Std. 12th Mathematics and Statistics - 1 Notes, Commerce

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Written as per the revised syllabus prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

Printed at: Jasmine Art Printers Pvt. Ltd., Navi Mumbai

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

P.O. No. 151398 Balbharati Registration No.: 2018MH0022 TEID: 13194

MATHEMATICS AND STATISTICS – ISTD. XII COMMERCE

PERFECT

Salient Features : • Precise Theory for every Topic. • Exhaustive coverage of entire syllabus. • Topic-wise distribution of all textual questions and practice problems at the

beginning of every chapter. • Relevant and important formulae wherever required. • Covers answers to all Textual Questions. • Practice problems based on Textual Exercises and Board Questions

(March 08 July 18) included for better preparation and self evaluation. • Multiple Choice Questions at the end of every chapter. • Two Model Question papers based on the latest paper pattern. • Includes Board Question Papers of 2017 and 2018 and March 2019.

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Mathematics is not just a subject that is restricted to the four walls of a classroom. Its philosophy and applications are to be looked for in the daily course of our life. The knowledge of mathematics is essential for us, to explore and practice in a variety of fields like business administration, banking, stock exchange and in science and engineering. With the same thought in mind, we present to you "Std. XII Commerce: Mathematics and Statistics-I" a complete and thorough book with a revolutionary fresh approach towards content and thus laying a platform for an in depth understanding of the subject. This book has been written according to the revised syllabus and includes two model question papers based on the latest paper pattern. At the beginning of every chapter, topic-wise distribution of all textual questions including practice problems have been provided for simpler understanding of various types of questions. Every topic included in the book is divided into sub-topics, each of which are precisely explained with the associated theories. We have provided answer keys for all the textual questions and miscellaneous exercises. In addition to this, we have included practice problems based upon solved exercises which not only aid students in self evaluation but also provide them with plenty of practice. We've also ensured that each chapter ends with a set of Multiple Choice Questions so as to prepare students for competitive examinations.

We are sure this study material will turn out to be a powerful resource for students and facilitate them in understanding the concepts of Mathematics in the most simple way.

The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on: [email protected]

Best of luck to all the aspirants! Yours faithfully Publisher Edition: Second

Disclaimer This reference book is transformative work based on textual contents published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

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BOARD PAPER PATTERN Time: 3 Hours Total Marks: 80 1. One theory question paper of 80 marks and duration for this paper will be 3 hours. 2. For Mathematics and Statistics, (Commerce) there will be only one question paper and two answer papers.

Question paper will contain two sections viz. Section I and Section II. Students should solve each section on separate answer books.

Section – I Q.1. This Question will have 8 sub-questions, each carring two marks. [12 Marks] Students will have to attempt any 6 out of the given 8 sub-questions. Q.2. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. Q.3. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions.

Section – II Q.4. This Question will have 8 sub-questions, each carring two marks. [12 Marks] Students will have to attempt any 6 out of the given 8 sub-questions. Q.5. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions. Q.6. This Question carries 14 marks and consists of two sub parts (A) and (B) as follows: [14 Marks] (A) It contains 3 sub-questions of 3 marks each. Students will have to attempt any 2 out of the given 3 sub-questions (B) It contains 3 sub-questions of 4 marks each. Students will have to attempt any 2 out of the given 3 sub-questions.

Evaluation Scheme for Practical i. Duration for practical examination for each batch will be one hour. ii. Total marks : 20

MARKWISE DISTRIBUTION

Unitwise Distribution of Marks Section - I

Sr.No. Units Marks with Option 1 2 3 4 5 6 7

Mathematical Logic Matrices Continuity Differentiation Application of Derivative Integration Definite Integrals

08 08 08 08 10 08 08

Total 58

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Section - II Sr. No. Units Marks with Option

1.

Commercial Arithmetic: Ratio, Proportion, Partnership Commission, Brokerage, Discount Insurance, Annuity

13

2. Demography 08 3. Bivariate Data Correlation 08 4. Regression Analysis 07 5. Random Variable and Probability Distribution 08 6. Management Mathematics 14 Total 58

Weightage of Objectives Sr. No. Objectives Marks Marks with Option Percentage

1 2 3 4

Knowledge Understanding Application Skill

08 22 32 18

13 32 45 26

10.00 27.50 40.00 22.50

Total 80 116 100.00

Weightage of Types of Questions Sr. No. Types of Questions Marks Marks with Option Percentage

1 2 3

Objective Type Short Answer Long Answer

24 24 32

32 36 48

30 30 40

Total 80 116 100.00

No. Topic Name Page No. 1 Mathematical Logic 1 2 Matrices 42 3 Continuity 122 4 Differentiation 151 5 Applications of Derivative 189 6 Integration 219 7 Definite Integrals 283 Model Question Paper - I 326 Model Question Paper - II 328 Board Questions Paper – March 2017 330 Board Questions Paper – July 2017 332 Board Questions Paper – March 2018 334 Board Questions Paper – July 2018 336 Board Questions Paper – March 2019 338

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Std. XII : Commerce (Maths - I)

Type of Problems Exercise Q. Nos.

Continuity of Standard Function 3.1 Q.1

Practice Problems (Based on Exercise 3.1)

Q.1

Examine the Continuity of a Function at a given point

3.1 Q.2, 3, 10


Q.2, 3, 9, 12, 13, 14, 15, 16, 17, 27

Miscellaneous Q.1

Practice Problems (Based on Miscellaneous)

Q.1, 2, 8

Types of Discontinuity (Removable Discontinuity/ Irremovable Discontinuity)

3.1 Q.4 Practice Problems

(Based on Exercise 3.1) Q.4

Miscellaneous Q.2, 10 Practice Problems

(Based on Miscellaneous) Q.3

Find the value of Function if it is Continuous at given point

3.1 Q.5, 7 Practice Problems

(Based on Exercise 3.1) Q.5, 7, 10, 24, 28

Miscellaneous Q.3, 4 Practice Problems

(Based on Miscellaneous) Q.4

Find the value of k/a/b if the Function is Continuous at a given point/points.

3.1 Q.6, 8, 9


Q.6, 8, 11, 18, 19, 20, 21, 22, 23, 25, 26

Miscellaneous Q.5, 6, 7, 8

Practice Problems (Based on Miscellaneous)

Q.5, 6, 7

Find the points of Discontinuity for the given Functions Miscellaneous Q.9

Continuity 03

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Chapter 03: Continuity

Syllabus: 3.1 Continuity of a function at a point 3.2 Algebra of continuous functions 3.3 Types of discontinuity 3.4 Continuity of some standard functions Introduction Continuity is ‘the state of being continuous’ and continuous means ‘without any interruption or disturbance’. For example, the price of a commodity and its demand are inversely proportional. The graph of demand curve of a commodity is a continuous curve without any breaks or gaps. Note: A graph consisting of jumps is not a graph of continuous function. 3.1 Continuity of a function at a point Definition: A function f is said to be continuous at a point x = a in the domain of f, if

alimx

f(x) exists and a

limx

f(x) = f (a).

i.e. if a

limx

f(x) = a

limx

f(x) = f (a)

If any of the above conditions is not satisfied by the function, then it is discontinuous at that point. The point is known as point of discontinuity. eg., Consider the function, f(x) = 2x + 7, x 4 = 5x 5, x 4 Since, f(x) has different expressions for the value of x left hand and right hand limits have to be found

out.

4lim

xf(x) =

4limx

(5x 5) = 5 4 5 = 15

Also, f (4) = 5 (4) 5 = 15 and

4lim

x f(x) =

4limx

(2x + 7) = 2 4 + 7 = 15

4

limx

f(x) = 4

limx

f(x) = f (4)

f(x) is continuous at x = 4. Continuity of a function on its domain Definition: A real valued function f : D R where D R is said to be a continuous function on D, if it is continuous at every point in the domain D. eg., Consider the functions, i. f(x) = 3x4 + x2 + 3x ii. f(x) = sin x These two functions are continuous on every domain D, where D R. 3.2 Algebra of continuous functions If f and g are two real valued functions defined on the same domain, which are continuous at x = a, then 1. the function kf is continuous at x = a, for any

constant k R. 2. the function f g is continuous at x = a 3. the function f . g is continuous at x = a 4. the function f

g is continuous at x = a, when

g (a) 0 5. composite functions, f[g(x)] and g[f(x)], if

well defined are continuous functions at x = a. 3.3 Types of discontinuity 1. Removable discontinuity: A real valued

function f is said to have removable discontinuity at x = a in its domain, if

alimx

f(x) exists but a

limx

f(x) f (a)

i.e. a

limx

f(x) =a

limx

f(x) f(a)

This type of discontinuity can be removed by redefining the function f at x = a as

f (a) = a

limx

f(x).

eg., Consider the function, f(x) = 5 6

2

x xx

, x 2

= 2 , x = 2 Here,

2limx

f(x) = 2

limx

2 5 6

2

x xx

= 2

limx

3 2

2

x x

x

= 2

limx

x 3

.... [ x 2, x 2, x 2 0] = 2 3 = 1

2limx

f(x) exists

Y

XO

X

YDemand

Pric

e

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Also, f (2) = 2 …. (given)

2limx

f(x) f (2)

function f is discontinuous at x = 2, This discontinuity can be removed by

redefining f as follows:

f(x) = 2 5 6

2

x xx

, x 2

= 1 , x = 2 x = 2 is a point of removable discontinuity. 2. Irremovable discontinuity: A real valued

function f is said to have irremovable discontinuity at x = a, if

alimx

f(x) does not

exist i.e. a

limx

f(x) a

limx

f(x) or one of the

limits does not exist. Such a function can not be redefined as

continuous function. eg., Consider the function, f(x) = x2 + 2x + 3 , x 3 = x2 1 , x 3 Here,

3lim

x f(x) =

3limx

x2 + 2x + 3

= (3)2 + 2(3) + 3 = 18 and

3lim

x f(x) =

3limx

x2 1= (3)2 1 = 8

3

limx

f(x) 3

limx

f(x)

limit of the function does not exist. f has irremovable discontinuity at x = 3 3.4 Continuity of some standard functions 1. Constant function: The constant function

f(x) = k (where k R is a constant). The function is continuous for all x belonging to its domain.

eg., f(x) = 10, f(x) = log10 100 , f(x) = e7 2. Polynomial function: The function

f(x) = a0 + a1x + a2x2 + …. + anxn, where n N, a0, a1 …. an R is continuous for all x belonging to domain of x.

eg., f(x) = x2 + 5x + 9, f(x) = x3 5x + 9,

f(x) = x4 16, x R. 3. Rational function: If f and g are two

polynomial functions having same domain then the rational function f

g is continuous in its

domain at points where g(x) 0.

eg.,

Consider the function, 2

2

5 69

x xx

Here, f(x) = x2 + 5x + 6 and g(x) = x2 9 Given function is continuous on its domain, where x2 9 0 i.e., (x + 3) (x 3) 0 i.e., x + 3 0, x 3 0 i.e., x 3, x 3 The function is continuous on its domain

except at x = 3, 3. 4. Trigonometric function: sin (ax + b) and

cos (ax + b), where a, b R are continuous functions for all x R.

eg., sin (5x + 2), cos (7x 11) x R. Note: Tangent, cotangent, secant and cosecant

functions are continuous on their respective domains.

5. Exponential function: f(x) = ax , a > 0, a 1, x R is an exponential function, which is continuous for all x R.

eg.,

f(x) = 3x , f(x) = 12

x

, f(x) = ex x R,

where a > 0, a 1. 6. Logarithmic function: f(x) = loga x where

a > 0, a 1 is a logarithmic function which is continuous for every positive real number i.e. for all x R+

eg., f(x) = loga 7x , f(x) = loga 9x2 x R, where

a > 0, a 1. Some Important Formulae Algebra of limits: If f(x) and g(x) are any two functions, 1.

alimx

[f(x) + g(x)] = a

limx

f(x) + a

limx

g(x)

2. a

limx

[f(x) g(x)] = a

limx

f(x) a

limx

g(x)

3. a

limx

[f(x)g(x)] = a

limx

f(x)a

limx

g(x)

4. a

f ( )limg( )

x

xx

= a

a

lim f ( )

lim g( )

x

x

x

x, where

alimx

g(x) 0

5. a

limx

[k.f(x)] = ka

limx

f(x), where k is a constant.

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Limits of Algebraic functions: 1.

alimx

x = a

2. a

limx

k = k, where k is a constant.

3. a

limx

n naa

xx

= nan 1 Limits of Trigonometric functions:

1. 0

limx

sin xx

= 1

2. 0

limx

tan xx

= 1

3. 0

limx

cos x = 1 Limits of Exponential functions:

1. 0

limx

a 1x

x= log a, where (a > 0, a 1)

2. 0

limx

(1 + x)1x = e

Limits of Logarithmic functions:

1. 0

limx

log 1 xx

= 1 Exercise 3.1 1. Are the following functions continuous on

the set of real numbers? Justify your answers. i. f(x) = 7 Solution: Given, f(x) = 7 It is a constant function. f(x) is continuous on the set of real

numbers i.e., x R ii. f(x) = e Solution: Given, f (x) = e It is a constant function …. [ e = 2.71828]

f(x) is continuous on the set of real numbers i.e., x R.

iii. f (x) = log 19 Solution: Given, f(x) = log 19 Here, log 19 is a constant f(x) is a constant function f(x) is continuous on the set of real

numbers i.e., x R.

iv. f(x) = 7x4 5x3 3x + 1 Solution: Given, f(x) = 7x4 5x3 3x + 1 It is a polynomial function f(x) is continuous on the set of real

numbers i.e., x R v. g(x) = sin (4x 3) Solution: Given, g(x) = sin (4x 3) It is a sine function f(x) is continuous on the set of real

numbers i.e., x R

vi. h(x) =2

3 25 +7 +2

+ + +3x x

x x x

Solution:

Given, h(x) = 2

3 2

5 7 23

x + x

x x x+

It is a rational function and is discontinuous if x3 + x2 + x + 3 = 0 But, x R, x3 + x2 + x + 3 0 h(x) is continuous on the set of real

numbers, except when x3 + x2 + x + 3 = 0

vii. g(x) = 2

213 16 19

2 1x x

x

Solution:

Given, g(x) = 2

2

13 16 192 1

x x

x

It is a rational function and is discontinuous, if 2x2 + 1 = 0 But x R, 2x2 + 1 0 g (x) is continuous on the set of real

numbers i.e., x R viii. f(x) = 5x Solution: Given, f(x) = 5x It is an exponential function It is continuous on the set of real

numbers i.e., x R ix. f(x) = 32x 15x Solution: Given, f(x) = 32x 15x It is the difference of two exponential functions It is continuous on the set of real

numbers i.e., x R

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x. f(x) = e(5x + 7) Solution: Given, f(x) = e(5x + 7) It is an exponential function It is continuous on the set of real

numbers i.e., x R 2. Examine the continuity of the following

functions at the given point. (All functions are defined on R R) i. f(x) = x2 – x + 9, for x 3 = 4x + 3, for x > 3; at x = 3.

[Mar 15, 18] Solution:

3lim

x f(x) =

3limx

(x2 x + 9)

= (3)2 3 + 9 = 15 …. (i) and

3lim

x f(x) =

3limx

(4x + 3)

= 4(3) + 3 = 15 …. (ii) Also, f (3) = (3)2 3 + 9 = 15 …. (iii)

3lim

x f(x) =

3lim

x f(x) = f(3)

…. [From (i), (ii) and (iii)] f is continuous at x = 3. ii. f(x) =

2 164

xx

, for x 4

= 8, for x = 4; at x = 4. [Oct 15] Solution:

4

limx

f(x) = 4

limx

2 164

xx

= 4

limx

2 244

xx

= 4

limx

( 4) ( 4)( 4)

x x

x

=

4limx

(x + 4)

….[ x 4, x 4, x 4 0]

4limx

f(x) = 4 + 4 = 8 …. (i)

Also, f (4) = 8 …. (ii)(given)

4limx

f(x) = f (4) …. [From (i) and (ii)]

f is continuous at x = 4.

iii. f(x) =x xx x

2

3

3 2 47 9 2

, for x 2

= 113

, for x = 2 ; at x = 2

Solution: Consider, x3 + 7x 9 2 By synthetic division, we get

2 1 0 7 9 2

2 2 9 2

1 2 9 0

x3 + 7x 9 2 = (x 2 ) (x2+ 2 x+ 9)

2

limx

f(x) =2

limx

2

2

2 2 2 42 2 9

x x xx x x

=2

limx

2

2 2 2 2

2 2 9

x x x

x x x

=2

limx

2

2 2 2

2 2 9

x x

x x x

=2

limx 2

2 22 9

x

x x

[x 2 , x 2 , x 2 0]

= 2

2 2 2( 2) 2 2 9

= 22 2 9

2

2lim f13

xx ….(i)

Also, f( 2 ) = 113

….(ii)(given)

2

limx

f(x) f ( 2 )

….[From (i) and (ii)] f is discontinuous at x = 2 . iv. f(x) = sin 5 x

x, for x 0

= 1, for x = 0; at x = 0. Solution:

0

limx

f(x) = 0

limx

sin5xx

= 0

limx

sin55

xx

5

= 5 0

limx

sin55

xx

= 5 (1)

.…[ x0, 5x0, 0

limx

sin xx

= 1]

0

limx

f(x) = 5

Also, f(0) = 1 …. (given)

0limx

f(x) f (0)

f is discontinuous at x = 0.

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v. f(x) =

3 2 7

1

x

x, for x 1

= – 13

, for x = 1; at x = 1. [Oct 14]

Solution:

1

limx

f(x) = 1

limx

3 2 71

xx

= 1

limx

3 2 7

1

x

x

3 2 7

3 2 7

x

x

= 1

limx

22(3) 2 7

1 3 2 7

x

x x

= 1

limx

9 2 7

1 3 2 7

x

x x

= 1

limx

2 21 3 2 7

xx x

= 1

limx

2 1

1 3 2 7

x

x x

= 1

limx

23 2 7

x

…. [x 1, x 1, x 1 0]

= 23 2 1 7

= 26 =

1

3

1

limx

f(x) = 13

…. (i)

Also, f(1) = 13

…. (ii)(given)

1

limx

f(x) = f(1) ….[From (i) and (ii)]


vi. f(x) =3 51 5

xx

, for x 4

= 18

, for x = 4; at x = 4.

Solution:

4

limx

f(x) = 4

limx

3 51 5

xx

= 4

limx

3 51 5

xx

1 51 5

xx

3 53 5

xx

= 4

limx

22

22

3 5 1 5

1 5 3 5

x x

x x

= 4

limx

9 5 1 5

1 5 3 5

x x

x x

= 4

limx

4 1 5

4 3 5

x x

x x

= 4

limx

4 1 5

4 3 5

x x

x x

= 4

limx

1 5

3 5

x

x

....[x 4, x 4, x 4 0]

= 1 5 4

3 5 4

= 1 1

3 3

= 2

6

4

limx

f(x) = 13

…. (i)

Also, f(4) = 18

…. (ii)(given)

4

limx

f(x) f(4) …. [From (i) and (ii)]

f is discontinuous at x = 4.

vii. f(x) = x2 cos

1x

, for x 0

= 0 , for x = 0; at x = 0 [Oct 15]

Solution:

0

limx

f(x) = 0

limx

x2cos 1 x

cos x [ 1, 1] for all x R,

also, when x 0, x 0 cos 1 x

exists.

Let 1cos x

= finite number = k (say)

0

limx

x2 1cos x

= 0

limx

x2k

where k [1, 1]

0

limx

f(x) = 0 …. (i)

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128


Also, f(0) = 0 …. (ii)(given)

0limx

f(x) = f(0) …. [From (i) and (ii)]


viii. f(x) =

tan sinsin 3 3sin

x xx x

, for x < 0

= 2 2

2

3 sin 2 sin3

x xx

,

for x 0; at x = 0 Solution: Consider,

0

limx

f(x) = 0

limx

tan sinsin 3 3 sin

x xx x

= 0

limx

3tan sin

3sin 4sin 3sin

x x

x x x

…[sin3 = 3sin – 4sin3]

= 0

limx

3tan sin

4sin

x x

x

= 0

limx

3

sin sincos

4sin

x xx

x

= 0

limx

3sin sin cos

4sin

x x xx

= 0

limx

3

sin 1 cos4 sin

x x

x

= 0

limx

2

3

sin 2sin2

4sin

xx

x

…[1 – cos x = 2sin2

2x]

= 12

0limx

2

3

3

3

sin .sin2

sin

xx

xx

x

=

2

20

3

30

sin1 sin 2lim2 4

4sinlim

x

x

xx

xx

xx

=

2

0 0

3

0

sin1 sin 1 2lim lim2 4

2sinlim

x x

x

xx

xx

xx

=

2

3

1 11 12 4

1

= 18

…[x0,2x 0,

0limx

sin xx

= 1]

0

limx

f(x) = 18

…. (i)

Also, 0

limx

f(x) = 0

limx

2 2

2

3sin 2sin3

x x

x

= 0

limx

22

2 2

2sin3sin3 3

xxx x

=

0limx

22

2 20

sinsin 2 lim3 x

xxx x

=0

limx

22

20

sinsin 2 lim3 x

xxx x

= (1)2 – 2

3 (1)

.…[0

limx

sin 1xx

]

= 1 – 23

0

limx

f(x) = 13 …. (ii)

0

limx

f(x) 0

limx

f(x) ….[From (i) and (ii)]

f(x) does not exist

f is discontinuous at x = 0

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ix. f(x) =

x x xx x

3 2

39 2

6

, for x < 2

=

3 2

1 42 2x x x

, for x > 2

= 4 for x = 2; at x = 2 Solution: Consider,

2

limx

f(x) =2

limx

3 2

39 2

6

x x x

x x

=

3 2

3

2 + 2 9 2 22 2 6

= 80

2

limx

f(x) does not exist.

But f(2) = 4

2lim

xf(x) f(2)

f is discontinuous at x = 2. x. f(x) = 6 3 2 1x x x

x, for x < 0

= 2

4 4 2x x

x, for x > 0

= 1, for x = 0, at x = 0 Solution:

0

limx

f(x) = 0

limx

6 3 2 1 x x x

x

= 0

limx

6 3 2 1 1 1 x x x

x

= 0

limx

6 1 3 1 2 1 x x x

x

= 0

limx

6 1 3 1 2 1 x x x

x

=0

limx

6 1 3 1 2 1

x x x

x x x

= 0 0 0

6 1 3 1 2 1lim lim lim

x x x

x x xx x x

= log 6 + log 3 – log 2

…[0

limx

a 1 log a

x

x]

= log 6 32

0

limx

f(x) = log 9 …. (i)

0

limx

f(x) =0

limx 2

4 4 2 x x

x

=0

limx 2

14 24

xx

x

=0

limx

2

2

4 2 4 1

4

x x

xx

=0

limx

2

2

4 1

4

x

xx

=0

limx

2

0

4 1 1lim4

x

xxx

= (log 4)2 . 014

…[0

limx

a 1 log a

x

x]

0

limx

f(x) = (log 4)2 .... (ii)

0

limx

f(x) 0

limx

f(x) .... [From (i) and (ii)]

0

limx

f(x) does not exist

f is discontinuous at x = 0 3. Discuss the continuity of the following

functions.

i. f(x) = 2a 1x

x, x 0, a 0 & a 1

= 2 log a, x = 0; at x = 0. Solution:

0

limx

f(x) =0

limx

2a 1x

x

=0

limx

2a 1 22

x

x=

02lim

x

2a 12x

x

0

limx

f(x) = 2 log a …. (i)

….[x0,2x0, 0

limx

a 1x

x = log a]

Also, f(0) = 2 log a …. (ii)(given)

0limx

f(x) = f(0) ….[From (i) and (ii)]

f is continuous at x = 0 ii. f(x) =

5 34 3

x x

x x , x 0

= log 54

, x = 0; at x = 0.

Solution:

0limx

f(x) =0

limx

5 34 3

x x

x x

SAMPLE C

ONTENT

130


=0

limx

5 3

4 3

x x

x xx

x

= 0

0

5 3lim

4 3lim

x x

xx x

x

x

x

= 0

0

5 1 3 1lim

4 1 3 1lim

x x

xx x

x

x

x

=0

0

5 1 3 1lim

4 1 3 1lim

x x

x

x x

x

x x

x x

= 0 0

0 0

5 1 3 1lim lim

4 1 3 1lim lim

x x

x xx x

x x

x x

x x

= log 5 log 3log 4 log 3

....[0

limx

a 1x

x= log a]

0

limx

f(x) =

5log34log3

…. (i)

Also, f (0) = log 54

…. (ii) (given)

0

limx

f(x) f(0)

f is discontinuous at x = 0

iii. g(x) =

2512

xx , x 0

= e5/2, x = 0; at x = 0. Solution:

0

limx

g(x) =0

limx

2512

xx

=0

limx

52551

2

xx

0

limx

g(x) = e5 …. (i)

....[ x 0, 52x 0

0limx

1

1 xx = e]

Also g(0) = 52e …. (ii)(given)

0

limx

g(x) g(0) ….[From (i) and (ii)]

g is discontinuous at x = 0

iv. h(x) = log 1 + 2 xx

, x 0

= 2, x = 0; at x = 0. Solution:

0

limx

h(x) =0

limx

log 1 2 xx

=0

limx

log 1 22

2

x

x

=0

2 limx

log 1 22 xx

= 2(1)

....[x0, 2x0, 0

limx

log 11

xx

]

0

limx

h(x) = 2 …. (i)

Also, h(0) = 2 …. (ii)(given)

0limx

h(x) = h(0) ….[From (i) and (ii)]

h is continuous at x = 0

v. f(x) = 32 1tan

x

x, x < 0

=

2

e e 2x xx , x > 0

= 2, x = 0 at x = 0. Solution:

0

limx

f(x) =0

limx

32 1tan

x

x=

0limx

32 1 33tan

x

xx

x

=

3

0

0

2 13lim3

tanlim

x

x

x

xx

x

= 3 log 21

....[x0, 3x0,

0limx

a 1x

x= log a,

0

tanlimx

xx

= 1]

0

limx

f(x) = 3 log 2 …. (i)

0

limx

f(x) =0

limx

2

e e 2 x x

x

=0

limx

2

1e 2e

xx

x

SAMPLE C

ONTENT

131


=0

limx

2

2e 2e 1

e

x x

x

x

=0

limx 2

2

1

e 1 1e

x

xx

= 2

0 0

1

e 1 1lim lime

x

xx xx

= 2

0

11logee

....[0

limx

e 1x

x= log e]

0

limx

f(x) = 1 .... (ii)

0

limx

f(x) 0

limx


f(0) does not exist f is discontinuous at x = 0 4. Discuss the continuity of the following

functions at the points given against them. If the function is discontinuous, determine whether the discontinuity is removable. In that case, redefine the function, so that it becomes continuous.

i. f(x) = x2 – 2x – 1, for x < 2 = 3x – 1, for x ≥ 2; at x = 2. Solution:

2lim

xf(x) =

2limx

(x2 – 2x – 1)

= (2)2 – 2(2) – 1 = 4 – 4 – 1 = 1

2lim

xf(x) = 1 …. (i)

and2

limx

f(x) =2

limx

(3x – 1)

= 3(2) – 1 = 6 1

2lim

xf(x) = 5 …. (ii)

2

limx

f(x) 2

limx


limit of the function does not exist. f has irremovable discontinuity at x = 2

ii. f(x) =

3 273

xx

, for x < 3

= 8x, for x 3; at x = 3. Solution:

3

limx

f(x) =3

limx

3 273

xx

=3

limx

3 333

xx

=3

limx

23 3 93

x x xx

....[a3 – b3 = (a – b)(a2 + ab + b2)]

=3

limx

(x2 + 3x + 9)

[ x 3, x 3 x – 3 0] = (3)2 + 3(3) + 9

3lim

xf(x) = 27 ….(i)

and3

limx

f(x) =3

limx

8x = 8(3)

3

limx

f(x) = 24 ….(ii)

3

limx

f(x) 3

limx


limit of the function does not exist f has irremovable discontinuity at x = 3 iii. f(x) = sin9

2x

x, for x 0

= 12

, for x = 0; at x = 0.

Solution:

0

limx

f(x) =0

limx

sin92

xx

=0

1 sin 9lim 92 9

x

xx

=0

9 sin 9lim2 9x

xx

= 9 12

[x0, 9x0, 0

limx

sin xx

= 1]

0

limx

f(x) = 92

…. (i)

Also, f(0) = 12

…. (ii) (given)

0

limx

f(x) f(0) ….[From (i) and (ii)]

SAMPLE C

ONTENT

132


f has removable discontinuity at x = 0 This discontinuity can be removed by

redefining the function as:

f(x) = sin92

xx

, for x 0

= 92

, for x = 0 ; at x = 0 iv. f(x) = 2e 1

5

x

x, for x 0

= 2, for x = 0; at x = 0 Solution:

0

limx

f(x) =0

limx

2e 15x

x

=2

0

1 e 1lim 25 2

x

x x

=2

0

2 e 1lim5 2

x

x x

= 2 log e5

[ x 0, 2x 0,0

limx

e 1 log e

x

x]

0

limx

f(x) = 25

…. (i)


0limx




f(x) =2e 15x

x, for x 0

= 25 , for x = 0 ; at x = 0

v. f(x) =

x

x3

3 2

1, for x 1

= 5, for x = 1; at x = 1 Solution:

1

limx

f(x) =1

limx 3

3 21

xx

=1

limx

3 3

3 2 3 2

1 3 2

x x

x x

=1

limx

2 2

2

3 2

1 1 3 2

x

x x x x

=1

limx 2

3 41 1 3 2

xx x x x

=1

limx

2

1

1 1 3 2

x

x x x x

=1

limx 2

11 3 2 x x x

....[x1, x 1 x–1 0]

= 2

11 1 1 1 3 2

=

13 2 2

1

limx

f(x) = 112

…. (i)


1limx




f(x) = 3

3 21

xx

, for x 1

= 112

, for x = 1 ; at x = 1 5. If f is continuous at x = 0, then find f(0)

i. f(x) = 2

5 5 2x x

x, x 0

Solution: Given, f is continuous at x = 0 f(0) =

0limx

f(x)

=0

limx 2

5 5 2 x x

x=

0limx 2

15 25

xx

x

=0

limx

2

2

5 2 5 15 x x

x x=

0limx

2

2

5 15x

x x

=0

limx

2

0

5 1 1lim5

x

xxx

= 2 1log 55

....[0

limx

a 1 log a

x

x]

f(0) = (log 5)2

SAMPLE C

ONTENT

133


ii. f(x) =

2sin3 1log 1

x

x x, x 0

[Mar 12; July 18] Solution: Given, f is continuous at x = 0 f(0) =

0limx

f(x)

=0

limx

2sin3 1log 1

x

x x

=

0limx

2sin 2

2

3 1 sin 1 1log( 1)sin

x xxx x xx

=

0limx

2 2sin3 1 sinsin

log 1

x xx x

+ xx

=

2 2sin

0 0

0

3 1 sinlim limsin

log(1 + )lim

x

x x

x

xx x

xx

= 2 2log3 1

1

.... [x0, sinx0, 0

limx

sinxx

=1,0

limx

a 1x

x= loga]

iii. f(x) = 15 3 5 1tan

x x x

x x , x 0 [Mar 15, 17]


0limx

f(x)

=0

limx

15 3 5 1tan

x x x

x x

=0

limx

5 3 3 5 1tan

x x x

x x

=0

limx

5 3 3 5 1tan

x x x x

x x

=0

limx

3 5 1 1 5 1tan

x x x

x x

=0

limx

5 1 3 1tan

x x

x x

=0

limx

2

2

5 1 3 1

tan

x x

xx x

x

=0

limx

5 1 3 1

tan

x x

x xx

x

= 0 0

0

5 1 3 1lim lim

tanlim

x x

x x

x

x xx

x

=

log5 log3

1

….[x 0, x 0,0

limx

a 1x

x = log a,

0limx

tan xx

= 1]

f(0) = (log 5) (log 3) iv. f(x) = 2

cos3 cosx xx , x 0 [Mar 16]


0limx

f(x)

=0

limx 2

cos3 cosx xx

=0

limx

3

24cos 3cos cos x x x

x

….[cos3 = 4cos3 – 3cos]

=0

limx

3

24 cos 4 cosx x

x

=0

limx

2

2

4cos cos 1x xx

=0

limx

2

2

4cos 1 cos x xx

=0

limx

2

24cos sin x x

x

= –40

limx

cos x . 0

limx

2sin

xx

= –4. cos(0) . (1)2

….[0

limx

sin 1xx

]

f (0) = –4

SAMPLE C

ONTENT

134


6. Find the value of k, if the function

i. g(x) =12 1

1xx

, for x 1

= k, for x = 1 is continuous at x = 1 Solution: Given, g is continuous at x = 1 and g(1) = k g(1) =

1limx

g(x)

k =1

limx

12 11

xx

=1

limx

12 1211

xx

=12(1)12–1

.... [a

limx

n nn 1a n a

a

xx

]

k = 12 ii. h(x) = x2 + 1, for x < 0 = 5 2 1x + k, for x 0 is continuous at x = 0 Solution: Given, h is continuous at x = 0

0lim

xh(x) =

0lim

xh(x) = h (0) ….(i)

Now,

0lim

xh(x) =

0limx

x2 + 1

= (0)2 + 1

0lim

xh(x) = 1

and0

limx

h(x) = 0

limx

25 1 kx

= 5 0 1 k

0lim

xh(x) = 5 + k

1 = 5 + k ....[From (i)] k = –4 iii. f(x) = tan 7

2x

x, for x 0

= k, for x = 0 is continuous at x = 0 [Mar 15] Solution: Given, f is continuous at x = 0 and f (0) = k f(0) =

0limx

f(x)

k =0

limx

tan72

xx

= 12 0

limx

tan 7 77

x

x

= 72 0

limx

tan77

xx

= 72

(1)

….[x0,7x0, 0

limx

tan xx

= 1]

k = 72

iv. h(x) = |x + k|, for x 17 = 20, for x = 17 is continuous at x = 17 Solution: Given h is continuous at x = 17, h(17) = 20 h(17) =

17limx

h(x)

= 17

limx

| x + k |

= 17

limx

(x + k)

h(17) = (17 + k) 17 + k = 20 or – (17 + k) = 20 k = 20 17 or 17 k = 20 k = 3 or k = 20 17 k = 3 or k = 37. 7. If f(x) =

21 sin

2

xx

, for x 2 , is continuous

at x =2 , then find f

2

. [Oct 15]

Solution:

Given, f is continuous at x = 2

f 2

lim f2

x

x = 2

2

1 sinlim2

x

xx

Put x = h2

h = x2

as , 0,h 02 2

x x

f2h 0

1 sin h2lim

22 h

2

= 2h 0

1 cos hlim2h

.... [sin cos2

]

SAMPLE C

ONTENT

135


=h 0lim 2

1 cosh 1 cosh1 cosh2h

=h 0lim

2

21 cos h

4h 1 cos h

= h 0lim

2

2sin h

4h 1 cosh

= 14 h 0

lim

2sin hh

h 0

lim

11 cos h

= 14

(1)2 1

1 cos 0….[

0limx

sinxx

= 1]

= 1 14 1 1

= 1 1

4 2

f 12 8

8. If f(x) =

2e 1a

x

x for x < 0, a 0

= 1 for x = 0

= log 1 7b

xx for x > 0, b 0

is continuous at x = 0, then find a and b. [Mar 16, 17; July 16] Solution: Given, f is continuous at x = 0 and f(0) = 1

0lim

xf(x) =

0lim

xf(x) = f(0)

0

limx

f(x) =0

limx

f(x) = 1 ….(i)

Now,

0

limx

f(x) =0

limx

2e 1ax

x

=2

0

1 e 1lim 2a 2

x

x x

=2

0

2 e 1lima 2

x

x x

= 2logea

….[0

a 10, 2 0, lim log a

x

xx x

x]

0

limx

f(x) = 2a

....(ii)

Also,0

limx

f(x) =0

limx

log 1 7b xx

= 0

log 1 71 lim 7b 7

x

xx

= 0

log 1 77 limb 7

x

xx

= 7 1b

.... [ 0

log 10,7 0, lim 1

x

xx x

x]

0

limx

f(x) = 7b

....(iii)

Now, 2 1a ….[From (i) and (ii)]

a = 2

and 7 1b ….[From (i) and (iii)]

b = 7 a = 2, b = 7 9. If f is continuous at x = 0 and

f(x) = 2 3 1x + a, for x < 0 = x3 + a + b, for x 0 and f(1) = 2, then find a, b. [July 17] Solution: Given, f is continuous at x = 0

0lim

xf(x) =

0lim

xf(x) …. (i)

Now,

0

limx

f(x) =0

limx

32 1x + a

= 2 0 1 a

0lim

xf(x) = 2 + a

and0

limx

f(x) =0

limx

x3 + a + b

= 0 + a + b

0lim

xf(x) = a + b

2 + a = a + b ….[From (i)] b = 2 Also, f(x) = x3 + a + b, for x 0 and f(1) = 2 f(1) = (1)3 + a + b 2 = 1 + a + b a + b = 1 …. (ii) Substituting b = 2 in (ii) we get a + 2 = 1 a = –1 a = –1, b = 2

SAMPLE C

ONTENT

136


10. Is the function f(x) = x3 + 2x2 – 5 cos x + 3

continuous at x =2 ? Justify.

Solution: Given, f(x) = x3 + 2x2 – 5 cos x + 3 f(x) = (x3 + 2x2 + 3) – 5 cos x Let x3 + 2x2 + 3 = p(x) and 5 cos x = q(x) f(x) = p(x) – q(x) ....(i) Now, p(x) = x3 + 2x2 + 3 p(x) is a polynomial function It is continuous at each value of x and q(x) = 5 cos x Here, 5 is constant function and cos x is cosine

function which is continuous. q(x) is continuous at each value of x f(x) is a difference of two continuous function

which is always continuous

f(x) is continuous at x = 2

Miscellaneous Exercise – 3 1. Discuss the continuity of the function.

f(x) =3

2

649 5

xx

, for x 4

= 10, for x = 4; at x = 4 [Mar 18]

Solution:

4limx

f(x) =4

limx

3

2

649 5

xx

=4

limx

23 3

2 2

9 54

9 5 9 5

xx

x x

=4

limx

2 2

2 22

4 4 16 9 5

9 5

x x x x

x

=4

limx

2 2

2

4 4 16 9 5

9 25

x x x x

x

=4

limx

2 2

2

4 4 16 9 5

16

x x x x

x

=4

limx

2 24 4 16 9 5

4 4

x x x x

x x

=4

limx

2 24 16 9 5

4

x x x

x

[ x 4, x 4, x – 4 0]

= 2 24 4 4 16 4 9 5

4 4

= 48 10

8

4

limx

f(x) = 60 …. (i)


4limx


f is discontinuous at x = 4 2. Examine the continuity of the function

f(x) =

23e 1log 1 3

x

x x , for x 0

= 10, for x = 0; at x = 0. If discontinuous, then state whether the

discontinuity is removable. If so, redefine and make it continuous.

Solution:

0limx

f(x) = 0

limx

23e 1log 1 3

x

x x=

0limx

23

2

e 19 log 1 3

9

x

x xx

=0

limx

23

2

e 1 1log 1 39

9

x

xxx

=0

limx

230

0

lim1e 11 log(1 3 )3 lim3 3

xx

x

xxx

=

2 1loge 1 13

....

0

e 10,3 0,lim loge,

x

xx x

x

0

log 1lim 1

x

xx

= 3(1)2

0limx

f(x) = 3 …. (i)


SAMPLE C

ONTENT

137


0

limx




f(x) = 23e 1log(1 3 )

x

x x

, for x 0

= 3 , for x = 0 ; at x = 0 3. The function f is defined as

f(x) = 7

512832

xx

, for x 2

=1/ 5 1/ 5

1/ 2 1/ 222

xx

, for x > 2

f(2) = 3 Examine, if f is continuous at x = 2. Solution:

2lim

xf(x) =

2limx

7

512832

xx

=2

limx

7 7

5 522

xx

=2

limx

7 7

5 5

2222

xx

xx

=

7 7

25 5

2

2lim22lim2

x

x

xx

xx

=

7 1

5 1

7 25 2

....

n nn 1

a

alim naa

x

xx

=6

47 25 2

=27 2

5

2

limx

f(x) = 285

and2

limx

f(x) = 2

limx

1 15 5

1 12 2

2

2

x

x

=2

limx

1 15 5

1 12 2

22

22

xx

xx

=

1 15 5

21 12 2

2

2lim2

2lim2

x

x

xx

xx

=

1 15

1 12

1 251 22

....[n n

n 1

a

alim naa

x

xx

]

=

45

12

1 251 22

=4 1

5 22 25

2

limx

f(x) = 25

(2)3

10

2

limx

f(x) 2

limx

f(x)

f(x) does not exist f is discontinuous at x = 2 4. The function f defined as

f(x) = 8 8 15 7 3

x xx x

, for x 1

is continuous at x = 1. Find f(1) Solution: Given, function is continuous at x = 1

f(1) = 1

limx

f(x)=1

limx

8 8 15 7 3x x

x x

=1

limx

8 8 1

5 7 3

x x

x x

5 7 3

5 7 3

x x

x x

8 8 1

8 8 1

x x

x x

=1

limx

2 2

2 2

8 8 1 5 7 3

5 7 3 8 8 1

x x x x

x x x x

=1

limx

8 8 1 5 7 3

5 7 3 8 8 1

x x x x

x x x x

=1

limx

7 7 5 7 3

8 8 8 8 1

x x x

x x x

=1

limx

7 1 5 7 3

8 1 8 8 1

x x x

x x x

SAMPLE C

ONTENT

138


=1

limx

7 5 7 3

8 8 8 1

x x

x x

....[ x1, x 1 x 1 0]

=

7 5 1 7 1 3

8 1 8 8 1 1

=

7 2 28 3 3

= 2848

f (1) = 712

5. Find k if the function given below is

continuous at x =2

f(x) = 3

2cos sin22

x xx

, for x 2

= k, for x =2

Solution:

Given, f is continuous at x = 2

f2

=2

lim

xf(x)

k = 2

lim

x 32cos sin 2

2

x xx

= 2

lim

x 32cos 2sin cos

2

x x xx

[sin2 = 2sin cos]

= 2

lim

x

3

2cos 1 sin2

x xx

Put x = h2

h = x – 2

as x 2 , x –

2 0, h 0

h = h 0lim 3

2 cos h 1 sin h2 2

2 h2

= h 0lim

3

2sin h 1 cos h2h

[cos sin , sin cos2 2

]

= h 0lim

2

3

h2sin h 2sin2

8h

= 12 h 0

lim

2

2

hsinsin h 2hh 44

=

2

h 0 h 0

hsin1 sin h 2lim .lim h8 h2

= 21 1 18

….h 0

h sin hh 0, 0, lim 12 h

6. If the function given below is continuous at

x = 2 as well as at x = 4 , then find the values of a and b.

f(x) = x2 + ax + b, x 2 = 3x + 2, 2 x 4 = 2ax + 5b, 4 x [July 18; Oct 14] Solution: Given, f is continuous at x = 2

2lim

xf(x) =

2lim

xf(x) ….(i)

Now,2

limx

f(x) =2

limx

x2 + ax + b

= (2)2 + a(2) + b

2lim

xf(x) = 4 + 2a + b

and2

limx

f(x) =2

limx

(3x + 2)

= 3(2) + 2

2lim

xf(x) = 8

4 + 2a + b = 8 ….[From (i)] 2a + b = 4 ….(ii) Also, f is continuous at x = 4

4lim

xf(x) =

4lim

xf(x) ….(iii)

Now,4

limx

f(x) = 4

limx

3x + 2 = 3(4) + 2

4

limx

f(x) = 14

and4

limx

f(x) = 4

limx

2ax + 5b = 2a(4) + 5b

4

limx

f(x) = 8a + 5b

14 = 8a + 5b ….[From(iii)] 8a + 5b = 14 ….(iv)

SAMPLE C

ONTENT

139


By eq. (iv) – 5 eq. (iii), we get

8a 5b 1410a 5b 20

2a 6

a = 3 Substituting a = 3 in eq. (ii), we get 2 3 + b = 4 b = 4 – 6 b = –2 a = 3 or b = 2 7. Find a and b if f is continuous at x = 1,

where f(x) =

sin

1

xx

+ a, x 1

= 2, x = 1

= 2

1 cos1

xx

+ b, x 1

Solution: Given, f is continuous at x = 1 and f (1) = 2

1lim

xf(x) =

1lim

xf(x) = f (1)

1

limx

f(x) = 1

limx

f(x) = 2 ….(i)

Now, 1

limx

f(x) = 1

limx

sin a1

xx

Put x = 1 + h h = x – 1 as x1, x 10, h0

1

limx

f(x) = h 0lim

sin 1 ha

1 h 1

= h 0lim

sin ha

h

= h 0lim

sin h ah

….[sin(+) = –sin]

= h 0lim

sin h ah

= –h 0 h 0

sin hlim lim ah

= –(1) + a

….[h0, h0,h 0lim

sin h 1h

]

1

limx

f(x) = – + a ….(ii)

Also 1

limx

f(x) =1

limx 2

1 cos b1

xx

Put x = 1 + h h = x – 1 as x1, x – 10, h0

1

limx

f(x) = 2h 0

1 cos 1 hlim b

1 1 h

= h 0lim

2

1 cos hb

1 1 h

= h 0lim 2

1 cos h bh

….[cos(+) = –cos]

= h 0lim

2

2

h2sin2 b

h

….[1 – cos = 2sin2

2 ]

= h 0lim

2

2

h2sin2 b

h 44

= h 0lim

2

2 2

hsin2 2 b4 h4

=

2

h 0 h 0

hsin2lim lim bh2

2

= 21 b2

.... [0

h sinh 0, 0, lim 12

x

xx

]

1

limx

f(x) = b2 ….(iii)

– + a = 2 ….[From (i) and (ii)] a = 2 + a = 3

and b 22 ….[From (i) and (iii)]

SAMPLE C

ONTENT

140


b = 2 – 2

b = 32

a = 3, b = 32

8. Find k, if the function f is continuous at

x = 0, where

i. f(x) =

2

e 1 sinx xx

, for x 0

= k, for x = 0 [July 16]

Solution: Given, f is continuous at x = 0 and f(0) = k f(0) =

0limx

f(x)

k = 0

limx

2

e 1 sinx xx

= 0

limx

e 1 sin

x xx x

= 0

limx

e 1x

x .

0limx

sinxx

= log e . (1)

….[0

limx

e 1x

x= log e,

0limx

sinxx

= 1]

k = 1

ii. f(x) = 27 3k 1

x x

x , for x 0

= 2, for x = 0 Solution: Given, f is continuous at x = 0 and f(0) = 2 f(0) =

0limx

f(x)

2 = 0

limx

27 3k 1

x x

x=

0limx

3 9 3k 1

x x

x =

0limx

3 9 3k 1

x x x

x = 0

limx

3 9 1k 1

x x

x

=

0limx

3 9 1

k 1

x x

xx

x

= 0 0

0

9 1lim3 .lim

k 1lim

xx

x xx

x

x

x

2 = 03 log 9log k

.... [0

limx

a 1 log a

x

x]

log k = log92

log k = log 129 .... [n log a = log an]

log k = log 3 k = 3

iii. f(x) = log 1 35 xx

, for x 0

= k, for x = 0 [Mar 18] Solution: Given, f is continuous at x = 0 and f(0) = k f(0) =

0limx

f(x)

k = 0

limx

log 1 35 xx

= 0

limx

log 1 31 35 3

xx

= 35 0

limx

log 1 33 xx

= 3 15

....[

0

log 10,3 0, lim 1

x

xx x

x]

k = 35

9. Find the points of discontinuity, if any, for

the following:

i. f(x) =2

2cos

1

x xx

Solution:

Given, f(x) = 2

2cos

1

x xx

Let, x2 + cosx = p(x) and x2 + 1 = q(x) Consider, p(x) = x2 + cosx Here, x2 is always continuous for all real

values of x and cosine is a continuous function p(x) is a continuous function and q(x) = x2 + 1 It is a polynonimal function It is continuous for all real values of x f(x) is a continuous function.

SAMPLE C

ONTENT

141


ii. f(x) = 25 4

4

xx

Solution:

Given, f(x) = 25 4

4

xx

f(x) =

5 42 2

x

x x f(x) is a rational function f(x) will be discontinuous if (x + 2) (x – 2) = 0 i.e., x + 2 = 0 or x – 2 = 0 i.e., x = –2 or x = 2 f(x) is discontinuous at x = –2 and x = 2 iii. f(x) =

2

23 4 9

6 10

x xx x

Solution:

Given, f(x) = 2

23 4 9

6 10

x xx x

f(x) is a rational function f(x) will be discontinuous if x2 – 6x + 10 = 0

i.e., x = 26 6 4 1 10

2

= 6 36 402

6 42

x

6 2i2

x

Value of x is a complex number f(x) is continuous for all real values of x

iv. f(x) =2 9

sin 9

xx

[Mar 17]

Solution:

Given, f(x) = 2 9

sin 9

xx

Let x2 – 9 = p(x) and sinx – 9 = q(x) Consider, p(x) = x2 – 9 It is a polynomial function It is continuous function and q(x) = sinx – 9

Here, sine is a continuous function and 9 is a constant function

q(x) is continuous as –1 sinx 1 f(x) is continuous function.

10. If possible, redefine the function to make it continuous.

i. f(x) =1

1 xx , for x 1

= e2, for x = 1; at x = 1. Solution:

1

1

1 1lim f lim

x

x xx x

Put x = 1 + h h = x – 1 as x1, x–10, h0

1limx

f(x) = h 0lim

1

1 h 11 h

=h 0lim

1h1 h

1

limx

f(x) = e

....[ 1

0lim 1 e

xx

x ]....(i) Also, f(1) = e2 …. (ii) (given)

1

limx




f(x) = 1

1 xx , for x 1

= e, for x = 1 ; at x = 1 ii. f(x) =

tan6 1sin

x

x, for x 0

= log 50, for x = 0; at x = 0. Solution:

0

limx

f(x) =0

limx

tan6 1sin

x

x

=0

limx

tan6 1sin coscos

x

x xx

= 0

limx

tan6 1tan cos

x

x x

= 0

limx

tan

0

6 1 1.limtan cos

x

xx x

= log 6 1

cos 0

.... [x0, tanx0, 0

limx

a 1 logax

x

]

0

limx

f(x) = log 6

Also, f(0) = log 50 ….(given)

0limx

f(x) f(0)

f has removable discontinuity at x = 0

SAMPLE C

ONTENT

142


This discontinuity can be removed by redefining the function as:

f(x) = tan6 1sin

x

x, for x 0

= log 6, for x = 0 ; at x = 0 iii. f(x) =

2

2sin 5 x

x, for x 0

= 5, for x = 0; at x = 0. [Oct 14]

Solution:

0

limx

f(x) = 0

limx

2

2sin 5x

x

=0

limx

2

2sin 5 2525

x

x

= 250

limx

2sin55

xx

= 25(1)2

[x0, 5x0, 0

limx

sin 1xx

]

0

limx

f(x) = 25

Also, f(0) = 5 …. (given)

0limx

f(x) f(0)



f(x) = 2

2sin 5x

x, for x 0

= 25, for x = 0 ; at x = 0 iv. f(x) =

2

3cos

1 sinx

x, for x <

2

=2

2 1 sincos x

x, for x >

2

= 23

, for x = 2 ; at x =

2 .

Solution:

2

lim

x

f(x) =2

lim

x

2

3cos

1 sinx

x

=2

lim

x

2

31 sin1 sin

xx

=2

lim

x

2

1 sin 1 sin1 sin 1 sin sin

x xx x x

....[a3 – b3 = (a – b) (a2 + ab + b2)]

=2

lim

x2

1 sin1 sin sin

xx x

….[x2 sinxsin

2 sinx1, 1–sinx0]

=2

1 sin2

1 sin sin2 2

= 2

1 11 1 1

2

lim f ( )x

x

= 23

and2

lim

x

f(x) =2

lim

x2

2 1 sincos x

x

=2

lim

x2

2 1 sin 2 1 sincos 2 1 sin

x xx x

=2

lim

x

2 2

2

2 1 sin

cos 2 1 sin

x

x x

=2

lim

x

2

2 1 sin

1 sin 2 1 sin

x

x x

=2

lim

x

1 sin

1 sin 1 sin 2 1 sin

x

x x x

=2

lim

x 1

1 sin 2 1 sin x x

….[x2 sinxsin

2 sinx1, 1–sinx0]

=1

1 sin 2 1 sin2 2

=

11 1 2 1 1

=

12 2 2

2

lim f ( )x

x

= 14 2

2

lim f ( )x

x

2

lim

x

f(x)

limit of the function does not exist

f has irremovable discontinuity at x = 2

SAMPLE C

ONTENT

143


v. f(x) = 2

2

21

x xx

, for x 1

=3 2

2 1

xx

, for x > 1


1

limx

f(x) =1

limx

2

2

21

x xx

=1

limx

2

2

1 1

1

x x

x

=1

limx

2

2 21 11 1

x xx x

=1

limx 2

1 111 1

x xx x

=1

limx

2 2

2 2

11

1 1

x

x x

=1

limx

11

1 1 1

x

x x x

=1

limx

111 1

x x

....[x1, x–10]

=1

limx

1 + 1

limx

11 1 x x

=1 +

11 1 1 1

= 1 + 12(2)

= 1 + 14

1

limx

f(x) = 54

and 1

limx

f(x) =1

limx

3 22 1x

x

=1

limx

3 2 3 2 2 12 1 3 2 2 1

x x xx x x

=1

limx

2 2

2 2

3 2 2 13 22 1

x xxx

=1

limx

3 4 2 12 1 3 2

x xx x

=1

limx

1 2 1

1 3 2

x x

x x

=1

limx

1 2 1

1 3 2

x x

x x

=1

limx

2 1

3 2

x

x ….[x1, x 1 x–10]

= 2 1 1 241 3 2

1

limx

f(x) = 12

1

limx

f(x) 1

limx

f(x)

limit of the function does not exist. f has irremovable discontinuity at x = 1

vi. f(x) = 2 3 4 4

1

x x x x

x , for x 1.


1

limx

f(x) =1

limx

2 3 4 41

x x x xx

=1

limx

2 3 41 1 1 11

x x x xx

=1

limx

2 3 41 1 1 11 1 1 1

x x x xx x x x

=1

limx

11

xx

+1

limx

2 211

xx

+1

limx

3 311

xx

+1

limx

4 411

xx

= (1) (1)1–1 + (2) (1)2–1 + (3) (1)3–1 + (4) (1)4–1

….[a

limx

n nn 1a na

a

xx

]

= 1 + 2 + 3 + 4

1limx

f(x) = 10 …. (i)


1limx

f(x) f(1)

f has removable discontinuity at x = 1

SAMPLE C

ONTENT

144


This discontinuity can be removed by redefining the function as:

f(x) = 2 3 4 4

1

x x x x

x, for x 1

= 10, for x = 1 ; at x = 1 Additional Problems for Practice Based on Exercise 3.1 1. Are the following functions continuous on the

set of real numbers? Justify your answer. (All functions are defined on R R)

i. f(x) = 8x2 + 9 ii. f(x) = e50

iii. f(x) = log 2319

iv. f(x) = 8x5 2x4 + 5x3 + 2x2 + x + 9 v. h(x) = cos(9x + 5)

vi. g(x) = 2

45

29 2

x x

x x vii. g(x) = 3x + 7x 2. Examine the continuity of the following

funtions at the given point:

i. f(x) = sinx

x + cos x, for x 0 = 2, for x = 0; at x = 0

ii. f(x) = 12

sin x2, for x 0

= 0, for x = 0; at x = 0 iii. f(x) = (1 + 2x)1/x, for x 0 = e2, for x = 0; at x = 0

iv. f(x) = 2 6

3

x xx , for x 3

= 7, for x = 3; at x = 3 v. f(x) = x2 + 6x + 10, for x 4 = x2 x + 38, for x > 4; at x = 4

vi. f(y) = 2

2

e 1 .siny yy , for y 0

= 4, for y = 0; at y = 0

vii. f(x) =

1415

xx, for x 0

= 45e , for x 0 at x = 0

viii. f(x) = 12 sin 2

(x + 1), for x 0

= 3tan sinx x

x , for x > 0; at x = 0

ix. f(x) = 3 2

3 22 2 53 3 1

x x xx x x , for x < 1

= 4 31 1

1 x x x , for x 1; at x = 1

x. f(x) = 3

3 21

xx

, for x 1

= 112

, for x = 1; at x = 1 3. Discuss the continuity of the following functions:

i. f(x) = 3 5a ax x

x, for x 0

= log a, for x = 0; at x = 0

ii. f(x) = 1

1a

xx , for x 0

= 1ae , for x = 0; at x = 0

iii. g(x) =

5log 12

x

x, for x 0

= 52

, for x = 0; at x = 0

iv. f(x) = 5 esin 2x x

x, for x 0

= 12

(log 5 + 1), for x = 0; at x = 0

v. f(x) = 2

2sin ax

x, for x 0

= 1, for x = 0; at x = 0

vi. f(x) = x2 sin 1x

, for x 0

= 0, for x = 0; at x = 0

vii. f(x) = 21 cos x

x , for x 0

= 0, for x = 0; at x = 0

viii. f(x) = 4 23

xx

, for x ≠ 0

= 112

, for x = 0; at x = 0

[Mar 16]

ix. If f(x) = 32 1tanx

x, for x ≠ 0

= 1 , for x = 0 [July 17] 4. Discuss the continuity of the following

functions at the points given against them. If the function is discontinuous, determine whether the discontinuity is removable. In that case, redefine the function, so that it becomes continuous.

SAMPLE C

ONTENT

145


i. f(x) = 1 cos3tan

xx x

, for x 0

= 9, for x = 0; at x = 0

ii. f(x) = sin5xx

, for x 0

= 5

, for x = 0; at x = 0

iii. f(x) = 2 4sin 2 x

x, for x 0

= 8, for x = 0; at x = 0

iv. f(x) = 2sin( )x xx

, for x 0

= 2, for x = 0; at x = 0 5. If f is continuous at x = 0, then find f(0).

i. f(x) = 2sin4 1log (1 2 )

x

x x, x 0

ii. f(x) = log(1 a ) log(1 b ) x xx

iii. f(x) = log(2 ) log(2 )tan

x xx

iv. f(x) = 2 2

2

cos sin 11 1

x xx

6. Find the value of k, if the function i. g(x) = |x 3|, for x 3 = k, for x = 3 is continuous at x = 3

ii. f(x) = 8 2k 1

x x

x , for x 0

= 2, for x = 0 is continuous at x = 0

iii. f(x) = log(1 k )sin xx

, for x 0

= 5, for x = 0 is continuous x = 0 iv. f(x) = x2 + k, for x 0 = x2 k, for x < 0 is continuous at x = 0

v. f(x) = 2

2

3 k2( 1)

x x

x, for x 1

= 54

, for x = 1

is continuous at x = 1

7. If f(x) = 2

1 cos[7( )]5( )

xx

, for x is

continuous at x = , find f()

8. If the function f(x) = k cos2

xx

, for x 2

= 3, for x = 2

be continuous at x = 2 , then find k

9. Is the function f(x) = 2x3 + 3x2 + 3x cos x + sin 5x + 3

continuous at x = 4 ? Justify

10. If the function f is continuous at x = 1, then

find f(1). Where f(x) = 2 3 2

1

x xx

for x ≠ 1.

[Mar 14] 11. If the function f is continuous at x = 2, then

find ‘k’ where f(x) = 2 5

1

xx

, for 1 < x 2

= kx + 1, for x > 2 [Mar 14]

12. Discuss the continuity of the function f

defined as

f (x) = 3

8 31

xx

, x ≠ 1

= 13 , x = 1; at x = 1 [Mar 08]

13. Discuss the continuity of the function f

defined as:

f (x) = 1

513

xx

, if x ≠ 0

= 53e , if x = 0; at x = 0 [Oct 08]

14. Discuss the continuity of f (x) at x = 2, where

f (x) = 2 4

2xx

, for x ≠ 2

= 4, for x = 2 [Mar 09] 15. Discuss the continuity of the function f

defined as, f(x) = 2x + 3, if 1 ≤ x ≤ 2 = 6x 1, if 2 < x ≤ 3; at x = 2 [Oct 10]

SAMPLE C

ONTENT

146


16. If f(x) = 4 2xx , for x ≠ 0

= 14, for x = 0

Discuss the continuity of f(x) at x = 0 [Mar 11, Oct 11 ]

17. Discuss the continuity of the following function:

1

f 1 3 xx x , for x ≠ 0 = e3, for x = 0; at x = 0 [Oct 12] 18. If f is continuous at x = 0, where f (x) = x2 + a, for x ≥ 0 = 2 2 1 bx , for x < 0 Find a, b given that f (1) = 2. [Mar 08] 19. Find k, if the function f defined as:

f (x) = 2

2 3 cos k xx

, x ≠ 0

= 2, x = 0 is continuous at x = 0 [Oct 08] 20. Find k, if the function

f (x) = 3 64

4xx

, for x ≠ 4

= k, for x = 4 is continuous at x = 4 [Oct 09] 21. If f(x) is continuous at x = 0 and it is defined as

a afx x

xx

, x ≠ 0

= k, x = 0 find k. [Mar 10] 22. The function f defined as

f(x) = sin pxx

, if x > 0

= q + 25 16x , if x ≤ 0 is continuous at x = 0. Find the values of p and

q, given that f (2) = 3. [Oct 10] 23. If f(x) = tan2

3x

x+ a, for x < 0

= 1, for x = 0 = x + 4 b, for x > 0 is continuous at x = 0, then find the values of a

and b. [Mar 11] 24. Find f(3) if f(x) =

2 93

xx

, x ≠ 3 is contiuous at

x = 3. [Oct 11]

25. If f is continuous at x = 0 where

f(x) = 3e 1ax

x, x ≠ 0

= 1 , x = 0 then find a. [Mar 12] 26. If f is continuous at x = 0 where f(x) = x2 + a, x 0 = 2 2 1 bx , x < 0, find a and b. Given that f(1) = 2 [Oct 12] 27. Examine the continuity of f at x = 1, if f(x) = 5x 3, for 0 x 1 = x2 + 1, for 1 x 2 [July 16] 28. If the function f is continuous at x = 2,

then find f (2)

where f (x) = 5 32

2xx

, for x ≠ 2. [July 17] Based on Miscellaneous Exercise - 3 1. Examine the continuity of the following

funtions at the given point:

i. f(x) = 10 7 14 5

1 cos

x x x x

x , for x 0

= 107 , for x = 0; at x = 0

ii. f(x) = sin3tan2

xx , for x < 0

= 32, for x = 0

= 2log(1 3 )

e 1x

x, for x > 0

iii. f(x) = 23 4 1

( 6)

xx , for x ≠ 6

= 15, for x = 6; at x = 6

iv. f(x) = 1 2 1 2 x xx

, for x < 0

= 2x2 + 3x 2, for x 0; at x = 0

v. f(x) = 3 2

2

16 20( 2)

x x xx , for x 2

= 7, for x = 2; at x = 2 2. Discuss the continuity of the following functions:

i. f(x) = 2 54 3

x x

x x , for x 0

= log 310

, for x = 0; at x = 0

SAMPLE C

ONTENT

147


ii. f(x) = 2(2 1)

tan .log(1 )

x

x x, for x 0

= log 4, for x = 0 3. Discuss the continuity of the following

functions at the points given against them. If the function is discontinuous, determine whether the discontinuity is removable. In that case, redefine the function, so that it becomes continuous:

i. f(x) = 4 e6 1

x x

x , for x 0

= log 23

, for x = 0; at x = 0

ii. f(x) = 23 3 2 x x

x, for x 0

= 2 log3, for x = 0; at x = 0

iii. f(x) =

6

3

16418

x

x, for x 1

2

= 13, for x = 1

2; at x = 1

2 iv. f(x) =

2(8 1)

sin log 14

x

xx, for x 0

= 8 log 8, for x = 0; at x = 0 4. If f is continuous at x = 0, then find f(0).

i. f(x) = 14 2 1

1 cos

x x

x, x 0

ii. f(x) = 5 2e esin 3x x

x

5. Find the value of k, if the function

f(x) = 2

2

3sin2x

x, for x 0

= k, for x = 0 is continuous at x = 0

6. If f(x) = sin45

xx

+ a, for x > 0

= x + 4 b, for x < 0 = 1, for x = 0 is continuous at x = 0, find a and b.

[Mar 09, 10; Oct 09]

7. If f(x) = 21 cos4 x

x, for x < 0

= a, for x = 0

= 16 4

x

x, for x > 0

is continuous at x = 0, then find the value of ‘a’. 8. Discuss the continuity of the function f at

x = 0, where f(x) = 5 5 2 ,cos 2 cos 6

x x

x x for x ≠ 0

= 21 log58

, for x = 0

[Mar 14] Multiple Choice Questions

1. If f(x) = 2 , 0 1c 2 , 1 2

xx x

is continuous at

x = 1, then c = (A) 2 (B) 4 (C) 0 (D) 1

2. If f(x) = 1 , if 3a b , if 3 57 , if 5

xx x

xis continuous,

then the value of a and b is (A) 3, 8 (B) –3, 8 (C) 3, –8 (D) –3, –8 3. The sum of two discontinuous functions (A) is always discontinuous. (B) may be continuous. (C) is always continuous. (D) may be discontinuous. 4. For what value of k the function

f(x) = 5 2 4 4 , if 2

2k ,if 2

x x xx

x

is

continuous at x = 2? (A) 1

4 3 (B) 1

2 3 (C) 1

4 3 (D) 1

2 3

5. The function f(x) = log (1 a ) log (1 b )x x

x

is

not defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is

(A) a b (B) a + b (C) log a + log b (D) log a log b

SAMPLE C

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148


6. In order that the function f(x) = (x + 1) cot x is continuos at x = 0, f(0) must be defined as

(A) f(0) = 1e (B) f(0) = 0

(C) f(0) = e (D) None of these

7. If f(x) =sin3 , 0sin

k, 0

x xx

xis a continuous

function, then k = (A) 1 (B) 3

(C) 13 (D) 0

8. A function f is continuous at a point x = a in the domain of ‘f’ if

(A) a

limx

f(x) exists (B) a

limx

f(x) = f(a)

(C) a

limx

f(x) f(a) (D) both (A) and (B). 9. Which of the following function is

discontinuous? (A) f(x) = x2 (B) g(x) = tan x

(C) h(x) = 23

1x

x (D) none of these

10. If the function f(x) =

k cos , when2 2

3, when2

x xx

x is

continuous at x = 2

, then k =

(A) 3 (B) 6 (C) 12 (D) None of these 11. The points at which the function

f(x) = 21

12

x

x x is discontinuous, are

(A) –3,4 (B) 3,–4 (C) –1,–3,4 (D) –1,3,4 12. Which of the following statement is true for

graph f(x) = log x (A) Graph shows that function is continuous (B) Graph shows that function is

discontinuous (C) Graph finds for negative and positive

values of x (D) Graph is symmetric along x-axis

13. If f(x) =

2 1, when 11

2, when 1

x xx

x, then

(A) 1

lim x

f(x) = –2

(B) 1

lim x

f(x) = –2

(C) f(x) is continuous at x = –1 (D) All the above are correct

14. If f(x) =

a, when a

a1, when a

xx

xx

, then

(A) f(x) is continuous at x = a (B) f(x) is discontinuous at x = a

(C) 0limx f(x) = 1

(D) None of these

15. If f(x) =

2

1 cos 4 , when 0

a when 0,

, when 016 4

x x <x

x =

x xx

is continuous at x = 0, then the value of ‘a’ will be (A) 8 (B) –8 (C) 4 (D) None of these

16. If f(x) =

4 16 , when 2216,when 2

x xx

x, then

(A) f(x) is continuous at x = 2 (B) f(x) is discountinuous at x = 2

(C) 2limx f(x) = 16

(D) None of these 17. The values of A and B such that the function

f(x) =

2sin ,2

Asin B, ,2 2

cos2

x x

x x

x, x

is continuous

everywhere are (A) A = 0, B = 1 (B) A = 1, B = 1 (C) A = –1, B = 1 (D) A = –1, B = 0

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149


18. If f(x) = 2

1 k 1 k ,for 1 0

2 3 2 ,for 0 1

x x x <x

x x x

, is

continuous at x = 0, then k = (A) –4 (B) –3 (C) –2 (D) –1 19. The function f(x) = sin |x| is (A) Continuous for all x (B) Continuous only at certain points (C) Differentiable at all points (D) None of these

20. The function f(x) = 1 sin cos1 sin cos

x xx x is not

defined at x = . The value of f(), so that f(x) is continuous at x = , is

(A) 12

(B) 12

(C) –1 (D) 1

21. The function f(x) = 2

3 2

2 73 3x

x x x

is discontinuous for

(A) x = 1 only (B) x = 1 and x = –1 only (C) x = 1, x = –1, x = –3 only (D) x = 1, x = –1, x = –3 and other values of x 22. The function ' f is defined by f(x) = 2x – 1, if

x > 2, f(x) = k if x = 2 and x2 –1, if x < 2 is continuous, then the value of k is equal to

(A) 2 (B) 3 (C) 4 (D) –3

23. Function f(x) = 21 cos4

8 x

x , where x 0 and f(x) = k, where x = 0 is a continous function at x = 0 then the value of k will be?

(A) k = 0 (B) k = 1 (C) k = –1 (D) None of these

24. If f(x) =

, when0 1/ 21, when 1/ 21 ,when1/ 2 1

x xx

x x

, then

(A) 1/2lim x f(x) = 2

(B) 1/2lim x f(x) = 2

(C) f(x) is continuous at x = 12

(D) f(x) is discontinuous at x = 12

25. If f(x) = 2

210 257 10

x xx x for x 5 and f is

continuous at x = 5, then f(5) = (A) 0 (B) 5 (C) 10 (D) 25

Answers to Additional Practice Problems Based on Exercise 3.1 1. i. Polynomial function continuous ii. Constant function continuous iii Constant function continuous iv. Polynomial function continuous v. Cosine function continuous vi. Rational function continuous for all x R, except when

x4 + 29x + 2 = 0 vii. Addition of exponential functions continuous 2. i. Continuous ii. Continuous iii. Continuous iv. Discontinuous v. Continuous vi. Discontinuous vii. Continuous viii. Continuous ix. Discontinuous x. Continuous 3. i. Discontinuous ii. Continuous iii. Continuous iv. Discontinuous v. Discontinuous vi. Continuous vii. Discontinuous viii. Continuous ix. Discontinuous 4. i. Discontinuous, removable ii. Discontinuous, removable iii. Discontinuous, removable iv. Discontinuous, removable

5. i. 2(log 4)

2 ii. a + b

iii. 1 iv. 4 6. i. 0 ii. 2 iii. 5 iv. 0 v. 4 7. 49

10

SAMPLE C

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150


8. 6 9. Addition of continuous functions. f(x) is continuous. 10. f(1) = 1 11. k = 4 12. Discontinuous 13. Discontinuous 14. Continuous 15. Discontinuous 16. Continuous 17. Discontinuous 18. a = 1, b = 1 19. k = ± 4 20. k = 48 21. 2 log a 22. p = 1, q = 3 23. a = 1

3, b = 3

24. 6 25. a = 3 26. a = 1, b = 3

4

27. Continuous at x = 1 28. 80 Based on Miscellaneous Exercise - 3 1. i. Discontinuous ii. Continuous iii. Discontinuous iv. Continuous v. Continuous 2. i. Discontinuous ii. Discontinuous 3. i. Discontinuous, removable ii. Discontinuous, removable iii. Discontinuous, removable iv. Discontinuous, removable 4. i. 2(log 2)2 ii. 1 5. 9

4 6. a = 1

5, b = 3

7. 8 8. Discontinuous

Answers to Multiple Choice Question

1. (B) 2. (C) 3. (B) 4. (C) 5. (B) 6. (C) 7. (B) 8. (D) 9. (B) 10. (B) 11. (B) 12. (A) 13. (D) 14. (B) 15. (A) 16. (B) 17. (C) 18. (C) 19. (A) 20. (C) 21. (C) 22. (B) 23. (B) 24. (D) 25. (A)

SAMPLE C

ONTENT

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Std. 12th Mathematics and Statistics - 1 Notes, Commerce