8122416845_Math2

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Copyright © 2005, New Age International (P) Ltd., PublishersPublished by New Age International (P) Ltd., Publishers

All rights reserved.No part of this ebook may be reproduced in any form, by photostat, microfilm,xerography, or any other means, or incorporated into any information retrievalsystem, electronic or mechanical, without the written permission of the publisher.All inquiries should be emailed to [email protected]

PUBLISHING FOR ONE WORLD

NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS4835/24, Ansari Road, Daryaganj, New Delhi - 110002Visit us at www.newagepublishers.com

ISBN (13) : 978-81-224-2535-2

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Mathematics is considered as the mother of Science as well as the mother of Philosophy. In fact, it isthe basic knowledge on which any human endeavour in the field of intellectual pursuit is based. Butsomehow, in our academic field Mathematics is considered as a subject tough to comprehend and tomaster. The otherwise enjoyable subject becomes sometimes to the students a frightening nightmareand in the course of my long career in the field of education, I have found many students even becomingnervous at the mention of Mathematics. That is, in my experience, mostly because of the way it istaught by some teachers and the way some of the books are written.

My esteemed colleague Smt. Veena, G.R. is not only a good teacher who would make the studentsto enjoy Mathematics, but also a good “Narrator” of Mathematics. Extremely resourceful and com-pletely committed, Veena has brought into this book her direct experience in the class room as a teacher.In lucid style, she has written this book making it students-friendly, teacher-friendly and teaching-learning focused.

While congratulating her for this good work, I wish that both the students and the teachers wouldwelcome this book and make use of it.

PPPPPRRRRROFOFOFOFOF. K.E. R. K.E. R. K.E. R. K.E. R. K.E. RADHAKRISHNADHAKRISHNADHAKRISHNADHAKRISHNADHAKRISHNAAAAA

Principal, Surana College,Bangalore-560004.


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This book is written as per the Syllabus of Basic Mathematics for II Year Pre-University Course ofKarnataka. Each topic has been discussed exhaustively as per the requirements of the latest syllabus.The book has been written in a very simple style, to enable students to understand the subject effec-tively. More focus is given for a systematic approach to enhance the grasping of the subject so that alltypes of questions could be answered well in each chapter. I hope this book will satisfy all the require-ments of the students for learning the subject successfully and getting through the examination withflying colours.

My sincere thanks to Prof. K.E. Radhakrishna, Principal, Surana College, an eminent academicianand educationist for his foreword. I also thank Ms. Sudha S., Department of Mathematics, SuranaCollege for reading the manuscript and identifying the unforeseen computational errors.

I am also thankful to Mr. R.K. Gupta, Chairman, Mr. Saumya Gupta, Managing Director, New AgeInternational Pvt. Ltd., New Delhi and Mr. Vincent D’souza, Branch Manager, Mr. Babu V.R., MarketingManager, Bangalore branch for accepting to publish the book.

I sincerely welcome criticism, views and suggestions from readers.

GGGGG.R. .R. .R. .R. .R. VVVVVEENEENEENEENEENAAAAA

[email protected]


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Foreword vii

Preface ix

1.1.1.1.1. Mathematical LogicMathematical LogicMathematical LogicMathematical LogicMathematical Logic 11111

1.1 Introduction 11.2 Propositions 11.3 Logical Connectives and Compound Propositions

Conjunction, Disjunction, Conditional, Biconditional, Negation 21.4 Tautology and Contradiction 91.5 Logical Equivalence 121.6 Converse, inverse and Contrapositive of a Conditional 14

2.2.2.2.2. PPPPPererererermmmmmutautautautautation and Combination and Combination and Combination and Combination and Combinationtiontiontiontion 2121212121

2.1 Introduction 212.2 Fundamental Principle 212.3 Permutation and Combination 222.4 Factorial of a Positive Integer 222.5 Permutation 22

2.5.1 Linear Permutation 232.5.2 Value of npr 232.5.3 Value of npn 23

2.5.4 Value of 0 242.6 Permutation of Things of which Some are Alike 282.7 Circular Permutation 312.8 Combination 33

2.8.1 Value of ncr 332.8.2 Complementary Combinations 34

3.3.3.3.3. PrPrPrPrProbaobaobaobaobabilitybilitybilitybilitybility 4646464646

3.1 Introduction 463.2 Terminology 463.3 Definition of Probability 473.4 Addition Rule of Probability 523.5 Conditional Probability 56

xii Contents

3.6 Multiplication Rule 56

4.4.4.4.4. Binomial Binomial Binomial Binomial Binomial TTTTTheorheorheorheorheorememememem 6565656565

4.1 Introduction 654.2 Statement of Binomial Theorem 65

5.5.5.5.5. PPPPPararararartial tial tial tial tial FFFFFrrrrractionsactionsactionsactionsactions 8888888888

5.1 Definitions 885.2 Partial Fractions 89

6.6.6.6.6. Matrices & DeterminantsMatrices & DeterminantsMatrices & DeterminantsMatrices & DeterminantsMatrices & Determinants 105105105105105

6.1 Introduction 1056.2 Matrix 1056.3 Types of Matrices 1056.4 Algebra of Matrices 1076.5 Transpose of a Matrix 1086.6 Determinants 1176.7 Properties of Determinants 1206.8 Minor, Co-factor, Adjoint and Inverse of a Square Matrix 1366.9 Characteristic Equation of a Square Matrix 1446.10 Cayley Hamilton Theorem 1466.11 Solution of Linear System of Equations 1536.12 Application of Matrices in Business Problems 164

7.7.7.7.7. RaRaRaRaRatio and Prtio and Prtio and Prtio and Prtio and Proporoporoporoporoportions,tions,tions,tions,tions, VVVVVararararariaiaiaiaiationstionstionstionstions 182182182182182

7.1 Introduction 1827.2 Ratio 1827.3 Proportion 1907.4 Direct Proportion or Direct Variation 1927.5 Problems on Time and Work 2027.6 Problem on Time and Distance 2097.7 Problems on Mixture 213

8.8.8.8.8. AAAAAvvvvverererereraaaaagggggeseseseses 221221221221221

8.1 Introduction 2218.2 Arithmetic Average or Mean 2218.3 Combined Average 222

9.9.9.9.9. Bill Bill Bill Bill Bill DDDDDiscountingiscountingiscountingiscountingiscounting 234234234234234

9.1 Introduction 2349.2 Terminology 2349.3 Formulae 235

Contents xiii

10.10.10.10.10. StocStocStocStocStocks and Sharks and Sharks and Sharks and Sharks and Shareseseseses 246246246246246

10.1 Stock 24610.2 Shares 24610.3 Distinction between Stock and Shares 24610.4 Terminology 247

11.11.11.11.11. LearLearLearLearLearning Curning Curning Curning Curning Curvvvvveeeee 255255255255255

11.1 Introduction 25511.2 Learning Curve 25511.3 The Learning Curve Ratio 25511.4 Graphical Representation of Learning Curve 25511.5 Learning Curve Equation 257

12.12.12.12.12. Linear PrLinear PrLinear PrLinear PrLinear Prooooogggggrrrrrammingammingammingammingamming 263263263263263

12.1 Introduction 26312.2 Linear Programming 26312.3 Solution to Linear Programming Problem 264

13.13.13.13.13. CirCirCirCirCircccccleslesleslesles 278278278278278

13.1 Definitions 27813.2 Equation of a Circle 27813.3 Point of Intersection of a Line and a Circle 29213.4 Equation of Tangent to the Circle

x2 + y2 + 2gx + 2fy + c = 0 at the Point (x1, y1) on it 29813.5 Length of the Tangent from the Point (x1, y1) to the

Circle x2 + y2 + 2gx + 2fy + c = 0. 29913.6 Condition for the Line y = mx + c to be a Tangent to

the Circle x2 + y2 = a2 and point of contact 30013.7 Condition for the Line lx + my + n = 0 to be a Tangent

to the Circle x2 + y2 + 2gx + 2fy + c = 0. 301

14.14.14.14.14. PPPPParararararaaaaabolabolabolabolabola 317317317317317

14.1 Introduction 31714.2 Parabola 31714.3 Equation of the Parabola in the Standard Form 31814.4 Different Forms of Parabola with Vertex (0, 0) 32014.5 Different Forms of Parabola with Vertex (h, k) 321

15.15.15.15.15. Limits and ContinuityLimits and ContinuityLimits and ContinuityLimits and ContinuityLimits and Continuity 343343343343343

15.1 Introduction 34315.2 Constants and Variables 34315.3 Function 343

xiv Contents

15.4 Limits 34415.5 Standard Limits 34515.6 Continuous Functions 356

16.16.16.16.16. DifDifDifDifDifffffferererererential Calculusential Calculusential Calculusential Calculusential Calculus 366366366366366

16.1 Introduction 36616.2 Derivative of a Function 36616.3 Derivative of Some Standard Functions from First Principles 36716.4 Rules of Differentiation 37116.5 Differentiation of Composite Functions 38616.6 Differentiation of Implicit Functions 39016.7 Differentiation of Parametric Functions 39316.8 Logarithmic Differentiation 39716.9 Successive Differentiation 401

17.17.17.17.17. AAAAApplicapplicapplicapplicapplication of Dertion of Dertion of Dertion of Dertion of Deriiiiivvvvvaaaaatititititivvvvveseseseses 429429429429429

17.1 Derivative as a Rate Measure 42917.2 Maxima and Minima 438

18.18.18.18.18. IntegrationIntegrationIntegrationIntegrationIntegration 455455455455455

18.1 Standard Integrals 45518.2 Algebra of Integrals 45618.3 Substitution Method 46118.4 Integration by Partial Fraction Method 46618.5 Integration by Parts 474

18.6 Integrals of the Type e f x f x dxx � � � �+ ′� 477

19.19.19.19.19. DefDefDefDefDefinite inite inite inite inite IIIIIntententententegggggrrrrralsalsalsalsals 484484484484484

19.1 Introduction 48419.2 Properties of Definite Integrals 49019.3 Application of Definite Integrals to Find Area 494

20.20.20.20.20. AAAAApplicapplicapplicapplicapplication of Calculus in Businesstion of Calculus in Businesstion of Calculus in Businesstion of Calculus in Businesstion of Calculus in Business 504504504504504

20.1 Terminology 504

ExaminaExaminaExaminaExaminaExamination Cortion Cortion Cortion Cortion Cornernernernerner 515515515515515

• Blue Print 516• Model Question Paper 1 517• Model Question Paper 2 520• Model Question Paper 3 522• Chapterwise Arranged Question Bank 526• Gist and Formulae 546

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Logic is the science dealing with principles of reasoning. We can find all the different ways of solvinga problem by logical reasoning. The English Mathematician George Boole is the founder of mathemati-cal logic. To express the principles of reasoning, a symbolic language has been developed. This sym-bolic language is called mathematical logic or symbolic logic.

Mathematical logic finds application in switching circuits, digital computers and other digital devices.

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A proposition is a statement which in the given context is either true or false but not both. The propo-sitions are denoted by small letters p, q, r...

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1. Sum of two even integers is even integer.

2. 3 is a rational number.

3. Earth is flat.

4. Delhi is the capital of Karnataka.

5. 7 is a prime number.

6. 5 − 7 = −2.

Note:Note:Note:Note:Note: The statements involving opinions, question marks, exclamatory mark, command, wish are notpropositions.

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1. Logic is interesting.

2. What a beautiful weather!

3. Where are you going?

4. Please sit down.

5. May God bless you.

2 Basic Mathematics

TRTRTRTRTRUTH UTH UTH UTH UTH VALUE:VALUE:VALUE:VALUE:VALUE: The truthness or falsity of a proposition is called its truth value. If a proposition istrue it is denoted by ‘T’ and if it is false it is denoted by ‘F’.

Example:Example:Example:Example:Example: The truth value of

1. 5 + 6 = 11 is ‘T’.

2. ‘Asia is in India’ is F.

3. ‘Today is Sunday’ is either ‘T’ or ‘F’ in the given context i.e., on a particular day it is only oneof ‘T’ or ‘F’.

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Two or more simple propositions are connected by using the words ‘and’, ‘or’, ‘if ... then’, ‘if andonly if’. These words or phrases are called logical connectives. Any proposition containing one or moreconnectives is called a compound proposition. The simple propositions occurring in a compound propo-sition are called its components.

TRTRTRTRTRUTH UTH UTH UTH UTH TTTTTABLEABLEABLEABLEABLE::::: The truth values of the compound proposition for all possible truth values of itscomponents is expressed in the form of a table called truth table.

For a compound proposition with only one proposition, Truth table consists of 2 possibilities (eitherT or F).

For a compound proposition with two propositions truth table consists of 22 = 4 possibilities. For acompound proposition with 3 propositions truth table consists of 23 = 8 possibilities.

CONJUNCTIONCONJUNCTIONCONJUNCTIONCONJUNCTIONCONJUNCTION::::: If p and q are 2 simple propositions. Then the proposition ‘p and q’ is called theconjunction of p and q. It is denoted by p ∧ q.

Example:Example:Example:Example:Example:

If p : 7 is a prime number.

q : 6 is an even number, then

p ∧ q : 7 is a prime number and 6 is an even number.

The truth value of the compound proposition p ∧ q depends on the truth values of p and q. Note thatthe conjunction of p and q is true only when both p and q are true, otherwise it is false.

TTTTTrrrrruth uth uth uth uth TTTTTaaaaabbbbblelelelele

p q p q

T T T

T F F

F T F

F F F

∧

DISJUNCTIONDISJUNCTIONDISJUNCTIONDISJUNCTIONDISJUNCTION::::: If p and q are 2 simple propositions, then proposition ‘p or q’ is called the disjunc-tion of p and q. It is denoted by p ∨ q.

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If p : 2 is rational number.

Mathematical Logic 3

q : 2 is odd number, then

p ∨ q = 2 is rational or 2 is odd number.

The truth value of p ∨ q depends on the truth values of p and q. Note that the disjunction of p andq is false only when both p and q are false. Otherwise it is true.


p q p q

T T T

T F T

F T T

F F F

∨

CONDITIONCONDITIONCONDITIONCONDITIONCONDITIONAL (IMPLICAAL (IMPLICAAL (IMPLICAAL (IMPLICAAL (IMPLICATION)TION)TION)TION)TION)::::: If p and q are two simple proposition, then the proposition if ‘p... then q’ is known as conditional or implication. It is denoted by p → q or p ⇒ q.

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If p : 6 is an even number.

q : 6 is divisible by 2, then

p → q : If 6 is an even number then 6 is divisible by 2.

The truth value of p → q depends on the truth values of p and q. Note that p → q is false only whenp is true and q is false.


p q p q

T T T

T F F

F T T

F F T

→

BICONDITIONBICONDITIONBICONDITIONBICONDITIONBICONDITIONAL (DOUBLE IMPLICAAL (DOUBLE IMPLICAAL (DOUBLE IMPLICAAL (DOUBLE IMPLICAAL (DOUBLE IMPLICATION OR EQTION OR EQTION OR EQTION OR EQTION OR EQUIVUIVUIVUIVUIVALENCE):ALENCE):ALENCE):ALENCE):ALENCE): If p and q are simple propo-sitions, then the proposition ‘p if and only if q’ is called biconditional or double implication. It isdenoted by p ↔ q.

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If p : k is odd number.

q : k2 is odd number, then

p ↔ q : k is odd number if and only if (iff) k2 is odd number.

Note that p ↔ q involves both the conditionals p → q and q → p.

4 Basic Mathematics

∴ p q p q q p↔ → ∧ → is � � � �

The biconditional p → q is true if p and q are both true or both false i.e., if p and q have same truthvalues. Otherwise it is false.

TTTTTrrrrruth tauth tauth tauth tauth tabbbbblelelelele

p q p q q p p q i e p q q p

T T T T T

T F F T F

F T T F F

F F T T T

→ → ↔ → ∧ →. ., � � � �

NEGANEGANEGANEGANEGATION:TION:TION:TION:TION: If ‘p’ is a proposition then the proposition ‘not p’ is called negation of p. It is denotedby ~p.

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If p : 6 is odd number then

~p : 6 is not an odd number.

If p is true then ~p is false and if p is false then ~p is true.

TTTTTrrrrruth tauth tauth tauth tauth tabbbbblelelelele

p p

T F

F T

~

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1.1.1.1.1. An integer is even if and only if it is divisible by 2.

Solution:Solution:Solution:Solution:Solution: Let p : An integer is even.

q : It is divisible by 2.

∴ The given proposition is p ↔ q.

2. 2. 2. 2. 2. If 6 + 3 = 7, then 7 − 3 = 6

Solution:Solution:Solution:Solution:Solution: Let p : 6 + 3 = 7

q : 7 − 3 = 6

Then the given proposition is p → q

3.3.3.3.3. I play chess or I study at home.

Solution:Solution:Solution:Solution:Solution: Let p : I play chess

q : I study at home.


The given proposition: p ∨ q.

4.4.4.4.4. It is raining and the ground is wet.

Solution:Solution:Solution:Solution:Solution: Let p : It is raining.

q : The ground is wet.

The given proposition : p ∧ q.

5.5.5.5.5. If it rains today then government declares a holiday and we are happy.

Solution:Solution:Solution:Solution:Solution: Let p : It rains today

q : Government declares a holiday

r : we are happy.

Given proposition: p → (q ∧ r).

6.6.6.6.6. If a number is not real then it is complex.

Solution:Solution:Solution:Solution:Solution: Let p : A number is real.

q : It is complex.

Given proposition in symbols: ~p → q.

7.7.7.7.7. If Rama is intelligent or hardworking then logic is easy.

Solution:Solution:Solution:Solution:Solution: Let p : Rama is intelligent

q : Rama is hardworking

r : Logic is easy.

Given proposition: (p ∨ q) → r.

8.8.8.8.8. If 3 is not odd and 2 is not even then 7 is not odd or 8 is not even.

Solution:Solution:Solution:Solution:Solution: Let p : 3 is odd

q : 2 is even

r : 7 is odd

s : 8 is even.

Given proposition: (~p ∧ ~q) → (~r ∨ ~s)

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1. 1. 1. 1. 1. p ∧ ~q

Solution:Solution:Solution:Solution:Solution: The question paper is difficult and I do not get good marks.

2.2.2.2.2. q → r

If I get good marks, then I can go abroad.

3.3.3.3.3. (~p ∧ q) → r

Solution:Solution:Solution:Solution:Solution: If the question paper is not difficult and I get good marks then I can go abroad.

4.4.4.4.4. r ↔ q

Solution:Solution:Solution:Solution:Solution: I can go abroad iff I get good marks.

5.5.5.5.5. (p ∧ ~q) → ~r

6 Basic Mathematics

Solution:Solution:Solution:Solution:Solution: If the question paper is difficult and I do not get good marks then I can not go abroad.

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1.1.1.1.1. p ∧ ~q

p q q p q

T F T T

~ ~∧

Given Given

So Truth value of p ∧ ~q is T when p is T and q is F.

2.2.2.2.2. p → (q ∧ r)

p q r q r p q r

T F T F F

∧ → ∧� �

3.3.3.3.3. (~p ∧ q) ↔ r

p q p p q r p q r

T F F F T F

~ ~− ∧ ∧ ↔� �

4.4.4.4.4. (p ∨ ~q) ↔ (q ∧ ~r)(1) (2) (1) ↔ (2)

p q q p q r r q r p q q r

T F T T T F F F

~ ~ ~ ~ ~ ~∨ ∧ ∨ ↔ ∧� � � �

Given Given Given

5.5.5.5.5. p → (q → ~r)

p q r r q r p q r

T F T F T T

~ ~ ~→ → →� �

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1. p ∧ (q ∨ r)

Given p : 2 is even number – True proposition.

So Truth value of p : T.

Similarly Truth value of q : F

Truth value of r : F


p q r q r p q r

T F F F F

∨ ∧ ∨� �

∴ Truth value of p ∧ (q ∨ r) is F.

2.2.2.2.2. p → (q ∧ ~ r)

p q r r q r p q r

T F F T F F

~ ~ ~∧ → ∧� �

∴ Truth value of p → (q ∧ ~ r) is F.

3.3.3.3.3. p ↔ (~ q ∧ ~ r)

p q q r r q r p q r

T F T F T T T

~ ~ ~ ~ ~ ~∧ ↔ ∧� �

∴ Truth value of p ↔ (~ q ∧ ~ r) is T.

V. 1.1.1.1.1. A certain compound proposition (p ∧ q) → r is known to be false. Find the truth values of p, qand r.

Given: p q r F∧ →� � is

This implies p ∧ q is T and r is F [��T → F is F]

⇒ p is T, q is T and r is F � is T T T∧

2.2.2.2.2. A certain compound proposition (p ∧ q) → (r ∨ ~s) is known to be false. Find the truth values ofp, q, r and s.

Given: p q r s F∧ → ∨� � � �~ is

⇒ p q T r s F∧ ∨ is and is ~ � T F F→ is

⇒ p T q T r F s F is , is and is , ~ is � T T T F F F∧ ∨ is and is

⇒ p T q T r F s T is , is , is and is . � ~ is F T .

3.3.3.3.3. A certain compound proposition ~ ~p q r s∧ ∧ ∧� � � � is given to be True. Find the truth values of

p, q, r and s.

Given: ~ ~p q r s T∧ ∧ ∧� � � � is

⇒ ~ ~p q T r s T∧ ∧ is and is

⇒ ~ p T q T r T s T is , ~ is and is , is � T T T∧ is

⇒ p F q F r T s T is , is ; is ; is . � ~ is T F .

8 Basic Mathematics

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1.1.1.1.1. p → ~p

p p p p

T F F

F T T

~ ~→

ExplanaExplanaExplanaExplanaExplanation:tion:tion:tion:tion: A compound proposition with one component will have 2 possibilities either T or F.Write ~p finally p → ~p.

2.2.2.2.2. ~p → ~q.

p q p q p q

T T F F T

T F F T T

F T T F F

F F T T T

~ ~ ~ ~→

ExplanaExplanaExplanaExplanaExplanation:tion:tion:tion:tion: A compound proposition with 2 components will have 4 possibilities. Write 2 T and 2Funder p, alternatively T, F under q to get all possible combinations. Now

~p is T when p is F and vice versa.

~q is T when q is F and vice versa.

~p → ~q is F only when ~p is T and ~q is F. Otherwise it is T.

3.3.3.3.3. p ↔ (p ∧ q)

p q p q p p q

T T T T

T F F F

F T F T

F F F T

∧ ↔ ∧� �

4.4.4.4.4. p ∨ ~(~p ∨ q)

p p q p q p q p p q

T F T T F T

T F F F T T

F T T T F F

F T F T F F

~ ~ ~ ~ ~ ~∨ ∨ ∨ ∨� � � �


5.5.5.5.5. p ∧ (q ∨ r)

p q r q r p q r

T T T T T

T T F T T

T F T T T

T F F F F

F T T T F

F T F T F

F F T T F

F F F F F

∨ ∧ ∨� �

ExplanaExplanaExplanaExplanaExplanation:tion:tion:tion:tion: A compound proposition having 3 components will have eight possibilities. Write 4T and4F under p, 2T, 2F, 2T, 2F under q and alternately T and F under r to get all the possible combinationsof truth values.

Now q ∨ r is F only when both q and r are F. Otherwise it is T.

p ∧ (q ∨ r) is T only when p is T and (q ∨ r) is T otherwise it is F.

6.6.6.6.6. (p → ~q) ↔ (q ∧ ~ r)

p q r q p q r q r p q q r

T T T F F F F T

T T F F F T T F

T F T T T F F F

T F F T T T F F

F T T F T F F F

F T F F T T T T

F F T T T F F F

F F F T T T F F

~ ~ ~ ~ ~ ~→ ∧ → ↔ ∧� � � �

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A tautology is a compound proposition which is always true irrespective of the truth values of itscomponents.

A contradiction is a compound proposition which is always false irrespective of the truth values ofits components.

Note:Note:Note:Note:Note: To determine whether a given proposition is a tautology or a contradiction, construct its truthtable. If its truth value for all possibility is True, then it is tautology. If its truth value for allpossibility is False then it is contradiction. Otherwise it is neither tautology nor contradiction.

10 Basic Mathematics

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1.1.1.1.1. Prove that p ∧ ~p is a contradiction.

p p p p

T F F

F T F

~ ~∧

From the last column p ∧ ~p is a contradiction.

2.2.2.2.2. Prove that p ∨ ~p is a tautology.

p p p p

T F T

F T T

~ ~∨

From the last column p ∨ ~p is a tautology.

3.3.3.3.3. Find whether p → ~p is tautology or contradiction or neither

p p p p

T F F

F T T

~ ~→

From the last column, clearly p → ~p is neither tautology nor contradiction.

4.4.4.4.4. Prove that (p ∧ ~q) ↔ q is neither tautology nor contradiction.

p q q p q p q q

T T F F F

T F T T F

F T F F F

F F T F T

~ ~ ~∧ ∧ ↔� �

From the last column it is clear that (p ∧ ~q) ↔ q is neither tautology nor contradiction.

5.5.5.5.5. Prove that (p → q) ↔ (~q → ~p) is a tautology.

x y

p q q p p q q p X Y

T T F F T T T

T F T F F F T

F T F T T T T

F F T T T T T

� � � �

~ ~ ~ ~→ → ↔

From the last column it is clear that (p → q) ↔ (~q → ~p) is a tautology.


6. Prove that (p ∨ q) ∧ (~p ∧ ~q) is a contradiction.

p q p q p q p q p q p q

T T F F T F F

T F F T T F F

F T T F T F F

F F T T F T F

~ ~ ~ ~ ~ ~∨ ∧ ∨ ∧ ∧� � � �

From the last column, the given proposition is contradiction.

7. Examine whether (p → q) ∧ (q → r) is a tautology or contradiction or neither

p q r p q q r p q q r

T T T T T T

T T F T F F

T F T F T F

T F F F T F

F T T T T T

F T F T F F

F F T T T T

F F F T T T

→ → → ∧ →� � � �

From the last column it is clear that the given proposition is neither tautology nor contradiction.

8. Prove that p q q r p r→ ∧ → → →� � � � is a tautology

p q r p q q r p q q r p r p q q r p r

T T T T T T T T

T T F T F F F T

T F T F T F T T

T F F F T F F T

F T T T T T T T

F T F T F F T T

F F T T T T T T

F F F T T T T T

→ → → ∧ → → → ∧ → → +� � � � � � � �

From the last column clearly the given proposition is tautology.


��6 ��7��

Two propositions X and Y are said to be logically equivalent if and only if they have same truth valuesand we write X ≡ Y. Note that X ≡ Y if and only if X ↔ Y is always true or X ↔ Y is tautology.

��7��

1.1.1.1.1. Prove that p ≡ ~(~p).

1 2 3

p p p

T F T

F T F

~ ~ ~� �

from column (1) and (3) p ≡ ~(~p).

2.2.2.2.2. Prove that ~(p ∧ q) ≡ ~p ∨ ~ q

1 2 3 4 5 6 7

p q p q p q p q p q

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

∧ ∧ ∨~ ~ ~ ~ ~� �

From columns 4 and 7, ~ (p ∧ q) ≡ ~p ∨ ~q.

3.3.3.3.3. Prove that ~(p ∨ q) ≡ ~p ∧ ~ q

1 2 3 4 5 6 7

p q p q p q p q p q

T T F F T F F

T F F T T F F

F T T F T F F

F F T T F T T

~ ~ ~ ~ ~∨ ∨ ∧� �

From column 6 and 7, ~ (p ∨ q) ≡ ~p ∧ ~ q.


4.4.4.4.4. Prove that ~ (p → q) ≡ p ∧ ~ q

1 2 3 4 5 6

p q p q p q q p q

T T T F F F

T F F T T T

F T T F F F

F F T F T F

→ → ∧~ ~ ~� �

From columns 4 and 6, the given propositions are logically equivalent.

5.5.5.5.5. Prove that ~ (p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)

1 2 3 4 5 6 7 8 9

p q p q p q p q p q q p p q q p

T T T F F F F F F

T F F T F T T F T

F T F T T F F T T

F F T T T T F T T

↔ ↔ ∧ ∧ ∧ ∨ ∧~ ~ ~ ~ ~ ~ ~� � � � � �

From columns 4 and 9, ~ ~ ~p q p q q p↔ ≡ ∧ ∨ ∧� � � � � �

Note:Note:Note:Note:Note: The results

~ ~ ~p q p q∧ ≡ ∨� �

~ ~ ~p q p q∨ ≡ ∧� �

~ ~p q p q→ ≡ ∧� �

~ ~ ~p q p q p q↔ ≡ ∧ ∨ ∧� � � � � �

are used to find the negation of compound propositions.

!��"��#��

��% &�'��

1.1.1.1.1. 6 is an odd number or 3 is an even number.

Let p : 6 is an odd number.

q : 3 is an even number.

Given proposition in symbol is

p q∨


We know ~ ~ ~p q p q∨ ≡ ∧� �

∴ Negation: 6 is not an odd number and 3 is not an even number.

2.2.2.2.2. He is rich and He is not happy

Let p : He is rich.

q : He is happy.

Given: p ∧ ~ q

We know ~ ~ ~ ~ ~p q p q∧ ≡ ∨� � � �

≡ ∨~ p q

∴ Negation: He is not rich or He is happy.

3.3.3.3.3. If the cow is big, then it is healthy.

Let p : Cow is big.

q : It is healthy.

Given proposition in symbols: p → q

We know ~ ~p q p q→ ≡ ∧� �

∴ Negation: Cow is big and it is not healthy.

4.4.4.4.4. If the triangles are not equiangular then the sides are not proportional.

Let p : The triangles are not equiangular.

q : The sides are not proportional.

Given proposition: p → q

Its negation: ~ ( p → q)

But ~ ~p q p q→ ≡ ∧� �

Negation: The triangles are not equiangular and the sides are proportional.

5.5.5.5.5. 6 is even if and only if it is divisible by 2.

Let p : 6 is even.

q : 6 is divisible by 2.

Given: p ↔ q.

Its negation is ~(p ↔ q)

But ~ ~ ~p q p q p q↔ ≡ ∧ ∨ ∧� � � � � �

∴ Negation: 6 is even and it is not divisible by 2 or 6 is not even and it is divisible by 2.

��8 �� 1�� 9��

Let p → q be the given conditional then the conditional q → p is called converse of p → q. Theconditional ~p → ~q is called inverse of p → q. The conditional ~q → ~p is called contrapositive ofp → q.

Note that contrapositive is converse of inverse.


!��"��#��

WrWrWrWrWrite the conite the conite the conite the conite the convvvvvererererersesesesese,,,,, in in in in invvvvverererererse and contrse and contrse and contrse and contrse and contraaaaapositipositipositipositipositivvvvve of the fe of the fe of the fe of the fe of the folloolloolloolloollowing conditionals:wing conditionals:wing conditionals:wing conditionals:wing conditionals:

1.1.1.1.1. If x = 3, then x2 = 9.

Solution:Solution:Solution:Solution:Solution: Let p : x = 3, q : x2 = 9.

Given: p → q

Converse: q → p

If x2 = 9, Then x = 3

Inverse: ~p → ~q

If x ≠ 3. Then x2 ≠ 9.

Converse: ~q → ~p

If x2 ≠ 9. Then x ≠ 3.

2. 2. 2. 2. 2. If two triangles are congruent then they are similar.

Let p : 2 triangles are congruent.

q : 2 triangles are similar.

Given: p → q

Converse: q → p

If 2 triangles are similar then they are congruent.

Inverse: ~p → ~q

If 2 triangles are not congruent then they are not similar.

Converse: ~q → ~p

If 2 triangles are not similar then they are not congruent.

3.3.3.3.3. If cows can fly then birds cannot fly.

Let p : cows can fly.

q : Birds can fly.

Given: p → ~q

Converse of p → q is q → p

Converse of p → ~q is ~q → p

i.e. If birds cannot fly then cows can fly.

Inverse of p → q is ~p → ~q

So inverse of p → ~q is ~p → ~(~q)

~p → q

i.e. If cows cannot fly then birds can fly.

Contrapositive of p → q is ~q → ~p

∴ Contrapositive of p → ~q is ~(~q) → ~p

i.e., q → ~p.

If birds can fly then cows cannot fly.


4.4.4.4.4. If I work hard then I can score 90% and I can go for engineering.

Solution:Solution:Solution:Solution:Solution: Let p : I work hard.

q : I can score 90%

r : I can go for engineering.

Given p → (q ∧ r)

Converse: q ∧ r → p

i.e. If I can score 90% and I can go for engineering then I work hard.

Inverse is ~p → ~(q ∧ r) i.e., ~p → ~ q ∨ ~r

If I do not work hard then I can not score 90% or I cannot go for engineering.

Contrapositive is ~(q ∧ r) → ~p i.e., (~q ∨ ~r) → ~p

If I cannot score 90% or I cannot go for engineering then I do not work hard.

5.5.5.5.5. If e is not irrational and π is rational then 6 is not even or 2 is odd.

Let X : e is not irrational and π is rational

Y : 6 is not even or 2 is odd.

Given X → Y

Converse: Y → X

If 6 is not even or 2 is odd then e is not irrational and π is rational.

Inverse: ~X → ~Y

If e is irrational or π is not rational

Then 6 is even and 2 is not odd.

Contrapositive: ~Y → ~X

If 6 is even and 2 is not odd then e is irrational or π is not rational.

��:��

• p ∧ q is T only when both p and q are true otherwise it is false i.e., T ∧ T is T otherwise it is F.

• p ∨ q is F only when both p and q are false. Otherwise it is True i.e., F ∨ F is F otherwise it isT.

• p → q is F only when p is true and q is false i.e., T → F is F otherwise it is T.

• p ↔ q is T only when both p and q are together True or False, i.e., T ↔ T and F ↔ F is Totherwise it is F.

• ~p is T when p is F and vice-versa.

• Tautology is a compound proposition which is always ‘true’ for all possible combinations of thetruth values of its components.

• Contradiction is a compound proposition which is always ‘False’ for all possible combinations ofthe truth values of its components.

• 2 propositions X and Y are logically equivalent if and only if they have identical truth values. Itis denoted by X ≡ Y.

• ~ (~p) ≡ p


• ~ (p ∧ q) ≡ ~p ∨ ~q

• ~ (p ∨ q) ≡ ~p ∧ ~q

• ~ (p → q) ≡ p ∧ ~ q

• ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (~p ∨ q)

• Converse of the conditional p → q is q → p

• Inverse of the conditional p → q is ~p → ~q

• Contrapositive of conditional p → q is ~q → ~p.

�#��

I. WrWrWrWrWrite the fite the fite the fite the fite the folloolloolloolloollowing compound prwing compound prwing compound prwing compound prwing compound propositions in symbols.opositions in symbols.opositions in symbols.opositions in symbols.opositions in symbols.

1. If a triangle is equilateral then all the sides of the triangle are equal.

2. If I don’t go to picnic then I will study at home.

3. Sun rises in the east and earth is not flat.

4. 2 is irrational or 5 is real.

5. A number is prime if and only if it is not composite.

6. If 2 + 2 ≠ 4 and 6 + 6 ≠ 12, then 4 + 7 = 6 or 5 + 3 = 9.

7. ABC is a right angled triangle if and only if one of the angle = 90° and square on the hypotenuse= sum of the squares on other 2 sides.

8. a + ib = x + iy iff a = x and b = y.

II. If If If If If ppppp,,,,, q q q q q and and and and and r r r r r ararararare 3 pre 3 pre 3 pre 3 pre 3 propositions with tropositions with tropositions with tropositions with tropositions with truth vuth vuth vuth vuth values alues alues alues alues TTTTT,,,,, F F F F F and and and and and T T T T T rrrrrespectiespectiespectiespectiespectivvvvvelelelelely then fy then fy then fy then fy then find the trind the trind the trind the trind the truthuthuthuthuthvvvvvalues of the falues of the falues of the falues of the falues of the folloolloolloolloollowing:wing:wing:wing:wing:

1. p q r∨ →� � ~ 2. ~ p q r→ ∧� � 3. p q q r∧ ↔ ∨� � � �~ 4. p q r p∧ ∨ ↔� �

5. p p q→ ∨� � 6. p r q↔ ∨� � ~

III. ConstrConstrConstrConstrConstruct the truct the truct the truct the truct the truth tauth tauth tauth tauth tabbbbble fle fle fle fle for the for the for the for the for the folloolloolloolloollowing prwing prwing prwing prwing propositions:opositions:opositions:opositions:opositions:

1. ~ ~p q∧ 2. p q∧ ~ 3. ~ ~p q→ 4. ~ ~p q→� �

5. p q q r∧ ↔ ∧� � � �~ 6. p p q∨ ∧~ � � 7. ~ p q r→ ∧� �

IV. VVVVVerererererify wify wify wify wify whether the fhether the fhether the fhether the fhether the folloolloolloolloollowing compound prwing compound prwing compound prwing compound prwing compound propositions aropositions aropositions aropositions aropositions are tautoloe tautoloe tautoloe tautoloe tautologggggies or contries or contries or contries or contries or contradictions oradictions oradictions oradictions oradictions orneither:neither:neither:neither:neither:

1. p p q→ ∨� � 2. p p∨ ~ 3. p p∧ ~ 4. p p q∧ ∨~� �

5. ~ ~ p p� �↔ 6. ~ p q q p∧ ∧ →� � � � 7. p q p∧ ∨� �

8. p q q r p r→ ∧ → → →� � � � � �~ 9. ~ ~p p q q∧ ∨ ∧� �

10. p q r p q p r∨ ∧ ↔ ∨ ∧ ∨� � � � � �


V. FFFFFind wind wind wind wind whether the fhether the fhether the fhether the fhether the folloolloolloolloollowing compound prwing compound prwing compound prwing compound prwing compound propositions aropositions aropositions aropositions aropositions are loe loe loe loe logggggicallicallicallicallically equiy equiy equiy equiy equivvvvvalent:alent:alent:alent:alent:

1. p q p q p q p q∧ → ∨ ∨ → ∧� � � � � � � �; 2. p q r p q r∨ ∨ ∨ ∨� � � �;

3. p q r p q p r→ ∧ → ∧ →� � � � � �; 4. p q p q→ →; ~ ~

5. p q r p q p r∧ ∨ ∧ ∨ ∧� � � � � �; 6. ~ ~ ; ~p q r p q r→ ∧ ∧ ∨� � � �

VI. NeNeNeNeNegggggaaaaate the fte the fte the fte the fte the folloolloolloolloollowing compound prwing compound prwing compound prwing compound prwing compound propositions:opositions:opositions:opositions:opositions:

1. 5 is odd and 6 is even.

2. Cow is not big or it is black.

3. If 2 lines are parallel then they do not intersect.

4. I will pass the examination iff the questions are easy.

5. p q∧ ~

6. p q→~

7. p q r∧ →� �

8. p q↔~

VII. FFFFFind the inind the inind the inind the inind the invvvvverererererse conse conse conse conse convvvvverererererse and contrse and contrse and contrse and contrse and contraaaaapositipositipositipositipositivvvvve of the fe of the fe of the fe of the fe of the folloolloolloolloollowing:wing:wing:wing:wing:

1. If x is even then x2 is even.

2. If a2 + b2 = c2 and a2 = c2 then b2 = 0.

3. If a number is real then it is rational or it is irrational.

4. If Smitha gets a first class then she is either intelligent or hard working.

5. If 3 is not prime and 7 is not an odd number then 37 is not an even number or 73 is an odd number.

6. ~p → ~q

7. p q p q∧ → ∨� � � �

8. p q r→ →� �

9. p q r→ →~ ~� �

10. p q r∨ →� � .

��!��

I. 1. p q→ 2. ~ p q→ 3. p q∧ ~

4. p q∨ 5. p q↔~ 6. ~ ~p q r s∧ → ∨� � � �

7. p q r↔ ∧� � 8. p q r↔ ↔� �

II.

1. F 2. T 3. F 4. F 5. F 6. T


III.

1.

~ ~p q

F

F

F

T

∧

2.

p q

F

T

F

F

∧ ~

3.

~ ~p q

T

T

F

T

→

4.

~ ~p q

T

F

T

F

→� �

5.

p q q r

F

T

T

T

T

F

T

T

∧ ↔ ∧� � � �~

6.

p p q

T

T

T

T

∨ ∧~ � �

7.

p q r

T

T

T

T

T

F

F

F

→ ∧

IV.

1. Tautology 2. Tautology 3. Contradiction 4. Neither

5. Tautology 6. Contradiction 7. Neither 8. Neither

9. Contradiction 10. Tautology

V.

1. No, not logically equivalent. 2. Logically equivalent.

3. Logically equivalent. 4. No, not logically equivalent.

5. Logically equivalent. 6. Logically equivalent.

VI.

1. It is not odd or 6 is not even

2. Cow is big and it is not black.

3. 2 lines are parallel and they intersect

4. I will pass the examination and the questions are not easy or I will not pass the examination andthe questions are easy.

5. ~p ∨ q

6. p ∧ q

7. ~p ∨ (q ∧ ~r)

8. (p ∧ q) ∨ (~p ∧ ~q)


VII.

1. Inverse: If x is not even then x2 is not even

Converse: If x2 is even then x is even.

Contrapositive: If x2 is not even then x is not even.

2. Inverse: If a2 + b2 ≠ c2 or a2 ≠ c2 Then b2 ≠ 0.

Converse: If b2 = 0 then a2 + b2 = c2 and a2 = c2.

Contrapositive: If b2 ≠ 0 then a2 + b2 ≠ c2 or a2 ≠ c2.

3. Inverse: If a number is not real then it is not rational and it is not irrational.

Converse: If a number is rational or it is not irrational then it is real.

Contrapositive: If a number is not rational and it is not irrational then it is not real.

4. Contrapositive: If Smitha is neither intelligent nor hardworking then she doesn’t get a first class.

Converse: If Smitha is either intelligent or hardworking then she gets a first class.

Inverse: If Smitha does not get first class then she is neither intelligent nor hardworking.

5. Contrapositive: If 37 is even and 73 is not odd then 3 is prime or 7 is odd.

Converse: If 37 is not even or 73 is odd then 3 is not prime and 7 is not odd.

Inverse: If 3 is prime or 7 is odd then 37 is even and 73 is not odd.

6. Inverse: p → q

Converse: ~q → ~p

Contrapositive: q → p

7. Inverse: ~ ~ ~ ~p q p q∨ → ∧� � � �

Converse: p q p q∨ → ∧� � � �

Contrapositive: ~ ~ ~ ~p q p q∧ → ∨� � � �

8. Inverse: ~ ~p q r→ ∧� �

Converse: q r p→ →� �

Contrapositive: q r p∧ →~ ~� �

9. Inverse: ~ ~p q r→ ∧� �

Converse: ~ ~q r p→ →� �

Contrapositive: ~ ~q r p∧ →� �

10. ~ ~ ~p q r∧ →

r p q→ ∨

~ ~ ~r p q→ ∧

�

��

��

In our daily life we come across situations where we have to select or arrange certain things out of agiven number of things. This selection or arrangement involves a principle known as fundamentalprinciple which is illustrated by the following example.

Suppose that in a auditorium there are 4 differententrance doors (say I1, I2, I3 and I4) and there are 5different exit doors (say O1, O2, O3, O4 and O5). Inhow many ways can a person enter and leave the au-ditorium?

If a person enters the auditorium through the doorI1, he can go out by any one of the exit doors O1 O2O3 O4 O5. So there are 5 ways of leaving the auditoriumif the person enters it through door I1.

Similarly corresponding to the entrance door I2 thereare 5 ways of leaving the auditorium. Altogether 5 +5 + 5 + 5 = 20 different ways. In general if there arem different entrance doors and n different exit doors,a person can enter and leave the auditorium in mnways. This is fundamental principle.

��

If one event can be done in m different ways and after it has been done in one of these ways, a secondevent (which is independent of the first) can be done in ‘n’ different ways then the two events togethercan occur in mn ways.

The extension of this principle (also called the mnp ... principle) to the case of more than 2 eventsis obvious.

��

1.1.1.1.1. A boy and a girl have to be selected from a group of 5 boys and 6 girls. In how may ways canthe selection be made?

Fig. 2.1


Solution: Solution: Solution: Solution: Solution: Here First operation is selecting a boy from a group of 5 boys. This can be done in 5 ways.After this is done, the second operation is selecting a girl from 6 girls. This can be done in 6 ways.

By fundamental principle, the total number of selections = 5 × 6 = 30 ways.

2.2.2.2.2. There are 4 candidates for the post of manager. 3 candidates for the post of officer and 5 for thepost of clerk. In how many ways can these posts be filled?

Solution:Solution:Solution:Solution:Solution: A manager may be selected in 4 ways. An officer may be selected in 3 ways and A clerkmay be selected in 5 ways.

By fundamental principle, the 3 posts together can be filled in 4 × 3 × 5 = 60 ways.

��

Suppose 3 members (say a, b, c) went to a cinema where they get only 2 tickets. So it is required toselect 2 members out of 3 members (a, b, c). The following selections are possible: ab, bc, ca. So thereare 3 ways of selecting 2 members out of 3 members. In symbols this can be written as 3c2. It is callednumber of combination of 3 members taken 2 at a time.

Now consider 3 symbols α, β and γ. If we wish to arrange these letters taken 2 at a time, we getthe following arrangements αβ, βγ, γα, βα, γβ, αγ. These arrangements are called number ofpermutations of 3 symbols taken 2 at a time. In symbols this can be written as 3p2.

Note that in arrangement order is important. Therefore αβ is different from βα, whereas in selectionorder is not important. Hence ab is same as ba and so it is regarded as only one selection.

��! ��"��#��$��

The product of first n natural numbers i.e. 1. 2. 3. ... (n − 1). n is called factorial n. It is represented

by the symbol n or n!.

∴ n n n n n! ... ...= ⋅ ⋅ − ⋅ = − ⋅ ⋅1 2 3 1 1 3 2 1� � � �

So 1 1=

2 2 1= ×

3 3 2 1= × ×

4 4 3 2 1 4 3 4 3 2= × × × = = × × and so on.

In general

n n n n n n= − = − ⋅ −1 1 2� � and so on.

��% ��

An arrangement of all or part of a set of objects in some order is called permutation.

If the objects are arranged along a straight line it is called a linear permutation. If the objects arearranged around a circle then it is called circular permutation.

Permutation and Combination 23

��%��

Each of the different arrangements in a straight line that can be made by taking some or all of a numberof things at a time is called linear permutation.

The number of permutation of n distinct things taken r at a time is denoted by npr.

��%�� #�&��'��

The number of permutation of n distinct things taken r at a time i.e. npr will be same as the numberof ways in which r blank places can be filled up with n given objects.

As the first place can be filled in by any one of the n objects, there are n ways of filling the firstplace.

After having filled in the first place by any one of the n things, there are (n − 1) objects left. Hence2nd place can be filled in (n − 1) ways. Similarly 3rd place can be filled in (n − 2) ways and so on.Hence rth place can be filled in (n − (r − 1)) ways i.e., (n − r + 1) ways.

Position of the object 1

Number of ways

st nd rd th2 3

1 2 1

......

......

r

n n n n r− − − +� � � � � �

By fundamental principle, r places can be filled by any r of n objects in n (n−1) (n−2) ... (n−r +1)ways.

∴ nrp n n n n n r= − − − − +1 2 3 1� �� ...

Multiplying and Dividing by (n − r) (n − r − 1) ... 2 ⋅ 1 in RHS we get

nrp

n n n n n r n r n r

n r n r= − − − − + − − − ⋅

− − − ⋅1 2 3 1 1 2 1

1 2 1

� ��

... ...

...

nrp

n

n r=

−

Hence value of n

rpn

n r=

− .

��%�� #�&��'��

The number of permutation of n distinct objects taken all at a time is npn. It is same as the number ofways in which n blank spaces can be filled up with n given objects.

Position of the object 1Number of ways

st nd rd th th2 3 11 2 2 1

......

......n n

n n n−

− −

As the first place can be filled in by any one of the n objects there are n ways of filling the firstplace. After having filled in the first place, 2nd place can be filled in (n − 1) ways and so on.

∴ By fundamental principle,

nnp n n n= − − ⋅1 2 2 1� �� ...


∴ nnp n=

∴ Value of nnp n= .

��%�! #�&��'� 0 �

We haven

rpn

n r=

−

Put r = n.

nnp

n

n n=

−

But nnp n=

∴ nn

=0

Cross-multiplying,

0 =n

n

⇒ 0 1=

(��)��*��"�

1.1.1.1.1. Evaluate:

(a) 52p (b) 6

3p (c) 77p

We have n

rpn

n r=

−

52

5

5 25 4 20p =

−= × = .

63 6 5 4 120p = × × = .

77 7 5040p = = .

2.2.2.2.2. If npn = 720 find n.

We have nnp n= = 720 given� �


n = × × × × ×6 5 4 3 2 1

n = 6

⇒ n = 6.

3.3.3.3.3. If np2 = 72, find n.

We have np2 = n (n − 1) = 72.

n (n − 1) = 9 × 8 [By inspection: 72 = 9 × 8].

⇒ n = 9.

4.4.4.4.4. In how many of the permutations of 7 things taken 4 at a time will (a) One thing always occur(b) One thing never occur?

Solution:Solution:Solution:Solution:Solution: (a) Keeping aside the particular thing which will occur, the number of permutation of 6things taken 3 at a time

= 6p3 = 6 × 5 × 4 = 120.

Now this particular thing can take up any one of the four places and so the total number of ways= 120 × 4 = 480 ways.

(b) Leaving aside the particular thing which has never to occur, the number of permutation of 6 thingstaken 4 at a time

= 6p4 = 6 × 5 × 4 × 3 = 360 ways.

5.5.5.5.5. How many 4 digit numbers can be formed with the digits 2, 4, 5, 7, 9. (Repetitions not beingallowed). How many of these are even?

Number of 4 digit numbers that can be formed with the digits 2, 4, 5, 7, 9 (without repetitions)= 5p4 = 5 × 4 × 3 × 2 = 120.

Since we require an even number, we musthave 2 or 4 in the unit’s place. After filling theunit’s place by 2 or 4, the remaining 3 places(Ten’s, Hundred’s and Thousand’s) can befilled by remaining digits 5, 7, 9, 2 or 4 in 4p3ways 4 × 3 × 2 = 24 ways.

∴ Number of even numbers that can be formed

= 2 2 24 4843× = × =p

6.6.6.6.6. How many numbers can be formed by using any number of digits 3, 1, 0, 5; no digit beingrepeated in any number:

Solution:Solution:Solution:Solution:Solution: The number of single digit numbers = 3p1 (excluding zero) = 3.

The permutation of 4 digits taking 2 at a time are 4p2 but 3p1 of these have zero in ten’s place soreduce to single digit number.

∴ Number of 2 digit numbers = 4p2 − 3p1.

Similarly number of 3 digit numbers = 4p3 − 3p2. Number of 4 digit numbers = 4p4 − 3p3.

2 7203 3604 1205 306 6

1

Th H T U

4p3

2 or 4


Total number of numbers

= + − + − + −3 42

31

43

32

44

33p p p p p p� � � � � �

= + × − + × × − × + × × × − × ×3 4 3 3 4 3 2 3 2 4 3 2 1 3 2 1� � � � � �

= 3 12 3 24 6 24 6+ − + − + −� � � � � � = 3 + 9 + 18 + 18 = 48.

7.7.7.7.7. There are 4 Kannada books, 3 Hindi books and 5 English books. In how many ways can thesebe placed on a shelf if the books of the same language are to be together?

Solution:Solution:Solution:Solution:Solution: Since the books of the same language are to be together. Let us consider the 4 Kannada booksas 1 unit, 3 Hindi books as another unit and 5 English books as a different unit. Then we have to arrange

3 different units. This can be done in3

3 3 6p = = ways.

But the 4 Kannada books remaining together can be arranged in4

4 4p = ways. Similarly, 3 Hindi

books can be arranged in 3 ways and 5 English books can be arranged among themselves in 5 ways.

∴ Number of permutations = × × ×3 4 3 5

= 3 × 2 × 4 × 3 × 2 × 1 × 3 × 2 × 5 × 4 × 3 × 2 × 1

= 6 × 24 × 6 × 120

= 103680 ways.

8.8.8.8.8. In how many different ways can 5 examination papers be arranged in a row so that the best andthe worst papers may never come together.

Solution:Solution:Solution:Solution:Solution: Without any restrictions the 5 exam papers can be arranged among themselves in 5p5 =

5 = 120 ways.

Considering now, the best and the worst papers as a single paper, we have only 3 + 1 = 4 papers.

Now these 4 papers can be arranged taking all at a time in 43 4 3 2 24p = × × = ways.

But in each of the 24 ways, the best and worst papers can be arranged among themselves in 2 =

2 ways.

∴ Total number of ways, where the best and the worst paper always come together

= 24 × 2 = 48.

Hence required number of arrangements where best and the worst paper never come together =120 − 48 = 72 ways.

9.9.9.9.9. In how many ways can 5 boys and 3 girls be seated in a row so that (a) each girl in between 2boys (b) no two girls sit together (c) all the girls are together.

Solution:Solution:Solution:Solution:Solution: (a) First we arrange 5 boys in a row. This can be done in 5p5 = 5 120= ways. .

Consider one such arrangement:

B1 ↓ B2 ↓ B3 ↓ B4 ↓ B5


There are 4 places available for 3 girls, so that each girl is between 2 boys. The 4 places can be filledby 3 girls in 4p3 ways = 4 × 3 × 2 = 24 ways.

For one arrangement of boys, there are 24 ways of arranging girls.

∴ For 120 arrangement of boys

= 120 × 24 = 2880 ways.

(b) Again we arrange 5 boys in a row. This can be done in 5 = 120 ways. Considering 1 such

arrangement

*B1*B2*B3*B4*B5*

There are 6 places available for the girls so that no two girls are together. The 6 places can befilled by 3 girls in 6p3 ways = 6 × 5 × 4 = 120 ways.

∴Required number permutations = 120 × 120 = 14400 ways.

(c) We require all the 3 girls to be together. So we consider 3 girls as one unit.

Now 5 boys and 1 unit (of 3 girls) can be arranged in 5 1 6 720+ = = ways.

Again 3 girls, wherever they are together can be arranged in 3 6= ways.

∴ Number of permutations = 720 × 6 = 4320 ways.

10.10.10.10.10. How many different words can be formed with the letters of the word ‘ORDINATE’.

(a) Beginning with O (b) beginning with O and ending with E. (C) So that vowels occupy oddplaces.

(a) Keeping ‘O’ in the first place. We can arrange the remaining 7 letters in 7 places in

77 7 5040p = = ways ways.

7p7

O

∴ Number of words that can be formed with the letters of word ‘ORDINATE’ that begins with0 = 5040.

(b) Keeping O in the first place and E in the last place, the remaining 6 letters can be arranged

in 6 place in 6p6 = 6 720= ways.

6p6

O E

(c) In the word ‘ORDINATE’ O, I, A, E are vowels.

There are 4 odd places and 4 vowels. The 4 vowels can be arranged in 4 odd places in 4p4

= 4 ways = 4 × 3 × 2 × 1 = 24 ways.


1st

3rd

5th 7

th

Remaining 4 even places can be occupied by remaining 4 letters in 4p4 ways

= =4 24ways ways.

∴Total number of permutations = 24 × 24 = 576.

11.11.11.11.11. If the letters of the word MAKE be permuted and the words so formed be arranged as indictionary what is the rank of the word?

Solution:Solution:Solution:Solution:Solution: The alphabetical order of the letters is AEKM.

If A is fixed in 1st place, the other 3 letters can be permuted in 3 3 2 6= × = ways. .

The number of words begin with A = 6.

Similarly the number of words which begin with E and K = 3 6= each.

The words beginning with M are

MAEK and MAKE.

∴ Rank of the word = 6 + 6 + 6 + 1 + 1

= 20.

12.12.12.12.12. Prove that the number of ways in which n books can be placed on a shelf when 2 particular books

are never together is n n− ⋅ −2 1� � .

Solution:Solution:Solution:Solution:Solution: Regarding the 2 particular books as one book, there are (n − 1) books now, which can be

arranged in nnp n−

− = −11 1 ways. Now these two books can be arranged in 2 ways. Therefore the

number of permutations in which 2 particular books are always together = 2 1. n − .

The number of permutations of n books without any restriction = n .

∴ The number of permutations in which 2 particular books never occur together

= − −n n2 1.

= ⋅ − − −n n n1 2 1.

= − −n n1 2� �

= − −n n2 1� �Hence proved.

��+ ��,��$"��(,��,�"��)�

TheoremTheoremTheoremTheoremTheorem: If x is the number of permutations of n things taken all at a time of which one set p are alike,another set q are alike and so on. Then prove that

xn

p q=

⋅ ...


Proof:Proof:Proof:Proof:Proof: Replacing p like objects by ‘p’ unlike objects, q like objects by ‘q’ unlike objects and so on,we arrive at a stage where all n objects are distinct. By permuting these unlike objects amongst

themselves each of the x permutations would give rise to p q⋅ ... permutations. Hence x permutations

give rise to x p q⋅ ⋅ ... . But the number of permutations of n distinct objects taken all at a time = n .

∴ x p q n⋅ ⋅ =...

⇒ xn

p q=

⋅ ....

(��)��*��"�

1.1.1.1.1. In how many ways can letters of the word ‘INDIA’ be arranged?

Solution:Solution:Solution:Solution:Solution: There are 5 letters in the word ‘INDIA’ of which I occur 2 times.

∴ Number of words possible = 5

2

= 5 × 4 × 3 = 60.

2.2.2.2.2. In how many ways can the letters of the word ‘PERMANENT’ be arranged so that 2E’s are alwaystogether.

Solution:Solution:Solution:Solution:Solution: There are 9 letters in the word ‘PERMANENT’ of which E occurs 2 times, N occurs 2 times.If 2 E’s are together, taking them as one letter we have to arrange 8 letters in which N occurs two times.

∴ Number of arrangements = 8

2

= 8 × 7 × 6 × 5 × 4 × 3 = 20,160

3.3.3.3.3. In how many ways can be letters of the word ‘HOLLOW’ be arranged so that the 2L’s do notcome together.

Solution:Solution:Solution:Solution:Solution: There are 6 letters in the word HOLLOW in which L occurs 2 times, O occurs 2 times.

∴ Number of arrangement =⋅6

2 2

= × × × ×× × ×

6 5 4 3 2

2 1 2 1

= 180.

If 2 L’s are together, taking them as one letter, we have to arrange 5 letters in which O occurs2 times.

∴ Number of arrangements in which 2 L’s are together = = × ×5

25 4 3

= 60.

∴ Number of arrangements in which 2L’s do not come together = 180 − 60 = 120.


4.4.4.4.4. Find the number of permutations of the word ‘EXCELLENCE’. How many of these permuta-tions (i) begin with E (ii) begin with E and end with C (iii) begin with E and end with E (iv)do not begin with E.

Solution:Solution:Solution:Solution:Solution: The word ‘EXCELLENCE’ contains 10 letters of which E occurs 4 times C occurs 2 times,L occurs 2 times.

Number of permutations =⋅ ⋅10

4 2 2

= × × × × ×× × ×

=10 9 8 7 6 5

2 1 2 137800.

(i) Put E in the first place and arrange the rest. Now there are 9 letters of which E occurs 3 times,L occurs twice and C occurs twice.

∴ Number of permutations =⋅ ⋅

9

3 2 2

= × × × × ×× × ×

9 8 7 6 5 4

2 1 2 1

= 15,120.

(ii) Put E in the first place and C in the last place, and arrange the rest. There are 8 letters of whichE occurs 3 times, L occurs twice.

∴ Number of permutations =⋅8

3 2

= × × × × =8 7 6 5 4

23360.

(iii) Put one E in the first place and another E in the last place and arrange the rest. There are 8 lettersof which E occurs 2 times, L occurs twice and C occurs twice.

∴ Number of permutations =⋅ ⋅

8

2 2 2

= × × × × ×× × ×

8 7 6 5 4 3

2 1 2 1

= 5040.

(iv) Number of permutations that do not begin with E = Total number of permutations.

— Number of permutations that begin with E.

= 37800 − 15120

= 22680.

5.5.5.5.5. How many numbers less than 3 millions can be formed using the digits of the number 2123343?


Solution:Solution:Solution:Solution:Solution:

Million HTh TTh Th H T U

1 or 2

Since the number should be less than 3 millions, we can have either 1 or 2 in the million’s place.

Keeping 1 in the million’s place and arranging the rest. There are 6 numbers 2, 2, 3, 3, 3, 4 of which2 repeats twice, 3 occurs thrice.

∴ Number of numbers = 6

2 3

6 5 4

2⋅= × ×

= 60.

Now keeping 2 in the million’s place and arranging the rest. There are 6 numbers 1, 2, 3, 3, 3, 4in which 3 occurs thrice.

∴ Number of numbers = 6

36 5 4= × ×

= 120.

Total number of numbers less than 3 millions.

= 60 + 120 = 180.

��- ��

If we have to arrange 4 letters A, B, C, D in a row. Then 2 of the arrangements would be ABCD, BCDAand we treat these two arrangements as different. But if we arrange along the circumference of the

circle, the two arrangements and are one and the same.

So we conclude that circular permutations are different only when the relative order of the objectsis changed; otherwise they are same.

In circular permutation of n different things one thing is kept fixed and the balance (n − 1) things

are arranged relative to it in n − 1 ways.

If the clockwise and anticlockwise orders are distinguished, the required number of permutations =

n − 1 .

∴ Number of ways in which n persons can occupy the chairs in a round table = n − 1 .

If the clockwise and anticlockwise orders are not distinguished then required number of permuta-

tions =−n 1

2.


∴ Number of ways in which n flowers or n beads are strung to form garland or necklace =−n 1

2.

(��)��*��"�

1.1.1.1.1. In how many ways 5 people sit around a table?

Solution:Solution:Solution:Solution:Solution: Fixing the position of one person, the remaining 4 persons can sit around a table in 4 ways

= 4 × 3 × 2 × 1 = 24.

∴ 5 people can sit round a table in 24 ways.

2.2.2.2.2. In how many ways can 7 different jewels be strung into a necklace?

Solution:Solution:Solution:Solution:Solution: Keeping one jewel fixed, remain 6 jewels can be arranged in 6 = 720 ways.

Since clockwise and anticlockwise arrangements are same, Required number of permutations.

1

2 ways.× =720 360

3.3.3.3.3. In how many ways can 7 people be arranged at a round table so that 2 particular persons alwayssit together.

Solution:Solution:Solution:Solution:Solution: First, the two particular persons can be arranged in 2 2= ways. .

Considering them as one fixed person, the remaining 5 persons can be arranged in 5 = 120 ways.

∴ The required number of permutation = 120 × 2 = 240.

4.4.4.4.4. A round table conference is to be held between delegates of 9 countries. In how many ways canthey be seated if 2 particular delegates must not sit next to each other?

Solution:Solution:Solution:Solution:Solution: The number of ways in which 2 particular delegates must not sit next to each other = Totalnumber of permutations — Number of permutations in which 2 particular delegates sit next to eachother.

Now Total number of circular permutations of 9 delegates = 9 1 8− =

Considering 2 delegates as one fixed person, the remaining 7 delegates can be arranged in 7 ways.

2 delegates again can be arranged in 2 ways.

∴ Number of permutations in which 2 particular delegates sit next to each other = 7 × 2

∴ Required number of permutation

= − × − × − = =8 7 7 7 7 8 2 6 7 302402 = 8 2 = � �5.5.5.5.5. In how many ways can 6 persons sit around a table so that all shall not have the same neighbours

in any 2 arrangements?

Solution:Solution:Solution:Solution:Solution: 6 persons can sit round a table in 5 = 120 ways. But each person will have the same

neighbours in clockwise and anticlockwise arrangements.


∴ Required number of ways = 5

2

120

260= = .

6.6.6.6.6. In how many ways can 4 gentlemen and 4 ladies sit down together at a round table so that notwo ladies may come together.

Solution:Solution:Solution:Solution:Solution: Let the gentlemen first take up their seats. They can sit in 3 = 6 ways. When they have

been seated, there remain 4 places for the ladies each between 2 gentlemen. Therefore the 4 ladies can

sit in 4 places in 4 = 24 ways.

∴ Required number of ways = 3 4×

= 6 × 24 = 144.

��. ��

Each of the different groups or selection which can be made by taking some or all of a number of thingsat a time (irrespective of the order) is called a combination.

The number of ways of selection of n different things taken r at a time is called the number ofcombination of n different things taken r at a time. It is written as ncr.

��.�� #�&��'��/��

The number of combinations of n different things taken r at a time can be arranged in r ways.

∴ ncr combinations will produce nrc r× permutations.

Now nrc r× = Number of permutations of n different things taken r at a time = npr.

∴ nr

nrc r p× =

⇒n

r

nrc

p

r=

Butn

rpn

n r=

−

∴ nrc

n

n r r=

− ⋅.

Note:Note:Note:Note:Note: (1) When r = n.

1. When r = n.

nnc

n

n n n

n

n=

− ⋅=

⋅=

01

� 0 1=

∴ =nnc 1


2. When r = 0

nr

nc cn

n= =

− ⋅=0 0 0

1

∴ nc0 1=

3. When r = 1.

ncn

n

n n

nn1 1 1

1

1=

− ⋅=

−−

=

∴ nc1 = n.

��.�� 0&��1��2�

1. Prove that ncr = ncn−r

Proof:Proof:Proof:Proof:Proof: 1. By using formula for ncr:

LHS: n

rcn

n r r=

− ⋅ ...(1)

RHS:n

n rcn

n n r n r

n

r n r− =− + ⋅ −

=⋅ −

nn rc

n

n r r− =− ⋅ ...(2)

From (1) and (2)

nr

nn rc c= −

Proof:Proof:Proof:Proof:Proof: By analytic method:ncr is selecting r things from n things. If we select r things from n things then (n − r) things are left.

∴ For every combination of (n − r) things, there corresponds a combination of r things.

∴ ncr = ncn − r. Hence proved.

2.2.2.2.2. Prove that ncr + ncr − 1 =n + 1cr

Proof:Proof:Proof:Proof:Proof: By using formula for ncr

LHS: ncr + ncn − r

=− ⋅

+− − ⋅ −

n

n r r

n

n r r1 1� �

=− ⋅ ⋅ −

+− + ⋅ −

n

n r r r

n

n r r1 1 1

=− ⋅ ⋅ −

+− + ⋅ − ⋅ −

n

n r r r

n

n r n r r1 1 1� �


=− ⋅ −

+− +

��

�

n

n r r r n r1

1 1

1

=− ⋅ −

− + +− +

�

��

�

n

n r r

n r r

r n r1

1

1� �

=⋅ +

− + ⋅=

++ − ⋅

n n

n r r

n

n r r

1

1

1

1

� �

= n + 1cr = RHS.

Proof by analytic method:Proof by analytic method:Proof by analytic method:Proof by analytic method:Proof by analytic method: n + 1cr is the total number of combination of n + 1 things taken r at a timewhich is nothing but the combination that contain a particular thing (ncr) plus the combination that donot contain a particular thing (ncr − 1).

i.e., nr

nr

nrc c c+−= +1

1

(��)��*��"�

1.1.1.1.1. Find the value of (i) 7c3 (ii) 5c2 + 5c1

(i) We have

nrc

n

n r r=

− ⋅

∴7

37

7 3 3c =

− ⋅

=× × ×× × ×

=7 6 5 4

4 3 2 135.

(ii) 52

51

5 12

62c c c c+ = =+

�n

rn

rn

rc c c+ =−+

11

Now6

26

6 2 2

6 5 4

4 215c =

− ⋅=

× ×⋅

= .

OR

5

25

15

5 2 2

5

5 1 1c c+ =

− ⋅+

− ⋅

=× ×

×+

×⋅

5 4 3

3 2

5 4

4 1

= 10 + 5 = 15.


2.2.2.2.2. Find n if nc20 = nc6

Solution:Solution:Solution:Solution:Solution: In the formula ncr = ncn − r

We observe r + n − r = n

∴ n nc c20 6=

⇒ 20 + 6 = n

⇒ n = 26.

3.3.3.3.3. If nc7 = nc23, find nc29.

n nc c7 23= �n

rn

n rc c r n r n= ⇒ + − =−� �

⇒ n = 7 + 23n = 30

Now,n c c29

3029

30

30 29 29= =

− ⋅

= 30.

4.4.4.4.4. If 20cr + 2 = 20c2r − 6, then find r.

Given 202

202 6c cr r+ −=

⇒ r r+ = −2 2 6

2 6 2+ = −r r

⇒ r = 8

OR

202

202 6c cr r+ −=

⇒ r r+ + − =2 2 6 20n

rn

n rc c r n r n= ⇒ + − =−

3r − 4 = 20

3r = 24

⇒ r = 8.

5.5.5.5.5. If nc2 = 36, find n.

Solution:Solution:Solution:Solution:Solution:nc

n

n2 2 236=

− ⋅=

n n n

n

− ⋅ −− ⋅

=1 2

2 236

� �

⇒n n −

=1

236

� �


n (n − 1) = 72

n (n − 1) = 9 × 8 (By inspection)

⇒ n = 9.

6.6.6.6.6. If nc5 = 24 nc4, find n.

n nc c5 424= ⋅

n

n

n

n− ⋅= ⋅

− ⋅5 524

4 4

⇒n

n

− ⋅− ⋅

=4 4

5 524

n n

n

− ⋅ − ⋅− ⋅ ⋅

=4 5 4

5 5 424

� �

n − 4 = 120

n = 120 + 4

n = 124.

7.7.7.7.7. If npr = 60 and ncr = 10, then find n and r.

Solution:Solution:Solution:Solution:Solution: We have nr

nrc

p

r=

∴ 1060=r

⇒ r = =60

106

r = 3

⇒ r = 3.

Given: npr = 60np3 = 60

n

n −=

360

n n n n

n

− − −−

=1 2 3

360

� ��

n (n − 1) (n − 2) = 60

n (n − 1) (n − 2) = 5 × 4 × 3. (By inspection)

⇒ n = 5.

2235

6030155

1


8.8.8.8.8. If 18cr =18cr + 2, then find rc5.

Given 18cr = 18cr + 2

� r r≠ + 2

nr

nn rc c

r n r n

=⇒ + − =�

��

�

−

r r+ + =2 18

2 16r =

⇒ r = 8.

rc c58

58

8 5 5= =

− ⋅

=× × ×× × ×

=8 7 6 5

3 2 1 556.

9.9.9.9.9. In how many ways can 4 persons be selected from amongst 9 persons? How many times will aparticular person be always selected?

Solution:Solution:Solution:Solution:Solution: The number of ways in which 4 persons can be selected from amongst 9 persons

= =− ⋅

= × × ×× × ×

=94

9

9 4 4

9 8 7 6

4 3 2 1126c .

Let a particular person is selected always. Then we have to select 3 persons from the remaining 8

persons. This can be done in 8c3 ways = 8

8 3 3− ⋅ = 56.

10.10.10.10.10. A student has to answer 7 out of 10 questions in an examination. How many choices has he, ifhe must answer the first three questions.

Solution:Solution:Solution:Solution:Solution: There are 10 questions of which a student must answer first 3 questions. Remaining 4questions (� he has to answer 7 questions) can be selected among 10 − 3 = 7 questions in 7c4 ways.

∴ Number of combinations = =− ⋅

74

7

7 4 4c

=× × ×× × ×

7 6 5 4

3 2 1 4

= 35 ways.

11.11.11.11.11. In how many ways 5 red and 4 green balls can be drawn from a bag containing 7 red and 8 greenballs.

Solution:Solution:Solution:Solution:Solution: Number of ways of drawing 5 red balls from 7 red balls = =− ⋅

75

7

7 5 5c

=× ×

×=

7 6 5

2 521.


Number of ways of drawing 4 green balls from 8 green balls = =− ⋅

84

8

8 4 4c

= × × ×× × ×

=8 7 6 5

4 3 2 170.

∴ Total number of ways (By fundamental principle)

= 70 × 21

= 1470.

12.12.12.12.12. Find the number of (a) Straight lines (b) triangles that can be drawn from 20 points of which4 are collinear.

Solution:Solution:Solution:Solution:Solution: Two points are needed for a straight line. If none of the 20 points are collinear then we wouldget 20c2 straight lines. But 4 points are given to be collinear. So we would not get 4c2 lines, instead weget only one straight line containing all the 4 points.

∴ Number of straight lines = 202

42 1c c− +

=− ⋅

−− ⋅

+20

20 2 2

4

4 2 21

= × − × +20 19

2

4 3

21

= 190 − 6 + 1 = 185.

(b) We need 3 non-collinear points for a straight line. If none of the 20 points are collinear then wewould get 20c3 triangles. Since 4 points are given to be collinear, we would not get 4c3 trianglesfrom these points.

∴ Number of triangles = 20c3 − 4c3

=− ⋅

−− ⋅

20

20 3 3

4

4 3 3

= 1140 − 4 = 1136.

13.13.13.13.13. A committee of 10 members is to be chosen from 9 teachers and 6 students. In how many waysthis can be done if

(i) The committee contains exactly 4 students.

(ii) There is to be a majority of teachers.

(iii) There are atleast 4 students.

(iv) There are at most 7 teachers.


(i) The committee contains exactly 4 students and 10 − 4 = 6 teachers.

4 students can be selected out of 6 students in 6c4 ways and 6 teachers out of 9 teachers can beselected in 9c6 ways.


∴ The number of selections =6c4 × 9c6 (By fundamental principle)

=− ⋅

×− ⋅

6

6 4 4

9

9 6 6

=× ×× ×

×× × ×× × ×

6 5 4

2 1 4

9 8 7 6

6 3 2 1

= 15 × 84 = 1260.

(ii) As there is to be majority of teachers, the committee may consist of

(a) 6 teachers 4 students

(b) 7 teachers 3 students

(c) 8 teachers 2 students

(d) 9 teachers 1 student.

∴ Number of selection =

(a) 96

64 1260c c× =

(b) 97

63 720c c× =

(c) 98

62 135c c× =

(d) 99

61 6c c× =

∴ Total number of selections = 1260 + 720 + 135 + 6

= 2121.

(iii) As there is to be at least 4 students, the committee may consist of

(a) 4 students 6 teachers

(b) 5 students 5 teachers

(c) 6 students 4 teachers.

Number of selections =

(a) 64

96 1260c c× =

(b) 65

95 756c c× =

(c) 66

94 126c c× =

So Total number of selections =

1260 + 756 + 126 = 2142.

(iv) As there is to be atmost 7 teachers, the committee may consist of

(a) 7 teachers and 3 students

(b) 6 teachers and 4 students

(c) 5 teachers and 5 students

(d) 4 teachers and 6 students.


∴ Number of selections =

(a) 97

63 720c c× =

(b) 96

64 1260c c× =

(c) 95

65 756c c× =

(d) 94

66 126c c× =

∴ Total number of selections

= 720 + 1260 + 756 + 126 = 2862.

14.14.14.14.14. A team of eleven is to be chosen out of 16 cricketers of whom 4 are bowlers and 2 others arewicket keepers. In how many ways can the team be chosen so that there are at least 3 bowlersand at least one wicket keeper.

"&��

4 Bowlers 2 Wicket keepers 10 others

(a) 3 1 7

(b) 4 1 6

(c) 3 2 6

(d) 4 2 5

Number of ways

(a) 43

21

107 960c c c× × =

(b) 44

21

106 420c c c× × =

(c) 43

22

106 840c c c× × =

(d) 44

22

105 252c c c× × =

∴ Total number of ways = 960 + 420 + 840 + 252

= 2472.

15.15.15.15.15. Arun has 7 friends, 4 of them are boys and 3 are girls. His sister, Aalekya has 7 friends, 4 of themare girls and 3 of them are boys. In how many ways can they invite for a party of 3 girls and3 boys. So that there are 3 of Arun’s friends and 3 of Aalekya’s friends.



Arun's friends Aalekya's friends

4 boys 3 girls 3 boys 4 girls

(a) 3 – – 3

(b) 2 1 1 2

(c) 1 2 2 1

(d) – 3 3 –

∴ Number of selections:

(a) 43

30

30

43 16c c c c× × × =

(b) 42

31

31

42 324c c c c× × × =

(c) 41

32

32

41 144c c c c× × × =

(d) 40

33

33

40 1c c c c× × × =

∴ Total number of selections = 16 + 324 + 144 + 1

= 485.

16.16.16.16.16. How many diagonals are there in a octagon?

Solution:Solution:Solution:Solution:Solution: Number of diagonals in octagon = 82 8c − [� octagon has 8 sides]

= 20.

��

•n

rpn

n r=

−

• n p0 1= , • nnp n= • n p n1 =

• Permutation of n objects of which p objects are of one kind, q are of another kind and so on is

n

p q⋅ ....

• Number of circular arrangement of n persons round a table = n − 1

• Number of circular arrangement of n beads or n flowers to form a necklace or garland = n − 1

2.

•n

rcn

n r r=

− ⋅


• nnc = 1

• nc n1 =

• nc0 1=

• nr

nn rc c= −

• nr

nr

nrc c c+ =−

+1

1

• Number of straight lines that can be drawn from n points of which p points are collinear =nc2 − pc2 + 1.

• Number of triangles that can be drawn from n points of which p points are collinear = nc3 − pc3.

• Number of diagonals in a polygon of n sides = nc2 − n.

�*��"�

I. Find the value of:

1. 10p3 2. 12p3 3. 15c8 4. 8c5 5. 14c10

II. Find n if

1. np2 = 90 2. nc3 = 20

III. Find r if

1. 6pr = 360 2. 13pr = 156

IV. Find n and r if

1. npr = 240 and ncr = 120 2. npr = 336 and ncr = 56

V.

1. If np4 = 12, np2 = 120, find n.

2. If np4 = 56, np2 = 120, find n.

VI.

1. How many 3 digit numbers can be formed by using the digits 9, 7, 6, 5, 3, 2 (repetitions notallowed)?

(a) How many of these are less than 400?

(b) How many of these are multiples of 5?

(c) How many of these are multiples of 2?

2. In how many ways can the letters of the word ‘STRANGE’ be arranged so that

(a) The vowels never come together.

(b) The vowels are never separated.

3. A shelf contains 6 Hindi books, 5 Kannada books and 8 English books. In how many ways canthey be arranged so that

(a) Hindi books are together?

(b) Hindi books are together and Kannada books are together.


(c) Books of the same languages are together.

(d) No two English books are together.

4. How many arrangements of the letters of the word SUNDAY can be made if the vowels are toappear only in the odd places.

5. How many numbers of four different digits can be formed using the digits 0, 1, 2, 3, 4, 5? Howmany of them are even?

6. If the letters of the word GATE be permuted and the words so formed be arranged as in adictionary what will be the rank of the word?

7. Find the number of permutations of the letters of the word INSTITUTION, when all the lettersare taken at a time. How many of them (i) Have 3T’s together (ii) begin with 2N’s.

8. Find the number of permutations of the letters of the word ASSASSINATION. How many ofthem (i) have 3 A’s together (ii) begin with 2 N’s.

9. Find the number of permutation of letters of the word TOMORROW. How many of them have(i) 3O’s together. (ii) End with 2R’s.

10. Find the number of ways in which 6 different beads can be arranged to form a necklace.

11. In how many ways can 5 persons sit around a table.

12. In how many ways can 4 boys and 4 girls be seated round a table so that no two boys are adjacent.

13. In how many ways can 7 persons sit around a table so that all shall not have the same neighboursin any two arrangements?

14. A round table conference is to be held between delegates of 20 countries. In how many ways canthey be seated if 2 participants may wish to sit together always.

15. (i) If nc10 = nc6, find n. (ii) If 43cr − 6 = 43c3r + 1, then find r.

16. From 8 lecturers and 4 students a committee of 6 is to be formed. In how many ways can thisbe done so that the committee contains (i) exactly 2 students (ii) atleast 2 students.

17. How many (i) straight lines (ii) triangles are determined by 12 points, no three of which lie onthe same straight line.

18. How many (i) straight lines (ii) triangles are determined by joining 20 points in a plane of which6 are collinear.

19. In how many ways a student can choose 8 questions from a set of 12 questions if the questions1 and 10 are compulsory.

20. Find the total number of diagonals of a hexagon.

21. Out of 3 books on maths, 4 books on Physics and 5 books on Chemistry, how many collectionscan be made, if each collection consists of

(i) exactly one book on each subject

(ii) at least one book on each subject.

��"(��"

I. 1. 720 2. 132 3. 6435 4. 56 5. 1001

II. 1. n = 10 2. n = 6.


III. 1. r = 4 r = 2

IV. 1. n =16, r = 2 2. n = 8, r = 3

V. 1. n = 4 2. n = 8

VI. 1. 120 (a) 40 (b) 20 (c) 40

2. (a) 3600 (b) 1440

3. (a) 13 6 (b) 10 6 5⋅ ⋅ (c) 3 6 5 8⋅ ⋅ ⋅ (d) 11 128⋅ p

4. 144

5. 300, 156

6. 14

7.11

3 3 2⋅ ⋅ (i) 30240 (ii) 10080

8.12

144(i)

10

24(ii)

10

72

9. 3360 (i) 360 (ii) 120

10. 60

11. 24

12. 144

13. 360

14. 2 18×

15. (i) 16 (ii) 12

16. (i) 420 (ii) 672

17. (i) 12c2 (ii) 12c3

18. (i) 176 (ii) 1120

19. 210

20. 9

21. (i) 60 (ii) 3255.


�

��

��

The term probability refers to the chance of happening or not happening of an event. The theory ofprobability provides a numerical measure of the elements of uncertainity. It enables us to take decisionunder conditions of uncertainity with a calculated risk. The theory of probability has its origin in thegames of chance, related to gambling for instance throwing a dice or tossing a coin.

Generally speaking, the probability of an event denotes the likelihood of its happening. The value ofprobability ranges between zero and one. If an event is certain to happen its probability would be 1 andif it is certain that the event wouldn’t take place, then the probability of its happening is zero. Ordinarilyin social sciences probability of the happening of an event is rarely 1 or 0. The reason is that in socialsciences we deal with situation where there is always an element of uncertainity about the happeningor not happening of an event.

��

Before we give definition of probability, it is necessary that we familiarise ourselves with certain termsthat are used in this context.

(i) Random eRandom eRandom eRandom eRandom experxperxperxperxperiment:iment:iment:iment:iment: It is an experiment which if conducted repeatedly under homogeneousconditions doesn’t give the same result. The result may be any one of the various possible out-comes.

For example: If a die is thrown it wouldn’t always fall with number 3 up. It would fall in any oneof six ways which are possible.

(ii) TTTTTrrrrrial and eial and eial and eial and eial and evvvvvent:ent:ent:ent:ent: The performance of a random experiment is called a trial and the outcome – anevent.

Event could be either simple or compound (or composite). An event is called simple if it corre-sponds to a single possible outcome. Thus in tossing a die, the chance of getting 3 is a simpleevent (� 3 occurs in a die only once). However the chance of getting an odd number is compound(� odd numbers are more than one — 1, 3 and 5).

(iii) ExhaustiExhaustiExhaustiExhaustiExhaustivvvvve cases:e cases:e cases:e cases:e cases: All possible outcomes of an event are known as exhaustive cases. In the throwof a single die the exhaustive cases are six, as the die has only 6 faces each marked with different

Probability 47

numbers. Similarly the number of exhaustive cases in tossing 2 coins would be 4: HH, HT ,THand TT (H-Head, T-tail).

(iv) FFFFFaaaaavvvvvourourourourouraaaaabbbbble cases:le cases:le cases:le cases:le cases: The number of outcomes which result in the happening of a desired event arecalled favourable cases. Thus in a single throw of a die the number of favourable cases of gettingan odd number are 3 (i.e. 1, 3 and 5).

(v) MutuallMutuallMutuallMutuallMutually ey ey ey ey excxcxcxcxclusilusilusilusilusivvvvve cases:e cases:e cases:e cases:e cases: Two or more cases are said to be mutually exclusive if the happeningof any one of them excludes the happening of all others in a single experiment. Thus in a throwof a single die, the events 5, 4 and 3 are mutually exclusive.

(vi) EquallEquallEquallEquallEqually liky liky liky liky likelelelelely cases:y cases:y cases:y cases:y cases: Two or more events are said to be equally likely if the chance of theirhappening is equal, i.e., there is no preference of any one event over the other. Thus in the throwof a die, the coming up of 1, 2, 3, 4, 5 or 6 is equally likely.

(vii) IndeIndeIndeIndeIndependent and dependent and dependent and dependent and dependent and dependent ependent ependent ependent ependent evvvvvents:ents:ents:ents:ents: An event is said to be independent if its happening is notaffected by the happening of the other events. So in the throw of a die repeatedly coming up of5 on the first-throw is independent of coming up of 5 again in the second throw. However we aresuccessively drawing cards from a pack without replacement, the event would be dependent.

��

We can define probability in 3 ways.

(i) Mathematical or classical definition.

(ii) Statistical or Empirical definition.

(iii) Subjective approach or set theoretic approach definition.

�� !"��#��#��$$�#��%!&�'��'�

If there are ‘n’ mutually exclusive, exhaustive and equally likely simple events in a trial, ‘m’ of themare favourable to the occurrence of an event A. Then probability or chance of occurrence of A equal to

P A� � =Number of favourable cases

Total number of all possibleequally likely cases

P Am

n� � = .

Note:Note:Note:Note:Note:

1. Since we have 0 0 1≤ ≤ ≤ ≤m nm

n.

or 0 1≤ ≤P A� � .

2. When m = n, P (A) = 1 and when m = 0, P (A) = 0, i.e. when m = n, the event A is certain andwhen m = 0, the occurrence of event A is an absolute impossibility.

3. The probability of non-occurrence of A is denoted by


P A� � =Number of unfavourable cases

Total number of all possible cases.

P An m

n� � =

−

= −n

n

m

n

= − = −1 1m

nP A� �

∴ P A P A� � � �= −1

⇒ P A P A� � � �+ = 1.

4. The main disadvantage of mathematical method is that it fails when there are infinite number ofpossible outcomes and it cannot be applied to trials where the outcomes are not equally likely.

��(��$�#��")��#��%!&�'��'��&�)��

If a random experiment is repeated for an indefinitely large number of times under identical conditions,then the limiting value of the ratio of the number of times an event occur to the total number of trialsis said to be the probability of occurrence of the event, provided the limit is a definite finite number.

If T is the number of trials and event A occurs f times in ‘T’ trials, then the probability of occurrence

of event A is given by probability = P Af

TT� � = �

��→∞

lim .

We use this method when the elementary events are not equally likely and the exhaustive number ofcases in a trial is infinite. The limitation of this method is that in practice an identical experimentalcondition doesn’t exist while repeating a random experiment for a large number of times. Moreover the

relative frequency i.e.f

T may not attain a unique limiting value when T → ∞.

��(*�+!#�,!�)��$!� !��!�#��))��# �

A set of points representing all possible elementary outcomes of a random experiment is called thesample space (S). The number of all possible sample points in the sample space S is represented by n(S).

The definition of probability is based on the following assumptions. (i) Total number of elementaryevents in the sample space (S) is finite say N (ii) N elementary events of the experiment are equallylikely.

P AA

S� � =

Number of elementary events favourable to event

Total number of equally elementary events in .

Probability 49

i.e., P An A

n S

A

S� �

� �

� �= = Number of sample points in .

Total number of sample points in .

Note:Note:Note:Note:Note: A pack of cards contain 52 cards, 26 red cards and 26 black cards.

Among 26 black cards, 13 are calavar and 13 are spade. Among 26 red cards, 13 are Diamond and13 are hearts.

Each symbol contain A, Q, K, J, 2, 3, 4, 5, 6, 7, 8, 9, 10. (13 cards)

So there are 4 A’s, 4 Q’s, ... 4-10’s.

A is also known as Ace.

Face cards are A, Q, K and J.

-��.��/��(�

1.1.1.1.1. If one card is drawn at random from a well shuffled pack of 52 cards. Then find the probabilityof each of the following.

(a) Drawing an ace card,

(b) Drawing a face card,

(c) Drawing a diamond card,

(d) Drawing either spade or hearts,

(e) Not drawing an ace of hearts.

Solution:Solution:Solution:Solution:Solution: (a) One card can be drawn out of 52 cards in 52c1 = 52 ways = n (S).

One ace card can be drawn out of 4 ace cards in 4c1 = 4 ways = n (A).

∴Probability of drawing an ace card = Number of favourable cases

Total number of all possible equally likely cases= n A

n S

� �

� �

= =4

52

1

13.

(b) A face card can be drawn out of 12 face cards in 12c1 = 12 ways.

∴Number of favourable cases = 12

Total number of all possible equally likely cases = 52c1 = 52.

∴Probability of drawing a face card = =12

52

3

13.

(c) A diamond card can be drawn out of 13 diamond cards in 13c1 = 13 ways.

∴Probability of drawing a diamond card

= =13

52

1

4.

(d) There are 13 spade and 13 hearts cards in a pack of cards. Either a spade or a heart can be drawnin 26c1 = 26 ways.


∴Probability of drawing either a spade or a hearts card = =26

52

1

2.

(e) There is one ace of hearts.

∴ Probability of drawing an ace of hearts = 1

52.

∴ Probability of not drawing an ace of hearts = 11

52

51

52− = .

2.2.2.2.2. Three balls are drawn at random from a bag containing 6 blue and 4 red balls. What is theprobability that two balls are blue and one is red?

Solution:Solution:Solution:Solution:Solution: The bag contains (6 + 4) = 10 balls.

3 balls can be drawn out of 10 balls in 10c3 = 10

10 3 3− ⋅ = 120 ways.

∴Total number of cases = 120.

Now 2 blue balls can be drawn out of 6 in 6c2 = 6

6 2 2− ⋅ = 15 ways.

1 red ball can be drawn out of 4 in 4c1 = 4 ways.

∴2 blue balls and 1 red ball can be drawn in 15 × 4 = 60 ways.

∴Number of favourable cases for the event = 60.

∴Probability of drawing 2 blue balls and 1 red ball = 60

120

1

2= . .

3.3.3.3.3. Three unbiased coins are tossed. What is the probability of obtaining (a) all heads (b) two heads(c) one head (d) atleast one head (e) atleast two heads (f) All tails.

Solution:Solution:Solution:Solution:Solution: There are 23 = 8 mutually exclusive exhaustive and equally likely cases

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

(a) Probability of all heads = 1

8 [There is only one HHH amongst 8 possiblities]

(b) Probability of 2 heads = 3

8→

��

�

��

HHTHTHTHH

(c) Probability of 1 head = 3

8→

��

�

��

HTTTHTTTH

(d) Probability of atleast one head = 7

8

Probability 51

(e) Probability of atleast 2 heads = 4

8

1

2= →

��

�

��

HHHHHTHTHTHH

(f) Probability of all tails = 1

8.

4.4.4.4.4. The marks obtained by 100 students are given below.

Marks

No. of students0 10 11 20 21 30 31 40 41 50 51 60 61 70 71 80 81 90 91 100

5 10 13 14 17 7 11 8 9 6− − − − − − − − − −

If a student is selected at random from the entire group of 100 students, find the probability that hismarks (i) is under 40. (ii) above 50 (iii) either between 31 to 40 or 41-50.

(i) Total number of students = 100.

Number of students obtaining marks less than 40 = 5 + 10 + 13 + 14 = 42.

∴ Required probability = 42

100

21

50= .

(ii) Number of students scoring above 50

= 7 + 11 + 8 + 9 + 6 = 41


100

(iii) Number of students obtaining marks between 31 to 40 = 14 and number of students obtainingmarks between 41 to 50 = 17.

∴ Total number of students whose score is either between 31 to 40 or 41 to 50 = 14 + 17 = 31.


100

5.5.5.5.5. If a pair of dice is thrown, find the probability that the sum of digits is neither 7 nor 11.

Solution:Solution:Solution:Solution:Solution: A pair of dice is thrown.

∴ n (S) = 6 × 6 = 36.

Let A be an event of getting the sum 7 and B be an event of getting the sum 11.

Then A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

∴ n (A) = 6

B = {(5, 6) (6, 5)}

n (B) = 2.

Probability of getting 7 = n A

n S

� �

� �= 6

36.


Probability of getting 11 = n B

n S

� �

� �= 2

36.

Probability of getting either 7 or 11 = 6

36

2

36

8

36+ = .

∴ Probability of getting neither 7 nor 11 = 18

36

28

36

7

9− = = .

��0 ��

StaStaStaStaStatement:tement:tement:tement:tement: If A and B are 2 events, then probability that at least one of them occurs is given by

P A B P A P B P A B∪ = + − ∩� � � � � � � �

PrPrPrPrProof:oof:oof:oof:oof: Consider the Venn diagram. The shaded portion denotes A∪B, i.e., set of all outcomes wheresome of the outcomes are common to both A as well as B. So P (A∪B) is the probability of happeningof atleast one of the events A and B.

A B

U

A∩B

Fig. 3.1

P (A) + P (B) is the sum of all the probabilities in A and all the probabilities in B.

So the probability in A ∩ B has been added twice in P (A) as well as in P (B). So we must subtractP (A ∩ B) once from P (A) + P (B) to obtain probabilities in A∪B.

∴ P A B P A P B P A B∪ = + − ∩� � � � � � � �

CorCorCorCorCorollarollarollarollarollary:y:y:y:y: If A and B are mutually exclusive, then P (A∪B) = P (A) + P (B)

PrPrPrPrProof:oof:oof:oof:oof: From addition rule we have

P A B P A P B P A B∪ = + − ∩� � � � � � � �

Since A and B are mutually exclusive, A and B are disjoint sets.

∴ A B∩ = φ

⇒ P A B∩ =� � 0

∴ P A B P A P B∪ = + −� � � � � � 0

⇒ P A B P A P B∪ = +� � � � � �

Probability 53

Note:Note:Note:Note:Note: 1. P (A∪B) means P (A or B) i.e., probability of happening of atleast one of the events A and B.P (A ∩ B) means P (A and B) i.e., probability of happening of both the events A and B.

2. P (A∪B∪C) is probability of happening of atleast one of the events A, B and C. It is given byaddition rule as

P A B C P A P B P C P A B∪ ∪ = + + − ∩� � � � � � � � � �

− ∩ − ∩ + ∩ ∩P B C P C A P A B C� � � � � �

This can be proved by writing the Venn diagram.

A B

U

C

Fig. 3.2

-��.��/��(�

1.1.1.1.1. A ticket is drawn from a bag containing 25 tickets bearing number 1, 2, 3, ..., 24, 25. Find theprobability of its bearing a number which is either even or a multiple of 3.

Solution:Solution:Solution:Solution:Solution: The events ‘even number’ and ‘a multiple of 3’ are not mutually exclusive as there are somenumbers which are even as well as multiples of 3. Ex: 6, 12, 24.

∴ P (an even number or a multiple of 3)

= P (an even number) + P (a multiple of 3) − P (an even number and a multiple of 3)

P P P2 4 6 8 24 3 6 9 24 6 12 24, , , , ..., , , , ..., , ,� �� + −

= + − =12

25

8

25

3

25

17

25.

[Since there are 12 even numbers from 1 to 25, 8 multiples of 3 and 3 numbers which are even aswell as multiples of 3 from 1 to 25].

2.2.2.2.2. What is the probability of getting either total of 7 or 11 when a pair of dice is tossed?

Solution:Solution:Solution:Solution:Solution: Total outcomes when a pair of dice is tossed = 6 × 6 =36.

The events ‘a total of 7’ and ‘a total of 11’ are mutually exclusive events.

∴ P (a total of 7 or 11)

= P (a total of 7) + P (a total of 11).


= P {(6, 1), (5, 2), (4, 3) (3, 4) (2, 5) (1, 6)} + P {(6, 5) (5, 6)}

= + = =6

36

2

36

8

36

2

9.

3. The probability that a contractor will get a plumbing contract is 2/3 and the probability that hewill not get an electric contract is 5/9. If the probability of getting atleast one contract is 4/5, whatis the probability that he will get both the contracts.

Solution:Solution:Solution:Solution:Solution: Let A be an event that a contractor gets plumbing contract.

B be an event that a contractor gets electrical contract.

Then Given:

P A� � = 2

3

P B� � = 5

9.

P B P B� � � �= − = − =1 15

9

4

9.

P A B∪ =� �4

5

P A B∩ =� � ?

From addition rule,

P A B P A P B P A B∪ = + − ∩� � � � � � � �

4

5

2

3

4

9= + − ∩P A B� �

⇒ P A B∩ = + −� �2

3

4

9

4

5

= + −30 20 36

45

P A B∩ =� �14

45.

∴ Probability that a contractor will get both the contracts = 14

45.

4.4.4.4.4. One card is drawn from a pack of 52 cards, what is the probability that the card drawn is neitherred nor king.

Solution:Solution:Solution:Solution:Solution: The event ‘card drawn is red’ and ‘card drawn is king’ is not mutually exclusive becausethere are two cards in the pack which are red as well as king.

∴ P (card drawn is red or king)

Probability 55

= P (card drawn is red) + P (card drawn is king) − P (card drawn is red and king)

Since there are 26 red cards, 4 king cards and 2 cards which are red as well as king,

P (Cards drawn is red or king),

= + − =26

52

4

52

2

52

28

52.

∴ P (card drawn is neither red not king)

= 1 − P (card drawn is red or king)

= − = − =128

52

52 28

52

24

52.

= 6

13.

5.5.5.5.5. A card is drawn at random from a well shuffled pack of 52 cards. What is the probability that itis a heart or a queen or black card.

Solution:Solution:Solution:Solution:Solution: Let A be an event that ‘the card drawn is heart’.

B be an event that ‘the card drawn is queen’ and C be an event that ‘the card drawn is black’.

A, B and C are not mutually exclusive events. So

P A B C P A P B P C P A B∪ ∪ = + + − ∩� � � � � � � � � �

− ∩ − ∩ + ∩ ∩P B C P C A P A B C� � � � � �.

There are 13 heart cards,

4 queen cards,

26 black cards,

1 heart queen card

2 queen black cards and

No heart black card.

No card which is heart, queen and black.

∴ P A B C∪ ∪ = + + − − − +� �13

52

4

52

26

52

1

52

2

520 0

P A B C∪ ∪ = + + − −� �

13 4 26 1 2

52

= =40

52

10

13.

∴ P (Card drawn is heart or queen or black)

= 10

13.


This can be illustrated by the following Venn diagram.

12Hearts 1 1

224

Black

12 othercardsQueen

Fig. 3.3

��1 ��

Let us consider the following example. A fair dice is thrown and the number that appeared is even.What is the probability that the number 4 has appeared? Since it is given that ‘the number appeared iseven’. Possible outcomes are no longer {1, 2, 3, 4, 5, 6} but only {2, 4, 6}. Out of these 3 possibleoutcomes there is 1 outcome in favour of appearance of 4. ∴ Probability of appearance of 4, given that

an even number has appeared = 1

3.

Formally, if E is the event ‘The number that appeared is 4’ and F the event ‘Number that appearedis even’ then P (E/F) denotes the probability of E given that F has happened.

∴ P E F� � =1

3.

The probability that an event B occurs subject to the condition that A has already occurred is calledthe conditional probability of occurrence of the event B. It is denoted by P (B/A).

��2 ��

StaStaStaStaStatement:tement:tement:tement:tement: The probability of the simultaneous occurrence of 2 events is the product of the probabilitythat one of the events will occur and the conditional probability that the other event will occur given thatthe first event has occurred.

If A and B are two events then P (A ∩ B) is the probability of their simultaneous occurrence and itis given by

P A B P A P B A∩ = ⋅� � � � � �

or P A B P B P A B∩ = ⋅� � � � � �

PrPrPrPrProof:oof:oof:oof:oof: Let A be any event with sample points n (A), i.e., P (A) > 0. If n (S) is the total number of samplepoints in S and B = another event such that A and B are not disjoint sets. (i.e., A∩B ≠ φ). Let the sample

points in A∩B be n (A∩B). Then from definition P An A

n S� �

� �

� �=

Probability 57

P A Bn A B

n S∩ = ∩

� ��

� �

S

A BA

B∩

Fig. 3.4

Now Let P (B/A) = Conditional probability of event B given A has occurred =

= ∩ = ∩ = ∩n A B

n A

n A B n S

n A n S

P A B

P A

� �

� �

� � � �

� � � �

� �

� �

i.e., P B AP A B

P A� �

� �

� �=

∩

⇒ P A B P A P B A∩ = ⋅� � � � � �

Similarly we can establish

P A B P B P A B∩ = ⋅� � � � � �

��!�'��,!��&�

If A and B are 2 mutually dependent events then conditional probability of B when A has occurred isproportional to P (A∩B).

i.e, P B A P A B� � � �∝ ∩

P B A K P A B� � � �= ⋅ ∩ where K is the constant of proportionality

To find K,

Conditional probability of A when A has occurred is 1

∴ P A A KP A A� � � �= ∩

1 = KP A� �

⇒ KP A

= 1

� �


Substituting KP A

P B A KP A B= = ∩1

� �� in

We get

P B AP A

P A B� ��

� �= ⋅ ∩1

⇒ P A B P A P B A∩ = ⋅� � � � � �

CorCorCorCorCorollarollarollarollarollary:y:y:y:y: If A and B are independent events then P (A ∩ B) = P (A) ⋅ P (B).

PrPrPrPrProof:oof:oof:oof:oof: If A and B are independent events then P (B/A) = P (B) and P (A/B) = P (A).∴ By multiplication theorem,

P A B P A P B A∩ = ⋅� � � � � �

Substituting P (B/A) = P (B) we get

P A B P A P B∩ = ⋅� � � � � �

Hence proved.

Note:Note:Note:Note:Note: 1. The multiplication theorem can be extended. For 3 dependent events A, B and C

P A B C P A P B A P C A B∩ ∩ = ⋅ ⋅ ∩� � � � � � � �

2. If there are n independent events A1, A2, .... An,

P A A A P A P A P A P An n1 2 1 2 3∩ ∩ = ⋅ ⋅... ... .� � � � � � � � � �

-��.��/��(�

1.1.1.1.1. A pair of dice is thrown and sum of the numbers on the two dice comes to be 7. What is theprobability that the number 4 has come on one of the dice?

Solution:Solution:Solution:Solution:Solution: Let the events A and B be such that

Event B: Sum of numbers on the two dice is 7.

Event A: The number 4 has come.

Total outcomes when a pair of dice is thrown = 36 = n (S)

P B P� � � � � � � � � � � � � �� = 6 1 5 2 4 3 3 4 2 5 1 6, , , , , , , , , , ,

P B� � =6

36.

To get P (A∩B) select the outcomes favourable to A from the outcomes that are favourable to B.

∴ P A B∩ = =� � � � � �� 4 3 3 42

36, , , .

From Multiplication theorem,

P A B P A P A B∩ = ⋅� � � � � �

Probability 59

⇒ P A BP A B

P A� �

� �

� �= ∩

P A B� � = = =

2

366

36

2

6

1

3.

∴ Probability that the number 4 has come on one dice given that sum of numbers on 2 dice is 7 = 1

3.

2. Two cards are drawn from a pack of 52 cards with replacement (i.e., the second card is drawnafter replacing the first card in the pack). Find the probability that (a) Both are ace, (b) First cardis jack and second card is king, (c) One is king and other is queen.

Solution:Solution:Solution:Solution:Solution: (a) In a pack of 52 cards, there are 4 ace cards.

∴ P (both cards are ace)

= × =4

52

4

52

1

169.

(b) P (First card is jack and second card is king)

= × =4

52

4

52

1

169.

[� There are 4 jack and 4 king cards in a pack of 52 cards]

(c) P [One card is king and other is queen]

= P [First is king and 2nd is queen] ∪ P [First is queen and 2nd is king]

= × + ×4

52

4

52

4

52

4

52 [Using addition theorem P (A∪B) = P (A) + P (B)]

= + =1

169

1

169

2

169.

3. Two cards are drawn without replacement from a pack of 52 cards. What is the probability that

(i) both are queen.

(ii) both are diamond cards.

(iii) one is king and the other is ace.

Solution:Solution:Solution:Solution:Solution: We are taking 2 cards from 52 cards. This can be done in 52c2 ways. There are 4 queen cardsfrom which we require 2 queen cards. This can be done in 4c2 ways.

∴ Required Probability = 4

252

2

4 3

52 51

c

c= ×

×

= 1

221.


OR

There are 4 queen cards in a pack of 52 cards.

∴ Probability of drawing first queen card = 4

52

Since the card drawn is not replaced, we are left with 51 cards and 3 queen cards.

∴ Probability of drawing 2nd queen card = 3

51

∴ Probability of both queen cards = × =4

52

3

51

1

221.

(ii) Required probability =13

252

2

c

c

=

××××

=

13 122 1

52 512 1

1

17.

(Since there are 13 diamond cards).

(iii) There are 4 favourable choices to take out a king and 4 favourable choices to take out an ace.

∴ Number of favourable cases = 4c1 × 4c1

In total, there are 52 cards out of which any 2 cards can be taken.

∴ Required probability = ×41

41

522

c c

c

= × ××

=4 4 2

52 51

8

663.

4. A lot contains 10 items of which 3 are defective. 3 items are chosen from the lot at random oneafter another without replacement. Find the probability that all the 3 are defective.

Solution:Solution:Solution:Solution:Solution: Let A, B and C be the events of drawing defective items in the first, second and third drawingrespectively. Hence Probability of all the three items being defective is given by

P A B C P A P B A P C A B∩ ∩ = ⋅ ⋅ ∩� � � � � � � �

= × ×3

10

2

9

1

8

= 1

120.

OR

3 defective items can be picked from 3 defective items in 3c3 ways.

Probability 61

3 items can be picked from 10 items in 10c3 ways.


310

3

c

c

110 9 83 2 1

6

10 9 8× ×× ×

=× ×

= 1

120.

5. Anil and Bharath appear in an interview for 2 vacancies. The probability of their selection being

1

7

1

5 and respectively.

Find the probability that (i) both will be selected (ii) only one is selected (iii) none will be selected(iv) atleast one of them will be selected.

Solution:Solution:Solution:Solution:Solution: Let A: Anil be selected.

B: Bharath be selected.

Given P A P B� � � �= =1

7

1

5 and

P (both will be selected)

= = ∩P A B P A B and � � � �

= ⋅P A P B� � � � [� A and B are independent events]

= × =1

7

1

5

1

35.

(ii) P (Only one will be selected)

= P A B A B and or and

= ∩ ∪ ∩P A B A B� � � �

= ⋅ + ⋅P A P B P A P B� � � � � � � �

= ⋅ −��

��

+ −��

��

⋅1

71

1

51

1

7

1

5

= ⋅ + ⋅1

7

4

5

6

7

1

5

= + = =4

35

6

35

10

35

2

7.


(iii) P (none will be selected)

= P A B and � �

= ∩ = ⋅P A B P A P B� � � � � �

= −��

��

−��

��

11

71

1

5

= × =6

7

4

5

24

35.

(iv) P (at least one of them will be selected)

= 1 − P (none of them will be selected)

124

35

11

35− = .

��

• P An A

n SP A� �

� �

� �� = ≤ ≤0 1.

• P A P A� � � �= −1

• P (A∪B) means P (A or B) and P (A∩B) means P (A and B)

• Addition rule: P (A∪B) = P (A) + P (B) − P (A∩B)

• For mutually exclusive cases

P (A∪B) = P(A) + P(B)

• P A B C P A P B P C P A B P B C P C A P A B C∪ ∪ = + + − ∩ − ∩ − ∩ + ∩ ∩� � � � � � � � � � � � � � � �

• P A B P A P B A∩ = ⋅� � � � � �

• P A B P B P A B∩ = ⋅� � � � � �

• P A B P A P B A B∩ = ⋅� � � � � � if and are independent events.

• P A B C P A P B A P C A B∩ ∩ = ⋅ ⋅ ∩� � � � � � � �

• If A1, A2, A3, ... An are independent events P A A A P A P A P An n1 2 1 2∩ ∩ ∩ =... ...� � � � � � � �

�/�� (�

1. A bag contains 100 tickets each bearing a distinct number from 1 to 100. A ticket is drawn fromthe bag. Find the probability that

Probability 63

(a) the ticket bears an odd number.

(b) the ticket bears a number divisible by 3.

(c) the ticket bears a number divisible by 3 or 5.

2. A pair of dice is thrown. Find the probability that the sum will be (a) equal to 4 (b) less than 4(c) greater than 4.

3. A bag contains 4 white, 5 red and 6 green balls. 3 balls are drawn at random. What is theprobability that (a) All are green (b) All are white.

4. A card is drawn from a pack of playing cards. What is the probability of drawing

(a) black card (b) king card (c) diamond card.

5. A box has 3 silver and 2 gold coins. 2 coins are drawn at random. Find the probability that (a)both the coins are silver ones. (b) both are gold coins (c) one of them is a silver coin.

6. A problem in statistics is given to 3 students A, B and C. Their probabilities in solving it are

1

8

1

6, and

1

4 respectively. What is the probability that the problem would be solved?

7. The following table gives a distribution of wages of 1000 workers.

Wages Rs. 120

No. of workers

� � − − − − − − −140 141 160 161 180 181 200 201 220 221 240 241 260

9 118 478 200 142 35 18

An individual is selected at random from the above group. What is the probability that his wagesare

(a) under Rs. 160 (b) above Rs. 200 (c) between Rs. 160 and 200.

8. One card is drawn from a pack of 52 cards. What is the probability that the card drawn is

(a) either red or king

(b) either king or queen

9. A ticket is drawn from a bag containing tickets bearing numbers 1 to 25. Find the probability thatthe number is either even or a multiple of 3.

10. A coin and a die are thrown. What is the probability of getting a head or an even number.

11. A box contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If one itemis chosen at random, what is the probability that it is rusted or a bolt?

12. A die is rolled. If the outcome is an odd number, what is the probability that it is prime.

13. A pair of dice is rolled. If the sum on the 2 dice is 9. Find the probability that one of the diceshowed 3.

14. If A and B are 2 events in a sample space S such that P A P B P A B� � � � � �= = ∪ =1

2

5

8

3

4, , .

Find (a) P (A∩B) (b) P A B∩� � .

15. 3 children are randomly selected from a class. What is the probability that (a) all 3 were born onMonday (b) 2 were born or Friday and the other on Tuesday (c) none were born on Wednesday.


��(-��(

1. (a)1

2(b)

33

100(c)

47

100

2. (a)3

36(b)

3

36(c)

30

36

3. (a)4

91(b)

4

455

4. (a)1

2(b)

1

13(c)

1

4

5. (a)3

10(b)

1

10(c)

3

5

6.290

64

7. (a)127

1000(b)

195

1000(c)

678

1000

8. (a)28

52(b)

2

13

9.17

2510.

3

411.

5

8

12.2

313.

1

214. (a)

1

8(b)

1

4

�

��

��

We know

x a x a+ = +� �1

x a x ax a+ = + +� �2 2 22

x a x a ax x a+ = + + +� � � �3 3 3 3

= + + +x x a xa a3 2 2 33 3 .

The general expansion for (x + a)n where n is positive integer is given by Sir Isaac Newton and isknown as Binomial Theorem.

��

If x and a are real numbers and n is any positive integer then

x a x c x a c x a c x a c x an n n n n n n n nn

n n n+ = + + + ⋅ +− − − −� � 11

22 2

33 3 ... .

= + ⋅ +−

⋅ ⋅ +− −

⋅ + +− − −x nx an n

x an n n

x a an n n n n1 2 2 3 31

2

1 2

3

� � � �� ! !

... .

Since,n

rcn

n r r=

− ⋅

ncn

n

n n

nn1 1 1

1

1=

− ⋅=

⋅ −

−= .

n ncn n

cn n n

2 31

2

1 2

3=

−=

− −� � � �� !

,!


and so on ... ncn = 1.

By taking 1 = C0 , nc1 = C1, n n

n nc C c C2 2 1= = = and so on ... .

We can write,

x a C x C x a C x a C an n n nn

n+ = + + +− −� � 0 11 2

23 3 ... .

��

(1) x a x a+ = +� �1 : Power : 1, Number of terms : 2

x a x ax a+ = + +� �2 2 22 3: .Power : 2, Number of terms

x a x x a xa a+ = + + +� �3 3 2 2 33 3 . Power : 3, No. of terms : 4.

The number of terms in the expansion of (x + a)n is n + 1.

(2) In the expansion of (x + a)n, from left to right the power of x decreases by 1 from one term to thenext term and the power of a increases by 1.

(3) In any term, the power index of x plus the power index of a = n.

(4) The co-efficients C0, C1, C2 ... Cn are called binomial co-efficients.

(5) The general term in the expansion is nr

n r rc x a− ⋅ which is (r + 1)th term. Hence T c x arn

rn r r

+−= ⋅1 .

(6) Since nr

nn rc c= − , the binomial co-efficients of first and last term are equal: Second and penul-

timate term are equal and so on.

(7) If n is even then the number of terms in (x + a)n is (n + 1) which is odd. So there will be only one

middle term and it is T n2

1+��

. If n is odd then the number of terms in (x + a)n is (n + 1) which is

even. So there will be two middle terms. They are T Tn n+��

��

+ +��

��

12

12

1 and

!��"��#��$��

1.1.1.1.1. Expand: xy

+��

��

22

5

xy

x c xy

c xy

+��

��

= + ⋅ ⋅��

+ ⋅��

2 2 22

55 5

14

2

15

23

2

2

+��

+��

+��

53

22

35

41

2

45

50

2

52 2 2

c xy

c xy

c xy

Binomial Theorem 67

= + ⋅ + × ⋅ ⋅ + × × ⋅ ⋅ + × × × ⋅ ⋅��

+x xy

xy

xy

xy y

5 42

34

26 8 105

2 5 4

2

4 5 4 3

3

8 5 4 3 2

4

16 32

! !

= + + + + +xx

y

x

y

x

y

x

y y5

4

2

3

4

2

6 8 10

1040

80 80 32.

2.2.2.2.2. Expand a

b

b

a+�

��

4

.

a

b

b

a

a

b

a

b

b

a

a

b

b

a

a

b

b

a

b

a+�

�� = ��

�� + �

�� + × �

�� + × × �

�� + ��

��

4 4 3 2 2 3 4

44 3

2

4 3 2

3!

= �� + �

�� + + ⋅ ��

�� + ��

��

a

b

a

b

b

a

b

a

4 2 2 4

4 6 4 .

3.3.3.3.3. Expand xy

xy

24

2

41 1

−��

��

= + −��

��

��

= + −��

+ ⋅ −��

+ ⋅ ⋅ −��

+ −��

x xy

xy

xy y

2 4 2 3 2 22

23 4

41 4 3

2

1 4 3 2

3

1 1

!

= − + − +xx

y

x

y

x

y y8

6 4

2

2

3 4

4 6 4 1.

4.4.4.4.4. Simplify: 3 2 3 25 5

+ + −

Solution:Solution:Solution:Solution:Solution: From Binomial theorem we have

x a x x a x a x a xa a+ = + ⋅ + ⋅ ⋅ + × × ⋅ ⋅ + × × × +� �5 5 4 3 2 2 3 4 555 4

2

5 4 3

3

5 4 3 2

4! !.

x a x x a x a x a xa a+ = + + + + +� �5 5 4 3 2 2 3 4 55 10 10 5 . ...(1)

Changing a to −a we get

x a x x a x a x a xa a− = − + − + −� �5 5 4 3 2 2 3 4 55 10 10 5 . ...(2)

Adding (1) and (2) we get

x a x a x x a xa+ + − = + +� � � �5 5 5 3 2 42 20 10 .

= + +2 10 54 2 2 4x x x a a


Putting x = 3 and a = 2 we get

3 2 3 2 2 3 3 10 3 2 5 25 5 4 2 2 4

+ + − = ⋅ + ⋅ +��

��

= + +6 81 180 5 4� �

= + = +6 261 20 1566 120

= 1686.

5.5.5.5.5. Simplify: 1 2 1 24 4

+ − −

We have from Binomial theorem

x a x x a x a x a a+ = + + × + × × +� �4 4 3 2 2 3 444 3

2

4 3 2

3!

x a x x a x a x a a+ = + + + +� �4 4 3 2 2 3 44 6 4 ...(1)

Replacing a by −a.

x a x x a x a x a a− = − + − +� �4 4 3 2 2 3 44 6 4 ...(2)

Subtracting (2) from (1)

x a x a x a xa+ − − = +� � � �4 4 3 38 8

Putting x = 1 and a = 2 we get

1 2 1 2 8 1 2 8 1 24 4 3 3

+ − − = ⋅ ⋅ + ⋅ ⋅

= + ⋅8 2 8 2 2

= + =8 2 16 2 24 2.

6.6.6.6.6. Find the middle term in the expansion of a

xbx+�

��

8

.

Solution:Solution:Solution:Solution:Solution: The expansion has 8 + 1 = 9 terms.

So T8

21+ is the middle terms i.e., T5 is the middle term.

We have

T c x arn

rn r r

+−= ⋅ ⋅1 .

Comparing a

xbx+�

��

8

with (x + a)n,

Binomial Theorem 69

We get xa

x=

a bx=

n = 8

r + =1 5

r = 4Substituting in

T c x arn

rn r r

+−= ⋅ ⋅1

T ca

xbx5

84

8 45= �

�� ⋅

−

� �

= �� ⋅8

4

45c

a

xbx� �

= ⋅ ⋅84

4

45 5c

a

xb x

= 84

4 5 5

4ca b x

x

=− ⋅

⋅8

8 4 44 5a b x

=× × × × /

/⋅

8 7 6 5 4

4 44 5a b x

= × × ×× × ×

×8 7 6 5

4 3 2 14 5a b x

= ×2 35 4 5a b x

= 70 4 5a b x.

7.7.7.7.7. Find the middle term in the expansion of 211

xy

x+�

�� .

Solution. Solution. Solution. Solution. Solution. The expansion has 11 + 1 = 12 terms.

So there are 2 middle terms.

T Tn n+ + +1

2

1

21

and are middle terms i.e., T T T T12

2

12

21

6 7 and and +

= are middle terms.


Comparing 211

xy

xx a n+�

�� + with � � we get x = 2x, a

y

x= and n = 11.

For T r r6 1 6 5, .+ = ⇒ =

T c x ann

rn r r

+−= ⋅1 formula

T c xy

x611

511 5

5

2= ⋅ ⋅ ��

−� �

= ⋅ ⋅ ��

115

65

2c xy

x� �

=− ⋅

⋅ ⋅ ⋅11

11 5 526 6

5

5xy

x

T xy66 511 10 9 8 7

5 4 3 2 12= × × × ×

× × × ×⋅ ⋅

T xy6529568= ⋅ .

For, T r r7 1 7 6, + = ⇒ =

T c x arn

rn r r

+−= ⋅1 .

T c xy

x711

611 6

6

2= ⋅ ⋅ ��

−� �

T xy

x75 5

6

6

11

11 6 62=

− ⋅⋅ ⋅ ⋅

Ty

x75

611 10 9 8 7

5 4 3 2 12= × × × ×

× × × ×⋅ ⋅ .

Ty

x7

6

14784= ⋅ .

8.8.8.8.8. Find the middle term in the expansion of x

x2

12

9

−��

�� .

The expansion has 9 + 1 = 10 term.

So there are 2 middle terms T Tn n+ + +1

2

1

21

and are middle terms.

Binomial Theorem 71

T T10

2

10

21

and +

are middle terms.

T5 and T6 are middle terms.

Comparing x

x2

12

9

−��

�� with (x + a)n we get

xx

ax

n= = − =2

192, , .

For T5, r + 1 = 5 ⇒ r = 4.

T c x arn

rn r r

+−= ⋅ ⋅1 .

T cx

x59

4

9 4

2

4

2

1= ⋅ �� ⋅ −��

��

−.

Tx

x5

5

5 8

9 8 7 6

4 3 2 1 2

1= × × ×× × ×

⋅ ⋅

Tx

x5

5

8

126

32=

×

Tx

5 3

63

16= .

For T6, r + 1 = 6 ⇒ r = 5.

T cx

x69

5

9 5

2

5

2

1= ⋅ ��

�� ⋅ −��

��

−

= − ⋅ ⋅1262

14

4 10

x

x

= − = −126 63

86 6x x.

9.9.9.9.9. Find the middle term in the expansion of 33

210

xy−�

�� .

The expansion has 10 + 1 = 11 terms.

So T T10

21

6+= is the middle term.

Comparing 33

210

xy−�

�� with (x + a)n we get


x x= 3 2

ay= −3

n = 10

r + =1 6

r = 5

T c x arn

rn r r

+−= ⋅1 .

T c xy

610

52 10 5 5

33

= ⋅ ⋅ −��

−

T xy

65 10

5

5

10

5 53

3=

−⋅

⋅ ⋅ ⋅

T x y610 510 9 8 7 6

5 4 3 2 1= − × × × ×

× × × ×⋅ ⋅

T x y610 5252= − ⋅ ⋅ .

10.10.10.10.10. Find the middle term in the expansion of xx

−��

��

2 10

.

The expansion has 10 + 1 = 11 terms.

So T102

1+ is the middle term

T6 is the middle term.

Comparing xx

−��

��

2 10

with (x + a)n we get

n = 10,

x x=

ax

= − 2

r + 1 = 6

r = 5

T c x arn

rn r r

+−= ⋅ ⋅1 formula� �

T c xx6

105

10 552= ⋅ −��

−

Binomial Theorem 73

T c xx

610

5

52

5

5

2= ⋅ ⋅

−� �

= −− ⋅

⋅ ⋅−10

10 5 52

5

25

5x

= − × × × ×× × × ×

⋅ ⋅−10 9 8 7 6

5 4 3 2 125 2 5x

= − ⋅

= −

−

−

252 2

8064

5 2 5

5 2

x

x

11.11.11.11.11. Find the co-efficient of x2 in the expansion of 31

2

8

xx

+��

�� .

Comparing 31

2

8

xx

+��

�� with (x + a)n we get x = 3x, a

x= 1

2 and n = 8.

T c x arn

rn r r

+−= ⋅1 formula

T c xxr r

rr

+−= ⋅ ��

��1

8 831

2� �

T c xxr r

rr

+−= ⋅ ⋅ ��

��1

8 831

2� �

= ⋅ ⋅ ⋅⋅

− −8 8 831

2c x

xr

r rr r

= ⋅ ⋅ ⋅− − − −8 8 83 2c xrr r r r

T c xr rr r r

+− − −= ⋅ ⋅ ⋅1

8 8 8 23 2 .

To get the co-efficient of x2, equating power index of x to 2.

8 2 2− =r

8 2 2− = r

6 2= r

r = 3.

Substituting r = 3 in (1),

T c x3 18

38 3 3 8 2 33 2+

− − −= ⋅ ⋅ ⋅ � �


T x45 3 28

8 3 33 2=

− ⋅⋅ ⋅ ⋅−

T x4

5

328 7 6

3 2

3

2= × ×

×⋅ ⋅

T x4

5

3256

3

2= × × .

∴ Co-efficient of x2 is 35 × 7.

12.12.12.12.12. Find the co-efficient of x25 in the expansion of xx

43

151−��

�� .

Solution.Solution.Solution.Solution.Solution. Comparing xx

43

151−��

�� with (x + a)n we get x = x4, a

x= − 1

3 and n = 15.

T c x arn

rn r r

+−= ⋅1 formula

T c xxr r

rr

+−

= ⋅ ⋅ −��1

15 4 15

3

1

T c x xr rr r

+− −= ⋅ ⋅ − ⋅1

15 60 4 2 31� �

= ⋅ ⋅ −− −15 60 4 3 1c xrr r r� �

T c xr rr r

+−= ⋅ ⋅ −1

15 60 7 1� � ...(1)

To get the coefficient of x25, equating power index of x to 25.

60 7 25− =r

60 25 7− = r

35 7= r

⇒ r = 5.

Substituting r = 5 in (1)

T c x5 115

525 51+ = ⋅ ⋅ −� �

T c x615

525 51 1= − ⋅ − = −� � �

∴ Co-efficient of x25 = −15c5.

13.13.13.13.13. Find the co-efficient of x60 in 2323

25

xx

−��

�� .

Binomial Theorem 75

Comparing 2323

25

xx

−��

�� with (x + a)n we get x = 2x2, a

x= − 3

3 and n = 25.

T c x arn

rn r r

+−= ⋅ ⋅1 formula

T c xxr r

rr

+−

= ⋅ ⋅ −��1

25 2 25

323

= ⋅ ⋅ ⋅ − ⋅− − −25 25 50 2 32 3c x xrr r r r� �

= ⋅ ⋅ ⋅ −− − −25 25 50 2 32 3c xrr r r r� �

= ⋅ ⋅ ⋅ −− −25 25 50 52 3c xrr r r� �

To get the co-efficient of x60, equating power index of x to 60.

i.e. 50 5 60− =r

⇒ 5 50 60r = −

5 10r = −

r = −2.

Since r is negative, there is no x60 term in the expansion of 2323

25

xx

−��

�� .

14.14.14.14.14. Find the constant term in the expansion of 3

2

1

3

2 9x

x−

��

��

.

Comparing 3

2

1

3

2 9x

xx a n−

��

��

+with � � .

We get xx

ax

n= = − =3

2

1

39

2

, , .

T c x arn

rn r r

+−= ⋅1 formula

T cx

xr r

r r

+

−

=��

��

⋅ −��1

92 9

3

2

1

3

= ⋅ ⋅−

⋅

− −

−9

9 18 2

9

3

2

1

3c

x

xr

r r

r

r

r r

� �

= ⋅ ⋅ ⋅ − ⋅− − − − − +9 9 18 2 93 1 2c xrr r r r r r� � ...(1)


To get the constant term, equating power index of x to zero

∴ 18 2 0− − =r r

18 3 0− =r

⇒ 3 18r =

r = 6.

Substituting r = 6 in (1)

T c x6 19

69 6 6 0 6 9 63 1 2+

− − − += ⋅ ⋅ − ⋅� �

=− ⋅

⋅ ⋅ − ⋅− −9

9 6 63 1 23 6 3� �

9 8 7

3 2 1

1

27

1

8

7

18

× ×× ×

⋅ ⋅ = .

15.15.15.15.15. Find the constant term in the expansion of 21

23

xx x

−��

��

.

Comparing 21

23

xx x

x a n−��

��

+ with � � .

We get x x ax x

n= = − =21

23, . and

T c x arn

rn r r

+−= ⋅1

T c xx xr r

rr

+−

= ⋅ ⋅ −��1

23 232

1

= ⋅ ⋅ ⋅ ⋅ −−−

− −23 2323

2 22 1c x x xrr

rr

rr� �

= ⋅ ⋅ ⋅ −−− − −23 23

232 22 1c xr

rr

rr

r� �

= ⋅ ⋅ ⋅ −−− − −

23 2323 2

22 1c xrr

r r rr� �

= ⋅ ⋅ ⋅ −−−

23 2323 4

22 1c xrr

rr� � .

To get the constant term equating power index of x to zero.

23 4 0− =r

Binomial Theorem 77

⇒ 4r = 23

r = 23

4.

Since r is a fraction there is no term independent of x or there is no constant term.

16.16.16.16.16. Prove that the constant term in the expansion of x

x2

22

10

−��

��

is 45

64.

PrPrPrPrProof:oof:oof:oof:oof: Comparing

x

xx a n

2

22

10

−��

��

+ with � �

We get xx

ax

n= = − =2

2102, . and

T c x arn

rn r r

+−= ⋅1 formula

T cx

xr r

r r

+

−

= ⋅��

⋅ −��1

1010

22

2

= ⋅ ⋅ ⋅ − ⋅− − + −10 5

2 10 22 2c x xr

rr r r� �

T c xr r

rr r r

+− − − += ⋅ ⋅ ⋅ −1

10 52

2 102 2� � ...(1)

To get the constant term equating power index of x to zero.

52

2 0− − =rr

10 4

20

− − =r r

10 5 0− =r

⇒ 5 10 2r r= ⇒ = .

Substituting r = 2 in (1) we get

T c x2 110

20 10 2 2 22 1 2+

− += ⋅ ⋅ ⋅ − ⋅� �

= ××

⋅ ⋅ ⋅−10 9

2 12 1 48

= × ××

=10 9 4

2 2

45

648 .


Hence the constant term in the expansion of x

x2

2 45

642

10

−��

��

is .

17.17.17.17.17. Find the value of (0.99)5 correct to 4 decimal places.

Solution:Solution:Solution:Solution:Solution: We know

0 99 1 0 01. .= −

∴ 0 99 1 0 015 5. .� � � �= −

We have from Binomial theorem,

x a x x a x a x a xa a+ = + ⋅ + × ⋅ ⋅ + × × + × × × ⋅ +� �5 5 4 3 2 2 3 4 555 4

2

5 4 3

3

5 4 3 2

4! !

Replacing a by −a.

x a x x a x a x a xa a− = − + − + −� �5 5 4 3 2 2 3 4 55 10 10 5 .

Taking x = 1 and a = 0.01, we get

1 0 01 1 5 1 0 01 10 1 0 015 5 4 3 2− = − +. . .� � � � � � � � � �

− + −10 1 0 01 5 1 0 01 0 012 3 4 5� � � � � �� . . . .

= − + − + × − ×− −1 0 05 0 001 0 00001 5 10 1 108 10. . .

≈ 0 9414801. .

≈ 0 9415. .Correct to 4 decimal places� �

18.18.18.18.18. Prove that sum of Binomial co-efficients of order n = 2n. Also prove the sum of odd binomial co-efficients = sum of even Binomial co-efficients = 2n − 1.

PrPrPrPrProof:oof:oof:oof:oof: We have from Binomial theorem.

x a C x C x a C x a C an n n nn

n+ = + ⋅ + + +− −� � 0 11

22 2 ... ...(1)

To get the sum of binomial co-efficient,

Taking x = a = 1 we get

1 1 1 1 1 1 10 11

22+ = ⋅ + ⋅ + ⋅ ⋅ + +− −� �n n n n

nC C C C... .

2 0 1 2n

nC C C C= + + + +... .

i.e., C C C Cnn

0 1 2 2+ + + + =... . ...(2)

Now,

Taking x = 1 and a = − 1 in (1)

1 1 1 1 1 1 10 11

22 2− = + − + − + +− −� � � � � �n n n n nC C C C. ...

Binomial Theorem 79

0 0 1 2 3= − + − − +C C C C Cn... ...(3)

� − = − − = − = −1 1 1 1 1 11 2 3� � � � � �, , & so on.

Adding (2) and (3) we get

2 2 2 20 2 4n C C C= + + + ...

2 2 0 2n C C= + + ...

2

2 0 2 4

n

C C C= + + + ...

⇒ C C C n0 2 4

12+ + = −...

Hence sum of even binomial co-efficients = 2n − 1.

Subtracting (3) from (2) we get

2 2 2 21 3 5n C C C= + + + ...

2

2 1 3 5

n

C C C= + + + ...

⇒ C C C n1 3 5

12+ + + = −... .

Hence sum of odd binomial co-efficients = 2n − 1.

19.19.19.19.19. The 1st, 3rd and 5th term in the expansion of (a + b)n are respectively 32, 240 and 90. Find a, b andn.

Given: In the expansion of (a + b)n,

T1 32= , T T3 5240 90= = and .

We know

T c x arn

rn r r

+−= ⋅1 formula

Comparing a b x an n+ +� � � � with we get x = a, a = b and n = n.

For 1st term r + 1 = 1 ⇒ r = 0.

For 3rd term r + 1 = 3 ⇒ r = 2.

For 5th term r + 1 = 5 ⇒ r = 4.

T c a bn n1 0

0 0 32= ⋅ =− .

1 32⋅ =an

an = 32 ...(1)

T c a bn n3 2

2 2 240= ⋅ ⋅ =−


n na bn−

⋅ ⋅ =−1

22402 2� �

n n a bn− ⋅ ⋅ =−1 4802 2� � ...(2)

T c a bn n5 4

4 4 90= ⋅ ⋅ =−

n n n na bn− − − ⋅ ⋅ =−1 2 3

4904 4� ��

!

n n n n a bn− − − ⋅ ⋅ =−1 2 3 21604 4� �� ...(3)

From (1) an = 32.

Since n is a +ve integer.

By inspection, 25 = 32.

∴ a = 2 and n = 5.

Substituting a = 2 and n = 5 in (2).

5 5 1 2 4805 2 2− ⋅ ⋅ =−� � b

5 4 2 4803 2× × ⋅ =b

b2 480

20 8=

×

b2 3=

⇒ b = 3.

%� �&�'� ��

By n n n n a bn− − − ⋅ =−1 2 3 21604 4� ��

5 4 3 2 2 3 21605 4 4× × × × ⋅ =−

120 2 9 2160× × =

2160 2160= .

Hence a = 2, n = 5 and b = 3 .

20.20.20.20.20. Using Binomial theorem prove that 6n − 5n always leaves the remainder 1 when divided by 25.

We know,

6n = (1 + 5)n

We have from Binomial theorem.

Binomial Theorem 81

x a x nx an n

x a an n n n n+ = + ⋅ + − ⋅ ⋅ + +− −� � � �1 2 21

2!... .

1 5 1 51

25

1 2

35 52 3+ = + ⋅ + − ⋅ + − − + +� � � � � �� n nn

n n n n n

!... .

6 1 51

25 52n nn

n n= + +

−⋅ + +

� �... .

6 5 11

25

1 2

35 52 3n nn

n n n n n− = + − + − − ⋅ + +� � � ��

!... .

Now right hand side contains all terms containing 5n except 1st term 1.

∴ when RHS is divided by 25, it leaves the remainder 1.

21.21.21.21.21. Find the greatest term in the expansion of (x − y)20 when x = 12 and y = 4

x y xy

xx

y

x− = −��

��

��

��

= −��

��

� �2020

2020

1 1

Consider

T

T

r

r

y

xr

r

+ = − + ��

1 20 1

T

T

r

r

y

xr

r

+ = − ��

1 21

When y = 4 and x = 12

T

T

r

r

r

rr

r

+ = − �� = −1 21 4

12

21

3

If T T r rr r+ ≥ − ≥1 21 3. Then

21 3≥ +r r

21 4≥ r

⇒ 4 21r ≤Greatest value of r is 5

∴ Greatest term in the expansion is T c x ar

rrn

rn r r

+−= ⋅ ⋅ + =

=11 5

4 with

n x a= = = −20 12 4, . and

T c520

420 4 412 4= ⋅ ⋅ −− � �


T c520

416 412 4= ⋅ ⋅ .

22.22.22.22.22. Find the greatest term in the expansion of (2p + 2q)17 when p = 12 and q = 14.

2 3 2 13

217 17

17

p q pq

p+ = ⋅ +

��

��

� � � �

T

T

r

r

q

pr

r

+ = − + ⋅��

1 17 1 3

2

= − ��

18 3

2

14

12

r

r

T

T

r

rr

r

+ = − ��

1 18 7

4

If T Tr r+ ≥1

18 7 4− ≥r r� �

126 7 4− ≥r r

126 11≥ r

11 126r ≤ .

The greatest value of r is 11.

∴ Greatest term is 11th term.

T c x arn

rn r r

+−= ⋅ ⋅1

T c p q1117

1017 10 102 3= ⋅ ⋅−� � � �

= ⋅ ⋅1710

7 1024 42c � � � � .

23.23.23.23.23. In the expansion of (1 + x)43, the co-efficients of (2m + 1)th and (m + 2)nd terms are equal,find m.

Solution.Solution.Solution.Solution.Solution. Given T2m + 1 = Tm + 2

We have T c x arn

rn r r

+−= ⋅ ⋅1 . formula

Here n x a x= = =43 1, . and

T c xm mm m

2 143

243 2 21+

−= ⋅ ⋅ .

T c xm mm

2 143

22 1+ = ⋅ . ...( ) � 1 143 2− =m

Now

T c xm mm m

+ +− + += ⋅ ⋅2

431

43 1 11

Binomial Theorem 83

T c xm mm

+ ++= ⋅2

431

1 ...(2)

Given (1) = (2)

∴ 432

2 431

1c x c xmm

mm⋅ = ⋅+

+

432

2

431

1 1c x

c xm

m

mm

⋅⋅

=+

+ .

43

43 2 243

43 1 1

1

2

1

− ⋅⋅

− + ⋅ +⋅

=+

m mx

m mx

m

m

43 1 1

43 2 212 1− + ⋅ +

− ⋅⋅ =− +m m

m mx m m� � .

43 1 1

43 2 212 1− − ⋅ +

− ⋅⋅ =− −m m

m mx m m .

42 1

43 2 211 0− ⋅ +

− ⋅⋅ = =−m m

m mx xm .

⇒ m − 1 = 0 [By equation power index of x]

⇒ m = 1 .

%� �&�'� ��

By Substituting m = 1 we get

42 1 1 1

43 2 1 2 111 1− ⋅ +

− ⋅ ⋅⋅ =−

� �x

41 2

41 210⋅

⋅⋅ =x

1 = 1.

24.24.24.24.24. The 21st and 22nd terms in the expansion of (1 + x)44 are equal. Find x.

Solution:Solution:Solution:Solution:Solution: Given T21 = T22

We know T c x arn

rn r r

+−= ⋅ ⋅1 . Formula

Comparing (1 + x)44 with (x + a)n we get

x a x n= = =1 44, . and

T c x2144

2044 20 201= ⋅ ⋅−


=− ⋅

⋅44

44 20 2020x

T c x2244

2144 21 211= ⋅ ⋅−

=− ⋅

⋅44

44 21 2121x

Given T21 = T22

∴44

44 20 20

44

44 21 2120 21

− ⋅⋅ =

− ⋅⋅x x

⇒44 21 21

44 20 20

21

20

− ⋅− ⋅

= x

x

23 21

24 20

⋅⋅

= x

23 21 20

24 23 20

⋅ ⋅⋅ ⋅

= x

⇒ x = 21

24

⇒ x = 7

8.

��

• x a x nx an n

x an n n

x a an n n n n n+ = + ⋅ + − ⋅ + − − ⋅ + +− − −� � � � � �� 1 2 2 3 31

2

1 2

3! !... .

• There are (n + 1) terms in the expansion of (x + a)n. The power indices of ‘x’ go on decreasingby 1 and those of ‘a’ go on increasing by 1 at each stage so that the sum of power indices is n.

• The general term or (r + 1)th term is given by T c x arn

rn r r

+−= ⋅1 .

• If n is even the number of terms in the expansion of (x + a)n is (n + 1) which is odd.

∴ There will be only one middle term i.e. Tn

21+.

• If n is odd, the number of terms in the expansion of (x + a)n is (n + 1) which is even. ∴ there will

be 2 middle terms: T Tn n+ + +1

2

1

21

and .

Binomial Theorem 85

• To find the term containing xm in the expansion of (x + a)n i.e., to find the co-efficient of xm, writeTr + 1. Simplify and equate power index of x to m. Get r and substitute the value of r in Tr + 1.

• For getting the term independent of x or constant term, equate the power index of x to zero afterwriting Tr + 1 simplify then get the value of r. If r is a positive integer greater than 0, substitute inTr + 1.

• If r is negative or a fraction then conclude that there is no term independent of x in the expansion.

�#��

I.I.I.I.I. Expand bExpand bExpand bExpand bExpand by using y using y using y using y using BBBBBinomial theorinomial theorinomial theorinomial theorinomial theorem:em:em:em:em:

1. xx

+��

1 4

2. 1 7+ xy� �

3.2

2

6p

q

q

p−

��

�� 4. 2

3

6

ab−�

��

5. yy

+��

��

15

II.II.II.II.II. Find the indicated term in the expansions:Find the indicated term in the expansions:Find the indicated term in the expansions:Find the indicated term in the expansions:Find the indicated term in the expansions:

1. 4th term in (2 + a)7 2. 12th term in yy

+��

��

113

3. 3rd term in xx

24

1+��

�� 4. 8th term in

x

y2

310

−��

��

5. 10th term in 212

ab

a+�

�� .

III.III.III.III.III. FFFFFind the midind the midind the midind the midind the middle terdle terdle terdle terdle term(s) in the fm(s) in the fm(s) in the fm(s) in the fm(s) in the folloolloolloolloollowing:wing:wing:wing:wing:

1. xx

+��

��

1

2

20

2.x

x2

12

10

−��

��

3.a

b

b

a−�

��

14

4.a

bab−�

��

12

5.a

a2

32

19

+��

�� 6. 3

2

2 9

xx

−��

��


7. 212

11

xx

−��

�� 8.

x

x2

32

19

+��

��

IVIVIVIVIV..... FFFFFind the terind the terind the terind the terind the term indem indem indem indem independent of pendent of pendent of pendent of pendent of xxxxx in the f in the f in the f in the f in the folloolloolloolloollowing ewing ewing ewing ewing expansions:xpansions:xpansions:xpansions:xpansions:

1.4

3

3

2

2 9a

a−

��

�� 2. x

x−��

34

10

3.x

x2

22

10

−��

�� 4. 2

115

xx

−��

��

5. 212

10

xx

+��

�� 6. x

x−��

12

21

VVVVV..... FFFFFind the coefind the coefind the coefind the coefind the coefffffficient of:icient of:icient of:icient of:icient of:

1. x x x23 2 20 in − 2. x x

x4 4

3

151 in +��

��

3.1 117

43

15

xx

x in −��

�� 4. a b a

b6 39

3 in 2 −��

��

5. x xx

5 23

5

21

in +��

�� 6.

1 14

8

x x in 2 −��

��

VI.VI.VI.VI.VI. FFFFFind the vind the vind the vind the vind the value of:alue of:alue of:alue of:alue of:

1. 2 3 2 35 5

+ + − 2. 2 1 2 16 6

+ − −

3. 1 5 1 55 5

+ + −

VII.VII.VII.VII.VII.

1. Prove that in the expansion of ayb

y2

25

+��

��

. There is no term independent of y.

2. The second, 3rd and 4th term of expansion of (x + y)n are 108, 54 and 12 respectively. Find x, yand n.

3. Find the value of (1.01)5 correct to four decimal places.

4. Prove that the sum of odd binomial co-efficients of order n = 2n − 1.

5. Prove that the sum of binomial co-efficient of order n = 2n.

Binomial Theorem 87

��!��

I. 1. x xx x

4 22 44 6

4 1+ + + +

2. 1 7 21 35 35 21 72 2 3 3 4 4 5 5 6 6 7 7+ + + + + + +xy x y x y x y x y x y x y

3. 64 96 60 2015

4

3

4

3

4 64

6 4 2 2 4 4 6

6

p

q

p

q

p

q

q

p

q

p

q

p

q

p

��

−��

+��

− +��

−��

−��

+ .

4. 64 6480

3

160

27

20

27

4

81 7296 6 4 2 3 3 2 2 5

6

a a b a b a b a b abb− + − + + − + .

5.1

5 10 105 13 2

2yy y y

y y+ + + + +

��

��

II. 1. 560 3a 2. 132

9c y− 3. 6 3x

4. 107

3 7 3 72 3c x y− −⋅ ⋅ 5. 17609

6

b

a.

III. 1. 2010

102c − 2. − 63

8 5x

3. −147c 4. 12

612 63c b⋅

5. 199

9 10 8 1910

10 9 103 2 3 2c x c x⋅ ⋅ ⋅ ⋅− − − − and 6.15309

8

5103

1613 14x x and .

7. − ⋅ ⋅ ⋅ ⋅115

19 2 115

764 32c x c x and . 8. 199

9 10 9 199

10 9 113 2 3 2c x c x⋅ ⋅ ⋅ ⋅− − − − and .

IV. 1. 2268 2. 405

3.45

644. 15

1052c ⋅

5. No term independent of x. 6. −217c

V. 1. −1140 2. 15c8 3. −1365

4. 93

62

27c ⋅ . 5. 80 6. 24 8

4⋅ c .

VI. 1. 724 2. 140 2 3. 352

VII. 2. x = 3, y = 1 and n = 4. 3. 1.0510.

�

��

��

1.1.1.1.1. PPPPPolololololynomial:ynomial:ynomial:ynomial:ynomial: An expression of the form

a x a x a x a a a a an n nn n0 1

12

20 1 2+ + + +− − ... , , ... where

are constants, x is a variable is called a polynomial in x of degree n (provided a0 ≠ 0).

2.2.2.2.2. PrPrPrPrProper and improper and improper and improper and improper and improper froper froper froper froper fractions:actions:actions:actions:actions: If f (x) and g (x) are two polynomials in x, then f x

g x

� �� is called

a rational fraction. If the degree of f (x) is less than degree of g (x) then it is called proper fraction.Otherwise it is called an improper fraction. By division, an improper fraction can always bereduced to the sum of a polynomial and a proper fraction.

��

1.4 1

2 8 12

x

x x

−+ − is a proper fraction since degree of numerator (=1) is less than degree of denomi-

nator (=2).

2.x

x

2

2

1

1

−+� � is an improper fraction since degree of numerator = Degree of denominator = 2.

Note that x

x

x

x x

2

2

2

2

1

1

1

2 1

−+

= −+ +� �

By division

x x x

x x

x

2 2

2

2 1 1

1

2 1

2 2

+ + −

+ +− − −

− −( ) ( ) ( )

We getx

x

x

x x

x

x x

2

2 2 2

1

11

2 2

2 11

2 2

2 1

−+

= + − −+ +

= − ++ +� �

which is polynomial and a proper fraction.

��

Consider1

1

2

4

4 2 1

1 4x x

x x

x x++

−= − + +

+ −� �

� � � �

= − + ++ −

= −+ − −

x x

x x

x

x x x

4 2 2

1 4

3 2

4 42� ��

= −− −

3 2

3 42

x

x x.

Here we have expressed the sum of 2 proper fractions as a single proper fraction. The reverse processof expressing a single proper fraction as the sum of the two or more proper fractions is called as‘Resolving into partial fractions’. The following rules are used to resolve a proper fraction into partialfractions:

1. To each linear factor (ax + b) which occurs only once as a factor of the denominator, there

corresponds a partial fraction of the form A

ax b+ where A is constant.

Example:Example:Example:Example:Example:x

x x

A

x

B

x

2 1

1 2 3 1 2 3

−+ +

=+

++� ��

.

2. To each linear factor (ax + b) which occurs r times as a factor of the denominator, there corre-sponds r partial fractions of the form,

A

ax b

A

ax b

A

ax b

A

ax bA A Ar

r r1 2

23

3 1 2++

++

++ +

+� � � � � �... , , where are constants.


x x

A

x

B

x

C

x

D

x

2

3 2 42 3 2 3 2 3 2 3+= +

++

++

+� � � � � � � �.

Note, here corresponding to linear factor x which occur only once in denominator, we have takenonly one constant A and corresponding to linear factor 2x + 3 which occur 3 times, we have taken 3constants B, C and D.

3. To each non factorisable quadratic factor ax2 + bx + c which occur only once as a factor of

denominator, there corresponds a partial fraction of the form Ax B

ax bx c

++ +2 where A and B are

constants.

Example:Example:Example:Example:Example:3 7

1 1 1 12 2

x

x x

Ax B

x

C

x

−+ −

= ++

+−� ��


Observe here, corresponding to x2 + 1, which is quadratic and non-factorisable we have taken Ax + Band corresponding to x − 1 which is linear we have taken one constant C.

4. To each repeated non-factorisable quadratic factor (ax2 + bx + c) which occurs r times as a factorof the denominator, there corresponds r partial fractions of the form.

A x B

ax bx c

A x B

ax bx c

A x B

ax bx c

r rr

1 12

2 2

2 2 2

++ +

++

+ ++

+

+ +� � � �...

where A1, A2, ..., Ar and B1 B2, ... Br are constants.


x

Ax B

x

Cx D

x

Ex F

x

2

2 3 2 2 2 2 3

1

2 1 2 1 2 1 2 1

−

+= +

++ +

++ +

+� � � � � �.

��

��! ��"!��#��$��

1.1.1.1.1.x

x x

−+ −

1

3 4� ��

This is a proper fraction. Resolve into partial fractions:

x

x x

A

x

B

x

−+ −

=+

+−

1

3 4 3 4� �� ...(1)

Multiplying both sides by (x + 3) (x − 4)

x A x B x− = − + +1 4 3� � � � ...(2)

Put x − 4 = 0

⇒ x = 4 in (2)

4 1 0 4 3− = + +A B� � � �

3 73

7= ⇒ =B B� � .

Put x + 3 = 0

x = −3 in (2)

− − = − − +3 1 3 4 0A B� � � �

− = − ⇒ = −−

⇒ =4 74

7

4

7A A A

Substituting A B= =4

7

3

7 and in 1� �

Partial Fractions 91

x

x x x x

−+ −

=+

+−

1

3 4

47

3

37

4� ��

=+

+−

4

7 3

3

7 4x x� � � �

2.2.2.2.2.4 1

12

x

x

−−

4 1

1

4 1

1 12

x

x

x

x x

−−

= −− +� ��

This is a proper fraction. Resolve into partial fractions.

4 1

1 1 1 1

x

x x

A

x

B

x

−− +

=−

++� �� ...(1)

Multiplying by (x − 1) (x + 1)

4 1 1 1x A x B x− = + + −� � � � ...(2)

Put x + 1 = 0

x = −1 in (2)

4 1 1 0 1 1− − = + − −� � � � � �A B

− = − ⇒ =5 25

2B B� �

Put x − 1 = 0

i.e. x = 1 in (2)

4 1 1 1 1 0� � � � � �− = + +A B

3 23

2= ⇒ =A A� �


2

5

2 and in (1)

4 1

1

4 1

1 1

32

1

52

12

x

x

x

x x x x

−−

= −− +

=−

++� ��

=−

++

3

2 1

5

2 1x x� � � �.


3.3.3.3.3.x

x x2 5 6− +

x

x x

x

x x x

x

x x x2 25 6 3 2 6 3 2 3− +=

− − +=

− − −� � � �

=− −

x

x x3 2� �� .


x

x x

A

x

B

x− −=

−+

−3 2 3 2� �� ...(1)

Multiplying by (x − 3) (x − 2)

x A x B x= − + −2 3� � � � ...(2)

Put x − 2 = 0

x = 2 in (2)

2 0 2 3= + −A B� � � �

2 1 2= − ⇒ = −B B� � .

Put x − 3 = 0

x = 3 in (2)

3 3 2 0= − +A B� � � �

3 3= ⇒ =A A .

Substituting A = 3 and B = −2 in (1)

x

x x x x x x− −=

−+ −

−=

−−

−3 2

3

3

2

2

3

3

2

2� �� .

4.4.4.4.4.2 1

1 2 2 1

x

x x x

−+ + +� ��


2 1

1 2 2 1 1 2 2 1

x

x x x

A

x

B

x

C

x

−+ + +

=+

++

++� �� ...(1)

Multiplying by (x + 1) (x + 2) (2x + 1)

2 1 2 2 1 1 2 1 1 2x A x x B x x C x x− = + + + + + + + +� �� ...(2)

Put x x+ = ⇒ = −1 0 1 in (2)

2 − − = − + − + + +1 1 1 2 2 1 1 0 0� � � � � �� A B C


− = + − ⇒ = +3 1 1 3A A� �� .

Put x x+ = ⇒ = −2 0 2 in (2)

2 − − = + − + + − + +2 1 0 2 1 2 2 1 0� � � � � � � �� A B C

− = − − ⇒ = −5 1 3

5

3B B� �� .

Put 2 1 01

2x x+ = ⇒ = − in 2� �

21

21 0 0

1

21

1

22−��

− = + + − +�

� − +��

A B C� � � �

− − = ��1 1

1

2

3

2C

− = �� ⇒ = −2

3

4

8

3C C

Substituting A B C= = − = −3

5

3

8

3, and in 1� �

2 1

1 2 2 1

3

1

5 3

2

8 3

2 1

x

x x x x x x

−+ + +

=+

+ −+

+ −+� ��

.

5.5.5.5.5.2 1

2 3 2

x

x x

−+ −� ��

This is a proper fraction. Resolve into partial fraction.

2 1

2 3 2 3 32 2

x

x x

A

x

B

x

C

x

−+ −

=+

+−

+−� �� ...(1)

Multiplying by (x + 2) (x − 3)2

2 1 3 2 3 22x A x B x x C x− = − + + − + +� � � �� ...(2)

x − =3 0

⇒ x = 3 (Put x = 3 in Eq. (2))

2 3 1 0 0 3 2� � � � � � � �− = + + +A B C

5 5 1= ⇒ =C C� �

x + =2 0


∴ x = −2 (Put x = −2 in Eq. (2))

2 2 1 2 3 0 02− − = − − + +� � � � � � � �A B C

− = − ⇒ = − = −5 55

25

1

52A A� �

Put x A C= = − =01

51, and in (2)

2 0 11

53 2 3 1 22� � � � � �� − = − − + − +B

− = − − +19

56 2B

− − + = −1 29

56B

− + = − ⇒ = −× −

=39

56

6

5 6

1

5B B

� �

Substituting A B C= − = =1

5

1

51, and in 1� �

2 1

2 3

1 5

2

1 5

3

1

32 2

x

x x x x x

−+ −

= −+

+−

+−� ��

= −+

+−

+−

1

5 2

1

5 3

1

3 2x x x� � � � � �.

6.6.6.6.6.x

x x

2

2

1

2 4

+− −� ��

x

x x

x

x x x

x

x x

2

2 2

2 2

2

1

2 2

1

2 2 2

1

2 2

+− −

= +− − +

= ++ −� ��


x

x x

A

x

B

x

C

x

2

2 2

1

2 2 2 2 2

++ −

=+

+−

+−� �� ...(1)

Multiplying with (x + 2) (x − 2)2

x A x B x x C x2 21 2 2 2 2+ = − + + − + +� � � �� ...(2)


Put x − =2 0 in (2)

⇒ x = 2

2 1 0 0 2 22 + = + + +A B C� � � � � �

5 45

4= ⇒ =C C� � .

Put x +2 = 0 in (2)

⇒ x = −2

− + = − − + +2 1 2 2 0 02 2� � � � � � � �A B C

5 165

16= ⇒ =A A� � .

Put x A C= = =05

16

5

4, and in 2� �

15

162 2 2

5

422= − + − +� � � �� B

15

164

4 45

2= × / + − +� �B

15

4

5

24− − = − B

4 5 10

44

11

4 4

11

16

− − = − ⇒ = −−

=B B� �

B = 11

16.

Substituting, A B C= = =5

16

11

16

5

4, and in 1� �

x

x x x x x

2

2 2

1

2 2

5

162

11

162

5

42

++ −

=+

+−

+−� ��

=+

+−

+−

5

16 2

11

16 2

5

4 2 2x x x� � � � � �.


7.7.7.7.7.2 3

9 32

x

x x

++ −� � � �


2 3

9 3 9 32 2

x

x x

Ax B

x

C

x

++ −

= ++

+−� � � � ...(1)

Multiplying by (x2 + 9) (x − 3)

2 3 3 92x Ax B x C x+ = + − + +� ��

2 3 3 3 92 2x Ax Bx Ax B Cx C+ = + − − + + .

2 3 3 3 92 2x Ax Cx Bx Ax B C+ = + + − − + .

Equating the co-efficients of like powers.

x A C A C2 0: = + ⇒ = − ...(2)

x B A: 2 3= − ...(3)

1 3 3 91 3

: = − +⇒ = − +

B CB C ...(4)

Substituting Eq. (2) i.e., A = −C in (3)

Equation (4):

B C

B C

B C

C C

− − =+ =

− + =

= ⇒ =

3 2

3 2

3 1

6 31

2

� �

A C A= − ⇒ = − 1

2

− + =B C3 1

− + �� =B 3

1

21

⇒ = − =B3

21

1

2

∴ B = 1

2


Substituting A B C= − = =1

2

1

2

1

2, and in 1� �

2 3

9 3

12

12

9

12

32 2

x

x x

x

x x

++ −

=− +

++

−� ��

= − ++

+−

x

x x

1

2 9

1

2 32� � � �.

8.8.8.8.8.8 1

1

2

3

x

x

+−

8 1

1

8 1

1

8 1

1 1

2

3

2

3 3

2

2

x

x

x

x

x

x x x

+−

= +−

= +− + +� ��

Resolve into partial fractions:

8 1

1 1 1 1

2

2 2

x

x x x

A

x

Bx C

x x

+− + +

=−

+ ++ +� �� ...(1)

Multiplying by x x x− + +1 12� ��

8 1 1 12 2x A x x Bx C x+ = + + + + −� � � ��

8 12 2 2x Ax Ax A Bx Bx Cx C+ = + + + − + −Equating the co-efficients of like powers:

x A B2 8: = + ...(2)

x A B C: 0 = − + ...(3)

1 1: = −A C ...(4)

Solving (3) and (4)

A B C

A C

A B

− + =− =− =

0

1

2 1...(5)

Solving (2) and (5)

A B

A B

A A

+ =− =

= ⇒ =

8

2 1

3 9 3


Substituting A = 3 in (2)

3 8+ =B

B = −8 3

B = 5

Substituting A = 3 in (4)

A C− = 1

3 1− =C

3 1 2− = ⇒ =C C

Substituting A = 3, B = 5 and C = 2 in (1)

8 1

1 1

3

1

5 2

1

2

2 2

x

x x x x

x

x x

+− + +

=−

+ ++ +� ��

9.9.9.9.9.2 1 2 1

13 2

x

x x

x

x x

++

= ++� �


2 1

1 12 2

x

x x

A

x

Bx C

x

++

= + ++� � ...(1)

Multiplying by x (x2 + 1)

2 1 12x A x Bx C x+ = + + +� � � ��

2 1 2 2x Ax A Bx Cx+ = + + +Equating the co-efficients of like powers:

x A B2 0: = +

x C: 2 =

1 1: = A

Now A B B B+ = ⇒ + = ⇒ = −0 1 0 1

Substituting A B C= = − =1 1 2 1, ( ) and in

2 1

1

1 2

12 2

x

x x x

x

x

++

= + − ++� �

.


10.10.10.10.10.x

x x

2

2

2

12

−+ −

This is an improper fraction. Since degree of numerator = degree of denominator = 2.

x x x

x x

x

2 2

2

12 2

1

12

10

+ − −

+ −− − +

− +( ) ( ) ( )

∴x

x x

x

x x

2

2 2

2

121

10

12

−+ −

= + − ++ −

...(1)

Consider

− ++ −

= − ++ − −

= − ++ − +

x

x x

x

x x x

x

x x x

10

12

10

4 3 12

10

4 3 42 2 � � � �

= − ++ −

x

x x

10

4 3� � � �.


− ++ −

=+

+−

x

x x

A

x

B

x

10

4 3 4 3� � � � ...(2)

Multiplying by (x + 4) (x − 3) we get

− + = − + +x A x B x10 3 4� � � �

Put x x− = ⇒ =3 0 3

− + = + +3 10 0 3 4A B� � � �

7 7 1= ⇒ =B B� � .

Put x + =4 0

x = −4

− − + = − − +4 10 4 3 0� � � � � �A B

14 7 2= − ⇒ = −A A� � .

Substituting A = −2 and B = 1 in (2)

− ++ −

= −+

+−

x

x x x x

10

4 3

2

4

1

3� ��


Substituting this in equation (1)

x

x x x x

2

2

2

121

2

4

1

3

−+ −

= + −+

+−

= −+

+−

12

4

1

3x x.

11.11.11.11.11.8 2

4 5 1

2

2

x x

x x

+ −+ +

This is an improper fraction. Since degree of numerator = Degree of denominator = 2.

4 5 1 8 2

2

8 10 2

9 4

2 2

2

x x x x

x x

x

+ + + −

+ +− − −

− −( ) ( ) ( )

∴8 2

4 5 12

9 4

4 5 1

2

2 2

x x

x x

x

x x

+ −+ +

= + − −+ +

8 2

4 5 12

9 4

4 5 1

2

2 2

x x

x x

x

x x

+ −+ +

= − ++ +

...(1)

Consider9 4

4 5 1

9 4

4 4 1 12 2

x

x x

x

x x x

++ +

= ++ + +

= ++ + +

= ++ +

9 4

4 1 1 1

9 4

1 4 1

x

x x x

x

x x� � � � � ��


9 4

1 4 1 1 4 1

x

x x

A

x

B

x

++ +

=+

++� �� ...(2)

Multiplying by (x + 1) (4x + 1)

9 4 4 1 1x A x B x+ = + + +� � � �

Put 4 1 0x + =

4 11

4x x= − ⇒ = −

91

44 0

1

41−��

+ = + − +�

�A B� �


− + = +��

9

44

3

4B

7

4

3

4

7

3= �� ⇒ =B B .

Put x x+ = ⇒ = −1 0 1

9 1 4 4 1 1 0− + = − + +� � � �� A B

− = − ⇒ =5 35

3A A� � .


3

7

3 and in (2)

9x

x x x x

++ +

=+

++

4

1 4 1

53

1

73

4 1� ��


8 1

4 5 12

5 3

1

7 3

4 1

2

2

x x

x x x x

+ −+ +

= −+

++

��

= −+

−+

25

3 1

7

3 4 1x x� � � �.

12.12.12.12.12.x

x x

3

3

1−+

.

This is an improper fraction since degree of numerator = Degree of denominator = 3.

x x x

x x

x

3 3

3

1

1

1

+ −

+− −

− −( ) ( )

∴x

x x

x

x x

x

x x

3

3 3 3

11

11

1−+

= + − −+

= − ++

...(1)

Consider

x

x x

x

x x

++

= ++

1 1

13 2� �This is a proper fraction.


Resolve into partial fractions

x

x x

A

x

Bx C

x

++

= + ++

1

1 12 2� � ...(2)

Multiplying by x (x2 + 1)

x A x Bx C x+ = + + +1 12� � � ��

x A x A Bx Cx+ = + + +1 2 2

Equating the co-efficients of like powers

x A B2 0: = +

x C: 1 =

1 1: = A

A B B B+ = ⇒ + = ⇒ = −0 1 0 1

Substituting A = 1, B = −1 and C = 1 in (2)

x

x x x

x

x

++

= + − ++

1

1

1 1

12 2� �


x

x x x

x

x

3

3 211 1 1

1+= − + − +

+��

��

= − − − ++

11 1

12x

x

x.

�� %��

Proper fraction Partial fraction

•Nr

a x b a x b a x b1 1 2 2 3 3+ + +� �� ...A

a x b

B

a x b

C

a x b1 1 2 2 3 3++

++

++ ...

•Nr

ax b cx d+ +� � � �2

A

ax b

B

ax b

C

cx d++

++

+� � � �2

•Nr

ax bx c dx e2 + + +� � � �Ax B

ax bx c

C

dx e

++ +

++2� �

Provided ax2 + bx + c is non-factorisable. If ax2 + bx + c is factorisable as (a1x + b1) (a2x + b2), thenreplace ax2 + bx + c by (a1x + b1) (a2x + b2). Then resolve,


Nr

a x b a x b dx e

A

a x b

B

a x b

C

dx e1 1 2 2 1 1 2 2+ + + ++

++

+� �� as

If an improper fraction is given to resolve, first by dividing the numerator by denominator write thegiven fraction as the sum of the polynomial and the proper fraction. Then the proper fraction is resolvedinto partial fractions.

��


1.8 1

2 3

x

x x

−− −� �� 2.

4 6

12

x

x

+− 3.

2 3

3 22

x

x x

+− +

4.1

1 2 3x x x+ + +� �� 5.x x

x x x

2

2

10 13

1 5 6

− +− − +� �� 6.

3 5

2 32

x

x x

++ −� � � �

7.9

1 2 2x x+ +� �� 8.4 3 1

2 1 1

2

2

x x

x x x

+ −+ + −� �� 9.

x

x x

2

2

1

2 4

+− −� ��

10.1

1 1 2x x− +� �� 11.3 2

3

x

x x

++ 12.

5 1

1

2

3

x

x

+−

13.3 1

2 1 2

x

x x x

−+ − +� �� 14.

x x

x x

2

2

2 4

2 2 3

+ +− +� �� 15.

x

x x

−+ +

1

1 12� ��

16.x

x x

2 2

4 3

−− +� �� 17.

2 3 2

2

2

2

x x

x x

+ +− −

18.x x x

x x

3 2

2

7 17 11

5 6

+ + ++ +

19.2 3 4

6

2

2

x x

x x

− −− −

20.x x x

x x

4 3 2

2

3 3 10

1 3

− − ++ −� � � �

��

1.3

2

5

3x x−+

+ 2.5

1

1

1x x−−

+ 3.7

2

5

1x x−−

−

4.1

2 1

1

2

1

2 3x x x+−

++

+� � � � 5.2

1

3

2

4

3x x x−+

−−

−

6.−

++

++

−14

25 2

1

5 2

14

25 32x x x� � � � � �


7.9

1

9

2

9

2 2x x x+−

+−

+� � 8. −+

++

+−

5

2 1

0

1

3

2 12x x x� � � � � �

9.5

16 2

11

16 2

5

4 2 2x x x++

−+

−� � � � � �

10.1

4 1

1

4 1

1

2 1 2x x x−+

++

+� � � � � �

11.2 2 3

12x

x

x+ − +

+ 12.2

1

3 1

12x

x

x x−+ +

+ + 13.−+

+− +

1

2 1 2x

x

x x

14.12

11 2

13 4

11 2 32x

x

x−− +

+� � � � 15.−+

++

1

1 12x

x

x16. 1

1

3

2

4−

++

−x x

17. 216

3 2

1

3 1+

−−

−x x� � � �18. x

x x+ +

++

+2

1

2

2

3� � 19. 2

2

2

1

3−

++

−x x

20. xx x x

− ++

−+

−−

217

16 1

11

4 1

17

16 32� �� .

�

��

��

The theory of matrices was developed in 1857 by the French mathematician Cayley. It was not welladvanced till 20th century. But now a days matrices are powerful tool in modern mathematics havingwide applications.

��

A matrix is an arrangement of numbers in rows (horizontal lines) and columns (vertical lines). Thearrangement is usually enclosed between square brackets [ ] or curved brackets ( ) or pairs of verticallines || ||. The matrix is usually denoted by a capital letter.

Order of a matrix = Number of rows × Number of columns.

If a matrix has m rows and n columns, then

Order = m × n (read as m by n)

Example:Example:Example:Example:Example: Matrix has the order 2 3.A = ��

��

×1 2 34 5 6

�� !��

1.1.1.1.1. Rectangular maRectangular maRectangular maRectangular maRectangular matrtrtrtrtrix:ix:ix:ix:ix: If the number of rows is not equal to number of columns in a matrix, thenthat matrix is called rectangular matrix.

Example:Example:Example:Example:Example: Aa a a a ab b b b bc c c c c

=�

��

�

��

×

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5 3 5

(a) RoRoRoRoRow maw maw maw maw matrtrtrtrtrix or Roix or Roix or Roix or Roix or Row vw vw vw vw vector:ector:ector:ector:ector: If a matrix has only one row, then it is called row matrix.

Example:Example:Example:Example:Example: X = ×1 2 3 1 3

(b) Column maColumn maColumn maColumn maColumn matrtrtrtrtrix or column vix or column vix or column vix or column vix or column vector:ector:ector:ector:ector: If a matrix has only one column then it is called columnmatrix.


Example:Example:Example:Example:Example: Y =�

��

�

��

×

123 3 1

2.2.2.2.2. SquarSquarSquarSquarSquare mae mae mae mae matrtrtrtrtrix:ix:ix:ix:ix: If the number of rows is equal to number of columns in a matrix, then it is calledsquare matrix.

Example:Example:Example:Example:Example: Aa a ab b bc c c

=�

��

�

��

×

1 2 3

1 2 3

1 2 3 3 3

In a square matrix, the entries from the left top corner to the right bottom corner are called principaldiagonal elements. So in the above example a1 b2 c3 are Principal diagonal elements.

(a) Diagonal matrix:Diagonal matrix:Diagonal matrix:Diagonal matrix:Diagonal matrix: If in a square matrix all the non-diagonal elements are zero, then it is calleddiagonal matrix.

Example:Example:Example:Example:Example: A B= ��

��

= −�

��

�

��

1 00 2

2 0 00 1 00 0 3

,

(b) Scalar matrix:Scalar matrix:Scalar matrix:Scalar matrix:Scalar matrix: If in a diagonal matrix, all the diagonal elements are equal, then it is called scalarmatrix.


��

=�

��

�

��

5 00 5

6 0 00 6 00 0 6

,

(c) Identity matrix or unit matrix:Identity matrix or unit matrix:Identity matrix or unit matrix:Identity matrix or unit matrix:Identity matrix or unit matrix: If in a scalar matrix all the diagonal elements are equal to 1, thenit is called unit matrix or identity matrix.

Example:Example:Example:Example:Example: I I2 31 00 1

1 0 00 1 00 0 1

= ��

��

=�

��

�

��

,

(d) Upper trUpper trUpper trUpper trUpper triangular maiangular maiangular maiangular maiangular matrtrtrtrtrix:ix:ix:ix:ix: A square matrix is called upper triangular if all the non-diagonalelements below the principal diagonal are zeroes.


a a ab b

c

1 2 3

1 2

1

00 0

�

��

�

��

(e) LoLoLoLoLowwwwwer trer trer trer trer triangular maiangular maiangular maiangular maiangular matrtrtrtrtrix:ix:ix:ix:ix: A square matrix is called lower triangular if all the non-diagonalelements above the principal diagonal are zeroes.


ab bc c c

1

1 2

1 2 3

0 00

�

��

�

��

Matrices & Determinants 107

Null matrix:Null matrix:Null matrix:Null matrix:Null matrix: If each element of a matrix is zero then it is called null matrix or zero matrix.


��

= ��

��

0 0 00 0 0

0 00 0

,

��" �#$�%��!��

1. Equality of maEquality of maEquality of maEquality of maEquality of matrtrtrtrtrices:ices:ices:ices:ices: Two matrices of same order are said to be equal iff the correspondingelements are equal,

i.e., a ab b

a b a b1 2

1 21 1 2 2

3 41 0

3 1 4 0��

��= −��

��

= = − = = iff , , , .

2. AdAdAdAdAddition and subtrdition and subtrdition and subtrdition and subtrdition and subtraction of maaction of maaction of maaction of maaction of matrtrtrtrtrices:ices:ices:ices:ices: Two or more matrices of the same order can be added orsubtracted. It is the matrix obtained by adding or subtracting the corresponding elements i.e.,

If Aa a aa a a

Bb b bb b b

= ��

��

= ��

��

1 2 3

4 5 6

1 2 3

4 5 6 and

Then A Ba b a b a ba b a b a b

+ = + + ++ + +

��

��

1 1 2 2 3 3

4 4 5 5 6 6

A Ba b a b a ba b a b a b

− = − − −− − −

��

��

1 1 2 2 3 3

4 4 5 5 6 6

Illustration: If A B= −��

��

= ��

��

1 20 1

3 14 6

and then,

A B+ = + ++ − +

��

��= ��

��

1 3 2 10 4 1 6

4 34 5

A B− = − −− − −

��

��= −

− −��

��

1 3 2 10 4 1 6

2 14 7

Scalar mScalar mScalar mScalar mScalar multiplicaultiplicaultiplicaultiplicaultiplication:tion:tion:tion:tion: If A is any matrix and k is any scalar or constant, then kA is a matrix obtainedby multiplying every element of A by k.

i.e., if thenAa a aa a a

= ��

��

1 2 3

4 5 6,

kAka ka kaka ka ka

= ��

��

1 2 3

4 5 6

Illustration: If A A= ��

��

= ��

��= ��

��

1 23 4

31 23 4

3 69 12

. . Then 3

MaMaMaMaMatrtrtrtrtrix mix mix mix mix multiplicaultiplicaultiplicaultiplicaultiplication:tion:tion:tion:tion: The product of 2 matrices exists only when the number of columns in 1stmatrix is equal to the number of rows in the 2nd matrix.


If A is a matrix of order m × n and B is a matrix of order n × p. Then AB exists and is of orderm × p. To get the elements of AB, the elements of 1st row of A multiplied by the corresponding elementsof the first column of B and the products are added. The sum is the element in the first row, first columnsof AB. Similarly other elements are obtained.

i.e., If A a Bbbb

= =�

��

�

��×

×

1 1 3

1

2

3 3 1

a a2 3 & , then AB exists and is of order 1 × 1.

AB a b a b a b= + +1 1 2 2 3 3

2. If Aa a aa a a

Bb bb bb b

= ��

��

=�

��

�

��

11 12 13

21 22 23

11 12

21 22

31 33

&

Then AB exists � 2 × ×3 3 2� � and is of order 2 × 2.

ABa b a b a b a b a b a ba b a b a b a b a b a b= + + + +

+ + + +��

��

11 11 12 21 13 31 11 12 12 22 13 33

21 11 22 21 23 31 21 12 22 22 23 33.

Illustration:

If A B= ��

��

=�

��

�

��×

×

1 2 34 5 6

1 42 53 62 3

3 2

& , then

AB exists and is of order 2 × 2.

AB = + + + ++ + + +

��

��

1 4 9 4 10 184 10 18 16 25 36

= ��

��

14 3232 77

.

Note:Note:Note:Note:Note:(1) A + B = B + A i.e., commutative law with respect to addition holds good.

(2) AB ≠ BA in general i.e., commutative law with respect to multiplication doesn’t hold good ingeneral.

��& �� !��

If A is any matrix then the matrix obtained by interchanging rows and columns of A is called transposeof A. It is denoted by A′ or AT.

For example: If A AT= ��

��

= ��

��× ×

1 23 4

1 32 42 2 2 2

then


If A AT= ��

��

=�

��

�

��×

×

1 2 34 5 6

1 42 53 62 3

3 2

then

If A is of order m × n then AT is of order n × m.

'��(��#� �

1.1.1.1.1. If A B= −��

��

= − −��

��

1 04 3

2 31 6

and Then find (i) A + B (ii) 2A − 3B (iii) A + 2A′

(i) A B+ = −��

��+ − −��

��= + +

− + − + −��

��

1 04 3

2 31 6

1 2 0 34 1 3 6

= − −��

��

3 35 3

(ii) 2 3 21 04 3

32 31 6

A B− = −��

��− − −��

��

= −��

��− − −��

��= − −

− − − − −��

��

2 08 6

6 93 18

2 6 0 98 3 6 18

= − −−��

��

4 95 24

(iii) A A+ ′ = −��

��+ −��

��= −��

��+ −��

��

21 04 3

21 40 3

1 04 3

2 80 6

=+ + −

− + +��

��= −

−��

��

1 2 0 84 0 3 6

3 84 9

.

2.2.2.2.2. If A B=−−

−

�

��

�

��

= −�

��

�

��

1 2 14 0 31 1 5

1 3 21 1 56 2 0

and

Then find (i) (A + B)′ (ii) 2A′ − 3B.

(i) A B+ =−−

−

�

��

�

��+ −�

��

�

��

1 2 14 0 31 1 5

1 3 21 1 56 2 0

A B+ =+ + − ++ + − − ++ − + +

�

��

�

��= −�

��

�

��

1 1 2 3 1 24 1 0 1 3 51 6 1 2 5 0

2 5 15 1 27 1 5


⇒ A B+ ′ = −�

��

�

��

2 5 75 1 11 2 5

(ii) 2 3 21 4 12 0 11 3 5

31 3 21 1 56 2 0

′ − = −− −

�

��

�

��− −�

��

�

��

A B

= −− −

�

��

�

��− −�

��

�

��

2 8 24 0 22 6 10

3 9 63 3 15

18 6 0

2 32 3 8 9 2 64 3 0 3 2 15

2 18 6 6 10 0

1 1 41 3 17

20 12 10′ − =

− − −− − − − −

− − − − −

�

��

�

��=

− − −−

− −

�

��

�

��

A B

3.3.3.3.3. If A B=−

�

�

��

�

�

��

= −4

0

1

1 2 3 and , then find AB and BA. Is AB = BA?

AB =−

�

�

��

�

�

��

− �

→

×

×

4

0

1

1 2 3 1 3

3 1

AB =× × × −× × × −

− × − × − × −

�

��

�

��=

−

− − +

�

��

�

��

4 1 4 2 4 30 1 0 2 0 3

1 1 1 2 1 3

4 8 120 0 01 2 3

BA = −−

�

�

��

�

�

��

→1 2 3

4

0

1

BA = × + × + − −1 4 2 0 3 1

BA = + + =4 0 3 7 .

∴ AB ≠ BA.

4.4.4.4.4. If A A=�

��

�

��

1 1 11 1 11 1 1

2, .then find


A A A2 = ⋅

A21 1 11 1 11 1 1

1 1 11 1 11 1 1

=�

��

�

��

�

��

�

��

=+ + + + + ++ + + + + ++ + + + + +

�

��

�

��

1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1

A A23 3 33 3 33 3 3

31 1 11 1 11 1 1

3=�

��

�

��=�

��

�

��= .

Hence A2 = 3A.

5.5.5.5.5. Find x and y if x y34 5

23 1

5 77 4

��

��+ −��

��= ��

��

Given

x y34 5

23 1

5 77 4

��

��+ −��

��= ��

��

= + ++ + −

��

��= ��

��

x y2 34 3 5 1

5 77 4

⇒ x y+ = + =2 5 3 7,

⇒ x y= − = −5 2 7 3,

⇒ x y= =3 4 and .

6. Find x and y if2 11 0

72−

��

��= ��

xy

2 11 0

72−

��

��= ��

xy

⇒2

1 0

7

2

2 7

0 2

x y

x y

x y

x

+− +��

�� =��⇒

+ =− + =

.

− = ⇒ = −x x2 2.

2 7x y+ =

2 2 7− + = y


y = +7 4

y =11.

So, x = −2 and y = 11

7.7.7.7.7. Find x and y given that 1 2

3 7

2

1−��

��= ��

x

y

Given1 23 7

21−

��

�� = ��

→xy

x y

x y

+− +��

��= ��

2

3 7

2

1

⇒ x y+ =2 2 ...(1)

− + =3 7 1x y ...(2)

⇒ (1) × 3 + (2)

3 6 6

3 7 1

13 7

x y

x y

y

+ =− + =

=

⇒ y = 7

13

Substituting y = 7

13 in (1)

x + � �� =2

7

132

x = −214

13

x = −26 14

13

x = 12

13.

8.8.8.8.8. Find A and B if A + B = 7 0

2 5

3 0

8 3��

��

− = ��

��

and A B .


A B+ = ��

��

7 0

2 5

A B− = ��

��

3 0

8 3

Adding A B A B+ + − = ��

��+ ��

��

7 02 5

3 08 3

210 0

10 8A = ��

��

⇒ A = ��

��= ��

��

1

2

10 0

10 8

5 0

5 4

Given A B+ = ��

��

7 02 5

5 0

5 4

7 0

2 5��

��+ = �

��

B

⇒ B = ��

��− ��

��

7 0

2 5

5 0

5 4

B =−��

��

2 03 1

VVVVVerererererififififificaicaicaicaication:tion:tion:tion:tion:

A B+ = ��

��+ −��

��= ��

��

5 05 4

2 03 1

7 02 5

9.9.9.9.9. If A =−

��

��

4 0

1 2, then prove that A2 − 2A − 8I = 0 where 0 is the null matrix.

Given: A =−

��

��

4 0

1 2

A A A2 4 0

1 2

4 0

1 2= ⋅ =

−��

�� −��

��

→

=× + × × + −+ − + − −

��

��

4 4 0 1 4 0 0 2

1 4 2 1 1 0 2 2


=+ +− +

��

��= ��

��

16 0 0 0

4 2 0 4

16 0

2 4

2 24 0

1 2

8 0

2 4A =

−��

��=

−��

��

8 81 00 1

8 00 8

I = ��

��= ��

��

Consider LHS,

A A I2 2 816 0

2 4

8 0

2 4

8 0

0 8− − = �

��−

−��

��− ��

��

=− − − −− − − − −

��

��= ��

��=

16 8 8 0 0 0

2 2 0 4 4 8

0 0

0 0 R.H.S.

Hence proved.

10.10.10.10.10. If A B C=−

��

��

= ��

��

=−��

��

1 2

3 1

1 0

2 1

2 1

1 1, and ,

then prove that A (B + C) = AB + AC.

Consider B C+ = ��

��+

−��

��= ��

��

1 02 1

2 11 1

3 11 2

A B C+ =−

��

��

�� =

+ +− −

��

��

→1 2

3 1

3 1

1 2

3 2 1 4

9 1 3 2

A B C+ = ��

��

5 5

8 1 ...(1)

Now

AB =−

��

��

�� =

+ +− −

��

��=

−��

��

→1 2

3 1

1 0

2 1

1 4 0 2

3 2 0 1

5 2

1 1

AC =−

��

�� −��

�� = − +

+ −��

��= ��

��

→1 2

3 1

2 1

1 1

2 2 1 2

6 1 3 1

0 3

7 2

AB AC+ =−

��

��+ ��

��= ��

��

5 2

1 1

0 3

7 2

5 5

8 1 ...(2)

From (1) and (2) A (B + C) = AB + AC


11.11.11.11.11. If A B C= ��

��

=�

�

��

�

�

��

=1 3 21 0 2

162

2 1, and , then prove that A (BC) = (AB) C

Consider BC =�

�

��

�

�

��

� =× ×× ×× ×

�

�

��

�

�

��

→1

6

2

2 1

1 2 1 1

6 2 6 1

2 2 2 1

BC =�

�

��

�

�

��

2 1

12 6

4 2

A BC = ��

��

�

��

�

�

�� =

× + × + × × + × + ×× + × + × × + × + ×��

��

→1 3 2

1 0 2

2 1

12 6

4 2

1 2 3 12 2 4 1 1 3 6 2 2

1 2 0 12 2 4 1 1 0 6 2 2

A BC =+ + + ++ + + +

��

��= ��

��

2 36 8 1 18 4

2 0 8 1 0 4

46 23

10 5 ...(1)

Now

AB = ��

��

�

��

�

�

�� =

× + × + ×× + × + ×��

��=

+ ++ +

��

��=

+ ++ +

��

��= ��

→1 3 2

1 0 2

1

6

2

1 1 3 6 2 2

1 1 0 6 2 2

1 18 4

1 0 4

1 18 4

1 0 4

23

5

AB C = ��

� =× ×× ×

��

��= ��

��

→23

52 1

23 2 23 1

5 2 5 1

46 23

10 5...(2)

From (1) and (2) A (BC) = (AB) C.

Hence proved.

12.12.12.12.12. If A B A B A B= ��

��

=−�

��

+ ′ = ′ + ′3 4

0 5

1 1

2 1 and then prove that ,

Consider

A B+ = ��

��+

−��

��= ��

��

3 4

0 5

1 1

2 1

4 3

2 6

A B+ ′ = ��

��

4 2

3 6 ...(1)


′ = ��

��

′ =−��

��

A B3 0

4 5

1 2

1 1 and

′ + ′ = ��

��+

−��

��= ��

��

A B3 0

4 5

1 2

1 1

4 2

3 6...(2)

From (1) and (2)

A B A B+ ′ = ′ + ′ Hence proved.

13.13.13.13.13. If A =−��

��

1 2

1 7, then find X such that AX = I where I is identity matrix of order 2 × 2.

Given A X AX I=−��

��

=1 2

1 7, To find such that

Let Xa b

c d= ��

��

� A

I

X

is of order 2 2 is of order 2 2, So is of order 2 2.

×××

�

�

��

�

�

��

Consider

AX = I

1 2

1 7

2 2

7 7

1 0

0 1−��

��

�� = + +

− + − +��

��= ��

��

→a b

c d

a c b d

a c b d

⇒

⇒

a c

a c

c

c

b d

b d

d

d

+ =− + =

==

+ =− + =

==

2 1

7 0

9 1

1 9

2 0

7 1

9 1

1 9.

;

;

.

Substituting c = 1/9 Substituting d = 1/9

a + � �� =2

1

91 b + �

�� =2

1

90

a = −12

9b = − 2

9

a = 7

9.


Hence Xa b

c d= ��

��=

−�

�

��

�

�

��=

−��

��

7

9

2

91

9

1

9

1

9

7 2

1 1

14.14.14.14.14. If A B= ��

��

=−�

��

3 4

0 5

1 1

2 4, , then find X such that A + 2X = B.

Solution.Solution.Solution.Solution.Solution. Since A is of order 2 × 2. B is of order 2 × 2. X is of order 2 × 2.

Now

A + 2X = B

2X = B − A

X B A= −1

2

X =−�

��− ��

��

1

2

1 12 4

3 40 5

X =− −

−��

��

1

2

2 5

2 1

X =− −

−��

��

1 5 2

1 1 2.

��

Every square matrix is associated with a unique real number called its determinant value.

�� )*� +,��-)*��.��

If Aa a

a a= ��

��

1 2

3 4, then determinant A is denoted by |A| or det A and its value is a a a a1 4 2 3− .

Example:Example:Example:Example:Example: If A A= ��

��

= × − × = − =1 2

3 71 7 2 3 7 6 1, . then

�� )*� +,��-)*��.��

If A

a a a

a a a

a a a

=�

�

��

�

�

��

1 2 3

4 5 6

7 8 9

, then


A aa a

a aa

a a

a aa

a a

a a= − +1

5 6

8 92

4 6

7 93

4 6

7 9

A a a a a a a a a a a a a a a a= − − − + −1 5 9 6 8 2 4 9 7 6 3 4 9 6 7� � � � � �For example,

If A = −�

��

�

��

1 2 32 1 13 1 2

, then

A =−

−−

+11 1

1 22

2 1

3 23

2 1

3 1

= + − + + −1 2 1 2 4 3 3 2 3

1 3 2 7 3 1 − + −

3 14 3 14− − = − .

Singular and non-singular maSingular and non-singular maSingular and non-singular maSingular and non-singular maSingular and non-singular matrtrtrtrtrices:ices:ices:ices:ices: If A = 0 for a matrix A then A is said to singular matrix. If

A ≠ 0 then A is called non-singular matrix.

'��(��#� �

1.1.1.1.1. If A A=−

��

��

1 2

4 1, .Then find

Solution:Solution:Solution:Solution:Solution: Given A =−

��

��

1 24 1

A = − −1 1 2 4

− − = −1 8 9.

2.2.2.2.2. If A B AB= ��

��

=−��

��

1 2

0 1

2 1

1 1 and then find , .

Solution:Solution:Solution:Solution:Solution: AB = ��

�� −��

��

→1 20 1

2 11 1

=+ − ++ − +

��

��

1 2 2 1 1 1 2 1

0 2 1 1 0 1 1 1


AB =− +− +

��

��=

−��

��

2 2 1 2

0 1 0 1

0 3

1 1

AB = − − =0 3 3 .

OR

AB A B= ⋅ = ⋅−

1 2

0 1

2 1

1 1

= − ⋅ − −1 1 2 0 2 1

1 2 1 1 3 3 + = = .

3.3.3.3.3. Find x if the matrix 1 2 1

0 1

1 2 6

x�

�

��

�

�

��

is singular

Given

1 2 10 11 2 6

x�

�

��

�

�

��

is singular.

∴

1 2 1

0 1

1 2 6

0x = .

11

2 62

0 1

1 61

0

1 20

x x− + =

1 6 2 2 0 1 1 0 0x x− − − + − =

6 2 2 1 1 0x x− − − + − =

6 2 2 0x x− / + / − =

5 0 0x x= ⇒ = .

4.4.4.4.4. If

1 2 2

1 3

0 6 2

2x x= , .then find


Given

1 2 21 30 6 2

2x =

13

6 22

1

0 22

1 3

0 62

x x− + =

1 6 6 2 2 0 2 6 2− − − + =x

6 6 2 2 12 2− − + =x

6 6 4 12 2− − + =x

− + =6 14 2x

− = − ⇒ =6 2 14 2x x .

5.5.5.5.5. Solve for x

x

x

x

: .2 1

2 51 2

0−

−=

x x x x x5 2 2 2 1 4 5 0− − + − + =

x x x3 2 3 9 0 − − =

3 6 9 02x x− − =

3 9 3 9 02x x x− + − =

3 3 3 3 0x x x− + − =

3 3 3 0x x+ − =

3 3 0 3 0x x+ = − = or

3 3x = −

x x= − = +1 3 or .

Hence x = −1 or x = 3.

��/ �� !��

1. Prove that the value of a determinant remains same when its rows and columns are interchanged

i.e., prove that A A= ′ .

PrPrPrPrProof:oof:oof:oof:oof: Let A

a a a

b b b

c c c

=�

�

��

�

�

��

1 2 3

1 2 3

1 2 3

27

9

3

6

2x

x

x

x

<−+−


Then ′ =�

�

��

�

�

��

A

a b c

a b c

a b c

1 1 1

2 2 2

3 3 3

To prove: A A= ′ .

Consider A

a a a

b b b

c c c

=1 2 3

1 2 3

1 2 3

= − +ab b

c ca

b b

c ca

b b

c c12 3

2 32

1 3

1 33

1 2

1 2

a b c b c a b c b c a b c b c1 2 3 3 2 2 1 3 3 1 3 1 2 2 1− − − + −� � � � � �

a b c a b c a b c a b c a b c a b c1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1− − + + − ...(1)

Now

′ = = − − − + −A

a b c

a b c

a b c

a b c b c b a c a c c a b a b1 1 1

2 2 2

3 3 3

1 2 3 3 2 1 2 3 3 2 1 2 3 3 2� � � � � �

′ = = − − + + −A

a b c

a b c

a b c

a b c a b c a b c a b c a b c a b c1 1 1

2 2 2

3 3 3

1 2 3 1 3 2 2 1 3 3 1 2 2 3 1 3 2 1 ...(2)

From (1) and (2) |A| = |A′|. Hence proved.

2. Prove that a determinant changes its sign when 2 of its rows are interchanged.


a a a

b b b

c c c

=�

�

��

�

�

��

1 2 3

1 2 3

1 2 3

B

b b b

a a a

c c c

=�

�

��

�

�

��

1 2 3

1 2 3

1 2 3

is the matrix obtained by interchanging 1st and 2nd rows.

To prove: A B= − .

Now


A

a a a

b b b

c c c

=1 2 3

1 2 3

1 2 3

= − − − + −a b c b c a b c b c a b c b c1 2 3 3 2 2 1 3 3 1 3 1 2 2 1� � � � � �

= − − + + −a b c a b c a b c a b c a b c a b c1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 ...(1)

B

b b b

a a a

c c c

b a c a c b a c a c b a c a c= = − − − + −1 2 3

1 2 3

1 2 3

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1� � � � � �

= − − + + −a b c a b c a b c a b c a b c a b c2 1 3 3 1 2 1 2 3 3 2 1 1 3 2 2 3 1

= − − + + − − +a b c a b c a b c a b c a b c a b c2 1 3 3 1 2 1 2 3 3 2 1 1 3 2 2 3 1 ...(2)

From (1) and (2)

B A= −

⇒ A B= − .

Hence proved.

3.3.3.3.3. Prove that the value of a determinant is zero if any two of its rows are identical.


a a a

a a a

b b b

=1 2 3

1 2 3

1 2 3

be the determinant whose 2 rows, (1st and 2nd) are identical.

To prove A = 0

Now A

a a a

a a a

b b b

a a b a b a a b a b a a b a b= = − − − + −1 2 3

1 2 3

1 2 3

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1� � � � � �

= − − + + −a a b a a b a a b a a b a a b a a b1 2 3 1 3 2 1 2 3 2 3 1 1 3 2 2 3 1

= 0.

Hence proved.

4.4.4.4.4. If every element of any row of a determinant is multiplied by a non zero constant k then prove thatthe whole determinant is multiplied by k.


a a a

b b b

c c c

B

a a a

b b b

kc kc kc

=�

�

��

�

�

��

=�

�

��

�

�

��

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

and


which is obtained by multiplying 3rd row of A by k.

To prove B = kA.

Now A

a a a

b b b

c c c

a b c b c a b c b c a b c b c= = − − − + −1 2 3

1 2 3

1 2 3

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1� � � � � � ...(1)

B

a a a

b b b

kc kc kc

a kb c kb c a kb c kb c a kb c kb c= = − − − + −1 2 3

1 2 3

1 2 3

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1� � � � � �

= − − − + −k a b c b c a b c b c a b c b c1 2 3 3 2 2 1 3 3 1 3 1 2 2 1� � � � � �

B k A= . from 1 Hence proved.

5.5.5.5.5. If each element of any row of a determinant is the sum of two terms, then prove that the determi-nants can be expressed as the sum of two determinants i.e.,

Prove that

a x a y a y

b b b

c c c

a a a

b b b

c c c

x y z

b b b

c c c

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

+ + += +

PrPrPrPrProof:oof:oof:oof:oof: L.H.S.:

a x a y a z

b b b

c c c

1 2 3

1 2 3

1 2 3

+ + +

= + − + + +a xb b

c ca y

b b

c ca z

b b

c c12 3

2 32

1 3

1 33

1 2

1 2� � � � � �

= + − +��

��+ +a

b b

c cx

b b

c ca

b b

c cy

b b

c ca

b b

c cz

b b

c c12 3

2 3

2 3

2 32

1 3

1 3

1 3

1 33

1 2

1 2

1 2

1 2

Rearranging,

= − + + − +ab b

c ca

b b

c ca

b b

c cx

b c

c cy

b b

c cz

b b

c c12 3

2 32

1 3

1 33

1 2

1 2

2 3

2 3

1 3

1 3

1 2

1 2

= + =a a a

b b b

c c c

x y z

b b b

c c c

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

RHS.

Hence proved.


6.6.6.6.6. Prove that the value of a determinant is not altered if to the elements of any row the samemultiples of the corresponding elements of any other row are added.

i.e., prove that

a a a

b b b

c c c

a kb a kb a kb

b b b

c c c

R R kR1 2 3

1 2 3

1 2 3

1 1 2 2 3 3

1 2 3

1 2 3

1 1 2=+ + +

′ = +here� �

PrPrPrPrProof:oof:oof:oof:oof: R.H.S.:

a kb a kb a kb

b b b

c c c

1 1 2 2 3 3

1 2 3

1 2 3

+ + +

= +a a a

b b b

c c c

kb kb kb

b b b

c c c

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

using property 5 .

= +a a a

b b b

c c c

k

b b b

b b b

c c c

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

using property 4 .

= +a a a

b b b

c c c

k1 2 3

1 2 3

1 2 3

0 � 2 rows are identical .

= =a a a

b b b

c c c

1 2 3

1 2 3

1 2 3

L.H.S.

Hence proved.

7.7.7.7.7. Prove that if all the elements in a row of a determinant is zero then the determinant value is zero.


a a a

b b b=1 2 3

1 2 3

0 0 0 whose 3rd row is zero row.

A ab b

ab b

ab b

= − +12 3

21 3

31 2

0 0 0 0 0 0

= − + =a a a1 2 30 0 0 0 . Hence proved.


8.8.8.8.8. In a determinant if all the elements on one side of the principal diagonal are zeros, then prove thatthe value of the determinant is equal to the product of the elements in the principal diagonal.


a a a

b b

c

=1 2 3

1 2

1

0

0 0 be the determinant in which all the elements to the left of prin-

cipal diagonal are zeros.

To prove: A a b c= 1 1 1

Consider

a a a

b b

c

ab b

ca

b

ca

b1 2 3

1 2

1

11 2

12

2

13

10

0 00

0

0

0

0 0= − +

= − − +a b c a a1 1 1 2 30 0 0� �

= a b c1 1 1 . Hence proved.

Note:Note:Note:Note:Note: The above properties which are proved for the rows also holds good for columns because|A| = |A′| from property 1. So

• The determinant changes its sign when 2 of its columns are interchanged i.e.,

a a a

b b b

c c c

a a a

b b b

c c c

1 2 3

1 2 3

1 2 3

2 1 3

2 1 3

2 1 3

= −

• The value of the determinant is zero when 2 of its columns are identical i.e.,

a a a

b b b

c c c

1 1 3

1 1 3

1 1 3

0= .

• If every element of any column of a determinant is multiplied by constant k. Then the whole

determinant is multiplied by k i.e.,

ka a a

kb b b

kc c c

k

a a a

b b b

c c c

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

=

• If each element of any column of a determinant is sum of two terms then the determinant can beexpressed as the sum of 2 determinants.

i.e.,

a x a a

b y b b

c z c c

a a a

b b b

c c c

x a a

y b b

z c c

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

2 3

2 3

2 3

+++

= +

• The value of a determinant is not altered if to the elements of any column the same multiples ofthe corresponding element of any other column are added.


i.e.,

a a a

b b b

c c c

a a ka a

b b kb b

c c kc cc c kc

1 2 3

1 2 3

1 2 3

1 2 3 3

1 2 3 3

1 2 3 32 2 3

=+++ ′ = +

• If all the elements in a column of a determinant are zero then the determinant is zero.

i.e.,

a a

b b

c c

1 2

1 2

1 2

0

0

0

0=

'��(��#� �

1.1.1.1.1. Find the value of: 3860 3861

3862 3863

3860 3861

3862 386312 2 1′ = − = −R R R kUsing property 6 with

= =3860 3861

2 22

3860 3861

1 1

= − = − = −2 3860 3861 2 1 2

2.2.2.2.2. Evaluate: 81 82 83

84 85 86

87 88 89


81 82 83

84 85 86

87 88 89

2 2 1

3 3 2

′ = −′ = −

R R R

R R R

= =81 82 833 3 33 3 3

0 2� rows are identical

3. Without expanding prove that

x y y z z x

y z z x x y

z x x y y z

− − −− − −− − −

= 0


LHS:

x y y z z x

y z z x x y

z x x y y z

R R R R

− − −− − −− − −

′ = + +1 1 2 3

0 0 0

0 1y z z x x y

z x x y y z

− − −− − −

= � st row is zero.

4.4.4.4.4. Evaluate: 99 101 104

100 102 105101 103 106

′ = −R R R2 2 1

′ = −R R R3 3 1

= = =99 101 104

1 1 1

2 2 2

2

99 101 104

1 1 1

1 1 1

0 Since 2 rows are equal.

5.5.5.5.5. Solve for x:

x

x

x

++

+=

2 3 42 3 42 3 4

0

Taking ′ = + +C C C C1 1 2 3

x

x x

x x

++ ++ +

=9 3 4

9 3 4

9 3 4

0

Taking x + 9 common

x x

x

+ ++

=9

1 3 4

1 3 4

1 3 4

0

⇒+ == −

=x

xx

x

9 0

9

1 3 4

1 3 4

1 3

0or +

+ 4

′ = −R R R2 2 1


and ′ = −R R R3 3 1

1 3 4

0 0

0 0

0x

x

=

⇒ ⋅ ⋅ =1 0x x

x2 0=

⇒ =x 0

Hence

x x= − =9 0 or .

6.6.6.6.6. Prove that 1

1

1

1

++

+= + + +

x y z

x y z

x y z

x y z

LHS:

1

1

11 1 2 3

++

+′ = + +

x y z

x y z

x y z

C C C C

=+ + ++ + + ++ + + +

1

1 1

1 1

x y z y z

x y z y z

x y z y z

111 11 1

2 2 1

3 3 1+ + + +

+

′ = −′ = −

x y z

y z

y z

y z

R R R

R R R

110 1 00 0 1

+ + +x y z

y z

1 1 1 1 1+ + + = + + + =x y z x y z R.H.S.

7.7.7.7.7. Prove that 1

11

2

2

2

a a

b b

c c

a b b c c a= − − −


LHS:

1

11

2

2

2

2 2 3

3 3 1

a a

b b

c c

R R R

R R R

′ = −′ = −

= − −− −

= − − +− − +

1

00

1

00

2

2 2

2 2

2a a

b c b c

c a c a

a a

b c b c b c

c a c a c a

Taking (b − c) and (c − a) common from R2 and R3

b c c a

a a

b c

c a

R R R− − ++

′ = − 10 10 1

2

3 3 2

b c c a

a a

b c

a b

− − +−

10 10 0

2

Taking (a − b) common from R3

a b b c c a

a a

b c− − − + 10 10 0 1

2

a b b c c a− − − 1 1 1

= − − − =a b b c c a RHS. Hence proved.

8.8.8.8.8. Prove that

x p q

p x q

p q x

x p x q x p q= − − + +

LHS:

x p q

p x q

p q x

C C C C′ = + +1 1 2 3

=+ ++ ++ +

= + +′ = −′ = −

x p q p q

x p q x q

x p q q x

x p q

p q

x q

q x

R R R

R R R

1

1

1

2 2 1

3 3 2


= + + −− −

x p q

p q

x p

q x x q

10 00

= + + −− − −

x p q

p q

x p

x q x q

1

0 00

Taking (x − p) common from R2 and (x − q) from R3

x p x q x p q

p q

R R R− − + +−

′ = + 1

0 1 0

0 1 13 3 2

= − − + +x p x q x p q

p q

10 1 00 0 1

= − − + + ⋅ ⋅x p x q x p q 1 1 1

= − − + + =x p x q x p q RHS.

9.9.9.9.9. Prove that

1 1 1

3 3 3x y z

x y z

x y y z z x x y z= − − − + +

LHS:

1 1 1

3 3 3

1 1 2

2 2 3x y z

x y z

C C C

C C C′ = −′ = −

= − −− −

0 0 1

3 3 3 3 3

x y y z z

x y y z z

= − −− + + − + +

0 0 1

2 2 2 2 3x y y z z

x y x xy y x y y yz z z � � � �

= − −+ + + +

x y y z z

x xy y y yz z z

0 0 1

1 12 2 2 2 3


Expanding

x y y z y yz z x xy y− − − + + + − + + � � � �0 0 1 2 2 2 2

x y y z y yz z x xy y− − + + − − − 2 2 2 2

x y y z z x z x y z x− − − + + −

x y y z z x x y z− − − + +

= − − − + +x y y z z x x y z .Hence proved.

10.10.10.10.10. Prove that

a b c a b

c b c a b

c a c a b

a b c

+ ++ +

+ += + +

22

22 3

LHS:

a b c a b

c b c a b

c a c a b

C C C C

+ ++ +

+ +′ = + +

2

2

21 1 2 3

=+ ++ + + ++ + + +

2 2 2

2 2 2 2

2 2 2 2

a b c a b

a b c b c a b

a b c a c a b

= + + + ++ +

′ = −′ = −

2

1

1 2

1 2

2 2 1

3 3 1a b c

a b

b c a b

a c a b

R R R

R R R

= + + + ++ +

210 00 0

a b c

a b

a b c

a b c

= + + + + + +2 1a b c a b c a b c = 2 (a + b + c)3. Hence proved.

11.11.11.11.11. Prove that −

−−

=a ab ac

ab b bc

ac bc c

a b c

2

2

2

2 2 24

LHS:

−−

−

a ab ac

ab b bc

ac bc c

2

2

2


Taking a, b, c common from C1, C2 and C3 respectively we get

abc

a a a

b b b

c c c

−−

−

Taking a, b, c common from R1, R2 and R3 respectively

abc abcR R R

R R R ⋅

−−

−

′ = +′ = +

1 1 11 1 11 1 1

2 2 1

3 3 1

a b c2 2 21 1 1

0 0 2

0 2 0

⋅−

Expanding

a b c2 2 2 1 0 4 1 0 1 0− − − +

a b c a b c2 2 2 2 2 24 4 = = RHS.

12.12.12.12.12. Prove that 1 1 1

1 1 11 1 1

11 1 1

++

+= + + +�

��

a

b

c

abca b c

LHS:

1 1 1

1 1 1

1 1 1

++

+

a

b

c

Taking a, b, c common from R1, R2 and R3 respectively we get

abc

a

a a a

b

b

b b

c c

c

c

⋅

+

+

+

1 1 1

1 1 1

1 1 1

abc

a a a

b b b

c c c

⋅

+

+

+

11 1 1

11

1 1

1 11

1


abc

a b c a b c a b c

b b b

c c c

⋅

+ + + + + + + + +

+

+

11 1 1

11 1 1

11 1 1

11

1 1

1 11

1

′ = + +R R R R1 1 2 3

abca b c b b b

c c c

C C CC C C 1

1 1 1

1 1 1

11

1 1

1 11

1

2 2 1

3 3 1+ + +�

�� +

+

′ = −′ = −

abc

a b c b

c

11 1 1

1 0 0

11 0

10 1

+ + +�

��

abca b c

11 1 1

1 1 1+ + +�

�� ⋅ ⋅ ⋅

= + + +�

��abc

a b c 1

1 1 1Hence proved.

13.13.13.13.13. Prove that

a b c

a b c

bc ca ab

ab bc ca b c c a a b2 2 2 = + + − − − .

LHS:

a b c

a b c

bc ca ab

C C C

C C C2 2 2 1 1 2

2 2 3

′ = −′ = −

a b b c c

a b b c c

bc ca ca ab ab

− −− −− −

22 2 2 2

a b b c c

a b a b b c b c c

c a b a b c ab

− −− + − +− − − −

2

Taking (a − b) common from C1, (b − c) common from C2


= − − + +− −

′ = −a b b c

c

a b b c c

c a ab

C C C 1 1

21 1 2

= − − − +− −

a b b c

c

a c b c c

a c a ab

0 12

Taking (c − a) common from C1

= − − − − +− −

′ = −a b b c c a

c

b c c

a ab

R R R 0 1

1

1

22 2 3

= − − − + + −− −

a b b c c a

c

a b c c ab

a ab

0 101

2

Expanding

= − − − − + − + + + +a b b c c a c ab c a b c � � 0 1 0 02

= − − − − / + + + + /a b b c c a c ab ac bc c 2 2

a b b c c a ab bc ca− − − + + . Hence proved.

14.14.14.14.14. Prove that

11

11

2

2

2

3 2x x

x x

x x

x= −� � .

LHS:

11

1

2

2

21 1 2 3

x x

x x

x x

C C C C′ = + +

=+ ++ ++ +

1

1 11 1

2 2

2

2 2

x x x x

x x x

x x x

= + +111 11 1

2

2

2x x

x xx

x� �


= + + − −− −

1

1

0 1

0 1

2

2

2

2 2

x x

x x

x x x

x x x� � ′ = −

′ = −R R RR R R

2 2 1

3 3 1

= + + − −− − −

110 1 1

0 1 1

2

2

2

x x

x x

x x x

x x x� �

� �

Taking (1 − x) common from R2, (1 − x) from R3

� 1 1 1 12 2 2− = − = − +x x x x

= + + − ⋅ −− +

1 1 110 10 1

2

2

x x x x

x x

x

x x� �

expanding, we get

= + + − + +1 1 12 2 2x x x x x� � � �

= + + − ⋅ + +1 1 12 2 2x x x x x� � � �

1 12 2 2+ + −x x x� � � 1 1 1 12 3 3 3− + + = − = −x x x x x � �

= − =1 3 2x� � RHS.

Hence proved.

15.15.15.15.15. If x, y, z are all different and

x x x

y y y

z z z

2 3

2 3

2 3

1

1

1

0

+++

= , then prove that 1 +xyz = 0.

Given

x x x

y y y

z z z

2 3

2 3

2 3

111

0+++

= .

⇒x x

y y

z z

x x x

y y y

z z z

2

2

2

2 3

2 3

2 3

1

11

0+ = (using property of determinants)


⇒x x

y y

z z

xyz

x x

y y

z z

2

2

2

2

2

2

1

11

1

11

0+ =

⇒

x x

y y

z z

xyz

x x

y y

z z

R R

R R

2

2

2

2

2

2

1 2

2 3

1

11

1

11

0+ = ��

��

By interchanging and Then and

⇒ 1111

0

2

2

2+ =xyz

x x

y y

z z

⇒ 1 + xyz = 0. � x, y and z are all different,

x x

y y

z z

2

2

2

111

0≠ .

Hence proved.

��0 ��1��2!��1��3��4�� !� 5��

Let [aij] be a square matrix of order n × n. Then minor of an element aij is the determinant obtained bydeleting the row and the column containing it (i.e. ith row & jth column). If minors are multiplied by(−1)i + j i.e., with proper signs + or − we get co-factors. Adjoint of a matrix is the transpose of co-factormatrix.

Illustration:

1.1.1.1.1. If A = ��

��

1 2

3 4, then

minor of 1 = 4

minor of 2 = 3

minor of 3 = 2

minor of 4 = 1

Co-factor matrix = 4 32 1

−−��

��

+ −− +��

��

Multiplying by

Adjoint of A =−

−��

��

4 2

3 1 Taking transpose of co-factor matrix i.e., interchanging rows and

columns.


2.2.2.2.2. If A =�

�

��

�

�

��

1 2 32 3 13 1 2

, then

minor of 1 = 3 1

1 26 1 5

1 2 3

2 3 1

3 1 2

= − =�

�

��

�

�

��

.

minor of 2 = 2 13 2

4 3 11 2 32 3 13 1 2

= − =�

�

��

�

�

��

.

minor of 3 = 2 33 1

2 9 71 2 32 3 13 1 2

= − = −�

�

��

�

�

��

.

Similarly minor of 2 : 2 31 2

4 3 11 2 32 3 13 1 2

= − =�

��

�

��.

minor of 3 : 1 33 2

2 9 7= − = − .

minor of 1 = 1 2

3 11 6 5= − = − .

minor of 3 = 2 33 1

2 9 71 2 32 3 13 1 2

= − = −�

�

��

�

�

��

.

minor of 1 = 1 32 1

1 6 5= − = − .

minor of 2 = 1 2

2 33 4 1= − = − .

Hence matrix of minors = 5 1 71 7 57 5 1

−− −

− − −

�

�

��

�

�

��


Multiplying with + − +− + −+ − +

�

�

��

�

�

��

the corresponding elements

We get co-factor matrix = 5 1 71 7 57 5 1

− −− − +− + −

�

��

�

��

Taking transpose, we get Adjoint matrix

∴ Adjoint of A = 5 1 7

1 7 5

7 5 1

− −− −− −

�

�

��

�

�

��

��

If A is a non-singular square matrix of order n × n then there exists a square matrix B of order n × n suchthat AB = BA = I where I is the identity matrix of order n × n. Here B is called inverse of A. It is denotedby A−1.

∴ A A A A I⋅ = ⋅ =− −1 1

Inverse of A can be found by using the formula

AA

A− =1 adjoint of

.

i.e., AA

A− =1 adj

.

Illustration:

��

1 12 2

−��

��

Let A =−�

��

1 1

2 2

A = − −1 2 1 2� � � ��

= + = ≠2 2 4 0.

∴ A is non-singular. Hence A−1 exists.

AA

A− =1 adj

formula


Now minor of 1 = 21 1

2 2

−��

��

minor of −1 = 2

minor of 2 = −1

minor of 2 = 1

∴ Co-factor matrix = 2 2

1 1

−��

��

. By multiplying minors with + −− +��

��

Taking transpose we get

Adjoint of A = 2 1

2 1−��

��

∴ A− = ⋅−��

��

1 1

4

2 1

2 1=

−

�

�

��

�

�

��

1

2

1

41

2

1

4

.

Note:Note:Note:Note:Note: We can write adjoint of a square matrix of order 2 × 2 directly by interchanging the principaldiagonal elements and changing the signs of secondary diagonal elements.

��

(i) If A = ��

��

1 7

2 3, then

Adj interchange 1 and 3, change the signs of 2 and 7.A =−

−��

��

3 7

2 1

(ii) If A A=−�

��

=−��

��

1 7

2 6

6 7

2 1, then adj

(iii) If A A=−−��

��

=−−

��

��

1 5

2 7

7 5

2 1, then adj

(iv) If A A=−

− −��

��

=−��

��

0 5

6 2

2 5

6 0, then adj

(v) If A A=− −− −��

��

=−

−��

��

1 6

8 7

7 6

8 1, then adj

��!"��#�

Let A =�

�

��

�

�

��

1 3 2

1 2 2

2 0 0


A = = − +1 3 21 2 22 0 0

12 20 0

31 22 0

21 22 0

= − − + −1 0 3 0 4 2 0 4� � � � � �

= − − + −1 0 3 0 4 2 0 4� � � � � �

0 12 8 4 0+ − = ≠Hence A−1 exists.

AA

A− =1 adj

.

Now A =�

�

��

�

�

��

1 3 21 2 22 0 0

minor of 1 = 2 2

0 00=

minor of 3 = 1 2

2 00 4 4= − = −

minor of 2 = 1 2

2 00 4 4= − = −

Similarly minor of 1 = 3 2

0 00=

minor of 2 = 1 2

2 00 4 4= − = −

minor of 2 = 1 32 0

0 6 6= − = −

Similarly minor of 2 = 3 2

2 26 4 2= − =

minor of 0 = 1 2

1 22 2 0= − =

minor of 0 = 1 3

1 22 3 1= − = −


∴Matrix of minors = 0 4 40 4 62 0 1

− −− −

−

�

�

��

�

�

��

+ − +− + −+ − +

�

�

��

�

�

��

Multiplying with

We get

Co-factors matrix = 0 4 40 4 62 0 1

−−

−

�

�

��

�

�

��

Taking transpose we get

Adj A = −− −

�

�

��

�

�

��

0 0 24 4 04 6 1

∴ AA

A− = = −

− −

�

�

��

�

�

��

1 1

4

0 0 24 4 04 6 1

Adj

A− = −− −

�

�

��

�

�

��

10 0 1 2

1 1 0

1 3 2 1 4

Note:Note:Note:Note:Note: We can write the adjoint of a square matrix directly by writing the minors multiplied by propersigns (+ or −) column wise.

i.e. If A =�

�

��

�

�

��

1 3 2

1 2 2

2 0 0

Then Adj A =

2 20 01 22 01 22 0

3 20 0

3 22 2

1 22 0

1 21 2

1 32 0

1 31 2

−

−

−

−

�

�

��

�

�

��

Adj A = 0 0 24 4 04 6 1

−− −

�

�

��

�

�

��


$��%�&�� '(��

1.1.1.1.1. Find A−1 if A =−

��

��

1 7

2 6

Solution:Solution:Solution:Solution:Solution: AA

A− =1 adj

formula� �

Now adj A =− −−��

��

6 7

2 1

By interchanging principal diagonal

elements and changing signs of

secondary diagonal elements.

��

�

��

A =−

= − − = −1 72 6

6 14 20

∴ A− =−

− −−��

��

1 1

20

6 7

2 1

A− =−

�

�

��

�

�

��

=−

�

�

��

�

�

��

1

6

20

7

202

20

1

20

3

10

7

201

10

1

20.

2.2.2.2.2. Verify A. adj A = |A|.I for the matrix A =−

��

��

1 2

3 1

Given: A =−

��

��

1 2

3 1

Hence adj A =− −−��

��

1 2

3 1

Consider A A⋅ =−

��

��

=− −−��

��

→

adj1 2

3 1

1 2

3 1

=− + − − +

− + − − − + −��

��

1 1 2 3 1 2 2 13 1 1 3 3 2 1 1� � � � � � � �

� � � ��

=− − − +− + − −��

��

=−

−��

��

1 6 2 2

3 3 6 1

7 0

0 7


A A I⋅ = −�

��

�

��

= −adj 71 0

0 17

But A =−

= − − = − − = −1 2

3 11 1 3 2 1 6 7� � � � .

∴ A.adjA = |A|.I.

3.3.3.3.3. Find A−1 if A = −�

�

��

�

�

��

1 2 3

0 1 1

1 1 2

Solution:Solution:Solution:Solution:Solution: AA

A− =1 adj

Now A = −1 2 3

0 1 1

1 1 2

= 1 (−2 −1) − 2 (0 − 1) + 3 (0 + 1)

= −3 + 2 + 3 = 2 ≠ 0.

Hence A−1 exists.

Now A = −�

�

��

�

�

��

1 2 3

0 1 1

1 1 2

Adj A =

−−

−

− −

−−

−

�

�

��

�

�

��

1 11 2

2 31 2

2 31 1

0 11 2

1 31 2

1 30 1

0 11 1

1 21 1

1 20 1

=− − − − +− − − −

+ − − −

�

�

��

�

�

��

2 1 4 3 2 3

1 2 3 1

0 1 1 2 1

� ��

� �

Adj A =− −

− −−

�

�

��

�

�

��

3 1 51 1 11 1 1


∴ AA

A− = =

− −− −

−

�

��

�

��

1 1

2

3 1 51 1 11 1 1

adj

4.4.4.4.4. Verify AA A A I A− −= = =−��

��

1 1 1 2

1 3 if .

Solution:Solution:Solution:Solution:Solution: Given A =−��

��

1 2

1 3

A =−

= − − =1 2

1 33 2 5� � .

∴ AA

A− = =

−��

��

1 1

5

3 2

1 1adj

Now AA− →

= −��

��

−��

�� = + − +

− + +��

��

1 1 21 3

1

5

3 21 1

1

5

3 2 2 23 3 2 3

AA I− = ��

��

= ��

��

=1 1

5

5 0

0 5

1 0

0 1

Similarly, A A− →

⋅ = −��

��⋅ −��

��1 1

5

3 21 1

1 21 3

=− − −+ − +

��

��

1

5

3 2 6 61 1 2 3� ��

= ��

��

= ��

��

=1

5

5 0

0 5

1 0

0 1I

Hence AA−1 = A−1A = I

)�* +,�+��+��

If A is any square matrix of order n × n and I is identity matrix of the same order then |A − λI| = 0 whereλ is a constant is called characteristic equation of a square matrix A. The roots of the equation |A − λI|= 0 (i.e., value of λ which satisfies this equation) are called characteristic roots or eigen values.

��

1.1.1.1.1. If A = ��

��

1 4

2 3 then


A I− = ��

��

− ��

��

λ λ1 4

2 3

1 0

0 1

= ��

��

− ��

��

=−

−��

��

1 4

2 30

0

1 4

2 3

λλ

λλ

A I− =−

−λ

λλ

1 42 3

Characteristic equation A I− =λ 0

∴1 4

2 30

−−

=λ

λ

1 3 8 0− − − =λ λ� � � �

3 3 8 02− − + − =λ λ λ

λ λ2 4 5 0− − = . This is the characteristic equation.

λ λ λ− + − =5 1 5 0� � � �

λ λ− + =5 1 0� ��

λ λ= = −5 1 or .

Hence the characteristic roots (or eigen values) are 5 and −1.

2.2.2.2.2. If A = −−

�

�

��

�

�

��

4 0 1

2 1 0

2 0 1

Then A I− = ⇒−− −− −

=λλ

λλ

04 0 1

2 1 02 0 1

0.

41 0

0 10 1

2 1

2 00−

−−

− +− −−

=λλ

λλ

� �

4 1 1 0 2 1 0− − − + + − =λ λ λ λ� ��

1 4 1 2 0− − − + =λ λ λ� � � ��

1 0 4 1 2 0− = − − + =λ λ λ or � ��


λ λ λ λ= − − + + =1 4 4 2 02 or

λ λ2 5 6 0− + =

λ λ λ2 3 2 6 0− − + =

λ λ λ− − − =3 2 3 0� � � �

λ λ− − =3 2 0� ��

λ λ λ= = =1 3 2 or or .

∴ Characteristic roots or eigen values are 1, 2 and 3.

)��- +.(�.�, �(��,��

Statement: Every square matrix A satisfies its characteristic equation, A I− =λ 0.

Illustration: If A = ��

��

1 22 3

Then characteristic equation is A I− =λ 0.

i.e.,1 2

2 30

−− =λ

λ

1 3 4 0− − − =λ λ� ��

3 3 4 02− − + − =λ λ λ

λ λ2 4 1 0− − = .

This is characteristic equation.

By Cayley Hamilton theorem ‘A’ satisfies this equation. So A A I2 4 0− − = ...(i)

where I is identity matrix of same order as that of A.

Now to verify Cayley Hamilton theorem,

We have to verify A2 − 4A − I = 0.

Now A2 1 22 3

1 22 3

= ��

��

��

→

A2 1 4 2 6

2 6 4 9

5 8

8 13=

+ ++ +

��

��

= ��

��

4 41 2

2 3

4 8

8 12A = �

��

= ��

��


I = ��

��

1 0

0 1

A A I2 45 8

8 13

4 8

8 12

1 0

0 1− − = �

��

− ��

��

− ��

��

=− − − −− − − −

��

��

= ��

��

5 4 1 8 8 0

8 8 0 13 12 1

0 0

0 0.

Hence verified.

Note: Note: Note: Note: Note: From (1) A A I2 4 0− − = .

Operating by A−1

A A A A A I− − −⋅ − ⋅ − =1 2 1 14 0.

A I A− − =−4 01

⇒ A A I− = −1 4 .

A− = ��

��

− ��

��

1 1 2

2 34

1 0

0 1

A− = ��

��

− ��

��

1 1 2

2 3

4 0

0 4

A− =−

−��

��

1 3 2

2 1

Hence we can find inverse of a square matrix by using Cayley Hamilton theorem.

$��%�&�� '(��

1.1.1.1.1. Find the eigen values of the matrix 1 4

3 2��

��

.

Solution:Solution:Solution:Solution:Solution: Characteristic equation is

A I− =λ 0

1 43 2

0−

−=

λλ

1 2 4 3 0− − − =λ λ� ��

2 2 12 02− − + − =λ λ λ


λ λ2 3 10 0− − =

λ λ λ2 5 2 10 0− + − =

λ λ λ− + − =5 2 5 0� � � �

λ λ= = −5 2 or .

Hence the eigen values are λ = 5 and λ = −2.

2.2.2.2.2. Find the characteristic roots of the matrix 1 1

3 1−��

��.

Solution:Solution:Solution:Solution:Solution: Characteristic equation is A I− =λ 0.

1 1

3 10

−− −

=λ

λ.

1 1 3 0− − − − =λ λ� ��

− − + − =1 1 3 0λ λ� ��

− − − =1 3 02 2λ� �

− + − =1 3 02λ

λ λ λ2 24 0 4 2− = ⇒ = ⇒ = ± .

Hence characteristic roots are +2 and −2.

3.3.3.3.3. Verify Cayley Hamilton theorem for the matrix 1 1

2 6

−��

��

.

Solution:Solution:Solution:Solution:Solution: Characteristic equation is A I− =λ 0

1 1

2 60

− −−

=λ

λ.

1 6 2 1 0− − − − =λ λ� ��

6 6 2 02− − + + =λ λ λ

λ λ2 7 8 0− + = .


By Cayley Hamilton theorem every square matrix obeys its characteristic equation.

Hence it is required to verify A2 − 7A + 8I = 0


Now A A A2 1 12 6

1 12 6

= ⋅ =−�

��

−��

��

→

1 2 1 6

2 12 2 36

1 7

14 34

− − −+ − +

��

��

=− −��

��

7 71 12 6

7 714 42

A = −��

��

= −��

��

8 81 0

0 1

8 0

0 8I = �

��

= ��

��

Now LHS: A A I2 7 8− +

=− −��

��

−−�

��

+ ��

��

1 7

14 34

7 7

14 42

8 0

0 8

=− − + − − − +

− + − +��

��

= ��

��

=1 7 8 7 7 0

14 14 0 34 42 8

0 00 0

� �R.H.S.

Hence verified.

4.4.4.4.4. Verify Cayley Hamilton theorem for the matrix A =−

−

�

�

��

�

�

��

1 1 0

2 1 0

1 2 1

Solution:Solution:Solution:Solution:Solution: Characteristic equation is A I− =λ 0

1 1 0

2 1 0

1 2 1

0

− −−

− −=

λλ

λ

11 0

2 11

2 01 1

0 0−−

− −− −

− −+ =λ

λλ λ� � � �

1 1 1 0 1 2 1 0 0 0− − − − − + − − − + =λ λ λ λ� � � ��

− − − − + =1 1 2 1 02 2λ λ λ� � � �

− − − − − =1 1 2 2 02λ λ λ� ��

− − − + − − =1 2 2 02 3λ λ λ λ� �


− + − − =λ λ λ3 2 3 0.


To verify Cayley Hamilton theorem, we have to verify − + − − =A A A I3 2 3 0

Now A A A2

1 1 02 1 01 2 1

1 1 02 1 01 2 1

= ⋅ =−

−

�

�

��

�

�

��

−

−

�

�

��

�

�

��

→

=− + − − + + ++ + − + + + ++ − − + − + +

�

�

��

�

�

��

1 2 0 1 1 0 0 0 0

2 2 0 2 1 0 0 0 0

1 4 1 1 2 2 0 0 1

A21 2 0

4 1 0

4 1 1

=− −

−−

�

�

��

�

�

��

A A A3 2

1 2 04 1 04 1 1

1 1 02 1 01 2 1

= ⋅ =− −

−−

�

�

��

�

�

��

−

−

�

�

��

�

�

��

→

=− − + − + + +

− + − − + + +− + − − + + −

�

�

��

�

�

��

1 4 0 1 2 0 0 0 04 2 0 4 1 0 0 0 04 2 1 4 1 2 0 0 1

A35 1 02 5 03 3 1

=− −

−− −

�

�

��

�

�

��

Now A I=− −

−− −

�

�

��

�

�

��

=�

�

��

�

�

��

=�

�

��

�

�

��

5 1 02 5 03 3 1

3 31 0 00 1 00 0 1

3 0 00 3 00 0 3

and

Consider LHS,

− + − −A A A I3 2 3

= −− −

−− −

�

�

��

�

�

��

+− −

−−

�

�

��

�

�

��

−−

−

�

�

��

�

�

��

−�

�

��

�

�

��

5 1 0

2 5 0

3 3 1

1 2 0

4 1 0

4 1 1

3 1 0

2 1 0

1 2 1

3 0 0

0 3 0

0 0 3


=− − − − + − + − −

− + − − − − − + − −− + − + − − + + + −

�

�

��

�

�

��

5 1 1 3 1 2 1 0 0 0 0 0

2 4 2 0 5 1 1 3 0 0 0 0

3 4 1 0 3 1 2 0 1 1 1 3

=�

�

��

�

�

��

0 0 00 0 00 0 0

Hence Cayley Hamilton theorem is verified.

5.5.5.5.5. By using Cayley Hamilton theorem find A−1 if A =−��

��

1 7

6 5

Characteristic equation = A I− =λ 0

1 76 5

0−− −

=λ

λ

1 5 6 7 0− − − − =λ λ� ��

5 5 42 02− − + + =λ λ λ

λ λ2 6 47 0− + = .

By Cayley Hamilton theorem, A2 − 6A + 47I = 0

Operating by A−1,

A A A A A I− − −− ⋅ + ⋅ =1 2 1 16 47 0

A I A− + =−6 47 01

⇒ 47 61A I A− = −

A I A− = −1 1

476

Now 6 61 00 1

6 00 6

I = ��

��

= ��

��

66 0

0 6

1 7

6 5

5 7

6 1I A− = �

��

−−��

��

=−�

��

∴ A− =−�

��

1 1

47

5 7

6 1.


6.6.6.6.6. Find A−1 if A = −�

�

��

�

�

��

3 2 14 1 01 3 1

by using Cayley Hamilton theorem.

Solution:Solution:Solution:Solution:Solution: Characteristic equation: A I− =λ 0

3 2 1

4 1 0

1 3 1

0

−− −

−=

λλ

λ

31 03 1

24 01 1

14 11 3

0−− −

−−

−+

− −=λ

λλ λ

λ� �

3 1 1 0 2 4 1 0 1 12 1 0− − − − − − − − + − − − =λ λ λ λ λ� � � ��

− − − − − + + + =3 1 8 1 12 1 02λ λ λ λ� ��

− + + − − + + + =3 3 8 8 13 02 3λ λ λ λ λ

− + + + =λ λ λ3 23 10 2 0

According to Cayley Hamilton theorem,

− + + + =A A A I3 23 10 2 0

Operating by A−1

− ⋅ + ⋅ + + ⋅ =− − − −A A A A AA A I3 1 2 1 1 13 10 2 0

− + + + =−A A I A2 13 10 2 0

⇒ A A A I− = − −1 21

23 10 .

Now,

A A A2

3 2 14 1 01 3 1

3 2 14 1 01 3 1

= ⋅ = −�

�

��

�

�

��

−�

�

��

�

�

��

→

=+ + − + + +− + + + + ++ + − + + +

�

�

��

�

�

��

9 8 1 6 2 3 3 0 112 4 0 8 1 0 4 0 03 12 1 2 3 3 1 0 1

A218 7 4

8 9 4

16 2 2

=�

�

��

�

�

��


3 33 2 14 1 01 3 1

9 6 312 3 0

3 9 3A = −

�

�

��

�

�

��

= −�

�

��

�

�

��

10 10

1 0 0

0 1 0

0 0 1

10 0 0

0 10 0

0 0 10

I =�

�

��

�

�

��

=�

�

��

�

�

��

A A I2 3 10

18 7 4

8 9 4

16 2 2

9 6 3

12 3 0

3 9 3

10 0 0

0 10 0

0 0 10

− − =�

�

��

�

�

��

− −�

�

��

�

�

��

−�

�

��

�

�

��

=−−

− −

�

�

��

�

�

��

1 1 14 2 4

13 7 11

∴ A A A I− = − − =−−

− −

�

�

��

�

�

��

1 21

23 10

1

2

1 1 14 2 4

13 7 11

)�� (��(��.��

Consider 3 equations in 3 variables:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3.

The solution is the value of x, y and z which simultaneously satisfy the above equations. The solutioncan be obtained by various methods. Cramer’s rule and matrix method are 2 such methods of solvingthe system of equations.

��+��/��"0��

Let ∆ =a b c

a b c

a b c

1 1 1

2 2 2

3 3 3

Multiplying both sides by x

x x

a b c

a b c

a b c

∆ =1 1 1

2 2 2

3 3 3


x

a x b c

a x b c

a x b c

C C C y C z∆ = ′ = + +1 1 1

2 2 2

3 3 3

1 1 2 3

x

a x b y c z b c

a x b y c z b c

a x b y c z b c

∆ =+ ++ ++ +

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

x

d b c

d b c

d b cx∆ ∆= =

1 1 1

2 2 2

3 3 3

say� �

∴ x x∆ ∆=

⇒ x x= ∆∆

.

Similarly, ∆ =a b c

a b c

a b c

1 1 1

2 2 2

3 3 3

Multiplying by y

y

a b y c

a b y c

a b y c

C C xC zC∆ = ′ = + +1 1 1

2 2 2

3 3 3

2 2 1 3

y

a b y a x c z c

a b y a x c z c

a b y a x c z c

∆ =+ ++ ++ +

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

y

a d c

a d c

a d cy∆ ∆= =

1 1 1

2 2 2

3 3 3

say� �

⇒ y y∆ ∆=

y y=∆∆


Similarly z z= ∆∆

.

Note:Note:Note:Note:Note: If there are 2 equations in 2 variables,

Say a1x + b1y = d1

a2x + b2y = d2

Then assume ∆ =a b

a b1 1

2 2

∆ x

d b

d b= 1 1

2 2 which is obtained by replacing the 1st column by d1 and d2

∆ y

a d

a d= 1 1

2 2 which is obtained by replacing 2nd column by d1 and d2.

By Cramer’s rule,

x yx y= =∆∆

∆∆

and .

�� 1��

To solve

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

Assume A

a b c

a b c

a b c

X

x

y

z

D

d

d

d

=�

�

��

�

�

��

=�

�

��

�

�

��

=�

�

��

�

�

��

1 1 1

2 2 2

3 3 3

1

2

3

, and

Then matrix equation is AX = D

Operating by A−1 [��A is non-singular, A−1 exists]

A AX A D− −⋅ =1 1

IX A D= −1

X A D= −1

i.e.,

x

y

z

A D�

�

��

�

�

��

= −1 .


Here A−1 can be calculated either by using Cayley Hamilton theorem or by using the formula

AA

A− =1 adj

.

$��%�&�� '(��

��0��1��00�2��3�45�+��/��"0��

1.1.1.1.1. 3x + 4y = 11

2x + 3y = 8

Let ∆ = = − = − =3 4

2 33 3 2 4 9 8 1� � � � .

∆ x = = − = − =11 48 3

11 3 4 8 33 32 1� � � �

∆ y = = − = − =3 11

2 83 8 11 2 24 22 2� � � �

x yx y= =∆∆

∆∆

and .

x y= =1

1

2

1 and

x = 1 and y = 2.

2.2.2.2.2. 2x + 4y = 7

x − 7y = 6.

Let ∆ =−

= − − = − − = −2 4

1 72 7 4 1 14 4 18� � � � .

∆ x =−

= − − = − − = −7 4

6 77 7 4 6 49 24 73� � � � .

∆ y = = − = − =2 7

1 62 6 7 1 12 7 5� � � � .

x yx y= =∆∆

∆∆

and

x y= −−

=−

73

18

5

18 and

x y= = −73

18

5

18 and .


3.3.3.3.3. 2x −4y + 3z = 3

3x + 3y + 2z = 15

5x − 2y + 2z = 7

Let ∆ =−

−

2 4 33 3 25 2 2

=−

− − +−

23 2

2 24

3 2

5 23

3 3

5 2� �

2 6 4 4 6 10 3 6 15+ + − + − −� � � � � �

2 10 4 4 3 21� � � � � �+ − + −

20 16 63− −

∆ = − = −20 79 59.

∆ x =−

−

3 4 315 3 27 2 2

=−

− − +−

33 2

2 24

15 2

7 23

15 3

7 2� �

= 3 (6 + 4) + 4 (30 − 14) + 3 (−30 − 21)

3 (10) + 4 (16) + 3 (−51)

30 + 64 − 153

∆x = 94 − 153 = −59.

∆ y =2 3 3

3 15 2

5 7 2

= − +215 2

7 23

3 2

5 23

3 15

5 7

= 2 (30 − 14) − 3 (6 − 10) + 3 (21 − 75)

= 2 (16) − 3(−4) + 3(−54)

= 32 + 12 − 162

∆y = 44 − 162 = −118.


∆ z =−

−

2 4 33 3 155 2 7

=−

− − +−

23 15

2 74

3 15

5 73

3 3

5 2� �

= 2 (21 + 30) + 4 (21 − 75) + 3 (−6 − 15)

= 2 (51) + 4 (−54) + 3 (−21)

102 − 216 − 63

∆z = 102 − 279 = −177.

x x= = −−

=∆∆

59

591

y y= = −−

=∆∆

118

592

z z= = −−

=∆∆

177

593

∴ x y z= = =1 2 3, . and

4.4.4.4.4. x − y − 2z = 3

2x + y + z = 5

4x − y − 2z = 11

Let ∆ =− −

− −

1 1 2

2 1 1

4 1 2

=− −

− −−

+ −−

11 1

1 21

2 1

4 22

2 1

4 1� � � �

= 1 (−2 − (−1)) + 1(−4 − 4) − 2 (−2 − 4)

= 1 (−2 + 1) + 1(−8) − 2 (−6)

= −1 −8 + 12 = +3.

∆ x =− −

− −

3 1 2

5 1 1

11 1 2

=− −

− −−

+ −−

31 1

1 21

5 1

11 22

5 1

11 1� � � �


3 2 1 1 10 11 2 5 11− − − + − − − − −� ��

3 2 1 1 21 2 16− + + − − −� � � � � �

3 1 1 21 2 16− + − − −� � � � � �

− − +3 21 32

− + =24 32 8.

∆ y =−

−

1 3 2

2 5 1

4 11 2

=−

−−

+ −15 1

11 23

2 1

4 22

2 5

4 11� �

=1 10 11 3 4 4 2 22 20− − − − − − −� � � � � �

= − − − −21 3 8 2 2� � � �

= − + −21 24 4

= − + = −25 24 1.

∆ z =−

−=

−− − +

−

1 1 3

2 1 5

4 1 11

11 5

1 111

2 5

4 113

2 1

4 1� �

1 11 1 5 1 22 20 3 2 4− − + − + − −� ��

11 5 1 2 3 6+ + + −� � � �

16 2 18 0+ − = .

∴ x y zx y z= = =∆∆

∆∆

∆∆

, and

x y z= = − =8

3

1

3

0

3, and

x y z= = − =8

3

1

30, . and

��3��1��0��1��00�2��3��5��!"��

1.1.1.1.1. 2x − 3y = 4

3x + 2y = 5


Let A Xx

yD=

−��

��

= ��

= ��

2 3

3 2

4

5, and

Matrix equation AX = D

X = A−1D.

Now AA

A− =1 adj

Adj and A A=−��

��

=−

= − − = + =2 3

3 2

2 3

3 24 9 4 9 13� �

∴ AA

A− = =

−��

��

1 1

13

2 3

3 2adj

Now X A D= =−��

��−

→

1 1

13

2 3

3 2

4

5

X =+

− +��

��

1

13

8 15

12 10

X =−��

1

13

23

2

x

y��

=−

�

�

��

�

�

��

23

132

13

⇒ x y= = −23

13

2

13; .

2.2.2.2.2. 2x − 3y = 4

4x − 5y = 10.

Let A Xx

yD=

−−

��

��

= ��

= ��

2 3

4 5

4

10, and

Matrix equation: AX = D

⇒ X = A−1D.

Now AA

A− =1 adj


Adj and A A=−−��

��

= − − − = − + =5 3

4 22 5 3 4 10 12 2� � � � .

AA

A− = =

−−��

��

1 1

2

5 3

4 2adj

X A D= =−−��

��−

→

1 1

2

5 3

4 2

4

10

=− +− +��

��

1

2

20 30

16 20

X = ��

= ��

1

2

10

4

5

2

x

y��

= ��

5

2

⇒ x y= =5 2 and .

3.3.3.3.3. x + y = 1

y + z = 7

z + x = 2

Given equations: x + y + 0z = 1

0x + y + z = 7

x + 0y + z = 2

Let A X

x

y

z

D=�

�

��

�

�

��

=�

�

��

�

�

��

=�

�

��

�

�

��

1 1 00 1 11 0 1

172

, . and


⇒ X = A−1D

Now AA

A− =1 adj

Consider A =�

�

��

�

�

��

1 1 00 1 11 0 1

A = − − = + =1 1 1 1 1 1 2� � � � .


Adj A =

−

− −

−

�

�

��

�

�

��

1 10 1

1 00 1

1 01 1

0 1

1 1

1 0

1 1

1 0

0 10 11 0

1 11 0

1 10 1

adj A =−

− − −− − −

�

�

��

�

�

��

=−

−−

�

�

��

�

�

��

1 1 1

1 1 1

1 1 1

1 1 1

1 1 1

1 1 1� �

� �

AA

A− = =

−−

−

�

�

��

�

�

��

1 1

2

1 1 1

1 1 1

1 1 1

adj

X A D= =−

−−

�

�

��

�

�

��

�

�

��

�

�

��

=− ++ −

− + +

�

�

��

�

�

��

−1 1

2

1 1 11 1 11 1 1

172

1

2

1 7 21 7 21 7 2

X =−�

�

��

�

�

��

=−�

�

��

�

�

��

1

2

468

234

x

y

z

x y z�

�

��

�

�

��

=−�

�

��

�

�

��

⇒ = − = =2

3

4

2 3 4, .and

x = −2, y = 3 and z = 4.

4.4.4.4.4. x + y + z = 3

x + 2y + 3z = 4

x + 4y + 9z = 6.

Let A X

x

y

z

D=�

�

��

�

�

��

=�

�

��

�

�

��

=�

�

��

�

�

��

1 1 11 2 31 4 9

346

, and


⇒ X = A−1D.

Now AA

A− =1 adj


Consider A =�

�

��

�

�

��

1 1 11 2 31 4 9

A = − − − + −1 18 12 1 9 3 1 4 2� � � � � �

A = − + =6 6 2 2.

adj A =

+ − +

− + −

+ − +

�

�

��

�

�

��

2 34 9

1 14 9

1 12 3

1 31 9

1 11 9

1 11 3

1 21 4

1 11 4

1 11 2

adj A =− − − −

− − − − −− − − −

�

�

��

�

�

��

18 12 9 4 3 2

9 3 9 1 3 1

4 2 4 1 2 1

� � � ��

� �

adj A =−

− −− +

�

�

��

�

�

��

6 5 16 8 2

2 3 1

AA

A− = =

−− −

− +

�

�

��

�

�

��

1 1

2

6 5 16 8 22 3 1

adj

Now X A D= =−

− −− +

�

�

��

�

�

��

�

�

��

�

�

��−

→

1 1

2

6 5 16 8 22 3 1

346

X =− +

− + −− +

�

�

��

�

�

��

=�

�

��

�

�

��

1

2

18 20 618 32 126 12 6

1

2

420

X

x

y

z

=�

�

��

�

�

��

=�

�

��

�

�

��

2

1

0

.

⇒ x = 2; y = 1; z = 0.


��

1.1.1.1.1. A company sold 22 scooters, 15 bikes and 10 mopeds in January and 20 scooters, 20 bikes and12 mopeds respectively in June. Represent the data in matrix form.

Scooters Bikes Mopeds

January 22 15 10

June 20 20 12

∴ Corresponding matrix = 22 15 10

20 20 12�

��

��

2.2.2.2.2. Suppose the matrices A and B represent the number of items of different kinds produced by 2manufacturing units in one day.

A B=�

�

��

�

�

��

=�

�

��

�

�

��

2

5

7

1

3

6

and Compute 2A + 5B what does 2A + 5B represent?

Solution:Solution:Solution:Solution:Solution: 2A + 5B

=�

�

��

�

�

��

+�

�

��

�

�

��

2257

5136

=�

�

��

�

�

��

+�

�

��

�

�

��

=+

++

�

�

��

�

�

��

4

10

14

5

15

30

4 5

10 15

14 30

=�

�

��

�

�

��

92544

.

2A + 5B represents the number of items produced by one unit in two days and another in 5 daystogether.

3.3.3.3.3. A man buys 3 kgs of dal, 2 kgs of rice and 5 kgs of oil. If the cost of each kg is Rs. 35, Rs. 20and Rs. 75 respectively. Then find the total cost by matrix method.

Solution:Solution:Solution:Solution:Solution: Matrix representing the commodities

= [3 2 5]

Matrix representing the costs =�

�

��

�

�

��

352075


Total cost =�

�

��

�

�

��

� →

3 2 5352075

= × + × + ×3 35 2 20 5 75

= + + =105 40 375 520

∴ Total cost = Rs. 520.

4.4.4.4.4. A company sold 30 metal chairs, 40 wooden chairs and 25 plastic chairs in February and 60, 50and 75 respectively in March. The selling price of a metal chair is Rs. 150, that of wooden chairis Rs. 500 and plastic chair is Rs. 300. Find the total revenue in February and March using matrixmethod.


Metal chair Wooden Chair Plastic chair

February: 30 40 25

March: 60 50 75

∴ Corresponding matrix = 30 40 25

60 50 75�

��

��

Matrix representing the price = 150

500

300

�

�

��

�

�

��

∴ Total revenue = 30 40 25

60 50 75

150

500

300

�

��

�

��

�

�

��

�

�

��

�

→

=× + × + ×× + × + ×

�

��

��30 150 40 500 25 300

60 150 50 500 75 300

=+ +

+ +�

��

��4500 20000 7500

9000 25000 22500

= ��

�

��32 000

56 500

,

,

∴ Total revenue in February = Rs. 32,000/-

and Total revenue in March = Rs. 56,500/-

∴ Total revenue = 32,000 + 56,500

= Rs. 88,500/-


5.5.5.5.5. Matrix A and matrix B give the daily sales and selling price of soft drinks for a shopkeeper.

Pepsi Coke Thumbsup

MonTue

WedThu

PepsiCoke

Thumbsup

76

6

A B=

�

�

��

�

�

��

=�

�

��

�

�

��

6 3 12 3 0

1 1 72 1 5

Find the total revenue from four days.

Solution:Solution:Solution:Solution:Solution: Revenue matrix = AB.

=

�

�

��

�

�

��

�

�

��

�

�

��

�

→6 3 12 3 01 1 72 1 5

766

=

× + × + ×× + × + ×× + × + ×× + × + ×

�

�

��

�

�

��

=

+ ++ +

+ ++ +

�

�

��

�

�

��

=

�

�

��

�

�

��

6 7 3 6 1 6

2 7 3 6 0 6

1 7 1 6 7 6

2 7 1 6 5 6

42 18 6

14 18 0

7 6 42

14 6 30

66

32

55

50

Mon

Tue

Wed

Thu

− Total revenue for 4 days = 66 + 32 + 55 + 50

= Rs. 203.

5.5.5.5.5. In a certain town there are 4 colleges and 12 schools. Each school has 8 peons, 5 clerks and 2cashiers. Each college has 10 peons, 7 clerks and 3 cashiers. In addition, each college has 1section officer and one librarian. The monthly salary of each of them is as follows: Peon: Rs.2000; Clerk Rs. 3000, Cashier Rs. 5000, Section Officer Rs. 6000 and Librarian Rs. 4500. Usingmatrix notation find (1) Total number of posts of each kind in schools and colleges taken together(2) Monthly salary bill of all the schools and colleges taken together.

Solution:Solution:Solution:Solution:Solution: Let A = [5 12] represents the number of colleges and schools in that order.

Let

Peon Clerk Cashier S. off. Librarian

CollegeSchool

10B = �

��

��7 3 1 1

8 5 2 0 0

C =

�

�

��

�

�

��

PeonClerkCashierS. off.

Librarian

2000300050006000

4500


Number of employees = AB

i.e., 5 1210 7 3 1 18 5 2 0 0

→ �

��

��

= × + × × + × × + × × + × × + ×5 10 12 8 5 7 12 5 5 3 12 2 5 1 12 0 5 1 1 0

= + + + + +50 96 35 60 15 24 5 0 5 0

= 146 95 39 5 5

1st element i.e. 146 represent number of peons, 2nd element, 95 represent number of clerks and soon.

6.6.6.6.6. Total monthly salary bill of each school and college is given by matrix BC.

10 7 3 1 1

8 5 2 0 0

2000

3000

5000

6000

4500

�

��

�

��

�

�

��

�

�

��

�

→

10 2000 7 3000 3 5000 1 6000 1 4500

8 2000 5 3000 2 5000 0 0

× + × + × + × + ×× + × + × + +

�

��

��

=+ + + +

+ + + +�

��

��20 000 21 000 15 000 6 000 4 500

16 000 15 000 10 000 0 0

, , , , ,

, , ,

= ��

�

��66 500

41 000

,

,

∴ Total monthly salary bill of all colleges and schools taken together

= = �

��

��A BC� 5 12

66 500

41 000

,

,

= × + ×5 66 500 12 41 000, ,

= +332500 492000

= [8,24,500]

∴ Total salary = Rs. 8,24,500.

��

Total monthly salary of all colleges and schools


= Total number of employees ⋅ salary

=

�

�

��

�

�

��

� →

146 95 39 5 5

2000

3000

5000

6000

4500

= × + × + × + × + ×146 2000 95 3000 39 5000 5 6000 5 4500

= + + + +292000 285000 195000 30000 22500

= 8 24 500, ,

Hence total salary = Rs. 8,24,500.

6.6.6.6.6. A salesman has the following record of sales during 3 months for 3 items A, B, C which havedifferent rates of commission.

Month Sales of units Total Commission

A B C

Jan 100 100 200 900Feb 300 200 100 1000Mar 100 200 300 1400

Find out the rates of commissions on items A, B and C.

Solution:Solution:Solution:Solution:Solution: Let x, y and z denote the rates of commission in Rupees per unit for A, B and C itemsrespectively. Then the data given can be expressed as a system of linear equation.

100 100 200 900x y z+ + =

300 200 100 1000x y z+ + =

100 200 300 1400x y z+ + =

or

x y z+ + =2 9

3 2 10x y z+ + =

x y z+ + =2 3 14

Solving these equations by matrix method:

Let A X

x

y

z

D=�

�

��

�

�

��

=�

�

��

�

�

��

=�

�

��

�

�

��

1 1 2

3 2 1

1 2 3

9

10

14

, and


AX = D

⇒ X A D= −1

Now AA

A− =1 adj

A = = × − − × − + −1 1 2

3 2 1

1 2 3

1 2 3 2 1 3 3 1 2 6 2� � �

= − − − +6 2 9 1 2 4� �

4 8 8 4 0− + = ≠ . ∴ A−1 exist.

Now adj A =

−

− −

−

�

�

��

�

�

��

2 12 3

1 22 3

1 22 1

3 11 3

1 21 3

1 23 1

3 21 2

1 11 2

1 13 2

AX = D

⇒ X = A−1D

Now adj A =−

− −− −

�

�

��

�

�

��

4 1 38 1 154 1 1

AA

A− = =

−− +

− −

�

�

��

�

�

��

1 1

4

4 1 38 1 54 1 1

adj

X A D= =−

− +− −

�

�

��

�

�

��

�

�

��

�

�

��

−1 1

4

4 1 3

8 1 5

4 1 1

9

10

14

X =× + × − ×

− × + × + ×× − × − ×

�

�

��

�

�

��

1

4

4 9 1 10 3 14

8 9 1 10 5 14

4 9 1 10 1 14

X =+ −

− + +− −

�

�

��

�

�

��

=�

�

��

�

�

��

1

4

36 10 42

72 10 70

36 10 14

1

4

4

8

12


X =�

�

��

�

�

��

123

x

y

z

�

�

��

�

�

��

=�

�

��

�

�

��

123

⇒ x y z= = =1 2 3, . and

i.e., Rate of commission per unit for the 3 items is Re. 1, Rs. 2 and Rs. 3 respectively.

7.7.7.7.7. The prices of 3 commodities X, Y and Z are x, y and z respectively. A sells 1 unit of X, 1 unit ofY and 1 unit of Z. B sells 3 units of X, 1 unit of Y and purchases 1 unit of z. C sells 1 unit of X,3 units of Y and purchases 1 unit of Z. In the process A, B and C earns Rs. 9000, Rs. 1000 and Rs.5000 respectively. Using matrices find the prices per unit of the commodities. (Note that sellingthe units is positive earning and buying the units is negative).

Solution.Solution.Solution.Solution.Solution. The above data can be written in the form of simultaneous equations:

A: x + y + z = 9000

B: 3x + y − z = 1000

C: x + 3y − z = 5000.

Solving these equations, by Cramer’s rule:

Let ∆ = −−

= − + − − + + −1 1 13 1 11 3 1

1 1 3 1 3 1 1 9 1� � �

2 1 2 1 8− − +� �

2 2 8 12+ + = .

∆ x = −−

9000 1 11000 1 15000 3 1

9000 1 3 1 1000 5000 1 3000 5000− + − − + + −� � �

9000 2 1 4000 1 2000� � � − + −

18000 4000 2000− −

= 12000

∆ y = −−

1 9000 1

3 1000 1

1 5000 1


= − + − − + + −1 1000 5000 9000 3 1 1 15000 1000� � �

4000 9000 2 1 14000− − +� �

4000 18000 14000+ +

= 36000.

∆ z =1 1 90003 1 10001 3 5000

= − − − + −1 5000 3000 1 15000 1000 9000 9 1� � �

2000 14000 72000− +

∆ z = 60000.

Hence x x= = =∆∆

12000

121000

y y= = =∆∆

36000

123000

z z= = =∆∆

60000

125000

Hence price per unit of x = Rs. 1000, that for y = Rs. 3000 and for z = Rs. 5000.

8.8.8.8.8. A company is considering which of the 3 methods of production it should use in producing 3products X, Y and Z. The amount of each product and produced by each method is as shownbelow.

Method Product X Product Y Product Z

I 4 8 2II 5 7 2III 3 6 5

Further information relating to profit per unit is as follows:

Product Profit/Unit

X 10Y 5Z 7

Using matrix multiplication find which method maximises the total profit:


Let A B=�

�

��

�

�

��

=�

�

��

�

�

��

4 8 25 7 23 6 5

1057

Total profit = AB

=�

�

��

�

�

��

�

�

��

�

�

��

4 8 25 7 23 6 5

1057

=× + × + ×× + × + ×× + × + ×

�

�

��

�

�

��

4 10 8 5 2 7

5 10 7 5 2 7

3 10 6 5 5 7

A =+ ++ ++ +

�

�

��

�

�

��

=�

�

��

�

�

��

40 40 1450 35 1430 30 35

949995

∴ Total profit in method I = 94.

That in method II = 99 and in method III = 95.

∴ Method II maximises the profit.

��

• Matrix is an arrangement of numbers in horizontal rows and vertical columns.

• Two matrices of the same order are said to be equal if and only if the corresponding elements areequal.

• Two matrices can be added subtracted if they have same order. It is obtained by adding/subtract-ing the corresponding elements.

• Multiplication of a matrix by a scalar is obtained by multiplying each and every element by ascalar.

• It A is of order m × n and B is of order n × p then AB is of order m × p, i.e., matrix multiplicationis possible only when number of column in 1st matrix is equal to number of rows in the 2nd

matrix.

• Transpose of a matrix is obtained by interchanging rows and columns.

• A unique value associated with every square matrix is called its determinant value.

• If det A = 0, i.e., |A| = 0 for a matrix A, then A is called singular matrix. Otherwise it is called non-singular matrix.

• The value of the determinant is unaltered if its rows and columns are interchanged.

• If 2 rows or columns are interchanged the value of the determinant changes its sign.

• If in a determinant 2 rows or columns are identical then the value of the determinant is zero.


• If the elements of any row (or column) is multiplied by k, the value of the determinant is multi-plied by k.

• If to the elements of any row (or column) of a determinant the same multiples of the correspond-ing elements of other rows (or columns) of the determinant are added the value of the determinantremains the same.

• The determinant obtained by deleting the row and column containing the element is called minorof that element. If the minors are multiplied with (−1)i + j [where i = row number and j = columnnumber of the element] we get co-factors. The transpose of the co-factor matrix is called adjointof the matrix.

• For a non-singular matrix A.

AA

A− =1 adj

.

• If A is a square matrix and I is the identity matrix of the same order.

• The characteristic equation: |A − λI| = 0. The values of λ obtained is called eigen values orcharacteristic roots.

• Every square matrix satisfies its characteristic equation, |A − λI| = 0. This is Cayley Hamiltontheorem.

• The solution of system of equations

a x b y c z d1 1 1 1+ + =

a x b y c z d2 2 2 2+ + =

a x b y c z d3 3 3 3+ + =

By Cramer’s rule:

Let ∆ ∆= =a b c

a b c

a b c

d b c

d b c

d b cx

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

,

∆ ∆y z

a d c

a d c

a d c

a b d

a b d

a b d

= =1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

,

Then x y zx y z= = =∆∆

∆∆

∆∆

, . and

By matrix method:

Let A

a b c

a b c

a b c

X

x

y

z

D

d

d

d

=�

�

��

�

�

��

=�

�

��

�

�

��

=�

�

��

�

�

��

1 1 1

2 2 2

3 3 3

1

2

3

, and


Then matrix equation

AX = D

X = A−1D.

and AA

A− =1 adj

or A−1 can be found by using Cayley Hamilton theorem.

��

1. If A B A B=−

�

��

��= ��

�

��+ ′

1 3

2 1

6 5

1 0 and find , .

2. If A B=−

�

��

��=

−�

��

��1 6 0

1 7 1

0 1 7

6 5 2 and ,

find A + 3B.

3. If A B=−−

�

��

��=

−�

��

��1 1

2 1

0 1

1 1 and , find matrix X such that A + X = B.

4. Find x and y if 4 7

7

2 1

5 0

1 8

6 7x

y�

��

��+

+��

�

��= ��

�

��.

5. If A B=−�

��

��=

−�

��

��1 2

7 6

1 3

7 8 and , verify that (A + B)′ = A′ + B′.

6. Solve for x: x x2 4

1 7

2 5

8 3

3 9

9 10�

��

�

�� + ��

�

��= ��

�

��

7. If A is of order 4 × 5 and B is of order 5 × 3, does AB and BA exists? If so what are their order?

8. If A B= ��

�

��=

−−

�

��

��1 7

5 2

1 0

6 5 and , then find AB.

9. If A = ��

�

��1 2

2 4, find AA′ and A′A. Is AA′ = A′A?

10. If 1 0 0

0 1 0

0 0 1

1

2

3

�

�

��

�

�

��

�

�

��

�

�

��

=�

�

��

�

�

��

x

y

z

, find x, y and z.

11. If A =−�

��

�

��

1 0

0 1, then prove that AA′ = A′A = I.


12. If Ai

i= ��

�

��0

0, then find A2.

13. If 1 3

7 1

2

3�

��

�� −�

��

��= ��

��x

y, then find x and y.

14. If A B= −�

�

��

�

�

��

= −−

�

�

��

�

�

��

1 2 31 5 31 2 7

1 7 21 0 65 2 1

, , then verify A (B + C) = AB + AC.

15. If A B C= ��

�

��=�

�

��

�

�

��

=1 2 31 5 2

132

2 1, , and then verify A (BC) = (AB) C.

16. If A = −�

�

��

�

�

��

1 2 3

0 1 1

1 1 2

, then find A3.

17. If A =−�

��

��1 2

7 3, then prove that A2 − 4A + 17I = 0. Where I is identity matrix of order 2 × 2.

18. If A = −−

�

�

��

�

�

��

1 2 02 1 61 3 6

, then prove that A3 − A2 + 13A − 9I = 0.

19. If A B=−

�

��

��=

−�

��

��1 2

3 1

2 1

3 2, , then prove that (AB)′ = B′A′.

20. Prove that A =�

�

��

�

�

��

1 2 22 1 22 2 1

satisfies A2 − 4A − 5I = 0.

21. Evaluate:

(a)1 2

7 5 (b)−1 3

6 5 (c)−

−1 7

2 6

(d)5 3

6 7

−− − (e)

−− −

7 5

5 2


22. Evaluate:

(a)

1 2 42 3 53 7 8

(b)

−−

−

1 2 73 5 42 0 8

(c)

1 1 22 1 14 1 2

−

− −

23. Find x if the matrix A is singular.

(a) Ax

x= ��

�

��

5

20 (b) A x= − −�

�

��

�

�

��

2 3 4

4 8

5 6 7

24. Solve for x:

(a)5

2 73

x= (b)

3

58

x

x= (c)

x

x

x

2 1

2 5

1 2

0

−

−=

25. Using the properties of determinants find

(a)3485 3486

3487 3488 (b)6001 6002

6004 6005

(c)

241 242 243

244 245 246

247 248 249(d)

2100 2101 2102

2110 2111 2112

2120 2121 2122

26. Prove that a b b c c a

b c c a a b

c a a b b a

− − −− − −− − −

= 0.

27. Prove that

111

3

3

3

a a

b b

c c

a b b c c a a b c= − − − + +� � � �

28. Prove that

1

11

2

2

2

a a

b b

c c

a b b c c a= − − −� � �

29. Prove that

−−

−=

a ab ac

ab b bc

ac bc c

a b c

2

2

2

2 2 24


30. Prove that

x p q

p x q

p q x

x p x q x p q= − − + +� � �

31. Prove that

11

11

++

+= + + +

a b c

a b c

a b c

a b c

32. Prove that x = −9 is a root of equation

x

x

x

++

+=

1 3 5

2 3 5

2 3 4

0

33. Prove that

a b c a b

c b c a b

c a c a b

a b c

+ ++ +

+ += + +

22

22 3�

34. Find the adjoint of:

( ) ( ) ( ) ( ) ( )a b c d e1 2

3 4

1 2

7 6

1 7

5 6

6 5

0 4

5 3

2 1�

��

��−

−�

��

��−

−�

��

��−��

�

��− −− −�

��

��

35. Find the inverse of:

(a)−��

�

��4 31 1 (b)

6 5

0 7�

��

��(c)

−−

�

��

��5 1

6 2

36. If A B= ��

�

��=

−��

�

��1 3

4 7

1 2

2 7 and , verify (AB)−1 = B−1A−1.

37. Find the adjoint of A =�

�

��

�

�

��

3 2 54 5 61 3 5

and hence find A−1.

38. Find the inverse of

0 1 1

4 2 0

3 1 4

− −−

�

�

��

�

�

��

.

39. Find the characteristic polynomial and characteristic roots of the matrix 1 4

2 3�

��

��.


40. Verify Cayley Hamilton theorem for the matrix 1 7

5 2

−�

��

��.

41. Verify Cayley Hamilton theorem for the matrix

1 2 01 3 21 6 5

−−

�

�

��

�

�

��

.

42. Find A−1 using Cayley Hamilton theorem for the matrix 1 0 2

2 1 0

3 2 1

�

�

��

�

�

��

.

43. Find A3 using Cayley Hamilton theorem if A = −�

�

��

�

�

��

1 2 30 1 11 1 2

.

44. Solve the following system of equations (a) By Cramer’s rule (b) By matrix method:

(i) 7 16

2 3 2

x y

x y

− =+ = −

(ii) 3 4 10

4 5 3

x y

x y

+ =− =

(iii) 2 3 4

3 2 5

x y

x y

− =+ =

(iv) x y

x y

+ =+ =

7

2 8

(v) 3 4 7

7 6

x y

x y

+ =− =

(vi) x y z

x y z

x y z

+ + =+ + =

+ + =

2 9

3 2 10

2 3 14

(vii) x y z

x y

x z

− − =+ =− =

4 3 9

5 19

2 5 3

(viii) x y z

x y z

x y z

− − =+ + =− − =

2 3

2 5

4 2 11

(ix) x y z

x y z

x y z

+ + =+ + =

+ + =

2 9

3 2 10

2 3 14

(x) x y

y z

z x

+ =+ =+ =

1

7

2

45. A man buys 8 dozens of mangoes; 10 dozens of apples and 4 dozen of bananas. Mangoes costRs. 18 per dozen, apple Rs. 9 per dozen and bananas Rs. 6 per dozen. Represent the quantitiesbought by a row matrix and prices by column matrix and hence find the total cost.

46. A company is considering which of the 3 methods of production it should use in producing 3goods X, Y and Z. The amount of each good produced by each method is shown in the matrix.


X Y Z

I

IIIII

4 8 2

5 7 15 3 9

�

�

��

�

�

��

The vector [10 4 6] represents the profit per unit for the goods X, Y and Z in that order. Findwhich method maximises profit.

47. Matrix A and B give the daily sales and sale price of chocolates for a shopkeeper.

Bar one Dairy milk Five star

Mon

TueWedThu

Bar oneDairy milk

Five star

A B=

�

�

��

�

�

��

=�

�

��

�

�

��

6 3 1

2 3 01 2 41 4 3

56

7

Find the total revenue for four days.

48. A salesman has the following record of sales during 3 months for 3 items, A, B and C which havedifferent rates of commission.

Months Sales of units Total commissionin Rs.

A B C

January 90 100 20 800

February 130 50 40 900

March 60 100 30 850

Find out the rates of commission on A, B and C.

49. The prices of the 3 commodities X, Y and Z are x, y and z per unit respectively. A purchases 4 unitsof z and 3 units of x and 5 units of y. B purchases 3 units of y and sells 2 units of x and 1 unitof z. C purchases 1 unit of x and sell 4 units of y and 6 units of z. In the process A, B and C earnRs. 6000, 5000 and 13,000 respectively. Using matrices, find the prices per unit of the 3 com-modities. (Note that selling the unit is positive earnings and buying the unit is negative earning).

50. Suppose the matrices X and Y represent the number of items of different kinds produced by 2manufacturing units in one day.

X Y=�

�

��

�

�

��

=�

�

��

�

�

��

4

5

3

4

56

and .

Compute 2X + 3Y, what does 2X + 3Y represent?


��

1.7 4

7 1−�

��

��2.

1 9 21

19 22 5

−�

��

��3.

−−�

��

��1 2

1 0

4. x y= − = −1 5, 6. x x= = −1 3or

7. AB exists and is of order 4 × 3. BA does not exist.

8.1 0

1 0�

��

��9. AA A A′ = �

��

��= ′

5 10

10 2010. x = 1, y = 2 and z = 3.

12.−

−�

��

��1 0

0 1 13. x y= − =7 11, 16.

11 22 194 5 99 17 16

− − −�

�

��

�

�

��

21. ( ) ( ) ( ) ( ) ( ) .a b c d e− − − −9 23 8 53 39 22. ( ) ( ) ( )a b c7 94 3

23. ( ) , ( )a x b x= = −10 6 24. ( ) ( ) ( )a b c19 4 3 1− − or

25. ( ) ( ) ( ) ( )a b c d− −2 3 0 0

34. ( ) ( ) ( ) ( ) ( )a b c d e4 2

3 1

6 2

7 1

6 7

5 1

4 5

0 6

1 3

2 5

−−�

��

��

��

��− −− −�

��

��−−

�

��

��−

−�

��

��

35. ( ) ( ) ( )a b c1

7

1 3

1 41

42

7 5

0 61

4

2 1

6 5−−

− −�

��

��−�

��

��− −− −�

��

��

37.1

28

7 5 13

14 10 2

7 7 7

−−

−

�

�

��

�

�

��

38.1

11

8 5 2

4 3 1

7 3 1

−−

−

�

�

��

�

�

��

39. λ λ2 4 5 0 5 1− − = −, .roots : and

42.1

3

1 4 2

2 5 4

1 2 1

−− −

−

�

�

��

�

�

��

43.

11 22 19

4 5 9

9 17 16

− − −+

�

�

��

�

�

��

44. ( ) , ( ) , ( ) ,i x y ii x y iii x y= = − = = = = −2 2 2 1

23

13

2

13

( ) , ( ) , ( ) , ,iv x y v x y vi x y z= = = = = = =1 6 1 1 1 2 3

( ) , , ( ) , ,vii x y z viii x y z= = − = = = − =4 1 1 8 3 1 3 0


( ) , , ( ) , , .ix x y z x x y z= = = = − = =1 2 3 2 3 4

45. 8 10 41896

258�

�

��

�

�

��

= .

Cost = Rs. 258.

46. Method III. Total profit 116.

47. Rs. 178.

48. Rs. 2, 4 and 11 respectively.

49. Price per unit of X = Rs. 3000

Y = Rs. 1000 and Z = Rs. 2000.

50.17

22

27

�

�

��

�

�

��

It represent the number of items produced by one unit in 2 days and another in 3 days

together.


�

��

��

Two quantities of the same kind can be compared either by subtraction method or by division method.In subtraction method we find how much more (or less) is one quantity than the other, and in divisionmethod, we find how many times (or what fractional part) is one quantity of the other. The quotient hereis nothing but the ratio of the two quantities. For example, if I have Rs. 100 and you have Rs. 600 thenwe can compare the money by subtraction method and say ‘You have Rs. 500 more than what I have’.Or we can compare by division method and say you have 6 times the money what I have.

If the ratio of 2 mutual quantities are equal then they are said to be proportional.

��

A ratio is a relation or comparison between two quantities of the same kind. The comparison is madeby considering what multiple, part or parts the first quantity is of the second.

The ratio of 2 quantities x and y is denoted by x : y or �

�� The first term x is called antecedent and

the second term y is called consequent.


1. A ratio is a pure number. Hence it has no units.

2. When the terms of the ratio are multiplied or divided by the same quantity the ratio is notaltered.

For instance 2 : 3 = 4 : 6 = 40 : 60 = 80 : 120...

3. If a : b and c : d are two ratios, then the ratio ac : bd is called their compound ratio.

Example : The compound ratio of 5 : 2 and 3 : 7 is 5 × 3 : 2 × 7 i.e., 15 : 14.

4. If a : b is the given ratio then the ratio

Ratio and Proportions, Variations 183

(i) a2 : b2 is called its duplicate ratio.

(ii) � �� is called its subduplicate ratio.

(iii) a3 : b3 is called its triplicate ratio.

(iv) � �� is called its sub-triplicate ratio.

�� !�"�

1.1.1.1.1. Express the following ratios in their simplest form

(a) 16 : 26 (b) 16 : 64 (c) 6 : 90 (d) 90 : 10


(a) ��

��

�

�� = = =

(b) ��

��

�

�� = = =

(c) � �

�

�

�

�� = = = =

(d) �

�

� �� = = =

2.2.2.2.2. Compare the following ratios :

(a) 2 : 5 and 3 : 7 (b) 12 : 13 and 13 : 2

(c) 5 : 11 and 11 : 6 (d) 8 : 3 and 4 : 7

Solution:Solution:Solution:Solution:Solution: (a) To compare 2 : 5 and 3 : 7

2 52

5: = and

3 73

7: =

To compare 2 fractions first we make denominators equal by multiplying with suitable numbers.

2

5

7

7

14

35× =

3

7

5

5

15

35× =

Clearly is less than 14

35

15

35

i.e. 14 : 35 < 15 : 35

i.e., 2 : 7 < 3 : 7


(b) 12 : 13 and 13 : 2

12 1312

13

2

2

24

26: = × =

13 213

2

13

13

169

26: = × =

Clearly24

26

169

26<

⇒ 12 : 13 < 13 : 2

(c) 5 : 11 and 11 : 6

5 115

11

6

6

30

66: = × =

11 611

6

11

11

121

66: = × =

Clearly30

66

121

6630 66 121 66< <ie, : :

∴ 5 : 11 < 11 : 6

(d) 8 : 3 and 4 : 7

8 38

3

7

7

56

21: = × =

4 74

7

3

3

12

21: = × =

Clearly56

21

12

21>

∴ 8 : 3 > 4 : 7.

3.3.3.3.3. Find the ratio between

(a) 1 hr 10 min and 140 min.

(b) 3 kg 30 gm and 1 kg 260 gms.

(c) 6 Rs. 50 ps. and 8 Rs. 75 ps.

Solution:Solution:Solution:Solution:Solution: (a) 1 hr 10 min = 60 + 10 min. = 70 min.

∴ Ratio between 70 min. and 140 min. = 70

140

= 70 : 140 = 1 : 2

(b) 3 kg 30 gm and 1 kg 260 gms.

3 kg 30 gm = 3000 + 30 gm

= 3030 gm


1 kg 260 gms = 1000 + 260 = 1260 gm

∴ Ratio between 3030 and 1260

= =3030

1260

101

42

= 101 : 42.

(c) 6 Rs. 50 ps. and 8 Rs. 75 ps.

6 Rs. 50 ps = 650 ps.

8 Rs. 75 ps = 875 ps.

Ratio between 650ps and 875 ps

= = =650

875

26

3526 35: .

4.4.4.4.4. Write the duplicate and subduplicate of the ratio 1 : 9

Solution:Solution:Solution:Solution:Solution: Duplicate of a : b is a2 : b2

∴ Duplicate of 1 : 9 is 12 : 92 = 1 : 81

Subduplicate of a : b = a b:

∴ Subduplicate of 1 : 9 = 1 9 1 3: : .=

5.5.5.5.5. Write the triplicate and subtriplicate of the ratio 1 : 8.

Solution:Solution:Solution:Solution:Solution: Triplicate of a : b is a3 : b3

∴ Triplicate of 1 : 8 is 13 : 83 = 1 : 512.

Subtriplicate of a : b is a b3 3:

Subtriplicate of 1 : 8 is 1 83 3:

= 1 : 2.

6.6.6.6.6. Raju’s monthly salary is Rs. 3000 and Rama’s annual income is Rs. 60,000. What is the ratio oftheir incomes?

Solution:Solution:Solution:Solution:Solution: Raju’s monthly salary = Rs. 3000

Raju’s annual income = 3,000 × 12

= 36,000

Given Rama’s annual income = Rs. 60,000

∴ Ratio of their income

= 36,000 : 60,000

= 6 : 10

= 3 : 5

OR

Given Raju’s monthly salary = Rs. 3000


Rama’s annual income = Rs. 60,000

∴ Rama’s monthly income = 60,000/12 = Rs. 5,000

∴ Ratio of their income

= 3000 : 5000

= 3 : 5.

7.7.7.7.7. A number is divided into 3 parts in the ratio 2 : 3 : 4. If the 3rd part is 32. Find the other 2 parts.

Solution:Solution:Solution:Solution:Solution: Let 2x, 3x and 4x be the parts of the number.

Given 3rd part = 32

4x = 32

⇒ x = 8

∴ 1st part = 2x = 2 (8) = 16

2nd part = 3x = 3 (8) = 24.

8.8.8.8.8. A bag contains rupee, 50 paise and 25 paise coins in the ratio 5 : 6 : 8. If the total amount isRs. 840, find the number of coins of each type.

Solution:Solution:Solution:Solution:Solution: Ratio of 1 Re., 50 ps., and 25 ps. coins = 5 : 6 : 8

∴ Ratio of values = 5

1

6

2

8

4: :

Ratio of values = 5 : 3 : 2

Now Divide Rs. 840 in the ratio 5 : 3 : 2

Sum of the terms of ratio = 5 + 3 + 2 = 10

∴ 1st part = Rs. Rs. 420.5

10840× =

2nd part = Rs. Rs. 252.3

10840× =

3rd part = Rs. Rs. 168.2

10840× =

∴ Number of 1 Re. coins = 420

Number of 50 ps. coins = 252 × 2 = 504

Number of 25 ps. coins = 168 × 4 = 672.

9.9.9.9.9. In a mixture of 35 litres, the ratio of milk and water is 4 : 1. If 7 litres of water is added to themixture, then find the ratio of milk and water in the new mixture.

Solution:Solution:Solution:Solution:Solution: Given milk : water = 4 : 1

Sum of terms = 4 + 1 = 5

∴ Milk in 35 litres mixture = 4

535×

= 28 litres.


Water in mixture = 35 − 28 = 7 litres.

Now if 7 litres of water is added.

Water in mixture = 7 + 7 = 14.

Milk in the mixture = 28.

∴ Milk : Water = 28 : 14 = 2 : 1.

10.10.10.10.10. A mixture contains nuts and screws in the ratio 4 : 3. If 7 screws are added to the mixture, the ratiobecomes 3 : 4. Find the number of nuts in the mixture.

Solution:Solution:Solution:Solution:Solution: Let the number of nuts and screws be 4x and 3x.

Given : If 7 screws are added the ratio becomes 3 : 4.

i.e., 4x : 3x + 7 = 3 : 4

i.e.,4x

x3 7

3

4+=

Cross multiplying

16x = 3 (3x + 7)

16x = 9x + 21

16x − 9x = 21 ⇒ 7x = 21 ⇒ x = 3.

∴ Number of nuts = 4x = 4 (3) = 12.

11.11.11.11.11. 5 years ago, Arun’s father’s age was 5 times his son’s age. After 2 years he will be 3 times Arun’sage. Find the ratio of their present ages.

Solution:Solution:Solution:Solution:Solution: Let Arun’s age 5 years ago be x yrs.

Then his father’s age = 5x years.

After 2 years,

Arun’s age = x + 5 + 2 = x + 7

His father’s age = 5x + 5 + 2 = 5x + 7

Given : 5x + 7 = 3 (x + 7)

5x + 7 = 3x + 21

5x − 3x = 21 − 7

2x = 14 ⇒ x = 7.

∴ Father’s age now = 5x + 5 = 5 (7) + 5 = 40 yrs.

Arun’s age now = x + 5 = 7 + 5 = 12 yrs.

∴ Ratio of their ages = 40 : 12 = 10 : 3.

12.12.12.12.12. If 4 kgs of tea is worth 3 kgs of sugar, 5 kgs of sugar is worth 16 kgs of flour and 7 kgs of flouris worth 2 kgs of coffee. How many kgs of tea is worth 24 kgs of coffee.

Solution:Solution:Solution:Solution:Solution: Given

Tea : Sugar = 4 : 3

Sugar : Flour = 5 : 16


Flour : Coffee = 7 : 2

Let Tea : Coffee = x : 24

NowTea

Coffee

Tea

Sugar

Sugar

Flour

Flour

Coffee= × ×

x

24

4

3

5

16

7

2= × ×

∴ x = × × ×× ×

=24 4 5 7

3 16 235

∴ x = 35

∴ Tea : Coffee = 35 : 24.

∴ 35 kgs of tea is worth 24 kgs of coffee.

13.13.13.13.13. A, B and C starts a business with a capital of Rs. 10,500 of this Rs. 4400 is contributed by A. Rs.3700 is contributed by B and the rest by C. After 5 months C withdraws Rs. 800 capital, whileA and B each adds Rs. 400. At the end of the year, profit of the original capital is shared in theratio of capitals. Find to the nearest rupee the amount to be received by each


Total capital : Rs. 10,500

A’s contribution : Rs. 4400

B’s contribution : Rs. 3700

C’s contribution : Rs. 10,500 − (4400 + 3700)

= Rs. 2400.

A’s share in capital

= Rs. 4400 used for 12 months and Rs. 400 used for (12 − 5) months.

i.e., 4400 × 12 + 400 × 7 = Rs. 55,600.

Similarly B’s share in capital

Rs. 3700 used for 12 months and Rs. 400 used for (12 − 5) months

Rs. 3700 × 12 + 400 × 7 = Rs. 47,200.

C’s share in capital :

Rs. 2400 used for 5 months and (Rs. 2400 − Rs. 800) used for (12 − 5) months

i.e., 2400 × 5 + 1600 × 7 = Rs. 23,200

Now Ratio of capital of A, B and C

= 55600 : 47200 : 23200

= 139 : 118 : 58

Total : 139 + 118 + 58 = 315

Now profit = 13¾% of capital


=×

× =55

4 10010 500 1443 75, .

A’s share in profit = 139

3151443 75× =. Rs. 637

B’s share in profit = 118

3151443 75× =. Rs. 541

C’s share in profit = 58

3151443 75× =. Rs. 266.

14.14.14.14.14. Three utensils contains equal mixture of milk and water in the ratio 6 : 1, 5 : 2, and 3 : 1respectively. If all the solutions are mixed together, find the ratio of milk and water in the finalmixture.

Solution:Solution:Solution:Solution:Solution: In 1st utensil milk : water = 6 : 1

∴ Quantity of milk in 1st utensil = 6

7

and Quantity of water in 1st utensil = 1

7

Similarly 2nd utensil contains 5

7 milk and

2

7 water.

Similarly quantity of milk in 3rd utensil = 3

4

and water = 1

4.

∴ If all solutions are mixed,

quantity of milk = 6

7

5

7

3

4+ +

= + + =24 20 21

28

65

28

quantity of water = 1

7

2

7

1

4+ +

= + + =4 8 7

28

19

28

∴ Milk : Water = 65

28

19

28:

i.e., Milk : Water = 65 : 19.


��# ��

If 2 ratios are equal then the 4 quantities comprising them form a proportion i.e. if the ratio a : b is equalto c : d, then 4 quantities a, b, c, d are in proportion.

ExampleExampleExampleExampleExample: 1, 2, 4, 8 are in proportion since 1 : 2 = 4 : 8.

Note Note Note Note Note : 1. a : b = c : d is also denoted by a : b : : c : d.

2. In a proportion a : b = c : d, the first andthe last terms i.e., a and d are called ex-tremes and the second and 3rd terms i.e. band c are called means.

3. In every proportion, the product of themeans is equal to the product of the ex-tremes. i.e., ad = bc

Conversely, if 4 quantities a, b, c, d are such that ad = bc, then they are said to be in proportion.

a : b = c : d ⇔ ad = bc is called rule of 3. This rule is used to solve a proportion when one of theterms is unknown.

For example: 4 : 5 = x : 15

⇒ 5 4 15x = ×

x = × =4 15

512.

1. InInInInInvvvvvererererertendo:tendo:tendo:tendo:tendo: If a : b = c : d, then prove that b : a = d : c

PrPrPrPrProof:oof:oof:oof:oof: Given a : b = c : d

⇔ =bc ad

⇔ =b

a

d

c

⇔ =b a d c: :

Hence proved.

2. AlterAlterAlterAlterAlternendo:nendo:nendo:nendo:nendo: If a : b = c : d, then prove that a : c = b : d


⇔ =bc ad

⇔ =b

d

a

c

⇔ =a c b d: :

Hence proved.

3. Componendo:Componendo:Componendo:Componendo:Componendo: If a : b = c : d, then prove that a + b : b = c + d : d


⇔ bc = ad

Fig. 7.1

Extremes

a : b = c : d

means


Adding bd to both sides

bc + bd = ad + bd

⇔ b c d d a b+ = +� � � �

c d

d

a b

b

+ = +

i.e, a b

b

c d

d

+ = +

⇔ a b b c d d+ = +: :

Hence proved.

4. DiDiDiDiDividendo:videndo:videndo:videndo:videndo: If a : b = c : d, then prove that a − b : b = c − d : d.


⇔ ad = bc

Subtracting bd on both sides.

ad bd bc bd− = −

d a b b c d− = −� � � �

⇔ a b

b

c d

d

− = −

⇔ a b b c d d− = −: :

Hence proved.

5. Componendo and DiComponendo and DiComponendo and DiComponendo and DiComponendo and Dividendo:videndo:videndo:videndo:videndo: If a : b = c : d, then prove that a + b : a − b = c + d : c − d


⇒ a + b : b = c + d : d (Componendo)

a b

b

c d

d

+ = +...(1)

Also a b b c d d− = −: : (Dividendo)

i.e.,a b

b

c d

d

− = −...(2)

Dividing (1) by (2) we get

a b

ba b

b

c d

dc d

d

+

− =

+

−

⇒ a b

a b

c d

c d

+−

= +−


⇒ a b a b c d c d+ − = + −: :

Hence proved.

��

Quantities a, b, c, d, ... are said to be in continued proportion if a

b

b

c

c

d= = ... .

i.e., a : b = b : c = c : d = ...

Three quantities a, b, c are in continued proportion if a : b = b : c. Here ‘b’ is called mean propor-tional and c is called 3rd proportional.

For instance 7, 14, 28, 56, ... are in continued proportion since 7 : 14 = 14 : 28 = 28 : 56 ...

��$ ��

Quantities are said to be in direct proportion when an increase (or decrease) in one kind is accompaniedby an increase (or decrease) in the other.

Example:Example:Example:Example:Example: Number of chocolates 1 2 3 4 ...

Cost of chocolates 3 6 9 12 ...

Here we notice as the number of chocolates increases cost of chocolates also increases. So numberof chocolates is directly proportional to cost of chocolates.

��"��"��

Quantities are said to be in inverse proportion when an increase (or decrease) in one kind is accompa-nied by decrease (or increase) in the other.

Example:Example:Example:Example:Example: Number of workers 1 2 3 4...

Number of days 12 6 4 3...

Here we notice as the number of workers increase, number of days required to finish the workdecreases and vice versa. So it is inverse proportion.

Note:Note:Note:Note:Note: If a : b = c : d represent a direct proportion, then a : b = d : c or (b : a = c : d) represent aninverse proportion.

�� !�"�

1.1.1.1.1. Find the missing term in the proportion:

(a) x : 4 = 27 : 12

(b) 10 : 50 = ? : 250.

Solution:Solution:Solution:Solution:Solution: (a) Given: x : :4 27 12=

12 27 4x = ×


x = ×27 4

12

x = 9.

(b) Given: 10 50 250: := x

50 10 250x = ×

x = ×10 250

50

x = 50.

2.2.2.2.2. Find the 3rd proportional to (a) 5 : 10 (b) 3 : 27

(a) Let the 3rd proportional be x.

Then 5 10 10: := x

5 10 10x = ×

x = × =10 10

520.

(b) Let the 3rd proportional be y.

Then 3 27 27: := y

3 27 27y = ×

y = × =27 27

3243.

3.3.3.3.3. Find the fourth proportional to

(a) 1 : 2 : 3

(b) 9 : 8 : 18

Solution:Solution:Solution:Solution:Solution: (a) Let the 4th proportional be x.

Then 1 2 3: := x

x = × =2 3 6.

(b) Let the 4th proportional be y.

Then 9 8 18: := y

9 8 18y = ×

y = × =8 18

816.

4.4.4.4.4. Find the mean proportional to (a) 2 : 8 (b) 5 : 45.


Solution:Solution:Solution:Solution:Solution: (a) Let the mean proportional be x.

Then 2 8: :x x=

x2 8 2= ×

x2 16=

x = 4 .

(b) Let the mean proportional be y.

Then 5 45: :y y=

y2 5 45= ×

y2 225=

y = 15.

5.5.5.5.5. If 15x = 12y. Then what is x : y?

Given: 15x = 12y

⇒x

y= 12

15

x

y= 4

5

⇒ x y: := 4 5 .

6.6.6.6.6. If 4x − 7y : 3x + y = 2 : 3. Then find x : y.

Given: 4 7 3 2 3x y x y− + =: :

3 4 7 2 3x y x y− = +� � � �

12 21 6 2x y x y− = +

12 6 2 21x x y y− = +

6 23x y=

⇒x

y= 23

6

⇒ x y: : .= 23 6

7.7.7.7.7. If x : y = 7 : 5. Then find 4x − 2y : x + 3y

Given : x : y = 7 : 5


⇒x

y= 7

5

Consider 4x − 2y : x + 3y

=−

+=

��

−

�

�

�

+

�

�

�

4 2

3

4 2

3

x y

x y

yx

y

yxy

=

�� −

+

475

2

75

3

=−

+=

−

+

28

52

7

53

28 10

57 15

5

= =18

22

9

11.

4 2 3 9 11x b a b− + =: : .

8.8.8.8.8. What number must be subtracted from each of 9, 11, 15 and 19, so that the difference will beproportional.

Solution:Solution:Solution:Solution:Solution: Let the number subtracted be x,

So that 9 − x, 11 − x, 15 − x and 19 − x will be in proportion.

∴ 9 11 15 19− − = − −x x x x: :

9 19 11 15− − = − −x x x x� � � � � � � �

171 19 9 165 15 112 2− − + = − − +x x x x x x

171 28 165 26 02 2− + − + − =x x x x

⇒ 6 2 0− =x

2 6 3x x= ⇒ = .

9.9.9.9.9. What must be added to each of 9, 42, 3 and 18 so that the sums will be in proportion.

Solution:Solution:Solution:Solution:Solution: Let the number added be x.

∴ 9 + x, 42 + x, 3 + x and 18 + x are in proportion.


9 42 3 18+ + = + +x x x x: :

9 18 42 3+ + = + +x x x x� � � � � ��

162 18 9 126 3 422 2+ + + = + + +x x x x x x

162 27 126 45 02 2+ + − − − =x x x x

⇒ 36 18 0− =x

⇒ x = 2.

10.10.10.10.10. If a : b = 2 : 1, b : c = 3 : 2, then find a : b : c.

Given: a b

b c

: :

: :

=

=

2 1

3 2

Multiplying 1st ratio by 3 and 2nd ratio by 1 to make the value of b same.

∴a b

b c

: :

: :

=

=

6 3

3 2

⇒ a b c: : : : .= 6 3 2

11.11.11.11.11. If x : y = 1 : 2, y : z = 3 : 4 and z : w = 5 : 1, then find x : y : z : w.

x y

y z

: :

: :

=

=

1 2

3 4

To make the value of y same, multiplying 1st ratio by 3 and 2nd ratio by 2.

x y

y z

: :

: :

=

=

3 6

6 8

∴ x y z: : : := 3 6 8

Given: z : w = 5 : 1.

To make the value of z same, multiplying first ratio by 5 and 2nd ratio by 8.

x y z: : : := 15 30 40

z : w = 40 : 8

∴ x : y : z : w = 15 : 30 : 40 : 8.

12.12.12.12.12. Divide Rs. 2360 among A, B, C so that A : B = 3 : 4, B : C = 5 : 6.

Given: A B

B C

: :

: :

=

=

3 4

5 6

⇒ A B C: : : : .= 15 20 24


Sum = 15 + 20 + 24 =59.

Out of 59. A’s share = 15

∴ Out of 2360 A’s share = 2360 15

59

×

= Rs. 600.

Out of 59, B’s share = 20

Out of 2360 B’s share = 20

592360×

= Rs. 800.Similarly

C’s share = 24

592360× = Rs. 960.

�%��&�'��

A’s share + B’s share + C’s share

= Rs. 600 + Rs. 800 + Rs. 960 = Rs. 2360 = Total.

13.13.13.13.13. Divide 166 into 3 parts such that 4 times the first part, 5 times the 2nd part and 7 times the 3rdpart are equal.

Solution:Solution:Solution:Solution:Solution: Let 1st, 2nd and 3rd part be a, b and c.

Given: 4a = 5b = 7c = x (say)

Then 4a = x, 5b = x

⇒ ax

bx

cx= = =

4 5 7, . and

∴ a b cx x x

: : : : .=4 5 7

i.e., a : b : c = 35 : 28 : 20.

Sum = 35 + 28 + 20 = 83.

∴ 1st part = 35

83166 70× =

2nd part = 28

83166 56× =

3rd part = 20

83166 40× = .

VVVVVerererererififififificaicaicaicaication:tion:tion:tion:tion: 70 + 56 + 40 = 166.

14.14.14.14.14. If the cost of 10 metres of cloth is Rs. 225, find the cost of 22 metres of cloth.



Length mts. Cost Rs.� � � �

10

22

225

x

As the length of the cloth increases. Cost also increases. ∴ Length and Cost are directly propor-tional. To denote direct proportion, we use 2 arrows, with same direction.

∴ 10 22 225: := x

10 22 225x = ×

x = ×22 225

10

x = 495.

∴ 22 mts of cloth costs Rs. 495.

15.15.15.15.15. If 60 men can complete a job in 12 days, how many days will 36 men take to complete the samejob?

Solution:Solution:Solution:Solution:Solution: Men days

60

36

12

x

As the number of men increases, the days required to complete the job decreases ∴ Men and daysare inversely proportional. To denote this we use 2 arrows with opposite direction.

60 36 12: := x

36 60 12x = ×

x = × =60 12

3620.

∴ 36 men can complete a job in 20 days.

16.16.16.16.16. If 10 men can earn Rs. 105 in 7 days, in how many days will 15 men earn Rs. 225?

Solution:Solution:Solution:Solution:Solution: Men Money days

10

15

105

225

7

x

Here number of days is unknown.

Leaving money, or keeping money constant, let us first consider men and days. As the number ofmen increases, days required to complete the job decreases. ∴ Men and days are inversely proportional.

Now, leaving men, let us consider money and days. As the days increase, money earned also in-creases. ∴ Money and days are directly proportional.


∴ Corresponding compound proportion is

10 15225 105

7::

:= x

15 105 7 10 225× × = × ×x

x = × ××

7 10 225

15 105

x = 10

∴ 15 men earn Rs. 225 in 10 days.

17.17.17.17.17. 5 carpenters can earn Rs. 3600 in 6 days, working 9 hrs. a day. How much will 8 carpenters earnin 12 days working 6 hrs. a day.


Carpenter days hrs. Money

5

8

6

12

9

6

3600

x

Here money is unknown.

Taking carpenter and money alone, as the carpenters increase, the money earned by them alsoincreases. So it is direct proportion. Taking days and money, as the day increases money also increases.So it is direct proportion. Now taking hours and money, as the number of hours increases, money alsoincreases. ∴ It is direct proportion.

So corresponding compound proportion is

5 86 129 6

3600:::

:= x

5 6 9 8 12 6 3600× × × = × × ×x

x = × × ×× ×

8 12 6 3600

5 6 9

x = 7680.

∴ 8 carpenters earn Rs. 7680 in 12 days working 6 hours a day.

18.18.18.18.18. A contractor undertook to make 15 kms of roadways in 40 weeks. In 10 weeks 3 kms werecompleted by 180 men working 8 hours a day. Then the men agreed to work 1 hour a day overtimeand some boys were engaged to assist them. The work was finished in the stipulated time (that is40 weeks). How many boys were employed if the work of 3 boys is equal to that of 2 men.


Men weeks hrs. kms.

180 10

40 10 30

8

8 1 9

3

15 3 12x − = + = − =� � � � � �


men and Kms are directly proportional.

men and hrs. are inversely proportional. men and week are inversely proportional.

∴ Required compound proportion is

180 30 109 83 12

: :::

x =

x × × × = × × ×30 9 3 180 10 8 12

x = × × ×× ×

=180 10 8 12

30 9 2

640

3.

Already 180 men are there,

∴640

3180

100

3− =

Given 3 boys ≡ 2 men.

⇒ 1 man ≡ 3

2boys

∴100

3

100

3

3

2men boys≡ ×

= 50 boys.

∴ 50 boys were engaged to assist them.

19.19.19.19.19. If 15 men build a wall 40 ft long, 2 and 1/2 ft. thick and 21 ft. height in 18 days working 10 and1/2 hrs. each day. In how many days working 15 hrs. a day will 45 men build a wall 200 ft. long,5 ft thick and 20 ft. height.

Solution:Solution:Solution:Solution:Solution: Men Length Breadth Height hours days(thickness)

15

45

40

200

212

5

21

20

1012

15

18

x

As the men increase, days decrease, so inverse proportion.

As length increase, days also increase, as breadth increase, days also increase.

As height increase, days also increase.

So, It is direct proportion.

As the number of hrs. increase, days required to construct decrease. So it is inverse proportion.

∴ Corresponding Compound proportion is:


15 45

200 40

5 21

220 21

101

215

18

:

:

:

:

:

:

�

�

��

�

��

= x

x × × × × × = × × × × ×45 40 21

221 15 18 15 200 5 20 10

1

2.

∴ x =× × × × ×

× × × ×

18 15 200 5 20 1012

45 40 212

21 15

x = 40

∴ 40 days are required.

20.20.20.20.20. If 12 pumps working 6 hours a day can draw 2000 gallons of water in 20 days, find in how manydays will 20 pumps working 9 hours a day draw 3000 gallons of water?

Solution:Solution:Solution:Solution:Solution: Pumps Hours Quantity Days12

20

6

9

2000

3000

20

x

As the number of pumps increase, the days required decreases. So it is inverse proportion. As thenumber of hours increases the days required to draw water decreases. So it is inverse proportion. As thenumber of gallons increase, number of days required also increases, so it is direct proportion.

∴ Corresponding compound proportion is

12 206 9

3000 200020

:::

:��

��= x

x × × × = × × ×20 9 2000 20 12 6 3000

x = × × ×× ×

20 12 6 3000

20 9 2000

x = 12

∴ 12 days are required.


��( ��)!� "��

Note:Note:Note:Note:Note: 1. If A can do a piece of work in n days. Then work done by A in 1 day = 1

n.

2. If B’s 1 day’s work = 1

x. Then B can finish the work in x days.

3. If A is twice as good a workman as B,

then

Ratio of work done by A and B = 2 : 1

Ratio of time taken by A and B = 1 : 2

4. If A can do a piece of work in x days, and B can do it in y days, then A and B working together

will do the same work in xy

x y+ days.

5. If A and B together can do a piece of work in z days and A alone can do it in x days then B

alone can do it in zx

x z− .

"�!�� !�"�

1.1.1.1.1. Ram can reap a field in 6 days which Raju alone can reap in 8 days. In how many days bothtogether can reap this field?

Solution:Solution:Solution:Solution:Solution: Ram’s 1 day’s work = 1

6

Raju’s 1 day’s work = 1

8

Ram and Raju’s 1 day’s work = 1

6

1

8+

= + =4 3

24

7

24.

∴ Both together can reap the field in 24

73

3

7= days.

OR

We know, if A can do a piece of work in x days and B can do it in y days then A and B together

can do it in xy

x y+ days.


∴ Ram and Raju together can reap the field in 6 8

6 8

×+ days.

= = =48

14

24

73

3

7 days.

2.2.2.2.2. X and Y together can dig a trench in 10 days which X alone can dig in 30 days. In how many daysY alone can dig it?

Solution:Solution:Solution:Solution:Solution: X and Y’s 1 day’s work = 1

10

X’s 1 day’s work = 1

30

∴ Y’s 1 day’s work = 1

10

1

30−

3 1

30

2

30

− =

∴ Y alone can dig a trench in 30

2 = 15 days.

OR

We know, if A and B together can do a piece of work in z days and A alone can do it in x days then

B alone can do it in zx

x z− days.

∴ Y alone can dig a trench in 30 10

30 10

×−

= =300

2015 days.

3. A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. In how manydays will they finish it together and separately?

(A + B)’s 1 day’s work = 1

12

(B + C)’s 1 day’s work = 1

15

(C + A)’s 1 day’s work = 1

20


Adding

2 [A + B + C]’s 1 day’s work = 1

12

1

15

1

20+ +

2 (A + B + C)’s 1 day’s work = 5 4 3

60

+ +

2 (A + B + C)’s 1 day’s work = 1

5

(A + B + C)’s 1 day’s work = 1

10

So A, B and C together finish the work in 10 days.

Now, A’s 1 day’s work = (A + B + C)’s 1 day’s work − (B + C)’s 1 day’s work.

= − =1

10

1

15

1

30

∴A alone can finish the work in 30 days.

Similarly,

B’s 1 day’s work = − =1

10

1

20

1

20

∴ B alone can finish the work in 20 days.

Similarly,

C’s one day’s work = − =1

10

1

12

1

60

∴ C alone can finish the work in 60 days.

OR

A, B and C can do together in

2xyz

xy yz zx+ + days

=× ×

× + × + ×2 12 15 20

12 15 15 20 20 12

� �

=+ +

= =2 3600

180 300 240

2 3600

72010

� � � ��

days.

A, B and C can together finish the work in 10 days.

4.4.4.4.4. A can do a piece of work in 25 days which B alone can finish in 20 days. Both work for 5 daysand then A leaves off. How many days will B take to finish the remaining work.


Solution:Solution:Solution:Solution:Solution: A’s work in 1 day = 1

25

B’s work in 1 day = 1

20

∴ (A + B)’s 1 day’s work = 1

25

1

20+

(A + B)’s 5 51

25

1

20 day' s work = +

��

= 9

20

Remaining work = 19

20− Note this step� �

= 11

20

Now 1

20 work is done by B in 1 day.

∴11

20 work will be done by B in

11

201

1

20

× = 11 days.

∴ B takes 11 days to finish the remaining work.

5.5.5.5.5. X is thrice as good a work man as Y and is therefore able to finish the piece of work in 60 daysless than Y. Find the time in which they can do it, working together.

Solution:Solution:Solution:Solution:Solution: Ratio of work done by x and y in same time = 3 : 1

∴ Ratio of time taken = 1 : 3

If Y takes y days to finish a work.

Then X takes y − 60 days to finish.

Now

y y− =60 1 3: :

3 60y y− =� �

3 180y y− =

2 180 90y y= ⇒ =


∴ Time taken by Y to finish the work = 90 days and time taken by X to finish the work = 90 − 60 =30 days.

∴ X’s 1 day’s work = 1

30

Y’s 1 day’s work = 1

90

(X + Y)’s 1 day’s work = 1

90

1

30

2

45+ =

∴ Both X and Y can finish the work in 45

222

1

2= days.

6.6.6.6.6.A can build a wall in 30 days which B alone can build in 40 days. If they build it together and geta payment of Rs. 1400 what is A’s share and B’s share?

Solution:Solution:Solution:Solution:Solution: A’s 1 day’s work = 1

30

B’s 1 day’s work = 1

40

∴ Ratio of their work = 1

30

1

40:

= 4 : 3

Total = 4 + 3 =7

A’s share = 4

71400 800× = Rs.

B’s share = 3

71400× = Rs. 600

7.7.7.7.7. A can do a piece of work in 10 days, while B alone can do it in 15 days. They work together for5 days and the rest of the work is done by C in 2 days. If they get Rs. 1200 for the whole workhow should they divide the money?


A’s 1 day’s work = 1

10

B’s 1 day’s work = 1

15


10

1

15

1

6+ =



6

5

6�� =

Remaining work = 15

6

1

6− = (Note this step)

Given: C completes rest of the work in 2 days.

∴ C’s 2 day’s work = 1

6

Now Consider

A’s 5 day’s work : B’s 5 day’s work : C’s 2 day’s work.

51

105

1

15

1

6��: :

1

2

1

3

1

6: :

= 3 : 2 : 1

Total = 3 + 2 + 1 = 6.

A’s share = 3

61200× = Rs. 600

B’s share = 2

6Rs. 400× =1200

C’s share = 1

6Rs. 200.× =1200

8.8.8.8.8. A certain number of men complete a piece of work in 60 days. If there were 8 men more the workcould be finished in 10 days less. How many men were originally there?

Solution:Solution:Solution:Solution:Solution: Let number of men = x

8 men more means x + 8

Given: Men daysx

x + − =8

60

60 10 50

As the number of men increases, the days required to complete the job decreases.

∴ It is inverse proportion.

x x: :+ =8 50 60

x x60 8 50� � � �= +

60 50 400x x− =


10 400 40x x= ⇒ = .

∴ 40 men were originally there.

9.9.9.9.9. 16 men or 28 boys can fence a farm in 40 days. In how many days will 24 men and 14 boyscomplete the same work?

Solution:Solution:Solution:Solution:Solution: Given,

16 men’s work ≡ 28 boy’s work.

i.e., 16 : 28

i.e., 8 : 14

∴ 8 men ≡ 14 boys.

Now 24 men and 14 boys ≡ 24 men + 8 men

≡ 32 men.

Given:

Men days16

32

40

x

As the number of men increases, the days required to fence a farm decreases. ∴ It is inverseproportion.

∴ 16 32 40: := x

32 16 40× = ×x

x = × =16 40

3220.

∴ 20 days are required to complete the work.

10.10.10.10.10. 2 men and 4 boys can do a work in 33 days. 3 men and 5 boys can do the same work in 24 days.How long shall 5 men and 2 boys take to finish it?

Given: 2 men and 4 boys can do the work in 33 days.

⇒ 2 × 33 men and 4 × 33 boys can do it in 1 day.

∴ 66 men and 132 boys can do it in 1 day ...(1)

Similarly

3 men and 5 boys can do the work in 24 days.

⇒ 3 × 24 men and 5 × 24 boys can do it in 1 day

i.e., 72 men and 120 boys can do it in 1 day ...(2)

From (1) and (2)

66M 132B 72M 120B+ ≡ +

132B 120B 72M 66M− ≡ −


12B 6M≡

M 2B.≡∴ One man’s work is equivalent to 2 boy’s work.

Now given: 2 men and 4 boys can finish the work in 33 days.

i.e., 2M + 2M can finish the work in 33 days.

4 men can finish the work in 33 days.

Now,

5 men and 2 boys ≡ 5M + 1M = 6 Men.

Men days4

6

33

x

As men increases, days required to finish the work decreases so it is inverse proportion.

∴ 4 6 33: := x

6 33 4x = ×

x = × ==33 4

622.

∴ 5 men and 2 boys can finish the work in 22 days.

��* ��)!� �� "��

Note :Note :Note :Note :Note :

(1) SpeedDistance

Time=

(2) x x km hr mts= ×��

��

5

18sec

(3) x x mts sec km hr.= ×��

��

5

18

�� !�"�

1.1.1.1.1. The distance between 2 stations A and B is 450 kms. A train starts at 4 p.m. from A and movestowards B at an average speed of 60 km/hr. Another train starts from B at 3 : 20 p.m. and movestowards A at an average speed of 80 km/hr. How far from A will the two trains meet and at whattime?


Solution:Solution:Solution:Solution:Solution: Let the trains meet at a distance x kms. from A.

Let trains from A to B and B to A be X and Y respectively.

Given: Speed of X = 60 km/hr

Speed of Y = 80 km/hr.

Time required to cover x kms. by X = x

60

Time required to cover (450 − x) kms. by Y = 450

80

− x

Difference between time

= 450

80 603 20 4

− − =x x: p.m. to p.m.

450

80 6040

− − =x x min.

450

80 60

40

60

− − =x x hrs.

4 − − =50

80 60

40

60

x x

3 450 4

240

2

3

− −=

x x� �

cross multiplying,

9 450 12 480− − =x x� �

4050 9 12 480 0− − − =x x

3570 21 0− =x

x = =3570

21170.

∴ The trains meet at a distance of 170 kms from A.

Time taken by X to cover 170 kms. = 170

60 hrs.

= 2hrs. 50 min.

So the trains meet at 4 pm + 2 hrs. 50 min.

= 6 : 50 pm.

2.2.2.2.2. Cycling 3

4 of his usual speed, a student is 10 min. late to his class. Find his usual time to cover

the distance.


Let the usual time taken be x min.

Time taken at 3

4 of the usual speed = ��

��

4

3x min. �

Distance

TimeSpeed, Time

Distance

Speed= =

�

�

�

Given:4

310x x− =

⇒ 4 3 30x x− =

x = 30.

∴ Usual time taken = 30 min.

3.3.3.3.3. A bullock cart has to cover a distance of 80 km in 10 hrs. If it covers half of the journey in 3

5��

th

time. What should be its speed to cover the remaining distance in the time left?

Solution:Solution:Solution:Solution:Solution: Total distance = 80 kms.

Total time = 10 hrs.

Distance left = 80

240= kms.

Given: Time taken to cover 1

2 distance =

3

510 6× =hrs. hrs.

Remaining time = 10 − 6 = 4 hrs.

SpeedDistance

Timekm hr= = 40

4

Speed = 10 km/hr.

4.4.4.4.4. A man travels 360 km in 4 hrs. partly by air and partly by train. If he had travelled all the way by

air, he would have saved 4

5 of the time he was in train and would have arrived at his destination

2 hrs. early. Find the distance he travelled by air and train.

Solution:Solution:Solution:Solution:Solution: Total time = 4 hrs.

Given: 4

5 of total time in train = 2 hrs.

Total time in train = 2 5

4

5

2

× = hrs.

Given: If 360 km is covered by air then time taken is 4 − 2 = 2 hrs.

∴ When 3

2 is spent in air,


Distance covered = 360

2

3

2270× = kms.

Distance covered in train = 360 − 270 = 90 km.

5. An aeroplane started 30 minutes later than the scheduled time from a place 1500 km away fromits destination. To reach the destination at the scheduled time, the pilot had to increase the speedby 250 km/hr. What was the speed of the aeroplane per hour during the journey?

Solution:Solution:Solution:Solution:Solution: Let the time taken by aeroplane in later case = x hrs.

We know SpeedDistance

Time=

Distance = 1500 kms.

Given:1500 1500

12

250x x

=+

+

1500 2 1500

2 1250

x x=

++

� �

1500 3000 250 2 1

2 1x

x

x=

+ ++� �

Cross multiplying

1500 2 1 3000 500 250x x x+ = + +� � � �

3000 1500 3000 500 250 02x x x x+ − − − =

500 250 1500 02x x+ − =

÷ by 250

2 6 0

2 4 3 6 0

2 2 3 2 0

2

2

x x

x x x

x x x

+ − =

+ − − =

+ − + =� � � �

−+

−

+

124

32x

x

x

x

2 2 3 2 0x x x+ − + =� � � �

2 3 2 0x x− + =� ��

2 3x =

x x= = −3

22 or

� x cannot be negative, x = 3

2


∴ The plane takes 3

21

1

2hrs. hrs.= in later case

So in Normal case it takes 11

2

1

22+ = hrs.

∴ Normal speed = 1500

2 km hr.= 750

�� )!� "��

Note:Note:Note:Note:Note: 1. The word Alligation literally means linking. The rule takes its name from the lines or linksused in working out questions on mixture.

2. Alligation method is applied for percentage value, ratio, rate, prices, speed etc. and not forabsolute values.

3. Alligation is the rule that enables us to find the proportion in which two or more ingredients atthe given price must be mixed to produce a mixture at a given price.

Cost price of unit quantity of the mixture is called the mean price.

4. Rule of alligation: If 2 quantities are mixed in a ratio, then

Quantity of cheaper

Quantity of dearer

C.P. of dearer Mean Price

Mean Price C.P. of Cheaper=

−−

� � � ��

i.e., cheaper quantity: dearer quantity = d − m: m − c.

We represent it as

Cost price of C.P. of dearercheaper (c) (d)

Mean Price (m)

(d − m) m − c

�� !�"�

1.1.1.1.1. The price of first quality of rice is Rs. 16 per kg and that of second quantity rice is Rs. 10. In whatratio these two should be mixed so that the mixture can be sold for Rs. 12 per kg.


Cost price of 1 kg C.P. of 1 kgcheaper Rice dearer rice

10 (c) 16 (d)

Mean Price12 (m)

(d − m) = 16 − 12 m − c= 4 = 12 − 10

= 2


Quantity of cheaper rice : Quantity of dearer rice.

= 4 : 2

= 2 : 1

OR

Let us mix these 2 types of rice in the ratio x : y.

The price of x kgs. cheaper rice = 10 x

Price of y kgs. dearer rice = 16 y

The price of (x + y) kgs. of mixture = 10x + 16y

Price of 1 kg of mixture = 10 16x y

x y

++

Given: Price of 1 kg of mixture = 12

∴10 16

12x y

x y

++

=

10 16 12 12x y x y+ = +

16 12 12 10y y x x− = −

4 2y x=

4

2= x

y

⇒ x y: := 4 2

∴ x y: := 2 1 .

2.2.2.2.2. Arjun travelled a distance of 80 km in 7 hrs. partly in bullock cart at the rate of 8 km/hr and partlyin tonga at 16 km/hr. Find the distance travelled in bullock cart.

Solution:Solution:Solution:Solution:Solution: Average distance travelled in

180 80

7 hr

km

7 hrs.km hr.= =

Distance travelled Distance travelledin 1 hr in bullock cart in 1 hr in tonga

8 km (c) 16 km (d)

Average in 1 hr

80

7m� �

(d − m) (m − c)

1680

7

32

7− =

80

78

24

7− =


Time taken by bullock cart

Time taken by tonga= −

−d m

m c

=

327

247

32

244 3= : .

∴ Out of 7 hrs., he took 4 hrs. to travel in bullock cart and 3 hrs. in tonga.

Distance covered by bullock cart = 4 × 8 km = 32 km.

3.3.3.3.3. In what ratio must a person mix three kinds of rice costing Rs. 12.00, Rs. 14.00 and Rs. 17.40 perkg. So that mixture may be worth Rs. 14.10 per kg.

Solution:Solution:Solution:Solution:Solution: Step 1: Mix 1st and 3rd kind of rice to get a mixture worth Rs. 14.10.

C.P. of 1 kg rice C.P. of 1 kg of riceof 1st kind of 3rd kind

Rs. 12.00 (c) Rs. 17.40 (d)

Mean PriceRs. 14.10

(d−m) = Rs. 3.30 (m−c) Rs. 2.10

By alligation rule

Quantity of 1st kind of rice

Quantity of 3rd kind of rice= −

−d m

m c

= =3 30

2 10

11

7

.

..

∴ They must be mixed in the ratio 11 : 7.

Step II: Mix rice of 1st and 2nd kind to obtain a mixture worth Rs. 14.10 per kg.

C.P. of 1 kg rice C.P. of 1 kg of riceof 1st kind of 2nd kind

Rs. 12.00 (c) Rs. 14.40 (d)

Mean Price (m)Rs. 14.10

(d−m) = Rs. 0.30 (m−c) = Rs. 2.10


By alligation rule,

Quantity of 1st kind of rice

Quantity of 2nd kind of rice= −

−=d m

m c

0 30

2 10

.

.

= 1

7.

∴ They must be mixed in the ratio 1 : 7.

Now

1st kind : 2nd kind = 1 : 7

1st kind : 3rd kind = 11 : 7

To make the value of 1st kind equal,

Multiply 1st ratio by 11 and 2nd ratio by 1.

∴ 1st kind : 2nd kind = 11 : 77

1st kind : 3rd kind = 11 : 7

∴ 1st kind : 2nd kind : 3rd kind = 11 : 77 : 7.

4.4.4.4.4. Gauri possessing Rs. 84,000 lent a part of it at 8% simple interest and the remaining at 6 and2/3% simple interest. Her total income after 1 and 1/2 years was Rs. 8820. Find the sum lent atdifferent rates.

Solution:Solution:Solution:Solution:Solution: Given:

P = 84000

I = 8820

T = 11

2 yrs.=

3

2yrs.

R = ?

We know,

I RI= ⇒ =PRT

100 PT

100

R = ×

×=100 8820

8400032

7%

Rate of Interest Rate of Interestcheaper (c) dearer (d)6 and 2/3% 8%

(m) Average rate7%

(d − m) = 8 − 7 = 1 m − c 7 62

3

1

3− =


By alligation rule,

Money given at 623

SI

Money given at 8% SI

%.= =1

13

3

1

= 3 : 1

Sum = 3 + 1 = 4.

∴ Money lent at 6 and 2/3% SI = 3

484000×

= Rs. 63000.

Money lent at 8% SI = 1

484000×

= Rs. 21000.

5.5.5.5.5. Adarsh buys 2 horses for Rs. 1350 and sells one at 6% loss and other at 7.5% gain and on thewhole, he neither gains, no loses. What does each horse cost?


Cheaper horse Dearer horse−6% (c) 7.5% (d)

MeanO

(d − m) (m − c)7.5 6

Cost of 1st horse

Cost of 2nd horse= = =7 5

6

75

60

5

4

.

1st horse : 2nd horse = 5 : 4

Sum = 5 + 4 = 9

∴ Cost of 1st horse = 5

91350× = Rs. 750

Cost of 2nd horse = 4

9Rs. 600.× =1350

�� )��

• For the given ratio a : b, duplicate ratio is a2 : b2, subduplicate ratio is a b: , Triplicate ratio

is a3 : b3 and subtriplicate ratio is a b3 3: .


• a : b : : c : d iff ad = bc.

• If a : b = b : c. Then b is called mean proportional and c is called 3rd proportional.

• If a : b = c : d. Then d is called fourth proportional.

• If a : b = c : d. Then

(i) b : a = d : c (Invertendo)

(ii) a : c = b : d (Alternendo)

(iii) a + b : b = c + d : d (Componendo)

(iv) a − b : b = c − d : d (Dividendo)

(v) a + b : a − b = c + d : c − d

(Componendo and dividendo)

• If a : b = c : d represent a direct proportion then a : b = d : c or b : a = c : d represent an inverseproportion and vice-versa.

• Rule of alligation: Quantity of cheaper

Quantity of dearer

C.P. of Dearer Mean

Mean C.P. of cheaper= −

−

c

d m

m

d

m c

� �

� �

� �

− −

i.e., cheaper : dearer = d − m : m − c.

��"�

1. Express the following ratios in their simplest form :

(a) 8 : 24 (b) 24 : 88 (c) 9 : 900

2. Compare the following ratios :

(a) 6 : 5 and 3 : 2 (b) 1 : 5 and 6 : 7 (c) 3 : 2 and 2 : 7

3. Find the ratio between :

(a) 1 hr 15 min and 105 min

(b) 3 kg 50 gm and 4 kg 500 gm

4. A number is divided into 3 parts in the ratio 2 : 3 : 4. If the 3rd part is 32. Find the other parts.

5. Find the third proportional to 2 : 6.

6. Find the mean proportional between 49 and 64.

7. Find the fourth proportional to 4, 5 and 12.

8. What number must be subtracted from each of 9, 11, 15 and 19. So that the difference will beproportional.

9. (a) If x : y = 2 : 3, y : z = 4 : 5. Then find z : x.

(b) If x : y = 2 : 3, y : z = 4 : 5 and z : w = 6 : 7. Then find x : w.

(c) If a : b = 2 : 3, b : c = 4 : 5; c : d = 6 : 7. Then find a : b : c : d.


10. Two numbers are in the ratio 5 : 8. If 9 is added to each then they are in the ratio 8 : 11. Find thenumbers.

11. The ratio between the ages of Khan and Ranjith is 6 : 5 and the sum of their ages is 44 years. Findthe ratio of their ages after 8 years.

12. One year ago the ratio of between Sarala and Saraswathi’s salary was 3 : 4. The ratio of theirindividual salaries between last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. Atpresent the total of their salary is Rs. 4160. Find the salary of Sarala now.

13. The ratio between Sumit’s and Prakash’s age at present is 2 : 3. Sumit is 6 years younger thanPrakash. Find the ratio of Sumit’s age to Prakash’s age after 6 years.

14. 8 labourers can build a wall in 6 days. In how many days, 12 labourers can do the same work?

15. 6 carpenters working 7 hrs. a day can complete 24 tables in 20 days. How many days will 12carpenters working 6 hrs. a day take to complete 36 tables.

16. A, B, C start a business with investments of Rs. 25,000, Rs. 16,000 and Rs. 12,000 respectively.If the profit for the year amounts to Rs. 6850. Find the share of each partner.

17. Ram can reap a field in 9 days which Deepak alone can reap in 12 days. In how many days bothtogether can reap this field.

18. A can do 1

3��

rd

of the work in 5 days and B can do 2

5��

th

of the work in 10 days. In how many

days both A and B together can do the work?

19. Sunil completes a work in 4 days, whereas Dinesh completes the work in 6 days. Ramesh works1 and 1/2 times as fast as Sunil. How many days it will take for the 3 together to complete thework.

20. A can complete a job in 9 days, B in 10 days and C in 15 days. B and C start the work and areforced to leave after 2 days. Find the time taken to complete the remaining work.

21. If 3 men or 4 women can construct a wall in 43 days. Then find the number of days that 7 menand 5 women take to construct it.

22. 8 men can dig a pit in 20 days. If a man works half as much again as a boy then in how many days4 men and 9 boys can dig a similar pit.

23. A train leaves Meerut at 6 a.m. and reaches Delhi at 10 a.m. Another train leaves Delhi at 8 am.and reaches Meerut at 11 : 30 a.m. At what time do the 2 trains cross each another.

24. A boy goes to school with a speed of 3 km/hr and returns to the village with a speed of 2 km/hr.If he takes 5 hrs. in all, find the distance between village and the school.

25. In what proportion must Ragi at Rs. 3.10 per kg be mixed with Ragi at Rs. 3.60 per kg so that themixture be worth Rs. 3.25 a kg.

26. A man possessing Rs. 1000 lent a part of it at 6% S.I. and the other at 8% SI the yearly incomeis Rs. 75. Find the sum lent at 8% SI.

27. Kantilal mixes 80 kgs of sugar worth Rs. 6.75 per kg with 120 kg worth Rs. 8 per kg. At what ratehe sell the mixture to gain 20%.

28. A jar contains a mixture of 2 liquids A and B in the ratio 7 : 5. When 9 litres of mixture is drawnoff and the jar is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid Awas contained by the jar initially?


29. A sum of Rs. 41 was divided among 50 boys and girls each boy gets 90 paise and a girl 65 paise.How many boys are there?

��"��"

1. (a) 1 : 3 (b) 3 : 11 (c) 1 : 100

2. (a) 3 : 2 is greater than 6 : 5 (b) 6 : 7 is greater than 1 : 5

(c) 3 : 2 is greater than 2 : 7.

3. (a) 5 : 7 (b) 61 : 90

4. 16 and 24 5. 18 6. 56 7. 15 8. 3

9. (a) 15 : 8 (b) 16 : 35 (c) 16 : 24 : 30 : 35

10. 15 and 24 11. 8 : 11 12. Rs. 1600 13. 3 : 4

14. 4 15. 105 16. Rs. 2650, Rs. 2200, Rs. 2000.

17. 51

7 days 18. 9

3

8days 19. 1

5

19 days 20. 6 days 21. 12 days

22. 16 days 23. 8.56 a.m. 24. 6 km. 25. 7 : 3 26. Rs. 750

27. Rs. 9 per kg 28. 21 litres 29. 34.

�

��

� ��

Condensation of data is necessary in statistical analysis because a large number of big figures are notonly confusing to mind but also difficult to analyse. In order to reduce the complexity of data and tomake them comparable it is essential that the various phenomena which are being compared are re-duced to one figure each. It is obvious that a figure which is used to represent a whole series shouldneither have the lowest value in the series nor the highest value but a value somewhere between thesetwo limits, possibly in the centre where most of the items of the series cluster. Such figures are calledmeasures of central tendencies or averages.

Averages are usually of the following types:

(a) Mathematical average

(i) Arithmetic average or mean

(ii) Geometric mean and

(iii) Harmonic mean.

(b) Average of position:

(iv) Median and

(v) Mode.

Of the above mentioned five important averages, we are going to discuss in this book, about arith-metic average or Mean in detail.

��

Mean of the certain number of quantities which are all of equal weightage or importance is the figureobtained by dividing the total values of various items by their number.

If the marks obtained by a student in five subjects are 56, 67, 82, 43 and 52, then to find the meanof these marks we shall add these marks and divide the total so obtained by the number of items whichis 5.

∴ Mean = + + + +56 67 82 43 52

5


Mean = =300

560.

In general, if x1, x2, x3, ... xn are the values in a data containing n items then their mean, denoted by

X Xx x x x

nn is = + + + +1 2 3 ...

i.e., X

x

n

ii

n

= =∑

1

If all the quantities are not of equal importance i.e. equal weight, we compute the weighted averageas follows:

If w1, w2, w3, ... wn are the weights associated to the values x1, x2 ... xn respectively the weighted

average Xw is given by

Xx w x w x w

w w wwn n

n

=+ + ++ + +

1 1 2 2

1 2

...

...

X

x w

ww

i ii

n

ii

n= =

=

∑

∑1

1

For example if Rama scores 35 in 2 subjects, 42 in 4 subjects and 72 in remaining 4 subjects. Thenaverage marks scored by Rama

= × + × + ×+ +

35 2 42 4 72 4

2 4 4

= + +70 168 288

10

= =526

1052 6. .

Note:Note:Note:Note:Note: To compute arithmetic mean in a continuous series, the midpoints of the various class inter-vals are written down to replace the class intervals. Once it is done, there is no differencebetween a continuous series and discrete series.

��

It is nothing but average of averages. Combined average is computed when the data set consists ofdifferent groups and average for each group is known.

Averages 223

If X X Xn1 2, ... , are the average of groups 1, 2, ..., n and N1, N2, ... Nn are number of quantities in

groups 1, 2, 3, ... n. Then combined average X n1 2 3, , ... is given by

XX N X N X N

N N Nnn n

n1 2 3

1 1 2 2

1 2, , ... ...

=+ ++ + +

For example if the average marks of group of 5 students is 56 and average marks of another groupof 6 students is 72 then combined average marks of 11 students.

= × + ×+

56 5 72 6

5 6

= + = ≈280 432

11

712

1164 72. .

�� !�"�

1.1.1.1.1. Find the arithmetic mean of 56, 87, 36, 72 and 44:

Solution:Solution:Solution:Solution:Solution: Arithmetic mean Xx x x x x= + + + +1 2 3 4 5

5

= + + + +56 87 36 72 44

5

= =295

559.

2.2.2.2.2. A market survey on demands of chocolates at a local shop provided the following distribution ofdaily demand:

Daily demand Frequency

(Number of chocolates) (Number of days)

20 5

30 27

40 62

50 4

60 2

Total: 100

Find the average demand of chocolates in numbers/day.

Solution:Solution:Solution:Solution:Solution: Mean =+ + + ++ + + +

x w x w x w x w

w w w wn n

n

1 1 2 2 3 3

1 2 3

...

...


Mean = × + × + × + × + ×+ + + +

20 5 30 27 40 62 50 4 60 2

5 27 62 4 2

Mean = + + + +100 810 2480 200 120

100

X = 37.1.

3.3.3.3.3. Find the arithmetic mean of the following data:

Class Interval: 0-20 20-50 50-90 90-140 140-200

Frequency: 10 20 40 15 15

C.I. Mid value Frequency X.W

(x) (w)

0-20 10 10 100

20-50 35 20 700

50-90 70 40 2800

90-140 115 15 1725

140-200 170 15 2500

ΣW =100 ΣXW = 7825

Xx w x w x w

w w wwn n

n

=+ + ++ + +

1 1 2 2

1 2

...

...

XXw

ww = = =ΣΣ

7825

10078 25. .

4.4.4.4.4. If 10 books are purchased at the rate of Rs. 7 each, 15 books are purchased at the rate of Rs. 9.50and 20 books are purchased at the rate of Rs. 13.50, find the average price of a book.

Solution:Solution:Solution:Solution:Solution: Given: N1 = 10, N2 = 15 and N3 = 20

X1 = 7, X2 = 9.50 and X3 = 13.50.

Combined average = X N X N X N

N N N1 1 2 2 3 3

1 2 3

+ ++ +

= × + × + ×+ +

7 10 9 50 15 13 50 20

10 15 20

. .

= + +70 142 50 270

45

.

= 10.7222.

∴ Average price of a book = Rs. 10.72.

5.5.5.5.5. The average marks of 15 students of class is 45. A student who has secured 17 marks leaves theclass. Find the average marks of the remaining 14 students.

Averages 225

Solution:Solution:Solution:Solution:Solution: Average marks of 15 students = 45

∴ Total marks = 45 × 15 = 675

If the one student with 17 marks leaves the class then total marks of remaining 14

Students = 675 − 17 = 658

Average marks of 14 students = =658

1447.

6.6.6.6.6. The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find theage of the 10th student.

Solution:Solution:Solution:Solution:Solution: Given: Average age of 10 students = 6 years

Total age of 10 students = 10 × 6 = 60 years

Given, the sum of ages of 9 of them = 52 years.

∴ Age of the 10th student = 60 − 52 = 8 years.

7.7.7.7.7. The average age of 10 students is 14 years. Among them, the average age of 5 students is 12 years.Find the average age of remaining students:

Solution:Solution:Solution:Solution:Solution: Given: Combined Average X = 14.

N1 + N2 = 10

N1 = 5, N2 = 5, X X1 212= =, ?

We have

XX N X N

N N=

++

1 1 2 2

1 2

1412 5 5

5 52= × + ×

+X

1460 5

102= + X

140 60 5 2= + X

5 140 60 802X = − =

X280

516= = .

∴ Average age of remaining 5 students = 16 years.

8. A shopkeeper purchased a certain number of dress materials at an average price of Rs. 190 each.The average price of 10 dress materials was Rs. 175 and that of remaining dress materials wasRs. 200. Find the total number of dress materials purchased.

Solution:Solution:Solution:Solution:Solution: Combined mean X = 190 (Given)


N X1 110 175= =,

X2 200=

Let total number of dress materials purchased = x

N N x1 2+ =

10 2+ =N x

N x2 10= −

We have

XN X N X

N N=

++

1 1 2 2

1 2

19010 175 10 200

=+ −� � � �x

x

190 1750 200 2000x x= + −

190 250 200x x= − + 250 200 190= −x x

250 10= x

⇒ x = =250

1025

∴ Total number of dress materials purchased = 25.

9.9.9.9.9. The average weight of a group containing 26 persons is 70 kg. 6 persons with average weight 67kg leave the group and 5 persons with weights 68, 72, 82, 56 and 54 kgs. joins the group. Findthe average weight of the group now.

Solution:Solution:Solution:Solution:Solution: Average weight of 26 persons = 70 kgs. (Given)

Total weight of 26 persons = 26 × 70 = 1820

6 persons with average weight 67 kgs leave the group. (Given)

i.e., Total weight of 6 persons = 67 × 6

= 402

Total weight of the remaining 20 persons

= 1820 − 402 = 1418

5 persons with weight 68, 72, 82, 56 and 54 kgs. join the group.

∴ Total weight of (20 + 5) persons

= 1418 + 332 = 1750

Average weight of the group now = 1750

2570= kgs.

Averages 227

10.10.10.10.10. A batsman realises that by scoring a century in the 11th innings of his test matches he has betteredhis average of the previous 10 innings by 5 runs. What is his average after the 11th inning:

Solution.Solution.Solution.Solution.Solution. Let the average runs in 10 innings be x

Then total runs in 10 innings = 10x

Average runs after 11th innings = x + 5 (Given)

Total runs in 11th innings = (x + 5) 11

Also given a batsman scores a century in the 11th innings.

∴ Runs in 11 innings — Runs in 10 innings.

= Runs in 11th innings = 100.

x x+ − =5 11 10 100� �

11 55 10 100x x+ − =

x = −100 55

x = 45.

∴ Average runs after 11th innings = x + 5 = 45 + 5 = 50.

11.11.11.11.11. Ms. Vani bought 17 books in a discount sale. The average price of books being Rs. 53. Theaverage price of the eleven Kannada books is Rs. 71. If the prices of the remaining 6 Englishbooks form an increasing arithmetic progression with last term Rs. 25. Find the price of cheapestEnglish book.

Solution.Solution.Solution.Solution.Solution.

Given: X = 53

X N1 171 11= =,

X N2 2 6= =?,

We have

XN X N X

N N=

++

1 1 2 2

1 2

5311 71 6

11 6

2=+

+

� � � �X

53781 6

172= + X

⇒ 781 6 53 172+ = ×X

6 901 7812X = −

6 1202X =

X2 20= .


∴ Total cost of 6 English books = 6 × 20

= Rs. 120.

Given the cost of 6 English books form an increasing A.P.

∴ Costs are a, a + d, a + 2d, a + 3d, a + 4d and a + 5d.

Given last term = 25

a + 5d = 25

5d = 25 − a

Also sum of first n terms in an A.P. is

Sn

a n dn = + −2

2 1� �

∴ S a d66

22 6 1= + −� �

S a d6 3 2 5= +

Substituting we get

120 3 2 25= + −a a

120

325= +a

40 25= +a

40 25− = a

⇒ a = 15.

∴ Cost of the cheapest English book = Rs. 15.

12.12.12.12.12. At a place, the average temperatures from Monday to Thursday was 35°C and from Tuesday toFriday was 38°C. Find the day temperatures on Monday and Friday if the ratio of temperatureson Monday and Friday is 5:7.

Solution:Solution:Solution:Solution:Solution: Average temperatures from Monday to Thursday = 35° (Given)

i.e.,Temp. on Mon Tue Wed Thu+ + + = °

435

∴ Temperature on Mon + Tue + Wed + Thu = 35° × 4 = 140° ...(1)

Average temperature from tuesday to friday = 38° (Given)

Temp. on Tue Wed Thu Fri+ + += °

438

∴ Temperature on Tue + Wed + Thu + Fri = 38° × 4 = 152° ...(2)

Equation (2) − Equation (1)

= Temp on Fri − Temp. on Monday = 152° − 140°

Averages 229

Temp. on Fri − Temp. on Monday = 12°

∴ Temp. on Fri = 12° + Temp. on Monday ...(3)

Also given ratio of temperature on Monday by Friday = 5:7

∴Temp. on Mon.

Temp. of Fri.= 5

7.

7 (Temperature on Monday) = 5 (Temperature on Friday)

7 (Temperature on Monday) = 5 (12 + Temperature on Monday)

7 60 5x x= +

7 5 60 2 60 30x x x x− = ⇒ = ⇒ =∴ Temperature on Monday = 30°

From (3) Temperature on Friday = 12 + Temperature on Monday

= 12° + 30° = 42°.

13.13.13.13.13. In a hostel of 80 students, the total monthly expenses were increased by Rs. 8000 when 20 morejoin the hostel. If the average monthly expenses thereby reduced by Rs. 40 per head, Find theaverage monthly expenses per student.

Solution.Solution.Solution.Solution.Solution. Let the average monthly expenses per student be x.

∴ Total monthly expenses for 80 students = 80 × x = 80x

When 20 more join the hostel, monthly expenses increases by Rs. 8000.

∴ Total monthly expenses of 100 students = 80x + 8000

Monthly average is reduced by 40 ⇒ x − 40.

∴ Monthly expenses of 100 students = 100 (x − 40)

∴ 100 40 80 8000x x− = +� �

100 4000 80 8000x x− = +

100 80 8000 4000x x− = +

20 12000x =

⇒ = =x12000

20600.

∴ Monthly expenses per student = Rs. 600.

14.14.14.14.14. Out of 3 numbers, the first is twice the second and is half of the third. If the average of the threenumbers is 56, Find the 3 numbers.

Solution.Solution.Solution.Solution.Solution. Let the 3 number be x, y and z.

Given: x yz x y

z y= = =

=22

24

i.e.,

Also given average = 56.

∴x y z+ + =

356


x y z+ + = × =56 3 168

2 4 168y y y+ + =

7 168y =

⇒ y = =168

724

∴ The 3 numbers are x = 2y = 2 (24) = 48

y = 24

and z = 4y = 4 (24) = 96.

15.15.15.15.15. The average expenditure of a man for the first five months is Rs. 1200 and for the next sevenmonths it is Rs. 1300. If he saves Rs. 2900 in that year, then find his average monthly income.

Solution.Solution.Solution.Solution.Solution. Given average expenditure for 5 months = Rs. 1200

∴ Total expenditure for 5 months = 5 × 1200

= Rs. 6000.

Also given average expenditure for 7 months = Rs. 1300

∴ Total expenditure for 7 months = 7 × 1300

= Rs. 9100

∴ Total expenditure for 12 months = Rs. 6000 + Rs. 9100

= Rs. 15,100

Savings = Rs. 2900 (Given)

We know Saving = Income − Expenditure

⇒ Income = Expenditure + Saving

∴ Total income = Rs. 15100 + 2900

= Rs. 18,000

Average monthly income = 18000

12= Rs. 1500.

16.16.16.16.16. Ten years ago, the average age of a family of 4 members was 24 years. Two children having beenborn, the average age of the family is same today. What is the present age of the youngest childif they differ in age by 2 years.

Solution:Solution:Solution:Solution:Solution: Let the age of youngest child be x yrs.

∴ Age of next child = x + 2 years.

10 years ago

Average age of 4 members = 24

Total age of 4 members = 24 × 4 = 96.

After 10 years,

Total age of 4 members = 96 + 4 × 10 [Each member’s age increases by 10 years]

= 96 + 40 = 136.

Averages 231

Now average of 6 members

= + + + =136 2

624

x xgiven� �

136 2 2 24 6+ + = ×x

138 2 144+ =x

2 144 138x = −

2 6x =

x = 3

∴ Present age of the youngest child = 3 yrs.

17.17.17.17.17. The average ages of A and B is 42 yrs., that of B and C is 28 yrs. and that of C and A is 40 yrs.Find the ages of A, B and C.

Solution:Solution:Solution:Solution:Solution: Let the ages of A, B and C be a, b and c. Given average of A and B = 42

⇒a b+ =

242

a + b = 84 ...(1)

Given: Average age of B and C = 28 years.

b c+ =2

28

b + c = 56 ...(2)

Also given average age of C and A is

i.e.,c a+ =

220

⇒ c + a = 40 ...(3)

Solving (1) and (2) we get

a bb c

a c

+ =+ =

− − −− =

8456

28� � � � � � ...(4)


⇒

+ =− =

=

c a

a c

a

40

28

2 68

⇒ a = 34

a c c c+ = ⇒ + = ⇒ =40 34 40 6


b c b b+ = ⇒ + = ⇒ =46 6 46 40

∴ Age of A = 34 yrs. B = 40 yrs. and C = 6 yrs.

��

• Mean Xx x x

nn=

+ + +1 2 ...

• Xw x w x w x

w w wwn n

n

=+ + ++ + +

1 1 2 2

1 2

...

...

• XX N X N X

N N NnN

n

n123

1 1 2 2

1 2

=+ + ++ + +

...

...

��"�"

1. A cricketer makes 72, 59, 101, 18 and 10 runs respectively in 5 matches played by him. Find hisaverage score.

2. The average weight of a class of 24 students is 35 kgs. If the weight of the teacher is included,the average rises by 400 gms. Find the weight of the teacher.

3. The average marks of 15 boys of a class is 65 and 11 girls of the same class is 78. Find the averagemarks of the students.

4. 3 tests in English, 2 in Hindi, 4 in Kannada and 5 in Sociology are conducted. The average marksscored by Raju in English is 60, that in Hindi is 56 and in Kannada is 45. If the average marksof all the subjects taken together is 48. Then find the average marks scored by him in Sociology.

5. 10 shirts and 5 pants were bought for Rs. 6000. If the average price of a shirt is Rs. 450, then findthe average price of a pant.

6. The average of 25 results is 18, that of first 12 is 14 and of the last 12 is 17. Then find the 13thresult.

7. The average age of A, B, C, D five years ago was 45 years. By including X, the present age ofall the five is 49 years. Find the present age of X.

8. A batsman makes a score of 87 runs in the 17th innings and thus increased his average by 3. Findhis average after 17th inning.

9. Miss Radha bought 51 dress materials in a discount sale. The average price of a dress materialbeing Rs. 318. The average price of 33 polyster dress materials is Rs. 426. If the prices of theremaining cotton dress materials form an increasing arithmetic progression with last term 150.Find the price of the cheapest cotton dress material.

10. The average temperature for Monday, Tuesday and Wednesday was 40°C. The average tempera-ture for Tuesday, Wednesday and Thursday was 41°C. If the temperature on thursday was 42°C.Then find the temperature on Monday.

Averages 233

11. The average age of a husband and a wife was 23 years when they were married 5 years ago. Theaverage age of the husband, the wife and a child who was born during the interval is 20 years now.How old is the child now?

12. The average age of 5 members of a committee is the same as it as 3 years ago, because an oldmember has been replaced by a new member. Find the difference between the ages of old and newmember.

13. There were 35 students in a hostel. If the number of students is increased by 7, the expenses ofthe mess were increased by Rs. 42 per day while the average expenditure per head diminished byRe 1. Find the original expenditure of the mess.

14. The average weight of A, B and C is 45 kg. If the average weight of A and B is 40 kgs and thatof B and C is 43 kg. Then find the weight of B.

15. The average salary of 20 workers in an office is Rs. 1900 per month. If the manager’s salary isadded, the average salary becomes Rs. 2000 per month. What is the manager’s annual salary?

��"��"

1. 52 runs 2. 45 kgs 3. 70.5 4. 40 5. Rs. 300 6. 78

7. 45 years 8. 39 9. Rs. 39 10. 39°C 11. 4 years 12. 15 years

13. Rs. 420 14. 31 kg 15. Rs. 48,000.


�

��

��

Suppose a merchant A purchases goods worth say Rs. 50,000 from another merchant B at a credit forcertain period say 6 months. Then B draws up a draft i.e. prepares a special bill called ‘Hundi’ or billof exchange. On the receipt of the goods, A gives an agreement dually signed on the bill stating that hehas accepted the bill and the money can be withdrawn from his bank account after 6 months of the dateof bill. On this bill, there is an order from A to his bank asking to pay Rs. 50,000 to B after 6 months.

If B needs the money of this bill earlier than 6 months, then B can sell the bill to a banker or a brokerwho pays him the money against the bill but somewhat less than the face value.

Bill discounting is essentially lending by the banker against the bills and the bankers charges acertain interest for doing this service.

��

Bill:Bill:Bill:Bill:Bill: The document which required an individual to pay a fixed amount after a fixed later date is calleda bill.

Discount:Discount:Discount:Discount:Discount: When a bill is cashed in advance of its date of maturity an amount is deducted by the moneylender or bank from the amount of the bill due. This amount deducted is called Discount.

TTTTTrrrrrue discount:ue discount:ue discount:ue discount:ue discount: The present value of a sum of money is that principal which, if placed on stipulated ratefor a specified period will amount to that sum of money at the end of the specified period. The intereston the present value of the bill is called true discount.

BankBankBankBankBanker’er’er’er’er’s discount:s discount:s discount:s discount:s discount: The interest on the face value or amount of the bill is called Banker’sdiscount.

BankBankBankBankBanker’er’er’er’er’s gs gs gs gs gain:ain:ain:ain:ain: The difference between banker’s discount and true discount is called Banker’sgain.

Bill Discounting 235

�� !�"�� #�

Stamp No. 1843, K.R. Road,Bangalore

To, 10.2.2005

Anil,

BTM Layout, Bangalore

Six months later pay me or my order the sum of Rs. 50,000 (Rupees fifty thousand) for valuereceived.

Anil Akram

In the above bill of exchange.

Drawer is Mr Akram so payee is Akram.

Drawee or Acceptor is Mr. Anil.

Drawing date: 10.2.2005

Bill amount or Face value of Bill = Rs. 50,000 = Nominal value of the bill.

Legally due date: drawing date + Bill period + grace period of 3 days

(It is customary to give 3 days grace period).

So In the above bill,

Legally due date is 13.8.2005

�

Date of drawing : 10 - 2 - 2005

Bill period : 0 - 6 - 0

Grace period : 3 - 0 - 0

Legally due date : 13- 8 - 2005

��$ %��&��

If F is the face value of the bill, BD is the banker’s discount, t is the time in years, R = Rate of interest

rR=

100; TD = True discount; BG = Banker’s gain, P = Present value of the bill, then

1. BD = Ftr

2. TD = Ptr

3. BG = BD − TD

4. Banker’s present worth or the discounted value of bill = F − BD

F Ftr−

F tr1−� �

5. True present worth PF

trw =+1


6. FBD TD

BG= ×

7. BG TD t r= ⋅ ⋅

8. BGTD

Pw

= � ��

2

where Pw = True present worth.

9. TD P w B G= ×. . . .� � � �

'��(��)&�*��+�

1.1.1.1.1. Find the present value, true discount, Banker’s discount and Banker’s gain on a bill of Rs. 10,450due in 9 months at 6% per annum.


Given: F = Rs. 10,450

t = 9 months = =9

12

3

4 yrs. yrs.

R r= ⇒ = =6%6

1000 06.

Banker’s discount = BD = Ftr

BD = × ×10 4503

40 06, .

BD = 470 25.

∴ Banker’s discount = Rs. 470.25.

True present worth, PF

tr=

+1,

P =+ ×

10 450

134

0 06

,

.

P = ×+

10 450 4

4 0 18

,

.

P = 10,000

∴ Present worth = Rs. 10,000.

True discount = Ptr

= × ×10 0003

40 06, .

= 450.


TD = Rs. 450.

∴ Banker’s gain = BD − TD

= 470.25 − 450

= 20.25.

∴ Banker’s gain = Rs. 20.25.

2.2.2.2.2. A bill for Rs. 13,000 was drawn on 3rd Feb 2005 at 6 months date and discounted on 13th Mar2005 at the rate of 8% p.a. For what sum was the bill discounted and how much did the bankergain on this?

Solution:Solution:Solution:Solution:Solution: Date of Bill : 3 - 2 - 2005Bill period : 0 - 6 - 0

Grace period : 3 - 0 - 0Legally due date : 6 - 8 - 2005

But bill was discounted on 13th Mar 2005.

∴ Period days

March April May June July Aug

=�

+�

+�

+�

+�

+�

18 30 days 31 30 31 6

� � � � � � � � � � � �

t = =146146

365 days yrs.

R r= ⇒ = =8%8

1000 08.

F = Rs. 13,000

BD Ftr= = × ×13000146

3650 08.

BD = 416.

∴ Banker’s discount = Rs. 416.

Present worth, PF

tr=

+1

P =+ ×

13000

1146365

0 08.

P = =13000

1 03212596 899

..

P ≈ 12596 90.

True discount = = × ×Ptr 12596 90146

3650 08. .


TD = 403.10

True discount = Rs. 403.10

Banker’s gain = BD − TD

= 416 − 403.10

= 12.90

Banker’s gain = Rs. 12.90.

3.3.3.3.3. A banker pays Rs. 2440 on a bill of Rs. 2500, 73 days before the legally due date. Find the rateof discount charged by the banker?

Solution:Solution:Solution:Solution:Solution: Given: F = Rs. 2500

Banker’s discount = Rs. 2500 − 2440

BD = Rs. 60

t = 73

365 yrs.

R = ?

We have BD = Ftr

60 = × ×250073

365r

⇒ r = ××

60 365

2500 73

r = 0.12

R = 100r = 0.12 × 100

R = 12%.

∴ Rate of discount charged by banker = 12%.

4.4.4.4.4. The banker’s gain on a bill is 1

5 of the banker’s discount and the rate of interest is 20% p.a. Find

the unexpired period of the bill.

Solution:Solution:Solution:Solution:Solution: Given: B G BD. .= ×1

5

R r= ⇒ = =20%20

1000 2.

BG BD TD= −

1

5BD BD TD= −

TD BD BD= − 1

5


TD BD= 4

5� P

F

tr=

+1

F

trtr Ftr

1

4

5+��

�

= ⋅

⇒1

1

4

5+=

tr

⇒1

1 0 2

4

5+=

.� �t

Cross multiplying 5 = 4 + 0.8t

5 4 0 8− = . t

⇒ t = =1

0 81 25

.. yrs.

t = 12 + 3 = 15 months.

5.5.5.5.5. The difference between the banker’s discount and true discount on a certain sum of money duein 4 months is Rs. 10. Find the amount of the bill if the rate of interest is 3% p.a.

Solution:Solution:Solution:Solution:Solution: Given BD − TD = Rs. 10

t = = =44

12

1

3 months yrs. yrs.

R r= ⇒ = =3%3

1000 03.

Now BD − TD = 10

FtrF

trtr−

+��

�

=1

10

Ftrtr

11

110−

+��

��=

F ⋅ × −+ ×

�

��

�

��=0 03

1

31

1

1 0 0313

10..

F 0 01 11

1 0110.

.� � −

�� =

F0 0001

1 0110

.

.��

��=


F = ×10 1 01

0 0001

.

.

F = 1 01 000, ,

∴ Amount of the bill = Rs. 1,01,000.

6.6.6.6.6. A bill for Rs. 2920 drawn at 6 months was discounted on 10.4.2000 for Rs. 2916. If the discountrate is 5% p.a. On what date was the bill drawn.

Solution:Solution:Solution:Solution:Solution: Given: F = Rs. 2920.

Discounted value Rs. 2916.

Rate = 5% ⇒ = =r5

1000 05.

Discounted values = F (1 − tr)

2916 2920 1 0 05= − t .� ��

2916

29201 0 05= − t .� �

t 0 05 12916

2920.� � = −

⇒ t = 0 027398. yrs.

⇒ t = ×0 027398 365. days

t = 10 days

∴ Number of days from legally due date to discounted date = 10 days.

Now

Bill was discounted on 10.4.2000

Legally due date = 10 10.4.2000 days+

= 20.4.2000

Bill period months= 6

Grace period days.= 3

∴

Bill drawing date .4.2000

3.6.0

Bill drawing date .10.1999

=−

=

20

17

� �

7.7.7.7.7. A bill was drawn on March 8th at 7 months period and was discounted on May 18th at 5%. If thebanker’s gain is Rs. 3 find the true discount, the banker’s discount, and the sum of the bill.

Solution:Solution:Solution:Solution:Solution: Date on which the bill was drawn = March 8th

i.e., 8 − 3

Period (7 months): 7

Grace period: 3 − 0


Legally due date: 11 − 10 i.e., Oct. 11th.

Date on which the bill was discounted: May 18th

Time for which the bill has yet to run:

May June July Aug Sep Oct13 30 31 31 30 11 days.

+ + + + += 146

∴ t = =146

365

2

5 yrs.

Now Banker’s gain = Simple interest on true discount i.e. Rs. 3 is the S.I. on T.D. for 2

5 at 5%

Then 3 =× ×P 5

25

100

⇒ P = × =100 3

2Rs. 150

Banker’s discount = T.D. + S.I. on TD

= +Rs SI on 150 for 2

5 yrs. at . 150 5%

150150

25

5

100+

× ×

= + =150 3 Rs. 153.

OR

B D TD B G. .= + . .

BD = + =150 3 Rs. 153.Now

Sum Rs. 7650.= ×−

= ×−

=BD TD

BD TD

153 150

153 150

8.8.8.8.8. A bill was drawn on the 10th July 1960 at 3 months after signed and was accepted on presentationon 1st Aug. 1960. It was discounted on 23rd Aug ’69 at 5% p.a. simple interest to realise Rs. 2475.Find the face values of thee bill and banker’s discount.


Drawing date : 10 - 7 -1960

Bill period : 0 - 3- 0

Grace period : 3 - 0 - 0

Legally due date : 13-10 -1960


Given: Bill was present on 1st Aug. 1969.

Number of days for which banker’s discount is paid =

30 30 13 73 days days.Aug Sept Oct

+ + =� � � � � �

t = =73

3650 2. yrs.

R r= ⇒ =5% 0 05.

Let the face value of bill be Rs. 100.

Then BD = Ftr

BD = 100 × 0.2 × 0.05

BD = 1.

For Rs. 100 face value bill, BD = Re 1

i.e., Rs. 99 is realised.

i.e., Rs 99 is realised when face value is Rs. 100

Rs. 2475 is realised when face value is x.

∴ x = ×2475 100

99

x = 2500.

∴ Face value of the bill = Rs. 2500.

Now, BD = Ftr

= 2500 × 0.2 × 0.05

BD = 25.

OR

For 100, BD = 1

For 2500, BD = × =2500 10

100Rs. 25.

9.9.9.9.9. The present worth of a bill due sometime hence is Rs. 1100 and the true discount on the bill isRs. 110. Find the banker’s discount and the extra gain the banker would make in the transaction.

Solution:Solution:Solution:Solution:Solution: Here time and rate of interest are not given.

Given: PW = Rs. 1100

TD = Rs. 110

We have TD PW BG= ×� � � �

Squaring TD PW BG� �2 = ×


⇒ BGTD

Pw= = =� � � �2 2110

1100Rs. 11.

Now BG BD TD= −

⇒ BD BG TD= +

BD = +Rs. 11 110

BD = Rs. 121.

10.10.10.10.10. The true discount on a bill of Rs. 1860 due after 8 months is Rs. 60. Find the rate, the banker’sdiscount and the banker’s gain.

Solution:Solution:Solution:Solution:Solution: Given: F = Rs. 1860

TD = Rs. 60

t = = =88

12

2

3 months yrs.

r = ?

BD BG= =? ? and

Present worth, P = 1860 − 60 = Rs. 1800.

TD = Ptr

60 18002

3= × × r

60 3

1800 2

××

= r

⇒ r = 0.05

Hence R = 5%.

BD = Ftr

BD = × ×18602

30 05.

BD = Rs. 62.

BG = BD − TD

= 62 − 60 = Rs. 2.

OR

BGTD

Pw= = =� � � �2 260

1800Rs. 2.

BD = TD + BG = 60 + 2 = Rs. 62.


��

• Legally due date = Bill Drawing date + Bil period + 3 days (Grace period).

• BD = Ftr [Simple interest on face value of the bill]

• TD = Ptr [Simple interest on present worth of the bill]

• BG = BD − TD

• Present worth = F

tr1+• BG = TD × tr

• BGTD

Pwr t= � �2

If and are not given

• TD B G Pw= ⋅ ⋅

�)��+�

1. A bill for Rs. 3500 due for 3 months was drawn on 27th March 2000 and was discounted at therate of 7% on 18th April 2000. Find the banker’s discount and discounted value of the bill.

2. The banker’s discount and true discount on a sum of money due four months are respectivelyRs. 510 and Rs. 500. Find the seem and the rate of interest.

3. The difference between BD and TD on a bill due after 6 months at 4% interest per annum isRs. 20. Find the true discount bill discount and face value of the bill.

4. The banker’s gain on a certain bill due 6 months hence is Rs. 10, the rate of interest being 10%p.a. Find the face value of the bill.

5. A banker pays Rs. 2340 on a bill of Rs. 2500, 146 days before the legally due date. What is therate of discount charged by banker?

6. A bill for Rs. 1460 drawn at 3 months was discounted at 4% p.a. on 9th November for 1454.40.On what date the bill was drawn?

7. A bill was drawn on April 14th at 8 months after date and was discounted on July 24th at 5% p.a.If the banker’s gain is Rs. 2, what is the face value of the bill.

8. Find the banker’s discount and cash value of a bill for Rs. 3400/- drawn on April 25th 1996 at 7months and discounted on September 16th, 1996 at 5%.

9. The banker’s gain of a certain sum due 2 years hence at 5% per annum is Rs. 8. Find the presentworth.

10. The present worth of a sum due sometimes hence is Rs. 576 and banker’s gain is Re. 1. Find thetrue discount.

11. The banker’s gain on a sum due 3 years hence at 5% is Rs. 90. Find the banker’s discount.

12. The banker’s discount on a bill due 1 year 8 months hence is Rs. 50 and true discount on the samesum at the same rate percent is Rs. 45. Find the rate of interest.


13. A bill for Rs. 3500 due for 3 months was drawn on 27th March 2000 and was discounted on 18thApril 2000 at 7% rate of interest. Find the banker’s discount and discounted value of the bill.

14. A bill for Rs. 2920 drawn at 6 months was discounted on 10.4.97 for Rs. 2916. If the discount rateis 5% per annum, on what date was the bill drawn?

15. If the difference of simple interest and true discount of a sum due one year 6 months, hence at 8%p.a. is Rs. 81.45. Find the sum.

&�+'��+

1. Rs. 3451, Rs. 49 2. 6%, Rs. 255

3. T.D. = Rs. 1000, B.D. Rs. 1,020 F = 51,000 4. Rs. 4200

5. 16% 6. September 11th

7. Rs. 5,100/- 8. Rs.34, Rs. 3,366

9. Rs. 800 10. Rs, 24

11. Rs. 690 12. 6 and 2/3%

13. Rs. 49, Rs. 3451 14. 17.10.96

15. Rs. 6335.


��

��

��

In order to meet the expenses of a certain plan or a big project. Loan is raised from the public at acertain fixed rate of interest. Bonds or promissory notes of a fixed value are issued for sale to the public.

If a man purchases a bond of Rs. 1000 at which 5% interest has been fixed. Then the holder of suchbond is said to have ‘a Rs. 1000 stock at 5%’. Here Rs. 1000 is called the face value of the stock.Usually a period is fixed for the repayment of the loan i.e., the stock matures at a fixed date only. Nowif the person holding a stock is in need of the money before the date of maturity of stock, he can sellthe bond to some other person, where by the claim of interest is transferred to that person.

��

To start a big concern or a business a large amount of money is needed. This is usually beyond thecapacity of one or two individuals. Therefore a group of individuals get together and form the company.The company issues a prospectus and invites the public to subscribe. The required capital is dividedinto equal small parts called shares, each of a particular fixed value. The person who possesses one ormore share is called a share holder. Sometimes the company asks its share holders to pay some amountimmediately and balance after some period. The total money raised immediately is called the paid upcapital.

��

The main distinction between stocks and shares is

(i) Shares can be issued directly, but stocks cannot be issued directly. Firstly amount is collected onshares from the public. Stock certificates are issued only after collecting full amount of shares.Stocks are never issued unless the shares are issued and subscribed in full by the public.

(ii) Shares need not be fully paid, but stocks must be fully paid. For example the value of a share isRs. 100. This is known as face value of the share. A share of Rs. 100 each may be called up atRs. 50 each by the company and the balance of Rs. 50 may be paid at later stage by the public asand when demanded by the company. Stock must be fully paid means face value of Rs. 100 oneach share must be paid in full by the public in order to get the stock certificate from the company.

Stocks and Shares 247

��! ��"��#�$%�

DeDeDeDeDebenturbenturbenturbenturbentures:es:es:es:es: Debentures are long term loans taken by the company from the public. Every person wholends such an amount is given a certificate of loan called debentures.

FFFFFace ace ace ace ace VVVVValue of Sharalue of Sharalue of Sharalue of Sharalue of Shares:es:es:es:es: It is the price at which shares are first issued by a company. It is the price printedon the share certificate.

MarMarMarMarMarkkkkket et et et et PPPPPrrrrrice:ice:ice:ice:ice: It is the price at which the share can be brought or sold on the stock/share market.

PPPPPar ar ar ar ar VVVVValue of Sharalue of Sharalue of Sharalue of Sharalue of Shares:es:es:es:es: When the shares are issued to the public at the face value, it is called par valueof share.

Example:Example:Example:Example:Example: When Rs. 10 share is issued at its face value, it is called par value.

AboAboAboAboAbovvvvve e e e e PPPPPar:ar:ar:ar:ar: If the market value of the share is more than the face value, it is said to be above par orat premium.

Example:Example:Example:Example:Example: When Rs. 10 shares are issued at Rs. 12 the shares are said to be issued at Rs. 2 premium.

BeloBeloBeloBeloBelow w w w w PPPPPar:ar:ar:ar:ar: If the market value of the share is less than the face value, it is said to be below par or itis said to be at discount.

Example:Example:Example:Example:Example: When Rs. 100 shares are issued at Rs. 90. Then the shares are said to be issued at 10%discount.

DiDiDiDiDividend:vidend:vidend:vidend:vidend: It is the portion of the profit of the company which is distributed to the share holders. Thedividend is always calculated on the face value of the share. Dividends may be cash dividends or sharedividends. Bonus shares are known as stock dividends.

Ex-DiEx-DiEx-DiEx-DiEx-Dividend and Cum-dividend and Cum-dividend and Cum-dividend and Cum-dividend and Cum-dividend vidend vidend vidend vidend PPPPPrrrrrices:ices:ices:ices:ices: Interest on bond is payable on pre-determined dates. If thebond is bought or sold on a date closer to the interest due date, the prices may be quoted Ex-interest orCum-interest. If the price is ex-interest the selling price of it is not inclusive of interest. If it is quotedcum interest, the buyers will receive the interest amount. In the case of shares, share dividend are paidinstead of interest. Cum dividend price quotations are usually higher than the Ex-dividend quotations.

YYYYYield:ield:ield:ield:ield: Actual dividend received by the actual amount invested in a stock or shares called yield.

i.e., YieldDividend

Amount invested=

i.e., = Nominal interest

Amount invested

BrBrBrBrBrokokokokokererererers and Brs and Brs and Brs and Brs and Brokokokokokerererereraaaaaggggge:e:e:e:e: Buying and selling of stocks or shares is done through the person called‘brokers’ at stock exchange. They charge certain amount called brokerage. Note that when stock/shareis purchased, brokerage is added to the cost price and when stock or share is sold brokerage is sub-tracted from the selling price.

Kinds of Kinds of Kinds of Kinds of Kinds of SSSSSharharharharhares:es:es:es:es: A company may issue two kinds of shares. They are

(i) Preference shares (ii) Equity shares

(i) PrPrPrPrPrefefefefeferererererence sharence sharence sharence sharence shares:es:es:es:es: A preference share holder enjoys a preferential claim with regard to thepayments of dividend and repayment of capital. The rate of dividend is fixed, but it is paid beforeprofit is distributed to other members.


(((((iiiiiiiiii))))) Equity sharEquity sharEquity sharEquity sharEquity shares:es:es:es:es: An equity share holder has no special rights. The rate of dividend is not fixed. Itvaries from year to year. An equity share holder is paid dividend only after the claims of prefer-ence share holders are satisfied.

&'��(��

Consider the statement “Government paper mills 11% shares at 110”. This is a quotation. This meansa share of the mill having face value Rs. 100 is available for sale at Rs. 110. This share fetches him adividend of Rs. 11 every year.

��)�"*#��

1.1.1.1.1. Find the cost of 80 shares at 5% if the market value of the share is 93 and its par-value is Rs. 100.If a person invests Rs. 37200 in such shares then find his annual income.

Solution.Solution.Solution.Solution.Solution. Cost of 1 share = Rs. 93

Cost of 80 such shares = 80 × 93 = Rs. 7440.

Now, the person has invested Rs. 37,200.

By investing Rs. 93, the person gets 1 share.

∴ By investing Rs. 37200 the person gets

= × =37200 1

93400.

∴ The person possesses 400 shares.

Face value of 1 share = Rs. 100

Face value of 400 shares = 400 × Rs. 100 = Rs. 40000

Annual income = 5% of face value of shares

= ×5

10040 000,

= Rs. 2,000.

2.2.2.2.2. Find the yield by investing Rs. 1140 on 15% stock quoted at Rs. 95.

Solution.Solution.Solution.Solution.Solution. YieldNominal interest

Amount invested=

For Rs. 100 stock, Rs. 95 is the amount invested and Rs. 15 is the nominal interest.

∴ Yield = = ≈15

9501578 016. . .

OR

By investing Rs. 95, 1 stock is obtained

By investing Rs. stocks are obtained.11401140 1

9512,

× =


1 stock has face value 100/-

12 stock has face value 1200/-

Nominal interest = 15% of Face value of stock

= 15% of 1200

= × =15

1001200 180.

YieldNominal interest

Amount invested= = 180

1140

= 0 1578.

≈ 0 16. .

3.3.3.3.3. Which of the following is better investment? 5 ½% stock at 102 or 4 and 3/4% stock at 106.

Solution.Solution.Solution.Solution.Solution. Yield from 5 and ½% stock at 102 = 5

12

102

=×

= =11

2 102

11

2040 05392. .

Yield from 4 and 3/4% stock at 106 = 4

34

106

=×

=19

4 106

19

424

= 0.044811.

So yield from 5 and ½% stock at 102 is greater. Hence it is a better investment.

4.4.4.4.4. Find the cash required to purchase Rs. 20,000 stock at 105 (brokerage ½%). Also find the annualdividend received if the company declares dividend of 8 and ½%.

Solution.Solution.Solution.Solution.Solution. Cash required to purchase Rs. 100 stock.

= +��

��=105

1

2

211

2

∴ Cash required to purchase Rs. 20,000 stock =×20 000

2112

100

,

= ××

=20 000 211

2 100

,Rs. 21,100.

Annual dividend for Rs. 100 stock = 8 and ½ Rs.


Annual dividend for Rs. 20,000 stock =×20 000 8

12

100

,

= × =20017

2Rs. 1,700.

5.5.5.5.5. A person has invested a certain sum of money in 13% stock at 96. He sold the investment whenthe market value went up to 101.5. He gained Rs. 1470 in this process. If he has paid thebrokerage at 2% for all the transaction, what was the amount of cash investment and what was thestock value of the investment in the first instance.

Solution.Solution.Solution.Solution.Solution. Let the amount invested = Rs. x

In the first instance,

Cost of 1 share = 96 + 2 (Brokerage)

i.e., Cost of 1 share = Rs. 98.

Now

For Rs. 100 share — Rs. 98 is the amount received.

For Rs. x share — 98

100

x is the amount received.

When the market value went upto Rs. 101.5.

Cost of 1 share = 101.5 − 2 (Brokerage)

= 99.5.

Amount received from Rs. 100 − 99.5

Amount received for Rs. x — 99 5

100

. x

The person gained Rs. 1470 in this process. (Given)

∴ − + =98

100

99 5

1001470

x x.

15

1001470

. x =

x = ×470 100

1 5.

x = 98,000

∴ Amount invested by a person = Rs. 98,000.

In the first instance,

For Rs. 98 he gets Rs. 100 worth stock


For Rs. 98000 he gets 98000 100

98

×

Rs. 1,00,000.

∴ Stock value of the investment in 1st instance = Rs. 1,00,000 and cash investment = Rs. 98,000.

6.6.6.6.6. Vivek has Rs. 16,500 stock in 3%. He sells it out at 1011

8 and invests the proceeds in 4% railway

debentures at 1317

8. Find the change in his income a brokerage of 1

8% being charged on each

transaction.

Vivek has Rs. 16,500 stock in 3%.

Income from Rs. 100 stock = Rs. 3

∴ Income from Rs. 16,500 stock = 16500 3

100

×

= Rs. 495.

He sells it out at 101 and 1/8 (brokerage 1/8%)

∴ S.P. of Rs. 100 stock = 1011

8

1

8−

S.P. of Rs. 100 stock = Rs. 101

∴ Selling price of Rs. 16500 stock = 16500 101

100

×

= 16,665.

Now he invests this Rs. 16665 in 4% railway debentures at 1317

8. (Brokerage = 1

8% )

i.e,

By investing Rs. 1317

8

1

8+�

��

, income derived = Rs. 4

∴ By investing Rs. 16665, income derived =×16665 4

132

= Rs 505.

∴ Change in income = Rs. 505 − Rs. 495 = Rs. 10.

∴ Income is increased by Rs. 10.

7.7.7.7.7. Tulasi has invested Rs. 1,00,000 partly in 12% stock at 120 and partly in 15% stock at 75. If thetotal income from both is Rs. 15,000. Find the amount invested in 2 types of stocks.

Solution.Solution.Solution.Solution.Solution. Let the amount invested in 12% stock at 120 be x.

Then Amount invested in 15% stock at 75

= 1,00,000 − x


Income from 12% stock at 120 = 12

120 10

x x=

Income from 15% stock at 75 = 15

751 00 000, , − x� �

= −1 00 000

5

, , x

Given total income = 15,000

∴ x x

10

1 00 000

515 000+ − =, ,

,

1 2 00 000 2

1015 000

x x− − =, ,,

− x + 2,00,000 = 1,50,000

⇒ x = 2,00,000 − 1,50,000

x = 50,000

∴ Amount invested in 12% stock at 120

= Rs. 50,000 and that invested in 15% stock at 75 = Rs. 1,00,000 − 50,000

= Rs. 50,000.

8.8.8.8.8. Mr. Gauriprasad sold Rs. 2250 stock at 75 and bought stock at 88.50 with proceeds. How muchstock worth does he buy if the brokerage is 2% for selling and 1.5% for buying.


Cost of 1 stock = Rs. 75 − 2 (Brokerage)

Cost of 1 stock = Rs. 73

∴ Amount received by selling stock worth Rs. 100 is Rs. 73.

∴ Amount received by selling stock worth Rs. 2250 = Rs. 2250 73×

100

= Rs. 1642.50.

Now Brokerage for buying = 1.5%

∴ Cost of 1 stock = 88.50 + 1.50 = 90

By investing Rs. 90, stock worth Rs. 100 can be purchased.

∴ By investing Rs. 1642.50, Stock worth − ×1642 50 100

90

. can be purchased.

= Rs. 1825.

∴ Mr. Gauriprasad bought stock worth Rs. 1825.

9.9.9.9.9. Pusphak buys Rs. 2,000 shares paying 9% dividend. If he wants to have an interest of 12% on hismoney, then find the market value of each share.


Solution.Solution.Solution.Solution.Solution. Dividend on Rs. 2000 share in 9% of 2000.

= ×9

1002000 = 180.

He wants to have 12% on his money.

Rs. 12 is an income on Rs. 100 share.

Rs. 180 is income on 180 100

121500

× = .

∴ Market value of each share must be Rs. 1500 in order to have interest of 12% on his money.

10.10.10.10.10. A man invests some money partly in 3% stock at 96 and partly in 4% stock at 120. What is theratio of money he must invest in order to get equal dividends from both.

Solution.Solution.Solution.Solution.Solution. For an income of Re 1 in 3% stock at 96, investment = =Rs. Rs.32.96

3

For an income of Re 1 in 4% stock at 120 investment = Rs.120

4

= Rs. 30.

∴ Ratio of investments = 32:30

= 16:15.

��"�"��

• Yield =Nominal interest

Amount invested

• When stock is purchased, brokerage is added to cost price.

• When stock is sold, brokerage is subtracted from selling price.

• Interest or dividend is paid on the face value of the stock or share not the market value.

• 4 and ½% stock at 96 means a stock whose face value is Rs. 100 is available at Rs. 96. Interestearned in 4 and ½.

• Shares need not be fully paid but stock must be fully paid.

�)��

1. Find the cost of

(a) Rs. 8750 8 and 3/4% stock at 92.

(b) Rs. 8500, 9 and ½% stock at 6 premium.

(c) Rs. 7200, 10% stock at 7 discount.

(d) Rs. 6400, 8% stock at par (brokerage 1/8%).

2. Find the cash required to purchase Rs. 1600, 8 and ½% stock at 105 (brokerage ½%).


3. Find the cash realised by selling Rs. 2400, 5 and ½% stock at 5 premium, brokerage being 1/4%.

4. Which of the following is better investment? 6% at 94 or 8% at 110.

5. Ramu possesses 150 shares of Rs. 25 each, the dividend declared by the company is 12%. Whatis the dividend earned by him. If he sells the shares at Rs. 40 and reinvest the proceeds in 7%shares of par value Rs. 100 at Rs. 80, Find the change in his dividend income.

6. A man invested Rs. 6750 partly in 6% stock at 140 and partly in 5% stock at 125. Find hisinvestment in cash if the income derived from both the investments is Rs. 280.

7. Mr. Vivek invested Rs. 2200 partly in 10% stock at 120 and partly in 12% stock at 96. Find hisinvestment in each if the income derived from both the investments is Rs. 200.

8. A man invests some money partly in 6% stock at 96 and partly in 5% stock at 120. In what ratio,he must invest the money so as to get equal dividends from both.

9. Rs. 2780 is invested partly in 4% stock at 75 and 5% stock at 80. If the income from bothinvestments are equal, find the investment in 5% stock.

10. Mr. Harish has invested a certain amount of money in 13% stock at 101. He sold it when marketvalue went down to 96.5. He lost Rs. 3564 on this process. If he has paid the brokerage at1 and ½% for all transactions, what was the amount of cash investment? What was the stock valueof the investment in first instance.

��

1. (a) Rs. 8050 (b) Rs. 9010 (c) Rs. 6696 (d) Rs. 6408

2. Rs. 1688

3. Rs. 2514

4. 8% at 110

5. Rs. 75

6. Rs. 3500 and Rs. 3250

7. Rs. 1800 and Rs. 400

8. 2 : 3

9. Rs. 1280

10. Rs. 73,062 and Rs. 72338.60.

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When an individual performs the same task repeatedly the second and subsequent time will provideexperience to the individual and there by there will be an increase in the degree of efficiency inperformance. This is due to the learning process.

The learning effect improves productivity of individuals especially that of workers engaged in fac-tories and industries.

��

It is the curvilinear relationship between the decrease in average labour hours per unit with increase inthe cumulative output. The assumption of learning curve theory is that every time total output of aproduct doubles, the cumulative average time taken to produce one unit decreases by a constant per-centage. For example: 80% learning effect means that when the cumulative output is doubled, thecumulative average hours per unit will be 80% of the previous level.

��

It is the ratio of average labour cost of first 2N units and average labour cost of first N units. It is alsoknown as experience ratio or improvement ratio or efficiency ratio.

∴ Experience ratioAverage labour cost of first 2 N units

Average labour cost of N units=

��

We know 80% learning effect means that when the cumulative output is doubled the cumulative aver-age hours per unit will be 80% of the previous level. Let us consider the effect of 80% learning in theform of a table taking cumulative time per unit as 100.


Units Total Cumulative Total number Average hour perProduced Output time percent of hours additional unit

1 1 unit 100 100 100

1 2 units 80% of 100 = 80 80 × 2 = 160160 100

160

−=

2 4 units 80% of 80 = 64 64 × 4 =256256 160

248

−=

4 8 units 80% of 64 = 51.2 51.2 × 8 = 409.6409 6 256

438

..4

−=

8 16 units 80% of 51.2 = 40.96 40.96 × 16 = 655.36655 36 409 6

830 72

. ..

−=

16 32 units 80% of 40.96 = 32.76 32.76 × 32 = 1048.571048 57 655 36

1624 57

. ..

−=

32 64 80% of ... ... ...

� � � � �

Taking the total output on x-axis and cumulative average time per unit on y-axis, we get the learningcurve as shown in the figure.

x

y

1 2 4 8 16 32

100 60 48 38 4 30 72 24 57. . .

100

80

60

40

20

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Scale:x - axis : 1 Unit = 1 cmy - axis : 20 units = 1 cm

y

x

Fig. 11.1

The curve clearly indicates that there will be fast learning effect in the initial stages and aftersometime there will be a steady state phase in which there is not any significant learning effect.

Learning Curve 257

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If b = logarithmic of learning ratio to base 2, y = cumulative average time per unit, x = cumulative totalnumber of units produced a = time for first unit, then the learning curve equation is given by

y = axb.

By considering log on both sides we get

log logy axb= � �

log log logy a xb= +

log log log .y a b x= +

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1.1.1.1.1. Find the index of learning for 60% learning effect.

Solution:Solution:Solution:Solution:Solution: Index of learning b = log260%

b =log

log10

10

60%

2

b = =log .

log

.

.10

10

0 6

2

1 7782

0 3010

=− + = −1 0 7782

0 3010

0 2218

0 3010

.

.

.

.

b = −0.7368.

2.2.2.2.2. A worker takes 10 hrs. to produce the first unit of product. What is the cumulative average timeper unit taken by him for the production of first two units? (Assuming the learning effect to be80%). Also find the total time for producing the first two units.

Solution:Solution:Solution:Solution:Solution: Given: a = 10 hrs.

x = 2

80% learning effect.

We have learning curve equation

y = axb

where b = =loglog .

log2 80%0 8

2

b = = − +1 9031

0 3010

1 0 9031

0 3010

.

.

.

.

b = − = −0 0969

0 30100 3219

.

..


Now y = × −10 2 0 3219.

y = 10

20 3219.

y = 10

1249.

y = ≈8 0064 8.

∴ Cumulative average time = 8 hrs.

Let x = 20.3219

Consider log on b.s.

log log .x = 20 3219

log . logx = 0 3219 2

log . .x = ×0 3219 0 3010

log .x = 0 09689

x = Antilog0.09689

= 1.249.

∴ Time taken for producing the first 2 units = 8 × 2 = 16 hrs.

3.3.3.3.3. The time required to produce the first unit of a product is 1000 hrs. If the manufacturers experi-ences 80% learning effect, calculate the average time per unit and the time taken to producealtogether 8 units. Also find the total labour charges for the production of 8 units at the rate ofRs. 12.50 per hour.

Solution:Solution:Solution:Solution:Solution: Given: a = 1000

x = 8

b = log280% = = −log .

log.

0 8

20 3219

Formula: y = axb

= × −1000 8 0 3219.

=1000

80 3219.

y = =1000

1 953512

..

∴ Average time/unit = 512 hrs.

To find x = 80.3219

log x = log80.3219

Learning Curve 259

0 3219 8. log×

0 3219 0 9031. .×

log .x = 0 2907

x = antilog 0 2907.

= 1 953. .

∴ Average time/unit = 512 hrs.

Total time to produce

8 units = 512 × 8 = 4096 hrs.

∴ Labour charges for 4096 hrs.

= 4096 × 12.50 [� Labour charges per hour = Rs. 12.50 (given)]

= Rs. 51,200.

OR

Units Total Cumulative Total number Average hour perProduced Output time percent of hours additional unit

1 1 1000 1000 1000

1 2 80% of 1000 = 800 800 × 2 = 16001600 1000

2300

−=

2 4 80% of 800 = 640 640 × 4 = 25602560 1600

4240

−=

4 8 80% of 640 = 512 512 × 8 = 40964096 2560

8192

−=

Total time to produce 8 units = 4096 hrs.

Cumulative average time per unit = 512

Total time to produce 8 units = 512 × 8 = 4096.

Given Labour charges for 1 hr. = Rs. 12.50

∴ Labour charges for 4096 hrs.

= 4096 × 12.50

= Rs. 51,200.

4.4.4.4.4. An engineering company has won the contract for supplying aircraft engines of a new type. Theprototype constructed to win the contract took 500 hrs. It is expected that there will be 90%learning effect. Estimate the labour cost of producing 8 engines of new order, if the labour costis Rs. 40 per hour.



Units Total Cumulative Total hours Average hour perOutput average time additional unit

1 1 500 500 500

1 2 90% of 500 = 450 450 × 2 = 900900 500

2200

−=

2 4 90% of 450 = 405 405 × 4 = 16201620 900

4180

−=

4 8 90% of 405 = 364.5 364.5 × 8 = 2916 2916 − 1620/8 = 162

Total time to produce 8 units = 2916

Given Labour charges per hour = Rs. 40

Labour charge for 2916 hrs.

= 2916 × 40 = Rs. 1,16,640.

OR

Given: a = 500

x = 8

b = log290% = log .

log

0 9

2

b = = − +1 9542

0 3010

1 0 9542

0 3010

.

.

.

.

b = − = −0 0458

0 30100 1521

.

.. .

Formula: y = axb

= × −500 8 0 1521.

= =500

8

500

13720 1521. .

y = 364.43

To find 80.1521

x = 80.1521

logx = 0.1521 × log8

= 0.152 × 0.9031.

Average time/unit = 364.43.

Total time to produce 8 units.

Learning Curve 261

= 364.43 × 8 = 2915.4518

≈ 2916.

Given labour charges/hr = Rs. 40

∴ Labour charges/2916 hrs.

= 2916 × 40

= 1,16,640.

��&�&'��

• The learning curve ratio = Average labour cost of first 2N units

Average labour cost of first N units

� �

� �

• The learning curve equation is

y = axb

where y = cumulative average time/unit.

a = Time for first unit.

b = log of learning ratio to base 2.

x = Cumulative total number of units produced.

• Index of learning, b = log of learning ratio to base 2.

�%��

1. Define the term learning curve ratio.

2. What is learning curve?

3. What do you mean by 80% learning effect.

4. Find the index of learning for 80% learning effect.

5. A worker requires 40 hrs. to produce first unit of a product. How much time is required for himto produce a total of 4 units.

6. A worker requires 20 hrs. to produce the first unit of a product. If the learning effect is 80%calculate how much time is required to produce 4 units. Also find the labour charges for theproduction of 4 units at the rate of Rs. 20 per hr.

7. A company requires 2134.2 hrs. to produce the first 40 machines. If the learning effect is 80%,find the number of hrs. required to produce next 120 machines.

8. The first unit of a product took 80 hrs. to manufacture. If the workers show 80% learning effect,find the total time taken to produce 8 units.

��#��

1.Average labour cost of first 2N units

Average labour cost of first N units.


2. Curvilinear relationship between the decrease in average labour hrs. per unit with increase incurve ratio.

3. When the cumulative output is doubled, cumulative average labour hrs./unit will be 80% of theprevious level.

4. −0.3219

5. 108.4 hrs.

6. 48.8 hrs.

7. 3329.35 hrs.

8. 32.768 hrs.

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Linear programming is a powerful technique which can indicate a definite conclusion as to the bestutilisation of available resources under given circumstances. Any industrial process may consists of anumber of activities relating to capital to be employed, products to be made and sold, materials to beused: machines to be run; inventories to the stored and consumed or a combination of the above. Sinceutilisation of one affects the utilisation of another and due to the limitation of the total availableresources, these activities are interdependent or interlocking. In such a situation, a large number ofways exists in which the available resources can be allocated to the competing demands. Linear pro-gramming enables us to arrange for that combination of resources which optimise the cost, production,profit etc.

This technique was evolved by George B. Dantzig as a tool for planning the diversification activitiesof U.S. Airforce in 1947.

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DefDefDefDefDefinition:inition:inition:inition:inition: Linear programming is a mathematical technique for determining the optimal allocation ofresources and obtaining a particular objective when there are alternative uses of the resources: money,manpower, material, machine and other facilities. The objective in resource allocation may be costminimisation or profit maximisation. The technique of linear programming is applicable to problems inwhich the total effectiveness can be expressed as a linear function of individual allocations and thelimitations on resources give rise to linear equalities or inequalities of the individual allocations.

ObjectiObjectiObjectiObjectiObjectivvvvve function:e function:e function:e function:e function: The objective function is a quantified statement in linear programming model ofwhat the best results or the best advantage which is aimed for as the objective of the resource allocationdecision. Objective function will either be to maximise a value or to minimise a value.

The objective function is expressed as a linear function of decision variables. It is usually stated asmaximise or minimise a1x1 + a2x2 + ... +anxn where x1 x2 ... xn are the decision variables in the problemand a1, a2, ..., an are constant values for each variable.

Decision vDecision vDecision vDecision vDecision vararararariaiaiaiaiabbbbbles:les:les:les:les: The decision variables in any activity is a variable which is competing with otheractivities for limited resources. These variables are inter-related linearly in terms of utilisation ofresources.


ConstrConstrConstrConstrConstraints:aints:aints:aints:aints: The resources like production capacity, manpower, time, space, technology, etc. arescarce and there are limitations on what can be achieved. These restrictions are a set of conditionswhich an optimal solution must satisfy. They are known as constraints. These are expressed as linearinequalities or equalities in terms of decision variables.

Non-neNon-neNon-neNon-neNon-negggggaaaaatititititivity conditions:vity conditions:vity conditions:vity conditions:vity conditions: All decision variables must assume non-negative values. If any of thevariable is unrestricted in sign, a trick can be employed which will enforce the non-negativity withoutchanging the original information of the problem.

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A set of values of decision variables x1 x2 ... xn which satisfies the constraints of linear programmingproblem is called solution.

FFFFFeasibeasibeasibeasibeasible solution:le solution:le solution:le solution:le solution: Any solution to a linear programming problem which satisfies the non-negativityrestriction of the problem is called a feasible solution to the Linear programming problem.

Optimal solution:Optimal solution:Optimal solution:Optimal solution:Optimal solution: Any feasible solution which optimises (maximises or minimises) the objectivefunction of a linear programming problem is called an optimal solution.

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If the objective function is a function of 2 variables only then the LPP can be solved by graphicalmethod.

One variable is taken along x-axis and another along y-axis. Since negative values are not allowed,the graph contains only first quadrant. That is, 0 and positive values of x and y are considered.

1

1

2

2

3

3

I quadrant

Fig. 12.1

�� )!��!��!�"��("��)�$��

Consider the graph of y = 2.

1

1

2

2

3

3

0

Fig. 12.2

Linear Programming 265

This is a straight line parallel to x-axis. All points below this line are represented by the inequalityy < 2 and all points above this line are represented by y > 2 and the corresponding graphs are

1

1

2

2

0

y > 2

1

1

2

2y = 2

0x x

y < 2

Fig. 12.3

Now consider the linear equation 3x + 2y = 6

To plot the graph, put x = 0

3 0 2 6� � + =y

2 6y =

y = 3

and put y x= + =0 3 2 0 6, � �

3 6x =

x = 2

Hence consider the points (0, 3) and (2, 0) and join them to get the graph of linear equation3x + 2y = 6.

1

1

2

2

3

3

0

(0, 3)

(2, 0)

3x + 2y = 6

Fig. 12.4

Any value (x, y) which will fall in the shaded area in figure (a) is represented by the inequality3x + 2y ≤ 6 and any value which will fall in the shaded area in figure (b) is represented by the inequality3x + 2y ≥ 6.


1

1

2

2

3

3

0 1

1

2

2

3

3

0

(3, 0)

(2, 0)

(a)

(0, 3)

(2, 0)(b)

3x + 2y 6≤ 3x + 2y 6≥

Fig. 12.5

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(1) When there are several inequalities, which are true at the same time, the feasible region ofcombinations of values x and y must be a region where all the inequalities are satisfied.

(2) Optimum solution to linear programming problem lies at one of the vertices only. Hence find theco-ordinates of each vertex and substitute in the objective function. The value for which theobjective function is highest/least is the optimum solution for maximisation/minimisation.

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1.1.1.1.1. Write the following data in the form of LPP:

Machine Chair Table Time

M1 5 2 50 min

M2 2 7 130 min

The price of each chair is Rs. 120 and each table is Rs. 130.

Solution:Solution:Solution:Solution:Solution: Let the number of chairs be x and number of tables be y.

Time availability for M1 ≤ 50 and

Time availability for M2 ≤ 130.

∴ 5 2 50x y+ ≤ and

2 7 130x y+ ≤

Objective function is Maximise Z = 120x + 130y

x ≥ 0 and y ≥ 0.

Hence LPP is given by

Maximise Z = 120x + 130y. Subject to constraints

5 2 50 2 7 130 0 0x y x y x y+ ≤ + ≤ ≥ ≥; , . and

2.2.2.2.2. A company manufacturers 2 types of bulbs A and B by using 2 machines M1 and M2. One bulbof type A require 2 hrs. at machine M1 and 1 hr. at machine M2 and one bulb of type B requires


one hr. at M1 and 2 hrs. at M2. The profit from each bulb of type A is Rs. 2 and that of type B isRs. 3. The number of hrs. available per week on machines M1 and M2 are 20 hrs. and 30 hrs.respectively. Formulate the above problem as LPP. The aim of the company is to maximise theprofit.

Solution:Solution:Solution:Solution:Solution: Let the number of bulbs of type A = x and that of type B = y

At machine M1, one bulb of type A requires 2 hrs. and one bulb of type B requires 1 hr. Also numberof hrs. available per week on machine M1 is 20.

∴ 2 1 20x y+ ≤

Similarly at machine M2, one bulb of type A requires 1 hr., B requires 2 hrs. Also number of hoursavailable per week = 30 hrs.

∴ 1 2 30x y+ ≤

The profit from each bulb of type A is Rs. 2 and that from type B = Rs. 3

∴ Profit function = 2x + 3y

∴ Mathematical model of LPP is

Maximise, Objective function, Z = 2x + 3y

Subject to constraints

2 1 20x y+ ≤

1 2 30x y+ ≤

x y≥ ≥0 0, .

3.3.3.3.3. Maximise Z = 20x + 30y subject to x + 2y ≤ 20, x + 5y ≤ 35 and x ≥ 0 and y ≥ 0.

Indicate the feasible region on the graph:

Solution:Solution:Solution:Solution:Solution: To draw x + 2y = 20

Put x = 0; 2y = 20 ⇒ y = 10

Put y = 0; x + 2 (0) = 20 ⇒ x = 20.

∴ (0, 10) and (20, 0) are points on straight line x + 2y = 20.

To draw x + 5y = 35

Put x = 0, 5y = 35

y = 7

Put y = 0, x = 35.

∴ (0, 7) and (35, 0) are points on the straight line x + 5y = 35.

The co-ordinates of the vertices are obtained by solving

x yx y

y

+ =+ =

− − −− = −

2 205 35

3 15� � � � � �

⇒ y = 5


Substituting y x y= + =5 2 20 in

x + =2 5 20� �

x + =10 20

x = 10 .

∴ (x, y) =(10, 5)

10 15 20 25 30 35 40

5

10

15

20

25

30

A

B

C

D

P

5

Scale:x-axis : 1 cm = 5 unitsy-axis : 1 cm = 5 units

0

Fig. 12.6

The shaded area OCPB in the figure is the feasible region.

The co-ordinates of vertices: O (0, 0), B ≡ (0, 7), P ≡ (10, 5) and C ≡ (20, 0).

Objective function: Z = 20x + 30y

for (0, 0) Z = 0.

for (7, 0) Z = 20 (0) + 30 (7) = 210

for (10, 5) = Z = 20 (10) + 30 (5) = 200 + 150 = 350

for (20, 0) = Z =20 (20) + 30 (0) = 400

∴ Z is maximum when x = 20 and y = 0.

i.e., Optimum solution is Z = 400 when x = 20 and y = 0.

4.4.4.4.4. Solve the following LPP graphically:

Maximise: Z = 5x + 6y Subject to x + y ≤ 300 and x + y ≤ 500 x ≥ 0 and y ≥ 0.

Solution:Solution:Solution:Solution:Solution: To plot the graph of x + y = 300

Put x = 0 ⇒ y = 300

Put y = 0 ⇒ x = 300

∴ (0, 300) and (300, 0) are points on the straight line x + y = 300

To plot graph of x + y = 500

Put x = 0 ⇒ y = 500 and

y = 0 ⇒ x = 500


∴ (0, 500) and (500, 0) are points on the straight line x + y = 500.

100

100

200

200

300

300

400

400

500

500

600

600

0A C

B

D

Fig. 12.7

The shaded region OAB in the figure is the feasible region.

To find the optimum solution:

Z = 5x + 6y

At (0, 0) Z = 0.

At 0 300 5 0 6 300 1800, ,� � � � � �Z = + =

At 300 0 5 300 6 0 1500, ,� � � � � �Z = + =

∴ Zmax = 1800 at x = 0 and y = 300.

5.5.5.5.5. A radio factory produces 2 different types of transistor radios (a) cheaper ordinary model and (b)an expensive special model. For greater efficiency the assembly and finishing operation andperformed in 2 workshop the ordinary model requires 3 hrs. of workshop A and 4 hrs. of work-shop B. While special model requires 6 hrs. of workshop A and 4 hrs. of workshop B. Due tolimited resources and skilled labour only 180 hrs. of workshop A and 200 hrs. in workshop B. Thefactory makes a profit of Rs. 300 on each ordinary model and Rs. 400 on each special model.Formulate the above as a LPP and solve by graphical method, assuming the company expectsmaximum profit.

Solution:Solution:Solution:Solution:Solution: Let the number of ordinary models = x and the number of special model = y.

Ordinary Special Total/hrs.

Workshop A: 3 6 180

Workshop B: 4 4 200

Objective function maximise Z = 300x + 400y

Subject to constraints,


3 6 180x y+ ≤

4 4 200 0 0x y x y+ ≤ ≥ ≥, and

To plot 3x + 6y = 180

Put x y y= = ⇒ =0 6 180 30,

Put y x x= = ⇒ =0 3 180 60,

∴ (0, 30) and (60, 0) are points on the straight line 3x + 6y = 180

To plot 4x + 4y = 200

⇒ x + y = 50

Put x = 0, y = 50

y = 0, x = 50

∴ (0, 50) and (50, 0) are points on the straight line 4x + 4y = 200

To get point of intersection

Solving 3x + 6y = 180 and 4x + 4y = 200

i.e.,

x yx y

y

+ =+ =

− − −=

2 6050

10� � � � � �

Substituting y = 10 in x + y = 50 we get x = 40.

∴ Point of intersection = (40, 10).

10

10

20

20

30

30

40

40

50

50

60 70A

B

C

D

P

0

Scale : 1 cm = 10 units for both x-axis and y-axis

Fig. 12.8

The shaded area OABC represent the feasible region.


At 0 (0, 0), Z = 0

At A (50, 0), Z = 300 (50) + 400 (0) = 15000


At B (40, 10), Z = 300 (40) + 400 (10) = 16000

At C (0, 30), Z = 300 (0) + 400 (30) = 12000

Hence optimum solution, i.e.,

Zmax = 16,000 and at x = 40 and y = 10.

i.e., Max profit is obtained if number of ordinary models = 40 and special models = 10.

6.6.6.6.6. Minimise Z = 3x + 2y subject to x y x y x y+ ≥ + ≥ ≥7 2 10 0, , by graphical method.

Solution:Solution:Solution:Solution:Solution: To plot x + y = 7

Put x = 0, y = 7

y = 0, x = 7

∴ (0, 7) and (7, 0) are points in x + y = 7

Now to plot 2x + y = 10

Put x = 0, y = 10

Put y = 0, 2x = 10 ⇒ x = 5

∴ (0, 10) and (5, 0) are points on 2x + y = 10

2

2

4

4

6

6

8

8

10

10

12

12

14

14

160 A

C

B

Scale :1 cm = 2 units

Fig. 12.9

To get point of intersection solving

x yx y

x

+ =+ =

− − −− = −

710

3� � � � � �

⇒ x = 3

Substituting x = 3 in x + y = 7 ⇒ y = 7 − 3 = 4


∴ (3, 4) is the point of intersection.

The shaded region in the figure is feasible region.

To minimise Z = 3x + 2y

At A (7, 0) : Z = 3 (7) + 2 (0) = 21

At B (3, 4) : Z = 3 (3) + 2 (4) = 9 + 8 = 17

At C (0, 10) : Z = 3 (0) + 2 (10) = 0 + 20 = 20

∴ Minimum value of Z = Zmin = 17

It is attained when x = 3 and y = 4

7.7.7.7.7. Food X contains 20 units of Vitamin A and 40 units of vitamin B. Food Y contains 30 units of eachvitamin A and vitamin B. The daily minimum human requirement of vitamin A and vitamin B are900 and 1200 units respectively. How many grams of each type of food should be consumed soas to minimise the cost if food X costs 60 paise per gram and food Y costs 80 per paise per gram.


Food A B Cost

Requirement:

X

Y

20 40 60

30 30 80

900 1200


To minimise 60x + 80y. Subject to the conditions

20x + 30y ≥ 900 and 40x + 30y ≥ 1200

i.e., 2x + 3y ≥ 90 and 4x + 3y ≥ 120

x y, ≥ 0

To plot 20x + 30y = 900

i.e., 2x + 3y = 90

Put x y y= = ⇒ =0 3 90 30, .

Put y x x= = ⇒ =0 2 90 45, .

∴ (0, 30) and (45, 0) are points on the straight line 2x + 3y = 90.

To plot

40x + 30y = 1200

i.e., 4x + 3y = 120

Put x y y= = ⇒ =0 3 120 40, .

Put y x x= = ⇒ =0 4 120 30, .

∴ (0, 40) and (30, 0) are points on the straight line 4x + 3y = 120.

To find the point of intersection solving


2x + 3y = 90 and 4x + 3y = 120

2 3 904 3 120

2 30

x yx y

x

+ =+ =

− − −− = −

� � � � � �

x = 15

Putting x in 2x + 3y = 90

2 15 3 90� � + =y

30 3 90+ =y

3 60y =

y = 20

∴ Point of intersection = (x, y) = (15, 20)

A0

Scale :10 units = 1 cm

10

10

20

20

30

30

40

40

50 60

P

B

D

C

x

y

Fig. 12.10

The shaded region in the figure is the feasible region. To minimise Z = 60x + 80y

At A Z45 0 60 45 80 0 2700, :� � � � � �= + =

At P Z15 0 60 15 80 20 900 1600 2500, :� � � � � �= + = + =

At D Z0 40 60 0 80 40 0 3200 3200, :� � � � � �= + = + =

∴ Z is minimum when x = 15 and y = 20

∴ Zmin = 2500.

8.8.8.8.8. A certain company wishes to plan its advertising strategy to reach certain minimum percentageof high and low income groups. Two alternatives are considered namely cinema commercials andmagazines. Magazines advertising has an exposure for the high income group of 2% per page butonly 1% per page exposure for the low income group. Cinema commercials exposes 3% of thelow income group per show and only 1% of high income group per show. Magazines advertising


costs Rs. 1000 per page and cinema commercial Rs. 4000 per show. If the firm wants a minimalexposure of 50% of high income group and 30% of low income group, what strategy should it useto minimize the advertisement cost? Also find the minimum advertisement cost by graphicalmethod:


High income group Low income group Cost

Magazine 2% 1% 1000/page

Cinema 1% 3% 4000/show

Minimal exposure 50% 30%

It is required to minimise

Z = 1000x + 4000y

Subject to conditions 2x + y ≥ 50

1x + 3y ≥ 30

x ≥0 and y ≥ 0.

To plot 2x + y = 50

Put x = 0, y = 50

y = 0, 2x = 50 ⇒ x = 25

∴ (0, 50) and (25, 0) are points on straight line 2x + y = 50.

To plot 1x + 3y = 30

Put x = 0, 3y = 30 ⇒ y = 10

Put y = 0, x = 30

∴ (0, 10) and (30, 0) are points on the straight line x + 3y = 30.

10

10

20

20

30

30

40

40

50

50

60

60

70

70

A

B

D

P

0

Scale :10 units = 1 cm

C

Feasibleregion

y

x

Fig. 12.11


To find point of intersection solving

2x + y = 50 and x + 3y = 30

2 502 6 60

5 10

x yx y

y

+ =+ =

− − −− = −

� � � � � �

⇒ y = 2

Put y = 2 in 2x + y = 50

2x + 2 = 50

2x = 48 ⇒ x = 24

∴ (24, 2) is the point of intersection.

The shaded region in the graph is the feasible region.

To minimise Z = 1000x + 4000y

At A Z30 0 1000 30 4000 0 30000,� � � � � �= + =

At P Z24 2 1000 24 4000 2 32000,� � � � � �= + =

At D Z0 50 1000 0 4000 50 200000,� � � � � �= + =

∴ Zmin = 30,000

Minimum advertisement cost = Rs. 30,000.

��

• The objective function is a quantified statement in linear programming problem of what the bestresult is aimed for Objective function will either to maximise or to minimise a value.

• If the objective function is a function of 2 variables only, the linear programming problem can besolved by graphical method.

• Optimum solution to linear programming problem lies at one of the vertices. Hence find the co-ordinates of each vertex and subject in the objective function. The values for which the objectivefunction is highest/least is the optimum solution for maximisation/minimisation.

�-��

1. Write the following data in the form of LPP:

Machine Product A Product B Time

M1 2 3 60 min.

M2 4 5 100 min.

The price of product A is Rs. 1000 each and product B is Rs. 1500 each.

2. Old hens can be brought at Rs. 200 each and the young ones at Rs. 500 each. The old hen lays3 eggs and young one lays 5 eggs per week: If there are only Rs. 8000 available to spend on


purchase, the hens should be bought in order to have a maximum profit per week, assuming thatthe house cannot accommodate more than 20 hens at a time.

3. Maximise: 30x + 20y subject to

10 6 1000 5 4 600x y x y+ ≤ + ≤,

x y, ≥ 0

4. Maximise 5x + 8y subject to

2 100 2 200x y x y+ ≤ + ≤;

x y, .≥ 0

5. A company produces 2 products x and y each of the product require two operations one onmachine A and the other on machine B. The machine hours required by these two products andthe total hours available are given as follows:

Machine hrs. required Product Total machine hrs. available

X Y

A 2 5 19

B 4 3 17

Each unit of the product x and y makes a profit of Rs. 3 and Rs. 4. Find the optimal solution ofthe product to obtain the maximum profit.

6. Minimise Z = 4x + 4 subject to 3 4 20 5 15 0x y x y x y+ ≥ + ≥ ≥, , , . .

7. Minimise Z = 10x + 6y subject to 2 60 4 80 01x y x y x y+ ≥ + ≥ ≥, , . .

8. Minimise 2 = 3x + 5y subject to 5 2 0 4 3 6 01x y x y x y x y+ ≥ + ≥ + ≥ ≥, , , . .

9. An animal feed company must produce 200 kgs of mixture consisting of ingredients x1 and x2daily. x1 costs Rs. 3 per kg and x2 Rs. 8 per kg. Not more than 80 kgs. of x, can be used and atleast60 kgs. of x2 must be used. Find how much of each ingredients should be used if the companywants to minimise the cost.

10. A chemist provides his customers at least cost, the minimum daily requirement of 2 vitamins Aand B by using a mixture of 2 products M and N. The amount of each vitamin in one gram of eachproduct, the cost per gram of each product and minimum daily requirements are given below:

Vitamin A Vitamin B Cost per gram

M 6 2 20 ps.

N 2 2 16 ps.

Minimum requirement 12 8

Formulate the problem of finding the least expensive combination which provide the minimumrequirement of the vitamins.


��+��

1. Maximise Z = 1000 x + 1500 y subject to

2 3 60x y+ ≤

4 5 100 0 0x y x y+ ≤ ≥ ≥, , .

2. Maximise Z = 3x + 5y

Subject to condition

3. x = 40, y = 100, Zmax = 3,200

4. x = 0, y = 100, Zmax = 800

5. x = 2 and y = 3 Max Profit = 18 Rs.

6. x = 0, y = 5, Zmin = 5

7. x = 10, y = 40, Zmin = 340

8. x = 3, y = 1, Zmin = 14

9. Minimise 3x + 8y subject to

x y x y+ ≥ < ≥200 0 601 1,

x y1 0≥

Optimum solution x = 80, y = 120.

Zmin = 1200 Rs.

10. Minimise Z = 20x + 16y subject to

6 2 12x y+ ≤

2 2 8x y+ ≤

x y≥ ≥0 0,


��

��

��

Circle is a locus of point which moves such that its distance from a fixed point is constant. The fixedpoint is called centrcentrcentrcentrcentreeeee and the distance is called rrrrradiusadiusadiusadiusadius of the circle.

The boundary line of a circle is called circircircircircumfcumfcumfcumfcumferererererenceenceenceenceence.


(1) The distance between 2 points (x1, y1) and (x2, y2) is given by distance formula

x x y y2 12

2 12− + −� � � � .

(2) Slope of the line joining (x1, y1) and (x2, y2) is y y

x x2 1

2 1

−−

(3) 2 lines are perpendicular if and only if the product of their slopes = −1

(4) Angle in a semi circle = 90°.

��

1.1.1.1.1. EquaEquaEquaEquaEquation of the cirtion of the cirtion of the cirtion of the cirtion of the circccccle wle wle wle wle whose centrhose centrhose centrhose centrhose centre is ae is ae is ae is ae is at the ort the ort the ort the ort the origigigigiginininininand radius = and radius = and radius = and radius = and radius = rrrrr units: units: units: units: units:

Let O (0, 0) be the centre of the circle and r be the radiusof the circle. Let P (x, y) be any point on the circle.

OP = Radius = Distance between O (0, 0) and P (x, y).

∴ r x y= − + −0 02 2� � � �

r x y= +2 2

⇒ x y r2 2 2+ = .Fig. 13.1

x

y

r

P (x, y)

O (0, 0)x′

y′

Circles 279

This is the equation of the circle whose centre is at the origin and radius = r units.

2.2.2.2.2. EquaEquaEquaEquaEquation of the cirtion of the cirtion of the cirtion of the cirtion of the circccccle wle wle wle wle whose centrhose centrhose centrhose centrhose centre is ae is ae is ae is ae is at (t (t (t (t (hhhhh,,,,, kkkkk) and r) and r) and r) and r) and radius = adius = adius = adius = adius = rrrrr units: units: units: units: units:

Let C (h, k) be the centre of the circle. ‘r’ be the radius of thecircle. P (x, y) be any point on the circle.

CP = Radius = Distance between C (h, k) and P (x, y)

r x h y k= − + −� � � �2 2

Squaring,

r x h y k2 2 2= − + −� � � �

i.e., x h y k r− + − =� � � �2 2 2.

This is the equation of the circle whose centre is at the origin and radius = r units.

3.3.3.3.3. EquaEquaEquaEquaEquation of the cirtion of the cirtion of the cirtion of the cirtion of the circccccle wle wle wle wle whichichichichich is descrh is descrh is descrh is descrh is described on line joining (ibed on line joining (ibed on line joining (ibed on line joining (ibed on line joining (xxxxx11111,,,,, yyyyy11111) and () and () and () and () and (xxxxx22222,,,,, yyyyy22222) as the diameter:) as the diameter:) as the diameter:) as the diameter:) as the diameter:

Let A (x1, y1) and B (x2, y2) be the ends of the diam-eter of a circle. Let P (x, y) be any point on the circle.Join AP and BP.

We know,

Angle in a semicircle = 90°

∴ APB = °90

So AP BP⊥

Slope of AP × Slope of BP = −1 ...(1)

Slope of line joining A (x1, y1) and P (x, y) =−−

y y

x x1

1

Slope of line joining B (x2, y2) and P (x, y) =−−

y y

x x2

2

Substituting in (1), we get

y y

x x

y y

x x

−−

×−−

= −1

1

2

2

1

y y y y

x x x x

− −− −

= −1 2

1 2

1� ��

y y y y x x x x− − = − − −1 2 1 2� ��

⇒ x x x x y y y y− − + − − =1 2 1 2 0� �� .

This is the equation of the circle which has (x1, y1) and (x2, y2) as ends of a diameter.

Fig. 13.2

Fig. 13.3

P (x, y)

C (h, k)

r

x

y

0

P (x, y)

A (x , y )1 1 A (x , y )2 2

90°


�� !��!"��

The equation x2 + y2 + 2gx + 2fy + c = 0 always represent a circle.

Consider

x y gx fy c2 2 2 2 0+ + + + =Rearranging,

x gx y fy c2 22 2 0+ + + + =

Adding and subtracting g2 and f 2

x gx g g y fy f f c2 2 2 2 2 22 2 0+ + − + + + − + =

x g g y f f c+ − + + − + =� � � �2 2 2 2 0

x g y f g f c− − + − − = + −� � � �2 2 2 2

x g y f g f c− − + − − = + −� � � � � �2 2 2 2

2

Comparing this with the equation of circle

x h y k r− + − =� � � �2 2 2

We get

Centre = = − −h k g f, ,� � � �

Radius = = + −r g f c2 2 .

Hence

x2 +y2 + 2gx + 2fy + c = 0 represent a circle with centre (−g, −f) and

radius = g f c2 2+ − .

�! ��

1. In the equation x2 + y2 + 2gx + 2fy + c = 0 co-efficient of x2 = co-efficient of y2 = 1 and there isno xy term.

2. Centre is the mid point of diameter.

3. Co-ordinates of mid point of line joining (x1, y1) and (x2, y2) is x x y y1 2 1 2

2 2

+ +��

, .

4. The point of intersection of 2 diameters of a circle = Centre

#��$��%�&'��

1.1.1.1.1. Find the equation of the circle with centre at the origin and radius = 3 units.

Circles 281

Solution:Solution:Solution:Solution:Solution: Equation of the circle with centre (0, 0) and radius = r units = x2 + y2 = r2.

∴ Equation of the circle with centre (0, 0) and radius r units = x2 + y2 = 32

x2 + y2 = 9.

2.2.2.2.2. Find the equation of circle with centre (3, 4) and radius 5 units.

Solution:Solution:Solution:Solution:Solution: Equation of circle with centre (h, k) and radius = r units = x h y k r− + − =� � � �2 2 2 . .

x y− + − =3 4 52 2 2� � � � .

x x y y2 29 6 16 8 25+ − + + − = .

x y x y2 2 6 8 25 25 0+ − − + − =

x y x y2 2 6 8 0+ − − = .

3.3.3.3.3. Find the equation of circle whose ends of diameter are (3, 1) and (−4, 2).

Solution:Solution:Solution:Solution:Solution: Equation of circle whose ends of diameter are (x1, y1) and (x2, y2) =

x x x x y y y y− − + − − =1 2 1 2 0� �� ∴ Equation of circle

= − − − + − − =x x y y3 4 1 2 0� � � ��

x x y y− + + − − =3 4 1 2 0� ��

x x x y y y2 23 4 12 2 2 0− + − + − − + =

x y x y2 2 3 10 0+ + − − = .

4.4.4.4.4. Find the equation of the circle with centre (4, 3) and which passes through (0, 0).

Solution:Solution:Solution:Solution:Solution: Centre = (4, 3) (Given)

Circle passes through (0, 0).

Distance between centre and any point on the circle = Radius.

∴ Distance between (4, 3) and (0, 0) = Radius

0 4 0 32 2− + − =� � � � Radius

16 9+ = Radius

⇒ Radius = 5 units.

Equation of circle with centre (4, 3) and radius 5 units = x y− + − =4 3 52 2 2� � � �

x x y y2 216 8 9 6 25+ − + + − =

x y x y2 2 8 6 0+ − − =

(4, 3)

(0, 0)

Fig. 13.4


��

Given: Centre = 4 3, ,� � � �= − −g f

⇒ g f= − = −4 3 and .

Let the equation of circle be x2 + y2 + 2gx + 2fy + c = 0

Given it passes through (0, 0)

0 + 0 + 2g (0) + 2f (0) + c = 0

⇒ c = 0

∴ Equation of circle

= + + − + − + =x y x y2 2 2 4 2 3 0 0� � � �

x y x y2 2 8 6 0+ − − = .

5.5.5.5.5. Find the equation of the circle whose two diameters are x + 2y = 3 and x − y = 6 and radius 6 units.

Solution:Solution:Solution:Solution:Solution: The point of intersection of 2 diameters = centre.

Solving

x y

x y

y

+ =− =

− + −= −

2 3

6

3 3

( ) ( ) ( )

y = −1.

Substituting

y x y= − + =1 2 3 in

x + − =2 1 3� �

x − =2 3

x = +3 2

x = 5.

Centre = (x, y) = (5, −1) = (h, k)

Radius = 6 units.

Equation of circle = x h y k r− + − =� � � �2 2 2 .

x y− + − − =5 1 62 2 2� � � �� .

x x y y2 2 25 10 1 2 36+ − + + + = .

x y x y2 2 10 2 26 36 0+ − + + − =

Fig. 13.5

Circles 283

x y x y2 2 10 2 10 0+ − + − = .

6.6.6.6.6. If one end of the diameter of the circle x2 + y2 − 2x − 4y + 4 = 0 is (1, 1), find the other end.

Solution:Solution:Solution:Solution:Solution: Centre is the mid point of the diameter.

Given: Equation of circle:

x y x y2 2 2 4 4 0+ − − + =

Comparing with general equation

x y gx fy c2 2 2 2 0+ + + + =

2 2 2 4 4g f c= − = − =, ,

⇒ g f= − = −1 2, .

Centre = − − =g f, , .� � � �1 2

Mid point formula:x x y y1 2 1 2

2 2

+ +��

,

∴ 1 21

2

1

22 2, ,� � = + +�

�

x y� 1 1,� � is one end of the diameter.

⇒ 1 and = + = +1

22

1

22 2x y

2 1 2 2 12 2= + × = +x y

x y2 21 4 1= − =

⇒ y2 3= .

∴ Other end of the diameter = (1, 3).

7.7.7.7.7. Find the centre and radius of the following circles:

(a) x y x y2 2 2 4 7 0+ + − − =

(b) 2 2 3 6 8 02 2x y x y+ − + − =

(c) 3 3 2 33 02 2x y x+ − − =

Solution:Solution:Solution:Solution:Solution: (a) x y x y2 2 2 4 7 0+ + − − = .

Comparing with general equation,

x y gx fy c2 2 2 2 0+ + + + =

We get 2 2 2 4 7g f c= = − = −, ,

⇒ g f c= = − = −1 2 7, ,


Centre = − − = −g f, ,� � � �1 2

Radius = g f c2 2 2 21 2 7+ − = + − − −� � � �

= + + = =1 4 7 12 2 3 units.

(b) 2 2 3 6 8 02

2 2x y x y+ − + − =÷ by

x yx

y2 2 3

23 4 0+ − + − =

Comparing with general equation,

23

22 3 4g f c= − = = −, and

g f c= − = = −3

4

3

24, and

Centre = − − = −��

g f, ,� � 3

4

3

2

Radius = + − = �� + −��

− −g f c2 2

2 23

4

3

24� �

= + + = + +9

16

9

44

9 36 64

16

Radius units.= =109

16

109

4

(c) 3 3 2 33 02 2x y x+ − − =÷ by 3.

x y x2 2 2

311 0+ − − =

Comparing with general equation we get

22

32 0 11g f c= − = = −, , .

g c= − = −1

311, .

Centre = − − = ��g f, ,� � 1

30

Circles 285

Radius = + − = −�� + − −g f c2 2

21

30 11� �

= + = + = =1

911

1 99

9

100

9

10

3 units.

8.8.8.8.8. Find the equation of the circle whose centre is same as the centre of the circle x2 + y2 − 2x +4y − 7 = 0 and which has the radius 3 units.

Solution:Solution:Solution:Solution:Solution: Centre of the circle,

x y x y2 2 2 4 7 0+ − + − = is

− − = + −= − == − =

��

�g f

g f

g f, ,� � � �1 2

2 2 2 4

1 2

�

Given: Centre = (1, −2) and radius = 3 units.

Equation of the circle = (x − h)2 + (y − k)2 = r2

x y− + − − =1 2 32 2 2� � � ��

x x y y2 2 21 2 4 4 9+ − + + + =

x y x y2 2 2 4 4 0+ − + − = .

9.9.9.9.9. Find the equation of the circle which is concentric with the circle 2x2 + 2y2 + 3x − 4y + 8 = 0 andhaving unit radius.

Solution:Solution:Solution:Solution:Solution: Circles having same centre are called concentric circles.

Now centre of the circle: 2 2 3 4 8 02 2x y x y+ + − + =

÷ by 2.

x y x y2 2 3

22

8

20+ + − + =

23

22 2g f= = −

g f= = −3

41

Centre = − − = − +��

g f, ,� � 3

41

Radius = 1 unit.

Equation of circle = x h y k r− + − =� � � �2 2 2

x y− −��

��

��

+ − =3

41 1

22 2� �


x y+�� + − =3

41 1

22� �

x x y y2 29

162

3

41 2 1+ + ⋅ + + − =

xx

y y2 29

16

3

21 2 1 0+ + + + − − =

x yx

y2 2 3

22

9

160+ + − + =

or 16 16 24 32 9 02 2x y x y+ + − + = .

10.10.10.10.10. Find the equation of the circle whose centre is same as the centre of the circle x y x2 2 4+ − +

8y − 7 = 0 and radius same as that of the circle x y y2 2 8 9 0+ + − = .

Solution:Solution:Solution:Solution:Solution: Centre of the circle x y x y2 2 4 8 7 0+ − + − = is (−g, −f)

= −2 4,� �� 2 4 2 8

2 4

g f

g f

= − == − =

��

��

Radius of the circle: x2 + y2 + 8x − 9 = 0

is g f c2 2+ −� 2 8

4

g

g

==

4 92 − −� �

2 0

0

9

f

f

c

=== −

= + = =16 9 25 5 units.

∴ Equation of the required circle

= x h y k r− + − =� � � �2 2 2

x y− + − − =2 4 52 2 2� � � ��

x x y y2 2 22 4 16 8 25+ − + + + =

x y x y2 2 4 8 5 0+ − + − = .

11.11.11.11.11. A circle touches x-axis at (1, 0) and passes through (5, 2) find its equation.

Solution:Solution:Solution:Solution:Solution: Let the equation of circle be x2 + y2 + 2gx + 2fy + c = 0 ...(1)

Since the circle touches x-axis, y-co-ordinate of the centre = Radius.

Circles 287

∴ − = + −f g f c2 2

Squaring,

f g f c2 2 2= + −

⇒ g c2 = . ...(2)

Given (1) passes through (1, 0)

1 0 2 1 2 0 02 2+ + + + =g f c� � � � ⇒ 2 0 1g c+ = −

2 1g c+ = − ...(3)


5 2 2 5 2 2 02 2+ + + + =g f c� � � �

25 4 10 4 0+ + + + =g f c

10 4 29g f c+ + = − ...(4)

From (2) and (3)

g c g c2 2 1= + = − and

2 12g g+ = −

g g2 2 1 0+ + =

g + =1 02� �

g = −1.

c g= 2

c c= − ⇒ =1 12� � .

Substituting g = −1, c = 1 in (4)

10 4 29g f c+ + = −

− + + = −10 4 1 29f

4 29 9f = − +

4 20f = −

f = −5.

Fig. 13.6

f

(5, 2)

(1, 0)x

y

c ( g, f)− −


Equation of required circle

x y gx fy c2 2 2 2 0+ + + + =

i.e. x y x y2 2 2 1 2 5 1 0+ + − + − + =� � � �

⇒ x y x y2 2 2 10 1 0+ − − + = .

12.12.12.12.12. Find the equation of circle with centre (4, 6) and whichtouches the y-axis.

Since the circle touches y-axis, radius = x-co-ordinates ofthe centre.

∴ Radius = 4 units. � centre = (4, 6).

Equation of circle with centre (h, k) and radius r units

= x h y k r− + − =� � � �2 2 2 .

x y− + − =4 6 42 2 2� � � � .

x x y y2 216 8 36 12 16+ − + + − =

x y x y2 2 8 12 36 0+ − − + = .

��

Let the equation of circle be x2 + y2 + 2gx + 2fy + c = 0 ...(1)

Given: Centre = 4 6, ,� � � �= − −g f

⇒ g f= − = −4 6 and .

Given: Circle touches y-axis.

∴ − = + −g g f c2 2

Squarring,

g g f c2 2 2= + −

⇒ f c2 = .

⇒ c f= = =2 26 36� � .

c = 36.

Equation of circle,

= + + − + − + =x y x y2 2 2 4 2 6 36 0� � � �

x y x y2 2 8 12 36 0+ − − + = .

x

y

0

4

(4, 6)

Fig. 13.7

Circles 289

13.13.13.13.13. A circle has its centre on the y-axis and passes through (−1, 3) and (2, 5). Find its equation.

Solution:Solution:Solution:Solution:Solution: Let the equation of the circle be

x2 + y2 + 2gx + 2fy + c = 0 ...(1)

Given: (1) has centre on Y-axis.

On y-axis, x-co-ordinate = 0.

∴ X-co-ordinates of the centre (−g, −f) = 0

i.e., − = ⇒ =g g0 0.

Given: (1) passes through (−1, 3)

− + + − + + =1 3 2 1 2 3 02 2� � � � � �g f c

1 9 2 0 6 0+ + + + =� � f c � g = 0� �⇒ 6 10f c+ = − ...(2)


2 5 2 2 2 5 02 2+ + + + =g f c� � � �4 25 0 10 0+ + + + =f c

10 29f c+ = − ...(3)

Solving (2) and (3)

6 10

10 29

4 19

f c

f c

f

+ = −+ = −

− − +− =

( ) ( ) ( )

f = − 19

4.

Substituting f = − 19

4 in (2)

619

410−��

+ = −c

− + = −57

210c

c = − +1057

2

c = − + =20 57

2

37

2


c = 37

2.

∴ Equation of the circle

= + + + −�� + =x y x y2 2 2 0 2

19

4

37

20� �

Multiplying by 2 we get

2 2 19 37 02 2x y y+ − + = .

14.14.14.14.14. Find the equation of circle which touch both the co-ordinate axis and passes through the point(2, 1).

Solution:Solution:Solution:Solution:Solution: Let the equation of circle bex2 + y2 + 2gx + 2fy + c = 0 ...(1)

Given (1) touches both co-ordinate axes.

[X-co-ordinate of centre = Radius = Y-co-ordinate of centre]

∴ − = + − − = + −g g f c f g f c2 2 2 2 and

∴ − = −g f

⇒ g f= .

Also g g f c2 2 2= + −

∴ f c g c2 2= =. .Similarly

∴ g f c2 2= = .


∴ 2 1 2 2 2 1 02 2+ + + + =g f c� � � �4 1 4 2 0+ + + + =g f c

5 4 2 0+ + + =g f c

5 4 2 0+ + + =f f c � g f=

5 6 0+ + =f c

5 6 02+ + =f f

f f2 6 5 0+ + = � c f= 2

f f f2 5 1 5 0+ + + =

f f f+ + + =5 1 5 0� � � �

r

r

Fig. 13.8

Circles 291

f f= − = −5 1 or

Since g f= ,

g g= − = −5 1 or

� c g= 2

c c= − = −5 12 2� � � � or

c c= =25 1 or .

∴ Equation of required circle

= + + − + − + =x y x y2 2 2 5 2 5 25 0� � � �

x y x y2 2 10 10 25 0+ − − + =

or

x y x y2 2 2 1 2 1 1 0+ + − + − + =� � � �

x y x y2 2 2 2 1 0+ − − + = .

15.15.15.15.15. A circle cuts off positive intercepts 5 and 6 on x and y-axes respectively, and passes through theorigin. Find its equation.

Solution:Solution:Solution:Solution:Solution: Let the equation of circle be

x2 + y2 + 2gx + 2fy + c = 0 ...(1)

Given (1) makes an intercept 5 on x-axis.

⇒ (1) passes through (5, 0)

∴ 5 0 2 5 2 0 02 + + + + =g f c� � � �25 10 0+ + =g c ...(2)

Given (1) makes an intercept 6 on y-axis.

⇒ (1) passes through (0, 6).

∴ 0 6 2 0 2 6 02+ + + + =g f c� � � �36 12 0+ + =f c ...(3)


∴ 0 0 2 0 2 0 0+ + + + =g f c� � � �

⇒ c = 0.

Substituting c = 0 in (2) we get

25 10 0+ =g


g = − = −25

10

5

2

Substituting c = 0 in (3)

36 12 0+ =f

⇒ f = −3.


x y x y2 2 25

22 3 0 0+ + −��

+ − + =� �

x y x y2 2 5 6 0+ − − = .

�� '��

ChorChorChorChorChord:d:d:d:d: A line which divides the circle into 2 parts is called chord. So chord is a linejoining any 2 distinct points on the circle.

TTTTTangangangangangent:ent:ent:ent:ent: A line which touches the circle at only one point and which is perpendicularto the radius drawn from the point of contact is called tangent.


If ax + by + c = 0 is the equation of the chord AB of the circle x2 + y2 + 2gx + 2fy + c = 0 whose centreis C.

Then from figure,

CB2 = CD2 + BD2

Now CB = Radius of the circle = g f c2 2+ −

CD = Length of the perpendicular from the centre(−g, −f) to the line ax + by + c = 0.

=− + − +

+

a g b f c

a b

� � � �2 2

BD AB= 1

2.

∴ Length of the chord AB can be found if we know the equation of the circle and equation of thechord.

#��$��%�&'��

1.1.1.1.1. Find the length of the chord 4x − 3y = 5 of the circle x2 + y2 + 3x − y − 10 = 0.

Centre: − −g f,� �

chord

AD

C

B

Fig. 13.9

Circles 293

= − +��

3

2

1

2,

Radius of the circle = g f c2 2+ −

= −�� + ��

− −3

2

1

210

2 2

� �

= + +9

4

1

410

= = = ⋅ ×50

4

1

250

1

225 2

= =5 2

2

5

2 units.

Perpendicular distance from C −��

3

2

1

2, , to the line 4x − 3y − 5 = 0 is

CD =−�� − �

� −

+ −

432

312

5

4 32 2� �

CD =− − −12

232

5

5

= − ⋅ =25

2

1

5

5

2.

CD = 5

2 units.

From figure,

CB CD DB2 2 2= +

⇒ DB CB CD2 2 2= −

DB = �� − ��

5

2

5

2

2 2

AD

C

B

Fig. 13.10


DB = − = −25

2

25

4

50 25

4

DB = =25

4

5

2.

∴ Length of the chord AB = 2 × DB

= × =25

25 units.

2. If x + 2y − 4 = 0 is the equation of the chord to the circle x2 + y2 = 4, then find its length.

Solution:Solution:Solution:Solution:Solution: Centre of the circle = (0, 0)

Radius = 4 units.= 2

Perpendicular distance from centre (0, 0) to the line x + 2y − 4 = 0 is

0 2 0 4

1 2

4

5

4

52 2

� � � �+ −

+= − = units.

From figure,

CB CD BD2 2 2= +

⇒ BD CB CD2 2 2= −

BD2 22

24

5= − ��

�

BD2 416

5= −

BD2 20 16

5= −

BD2 4

5=

BD = =4

5

2

5 units.

∴ Length of the chord AB = 2 × BD

= × =22

5

4

5 units.

3. If 3x + y − 5 = 0 is a chord to the circle x2 + y2 − 22x − 4y + 25 = 0, then find its length.

AD

C

B

Fig. 13.11

Circles 295


Centre = (−g, −f)

= 11 2,� �

Radius = + − = + −g f c2 2 2 211 2 25

= + − = =121 4 25 100 10 units.

Perpendicular distance from the centre (11, 2) to the line 3x + y − 5 = 0 is

3 11 2 5

3 1

33 2 5

102 2

� � + −

+= + −

CD = 30

10 units.

From figure,

CB CD BD2 2 2= +

BD CB CD2 2 2= −

⇒ BD = − ��10

30

102

2

= −100900

10

BD = 10 units.

AB BD= = ×Length of the chord 2

= ×2 10

= 2 10 units.

4.4.4.4.4. Find the point of intersection of the line 4x − 3y = 10 and the circle x2 + y2 − 2x + 4y − 20 = 0and find the length of chord intercepted by the line.

Equation of circle: x y x y2 2 2 4 20 0+ − + − =

Equation of line: 4 3 10x y− =

4 10 3x y= +

xy= +10 3

4.

AD

C

B

Fig. 13.12


Substituting xy= +10 3

4 in the equation of circle we get the point of intersection.

10 3

42

10 3

44 20 0

22+�

� + − +�

� + − =y

yy

y

10 3

162

10 3

44 20 0

22+

+ −+

+ − =y

yy

y� � � �

Multiplying by 16

10 3 16 8 10 3 64 320 02 2+ + − + + − =y y y y� � � �

⇒ y y2 4 12 0+ − =

y y y2 6 2 12 0+ − − =

y y y+ − + =6 2 6 0� � � �

y y− + =2 6 0� � � �

∴ y y= = −2 6 or

� xy= +10 3

4

x x=+

=+ −10 3 2

4

10 3 6

2

� � � � or

x x= = −4 4 or

The point of intersection are (4, 2) and (−4, −6).

Length of the chord = Distance between (4, 2) and (−4, −6).

= − + −x x y y2 12

2 12� � � �

= − − + − −4 4 6 22 2� � � �

= + = +8 8 64 642 2

= =128 8 2 units.

5.5.5.5.5. If x + y = 1 is the chord to the circle x2 + y2 − 2x − 4y − 29 = 0, then find its length. Also find theco-ordinates of middle point of chord intercepted.


Equation of the line = x + y = 1

y = 1 − x

Fig. 13.13

B (4, 2)A( 4, 6)− −

Circles 297

Substituting y = 1 − x in the equation of the circle we get

x x x x2 21 2 4 1 29 0+ − − − − − =� � � �

x x x x x2 21 2 2 4 4 29 0+ + − − − + − =

2 32 02x − =

2 322x =

x2 32

2=

x2 16=

⇒ x = ± 4.

y x= −1

y y y= ± ∴ = = −1 4 5 3 or .

∴ Point of intersection of the circle with the line = (4, −3) and (−4, 5)

Length of the chord = − + −x x y y2 12

2 12� � � �

= − − + − −4 4 5 32 2� � � ��

= +64 64

= 8 2 units.

Co-ordinates of the middle point of the chord whose end points are (4, −3) and (−4, 5)

= + +��

x x y y1 2 1 2

2 2,

=+ − − +�

��

4 4

2

3 5

2

� �,

= (0, 1).

6.6.6.6.6. If the straight line x

a

y

b+ = 1 cuts the circle x y r2 2 2+ = in 2 points, then prove that the length

of the chord

=+ −

+2

2 2 2 2 2

2 2

r a b a b

a b

� �

PrPrPrPrProof:oof:oof:oof:oof: Equation of line

= + =x

a

y

b1

Fig. 13.14

A( 4, 5)−

B (4, 3)−


bx ay

ab

+ = 1

⇒ bx ay ab+ − = 0

The equation x2 + y2 = r2 represent the circle whose centre = (0, 0) and radius = r units.

Length of the perpendicular from (0, 0) to the line bx + ay − ab = 0.

ODb a ab

b a

ab

a b=

+ −

+=

+

0 02 2 2 2

� � � �.

From figure,

OB OD BD2 2 2= +⇒ BD OB OD2 2 2= −

BD rab

a b

2 2

2 2

2

= −+

��

�

BD ra b

a b2 2

2 2

2 2= −+

BD ra b

a b

r a b a b

a b= −

+=

+ −

+2

2 2

2 2

2 2 2 2 2

2 2

� �

Length of the chord AB = 2 × BD

ABr a b a b

a b=

+ −

+2

2 2 2 2 2

2 2

� �.

Hence proved.

��( �� )��xxxxx22222 + + + + + yyyyy22222 + 2 + 2 + 2 + 2 + 2gxgxgxgxgx + 2 + 2 + 2 + 2 + 2fyfyfyfyfy + + + + + ccccc = 0 = 0 = 0 = 0 = 0 ��)��'��*xxxxx11111,,,,, yyyyy11111

+��

Let P (x1, y1) be any point on the circle x2 + y2 + 2gx + 2fy + c = 0. Let C (−g, −f) be the centre. PTbe the tangent. Joint CP.

We know,

Slope of the line joining 2 points (x1, y1) and (x2, y2) = y y

x x2 1

2 1

−−

∴ Slope of line joining C (−g, −f) and P (x1, y1) = y f

x g1

1

− −− −� ��

=++

y f

x g1

1

Fig. 13.15

P (x , y )1 1

T

C ( g, f)− −

Fig. 13.16

bx + ay ab = 0−

Circles 299

Since PT perpendicular CP

Slope of PT × Slope of CP = −1

∴ Slope of PT = −1

Slope of CP

= −++

1

1

1

y f

x g

Slope of PTx g

y f= −

++

��

�

1

1

Equation of a straight line passing through (x1 y1) and having slope m is

y y m x x− = −1 1� �∴ Equation of tangent PT

= − = −++

��

�

−y yx g

y fx x1

1

11� �

y y y f x g x x− + = − + −1 1 1 1� ��

yy y yf y f xx gx x gx1 12

1 1 12

1− + − = − + − −

yy y yf y f xx gx x gx1 12

1 1 12

1 0− + − + + − − =

⇒ xx yy gx fy x y gx fy1 1 12

12

1 1+ + + = + + +

adding gx fy c1 1+ + to both sides

∴ xx yy gx fy gx fy c x y g x fy c1 1 1 1 12

12

1 12 2+ + + + + + = + + ⋅ + +

Since (x1, y1) lie on the circle,

= + + + + =x y gx fy c12

12

1 12 2 0

∴ xx yy g x x f y y c1 1 1 1 0+ + + + + + =� � � � .

This is the equation of the tangent to the circle.

��, ��)�� )�� &��)��'��(((((xxxxx11111,,,,, yyyyy11111)))))��)��xxxxx22222 + + + + + yyyyy22222 + 2 + 2 + 2 + 2 + 2gxgxgxgxgx + 2 + 2 + 2 + 2 + 2fyfyfyfyfy + + + + + ccccc = 0. = 0. = 0. = 0. = 0.

Let C (−g, −f) be the centre of the circle x2 + y2 + 2gx + 2fy + c = 0. Let P (x1, y1) be any point outside

the circle. Let PT be the tangent drawn to the circle from P. Join CT and CP.


P (x , y )1 1T

C ( g, f)− −

Since PT is a tangent,

CT ⊥ PT.

∴ CTP is a right angled triangle.

∴ By Pythogoras theorem,

CP2 = CT2 + PT2

⇒ PT2 = CP2 − CT2.

Now CP = Distance between C (−g, −f) and P (x1, y1)

CP x g y f= − − + − −12

12� � � �

CP x g y f= + + +12

12� � � �

CT g f c= = + −Radius 2 2

∴ PT x g y f g f c21

21

22

2 22

= + + + ��

��

− + −� � � �

PT x g y f g f c21

21

2 2 2= + + + − + −� � � � � �

PT x g gx y f fy g f c212 2

1 12 2

12 22 2= + + + + + − − +

PT x y gx fy c212

12

1 12 2= + + + + .

∴ PT x y gx fy c= + + + +12

12

1 12 2 .

This is the length of the tangent from P (x1, y1) to the circle.

��- �� )��y y y y y = = = = = mx mx mx mx mx + + + + + ccccc��.��)��xxxxx22222 + + + + + yyyyy22222 = = = = = aaaaa22222��'��

The line y = mx + c is a tangent to the circle x2 + y2 = a2 if and only if length of the perpendicular fromthe centre to the line is equal to radius.

Now centre of the circle = (0, 0)

Radius = a

∴ Length of perpendicular from (0, 0) to the line y = mx + c = Radius

(i.e. mx − y + c = 0)

⇒m c

ma

0 0

12

� � � �− +

+=

c

ma

2 1+=

Fig. 13.17

Circles 301

c

ma

2 1+= ±

⇒ c a m= ± +2 1

Squaring,

c a m2 2 2 1= +� �.

This is the condition for the line y = mx + c to be a tangent to thecircle x2 + y2 = a2.

∴ Equation of tangent = = ± +y mx a m2 1 ...(1)

Also we know that equation of tangent at (x1, y1) on x2 + y2 = a2 is

xx yy a1 12+ =

⇒ yy xx a1 12= − + ...(2)

Comparing (1) and (2) we get

y x

m

a

a m

1 12

21 1=

−=

± +Taking 1st and 3rd ratio,

ya

m1 2 1

= ±+

Taking 2nd and 3rd ratio.

− = ±+

x

m

a

m

1

2 1

xam

m1 2 1

=+

�

∴ The points of contact are

am

m

a

m

am

m

a

m2 2 2 21 1 1 1+

−

+

��

�

−

+ +

��

�

, , or

out of the 2 points of contact, one will satisfy the equation of tangent.

��/ �� )��lx lx lx lx lx + + + + + mmmmmy y y y y + + + + + n n n n n = 0= 0= 0= 0= 0��.��)��xxxxx22222 + + + + + yyyyy22222 + 2 + 2 + 2 + 2 + 2gxgxgxgxgx + 2 + 2 + 2 + 2 + 2fyfyfyfyfy + + + + + ccccc = 0 = 0 = 0 = 0 = 0

lx + my + n = 0 may be a tangent to the circle iff length of the perpendicular from the centre = Radius.

Centre = (−g, −f) and radius = g f c2 2+ −

y = mx + c

(0, 0)

Fig. 13.18


i.e.,l g m f n

l mg f c

− + − +

+= + −

� � � �2 2

2 2

− − +

+= + −lg mf n

l mg f c

2 2

2 2

Squaring

n mf

l mg f c

− −+

= + −lg� �2

2 22 2

n mf l m g f c− − = + + −lg� � � �� 2 2 2 2 2

This is the required condition.


The length of the tangent from the point P (x1, y1) tothe circle x2 + y2 + 2gx + 2fy + c = 0 is given by

x y gx fy c12

12

1 12 2+ + + + .

If P lie out the circle, then AP is positive or

x y gx fy c12

12

1 12 2 0+ + + + >

or x y gx fy c12

12

1 12 2 0+ + + + > .

If P coincide with A, then AP = 0 and the point P lie on the circle.

∴ x y gx fy c12

12

1 12 2 0+ + + + = .

If P lie inside the circle then

x y gx fy c12

12

1 12 2 0+ + + + < .

#��$��%�&'��

1.1.1.1.1. Find the equation of tangent to the circle 2 2 3 12 02 2x y x y+ + − − = at (1, 3).

Solution:Solution:Solution:Solution:Solution: Equation of circle:

2 2 3 12 02 2x y x y+ + − − =÷ by 2.

x yx

y2 2

2

3

26 0+ + − − =

21

22

3

26g f c= = − = −

lx + my + n = 0

( g, f)− −

Fig. 13.19

Fig. 13.20

P (x , y )1 1A

Circles 303

⇒ g f= = −1

4

3

4.

x y1 1 1 3, , .� � � �=

Equation of tangent

xx yy g x x f y y c1 1 1 1 0+ + + + + + =� � � �

x y x y1 31

41

3

43 6 0� � � � � � � �+ + + + −�

� + − =

x y xy+ + + − − − =3

1

4

1

4

3

4

9

46 0

4 12 1 3 9 24

40

x y x y+ + + − − − =

Crossmultiplying

5 9 32 0x y+ − = .

2.2.2.2.2. Find the equation of the tangent to the circle x y x y2 2 8 6 11 0+ + − − = parallel to 4x − 3y + 1 = 0.

Solution:Solution:Solution:Solution:Solution: Any line parallel to 4x − 3y + 1 will have the equation 4x − 3y + k = 0.

Now 4x − 3y + k = 0 will be a tangent to the circle if the length of the perpendicular from the centreto the line = Radius.

Now Equation of circle = x y x y2 2 8 6 11 0+ + − − =

Centre = (−g, −f) = (−4, 3)

Radius of x y x y2 2 8 6 11 0+ + − − = is 4 3 112 2+ +

= + =25 11 6 units.

Length of perpendicular from (−4, 3) to 4x − 3y + k = 0 = Radius

4 − − +

+=

4 3 3

4 36

2 2

� � � � k

− − + =16 9

56

k

− + =25 30k [Note: We get 2 tangents parallel to given line]

− + = ±25 30k

4x 3y + 1−

Fig. 13.21


k = ± +30 25

∴ k k= = −55 5 or .

∴ Equations of tangents

= − + = − − =4 3 55 0 4 3 5 0x y x yand

3.3.3.3.3. Find the equation of tangent to the circle x y x y2 2 2 4 4 0+ − − − = which are perpendicular to

the line 4x + 3y − 7 = 0.

Any line perpendicular to 4x + 3y − 7 will have the equation 3x − 4y + k = 0.

3x − 4y + k = 0 will be a tangent to the circle if the length of the perpendicular from the centre =Radius.

Now equation of circle = x y x y2 2 2 4 4 0+ − − − =

Centre = (−g, − f) = (+1, 2)

Radius = g f c2 2 2 21 2 4+ − = + − −� �

= + + =1 4 4 3 units.

Length of the perpendicular from (1, 2) to 3x − 4y + k = 0 (Radius)

i.e.,3 1 4 2

3 43

2 2

� � � �− +

+=

k

= − + =3 8

53

k

= − + =5 15k

− + = ±5 15k

k = ± +15 5

k k= = −20 10 or .

∴ Equation of tangents are

3 4 20 0 3 4 10 0x y x y− + = − − = and .

4.4.4.4.4. Find the value of k if x + y + k = 0 touches the circle x y x y2 2 7 5 18 0+ − − + = .

Solution:Solution:Solution:Solution:Solution: x + y + k touches the circle if the length of the perpendicular from the centre to the line =Radius.

Now Equation of circle = x y x y2 2 7 5 18 0+ − − + =

Centre = − − = ��g f, ,� � 7

2

5

2

Fig. 13.22

4x + 3y 7 = 0−

[Note: We get 2 tangents perpendicular to given line]

Circles 305

Radius = + − = �� + ��

−g f c2 2

2 27

2

5

218

= + −49

4

25

418

Radius units.= − = =74 72

4

2

4

1

2

Length of the perpendicular from 7

2

5

2,�� to x + y + k = 0 = Radius

72

52

1 1

1

22 2

+ +

+=

k

122

1 1

1

22 2

+

+=

k

122

2

1

2

+=

k

6 1+ =k

6 1+ = ±k

k = ± −1 6

k k= − = −7 5 or

Hence k = −7 or −5.

5.5.5.5.5. Find the length of the tangent to the circle 3 3 7 6 12 02 2x y x y+ − − − = from (6, −7)

Solution:Solution:Solution:Solution:Solution: Equation of the circle

3 3 7 6 12 03

2 2x y x y+ − − − =÷ by .

x y x y2 2 7

32 4 0+ − − − = .


Length of the tangent from (x1, y1)

= + + + +x y gx fy c12

12

1 12 2

Length of tangent from (6, −7)

= + − − − − −6 77

36 2 7 42 2� � � � � �

= + − + −36 49 14 14 4

= =81 9 units.

6.6.6.6.6. Find whether the origin is inside or on or outside the circle

(i) x y x y2 2 4 7 8 0+ + − + =

(ii) x y2 2 7+ =

(iii) x y y x2 2 2 6 0+ + + =

Solution:Solution:Solution:Solution:Solution: (x1, y1) lie inside the circle

x y gx fy c2 2 2 2 0+ + + + = if

x y gx fy c12

12

1 12 2 0+ + + + <

Outside if x y gx fy c12

12

1 12 2 0+ + + + > and on the circle if

x y gx fy c12

12

1 12 2 0+ + + + =

(i) Clearly (0, 0) lie outside the circle x y x y2 2 4 7 8 0+ + − + =

� 0 0 4 0 7 0 8 8 0+ + − + = >� � � �

(ii) Clearly (0, 0) lie inside the circle x y2 2 7 0+ − = . Since 0 + 0 − 7 = −7 < 0

(iii) (0, 0) lie on the circle x y y x2 2 2 6 0+ + + =

� 0 0 2 0 6 0 0+ + + =� � � �7.7.7.7.7. Find the equation of the circle so that the lengths of the tangents from (−1, 0), (0, 2) and (−2, 1)

are respectively 3, 10 and 3 3.

Solution:Solution:Solution:Solution:Solution: Let the equation of circle be

x y gx fy c2 2 2 2 0+ + + + = ...(1)

Given: length of tangent from (−1, 0) to circle is 3 units.

∴ − + + − + + =1 0 2 1 2 0 32 2� � � � � �g f c

Circles 307

1 2 3− + =g c

Squaring

1 2 9− + =g c

− + =2 8g c ...(2)

Given: Length of tangent from (0, 2) to the circle = 10 units.

0 2 2 0 2 2 102+ + + + =g f c� � � �Squaring

4 4 10+ + =f c

4 6f c+ = ...(3)

Given: Length of the tangent from (−2, 1) is 3 3

∴ − + + − + + =2 1 2 2 2 1 3 32 2� � � � � �g f c

Squaring,

4 1 4 2 3 32

+ − + + =g f c � �

5 4 2 27− + + =g f c

− + + =4 2 22g f c ...(4)

Solving (3) and (4)

4 6

4 2 22

4 2 16

2

f c

g f c

g f

+ =− + + =+ − − −

+ = −÷

� � � � � � � �

by

2 8g f+ = − ...(5)

Solving (2) and (3)

− + =+ =

− − −− − =

2 8

4 6

2 4 2

g c

f c

g f

( ) ( ) ( ) ...(6)


Solving (5) and (6)

( ):

( ):

5

6

2 8

2 4 2

3 6

g f

g f

f

+ = −− − =

− = −

⇒ f = 2.

Substituting f = 2 in (5)

2 2 8g + = −

2 8 2g = − −

2 10g = −

⇒ g = −5

Substituting g = −5, f = 2 in (2)

− + =2 8g c

− − + =2 5 8� � c

c = −8 10

c = −2

Substituting g = −5, f = 2 and c = −2 in (1).

Equation of circle

= + + − + + − =x y x y2 2 2 5 2 2 2 0� � � � � �

x y x y2 2 10 4 2 0+ − + − = .

8.8.8.8.8. Find the equation of the circle passing through the points (1, 2) and (3, 4) and touching the line3x + y − 3 = 0.

Solution:Solution:Solution:Solution:Solution: Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 ...(1)

Given: (1) passes through (1, 2)

∴ 1 2 2 1 2 2 02 2+ + + + =g f c� � � �5 2 4 0+ + + =g f c

2 4 5g f c+ + = − ...(2)


3 4 2 3 2 4 02 2+ + + + =g f c� � � �25 6 8 0+ + + =g f c

6 8 25g f c+ + = − ...(3)

Circles 309


2 4 5

6 8 25

4 4 20

g f c

g f c

g f

+ + = −+ + = −

− − − +− − =

÷ −

� � � � � � � �

by 4

g f+ = −5

f g= − −5 ...(4)

Given: 3x + y − 3 = 0 touches the circle.

∴ Length of the perpendicular from centre (−g, −f) to 3x + y − 3 = 0 (Radius)

∴3 3

3 12 2

− + − −

+=

g f� � � �Radius

− − − = + + +3 3

101 22 2g f

g f� � � � �Radius Distance between centre

and any point on circumference

= ��

��

Substituting f = −5 −g we get

− − − − −= + + − −

3 5 3

101 2 52 2g g

g g� � � � � �

− − − − −= + + + + +

3 5 3

101 2 9 62 2g g

g g g g� �

Squaring,

− +��

� = + +2 2

1010 2 8

22g

g g

4

101 2 8 102 2g g g− = + +� �

gg g

−= + +

1

54 5

22� �

g g g g2 22 1 5 20 25− + = + +

⇒4 22 24 02g g+ + =

÷ by 2

(1, 2)

C ( g, f)− −

Fig. 13.23


2g g2 11 12 0+ + =

2 8 3 12 02g g g+ + + =

2 4 3 4 0g g g+ + + =� � � �

2 33

24

g

g g= −

= − = − or

when g = −4,

Substituting g = −4 in (4)

f g= − −5

f = − − −5 4� �

f = −1.

Substituting g = −4, f = −1 in (2)

2 4 5g f c+ + = −

2 4 4 1 5− + − + = −� � � � c

c = − +5 12

c = 7

Similarly we can get f c= − =7 2 12 and when g = −3/2.


x y x y2 2 2 4 2 1 7 0+ + − + − + =� � � �

x y x y2 2 8 2 7 0+ − − + = or

x y x y g f c2 2 3 7 12 03

2

7

212+ − − + = = − = − =

��

. ,By putting and

9.9.9.9.9. Find the equation of the circle which is concentric with the circle x y x y2 2 8 12 20 0+ − + − =and which touches the line 4x − 3y − 14 = 0.

Solution:Solution:Solution:Solution:Solution: Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 ...(1)

Given (1) is concentric with

x y x y2 2 8 12 20 0+ − + − =

∴ Centre of (1) = Centre of x y x y2 2 8 12 20 0+ − + − =

248

3

11

2gg

g

g

+++

Circles 311

− − = −g f, ,� � � �4 6

∴ g f= − =4 6 and .

Also given (1) touches 4x − 3y − 14 = 0

∴ Length of the perpendicular from centre = Radius

∴4 4 3 6 14

4 32 2

2 2� � � �− − −

+= + −g f c

= + − = + −16 18 14

54 62 2 c

20

516 36= + − c

Squaring,

4 522 = − c

16 52= − c

c = − =52 16 36


x y x y2 2 2 4 2 6 36 0+ + − + + =� � � �x y x y2 2 8 12 36 0+ − + + = .

10.10.10.10.10. Find the equation of the circle passing the points (4, 1) and (6, 5) and having its centre on the line4x + y = 16.

Solution:Solution:Solution:Solution:Solution: Let the equation of the circle be

x2 + y2 + 2gx + 2fy + c = 0 ...(1)


4 1 2 4 2 1 02 2+ + + + =g f c� � � �16 1 8 2 0+ + + + =g f c

8 2 17g f c+ + = − ...(2)


6 5 2 6 2 5 02 2+ + + + =g f c� � � �12 10 61g f c+ + = − ...(3)

Also given (1) has centre on 4x + y = 16 i.e., (−g, −f) lie on 4x + y = 16.

∴ 4 16− + − =g f� � � �− − =4 16g f ...(4)


Solving (2) and (3)

8 2 17

12 10 61

4 8 44

g f c

g f c

g f

+ + = −+ + = −

− − − +− − =( ) ( ) ( ) ( ) ...(5)

Solving (4) and (5)

− − =− − =

+ + −= −

4 16

4 8 44

7 28

g f

g f

f

( ) ( ) ( )

f = − 4

Substituting f = −4 in (4)

− − − =4 4 16g � �

− + =4 4 16g

− = −4 16 4g

− =4 12g

g g= − ⇒ = −12 4 3

Substituting g = −3 and f = −4 in (2)

8 3 2 4 17− + − + = −� � � � c .

− − + = −24 8 17c

c = − +17 32

c = 15


= + + − + − + =x y x y2 2 2 3 2 4 15 0� � � �

x y x y2 2 6 8 15 0+ − − + = .

��&�&.��

• Distance between 2 points (x1, y1) and (x2, y2)

x x y y2 12

2 12− + −� � � � .

Circles 313

• Slope of the line joining (x1, y1) and (x2, y2)

=−−

y y

x x2 1

2 1

.

• 2 lines are perpendicular iff product of their slopes = −1.

• Equation of circle with centre (0, 0) and radius r units: x2 + y2 = r2.

• Equation of circle with centre (h, k) and radius r units: x h y k r− + − =� � � �2 2 2 .

• Equation of circle which is described on line joining (x1, y1) and (x2, y2) as diameter =

x x x x y y y y− − + − − =1 2 1 2 0� � � � � � � � .

• General equation of the circle x2 + y2 + 2gx + 2fy + c = 0

Centre = (−g, −f), Radius = g f c2 2+ − .

• Centre is the mid point of diameter.

• Co-ordinates of mid point of line joining (x1, y1) and (x2, y2) = x x y y1 2 1 2

2 2

+ +��

,

• The point of intersection of 2 diameters of a circle = centre.

• If a circle x2 + y2 + 2gx + 2fy + c = 0 touches x-axis, then, radius = − f and g2 = c.

• If a circle x2 + y2 + 2gx + 2fy + c = 0 touches y-axis, then Radius = − =g f c and 2 .

• Length of the perpendicular from (x1, y1) to the line ax + by + c = 0 is

ax by c

a b

1 1

2 2

+ +

+

• Equation of tangent at (x1, y1) on the circle x2 + y2 + 2gx + 2fy + c = 0 is

xx yy g x x f y y c1 1 1 1 0+ + + + + + =� � � � .

• Length of the tangent from (x1, y1) to the circle x2 + y2 + 2gx + 2fy + c = 0 is

x y gx fy c12

12

1 12 2+ + + + .

• Any line is a tangent to the given circle if the length of the perpendicular from centre to the line= Radius of the circle.

• Condition for the line y = mx + c to be a tangent to the circle x2 + y2 = a2 is c a m2 2 2 1= ± +� � and

point of contact is am

m

a

m

am

m

a

m2 2 2 21 1 1 1+

−

+

��

�

−

+ +

��

�

, , . or .

• Condition for the line lx + my + n = 0 to be a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 is

n mf l m g f c− − = + + −lg� � � �� 2 2 2 2 2 .


• The point (x1, y1) lie inside the circle x2 + y2 + 2gx + 2fy + c = 0 if x y gx fy c12

12

1 12 2 0+ + + + < .

Outside the circle if x y gx fy c12

12

1 12 2 0+ + + + > .

On the circle if x y gx fy c12

12

1 12 2 0+ + + + = .

• Any line parallel to ax + by + c = 0 will have the equation ax + by + k = 0.

[Take x and y co-efficients same].

• Any line perpendicular to ax + by + c = 0 will have the equation bx − ay + k = 0 [interchange xand y co-efficients and change the sign of anyone].

�%��

I. FFFFFind the equaind the equaind the equaind the equaind the equation of cirtion of cirtion of cirtion of cirtion of circccccle in eacle in eacle in eacle in eacle in each of the fh of the fh of the fh of the fh of the folloolloolloolloollowing:wing:wing:wing:wing:

1. Centre: Origin, Radius: 3 Units.

2. Centre: (0, 2), Radius: 2 Units.

3. Centre: (3, −4), Radius: 5 Units.

4. Centre: 4

3

7

3,�� , Radius: 15 units.

5. Two diameters are x + y = 8 and 2x − y = 4 and radius 4 units.

6. Diameter is the line joining (5, 0) and (0, 5).

7. Centre: (3, 4), touching the x-axis.

8. Centre: (−5, 1), touching the y-axis.

9. Centre: (4, 4), touching the axes.

10. Centre: (−3, 2), passes through the origin.

II. FFFFFind the centrind the centrind the centrind the centrind the centre and re and re and re and re and radius of the fadius of the fadius of the fadius of the fadius of the folloolloolloolloollowing cirwing cirwing cirwing cirwing circccccles:les:les:les:les:

1. x y x y2 2 8 4 2 0+ − + + =

2. 2 2 6 10 15 02 2x y x y+ − − − =

3. 3 3 6 12 2 02 2x y x y+ − − − = .

III.

1. If one end of a diameter of a circle x y x y2 2 4 6 14 0+ + − − = is (−5, −1), find the other end.

2. Find the equation of the tangent to the circle 2 2 3 12 02 2x y x y+ + − − = at (1, 3).

3. Find the equation of tangent at the point (−3, −5) on the circle x y x y2 2 4 6 8 0+ + + + = .

4. Find the value of k if 3x + 4y = k touches the circle x2 + y2 = 16. Also find the point of contact.

5. Find the equation of tangent to the circle x y x y2 2 2 6 15 0+ − + − = which are parallel to the line

3x + 4y − 8 = 0.

Circles 315

6. Find the equation of tangent to the circle x y x y2 2 2 4 1 0+ − − + = which are perpendicular to

the line 3x − 4y + 8 = 0.

7. Find the length of the chord 4x − 3y = 5 of the circle x y x y2 2 3 10 0+ + − − = .

8. Find the length of the tangent to the circle 3 3 7 6 12 02 2x y x y+ − − − = from (6, −7).

9. Find whether (1, −2) lie inside or on or outside the circle x y x y2 2 4 8 8 0+ − + − = .

10. Find the equation of the circle so that the lengths of the tangents from (−1, 2), (−2, 3) and (2, 4)

are respectively 2 3 30, and 17 units.

11. Find the equation of the circle which passes through (1, 1), (2, 2) and whose radius is unity.

12. Find the equation of circle which has centre on x-axis, passes through (1, 3) and whose radius is

18 units.

13. Find the equation of circle which makes an intercepts −8 and 6 on X and Y-axes and which passesthrough the origin.

14. Find the equation of circle passing through the point (1, −2), (4, −3) and whose centre lies on3x + 4y − 7 = 0.

15. Find the equation of circle concentric with x y x y2 2 6 8 16 0+ − + + = and touching the line

5 12 15 0x y− + = .

16. Find the equation of circle which touches the y-axis at (0, 9) and cuts x-axis at (−3, 0) and(−9, 0).

��#��

I. 1. x y2 2 9+ = 2. x y y2 2 4 0+ − =

3. x y x y2 2 6 8 0+ − + = 4. 9 9 24 42 70 02 2x y x y+ − − − =

5. x y x y2 2 8 8 16 0+ − − + = 6. x y x y2 2 5 5 0+ − − = .

7. x y x y2 2 6 8 9 0+ − − + = 8. x y x y2 2 10 2 1 0+ − + + =

9. x y x y2 2 8 8 16 0+ − − + = 10. x y x y2 2 6 4 0+ + − =

II. 1. C r≡ − =4 2 3 2, ,� � units. 2. C r≡ �� =3

units.2

5

24, ,

3. C r≡ =1, 2 units.� �, 17 3

III. 1. 1, 7� � 2. 5 9 32 0x y+ − =


3. x y+ + =2 13 0 4. k = − − ��

��

2012

5

16

5, ,

5. 3 4 34 0 2 4 16 0x y x y+ + = + − = and . 6. 4 3 20 0 4 3 0x y x y+ − = + =;

7. 5 units. 8. 9 units.

9. Inside 10. x y x y2 2 6 4 7 0+ − + − =

11. x y x y2 2 4 2 4 0+ − − + = 12. x y x2 2 4 14 0+ + − =

13. x y x y2 2 8 6 0+ + − = 14. 15 15 94 18 55 02 2x y x y+ − + + =

15. x y x y2 2 6 8 11 0+ − + − = 16. x y x y2 2 12 18 27 0+ + − + = .

��

��

��

When a solid cone is cut by a plane, the curves which lies on the surface of the cone and the plane,are the curves – circle, parabola, ellipse and hyperbola. These curves are called as conic sections.

��

Let l be a fixed line and S be fixed point. A point P movesin a plane containing S and l such that its distance from Sbears a constant ratio to its distance from the line l, i.e.,

SP

PM = a constant.

The locus of the point P is called a conic.

A conic is the locus of the point which moves such thatthe ratio of its distance from a fixed point in the plane to itsdistance from a fixed line in a plane is constant.

The fixed point is called focus and the fixed line is called directrix and the constant ratio SP

PM is

called eccentricity of the conic.

If the eccentricity of the conic is less than 1 then the conic is called ellipse.

If the eccentricity is greater than 1 then it is called hyperbola.

If the eccentricity is equal to one then the conic is called parabola.

Circle is regarded as the conic of eccentricity zero. It is a particular case of an ellipse.

��

Parabola is the locus of point which moves in a plane such that its distance from a fixed point is equalto its distance from a fixed line. The fixed point is called focus. The fixed line is called directrix.

S

PM

l

Fig. 14.1


�� ! �"� �� "� #�� !� $�

Let S be the focus, l be the directrix.

Draw SZ ⊥ l.

Let ‘O’ be the midpoint of SZ. Let SZ = 2a. So that OS = OZ = a.

Choose O as origin the 2 mutually perpendicular lines OX and OY as co-ordinate axes.

Let P (x, y) be any point on the parabola.

Draw PM ⊥r to l and PN ⊥r X-axis. Join PS.

O N S

M

x

y

P

zx′

y′

l

Fig. 14.1

From definitionSP

PM= 1

SP PM= ...(1)

Since S is on X-axis and OS = a,

Co-ordinates of S = (a, 0)

SP = Distance between S (a, 0) and P (x, y).

Distance formula: x x y y2 12

2 12− + −�

��

SP x a y x a y= − + − = − +� � � � � �2 2 2 20 ...(2)

Now

PM = NZ = ON + OZ

PM = x + a ...(3)

Substituting (2) and (3) in (1) we get

x a y x a− + = +� �2 2

Parabola 319

Squaring,

x a ax y x a ax2 2 2 2 22 2+ − + = + +

⇒ y ax ax2 2 2= +

⇒ y ax2 4= .

This is the standard equation of parabola.

�"� �� yyyyy22222 = 4 = 4 = 4 = 4 = 4ax.ax.ax.ax.ax.

For the parabola y2 = 4ax we have the following:

1. Shape:

• Since (0, 0) satisfy the equation y2 = 4ax, the curve passes through the origin.

• The equation y2 = 4ax remains unchanged if we replace y by −y. So the curve is symmetricabout x-axis.

• For negative values of x, we get imaginary values for y. So the curve entirely lies on the rightside of y-axis if a is +ve or a > 0. (and if ‘a’ is −ve, the curve entirely lies on the left side ofy-axis).

Hence the shape of the parabola is

S

L

l

L′

Ox

Fig. 14.2

2. The point O(0, 0) is called vertex of the parabola.

3. The line OX, the +ve x-axis is called the axis of the parabola.

4. S is the focus and its co-ordinate = (a, 0).

5. The line l is the directrix and its equation is x = −a.

6. The line LSL′ which is perpendicular to axis and passing through the focus is called Latus rectum.L and L′ are ends of latus rectum. L and L′ has x-co-ordinate ‘a’. To get y-co-ordinate, since L andL′ lie on the parabola y2 = 4ax it satisfies the equation.

∴ y ax2 4=


∴ y a a x a2 4= =� � �

y a y a2 2 24 4= ⇒ = ±

y a= ±2 .

∴ Ends of latus rectum = (a, ±2a).

Equation of latus rectum: x = a

Length of latus rectum = 2a + 2a = 4a.

�� !!� �� !� $# �! �� %�" &� ��' ()* )+�

1.1.1.1.1. The parabola The parabola The parabola The parabola The parabola yyyyy22222 = = = = = −−−−−44444axaxaxaxax.....

Shape:

S x

y

lL

x′

y′

L′

Fig. 14.3

Vertex: (0, 0)

Axis: Negative x-axis.

Co-ordinates of focus = (−a, 0)

Equation of directrix = x = +a

Ends of latus rectum = (−a, ±2a)

Equation of latus rectum: x = −a

Length of latus rectum: 4a.

2.2.2.2.2. The parabola The parabola The parabola The parabola The parabola xxxxx22222 = 4 = 4 = 4 = 4 = 4ayayayayay

Shape:

Vertex: (0, 0)

Axis: Positive y-axis.

Co-ordinates of focus = (0, a)

Equation of directrix = y = −aFig. 14.4

S

y

L

l

x

y′

L′

x′

Parabola 321

Ends of latus rectum = (±2a, a)

Equation of latus rectum = y = a

Length of latus rectum = 4a.

3.3.3.3.3. The parabola The parabola The parabola The parabola The parabola xxxxx22222 = = = = = −−−−−44444ayayayayay

Shape:

Vertex: (0, 0)

Axis: Negative y-axis

Co-ordinate of focus = (0, −a)

Equation of directrix = y = a

Ends of latus rectum = (±2a, −a)

Equation of latus rectum = y =−a

Length of latus rectum = 4a

��, �!!� �� !� $# �! �� %�" &� ��' (�* �+�

1.1.1.1.1. y k a x h− = −� � � �2 4

O

y

x

(h, k) s

Fig. 14.6

Let OX and OY be x and y-axis respectively change the origin to the point (h, k) without changingthe direction. Then the new co-ordinates (X, Y) are given by

x X h y Y k= + = + and

⇒ X x h Y y k= − = − and

∴ y k a x h− = −� � � �2 4

⇒ Y2 = 4aX

which is the equation of parabola, which opens to the right.

For the parabola y k a x h− = −� � � �2 4 we have the following

Vertex = (h, k)

Axis: y = k

Fig. 14.5

Sx

y

L

l

L′

x′

y′


Co-ordinates of focus = (a + h, k)

Equation of directrix = x = −a + h.

Ends of Latus rectum = (a + h, ±2a + k)

Equation of Latus rectum = x = a + h

Length of Latus rectum = 4a.

2.2.2.2.2. y k a x h− = − −� � � �2 4

Shape:

h, k

Fig. 14.7

Parabola which opens to the left in any quadrant depending on value of h and k.

Vertex: (h, k)

Axis: y = k

Co-ordinates of focus = (−a + h, k)

Equation of directrix: x = a + h

Ends of latus rectum = (− a + h, ±2a +k)

Equation of Latus rectum: x = −a + h

Length of Latus rectum = 4a

3.3.3.3.3. x h a y k− = −� � � �2 4

Shape:

(h, k)

Fig. 14.8

Parabola which opens upward in any quadrant depending on value of h and k.

Parabola 323

Vertex : (h, k)

Axis: x = h

Focus: (h, a + k)

Ends of latus rectum = (±2a + h, a + k)

Equation of latus rectum = y = a + k


4.4.4.4.4. x h a y k− = − −� � � �2 4

Shape:

(h, k)

Fig. 14.8

Parabola which open downward is any quadrant depending on value of h and k.

Vertex : (h, k)

Axis : x = h

Focus = (h, −a + k)

Equation of directrix: y = a + k

Ends of Latus rectum: (±2a + h, −a + k)

Equation of latus rectum: y = −a + k

Length of latus rectum: 4a.

Note:Note:Note:Note:Note: In any parabola, the focus is inside the curve (i.e. on the axis of the parabola) and directrix isaway from the curve (i.e. directrix never meet the parabola).

2. The distance between the vertex and the focus is equal to the distance between the vertex andthe directrix is equal to a.

3. The distance of a point on the parabola from the focus is called the focal distance of the point.

4. A chord drawn through the focus is called focal chord and the focal chord perpendicular to theaxis is Latus rectum.

%� -�� '�$��#�

!��. �/� 0��1* �1��* ��23�* �43�� .��2��1* ��.� �� 3� ��2�35* ��6�/ �� 3��2�35* �43�� 3� ��2�35 �� /� ��7��6 8��

1.1.1.1.1. x2 = 8y


Comparing with x2 = 4ay we get

4a = 8 ⇒ a = 2.

Vertex = (0, 0)

Axis: y-axis or x = 0.

Focus: (0, a) = (0, 2)

Equation of directrix: y = −a

y = −2

Ends of latus rectum = (±2a, a)

= (±2(2), 2)

= (±4, 2).


= 4 (2) = 8.

Equation of latus rectum = y = a

y = 2.

2.2.2.2.2. y2 = −6x.

Comparing with y2 = −4ax we get

4 6a =

a = =6

4

3

2.

Vertex: (0, 0)

Axis: x-axis or y = 0

Focus = (−a, 0) = −��

3

20, .

Equation of directrix = x = a

x = 3

2.

Ends of Latus rectum: (−a, ±2a)

= − ± ��

��

��

3

22

3

2,

= − ±��

��

3

23,

Equation of latus rectum x = −a

x = − 3

2.

Parabola 325


= �� =4

3

26.

3.3.3.3.3. x2 = −y

Comparing with x2 = −4ay we get

4 11

4a a= ⇒ =

Vertex = (0, 0)

Axis: x-axis or x = 0.

Focus = (0, −a) = 01

4, −��

��

Equation of directrix : y = a

y = 1

4.

Ends of latus rectum = ± −2a a,� �

= ± �� −�

�� = ± −��

��2

1

4

1

4

1

2

1

4, ,

Equation of latus rectum: y = −a

y = − 1

4


= 41

41�

�� = .

4.4.4.4.4. y2 = 4 (x + 1)

⇒ y x− = − −0 4 12� � � �� Comparing with (y − k)2 = 4a (x − h)

We get (h, k) = (−1, 0)

4a = 4 ⇒ a = 1

Vertex = (h, k) = (−1, 0)

Axis: y = k

y = 0.

Focus: (a + h, k) = (1 + (−1), 0)

= (0, 0).


Equation of directrix x = −a + h

x = −1 −1

x = −2

Ends of latus rectum = (a + h, ±2a + k)

= (1 + (−1), ±2 (1) + 0)

= (0, ±2).

Equation of latus rectum = x = a + h

x = 1 + (−1)

x = 0.


= 4 (1) = 4.

5.5.5.5.5. y x+ = +2 3 12� �

y x+ = +��2 3

1

32� �

y x− − = − −��

��

��

2 31

32� �

Comparing with (y − k)2 = 4a (x − h)

We get

h k, ,� � = − −��

��

1

32

4 33

4a a= ⇒ =

Vertex = h k, ,� � = − −��

��

1

32

Axis: y = k, y = −2

Focus: (a + h, k)

= − −��

��

3

4

1

32,

= − −��

�� = −��

��

9 4

122

5

122, ,

Equation of directrix: x = −a + h

x = − −3

4

1

3.

Parabola 327

x = − 13

12.

Ends of latus rectum = (a + h, ±2a + k)

= − ± �� −�

��

3

4

1

32

3

42,

5

12

3

22

5

12

7

2

5

12

1

2, , , .± −�

�� = −��

�� −�

�� and

Equation of Latus rectum x = a + h

x = − =3

4

1

3

5

12.

Length of latus rectum = 4 43

43a = �

�� = .

6.6.6.6.6. x y+ = − −3 24 12� � � �

⇒ − − = − −x y3 24 12� ��

Comparing with (x − h)2 = −4a (y − k)

We get (h, k) = (−3, 1)

4 24 6a a= ⇒ = .

Vertex = (h, k) = (−3, 1)

Axis: x = h, x = −3.

Focus = (h, −a + k) = (−3, −6 + 1)

= (−3, −5)


y = 6 + 1 = 7

Ends of latus rectum = (±2a + h, −a + k)

= ± + − − +2 6 3 6 1� � � �� ,

= ± − −12 3 5,� �

= − − −15 5, .� � � � and 9, 5

Equation of latus rectum: y = −a + k

y = −6 + 1 = −5


= 4 (6) = 24.


7.7.7.7.7. x x y2 8 12 4 0+ + + =

Consider

x x y

x x y

2

2 2 2

8 12 4 0

8 4 4 12 4 0

+ + + =

+ + − + + =

Co - efficient of

Square it, Add & Subtract

x →

= →

�

�

�

�

��

8

8

24

x y+ − + + =4 16 12 4 02� �Comparing with (x − h)2 = −4a (y − k)

We get

h k, ,� � � �= −4 1

4 12 3a a= ⇒ = .

Vertex: h k, ,� � � �= −4 1

Axis: x h x= = −, .4

Focus: h a k, , ,− + = − − + = − −� � � � � �4 3 1 4 2


y = 3 + 1 = 4

Ends of latus rectum: (±2a + h, −a + k)

= ± + − − +2 3 4 3 1� � � �� ,

= ± − −6 4 2,� �

= − − −10 2 2 2, , .� � � � &

Equation of latus rectum, y = −a + k

y = −3 + 1

y = −2


= 4 (3) = 12.

8.8.8.8.8. y y x2 4 6 13 0− − + =

Consider

y y x2 4 6 13 0− − + =

y y x2 2 24 2 2 6 13 0− + − − + =

y x− − − + =2 4 6 13 02� �

Parabola 329

y x− − + =2 6 9 02� �

y x− = −2 6 92� � .

y x− = −��2 6

9

62� �

y x− = −��2 6

3

22� �

Comparing this with (y − k)2 = 4a (x − h)

We get h k, ,� � = ��

3

22

4 66

4

3

2a a= ⇒ = =

Vertex: h k, ,� � = ��

3

22

Axis: y k y= =, .2

Focus: a h k+ = +��

��, ,� � 3

2

3

22

= �� =6

22 3 2, , .� �

Equation of directrix: x = −a + h

x = − +3

2

3

2

x = 0.

Ends of Latus rectum = (a + h, ±2a + k)

= + ± ⋅ +��

��

3

2

3

22

3

22,

= ± + = −3 3 2 3 5 3 1, , & , .� � � � � �

Equation of latus rectum: x a h= +

x = + = =3

2

3

2

6

23.


Length of latus rectum: 4a

= �� =4

3

26.

9.9.9.9.9. 3 6 8 5 02y y x+ + − = .

Consider

3 6 8 5 02y y x+ + − = .

3 2 8 5 02y y x+ + − =� �

3 2 1 1 8 5 02 2 2y y x+ + − + − =

3 1 1 8 5 02y x+ − + − =� �

3 1 3 8 5 02y x+ − + − =� �

3 1 8 8 02y x+ + − =� �

3 1 8 82y x+ = − +� �

3 1 8 12y x+ = − −� � � �

y x+ = − −18

312� � � �

y x− − = − −18

31

2� ��

Comparing with (y − k)2 = −4a (x − h) we get (h, k) = (1, −1), 48

3

2

3a a= ⇒ =

Vertex = (h, k) = (1, −1)

Axis: y = k, y = −1

Focus: (−a + h, k) = − + −��

��

2

31 1,

= −��

��

1

31,

Equation of directrix : x = a + h

x = +2

31

x = 5

3.

Parabola 331

Ends of latus rectum:

− + ± +a h a k, 2� �

= − + ± �� −�

��

2

31 2

2

31,

= ± −��

�� = −��

��

��

��

1

3

4

31

1

3

7

3

1

3

1

3, , , . and

Equation of latus rectum

x = −a + h,

x = − + =2

31

1

3.

Length of latus rectum: 42

3

8

3�� = .

10.10.10.10.10. 2 5 3 4 02x x y− + + =

Consider 2 5 3 4 02x x y− + + =

25

23 4 02x x y−�

�� + + =

25

2

5

4

5

43 4 02

2 2

x x y− + �� − ��

��

�

�

�

��

+ + =

25

4

25

163 4 0

2

x y−�� −

�

�

�

��

+ + =

25

4

25

83 4 0

2

x y−�� − + + =

25

43

7

80

2

x y−�� + + =

25

43

7

8

2

x y−��

��

= − −

25

43

7

24

2

x y−��

��

= − +��

��


x y−�� = − − −��

��

��

��

5

4

3

2

7

24

2

Comparing with (x − h)2 = −4a (y − k) we get

h k a a, , , .� � = −��

�� = =5

4

7

244

3

2

3

8 and

Vertex = (h, k) = 5

4

7

24, −�

��

Axis: x = h

x = 5

4

Focus: (h, −a + k)

= − −��

��

5

4

3

8

7

24,

= − −��

��

5

4

9 7

24,

= −��

�� = −��

��

5

4

16

24

5

4

2

3, , .

Equation of directrix:

y = a + k

y = −3

8

7

24

y = − = =9 7

24

2

24

1

12.

Ends of latus rectum

± + − + = ± �� + − −�

��2 2

3

8

5

4

3

8

7

24a h a k, ,� �

= ± + −��

��

3

4

5

4

16

24, .

22

3

1

2

2

3, , .−��

�� −�

�� and

Parabola 333

Equation of latus rectum,

y a h= − + = − −3

8

7

24

y = − 2

3.


= �� =4

3

8

3

2.

!��. �/� �43�� 6�0�� /��

1.1.1.1.1. Vertex: (0, 0), Focus: (5, 0)

Solution: Solution: Solution: Solution: Solution: Since the focus lies inside the parabolaand vertex is given as origin. The parabola opens tothe right.

∴ Equation of the parabola with vertex (0, 0) andwhich opens to right = y2 = 4ax

Co-ordinates of focus = (a, 0) = (5, 0) [given]

⇒ a = 5.

Equation of the parabola = y2 = 4 (5) x

y2 = 20x

2.2.2.2.2. Vertex (0, 0), Focus (0, 4)

Solution:Solution:Solution:Solution:Solution: Since focus = (0, 4) and vertex: (0, 0) andfocus lies inside the parabola, the parabola opensupward.

Equation of parabola: x2 = 4ay

Given focus = (0, 4) = (0, a)

⇒ a = 4

∴ Equation of parabola: x2 = 4 (4) y

x2 = 16y

3.3.3.3.3. Vertex: (0, 0), Axis = Negative Y-axis andlength of latus rectum = 5.

Since axis = Negative Y-axis, and vertex = (0, 0)parabola opens downward as shown in fig.

Equation of parabola: x2 = −4ay

Given length of latus rectum = 5

4a = 5

Fig. 14.9

Fig. 14.10

Fig. 14.11

(5, 0)

(0, 4)


∴ Equation of required parabola.

x2 = −5y.

4.4.4.4.4. Vertex (0, 0), Directrix x = 4.

Since directrix x = 4, and directrix is away from the pa-rabola and vertex = (0, 0), Parabola opens to the left.

Equation of parabola: y2 = −4ax.

For y2 = −4ax parabola, Equation of directrix is

x = a.

a = 4 (� x = 4 is given)

∴ Equation of the parabola: y x y x2 24 4 16= − ⇒ = −� �5.5.5.5.5. Vertex (1, −2), Focus (−1, −1)

Since vertex = (1, −2) and focus = (1, −1) and focus lieinside the parabola, the parabola open upwards. Equation of

parabola = x h a y k− = −� � � �2 4

Given vertex (1, −2)

∴ (h, k) = (1, −2)

⇒ h k= = −1 2, .

Focus = (h, a + k) = (1, −1)

∴ a + k = −1

a + (−2) = −1

a = −1 + 2

a = 1

⇒ a = +1.

∴ Equation of parabola

x y− = − −1 4 1 22� � � � � ��

x y− = +1 4 22� � � �

x x y2 1 2 4 8+ − = +

x x y2 2 4 7 0− − − = .

��

Given: Vertex = (1, −2) = (h, k)

⇒ = = −h k1 2,

Fig. 14.12

Fig. 14.13

S

x

Y

0

Y′

x′

V

x = 4

Y�

x

Parabola 335

We know,

Distance between vertex and focus = a

i.e. distance between (1, −2) and (1, −1) = a

1 1 1 22 2− + − + =� � � � a

1 12 = ⇒ =a a .

∴ Equation of parabola

x h a y k− = −� � � �2 4

x y− = +1 4 1 22� � � ��

⇒ x x y2 2 4 7 0− − − = .

6.6.6.6.6. Focus (1, 1), directrix x + 8 = 0.

Let S (1, 1) be the focus and l: x + 8 = 0 be the directrix.

Let P (x, y) be any point on the parabola. Join SP, drawPM ⊥ l. We know from definition,

SP

PM= 1

SP PM=SP = Distance between S (1, 1) and P (x, y)

= − + −x y1 12 2� � � �[Using distance formula]

PM = perpendicular distance of P (x, y) from x + 8 = 0

= +

+

x 8

1 02

∴ x yx− + − = +

1 18

12 2� � � �

Squaring,

x yx

− + − =+

1 18

12 2

2

� � � � � �

x x y y x x2 2 21 2 1 2 64 16+ − + + − = + +

y y x2 2 18 62 0− − − = .

Fig. 14.14

S

PM

�


7.7.7.7.7. Focus (4, 0) and directrix x = 6.

Let P (x, y) be any point on the parabola whose focus is S (4, 0) and directrix is x − 6 = 0. Join, SP anddraw PM ⊥ l.

From definition

SP

PM= 1

SP PM=

SP S P x y= Distance between and 4 0, ,� � � �

= − + − = − +x y x y4 0 42 2 2 2� � � � � �

PM = perpendicular distance of P (x, y) from x − 6 = 0.

= − = − = −x xx

6

1

6

16

2.

∴ x y x− + = −4 62 2� �Squaring,

x y x− + = −4 62 2 2� � � �

x x y x x2 2 216 8 36 12+ − + = + −

y x2 4 20 0+ − = .

8.8.8.8.8. Ends of latus rectum (−5, 2) and (−3, 2)

SLL′

( 5, 2)−

x

Y′

Y

x′

( 3, 2)−

Fig. 14.16

Since the ends of latus rectum are (−5, 2) and (−3, 2). There are 2 parabolas, open upwards and opendownwards.

So equations are

x h a y k− = ± −� � � �2 4 .

Fig. 14.15

S

PM

l

(x, y)

(4, 0)x 6 = 0−

Parabola 337

To find h, k and a:

Length of latus rectum = 4a i.e., Distance between (−3, 2) and (−5, 2) = 4a.

− − − + − =5 3 2 2 42 2� �� a

− + + =5 3 0 42� � a

− =2 42� � a

4 4= a

a = 1

2.

Focus = Mid point of latus rectum

Focus = mid point of (−3, 2) and (−5, 2)

Focus =− + − +��

��

3 5

2

2 2

2

� �,

= −�� = −8

2

4

24 2, , .� �

But Focus = (h, ± a + k)

∴ (h, ± a + h) = (−4, 2)

⇒ h a k= − ± + =4 2,

± + =1

22k

k = ±21

2.

k k= =5

2

3

2 or

(for −ve a) (for +ve a)

∴ Equations of parabolas:

(i) x y− − = �� −��

��4 4

1

2

3

22� �

x y+ = −��4 2

3

22� �

x x y2 16 8 2 3+ + = −

x x y2 8 2 19 0+ − + = .


(ii) x y− − = − �� −��

��4 4

1

2

5

22� �

x y+ = − −��4 2

5

22� �

x y+ = − +4 2 52� �

x x y2 8 16 2 5 0+ + + − =

x x y2 8 2 11 0+ + + =

9.9.9.9.9. Vertex: (3, 4) Directrix y = 1

Given: Vertex = (3, 4)

∴ (h, k) = (3, 4) ⇒ h = 3 and k = 4.

Since directrix is away from the parabola and per-pendicular to axis, the parabola opens upwards.

Equation of parabola is (x − h)2 = 4a (y − k)

Equation of directrix: y = −a + k

� y = 1 is directrix.

1 = − +a k

1 4= − +a

a = −4 1

a = 3.

∴ Equation of parabola: x y− = −3 4 3 42� � � �.

x x y2 9 6 12 48+ − = −

x x y2 6 12 57 0− − + = .

10.10.10.10.10. Directrix: x + 3 = 0, axis y = 2 and length of latus rectum: 6.

Given: Axis y = 2, and directrix x + 3 = 0, x = −3.

From figure,

Since directrix is away from parabola, the parabolaopens to the right.

∴ Equation of parabola.

y k a x h− = −� � � �2 4

Axis: y k y= =, .� 2

∴ k = 2.

Fig. 14.17

Fig. 14.18

(3, 4)y = 1

laxisy = 2

x = 3

Parabola 339


4 6a = (Given)

a = 6

4

a = 3

2.

Directrix: x = −a + h

x h= − +3

2

− = − +33

2h

− + =33

2h

− + =6 3

2h

h = − 3

2.

∴ Equation of parabola:

y x− = �� − −��

��

��

��2 4

3

2

3

22� �

y x− = +��2 6

3

22� �

y y x2 4 4 6 9+ − = +

y y x2 4 6 5 0− − − =

340B

asic Mathem

atics �$�$��

Equation Figure Vertex Focus Equation Axis Ends Equation Lengthof of of of

directrix Latus rectum Latus rectum Latus rectum

y2 = 4ax (0, 0) (a, 0) x = −a x-axis (a, ±2a) x = a 4ay = 0

y2 = −4ax (0, 0) (−a, 0) x = a x-axis (−a, ±2a) x = −a 4ay = 0

x2 = 4ay (0, 0) (0, a) y = −a y-axis (± 2a, a) y = a 4ax = 0

x2 = −4ay (0, 0) (0, −a) y = a y-axis (± 2a, −a) y = −a 4ax = 0

(y − k)2 = 4a (x − h) (h, k) (a + h, k) x = −a + h y = k (a + h, ±2a + k) x = a + h 4a

(y − k)2 = −4a (x − h) (h, k) (−a + h, k) x = a + h y = k (−a + h, ±2a + k) x = −a + h 4a

(x − h)2 = 4a (y − k) (h, k) (h, a + k) y = −a + k x = h (± 2a + h, a + k) y = a + k 4a

(x − h)2 = −4a (y − k) (h, k) (h, −a + k) y = a + k x = h (± 2a + h, −a + k) y = −a + k 4a

(h, k)

(h, k)

(h, k)

(h, k)

Parabola 341

• In any parabola, focus is inside the curve and directrix is away from the parabola.

• Distance between vertex and focus = a.

• For the given ends of latus rectum, there are 2 possible parabolas.

• Focus is the mid point of latus rectum.

• Axis is perpendicular to the directrix. Distance between directrix and vertex = a.

�'� �#�

I. FFFFFind the vind the vind the vind the vind the vererererertetetetetex,x,x,x,x, F F F F Focus,ocus,ocus,ocus,ocus, equa equa equa equa equation of dirtion of dirtion of dirtion of dirtion of directrectrectrectrectrix,ix,ix,ix,ix, axis, axis, axis, axis, axis, length of la length of la length of la length of la length of latus rtus rtus rtus rtus rectum,ectum,ectum,ectum,ectum, ends of la ends of la ends of la ends of la ends of latustustustustusrrrrrectum and equaectum and equaectum and equaectum and equaectum and equation of lation of lation of lation of lation of latus rtus rtus rtus rtus rectum of the fectum of the fectum of the fectum of the fectum of the folloolloolloolloollowing parwing parwing parwing parwing paraaaaabolas:bolas:bolas:bolas:bolas:

1. y x2 4= 2. y x2 8= −

3. x y2 3= 4. x y2 7= −

5. y x− =1 42� � 6. x y+ = +1 6 12� � � �

7. x y2 12 7= − −� � 8. y x+ = − −3 8 72� � � �

9. y x y2 20 10 15 0− − + = 10. y y x2 4 6 8 0− + − =

11. 2 5 3 4 02y y x− + + = 12. x x y2 4 10 8 0+ + + =

13. x y x2 4 2 3 0+ − + = 14. x x y2 8 16 0− + =

15. x x y2 4 5 1 0− − − =

II. FFFFFind the equaind the equaind the equaind the equaind the equation of partion of partion of partion of partion of paraaaaabola gbola gbola gbola gbola giiiiivvvvven thaen thaen thaen thaen that:t:t:t:t:

1. Vertex: (0, 0), Focus (−3, 0)

2. Vertex: (0, 0), Directrix : y = 8

3. Vertex : (0, 0), Length of Latus rectum = 3, axis positive x-axis.

4. Vertex: (0, 0), Equation of latus rectum x = 4.

5. Vertex : (0, 0), Axis x-axis and passing through (3, −1)

6. Focus (0, 5) and directrix y = −4

7. Focus (4, 0), Directrix x = 6

8. Vertex: (1, 1), Focus (4, 1)

9. Ends of latus rectum (−3, 1) and (1, 1)

10. Directrix x + 2 = 0, axis y = 3 and length of latus rectum = 6.

11. Vertex: (3, 4), Directrix y = 1

12. Vertex: (3, −3), Directrix y = −7 and axis parallel to y-axis.

13. Directrix 4x − 1 = 0, Axis y = 2 and length of latus rectum 8.

14. Ends of latus rectum (8, 4) and (8, −2)

15. Directrix: 4x + 1 = 0, axis 2y + 1 = 0 and length of latus rectum 24.


��#%� #

I.

Question Vertex Focus Directrix Axis Length of Ends of Equation of No. Latus rectum Latus rectum Latus rectum

1 (0, 0) (1, 0) x = −1 x-axis y = 0 4 (1, ± 2) x = 1

2 (0, 0) (−2, 0) x = 2 x-axis y = 0 8 (−2, ± 4) x = −2

3 (0, 0) (0, 3/4) y = −3/4 y-axis x = 0 3 (±3/2, 3/4) y = 3/4

4 (0, 0) (0, −7/4) y = 7/4 y-axis x = 0 7 (±7/2, −7/4) y = −7/4

5 (0, 1) (1, 1) x = −1 y = 1 4 (1, ±2 + 1) x = 1

6 (−1, −1) (−1, 1/2) y = −5/2 x = −1 6 (−1±3, 1/2) y = 1/2

7 (0, 7) (0, 10) y = 4 x = 0 12 (±8, 10)

8 (7, −3) (5, −3) x = −5 y = −3 8 (5, ±4 −3) x = 5

9 (1/2, 5) (9/2, 5) x = −11/2 y = 5 20 (9/2, 25) x = 9/2(9/2, −15)

10 (2, 2) (1/2, 2) x = 7/2 y =2 6 (1/2, 5) x = 1/2(1/2, −1)

11 (−7/24, 5/4) (−2/3, 5/4) x = 1/12 y = 5/4 3/2 (−2/3, 2) x = −2/3(−2/3, 1/2)

12 (2, −2/5) (−2, −29/10) y = −21/10 x = 2 10 (3, −29/10) y = −29/10(−7, 29/10)

13 (1, −1/2) (1, −3/2) y = −1/2 x = 1 4 (3, −3/2) y = −3/2(−1, −3/2)

14 (4, 1) (4, −3) y = 5 x = 4 16 (12, 1) y = 1(−4, 1)

15 (2, −1) (2, 1/4) y = −9/4 x = 2 5 (9/2, 1/4) y = 1/4(−1/2, 1/4)

1. y x2 12= − 2. x y2 24= − 3. y x2 3=

4. y x2 16= 5. y x2 1

3= 6. x y2 18

1

2= −��

��

7. y x2 4 5= − −� � 8. y x− = −1 12 12� � � � 9. x y+ = − −1 4 22� � � �

10. y x− = +��3 6

1

22� � 11. x y− = −3 12 42� � � � 12. x y− = +3 16 32� � � �

13. y x− = +��2 8

7

42� � 14. y x− = −��

��1 6

13

22� � 15. y x− = +��

��2 8

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15

Limits and Continuity

15.1 INTRODUCTION:The discovery of calculus was done independently and almost during the same time by Sir IsaacNewton and Gottfried Wilhelm Leibnitz. Calculus is developed on the basis of a more fundamentalconcept called the Limit.

15.2 CONSTANTS AND VARIABLES:A quantity which remains the same throughout any mathematical discussion is called a constant.

A quantity which takes different values in any mathematical discussion is called a variable. Forinstance, if we increase the production by using more raw materials, the cost of the machine doesn’tchange, but the costs of raw materials, Labour sales change. So cost of the machine which remains thesame value throughout is constant and cost of raw materials and labour which assumes any numericalvalue out of given set of values are variables.

Constants are represented by a, b, c, d, e and variables are represented by u, v, w, x, y, z.

15.3 FUNCTION:Let A and B be 2 non empty sets. A rule f which associates each element x of A to an unique elementy of B is called a function. It is denoted by f : A → B. If x ∈ A is related to y ∈ B, then we write y = f (x).

Algebraic function:

A function y = f(x) is said to be an algebraic function if f(x) is a polynomial function (eg: x2 + 3x − 1)

or rational function 3 8

4 1

2

3x

x

+−

FHG

IKJ or irrational function (e.g.: (1 − x)2/3).

Note: The functions other than algebraic functions are called transcendental functions. They include,trigonometric functions, logarithmic function, inverse trigonometric functions, exponential func-tions, hyperbolic functions etc.


Exponential function:

The function which associates every real number x to the real number ex is called exponential function.Where e is the sum of the infinite series

111

12

13!

14

+ + + + +! ! !

...

Its value is 2.7182818284.... e is an irrational number.

Logarithmic function:

The function which associates every positive real number x to the real number loge x is called thelogarithmic function.

15.4 LIMITS:Consider the function

yxx

= −−

2 11

when x = 1, y = −−

=1 11 1

00

2

which is not defined.

Instead of giving x = 1, Let us give, x a value which is slightly less than 1 or slightly greater than 1.Then

when x = 0.9, y = 1.9

x = 0.99, y = 1.99

x = 0.999, y = 1.999

when x = 1.0001, y = 2.0001

x = 1.001, y = 2.001

x = 1.01, y = 2.01

x = 1.1, y = 2.1

From the above set of values, we can observe that the value of y is slightly less than 2 or slightlymore than 2 when the value of x is slightly less than 1 or slightly more than 1. But when x = 1, y hasno value. In otherwords, when x is very nearly equal to 1, y is very nearly equal to 2. In symbols this

is expressed as limx

xx→

−−

=1

2 11

2

i.e. limx

y→

=1

2

Read it as the limit of y as x tends to 1 is 2.

The statement simply imply when x ≈ 1 y ≈ 2 even though it doesn’t exist when x = 1.

Limits and Continuity 345

Definition of Limit:

A function f(x) is said to tend to a limit l as x tends to a if the numerical difference between f(x) andl can be made as small as we please by taking the numerical difference between x and a as very small.

In other words f(x) is very nearly equal to l when x is very nearly equal to a.

It is denoted by limx a

f x l→

=a f or

ltx a

f x l→

=a f .

Left Hand and Right Hand Limits:

When x → a, through the values which are smaller than ‘a’ then we write x → a − 0 this

lim or limx a h

f x f a h→ − →

−0 0a f a f is called Left Hand Limit [LHL] of a function f (x).

For example: if x → 3 by taking the values 2.9, 2.99, 2.999, 2.9999... then x → 3 − 0 or Simplyx → 3−

When x → a, through the values which are greater than ‘a’ then we write x → a + 0 this limx a

f x→ +0a f

or limh

f a h→

+0a f is called Right Hand Limit [RHL] of a function f(x).

For example: if x → 3 by taking the values 3.1, 3.01, 3.001... then x → 3 + 0 or Simply x → 3+

15.5 STANDARD LIMITS:

1. Prove that limx a

n nnx a

x ana

→−−

−= 1, where n is any rational number.

Proof. Case (i): Let n be a positive integer. We know

x a x a x x a x a an n n n n n− = − + ⋅ + ⋅ + +− − − −a fd i1 2 3 2 1...

lim lim...

x a

n n

x a

n n n nx a

x a

x a x x a x a a

x a→ →

− − − −−−

=− + ⋅ + ⋅ + +

−

a fd i1 2 3 2 1

= + ⋅ + ⋅ + +→

− − − −lim ...x a

n n n nx x a x a a1 2 3 2 1(n terms)

= + ⋅ + ⋅ + +− − − −a a a a a an n n n1 2 3 2 1...

= a a an n n− − −+ + +1 1 1 ... (n terms)

limx a

n nnx a

x ana

→−−

−= 1

Case (ii): Let n be a negative integer, say n = −m, where m is a +ve integer.


Consider

lim limx a

n n

x a

m mx ax a

x ax a→ →

− −−−

= −−

=−

−→limx a

m mx ax a

1 1As a

am

m− = 1

=

−

−→limx a

m m

m ma x

x ax a

= −⋅ −→

limx a

m m

m m

a x

x a x aa f

= − −−

LNM

OQP

⋅⋅→

limx a

m m

m m

x ax a x a

1 Using case (i) as is + ve integermLNM

OQP

limx a

n nm

m m

x ax a

maa a→

−−−

= −⋅

1 1

= −−ma

a

m

m

1

2

= − − −mam m1 2

= − − −ma m 1

∵ n m= −

limx a

n nnx a

x ana

→−−

−= 1

Case (iii) Let be a fraction, positive or negative.

Take np

q= where p is an integer and q ≠ 0.

Now

lim limx a

n n

x a

p q p qx ax a

x ax a→ →

−−

= −−

Taking x = yq and a = bq

⇒ x1/q = y and a1/q = b


So that

x a y bq q→ ⇒ →

y b→ .

lim limx a

p q p q

y b

p p

q q

x a

x a

y b

y b→ →

−−

= −−

Dividing both numerator and denominator by y − b

limx a

p p

q q

p

q

y by b

y by b

pb

qb→

−

−

−−−−

=1

1 ∵ limx a

n nnx a

x ana

=

−−→

=LNM

OQP

1

p

qbP q⋅ − − +1 1

p

qb p q⋅ −

But b a q=1

limx a

p q p qq p qx a

x a

p

qa

→

−−−

= ⋅ 1d i

limx a

p q p q pqx a

x a

p

qa

→

−−−

=1

∴ limx a

n nnx a

x ana

→−−

−= 1 for all rational values of n.

Some Standard Limits:

Here we state some standard limits (without proof) and use them while solving problems.

1. (a) limn

n

ne

→∞+FHIK =1

1(b) lim

nnn e

→+ =

0

11a f

2. limn

nkk

ne

→∞+FHIK =1

3. limx

xex→

− =0

11


4. lim log .x

x

ea

xa a

→

− = >0

10 where

Algebra of Limits:

If f(x) and g(x) are 2 functions of x and k is any scalar

Then

1. lim lim limx a x a x a

f x g x f x g x→ → →

± = ±a f a f a f a f

2. lim limx a x a

k f x k f x→ →

⋅ = ⋅a f a f

3. lim lim limx a x a x a

f x g x f x g x→ → →

⋅ = ⋅a f a f a f a f

4. limlim

limlim

x a

x a

x ax a

f g

g x

f x

g xg x

→

→

→→

= ≠a fa f

a fa f a f provided 0 .

Indeterminate Forms:

In mathematics the forms like 00

0 10, , , ...∞∞

∞ are called indeterminate forms.

Ex: 1. If yxx

= −−

2 11

then the value of y = 00

when x = 1 ∵ y = −−

=1 11 1

00

2

2. If y n n= +11a f then value of y = +1 0

10a f y = 1∞ when n = 0.

Evaluation of Limits:

To evaluate limx a

f x→a f , first find f(a).

(i) If f (a) ≠ an indeterminate form then f (a) itself is the limit.

(ii) If f (a) = an indeterminate form then suitable method / formula is applied to reduce f (x) so thatit will not take an indeterminate form when x is replaced by a.

WORKED EXAMPLES:

1. Evaluate: limx

x x

x x→

+ +− −0

2

24 3 1

3 4 1

Solution: By Putting x = 0. We get

limx

x x

x x→

+ +− −

=+ +− −

=−

= − ≠0

2

24 3 1

3 4 1

4 0 3 0 1

3 0 4 0 111

1a f a fa f a f an intermediate.


Hence limx

x x

x x→

+ +− −

= −0

2

24 3 1

3 4 11

2. Evaluate: limx

x x

x x→

− +− +1

2

25 4

4 3

Solution: By putting x = 1 we get 1 5 1 4

1 4 1 300

2

2

− +− +

= =a fa f an indeterminate

Hence consider

limx

x x

x x→

− +− +1

2

25 4

4 3 by factorising both numerator and denominator we get

lim limx x

x x x

x x x

x x x

x x x→ →

− − +− − +

=− − −− − −1

2

2 1

4 1 4

3 1 3

4 1 4

3 1 3a f a fa f a f

= lim limx x

x x

x x

x

x→ →

− −− −

= −−1 1

1 4

1 3

4

3a fa fa fa f

Now substituting x = 1 or applying limx→1

we get

1 41 3

32

32

−−

= −−

= .

∴ lim .x

x x

x x→

− +− +

=1

2

25 4

4 3

32

3. Evaluate: limx

x xx→

+ − −0

3 3

Solution: By putting x = 0 we get

3 0 3 0 3 3 00

+ − − = − = =x 0

an indeterminate

Consider

limx

x xx→

+ − −0

3 3 By rationalising the numerator,

limx

x x

x

x x

x x→

+ − − × + + −+ + −0

3 3 3 3

3 3

limx

x x

x x x→

+ − −

+ + −0

2 23 3

3 3

d i d id i


= + − ++ + −→

limx

x x

x x x0

3 3

3 3d i

=+ + −→

limx

x

x x x0

2

3 3d i

=+ + −→

limx x x0

23 3

Now applying limx→0

,

2

3 0 3 0

2

3 3+ + −=

+

= =2

2 3

1

3

∴ lim ,x

x x

x→

+ − − =0

3 3 1

3

4. Evaluate: limx

xx→

−−3

5 2433

Solution: lim limx x

xx

xx→ →

−−

= −−3

5

3

5 52433

33

Formula: limx a

n nnx a

x ana

→

−−−

=LNM

OQP

1

= ⋅ = ⋅ = =−5 3 5 3 5 81 4055 1 4 a f

∴ limx

xx→

−−

=3

5 2433

405

5. Evaluate: limx

x

x→

−−5

3 3 5

5

Consider

limx

x

x→

−−5

1 3 1 3

1 2 1 25

5

Dividing both numerator and denominator by x − 5.

limlim

limx

x

x

xx

xx

xx

xx

→

→

→

−−−−

=

−−−−

5

1 3 1 3

1 2 1 25

1 3 1 3

5

1 2 1 2

5555

5555

[using algebra of limits]


Now applying the formula,

limx a

n nnx a

x ana

→−−

−= 1 for numerator and denominator separately we get

13

5

12

5

13

5

12

5

13

1

12

1

23

12

−

−

−

−=

=− +2

35

23

12

23

523

52

3 5

4 36

16

6

− + −= =

⋅

∴ lim .x

x

x→

−−

=⋅5

3 3

6

5

5

2

3 5

6. Evaluate: limx

x

x→−

+−2

5

8 832

2

lim limx x

x

x

x

x→− →−

+−

=− −− −2

5

8 8 2

5 5

8 8

322

2

2

a fa f

Dividing both Numerator and Denominator by ∵ − = −2 325a fx − (−2) we get − = +2 28 8a f

limlim

limx

x

x

xx

xx

xx

xx

→−

→−

→−

− −− −− −− −

=

− −− −− −− −

2

5 5

8 8

2

5 5

2

8 8

22

22

22

22

a fa fa fa f

a fa fa fa f

=−−

=−−

−

−5 2

8 2

5 2

8 2

5 1

8 1

4

7

a fa f

a fa f

= ⋅ − = ⋅ −− −58

258

24 7 3a f a f

58

18

564

⋅−

=−a f


7. Evaluate: limn

n n

n n→∞

+ ++ −

2

21

3 2 1.

Solution: limn

n n

n n→∞

+ ++ −

= ∞∞

=2

21

3 2 1an indeterminate .

Consider

lim limn n

n n

n n

nn n

nn n

→∞ →∞

+ ++ −

=+ +FH

IK

+ −FH

IK

2

2

22

22

13 2 1

11 1

32 1

=+ +

+ −→∞limn

n n

n n

11 1

32 1

2

2

Now applying the limn→∞

we get

1 0 03 0 0

13

+ ++ +

=

∴ lim .n

n n

n n→∞

+ ++ −

=2

21

3 2 1

13

8. Evaluate: lim...

....

n

n n

n→∞

+ + +

+ + +

1 2 3

1 2 3

2 2 2 2

3 3 3 3

limn

nn n n

n n→∞

+ +LNM

OQP

+

1 2 16

14

2 2

a fa f

a f ∵1 2 31 2 16

2 2 2 2+ + + =+ +

...nn n na fa f

limn

n n n

n n→∞

+ +×

+

2

2 2

1 2 16

4

1

a fa fa f and 1 2

1

43 3 3

2 2

+ + =+

...nn na f

limn

n nn

nn

n nn

→∞

⋅ +FHIK +FH

IK

⋅⋅ +FH

IK

2

2 22

11

21

6

4

11


limn

n n

n

→∞

+FHIK +FH

IK

⋅+FHIK

11

21

64

11 2

Applying the limn→∞

1 0 2 06

4

1 0

26

41

432

+ +×

+= × =a fa f

a f

∴ lim...

....

n

n n

n→∞

+ +

+ +=

1 2

1 243

2 2 2

3 3 3

9. Evaluate: limn

n

n→∞

+

−

1

4 32

lim limn n

n

n

nn

nn

→∞ →∞

+

−=

+FHIK

−FHIK

1

4 3

11

432

22

=+FHIK

−→∞limn

nn

nn

11

432

Applying the limit, we get

1 0

4 0

1

4

12

+−

= = .

∴ limn

n

n→∞

+

−=1

4 3

1

22

10. Evaluate: limn

nn

n→∞

+FHIK

1 4

lim limn

n

n

nn

n n n→∞ →∞+FHIK = +FH

IK

11

14 4

= +FHIK

LNMM

OQPP

=→∞lim .n

n

ne1

14

4 ∵ limn

n

ne

→∞+FHIK =1

1


11. Evaluate: limn

nn

n→∞ +FHIK1

6

Dividing both numerator and denominator by n

= +

F

HGGG

I

KJJJ

=+

F

HGGG

I

KJJJ→∞ →∞

lim limn

n

n

nnn

nn

nn n

11

1

6 6

limn

n

n

n

n n

→∞ +

F

HGGG

I

KJJJ

=+FHIK

1

11

1

11

6

6

6

= =

+FHIK

LNM

OQP

= =→∞

−lim .n

n

n

n

ee

1

11

16

6 66

12. Evaluate: lim ... .n n→∞

+ + + +16

1

6

1

6

1

62 3

Solution: 16

1

6

1

6

1

62 3+ + + +... n is a G.P. with a r= = =1

6

1

616

1

6

2,

Sa r

rn

n

=−

−

1

1

d i [formula]

lim ... .n n→∞

+ + + +16

1

6

1

6

1

62 3

=− FHIK

LNM

OQP

−=

− FHIK

LNM

OQP =

− FHIK

→∞ →∞lim limn

n

n

n n16

116

116

16

116

56

116

5

Applying the lim, we get

1 05

15

15

− = = .


13. Evaluate: lim ...n

n→∞

− + − +112

14

18

terms.

Solution: 1 terms is a G.P.− + −12

14

18

...n

with a r= =

−

= −1

121

12

,

Sa r

rn

n

=−

−

1

1

d iFormula

∴ lim ...n

n→∞

− + − +112

14

18

terms.

lim limn

n

n

n

→∞ →∞

− −FHIK

FHG

IKJ

− −FHIK

=− −FHIK

+

1 112

112

112

112

Applying the lim .n→∞

− =1 032

23

14. Evaluate: limx

x xa bx→

−0

limx

x xa bx→

− + −0

1 1Adding and Subtracting 1

=− − −

→limx

x xa b

x0

1 1 1d i

limx

x xax

bx→

− − −0

1 1

lim limx

x

x

xax

bx→ →

− − −0 0

1 1

= −log loge ea b

= FHIKloge

a

b


15. Evaluate limx

x

x→

−+ −0

3 1

1 1

Consider

3 1

1 1

x

x

−+ − Rationalising the denominator,

3 1

1 1

1 1

1 1

x

x

x

x

−+ −

× + ++ +

=− + +

+ −

3 1 1 1

1 1

x x

x

d id i

3 1 1 1x x

x

− + +d id i

∴ lim limx

x

x

x

x

x

x→ →

−+ −

=− + +

0 0

3 1

1 1

3 1 1 1d id i

Applying the limx→0

loge 3 0 1 1+ +d i ∵ lim logx

x

ex→

− =0

3 13

loge 3 2a f= 2 3loge

15.6 CONTINUOUS FUNCTIONS:

A function y = f (x) is said to be continuous at x = a if lim limx a x a

f x f a f x→ →− +

= =a f a f a fi.e., Left Hand Limit = f (a) = Right Hand Limit

or lim limh h

f a h f a f a h→ →

− = = +0 0a f a f a f

In other words, y = f(x) is continuous at x = a if limx a

f x f a→

=a f a fGeometrically y = f (x) is continuous at x = a means that there is no break in the graph of y = f (x)

at x = a.

Note: A function which is not continuous at x = a is said to be discontinuous at x = a.


WORKED EXAMPLES:

1. Prove that the function f xx x

xx x

a f =+ ≤

=− ≥

RS|T|

3 25 2

3 1 2

when when

when is continuous at x = 2.

Proof. For a function y = f (x) to be continuous at x = a we have LHL = f (a) = RHL.

Here a = 2.

LHL = = + = + =→ →−lim lim

x xf x x

2 23 2 3 5a f

f(a) = f(2) = 5 (Given)

RHL = = − = − =→ →+lim lim

x xf x x

2 23 1 3 2 1 5a f a f

∴ LHL RHL= =f aa fHence f(x) is continuous at x = 2.

2. Verify whether the function f xx x

xx x

a f =+ >

=− <

RS|T|

3 12 1

2 1 1

when when

when is continuous at x = 1

Solution: For a function y = f (x) to be continuous at x = a we have LHL = f (a) = RHL.

Here a = 1

LHL = = − = − =→ →−lim limx x

f x x1 1

2 1 2 1 1 1a f a ff (a) = f (1) = 2 (Given)

RHL = = + = + =→ →+lim limx x

f x x1 1

3 1 3 4a f

∴ LHL RHL≠ ≠f aa fSo f (x) is discontinuous at x = 1

3. Prove that the function defined by

f xx

xx

x

a f =−−

≠

=

RS|T|

3

2

64

1664

26 4

when

when is not continuous at x = 4.

Proof. For a function f (x) to be continuous at x = a we have LHL = f (a) = RHL.

Here a = 4

LHL RHL= = −−→

limx

x

x4

3

264

16x x x≠ ⇒ < >4 4 4 or


=

−−−−

= = ××

= × =→

−

−lim..x

xx

xx

4

3 3

2 2

3 1

2 1

24444

3 42 4

3 42 4

3 42

6

Given f(a) = f(4) = 26

∴ LHL = RHL ≠ f(a).

∴ The function f(x) is not continuous at x = 4.

4. Prove that f x x xe x

xa f a f= + ≠=

RS|T|

1 2 00

1

2for

for is continuous at x = 0.

Proof. For a function to be continuous at x = a we have LHL = f(a) = RHL.

Here a = 0

LHL RHL= = +→

limx

xx0

11 2a f

limx

xx→

+0

221 2a f

= +LNMOQP→

limx

xx0

12

2

1 2a f

= e2 ∵ limx

nn e→

+ =LNM

OQP0

11a f

f a f ea f a f a f= =0 2 given

∴ = =LHL RHL f 0a fHence f (x) is continuous at x = 0.

5. If f (x) = x2 + 3x − 1, then prove that f (x) is continuous at x = 1.

Proof. For a function f (x) to be continuous at x = a we have LHL = f (a) = RHL.

Here a = 1.

LHL = = −→ →−lim limx h

f x f h1 0

1a f a f

= − + − −→

limh

h h0

21 3 1 1a f a f ∵ f x x xa f = + −2 3 1

= + − + − −→

limh

h h h0

21 2 3 3 1

= − +→

limh

h h0

2 5 3


Applying the limh→0

we get

LHL = 0 − 5(0) + 3 = 3.

RHL: lim limx h

f x f h→ →+

= +1 0

1a f a f

= + + + −→

limh

h h0

21 3 1 1d i a f

= + + + + −→

limh

h h h0

21 2 3 3 1

= + +h h2 5 3

Applying the limh→0

we get

RHL = 3.

f a fa f a f a f= = + − = + − =1 1 3 1 1 1 3 1 33 .

∴ LHL RHL.= =f aa fHence the function y = f(x) is continuous at x = 1.

6. Find K if the function f xx x

xx

K xa f =

− +−

≠

=

RS|T|

2 5 6

22

2

if

if is continuous at x = 2.

Given f (x) is continuous at x = 2.

LHL = f (a) = RHL.

i.e. limx

x xx

K→

− +−

=2

2 5 62

limx

x x

xK

→

− −−

=2

3 2

2a fa fa f

limx

x K→

− =2

3

2 3− = K

⇒ K = −1

7. Define f (0) so that f x x xxa f a f= + ≠1 3 01

is continuous at x = 0.

Given: f (x) is continuous at x = 0.

limx

f x f→

=0

0a f a f

limx

xx f→

+ =0

11 3 0a f a f


limx

xx f→

+ =0

331 3 0a f a f

limx

xx f→

+LNMOQP =

0

13

3

1 3 0a f a f

⇒ e f3 0= a f ∵ limn

nn e→

+ =0

11a f

8. Find K if f xx xx

x

K x

a f =+ − ≠

+ =

RS|T|

10

3 0

if


Given f (x) is continuous at x = 0

∴ limx

f x f→

=0

0a f a f

limx

xx

K→

+ − = +0

1 13

limx

x

x

x

xK

→

+ − × + ++ +

= +0

1 1 1 1

1 13

limx

x

x xK

→

+ −

+ += +

0

2 21 1

1 13

d id i

limx

x

x xK

→

+ −+ +

= +0

1 1

1 13

d i

limx

x

x xK

→ + += +

0 1 13

d iApplying the lim

x→0 we get

1

1 0 13

+ += +K

11 1

3+

= +K

12

3− = K


⇒1 6

2− = K

⇒ K = −52

9. Find a if f a x a xx xa f {= + ≤

− >2 2

1 2 if


Given: f(x) is continuous at x = 2.

LHL = RHL = f(a)

lim limx x

f x f f x→ →− +

= =2 2

2a f a f a flim limx x

x a x→ →

+ = −2 22 1

2 2 2 1a f + = −a

4 1+ =a

a = −1 4

a = −3

REMEMBER:

• limx a

n nnx a

x ana

→−−

−= 1

• limx

xex→

− =0

11

• lim logx

x

ea

xa

→

− =0

1

• limn

n

ne

→∞+FHIK =1

1

• limn

nn e→

+ =0

11a f

• A function y = f (x) is said to be continuous at x = a if lim limx a x a

f x f a f x→ →− +

= =a f a f a fi.e., LHL = f (a) = RHL.

• Limit of a function exists at x = a if


lim limx a x a

f x f x→ →− +

=a f a f∴ f(x) is continuous at x = a iff

limx a

f x f a→

=a f a f

EXERCISE

Evaluate:

1. limx

x x→

+ +2

2 1 3 2a fa f

2. limx

x x x

x x→

+ − +− −1

3 2

24 3 3

7 1

3. limx

x x

x x→

+ −− −0

2

23 8 9

4 9 3

4. limx

x x x

x x→

+ + −+ +1

3 2

3 23 2 1

3 6

5. limx

x x

x x→

− ++ −3

2

25 6

3 3 7

6. limx

xx→

+ −0

3 3

7. limx

a x a xx→

− − +0

8. limx

x

x x→

−− − −3

3

2 4

9. limx

xx→

− +−4

3 54

10. limx

x

x x→

−− − −3

3 27

3 4 2 1

11. limx

x

x x→

−− − +2

2 4

3 2 2

12. limx

x xx→

− − −−2

3 4 42

13. limx

xx→ −

++2

5 322

14. limx

xx→

−

−4

31 1

644

15. limx

xx→

−−25

1617

16. limx

xx→

−−3

4 813

17. limx

xx→

−−1

3 2 11

18. limx

x

x x→

−− +5

3

2125

6 5

19. limz

z z

z z→

− −− +3

2

2

26 5 6

8 14 3

20. limn

n n n

n→−

+ + +

−1

2

2

2 4 3

1

a fd i

21. limy

y y y

y y y→−

− − ++ − −1

3 2

3 22 3 3 2

3 2 11 10

22. limx a x a

a

x ax→ −−

−1

2

23. limx

x

x→

−

−3

3

4

1 127

1 181


24. limx

x

x x→

−−3

3 27

3 3

25. limx

x

x→

− −−5

53 32

5a f

26. limx

x xa bx→

−0

27. limx

x xe ex→

−−0

28. limx

x

x→

−0

2 1

29. limx

x

x→

−+ −0

2 1

1 1

30. limn

n n

n→∞

− +−

3 6 8

4 8

2

2

31. lim...

n

n

n→∞

+ + + +1 2 3

3 2

a f

32. lim...

n

n

n n→∞

+ + + +− +

1 2 3

3 8 1

2 2 2 2

3

33. limn

n n n→∞

− −2 4

34. lim ...n n→∞

+ + +13

1

3

1

32

35. lim...

n

n

n→∞

+ + ++

1 2

4

3 3 3

4a f

36. limn

n

n→∞

+

+++

3 1

3 2

1

2

37. If f x x xx xa f {= + <

− >3 2 27 6 2

if if , then find lim

xf x

→2a f .

38. Prove that limx

x

x→0 does not exist


39. Evaluate (a) limn

nn

→∞+1 2 3a f (b) lim

n

xx→

+0

41 3a f40. Examine the Continuity of the following functions:

(a) f(x) = x2 + x + 1 at x = 1

(b) f xx x

xx x

a f =+ <

=+ >

RS|T|4 1 0

1 02 3 0

if if

if at x = 0

(c) f xx

x xx

xa f =

−− +

≠

=

RS|T|

2

29

4 33

3 3

for

for at x = 3.

(d) f xx

xx

xa f = + −

≠

=

RS|T|

1 10

3 0

for

for at x = 0

(e) f xbx

xx

b xa f

a f=

+≠

=

RS|T|

log 10

0

when

when at x = 0

41. Prove that the function f(x) = |x| is continuous at x = 0.

42. Prove that f xx

xx

xa f = ≠

=

RS|T|

for

for

0

0 0 is discontinuous at x = 0.

43. Find K if the function f x x xe K x

xa f a f= + ≠=

RS|T|

1 for for

14 0

02 is continuous at x = 0.

44. Find K if the function f x

a x a x

xx

Kx

a f =+ − − ≠

=

RS||

T||

for

for

0

10

is continuous at x = 0.

45. Find a if f xx

x

x

K x

a f =−

−≠

=

R

S||

T||

1

for

for

4181

1 127

3

3

3 is continuous at x = 3.


ANSWERS

1. 40 2. 1 3. 3 4.12 5. 0 6.

12 3

7. − a

2 8. 1 9. − 16

10. 54 5 11. 8 12. 2

13. 80 14.−3256

15.12

16. 108 17.32

18.754

19.52 20. −1 21.

−32 22.

1a

23.94 24. 6 3

25. 80 26. logab

27. 2 28. log 2 29. log 4 30.34

31.16 32.

19 33. −2 34.

12 35.

14 36.

411

37. 8 39. (a) e6 (b) e12

40. (a) Continuous (b) discontinuous (c) Continuous (d) discontinuous (e) Continuous.

43. e2 44. a 45.49


16

Differential Calculus

16.1 INTRODUCTION:Differential calculus was discovered by Sir Isaac Newton of England and Wilhelm Leibnitz of Ger-many. It deals with the study of rate of change of one quantity with another.

16.2 DERIVATIVE OF A FUNCTION:Let y = f(x).

Let δx be an increment given to x.

δy be the corresponding change in y.

y + δy = f(x + δx)

δy = f(x + δx) − y

δy = f(x + δx) − f(x)

Dividing by δx and taking limδx→0

.

lim limδ δ

δδ

δδx x

yx

f x x f x

x→ →=

+ −0 0

a f a f

If limδ

δδx

f x x f x

x→

+ −0

a f a fexists and finite then the function y = f (x) is said to be differentiable at x

and limδ

δδx

f x x f x

x→

+ −0

a f a f is called derivative or differential co-efficient of y with respect to x. It is

denoted by dydx

or y′ or y1 or f′(x).

∴ lim limδ δ

δδ

δδx x

yx

dydx

f x x f x

x→ →= =

+ −0 0

a f a f

Differential Calculus 367

16.3 DERIVATIVE OF SOME STANDARD FUNCTIONS FROMFIRST PRINCIPLES:

1. xn.Let y = xn.


δy be the corresponding increment in y

y + δy = (x + δx)n

δy = (x + δx)n − y

δy = (x + δx)n − xn.

Divide by δx and take limδx→0

lim limδ δ

δδ

δδx x

n ny

x

x x x

x→ →=

+ −0 0

a f

Add and subtract x in the denominator of RHS.

lim limδ δ

δδ

δδx x

n nyx

x x x

x x x→ →=

+ −+ −0 0

a f

As lim δx → 0,

lim x + δx → x.

Also limx a

n nnx a

x ana

→−−

−= 1 [formula].

∴ lim limδ δ

δδ

δδx x

n nny

x

x x x

x x xnx

→ →

−=+ −

+ −=

0 0

1a f

⇒dydx

nxn= −1

i.e.,ddx

x nxn nd i = −1

Hence derivative or differential co-efficient of xn is nxn−1.

2. ex.Let y = ex.


δy be the corresponding increment in y.

y y ex x+ = +δ δ

δ δy e yx x= −+


δ δy e ex x x= −+

δ δy e e ex x x= ⋅ − ∵ a a am n m n+ = ⋅

δ δy e ex x= − 1e jDividing by δx and taking lim

δx→0.

lim limδ δ

δδδ δx x

x xy

x

e e

x→ →=

−

0 0

1e j

But we have limx

xex→

− =0

11 (formula)

∴ lim limδ δ

δδδ δx x

x xyx

e e

x→ →=

−

0 0

1e j

⇒dydx

ex= ⋅1

dydx

ex=

i.e.,ddx

e ex xd i = .

Hence derivative or differential co-efficient of ex is ex.

3. ax.Let y = ax.

Let δx be an increment given to x. δy be the corresponding increment in y.

y y ax x+ = +δ δ

δ δy a yx x= −+

δ δy a ax x x= −+

δ δy a a ax x x= ⋅ − ∵ a a am n m n+ = ⋅

δ δy a ax x= − 1e jDividing by δx and taking lim

δx→0


lim limδ δ

δδδ δx x

x xy

x

a a

x→ →=

−

0 0

1e j

∵ lim logx

x

ea

xa

→

− =0

1(formula)

∴ lim limδ δ

δδδ δx x

x xy

x

a a

x→ →=

⋅ −

0 0

1e j

⇒dydx

a axe= ⋅ log .

∴ddx

a a ax xed i = ⋅ log .

Hence derivative or differential co-efficient of ax is ax.logea.

4. logex.Let y = logex



y y x xe+ = +δ δlog a f δ δy x x ye= + −log a f

δ δy x x xe e= + −log loga f but log log logm nm

n− = F

HIK

∴ δ δy

x x

xe= +FHGIKJlog

δ δy

x

x

x

xe= +FHGIKJlog

δ δy

x

xe= +FHGIKJlog 1


lim limlog

δ δ

δδ

δ

x x

ey

x

xx

x→ →=

+FHGIKJ

0 0

1


lim lim logδ δ

δδ δ

δx x

ey

x x

x

x→ →= +FHG

IKJ0 0

11

Multiplying and dividing by x in RHS.

lim lim logδ δ

δδ δ

δx x

y

x x

x

x

x

x→ →= ⋅ ⋅ +FHG

IKJ0 0

11

But n m mnlog log=

lim lim logδ δ

δδδ

δx x

e

xxy

x x

x

x→ →= ⋅ +FHG

IKJ0 0

11

As limδ

δδx

xxx

xe

→+FHGIKJ =

01

lim lim logδ δ

δδδ

δx x

e

xxy

x x

x

x→ →= +FHG

IKJ0 0

11

⇒dydx x

ee= ⋅1log But logee = 1

∴dydx x

= ⋅11

dydx x

= 1.

i.e.,ddx

xx

loga f = 1

∴ Derivative or differential Co-efficient of logx is 1x

.

5. Constant function:Let y = c where c is a constant.

Let δx be an increment given to x. δy be the corresponding increment in y. But Since c is a constantfunction any change in x will not cause change in y. In other words δy = 0.

∴ limδ

δδ δx

yx x→

= =0

00

∴dydx

= 0


Hence ddx

constanta f = 0.

∴ derivative of a constant is zero.

Ex.:d

dx

d

dx

d

dxa a−FH

IK = = =5

20 5 0 0; a f a f if is constant.

16.4 RULES OF DIFFERENTIATION:

1. Derivative of Product of Constant and a Function:

Let y = Ku where K is a constant and u is a function of x.


δu be the increment in u and


∴ y y K u u+ = +δ δa fy y Ku K u+ = +δ δ

δ δy Ku K u y= + −

δ δy Ku K u Ku= + −

δ δy K u=


lim limδ δ

δδ

δδx x

y

xK

u

x→ →= ⋅

0 0

⇒ dydx

Kdudx

= ⋅

i.e.,ddx

Ku Kdudx

a f = ⋅

Hence derivative or differential co-efficient of constant multiple of a function is constant into deriva-tive of the function.

Examples:

(a)ddx

xddx

x4 4log loga f a f= ⋅

= ⋅ =41 4x x

.


(b)ddx

eddx

ex x8 8d i d i= ⋅

= ⋅8 e x

2. Derivative of Sum of 2 Functions:

Let y = u + v where u and v are functions of x.


δu, δv be the increments in u and v.


∴ y y u u v v+ = + + +δ δ δ

δ δ δy u u v v y= + + + −

δ δ δy u u v u u v= + + + − +a f δ δ δy u u v v u v= / + + / + − / − /

δ δ δy u v= +


lim limδ δ

δδ

δ δδx x

y

x

u v

x→ →= +

0 0

lim limδ δ

δδ

δδ

δδx x

yx

ux

vx→ →

= +0 0

⇒dydx

dudx

dvdx

= +

i.e.,ddx

u vdudx

dvdx

+ = +a f

∴ Derivative or differential co-efficient of sum of 2 functions is derivative of first function plus thederivative of the second function.

Examples:

(a) If y = x4 + ex, then dydx

ddx

x ex= +4d i

= +ddx

xddx

ex4d i d i


dydx

x ex= +−4 4 1

dydx

x ex= +4 3

(b)ddx

x xlog + 2d i

= +ddx

xddx

xloga f d i2

= + −12 2 1

xx

= +12

xx

3. Derivative of Difference of 2 Functions:

Let y = u − v where u and v are functions of x. Let δx be an increment given to x.



y y u u v v+ = + − +δ δ δa f δ δ δy u u v v y= + − + −a f

δ δ δy u u v v u v= + − + − −a f a f δ δ δy u u v v u v= + − − − +

δ δ δy u v= −


lim limδ δ

δδ

δ δδx x

y

x

u v

x→ →= −

0 0

lim limδ δ

δδ

δδ

δδx x

yx

ux

vx→ →

= −0 0

lim lim limδ δ δ

δδ

δδ

δδx x x

y

x

u

x

v

x→ → →= −

0 0 0

⇒ dydx

dudx

dvdx

= −


i.e.,ddx

u vdudx

dvdx

− = −a f∴ Derivative of difference of 2 functions is the derivative of first function minus the derivative of the

second function.

Examples:

(a) ddx

x x3 2−d i

= −ddx

xddx

x3 2d i d i

= −− −3 23 1 2 1x x

3 22x x−(b) If y = logx − ex, then

dydx

ddx

x ex= −logd i

= −ddx

xddx

exloga f d i

= −1x

ex .

4. Derivative of Product of 2 Functions:

Let y = uv where u and v are functions of x. Let δx be an increment given to x, δu, δv be the incrementsin u and v. Let δy be the corresponding increment in y.

y y u u v v+ = + +δ δ δa fa f δ δ δy u u v v y= + + −a fa f δ δ δ δ δy uv u v v u u v uv= + + + −

δ δ δ δ δy u v v u u v= + +


lim limδ δ

δδ

δ δ δ δδx x

yx

u v v u u vx→ →

= + +0 0

= + +→

limδ

δδ

δδ

δ δδx

uvx

vux

u vx0


As limδx→0

, δu and δv are small. Hence the product δu × δv is very very small. So the term δ δδu vx

can

be neglected.

∴ lim limδ δ

δδ

δδ

δδx x

yx

uvx

vux→ →

= +0 0

⇒dydx

udvdx

vdudx

= +

i.e.,ddx

uv udvdx

vdudx

a f = +

So Differential Co-efficient or derivative of product of 2 function is first function into derivative of2nd function plus second function into derivation of the first function.

i.e.,ddx

[I function ⋅ II function]

= (I function) ⋅ ddx

(II function) + (II function) ⋅ ddx

(I function).

This rule is known as product rule.

Examples:

(a)ddx

x e xddx

e eddx

xx x x3 3 3d i d i= ⋅ + ⋅

= x e e xx x3 3 13⋅ + ⋅ −

= x e e xx x3 23+ ⋅(b) If y = x4 log x, then

dydx

ddx

x x= 4 logd i

= ⋅ + ⋅xddx

x xddx

x4 4log loga f a f d i

= ⋅ + ⋅ −xx

x x4 4 114log

x x x3 34+ log

x x x3 34+ log


5. Derivative of Quotient of 2 Functions:

Let yuv

= where u and v are functions of x. Let δx be an increment given to x.



y yu uv v

+ = ++

δ δδ

δ δδ

yu u

v vy= +

+−

δ δδ

yu uv v

uv

= ++

−

δδ δ

δy

v u u u v v

v v v=

+ − ++

a f a fa fa f

δ δ δδ

yuv v u uv u v

v v v= + − −

+a f

δ δ δδ

yv u u v

v v v= −

+a fDivide by δx and take lim

δx→0

lim limδ δ

δδ

δ δδ δx x

y

x

v u u v

v v v x→ →= −

+ ⋅0 0 a f

lim limδ δ

δδ

δδ

δδ

δx x

y

x

vux

uvx

v v v→ →=

−

+0 0 a f (note this step)

As limδx→0

, δv → 0

∴ lim limδ δ

δδ

δδ

δδ

δx x

y

x

vux

uvx

v v v→ →=

−

+0 0 a f

⇒ dy

dx

vdudx

udvdx

v v=

−

+ 0a f


dydx

vdudx

udvdx

v=

−2

i.e., ddx

uv

vdudx

udvdx

vFHIK =

−2

Hence Derivative or differential co-efficient of quotient of 2 functions is Denominator into differen-tial co-efficient of Numerator, minus Numerator into differential co-efficient of Denominator, wholedivided by square of the denominator.

i.e.,d

dx

ddx

ddxNr.

Dr.

Dr. Nr. Nr. Dr.

Dr.FHIK =

−a f a f a f a fa f2

Nr.: Numerator

Dr.: Denominator

This rule is known as quotient rule.

Examples:

(a) d

dx

x

e

eddx

x xddx

e

ex

x x

x

log log logFHIK =

⋅ − ⋅a f a f d id i2

=⋅ − ⋅e

xx e

e

x x

x

1

2

log

d i

(b) If yx

x=

3

log, then

dy

dx

d

dx

x

x

xddx

x xddx

x

x=FHGIKJ =

⋅ − ⋅33 3

2log

log log

log

d i a fa f

=⋅ − ⋅log

log

x x xx

x

312 3

2a f

= ⋅ −log

log

x x x

x

3 2 2

2a f


Some particular cases of ddx

x nxn nd i = −1

1. When n = 0

ddx

x x0 0 10d i = −But x0 = 1.

∴ ddx

1 0a f =

Also we have ddx

K u Kdudx

⋅ = ⋅a f

∴ddx

K Kddx

K⋅ = ⋅ = ⋅ =1 1 0 0a f a f

∴ddx

Ka f = 0

2. When n = 1,

ddx

x x x1 1 1 01 1 1d i = = =−

∴ddx

xa f = 1

3. When n = 2,

ddx

x x x2 2 12 2d i = =−

∴ddx

x x2 2d i =4. When n = 3,

ddx

x x3 3 13d i = −

ddx

x x3 23d i =

Similarly, ddx

x xddx

x xddx

x x4 3 5 4 100 994 5 100d i d i d i= = =, ... and so on.

5. When n = 12

,


d

dxx x

12

12

112

FHGIKJ

=−

ddx

x xd i = −12

12

∵ x x12 =

d

dxx

xd i = 1

2a

am

m− = 1

d

dxx

x= 1

2

6. When n = −1

ddx

x x− − −= −1 1 11d i

d

dx xx

11 2F

HIK = − −

d

dx x x

1 12

FHIK = −

7. When n = −2

ddx

x x x− − − −= − = −2 2 1 32 2d i

ddx x x

1 22 3FHIK = −

Similarly

ddx x x

ddx x x

1 3 1 43 4 4 5FHIK = − F

HIK = −,

ddx x x

1 55 6FHIK = − and so on.


List of Formulae:

ydydx

(1) xnnx n−1

(a) x 1

(b) x2 2x

(c) x33 2x

(d) x44 3x

(e) x5 5 4x

(f) x1

2 x

(g)1x

− 12x

(h)12x

−23x

(i)13x

−34x

(j)14x

−45x

(k) K (Constant) 0

2. exex

3. ax a axelog

4. log x1x

5. u v±dudx

dvdx

±

6. (I function) (II function) I function II II I⋅ + ⋅ddx

ddx

a f a f

7.Nr.Dr.

Dr. Nr. Nr. Dr.

Dr

ddx

ddx

a f a f a fa f− ⋅

2


Note: Sum, Difference and product rule can be extended i.e.,

ddx

u v wdudx

dvdu

dwdu

± ± ± = ± ± ±... ...a f

ddx

uvwdudx

vwdvdx

uwdwdx

uv... ... ... ... ...a f = ⋅ + ⋅ + +

WORKED EXAMPLES:

1. Find dydx

if y = x3 − 3x + 7

Consider y x x= − +3 3 7, differentiate with respect to (w.r.t) x.

dydx

ddx

x x= − +3 3 7

= − +ddx

xddx

xddx

3 3 7d i a f a f

3 3 02xddx

x− +a f

3 3 1 02x − ⋅ +a fdydx

x= −3 32 .

2. Find y′ if y e xxe= −7 4 log .

Given: y e ex x= −7 4 log

diff. w.r.t.x. (differentiate with respect to x)

′ = −yddx

eddx

xxe7 4d i b glog

′ = ⋅ − ⋅y ex

x7 41

′ = −y ex

x74

3. Find f ′(x) if f(x) = 7x + 8 ex − 9

Consider f(x) = 7x + 8 ex − 9

diff. w.r.t. x.


′ = + ⋅ −f xddx

ddx

eddx

x xa f d i a f7 8 9

′ = + ⋅ −f x ex xa f 7 7 8 0log

′ = +f x ex xa f 7 7 8log

4. If y xx

= + 1, then find y1

y xx

= + 1

diff. w.r.t. x.

yd

dxx

d

dx x11= + FHGIKJd i

= + −ddx

xddx

x1 2 1 2d i d i

= + −FHG

IKJ

− − −12

12

12

112

1x x

y x x11 2 3 21

212

= −− −

yx x1 3 2

1

2

1

2= −

Alieter:

y xx

= + 1

yx

x= +1

diff. w.r.t. x. using quotient rule.

yx

ddx

x xddx

x

x1 2

1 1=

⋅ + − + ⋅a f a f d id i

x x

xx

1 0 11

2+ − + ⋅a f


x

xx xx

− −2

12

= −⋅

−⋅

x

x

x

x x x x2

1

2

= − −1 1

2

1

2x x x x

= − − +2 1

2

1

2 1 1 2x x

yx x1 3 2

1

2

1

2= − .

5. Find dydx

if y = x38x.

y x x= 3 8 .

diff. w.r.t. x using product rule.

dydx

xddx

ddx

xx x= ⋅ + ⋅3 38 8d i d idydx

x xx x= ⋅ + ⋅3 28 8 8 3log

6. If f xx

xa f = +

−

2

31

7, then find f ′(x)

f xx

xa f = +

−

2

31

7

diff. w.r.t. x using quotient rule.

′ =− ⋅ + − + ⋅ −

−f x

xddx

x xddx

x

xa fd i d i d i d i

d i

3 2 2 3

3 2

7 1 1 7

7

=− +LNM

OQP − + −L

NMOQP

−

xddx

xddx

xddx

xddx

x

3 2 2 3

3 2

7 1 1 7

7

d i d i a f d i d i a fd i


x x x x

x

3 2 2

3 2

7 2 1 3

7

− − +

−

d ia f d id id i

′ = − − −

−f x

x x x x

xa f

d i2 14 3 3

7

4 4 2

3 2

′ = − − −

−f x

x x x

xa f

d i4 2

3 23 14

7

7. If yx

xx= + −

22

2 2 , then find dydx

.

yx

xx= + −

22

2 2

diff. w.r.t.x.

dy

dx

d

dx

x d

dx x

d

dxx= F

HIK +FHIK −2

22 2d i

= + FHIK −

12

21

2 2d

dxx

d

dx x

d

dxxa f d i

= ⋅ + −FHIK − ⋅1

21 2

12 22x

xa f

dydx x

x= − −12

242 .

8. If yxe

x

x

=+log 7a f , then find

dydx

.

yxe

x

x

=+log 7

.


dydx

xddx

xe xeddx

x

x

x x

=+ ⋅ − ⋅ +

+

log log

log

7 7

7 2

a f d i d i a fa f

=+ ⋅ + ⋅LNM

OQP − +LNM

OQP

+

log log

log

x xddx

e eddx

x xeddx

xddx

x

x x x7 7

7 2

a f a f a f a fa f


=+ ⋅ + ⋅ − +LNM

OQP

+

log

log

x x e e xex

x

x x x7 11

0

7 2

a fa f

dydx

x xe e e

x

x x x

=+ + −

+

log

log

7

7 2

a fd ia f

9. If yx n

e

n x

x= −, then find

dydx

.

yx n

e

n x

x= −


dy

dx

eddx

x n x nddx

e

e

x n x n x x

x=

⋅ − − −d i d i d id i2

=−L

NMOQP − − ⋅e

ddx

xddx

n x n e

e

x n x n x x

x

d i d i d id i2

=− − −−e nx n n x n e

e

x n x n x x

x

1

2

log d id i

=− − +−e nx n n x n

e

x n x n x

x

1

2

log

d i

dydx

nx n n x n

e

n x n x

x= − − +−1 log.

10. If f(x) = (x2 + 1) (x3 + 7x + 8) ex, then find f ′(0).

f x x x x exa f d id i= + + + ⋅2 31 7 8

diff. w.r.t. x using extended product rule.

′ = + + + ⋅ + + ⋅ + + + + + ⋅ +f x x x xddx

e x eddx

x x x x eddx

xx x xa f d id i d i d i d i d i d i2 3 2 3 3 21 7 8 1 7 8 7 8 1


′ = + + + ⋅ + + ⋅ + + + + ⋅ ⋅f x x x x e x e x x x e xx x xa f d id i d i d i d i a f2 3 2 2 31 7 8 1 3 7 7 8 2

∴ ′ = + + + + + + + + ⋅f e e e0 0 1 0 8 0 1 0 7 0 0 8 2 00 0 0a f a fa f a f a f a f a f= 8 + 7 + 0

′ =f 0 15a f

16.5 DIFFERENTIATION OF COMPOSITE FUNCTIONS:Chain rule: If y = g(u) and u = f (x) are 2 differentiable functions, then the composite functiony = g[f (x)] can be differentiated by using chain rule, which can be stated as

dydx

dydu

dudx

= ⋅

i.e.dydx

g u f x= ′ ⋅ ′a f a f

i.e.ddx

g f x g f x f xa fb g a f a f= ′ ⋅ ′

Examples:

1. If y = log (x2 − 4x + 8), then

dydx x x

ddx

x x=− +

⋅ − +1

4 84 82

2d i

=− +

⋅ −1

4 82 42x x

xa f

= −− +

2 4

4 82

x

x x

Alieter: Consider y = log (x2 − 4x + 8)

Let y = logu where u = x2 − 4x + 8

diff. w.r.t. x.

dydx u

dudx

= ⋅1u x x= − +2 4 8 diff. w.r.t. x

dydx u

x= ⋅ −12 4a f ∴

dudx

x= −2 4

dydx u

x= ⋅ −12 4a f


dy

dx x xx=

− +⋅ −1

4 82 42 � �

dy

dx

x

x x= −

− +2 4

4 82

2. If f x ex� � = 2, then

′ = ⋅f x ed

dxxx� � � �

2 2

′ = ⋅f x e xx� � 2

2 .

Note:Note:Note:Note:Note: Chain rule can be extended.

i.e., If y = f[g(h(x))], then

dy

dxf g h x g h x h x= ′ ⋅ ′ ⋅ ′� ��

Example.Example.Example.Example.Example. If y ex=+2 8

4� � , then

dy

dxe

d

dxx

x= ⋅ ++2 8

4 2 84� � � �

dy

dxe x

d

dxx

x= ⋅ + ⋅ ++2 8

4 2 7 28 4 4� � � � � �

= ⋅ + ⋅+

e x xx2 8

4 2 78 4 2

� � � � .

��

1.1.1.1.1. Find dy

dx if y = log (xn −−−−− ex)

y x en x= −log� �diff. w.r.t. x.

dy

dx x e

d

dxx e

n xn x=

−⋅ −1� �

=−

⋅ −−1 1

x enx en x

n x

= −−

−nx e

x e

n x

n x

1

.


2.2.2.2.2. Find dy

dxy x x if = + −2 4 9 .

Consider y x x= + −2 4 9

diff. w.r.t x.

dy

dx x x

d

dxx x=

+ −⋅ + −1

2 4 94 9

2

2� �

dy

dx x xx=

+ −+ −

1

2 4 92 4 0

2

dy

dx x xx=

+ −⋅ +1

2 4 92 2

2� �

dy

dx

x

x x= +

+ −

2

4 92.

3.3.3.3.3. If yx x

=+ −

73 9 6� � , then find

dy

dx.

yx x= + −

73 9 6� �

diff. w.r.t. x.

dy

dx

d

dxx x

x x= ⋅ ⋅ + −+ −7 7 9 6

3 9 6 3� � � �log

= ⋅ + −+ −

7 7 3 9 1 03 9 6 2x x

x� � � �log

dy

dxxx x= ⋅ ⋅ ⋅ ++ −7 7 3 3

3 9 6 2log � � .

4.4.4.4.4. If f x x e x� � = +3 5 82

, then find f ′(x)

Consider f x x e x� � = +3 5 82


′ = ⋅ + ⋅+ +f x xd

dxe e

d

dxxx x� � � � � �3 5 8 5 8 32 2

= ⋅ ⋅ + + ⋅+ +x ed

dxx e xx x3 5 8 2 5 8 22 2

5 8 3� � � �

′ = ⋅ + + ⋅+ +f x x e x e xx x� � � �3 5 8 5 8 22 2

10 0 3


5.5.5.5.5. If yx

x x= +

+ −

1

8 92 3� �

, then find dy

dx.

Consider

yx

x x= +

+ −

1

8 92 3� �


dy

dx

x xd

dxx x

d

dxx x

x x

=+ − ⋅ + − + ⋅ + −

+ −�

� �

2 3 2 3

2 3 2

8 9 1 1 8 9

8 9

� � � � � � � �

� �

=+ − ⋅ + − + ⋅ + − ⋅ + −

��

+ −

x x x x xddx

x x

x x

2 3 2 2 2

2 6

8 9 1 0 1 3 8 9 8 9

8 9

� � � � � � � � � �

� �

dy

dx

x x x x x x

x x=

+ − − + + − +�

� �

+ −

2 3 2 2

2 6

8 9 1 3 8 9 2 8

8 9

� � � � � � � �

� �.

6.6.6.6.6. If yxe

x

x

=−

+3 7

7 6log� � , then find dy

dx.

yxe

x

x

=−

+3 7

7 6log� �diff. w.r.t. x using quotient rule.

dy

dx

xd

dxxe xe

d

dxx

x

x x

=− ⋅ − ⋅ −

−

+ +log log

log

7 6 7 6

7 6

3 7 3 7

2

� � � �

� �

=− + ⋅�

� �− ⋅

−⋅ −

��

−

+ + +log

log

7 61

7 67 6

7 6

3 7 3 7 3 7

2

x xd

dxe e

d

dxx xe

x

d

dxx

x

x x x� � � � � � � �

� �

=− ⋅ ⋅ + + ⋅�

� �−

−⋅

��

−

+ + +log

log

7 6 3 7 11

7 67

7 6

3 7 3 7 3 7

2

x x ed

dxx e xe

x

x

x x x� � � � � �

� �


dy

dx

x xe e xex

x

x x x

=− ⋅ + − ⋅

−−

+ + +log

log

7 6 37

7 67 6

3 7 3 7 3 7

2

� �

� �

��

Consider the function, y = x2 +ex −−−−− logx. Here y is expressed as a function of x. i.e., y = f (x). Suchfunctions are called explicit functions.

Now consider the function xy + log y + ex = 0. Here y is not expressed explicitly as a function of x.Such functions are called implicit functions. They are of the form f (x, y) = 0.

To find dy

dx, differentiate the given function shift all the terms containing

dy

dx to Left Hand Side and

the remaining terms to Right Hand Side. Take dy

dx common and shift the co-efficient of

dy

dx to Right

Hand Side.

��

1.1.1.1.1. Find dy

dx if x3 + y3 = a3.

Consider x3 + y3 = a3

diff. w.r.t. x.

3 3 02 2x ydy

dx+ ⋅ =

� ad

dxa

d

dxy y

d y

dx

is constant

using chain rule

3

3 2

0

3

� �

� ��

� �

=

= ⋅

��

�

��

3 3 02 2x ydy

dx+ =

3 32 2ydy

dxx= −

dy

dx

x

y= −3

3

2

2

dy

dx

x

y=− 2

2 .


2.2.2.2.2. If y + x2 + ey = 0, then find dy

dx.

Solution:Solution:Solution:Solution:Solution: Consider y + x2 + ey = 0

diff. w.r.t. x.

dy

dxx e

dy

dxy+ + ⋅ =2 0

dy

dxe

dy

dxxy+ ⋅ = −2 .

dy

dxe xy1 2+ = − .

dy

dx

x

ey= −+2

1.

3.3.3.3.3. If ex + ey = logx, then find dy

dx when x = 1 and y = 0


Consider ex + ey = logx.

diff. w.r.t. x.

e edy

dx xx y+ = 1

edy

dx xey x⋅ = −1

dy

dxx

e

e

x

y=−1

dy

dxx y

e

e when and = = =

−1 0

11

1

0

dy

dx

ee

1, 0� �= − = −1

11 .

4.4.4.4.4. If y x x x x= + + + + ∞...

Then find dy

dx.


Solution:Solution:Solution:Solution:Solution: y x x x x= + + + + ∞...

y x y= +

Squaring,

y x y2 = +

diff. w.r.t. x.

2 1ydy

dx

dy

dx⋅ = +

2 1ydy

dx

dy

dx⋅ − =

dy

dxy

dy

dx y2 1 1

1

2 1− = ⇒ =

−� �

5.5.5.5.5. If y ax ax ax= ∞... , then find dy

dx.

y ax ax ax= ∞...

y axy=

Squaring

y axy2 = .

diff. w.r.t. x (using product rule in RHS, Chain rule in LHS)

2ydy

dxa x

dy

dxy

d

dxx= ⋅ + ⋅

��

� �

2ydy

dxa x

dy

dxy= +

��

2ydy

dxax

dy

dxay= +

2ydy

dxax

dy

dxay− =

dy

dxy ax ay2 − =� �

dy

dx

ay

y ax=

−2.


��

If both x and y are expressed as a function of another variable say t then the function y = f (x) is saidto be in parametric form. The variable t is called a parameter.

To find dy

dx we use.

dy

dx

dy

dtdxdt

=

��

1.1.1.1.1. Find dy

dx if x = at2 and y = 2 at.

Consider y = 2 at

diff. w.r.t. t.

dy

dta= 2 1� �

dy

dta= 2

Now Consider x = at2

diff. w.r.t. t.

dx

dta t= 2� �

dx

dtat= 2

Now

dy

dx

dy

dtdxdt

a

at= = 2

2

dy

dx t= 1

.

2.2.2.2.2. If x = 7t + et and y = et −−−−− 7t, then find dy

dx when t = 0.

Consider y = et −−−−− 7t.


diff. w.r.t. t.

dy

dtet t= − 7 7log .

Now consider

x et t= +7

diff. w.r.t. t.

dx

dtet t= +7 7log

dy

dx

dy

dtdxdt

=

= −+

e

e

t t

t t

7 7

7 7

log

log

dy

dx

e

etwhen == −

+0

0 0

0 0

7 7

7 7

log

log

dy

dx twhen == −

+0

1 7

7 1

log

log

3.3.3.3.3. Differentiate ex3

with respect to log x.

Solution.Solution.Solution.Solution.Solution. Let u ex=3

and v = log x.

To find:du

dv

du

dv

du

dxdvdx

=

Consider u ex=3

diff. w.r.t. x.

du

dxe

d

dxxx= ⋅

3 3� �

du

dxe xx= ⋅

3

3 2


v x= log

diff. w.r.t. x

dv

dx x= 1

∴∴∴∴∴du

dv

e x

x

x

= ⋅3

31

2

du

dve xx= ⋅

3

3 3.

4.4.4.4.4. Differentiate (x2 + 8x −−−−− 1)4 with respect to ex2 9−

Solution.Solution.Solution.Solution.Solution. Let u = (x2 + 8x −−−−− 1)4 and v ex= −2 9

To find: du

dv

du

dv

dudxdv

dx

=

Consider u = (x2 + 8x −−−−− 1)4

diff. w.r.t. x.

du

dxx x

d

dxx x= + − ⋅ + −4 8 1 8 12 3 2� � � �

du

dxx x x= + − ⋅ +4 8 1 2 82 3� � � �

Now, v ex= −2 9

diff. w.r.t. x.

dv

dxe

d

dxxx= ⋅ −−2 9 2 9� �

dv

dxe xx= ⋅−2 9 2

Hencedu

dv

du

dxdvdx

x x x

e xx= =

+ − +

⋅−

4 8 1 2 8

2

2 3

92

� � � �.


5.5.5.5.5. Differentiate 4 + log x with respect to exlog 2 9+� �

Let u x v ex

= + =+

42 9

loglog

and � �

⇒ = + =v x e f xf h2 9 log � �� formula

To find: du

dv

du

dv

dudxdv

dx

=

Consider u x= +4 log

diff. w.r.t. x.

du

dx x

d

dxx=

+⋅ +1

2 44

loglog� �

du

dx x x=

+⋅1

2 4

1

log

du

dx x x=

+1

2 4 log

Now v = x2 + 9.

diff. w.r.t. x.

dv

dxx= +2 0

dv

dxx= 2

∴∴∴∴∴du

dv

dudxdv

dx

x x

x= =

+1

2 4

2

log

∴∴∴∴∴du

dv x x=

+1

4 42 log.


��

If it is required to differentiate (function)function or (constant)function then we consider log on both sides,apply logmn = nlogm and then differentiate.

��

1.1.1.1.1. Find dy

dx if y = xx

y x x=

Consider log on both sides

log logy x x=

log logy x x=

diff. w.r.t. x.

1

y

dy

dxx

d

dxx x

d

dxx⋅ = ⋅ + ⋅log log� � � �

1 11

y

dy

dxx

xx⋅ = ⋅ + ⋅log

11

y

dy

dxx⋅ = + log

dy

dxy x= +1 log� �

∴∴∴∴∴d

dxx x xx x� � � �= +1 log

2.2.2.2.2. Find dy

dxx yy x if = .

x yy x=


log logx yy x=

y x x ylog log=

diff. w.r.t. x.

yd

dxx x

dy

dxx

d

dxy y

d

dxx⋅ + ⋅ = ⋅ + ⋅log log log log� � � � � �


yx

xdy

dxx

y

dy

dxy⋅ + ⋅ = ⋅ ⋅ + ⋅1 1

1log log

y

xx

dy

dx

x

y

dy

dxy+ ⋅ = ⋅ +log log

log logxdy

dx

x

y

dy

dxy

y

x⋅ − = −

dy

dxx

x

yy

y

xlog log−��

�� = −

dy

dx

yy

x

xxy

=−

−

log

log.

3.3.3.3.3. If yx

x

dy

dx= +

−1

1, then find .

yx

x= +

−1

1

Consider log on both sides.

log logyx

x= 1+

−1

log logyx

x= 1+

−��

��1

12

log logyx

x= 1+

−��

��

1

2 1

log log logy x x= + − −1

21 1� � � �

diff. w.r.t. x.

1 1

2

1

1

1

11

y

dy

dx x x

d

dxx⋅ =

+−

−⋅ −

��

1 1

2

1

1

1

11

y

dy

dx x x⋅ =

+−

−−

��


dy

dx

y

x x=

++

−�

� �2

1

1

1

1

dy

dx

y x x

x x= − + +

+ −�

� �

2

1 1

1 1� ��

dy

dx

x

xx x

=

+− ⋅

+ −�

� �

112

2

1 1� ��

dy

dx

x

x x x x x= +

−⋅

+ −=

+ ⋅ −1

1

1

1 1

1

1 1 3 2� ��

��

yx

x= +

−1

1

diff. w.r.t. x.

dy

dx x

x

d

dx

x

x=

+−

⋅ +−

��

��

1

211

1

1

dy

dx xx

xddx

x xddx

x

x=

+−

⋅− ⋅ + − + ⋅ −

−1

211

1 1 1 1

1 2

� � � � � � � �

� �

1

2 1

1 1 1 1

1 2

−+

⋅− − + −

−x

x

x x

x

� ��

1

2 1

1 1

1 2

−+

⋅ − + +−

x

x

x x

x� �

1

2 1

2

1 2

−+

⋅−

x

x x� �

dy

dx x x x x=

+ ⋅ −=

+ ⋅ −−

1

1 1

1

1 1212

32� � � � � �


4.4.4.4.4. If yey = xx, then find dy

dx.

Consider yey = xx


log logye xy x� � = � log log logmn m n= +

log log logy e xy x+ =

log log logy y e x x+ =

log logy y x x+ =

diff. w.r.t. x. � loge = 1

1

y

dy

dx

dy

dxx

d

dxx x

d

dxx⋅ + = ⋅ +log log� �

1 1

1y

dy

dx

dy

dxx

xx⋅ + = ⋅ + ⋅log

dy

dx yx

11 1+

�

� �= + log

dy

dx

x

y

= +

+

11

1

log

dy

dx

y x

y=

++

1

1

log.

� �

5.5.5.5.5. If ey = ax+y, then find dy

dx.

e ay x y= +


log loge ay x y= +

y e x y alog log= +� �

y x y a⋅ = +1 � � log

diff. w.r.t.x.

dy

dxa

dy

dx= +

��

log 1 � log a is a constant.


dy

dxa a

dy

dx= + ⋅log log

dy

dxa

dy

dxa− ⋅ =log log .

dy

dxa a1− =log log .� �

dy

dx

a

a=

−log

log.

1

��! �� "��

If y = f (x), then

dy

dxy y f x= = ′ = ′1 � � is first order derivative of y with respect to x. It is a function of x. The deriva-

tive of this function with respect to x is d y

dxy y f x

2

2 2= = ′′ = ′′� � and is called second order derivative.

i.e., d

dx

dy

dx

d y

dxf x y y�

�� = = ′′ = ′′ =

2

2 2� � .

��

1.1.1.1.1. Find d y

dx

2

2 if y = x2 + 3x + 8.

Solution: Solution: Solution: Solution: Solution: y x x= + +2 3 8

diff. w.r.t. x.

dy

dxx= + +2 3 1 0� �

dy

dxx= +2 3

diff. again w.r.t. x.

d y

dx

2

2 2 1 0= +� �

d y

dx

2

2 2= .


2.2.2.2.2. If y x ed y

dxx= 2

2

2, . then find

y x ex= 2 .


dy

dxx

d

dxe e

d

dxxx x= ⋅ + ⋅2 2� � � �

dy

dxx e e xx x= ⋅ + ⋅2 2

dy

dxe x xx= +2 2� �

diff. again w.r.t. x using product rule,

d y

dxe

d

dxx x x x

d

dxex x

2

22 22 2= ⋅ + + +� � � �

= + + + ⋅e x x x ex x2 2 22� �

d y

dxe x x xx

2

222 2 2= + + +

d y

dxe x xx

2

22 4 2= + + .

3.3.3.3.3. If x = t2 and y = 4t, then find d y

dx

2

2 at t = 1.

Consider x = t2

diff. w.r.t. t.

dx

dtt= 2

Consider y = 4t.diff. w.r.t. t.

dy

dt= ⋅4 1� �

dy

dt= 4.

dy

dx

dydtdx

dtt

= = 4

2


dy

dx t= 2


d y

dx t

d

dxt

2

2 221= −�� ⋅ � � [Note this step]

d y

dx t t

2

2 2

2 1

2= − ⋅ �

dx

dtt= 2

d y

dx t

2

2 3

1= −dt

dx t= 1

2.

d y

dx t

2

21

3

1

1when == − .

d y

dx t

2

21

1when =

= − .

4.4.4.4.4. If y = e2t and x = log 3t, then

Find d y

dx

2

2 .

y e t= 2

diff. w.r.t. t

dy

dte

d

dttt= ⋅2 2� �

dy

dte et t= ⋅ =2 22 2

x = log 3t

diff. w.r.t. t.

dx

dt t

d

dtt= ⋅1

33� �

dx

dt t= ⋅1

33

dx

dt t= 1

∴∴∴∴∴dy

dx

dydtdx

dt

e

t

t

= = 21

2


dy

dxte t= 2 2

diff. w.r.t. t.

d y

dxt

d

dxe e

d

dxtt t

2

22 22= ⋅ + ⋅

��

d y

dxt e

d

dxt e

dt

dxt t

2

22 22 2= ⋅ ⋅ + ⋅

��

� �

d y

dxte

dt

dxe

dt

dxt t

2

22 22 2= ⋅ + ⋅

��

d y

dxe

dt

dxtt

2

222 2 1= ⋅ +

d y

dxe t tt

2

222 2 1= ⋅ +

5.5.5.5.5. If y ae bemx mx= + − , then prove that y m y22 0− =

y ae bemx mx= + −

diff. w.r.t. x.

y aed

dxmx be

d

dxmxmx mx

1 = ⋅ + ⋅−� � � �

y ae m be mmx mx1 = + −−� � � �

y m ae bemx mx1 = − −


y m aed

dxmx be

d

dxmxmx mx

2 = ⋅ − ⋅ −�

� �

−� � � �

y m ae m be mmx mx2 = ⋅ − −−� � � �

y m ae m be mmx mx2 = + −� �

y m ae bemx mx2

2= + −

y m y22=

�dx

dt tdt

dxt

=

=

1


y m y22 0− =

Hence proved.

� �� #��

��$��% &�� '(��)*�

1.1.1.1.1. Differentiate xe + ex −−−−− ee with respect to x.


Let y = xe + ex −−−−− ee

diff. w.r.t. x.

dy

dxex ee x= + −−1 0

dy

dxex ee x= +−1

2.2.2.2.2. Differentiate 2x + x2 −−−−− logx with respect to x.

Let y = 2x + x2 −−−−− logx

diff. w.r.t. x.

dy

dxx

xx= + −2 2 2

1log

3.3.3.3.3. Differentiate ex x2 2log +� � w.r.t. x.

Let y ex x= +2 2log� �

diff. w.r.t. x.

dy

dxe

d

dxx x

x x= ⋅ ++2 22

2log

log� � � ��

= ⋅ +��

��

+e

xx

x x2 2

21

2log� �

= +��

��

+e

xx

x x2 2 24

log.

� �

4.4.4.4.4. If y x x= + −52 4 7 , then find y1

y x x= + −52 4 7

diff. w.r.t. x.


yd

dxx xx x

14 7 25 5 4 7

2

= + −+ − log � �

y xx x1

4 75 5 2 42

= ⋅ ⋅ ++ − log � �

5.5.5.5.5. Differentiate ex xlog 2 4+� �

w.r.t. x.

Let y ex x

=+log 2 4� �

y x x= +2 4 � e f xf xlog � � � �=

diff. w.r.t. x.

dy

dxx= +2 4.

6.6.6.6.6. If yx

=1

23, then find

dy

dx.

yx x

x= = =−1 1

232

13

23

� �.

y x=−2

3 .

diff. w.r.t. x.

dy

dxx x= − ⋅ = −− − −

2

3

2

3

23

15

3

= − 2

3 5 3x.

7.7.7.7.7. If y = 3−8x, then find dy

dx

y x= −3 8

diff. w.r.t. x.

dy

dx

d

dxxx= ⋅ ⋅ −−3 3 88 log � �

dy

dxx= ⋅ ⋅ −−3 3 88 log � �


8.8.8.8.8. If y x ex= −2 4 , then find y′′′′′

y x ex= −2 4

diff. w.r.t. x.

′ =−

⋅ −yx e

d

dxx e

x

x1

2 44

2

2� �

′ =−

⋅ −yx e

x ex

x1

2 42 4

2

′ = −

−y

x e

x e

x

x

2

42

9.9.9.9.9. If f xx

� � =−4

92 , then find f ′(x)

f x x� � � �= −−

4 92 1

diff. w.r.t. x.

′ = − −�

� � ⋅ −

− −f x x

d

dxx� � � � � �4 1 9 92 1 1 2

′ = − −�

� �

−f x x x� � � �4 9 22 2

′ = −

−f x

x

x� �

� �8

92 2

10.10.10.10.10. If yx

= 1

33, then find y1

yx x

x= = = −1

3

1

331 3 3

3

� �

y x= −3 3

diff. w.r.t. x.

yd

dx

xx1

33 33

= ⋅ ⋅ −��

− log

y x1

33 31

3= ⋅ ⋅ −��

��

− log


y x x1 3 1 3

3

3 3

3

3= −

⋅= −− −

log log.

��+'�% &*�� '(��)*�

1.1.1.1.1. If x2 + y2 = 10, then find dy

dx at (1, −−−−−1)

Consider x2 + y2 = 10

diff. w.r.t. x.

2 2 0x ydy

dx+ ⋅ =

2 2ydy

dxx= − .

dy

dx

x

y= −2

2

dy

dx

x

y= −

dy

dx 1 1

1

1,−= −

−� � � �

dy

dx 1 11

,.

−=

� �

2.2.2.2.2. If x = 4t and y = 5t2, then find dy

dx.

Solution:Solution:Solution:Solution:Solution: dy

dx

dy

dtdxdt

=

Now

y = 5t2

diff. w.r.t. t.

dy

dtt= 5 2� �

dy

dtt= 10


Next, x = 4 t

diff. w.r.t. t.

dx

dt= ⋅ =4 1 4.

dy

dx

dy

dtdxdt

t t= = =10

4

5

2.

3.3.3.3.3. If f x ax

� � = log ,2 then find f ′(a)

f x aa

xxe

e

� � = =loglog

log2 2

=log

loge

e

a

x2

f xa

xe

e

� � = ⋅log

log2

1

diff. w.r.t. x.

′ = ⋅ −�

��

�

��f x

a

x

d

dxxe

e

� �

� �log

loglog

2

12

′ = − ⋅ ⋅f xa

x xe

e

� �

log

log2

1 12

′ = −⋅

f aa

a ae

e

� � log

log4

′ = −f aa

� � 1

4

4.4.4.4.4. If yx

x= −

+log ,

1

1 then find

dy

dx

yx

x= −

+��

��log

1

1

12

yx

x= −

+��

��

1

2

1

1log


y x x= − − +1

21 1log log� � � �

diff. w.r.t. x.

dy

dx x

d

dxx

x

d

dxx=

−⋅ − −

+⋅ +�

� ��

1

2

1

11

1

11� � � �

dy

dx x x=

−− −

+⋅�

� ��

1

2

1

11

1

11� �

dy

dx x x= −

−−

+��

��

1

2

1

1

1

1

=− − − +− +

��

��

1

2

1 1

1 1

x x

x x� ��

dy

dx x= −

−��

��

1

2

2

1 2

dy

dx x= −

−1

1 2 .

5.5.5.5.5. If y x x= + +log 1 2� � , then Prove that dy

dx x=

+

1

1 2.

PrPrPrPrProof:oof:oof:oof:oof: Consider y x x= + +log 1 2� �

diff. w.r.t. x.

dy

dx x x

d

dxx x=

+ +⋅ + +1

11

2

2� �

dy

dx x x x

d

dxx=

+ ++

+⋅ +

�

�

�

��

1

11

1

2 11

2 2

2� �

dy

dx x x xx=

+ ++

+⋅

�

�

�

��

1

11

1

2 12

2 2� �

dy

dx x x

x x

x=

+ +

+ +

+

�

�

�

��

1

1

1

12

2

2


dy

dx x=

+

1

1 2.

Hence proved.

6.6.6.6.6. If xmyn = am+n, then prove that

dy

dx

my

nx= − .

Consider xmyn = am+n.

Taking log on both sides log (xmyn) = log am+n

log log logx y am n m n+ = +

m x n y m n alog log log .+ = +� �diff. w.r.t. x.

mx

ny

dy

dx⋅ + ⋅ ⋅ =1 1

0 � m n a+� � log is constant

m

x

n

y

dy

dx+ ⋅ = 0

n

y

dy

dx

m

x= −

dy

dx

my

nx= − .

Hence proved.

7.7.7.7.7. If y = ax+y, then prove that dy

dx

y a

y a=

−log

log1

Consider y = ax+y


log logy ax y= +

log logy x y a= +� �diff. w.r.t. x.

11

y

dy

dxa

dy

dx= +�

� ��

log

1

y

dy

dxa a

dy

dx= + ⋅log log


1

y

dy

dxa

dy

dxa− ⋅ =log log

dy

dx ya a

1 −��

��=log log

dy

dx

a

ya

=−

log

log1

dy

dx

ay a

y

= −log

log1

dy

dx

y a

y a=

−log

log1

Hence proved.

8.8.8.8.8. If yey = x, then prove that dy

dx

y

x y=

+1� �Consider

yey = x.

taking log on both sides

log logy e xy⋅ =� �

log log logy e xy+ =

log log logy y e x+ ⋅ =

log logy y x+ = � loge = 1

diff. w.r.t. x.

1 1

y

dy

dx

dy

dx x+ =

dy

dx y x

11

1+��

�� =

dy

dx

y

y x

1 1+��

�� =


⇒dy

dx

y

x y=

+1� �Hence proved.

��

Consider yey = x.

diff. w.r.t. x. using product rule.

yd

dxe e

dy

dxy y⋅ + ⋅ =� � 1

y edy

dxe

dy

dxy y⋅ ⋅ + ⋅ = 1

dy

dxe y ey y⋅ + =� � 1

dy

dx e yy=+

1

1� � But ye xy = , ex

yy =

dy

dx x

yy

=+

1

1� �

dy

dx

y

x y=

+1� �Hence proved.

��

1.1.1.1.1. If x y y x x y1 1 0+ + + = ≠ and , prove that

dy

dx x= −

+1

1 2� �Consider

x y y x1 1 0+ + + =

x y y x1 1+ = − +

Squaring

x y y x1 12 2

+ = − +� � � �


x y y x2 21 1+ = +� � � �

x x y y y x2 2 2 2+ = +

x x y y y x2 2 2 2 0+ − − =

x y xy x y2 2 0− + − =� �

x y x y xy x y− + + − =� �� 0

x y x y xy− + + =� � 0

⇒ x y x y xy− = + + =0 0 or

� x y x y xy≠ + + =, 0

y xy x+ = −

y x x1+ = −� �

yx

x= −

+1

diff. w.r.t. x.

dy

dx

xddx

x xddx

x

x=

+ ⋅ − − − ⋅ +��

��

+

1 1

1 2

� � � � � � � �

� �

dy

dx

x x

x

x x

x x=

+ − ++

= − − ++

= −+

1 1 1

1

1

1

1

12 2 2

� ��

Hence proved.

2.2.2.2.2. If x = at2, y = 2at, then prove that

d y

dx at

2

2 3

1

2= −

Consider x = at2

diff. w.r.t. t.

dx

dta t= ⋅ 2� � ...(1)

y at= 2

diff. w.r.t. t.

dy

dxa= 2 1� �


dy

dta= 2 ...(2)

∴dy

dx

dydtdx

dt

a

at t= = =2

2

1

dy

dx t= 1

diff. w.r.t. x.

d y

dx t

dt

dx

2

2 2

1= − ⋅

d y

dx t at

2

2 2

1 1

2= − ⋅ ��

��

d y

dx at

2

2 3

1

2= −

Hence proved.

3.3.3.3.3. Differentiate x from 1st principles:

Let y x=



y y x x+ = +δ δ

δ δy x x y= + −

δ δy x x x= + −


lim limδ δ

δδ

δδx x

y

x

x x x

x→ →= + −

0 0

Add and subtract x in the Denominator of RHS.

lim limδ δ

δδ

δδx x

y

x

x x x

x x x→ →= + −

+ −0 0


lim limδ δ

δδ

δδx x

y

x

x x x

x x x→ →=

+ −+ −0 0

12

12� �

RHS is of the form limx a

n nx a

x a→

−−

with x + δx in place of x

x in place of a.

and 1/2 in place of n.

As δx → 0

x + δx → x.

∴ By applying the formula.

limx a

n nnx a

x ana

→−−

−= 1

we get

lim limδ δ

δδ

δδx x

y

x

x x x

x x x→ →=

+ −+ −0 0

12

12� �

⇒dy

dxx=

−1

2

12

1

dy

dxx

x= =

−1

2

1

2

1

2

∴d

dxx

x� � = 1

2.

4.4.4.4.4. If xmyn = (x + y)m+n, then prove that

dy

dx

y

x= .

PrPrPrPrProof:oof:oof:oof:oof: xmyn = (x + y)m+n


log logx y m n x ym n� � � � � �= + +

log log logx y m n x ym n+ = + +� � � �

m x n y m n x ylog log log+ = + +� � � �diff. w.r.t. x.


mx

ny

dy

dxm n

x y

d

dxx y⋅ + ⋅ = + ⋅

+⋅ +1 1 1� � � �

m

x

n

y

dy

dx

m n

x y

dy

dx+ ⋅ = +

++��

��

1

m

x

n

y

dy

dx

m n

x y

m n

x y

dy

dx+ ⋅ = +

++ +

+⋅

⇒n

y

dy

dx

m n

x y

dy

dx

m n

x y

m

x− +

+⋅ = +

+−

⇒dy

dx

n

y

m n

x y

m n x m x y

x x y− +

+��

��=

+ − ++

� � � ��

dy

dx

n x y m n y

y x y

mx nx mx my

x x y

+ − ++

��

�� =

+ − −+

� � � ��

dy

dx

nx ny my ny

y

nx my

x

+ − −��

��= −

dy

dx

nx my

y

nx my

x

−��

��= −

dy

dx

y

x=

Hence proved.

5.5.5.5.5. If y x xm

= + +2 1� � , then prove that 1 022 1

2+ + − =x y xy m y� � .

Consider

y x xm

= + +2 1� �

diff. w.r.t. x

y m x xd

dxx x

m

12

121 1= + + ⋅ + +

−

� � � �

y m x xx

d

dxx

m

12

1

2

21 11

2 11= + + ⋅ +

+⋅ +

�

�

�

��

−� � � �


y m x xx

xm

12

1

21 1

1

2 12= + + +

+⋅

�

�

�

��

−

� �

y m x xx x

x

m

12

1 2

21

1

1= + + ⋅ + +

+

�

��

�

��

−� �

y mx x

x

m

1

21 1

2

1

1=

+ +

+

− +

� �

ym x x

x

m

1

2

2

1

1=

+ +

+

� �But x x y

m

+ + =2 1� �

ymy

x1 2 1=

+

Cross multiplying

y x my12 1+ =� �

Squarring

y x m y12 2 2 21+ =� �

Diff. again w.r.t. x.

y x x y y m yy12 2

1 22

12 1 2 2⋅ + + ⋅ = ⋅� �

2 1 21 12

2 12y xy x y y ym+ + =� � � �

xy x y m y12

221+ + =� �

Rearranging

x y xy m y22 1

21 0+ + − =� �Hence proved.

6.6.6.6.6. If y = ex log x, then prove that

xy x y x y2 12 1 1 0− − + − =� � � �

Given: y e xx= log

diff. w.r.t. x.


y ex

x ex x1

1= ⋅ + ⋅log

ye

xy

x

1 = + � y e xx= log

ye xy

x

x

1 =+

Cross multiplying

xy e xyx1 = + ...(1)


xy y e x y yx2 1 11 1+ ⋅ = + ⋅ + ⋅� �

xy y e xy yx2 1 1 0+ − − − =

But from (1) xy1 = ex + xy

e xy xyx = −1

xy y xy xy xy y2 1 1 1 0+ − − − − =

xy y xy xy xy y2 1 1 1 0+ − + − − =

xy y xy xy y2 1 12 0+ − + − =

xy y x y x2 1 1 2 1 0− − + + − =� � � �

xy y x y x2 1 2 1 1 0− − + − =� � � �Hence proved.

7.7.7.7.7. Differentiate eax from first principles:

Let y = eax

Let δx be an increment give to x.


y y ea x x+ = +δ δ

δ δy e yax a x= −+

δ δy e eax a x ax= −+

δ δy e e eax a x ax= ⋅ −

δ δy e eax a x= −1



limδ

δδδ δx

ax a xy

x

e e

x→=

−

0

1� �

Multiply and divide by a in RHS.

� limx

xe

x→

− =0

11

limδ

δδδ δx

ax a xy

x

ae e

a x→=

⋅ −

0

1� �

⇒dy

dxa eax= ⋅ ⋅1

∴d

dxe aeax ax� � =

8.8.8.8.8. If y x a x= + +log 2 2� � , then prove that a x y xy2 22 1 0+ + =� � .

Consider y x a x= + +log 2 2� �

diff. w.r.t. x.

yx a x

d

dxx a x1 2 2

2 21=

+ +⋅ + +� �

yx a x a x

d

dxa x1 2 2 2 2

2 211

1

2=

+ ++

+⋅ +

�

�

�

��

� �

yx a x a x

x1 2 2 2 2

11

1

20 2=

+ ++

+⋅ +

�

�

�

��

� �

yx a x

a x x

a x1 2 2

2 2

2 2

1=+ +

+ +

+

�

�

�

��

ya x

1 2 2

1=

+.

a x y2 21 1+ =


Squaring

a x y2 212 1+ =� �

diff. again with respect to x.

a x y y y x2 21 2 1

22 2 0+ ⋅ + ⋅ =� � � �

2 012 2

2 1y a x y xy+ + =� �

a x y xy2 21 1 0+ + =� �

Hence proved.

9.9.9.9.9. If xy = ex−y, then prove that

dy

dx

x

x=

+log

log1 2� �

PrPrPrPrProof:oof:oof:oof:oof: x ey x y= −


log logx ey x y= −

y x x y elog log= −� �y x x ylog = − ...(1) � loge = 1

diff. w.r.t. x.

yx

xdy

dx

dy

dx⋅ + ⋅ = −1

1log

y

xx

dy

dx

dy

dx+ + =log 1

dy

dxx

y

xlog + = −1 1� �

dy

dxx

x y

xlog + = −

1� �

dy

dx

x y

x x= −

+1 log� �dy

dx

y x

x x=

+log

log1� �But x − y = y logx from (1)

∴dy

dx

x

x=

+log

log1 2� �Hence proved.

� x y y x

x y x y

x y x

x

yx

y

x x

− =

= +

= +

= +

=+

�

�

�

�

��

log

log

log

log

log

1

1

1

1

� �

� �


10.10.10.10.10. Differentiate x x x...∞

with respect to e8x.

Let u x v ex xx

= =∞...

. and 8

To find du

dv

du

dxdvdx

=

Now u x xx

=∞...

u xu=Consider log on both sides

log logu xu=

log logu u x=

diff. w.r.t. x.

1 1

u

du

dxu

xx

du

dx⋅ = ⋅ + ⋅log

1

u

du

dxx

du

dx

u

x⋅ − ⋅ =log

du

dx ux

u

x

1 −��

��=log .

du

dx

u

x

ux

u

x u x=

−=

−1 1

2

log log� �

Now v = e8x

diff. w.r.t. x.

dv

dxe

d

dxxx= ⋅8 8� �

dv

dxe x= ⋅8 8.

Hence

du

dv

dudxdvdx

ux u x

e x= =−

2

8

1

8

log� �


du

dv

u

x u x e x=

−

2

88 1 log� � where u x x x

=∞

�

��

ydy

dx

xn nxn−1

x 1

x2 2x

x3 3x2

x1

2 x

1

x− 1

2x

12x

−23x

13x

−34x

ex ex

ax ax loga

logx1

x

Cu Cdu

dx⋅

u v±du

dx

dv

dx±

I II⋅ I II II I⋅ + ⋅d

dx

d

dx� � � �

Nr.

Dr.

*

**

Dr Nr Nr Dr.

Dr.

.d

dx

d

dx� � � � � �

� �

− ⋅2

*: Numerator **: Denominator


• Chain rule: Chain rule: Chain rule: Chain rule: Chain rule: If y = g(u) & u = f(x)

Then y = g[f(x)] is differentiated by chain rule

dy

dx

dy

du

du

dx= ⋅

• Implicit difImplicit difImplicit difImplicit difImplicit difffffferererererentiaentiaentiaentiaentiation:tion:tion:tion:tion: Given function f(x, y) = 0.

diff. w.r.t. x., take dy

dx common among the terms containing

dy

dx and shift the remaining terms to

RHS. Then find dy

dx.

• PPPPParararararametrametrametrametrametric dific dific dific dific difffffferererererentiaentiaentiaentiaentiation:tion:tion:tion:tion:

Given x = f (t) and y = g(t).

Then to find dy

dx, use

dy

dx

dy

dtdxdt

=

• Second orSecond orSecond orSecond orSecond order derder derder derder derder deriiiiivvvvvaaaaatititititivvvvve:e:e:e:e:

If y = f (x), then by differentiating we get dy

dx or y′ or y1 or f ′(x). This is a function of x. By

differentiating this again with respect to x we get d y

dx

2

2 or y″ or f ″(x) or y2.

��

(1)(1)(1)(1)(1) DifDifDifDifDifffffferererererentiaentiaentiaentiaentiate the fte the fte the fte the fte the folloolloolloolloollowing functions frwing functions frwing functions frwing functions frwing functions from fom fom fom fom fiririririrst prst prst prst prst principles:inciples:inciples:inciples:inciples:

(a) xn (b) eax

(c) logax (d) x

(e) 7x

(2)(2)(2)(2)(2) DifDifDifDifDifffffferererererentiaentiaentiaentiaentiate the fte the fte the fte the fte the folloolloolloolloollowing with rwing with rwing with rwing with rwing with respect to espect to espect to espect to espect to xxxxx:::::

(a) x ex4 3 7+ − (b) 3 8 93x x+ +

(c) xx x

+ −1 1

34(d) e e ex e e+ − +π π

(e) a x ex a a+ − (f) 7 77 7x ex e+ + −


(3)(3)(3)(3)(3) Find Find Find Find Find dy

dx if if if if if

(a) y x x= + +3 8 5 72 3� �� (b) y x x= + −7 63� ��

(c) y x x= +��

��5

1

12 (d) yx x

x x= + +

− +

2

2

2

2 3

(e) ye e

e e

x x

x x= −+

−

− (f) yx

x= +

−1

1

2

2

(4)(4)(4)(4)(4) Find Find Find Find Find dy

dx if if if if if

(a) y e e ex x x= + +51

π (b) y e x x= + +−7 68 7π

(c) y e xx= − +6 3 8log � � (d) y x a x= + +log 2 2� �

(e) y e ex x= + −log � � (f) ya

x a

x a= −

+��

��

1

2log

(g) y ee

ex

x

x= +−

��

��

log1

1 (h) y x x a a x x a= + + + +2 2 2 2 2log� �

(i) yx x

x x= − −

+ −

�

��

�

��log

1

1

2

2(j) y e

x

xx= −

+��

��

��

��

log2

2

3 4

5.5.5.5.5. Find Find Find Find Find dy

dx if if if if if

(a) x y2 2 10+ = (b) y x x= + + + ∞log log log ...

(c) y x xx

=∞...

6.6.6.6.6. If (If (If (If (If (aaaaa) ) ) ) ) y a x x2 32 − =� � . . . . . TTTTThen prhen prhen prhen prhen prooooovvvvve thae thae thae thae that t t t t dy

dxa a at , .� � = 2

(b) If x2 + 3y2 − 7xy = 5 Then prove that dy

dx

x y

x y= −

−2 7

7 6

(c) If x y x y y x≠ + + + =, 1 1 0 . Then prove that dy

dx x= −

+1

1 2� �


(d) If ax2/3 + by2/3 = (a2 − b2)2/3

Then prove that dy

dx

y

x= −��

��

13.

7.7.7.7.7. Find Find Find Find Find dy

dx if if if if if

(a) x t y t= = +3 2 1, (b) xt

ty

t

t= −

+=

+1

1

2

1

2

2 2,

(c) xat

ty

at

t=

+=

+3

1

3

13

2

3,

8.8.8.8.8. DifDifDifDifDifffffferererererentiaentiaentiaentiaentiatetetetete

(a) loge x with respect to ex

(b) log10 x with respect to x2

(c) 10x with respect to 5x

(d) If xm⋅yn = (x + y)m+n, then prove that dy

dx

y

x= .

(e) If ex + ey = ex+y, then prove that dy

dxey x= − −

(f) If xy = ex−y, then prove that dy

dx

x

x=

+log

log1 2� � .

9.9.9.9.9. Find Find Find Find Find d y

dx

2

2 if if if if if

(a) x y a2 2 2+ = (b) y x ex= 2

(c)x

a

y

b

2

2

2

2 1+ = (d) x y a3 3 5=

10. (a) If y x xm

= + +1 2� � , then prove that 1 022 1

2+ + − =x y xy m y� � .

(b) If y x a x= + +log 2 2� � , prove that a x y xy2 22 1 0+ + =� � .

(c) If y = aemx + be−mx, then prove that y2 − m2y = 0.

(d) If y = exlogx, then prove that xy x y x y2 12 1 1 0− − + − =� � � � .

(e) If y = (x2 − 1)n, prove that x y x n y ny22 11 2 1 2 0− + − − =� � � � .


(f) If y a x= +2 2 6� � , prove that x a y xy y2 2

2 110 12 0+ − − =� � .

(g) If y axb

xn

n= ++1, then prove that x y n n y2

2 1 0+ + =� � .

(h) If y = (a + bt)e−nt, then prove that d y

dtn

dy

dtn y

2

222 0+ ⋅ + = .

(i) If y a x x b x xn n

= + − + − −2 21 1� � � � , then prove that x y xy n y22 1

21 0− + − =� � .

��

1. (a) nxn−1 (b) aeax (c) 1

x(d)

1

2 x (e) 7x log 7.

2. (a) 4x3 + 3ex (b) 9x2 + 8 (c) 1

2

1

2

3

4

1

4

5

4

x x x

x− +−

(d) ex

(e) ax loga + axa−1 (f) 7x log7 + 7x6.

(3) (a) 3 8 15 5 7 62 2 3x x x x+ + +� � � � (b) x x xx

+��

�� + − �

��

−7

1

36

1

2

23 3� � � �

(c) x xx

5 51

12

1

2� � + +�

�� (d)

x x x x x x

x x

2 2

2 2

2 3 2 1 2 2 2

2 3

− + + − + + −

− +

� ��

� �

(e) 4

2e ex x− −� � (f)

4

1 2 2

x

x−� �

4. (a) 5 5

1

2ee

xex

xx− + π π (b) − + ⋅ ⋅−7 6 8 87 6e x x log

(c) e

x

x

x

x6 2

32 66

3

8⋅ −

+(d)

12 2x a+

(e) e e

e e

x x

x x

+−

−

− (f)1

2 2x a−


(g) ee

e

e

ex

x

x

x

xlog+−

��

�� − −

1

1

2

1

2

2 (h) 2 2 2x a+

(i) 2

2 1 12 2x x− −� � (j)x

x

2

2

1

4

−−

5. (a) −x

y (b) 1

2 1x y −� � (c) y

x y x

2

1− log� �

7. (a) 2

3t(b)

t

t

x

y

2 1

2

−− or (c)

2

1 2

4

3

t t

t

−−

8. (a) e

x

x−

(b) 1

2 102x elog (c) 10 10

5 5

x

x

log

log

9. (a) −a

y

2

3 (b) e x xx 2 4 2+ +� � (c) −b

a y

4

2 3 (d) 15

4 2

y

x.

��

��

Derivatives have many applications. To mention a few, we use differentiation in finding 1. Equation oftangent and normal. 2. Length of subtangent and subnormal 3. Angle between the curves 4. Ratemeasure i.e., variation with respect to time 5. Maximum and minimum values of the function.

Derivatives are also used in finding marginal cost and marginal revenue.

��

VVVVVelocity:elocity:elocity:elocity:elocity: Rate of change of displacement is called velocity. If v is the velocity, ‘s’ is the displacementand s = f (t) Then

Velocity = = ′ds

dtf t� �

AccelerAccelerAccelerAccelerAcceleraaaaation:tion:tion:tion:tion: The rate of change of velocity is called acceleration. If ‘a’ is the acceleration v is thevelocity. Then

adv

dt

d

dt

ds

dt

d s

dt= = �

�� =

2

2


1. Rate means differentiation with respect to time, t.

Rate of increase of area = dA

dt.

Rate of decrease of volume = dV

dt and so on.

2. Formulae:

(a) Area of square = (side)2 = S2. (d) Volume of sphere = 4

33πr .

(b) Area of circle = π(radius)2 = πr2. (e) Volume of a cylinder = πr2h.

(c) Surface area of Sphere = 4πr2 (f) Volume of a cone = 1

32πr h.


�� !"��

1.1.1.1.1. If the displacement of a particle at time t seconds is s = 3t2 − 7t + 6, then find the velocity andacceleration at t = 1 second.

Given: s = 3t2 − 7t + 6

diff. w.r.t. t.

ds

dtt= −3 2 7 1� � � �

Velocity = = −ds

dtt6 7

Velocity when t = 1 sec. = 6(1) − 7.

= 6 − 7 = −1 unit/sec.

Now

Velocity = 6t − 7

v = 6t − 7

diff. w.r.t. t.

dv

dt= −6 1 0� �

dv

dt= 6

acceleration = 6 units/sec2

acceleration when t = 1 is 6 units/sec2.

2.2.2.2.2. When the brakes are applied to the moving car, the car travels a distance s mts in t seconds givenby s = 6t − 3t2, when and where does the car stops?

Solution:Solution:Solution:Solution:Solution: Car stops when velocity = 0.

Now s = 6t − 3t2.

diff. w.r.t. t.

ds

dtt= −6 1 3 2� � � �

ds

dtt= −6 6

Velocity = =ds

dt0

∴ 0 6 6= − t

⇒ 6 6 1= ⇒ =t t .

Application of Derivatives 431

∴ Car stops when t = 1 sec.

Distance travelled in 1 sec.

s t t= −6 3 2 when t = 1 sec.

s = −6 1 3 1 2� � � �s = −6 3

s = 3 mts.

So car stops after travelling a distance of 3 mts.

3.3.3.3.3. With usual notation, if s = at + b. Where a and b are constants then prove that velocity is constantand acceleration is zero.

Solution:Solution:Solution:Solution:Solution: Given s = at + b

diff. w.r.t. t.

ds

dta= +1 0� � � a and b are constants

ds

dta= = =velocity a constant.

diff. w.r.t. t.

d s

dta

2

2 0= is constant.�

∴ acceleration = 0

4.4.4.4.4. A square plate expands uniformly, the side is increasing at the rate of 6 mm/sec.

What is the rate of increase in area when side is 13 mm.

Solution:Solution:Solution:Solution:Solution: Given

Rate of increase of side = 6 mm/sec.

i.e., dS

dt= 6 mm sec.

To find: dA

dt when S = 13 mm.

We have Area of square = side × side

A S= 2

diff. w.r.t. t.

dA

dtS

dS

dt= ⋅2

dA

dt= 2 13 6� ��


dA

dt= 156 mm2 sec.

∴ Rate of increase of area = 156 mm2/sec.

5.5.5.5.5. A stone is dropped into a pond, waves in the form of circles are generated and the radius of theoutermost wave increase at the rate of 2 mm/sec. How fast is the area increasing (a) when theradius is 5 mm (b) after 3 sec.

Solution:Solution:Solution:Solution:Solution: If r is the radius at time t;

Then given: dr

dt= 2 mm sec

To find (a) dA

dtr when mm= 5

(b) dA

dtt when sec= 3

We have

Area of circle = πr2

A = πr2

diff. w.r.t. t.

dA

dtr

dr

dt= π 2� � ...(1)

(a)dA

dt= ⋅π 2 5 2� ��

dA

dt= 20π

dA

dt= 20π mm2 sec

(b)dA

dtr

dr

dt= π 2� �

Now we have radius is increasing at the rate of 2 mm/sec.

After 3 seconds, radius = 3 × 2 = 6 mm.

( sec�

dr

dt= 2 mm

For 1 sec radius is 2 mm.

For 3 sec. radius is 3 × 2 = 6 mm)

From (1)


dA

dtr

dr

dt= ⋅π 2� �

dA

dt after 3 seconds= × ⋅π 2 6 2� �

dA

dt after 3 sec= 24π

∴ After 3 seconds, Area is increasing at the rate of 36 π mm2/sec.

6.6.6.6.6. Water flows into a cylindrical tank of radius 4 mts at 80,000 cc/hr. How fast water level is raising?

Solution:Solution:Solution:Solution:Solution: If r is the radius, h is the height. The volume V is given by

V r h= π 2

GivendV

dt= 80 000, cc hr.

If h is the height of the water level, dh

dt represent rate of increase in water level.

r = radius = 4 mts = 400 cm = a constant while filling the water in the cylindrical tank.

Consider

V r h= π 2

diff. w.r.t. t.

dV

dtr

dh

dt= ⋅π 2

80 000 400 2, = ⋅π � � dh

dt

80 000

160000

,

π= dh

dt

= =1

2πdh

dt

∴ Water level is rising at the rate of 1

2π cms/hr.

7.7.7.7.7. Sand is poured at the rate of 30 cc/sec and it forms a conical pile in which the diameter of thecircular base is always equal to one third the height. At what rate height of the pile is increasingwhen the height is 30 cm.


2 × Radius = Diameter


Given Diameter = 1

3⋅ h

2 × radius = 1

3h

r h= 1

6 ...(1)

dV

dt= 30 cc sec

To find dh

dt when h = 30 cm.

We have volume of the cone = 1

32πr h.

V r h= 1

32π

V h h= ⋅ �� ⋅1

3

1

6

2

π [from (1)]

Vh h h= ⋅ ⋅ =1

3 36 108

2 3

π π

Vh= π 3

108

diff. w.r.t. t.

dv

dth

dh

dt= ⋅�

� ��

π108

3 2

30108

3 30 2= × ⋅��

��

π � � dh

dt

30 108

3 900× = ×π

dh

dt

30 108

3 900

×× ×

=π

dh

dt

⇒ dh

dt= 6

5π

3h

Fig. 17.1


Height of the pile is increasing at the rate of 6

5π cm/sec.

8.8.8.8.8. A spherical snow ball is forming so that its volume is increasing at the rate of 8 cm/sec. Find therate at which the radius is increasing when the snow ball is 2 cm in diameter. Also find the rateof increase in surface area.

Solution.Solution.Solution.Solution.Solution.

Given: dV

dt= 8 cm sec.

To find: dr

dt when diameter = 2 cm.

and ds

dt i.e. radius

2

21= cm .

We have for a sphere,

V r= 4

33π

diff. w.r.t. t.

dV

dtr

dr

dt=

/⋅ / ⋅4

33 2π

dV

dtr

dr

dt= ⋅4 2π

8 4 1 2= ⋅ ⋅π � � dr

dt

⇒dr

dt= =8

4

2

π π

∴dr

dt= 2

πcm sec.

Now surface area = 4 πr2.

diff. w.r.t. t.

ds

dtr

dr

dt= 4 2π � �

ds

dt= ⋅4 2 1

2ππ

� �

ds

dt= 16 sq cm sec.


9.9.9.9.9. A ladder 13 ft long rests with its ends on a horizontal floor and against a smooth vertical wall. Ifthe upper end is coming downwards at the rate of 1 ft/min. Find the rate at which the lower endmoves, when the upper end is 5 ft from the ground.

Solution:Solution:Solution:Solution:Solution: Let PQ be the ladder.

At time t, Let OP = y,

OQ = x

From figure OP2 + OQ2 = PQ2

y2 + x2 = 132

Given dy

dt= 1ft min.

To find dx

dt when y = 5ft.

i.e., To find dx

dt when y = 5 ft. and x = 12 ft.

y x2 2 213+ =

5 1692 2+ =x

x2 169 25= −

x2 144=

x = 144

x = 12

From fig.,

y x2 2 213+ =

diff. w.r.t. t.

2 2 0ydy

dtx

dx

dt+ =

2 2xdx

dty

dy

dt= −

dx

dt

y

x

dy

dt= − ⋅2

2

dx

dt

y

x

dy

dt= −

dx

dt= − ⋅5

121

Fig. 17.2

0 x

y

Q

P


dx

dt= − 5

12

∴ Lower end moves at the rate of 5

12 ft/min. Negative sign shows as x increases, y decreases.

10.10.10.10.10. An aeroplane at an altitude of 400 kms flying horizontally at 500 km/hr passes directly over anobserver. Find the rate at which it is approaching the observer when it is 500 kms away from him.

Solution.Solution.Solution.Solution.Solution. Let at time ‘t’ ‘A’ be the position of theaeroplane, B be the point on the path directly abovethe observer O. Let AB = x. OA = y.

From fig. AB OB OA2 2 2+ =

x y2 2 2400+ =

diff. w.r.t. t.

2 0 2xdx

dty

dy

dt+ = ...(1)

Givendx

dt= −500 km hr

[−ve Sign is taken since x is decreasing]

To find dy

dt when y = 500 kms.

From fig.

x y2 2 2400+ =

x2 2 2400 500+ = � y = 500� �

x2 2 2500 400= −

x2 250000 160000= −

x2 90000=

⇒ x = 300

Substituting x = 300, y = 500 and dx

dt= −500 in (1) we get

2 300 500 2 500� �� − = ⋅ dy

dt

− =300dy

dt

O

y 400 kms

A x B

Fig. 17.3


∴dy

dt= −300 kms hr.

∴ The aeroplane is approaching at the rate of 300 km/hr.

��# �� $��$��

Let y = f (x) be a continuous function. As x incrincrincrincrincreases,eases,eases,eases,eases, if y also incrincrincrincrincreaseseaseseaseseaseseases then y = f (x) is said to be anincrincrincrincrincreasing function.easing function.easing function.easing function.easing function.

Ex.: y = x + 3 is an increasing function

�

x

y

0 1 2 3

3 4 5 6

.......

.......

increase

increase

→

→

As x incrincrincrincrincreaseseaseseaseseaseseases if y decrdecrdecrdecrdecreases eases eases eases eases then the function y = f(x) is said to be a decrdecrdecrdecrdecreasing functioneasing functioneasing functioneasing functioneasing function.

Ex.: y = 7 − x is a decreasing function.

�

x

y

0 1 2 3

7 6 5 4

.......

.......

decrease

increase

→

→

Note that for an increasing function, dy

dx> 0 i.e. positive.

For a decreasing function, dy

dx< 0 i.e., negative.

y is said to be stationary at a point if it neither increases nor decreases. At such point dy

dx= 0 .

If a continuous function increases upto a certain value and then decreases from that value, then thatvalue is called a maximum value of a function.

Similarly if continuous function decreases to a value and then increases, then that value is calledminimum value of the function.

Note that a continuous function may attain maxima/minima at several points or it may neither havemaxima nor minima.

In the figure, the points P, Q, R, S, are points of maxima. The points A, B, C, are points of minima.

Between any 2 maxima there must be a minima and vice versa.

Fig. 17.4

Fig. 17.5


Y

X

S

Fig. 17.6

��%�� $��$��

Let y = f (x) be the given function. To find the maximum and / or minimum value of the function. We

find dy

dx and equate it to zero. Let x = a, x = b and x = c be the points.

We find d y

dx

d y

dxx a x a

2

2

2

2at at

, if = =

is greater than zero, then x = a is a point of minima. Minimum value

of the function is y = f (a).

Next we find d y

dx

d y

dxx b x b

2

2

2

2at at

if = =

is less than zero, then x = b is a point of maxima. Maximum

value of the function is y = f (b).

If d y

dx x c

2

2at =

is equal to zero, then at x = c, the function neither attains maxima, nor minima. The

point x = c is called a point of inflexion.

�� !"��

1.1.1.1.1. Find the maximum and minimum value of the function 2x3 − 15x2 + 36x + 10.

Solution:Solution:Solution:Solution:Solution: Let y = 2x3 − 15x2 + 36x + 10.

diff. w.r.t. x.

dy

dxx x= − + +2 3 15 2 36 1 02 � � � �

dy

dxx x= − +6 30 362 . ...(1)


At extremum, dy

dx= 0 .

0 6 30 362= − + ÷x x by 6.

x x2 5 6 0− + =

x x x2 3 2 6 0− − + =

x x x− − − =3 2 3 0� � � �

∴ x x− − =2 3 0� � � �

x x= =2 3 or .

Consider equation (1)

dy

dxx x= − +6 30 362

diff. w.r.t. x.

d y

dxx

2

2 6 2 30 1 0= − +� � � �

d y

dxx

2

2 12 30= −

d y

dx x

2

22

12 2 30at =

= −� �

= − < →6 0 negative.

∴ x = 2 is a point of maxima.

Maximum value of the function

2 15 36 103 2x x x− + +

is 2 2 15 2 36 2 103 2� � � � � �− + +

2 8 60 72 10� � − + +

16 60 72 10− + +

= − =98 60 38.

Now

d y

dxx

2

2 12 30= −

d y

dx x

2

23

12 3 30 6 0at

positive=

= − = > →� �

∴ x =3 is a point of minima.

−

−−

36

25

2x

xxx


Minimum value of the function

2x x x3 215 36 10− + + is

2 3 3 3� � � � � �3 215 36 10− + +

2 27 9� � � �− + +15 108 10

54 135 118− +

172 135 37− = .

2.2.2.2.2. Find two numbers such that their sum is 20 and their product is maximum.

Solution.Solution.Solution.Solution.Solution. Let the 2 numbers be x and y.

Given: Their sum = 20.

x y y x+ = ⇒ = −20 20

To find: x and y such that their product is maximum.

Their product = xy

P xy=

P x x= −20� �

P x x= −20 2

For P to be extremum, dP

dx= 0

P x x= −20 2

diff. w.r.t. x.

0 20 1 2= −� � x

2 20x =

x = 10.

So the numbers are x = 10 and y = 20 − 10 = 10. Reqd. Numbers are: 10 and 10.

3.3.3.3.3. Find two numbers whose sum is 16 and sum of their cubes is minimum.

Solution:Solution:Solution:Solution:Solution: Let the 2 numbers be x and y.

Given: Their Sum = 16.

x y+ = 16

⇒ y x= −16 ...(1)

To find: x and y such that sum of their cubes is minimum.

i.e., x y3 3+ Should be minimum.


Let P x y= +3 3

P x x= + −3 316� � [from (1)]

For P to be extremum,

dP

dx= 0

Now P x x= + −3 316� �diff. w.r.t. x.

dP

dxx x

d

dxx= + − ⋅ −3 3 16 162 2� � � �

dP

dxx x= + − −3 3 16 12 2� � � �

0 3 3 162 2= − −x x� �

0 3 16 322 2 2= − + −x x x

0 256 322 2= − − +x x x

0 256 32= − + x

32 256x =

x = = =256

32

16

28.

So the numbers are x = 8 and y = 16 − 8 = 8 i.e. 8 and 8.

4.4.4.4.4. Prove that logx do not have maxima or minima.

PrPrPrPrProof:oof:oof:oof:oof: Let y = log x.

For maxima or minima, dy

dx= 0

y x= log

dy

dx x= 1

01=x

1

0x

=


⇒ x has no finite value.

∴ logx do not have maxima or minima.

5.5.5.5.5. The product of 2 positive numbers is 36. Find the minimum value of their sum.

Solution:Solution:Solution:Solution:Solution: Let the 2 number be x and y.

Given: Their product = 36.

xy = 36

⇒ yx

= 36

Their sum = x + y.

S xx

= + 36

For S to be extremum dS

dx= 0

S xx

= + 36

diff. w.r.t. x.

dS

dx x= + −�

��1 36

12

0 136

2= −x

0362

2= −x

x

⇒ x2 36 0− =

x x2 36 6− ⇒ = ±

dS

dx x= −1

362

diff. w.r.t. x.

Nowd S

dxx

2

2336 2= − − −

d S

dx x

2

2 3

72=

d S

dx x

2

26

3

72

60

atpositive.

== > →


∴ x = 6 is a point of minima.

yx

= = =36 36

66.

∴ Minimum value of their sum = 636

6+

= + =6 6 12.

6.6.6.6.6. Prove that a maximum rectangle that can be drawn with a constant perimeter is a square.

PrPrPrPrProof:oof:oof:oof:oof: Let x be the length and y be the breadth of a rectangle.Given: Perimeter = constant.

Sum of all sides = constant.

x y x y k+ + + = 2 say� �

2 2 2x y k+ =

2 2x y k+ =� �

x y k+ =

y k x= −

Area of rectangle = Length × Breadth

A xy=

A x k x= −� �

A kx x= − 2

For area to be extremum, dA

dx= 0

A kx x= − 2

diff. w.r.t. x.

dA

dxk x= −1 2� �

0 2= −k x

⇒ 2x k=

xk=2

Consider dA

dxk x= − 2 diff. again w.r.t. x,

d A

dx

2

2 2 0= − < for all

values of x.

A B

CD

x

x

y y

Fig. 17.7


∴ A attains maxima for all values of x.

Now,

y k x kk k k k= − = − = − =2

2

2 2

Hence x yk= =2

Length 2 22

xk

k= ⋅ =

Breadth 2 22

yk

k= ⋅ =

∴ Rectangle has Length = Breadth.

Hence it is a square.

7.7.7.7.7. Prove that the maximum rectangle that can be inscribed in a circle is a square.

A B

CD

x

x

y y

Fig. 17.8

Let ABCD be the rectangle with length x and breadth y inscribed in a circle of radius say a.

∴ OB OD a= =

∴ BD a= 2

From right angled triangle, ABD,

AB AD BD2 2 2+ =

x y a2 2 22+ = � �

x y a2 2 24+ =

y a x2 2 24= −

y a x= −4 2 2


Now, Area of the rectangle = Length × Breadth

A x y= ×

A x a x= −4 2 2

A x a x= −4 2 2

For A to be maximum/minimum dA

dx= 0

A x a x= −4 2 2

diff. w.r.t. x.

dA

dxx

d

dxa x a x

d

dxx= ⋅ − + − ⋅4 42 2 2 2� � � �

dA

dxx

a x

d

dxa x a x= ⋅

−⋅ − + − ⋅1

2 44 4 1

2 2

2 2 2 2 � �

dA

dx

x

a xx a x=

−⋅ − + −

2 42 4

2 2

2 2� �

02

2 44

2

2 2

2 2=−

−+ −

x

a xa x

04

4

2 2 2

2 2= − + −

−

x a x

a x

0 2 42 2= − +x a

⇒ 2 42 2x a=

x a2 22=

x a= 2

Alsod A

dx x a

2

22at =

is < 0.

∴ A attains maxima at x a= 2 .

Now y a x a a= − = −4 4 22 2 2 2

y a a a= − =4 2 22 2


∴ Rectangle has maximum area when its length = 2 a and Breadth = 2 a , i.e., when its length =

Breadth = 2 a , i.e., when it is a square.

8.8.8.8.8. What is the largest size rectangle that can be inscribed in a semicircle of radius r unit so that 2vertices lie on the diameter.

Solution:Solution:Solution:Solution:Solution: Let AB be the diameter of a semicircle.Let 2 vertices P and Q of the rectangle PQRS lie onthe diameter.

Let PQ RS x= = 2

PS QR y= = 2

Join OR. Given: r is the radius of the semicircle.

From fig.

OQ QR OR2 2 2+ =

x y r2 2 22+ =� �

x y r2 2 24+ =

4 2 2 2y r x= −

y r x2 2 21

4= − ...(1)

Now,

Area of the rectangle = Length × Breadth

A x y= ×2 2

A xy= 4

A x r x= ⋅ −41

22 2

from (1)

A x r x= −2 2 2

Area is extremum when dA

dx= 0

Now A x r x= −2 2 2

diff. w.r.t. x.

dA

dxx

d

dxr x r x

d

dxx= −�

� ��

+ − ⋅2 22 2 2 2� � � �

Fig. 17.9

A BP Q2x

S R2x

2y2y

O


dA

dxx

r x

d

dxr x r x= ⋅

−⋅ − + − ⋅2

1

22

2 2

2 2 2 2

dA

dx

x

r xx r x=

−⋅ − + −

2 2

2 22 2� �

02 22 2 2

2 2=

− + −

−

x r x

r x

0 2 2 22 2 2= − + −x r x

4 22 2x r=

⇒ xr2

2

2=

xr=2

.

Also d A

dx xr

2

2

2at

is negative.=

Hence A attains maxima when x r= 2 .

∴ Area of rectangle is maximum when its length = 2 22

2xr

r= ⋅ =

Breadth = 2 21

2 22

2

y rr= ⋅ − ��

Breadth = − =rr r2

2

2 2.

Maximum area Length Breadth= ×

= ×rr

22

= r2.

9.9.9.9.9. A box is constructed from a square metal sheet of side 60 cm by cutting out identical squares fromthe four corners and turning up the sides. Find the length of the side of the square to be cut outso that the box is of maximum volume.


Solution:Solution:Solution:Solution:Solution: Let the side of the cut out square be x.

Now length of the box = 60 − 2x

Breadth of the box = 60 − 2x

Height of the box = x.

Volume = Length × Breadth × height.

V x x x= − −60 2 60 2� ��

V x x= − ⋅60 2 2� �

V attains extrema when dV

dx= 0

Consider

V x x= −60 2 2� �diff. w.r.t. x.

dV

dxx

d

dxx x

d

dxx= − ⋅ + ⋅ −60 2 60 22 2� � � � � �

dV

dxx x x

d

dxx= − ⋅ + ⋅ − ⋅ −60 2 1 2 60 2 60 22� � � � � �

dV

dxx x x= − + − −60 2 2 60 2 22� � � ��

dV

dxx x x= − − +60 2 240 82 2� �

dV

dxx x x x= + − − +60 4 240 240 82 2 2

0 3600 12 4802= + − ÷x x by 12.

300 + − =x x2 40 0

x x2 40 300 0− + =

x x x2 30 10 300 0− − + =

x x x− − − =30 10 30 0� � � �

x x− − =30 10 0� �� x x= =30 10 or

Now

dV

dxx x= + −3600 12 4802

−

−−

30300

1040

2x

xxx

x x

x x

x x

x x

60 2x−

60 2x−

Fig. 17.10


d V

dxx

2

2 24 480= − .

d V

dx x

2

210at

is negative.=

∴ V attains maxima when x = 10.

∴ Length of the square to be cut = 10 cm.

10.10.10.10.10. Prove that ±��

��

−1

2

12, e are the points of inflection for the curve y e x= − 2

.

PrPrPrPrProof:oof:oof:oof:oof: Consider

y e x= − 2

diff. w.r.t. x.

dy

dxe

d

dxxx= ⋅ −− 2 2

dy

dxe xx= ⋅ −− 2

2� �


d y

dxe

d

dxx x

d

dxex x

2

2

2 2

2 2= ⋅ − + −− −� � � � � �

d y

dxe x e xx x

2

2

2 2

2 2 2= ⋅ − − ⋅ ⋅ −− −� � � � � �

d y

dxe x ex x

2

222 4

2 2

= − +− −

y attains neither maxima nor minima when

d y

dx

2

2 0= i.e.

0 2 42 22= − +− −e x ex x

0 2 1 22 2= − +−e xx

e xx− = − + =2

0 1 2 02 or

− =

=

=

= ±xx

x

x

22

02 1

1

20

or


Now

when x y e e ex= = − = =−−��

�� −1

2

2

21

21

2,

when x y e ex= − = =− −1

2

212, .

∴ The points of inflection for the curve y e e ex=��

�� −

��

��− − −2 1

2

1

2

12

12 are and , , .

Hence proved.

��&��

• Velocity = ds

dt

• Acceleration = =dV

dt

d s

dt

2

2

• Rate means differentiation w.r.t. t.

∴ rate of change of area = dA

dt

• Area of Square = S2

• Area of circle = πr2

• Surface area of sphere = 4πr2

• Volume of Sphere = 4

33πr

• Volume of a cylinder = πr2h

• Volume of a cone = 1

32πr h

• For an increasing function dy

dx> 0 and for a decreasing function

dy

dx< 0.

• To find maximum and/or minimum value of the function y = f (x), find dy

dx, equate it to zero. Let

x = a, x = b, x = c be the points. Find d y

dx

d y

dx

d y

dxx a x b x c

2

2

2

2

2

2at at at

and = = =

,..


if d y

dx x a

2

2 0at =

> , then x = a is a point of minima. Minimum value of the function is y = f (a).

if d y

dx x b

2

2at =

is less than zero, x = b is a point of maxima. Maximum value of the function is

y = f (b).

if d y

dx x c

2

2at =

is equal to zero, then x = c is called point of inflection. At x = c the function neither

attains maxima nor minima.

� ��'��

1. The distance s is metres moved by a particle in ‘t’ seconds is given by s = 45t + 11 t2 − t3. Findthe time when the particle comes to rest?

2. The displacement s of a particle at time ‘t’ seconds is given by 2t3 − 3t2 − 36 t + 90. Find the (a)velocity after 4 seconds (b) displacement and acceleration when the velocity vanishes (c) accel-eration after 4 seconds.

3. With usual notation if.

If s2 = at2 + 2bt + c, then prove that

(a) the acceleration is inversely proportional to s3.

(b) the acceleration is a v

s

− 2

where v is the velocity.

4. The equation of motion of a particle is given by s = 9t2 − t3. Find the displacement when velocityis zero and velocity when the acceleration is zero.

5. If s = at3 + bt, find a and b given that when t = 3 velocity is zero and acceleration is 14 units.

6. When breaks are applied to the moving car, the car travels a distance S feet in t seconds given bys = 20t − 40 t2. When does the car stop?

7. The side of a square sheet metal is increasing at 3 mm/min. At what rate is the area increasingwhen the side is 10 mm long.

8. A circular patch of oil spreads on water, the area is growing at the rate of 2 sq cm/hr. How fastare the radius and the circumference increasing when the diameter is 24 cm.

9. A drop of ink spreads over a blotting paper so that the circumference of the blot which is circularincreases at the rate of 3 cm/min. Find the rate of increase of the radius and area when itscircumference is 4π cm.

10. A stone is dropped into a pond, waves in the form of circles are generated and the radius of theoutermost ripple increases at the rate of 2 mm/min. How fast is the area increasing when theradius is 5 mm, after 5 min?

11. A cylindrical tank is 10 mts in diameter, water is flowing in it at the rate of 24 m3/min. at whatrate height of the water is rising?


12. Water is flowing into a right circular cylindrical tank of radius 50 cm at the rate 500 π cc/min.Find how fast is the level of water rising?

13. A ladder 20 ft long rests with its ends on a smooth horizontal floor and against a smooth verticalwall. If the lower end is moved at the rate of 5 ft/min. Find the rate at which the upper end moveswhen the lower end is 12 ft. from the wall.

14. The radius of the sphere is decreasing at the rate of 3 cm/sec. Find the rate at which surface areais decreasing when radius is 12 cm.

15. The height of circular cone is 30 cm and it is constant. The radius of the base is increasing at therate of 0.25 cm/sec. Find the rate of increase of volume of the cone when the radius is 10 cm.

16. A man 6 ft tall is moving directly away from a lamp at a height of 10 ft above the floor. If he ismoving at the rate of 6 ft/sec find the rate at which the length of his shadow is increasing?

17. Find the maximum and minimum value of the function 4x3 − 15x2 + 12 x + 7.

18. Find the maximum and minimum value of the function x3 − 3x2 − 9x + 17.

19. Prove that the function xe−x attains maxima at x = 1 and its maximum value is 1

e.

20. Prove that xx is minimum when xe

= 1.

21. Prove that the maximum value of 1

x

x�� is e1/e.

22. Prove that x3 − 3x2 − 9x + 9 has a point of inflection at x = 1

23. Prove that y = a−(x − b)2/5 has no point of inflection.

24. Find two number whose sum is 24 and their product is maximum.

25. The sum of two numbers is 20. If the product of square of the first and cube of the 2nd ismaximum. Find the numbers.

26. Prove that the area of the right angled triangle of given hypotenuse is maximum when the triangleis isoceles.

27. Prove that among all rectangles of a given area the square has minimum perimeter.

28. Prove that among all rectangles of given perimeter, the square is the one with shortest diagonal.

29. What is the largest rectangle that can be inscribed in a semicircle of radius 1 so that two verticeslie on the diameter.

30. A rectangular field is to be fenced off along the bank of a river, no fencing is required along theriver. If the material for fence cost Rs. 8 per running foot for 2 ends and Rs. 12 per running footfor the side parallel to the river, find the dimension of the field of largest possible area that canbe enclosed with Rs. 3600 worth of fences.

�$��

1. 9 Seconds

2. (a) 0 (b) 9 cm; 30 cm/sec2 (c) 42 cm/sec2,


4. 108 units; 27 units/sec. 5. a = 7

9, b = −21

6.1

4sec. 7. 60 mm2/min.

8.1

12π cm/hr; 1

6 cm/hr. 9.

3

2π cm/min; 6 sq cm/min

10. 20 π mm2/m’ 40 πmm2/min 11.24

25π m/min

12.1

5 cm/min 13. − 15

4 ft/sec.

14. 288 π sq cm/sec. 15. 50 π cm3/sec.

16. 9 ft/sec. 17.39

43;

18. 22; −10 24. 12; 12

25. 8 and 12 30. 1121

2ft and 150 ft.

��

��

If d

dxf x g x� � � �= ,

d

dxf x c g x� � � �+ = . Then g(x) is called integrand and f(x) or f(x) + c is called

integral or primitive or antiderivative of g(x) with respect to x. It is denoted by g x dx� � .�∴ g x dx f x c� � � �� = +

��

By using the definition of integration as the reverse operation of differentiation, the following formulaemay be obtained.

1.1.1.1.1. x dxx

ncn

n

� =+

++1

1

Provided n ≠ −1

�

d

dx

x

nc

n x

nx

n nn

+ + −

++

��

�� =

++

+ =1 1 1

1

1

10

� �

if n = −1,

x dxx

dx x c−� �= = +1 1log

Sinced

dxx c

x xlog + = + =1

01

2.2.2.2.2. e dx e cx x� = +

�

d

dxe c ex x+ = �


3.3.3.3.3. a dxa

acx

x

� = +log

�

d

dx

a

ac

a a

a

x x

log

log

log+

��

�� = + 0

= ax

4.4.4.4.4. k dx kx c� = +

�

d

dxkx c k k+ = + =� � � �1 0

��

1. If K is constant, f(x) and g(x) are 2 integrable functions, then

(i) k f x dx k f x dx⋅ = ⋅� ��

i.e., Integral of constant multiplied by function is constant multiplied by integral of the function.

(ii) f x g x dx f x dx g x dx� � � �� ± = ±� ��i.e., Integral of sum or difference of 2 or more functions = Integral of the first function ± integralof the second function.

��

1.1.1.1.1. Evaluate: 4 13x dx−� �

Solution:Solution:Solution:Solution:Solution: 4 1 4 13 3x dx x dx dx− = −� � � �

= −� �4 13x dx dx

44

4

⋅ − +xx c.

where c is the constant of integration

4 13 4x dx x x c− = − +� � .

2.2.2.2.2. Integrate 1

xex+ with respect to x.

Integration 457

1 1

xe dx

xdx e dxx x+�

�� = +� � �

= + +log .x e cx

3.3.3.3.3. Evaluate: 1 1 4

2 3x x xdx− +�

��

Solution:Solution:Solution:Solution:Solution: 1 1 4 1 1 42 3 2 3x x x

dxx

dxx

dxx

dx− +��

�� = − +� � � �

log xx x

c−− +

+− +

+− + − +2 1 3 1

2 1

4

3 1

log xx x

c−−

+−

+− −1 2

1

4

2

log xx x

c+ − +1 22 .

4.4.4.4.4. Evaluate: xx

dx+�� 1 2

xx

dx xx

xx

dx+�� = + + ⋅ ⋅�

�� 1 1

212

22

= + +� � �x dxx

dx dx22

12

=+

+− +

+ ++ − +x x

x c2 1 2 1

2 1 2 12

= +−

+ +−x x

x c3 1

3 12

x

xx c

3

3

12− + + .

5.5.5.5.5. Evaluate: x x

xdx

2 3 1+ −��

��

Solution:Solution:Solution:Solution:Solution: x

x

x

x xdx

2 3 1+ −��

��


= + −− − −� � �x dx x dx x dx

212

112

123

x dx x dx x dx32

12

123� � �+ −

−

x x x

c

32

112

112

1

32

1

312

112

1

+ + − +

++

+−− +

+

= + − +x x xc

52

32

12

52

332

12

2

52 2

52 3

212x

x x c+ − + .

6.6.6.6.6. Integrate (x2 + 1) (2x3 − 6x) with respect to x.

x x x dx x x x x dx2 3 5 3 31 2 6 2 6 2 6+ − = − + −� � � � �

= − −� 2 4 65 3x x x dx �

= − −� � �2 4 65 3x dx x dx x dx

= − − +2

6

4

4

6

2

6 4 2x x xc

= − − +xx x c

64 2

33 .

7.7.7.7.7. Evaluate: 2 3 2x xe x dx− +� �

2 3 2x xdx e dx x dx� � �− +

2

2

3

3

3xxe

xc

log− + +

2

23

xxe x c

log.− + +

Integration 459

8.8.8.8.8. Evaluate: x

xdx

−��

�� 1

2

2

x

xdx

x

x xdx

−��

�� = −�

�� 1 1

2

2

2 2

2

= −��

�� 1 1

2

2

x xdx

= + − ⋅ ⋅��

�� 1 1

21 1

2 4 2x x x xdx

= + −� � �1 12

12 4 3x

dxx

dxx

dx

x x xc

− + − + − +

− ++− +

− ⋅− +

+2 1 4 1 3 1

2 1 4 12

3 1

x x xc

− − −

−+−

−−

+1 3 2

1 3

2

2

− − + +1 1

3

13 2x x x

c.

Note:Note:Note:Note:Note: If f x dx g x c� � � �� = + , then f ax b dx g ax ba

c+ = + ⋅ +� � � � � 1 where a and b are constants.

ExamplesExamplesExamplesExamplesExamples 1. e dx e cx x� = +

∴ e dx e cx x3 7 3 7 1

3+ +� = ⋅ +

2. x dxx

cx

c33 1 4

3 1 4� =+

+ = ++

∴ 6 46 4

4

1

43

4

− =−

⋅−

+� x dxx

c� � � �

=−−

+6 4

16

4xc

� �.

��

1.1.1.1.1. Evaluate: 1

8 6xdx

−� .


1

8 68 6

1

8xdx x c

−= − ⋅ +� log� �

=−

+log

.8 6

8

xc

� �

2.2.2.2.2. Evaluate: e dxx x3 67−� � .

= −� �e dx dxx x3 67

= − ⋅ +e

cx x3 6

3

7

7

1

6log.

3.3.3.3.3. Evaluate: 4 6 7x dx−� � �

4 64 6

7 1

1

47

7 1

x dxx

c− =−+

⋅ +�+

� � � �

=−

+4 6

32

8xc

� �.

4.4.4.4.4. Integrate 6 4− x with respect to x.

Solution:Solution:Solution:Solution:Solution: 6 4 6 412− = −� �x dx x dx� � .

=−

+⋅−

++6 4

12

1

1

4

12

1xc

� �

= −−

⋅ +6 4

32

1

4

32x

c� �

=−−

+6 4

6

32x

c� �

5.5.5.5.5. Integrate 1

8 9e x− with respect to x.

18 9

8 9

edx e dxx

x−

− −� �= � �

Integration 461

= − +� e dxx8 9�

1

aam

m= −

=−

+− +e

cx8 9

8.

��! "� ��"��#��

To evaluate integrals of the type f x f x dxn� � � �⋅ ′� , we put f (x) = t

Diff. w.r.t. x.

′ =f xdt

dx� �

′ =f x dx dt� � .

Substituting we get

f x f x dx t dtn n� � � �⋅ ′ = ⋅� �

t

nc

n+

++

1

1 provided n ≠ −1

if n = −1,

t dt t dt t cn� �= = +−1 log

∴ f x f x dxf x

nc n

nn

� � � � � �⋅ ′ =

++ ≠ −�

+1

11 if

log f x c n� � + = − if 1

Note:Note:Note:Note:Note: To evaluate ∫ef(x)⋅f ′(x)dx, put f(x) = t and proceed the same way.

��

1.1.1.1.1. Evaluate: x x x dx2 41 2 1+ − ⋅ +� � � �

Solution:Solution:Solution:Solution:Solution: x x x dx2 41 2 1+ − ⋅ +� � � �

Put x2 + x + 1 = t

diff. w.r.t. x.

2 1xdt

dx+ =


2 1x dx dt+ =� �Substituting we get

x x x dx t dt2 4 41 2 1+ + + = ⋅� � � � �

= +tc

5

5

=+ +

+x x

c2 5

1

5

�.

2.2.2.2.2. Evaluate: 4

9

3

4

x

xdx

−� .

Solution:Solution:Solution:Solution:Solution: Put x4 − 9 = t

diff. w.r.t. x.

4 3xdt

dx=

4 3x dx dt=

∴4

9

3

4

x

xdx

dt

tt c

−= = +� � log

= − +log .x c4 9 �


xdx

4 72 +� .

Solution:Solution:Solution:Solution:Solution: Put 4x2 + 7 = t

diff. w.r.t. x.

8xdt

dx=

8x dx dt=

x dx dt= 1

8.

Substituting,

x

xdx

tdt

4 7

1

82 +=� �= +1

8log t c

Integration 463

= + +1

84 72log .x c �

4.4.4.4.4. Evaluate: 9

x xdx

log�Solution:Solution:Solution:Solution:Solution: Put log x = t.

diff. w.r.t. x.

1

x

dt

dx=

dx

xdt= .

Substituting,

99 9

x xdx

dt

tt c

loglog� �= ⋅ = +

= +9log log .x c� �

5.5.5.5.5. Integrate e

e

x

x6 4− with respect to x.

Solution:Solution:Solution:Solution:Solution:e

edx

x

x6 4−�Put 6 − 4ex = t

diff. w.r.t. x.

0 4− =edt

dxx

− =4e dx dtx

e dxdtx =−4

Substituting,

e

edx

dt

tt c

x

x6 4 4

1

4−=

−= − +� � log

− − +1

46 4log .e cx �

6.6.6.6.6. Evaluate: e x dxx3 2⋅�Solution:Solution:Solution:Solution:Solution: Put x3 = t


diff. w.r.t. x.

3 2xdt

dx=

3 2x dx dt=

x dxdt2

3=

∴ e x dx e dt e cx t t3 2� �= ⋅ = +

= +e cx3

.

7.7.7.7.7. Evaluate: e x dxx x4 8 72

1+ − +� � �

Solution:Solution:Solution:Solution:Solution: Put 4x2 + 8x − 7 = t

diff. w.r.t. x.

8 8xdt

dx+ =

8 8x dx dt+ =� �

8 1x dx dt+ =� �

x dxdt+ =18

� �

Substituting,

e x dx edtx x t4 8 72

18

+ − ⋅ + = ⋅� ��

= ⋅ = +�1

8

1

8e dt e ct t

= ⋅ ++ −1

84 8 72

e cx x.

8.8.8.8.8. Evaluate: xe dxx2 2�Solution:Solution:Solution:Solution:Solution: Put 2x2 = t

diff. w.r.t. x.

4xdt

dx=

4x dx dt=

Integration 465

x dxdt=4

Substituting,

xe dx edt

e dtx t t2 2

4

1

4� � �= ⋅ =

= + = +1

4

1

42 2

e c e ct x .

9.9.9.9.9. Evaluate: x e

x edx

e x

e x

− −++�

1 1

Solution:Solution:Solution:Solution:Solution: Put xe + ex = t

diff. w.r.t. x.

ex edt

dxe x− + =1

e xe

e

dt

dxe

x− +

��

�� =1

e xe

edx dte

x− +

��

�� =1

x e dxdt

ee x− −+ =1 1 � .

Substituting

∴x e

x edx

dt

e t

e x

e x

− −++

=� �1 1

� �

= +1

et clog

1

ex e ce xlog + + � .

10.10.10.10.10. Evaluate: dx

x x x−� log

Solution:Solution:Solution:Solution:Solution:dx

x x x

dx

x x−=

−� �log log1� �Put 1 − logx = t


diff. w.r.t. x.

− =1

x

dt

dx

dx

xdt= −

Substituting

dx

x x

dt

tt c

1−= − = − +� �log

log� �

= − − +log log .1 x c� �

��$ ��%� ��&��#��

To evaluate integrals of the type px q

ax b cx d

px q

ax b cx d

++ +

++ +��

or 2 , first resolve into partial

fractions. i.e.,

Expresspx q

ax b cx d

A

ax b

B

cx d

++ +

=+

++� ��

orpx q

ax b x d

A

ax b

B

ax b

C

cx d

++ +

=+

++

++� � � � � � � �2 2

(where A, B and C are constants to be determined) and then integrate.

��

1.1.1.1.1. Evaluate: 3 1

3 1

x

x xdx

−− +� � ��

.

Consider 3 1

3 1

x

x x

−− +� ��

,

Resolve into partial fractions

3 1

3 1 3 1

x

x x

A

x

B

x

−− +

=−

++� �� ...(1)

3 1

3 1

1 3

3 1

x

x x

A x B x

x x

−− +

= + + −− +� ��

� � � ��

3 1 1 3x A x B x− = + + −� � � �Put x + 1 = 0

Integration 467

⇒ x = −1

3 1 1 0 1 3− − = + − −� � � � � �A B

− = −4 4B� �

⇒ B = 1

3 1 1 3x A x B x− = + + −� � � �Put x − 3 = 0

x = 3

3 3 1 3 1 0� � � � � �− = + +A B

8 4= A� �

⇒ A = 2.

Substituting A = 2 and B = 1 in (1) we get

3 1

3 1

2

3

1

1

x

x x x x

−− +

=−

++� ��

integrating with respect to x.

3 1

3 1

2

3

1

1

x

x xdx

xdx

xdx

−− +

=−

++� � ��

.

= − + + +2 3 1log log .x x c� � � �

= − + +log .x x c3 12� � � ��

2.2.2.2.2. Evaluate: 4 6

12

x

xdx

+−� .

Consider

4 6

1

4 6

1 12

x

x

x

x x

+−

= +− +� ��

Resolve into partial fractions.

4 6

1 1 1 1

x

x x

A

x

B

x

+− +

=−

++� ��

. ...(1)

⇒ 4 6 1 1x A x B x+ = + + −� � � �Put x + 1 = 0

x = −1


4 1 6 0 1 1− + = + − −� � � � � �A B

− + = −4 6 2B� �

2 2 1= − ⇒ = −B B� �Now

4 6 1 1x A x B x+ = + + −� � � �Put x − 1 = 0

x = 1

4 1 6 1 1 0� � � � � �+ = + +A B

10 2= A� �

⇒ A = 5

Substituting A = 5 and B = −1 in (1)

4 6

1 1

5

1

1

1

x

x x x x

+− +

=−

+ −+� ��

integrating,

4 6

1 1

5

1

1

1

x

x xdx

xdx

xdx

+− +

=−

+ −+� � ��

=−

− −+� �5

1

1

1

1xdx

xdx

4 6

1 15 1 1

x

x xdx x x c

+− +

= − − + +� � �� log log .


x x xdx

2 2

1 2 3 4 1

++ + −� � �� .


Consider

x

x x x

2 1

1 2 3 4 1

++ + −� ��

Resolve into partial fractions,

x

x x x

A

x

B

x

C

x

2 1

1 2 3 4 1 1 2 3 4 1

++ + −

=+

++

+−� �� ...(1)

Integration 469

⇒ x A x x B x x C x x2 1 2 3 4 1 1 4 1 1 2 3+ = + − + + − + + +� �� Put x + 1 = 1

x = −1

− + = − + − − + +1 1 2 1 3 4 1 1 0 02� � � �� A B C

2 1 5= −A� ��

⇒ A = −2

5

Put 2x + 3 = 0

2 3x = −

x = −3

2

−�� + = +

−+�

��

−�� −

��

�� +

3

21 0

3

21 4

3

21 0

2

A B C� � � �

9

41

3 2

2

12 2

2+ = − +�

��

− −��

��B

13

4

14

4= ��B

⇒ B= 13

14

Put 4x − 1 = 0

4x = 1

x = 1/4

1

41 0 0

1

41 2

1

43

2�� + = + + +�

�� ⋅ +��

��A B C� � � �

1

161

5

4

14

4+ = �

��C

17

16

70

16= ��C

⇒ C=17

70


Substituting the values of A, B and C in (1)

x

x x x x x x

2 1

1 2 3 4 1

251

13 14

2 3

17 70

4 1

++ + −

=−

++

++

−� �� integrating w.r.t. x.

x

x x xdx

xdx

xdx

xdx

2 1

1 2 3 4 1

2 5

1

13 14

2 3

17 70

4 1

++ + −

= −+

++

+−� � � ��

=−

+ ++

+−

+2

51

13 2 3

28

17 70 4 1

4log

log logx

x xc� � � � � �

4.4.4.4.4. Evaluate: 2 3

3 2

x

x x xdx

−+ −� � ��

Consider

2 3

3 2

x

x x x

−+ −� ��

, Resolve into partial fractions.

2 3

3 2 3 2

x

x x x

A

x

B

x

C

x

−+ −

= ++

+−� ��

2 3

3 2

3 2 2 3

3 2

x

x x x

A x x B x x C x x

x x x

−+ −

= + − + − + ++ −� ��

� ��

⇒ 2 3 3 2 2 3x A x x Bx x Cx x− = + − + − + +� �� Put x + 3 = 0

x = −3

2 3 3 0 3 3 2 0− − = + − − − +� � � � � �� A B C

− − = − −6 3 3 5B� ��

− =9 15B � �

⇒ B B= − ⇒ = −9

15

3

5

Put x − 2 = 0

x = 2

2 2 3 0 0 2 2 3� � � � � � � �� − = + + +A B C

4 3 2 5− = C � ��

Integration 471

1 10= C � �

⇒ C = 1

10

Put x = 0,

2 0 3 0 3 0 2 0 0� � � �� − = + − + +A B C

− = −3 6A� �

⇒ A A= −−

⇒ =3

6

1

2

Substituting A B C= = − =1

2

3

5

1

10, and in 1� �

2 3

3 2

1 2 3 5

3

1 10

2

x

x x x x x x

−+ −

= + −+

+−� ��

integrating with respect to x,

2 3

3 2

12

35

3

110

2

x

x x xdx

xdx

xdx

xdx

−+ −

= −+

+−� � � ��

= − + + − +1

2

3

53

1

102log log logx x x c� � � �

5.5.5.5.5. Evaluate: 3 4

2 12

x

x x

+− +� � � � �

Consider 3 4

2 12

x

x x

+− +� � � �


3 4

2 1 2 2 12 2

x

x x

A

x

B

x

C

x

+− +

=−

+−

++� � � � � � � � ...(1)

Multiplying by (x − 2)2 (x + 1)

3 4 2 1 1 2 2x A x x B x C x+ = − + + + + −� �� Put x − 2 = 0

x = 2

3 2 4 0 2 1 0� � � � � � � �+ = + + +A B C

6 4 310

3+ = ⇒ =B B� �


Put x + 1 = 0

x = −1

3 1 4 0 0 1 2 2− + = + + − −� � � � � � � �A B C

1 91

9= ⇒ =C C� �

Put and x B C= = =010

3

1

9,

4 2 110

31

1

92 2= − + + −A� ��

4 210

3

4

9= − + +A

410

3

4

92− − = A

36 30 4

92

− − = A

⇒ + = ⇒ =2

92

1

9A A

Substituting A B C= = =1

9

10

3

1

9, and in 1� �

3 4

2 1

1

92

10 3

2

1

912 2

x

x x x x x

+− +

=−

+−

++� � � � � �

integrating we get,

3 4

2 1

1 9

2

10 3

2

1 9

12 2

x

x xdx

xdx

xdx

xdx

+− +

=−

+−

++� � � ��

= − + ⋅−− +

+ + +− +1

92

10

3

2

2 1

1

91

2 1

log logxx

x c� � � � � �

= − −−

+ + +1

92

10

3 2

1

91log log .x

xx c� � � � � �

6.6.6.6.6. Evaluate: 4 8

2 1 32

x

x x

+− +� � � � �

Integration 473

Consider

4 8

2 1 32

x

x x

+− +� � � � Resolve into partial fractions

4 8

2 1 3 2 1 2 1 32 2

x

x x

A

x

B

x

C

x

+− +

=−

+−

++� � � � � � ...(1)

Multiplying by (2x − 1)2 (x + 3)

4 8 2 1 3 3 2 1 2x A x x B x C x+ = − + + + + −� �� Put 2x − 1 = 0

2x = 1

x = 1/2.

41

28 0

1

23 0�

�� + = + +��

�� +A B C� � � �

107

2

20

7= �� ⇒ =B B

Put x + 3 = 0

x = −3

4 3 8 0 0 2 3 12− + = + + − −� � � � � � � �� A B C

− + = −12 8 7 2C � �

− = ⇒ = −4 49

4

49C C� �

Substituting B C x= = − =20

7

4

490, and

we get

4 0 8 1 320

73

4

491 2+ = − + − −� � � �� A

32 360

7

4

49= − + −A

3 3260

7

4

49A = − + −

31568 420 4

49A = − + −


31152

49A = −

A = −×

= −1152

49 3

384

49

Substituting in (1)

4 8

2 1 3

384

492 1

20

72 1

4

4932 2

x

x x x x x

+− +

=

−

−+

−+

−

+� � � � � �integrating with respect to x.

4 8

2 1 3

384

492 1

20

72 1

4

4932 2

x

x x xdx

xdx

xdx

+− +

=

−

−+

−+

−

+� � � ��

= − ⋅−

+ ⋅−− +

⋅ + − ⋅ + +− +384

49

2 1

2

20

7

2 1

2 1

1

2

4

493

2 1loglog

x xx c

� � � � � �

= − ⋅ − −−

− + +384

49

2 1

2

20

14 2 1

4

493

loglog

x

xx c

� �� .

��' ��%� ��

From product rule in differentiation we have

d

dxuv u

dv

dxv

du

dx� � = ⋅ + ⋅

Integrating w.r.t. x. we get

d

dxuv u

dv

dxdx v

du

dxdx� �� = +

⇒ uv udv v du= +� � .

⇒ udv uv vdu

dxdx� �= − ��

By taking u as first function,

dv as second function

du as differential of 1st function

v as the integral of the 2nd function we get

Integration 475

I function II function I function II function II function I function� �� dx dxd

dxdx� � ��= − ⋅�

��

This is known as integration by parts.

The success of the method of integration by parts depends on choice of 1st function and 2nd function.Logarithmic, algebraic and exponential functions should be taken in the same order of priority. [LIATE:log, Inverse trig., Algebraic, Trigonometric and Exponential]

For example, to integrate xex, x is first function and ex is 2nd function where as to integrate xlogx, logxis first function and x is 2nd function.

��

1.1.1.1.1. Evaluate: x e dxx� .

x e dx x e ed

dxx dxx x x

I II� ��= ⋅ − ⋅��

��

= − ⋅�xe e dxx x 1

xe e cx x− + .

2.2.2.2.2. Evaluate: x x dxlog� .

x x dx x x dxlog log� �= ⋅I II

log log .x x dxx d

dxx dx� �− ⋅

��

��

2

2� �

= ⋅ − ⋅��

��log x

x x

xdx

2 2

2 2

1

log xx x

dx⋅ − �2

2 2

log xx x

c⋅ − ⋅ +2 2

2

1

2 2

log .xx x

c⋅ − +2 2

2 4


3.3.3.3.3. Evaluate: log .x dx�log logx dx x dx� �= ⋅

I II1

log logx dx xd

dxx dx⋅ − ⋅�

�� 1 � �

log x x xx

dx⋅ − ⋅� 1

log x x dx⋅ − �1log .x x x c⋅ − +

4.4.4.4.4. Evaluate: x e dxx3 2

.�Put x2 = t.

diff. w.r.t. x.

2xdt

dx=

2x dx dt=

x dxdt=2

Substituting we get

x e dx x e x dxx x3 22 2� �= ⋅

= ⋅�12 t e dtt

I II

1

2t e e

d

dtt dtt t− ⋅�

��

��

��

1

2te e dtt t− ⋅�

1

2te e ct t− +

=−

+x e e

cx x2 2 2

2.

Integration 477

5.5.5.5.5. Evaluate: x x dx nn log� ≠ − where 1

log x x dxn

I II⋅�

log logx x dxx

n

d

dxx dxn

n

⋅ −+

��

��

+1

1� �

= ⋅+

−+

⋅��

��

+ +

�log xx

n

x

n xdx

n n1 1

1 1

1

log xx

n

x

ndx

n n

⋅+

−+

+

�1

1 1

log xx

n n

x

nc n

n n

⋅+

−+ +��

�� + ≠ −

+ +1 1

1

1

1 11 if

= ⋅+

−+

++ +

log .xx

n

x

nc

n n1 1

21 1� �

��( �� #��% �� e f x f x dxx � � � �+ ′�Consider

e f x f x dxx � � � �+ ′�= ⋅ + ⋅ ′� �e f x dx e f x dxx x

I II� � � �

f x e dx ed

dxf x dx e f x dxx x x� � � � � �⋅ − �

��

��

+ ′� ��f x e e f x dx e f x dxx x x� � � � � �⋅ − ⋅ ′ + ⋅ ′� �

∴ = ⋅f x ex� �

∴ e f x f x dx e f x cx x� � � � � �+ ′ = ⋅ +� .

��

1.1.1.1.1. ex x

dxx 1

1

1

1 2+−

+

��

��


ex

cx ⋅+

+1

1

Sinced

dx x x x

1

1

1

1

1

12 2+��

�� = −

+= −

+� � � �.

2.2.2.2.2. ex x

xdx e

xx dxx x1 1+

= +��

�� log

log� �

= ⋅ +e x cx log

Since d

dxx

xlog .� � = 1

3.3.3.3.3. e xx x

dxx log loglog

� � +��

�� 1 = ⋅ +e x cx log log� �

Since d

dxx

x

d

dxxlog log

loglog� � � �= ⋅1

= ⋅ =1 1 1

log log.

x x x x

4.4.4.4.4. e x dx e x cx x+ = ⋅ +� 1� �

�� d

dxx� � = 1.

5.5.5.5.5. ex

xdxx 2 1

2

+��

�� = +�

�� e x

x xdxx 2

2

1

2

= +��

�� e x

xdxx 1

2

= ⋅ +e x cx

Since d

dxx

x � = 1

2.

��

• x dxx

ncn

n

� =+

++1

1

Provided n ≠ −1

Integration 479

• if n = −1

x dxx

dx x c−� �= = +1 1log .

• e dx e cx x� = + .

• a dxa

acx

x

� = +log

• k dx kx c� = +

• k f x g x dx k f x dx k g x dx� � � � � � � �± = ±� � �• If f x dx g x c� � � �� = + , then

f ax b dxg ax b

ac+ =

++� � � � �

• To evaluate f x f x dx f x tn� � � � � �⋅ ′ =� , put and proceed to get the answer.

f x

nc n

n� � +

++ ≠ −

1

11 for

and log [f (x)] + c for n = −1

• To evaluate ∫ef(x) ⋅ f ′(x) dx, put f (x) = t and proceed to get e f(x) + c as answer.

• To evaluate integrals of the type

px q

ax b cx ddx

px q

ax b cx d

++ +

++ +� ��

or 2

First resolve into partial fractions

px q

ax b cx d

A

ax b

B

cx d

++ +

=+

++� ��

or

px q

ax b cx d

A

ax b

B

ax b

C

cx d

++ +

=+

++

++� � � � � � � � � �2 2

Find A, B, C, then finally integrate.

• I function II function� � � � dx�


= − ��

�� I function II function II function I function

d

dxdx� �

I function and II function are kept by making use of LIATE rule.

• e f x f x dx e f x cx x� � � � � �+ ′ = ⋅ +� .

��&� �

I.I.I.I.I. EvEvEvEvEvaluaaluaaluaaluaaluate:te:te:te:te:

1. xx

dx3 4+��

��

2. 45 822x

x xdx− +�

��

3. axb

x

c

xd dx2

2+ + +��

��

4. x e e dxe x x+ −� �

5. x x e dxe eπ ππ+ − +� �

6. 6 26x x dx−� � �

7.6 8 94x x

xdx

+ −��

��

8. xx

x dx+�� −� 1

82 �

9. 6 4 3x dx+� � �

10.4 9 12x x

xdx

− +��

��

11.1

4 7

1

8 72x xdx

+−

+��

��

12.1

3 4

1

8 5 3−+

−� x xdx

� �

13.2 7

7 92

x

x xdx

++ −�

14. 2 9 9 62 2x x x dx− − +� � � �

15. x x x dx3 26 1 2+ − ⋅ +� �

16. ax bx c ax b dx2 2+ + +� ��

17.1

x xdx

log�

18.x

xdx

2

3 42 7−� �

19. 22

x e dxx�20.

e e

e edx

x x

x x

+−

−

−�

21.log 7 6

7 6

x

xdx

−−� � �

� �

22.1

6x

x dx−� log

23.6log x

xdx�

24.x

xdx

+� 6

Integration 481

25.1

7 122x xdx

+ +� 31.x

x xdx

− +� 1 22� � � �

26.1

2 82x xdx

− −� 32. x e dxx2�27.

2 1

6 8

x

x xdx

++ +2� 33. x e dxx4 2−�

28.3 7

3 7 62

x

x xdx

+− −� 34. x x dx−� 1 20� �

29.2 1

1 2 2

x

x x

++ +� � �� 35. x x dx6 log�

30.e

e edx

x

x x

2

4 24 3+ +� 36. log x x a dx+ +� 2 2� �

37. x e dxx5 2� Hint: x x x x t5 3 2 3= ⋅ =Put

38. x e dxx−� Hint: Put x t= 2

39.xe

x

x

+� 1 2� �

40. ex

xx

2

2

1

1

++

��

�� [Hint:Hint:Hint:Hint:Hint: Add and Subtract 1 in the Numerator and simplify]

41. log x dx−� 1

42.1

1

−+�

x

xdx

43.2 3

4 7

x

xdx

+−�

44. e xx x

dxx log loglog

� � +��

�� 1

45.log log logx

xe dxx� �

+��

��


��

1.x

x c4

44+ +log

2.4

3

58

3x

xx c+ + +log

3.ax b

xc x dx k

3

3− + + +log

4.x

ee e c

ex x

+

++ − +

1

1

5.x x

ex e x c

ee

ππ

ππ

+ +

++

+− + +

1 1

1 1

6. 43

4

32

43

xx

c− +

7.3

28 9

4xx x c+ − +log

8.x x

x c4 2

4

7

28− − +log

9.6 4

24

4xc

++� �

10. 8

56 2

52 3

212x

x x c− + +

11.−+

− + +1

4 4 7

8 7

8x

xc

� �� log

12.log 3 4

4

1

10 8 5 2

−−

+−

+x

xc

� ��

13. log x x c2 7 9+ − + �

14.x x

c2 2

9 6

2

− ++

�

15.2 6 1

9

33

2x xc

+ −+

�

16.ax bx c

c2 2

2

+ ++

17. log log x c+

18.1

63 2 7 3 3−

+x

c �

19. e cx2

+

20. log e e cx x− +−

21.log 7 6

14

2x

c−

+� �

22.− −

+2 6

3

32log x

c

23.6

6

log

log

x

c+

24. x c+ +62

�

25. logx

xc

++

��

�� +

3

4

26.1

6

2

4log

x

xc

−+

��

�� +

27.7

24

3

22log logx x c+ − + +� � � �

28. − + + − +5

33 2

16

113log logx x c� � � �

Integration 483

29.1

31

1

32

3

2log logx x

xc+ − − −

−+� � � � 38. e x x x cx− + + +

�

��

�

�+

32 3 6 6

30.1

4

1

3

2

2loge

ec

x

x

++

��

��+ 39.

e

xc

x

1++

31. − − + − − + +1

31

2

91

2

92x x x c� � � � � �log log 40. e

x

xcx −

+��

��+1

1

32. x e xe e cx x x2 2− + + 41.1

21

1

2

1

21x x x x clog log− − − − +� � � �

33. ex

xx x

cx− − − + − −��

�� +2

43

2

2

3

2

3

2

3

4 42. 2 1log + − +x x c� �

34. xx x

c−

−−

+1

21

1

462

21 22� � � �43.

1

2

13

4

7

4x x c+ −��

��

��

��+log

35.x

xx

c7 6

7 42log − + 44. e x cx log log� � +

36. x x x a x a clog + + − + +2 2 2 2� � 45. log log log log .x x xx

c⋅ − + +� �2

2

37.e

x cx3

313 − +


��

��

��

If f x dx g x c� � � �� = + . Then f x dx g x g b g aa

b

a

b

� � � � � � � �� = = − .

f x dx� �� is called indefinite integral and f x dxa

b

� �� [Read as integral from a to b f (x)dx] is called

definite integral of f (x) from a to b. Here a is called lololololowwwwwer limiter limiter limiter limiter limit and b is called upper limitupper limitupper limitupper limitupper limit.

To evaluate definite integral, integrate the given function as usual. In the final answer substitute theupper limit value for x − lower limit value for x.

��

I. EvI. EvI. EvI. EvI. Evaluaaluaaluaaluaaluate:te:te:te:te:

1.1.1.1.1. x dxx2

1

2 3

1

2 3 3

3

2

3

1

3

8

3

1

3

7

3� = = − = − = .

2.2.2.2.2. e dx e e e ex x

0

1

0

11 0 1� = = − = − .

3.3.3.3.3. xx

dx+�� 1 2

0

1

= + +��

�� x

xdx2

2

0

11

2

= +− +

+�

�

�

− +x xx

3 2 1

0

1

3 2 12

Definite Integrals 485

= − +�

�

�

x

xx

3

0

1

3

12

1

3

1

12 1

0

3

1

02 0

3 3

− +�

�

� − − +�

�

� � � � �

1

31 2 0− + −

1

31

4

3+ = .

4.4.4.4.4. 1 21

2

+� x dx� �

1 21

2

1

2

+� x dx� � =+

+⋅

+1 212

1

1

2

12

1

1

2x� �

1 232

1

2

32

1

2+⋅

x� �

1 2

3

3

2

1

2+ x� �

1 2 2

3

1 2 1

3

3

2

3

2+−

+� ��

1 4

3

1 2

3

32

32+

−+� �

= − − −5 3

3

5 5 3 3

3

32

32

.

5.5.5.5.5. x e dxx

−�1

1

x e ed

dxxx x⋅ − ⋅�

��

−1

1


xe ex x−−1

1

1 11 1 1 1e e e e− − − −− −

e e e e− − − −− −1 1

0 2 1− − −e

= =−221ee

.

6.6.6.6.6. x x dxe

II Ilog

1�

log logx xdxx d

dxx dx

e

� �− ⋅�

�

�

�

�

�

2

12

� �

log xx x

xdx

e

⋅ − ⋅�

�

� �

2 2

12 2

1

log xx x

dx

e

⋅ −�

�

� �

2

12 2

log xx x

e

⋅ − ⋅�

�

�

2 2

12

1

2 2

log xx x

e

⋅ −2 2

12 4

log logee e⋅ −

�

�

� − ⋅ −�

�

�

2 2 2 2

2 41

1

2

1

4

12 4

01

4

2 2e e−�

�

� − −�

��

e e e e e2 2 2 2 2

2 4

1

4

2 1

4

1

4− + = − + = +

.


7.7.7.7.7.x

x xdx

++ +� 2

4 82

0

1

.

Put x2 + 4x + 8 = t

diff. w.r.t. x.

2 4xdt

dx+ =

2 4x dx dt+ =� �

2 2x dx dt+ =� �

x dxdt+ =22

� � .

when x = 0, x x t2 4 8+ + =

0 4 0 8+ + =� � t

⇒ t = 8

when x = 1, x x t2 4 8+ + =

1 4 1 8+ + =� � t

⇒ t = 13

∴x

x x

dt

tt

++ +

= =� �2

4 8 2

1

22

0

1

8

13

8

13

log

1

213 8log log−

1

2

13

8log .��

8.8.8.8.8. x x dx1 2

1

2

+�Put 1 + x2 = t

diff. w.r.t. x.

2xdt

dx=

2x dx dt=


x dxdt=2

when x = 1, 1 2+ =x t

1 1+ = t

⇒ t = 2

when x = 2, 1 22+ = t

⇒ t = 5

Substituting,

x x dxt dt

12

2

1

2

2

5

+ =� �

=+

+1

2 1 2 1

1 2 1

2

5t

2

3 2 3

3 2

2

5 3 2

2

5t t

×=

1

35 2

1

35 5 2 23 2 3 2− = ⋅ −

= −5 5 2 2

3.

9.9.9.9.9.1

10

1−+�

x

xdx

−−+� x

xdx

1

10

1 � �

= − + − −+

= − ++

−+

�

�

� � � �x

xdx

x

xdx

xdx

1 1 1

1

1

1

2

10

1

0

1

0

1

− −+

�

�

� � �1 2

1

10

1

0

1

dxx

dx

− − +x x2 10

1log� �


− − + − − +1 2 1 1 0 2 0 1log log� � � ��

− − − −1 2 2 0 2 1log log� �

− + −1 2 2 0log

= −2 2 1log .

10.10.10.10.10.2 3

5 72

3x

xdx

+−�

2 3

5 7

232

575

2

3

2

3x

xdx

x

xdx

+−

=+��

−��

� �

=− + +

−�25

7

5

7

5

3

27

52

3 x

xdx

2

5

7575

75

3275

2

3

2

3x

xdx

xdx

−

−+

+

−

�

�

�

� �

2

51

2910

75

2

3

2

3

dxx

dx� �+−

�

�

�

2

5

29

10

7

5 2

3

x x+ −��

�

��

log

2

53

29

103

7

52

29

102

7

5+ −��

��

�

��

− + −��

�

��

��

��

log log

2

53

29

10

8

52

29

10

3

5+ �

�� − − �

��

��

��

log log


2

51

29

10

8535

+

�

�

��

�

�

��

�

�

�

log

2

51

29

10

8

3+ �

��

�

��

log

��! �� "��"��#��

1. f x dx f t dt f z dza

b

a

b

a

b

� � � � � �� = =

2. f x dx f x dxa

b

b

a

� � � �� = −

3. f x dx f x dx f x dx a c ba

b

a

c

c

b

� � � � � �� = + < < where .

4. f x dx f a x dxa a

� � � �0 0� �= −

5. f x dx f a b x dxa

b

a

b

� � � �� = + −

��

1.1.1.1.1. Evaluate: x x dx−� 1 5

0

1

� �

Let I x x dx= −� 1 5

0

1

� �

using property (4),

I x x dx= − − −� 1 1 1 5

0

1

� ��


I x x dx= − −� 1 5

0

1

� ��

I x x dx= − +� 5 6

0

1

� �

Ix x

= −+

++

+ +5 1 6 1

0

1

5 1 6 1

Ix x= − +

6 7

0

1

6 7

I = − + −1

6

1

70

I = − +7 6

42

I = − 1

42

∴ x x − = −� 11

425

0

1

� � .

2.2.2.2.2. Evaluate: x a x dxa

− ⋅� � �4

0

Let I x a x dxa

= − ⋅� � �4

0

using property (4),

I a x a a x dxa

= − − ⋅ −� � � � �4

0

I x a x dxa

= − −� � � � �4

0

I x a x dxa

= −� 4

0

� �


I x a x dxa

= ⋅ −� 4 5

0

Ix

ax

a

= ⋅ −5 6

05 6

Ia a a= ⋅ − −

5 6

5 60

Ia a a a a= − = − =

6 6 6 6 6

5 6

6 5

30 30

∴ x a x dxa

a

− =� � �4

0

6

30.


a xdx

a

−�0

.

Let Ix

a xdx

a

=−�

0

using property (4)

Ia x

a a xdx

a

= −− −� � �

0

= −− +

= −� �a x

a a xdx

a x

x

a a

0 0

Ia

x

x

xdx

a

= −��

��

0

= −� �a

xdx dx

a a

0 0

1

= −a x xalog 0

a a alog − − 0

= −a a alog .


��

x

a xdx

x

a x

a a

−= − −

−� �0 0

= − − − +−

�

�� x a a

a xdx

a

0

= − −−

+ −−

�

�

� � �a x

a x

a

a xdx

a a

0 0

− ⋅ −−

�

�

� � �1

0 0

dxa

a xdx

a a

− −−−

�

��

x aa x

a

log� �

1 0

− + −x a a xa

log� �0

− + − − + −a a a a a alog log� � � �� 0 0

− + −a a a alog log0

− +a a alog

= −a a alog .


xdx

+−�

4

40

4

Let I x

xdx

+−�

4

40

4

using property (4)

Ix

x= − +

− −� 4 4

4 40

4

Ix

xdx

xdx

x

xdx= −

−= − −�

�

� � � �8 8

0

4

0

4

0

4


− −80

4

log x x

− − = −8 4 4 4 8 4log log .� �

��

x

xdx

+−�

4

40

4

= + − +−� x

xdx

4 4 4

40

4

= −−

+−� �x

xdx

xdx

4

4

8

40

4

0

4

18

40

4

0

4

� �−−

dxx

dx

xx

−−

−8

4

1 0

4log� �

4 8 4 4 0 8 4 0+ − − + −log log� � � �

4 0 0 8 4− − − log

4 8 4− log .

��$ ��"��"��#�� "��

One of the application of definite integrals is to findarea.

(i) Area enclosed by the curve y = f (x), the x-axis and the lines x = a and x = b is given by

A y dx f x dxa

b

a

b

= =� � � � .

��

1.1.1.1.1. Find the area bounded by the curve y = x2 − x, x-axis and the ordinates x = 1 and x = 2.

y

0 x = a x = b

y = f (x)

x

Fig. 19.1


Solution:Solution:Solution:Solution:Solution: Area = � y dxa

b

A x x dx= −� 2

1

2

� �

Ax x= −

3 2

1

2

3 2

A = −��

��

− −��

��

2

3

2

2

1

3

1

2

3 2 3 2

= − − + = − − +8

32

1

3

1

2

16 12 2 3

6

Area square units.= 5

6

2.2.2.2.2. Find the area bounded by the curve y = 3x2 − 8 with x-axis and ordinates x = 0 and x = 3.

Solution:Solution:Solution:Solution:Solution: Area = � y dxa

b

= −� 3 82

0

3

x dx� �

= −33

83

0

3x

x

= −x x3

0

38

3 8 3 03 − −� �

27 24 3− = Sq. units.

3.3.3.3.3. Find the area bounded by the parabola x2 = 4y, x-axis and x = 1 and x = 2.

Solution:Solution:Solution:Solution:Solution: A ydxa

b

= �

Ax

dx= �2

1

2

4 � x y y x yx2 2

2

4 44

= = =, ,


Ax= ⋅1

4 3

3

1

2

A = −�

�

� = −�

��

=1

4

2 1

3

1

4

8 1

3

7

12

3 3

sq. units.

4.4.4.4.4. Find the area bounded by the x-axis and the curve y = x2 − 5x + 6

Solution.Solution.Solution.Solution.Solution. On X-axis, y = 0.

∴ − + =x x2 5 6 0

x x x2 3 2 6 0− − + =

x x x− − − =3 2 3 0� � � �

x x− − =2 3 0� ��

x x= =2 3 or .

So Area bounded by the x-axis and the curve y x x y dx= − + = �2

2

3

5 6

= − +� x x dx2

2

3

5 6� �

= − ⋅ +x x

x3 2

2

3

35

26

3 2

35

3 2

26 3 2

3 3 2 2− − ⋅ − + ⋅ −� �

27 8

35

9 4

26 1

− − −��

�� + � �

19

3

25

26

38 75 36

6

1

6− + = − + = −

� Area is +ve, A = 1

6 sq. units.

5.5.5.5.5. Find the area bounded by the curve y = x2 − x with x-axis.

Solution:Solution:Solution:Solution:Solution: On X-axis, y = 0.

x x2 0− =


x x − =1 0� �

x x= − =0 1 0 or

x x= =0 1 or

So Area bounded by the curve y = x2 − x with x-axis

= = −� �y dx x x dxa

b2

0

1

� �

= − = − −x x3 2

0

1

3 2

1

3

1

20

2 3

6

1

6

− = −

� Area is non-negative,

Required Area = 1

6 sq. units.

6.6.6.6.6. Find the area between the parabolas y2 = x and x2 = y.


Point of intersection: y x2 =

x y2 =

Squaring,

x y4 2=

x x4 = � y x2 =� �

x x4 0− =

x x3 1 0− =� �

x x= =0 1 or

Area bounded by the curve y2 = x, x-axis, x = 0 and x = 1 is y dxa

b

� .� y x

y x

2 =∴ =

= � x dx0

1

Fig. 19.2

0(0, 0)

y

x

(1, 1)

x = y2

y = x2


=+

+x

12

1

0

1

12

1

x3 2

0

1

3 2

1

3 20

2

3

3 2

− =

Area bounded by the curve x2 = y, x-axis x = 0 and x = 1 is y dxa

b

� . y x= 2

= � x dx2

0

1

= =x3

0

1

3

1

3.

Area between the parabolas = Area bounded by the parabola y2 = x − Area bounded by the parabolax2 = y with X-axis, x = 0 and x = 1.

2

3

1

3

1

3− = sq units.

∴ Reqd. area = 1

3 sq. units.

7.7.7.7.7. Find the area between the parabola y2 = 4x and the line y = x.

Point of intersection:

y2 = 4x and y = x

y x2 2=

4 2x x=

x x2 4 0− =

x x − =4 0� �

x x= =0 4 or .

Fig. 19.3

(0, 0)

y

x

y = 4x2

y = x

(4, 4)


Area bounded by the parabola y2 = 4x, the lines x = 0, x = 4 and X-axis = � y dxa

b

.

= � 20

4

x dxy x

y x x

2 44 2

== =

2 12

1

12

1

0

4

⋅+

+x

232

2 2

3

32

0

4

32

0

4

⋅ = × ⋅xx

4

34 03 2⋅ −� �

4

38

32

5⋅ = .

Area bounded by the line y = x, x = 0, x = 4 and x-axis = � y dxa

b

= = = = =� x dxx

0

4 2

0

4 2

2

4

2

16

28.

Required area = Area bounded by the parabola y2 = x − Area bounded by the line x =y with x-axisand lines x = 0 and x = 4

= −32

38

=32 24

3

8

3

− = sq. units.

8.8.8.8.8. Find the area between the curves x2 = 5y and y = 2x.

Solution:Solution:Solution:Solution:Solution: Required area = A1 ~ A2.

Now: x2 = 5y and y = 2x.

x x2 5 2= � �


x = 5y2

0 B

A

y =

2x

A1A2

Fig. 19.4

x x2 10=

x x2 10 0− =x x − =10 0� �x x= =0 10 or .

y x y y= ⇒ = =2 2 0 2 10� � � � or

y y= =0 20or .

∴ Point of intersection: (0, 0) and (10, 20).

A1 = Area bounded by x2 = 5y, x-axis and lines x = 0 and x = 10

y dxx

dxa

b

� �=2

0

10

5

� 5

5

2

2y x

yx

=

=

= 1

5 3

3

0

10x

1

1510 0

1000

15

200

33 3− = =

A2 = Area bounded by y = 2x, x-axis and lines x = 0 and x = 10.

∴ = = = = =� �ydx x dxx

a

b

22

210 100

0

10 2

0

102 .

Required area = A1 ∼ A2.

= = =200

3100

200 300

3

100

3~

~


∴ Required area = 100

3 sq. units.

��%��

• If f x dx g x c� � � �� = + , then f x dx g x g b g aa

b

a

b� � � � � � � �� = = −

• f x dx f t dt f z dza

b

a

b

a

b

� � � � � �� = = and so on.

• f x dx f x dxa

b

b

a

� � � �� = −

• f x dx f x dx f x dxa

b

a

c

c

a

� � � � � �� = +

• f x dx f a x dxa a

� � � �0 0� �= −

• Area enclosed by the curve y = f (x), X-axis and the lines x = a and x = b is given by

A ydx f x dxa

b

a

b

= =� � � � .

��

I. Evaluate:

1. x dxa

b

+� 1� � 2. x dx2

1

2

1−� � �

3. 3 2 12

1

3

x x dx− +� � � 4. 2 1 8

0

1

x dx−� � �

5. xx

dx+��

�� 1

1

4

6. x x x dx3 2

0

2

2 1+ − +� � �


7. x x dx3

1

1

4 2+ −−� � � 8. x

xx dx2

2

1

21

+ −��

��

9. x x x dx3 2

1

2

3+ +−� � � 10. 6 2 45 4

1

2

x x dx− + +� � �

11. x x x dx+ −� 1 10

1

� �� 12. x x dx+ −� 1 20

1

� ��

13.x

x xdx

+

+ +� 2

4 320

1

14.1

10

x xdx

e

+� log

15.log x

xdx

� �2

1

2

� 16. log x dxe

1�

17. 10

1

+ −� e dxx� � 18. x x25

2

0

1

1 −� � �

19. x e dxx2

1

1−

−� 20.

x

xdx

+� 10

1

21.dx

x x1 21

0

− +−� � �� 22.

x

xdx2

0

1

1+�

23.dx

x x2

2

5

4 3+ +� 24.x

edx

x

3

2

0

1

�

25. log x dxe

e

� �22

�

II. 1. Find the area bounded by the curve y = x3, the x-axis and the lines x = 1 and x = 2.

2. Find the area bounded by the curve y = x2 − 4x, x-axis and the lines x = 1 and x = 3.

3. Find the area enclosed by the curve yx

x=

+1 x-axis and ordinates x = 0 and x = 1.


4. Find the area bounded by the x-axis and the curve y = x2 − 7x + 10.

5. Find the area bounded by the curve y = 4x − x2 − 3 with x-axis.

6. Find the area bounded by the curve y = 4x − x2 and x-axis.

7. Find the area between the curves y2 = 4x and y = 2x.

8. Find the area between the curves y2 = 2x − 2 and the line y = x − 5.

9. Find the area between the parabolas y2 = 4x and x2 = 4y.

10. Find the area between the parabola y2 = 4ax and the line y = 2x.

11. find the area between x2 = y and y2 = 8x.

12. Find the area between the curves y = 11x − 2y − x2 and y = x.

��

1. b ab a− + +�

��

� � 2

2 2.4

33. 20 4.

1

95.

20

3

6.28

37. −4 8.

4

39.

57

410.

365

3211. − 1

4

12. − 4

313. 8 3− 14. 2 2 1−� � 15.

log2

3

3� �16. −1 17. 2

1−e

18.16

69319.

e

e

2 5−20. 1 2− log 21.

2

32log 22. log 2 23.

1

2

5

4log

24.− +19

8

3

82e25. 2 2e e−

II. 1.15

4 sq units 2.

22

3 sq units 3. 1 2− loge� � sq units

4.9

2 sq units 5.

4

3 sq units 6.

64

3 sq units

7.1

32a sq units 8. 18 sq units 9.

16

3 sq units

10.1

32a sq units 11.

8

3 sq units 12.

4

3 sq units


��

��

��

Cost Cost Cost Cost Cost FFFFFunction:unction:unction:unction:unction: The outflows usually raw materials, rent, utilities, pay of salaries and so forth form thetotal cost. It is sum total of all costs. Economists and accountants often define total cost as sum of 2components. Total variable cost and total fixed cost.

Total variable cost varies with the level of output. (Eg.: raw materials) whereas total fixed costremains the same (for example rent). Hence Total cost = Variable cost + Fixed cost

AAAAAvvvvverererereraaaaaggggge Cost:e Cost:e Cost:e Cost:e Cost: Average Cost is the Cost per unit of the output. It is obtained by dividing total cost bythe total quantity produced. If TC is the total cost of producing x units or q units, then average cost AC

is given by, ACTC

or TC=

x q.

MarMarMarMarMarggggginal Cost:inal Cost:inal Cost:inal Cost:inal Cost: Marginal cost is the additional cost incurred as a result of producing and selling onemore unit of the product.

If TC is the total cost of producing x units or q units then the derivative d

dx

d

dqTC or TC repre-

sents the instantaneous rate of change of total cost, given a change in number of units (x or q) produced.

∴ Marginal Cost, MC TC or TC= d

dx

d

dq.

ReReReReRevvvvvenenenenenue function:ue function:ue function:ue function:ue function: The money which flows into an organisation from either giving service or sellingproducts is called as revenue. The most fundamental way of computing total revenue from selling aproduct is

Total revenue = Price per unit × Quantity sold.

If TR is the total revenue p is the price per unit and q or x is the quantity sold then

TR = pq or TR = px.

Application of Calculus in Business 505

MarMarMarMarMarggggginal rinal rinal rinal rinal reeeeevvvvvenenenenenue:ue:ue:ue:ue: It is the additional revenue derived from selling one more unit of a product. It isobtained by differentiating total revenue with respect to quantity demanded. If MR is the marginalrevenue. TR is the total revenue and x or q is the No. of units produced. Then

MR TR or MR TR= =d

dx

d

dq� � .

PrPrPrPrProfofofofofit function:it function:it function:it function:it function: Profit for an organisation is the difference between total revenue and total cost. If TRis the total revenue and TC is the total cost then

Profit = TR − TC.

For many production situations, the marginal revenue exceeds the marginal cost, at lower level ofoutput. As the level of output increases, the amount by which marginal revenue exceeds marginal costbecomes smaller. Eventually a level of output is reached at which marginal revenue = Marginal cost.Beyond this point marginal revenue is less than marginal cost and the total profit begins to decreaseswith added output. So from theoretical stand point, the profit maximization level of output can beidentified by the criterion.

Marginal revenue = Marginal cost.

i.e., MR = MC.

i.e.,d

dx

d

dxTR TC� � =

or

d

dq

d

dqTR TC� � =

Note:Note:Note:Note:Note: We know

if d

dxf x g x� � � �=

Then g x dx f x c� � � �� = + where c is the constant of integration.

By applying this to,

d

dxTR MR and=

d

dxTC MC=

We get

MR TR and MC TC.� � � �dx dx� �= =

So if we know marginal cost/marginal revenue (MC/MR) we can calculate, total cost or Totalrevenue (TC/TR) by using


TC MC or MR= � �dx dq

TR MR or MR= � �dx dq

��!�"��#��$��%�

1.1.1.1.1. If the total cost function C(x) of a firm is given by C(x) = x3 − 3x + 7. Then find the average costand marginal cost when x = 6 units.

Solution:Solution:Solution:Solution:Solution: Given c(x) = x3 − 3x + 7.

We have Average Cost = Average CostTotal Cost=

x

= − +x x

x

3 3 7

= − + = − +x

x

x

x xx

x

323 7

37

.

Marginal Cost = d

dxC x� �

= − +d

dxx x3 3 7

= − +3 3 02x

MC = −3 32x

MC when x = 6 is 3(6)2 − 3

3(36) − 3

108 − 3 = 105.

2.2.2.2.2. If the marginal cost function is given by x3 + 3x − 7 and fixed cost is Rs 750 then find the totalcost function.

Solution:Solution:Solution:Solution:Solution: Given:

Marginal cost function = x3 + 3x − 7.

MC = + −x x3 3 7

We know Total cost = MC� dx .

= + −� x x dx3 3 7� �


= + −� � �x dx x dx dx3 3 7 .

TC = + − +x xx c

4 2

4

3

27

where c is the constant of integration.

Here it is fixed cost.

Given fixed cost = Rs. 750

∴ Total function cost = + − +x xx

4 2

4

3

27 750

3.3.3.3.3. The total revenue function is given by R = x + 3x2. Find the marginal revenue and demandfunction.

Solution:Solution:Solution:Solution:Solution: Given TR = x + 3x2.

diff. w.r.t. x.

d

dx

d

dxx xTR = + 3 2

d

dx

d

dxx

d

dxxTR = + ⋅� � � �3 2

⇒ Marginal revenue = 1 + 3.(2x)

MR = 1 + 6x

Demand function = Average revenue

= Total revenue

x

= +x x

x

3 2

= + = +x

x

x

xx

31 3

2

.

4.4.4.4.4. If the total revenue function is given by R qq

q� � = + +596

6 2 where q is the number of units

manufactured. Find the maximum value of total revenue.

Solution:Solution:Solution:Solution:Solution: Given R qq

q� � = + +596

6 2

diff. w.r.t. x.

′ = + +��

�

R qd

dq qq� � 5

966 2


′ = + −� �

��

+R qq

q� � � �0 961

6 22

R(q) attains maxima when R′(q) = 0.

096

122= − +q

q

96122q

q=

96 12 3= q

⇒ q3 96

12=

q3 8=

q = 2

∴ Maximum value of the revenue = R(2)

R .2 596

26 2 5 48 24 772� � � �= + + = + + =

5.5.5.5.5. The marginal revenue (in thousands of rupees) function for a particular, commodity is 4 + e−0.03x,where x is denotes the number of units sold. Determine the total revenue from the sale of 100 unitsgiven that e−3 = 0.05 (Approx).


We know Total revenue Marginal revenue= � � �dx

Total revenue from the sale of

1000

100

units MR= � � �dx

= + −� 4 0 03

0

100

e dxx.� �

= +−

−4

0 03

0 03

0

100

xe x.

.

= +−

�

��

� − +

−�

��

�

− ×4 100

0 034 0

0 03

0 03 100 0

� � � �e e.

. .


= −�

��

� − −��

�

−400

0 03

1

0 03

3e

. .

4000 05

0 03

1

0 03400 1 66 33 33 431 667− + = − + =.

. .. . .

∴ Total revenue = 431.667 thousands of rupees.

= Rs. 4,31,667

6.6.6.6.6. Find the total cost of producing 1000 electric bulbs if the marginal cost (in Rs. per unit) is given

by ′ = +C xx� �

10002 5. where x is the output.


We know, Total cost = ∫(Marginal cost)dx.

∴ Total cost of producing 1000 electric bulbs = � MC� �dx0

1000

= +�

�� x

dx1000

2 50

1000

.

= +��

�

1

1000 22 5

2

0

1000x

x.

= +�

��

� −1

1000

1000

22 5 1000 0

2

. � �

1000

22 500 500 2500+ = +,

= 3000.

∴ Total cost of producing 1000 bulbs = Rs 3000.

7.7.7.7.7. If the total revenue function and total cost function are given by TR = 40x − x2 and TC = 2 + 4xrespectively. Then find at what level of output profit is maximised.

Solution:Solution:Solution:Solution:Solution: We know, profit is maximised when marginal cost = marginal revenue. Now

Marginal cost = d

dx[Total cost]

MC = + = +d

dxx2 4 0 4

MC = 4


Marginal revenue = d

dx[Total revenue]

= −d

dxx x40 2

MR = 40 − 2x

Profit is maximised when, MR = MC.

∴ 40 − 2x = 4

⇒ 40 − 4 = 2x

36 = 2x

⇒ x = 18

∴ Profit is maximised when 18 units are produced.

8.8.8.8.8. Find the maximum profit that a company can make, if the profit function P(x) = 41 + 24x − 18x2.

Solution:Solution:Solution:Solution:Solution: Consider

P x x x� � = + −41 24 18 2

diff. w.r.t. x.

dP

dxx= + −0 24 1 18 2� � � �

dP

dxx= + −24 36

P attains extrema whendP

dx= 0

0 24 36= + − x

⇒ x = + = +24

36

2

3.

Again consider dP

dxx= + −24 36

diff. w.r.t. x.

d P

dx

2

2 36 0= − <

∴ P attains maxima at x = 2/3.

Maximum value of the profit

= + � �� − �

��41 24

2

318

2

3

2


= + − ⋅41 16 184

9

= − =57 8 49.

9.9.9.9.9. A T.V. manufacturer produces x sets per week at a total cost of Rs x2 + 1560x + 50,000.

He is a monopolist and the demand function for this product is xP= −12000

179 where P is the price

per set. What is the monopoly price in order to maximise the profit?

Given: Total cost = x2 + 1560 x + 50,000

diff. w.r.t. x.

d

dxxTC� � = +2 1560

i.e., Marginal Cost = 2x + 1560 ...(1)

Now, also given

Demand function = = −x

P12000

179

⇒ P x= −12000 179

Total revenue = Px.

TR = −12000 179 2x x

diff. w.r.t. x.

d

dxxTR� � � �= −12000 179 2

i.e., Marginal revenue = −12000 358x

MR = −12000 358x ...(2)

Profit is maximixed when

Marginal cost = Marginal revenue.

i.e., Equation (1) = (2)

2 1560 12000 358x x+ = −

2 358 12000 1560x x+ = −

360 10440x =

x = 10440

360

x = 29.

∴ Monopoly price P x= −12000 179


= −12000 179 29� �

= Rs. 6809.

10.10.10.10.10. The cost function of a company is given by C x xx= − +500 203

23

, where x stand for output.

Calculate the output when marginal cost is equal to average cost.

Given: C x xx= − +500 203

23

diff. w.r.t. x.

dC

dxx

x= − +500 20 23

3

2

� �

i.e., Marginal Cost = 500 − 40x + x2 ...(1)

Now,

average costTotal cost=

x

=− +500 20

32

3

x xx

x

= − +500 203

2

xx

...(2)

Equating (1) & (2)

500 40 500 203

22

− + = − +x x xx

− + + − =40 203

022

x x xx

− + − =203

022

x xx

− + − =60 3

30

2 2x x x

− + =60 2 02x x

2 30 0x x− + =� �

2 0 30 0x x= − + = or

x x= =0 30 or .


∴ The output when the marginal cost = average cost is 30.

��

• Total cost = Fixed cost + Variable cost.

• Marginal cost =d

dx

d

dq Total cost or total cost� � � �

• Total cost = ∫(Marginal cost) dx or ∫(Marginal cost) dq.

• Marginal revenue =d

dx (Total revenue) or

d

dq (Total revenue)

• Total revenue = ∫(Marginal revenue) dx or ∫(Marginal revenue) dq.

• Average cost = Total cost or

Total cost

x q

• Profit is maximised when marginal cost = Marginal revenue.

• Average revenue is nothing but demand function = Total revenue or

Total revenue

x q

�#��%�

1. If the marginal cost of a product is 3x2 − 4x where x is the output, find the total cost of producing10 units.

2. If the marginal revenue function is given by 2

2 1x + , find the total revenue function.

3. A company has revenue function given by R = 100q − q2. Find the marginal revenue function(where q is the output).

4. Find the marginal revenue for the demand function 3x − x2.

5. Find the average cost function for the marginal cost function x2 + 2x.

6. The demand function of a firm is given by P = 50 − 2x. Where P is the price per unit for x units.Determine the marginal revenue.

7. The cost function Cx

x= + +548

3 2 where x is the number of articles produced. Find the mini-

mum value of C.

8. If the marginal cost is given by x2 + 7x + 6 and fixed cost is Rs 2500, determine the total cost ofproducing 6 units.

9. If C is the total cost for producing x units of a product and the average cost function is given by

0.003 x2 − 0.04 x + 6 + 3000

x, find the marginal cost function.


10. The total cost function of a firm is given by C(x) = 2x3 − x2 + 5x. Find the average and marginalcost function.

11. If the marginal revenue is given by 3030

3

− x, then find the total revenue function.

12. The cost function of a firm is given by C x x x x� � = − +300 101

32 3. Where C is the total cost for

x units. Calculate the output at which the average cost is minimum.

13. The unit demand function is x = 25 − 2p where x is the number of units and p is the price. Let theaverage cost per unit be Rs 20. Find the price per unit that maximises the profit function.

14. The demand function of a company is given by P(q) = 800 − 0.4q and the total cost function isgiven by C(q) = 80 q + 8000 where q is the level of output,P is the price per unit. Find the levelof output and the price charged which maximises the profit.

15. Find the total revenue, marginal revenue, marginal revenue and average revenue when the de-mand function is given by Q = 30 − 4P + P2 where P is price and Q is the quantity demanded.Also calculate the marginal revenue when P = 3.

16. A manufacturer has a total cost function C x= +100 32 . He would like to know at what levelof output his marginal cost will be Rs. 2.00.

17. If the total cost function is C = 8x3 − 12x2 + 20x where x is the level of output, find x at whichthe average cost is minimised find the total cost for this output.

18. The marginal cost function of a firm is 150 − 10x + 0.2x2, where x is the output, find the total costfunction, if the fixed cost is Rs 750, what is the average cost?

19. If the total revenue (R) and total cost (C) functions are given by R = 30x − x2 and C = 20 + 6x,find x if marginal revenue = marginal cost. Also find average cost and average revenue.

20. If the marginal revenue is given by 1515

+ q, find the total revenue obtained from an output of

30 units.

��% ��%

1. 1200 2. log(2x + 1) + c 3. 100 − 2q 4. 6x − 3x2 5.x

xc

x

2

3+ +

6. 50 − 4x 7. 41 8. 2734 9. 0.009x2 − 0.08x + 6

10. 0 006 0 043000

2. .xx

− − 11. 6x2 − 2x + 5 12. 15 13. Rs 12.34

14. Rs 440 15. 30 − 4P + P2 16. 64 17.3

4

18. AC = − + +150 50 2

3

7502

xx

x

.19. x = 12 20. 480.

��

516B

asic Mathem

aticsQuestion PQuestion PQuestion PQuestion PQuestion Paaaaaper Blue Prper Blue Prper Blue Prper Blue Prper Blue Print fint fint fint fint for Second or Second or Second or Second or Second YYYYYear Prear Prear Prear Prear Pre-Unie-Unie-Unie-Unie-Univvvvvererererersity [90 Hoursity [90 Hoursity [90 Hoursity [90 Hoursity [90 Hours]s]s]s]s]

Sl. Objectives → No. of Hours Knowledge Understanding Application Skill

No. Content ↓ VSA SA ET VSA SA ET VSA SA ET VSA SA ET TOTAL

1. Mathematical Logic 8 1 1 1 7

2. Permutation Combination 15 1 1 1 1 1 1 15and Probability

3. Binomial Theorem 4 1 1 4

4. Partial Fractions 4 1 4

5. Matrices and Determinants 15 1 1 1 1 15

6. Ratio, Proportions and Variations 10 1 1 1 1 7

7. Averages 4 1 4

8. Bill Discounting 4 1 1 5

9. Stocks and Shares 4 1 1 3

10. Learning Curve 4 1 4

11. L.P.P. 6 1 1 5

12. Circles, Parabola 10 1 1 1 9

13. Limits and Continuity 8 1 1 1 7

14. Differentiation 10 1 1 1 1 1 13

15. Applications of Derivatives 8 1 4

16. Integration 10 1 2 10

17. Applications of Integration 6 1 4

Total 130 7 6 8 2 16 20 1 6 24 2 28 120

Examination Corner 517

��

�� !��"#

� ��$%��&&'(�$��$ �)�*+�*

1. Negate “if two triangles are similar, then their areas are equal”.

2. If nc7 = nc3, find nc10.

3. Find the value of 2400 24012402 2403

.

4. Find the fourth proportional of 2, 6, 5.

5. A bill drawn for 4 months was legally due on 10.10.2003. Find the date of drawing the bill.

6. Define yield.

7. What is feasible region in L.P.P.

8. Find the centre of the circle 4x2 + 4y2 − 15x − 8y + 11 = 0

9. Evaluate Ltx

xe

x→

−0

5 1.

10. Total cost of a commodity is given by C = 1/3 x3 − x2 + 3x + 10. Find the marginal cost.

�� $%��,�-�,�&&�%��./ 0)�*+0*

11. If p is 2 is prime, q is 3 is even, r is 2 3 5+ = , , find the truth value of (p → q) → r.

12. Find the number of words that can be formed using all the letters of the word “BOOKS”. Howmany of them begin with ‘B’.

13. There are 10 points out of which 3 are collinear. Find the number of straight lines that can beformed.

14. Two dice are thrown simultaneously. What is the probability of getting the sum 7?

15. Expand 212

5

xx

−��

�� using Binomial theorem.

16. Find the middle term or terms in the expansion of xx

3133+�

�� .

17. If A =

2 11 21 3

−−�

��

�

find AA1.


18. Find the interest earned on Rs. 2448.35 cash invested in 15% Stock at 81.5, given that brokerageis 0.125%.

19. Find the equation of the parabola whose focus is (4, 0) and directrix is x + 4 = 0. Find the lengthof latus rectum.

20. If f xx

x

� � = −2 1, is continuous at x = 0, find f (0).

21. If y x xe= + +log ,1 2 � prove that dy

dx x=

+

1

1 2.

22. Integrate w.r.t. x, x x

x

+ +3 42

.

�� $%��-��,�-�,�&&�%��./ 1)2+�0

23. Prove that p q p q→ ↔ ∨� � � �~ is a tautology.

24. Prove that ncr + ncr − 1 = n + 1cr, and verify this for n = 5 and n = 2.

25. Resolve into partial fractions 1

1x x2 +� �.

26. Prove that if each element of a row (or column) constant multiples of corresponding elements ofother rows (or columns) are added, then the value of the determinant is unaltered.

�3 ��$%��-��,�-�,�&&�%��./ 1)2+�0

27. If two men and four women can do a work in 33 days and 3 men and 5 women can do the samework in 24 days, how long shall 5 men and 2 women take to do the same work?

28. Calculate the total wages earned per week by 400 workers in a factory from the following data

Daily wages Number of Workers

12.50 – 17.50 51

17.50 – 22.50 38

22.50 – 27.50 42

27.50 – 32.50 59

32.50 – 37.50 80

37.50 – 42.50 60

42.50 – 47.50 50

47.50 – 52.50 20

29. A bill for Rs. 2725.75 was drawn on 03.06.1997 and made payable 3 months after due date. It wasdiscounted on 15.06.1997 at 16% p.a. What is the discounted value of the bill and how much hasthe banker gained in this transaction.

30. A business men buys and sells chairs and tables. He has Rs. 3000/- to invest. A chair costs him


Rs. 50/- and a table costs him Rs. 90/-. He has space which can accommodate at most 48 pieces.If he sells each chair for Rs. 200 and each table for Rs. 400/-. Find the number of chairs and tableshe has to buy to obtain maximum profit.

3 1

31. Find the equation of the circle passing through the origin and having centre at (−3, 4).

OR

Derive the equation of a parabola in the form y2 = 4ax.

3� ��$%��-��,�-�,�&&�%��./ 1)2+�0

32. Prove that Ltx a

n nnx a

x ana

→−−

−= 1. for all rational values of n.

33. If x y y x x ydy

dx x1 1 0

1

1 2+ + + = ≠ = −+

and prove that , .� �

34. If x = at2, y = 2 at prove d y

dx at

2

2 3

1

2= − .

35. Evaluate e x dxx2 3� .

3�� $%��%��,�-�,�&&�%��./ �*)0+0*

36. (a) Two number are in the ratio 4:7. If 12 is added to each of them then the new ratio is 8:11. Findthe numbers.

(b) In how many ways 3 boys and 5 girls be arranged in a row so that

(i) no two boys are together.

(ii) all the girls are together.

(c) Solve by matrix method.

2x + y + z = 7.

x − y + 2z = 5

3x − 2y + 2z = 5.

37. (a) Find the equation of the circle with (−3, 5) and (6, 1) as the extremities of a diameter.

(b) An engineering company has succeeded in winning a contract for supplying aircraft engines.The prototype constructed to win the contract took 4000 labour hours and the companyexperiences 80% learning effect. Find the total cost of 7 engines of new order if labour costis Rs. 30/- per hour.

(c) The radius of a circular blot of ink is increasing at the rate of 3 cm per minute. Find the rateof increases of its area when its radius is 2 cms. What is the rate of increases of its circum-ference?

38. (a) Find the range in which the function f (x) = x2 − 6x + 5 is (1) increasing, (2) decreasing.

(b) Find the area bounded by the curve y = x2 − 7x + 10 with x-axis.


(c) Prove without expanding 111

2

2

2

a ab bc c

a b b c c a= − − −� �� .

* * *

��0/

� ��$%��&&'(�$��$/ �)�*+�*

1. If the truth value of p and q are T and F respectively. Then what will be the truth value of

p q p q∧ → ∨� � � �~ ~ .

2. Find the number of permutation of the letters of the word BALLOON taken all together.

3. Find the mean proportional to 2:8.

4. Find x if 2 7

14x��

� is singular.

5. A bill was drawn on 14.3.1997 for 3 months. When does the bill fall legally due?

6. Find the yield by investing Rs. 1140 on 15% stock quoted at Rs. 95.

7. Write any one step to formulate an LP problem.

8. Find the radius of the circle 4x2 + 4y2 + 8x + 9y − 7 = 0.

9. Evaluate: limx

x x→

− −1

3 41 1� � � �

10. If the marginal revenue function is given by 3x2 − 8, then find the total revenue function.

�� $%��,�-�,�&&�%��./ 0)�*+0*

11. Construct the truth table for p q p∧ →~ ~� � .

12. How many four digits numbers greater than 2000 can be formed with the digits 0, 1, 2, 3, 4, 5.

13. Find the number of permutations of the letters of the word ‘BANANA’ taking all letters. Howmany of them begin with N.

14. Two cards are drawn at random from a pack of well shuffled 52 cards. What is the probability ofgetting king and queen cards.

15. Expand using binomial theorem: 32 4

+��x

.

16. Find the middle term or terms in the expansion of 1

22

10

x

x+��

�� .

17. If A = −�

��

�

2 0 11 6 21 3 2

, then find AA′.


18. Find the cash required to purchase Rs. 1600, 8 1/2 stock at 105, brokerage is 1/2%.

19. Find the equation of parabola whose vertex is (1, 1) and focus is (3, 1).

20. If f xe

x

x

� � = −3 1 is continuous at x = 0. Then find f (0).

21. If x2 + y2 = 16. Then find dy

dx when x = 3 and y = 4.

22. Integrate w.r.t. x: x xx

3 1

3 2+

+

�� $%��-��,�-�,�&&�%��./ 1)2+�0

23. Prove that p q p q∧ ∧ ∨~ ~� � � � is a contradiction.

24. Prove that ncr = ncn − r and hence find n if nc18 = nc10.

25. Resolve into partial fractions:

2 3 2

2

2

2

x x

x x

+ +− −

.

26. If the elements of any row (or column) is multiplied by non zero constant K. Then prove that thevalue of the determinant is multiplied by K.

�3 ��$%��-��,�-�,�&&�%��./

27. 16 men or 28 boys can fence a farm in 40 days. In how many days will 24 men and 14 boyscomplete the same work.

28. 3 tests in English, 2 in Hindi, 4 in Kannada and 5 in Sociology are conducted. The average marksscored by Rashmi in English is 60, that in Hindi is 56 and that in Kannada is 45. If the averagemarks of all the subjects and all the tests taken together is 48. Then find the averages marks scoredby her in Sociology.

29. The difference between the bill discount and true discount on a certain sum of money due in 4months in Rs. 10. Find the amount of the bill (or face value) if the rate of interest is 3% p.a.

30. Maximise Z = 7x + 4y subject to x y x y x y+ ≤ + ≤ ≥ ≥3 3 6 3 8 0 0, ; , . .

3

31. Derive the equation of circle which is described on the line joining (x1 y1) and (x2 y2) as the endsof diameter.

OR

Find the equation of parabola whose ends of latus rectum are (3, 2) and (3, −4).

3� ��$%��-��,�-�,�&&�%��./ 1)2+�0

32. Evaluate: (i) lim...

n

n

n→∞

+ + +−

1 2 3

6

2 2 2 2

3 (ii) limn

xx→∞

+1 31

� �


33. If y x x x y xy m ym

= + + + + − =1 022 1

2 � � �, . then prove that 1 2

34. If xy + 4y = 3x, then prove that d y

dx x

2

2 3

24

4= −

+� � .

35. Evaluate: (a) x x dxlog�

(b) x e

x edx

e x

e x

− −++

��

��

1 1

3�� $%��%��,�-�,�&&�%��./ �*)0+0*

36. (a) Divide Rs. 118 among A, B and C so that A:B = 3:4, B:C = 5:6.

(b) From 8 lecturers and 4 students a committee of 6 is to be formed. In how many ways can thisbe done so that the committee contains (i) Exactly 2 students (ii) atleast 2 students.

(c) Solve by matrix method:

5 4 4 4 2 12 3 3x y z x y z x y z− − = + + = − − =, , .

37. (a) A circle with centre (2, −1) passes through (−1, 3). Find the equation of the circle.

(b) The first sample batch is of 50 units of product. A took 80 hours to make. The company nowwishes to estimate the total time taken to make 100 additional units. Solve the problem using80% learning effect.

(c) If the base of the triangle is 3 times the height and the height is decreasing at the rate of 4 cm/sec. then find the rate of decrease of the area when height is 1 cm.

38. (a) Find 2 numbers whose sum is 20 and whose product is maximum.

(b) Find the area bounded by the curve y2 = 4x and the line y = x.

(c) Prove without expanding a ab acab b bcac bc c

a b c

2

2

2

2 2 24− −−

= .

* * *

��2/

� ��$%��&&'(�$��$/ �)�*+�*

1. Write the contrapositive of ‘If the triangle is not right angled then its two sides are not perpendi-cular’.

2. If nc2 = 105, then find n.

3. Find x and y if 234 0

38 0

0 132 3

x y x��

�

+ ��

�

= ��

� .


4. If a : b = 2 : 3 and b : c = 5:7, then find a : c.

5. The present worth of a bill due sometime hence is Rs. 1100 and the true discount is Rs. 110. Findthe banker’s gain.

6. Find the proceeds got by selling shares worth Rs. 3000 at 96 1/8 brokerage 1/8%.

7. What is an objective function in LPP.

8. Find the focus of the parabola x2 + 8y = 0.

9. Evaluate: limn

n n

n→∞

+

−

2

3

2

6 8

� �

10. If the marginal cost function is x3 + 2x − 1 and fixed cost is Rs. 2500, then find total cost function.

�� $%��'(�$��$/ 0)�*+0*

11. The compound proposition p q r s∧ → ∨� � � � has truth value false. Find the truth values of p, q,

r and s.

12. In how many ways can 3 girls and 4 boys sit round the table so that no two girls sit next to eachother.

13. Find the number of triangles that can be formed from 12 points of which 4 are given to becollinear.

14. 2 dice are thrown simultaneously what is the probability of getting the sum 8.

15. Expand a

x

x

b+�

��

5

using binomial theorem.

16. Find the term independent of x in the expansion of xx

−��

12

21

.

17. Solve by Cramer’s rule: 3x + y = 8 and 4x − 9y = 6.

18. Find the cash you would require to purchase a share worth Rs. 3000 of 4% at 102 8/9 Brokerage1/9%.

19. Find the vertex focus, directrix and length of latus rectum of the parabola x2 − 8y + 24 = 0.

20. If f xx xk x

� � = − ≥+ <

��4 3 1

1 if

2 if is continuous at x = 1, then find K.

21. If y a x y= + , then find dy

dx.

22. Evaluate: x e dxx2� .

�� $%��-��,�-�,�&&�%��./ 1)2+�0

23. Verify p q p q∧ ∧ →� � � �~ is a tautology or contradiction or neither.


24. Prove that n

rpn

n r=

− and hence find 5p2.

25. Resolve into partial fractions:

2 3

3 2

x

x x x

−+ −� ��

26. A salesman has the following records of sales during 3 months for 3 items A, B, C which hasdifferent rates of commission.

Months Sales of units Total Commissionin Rs.

A B C

Jan 90 100 20 800

Feb 130 50 40 900

Mar 60 100 30 850

Find out the rates of commission on items A, B and C.

�3 ��$%��-��,�-�,�&&�%��./ 1)2+�0

27. In a mixture of 35 litres, the ratio of milk and water is 6:1. If 2 litres of milk and 3 litres of waterare added to the mixture, what is the new ratio of milk and water?

28. The average age of A and B is 18 yrs., that of B and C is 17 yrs. and that of C and A is 20 yrs.Find the ages of A, B and C.

29. The banker’s gain on a certain bill due 6 months hence is Rs. 20. The rate of interest being 20%per annum. Find the face value of the bill.

30. A company produces 2 products x and y. Each of the product require two operations one onmachine A and other on machine B. The machine hours required by these two products and thetotal hours available are given as follows:

Machine hrs. Product Total machine required —–––––––———–––––––––––––––– hrs. available

x y

A 2 5 19

B 4 3 17

Each unit of the product x and y makes a profit of Rs. 3 and Rs. 4. Find the optimal solution ofthe product to be produced to obtain the maximum profit.

3 �1#

31. Find the equation of circle passing through (4, 0) and (0, 4) and its centre lies on 3x + 4y = 7.OR


Find the equation of parabola whose vertex is (0, 0) axis: Y-axis and passing through (−1, 4).

3� ��$%��2�,�-�,�&&�%��./ 1)2+�0


n nnx a

x ana

→−−


33. Find dy

dxx yy x if = .

34. If y ae bemx mx= + − , then prove that y m y22 0− = .

35. Evaluate: ex

xdxx

2

2

1

1

++

��

�� .

3�� $%��%��,�-�,�&&�%��./ �*)0+0*

36. (a) If a : b = 7 : 5, find 4a − 2b : a + 3b. (2)

(b) In how many ways can 5 members forming a committee but of 10 be selected so that

(i) 2 particular members must be included.

(ii) 2 particular members must not be included. (4)

(c) Solve by matrix method:

x y z− − = −1

y z+ =3 3

z z+ =4 2 (4)

37. (a) Prove that x + y = 2 touches the circle x y x y2 2 6 10 0+ − + + =� � . (2)

(b) What do you mean by 80% learning effect?

Find the index of learning for 80% learning effect. (4)

(c) A ladder 5 mts. long rests with its ends on a horizontal floor and against a smooth verticalwall. If the upper end is coming downwards at the rate of 40 cm/sec. Find the rate at whichthe lower end moves when the upper end is 3 mts. from the ground. (4)

38. (a) The area of square is increasing at the rate of 2 cm2/sec. Find the rate at which the perimeteris increasing when the side of the square is 16 cm. (2)

(b) Find the area between the parabolas.

x y y x2 24 4= = and . (4)

(c) Prove without expanding 1

11

1+

++

= + + +a b c

a b ca b c

a b c. (4)

* * *


��

��

� �� !"#$�!�%&�'�

1. Define a proposition.

2. Give the truth value of ‘4 + 3 ≥ 8’.

3. If p : 6 is an even number, ‘q : 4 is odd’, then write the true value of p ∨ q.

4. Write in symbols: The necessary and sufficient condition for a triangle to be equilateral is all itssides should be equal.

5. Write symbolically: ‘6 is even and 2 is not irrational’.

6. Negate: ‘Cow is not big or it is black.’

7. If the truth values of p and q are T and F respectively then what will be the truth value of

p q p q∧ → ∨� � � � .8. If the truth values of p, q, r are T, T, F respectively, find the truth value of p → (~r → q).

9. If the truth value of p → ~r is F, then find truth value of r.

10. Write the truth table for p ∧ ~p.

11. Write the truth table for p ∨ ~p.

12. Write the converse of ‘If a + 2 = 3, then a = 1.’

13. Write the converse of ‘If cows can fly then birds cannot fly’.

14. Write the inverse of ‘If he is rich then he is happy.’

15. Write the contrapositive of ‘If e is not irrational then 2 is rational’.

16. Construct the truth table for p → ~p.

17. Negate: ‘If 2 is even then 3 is odd’.

18. Negate: p → ~q.

19. Negate: ‘If 6 is even and 7 is odd, then 2 is prime.’

20. Prove that p ≡ ~(~p).

� �($��!�� !"#$�!�%�'�

1. Construct the truth table for ~p ∧ q.

2. Construct the truth table for p → (~q ∧ p).

3. Negate: ‘If a triangle is equilateral then its sides are equal and angles are equal.’

4. Negate: If x + y ≠ 6 and x ≠ 5, then y ≠ 0 or y ≠ 7.

5. Write the inverse and converse of:

If a triangle is equilateral then its sides are equal and angles are equal.

6. Write the converse and contrapositive of ‘If a number is real then it is either rational or irrational.’

7. Prove that ~[p → q] ≡ p ∧ ~q.


8. If p, q, r and s are false propositions, then find the truth value of (p ∧ ~q) → (~r ∨ s).

9. Define Tautology.

10. Define converse and inverse of a conditional.

11. Define logically equivalent propositions.

12. Define contradiction.

13. If p, q, r, s are propositions with truth values T, F, T, T respectively, then find the truth value of

p q q r s∨ ↔ ∧ →� � � �~ ~ .

14. Negate: (p → q) ∧ (p → r).

15. Negate: If it rains today then principal does not declare a holiday and we are not happy.

� )$ ��!�� !"#$�!�%��'�

1. Prove that ~ ~p q p q∨ ∨ ∧� � � � is a tautology.

2. Verify whether p q p q∧ ∧ ∨~ ~� � � � is a tautology or contradiction.

3. Prove that p q p q q∧ ∨ ∨ ↔� � � � is neither tautology nor contradiction.

4. Prove that the negation of disjunction of 2 proposition is logically equivalent to the conjunctionof their negation.

5. Prove that p q p p∨ ∧ →� � ~ is neither tautology nor contradiction.

6. Prove that ~ ~p p q q∧ ∨ ∧� � is a contradiction.

7. Prove that ~ ~ ~p q p q q p↔ ≡ ∧ ∨ ∧� � � � .8. Define converse and contrapositive of a conditional and prove that a conditional and its

contrapositive are logically equivalent.

9. Prove that the converse and inverse of a conditional are logically equivalent.

10. Prove that contrapositive of a conditional is converse of inverse of a conditional.

*� ��+��,�

� �� !"#$�!�%&�'�

1. If nc4 = nc6, then find nc7.

2. Find the number of permutations of the letters of the word BOOKS, taking all letters.

3. How many of the arrangements of the word ‘ABACUS’ begin with A.

4. If nc2 = 105, find n.

5. If np2 = 30, find n.

6. If np5 = 24 nc4, then find n.

7. How many of the arrangements of the word ‘CRICKET’ begin and end with C.

8. State addition rule of probability for mutually exclusive events.


9. A ticket is drawn from a bag containing tickets bearing numbers 1 through 25. Find the probabil-ity of its bearing a number which is a multiple of 3.

10. One card is drawn from a pack of 52 cards. What is the probability of getting a king card?

� �($��!�� !"#$�!�%�'�

1. In how many ways 4 boys and 3 girls may be arranged for a photograph so that all the 3 girls areshown together?

2. In how many ways can 5 members forming a committee out of 10 be selected so that 2 particularmembers must be included?

3. Six students have taken an examination. In how many ways can the first 3 positions be declared?

4. In how many ways can the letters of the word HEXAGON be arranged so that H appears in thebeginning.

5. There are 5 maths books, 4 accounts books and 3 economics books. In how many ways can thebooks be arranged so that the books of the same subject are together.

6. How many words can be formed by using all the letters of the word ‘ALLAHABAD’?

7. 20 persons were invited to a party. In how many ways can they and host be seated at a circulartable?

8. There are 12 points in a plane. 4 points are collinear. Find how many straight lines can be drawn?

9. A box contains 7 red, 6 white and 4 blue balls. How many selections of 3 balls can be made sothat none is red.

10. How many diagonals are there in an octagon?

11. Three coins are tossed. What is the probability of getting.

1. Exactly 2 heads.

2. at least 2 heads.

12. A pair of dice is thrown. What is the probability that the sum of the numbers obtained is more than10.

13. A card is drawn from a pack of 52 cards. What is the probability that it is either king card or a redcard.

14. A pair of dice is rolled. If the sum on the 2 dice is 9, find the probability that one of the dice isshowed 3.

15. Two dice are thrown, find the probability of a doublet.

16. Prove that 0 1=

� )$ ��!�� !"#$�!�%��'�

1. A student has to answer 8 out of 10 questions in an examination. How many choices has he? Howmany choices has he if he must answer the first 3 questions.

2. Prove that nrp

n

n r=

−.


3. Prove that nrc

n

n r r=

− ⋅.

4. Prove that ncr = ncn − r and hence find n if nc8 = nc4.

5. Prove that ncr + ncr − 1 = n + 1cr and verify this for n = 5 and r = 3.

6. A committee of 10 members is to be chosen from 9 teachers and 6 students. In how many waysthis can be done if

(1) the committee contains exactly 4 students.

(2) there are atmost 7 teachers.

7. How many 4 digits numbers greater than 3000 can be formed with the digits 1, 2, 3, 5, 7 (repeti-tions of digits is not allowed). How many of them are even?

8. How many (i) Straight lines (ii) Triangles are determined by 12 points. No three of which lie onthe same straight line.

9. There are 5 questions in part A and 4 in part B of a question paper. In how many ways can studentanswer 6 questions if he has to choose at least 2 from each part.

10. A man has 7 relatives. 4 of them are ladies and 3 are gentleman. His wife also has 7 relatives 3of them are ladies and 4 are gentlemen. In how many ways can they invite 3 ladies and 3gentleman to a dinner partly so that there are 3 of man’s relatives and 3 of wives’ relatives.

11. State and prove addition rule of probability.

12. What is the probability of getting neither 8 nor 11 when a pair of dice is tossed?

13. State and prove multiplication rule of probability.

14. Out of the numbers 1 to 120, one number is selected at random. What is the probability that it isdivisible by 8 or 10.

15. A card is drawn from a pack of 52 cards. What is the probability that it is neither a red card nora jack card.

16. If P A P B P A B P A B P B A� � � � � � � � � �= = ∪ =3

8

5

8

3

4, , ; . and find Are the events independent?

-� ��.��)��

� �� !"#$�!�%&�'�

1. The second term in the expansion of (x + 3)5 = 240. Find x.

2. Write the middle term in the expansion of a

x

x

a+�

��

4

.

3. Write the 5th term in the expansion of 21 8

−��x

.


4. Find the middle term in the expansion of xx

−��

1 8

.

5. State Binomial theorem.

� �($��!�� !"#$�!�%�'�

1. Find the middle term in the expansion of 312

9

xx

+��

�� .

2. Expand: a bx+� �4

3. Expand: 1 2 5− x� �

4. Find the middle term in the expansion of1

22

10

x

x−�

�� .

5. Find the co-efficient of x5 in the expansion of axx

23

51+��

�� .

6. Find the term independent of x in the expansion of axx

23

51+��

�� .

7. Prove that there is no term independent of x in the expansion of 212

10

xx

+��

�� .

8. Find the value of (1.01)3 correct to 3 decimal places.

9. Prove that the sum of Binomial co-efficient of order n is 2n.

10. Prove that the sum of odd Binomial co-efficient = sum of even Binomial coefficients of ordern = 2n − 1.

� )$ ��!�� !"#$�!�%��'�

1. State and Prove Binomial theorem.

2. Simplify: 2 3 2 35 5

+ + −� � � �

3. Find the middle term in the expansion of 212

11

xx

−��

�� .

4. Find the coefficient of a6b3 in the expansion 23

9

ab−�

�� .


5. Find the constant term in the expansion of 21

15

xx

−��

�� .

6. Prove that there is no term independent of x in the expansion of ayb

y2

25

+��

�� .

7. Resolve into partial fractions: 7 9

1 2

x

x x

−− −� ��

8. Resolve into partial fractions: x

x x

+− −

8

2 12

9. Resolve into partial fractions: 1

23 2x x x+ −

10. Resolve into partial fractions: x x

x x x

2

2

10 13

1 5 6

− +− − +� ��


2 12

x

x x

−− +� � � �


x x

−+ +

1

1 12� ��


x x

2 1

1 2

++ +� ��

14. Resolve into partial fractions: 2 3 3 2

1 1

3 2

2

x x x

x x

+ − +− +� ��


1

2

3

x

x

+−

/� ��

� �� !"#$�!�%&�'�

1. If A = [1 2 3], then find AA′.2. If A is of order m × n, B is of order n × p, then does AB exists? If so what is its order?


3. Find x if 1 2 3

1

7

3x

�

�

�

�

��

=

4. If and then find A B AB= ��

��

= ��

1 2

1 7

2

3, .

5. Write the adjoint of 1 7

0 6��

��

.

6. State whether 2 3

7 8��

�� is singular or non-singular.

7. If A = ��

��

2 0

0 2, then find A2.

8. Find the value of 99 100

101 102.

9. If D x x x=2 3 4

2 3 4

9 8 7

, then prove that D = 0.

10. If then find AA

=−��

��

7 7

6 5 2, .

11. If A

x

y

z

x

y

z

A

�

�

�

�

��

=�

�

�

�

��, .then find

12. Find x if

1 2 7

1 8

1 2 3

x

�

�

�

�

��

is singular.

13. If

1 0 0

0 1 0

0 0 1

2

3

4

�

�

�

�

��

�

�

�

�

��

=−

�

�

�

�

��

x

y

z

, then find x, y and z.

14. The following matrix shows the belongings of 3 friends Raju, Rakshith and Rama.


Shirts Trousers

Raju

Rakshith

Rama

5

7

8

3

4

9

�

�

�

�

��

Write the row matrix which represents the belongings of Rakshith.

15. A company sold 22 chairs, 15 tables in January and 16, 28 respectively in June. Represent the datain matrix form.

16. A business organisation has 2 distribution outlets. At the beginning of the month, the stock onhand of products A, B and C are given in matrix M and the quantities of A, B and C sold duringthe month are given in matrix N. What are the stocks on hands (in matrix form) at the end of themonth?

M N= ��

��

= ��

��

20 30 5

6 7 9

17 12 2

3 1 7,

17. If A B AB= ��

��

= ��

��

1 2

7 6

2 0

6 7 and then find , .

18. Find x and y if

x

y

7

4 0

7 1

9

8 8

13 8��

��

+ ��

��

= ��

��

19. Find the minor of 3 in 1 7

3 6��

��

.

20. If A =1 2 7

1 5 6−��

��

, then prove that (A′)′ = A.

� �($��!�� !"#$�!�%�'�

1. If A B= ��

��

=−�

� ��

1 7

6 5

2 1

3 2 and , then prove that (AB)′ = B′A′.

2. If 27 11

7 42

8 13

14 5A B A B+ =

−��

��

+ =−��

��

and , then find A.

3. Find AB if A B

1 7

2 6

5 2

2

7

�

�

�

�

��

=−−��

and .


4. If A = ��

��

1 0

2 3, then prove that A2 − 4A + 3I = 0.

5. Find A and B if A B A B+ = ��

��

− = ��

��

7 0

2 5

3 0

3 3 and .

6. Prove that

x y y z z x

y z z x x y

z x x y y z

− − −− − −− − −

= 0 .

7. Evaluate:

100 101 102

103 104 105

106 107 108

.

8. Without expanding the determinant prove that

1 2 3

2 3 4

3 4 5

0= .

9. If A B AB= �

� ��

= ��

��

−2 2

3 5

7 0

1 6 2 and then find , .

10. Solve by Cramer’s rule: 2x + y = 1, x − 3y = 4.

11. Solve by Matrix method: 3x + 4y = 7, 7x − y = 6.

12. Suppose the matrix A and B represent the number of items of different kinds produced by 2

manufacturing units in one day A B=�

�

�

�

��

=�

�

�

�

��

1

2

3

7

6

5

and . Compute 2A + 3B. What does 2A + 3B

represent?

13. Ajith buys 5 kgs of tomato, 2 kgs. of Beans and 8 kgs of carrot. If the cost of each is Rs. 5,Rs. 7 and Rs. 9 respectively, then find the total cost by matrix method.

14. A company sold 40 metal chairs, 20 wooden chairs and 10 plastic chairs. The selling price of ametal chair is Rs. 150 that of wooden chair is Rs. 500 and plastic chair is Rs. 300. Find the totalrevenue using matrix method.

15. Find the adjoint of 1 2

7 0

−��

��


� )$ ��!�� !"#$�!�%��'�

1. Solve the following using Cramer’s rule:

4x + y = 7, 3y + 4z = 2 3z + 5x = 2.


2. Solve by using Cramer’s rule: 3 2 18 2 3 3 5 8x y z x y z x y z− + = + − = + − = −, , .

3. Find the adjoint of

1 4 2

2 5 4

1 2 1

−−

−

�

�

�

�

��


4. Find the inverse of

1 1 2

0 2 2

1 1 3

−−

�

�

�

�

��.

5. Prove that A =�

�

�

�

��

1 2 2

2 1 2

2 2 1

satisfies the equation A2 − 4A − 5I = 0 .

6. Solve by matrix method: x + y + z = 6, 3 2 2 5x y z x z y+ − = + − =, .

7. Solve by matrix method: 4 7 3 4 5 3 5 2x y y z z x+ = + = + =, , .

8. Prove that

1 2 1

1 2 1

1 2 1

42

+ −− + −

− += +

x

x

x

x x� � .

9. Prove that

x p q

p x q

p q x

x p x q x p q= − − + +� �� .

10. Prove that

a b c a b

c b c a b

c a c a b

a b c

+ ++ +

+ += + +

2

2

2

2 3� � .

0� ��+��&��

� �� !"#$�!�%&�'�

1. Find the fourth proportional of 2, 6, 5.

2. Express 13:169 in the simplest form.

3. Compare 2:5 and 3:2.

4. Find the ratio between 1 hr 10 min and 575 min.

5. Write the subtriplicate of 1:27.

6. If a:b = c:d, then prove that b:a = d:c.


7. Find the missing term: x:4 = 27:12.

8. If a:b = 2:3, b:c = 4:5, then what is c:a?

9. If Ramu can do a piece of work in 8 days and Rithu can do the same work in 3 days, then in howmany days, both together can do the same work?

10. Divide Rs. 180 in the ratio 1:2:3.

� �($��!�� !"#$�!�%�'�

1. Two number are in the ratio 4:7. If 12 is added to each of them then the new ratio is 8:11. Findthe numbers.

2. The prices of a scooter and a T.V. are in the ratio 3:2. If the scooter costs Rs. 6000 more than theT.V., then find the cost of TV.

3. A certain amount was divided between Kavitha and Reena in the ratio 4:3. If Reena’s share wasRs. 2400, then find the amount divided.

4. What number should be added to each one of 6, 14, 18, 38 to make it equally proportionate.

5. Three friends divide Rs. 624 among themselves in the ratio 1

2

1

3

1

4: : . Find the share of the third

friend.

6. Gold is 19 times as heavy as water and copper 9 times as heavy as water. Find the ratio in whichthese two metals be mixed so that the mixture is 15 times as heavy as water.

7. The ages of Vivek and Sumit are in the ratio 2:3. After 12 years their ages will be in the ratio11:15. Find the age of Sumit.

8. If 60 men can complete a job in 12 days, how many days will 36 men take to complete the samejob?

9. Find 3 numbers in the ratio 2:3:5, the sum of whose squares is 608.

10. Cycling 3

4 of his usual speed, a student is 10 minute late to his class. Find his usual time to cover

the distance.

� )$ ��!�� !"#$�!�%��'�

1. If 2 men and 4 women can do a work in 33 days and 3 men and 5 women can do the same workin 24 days. How long shall 5 men and 2 women take to do the same work.

2. A can do a piece of work in 6 days and B alone can do it in 8 days. A and B under took to do itfor Rs. 320, with the help of C, they finished it in 3 days. How much is paid to C?

3. 12 men or 18 women can reap a field in 14 days. In how many days 8 men and 16 women canreap it?

4. 4 men and 6 women finish a job in 8 days while 3 men and 7 women finish it in 10 days. In howmany days 10 women working together will finish it?

5. 5 carpenters can earn Rs. 3600 in 6 days working 9 hours a day. How much will 8 carpenters earnin 12 days working 6 hours a day?


6. Two trains start at the same time from Aligarh and Delhi and proceed towards each other at 16km/hr and 21 km/hr respectively, when they meet, it is found that one train has travelled 60 kmmore than the other. Find the distance between 2 stations.

7. Kanthilal mixes 80 kgs of sugar worth Rs. 6.75 per kg with 120 kg worth of Rs. 8 per kg. At whatrate shall he sell the mixture to gain 20%?

8. A can contains a mixture of 2 liquids A and B in the ratio 7:5, when 9 litres of mixture is drawnoff and the can is filled with B, the ratio becomes 7:9. How many litres of the liquid A wascontained by the can initially?

9. Two men undertake to do a piece of work for Rs. 400. One alone can do it in 6 days and the otherin 8 days. With the help of a boy they finished it in 3 days. What is boy’s share?

10. Adarsh buys 2 horses for Rs. 1350 and sells one at 6% loss and other at 7.5% gain and on thewhole, he neither gains nor loses. What does each horse cost?

1� ��)�)��%�2��3�!+��#44��#!5$ �"#�3+�"$5�!�.�6��!+��#�3�� 2�+��'

� �� !"#$�!�%&�'�

1. A bill drawn for 4 months was legally due on 10-10-2003. Find the date of drawing the bill.

2. Define yield.

3. What is feasible region in LPP?

4. The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find theage of 10th student.

5. If 10 kgs of a commodity is purchased at Rs. 24 per kg and 5 kgs at Rs. 18 per kg, what is theaverage price of the commodity per kg?

6. Find the cost of Rs. 9100, 8 34 % stock at 92.

7. The present worth of a certain sum due sometime hence is Rs. 1600 and the true discount isRs. 160. What is the banker’s gain?

8. What is learning curve?

9. Find the index of learning for 80% learning effect.

10. What is the cost price of Rs. 100 share at 4 discount if the brokerage is 1/4%?

� �($�� !"#$�!�%�'�

1. Find the interest earned on Rs. 2448.35 cash invested in 15% stock at 81.5, given that brokerageis 0.125%.

2. Find the market value of 5 and 1/4% stock, in which an income of Rs. 756 is derived by investingRs. 14976, brokerage being 1/4%.

3. The average age of 10 students is 14 years. Among them, the average age of 4 students is 12 years.Find the average age of the remaining students.

4. Find the income derived from 44 shares of Rs. 25 each at 5 premium (brokerage 1/4 per share)the rate of dividend being 5%. Also find the rate of interest on the investment.


5. A worker takes 20 hours to produce the first unit of product. What is the cumulative average timeper unit taken by him for the production of first 2 units?

� )$ ��!�� !"#$�!�%��'�

1. A bill for Rs. 2725 was drawn on 3-6-1997 and made payable 3 months after due date. It wasdiscounted on 15-6-1997 at 16% p.a. What is the discounted value of the bill and how much hasthe banker gained in this transaction.

2. A business man buys and sells chairs and tables. He has Rs. 3000 to invest. A chair costs himRs. 50 and a table costs him Rs. 90. He has space which can accommodate at most 48 pieces. Ifhe sells each chair for Rs. 200 and each table for Rs. 400, find the number of chairs and tables hehas to buy to obtain maximum profit.

3. An engineering company has succeeded in winning a contract for supplying aircraft engines. Theprototype constructed to win the contract took 4000 labour hours and the company experiences80% learning effect. Find the total cost of 7 engines of new order if labour cost is Rs. 30 per hour.

4. A bill of Rs. 5000 drawn on 10-04-96 at 3 months was discounted on 1-5-96 at 7% per annum.Find the discounted value of the bill and banker’s gain.

5. A banker pays Rs. 2340 on a bill of Rs. 2500, 146 days before the legally due date. What is therate of discount charged by the banker?

6. A bill for Rs. 2920 drawn at 6 months was discounted on 10-4-97 for Rs. 2916. If the discountrate is 5% per annum on what date was the bill drawn?

7. A batsman realises that by scoring a century in the 11th innings of his test matches he has betteredhis average of the previous 10 innings by 5 runs. What is his average after the 11 innings?

8. Ms. Vidya bought 17 books in a discount sale. The average price of the books being Rs. 53. Theaverage price of the eleven maths books is Rs. 71. If the prices of the remaining 6 accounts booksform an increasing arithmetic progression with last term Rs. 25, find the price of the cheapestbook.

9. Maximise Z = 10x + 30y subject to x + 2y ≤ 20. x + 5y ≤ 35 and x, y ≥ 0. Indicate the feasibleregion on the graph sheet.

10. Minimise Z = 20x + 10y subject to x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x ≥ 0, y ≥ 0.

11. Mr. Harsha has invested a certain amount of money in 13% stock at 101. He hold the investmentwhen the market value went down to 96.5. He lost Rs. 3564 on this process. If he has paid thebrokerage of 1.5% for all the transactions, what was the amount of cash investment and what wasthe stock value of the investment in the first instance.

12. A person has invested Rs. 4300 partly in 4.5% stock at 72 and partly in 5% stock at 95. If the totalinterest is Rs. 250, find his investment in each type of stock.

13. The time required to produce the first unit of a product is 1000 hrs. If the manufacturer experi-ences 80% learning effect, calculate the average time per unit and the time taken to producealtogether 8 units. Also find the total labour charges for the production of 8 units at the rate of Rs.10 per hour.


7� ��+��

� �� !"#$�!�%&�'�

1. Find the radius of the circle 2 2 3 8 6 02 2x y y x+ − + − = .

2. Find the centre of the circle 3 3 6 7 8 02 2x y x y+ − + − = .

3. Find the equation of the tangent to the circle x2 + y2 = 2 at (1, 1).

4. Find the length of the tangent from (−1, 1) to the circle x y x y2 2 2 3 1 0+ + + + = . .

5. Find the distance of the centre of x y x y2 2 4 6 9 0+ + + + = from x-axis.

6. If the circle (x − 6)2 + (y − 6)2 = 36 touches both co-ordinate axes, find the distance of co-ordinateaxes from the centre.

7. Mention the formula to find the equation of tangent at (x1, y1) to the circle with centre origin andradius r units.

8. Write the general equation of the circle.

9. If 3 9 8 7 02 2x ay bxy x y+ + + − + = represent a circle, then what is the value of a and b?

10. Where do the point (−2, 3) lie with respect to the circle x y x y2 2 16 3 7 0+ + − − = .

11. Write the equation of the directrix of the parabola y2 = 12x.

12. Find the co-ordinates of ends of latus rectum of x2 = −4y.

13. Find the length of latus rectum of x2 + 8y − 7 = 0.

14. What is the eccentricity of the parabola x2 = 8y.

15. Find the vertex of the parabola 3x2 − 2y = 0.

� �($��!�� !"#$�!�%�'�

1. Find the equation of the circle whose 2 diameters are 2x + y =3 and x − y = 6 and radius 3 units.

2. Prove that x + y = 2 touches the circle x y x y2 2 6 6 10 0+ − − + = . .

3. If the circle x y gx fy c2 2 2 2 0+ + + + = touches x-axis then prove that g2 = c.

4. Find the length of the perpendicular from the centre of the circle x y x y2 2 4 3 0+ + − + = to the

chord 3x + 4y + 1 = 0.

5. A circle with centre (2, −1) passes through (−1, 3). Find its equation.

6. If one end of diameter of the circle x y x y2 2 2 0+ − + − = is (1, 1), then find the other end.

7. Find the vertex, and axis of the parabola x2 + 2x −y − 4 = 0.

8. Find the focus and length of latus rectum of (y − 2)2 = −8 (x + 2).

9. Find the ends of latus rectum for the parabola (y + 2)2 = 4 (x − 1).

10. Find the equation of parabola given that vertex is (0, 0) and focus is (0, 4).


11. Define a parabola: what is the eccentricity of the parabola.

12. Find the length of latus rectum for the conic 4 3 6 12y y x+ + = .

� )$ ��!�� !"#$�!�%��'�

1. Derive the standard equation of the circle. Find its centre and radius.

2. Derive the equation of tangent to the circle x y gx fy c x y2 21 12 2 0+ + + + = at ,� � on it.

3. Derive the length of the tangent to the circle x y gx fy c2 2 2 2 0+ + + + = from (x1, y1).

4. Find the condition for the line y = mx + c to be a tangent to the circle x2 + y2 = a2. Find the pointof contact also.

5. Derive the equation of circle which is described on the line joining (x1, y1) and (x2, y2) as diameter.

6. Derive the equation of parabola in the standard form.

7. Find the equation of the circle passing through (0, −2) having centre on x + y = 0 and having 6as the length of the tangent from (1, 1).

8. Find the equation of the circle passing through the points (1, 1), (−2, 2) and (−6, 0).

9. Find the equation of the circle given that the length of the tangents from the points (−1, 2),

(−2, 3) and (2, 4) are respectively 2 3 30 17, . and

10. Find the equations of the tangents to the circle x y x y2 2 2 2 7 0+ + + − = which are parallel to 3x

− 4y + 8 = 0.

11. Find the equations of tangents to the circle 5 5 4 2 60 02 2x y x y+ + + − = which are ⊥r to 16x −7y + 7 = 0.

12. Find the equation of the parabola given that vertex (3, 4) and directrix y = 1.

13. Find the equation of the parabola given that Focus = (−4, 0) and directrix x = 4.

14. Find the equation of the parabola given that vertex (0, 0) axis y-axis and passing through (−1, 4).

15. Find the focus, directrix, axis, length of latus rectum of the parabola y y x2 3 4 0+ + + = .

16. Find the vertex, axis, ends of latus rectum of the parabola x x y2 2 4 0+ − − = .

17. Find the equation of parabola given that ends of latus rectum (−2, 3), (−2, 6).

18. Prove that the circles x y x y x y x y2 2 2 22 4 3 0 14 16 81 0+ − − − = + − − + = and touch each other.

Also find the co-ordinate point of contact.

8� ��,�

� �� !"#$�!�%&�'�

1. Evaluate: limx

xe

x→

−0

3 1


2. Evaluate: limx

x

x→

−−3

4 4

3 3

3

3

3. Evaluate: lim...

n

n

n→∞

+ +1 22 2 2

3

4. Evaluate: limx

x x x

x x→

+ − +− +2

3 2

2

4 3 2

7 2

5. Evaluate: limx

x

x→

−0

38 1

6. Evaluate: limn

n

n→∞+��1

2 3

� �($��!�� !"#$�!�%�'�

1. Find f (0) if f xx x

x x� � = − <

− >��

3 2 0

2 0

if

4 if is continuous at x = 0.

2. If f xx

x x� � = −

− +

2

2

9

4 3 for x ≠ 3 is continuous at x = 3, find f (3).

3. Prove that limx

x

x→0 does not exist.

4. If f x x x� � � �= +1 21

is continuous at x = 0 find f (0).

5. Evaluate: limx

x

x→

+ −0

3 3

6. Evaluate: limx

x

x→

− −

− −

−

−3

3 3

4 4

3

3

� ��

� )$ ��!�� !"#$�!�%��'�


n nnx a

x ana

→−−


2. Prove that the function f (x) = |x| is continuous at x = 0.


3. Prove that f x x

x

x� � =−

−≠

��

��

1 181

1 127

34

3

for is continuous at x = 3.

4

9 for x = 3

4. Verify for continuity of f (x) at x = 0, if f x

x

xx

b x� �

� �=

+ ≠

=

��

��

log 1 60

0

for

for.

5. Evaluate:

(a) limn

n

n→∞

+

+++

3 1

3 2

1

2

(b) limn

n

n→∞

+��

��

1 2 4

6. Define a continuous function and prove that f (x) = 3x + 2 is a continuous function.

9� �))��)��&��&��

� �� !"#$�!�%&�'�

1. What is the derivative of 35 w.r.t. x?

2. Differentiate xe + ex + ee w.r.t. x.

3. If x y ady

dxa a2 2 2+ = find at , .� �

4. Find dy

dxy ex if = 2 2 .

5. If y x= +log 2 1 , find y1.

6. If y x= −2 , then find y′.

7. If f x xx

f� � � �= + ′22

11, . then find

8. If the total cost of a commodity is C x x x= + + +1

33 103 2 , then find marginal cost.

9. If the total cost function is C (x) = x3 − 7x + 8, then find average cost.


10. If the average cost function is 1 1

2x x− , then find marginal cost.

11. Differentiate 3ex w.r.t. logx.

12. If the total revenue function is given by R = 3q2 − q + 8, then find marginal revenue function.

� �($��!�� !"#$�!�%�'�

1. If y x x= + +log 1 2� � , then prove that dy

dx x=

+

1

1 2.

2. If y e emx mx= + − , then prove that y m y22 0− = .

3. If y xx

d y

dx= + 1 2

2, . then find

4. If x a ya= =θθ

, , then prove that dy

dx

y

x+ = 0.

5. Differentiate ex2

with respect to ex.

6. If y x a xdy

dx a x= + + =

+log , .2 2

2 2

1� � then prove that

7. If yey = x, then prove that dy

dx

y

x y=

+1� � .

8. With usual notation, S = at + b, where a and b are constants. Prove that velocity is constant andacceleration is zero.

9. Find 2 numbers whose sum is 10 and whose product is maximum.

10. If the total cost function c (x) of a firm is given by c (x) = x3 − 3x + 7, then find the average costand marginal cost where x = 6 units.

� )$ ��!�� !"#$�!�%��'�

1. If x y y x x y1 1 0+ + + = ≠ and , then prove that dy

dx x= −

+1

1 2� �.

2. If x = at2, y = 2at, then prove that d y

dx at

2

2 3

1

2= − .

3. The radius of a circular blot of ink is increasing at the rate of 3 cm per minute. Find the rate ofincrease of its area when its radius is 2 cm. What is the rate of increase of its circumference?

4. Prove that the largest rectangle that can be inscribed in a circle of given radius is a square.

5. Find the derivative of x3n from first principle.

6. State and prove quotient rule in differentiation.


7. If x y x ym n m n= + +� � , then prove that dy

dx

y

x= .

8. Find the derivative of x5 from first principles.

9. If y x edy

dxx x= +

2

, . then find

10. Find the maximum and minimum value of the function 2 15 36 103 2x x x− + + .

11. Prove that the area of the right angled triangle of given hypotenuse is maximum when the triangleis isosceles.

12. If total revenue function is R qq

q� � = + +596

6 2 where q is the number of units manufactured.

Find the maximum value of total revenue.

13. If total revenue function and total cost function are given by TR = 40x and TC = 2 + 4x respec-tively, then find at what level of output profit is maximised?

14. A TV manufacturers produces x sets per week at a total cost of Rs. x2 + 1560x + 50,000. He is

a monopolist and the demand function for this is xP= −12000

179 where P is price per unit, what

is the monopoly price in order to maximise the profit?

�:� �� )��

� �� !"#$�!�%&�'�

1. Evaluate: 2 2x xe x dx+ −� � �

2. Integrate e5x + 7 with respect to x.

3. Evaluate: log x dx�

4. Evaluate: x x dx2

1

2

3 7+ −� � �

5. Find the area bounded by the curve y = x2, x-axis and the ordinates x = 0 and x = 1.

6. If the marginal cost function is given by x2 + 7x − 8, then find total cost function.

� �($��!�� !"#$�!�%�'�

1. Integrate with respect to x: x x

x

+ +3 42

.

2. Integrate (x2 + 7) (2x3 − 7) with respect to x.


3. Evaluate: x

xdx

−� 1 2

3

� �

4. Evaluate: x x x dx2 51 2 1+ − +� � � � �

5. Evaluate: 2 1

1 2

x

x xdx

++ +� � ��

6. Evaluate: x x dx3 log�

7. Evaluate: 1

10

1−+�

x

xdx

8. Find the area bounded by the curve y = x2 − x with x-axis.

9. The marginal cost function of a firm is 150 − 10x + 0.2x2 where x is the output. Find the total costfunction, if the fixed cost is Rs. 750. What is the average cost?

10. The marginal revenue function of a firm is 60x − x2 where x is the output. Find the total revenuefunction.

� )$ ��!�� !"#$�!�%��'�

1. Evaluate: e x dxx2 3⋅�

2. Evaluate: x a x dxa

−� � �40

3. Find the area between the parabolas y2 = x and x2 = y.

4. Find the area bounded by the curve y x x= − +2 7 10 with x-axis.

5. Evaluate: ex

xdxx 2 1

2

+��

��

6. Evaluate: x x dx1 2

1

2

+�

7. Evaluate: 2 3

5 72

3x

xdx

+−�

8. Evaluate: 4 8

2 1 32

x

x x

+− +� � � �

.


��

��

• p ∧ q is T only when both p and q are true otherwise it is false i.e., T ∧ T is T otherwise it is F.

• p ∨ q is F only when both p and q are false. Otherwise it is True i.e., F ∨ F is F otherwise it isT.

• p → q is F only when p is true and q is false i.e., T → F is F otherwise it is T.

• p ↔ q is T only when both p and q are together True or False, i.e., T ↔ T and F ↔ F is Totherwise it is F.

• ~p is T when p is F and vice-versa.

• Tautology is a compound proposition which is always ‘true’ for all possible combinations of thetruth values of its components.

• Contradiction is a compound proposition which is always ‘False’ for all possible combinations ofthe truth values of its components.

• 2 propositions X and Y are logically equivalent if and only if they have identical truth values. Itis denoted by X ≡ Y.

• ~ (~p) ≡ p

• ~ (p ∧ q) ≡ ~p ∨ ~q

• ~ (p ∨ q) ≡ ~p ∧ ~q

• ~ (p → q) ≡ p ∧ ~ q

• ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (~p ∨ q)

• Converse of the conditional p → q is q → p

• Inverse of the conditional p → q is ~p → ~q

• Contrapositive of conditional p → q is ~q → ~p.

��

•n

rpn

n r=

−

• n p0 1= , • nnp n= • n p n1 =

• Permutation of n objects of which p objects are of one kind, q are of another kind and so on is

n

p q⋅ ....

• Number of circular arrangement of n persons round a table = n −1

• Number of circular arrangement of n beads or n flowers to form a necklace or garland = n −1

2.


•n

rcn

n r r=

− ⋅

• nnc = 1

• nc n1 =

• nc0 1=

• nr

nn rc c= −

• nr

nr

nrc c c+ =−

+1

1

• Number of straight lines that can be drawn from n points of which p points are collinear =nc2 − pc2 + 1.

• Number of triangles that can be drawn from n points of which p points are collinear = nc3 − pc3.

• Number of diagonals in a polygon of n sides = nc2 − n.

��

• P An A

n SP A� � � �

� �� = ≤ ≤0 1.

• P A P A� � � �= −1

• P (A∪B) means P (A or B) and P (A∩B) means P (A and B)

• Addition rule: P (A∪B) = P (A) + P (B) − P (A∩B)

• For mutually exclusive cases

P (A∪B) = P(A) + P(B)

• P A B C P A P B P C P A B P B C P C A P A B C∪ ∪ = + + − ∩ − ∩ − ∩ + ∩ ∩� � � � � � � � � � � � � � � �

• P A B P A P B A∩ = ⋅� � � � � �

• P A B P B P A B∩ = ⋅� � � � � �

• P A B P A P B A B∩ = ⋅� � � � � � if and are independent events.

• P A B C P A P B A P C A B∩ ∩ = ⋅ ⋅ ∩� � � � � � � �

• If A1, A2, A3, ... An are independent events P A A A P A P A P An n1 2 1 2∩ ∩ ∩ =... ...� � � � � � � �

��

• x a x nx an n

x an n n

x a an n n n n n+ = + ⋅ +−

⋅ +− −

⋅ + +− − −� � � � � �� 1 2 2 3 31

2

1 2

3! !... .


• There are (n + 1) terms in the expansion of (x + a)n. The power indices of ‘x’ go on decreasingby 1 and those of ‘a’ go on increasing by 1 at each stage so that the sum of power indices is n.

• The general terms or (r + 1)th term is given by T c x arn

rn r r

+−= ⋅1 .

• If n is even the number of terms in the expansion of (x + a)n is (n + 1) which is odd.

∴ There will be only one middle term i.e. Tn

21+.

• If n is odd, the number of terms in the expansion of (x + a)n is (n + 1) which is even. ∴ there will

be 2 middle terms: T Tn n+ + +12

12

1 and .

• To find the term containing xm in the expansion of (x + a)n i.e., to find the co-efficient of xm, writeTr + 1. Simplify and equate power index of x to m. Get r and substitute the value of r is Tr + 1.

• For getting the term independent of x or constant term, equate the power index of x to zero afterwriting Tr + 1 simplify then get the value of r. If r is a positive integer greater than 0, substitute inTr + 1.

• If r is negative or a fraction then conclude that there is no term independent of x in the expansion.

��

Proper fraction Partial fraction

•Nr

a x b a x b a x b1 1 2 2 2 3+ + +� �� ...A

a x b

B

a x b

C

a x b1

1 1 2 2 3 3++

++

++ ...

•Nr

ax b cx d+ +� � � �2

A

ax b

B

ax b

C

cx d++

++

+� � � �2

•Nr

ax bx c dx e2 + + +� � � �Ax B

ax bx c

C

dx e

++ +

++2� �

Provided ax2 + bx + c is non-factorisable. If ax2 + bx + c is factorisable as (a1x + b1) (a2x + b2), thenreplace ax2 + bx + c by (a1x + b1) (a2x + b2). Then resolve,

Nr

ax b a x b dx e

A

a x b

B

a x b

C

dx e+ + + ++

++

+1 2 2 1 1 2 2� �� as

If an improper fraction is given to resolve, first by dividing the numerator by denominator write thegiven fraction as the sum of the polynomial and the proper fraction, then the proper fraction is resolvedinto partial fractions.

��

• Matrix is an arrangement of numbers in horizontal rows and vertical columns.

• Two matrices of the same order are said to be equal if and only if the corresponding elements areequal.


• Two matrices can be added subtracted if they have same order. It is obtained by adding/subtract-ing the corresponding elements.

• Multiplication of a matrix by a scalar is obtained by multiplying each and every element by ascalar.

• If A is of order m × n and B is of order n × p then AB is of order m × p, i.e., matrix multiplicationis possible only when number of column in 1st matrix is equal to number of rows in the 2nd

matrix.

• Transpose of a matrix is obtained by interchanging rows and columns.

• A unique value associated with every square matrix is called its determinant value.

• If det A = 0, i.e., |A| = 0 for a matrix A, then A is called singular matrix. Otherwise it is called non-singular matrix.

• The value of the determinant is unaltered if its rows and columns are interchanged.

• If 2 rows or columns are interchanged the value of the determinant changes its sign.

• If in a determinant 2 rows or columns are identical then the value of the determinant is zero.

• If the elements of any row (or column) is multiplied by k, the value of the determinant is multi-plied by k.

• If to the elements of any row (or column) of a determinant the same multiples of the correspond-ing elements of other row (or column) of the determinant are added the value of the determinantremain the same.

• The determinant obtained by deleting the row and column containing the element is called minorof that element. If the minors are multiplied with (−1)i + j [where i = row number and j = columnnumber of the element] we get co-factors. The transpose of the co-factor matrix is called adjointof the matrix.

For a non-singular matrix A.

AA

A− =1 adj

.

If A is a square matrix and I is the identity matrix of the same order.

The characteristic equation: |A − λI| = 0. The values of λ obtained is called eigen values orcharacteristic roots.

Every square matrix satisfies its characteristic equation, |A − λI| = 0. This is Cayley Hamiltontheorem.

The solution of system of equations

a x b y c z d1 1 1 1+ + =

a x b y c z d2 2 2 2+ + =

a x b y c z d3 3 3 3+ + =

By Cramer’s rule:


Let ∆ ∆= =a b c

a b c

a b c

d b c

d b c

d b cx

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

,

∆ ∆y z

a d c

a d c

a d c

a b d

a b d

a b d

= =1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

,

Then x y zx y z= = =∆∆

∆∆

∆∆

, . and

By matrix method:

Let A

a b c

a b c

a b c

X

x

y

z

D

d

d

d

=�

�

�

��

=�

�

�

��

=�

�

�

��

1 1 1

2 2 2

3 3 3

1

2

3

, and

Then matrix equation

AX = D

X = A−1D.

and AA

A− =1 adj

or A−1 can be found by using Cayley Hamilton theorem.

��

• For the given ratio a : b, duplicate ratio is a2 : b2 Subduplicate ratio is a b: , Triplicate ratio

is a3 : b3 and subtriplicate ratio is a b3 3: .

• a : b : : c : d iff ad = bc.

• If a : b = b : c, then b is called mean proportional and c is called 3rd proportional.

• If a : b = c : d, then d is called fourth proportional.

• If a : b = c : d, then

(i) b : a = d : c (Invertendo)

(ii) a : c = b : d (Alternendo)

(iii) a + b : b = c + d : d (Componendo)

(iv) a − b : b = c − d : d (Dividendo)

(v) a + b : a − b = c + d : c − d

(Componendo and dividendo)

• If a : b = c : d represent a direct proportion, then a : b = d : c or b : a = c : d represent an inverseproportion and vice-versa.


• Rule of alligation: Quantity of cheaper

Quantity of dearer

C.P. of Dearer Mean

Mean C.P. of cheaper= −

−

c

d m

m

d

m c

� �

� �

� �

− −

i.e., cheaper : dearer = d − m : m − c.

!� � ��

• Mean Xx x x

nn=

+ + +1 2 ...

• Xw x w x w x

w w wwn n

n

=+ + ++ + +

1 1 2 2

1 2

...

...

• XX N X N X

N N NnN

n

n123

1 1 2 2

1 2

=+ + ++ + +

...

...

"� ��

• Legally due date = Bill Drawing date + Bil period + 3 days (Grace period).

• BD = Ftr [Simple interest on face value of the bill]

• TD = Ptr [Simple interest on present worth of the bill]

• BG = BD − TD

• Present worth = F

tr1+• BG = TD × tr

• BGTD

Pwr t= � �2

If and are not given

• TD B G Pw= ⋅ ⋅

�#� ��$��

• Yield =Nominal interest

Amount invested

• When stock is purchased, brokerage is added to cost price.

• When stock is sold, brokerage is subtracted from selling price.

• Interest or dividend is paid on the face value of the stock or share not the market value.


• 4 ½% stock at 96 means a stock whose face value is Rs. 100 is available at Rs. 96. Interest earnedin 4 ½.

• Shares need not be fully paid but stock must be fully paid.

��

• The learning curve ratio = Average labour cost of first 2N units

Average labour cost of first N units

� ��

• The learning curve equation is

y = axb

where y = cumulative average time/unit.

a = Time for first unit.

b = log of learning ratio to base 2.

x = Cumulative total number of units produced.

• Index of learning, b = log of learning ratio to base 2.

��

• The objective function is a quantified statement in linear programming problem of what the bestresult is aimed for Objective function will either to maximise or to minimise a value.

• If the objective function is a function of 2 variables only, the linear programming problem can besolved by graphical method.

• Optimum solution to linear programming problem lies at one of the vertices. Hence find the co-ordinates of each vertex and subject in the objective function. The values for which the objectivefunction is highest/least is the optimum solution for maximisation/minimisation.

��

• Distance between 2 points (x1 y1) and (x2 y2)

x x y y2 12

2 12− + −� � � � .

• Slope of the line joining (x1 y1) and (x2 y2)

=−−

y y

x x2 1

2 1

.

• 2 lines are ⊥r iff product of their slopes = −1.

• Equation of circle with centre (0, 0) and radius r units: x2 + y2 = r2.

• Equation of circle with centre (h, k) and radius r units: x h y k r− + − =� � � �2 2 2 .

• Equation of circle which is described on line joining (x1 y1) and (x2 y2) as diameter =

x x x x y y y y− − + − − =1 2 1 2 0� � � � � � � � .


• General equation of the circle x2 + y2 + 2gx + 2fy + c = 0

Centre = (−g, −f), Radius = g f c2 2+ − .

• Centre is the mid point of diameter.

• Co-ordinates of mid point of line joining (x1 y1) and (x2 y2) = x x y y1 2 1 2

2 2

+ + �

��,

• The point of intersection of 2 diameters of a circle = centre.

• If a circle x2 + y2 + 2gx + 2fy + c = 0 touch x-axis, then, radius = − f and g2 = c.

• If a circle x2 + y2 + 2gx + 2fy + c = 0 touches y-axis, then Radius = − =g f c and 2 .

• Length of the perpendicular from (x1 y1) to the line ax + by + c = 0 is

ax by c

a b

1 1

2 2

+ +

+

• Equation of tangent at (x1 y1) on the circle x2 + y2 + 2gx + 2fy + c = 0 is

xx yy g x x f y y c1 1 1 1 0+ + + + + + =� � � � .

• Length of the tangent from (x1 y1) to the circle x2 + y2 + 2gx + 2fy + c = 0 is

x y gx fy c12

12

1 12 2+ + + + .

• Any line is a tangent to the given circle if the length of the ⊥r from centre to the line = Radius ofthe circle.

• Condition for the line y = mx + c to be a tangent to the circle x2 + y2 = a2 is c a m2 2 2 1= ± +� � and

point of contact is am

m

a

m

am

m

a

m2 2 2 21 1 1 1+

−

+

��

��−

+ +

��

��, , . or .

• Condition for the line lx + my + n = 0 to be a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 is

n mf l m g f c− − = + + −lg� � � �� 2 2 2 2 2 .

• The point (x1 y1) lie inside the circle x2 + y2 + 2gx + 2fy + c = 0 if x y gx fy c12

12

1 12 2 0+ + + + < .

• Outside the circle if x y gx fy c12

12

1 12 2 0+ + + + > .

On the circle if x y gx fy c12

12

1 12 2 0+ + + + = .

• Any line parallel to ax + by + c = 0 will have the equation ax + by + k = 0.

[Take x and y co-efficients same].

• Any line perpendicular to ax + by + c = 0 will have the equation bx − ay + k = 0 [interchangex and y co-efficients and change the sign of anyone].

554B

asic Mathem

atics��

Equation Figure Vertex Focus Equation Axis Ends of Equation Lengthof of of of

directrix Latus rectum LR LR

y2 = 4ax (0, 0) (a, 0) x = −a x-axis (a, ±2a) x = a 4ay = 0

y2 = −4ax (0, 0) (−a, 0) x = a x-axis (−a, ±2a) x = −a 4ay = 0

x2 = 4ay (0, 0) (0, a) y = −a y-axis (±2a, a) y = a 4ax = 0

x2 = −4ay (0, 0) (0, −a) y = a y-axis (±2a, −a) y = −a 4ax = 0

(y − k)2 = 4a (x − h) (h, k) (a + h, k) x = −a + h y = k (a + h, ±2a + k) x = a + h 4a

(y − k)2 = −4a (x − h) (h, k) (−a + h, k) x = a + h y = k (−a + h, ±2a + k) x = −a + h 4a

(x − h)2 = 4a (y − h) (h, k) (h, a + k) y = −a + k x = h (±2a + h, a + k) y = a + k 4a

(x − k)2 = −4a (y − h) (h, k) (h, −a + k) y = a + k x = h (±2a + h, −a + k) y = −a + k 4a

(h, k)

(h, k)

(h, k)

(h, k)


• In any parabola, focus is inside the curve and directrix is away from the parabola.

• Distance between vertex and focus = a.

• For the given ends of latus rectum, there are 2 possible parabolas.

• Focus is the mid point of latus rectum.

• Axis is ⊥r to the directrix. Distance between directrix and vertex = a.

��

• limx a

n nnx a

x ana

→−−

−= 1

• limx

xe

x→

− =0

11

• lim logx

x

ea

xa

→

− =0

1

• limn

n

ne

→∞+ �� =1

1

• limn

nn e→

+ =0

11� �

• A function y = f(x) is said to be continuous at x = a if lim limx a x a

f x f a f x→ →− +

= =� � � � � �

i.e., LHL = f(a) = RHL.

• Limit of a function exists at x = a if

lim limx a x a

f x f x→ →− +

=� � � �

∴ f(x) is continuous at x = a iff

limx a

f x f a→

=� � � �

��

ydy

dx

xn nxn−1

x 1

x2 2x


x3 3x2

x1

2 x

1

x− 1

2x

12x

−23x

13x

−34x

ex ex

ax ax loga

logx1

x

Cu Cdu

dx⋅

u v±du

dx

dv

dx±

I II⋅ I II II I⋅ + ⋅d

dx

d

dx� � � �

Nr.

Dr.

*

**

Dr Nr Nr Dr.

Dr.

.d

dx

d

dx� � � � � �

� �

− ⋅2

• Chain rule: • Chain rule: • Chain rule: • Chain rule: • Chain rule: If y = g(u) & u = f (x),

then y = g[f(x)] is differentiated by chain rule

dy

dx

dy

du

du

dx= ⋅

• Implicit dif• Implicit dif• Implicit dif• Implicit dif• Implicit difffffferererererentiaentiaentiaentiaentiation:tion:tion:tion:tion: Given function f (x, y) = 0.

diff. w.r.t. x., take dy

dx common among the terms containing

dy

dx and shift the remaining terms to

RHS. Then find dy

dx.

*: Numerator **: Denominator


Parametric differentiation:

Given x = f(t) and y = g(t).

Then to find dy

dx, use

dy

dx

dy

dtdxdt

=

• Second or• Second or• Second or• Second or• Second order derder derder derder derder deriiiiivvvvvaaaaatititititivvvvve:e:e:e:e:

If y = f(x), then by differentiating we get dy

dx or y1 or y1 or f1(x). This is a function of x. By

differentiating this again with respect to x we get d y

dx

2

2 or y″ or f ″(x) or y2.

��

• Velocity = dS

dt

• Acceleration = =dv

dt

d S

dt

2

2

• Rate means differentiation w.r.t. t.

∴ rate of change of area =dA

dt

• Area of Square = S2

• Area of circle = πr2

• Surface area of sphere = 4πr2

• Volume of Sphere = 4

33πr

• Volume of a cylinder = πr2h

• Volume of a cone = 1

32πr h

• For an increasing function dy

dx> 0 and for a decreasing function

dy

dx< 0.

• To find maximum and/or minimum value of the function y = f(x), find dy

dx, equate it to zero. Let

x = a, x = b, x = c be the points. Find d y

dx

d y

dx

d y

dxx a x b x c

2

2

2

2

2

2at at at

and = = =

,..


if d y

dx x a

2

2 0at =

> , then x = a is a point of minima. Minimum value of the function is y = f(a).

if d y

dx x b

2

2at =

is less than zero, x = b is a point of maxima. Maximum value of the function is

y = f(b).

if d y

dx x c

2

2at =

is equal to zero, then x = c is called point of inflection. At x = c the function neither

attains maxima nor minima.

�!� ��

• x dxx

ncn

n

� =+

++1

1

Provided n ≠ −1

• if n = −1

x dxx

dx x c−� �= = +1 1log .

• e dx e cx x� = + .

• a dxa

acx

x

� = +log

• k dx kx c� = +

• k f x g x dx k f x dx k g x dx� � � � � � � �± = ±� � �• If f x dx g x c� � � �� = + , then

f ax b dxg ax b

ac+ =

++� � � � �

• To evaluate f x f x dx f x tn� � � � � �⋅ ′ =� , put and proceed to get the answer.

f x

nc n

n� � +

++ ≠ −

1

11 for

and log [f (x)] + c for n = −1


• To evaluate ∫ef(x) ⋅ f′(x) dx, put f(x) = t and proceed to get ef(x) + c as answer.

• To evaluate integrals of the type

px q

ax b cx ddx

px q

ax b cx d

++ +

++ +� ��

or 2

First resolve into partial fractions

px q

ax b cx d

A

ax b

B

cx d

++ +

=+

++� ��

or

px q

ax b cx d

A

ax b

B

ax b

C

cx d

++ +

=+

++

++� � � � � � � � � �2 2

Find A, B, C, then finally integrate.

• I function II function� � � � dx�= − �

�� I function II function II function I function

d

dxdx� �

I function and II function are kept by making use of LIATE rule.

• e f x f x dx e f x cx x� � � � � �+ ′ = ⋅ +� .

�"� ��

• If f x dx g x c� � � �� = + , then f x dx g x g b g aa

b

a

b� � � � � � � �� = = −

• f x dx f t dt f z dza

b

a

b

a

b

� � � � � �� = = and so on.

• f x dx f x dxa

b

b

a

� � � �� = −

• f x dx f x dx f x dxa

b

a

c

c

a

� � � � � �� = +

• f x dx f a x dxa a

� � � �0 0� �= −


• Area enclosed by the curve y = f(x), X-axis and the lines x = a and x = b is given by

A y dx f x dxa

b

a

b

= =� � � � .

�#� ��

• Total cost = Fixed cost + Variable cost.

• Marginal cost =d

dx

d

dq Total cost or total cost� � � �

• Total cost = ∫(Marginal cost) dx or ∫(Marginal cost) dq.

• Marginal revenue d

dx (Total revenue) or

d

dq (Total revenue)

• Total revenue = ∫(Marginal revenue) dx or ∫(Marginal revenue) dq.

• Average cost =Total cost

or Total cost

x q

• Profit is maximised when marginal cost = Marginal revenue.

• Average revenue is nothing but demand function =Total revenue

or Total revenue

x q


Chapter One mark Two mark Four mark Total marksquestions questions questions

(VSA) (SA) (ET)

Mathematical Logic 1 1 1 7

Permutation, Combination 1 3 2 15and Probability

Binomial Theorem and – 2 1 8Partial Fractions

Matrices and Determinants 1 1 3 15

Ratio, Proportion & Variation 1 1 1 7

Mathematics of Finance 3 1 4 21(Averages, B.D., Stock andShares, L.C., LPP)

Circles, Parabola 1 2 1 9

Limits and Continuity 1 1 1 7

Differentiation and Its 1 2 3 17Application

Integration and Its – 1 3 14Application

Total: 10 15 20 120

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8122416845_Math2