Study Material for B.Tech Mathematics -Code -BTC 11 Sharada Vikas Trust (R

Karnataka State Open University

Study Material for B.Tech

Mathematics - Code - BTC 11

by

K.S. SrinivasaRetd. Principal &

Professor of Mathematics Bangalore

Published by Sharada Vikas Trust (R)

Bangalore

BTC 11 MATHEMATICS

Syllabus

1. Complex Trigonometry

Revision of Plane Trigonometry - trigonometric ratios, expressions for relation between allied angles and trigonometrical ratios. Addition formulae for trigonometrical ratios and simple problems. Complex numbers and functions, definition,properties, De Moivre's Theorem (without proof), Roots of a complex number, expansions of sin ( n ), cos ( n ) inpowers of sin & cos , addition formulae for any number of angles, simple problems.

2. Matrix Theory :

Review of the fundamentals. Solution of linear equations by Cramers' Rule and by Matrix method, Eigen values and Eigen vectors, Cayley Hamilton's Theorem, Diagonalization of matrices, simple problems.

3. Algebraic Structures

Definition of a group, properties of groups, sub groups, permutation groups, simple problems, scalars & vectors, algebraof vectors, scalar & vector products, scalar triple product, simple problems.

4. Differential Calculus

Limits, continuity and differentiability (definition only), standard derivatives, rules for differentiation, derivatives offunction of a function and parametric functions, problems. Successive differentiation,n derivative of standard functions,t h

statement of Leibnitz's Theorem, problems, polar forms, angle between the radius vector and the tangent to a polar curve, (no derivation) angle between curves, pedal equation, simple problems, indeterminate forms, L' Hospital's rule,partial derivatives, definition and simple problems.

5. Integral Calculus

Introduction, standard integrals, integration by substitution and by parts, integration of rational, irrational and trigonometricfunctions, definite integrals, properties (no proof), simple problems, reduction formulae and simple problems.

6. Differential Equations of first order

Introduction, solution by separation of variables, homogeneous equations, reducible to homogeneous linear equation,Bernoulli's equation, exact differential equations and simple problems.

Text Books 1. Elementary Engineering Mathematics by Dr. B.S. Grewal, Khanna Publications

2. Higher Engineering Mathematics by B.S. Grewal, Khanna Publications

Reference Books 1. Differential Calculus by Shanti Narayan, Publishers S. Chand & Co.

2. Integral Calculus by Shanti Narayan, Publishers S. Chand & Co.

3. Modern Abstract Algebra by Shanti Narayan, Publishers S. Chand & Co.

CONTENTS

Page Nos.

1. Complex Trigonometry 01

2. Matrix Theory 27

3. Algebraic Structures 47

4. Differential Calculus 69

5. Integral Calculus 101

6. Differential Equations 123

COMPLEX TRIGONOMETRY

Trigonometric ratios of acute angles

A èConsider a right-angled triangle ABC , right angled at C. Let The side AC A BC = .opposite to angle is called ‘opposite side’. The side BC is called ‘adjacent side’. Theside AB is hypotenuse.

opposite side AC The ratio is defined as ‘sine of ’& written as sin .ie hypotenuse AB

adjacent side BC The ratio is defined as ‘cosine of ’& written as cos . ie hypotenuse AB

opposite side AC 90ºThe ratio is defined as ‘tangent of ’& written as tan . ie BC adjacent side BC sin è tanIt can be seen by the above definition that = ècos è

1 AB The reciprocal of sin ie is defined as ‘cosecant of ’and written as cosec .ie sin è AC

1 AB The reciprocal of cos ie is defined as ‘secant of ’and written as sec . ie cos è BC BC The Reciprocal of tan ie is defined as ‘cotangent of ’and written as cot .AC

Identities:- (1) sin + cos = 1 2 2

(2) sec = 1 + tan2 2 (3) cosec = 1 + cot 2 2

Proof:- From the right angled triangle ABC

AC + BC = AB 2 2 2

AC BC 2 2

ie sin + cos = 12 divide by AB ,we have + = 1 2 22 2AB AB 2 2 AC AB

ie tan + 1 = sec BC , + =2 divide by we have 1 2 2

BC BC 2 2 2 2 BC AB

ie 1 + cot = cosecAC , + =2divide by then 1 2 2

AC AC 2 2

Trigonometric ratios of 30° & 60° Consider an equilateral triangle of side 2 units AB = BC = AC = 2. , then BD = DC = 1. Then in theDraw AD to BC r triangle ABD , , further AD = AB – BD = 4 – 1 = 3, AD = 3A DB = 90 ° , A BD = 60 ° & B AD = 30 ° 2 2 2

2 KSOU Complex Trigonometry

In the triangle ABD A AD 3

sin 60° = = 30ºtake A BD AB 2 BD 1

cos 60° = =AB 2 2 2

3 AD 3 tan60° = =BD 1

2 1 90º60ºalso cosec 60 ° = , sec 60 ° = 2 & cot 60 ° =3 3 BDC 11

BD 1take 2B AD sin 30 ° = =

AB AD 3 cos 30° = =AB 2 BD 1

tan 30 ° = =AD 3

2also cosec 30 ° = 2 , sec 3 0 ° = & cot 3 0 ° = 3

3 From the above results, it can seen that sin60º = cos30º, cos60º = sin30º and tan60º = cot30º.

Trigonometric ratios of 45 º

A Consider a right-angled isosceles triangle ABC where A C B = 90 ° A B C = 45° = B AC

Let AC = BC = 1 unit then units AB 2 = 2

1 AC 1\ sin 45 ° = =

AB 2

BC 1 cos 45 ° = = B C 1AB 2

AC 1tan 45 ° = = = 1

BC 1

also, cosec45 ° = 2 , sec 45 ° = 2 and cot 45 ° = 1

Note :- Trigonometric ratios of 30º, 45º and 60º are called ‘Standard Trigonometric’ ratios which are always useful, hence these values have to be always remembered.

Trigonometric ratios of any angle (from 0 ºto 360 º)

Let XOX' & YOY' be co-ordinate axes where O is the origin. Consider a circle of radius r with centre O . Let P be any point on the cirlce whose co-ordinates are ( x , y ).

Draw PM perpendicular to OX .

BTC11

Then OM = x & PM = y Y Let M O P = è

y x y sin è = , cos è = , tan è =r r x (–, +) (+, +) P (x , y )

r r x y also cosec è = , sec è = , cot è =x y x y X' X O M

When ,M O P satisfy 0º < < 90º the print P will be in first quadrant(–, –) (+, –) of the circle, when it satisfy 90º < < 180º, P will in second quadrant,

when 180º < < 270º, P will be in third quadrant & finally when 270º < < 360º the point P will be in fourth quadrant, because of thesepositions, signs of the Trigonometric ratios changes.

Y' In the I quadrant bothx & y are +ve and r is always +ve.

Therefore sin , cos & tan are +ve, their reciprocals are also +ve.

In the II quadrantx is –ve, y is +ve

Therefore sin & cosec are +ve cos , tan , sec & cot are –ve. In the III quadrant both x & y are –ve

Therefore, tan & cot are +ve and sin , cos , cosec & sec are –ve.

In the IV quadrantx is +ve & y is –ve Therefore, cos & sec are +ve and sin , tan , cosec & cot are –ve.

Note :- The signs of the trigonometric ratios can be easily remembered with the help of the following diagram

Sine is +ve All are +ve SA

in shortTC

tan is +ve cos is +ve

P Trigonometric ratios of angles 0 º, 90 º, 180 º, 270 ºand 360 º(can be called border angles). y

x O M

Let be an angle whose measure is very close to zero (as in fig. 1)M O P As 0, y 0 , x r

0\ sin 0° = = 0

r fig.1


P r

cos 0 ° = = 1 r 0 y tan 0 ° = = 0r

x O M If ,M O P is very close to 90º as in fig. 2.

As 9 0°, x 0, y r r

\ sin 90° = =1 r fig.20

cos 90° = = 0 r r tan 90 ° = = 8

P 0 y

If is very close to 180º as in fig. 3 x M O As 18 0 °, x - r , y 0

0 \ sin 180 ° = = 0

r - r fig.3cos 180 ° = = - 1

r 0

tan 180 ° = = 0 r

If is very close to 270º as in fig. 4 M

As 270° , 0x , y - r x O - r y \ sin 270 ° = = - 1 r r 0

cos 270 ° = = 0r P fig.4- r

tan 270 ° = = -80

If is very close to 360º as in fig. 5 When 360 ° , x r , y 0

x M 0 \ sin 360 ° = = 0 y r r

P r cos 360 ° = = 1

r 0

tan 360 ° = = 0 fig.5r

BTC11

Thus we have the following

sin 0 ° = 0, cos 0° = 1, tan 0° = 0

sin 90 ° = 1, cos 90 ° = 0 , tan 90 ° = 8

sin 180 ° = 0 , cos 180 ° = - 1, tan 180 ° = 0sin 270 ° = - 1, cos 270 ° = 0 , tan270 ° = -8sin 360 ° = 0, cos 360 ° = 1 , tan 360 ° = 0

Note :- From the above derivations it can be seen that the values of sin & cos always be between –1 & 1. Whereas thevalue of tan will be between & and hence the graphs of the trigonometric functions -8 8

are as follows.y = sin x , y = cos x & y = tan x y = sin x

O x 90° 270°180° 360°

fig.6

y = cos x

O x 90° 270° 180° 360°

fig.7

y = tan x

O x 90° 270°180° 360°

fig.8


Rules for allied angles sin( 90 ° - ) = cos sin( 270 ° - ) = - cos

cos( 90 ° - ) = sin cos( 270 ° - ) = - sin tan( 90 ° - ) = cot tan( 270 ° - ) = cot

sin( 90 ° + ) = cos sin( 270 ° + ) = - cos

cos( 90 ° + ) = - sin cos( 270 ° + ) = sin

tan( 90° + ) = - cot tan( 270 ° + ) = - cot

sin(180 ° - ) = sin sin( 360 ° - ) = - sin

cos( 180 ° - ) = - cos cos( 360 ° - ) = cos

tan( 180 ° - ) = - tan tan( 360 ° - ) = - tan

sin(180 ° + ) = - sin sin( 360 ° + ) = sincos( 180 ° + ) = - cos cos( 360 ° + ) = cos

tan( 180 ° + ) = tan tan( 360 ° + ) = tan

Using the Trigonometric ratios of standard angles 30°, 45° & 60° and using the above rules for allied angles, we canfind the trigonometric ratios of other angles as follows.

3Eg.1 2sin 120 ° = sin(90° + 30°) = cos 30° =

3or 2sin120° = sin(180° - 60° ) = sin 60° =

3Eg.2 2cos 150° = cos(90° + 60°) = - sin 60° = -

3 or 2cos 150° = cos(180 ° - 30° ) = - cos 30° = -

1Eg.3 sin 315 ° = sin( 360 ° - 45° ) = - sin 45° = -

2 1

or sin 315 ° = sin( 270 ° + 45° ) = - cos 45° = -2

B Radian Measure r r Consider a circle of radius r with centre O . Let AB be a arc such that arc AB = r .

1 radian A O r Measure of the is called a 'radian' denoted as 1 .A O B C

To prove that radian is a Constant angle

Consider a circle of radius r with centre O . Let arc AB = r so that is 1 radian.A O B Let C be a point on the circle such that A O C = 90 ° .

Since the angle subtended at the centre of a circle is proportional to the corresponding arc.

MCA 11 - Mathematics SVT 7

C A O B arc AB =

B arc AC A O C 1 radian r 2 r = = r p 1 90 ° p (2 r )

4A O r 1 1= p Q AC = Circumfere nce ( 2 r )

4 4

p radians = 180 °

p ie = 180 ° imp. result.

p p p p = 60 °, = 30 ° .= 90 °, = 45 ° ,

2 4 3 6 Using the relation p = 180° measurement of an angle in degrees can be converted to radians and vice-versa.

Eg. (1) Qn. : Convert 40° to radians

p p 2Solution : 40 ° = 40 × = radius

180 9 p 2Eg. (2) Qn. : Convert radians in to degrees3

p 2 180Solution : × = 120 °

p 3

Some Problems

1. Show that2 2 2 (1 + cot A ) + (1 - cot A ) = 2 cosec A

2 2Solution : LHS = 1 + cot 2 A + 2 cot A + 1 + cot A - cot A 2= 2 cot+ 2 A

= 2(1 + cot A )2

2= 2cosec A = RHS.

2. Show that

1 + sin 4A 1 - sin A - = sec A tan A

1 - sin A 1 + sin A (1 + sin A ) - (1 - sin A )2 2

(taking LCM)Solution : LHS=(1 - sin A )(1 + sin A )

2 2(1+ sin A + 2 sin A ) - (1 + sin A - 2 sin A )=

21 - sin A 2 21 + sin A + 2 sin A - 1 - sin A + 2 sin A

=2cos A

4 sin A 1 sin A = = 4 ·

cos A cos A cos A 2

= 4 sec A · tan A = RHS


2a - 1 3. If sec A + tan A = a , then prove that = sin A a + 1 2

2 2 2a - 1 sec A + tan A + 2 sec A tan A - 1Solution : LHS = =

a + 1 sec A + tan A + 2 sec A tan A + 1 2 2 2

2 2 sec A - 1 + tan A + 2 sec A tan A =

sec A + tan A + 1 + 2 sec A tan A 2 2

22 tan A + 2 sec A tan A = (using the identity)

2 sec A + 2 sec A tan A 2

2 tan A( tan A + sec A) =

2sec A (sec A + tan A )tan A sin A cos A

= = × = sin A = RHSsec A cos A 1

4 sin + cos4. If find the value ofsin 5 = ,

tan - cot

4 then opposite side is 4 hypotenusis 5.Solution : If sin =5

3 4cos = & tan =adjacent side = 25 - 16 = 9 = 3

5 3

4 3 7 +

sin + cos 125 5 5= = = 4 3 7 tan - cot 5-3 4 12

p 3 5 cos + 8 tan5. If p tan = - , < < find 4 2 8sec - 3 cosec

Solution : Since lies in II Quadrant sine is +ve cosine & tangant are –ve.

3 4 sin = and cos = -

5 5 - 4 - 3

5 + 85 cos + 8 tan 5 4

=8sec - 3 cosec - 5 5

8 - 34 3

- 4 - 6 10 2= = =

- 10 - 5 15 3

6. Prove that

tan(180 sec° + ) sec( 180 ° - ) cosec (90 ° + )= 2

sec( 360 ° - ) cot( 90 ° + ) sin( 90 ° - )


tan (- sec ) sec= = 2Solution sec: LHS

sec ( - tan ) cos

7. Prove that

p -p p sin( + ) cos( 2 - ) cot2

= sinp p 3+ +tan cot sin( - )2 2

(- sin ) cos tanSolution : LHS =

( - cot ( - tan )( - sin )

= cos tan

sin= cos × = sin = RHS

cos

Addition Formulae

sin( A + B ) = sin A cos B + cos A sin B cos( A + B ) = cos A cos B - sin A sin B

tan A + tan B tan( A + B ) =

1 - tan A tan B replacing B by –B

sin( A - B ) = sin A cos B - cos A sin B cos( A - B ) = cos A cos B + sin A sin B

tan A - tan B tan( A - B ) =

1 + tan A tan B Using the above formulae we can find the trigonometric ratios of 15°, 75°, 105° etc.

sin 75 ° = sin( 45 ° + 30 °)

= ° ° + ° °sin 45 sincos 30 cos 45 30

1 3 1 1 3 +1 = × + × =2 2 22 2 2

cos 75 ° = cos( 45° + 30 ° )= cos 45 sin° cos 30 ° - sin 45 ° 30 °

1 3 1 1 3 - 1 = × - × =2 2 22 2 2

sin75° 3 +1tan75° = =cos 75° 3 - 1

sin 15 ° = sin( 45 ° - 30 ° )= sin 45 sin° cos 30 ° - cos 45 ° 30 °


1 3 1 1 3 - 1 = × - × =2 2 2 2 2 2

cos 15 ° = cos( 45° - 30 °)

= cos 45 sin° cos 30 ° + sin 45 ° 30 °

1 3 1 1 3 +1 = × + × =2 2 2 2 2 2

sin15° 3 - 1 tan15° = =

cos 15° 3 +1 sin 105 ° = sin( 60 ° + 45 °)

= sin 60 sin° cos 45 ° + cos 60 ° 45 ° 3 1 1 1 3 +1 = × + × =2 2 2 2 2 2

cos 105 ° = cos( 60 ° + 45 ° )= cos 60 sin° sin 45 ° - sin 60 ° 45 °

1 1 3 1 1 - 3= × - × =2 22 2 2 2 3 +1

tan105 ° =1- 3

Alternate method sin 15 cos° = sin( 90 ° - 75 °) = 75 °

3 - 1=2 2

cos 15 sin° = cos( 90 ° - 75 °) = 75 ° 3 +1

=2 2

sin 105 sin° = sin(180 ° - 75 °) = 75° 3 +1=

2 2

cos 105 cos ° = cos(180 ° - 75 ° ) = - 75 ° ()- 3 - 1 1 - 3

= =2 2 2 2


To find sin2 , cos2 & tan2 .

sin( A + B ) = sin A cos B + cos A sin B put A = = B sin 2 = sin cos + cos sin

= 2 sin cos cos( A + B ) = cos A cos B - sin A sin B put A = B =cos 2 = cos cos - sin sin

2 2= cos - sin

using cossin = 1-2 2

cos 2 = 2 cos - 12

2 2also sinusing cos = 1 -

cos 2 sin = 1 - 2 2

tan A + tan B tan( A + B ) =

1 - tan A tan B put A = = B

tan + tantan 2 =

1 - tan tan

2 tan=

1 - tan2

To find sin3 , cos3 & tan3 .

sin 3 = sin( 2 + )

= sin 2 2 cos + cos sin

2= 2sin cos cos + (1 - 2 sin )sin

2 3= 2 sinsin (1- sin ) + sin - 2

3 3= 2 sin sin - 2 sin + sin - 2

3= 3 sin sin - 4

cos 3 = cos( 2 + )

= cos 22 cos - sin sin

= (2 cos - 1) cos - 2sin cos2 2

3 2= 2cos - cos - 2 (1 - cos )cos

3 3= 2 coscos - cos - 2 cos + 2

3 = 4 cos 3 - cos


tan 3 = tan( 2 + )

tan 2 + tan=

1 - tan 2 tan

2 tan+ tan

1- tan2=

2 tan1 - × tan2 1- tan

22 tan + tan (1 - tan )2 1 - tan=

2 21 - tan - 2 tan2 (1 - tan )

tan + tan - tan32=

21 - 3 tan

tan - tan3 3=

2 1 - 3 tan

Thus we have,

sin 2 2 = sin cos

2 2 2 2 cos 2 sin = cos - = 2 sin cos - 1 = 1 - 2

2 tantan 2 =

21 - tan

3sin 3 sin = 3 sin - 4

3cos 33 = 4 cos - cos

33 tan - tantan 3 =

2 1- 3 tan

Problems

sin 3 A cos 3 A 1. Show that - = 2

sin A cos A 3 33sin 3 A - 4 sin A 4 cos A - cos A

Solution : LHS = -sin A cos A

2 2= 3 - 4sin A - 4 cos A + 3) = 6 - 4(sin A + cos A 2 2

= 6 - 4 = 2 = RHS

22. If find sin2 A & cos3 A .cos A =3

2 9 - 4 5 Solution : Given cos A = , sin A = =

3 3 3


5 2 4 5 sin 2 2 A = sin A cos A = 2 × × =

3 3 9 3cos 33 A = 4 cos A - cos A

8 2 32 32 - 54 - 22= 4 × - 3 × = - 2 = =

27 3 27 27 27

4 5 - 22& cos 3A =sin 2 A =

9 27

3. Prove that

1 + sin 2= tan (45 ° + )2

1 - sin 2

1 + 2sin cosSolution : LHS =

1 - 2 sin cos

2(cos + sin )=

(cos - sin )2 2 cos + sin

= (dividing Nr & Dr inside the bracket by cos )cos - sin

21 + tan=

1 - tan

= tan( 45 ° + ) = RHS

1 - cos4. Prove that and hence prove that 32 tan15° = 2 -= tan1 + cos 2

- 21 - 1 2 sin1 - cos 2 using cos2 A formulaSolution : =1 + cos - 21 + 2 cos 1

2

2 21 - 1 + 2 sin 2 sin2 2= = = tan2

21 + 2 cos - 1 2 cos2 2 2 2

1 - cos2 = tan

1 + cos 2

31 -

1 - cos 30 ° 2 - 32 put = 30 ° . tan 15 ° = = = 2 1 + cos 30 ° 3 2 + 3

1 + 2

()()()() 2 -()() 2

- 3 2 - 3 2 3 2= = = 2 - 3

4 - 32 + 3 2 - 3

tan = 2 - 3 2


Complex Number

Definition :A number of the form is defined as a Complex Number and usually denotedx + iy where x R , y R & i = - 1

as Z. x is called Real part & y is called Imaginary part. x – iy is called Conjugate. Complex number denoted as . Z Aor z complex number can be represented by a point on a plane by taking real part onx -axis & imaginary part ony -axis. The planeon which complex numbers are represented is called a Complex Plane. For every point in a plane there is a complex number& for every complex number there is a point in the plane. In a complex x -axis is called Real axis & y -axis Imaginary axis.

Properties

(1) Equality. Two complex numbers are said to be equal ifz = x + iy , z = x + iy x = x , y = y 1 1 1 2 2 2 1 2 1 2

(2) Addition. If z = x + iy , z = x + iy 1 1 1 2 2 2

z + z = ( x + x ) + i ( y + y )1 2 1 2 1 2

(3) Subtraction. z - z = ( x - x ) + i ( y - y )1 2 1 2 1 2

(4) Multiplication. z z = ( x + iy )( x + iy )1 2 1 1 2 2

2 = x x + ix y + ix y + i y y 1 2 2 1 1 2 1 2

2= x x - y y + i (x y + x y ) Q i = - 11 2 1 2 1 2 2 1

Note : i = - 1, i = - 1 , i = - i , i = 1 etc2 3 4

z ( x + iy )=1 1 1 (5) Division. z ( x + iy )

2 2 2

multiply numerator & denominator by the conjugate of x + iy ie x – iy , then 2 2 2 2

z ( x + iy )( x - iy )=1 1 1 2 2

z ( x + iy )( x - iy )2 2 2 2 2

x x + y y + i ( x y - x y )= 1 2 1 2 2 1 1 2

x + y 2 22 2

x x + y y x y - x y = + i 1 2 1 2 2 1 1 2

2 2 2 2 x + y x + y 2 2 2 2

which is a complex number.

Note :- Product of a complex number with its conjugate is always a positive real number.2 2 ie z z = (x + iy )( x - iy ) = x + y & z + z = 2 x , z - z = 2 iy

z + z z - z y x = & y =2 2i

Polar form the complex number

z = x + iy is called the Cartesian form.P (x ,y )

Let P (x , y ) be any point in the plane which represents a complex number. r y r Draw PM to x -axis & join PM . Then OM = x , MP = y . Let &M OP =

x OP = r. O x M From the triangle OPM ,


OM x cos = = x = r cos

OP y PM y

sin = = y = r sinOP r

z = x + iy = r cos + ir sin = r (cos + i sin )

This form of the complex number is called 'Polar form'. Where r is called Modulus & is called argument which aregiven by

y 2 2 - 1r = x + y & = tanx

r is always positive and argument varies from 0° to 360°. The value of argument satisfying – p < = p is definedas amplitude which is unique for a complex number.

Thus we have

2 2 Mod. z = z = r = x + y y and amp. z = where –p < = p - 1 arg z = tanx

x y while finding the amplitude of a complex number, we have to find satisfying ie cos = and sin =

y x

Examples

2 + 3 i 1. Express in the form x + iy

1 + i 22+ 3 i - 2i - 3i 2 + 3 i ( 2 + 3i )(1 - i )

=Solution : = 1 +1 1 + i (1 + i )(1 - i )

2 + i + 3 5 + i 5 1= = = + i

2 2 2 2

(1 + i )2

2. Express in the form x + iy 3 - i 2 2(1 + i ) 1 + i + 2 i 2i ( 13 + i )Solution : = 2= (using i = - )

3 - i (3 - i ) (3 - i )( 3 + i )26 i + 2 i - 2 + 6 i - 2 6 i 1 3i

= = = + = - +9 + 1 10 10 10 5 5

3. Find the modulus and amplitude of 1 + i .

Solution : z = r = x + y = 1 + 1 = 22 2

p x 1 y 1 cos = = and sin = = = 45 ° =

r 2 r 42

p Modulus is 2 and amplitude is

4


4. Find the modulus & amplitude of - 3 + i

Solution : z = 3 + 1 = 4 = 2

- 3 cos = 5p 2 =61 sin =

2 p 5

Modulus is 2 & amplitude is 6

5. Find the modulus & amplitude of 3 - 1 - i Solution : z = 1 + 3 = 4 = 2

1cos = -2 2p

= -3 3 sin = -

2

p 2 Modulus is 2 & amplitude is -

3

6. Find the modulus & amplitude of 1 – i .

Solution : z = 1 + 1 = 2

x 1cos = =

p r 2= -

y 1 4 sin = = -

r 2

p Modulus is 2 & amplitude is -

4

1a ß a ß 2 2 2 2If + i = ,then prove that ( + )( a + b ) = 1 7.a + ib

1 (a - ib )a ß Solution : + i = =

a + ib ( a + ib )( a - ib )a - ib a b

= = - i 2 2 2 2 2 2a + b a + b a + b

Equating real and imaginary parts separately

a - b a ß = and = a + b a + b 2 2 2 2


Squaring and adding

2 2 2 2a b (a + b ) 1a + ß = ()() + = () =2 2

a + b 2 2 2 2 22 2 2 2 2 2 a + b a + b a + b cross - multiplyin g 2 2 2 2(a + ß )(a + b ) = 1

DeMoivre's Theorem

Statement : If n is a +ve or –ve integer, then (cos + i sin )n = cos n + i sin n If n is a +ve or –ve fraction, one of the values of is(cos + i sin )n cos n + i sin n .

Proof : Case (i) when n is a +ve integer proof by Mathematical Induction.

When n = 1,1(cos + i sin ) = cos + i sin = cos 11. + i sin .

result is true for n = 1. Let us assume that the result is true for n = m

m ie (cos + i sin ) = cos m + i sin m multiply both sides by cos + i sin

(cos + i sin ) (cos + i sin )m = (cos m + i sin m )(cos + i sin )

+ie (cos + i sin )m 1

= cos m cos + i sin m cos + i cos m sin - sin m sin = cos m cos - sin m sin + i [sin m cos + cos m sin ]

= cos( 1 m + 1) + i sin( m + )

the result is true for n = m + 1. Thus if the result is true for n = m then it is true for n = m + 1. ie If it is true for one integer it is true for next integer, hence by Induction the result is true for all +ve integers.

p Case (ii) When where p & q are +ve integers.n =

q q + p p Consider cos i sin

q q

p p = (cos + i sin )p = cos · q + i sin · q = cos q + i sin p q q

q + p p p ie (cos + i sin ) = cos i sinq q

p p p taking q th roots on both sides. One of the values of (cos + i sin ) = cos + i sinq

q q

Case (iii) when n is –ve integer or –ve fraction. Let n = –m where m is a +ve integer or +ve fraction (cos + i sin ) = (cos + i sin )n - m


1 1= =

cos m + i sin m m (cos + i sin )

(cos m - i sin m ) cos m - i sin m = =(cos m + i sin m )(cos m - i sin m ) 2 2cos m + sin m cos m - i sin m

= = cos( - m ) + i sin( - m ) = cos n + i sin n .1

Important Results1

(i) If then = cos - i sinx = cos + i sin x

1 1x + = 2 cos and x - = 2 i sin .

x x (ii) If x = cos + i sin

x = (cos + i sin )n n = cos n + i sin n 1 1&= cos - i sin = (cos - i sin )n x x n

= cos n - i sin n 1 1&n n x + = 2 cos n x - = 2 i sin n

x x n n Note :- For convenience can be written ascos + i sin cis .

Roots of a complex number Let z = x + iy , express the complex number in the polar form.

ie z = r (cos + i sin )

[] p p = r cos( 2k + ) + i sin( 2 k + ) where k I [] p p 1 1 1 z = r cos( 2 )k + ) + i sin( 2k +n n n

p p 2k + ( 2k )+ = r cos + i sin1

n n n

where k = 0, 1, 2, ......... n – 1. Let us denote the n th roots of the complex number by z , z , z , .............. z 0 1 2 n – 1

Then, +

k = 0, z = r cos i sin = r cisfor 1 1n n

0 n n n p p p 2 + 2 2 ++k = 1 , z = r cos + i sin = r cis1 1

n n 1 n n n

p p p 4 + 4 4 + + k = 2, z = r cos + i sin = r cis1 1n n

2 n n n p p p 2( n - 1 ) + 2(n - 1) 2 (n - 1 ) ++

k = n - 1 , z = r cos + i sin = r cis1 1n n

n - 1 n n n the above n values gives n th roots of z = x + iy


Note :- If k = n , n + 1, n + 2 etc. The values will repeat. Hence these will be only n values of which are distinct. 1z n Using the polar form of the complex number we can plot the nth roots of the complex in the following way.

z 2Draw a circle of radius whose centre is O . Mark a point on the circle and taker 1n

z C 1 OA as intial line. Take a point B such that . Then B represent z . Take a A OB =

0n z p 2 +

point C such that then C represent z , like this all the n th roots can 0A OC = B n 1

be represented. This diagram is called 'Argand Diagram'. / n O A 1 r n

Problems :

z 2 - 3cos 5 - i sin 5 ) (cos 7 + i sin 7 )(1 ) Simplify n

9 5(cos 4 - i sin 4 ) (cos + i sin )

Solution : G.E. (given expression)

(cos + i sin ) (cos + i sin )- 1 0 - 2 1

= - 3 6 5(cos + i sin ) (cos + i sin )

- - + -= (cos + i sin ) = (cos + i sin ) = 110 21 36 5 0

- 5 - 3 (cos 2 + i sin 2 ) (cos 3 - i sin 3 )( 2) Simplify

4 10(cos 4 - i sin 4 ) (cos + i sin )

(cos + i sin ) (cos + i sin )- 10 9

Solution : G.E. =- 16 10(cos + i sin ) (cos + i sin )

- + + -= (cos + i sin ) = (cos + i sin )10 9 16 10 5 = cos 5 sin+ i 5

1 1f (3 ) If 2 cos = x + & 2 cos = y +

x y Show that

m n 1 x y f f m m &(i) x y + = 2 cos( m + n ) (ii) + = 2 cos( m - n )m m n m x y y x

21 x + 1Solution : 2 cos = x + =

x x x + 1 = 2cos · x 0 ie x - 2 cos x +1 =2 2

2cos ± (- 4)(1 - cos )22 cos ± 4 cos - 422 =x =

2

22 cos ± - 4 sin 2 cos ± 2 i sin= = = cos ± i sin

2 2

1If x = cos + i sin ,then = cos - i sin

x 1 1f f f f f Similarly if 2 cos = y + , y = cos + i sin & = cos - i siny y


(i) = (cos + sin )m m x i = cos m + i sin m y = (cos f + i sin f ) f f n n = cos n + i sin n x y = (cos m + i sin m )(cos n f + i sinn f )m n

f f (1) = cos( m + n ) + i sin( m + n )

1 f f = 2 cos( m + n ) - i sin( m + n ) (2) m n x y adding (1) & (2)

1 f x y + = 2 cos( m + n )m n x y m n

(ii) x = cos m + i sin m m 1 f f = cos n = i sin n

n y x y m n

f f & f f = cos( m - n ) + i sin( m - n ) = cos( m - n ) - i sin( m - n )n m y x

adding m n x y f + = 2 cos( m - n )n m y x

(4) Prove thatm b

2 2 - 1(a + ib ) + (a - ib ) = 2 tan(a + b ) cosm m m n n 2 n

n a b

2 2 - 1Solution : Let a + ib = r (cos + i sin ) where r = a + b & = tana

m m + (1) ( + )m = m (cos + sin )m a ib r i = r cos i sinm n n n n n n

a - ib = r (cos - i sin )

m m -(a - ib ) = r cos i sin m m (2) n n n n

adding (1) & (2) m

( 2a + ib ) + (a - ib ) = r cos m m m n n n

n m m b + 2 n

(a + ib )m + (a - ib )m = a b cos tan2 2 - 1 2n n n a

b b Q r = a + b & tan = = tan2 2 - 1

a a () m b m 2 2 - 1 (a + ib ) + (a - ib ) = 2 a + b 2 cos tanm m 2 n n n

n a


(5) Find the cube roots if 1 + i and represent them on Argand diagram. Solution ): Let z = 1+ i = r (cos + i sin say

r cos = 1 , r sin = 1Squaring& adding r (cos + sin ) = 1 +1 2 2 2

r = 2 r = 2 2

1 cos =

p 2=

1 4 sin =

2

p p + z = 2 cos i sin

4 4

p p + + p p = 2 cos 2 k + i sin 2 k for k I 4 4

p + p z = 1 + i = 2 cis 2 k 4

() 18k p p + p p 3 8k +1

z = (1 + i ) = 2 cis 1 1 3 1= 2 cis for k = 0, 1, 23 3 6

4 12

p 1 1when k = 0, z = 2 cis = 2 cis15 ° 6 60 12

p p 9 31 1 1k = 1 , z = 2 cis = 2 cis = 2 cis 135 °6 6 61 12 4

p p 17 171 1 1k = 2, z = 2 cis k = 2 , z = 2 cis = 2 cis255 ° 6 6 62 212 12

To represent them on Argand diagram. Draw a circle of radius with centre O . Let OA be the initial line.1 C 2 6

Take a point B on the circle such that , A OB = 15 ° take a point C on the circleB 135°such that & take a point D on the circle such that thenA OC = 135 ° A OD = 255 °

the points B , C , D represent z , z , z . A 15°255° O 0 1 2 12 6

1- 1 3 4

i (6) Find all the values of & find their continued product.2 2

1 3 Solution : Let z = - i = r (cos + i sin ) D 2 2

1 3 r cos = , r sin = -

2 2Squaring & adding

1 3 4r = + = = 1 2

4 4 4


1cos =2 p

= -3 3

sin = -2

p p - -z = 1 cos + i sin

3 3

p - p z = cis 2k for k I 3

1 - 1 1

1 3 - p - 6k p p 4 4 4

z = i = cis 2 k p = cis 14

2 2 3 3

p p 6 k -= cis , for k = 0 , 1 , 2 3

12

p p 5-for k = 0 , z = cis k = 1, z = cis0 112 12

p p 11 17k = 2, z = cis k = 3 , z = cis

2 312 12

Their continued product

p p p p 5 11 17-= ci s cis cis cis12 12 12 12

p p p p p p + 5 11 17 32 8= cis - + + = cis = cis

12 12 12 12 12 3 p p p p 1 3+ 2 2 2 2 p = - + i = cis 2 = cis = cos + i sin

2 23 3 3 3

Expansion of sin ( n ) and cos ( n ) in powers of sin & cos

n Consider (cos + i sin ) = cos n + i sin n n Expand (cos n + i sin n )

Using Binomial Theorem

( + ) = + + + ........ +n n n - 1 n - 2 2 n x a x nC x a nC x a nC a 1 2 n Equating real and imaginary parts separately we get the expressions of cos n & sin n

Eg. (1) Express in powers of & .cos 5 sin + i 5 sin cos

5Solution ): cos 5 + i sin 5 = (cos + i sin = cos + C cos · i sin + C cos (i sin ) + C cos (i sin ) + C cos ( i sin ) + C ( i sin )5 4 3 2 2 3 4 55 5 5 5 51 2 3 4 5

5 4 3 2 2 3 4 5= cos + 5 i cos sin - 10 cos sin - 10 i cos sin + 5 cos sin + i sin

= cos - 10 cos sin + 5 cos sin + i [5cos sin - 10 cos sin + sin ]5 3 2 4 4 2 3 5


Equating real & imaginary parts separately

and 5 3 2 4cos 5 sin= cos - 10 cos sin + 5 cos

4 2 3 5sin 5 sin = 5 cos sin - 10 cos sin +

Eq. (2) Express cos 6 & sin 6 in powers of cos + i sincos +i sin = (cos + i sin 6Solution ): 6 6 = cos + 6C cos (i sin ) + 6 C cos (i sin ) + 6 C cos (i sin ) + 6 C cos ( i sin )6 5 4 2 3 3 2 4

1 2 3 4 + 6 ) C cos ( i sin ) + 6C (i sin5 6

5 6 = cos + i cos sin - cos sin - i cos sin + cos sin + i cos sin - sin 6 5 4 2 3 3 2 4 5 66 15 20 15 6

6 4 2 2 4 6 5 3 3 5= cos - 15 cos sin + 15 cos sin - sin + i [6cos sin - 20 cos sin + 6cos sin ]


cos 6 = cos - 15 cos sin + 15 cos sin - sin and6 4 2 2 4 6

5 3 3 5 sin 6 sin = 6 cos sin - 20 cos sin + 6 cos

Addtion formulae for any number of anglesWe have,

cis cis cis ......... cis = cis( + + + .......... ... + )1 2 3 n 1 2 3 n

Now cis cis cis ......... cis1 2 3 n = cos (1 + i tan ) cos (1 + i tan )......... cos (1 + i tan )

1 1 2 2 n n = cos cos ......... cos (1 + i tan )(1 + i tan )......... .....( 1 + i tan )

1 2 n 1 2 n ( + + + .......... .. + )ie cis 1 2 3 n

2 3= cos cos cos .......... cos (1+ iS + i S + i S + .......... ....)1 2 3 n 1 2 3 where S = tan

1 1

S = tan tan2 1 2

S = tan tan tan3 1 2 3

.........................................

cis ( è + è + + è ).......... .. 1 2 n

= cos cos .......... cos ( + iS - S - iS + S .......... ....)11 2 n 1 2 3 4

= cos cos .......... cos ( - S + S .......... ....) + i cos cos .......... cos ( S - S .......... ..)11 2 n 2 4 1 2 n 1 3


cos( + + + .......... ... + ) = cos cos .......... cos (1 - S + S .......)1 2 3 n 1 2 n 2 4

and

sin( + + + .......... ... + ) = cos cos .......... cos (S - S .......)1 2 3 n 1 2 n 1 3

S - S .......... ....... è + è + è + .......... ... + è = 1 3tan( )

1 2 3 n 1 - S + S .......... ....... 2 4


Exercise

1. Find the continued product of fourth roots of unity.

2 32. If ' ' isthe cube root of unity, then findthe value of (4 - 3 - 3 ) .

8sin(p / 8) + i cos(p / 8)3. Findthe value of sin(p / 8) - i cos(p / 8 )

4. Write the conjugate of the multiplica tive inverse of the complex number ( sin + i cos ).

5. If z = x + iy then wha t does z + 1 =1 represents? 26. Simplify (sin 2 x + i cos 2 x ) using De Moivre's Theorem.

7. Find the realvalues of x and y which satisfy th e equation ( 02 x - y ) + i (3x + 2 y + 1 ) = . 8. Find the realvalues of x and y which satisfy th e equation ( 02 + i ) x + (i - 3) y - 4 = .

9. If 'n 'is anyinteger, find i + i + i + i . - n - n +1 - n + 2 - n + 3

10. What isthe multiplica tive inverse of i ?101

3 2 3 11. If ' ' in the cube root of '1', findthe value of (1 + ) - (1+ ) .

2 4 2 4 . + + + .12 Simplify

i i i i 2 3 4 p p + 3 3

13 . Express 2 cos i sin in the Cartesian form.4 4

p p - 1+ cos i sin .14. Find the real part of 1 + 5 5

a + ib a + b 2 2 2 2 215 . If x + iy = , prove that ( x + y ) = .

c + id c + d 2 2 n 1 + i

16 . Find the least positive integer 'n 'for which = 1 .1 - i

2 217 . If z = x + iy and z + 6 = 2z + 3 , show that x + y = 9 .2 218 . If z = a + ib and z - 2 = 2 z - 1 , show that a + b = 1.

p z - 12 219 . If z = x + iy be such that amp = , show that x + y - 2 y = 1 .

z + 1 4 20 . Express 1 - i in the polar form.

3 - i 21. Express in the polar form.

2 22 + i

22. Find the modulus and amplitude of .3 - i

(sin + i cos )3 23 ). Simplify

(cos a + i sina 2


(cos 7 + i sin 7 ) (cos 3 + i sin 3 )- 2 424. Simplify

- 4 (cos + i sin )

25. If is a cube root of unity, showthat (1 - + )(1 - + )(1- + )(1 - + ) =16.2 2 4 4 8 8 16

1 526. If x = cis ,then find x + .x 5

27. If i = - 1 , then find i + i + i + L + (2n )terms. 2 2 4 6

a ß x - y -a ß 28. If x = cis and y = cis , prove that = i tanx + y 2

p n a ß a ß 2 n n n + 129 . If & arethe roots of x - 2 x + 4 = 0 ,then prove that + = 2 cos .3

30. Provethe following()() 6 6 i) (1 + i ) + (1 - i ) = - 4 ii) 1 - i 2 + 1 + 2 = 46.3 3

1 1 n n 31 . If Z = cos + i sin , show that Z + = 2 cos n and Z - = 2i sin n .Z Z n n

a ß 32 . If x = cis , y = cis , prove that 1 1 a ß a ß i) xy + = 2 cos( + ) ii) xy - = 2 i sin( + ).

xy xy a ß 33 . If x = cis , y = cis , prove that

1 1a ß a ß 2 3 2 3i) x y + = 2 cos( 2 + 3 ) ii) x y - = 2 i sin( 2 + 3 ).

x y x y 2 3 2 3

x - 12 n 34 . If x = cis , prove that = i tan n .

2 n x + 1

8 8 9 835. Provethat (1+ cos + i sin ) + (1 + cos - i sin ) = 2 ·cos(4 ) ·cos ( / 2).

36. If cos a + 2cos ß + 3 cos = 0 = sin a + 2sin ß + 3 sin , provethati) cos 3 a + 8 cos 3 ß + 27 cos 3 = 18 cos(a + ß + ) ii ) sin 3a + 8 sin3ß + 27 sin 3 = 18 sin(a + ß + ).

1 1 137. If 2 cosa = x + , 2 cos ß = y + and 2 cos = z + , provethat

x y z 1 1i) xyz + = 2 cos(a + ß + ) ii ) xyz - = 2 i sin(a + ß + ).

xyz xyz []n n n - 38. Showthat (a + ib ) - (a - ib ) = 2 i (a + b ) · sin 2n · tan (b / a )2 2 2 2 1

n p p 1 + sin + i cos - n - n 39 . Prove that = cos n + i sin n

1 + sin - i cos 2 2

71 - cos - i sin 740 . Prove that = i tan · cis 7 .1 + cos - i sin 2

a n a 1 + i tan 1+ i tan41 . Prove that = .

a a 1 - i tan 1 - i tan


. Z = (p / ); r = , , L Z · Z · Z L L 8 = i .r 42 If cis 3 1 2 3 prove thatr 1 2 3

43. Findthe cube roots of - 1 + i 3 .

/ 344. Findallthe values of (1+ i ) .2 2 / 345. Findallthe values of (1 - i ) .

46 . Find the fourth roots of 1 + i 3 and represent them in the Argand diagram.

7 47. Solve x - x = 0. 5 48 . 0 Solve x + 1 = .

49. Express sin 7 & cos 7 in powers of sin & cos .

50 . Express cos 8 & sin 8 in powers of cos & sin .

–––––––––––––

MATRIX THEORY

Review of the fundamentals

A rectangular array ofmn elements arranged inm rows & n columns is called a 'Matrix' of a orderm ×n matrices are denotedby capital letters of The English Alphabet.

Examples

a b 1 1 a b Matrix of order 3 ×2 is 2 2

a b 3 3

a a a 1 2 3

b b b Matrix of order 4 ×3 is 1 2 3

c c c 1 2 3

d d d 1 2 3

a b c 1 1 1

a b c Matrix of order 3 ×3 is 2 2 2

a b c 3 3 3

Note :- Elements of Matrices are written in rows and columns with in the bracket ( ) or [ ]. Types of Matrices

(1) Equivalent Matrices : Two matrices are said to be equivalent if the order is the same.

(2) Equal Matrices : Two matrices are said to be equal if the corresponding elements are equal. (3) Rectangular & Square Matrices : A matrix of order m ×n is said to be rectangular if m n , square if m = n .(4) Row Matrix : A matrix having only one row is called Row Matrix.

(5) Column Matrix : A matrix having only one column is called Column Matrix. (6) Null Matrix or Zero Matrix : A matrix in which all the elements are zeros is called Null Matrix or Zero Matrix

denoted as O . [English alphabet O not zero where as elements are zeros]

(7) Diagonal Matrix : A diagonal matrix is a square matrix in which all elements except the elements in the principaldiagonal are zeros.

2 0 0 4 0

Example 0 1 0 0 6

0 0 4 are diagonal matrices of order 2 & 3.

(8) Scalar Matrix : A diagonal matrix in which all the elements in the principal diagonal are same.

28 KSOU Matrix Theory

8 0 0 4 0

Example 0 8 0 0 4

0 0 8 are Scalar Matrices of order 2 & 3.

(9) Unit Matrix or Identity Matrix : A diagonal matrix in which all the elements in the principal diagonal is 1 iscalled Unit Matrix or Identity Matrix denoted by I .

1 0 0 0

1 0 0 1 0 0Example , :

0 1 0 0 1 0

0 0 0 1

are unit matrices of order 2 & 4. (10) Transpose of a Matrix : If A is any matrix then the matrix obtained by interchanging the rows & columns of A is

called 'Transpose of A and it is written as A' or A . T

a b a c e

Example : If A = c d then A ' isb d f

e f

A is of order 3 ×2 but A' is of order 2 ×3.

Matrix addition Two matrices can be added or subtracted if their orders are same.

a b c c d e 1 1 1 1 1 1Example : If A = & B =

a b c c d e 2 2 2 2 2 2

a + c b + d c + e A + B = 1 1 1 1 1 1

a + c b + d c + e 2 2 2 2 2 2

a - c b - d c - e 1 1 1 1 1 1A - B =

a - c b - d c - e 2 2 2 2 2 2

Matrix MultiplicationIf A is a matrix of orderm × p and B is matrix of order p × n , then the productAB is defined and its order is m × n . (ie. for AB to be defined number of columns of A must be same as number of rows of B )

a ß 1 1 a b c A = & B = a ß 1 1 1Example : Let 2 2a b c 2 2 2 a ß 3 3

a a a ß ß ß a + b + c a + b + c 1 1 1 2 1 3 1 1 1 2 1 3then AB =a a a ß ß ß a + b + c a + b + c

2 1 2 2 2 3 2 1 2 2 2 3

which is of order 2 ×2.

Note :- If A is multiplied by A then AA is denoted as A , AAA .... as A etc.2 3


Scalar Multiplication of a Matrix

If A is a matrix of any order andK is a scalar (a constant), thenKA represent a matrix in which every element ofA is multipliedby K .

a b c Ka Kb Kc 1 1 1 1 1 1Example : If A = a b c then KA = Ka Kb Kc 2 2 2 2 2 2

a b c Ka Kb Kc 3 3 3 3 3 3

Symmetric and Skew Symmetric Matrices Let A be a matrix of ordern × n an element in i row and j column can be denoted as aij . Hence a matrix of order n × n can t h th

be denoted as (aij ) or [aij ] where i = 1, 2, ....... n , j = 1, 2, ....... n A matrix of ordern × n is said to be Symmetric if aij = aji and Skew Symmetric if aij = –aji or A is symmetric if A = A T

or A = A' , skew symmetric if A = –A or A = –A' also A + A' is symmetric & A – A' is skew symmetric.T

Note :- In a skew symmetric matrix the elements in principal diagonal are all zeros.

2 3 5 'Example : A = 3 7 6 is symmetricwhere A = A

5 6 8

0 - 2 7B = 2 0 6 is skew symmetricwhere B = - B '

- 7 - 6 0

Determinant A determinant is defined as a mapping (function) from the set of square matrices to the set of real numbers.

A If A is a square matrix its determinant is denoted as .

a b c a b c 1 1 1 1 1 1

Example: Let A = a b c then det. A or A = a b c 2 2 2 2 2 2

a b c a b c 3 3 3 3 3 3

Minors and Co-factors

Let ,A = ( aij ) i = 1 , 2 , 3 j = 1 , 2 3

a a a 11 12 13 ie A = a a a 21 22 23

a a a 31 32 33

a a a 11 12 13

A = a a a 21 22 23 a a a 31 32 33

a a 2 2 2 3Consider which is a determinant formed by leaning all the elements of row and column in which all lies. Thisa a 3 2 3 3

determinant is called Minor of a . Thus we can form nine minors. In general if A is matrix of order n × n then minor of aij is11


obtained by leaning all the elements in the row and column in whichaij lies in . A The order of this minor isn – 1where as theorder of given determinant is n if this minor is multiplied by (–1) then it is called Co-factors of aij .i + j

a a a 11 12 13 Example : Let A = a a a 21 22 23

a a a 31 32 33 a a

22 23 a =Minor of1 1 a a

32 33 a a a a

22 2 3 2 2 2 3Co - factor of a = ( - 1 ) =1 + 11 1 a a a a 3 2 3 3 3 2 3 3

a a 1 2 13 Minor of a is 21 a a 3 2 3 3

a a a a 1 2 13 1 2 13 Co - factor of a is (- 1) = -2 + 1

21 a a a a 3 2 3 3 3 2 3 3

Value of a determinant Consider a matrixA of ordern × n . Consider all the elements of any row or column and multiply each element by its correspondingco-factor. Then the algebraic sum of the product is the value of the determinant.

a b A = 1 1Example : Let

a b 2 2

Co-factor of a is b 1 2

Co-factor of b is –a 1 2

A = a b - b a 1 2 1 2 a b c 1 1 1

Let A = a b c 2 2 2

a b c 3 3 3

b c b c a ( - ) =1+ 1 2 2 2 2Co - factor of is 1

1 b c b c 3 3 3 3

a c a c 2 2 2 2 Co - factor of b is (- 1) = -1 + 2

1 a c a c 3 3 3 3

a b a b 1 + 3 2 2 2 2Co - factor of c is (- 1) =

1 a b a b 3 3 3 3

b c a c a b 2 2 2 2 2 2A = a - b + c

1 1 1b c a c a b 3 3 3 3 3 3

= a (b c - b c ) - b (a c - a c ) + c (a b - a b )1 2 3 3 2 1 2 3 3 2 1 2 3 3 2

= a b c - a b c ) - a b c - a b c + a b c - a b c 1 2 3 1 3 2 2 1 3 3 1 2 2 3 1 3 2 1


Properties of determinants (1) If the elements of any two rows or columns are interchanged then value of the determinant changes only in sign. (2) If the elements of two rows or columns are identical then the value of the determinant is zero. (3) If all the elements of any row or column is multipled by a constant K , then the value of the determinant is multipled

by K .(4) If all the elements of any row or column are written as sum of two elements then the determinant can be written as

sum of two determinants.

(5) If all the elements of any row or column are multiplied by a constant and added to the corresponding elements ofany other row or column then the value of the determinant donot alter.

Adjoint of a Matrix

a b c 1 1 1

Let A = a b c 2 2 2

a b c 3 3 3

Let us denoted the co-factors of a , b , c , a , b , c , a , b , c as A , B , C , A , B , C , A , B , C transpose of matrix of 1 1 1 2 2 2 3 3 3 1 1 1 2 2 2 3 3 3co-factors is called Adjoint of the Matrix.

A B C 1 1 1

Matrixof Co - factors = A B C 2 2 2 A B C 3 3 3

A A A 1 2 3

Adjoint of A = B B B 1 2 3C C C 1 2 3

Theorem A . adjadj.A = A I = .A .A a b c A A A 1 1 1 1 2 3

A .adj.A = a b c B B B 2 2 2 1 2 3

a b c C C C 3 3 3 1 2 3

a A + b B + c C a A + b B + c C a A + b B + c C 1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3 = a A + b B + c C a A + b B + c C a A + b B + c C 2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3

a A + b B + c C a A + b B + c C a A + b B + c C 3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3 b c a c a b

The value of the det. A .2 2 2 2 2 2Now a A + b B + c C = a - b + c =1 1 1 1 1 1 1 1 1b c a c a b

3 3 3 3 3 3

a A + b B + c C =Similarly2 2 2 2 2 2

a A + b B + c C =3 3 3 3 3 3

b c a c a b 1 1 1 1 1 1a A + b B + c C = - a + b - c

1 2 1 2 1 2 1 1 1b c a c a b 3 3 3 3 3 3

= - a (b c - b c ) + b ( a c - a c ) - c (a b - a b )1 1 3 3 1 1 1 3 3 1 1 1 3 3 1

= - a b c + a b c + a b c - a b c - a b c + a b c = 01 1 3 1 3 1 1 1 3 3 1 1 1 3 1 3 1 1


Similarly the other five elements of A adj.A is zero.

0 0 A .adj.A = 0 0 where = A

0 0 1 0 0

= 0 1 0 = .I 0 0 1

A . adjadj.A = A · I = .A

Singular and Non-singular Matrices

A square matrix A is said to be singular if 0A = and non-singular if .A 0

Inverse of a MatrixTwo non-singular matricesA & B of the same order is said to be inverse of each other ifAB = I = BA . Inverse of A is denoted as A . Inverse of B is denoted as B and further ( AB ) = B A .– 1 –1 –1 –1 – 1

To find the inverse of A

A .adj.A = A · I A , AA . .A = A A - 1 - 1 - 1multiply by adj

. A ie adj. adj A = A A A =- 1 - 1

A

1 4 - 2- 2 - 5 4Example : Find the inverse of1 - 2 1

1 4 - 2Let A = - 2 - 5 4

1 - 2 1

- 5 4 - 2 4 - 2 - 5-

- 2 1 1 1 1 - 2

4 - 2 1 - 2 1 4 Matrix of Co - factors = - -

- 2 1 1 1 1 - 2 4 - 2 1 - 2 1 4

-- 5 4 - 2 4 - 2 - 5

(- 5 + 8 ) - (- 2 - 4) (4 + 5 )= - (4 - 4) (1 + 2) - (- 2 - 4)

(16 - 10) - (4 - 4 ) (- 5 +8)

3 6 9= 0 3 6

6 0 3


3 0 6adj.A = 6 3 0

9 6 3

1 4 - 2 A = - 2 - 5 4 = 1( 5- 5 + 8 ) - 4( - 2 - 4 ) - 2( 4 + ) = 3 + 24 - 18 = 9

1 - 2 1

3 0 6 0 3 6 9 91 1

= 6 3 0 = 0 - 1A = adj.A 6 39 9 9 A

9 6 3 9 6 3 9 9 9

0 1 2 3 3

A = 0 - 1 2 1 3 3 1 2 1

3 3

Solutions of Linear equations

Cramer's RuleTo solve the equations

a x + b y + c z = d 1 1 1 1

a x + b y + c z = d 2 2 2 2

a x + b y + c z = d 3 3 3 3

a b c 1 1 1

Consider = a b c (1)2 2 2 a b c 3 3 3

first evaluate & if it is not zero then multiply both sides of (1) by x .a b c a x b c 1 1 1 1 1 1

x = x a b c = a x b c 2 2 2 2 2 2

a b c a b c 3 3 3 3x 3 3

multiply the elements of columns 2 & 3 by y & z and add to elements of column 1.

a x + b y + c z b c 1 1 1 1 1

then x = a x + b y + c z b c 2 2 2 2 2

a x + b y + c z b c 3 3 3 3 3

d b c 1 1 1 = d b c = (say) (2)2 2 2 1

d b c 3 3 3 multiply both sides of (1) by y

a b c a b y c 1 1 1 1 1 1

y = y a b c = a b y c 2 2 2 2 2 2a b c a b y c 3 3 3 3 3 3


multiply the elements of columns 1 & 3 by x & z and add to the elements of column 2.

a a x + b y + c z c a d c 1 1 1 1 1 1 1 1

= a a x + b y + c z c = a d c = (say) (3) 2 2 2 2 2 2 2 2 2

a a x + b y + c z c a d c 3 3 3 3 3 3 3 3

multiply both sides of (1) by z a b c a b c z 1 1 1 1 1 1

z = z a b c = a b c z 2 2 2 2 2 2 a b c a b c z 3 3 3 3 3 3

multiply the elements of columns 1 & 2 by x & y and add to the elements of column 3.

a b a x + b y + c z 1 1 1 1 1

= a b a x + b y + c z 2 2 2 2 2

a b a x + b y + c z 3 3 3 3 3

a b d 1 1 1= a b d = (say) (4) 2 2 2 3

a b d 3 3 3

then 1 x = from (2)

y = from (3) 2

z = from (4)3

Note :- Verification of values of x , y ,z can be done by substituting in the given equations.

Example - 1

Solve 2 x + y - z = 31 x + y + z =4 x - 2y - 3 y =

2 1 - 1Let = 1 1 1 (1)

1 - 2 - 3

= 2 ( 1- 3 + 2 ) - 1(- 3 - 1 ) - 1( - 2 - ) = - 2 + 4 + 3 = 5

multiply both sides of (1) by x 2 1 - 1 2 x 1 - 1

then x = x 1 1 1 = x 1 1 1 - 2 - 3 x - 2 - 3

multiply the elements of columns 2 & 3 by y andz and add to the elements of column 1.

2 x + y - z 1 - 1 x = x + y + z 1 1

x - 2 y - 3y - 2 - 3


3 1 - 1= 1 1 1 = 3 ( 4- 3 + 2 ) - 1 ( - 3 - 4 ) - 1 (- 2 - ) = - 3 + 7 + 6 = 10

4 - 2 - 3

10 10x = = = 2

5 multiply both sides of (1) by y

2 3 - 1 2 y - 1 y = y 1 1 1 == 1 y 1

1 4 - 3 1 - 2 y - 3 multiply the elements of column 1 by x & 3 by z and to the corresponding elements of column 2.

2 2x + y - z - 1 then y = 1 x + y + z 1

1 x - 2 y - 3 z - 3

2 3 - 1 = 1 1 1 = 2( 1- 3 - 4 ) - 3( - 3 - 1 ) - 1(4 - ) = - 14 + 12 - 3 = - 5

1 4 - 3 - 5 - 5

y = = = - 15

multiply both sides of (1) by z 2 3 - 1 2 3 -z

z = z 1 1 1 = 1 1 z 1 4 - 3 1 4 - 3 z

multiply the elements of column 1 by x & column 2 by y and to the corresponding elements of column 3.

2 1 2 x + y - z then z = 1 1 x + y + z

1 - 2 x - 2 y - 3z 2 1 3

= 1 1 1 = 2( 14 + 2) - 1(4 - 1) + 3( - 2 - ) = 12 - 3 - 9 = 0

1 - 2 4 0 0

z = = = 0 5

Thus solution is x = 2, y = –1 & z = 0 which can be verified by substituting in the given equations.

Example - 2

Solve 4 x + y = 7 5 3 y + 4 z =

5 x + 3 z = 2

4 1 0 = 5 3 4 = 4( 15 9 - 0) - 1 (0 - 20 ) + 0 ( 0 - ) = 36 + 20 = 56

2 0 3


7 1 0 = 5 3 4 = 7( 69 - 0) - 1 (15 - 8) + 0(0 - ) = 63 - 7 = 561

2 0 3

4 7 0= 0 5 4 = 4(15 - 8) - 7(0 - 20 ) + 0 ( 0 - 25) = 28 + 140 = 1682

5 2 3

4 1 7= 0 3 5 = 4( 156 - 0 ) - 1 (0 - 25 ) + 7( 0 - ) = 24 + 25 - 105 = - 563

5 0 2

561x = = = 1

56

168y = = = 3 2

56

- 56z = = = - 13

56

Solution of Linear equations by Matrix Method Given a x + b y + c z = d

1 1 1 1

a x + b y + c z = d 2 2 2 2

a x + b y + c z = d 3 3 3 3

a b c x d 1 1 1 1

Consider A = a b c X = y & B = d 2 2 2 2

a b c z d 3 3 3 3

then given equations can be written in Matrix form as AX = B . If 0A solution exists multiply both sides by A –1

- 1 - 1A ( AX ) = A B A AX = A B - 1 - 1 ie IX = A B - 1

- 1 X = A B

Example

Solve 3x - y + 2 z = 13

3 2x + y - z =x + 3 y - 5 z = -8

3 - 1 2 x 13Let A = 2 1 - 1 X = y , B = 3

1 3 - 5 z - 8

then given equations can be written as AX = B


X A B (1)-1 =

To find A –1

3 - 1 2 A = 2 1 - 1 = 3 (- 5 + 3 ) + 1(- 10 + 1 ) + 2( 6 - 1) = - 6 - 9 + 10 = - 5

1 3 - 5

1 - 1 2 - 1 2 1 -

3 - 5 1 - 5 1 3- 1 2 3 2 3 - 1

Matrix of Co - factors = - -3 - 5 1 - 5 1 3

- 1 2 3 2 3 - 1 -

1 - 1 2 - 1 2 1

(- 5 + 3 ) - (- 10 +1) (6 - 1) - 2 9 5

= - (5 - 6) (- 15 - 2 ) - (9 + 1) = 1 - 17 - 101- 2 - ( - 3 - 4) (3 + 2) - 1 - 7 5

- 2 1 - 1 - 2 1 - 11 1adj.A = 9 - 17 7 A = adj.A = - 9 - 17 7- 1A 5

5 - 10 5 5 - 10 5

Using this in (1) - 2 1 - 1 13 - 26 3 8 - 15 3

1 1 1 X = - 9 - 17 7 3 = - 117 - 51 - 56 = - 10 = - 25 5 5

5 - 10 5 - 8 65 - 30 - 40 - 5 1 x = 3, y = –2, z = 1 is the solution

Verification : Consider the first equation

3 x - y + 2z = 9 + 2 + 2 = 13

Characteristic equation, Eigen Values & Eigen Vectors

A - I =Let A & I be square matrices of same order and a scalar then 0 is called Characteristic equation and the roots ofthis equation ie values of are called Eigen Values or Characteristic roots. The matrixX satisfying AX = X is called EigenVector.

Example - 1

1 4 Find the eigen roots and eigen vectors of the matrix

2 3 1 4 1 0 1 4 - = 0Characteristic equation isLet A = 2 3 0 1 2 3

1- 4 ie = 0 (1 - )(3 - ) - 8 = 0

4 3 -

2 23 - 3 - + - 8 = 0 - 4 - 5 = 0


( - 5 ,)( + 1 ) = 0 = - 1 5

Eigen roots are –1 & 5.

To find eigen vector X , corresponding to –1,

AX = –1

1 4 x - x ie =

2 3 y - y x + 4 y - x

=2 x + 3 y - y x + 4 y = - x ix = - 4y x y

= 2 x + 3y = - y ie x = - 2 y - 2 1

Eigen vector corresponding to eigen value –1 is (–2, 1)

To find the eigen vector corresponding to 5.

1 4 x x x + 4 x 5 x 1 1 1 2 1ie = 5 =

2 3 x x 2 x + 3x 5 x 2 2 1 2 2

x + 4 x = 5 x x x 1 2 1 1 2ie x = x ie = 1 2x + x = x 2 3 5 1 1

1 2 2

Eigen vector is (1, 1)

Eigen vector corresponding to eigen root 5 is (1, 1)

Example - 2

6 - 2 2 - 2 3 - 1Find the eigen roots and eigen vectors of the matrix 2 - 1 3

6 - 2 2 6 - - 2 2 Let A = - 2 3 - 1 Characteristic equation is 0- 2 3- - 1 =

2 - 1 3 2 - 1 3-

[] [][] ie (6 - ) (3 - ) - 1 + 2 - 2(3 - ) + 2 + 2 2 - 2 (3 - ) = 0 2 [] [][]ie (6 - ) 9 + - 6 - 1 + 2 - 6 + 2 + 2 + 2 2 - 6 + 2 ) = 02

ie (6 - )( - 6 + 8) + 2(2 - 4 ) + 2 (2 - 4) = 02

ie 6 - 36 + 48 - + 6 - 8 + 4 - 8 + 4 - 8 = 02 3 2

ie - + 12 - 36 + 32 = 0 ie - 12 + 36 - 32 = 0 3 2 3 2

which is the characteristic equation, by inspection 2 is a root3 2dividing - 12 + 36 - 32 by - 2 ,we have

0( - 2)( - 10 + 16) =2

= 2 ,, 2 8ie ( - 2)( - 2 )( - 8) = 0


To find eigen vector or = 2

Consider AX = 2 X 6 - 2 2 x x x 1 1 1

ie - 2 3 - 1 x = 2 x where X = x 2 2 22 - 1 3 x x x 3 3 3

x - x + x = x 4 x - 2 x + 2x = 0 6 2 2 2 1 2 3 1 1 2 3

- x + x - x = x - 2x + x - x = 02 3 2 1 2 3 2 1 2 3

x - x + x = x 2 x - x + x = 0 2 3 2 1 2 3 3 1 2 3

the above three equations represent one equation 2x – x + x = 0. 1 2 3

x x x x x Let x = 0,then 2 x = x ie = ie = =1 2 1 2 3

3 1 2 1 2 1 2 0 Eigen Vector is (1, 2, 0)

To find the eigen vector for = 8 Consider AX = 8 X

6 - 2 2 x 8x 1 1

ie - 2 3 - 1 x = 8 x 2 2

2 - 1 3 x 8 x 3 3

ie 6 x - 2x + 2x = 8 x ie - 2 x - 2 x + 2 x = 0 1 2 3 1 1 2 3

- 2x + 3x - x = 8x - 2 x - 5 x - x = 01 2 3 2 1 2 3

2 x - x + 3x = 8 x 2 x - x - 5 x = 01 2 3 3 1 2 3

(1)ie x + x - x = 01 2 3

(2)2 x + 5 x + x = 0 1 2 3

(3)2 x - x - 5 x = 0 1 2 3

adding (1) & (2) we get 3 x + 6x = 01 2

x x ie x + 2 x = 0 ie x = - 2 x = = K (Say)1 2

1 2 1 2 - 2 1

then ,x = - 2 K x = K 1 2

subsitituting in (1)

- 2 K + K - x = 0 x = - K 3 3

x = - 2 &K , x = K x = - K 1 2 3

Eigen vect or is (- 2, 1, - 1) or (2, - 1, 1)

Eigen roots are 2, 2, 8 & 1Eigen vect ors are (1, 2, 0) & (2 , - 1 , )

Properties of Eigen values

(1) The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal.

(2) The product of the eigen values of a matrix is equal to the value of its determinant.

1 (3) If is an eigen value of A then is the eigen value of A . –1


Cayley - Hamilton Theorem Every square matrix satisfies its characteristic equation.

a - b a b = 0Characteristic equation isLet A =

c d -c d which on simplification becomes a quadric equation in in the form 0 + a + a = where a , a are constants.2

1 2 1 2

A + a A + a I =2 Cayley Hamilton Theorem states that 0 where I is a unit matrix of order 2 & 0 is a null matrix of order 2.1 2 a b c 1 1 1

If A = a b c 2 2 2 a b c 3 3 3

a - b c 1 1 1

then characteristic equation is 0 a b - c =2 2 2

a b c -3 3 3

which on simplification becomes 0 which is a cubic equation. 3 2 - + a + a + a =1 2 3 Then as per Cayley Hamilton Theorem 0- A + a A + a A + a I = where I is a unit matrix of order 3 & 0 is a null3 2

1 2 3matrix of order 3.

In general if A is a square matrix of ordern then characteristic equation will be of the form

- - (- 1 ) + a + a + ........ + a I = 0 n n n 1 n z 2 n

and by Cayley Hamilton Theorem

n n n - 1 n - z (- 1 ) A + aA + a A + ........ + a I = 02 n where I is a unit matrix of order n & 0 is a null matrix of order n .

Note :- If we put = 0 in the characteristic equation then a = A n If a = 0, matrix A is singular & the matrix A is non-singular & hence inverse exists and we can find thea 0

n n inverse of A using Cayley Hamilton Theorem.

Example - 1

a b Let A =

c d a - b

Characteristic equation is = 0 c d -

ie + a + a = 0 where a , a are constants. 21 2 1 2

By Cayley Hamilton Theorem

2 A + a A + a I = 0 1 2 multiply both sides by A –1

2 -1 - 1 - 1 then A A + a AA + a A = 01 2

- 1ie A + a I + a A = 0 1 2

a A = - ( A + a I )- 12 1

1 A = - ( A + a I )- 1

1a 2


Example - 2

The characteristic equation of a matrix A of order 2 is - 5 + 10 = 0 find A .2

Solution : put = 0 in C.E. then the constant 10 is . A

Example - 3

2 - 1Find the inverse of using Cayley Hamilton Theorem.- 3 4

2 - 1 2 - - 1Solution : Let A = C.E. is = 0

- 3 4 - 3 4 -ie 8 - 4 - 2 + - 3 = 02ie ( 2 - )( 4 - ) - 3 = 0

ie - 6 + 5 = 0 2

by Cayley Hamilton Theorem

2A - 6 A + 5 I = 0 multiply both sides by A –1

A - 6 I + 5 A = 0- 1

- 1 5A = - A + 6I 2 - 1 1 0 - 2 + 6 1 + 0 4 1

= - + 6 = = - 3 4 0 1 3 + 0 - 4 + 6 3 2

4 11 A = - 1

5 3 2

Diagonalisation of Matrices If A is a square matrix of order n where all the eigen values are linearly independent then a matrix P can be found such thatP AP is a Diagonal Matrix.–1

Let A be a square matrix of order 3 and let , , be the eigen values, corresponding to these. Let X , X , X be three1 2 3 1 2 3

vectors where

x y z 1 1 1 X = x , X = y , X = z

1 2 2 2 3 2 x y z 3 3 3

x y z 0 0 1 1 1 1- 1Let P = x y z Then P AP = 0 0 2 2 2 2

x y z 0 03 3 3 3

Example - 1

1 - 2 Let A =

- 5 4 1 - - 2

C.E. is = 0 (1 - )( 4 - ) - 10 = 0- 5 4 -


ie - 5 - 6 = 0 ( + 1 )( - 6 ) = 02 = - 1, 6 are eigen valu es

For , = - 1 let AX = - X 1 - 2 x - x

1 1=ie- 5 4 x - x

2 2

x - 2 x = - x 1 x x 1 2 1ie x = x = ....eigen vect or is 1 21 2- x + x = - x 5 4 1 1 1 1 2 2

For = 6 , AX = 6X 1 - 2 x 6x

1 1ie = - 5 4 x 6 x

2 2

x - 2 x = - x 1 2 1 5 x = - 2 x

1 2- 5 x + 4 x = - x 1 2 2 x x - 2

ie = eigen vect or is1 2 5 - 2 5

1 2 5 2- 1Let P = Then P =- 1

1 5 7 - 1 1

5 2 1 - 2 1 2 1 -- 1P AP =

7 - 1 1 - 5 4 1 5 5 - 10 - 10 + 8 1 2 - 5 - 2 1 2 1 - 1 -

= =7 - 1 - 5 2 + 4 1 5 7 - 6 6 1 5

- 5 - 2 10 - 10 7 0 1 0 1 1 - -= = =

7 - 6 + 6 12 + 30 7 0 42 0 6 1 2 1 - 2 -

Thus P = diagonaliz ethe matrix1 5 - 5 4

Example - 2

1 1 3Let A = 1 5 1

3 1 1

1- 1 3 1 5- 1 =Characteristic equation is 03 1 1 -

[][][]- - - - - - - + - - = ie (1 ) (5 )(1 ) 1 11 3 3 1 3( 5 ) 0 2ie (1- )( - 6 + 4 ) - ( - - 2 ) + 3 (- 14+ 3 ) = 0

ie - 6 + 4 - + 6 - 4 + + 2 - 42 + 9 = 02 3 2 ie - + 7 - 36 = 0 ie - 7 + 36 = 03 2 3 2

by inspection –2 is a root + 2 is a factor. equation becomes2( + 2 )( - 9 +18) = 0


6 = - 2 ,, 3 ie ( + 2)( - 3( - 6) = 0 ie characteristic roots are –2, 3, 6.

To find the eigen vector for = –2 Consider AX = - 2 X 1 1

x 1where X = x

1 2x 3

1 1 3 x - 2x 1 1

ie 1 5 1 x = - 2 x 2 2

3 1 1 x - 2 x 3 3

(1)ie x + x + 3x = - 2x ie 3x + x + 3x = 01 2 3 1 1 2 3

(2)x + 5x + x = - 2x x + 7 x + x = 0 1 2 3 2 1 2 3

(3)3x + x + x = - 2x 3x + x + 3x = 01 2 3 3 1 2 3

(1) & (3) are same.

Put x = 0 in (1) or (2), then x + x = 02 1 3

1 -x x eigen vector X = 0ie x = - x = 1 3

11 3 - 1 1 1

Let X be the eigen vector for = 3.2

ie AX = 3X 2 2

1 1 3 y 3 y y 1 1 1

ie 1 5 1 y = 3 y where X = y 2 2 2 23 1 1 y 3 y y 3 3 3

(1)y + y + y = y ie - 2 y + y + 3 y = 0ie 3 31 2 3 1 1 2 3

(2)y + y + y = y y + 2 y + y = 05 3 3 1 2 3 2 1 2 3

(3)y + y + y = y 3y + y - 2 y = 0 3 3 31 2 3 3 1 2 3

Let us eliminate y from (1) & (2)1

(1 ) × 1is - 2 y + y + 3 y = 01 2 3

(2) ×2 is 2 y + 4 y + 2 y = 01 2 3 adding 5 y + 5 y = 0

2 3 y y

(4)y + y = 0 ie y = - y = 2 3 2 3 2 3 - 1 1

Let us eliminate y from (2) & (3)2

(2) ×1 is y + 2 y + y = 01 2 3

(3) × 2 is 6 y + 2 y - 4 y = 0 1 2 3

subtractin g - 5 y + 5 y = 0 2 3

y y (5)5 y = 5 y y = y =1 3

1 3 1 3 1 1


1 y y y X = - 1 1 2 3 From (4) & (5) = = 21 - 1 1

1 Next, let X be the eigen vector for = 6

3

ie AX = 6 X 3 3

1 1 3 z 6 z z 1 1 1 ie 1 5 1 z = 6 z where X = z 2 2 3 2

3 1 1 z 6 z z 3 3 3

(1) z + z + z = z ie - 5 z + z + 3 z = 0ie 3 6 1 2 3 1 1 2 3

(2) z + z + z = z z - z + z = 05 6 1 2 3 2 1 2 3

(3) z + z + z = z 3 z + z - 5z = 03 6 1 2 3 3 1 2 3

adding (1) & (2), - 4 z + 4 z = 0 1 3

z z (4) 1 3ie z = z =

1 3 1 1

Let us eliminate z from (2) & (3)3

(2)× 5 is 5 z - 5 z + 5 z = 01 2 3

(3) ×1 is 3z + z + - 5 y = 01 2 3

adding 8z - 4 z = 01 2

z z 1 2 (5) ie 2 z = z =

1 2 1 2 1

z z z X = 2From (4) & (5), = = 1 2 331 2 1

1

- 1 1 1 - 2 0 0- 1 Let P = 0 - 1 2 Then P AP = 0 3 0

1 1 1 0 0 6

- 1 1 1 1 1 3P = 0 - 1 2 diagonalize 1 5 1

1 1 1 3 1 1

Exercise

1 2 3 1 . Evaluate - 1 2 3 .

2 3 1

a - b b - c c - a 2. Evaluate b - c c - a a - b .

c - a a - b b - c


1 2 3 3. Evaluate 2 3 4 .

3 4 4 2 0 4

4. If 6 x 5 = 0,then find x .- 1 - 3 1

1 p - q 5. Evaluate - p 1 r .

q - r 1

2 06. Find the adjoint of .

2 3

1 2 3 7. If A = 2 3 4 findthe co - factorof 1.

3 4 2

2 1 5' '8. If A = find A A & A A .

0 3 7

sec tan9. Find the inverse of .

tan sec

2 3 4 1 2 3-10 . If A = and B = find 5 A - 3B and 6 B - 7 A .

- 1 0 5 0 4 2

4 2 5 2 0 4-11 . If 3 A + B = and 2 B + A = find A and B .

3 7 6 - 1 2 3

1 x x x y 4 5+ -12 . If + = find x and y .

5 x 7 - y 12 4 x

5 6 7 () ''13 . If A = verify th at A = A .8 9 10

2 4 - 3 4 5 ' '14. If A = and B = - 4 3 find A + B and A - B .6 - 2 1

3 - 2

3 4 2 15 . 0If A = prove that A - 10 A + I = .

5 7 6 5

16 . Find the inverse of . 3 2

1 2 17 . Find the characteri stic equation of .

0 4


2 3 18 . Find the eigen valu es of .

0 4 19 . Solve 2 x + z = - 1 , 2 y + x = 5, z - y = - 2 by Cramer' s Rule.

- + =5 x y 4z 520. Solve

2 x + 3y + 5 z = 2

7 x - 2y + 6 z = 5by matrix method.

2 1-21. Find the characteri stic roots of .

0 1

- 1 0 122. Findthe characteristic roots of 1 2 1 .

2 2 3

1 223. Verify Cayley - Hamilton Theorem for the matrix .

3 4

2 024. Verify Cayley - Hamilton Theorem for the matrix .

1 - 1

1 2 25. Find the eigen vect ors for the matrix .

2 1 6 - 2 2

26. Findthe eigen values andeigen vectors for the matrix - 2 3 - 1 .2 - 1 3

BCA 21 / IMCA 21 / Mathematics SVT 21

ALGEBRAIC STRUCTURES

Abreviations used N : represent set of natural numbers.

Z or I : represent set of +ve and –ve integers including zero.

Z : represent non-negative integers ie. +ve integers including zero.+ Q : represent set of rational numbers.

R : represent set of real numbers.

C : represent set of complex numbers.

Z = {0, 1, 2, 3, .............. n – 1} ie. Z represent set of integers modulo n .n n

Q : represent set of +ve rational numbers.+ z - {o} : represent set of integers except 0.

Q - {o} : represent set of rational numbers except zero.

R - {o} : set of real numbers except zero.

A set in general is denoted by S .: for all

: belongs to

Binary OperationIf S is a non-empty set then a mapping (function) from S × S to S is defined as Binary Operation (in short B.O.) and denotedby * (read as star). ie. : S ×S S (Star maps S cross S to S )

Another DefinitionIf S is non-empty set then * (star) is said to be a Binary operation if a , b S , a * b S .

Examples

(1) on N + and ×(ie addition & multiplication) are B.O. 2 + 3 = 5 N 2 × 3 = 6 N

(2) on Z , +, – & ×are B.O.

4 + 5 Z , 3 - 4 = 1 Z , 4 ,- 3 = 1 Z 5× 6 = 30 Z a (3) On Q & R +, – & ×are B.O. but ÷ is not a B.O. on Q & R for 0, but on Q - {o} & R - {o} ÷is a B.O.Q Q & R 0

(4) on C , + and ×are B.O.

( x + iy ) + ( x + iy ) = ( x + x ) + i ( y + y ) C 1 1 2 2 1 2 1 2

( x + iy ) + ( x + iy ) = ( x x - y y ) + i ( x y + x y ) C 1 1 2 2 1 2 1 2 1 2 2 1

48 KSOU Algebraic Structures

Definitions (1 ) A non-empty set S with one or more binary operations is called an 'Algebraic Structure'.

(N , +), ( Z , +, ×), (Q , +, ×) are all algebraic structures.

(2) Closure Law : A set S is said to be closed under a B.O. * if a , b S , a * b S .(3) Associative Law : A B.O. * is said to be associative on S if a ,b ,c S

(a * b ) * c = a * ( b * c )(4) Commutative Law : A B.O. *is said to be commutative on S if a ,b S , a * b = b * a .(5) Identity Law : An element e S satisfying

a * e = a = e * a . a S is called an identity element for the B.O. *on S .

Examples

(i) + and ×(addition and multiplication) are associative and commutative on N , Z , Q & R .(ii) B.O. – (subtraction) is not associative & commutative.

(iii) 1 is an identity for B.O. ×on N but + has no identity on N . Where as O is an identity on Z , Q and R for the B.O. +.

1 0 (iv) If S is a set of 2 ×2 matrices and B.O. is matrix multiplication then is an identity element. I =

0 1

Group A non-empty set G together with a B.O. *ie ( G , *) is said to form a group if the following axioms are satisfied.

G . Closure Law : a , b G , a * b G 1

G . Associative Law : a , b , c G , (a * b ) * c = a * ( b * c )2

G . Identity Law : There exists an element e G such that a G , a * e = a = e * a .3

G . Inverse : a G , there exists an elementb such that a * b = e = b * a . This b is called inverse of a and usually denoted 4 as a –1

ie. a * a = e = a * a . –1 –1 In addition to the above four axions if a , b G , a * b = b * a . Then (G , * ) is called an'abelian group' or 'commutativegroup'.

Note (1 ) If for ( G , *) only G is satisfied it is called a 'groupoid'.1

(2 ) If for ( G , *) G & G are satisfied it is called a 'semi-group'.1 2

(3 ) If for ( G , *) G , G & G are satisfied it is called 'Monoid'.1 2 3

Examples

(i) ( N , +) is a groupoid and semigroup.

(ii) ( N , ×) is a groupoid, semigroup and Monoid (identity for ×is 1)

(iii) ( Z , +) is a group (identity is O and a = – a ) ie. a + (– a ) = 0 = – a +a . –1

Note :- Every group is a monoid but the converse is not true, (Z , +) is a group and also a monoid but (N , ×) is a monoid butnot a group.


Properties of Groups

1. Cancellation laws are valid in a group

ie if is a group then ( G , *) a , b , c G ,(i () a * b = a * c b = c left cancellati on law)

(ii () b * a = c * a b = c right cancellati on law)

Proof :- a * b = a * c , as , a G a G - 1

- -a *(a * b ) = a * (a *c )1 1 - 1 - 1 ie (a * a )* b = (a * a )* c

ie e * b = e * c where e is the identity.

b = c Similarly by considering

- 1 - 1(b *a) * a = (c * a) * a we get b = c

2. In a group G, the equations a * x = b and y * a = b have unique solutions, a , b G .Proof :- a * x = b

Operating on both sides by a –1

-1 - 1a *(a * x ) = a * b - 1 - 1ie )(a * a * x = a * b

- 1ie e * x = a * b x = a * b - 1

To prove that the solution is unique, let x & x be two solutions of a * x = b . 1 2

a * x = b a * x = b ie &1 2

a * x = a * x 1 2

Operating on both sides by a –1

- 1 - 1We get a * (a * x ) = a * (a * x )1 2

- 1 - 1 ie (a * a ) * x = (a * a )* x 1 2

ie e * x = e * x 1 2

x = x 1 2

solution is unique. 3. In a group the identity element and inverse of an element are unique.

Proof :- To prove identity is unique. If possible let e & e are two identities then1 2

(1)a G , a * e = a = e * a 1 1

(2)& a * e = a = e * a 2 2

From LHS of (1), (using (2))a * e = a = e * a 1 2

(using LHS of (2))ie a * e = e * a = a * e 1 2 2


by left cancellation, Thus the identity is unique. e = e . 1 2

To prove that inverse of an element is unique. - -Let if possible let b & c are inverses of a ie b = a and c = a 1 1a G ,

Now, a * b = e & a * c = e where e is the identity of the group

a * b = a * c by left cancellation law b = c . Thus inverse of an element is unique.

() - 14. In a group G , a = a a G - 1

Proof :- As a is the inverse of a – 1

- 1 - 1We have a * a = e = a * a () - 1it can be easily seen from above relation that inverse of a is a ie a = a . - 1–1

Note :- If b & c are elements of G , such that then each is the inverse of the other.b * c = e = c * b

5. In a group (G , *)

- 1 - 1 - 1a , b G , (a * b ) = b * a Proof :- ) - 1 - 1 Consider (a * b ) * (b * a

- 1 - 1 - 1 - 1= a * (b * b )* a = a * e *a = a * a = e (a * b ) = b * a - 1 - 1 - 1

6. A group of order 3 is abelian.

Proof :- Order of group means the number of elements in a group. If a groupG has n elements. The order of G is n , which is denoted as O (G ) = n .If n is finite it is called finite group and n is Infinite then it is called Infinite group.

be a group with a binary operation * .Let G = { e , a , b }e is the identity by definition of identity it commutes with every element Q a * e = a = e * a

So we have to prove that * = *a b b a Let a * b = a (a * a )* b = a * a - 1 - 1 e * b = e b = e

which is not possible. Let a * b = b

-1 - 1(a * b )*b = b * b a * e = e a = e which is not possible. cannote be equal to a or b *a b a * b = e

Similarly we can prove that

b * a = e a * b = b * a (G , *) is abelian.

BTC111

Subgroups A non-empty subset H of a group G is said to form a subgroup with respect to the same binary operation * if is a( H , * )group.

Eg. (1) ( z , +) is a subgroup of ( Q , +)

(2) is a subgroup of with B.O. H = { 40, 2, } G = { 50, 1 , 2 , 3, 4, } + mod 6 ie6

(3) is a subgroup of with respect to the B.O. multiplication. H = { 1- 1 , } G = {1, - 1, i , - i }

Theorem

A non-empty subset H of a group G is a subgroup of G if and only if . a , b H , ab H - 1

Proof :- case (i) Let H be a subgroup of G then H is a group- 1 - 1a , b G , b G (inverse axion) & ab G (closure law)

condition is satisfied. case (ii) Let H be a non-empty subset of G with the property . - 1a , b H , ab H

We have to prove that H is a subgroup.

Let ,a H , a , a H aa = e H & e a H ea H ie a H - 1 -1 - 1 - 1 Since all elements of H are elements of G , associative property is satisfied.

() - 1 Let b H , ie closure property is satisfied.b H & for a H , a b ie ab H - 1 -1

Hence H is a group and hence a subgroup.

Permutation group

Let S = {a , a , a , .......... a } 1 2 3 n Then a one-one and onto mapping or function from S onto itself is called a Permutation.

a a a .......... a Permutation is denoted as 1 2 3 n

f (a ) f (a ) f (a ) .......... f ( a )1 2 3 n

There will be n ! ie permutations the set of permutations is denoted by S .n n

There is a composite mapping for f & g denoted as this can be taken as binary operation. Then theLet f , g S . f o g ,n set S with binary operation 'O ' (ie composite mapping) will form a group. For convenience is denoted as gf .f o g

n 1 2 3 4 1 2 3 4

the B.O. composite function is given byLet f = & g = S 4 3 1 4 2 2 3 4 1

1 2 3 4 1 2 3 4 1 2 3 4f o g = o =

3 1 4 2 2 3 4 1 ? ? ? ?

to fill up the second row, following is the procedure.

in g : g (1) = 2 , g ( 2) = 3 , g (3 ) = 4, g (4) = 1 & in f : f (1) = 3 , f ( 2) = 1, f ( 3 ) = 4, f ( 4) = 2

Now f o g (1) = f [ g (1 )] = f ( 2 ) = 1

f o g (2) = f [ g ( 2)] = f (3) = 4

f o g ( 3) = f [ g ( 3)] = f ( 4 ) = 2 f o g ( 4) = f [ g ( 4)] = f (1 ) = 3


1 2 3 4 f o g =

1 4 2 3 for convenience is written as gf f o g

1 2 3 4 1 2 3 4 1 2 3 4 ie f o g = o =

3 1 4 2 2 3 4 1 1 4 2 3 1 2 3 4 1 2 3 4 1 2 3 4

& gf = = 2 3 4 1 3 1 4 2 1 4 2 3

the composite function is also called product function.

Eg. (1) Let S = { 21, }

1 2 1 2 then S = ,

2 1 2 2 1 Let B.O. be product permutation

1 2 1 2 1 2then =

1 2 2 1 2 1 1 2 1 2 1 2

= 1 2 1 2 1 2 1 2 1 2 1 2

= 2 1 2 1 1 2

1 2closure law is satisfied, associative law can be easily verified. inverse e =

1 2- 11 2 1 2

=1 2 1 2

- 11 2 1 2=

2 1 2 1

S forms a group.2

Eg. (2) Let S = { 31, 2 , }

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3S =

3 1 2 3 1 3 2 3 2 1 2 3 1 3 1 2 3 2 1

Let us denote the elements as respectively.f , f , f , f , f & f 1 2 3 4 5 6

{}ie S = f , f , f , f , f , f 3 1 2 3 4 5 6

The following is the multiplication table. f f f f f f 1 2 3 4 5 6

f f f f f f f 1 1 2 3 4 5 6 f f f f f f f 2 2 1 4 3 6 5 f f f f f f f 3 3 5 1 6 2 4 f f f f f f f 4 4 6 2 5 1 3 f f f f f f f 5 5 3 6 1 4 2 f f f f f f f 6 6 4 5 2 3 1

BTC113

It can be seen from the table that closure law is satisfied. For associative law

( f f ) f = f f = f Consider3 4 5 6 5 3

f ( f f ) = f f = f 3 4 5 3 1 3

hence associative law is satisfied. ( f f ) f = f ( f f )3 4 5 3 4 5

identity e = f 1

-1 - 1 - 1 - 1 - 1 - 1 f = f , f = f f = f , f = f f = f & f = f 1 1 2 2 3 3 4 5 5 4 6 6

inverse of all elements exists.

S forms a group, but it is not an abelian group3

Q f f = f but f f = f f f f f . 3 4 6 4 3 2 3 4 4 3

Examples

(1) Show that the set R - {o} with B.O. ×forms a group.

Solution : For any elements a , b R - {o}. a * b R - {o}

2, 3 R - {o}, 2 ×3 = 6 R - {o}

closure law is satisfied.

For any three elements a , b ,c R - {o}

(a × b )× c = a ×( b × c )(–3 ×4) ×5 = –12 ×5 = –60.

–3 ×(4 ×5) = –3 ×20 = –60. associative law is satisfied.

Identity element is 1,

ie. a R - {o}, a ×1 = a = 1 × a . Let a R - {o} then there exists

1 } 1 1R - { o such that a × = 1 = × a

a a a inverse exists for all elements R - {o}. ( R - {o}, ×) forms an abelian group.

(2) Show that forms an abelian group.( Z , )5 5

Solution : Let us construct table for (Z , )5 5

0 1 2 3 45

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3


From the above table it can be easily seen that closure law is satisfied.

2 (3 4) = 2 2 = 45 5 5

( 2 3 ) 4 = 0 4 = 45 5 5

ie 2 (3 4 ) = ( 2 3) 45 5 5 5

associative law can be versified.

identity element is 0.

inverse of 0 is 0 inverse of 1 is 4 inverse of 2 is 3 inverse of 3 is 2 inverse of 4 is 1

form a group, further it can be seen from the table that for any two element a , b Z (Z , )5 5 5

a b = b a 5 5

is an abelian group.(Z , )5 5

(3) Show that G = {1, 2, 3, 4} with B.O. multiplication mod 5 ie is an abelian group. 5

Solution : The following in the relevant table for elements

1 2 3 4 5

1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1

From the above table it can be seen that closure law is satisfied.

( 2 3) 4 = 1 4 = 45 5 5

2 (3 4) = 2 2 = 45 5 5

associate law is satisfied.

identity is 1.

inverse of 1 is 1 inverse of 2 is 3 inverse of 3 is 2 inverse of 4 is 4

forms a group and it can be seen from the table that it forms an abelian group. (G , )5

4. If R is the set of real numbers and *is a binary operation defined on . R as x * y = 1 + xy , x , y R Show that * is commutative but not associative.

Commutativ e property a * b = b * a , a , b R x * y = 1+ xy

BTC115

y * x = 1+ yx x * y = y * x Associativ e property ( x * y ) * z = x * ( y * z )LHS )= (1 + xy ) * z = P * z = 1 + pz = 1 + (1 + xy z = 1 + z + xyz RHS )= x * (1+ yz ) = x * Q = 1 + xQ = 1 + x (1 + yz = 1 + x + xyz

LHS RHS * is not a associativ e

5. Show that set of integers with a * b = a - b where a , b I is not a group

a * b = a - b I closure axiom is satisfied ( a * b ) * c a * (b * c ), a * b = a - b I , a , b I .LHS = (a - b ) * c = a - b - c RHS = a * ( b - c ) = a - b + c associativ e axiom is not satisfied

is not a group.( I , *)

When one of the axiom is not satisfied, it is not a group. Hence, we need not have to check the rest of the axioms.

ab 6. In a group Find the identity element, inverse of 4 and solve ( 2 G , *), a * b = . 4 * x = 5To find e : a * e = a = e * a

ae a * e = = a e = 2 i.e. identity element is 2.

2 a * a = e = a * a - 1 - 1

aa 4 - 1- 1 - 1a * a = 2 = 2 a =

2 a 4

4 = = 1. inverse of 4 is 1.- 14

4 x 10 54 * x = 5 = 5 x = =

2 4 2

{}G = cos + i sin is real7. Prove that is an abelian group under multiplication

{} G = cos + i sin is real

Let x , y , z be any three elements of G .a a ß ß Take x = cos + i sin , y = cos + i sin , z = cos + i sin .

a ß where , , are real numbers.

i) a a ß ß a ß a ß a ß xy = (cos + i sin )(cos + i sin ) = cos( + ) + i sin( + ) G Q ( + ) is real

Closure axiom is satisfied

ii) a a ß ß (xy ) z = [(cos + i sin )(cos + i sin )](cos + i sin )

a ß a ß = {cos( + ) + i sin( + )}(cos + i sin )

a ß a ß = cos{( + ) + } + i sin{( + ) + }

a ß a ß = cos{ + ( + )} + i sin{ + ( + )}

= x ( yz ) multiplication is associative on R Associative axiom is satisfied.Q iii) 1 = cos 0 + i sin 0 G is the identity element


iv) ( cos + i sin )(cos - i sin ) = 1 cos - i sin is the multiplica tive inverse of cos + i sin . v) a ß ß xy = (cos + i sin )(cos + i sin )

a ß a ß = cos( + ) + i sin( + )

ß a ß a = cos( + ) + i sin( + ) Q is commutativ e on R = yx

Commutative law is satisfied.

So all the axioms are satisfied. Hence, G is an abelian group under multiplication.

8. Show that the cube roots of unity form an abelian group under multiplication

We know that the cube roots of unity are }1 , and . Let G = {1, ,2 2

. 1 2

here = 1 31 1 2

4 3& = · = .12

12 2

i) All the entries in the table are the same as the elements of the set. This means the closure law is satisfied.

ii) 1 Consider 1· ( · ) =1 · 1=2

2 2(1 · ) · = · = 1 2 2 1· ( · ) = (1 · ) · . Associative law is satisfied.

iii) The row heading 1 is the same as the topmost row.

1 is the identity element.

iv) Inverse of 1 is 1, inverse of is and inverse of is .2 2

Every element has a inverse.

v) The table is symmetrical about the principal diagonal

commutative law holds good. So G is an abelian group under multiplication.

1 0 - 1 0 1 0 - 1 09. form an abelian group under matrix Show ,that the four matrices , and

0 1 0 1 0 - 1 0 - 1multiplication.

1 0 1 0 1 0 - 1 0-Take = I , = A , = B , = C ; G = {1 , A , B , C }

0 1 0 1 0 - 1 0 - 1

IA = AI = A , IB = BI = B , IC = CI = C 1 0 1 0 - 1 + 0 0 + 0 - 1 0-

AB = = = = C 0 1 0 - 1 0 + 0 0 - 1 0 - 1

Similarly, it can be shown that BA = C , AC = CA = B , BC = CB = A , A · A = I etc.

BTC117

The composition table is . I A B C I I A B C A A I C B B B C I A C C B A I

i) The entries in the table are the same as the elements of the set G . Closure law is satisfied.

ii) Associative law is satisfied. A ( BC ) = A ( A ) = I ; ( AB )C = (C )C = I iii) I is the identity element.

iv) Inverses of I , are respectivelyA , B I , is a group under matrix multiplicationA , B , C . G is a group.(G , ·)

v) Since the entries on either side of the leading diagonal are symmetric, is an abelian group.(G , ·)

10. If every element of a group G has its own inverse, show that G is abelian (1)-Given a = a , a , b G 1

(2)- 1 - 1 - 1 we knowthat, a , b G , (ab ) = b a - 1 -1 - 1from Using these in (2), (1), a = a , b = b and (ab ) = ab . ab = ba , a , b G

the commutative law is satisfied, so G is abelian.

11. In a group . (G , ·) if Prove that(ab ) = a b , a , b G is abelian and conversely.2 2 2 (G , ·)

(ab ) = (ab )(ab ) = (aa )(bb )2

a [ b (ab )] = a [a (bb )] using L.C.L.

it is abelian=b (ab ) = a ( bb ) (ba )b = (ab ) b using RCL ba ab Conversely

If ab = ba pre operating bya we get, a ( ab ) = a (ba ) ( a · a )b = ( ab )a post operating by b we get [(a · a ) ·b ] = [( ab )a ]b

(a · a )(b · b ) = (ab )(ab ) a b = (ab )2 2 2

{}12. Given Q , the set of non zero rational numbers is a multiplicative group and show that H is a subgroupn H = 2 n Z ,0

of Q under multiplication.{}{ }0

H = 2 n Z = ... 2 , 2 , 2 2 .....n - 2 - 1 0 , 1

i) 2 , 2 H , 2 · 2 = 2 H closure law is satisfiedm n m n m +n

ii) m n r m n r m + n r m n +r ( 22 ie )· 2 )2 = 2 ·(2 · ), m , n , r z (2 )2 = 2 (2

( m + n ) + r m + (n + r ) m +n + r m +n +r 2 = 2 i.e. 2 = 2 Associativ e law is satisfied.

iii) 2 is the identity element0

iv) 2 ; there exist 2 such that 2 · 2 = 2 Inverseof 2 is 2 .m - m m - m 0 m - m H is a group under multiplication and ie H is a subgroup of Q under multiplication. H Q

0 0


Exercise

1. If N = {1, 2, 3, ....}, which of the following are binary operation of N .a

(1 ) a * b = a + 2 b (2) a * b = 3 a - 4b (5) a * b =b

2. Which of the following operations on the given set are binary (1) on I , the set of integers, a * b = 3 a - 4 b (2) 2 2on R , a * b = a - b (3) on R , a * b = ab

3. If *is given by show that *is not a binary operation of Z .a b a b * = ,

4. Why the set of rationals does not form a group w.r.t. multiplication ?

5. If x , y , z are any three elements of a group G , find (xyz ) . –1

() - 16. In a group G , a , b G , a b .- 1 -1find

7. If the binary operation *on the set Z is defined by ,a * b = a + b + 5 find the identity element. 8. In the group of non zero integers mod 5. Find the multiplicative inverse of 4.

1 2 3 1 2 3 9. If f = and g = are permutatio ns in S , find f o g .31 2 3 2 3 1 10. If S = {1, 2, 3, 4, 5, 6} w.r.t. multiplication (mod 7), solve the equation 3 x = 5 in S .11. Show that S = {1, 2, 3} under multiplication (mod 4) is not a group.

1 2 3 4 1 2 3 412. If f = and g = find gf .

3 4 1 2 2 3 1 4

ab 13. The binary operation *is defined by on set of rational numbers, show that * is associative.a * b = ,7

ab 14. If *is defined by on the set of real numbers, show that *is both commutative and associative.a * b = ,

2

15. If *is defined by show that *is associative.2 2a * b = a + b ,

16. On the set of real numbers, Show that *is associative.R , a * b = a + b - 1, a , b R .17. On the set of real numbers, R , *is defined by , a * b = 2 examine whether a - 3b + ab * is commutative and associative.

18. In the set of rationals except 1, binary operation * is defined by Find the identity and inverse of 2. a * b = a + b - ab .ab 19. On the set of positive rational numbers Find the identity element and the inverse of 8.Q , a * b = , a , b Q .+ +

4

20. In a group of integers, an operation *is defined by Find the identity element. a * b = a + b - 1.

BTC119

Vectors and Scalars Vector : A physical quantity which has both direction and magnitude is called a 'Vector'. B Eg. Velocity, acceleration, force, weight etc.

Scalar : A physical quantity which has only magnitude and no direction is called a 'Scalar'. A Eg. speed, volume, mass, density, temperature etc.

Vectors are represented by directed line segments. LetAB be the line segment. Vector fromA to B is denoted by andAB vector from B to A is denoted by can also be represented by . The length of is magnitude of the vector denoteda BA . AB AB as or or simply a . For A is the initial point and B the terminal point.AB a AB ,

Scalar multiplication of a vector

Let be a scalar and a a vector then a represents a vector whose magnitude is times the magnitude of a and the direction is same as that of a if is positive but opposite to that of a if is negative. If it represents a null vector denoted by= 0'0 ie a' null vector is a vector of magnitude zero but its direction is arbitrary.

A vector whose magnitude is 1 is called a unit vector and a unit vector in the direction of a is written as

a a or or a (readas a cap)a a

Like and unlike vectors Vectors having same direction are called 'like vectors' and those having the opposite direction are called 'unlike vectors'.

Co-initial vectors : Vectors having the same initial point are called co-initial vectors.

Coplanar vectors : Vectors in the same plane are called 'coplanar vectors'.

Parallel vectors : Vectors having same direction but different initial C points are called 'parallel vectors'.

Triangle Law for addition of vectors

Let & represents two vectors then representsBC AC AB b AB + BC ie a + b

A B a Parallelogram Law

Let OA = a & OB = b Complete the parallelogram OACB .then OA + AC = OC by triangl e law

B C but AC = OB (parallel vectors)

OC = OA + OB = a + b b Note : - AB = OB - OA

& BA = OA - OB O A a


Properties

(i) Vector addition is commutative ie . a + b = b + a ( ) ( )

(ii) Vector addition is associative ie a + b + c = a + b + c .(iii) Set of vectors V , with binary operation vector addition will form a'Group' . The identity being 0 (null vector) and

inverse of . a is - a

Position vectorsY

(i) Let P be a point in a plane where O is the origin and OX & OY are co-ordinate axes. is called position vectors of P .OP

r Draw PQ to OX , then OQ = x &QP = y P (x , y )

j Let represents unit vectors in the direction ofi & j OX &OY . y Then OQ = x i , QP = y j

X x OQ i OP = OQ + QP = x i + y j

2 2 OP = x + y

Note :- A plane vector is an ordered pair of real numbers and the distance between O & P is the magnitude of .OP

(ii) Let P be a point in three dimensional space whereOX , OY & OZ are co-ordinate axes. Let ( x , y , z ) be the co-ordinates ofr P . Draw PQ to the plane XOY & QA & QB parallel toOY & OX respectives to meet OX atA & OY at B .

be unit vectors in the direction of OX , OY & OZ Z Let i , j & k

Then OA = x i , OB = y j , QP = z k

OQ = OA + OB (by parallelog ram law)P (x , y ,z )k = x i + y j

O B j OP = OQ + QP (by triang le law) Y i

= x i + y j + z k A

Q 2 2 2OP = x + y + z

This position vector OP of the point P is usuallyX denoted by r ie OP = r

x i + y j + z k unit vecto r in the direction of OP is given by

x + y + z 2 2 2


Z be any two points in 3-spaceLet P ( x , y , z ) & P ( x , y , z )1 1 1 2 2 2

to ,find PQ fromthe triangle OPQ Q ( x , y z )

2 2 , 2

OP + PQ = OQ

P ( x , y z )PQ = OQ - OP = x i + y j + z k - x i - y j - z k 1 1 , 12 2 2 1 1 1 = ( x - x ) i + ( y - y ) j + ( z - z )k

O Y 2 1 2 1 2 1 and unit vector in the direction of

( x - x ) i + ( y - y ) j + ( z - z ) k PQ is 2 1 2 1 2 1

(x - x ) + ( y - y ) + ( z - z )2 2 22 1 2 1 2 1

X Scalar product of two vectors

are two non-zero vectors, then is defined as 'scalar product'If a = a i + a j + a k and b = b i + b j + b k a b + a b + a b 1 2 3 1 2 3 1 1 2 2 3 3

of two vectors also known as 'dot product'.a & b , denoted as a ·b

Vector product of two vectors

are two non-zero vectors, then a vectorIf a = a i + a j + a k , b = b i + b j + b k 1 2 3 1 2 3

( a b - a b )i + ( a b - a b ) j + ( a b - a b )k 2 3 3 2 3 1 1 3 1 2 2 1

i j k ie a a a is defined as 'vector product'of two vector a & b , denoted as a ×b also known as cross product.1 2 3

b b b 1 2 3

i j k ie a × b = a a a = - (a b a b ) i 1 2 3 2 3 3 2

b b b 1 2 3

Geometrical Meaning of a · b and a × b B Let OA = a & letOB = b and A O B =

b then a ·b = ab cos

OA where a = a & b = b a when cos= 0° , a · b = ab 0° = ab when = 90 ° , a · b = ab cos 90 ° = 0 hence two vectors are said to be 'orthogonal' if .a · b = 0

Let ON represent a line which is to both OA & OB . ie ON is to the plane OAB . Let represent a unit vector inr r n the direction of ON .

Then ab sin n is a × b and - ab sin n is b × a if = 0 °, then sin = 0 is a null vector, hence two vectors are said to be parallel or coincident ifa × b or - b × a a × b = 0 (null vector)


Note :- N

B n b a × b = ab sin n represent anti - clockwise rotation.

O A a

B b b × a = - ab sin n represent clockwise rotation.

O A a

- n

N also for unit vectors i , j & k

i × j = k , k × i = j , j × k = i & j × i = - k , i × k = - j , k × j = - i .

Projection of b upon a B

r Let OA = a , OB = b & A OB = . Draw BD to OA then OD isthe projection of b upon a .b a ·b

Since a · b = ab cos , cos = ab O D A

a OD OD

fromthe OBD cos = =l e OB b

a · b OD = b cos = b ·

ab a

= · b = a ·b a

further area of parallelogram whose adjacent sides areOA & OB is given by

BD OA · BD = ab sin Q = sin but a × b = ab sin .

OB Area of parallelogram whose co-terminus edges are a & b is given by a × b

1l e and therefore area of OAB = a × b .2

1Also area of the parallelogram whose diagonals are d & d is d × d 1 2 1 22


Scalar Triple Product

a = a i + a j + a k , b = b i + b j + b k & c = c i + c j + c k Let 1 2 3 1 2 3 1 2 3

is called 'Scalar Triple Product'Then a · ( orb × c ) ( a × b ) · c both on computation gives a b c - a b c + a b c - a b c + a b c - a b c

1 2 3 1 3 1 2 3 1 2 1 3 3 1 2 3 2 1 a a a 1 2 3

which is b b b 1 2 3

c c c 1 2 3

Thus a · ( orb × c ) ( a × b ) ·c [ ]which is 'Scalar Triple Product' which is usually denoted as also called 'Box-Product'. a b c

[ ]Geometrical Meaning of a b c N

Consider a parallelopiped whose co-terminus edges areOA , OB , OC ie a , b , c

Area of parallelogram A OBDC is b × c P a Let OP bethe projection of OA ie a upon ON which is tor

c D the plane OBDC . C () O B b × c a · b × c b OP = · a =

b × c b × c Volume of parallelopiped = area of parallelogram × OP

( ) a · b ×c

= b × c b × c

( ) [ ]= a · b × c = a b c Volume of parallelopiped whose coterminus edges are given by isa , b , c

a a a 1 2 3[ ] = b b b a b c 1 2 3

c c c 1 2 3

[ ]If = 0 then the vectors a , b & c are coplanar.a b c

Vector Triple Product

If are three non-zero vectors thena , b , c ()()()

a × b × c = a ·c b - a ·b c () ()()also a × b × c = a ·c b - b · c a


Examples

1. If a = ( 2 , 5), b = (- 2 , 3) find a ·b .Solution : 11a ·b = - 4 +15 =

()()2. a ß a ß a ß If = i + 2 j - 3 k , = 3 i - j - 2 k , find 2 + · + 2 ()()Solution : a ß a ß 2 + · + 2 = ( 5i + 3 j - 8k ) · ( 7 i - 7 k ) = 35 + 0 + 56 = 91

3. Prove that the vectors a = 3i - 2 j - k , b = 2i + j - 4k are perpendicu lar to each other.

Solution : a ·b = 6 - 2 - 4 = 0 a & b are perpendicu lar to each other.

4. Find m such that 3 i + m j + k and 2 i - j - 8 k are orthogonal .

Solution : (3 i + m j + k ) · ( 2 i - j - 8 k ) = 0 6 - m - 8 = 0 - m = - 2

5. If a + b = a - b . Show that a and b are orthogonal to each other.

Solution : Given a + b = a - b ( ) ( ) ( ) ( )

Squaring both the sides, a + b · a + b = a - b · a - b ()a · a + a · b + b ·a + b ·b = a · a - a ·b - b · a + b · b i.e. 2 a · b = 0

a · b = 0 a is perpendicu lar to b .

6. Find the projection of a = i + 3 j + 5k on b = - 3i + j + k .a · b - 3 + 3 + 5 5 Solution : Projection of a on b = = = .

9 +1 +1 11b

7. If a = 2i + j + k and b = i - 2 j + 2 k and c = 3i - 4 j + 2 k , find the projection of a + c on b .()()( ) ( ) a + c · b 5i - 3 j + 3 k · i - 2 j + 2k 5 + 6 + 6 17Solution : Projection of a + c on b = = = =

3 31 + 4 + 4 b

8. Find the cosine of the angle between ve ctors a = 4 i - 3 j + 3k and b = 2 i + j - k a · b 8 - 3 - 3 2Solution : cos = = =

16 + 9 + 9 4+ 1 +1 34 6a b

9. Find the cross product of vectors a = 9 i - j + 4 k , b = 8 i - j + k i j k

Solution : a ×b = 9 - 1 4 = i (- 1 + 4 ) - j (9 - 32) + k (- 9 + 8 ) = 3 i + 23 j - k 8 - 1 1

10. If , a = i + 2 j + 2k and b = 2i + j - 2k then find a × b


ˆˆ ˆi j k ˆ ˆˆ ˆ ˆ ˆSolution : a b 1 2 2 i ( 4 2 ) j ( 2 4 ) k (1 4 ) 6 i 6 j 3 k × = = - - - - - + - = - + -

2 1 - 2

a × b = 36 + 36 + 9 = 81 = 9

211. Find the unit vecto r perpendicu lar to the pair of vectors a = 6 i - 2 j + k and b = 3 i + j - k i j k

Solution : Vector perpendicu lar toa & b is a × b = 6 - 2 1 = i (4 - 1 ) - j (- 12 - 3 ) + k (6 + 6) = 3i +15 j +12k 3 1 - 2

a × b 3i +15 j +12k 3 i + 15 j + 12k n = = =9 + 225 +144 378a × b

( ) ( ) ( )12. Prove that 2a + b × a + 2b = 3 a × b

()()()() ()() ()Solution : 2a + b × a + 2b = 2 a × a + 4 a × b + b × a + 2 b × b = 4 a × b + b × a = 3 a × b .

()()() 13. If a , b , c are non zerovectors prove that a × b + c + b × c + a + c × a + b = 0

Solution : a × b + a × c + b × c + b × a + c × a + c × b = a × b + a × c + b × c - a × b - a × c - b × c = 0.

214. Find the sine of the angle between th evectors a = i + j + k and b = 2 i - 3 j + k

i j k Solution : a ×b = 1 1 1 = i (2 + 3 ) - j (2 - 2) + k (- 3 - 2 ) = 5 i - 5 k , a × b = 5 + 5 ,2 2

2 - 3 2

2 2 5 + 5 5 2sin = =

1 + 1 + 1 4 + 9 + 4 3 17

15. If a + b + c = 0, then show that a × b = b × c = c × a Solution : Given a + b + c = 0

( ) ( ) [ ]a × a + b + c = a × a + a × b + a × c = 0 Q a × b = - a × c a × a = 0 a × b = c × a Similarly b × a = c × b or c × b = - a × b or a × b = b × c Similarly c × a = b × c a × b = b × c = c × a

5 16. Find the area of the triangle whose sides are a = 3 i - 2 j + k and b = i - 3 j + k 1

Solution : A = a × b 2

i j k a × b = 3 - 2 1 = i (- 10+ 3 ) - j (15 - 1 ) + k (- 9 + 2 ) = - 7i - 14 j - 7k

1 - 3 5


49 +196 + 49 294area = = sq.units

2 2

17. Find the area of trianglePosition v ectors of the points A , B and C are respective ly i - j + k , 2 i + j - k and 3i - 2 j + k .ABC .

Solution : AB = OB - OA = ( 2 i + j - k ) - ( i - j + k ) = i + 2 j - 2 k .AC = OC - OA = 3 i - 2 j + k - (i - j + k ) = 2 i - j .

i j k AB × AC = 1 2 - 2 = i (0 - 12 - j (4) + k (- 1 - 4 ) = - 2i - 4 j - 5 k .

2 - 1 0

1 45area = 4 + 16 + 25 = sq.units2 2

18. Find the area of a parallelog ram whose adjacent sides are a = 3 i + 2 j - k and b = i + 2 j + 3 k .

i j k Solution : a b 3 2 1 i 6 2 j 9 1 k 6 2 8i 10 j 4k × = - = ( + ) - ( + ) + ( - ) = - +

1 2 3

A = 64 + 100 + 16 = 180 sq.units

19. Find the area of a parallelog ram whose diagonals are d = 3 i + j + 2 k and d = i - 3 j + 4k . 1 2

1 Solution : A = d × d

1 2 2 i j k

d d 3 1 2 i 4 6 j 12 2 k 9 1 10i 10 j 10k × = = ( + ) - ( - ) + (- - ) = - -1 2 1 - 3 4

1 1 d × d = 100 + 100 + 100 ; d × d = 300 sq.units = 5 3 sq.units.

1 2 1 2 2 2

20. Find the scalar tri ple product of vectors a = 2 i - j + 3k , b = i + 2 j + 3k and c = 3 i + j - k .

( ) 2 - 1 3 Solution : a · b ×c = 1 2 3 = 2( 35- 2 - 3 ) +1( - 1- 9) + 3(1- 6 ) = - 10 - 10 - 15 = - .

3 1 - 1

[]Evaluate i - j , j - k , k - i 21.

1 - 1 0 Solution : 0 1 - 1 = 1 (1 ) + 1(- 1 ) + 0 = 0

- 1 0 1

22. Find the volume of parallelopiped whose coterminus edges are

a = i - j + k , b = 2i - 4 j + 5 k , c = 3 i = 5 j + k .


( ) 1 - 1 1 Solution : a · b ×c = 2 - 4 5 = 1 (- 4+ 25) +1(2- 15) +1(- 10 +12) = 21 - 13 + 2 = 10 cubic units

3 - 5 1

23. Find if the vectors a = ( 2 , - 3 4), b = (1 , 2, - 1) and c = ( , - 1 , 2) are coplanar.

[] 2 - 3 4Solution : a b c = 1 2 - 1 = 0

- 1 2

i.e., 2(4 - 1) + 3(2 + ) + 4(- 1 - 2 ) = 0 8

6 + 6 + 3 - 4 - 8 = 0 5 = 8 ; 5 = .

24. Show that the points A (4, 5, 1), B (0 , - 1, - 1), C (3, 9 , 4 ) and D (- 4, 4 , 4) are coplanar.

Solution : OA = 4 i + 5 j + k , OB = - j - k , OC = 3 i + 9 j + 4k , OD = - 4i + 4 j + 4 k AB = OB - OA = - 4i - 6 j - 2 k , AC = OC - OA = - i + 4 j + 3k , AD = OD - OA = - 8 i - j + 3k

Consider

( ) - 4 - 6 - 2

AB · A C × A D = - 1 4 3 = - 4 (12 + 3) + 6 (- 3 + 24) - 2(1 + 32) = - 60 + 126 - 66 = 0- 8 - 1 3

A , B , C and D are coplanar.

Exercise

1 . The position vectors of the points A, B and C are respective ly a , b and 2 a - 3 b . Express vectors BC , AC , AB in terms of

a and b .2. Position v ectors are A , B , C and D are 2a + 4c , 5a + 3 3 b + 4c , - 2 3 b + c and 2a + c respective ly.

3Show that AB || CD and AB = CD

2

3. If a = 3i - 4 j , b = 2i - 3 j , c = i + j , find (i) 2b - c + 2 a (ii) 2 a - 3c + 2b 4. Find the unit vecto r in the direction of a + b + c where a = i + 4 j + 2k , b = 3i - 3 j - 2k and c = - 2i + 2 j + 6 k .5. If a = 2 i + 3 j and b = - i + 2 j , find a ·b .6. Show that the vectors (–1, 2, 3) and (2, –5, 4) are orthogonal.

7. Find the values of such that i - 3 j + k and i + j + 2 k may be orthogonal . 8. If a and b are unit vecto r show that the vectors a + b and a - b are orthogonal .

9. If a = (1 , - 1, 3 ) and b = (2, 1 , 1 ), find a × b .10. Find the projection of a on b where a = i + 2 j - 3k and b = - 2i + 3 j .11. If a = 2 i + j - 3 k and b = i - 2 j + k . Find the projection of b on a .


12. If a = (2 , 3 ) and b = (8 , m ) and a × b is a null vector, find m .13. If a = (2 , - 1 , 3), b = (- 2 , 1, 4 ) and c = (2 , 1, - 7), find the unit vecto r in the direction of a + b + c .14. f f Show that cos sin i + sin j + cos k is a unit vecto r.

15. Show that points A (3, –2, 4), B (1, 1, 1) and C (–1, 4, –2) are collinear.

16. Show that points A (1, 1, 1), B (7, 2, 3), C (2, –1, 1) form a triangle.

17. respectively form aShow that points A , B and C whose position v ectors are 2 i + j + k , i + 3 j - 5 k and 3 i - 4 j - 4k right-angled triangle.

18. Find the cosine of the angle between th evectors a = i + j + k and b = 2i + 3 j - 2 k .19. Find the sine of the angle between th evectors 2i - 3 j + k and 3i + j - 2k .20. Find the unit vecto r perpendicu lar tothe vectors 3 i + j + 2k and 2 i - 2 j + 4k . 21. Find the volume of the parallelopiped whose co-terminal edges are represented by a = 2i - 3 j + 4k ; b = i + 2 j - k and

c = 3 i + j + 2 k .22. Show that vect ors 2 i - j + k , i + 2 j - 3k and 3 i - 4 j - 5k are coplanar.

23. If the vectors i + 2 j + 5k , 2 i + x j - 10k and 3i + 9 j - 2k are coplanar. find x .24. Show that points A (2, 3, –1), B (1, –2, 3), C (3, 4, –2) and D (1, –6, 6) are coplanar. 25. Find the scalar tri ple product of, a = i - 2 j + k , b = 2 i + j + k , c = i + 2 j + k .

( ) ( ) ( )26. Prove that 2 a + 3 b × a + 4b = 5 a × b 27. Find the area of the triangle whose 2 sides represente d by a = i - 3 j - 2 k , b = i + 2 j + k .28. Find the area of the triangle whose vertices are i - j + 2k , 2 j + k and j + 3k .29. Find the area of parallelog ramwhose diagonals are a = 3 i + j - 2k and b = i - 3 j + 4 k .

( ) ( )30. If a = (1, - 1 , 2), b = (1 , 2, 3) & c = (3 , - 2, 4). Find a × b × c and a × b × c .

BCA 21 / IMCA 21 / Mathematics SVT 65

DIFFERENTIAL CALCULUS

Limits of functions

x - 1 2Consider y = f ( x ) = the function is defined for all values of x except for x = 1.

x - 1

1 - 1 0 which is indeterminate.for x = 1, f ( x ) = = 1 - 1 0

Let us consider the values of f (x ) as x approaches 1 2x - 1

x f ( x ) = x - 1

.9 1.9

.99 1.99

.999 1.999

1.01 2.01

1.001 2.001

1.0001 2.0001

2 x - 1It can be seen from the above values that as x approaches 1, approaches 2.x - 1

2 2 2 x - 1 x - 1 x - 1This value 2 is called "Limit of as x approaches 1"which can bewritten as lim 2 or Lt = .

x - 1 x - 1 x - 1x 1 1x In general the limit of a function f ( x ) as x approaches a is denoted as l and which is written as

lim orf (x ) = l Lt f (x ) = l x a x a

Properties

[](1) lim f (x ) ± g (x ) = lim f (x ) ± lim g ( x )x a x a x a

[](2) lim f (x )g (x ) = lim f (x ) lim g (x )wherein particularlim kf (x ) = k · lim f (x ) k is a contant x a x a x a x a x a

lim f (x )f (x )(3) . lim 0 = provided lim g (x )x a g (x ) lim g (x )x a x a

x a

Standard Limits

n n x - a (1) lim = na n - 1 x - a x a

70 KSOU Differential Calculus

sin tan(2) lim = 1 ( in radians) also lim = 1 0 0

n + 1 1lim 1 ) = e 2 < e < 3 or lim (1 + x = e (3) x

n n 8 x 0

a - 1 e - 1x x (4) lim = log a ( a > 0) in particular lim = 1 .

e x x x 0 x 0

Examples

x + 4 x + 3 0 + 0 + 3 3 2 (1) lim = =

0 - 0 + 4 4 x - 5x + 4 2 0x 3 42 - +

2 - 0 + 0 22 x 2 x - 3x + 4 2x (2) 3= lim = =lim (dividing Nr & Dr by x )22 1 3 + 0 + 023x + 2 x + 1 x 8x 8 3 + +x 2x

x - 81 x - 34 4 4(3) lim = lim = 4 (3 ) = 1083

x - 3 x - 33 3 x x 7 7x - (- a )7 7x + a

x + a x - (- a ) 7 ( - a ) 7 7 7 6x + a 2(4) lim = lim = lim = = a 5 5 5 5 5 5 5 4x + a x + a x - (- a ) 5 - (- a )x - a x - a x - a

x + a x - (- a )

sin 7x sin 7 x (5) lim = lim × 7 = 1 × 7 = 7x x x 0 x 0

21 - cos 2 2 sin(6) lim = lim = 2 2 20 0

tan 3x - 1

tan 3 x - x 3 - 1x (7) lim (dividing Nr & Dr by x ) = lim = = 1sin x 3 x - sin x 3 - 1x 0 x 0 3 -

x ab 1b ab (8) lim(1 ) + ax ) = lim (1 + ax = e x ax

x 0 x 0

3n n 3 3+ + 3 3(9) lim 1 = lim 1 = e n n n 8 n 8

- 211 - lim(1 - 2 x ) = lim (1- 2 x ) = e - 2(10) x 2x

x 0 x 0

a - b (a - x ) - (b - x ) a - x b - x a x x x x x x (11) lim = lim = lim - lim = log a - log b = loge e e x x x x b x 0 x 0 x 0 x 0


x 2 - 1log 2 x 2 - 1 x (12) 2 lim = lim = = log e

e sin x sin x 1x 0 x 0

x

Continuity of a function

A function f (x ) is saidto be continuous at x = a if lim f (x ) = f (a )x a

A function f (x ) is saidto be continuous if lim f (x ) exists, f (a ) exists and they areequal.x 0

If these donot happen then the function is said to be not continuous or discontinuous. A function f ( x ) is said to be continuous in an interval if it is continuous at all points in the interval.

Examples

x - 1 x - 1 0 2 2(1) f ( x ) = is not continuous at x = 1 Q lim = 2 exists but f (0 ) = donot exists.

x - 1 x - 1 0 x 1 where as it is continuous at all other values of x .

4 x + 3 for x = 4(2) at x = 4.Discuss the continuity of function f ( x ) =

3 x + 7 for < 4

Solution : While finding the limit of a function f ( x ) as x approaches a , if we consider the limit of the function asx approachesa from left hand side, the limit is called 'Left Hand Limit' (LHL) and if x approaches a from right hand side the limit is called 'Right Hand Limit' (RHL) and the limit of the function is said to exists if both LHL & RHL existsand are equal, for convenience LHL & RHL are denoted as )lim f (x )and furtherand lim f (x

- +x a x a LHL= lim f (x ) = lim f (a - h ) & RHL = lim f (x ) = lim f (a + h )

x a - h 0 x a + h 0 For the given problem LHLat x = 4 is lim f (x ) = lim f (4 - h ) = lim 3 (4 - h ) + 7 = 19

x 4- h 0 h 0 RHL at x = 4 is lim f (x ) = lim f (4 + h ) = lim 4(4 + h ) + 3 = 19

h 0 h 0x 4 +and f ( 4) = 19

The function is continuous at x = 4 sin x

for x 0 (3) at x = 0.Examine the continuity of f ( x ) = x

2 for x = 0

sin x Solution : lim = 1 , but f ( 0 ) = 2

x x 0

The function is discontinu ous at x = 0.

5 x - 4 for 0 < x = 1 (4) 2Examine the continuity of the function f ( x ) = 4 x - 2 x for 1 < x < 2

4 x + 4 for x = 2 at x =1 and x = 2.


at x = 1, LHL = lim f ( x ) = lim f (1 - h ) = lim 5 (1 - h ) - 4 = 1-1 h 0 h 0x

RHL = lim f (1 + h ) = lim 4 (1 + h ) - 2(1 + h ) = 4 - 2 = 22

h 0 h 0

LHL RHL at x = 1

The function is discontinuous at x = 1.

at x = 2 , LHL = lim f ( 2 - h ) = lim 4( 2 - h ) - 2( 2 - h ) = 16 - 4 = 122

h 0 h 0

RHL = lim f (2 + h ) = lim 4(2 + h ) + 4 = 12h 0 h 0

and f (2) = 4 × 2 + 4 = 12

lim 2 f (x ) = f ( )x 2

The function is continuous at x = 2 .

Differentiability of a function

f ( a + h ) - f ( a )A function f ( x ) is said to be differenti able at a point a if lim exists and the derivative is denoted as f '(a ).

h h 0

d f ( x + x ) - f ( x )A function f ( x ) is said to be differenti able at x if exists andthe derivative is denoted as f '( x ).lim

d x d x 0

[d x is called the increment in x which is very very small].

dy If y = f ( x ), the derivative is denoted by or f '( x )

dx Note :- A function which is differentiable is always continuous but the converse is not always true.

- x for x < 0 is continuous for all x but not differentiable at x = 0.eg. y = f ( x ) = x =

x for x = 0

To find the derivates of x , log x , a , sin x , cos x and a constant C with respect to x .n x e

d n n dy ( x + x ) - x (1) Let y = x , = = nx n n - 1 limd dx x + x - x d x 0

()() ()d d d 9 9 7 6 - 4 - 5 - 1 Eg. (i) x = 7 x , ( ii ) x = - 4 x (iii) x = x = x 9 9 54 4 4

dx dx dx 4 4 ()() d 1d 5 5 ( v) x =- - - 1 -(iv) x = - x = - x 5 5 8

3 3 3

dx dx 3 3 2 x d dy log ( x + x ) - log x (2) Let y = log x = e e then limd e dx x d x 0

x d d 1 x x x 1 x 1 1+ + d x = lim · log = lim · log 1 = log e = .

d e x x x x x x x d d x 0 x 0 Let y = a x (3)

d d x + x x x dy a - a (a - 1 ) d x x in particular= lim 0 = lim a = a log a ( a > ) (e ) = e x x d d e dx x x dx d d x 0 x 0


(4) Let y = sin x d dy sin( x + x ) - sin x

= limd dx x d x 0

d d d x + x + x x + x - x x 2 cos sin sind + x 2 2 2= lim = lim cos x · = cos( x + 0 ) ·1 = sin x .

d d x x 2d d x 0 x 0

2

(5) Let y = cos x d dy cos( x + x ) - cos x

= limd dx x d x 0

d x sind + x 2= lim - sin x · = - sin( x + 0 ) ·1 = - sin x .

d x 2d x 0

2

dy c - c (6) Let (y = c a constant) = = 0 limd dx x d x 0

Thus derivative of a constant is zero.

Thus we have the following standard derivatives

Function y = f ( x ) Derivative

x nx n n - 1

1log x

x a log a x x a e cos x sin x

cos x - sin x constant zero

Rules for differentiation I Sum Rule

dy If &y = u + v whose u v are functions of x then to find

dx d d d d Give a small increment x to x , let the correspond ing increments in u , v & y be u , v & y respective ly.

d d d Then y + y = u + u + v + v Subtracting

y + d y - y = u + d u + v + d v - u - v d d d ie y = u + v d d d y u v d divide through out by x then = + d d d x x x

d take limits on both sides as x 0 .d d d y u v

Then lim = lim + limd d d x x x d d d x 0 x 0 x 0


dy du dv ie = +

dx dx dx dy du dv dw

Note : - If y = u + v - w then = + -dx dx dx dx

dy d d d Eg. (1) If y = x + sin x - cos x then = ( x ) + (sin x ) - (cos x )6 6

dx dx dx dx 5 5= 6 x + cos x - (- sin x ) = 6 x + cos x + sin x .

dy 1 ( 2) If y = 3 - log x + c then = 3 log 3 - + 0 .x x

e e dx x

II Product Rule

dy If y = uv where u & v are functions of x then to find .

dx d d d d Give a small increment x to x , let the correspond ing increments in u , v & y be u , v & y respective ly.

d d d d d d d then y + y = ( u + u )( v + v ) = uv + u v + v u + u · v d y = y + d y - y = u d v + v d u +d u ·d v

d divide through out by x ,d d d d y v u v d = u + v + u · d d d d x x x x

take limit on bothsides as d x 0,

d d d d y v u v d = u + v + u ·then lim lim lim limd d d d x x x x d d d d x 0 x 0 x 0 x 0

dy dv du dv ie = u + v + o

dx dx dx dx dy dv du

ie = u + v dx dx dx

dy dv Note (1) If y = kv where k is a constant then = k

dx dx dy dw dv du

(2 ,) If y = uvw then = uv + uw + vw dx dx dx dx

dy d d 3 3 3Eg. (1) If y = x sin x then = x (sin x ) + sin x x 2 2 2

dx dx dx 3 3 1

= x cos x + sin x · x 2 2

2 dy

( 82) If y = 8 cos x then = - sin x dx

dy d d d ( 3) If y = e sin x · log x then = e sin x (log x ) + e log x · (sin x ) + sin x · log x ( e )x x x x

dx dx dx dx 1

= e sin x · + e log x · cos x + sin x · log x · e .x x x x


III Quotient Rule

u dy If y = where u & v are functions of x then to find .

v dx d d d d Give a small increment x to x , let the correspond ing increments in u , v & y be u , v & y respective ly.

d u + u d then y + y =d v + v

d d d d d u + u u v (u + u ) - u (v + v ) vu + v u - uv - u v d y = - = =d d d v + v v v ( v + v ) v (v + v )

d divide through out by x d u d v v - u d y d x d x =

d x v (v +d v )d take limit on both sides as x 0

d u d v v - u d y d x d x ie lim = lim

d x v (v + d v )d x 0 d x 0 du dv

v - u dy dx dx ie =dx 2v

This rule can be easily remembered in the following manner u Nr

If y = = (say)v Dr

dy Dr ( derivative of Nr ) - Nr (derivative of Dr )Then =

dx 2( Dr )k dy k dv

Note : - If y = where k is a constant = -then v dx dx v 2

sin x Eg. (1) If y = tan x =

cos x d d

cos x (sin x ) - sin x (cos x )dy dx dx =then cosdx x 2

cos x · cos x - sin x (- sin x ) cos x + sin x 12 2 = = = = sec x .2

cos x cos x cos x 2 2 2

dy 2If y = tan x , then = sec x .dx

cos x ( 2) If y = cot x =

sin x dy sin x ( - sin x ) - cos x (cos x )

then sin =dx x 2

2 2- sin x - cos x - 1= = = - cosec x . 2

sin x sin x 2 2


dy If y = cot x , then - cosec x .2

dx

( 13) If y = sec x = cos x

dy 1 d then = - · (cos x )

dx dx 2cos x 1 1 1 sin x

= - ( - sin x ) = (sin x ) = · = sec x tan x .cos x cos x cos x 2cos x dy

If y = sec x , then = sec x tan x .dx

( 14) If y = cosec x = sin x

dy 1 then = - · cos x

dx sin x 2

1 cos x = - · = - cosec x · cot x .

sin x sin x dy

If y = cosec x , then = - cosec x ·cot x .dx

IV Chain Rule or function of a function rule d d d dy y y u

If , y = f ( u ) where u = g ( x ) to find consider = × d d d dx x u x

d d d y y u lim = lim ×

d d d x u x d d x 0 x 0 dy dy du dy dy du dv

ie = · also if y = f ( x ), u = g (v ) & v = h ( x ) then = · ·dx du dx dx du dv dx

2 n 2 Eg. ) (1) If y = (ax + bx + c then put u = ax + bx + c dy

y = u = nu n n - 1

du 2 u = ax + bx + c

du = 2 ax + b

dx dy dy du 2 n = · = (ax + bx + c ) ( 2 ax + b ).dx du dx

3 2 ( 72) If y = log( x - 2x + )

2dy 1 d 3x - 4 x then = × ( x - 2 x + 7 ) =3 2

dx dx 3 2 3 2( x - 2 x + 7 ) x - 2x + 7

( 33) If y = sin( 2x + 4 x - )2

dy then = cos( 32 x + 4 x - 3 )( 4 x + 4) = 4( x + 1 ) cos( 2 x + 4 x - )2 2

dx


2x - 1(4) If y = tan2x + 1

dy x - 1 d x - 1 2 22then = sec

dx dx x + 1 x +1 2 2

2 2 2 2 2 2x - 1 (x +1 )(2x ) - (x - 1 )2 x x - 1 2x (x +1 - x + 1 )2 2= sec × = sec ×2 2 2 2 2 2x +1 (x +1) x +1 (x +1 )2x - 1 4 x = sec ×2

x +1 (x +1 )2 2 2

Derivative of Hyperbolic functions (1 ) If y = sinh x ,

+dy d d e e e - e x - x x - x then = (sinh x ) = = = cosh x .

dx dx dx 2 2 ( 2) If y = cosh x ,

- x - x x - x dy d d e e e + e then = (cosh x ) = = = sinhx .dx dx dx 2 2

sinh x (3 ) If y = tanh x = ,

cosh x dy d sinh x

then = dx dx cosh x

2 2cosh x cosh x - sinh x sinh x cosh x - sinh x 1= = = = x . 2sech

cosh x cosh x cosh x 2 2 2

cosh x ( 4) If y = coth x = ,

sinh x dy d cosh x

then = dx dx x sinh

sinh x sinh x - cosh x cosh x sinh x - cosh x - 12 22= = = = - cosech x .

sinh x sinh x sinh x 2 2 2

1 (5 ) If y = sech x = ,

cosh x dy d 1 1 d

then = = - (cosh x )dx dx cosh x dx 2 cosh x

- 1 - 1 sinh x = · sinh x = · = - sech x tanh x .

sech x cosh x 2 cosh x


1(6 ) If y = cosech x = ,

sinh x dy - 1 d

= - (sinh x )then dx dx sinh x 2

- 1 - 1 cosh x = ·cosh x = · = - cosech x · coth x .

sinh x sinh x 2 sinh x

Implicit Functions

Function of the type is called Implicit function. f ( x , y ) = 0 dy

To &find treat y as a function of x use chain rule. dx

2 2Eg. (1) If ax + 2 hxy + by = 0

dy dy +then 2 ax + 2h x y + 2by = 0

dx dx dy

ie (2 hx + 2 by ) = - 2 ax - 2hy dx

dy - 2(ax + hy ) - (ax + hy )= =

dx 2( hx + by ) hx + by Eg. (2) If x sin y + y sin x = 10

differentiating w.r.t.x dy dy

x cos y + sin y + y cos x + sin x = 0dx dx

dy ie (x cos y + sin x ) = - sin y - y cos x

dx dy - (sin y + y cos x )

=dx ( x cos y + sin x )

Parametric functions

taken together is called Parametric function, where t is the paramter. ParametricFunctions of the type x = f ( t ), y = g (t )functions are also denoted as x = f ( ), y = f ( )where isthe parameter.

dy dx dy dx dy To &find , consider or &

dx dt dt d d dy dy dt dy dy d

then = or =dx dx dt dx dx d

dy 3 3Eg. (1) If x = a cos t , y = a sin t , find

dx dx dy 3 2 3 2Solution : x = a cos t , = - 3 a cos t sin t y = a sin t , = 3a sin t cos t ,dt dt

dy dy dt a sin t cos t 23 = = = - tan t dx dx dt 2- 3a cos t sin t


dy 3 3(2) If x = a (3 cos - 4 sin ) & y = a (3sin - 4 cos ), find dx

dy dy d a (3 cos + 12 cos sin )2

Solution : = = dx dx d 2a ( - 3 sin - 12 sin cos )

3 cos (1 + 4 sin cos )= = - cot

- 3 sin (1 + 4 sin cos )

Differentiation using Logarithms g ( x ) If y = f (x )

use logarithms on both sides

g ( x )log y = log f ( x ) = g (x ) log f (x )differentiating w.r.t.x 1 dy 1 g ( x ) f ' ( x )

= g ( x ) f '( x ) + g '( x ) log f ( x ) = + g '( x ) log f ( x )y dx f ( x ) f (x )

+ dy g ( x ) f '( x )g x = f ( x ) g ' ( x ) log f ( x )( )

dx f ( x )dy

Eg. (1) Find if y = x x dx

Solution : log y = x log x 1 dy 1

= x · + log x · 1y dx x dy x ie = x [1 + log x ]dx

dy s i n x ta n x (2) If y = x + (cos x ) , find .

dx Solution : Let y = u + v where u = x & v = (cos x )sin x tan x

log u = sin x log x 1 du sin x

= + cos x log x u dx x du sin x +ie = x cos x log x s i n x dx x

tan x v = (cos x )log v = tan x log cos x

1 dv 1 ( - sin x ) 2 = tan x + sec x log cos x v dx cos x dv tan x 2 2= (cos x ) [ - tan x + sec x log cos x ]dx dy du dv + sin x

= + = x cos x log x + (cos x ) [ - tan x + sec x log cos x ]si n x t an x 2 2

dx dx dx x


Derivatives of inverse Trigonometric functions

d 1 (1 ) (sin x ) = for x < 1-1 dx 21 - x d - 1(2) (cos x ) = for x < 1 - 1dx 2 1- x d 1

(3 ) (tan x ) =- 1

dx 21 + x d - 1- 1( 4) (cot x ) =

dx + x 21

d 1- 1(5 ) (sec x ) = for | x | > 1dx x x - 12

d - 1- 1(6 ) (cosec x ) = for x > 1 dx x x - 1 2

Derivatives of inverse hyperbolic functions d 1- 1(1 ) ) (sinh x = x dx 1+ x 2d 1- 1(2) (cosh x ) = for x > 1dx 2 x - 1

d 1- 1(3 ) (tanh x ) = for x < 1

dx 1 - x 2d - 11( 4) (coth x ) = for x > 1 -

dx 2x - 1

d - 1 (5 ) (sech x ) = for x < 1 - 1dx 2x 1 - x d - 1- 1(6 ) (cosech x ) =dx x 1+ x 2

Exercise

dy Find of the following

dx 2 21. y = x + a 2 . y = log (3 x + 2) 3 . y = log cos x e e x + 1. y = . y = x e . y = ( x + )2 x 24 5 6 2 3 x - 1

2 x 1/ 37 . y = ax + bx + c 8 . y = e 9 . y = (3 x + 5)

2 x 5 - 110. y = sin x + log x + x + e 11. y = cosec x - tanx + x 12. y = sinh x e 13. y = sinh x sinh x 14. y = sin x · sin x 15. y = sec x + cosec x - 1 - 1 - 1 - 1


2x + x + 1 2 -1 216. y = (1 + x )tan x 17. y = (1 + x ) sin x 18. y =x

19. y = (x +1 ) (x + 2) 20. y = 5e + 4 log x 21. y = e (sin x + cos x )2 x x e

x x - x e e + e 22 . y = 23 . y = 24 . y = sin x ·sin 2 x

22x 2x x + a - 1 - 125 . y = ( x + a )( x + b )( x + c ) 26 . y = sin 27 . y = tan

1 - ax 21 + x 1 - cos x

28 . y = 29 . y = e 30 . y = log sin x s in h x e 1 + cos x

1 + x 31 . y = log ( x tan x ) 32 . y = (1 - x ) cos x 33 . y = tan2 2 - 1 - 1

e 1 - x 1 - x - 1 2 - 1 - 134 . y = tan 35 . xy = c 36 . sin x + sin y = 0 1 + x

2 2 2 2 / 3 2 / 3 2 /3 237. x + y = a 38. x + y = a 39. x = at , y = 2at 1

40 . x = t , y = 41 . x = 4 cosh t , y = 4 sinh t 42 . x = a cos t , y = b sin t t

243. x = log sect , y = tan t 44. y = x cos xy 45. y = sin xy e 2 x x - x ( x +1 )e e - e 46. y = 47. y = 48. y = (1+ x ) tan x 2

log x ·cosec x x - x e + e e x 2 x - 1 xe x + e cos x 49. y = 50 . y = 51. y =

1+ cos x 1- x log x log x e e x e x + x - x 3 x 2 31 3 - 1 - 152. y = 53. y = sec 54. y = tan

- 1 2 2cos x x - 1 1 - 3x 35x cosh x 1 + sin x x + 355. y = 56. y = 57. y = sec x

cosech x 1 - sin x x - 1

58. y = (x - 1 ) cosec (x - 1 ) 59. sin(x + y ) = log (x + y ) 60. y sin x + x sin y = 0 2 - 1e

61. x = e (cos t + sin t ), y = e (cos t - sin t ) 62. x = a cos t , y = a sin t t t 4 4

a a 2 n 63 . x = log sec t , y = tan t 64 . x = , y = 65 . y = sin x · sin nx e 1 + cos t 1 - cos t

66 . y = (sin x ) 67 . y = (sin x ) 68 . y = (sin x )- 1 ta n x si n x ta n- 1 x l og x x + y x y 69 . y = (log x ) 70 . e = e + e e

x


Successive Differentiation

dy If y = f ( x )then = f '( x ) is also a function of x , hence further derivative s can be obtained.

dx 2d y '' 2The second derivative is denoted by or f ( x ) or y or D y .

2 2dx 3d y ''' 3The third derivative is denoted by or f ( x ) or y or D y .

3 3dx n d y t h n n In general n derivative is denoted as or f ( x ) or y or D y .( )

n n dx

Examples

2d y 2 - 11 . If y = (1 + x ) tan x find at x = 12dx

2 - 1Solution : y = (1 + x ) tan x dy 1

= ( + x ) × + tan x · ( x ) = + x tan x .2 - 1 - 1 1 2 1 2 dx 1 + x 2

2d y 1= 0 + 2 x × + 2 tan x - 1

2 2dx 1 + x p p d y 22

at x = 1, 2 = + 2 × = 1 + .1 + 1 42dx

x = 1

d y - 4 a 2. y = ax , = 22 If 4 show that

dx y 2 3

2Solution : y = 4ax differenti ating w.r.t. x

dy dy 2a 2 y = 4a =

dx dx y again differenti ating w.r.t. x d y a dy a a - a 2 22 2 2 4

= - = - = dx y 2 2 2 3dx y y y

t d y sin t 2+3. If x = a cos t log tan & y = a sin t , thenshow that =2 dx a cos t 2 4

+ t Solution : x = cos t log tan

2

dx 1 1 t = a - sin t + sec2 1 dt 2 2tan2


+1 1 1 - sin t 1 a cos t 2 2 += a - sin t + = a - sin t + = a - sin t = a =1 t t t sin t sin t sin t 22 tan cos 2 sin cos2 2 2 2

dy y = a sin t , = a cos t

dt cos dy dy dt a t

= = × sin t = tan t dx dx dt 2a cos t d y d dy d dy dt 1 sec t 2 2

= = = sec t × =2 dx dx dt dx dx dx dt dx a cos t 2 2

sin t 2d y sin t

=2 4dx a cos t

d y 22 4. If y = sin x , find

2 dx dy

Solution sin: = 2 sin x cos x = 2 x dx d y 2

= cos 2 x · 2 = 2 cos 2 x dx 2

d y 2 25. If y = x log x find e 2dx

dy 12Solution : = x · + log 1 x · 2x = x (2 log x + )e e dx x

d y 22 = x + ( 2 2 log x +1) = 3+ log x e e x dx 2

2d y 6. If y = e ·sin( bx + c ) find ax

2dx dy a x a x Solution : = e · cos( bx + c ) · b + sin( bx + c ) · ae dx dy {} = e b cos( bx + c ) + a sin( bx + c )ax dx

{}2d y {} = e - b sin( bx + c ) + ab cos( bx + c ) + ae b cos( bx + c ) + a sin( bx + c )ax 2 a x 2dx

{}ax = e 2ab cos(bx + c ) + (a - b ) sin(bx + c )2 2

d y 2

7. If x = a ( - sin ), y = b (1 - cos ) find2dx

dx dy Solution : = a (1 - cos ), = b sin

d d dy b sin 2 b · 2 sin( 2) cos( 2 ) b

= = = cot( )dx a (1 - cos ) a a · 2 sin ( 2 )2


2 d y b 1 d b 1 b = - cosec · · = - cosec · = - cosec ( ).2 2 4

a 2 2 dx 2a 2 a ( 2 1 - cos )2 2dx 4a

8. If y = e , then prove that (1 - x ) y - xy - m y = 0 m s i n-1 x 2 22 1

dy m m sin- 1 x Solution : = e ·dx 21 - x cross multiplying & squaring, we have

2 2 2 2(1- x ) y = m y 1

differentiating w.r.t. x (1- x ) · 2 y y + y (- 2 x ) = m · 2 yy 2 2 2

1 2 1 1

2 2 Dividing throughou t by 2 y , we get (1 - x )y - xy - m y = 0 1 2 1

- 1 2 29. If y = sin( m sin x ), then provethat (1- x ) y - xy + m y = 02 1 m -Solution : y = cos(m sin x ) ·1

1 - x 2 1

2m 1 - y y = (1 - x ) y = m (1 - y )2 2 2 2

1 1- x 21

differentiating w.r.t. x 2 2 2(1- x )2 y y + y (- 2x ) = m (- 2 yy )1 2 1 1

Dividing throughtout by 2 y we get ( 01 - x ) y - xy + m y = .2 2 1 2 1

10. If y = e tan x , provethat (1+ x ) y - 2(1- x + x )y + (1 - x ) y = 0 x - 1 2 2 2 2 1

dy 1 x x - 1 Solution : = e · + e tan x dx + x 2 1

(1 + x )[ y - y ] = e 2 x 1

differentiating w.r.t. x 2 x (1 + x )(y - y ) + 2x (y - y ) = e 2 1 1

(1 + x )y - (1 + x - 2 x )y - 2 xy = (1 + x )y - (1+ x )y 2 2 2 22 1 1

2 2 2 (1 + x )y - 2 (1 + x - x )y + (1 - x ) y = 0 2 1

Exercise

2 2 d y 16 d y - 2(1 ) If 4 x + 9 y = 36 show that = - ( 2 ) If x + 2xy + 3 y = 1 prove that =2 2 2 2

2 2 2 3dx 9 y dx ( x + 3 y )

d y 1 d y 2 2 ( ) x = a cos , y = a sin ( ) x = a tan , y = a sin3 3 3 If find 4 If 2 find

dx 2 dx 2 2


d y sec2 3( ) x = a (cos + sin ), y = a (sin - cos ) =5 If show that

a dx 2d y d y 2 2

( ) y = sin x , ( ) y = x log x ,- 1 6 If 2 then find 7 If then find e dx dx 2 2

d y d y 2 2( ) y = e · sec x , ( ) y = a ,4 x x 8 If 3 then find 9 If then find

dx dx 2 2

d y d y 2 2 ( ) x = at , y = at , ( ) x y = a ,2 3 3 x 10 If 2 then find 11 If then find

dx dx 2 2

2d y (12 ) If y = a cos mx + b sin mx , prove that + m y = 02

dx 2 m + 2 2 2(13) If y = x + x 1 , prove that (x +1 ) y + xy - m y = 02 1

n +1 - n 2 (14) If y = ax + bx , provethat x y - n (n +1 ) y = 02 (15) If x = sin t , y = sin pt , prove that (1 - x ) y - xy + p y = 0 2 2

2 1

2 2 3 (16) If x + xy + y = 1, then prove that (x + 2 y ) y + 6 = 0 2 ax 2 2(17) If y = e sin bx provethat y - 2ay + (a + b )y = 02 1

b 2(18 ) If y = ax + then show that x y + 2( xy - y ) = 0 2 12x

b (19 ) If y = ax + then show that x y = n (n + 1) y n +1 2

2x n 2d x

( 20 ) If x = a cos nt + b sin nt then show that + n x = 022dt

n derivative of Standard functionsth

1 ) . If y = (ax + b to find y m n

m - 1 m - 2 2 Solution : y = m (ax + b ) a ; y = m (m - 1)(ax + b ) a 1 2

differenti ating n times, we have y = m ( 1 m - 1 L) (m - n + )(ax + b ) a m - n n

n 1

in ,particular if m = - 1 ie y =ax + b

( L )( )( ) ( )( - 1- n n y = - 1 )- 2 - 3 - n ax + b a n n n ( - 1) n ! a

ie y = n n + 1(ax + b )2 ). If y = log( ax + b to find y n

1Solution : y = · a 1 ax + b


differenti )ating ( n - 1 times ( - ) a ( - ) a n - 1 n - 1 n - 1 n 1 1

y = · a =n n n (ax + b ) ( ax + b )3. If y = a to find y mx

n mx mx 2 2 Solution ): y = a · m (log a ) ; y = a · m (log a 1 2

n n mx Ingeneral, y = m (log a ) a n 4 ). If y = sin( ax + b to find y n

p +Solution : y = a cos( ax + b ) = a sin ax + b

1 2

again differenti ating w.r.t. x p p + ·

y = a cos ax + b = a sin ax + b + 2 2 22 2 2

p p · ·y = a cos ax + b + 2 = a sin ax + b + 33 3

3 2 2

p ·In general, y = a sin ax + b + n n

n 2

5 ). If y = cos( ax + b to find y n p +

Solution : y = - a sin( ax + b ) = a cos ax + b 1 2

again differenti ating w.r.t. x p p + ·

y = - a sin ax + b = a cos ax + b + 22 22 2 2

p ·In general, y = a cos ax + b + n n

n 2 6 ). If y = e cos(bx + c to find y ax

n Solution : differenti atingw.r.t. x

ax ax y = ae cos(bx + c ) - be sin(bx + c )1 a a put a = r cos , b = r sin

then y = re cos( bx + c ) cos a - re sin(bx + c )sin a ax ax 1

ax ax = re [cos(bx + c ) cosa - sin( bx + c ) sin a ] = re cos(bx + c +a )again &differenti ating w.r.t. x simplifyin g, we get

y = r e cos(bx + c + 2 a )2 ax 2

b a a n a x 2 2 - 1In general, y = r e cos( bx + c + n ) where r = a + b & = tann a


7 ). If = sin( + to findax y e bx c y n Solution : differenti ating w.r.t. x

y = ae sin(bx + c ) + be cos(bx + c )ax ax 1

a a put a = r cos , b = r sin then y = re [sin(bx + c ) cos a + cos(bx + c )sin a ] = re sin(bx + c + a )ax ax

1

again & differenti ating w.r.t. x simplifyin g, we get

2 ax y = r e sin(bx + c + 2a )2

b a a n a x In general, y = r e sin( bx + c + n ) where r = a + b & = tan2 2 - 1n a

8. Statement of Leibmitz's Theorem on n derivative of a productth

If u & v are functions of x , the n derivative of the productuv is given byth

(uv ) = u v + nc u v + nc u v + L L + nc uv n n 1 n - 1 2 2 n - 2 2 n n where suffixes of u & v represent order of the derivative s of u & v .

Examples

1t h1 . Find the n derivative of

x - 6 x + 82

1 1 A B Solution : Let y = = = + (Say)

( x - 2 )( x - 4) ( x - 2) ( x - 4)2x - 6 x + 8

multiplyin g throughou t by ( 4x - 2)( x - )

1 = A ( 2x - 4) + B ( x - )

1put x = 2, 1 = A (- 2) A = -

2

1put x = 4, 1 = 2 B B =

2

1 1 -2 2 y = +

( 4 x - 2 ) (x - )

1 1- ( - ) n ! ( - ) n !n n 1 1

2 2differenti ating n times, we have y = + n n +1 n +1( x - 2) ( x - 4)

t h 2 3 2. cos Find the n derivative of sin x x ( - cos ) cos + cos1 2 x 3x 3 x 2 3Solution : Let y = sin x cos x = ×

2 4

1 ie y = [cos 3x + 3 cos x - cos 3x cos 2 x - 3cos 2 x cos x ]

8 1 1 3 += cos 3 x + 3 cos x - (cos 5 x + cos x ) - (cos 3 x cos x )8 2 2 1 - 1 1 3 3

= cos 3 x + 3 cos x - cos 5x - cos x - cos 3 x cos x )8 2 2 2 2


1 cos 1 1-y = cos x - cos 3 x 5 x 8 2 2

p p p 1 1 1+ + + n n differenti ating n times, we have y = cos x - ·3 cos 3 x n - ·5 cos 5x n n 8 2 2 2 2 2 2x 23. If y = e sin x to find y n

1 1 12 x 2 x 2 x Solution : y = e (1 - cos 2x ) y = e - e cos 2x 2 2 2

1 1 a n 2 x n 2 x differenti ating n times, we have y = 2 e - · r e cos( 2 x + n - )n 2 2

p 2a where r = 4 + 4 = 8, = tan = tan 1 = - 1 - 1

2 4 () p 1 ·n n - 1 2 x y = 2 e - 8 cos 2 x + n n 2 2

44. If y = e sin 5x cos 3 x find y x n

1 1 1 4 x 4 x 4 x Solution : y = e [sin 8 x + sin 2x ] ie y = e sin 8 x + e sin 2 x

2 2 2 () ()1 8 1 2+ +n n differenti ating n times, we have y = 16 + 64 e sin 8 x n tan + 16 + 4 e sin 2 x n tan4 x - 1 4 x - 1

n 2 4 2 4

() ()1 1 1+n n ie y = 80 e sin( 8 x + n tan 2) + 20 e sin 2 x n tan4 x - 1 4 x - 1n 2 2 2

5. If y = x log 3 x find y 2n

2Solution : y = x log 3x Let u = log 3x = log 3 + log x

(- )n - 11 2 u = , v = x n n x differenti ating n times, using Leibnitz' s Theorem,

( - 1) ( - 1) ( - 1)n - 1 n - 2 n - 32y = ( uv ) = · x + nC 2x + nC · 2n n 1 2 n n - 1 n - 2x x x

[][]( - ) (-n - 3 n - 31 ) 12 1 2= ( - 1) + (- 1 ) 2 n + n ( n - 1 ) = 1 - 2 n + n - n n - 2 n - 2 x x

[](- 1 )n - 3 2ie y = n - 3n + 1n n - 2 x

2 x 26. If y = a cos(log x ) + b sin(log x ) show that x y + (2n + 1) y + (n +1 )y = 0n +2 n +1 n Solution : Let y = a cos(log x ) + b sin(log x )

differenti ating w.r.t. x 1 1

y = - a sin(log x ) + b cos(log x )1 x x


ie xy = - a sin(log x ) + b cos(log x )1

differenti ating w.r.t. x again 1 1

xy + y = - a cos(log x ) - b sin(log x )2 1 x x

2ie x y + xy = - a cos(log x ) - b sin(log x ) = - y 2 1

2ie x y + xy + y = 0 2 1

differenti ating n times, using Leibnitz' s Theorem, we have

x y + nC 2 xy + nC · 2 y 2n + 2 1 n + 1 2 n

+ xy + nC · 1 y n + 1 1 n

+ y = 0n adding, x y + ( 2n + 1) xy + [ n ( n - 1 ) + n + 1] y = 02

n + 2 n +1 n 2 2ie x y + (2n +1 )xy + (n + 1 )y = 0 n +2 n +1 n

7. If y = e , prove that (1 - x ) y - ( 2n + 1 ) xy - ( n + m ) y = 0 m co s - 1 x 2 2 2n + 2 n +1 n

m co s- 1 x Solution : y = e differenti ating w.r.t. x

- m y = e ·m cos x ie 1- x y = - my 21 1 21 - x

squaring both sides (1 - x ) y = m y 2 2 2 2 1

differenti ating again w.r. t. x ,2 2 2(1 - x )2 y y + y (- 2x ) = m 2yy 1 2 1 1

dividing by 2 y , we have 1

(1 - x )y - xy - m y = 02 22 1

differenti ating w.r.t. x , n times using Leibnitz' s Theorem,

2(1 - x ) y + nC y ( - 2 x ) + nC y (- 2 )n + 2 1 n + 1 2 n - xy + nC y (- 1)

+ 1 1n n 2- m y = 0n

2 2 2adding, (1 - x ) y - (2 n + 1) xy - ( n - n + n + m ) y = 0 n + 2 n +1 n

2 2 2ie (1 - x ) y - (2 n +1 )xy - (n + m )y = 0 n + 2 n +1 n 8. If y + y = 2x , prove that (x - 1)y + (2 n +1 )xy + (n - m )y = 0 1 m - 1 m 2 2 2

n +2 n +1 n 1 m - 1 m Solution : y + y = 2 x

() 2ie y + 1= 2xy 1 m 1 m () 21 m 1 m y - 2xy + 1 = 0


22 x ± 4x - 4which is a quadratic equation in y y = = x ± x - 11 m 1 m 2

2

m -1 m 2 2Consider, y = x + x - 1 y = x + x 1

differenti ating w.r.t. x + 2x - 1 x 2x m - 1 m - 1 - -y = m x + x 1 1 + = m x + x 1 2 2

1 2 22 x - 1 x - 1

ie x - 1 y = my 2 1

Squaring both sides (x - 1)y = m y 2 2 2 21

differenti ating w.r.t. x (x - 1 )2y y + y (2 x ) = 2m yy 2 2 2

1 2 1 1

dividing throughou t by 2 y , we have 1

2 2(x - 1 ) y + xy - m y = 02 1

differenti ating n times using Leibnitz' s Theorem

( - ) + + 2x 1 y nC y 2 x nC y 2n + 2 1 n +1 2 n + xy + nC y 1n + 1 1 n

2- m y = 0n

2 2 2adding, ( x - 1 ) y + (2 n + 1 )xy + (n - n + n - m ) y = 0 n + 2 n +1 n ie (x - 1)y + (2n +1 )xy + (n - m )y = 0 2 2 2

n + 2 n +1 n m - -We obtain the same result if y = x - x 1 ie y = x - x 11 m 2 2

Exercise

1 1 . Find the n derivative oft h

2 x - 5 x + 6 th 3 3 2. Findthe n derivative of (i ) sin x ( ii ) cos x (iii ) sin 4 x cos 3x (iv ) sin 8x sin 4 x (v ) cos 5 x cos x .th 3 x 2 2 x 2 x 3. Findthe n derivative of (i ) e sin x ( ii ) e cos x (iii ) e sin5 x cos 2 x

- 1 2 2 24. If y = sin(m sin x ), provethat (1 - x ) y - (2 n +1 )xy - (n - m )y = 0 n + 2 n +1 n 5. If y = e , prove that (1 - x )y - (2 n + 1) xy - (n + a ) y = 0a si n- 1 x 2 2 2

n + n + n 2 1 2 n 26. If y = (x - 1 ) , provethat (x - 1 )y + 2xy - n (n +1 )y = 0n +2 n +1 n


Polar Co-ordinates r = f ()q

Let O be a fixed point and OA the initial line. Let P be any point on the curve.

Let A O P = & OP = r P r (, ) qjthen )co - ordinates of P are (r , which are called Polar co - ordinates. r

r is called radiusvector, which is distance of P from q y A O O and isthe A O P measuredin anticlock direction. p

Q Equation t o the curve is taken as r = f ( )

Let PT be the tangent to the curver = f ( ) at P , then angle made by the tangent with radius vectorOP is denoted as and angleie O PT = made by the tangent with initial line OA is denoted by . It can be seen T from the figure that

= + (1)

Let OQ represent perpendicular from the poleO upon the tangent at P , it is denoted as p .OQ p f It can be seen fromthe triangle OPQ that sin = = OP r

f (2)p = r sin

rd f To prove that tan = (important result)

dr dy

If the Cartesian co - ordinates of P are ( x , y ) then x = r cos y = r sin and tan = dx

dr r cos + sin

dy d d tan = =dr dx d cos - r sind

dr using , (1) & dividing Nr & Dr of RHS by cos we have

d d d

r + tan tan + r f tan + tandr dr tan( + ) = ie =f d d 1 - tan tan1 - r tan 1 - tan · r

dr dr

d Comparing LHS & RHS, we have f (3)tan = r dr

[] 21 1 1 1 1 dr f f f from (2) p = r sin = cosec = 1 + cot = + using (3)2 2

p rd r r r r 2 2 2 22

2 1 1 1 dr = + (4)

p d r r 2 42

which is called Pedalusing r = f ( ) & ( 4) we can eliminate and obtain a relation between p & r for the curve,

Equation or (p , r ) equation of the curve.


r = g ()qTo find the angle between two curves r = f ( ) & r = g ( ) r = f ()qLet the two curves r = f ( ) & r = g ( ) intersect at P .

be the tangents to the two curves and letf f bet the anglesLet PT & PT &1 2 1 2 P made by the tangents with the radius vectorOP .

ff 1

f f Angle between tw o curves is given by - 21 2f -f

f f 1 2() tan - tanf f Now tan - = 1 2 (5)f f 1 2 1 + tan tan1 2

T f f and hence angle between two using the result in (3) we can find tan & tan T 1 2 1

curves at the point of intersection can be found out. 2

() f f f f If tan = tan then tan - = 01 2 1 2

f f ie = thetwo curves touch each other at P .1 2

p f f f f If tan tan = - 1 then - =

1 2 1 2 2 thetwo curves are said to intersect orthogonal ly.

Examples

(1) Showthat in the equiangula rspiral r = ae thetangent is inclined at a constant angleto the radiusvector.cota a co tSolution : r = ae

dr a a a cotdifferenti ating w.r.t. , = ae cot = r cotd

d r r f a f a hence the result.tan = r = = = tan =

a dr dr r cotd

p ( 2) Show that the tangent t othe Cardiod r = a (1 + cos ) at the point = is parallel to the initial line.

3 Solution : r = a (1 + cos )

dr differenti ating w.r.t. , = a (0 - sin )

d

22 cos p d r a (1 + cos ) +2f tan = r = = = = - cot = tandr dr - a sin 2 2 2- sin cosd 2 2

p p p p p p p f f p f when & = , = + = + = + + == +3 2 6 3 2 62 2

thetangent is parallel to the initial line.

2 a f p (3) For the curve = 1 - cos show that (i ) = - and (ii ) p = a cosec

r 2 2

2a Solution : = 1 - cos

r


22 a dr dr - r sindifferenti ating w.r.t. , - = sin =

d d 2a r 2

2 sin2 d r r 2a 1- cos 2Now tanf = r = = = - = = = - tan

dr dr r sin - sin 22- r sin - sin cosd 2 2 2a

- -f p f p f p ie tan = - tan = tan = - & p = r sin = r sin = r sin2 2 2 2 2 2a 2a a

ie p = sin · = sin · = p = a cosec2 1- cos 2 2 2 sin sin2

2 2 ( 4 ) Find the angle between th e curves r = 2 sin & r = sin + cos .

dr Solution : r = 2 sin 2 , = cos

d f Let bethe angle made by the tangent at the point of intersecti on with radius vector

1 d r 2 sin f tan = r = = = tan

dr 1 dr 2 cos d

f (1)= 1

for the curve r = sin + cos

dr = cos - sin

d f Let bethe angle w.r.t. this curve

2

d r sin + cos tan + 1 + p tanf = r = = = = tandr 2 dr cos - sin 1 - tan 4d

p f (2)= +2 4

p f f from (1) & (2) - = + -

1 2 4

p angle between tw o curves is

4 (5 ) Prove that the curves r = a (1 + cos ) & r = b (1 - cos ) intersect at right angles.

Solution : Consider r = a (1+ cos )

dr = - a sin

d d r a (1 + cos ) (1 + cos )

f tan = r = = = dr 1 dr - a sin - sin (1)

d


dr Consider r = b (1 - cos ), = b sin

d d r b (1 - cos ) (1- cos )

f tan = r = = = dr 2 (2) dr b sin sind

from (1) & (2)

(1 + cos ) (1 - cos ) 1 - cos sin2 2

f f tan tan = × = = = - 1 1 2 - sin sin 2 2- sin - sin

thetwo curves intersect orthogonal ly. 2a

(6 ) Find the pedal equation of = 1 + cos r

2a Solution : = 1 + cos

r differenti ating w.r.t.

2 a dr - = - sin

d 2r 1 dr sin

(1) ie =d 2 a r 2

2 21 1 1 dr 1 1 dr 1 sin 2

Now = + = + = + using (1) p d d r r r r r 4 a 2 4 2 2 2 2

2

[] 2 +1 1 1 1 2 a 1 1 4a 4a 2-2= + 1 - cos = + 1 - 1 = + 1- - 1r r r 4a r 4a r 4a r 2 2 2 2 2 2 2

+1 1 1 - 4 a 4 a 1 1 1 12ie = + = - + =

r ar ar 2 2 2 2 2 2p r 4 a r r r p = ar which isthe required pedal equation.2

Exercise

1. Find the length of the perpendicu lar from the pole on the tangent t othe curve. r = a (1 - cos ) Ans : 2a sin 3

2

2. Forthe curve r = a sin 2 showthat = 3 .2 2

p 3. Find the angle of intersecti on of the curves r = 2 sin 2 , r = cos Answer :

2

4. Show that the curves r = a (1 + sin ) & r = a (1 - sin ) intersect orthogonal ly.

m m +5. Findthe pedalequation of the curve r = a cos m Answer : pa = r m m 1

2 2 2 4 Answer : p (r + a ) = r 6. Find the pedal equation of r = a


Indeterminate Forms

[]f ( x ) 0 8 they - 8 - 8 ¥ °g ( x ) 8While evaluating lim or lim f ( x ) g ( x ) or lim f ( x ) when it takes the forms , , , 1 , , 0

g ( x ) 0 8 x a x a x a are called Indeterminate forms and to evaluate such forms the following rule known as L' Hospital's Rule is used.

L' Hospital's Rule

'f ( x ) 0 8 f ( x ) f ( x ) 0 8 If lim is of the form or then lim = lim again if this is of the form or then

g ( x ) 0 8 g ( x ) g '( x ) 0 8 x a x a x a ''f ( x ) f ( x ) 0 8

lim = lim whenever it is of the or This rule can be applied.g ( x ) g ''( x ) 0 8x a x a

[] To evaluate lim f (x ) - g (x ) when it is of the form 8 - 8 , thenx a

1 1 -

[] 0g ( x ) f ( x )Consider lim f ( x ) - g ( x ) = lim which is of the form & hence L' Hospital' s Rule can be applied.

1 0x a x a g ( x ) f ( x )

Consider ( lim f ( x ) = y say) g ( x )x a

log f ( x )then log y = lim log f ( x ) = lim g ( x ) log f ( x ) = limg ( x )

1x a x a x a g ( x )

0 8which is of the form or and hence L' Hospital' s rule can be applied.

0 8

Examples

log x (1) Evaluate lim

2 x - 3x + 2 x 1

log x 0Solution : lim this is of the form using L' Hospital' s rule

02x - 3 x + 2x 1

1 1x = lim = = - 1

2 x - 3 2 - 3x 1

2sin x - sin 2 x lim(2) Evaluate

x 3 x 0

sin x - sin x 02 2Solution : lim This is of the form applying L' Hospital' s rule

03x x 0

2 cos x - 2 cos 2 x 0= lim again it is of the form applying L' Hospital' s rule again

02 3x x 0

- 2 sin x + 4 sin 2 x + 2 sin x 4 sin 2 x 2 4 sin= lim = lim - · = - × 1 + × 1 Q lim = 1

6 x 6 x 3 2x 6 3x 0 x 0 0 1 4

= - + = 1 3 3


tan x - x (3 ) Evaluate lim

2 x tan x x 0

tan x - x tan x - x x tan x - x x Solution : lim = lim × = lim Q lim = 1

tan x tan x x tan x x x 2 3 3x 0 x 0 x 0 x 0

0This is of the form using L' Hospital' s rule

0

sec x - 1 0 2using the rule again= lim =

3x 0 2x 0 2 sec x · sec x tan x 2 sec x tan x 2 12

= lim = lim × = × 1 =6 x 6 x 6 3x 0 x 0

1 1 (4) Evaluate lim -

x e - 1 x 0x Solution : This is of the form 8 - 8

x 1 1 (e - 1 ) - x 0lim - = lim which is of the form using L' Hospital' s rule

x 0e - 1 ( e - 1) x x x x 0 x 0

e - 1x = lim

e - 1 + xe x x x 0 0

which is again of the form using the rule again0

x e 1 1= lim = =

1 + 0 + 1 2x x x e + xe + e x 0 - x

( 15 ) Evaluate limx - 1 log x x 1

Solution : This of the form 8 - 8

x - 1 log x -- x 1 x lim = lim

x - 1 log x x 1-x 1 x 1 log x x

0This is of the form using L' Hospital' s rule

0

1 1 -

x - 1 0x x 2 = lim = lim is form again applying the L' Hospital' s rule 1 x - 1 1 log x + x - 1 0x 1 x 1log x × + ×

x x x 2 1 1

= lim = 1 2x 1 + 1 x

(6 ) Evaluate lim (tan x )cot x p

x 2


0 Solution : This is of the form 8 y lim (tan x )cot x Let =

p x

2

log tan x x log y = lim log(tan x ) = lim cot x log tan x = lim using L' Hospital' s rulecot tan x p p p

x x x 2 2 2

1 2 · sec x 1 tanx = lim = lim = lim cot x = 0sec x tanx 2p p p x x x

2 2 2 0y = e = 1

1 sin x x

(7) Evaluate limx x 0

Solution : This is of the form 18 1

sin x x Let y = lim

x x 0

sin x log1 sin x x log y = lim log = limx x x x 0 x 0

0 This is of the form applying the L' Hospital' s rule

0 x x cos x - sin x

× x x cos x - sin x x cos x - sin x sin x 2 x = lim = lim × = 1 × lim

1 sin x x x 2 2x 0 x 0 x 0

0 This is of the form again applying the L' Hospital' s rule

0 cos x - x sin x - cos x - x sin x - sin x

= lim = lim = lim = 02 x 2 x 2x 0 x 0 x 0

ie log y = 0 y = 1

a sin x - sin 2x (8 ) If lim is finite, find the value of a and the limit.

3tan x x 0

0 Solution : The given limit is of the form applying the L' Hospital' s rule

0 a cos x - 2 cos 2 x

= lim3 tan x · sec x 2 2x 0

0limit exists if this is of the form

0

a cos x - 2 cos 2 x = 0 for x = 0 a = 2 2 2 cos x - 2 cos 2 x x 1 3cos x - 2 cos 2 x 0

= lim × × = lim × 1 × 1 is of form0 3x tan x sec x 3 x 2 2 2 2x 0 x 0


using L' Hospital' s rule

- 2 sin x + 4 sin 2 x - 2 sin x 4 sin 2 x 2 4 1 4= lim = lim + × = - + = - + = 1

6 x 6 x 3 2 x 6 3 3 3x 0 x 0

a = 2 and the limit is 1.

Exercise

Evaluate the following

cosh x - cos x e - e x s i n x (1) lim (2 ) lim

x sin x x - sin x x 0 x 0

x x e - sin x - 1 xe - log(1 + x )(3 ) lim (4) limlog(1+ x ) x 2 x 0 x 0

1 1 1 1 - -(5 ) lim (6 ) lim

x sin x sin x x 2 2 x 0 x 0

1 1(7 ) lim (cos x ) (8 ) lim ( x )x 2 1 - x

x 0 x 1

1 1 + x x x a + b c x

(9 ) lim (cot x ) (10 ) liml o g x 3x 0 x 0

3 1 1 1 1 1Answers : (1) 1 (2) 1 (3) 2 (4) (5 ) - (6) 0 ( 7) (8) ( 9 ) (10 ) ( abc ) 3

2 3 e e e


Partial Derivatives A function of two independent variables and a dependent variable is denoted as which is explicit function wherez = f ( x , y )x & y are independent variables and z a dependent variable. Implicit function is denoted by f ( x , y , z ) = C

d f ( x + x , y ) - f ( x , y ) z f If lim exists then it is called Partial derivative of z or f w.r.t. x and denoted by or

d x x x d x 0

d ( , + ) - ( , )f x y y f x y z f If lim exists then it is called Partial derivative of z or f w.r.t. y and denoted by or

d y y y d y 0

z z while obtaining the derivative differenti atethe given function w .r.t. x treating y as a constant and while finding

x y differentiate the given function with respect to y , treating x as a constant.

z z 2 2Eg. (1) If z = x + xy - y then = 2 x + y + 0 = 2 x + y & = x - 2 y x y

z z 2 2 2( 2) If z = x y - x sin xy then = 2 xy - x cos xy · y - sin xy & = x - x cos xy · x = x (1 - cos xy )x y

2 22 xy z 1 ( x - y ) 2 y - 2xy · 2 x - 1(3 ) If z = tan then = × x 2 2 2 2 2x - y 4 x y x - y 2 2

() ()1+2x - y 2 2

() 2x - y x y - y - x y - y - x y - y ( y + x ) - y 2 2 2 3 2 3 2 2 2 2 2 4 2 2 2 2 () () ()()= × = = =

x + y 2 2 2 2 2 2x - y + 4 x y x - y x + y x + y 2 2 2 2 2 2 2 2 2 2

z 1 (x - y )(2x ) - 2 xy (- 2 y )2 2 & = ×

y 4 x y 2 2 22 2x - y () ()1+2 2 2 x - y

() 2x - y 2 x y - 2 xy + 4 xy 2x + 2 xy 2 x ( x + y ) 2 x 2 2 3 2 2 3 2 2 2

() () ()()= × = = = x + y 2 2 2 2 2 22 2 2 2 2 2 2 2 2 2x - y + 4 x y x - y x + y x + y

Successive derivatives

z z For the function z = f ( x , y ) & are first order partial derivative s, the second order partial derivative s are

x y 2 2 2 2 2 2z z z z z z z z z z

, , , which are denoted as , , , but in general = . x x x y y x y y x y y x x y y x 2 2x y

z z z z 2 2 2 2

In example (1) = 2, = - 2 , = 1 & = 1x y y x 2 2x y

z z 2 22 2Inexample (2) = 2 x + x y sin xy - x cos xy - x cos xy & = 2x + x y sin xy - 2x cos xy

y x x y


2 2z z =x y y x

z - y x + y - + y · y - x - y + y - x - y 2 2 2 2 2 2 2 22 ( )( 2) 2 2 2 2 4 2( )() () ()In example (3) = = = =

y x y 2 2 2 2 2x + y x + y x + y x + y 2 2 2 2 2 2

2 2 2 2 2 2 2z 2x ( x + y ) 2 - 2 x · 2 x 2 y - 2 x - 2( x - y )() ()() and = = = =

x y x 2 2 2 2 2x + y x + y x + y x + y 2 2 2 2 2 2

2 2z z Thus in general, always =x y y x

Exercise

z z z z 2 22 2 (1 ) Find , for z = log( x + y ) andshowthat =

x y x y y x 2 2 z z f ( 2) If x = f ( x + ct ) + ( x - ct ) show that = c where c is a constant.2

2 2t x z z

(3) If z = e f ( ax - by )then show that b + a = 2abz a x + by x y

22 2x + y u u u u (4) If u = then show that - = 4 1 - -x + y x y x y

2 2y u u (5 ) If u = sin then showthat = .- 1x x y y x


INTEGRAL CALCULUS

dy 'Integration' and the resulting function is called 'Integral'. If g (x ) isGiven = f ( x ), the process of finding y is calleddx

f x dx g x the integral then ) ( ) = ( is the notation used to represent the process. d [] In the above notation f ( x ) is called 'Integrand' and further g ( x ) = f ( x ).

dx d [] f x dx g x c ( ) = ( ) +But g ( x ) + c = g '( x )when c is a constan t

dx Thus integral of a function is not unique and two integrals always differ by a constant.

Properties

[](1 ) f (x ) ± f (x ) dx = f (x )dx ± (x )dx

ò(2) Kf (x )dx = K f (x )dx where K is a constant

(3) 0 dx = c (a constant)

Standard Integrals

+n +1 n +1x d x 1. x dx = + c (n - 1 ) Q c = x n n n +1 dx n +1

1 d 12. Q dx = log x + c (log x + c ) =

e x dx x +x x a d a x x x x 3 log. Q a dx = + c c = a in particular e dx = e + c

log a dx a

4. sin x dx = - cos x + c

5. cos x dx = sin x + c

6. sec x tanx dx = sec x +1

7 cot. cosec x x dx = - cosec x + c

8. sinh x dx = cosh x + c

9. cosh x dx = sinh x + c

10 tanh . sech x x dx = - sech x + c

102 KSOU Integral Calculus

11 coth . cosech x x dx = - cosesh x + c

112 cot . dx = tan x + c or - x + c - 1 - 1

21 + x 113 cos. dx = sin x + c or - x + c - 1 -1

2 1 - x 114 sinh. dx = x + c - 1

1 + x 2 1 - 115 cosh . dx = x + c

x - 12 1 -1 - 1 16 sec. dx = x + c or - cosec x + c

x x - 12

1 17. dx = - sech x + c -1 2x 1- x

1 18. dx = - cosech x + c -1 2x 1 + x

Methods of Integration There are two methods (1) Integration by substitution & (2) Integration by parts.

1. Integration by substitution

dx Consider f x dx f f f ( ) put x = ( t ) then = '(t ) ie dx = '(t ) dt dt

[] f (x ) dx = f f (t ) f '( t ) dt now for the new integrand, we can use the standard forms, ie. we have to make a proper substitution so that the given

integrand reduced to a standard one.

Examples

sin x 1 . tan x dx = dx

cos x dt

put cos x = t , then - sin x = ie sin x dx = - dt dx

ò dt tan x dx = - = - log t = - log cos x = log sec x + c

t

ò tan x sec x or tan x dx = dx

sec x put ,sec x = t differenti ating w.r.t. x

dt sec x tan x = sec x tan x dx = dt

dx


ò dt tan x dx = = log t = log sec x + c

t ò cos x

2. cot x dx = dx sin x

dt put ,sin x = t differenti ating w.r.t. x cos x = ie cos x dx = dt

dx ò dt

= = = + cot x dx log t log sin x c t

3. tanhx dx = log cosh x + c

4. coth x dx = log sinh x + c

5. sec x dx = log(sec x + tan x ) + c

6. cosec cosec x dx = log( x - cot x ) + c

' n +1f (x ) [] [] f (x )n Ingeneral dx = log f (x ) + c also f (x ) f '(x ) dx = + c (n - 1 )

f (x ) n +1

2. Integration by parts

d dv du If u & v are functions of x , we know that, (uv ) = u + v

dx dx dx By definition of Integratio n

òdv du dv du += = + uv u v dx u dx v dx using property (1)dx dx dx dx

ò dv du u dx = uv - v dx

dx dx The result can be used as the standard result. Out of the two functions of the product, one has to be taken as u & another

dv then the RHS after evaluation gives the integral or if both functions have taken as u & v then the result is as follows dx

òdu ·= -uv dx u v dx v dx dx dx

any one formcan be used depending on convenience. The firstone can also bewritten as u v ' dx = uv - u 'v dx

Examples

1. 1 xe dx put u = v , v ' = e , u ' = , v = e x x x

'x x x x x xe dx = uv - u v dx = xe - 1 · e dx = xe - e + c

2. 1 x sin x dx = x sin x du - · sin x dx = - x cos x - - cos x dx = - x cos x + cos x dx = - x cos x + sin x + c

' '3. 1 log x dx put u = log x , v = 1, u = , v = x x


' 'u v dx = uv - u v dx

ò 1ie log 1x dx = x log x - · x dx = x log x - · dx = x log x - x + c x

1 ' '4. sin x dx put u = sin x , v = 1, u = , v = x - 1 - 1 1 - x 2

1 - -sin x dx = x sin x - x dx 1 1

1 - x 2 - x differenti atingw.r.t. x to evaluate dx put 1 - x = t 2 2

1 - x 2- 2 x dx = 2 t dt - x dx = t dt

- x dx t dt = = 1· dt = t = 1 - x 2

1 - x t 2 2

sin x dx = x sin x - - x + c - 1 - 1 21

Special Types of Integrals Type I

dx dx dx dx , ( ) , ( ) & ( )(1) 2 3 4

a + x x - a a - x Ax + Bx + C 2 2 2 2 2 2 2

to evaluate(1) put x = at , dx = a dt dx a dt 1 tan dt 1 1 x - 1 - 1 = = = tan t = + c

a a a a 2 2 2 2 2 2a + x a + a t 1 + t to evaluate(2) & (3) use partialfractions

1 1 A B = = + (Say)

( x + a )( x - a ) x + a x - a 2 2x - a multiply throughoutby x - a 2 2

1 = A (x - a ) + B (x + a )1 put x = a , 1 = 0 + B · 2 a B =

2 a - 1put x = - a , 1 = A (- 2a ) + 0 A =2 a

- 1 11 2 a 2 a = +

x + a x - a 2 2x - a 1 1 dx 1 dx 1 1 1 x - a

dx = - + = - log( 2x + a ) + log( x - a ) = logx - a 2 a x + a 2 a x - a 2a 2 a a x + a 2 2

dx 1 x - a = log + c

2 a x + a 2 2x - a


1 1 A B next, = = + (Say)

(a + x )( a - x ) a + x a - x x - a 2 2

multiplying throughout by a - x , then2 2

1 = A (a - x ) + B (a + x )1 put x = a , 1 = 0 + B (2 a ) B =

2a 1

put x = - a , A ( 2 a ) + 0 A =2 a

1 11 2 a 2 a = +

a + x a - x 2 2 a - x 1 1 1 1 1 1 1 1 a + x

dx = dx + dx = log( 2a + x ) - log( a - x ) = log2 a a + x 2 a a - x 2 a 2a a a - x 2 2a - x

1 1 a + x dx = log + c

2 a a - x 2 2a - x dx

to evaluate (4) Ax + Bx + c 2

1 dx 1 dx 1 dx G.I. = = =

B C A A A 2 2B B C B B - 4 AC 2 2+ +2x + x + x - + x -A A 2 A A 2 A 2 24 A 4 A This integral will take any one of (1), (2) or (3) and hence can be evaluated.

Examples

dx (1 ) Evaluate

3x - 2x + 42

dx 1 dx 1 dx 1 dx Solution : = = =

2 4- + 3 3 3 2 2 23x 2 x 4 - 1 4 4 - 1 - 4 + 12x - x +2 x - + x +3 3 3 9 3 3 9

1 x -1 dx 1 dx 1 1 1 - 3 x 1 3 = = = × tan = tan + c - 1 - 1

3 3 3 2 2 2 2 2 2 2 2 2 28 1 2 2 2 1 2- -+ x + x 3 33 3 3 3

dx ( 2) Evaluate

x - 10 x + 212 dx dx dx 1 x - 5 - 2 1 x - 7

Solution : = = = log = log + c 2 × 2 x - 5 + 2 4 x - 3 x - 10 x + 21 ( x - 5) - 25 + 21 ( x - 5) - 2 2 2 2 2

dx (3) Evaluate

26 - 4x - 2 x


dx dx dx dx 1 1 1 Solution : = = =

- - 2 - ( + ) 2 - ( + ) + 2 - ( + )2 2 2 2 2 6 4 x 2x 3 x 2x 3 x 1 1 2 x 1

1 log 1 2 + ( x + 1) 1 3 + x = × log = + c

2 2 × 2 2 - ( x + 1) 8 1 - x

Typ e II

dx dx dx dx (1) ) , (2 ) , (3 ) , (42 2 2 2 2 2 2 a - x a + x x - a Ax + Bx + C

dx to evaluate put x = a sin , dx = a cos d

a - x 2 2 dx a cos d a cos d x d sin - 1= = = 1· = =

a cos a 2 2 2 2 2a - x a - a sin

dx x = sin + c - 1a 2 2 a - x

dx to evaluate put x = a sinh , dx = a cosh d a + x 2 2

dx a cosh d a cosh d x - 1 = = = 1 · d = = sinha cosh a a + x a + a sinh2 2 2 2 2

dx x - 1 = sinh + c a a + x 2 2

dx to evaluate put x = a cosh , dx = a sinh d

2 2x - a dx x = cosh + c - 1

a 2 2 x - a dx a sinh d a sinh d x = = = 1 · d = = cosh - 1

a sinh a x - a a cosh - a 2 2 2 2 2

dx to evaluate

Ax + Bx + C 2

1 dx 1 dx 1 dx G.I. = = =

A B C A A 2 22 2+ B B C + B B - 4 AC x + x + 2x - + x -A A 2 A A A 2 A A 2 24 4

This will reduce to any one of (1), (2) & (3) and hence can be evaluated.

Examples

dx (1 ) Evaluate2 2 x - 5 x


1 x -dx 1 dx 1 dx 1 15 - 1 - 1Solution : = = = sin 1 = × sin (5 x - ) + c

11 2 5 5 5 5- 2 2 21 1 -2 x - x 2 - x x - x 55 5 5 5 5 dx

(2) Evaluatex - 2x + 52

dx dx dx x - 1 - 1 Solution : = = = sinh + c 22 2 2 2 2x - 2 x + 5 ( x - 1 ) + 2 2 + ( x - 1 )

dx (3 ) Evaluate24 x - 12x + 8

1 dx 1 dx 1 dx = = = Solution : G.I.

2 2 4 2 2 2x - 3 x + 2 3 9 3 9 - 8- -x - + 2 x -2 4 2 4

3x -1 dx 1 12- 1 - 1= = cosh 3 = cosh (2 x - ) + c

1 2 2 22 2 3 1 -x - 2

2 2

Type III

px + q px + q dx and

2 Ax + Bx + C Ax + Bx + C 2

2to evaluate put px + q = l ( 2 derivative of Ax + Bx + C ) + m = l ( Ax + B ) + m where l & m are the constants to be found out by equating the co-efficients of corresponding terms on both sides. ie. to

solve for m & n from the equations

2 Al = p and lB + m = q px + q 2 Ax + B dx dx

then dx = l dx + m = l · log( Ax + Bx + C ) + m 22 2 2 2Ax + Bx + c Ax + Bx + C Ax + Bx + C Ax + Bx + C

the second integral in RHS is Type I and hence can be evaluated.

ò px + q px + q dx dl 2 dx = l dx + m = 2l Ax + Bx + C + m Ax + Bx + C Ax + Bx + C Ax + Bx + C Ax + Bx + C 2 2 2 2

the second integral in RHS is Type II and hence can be evaluated.

Examples

2x + 3(1 ) Evaluate dx

3x - 4 x + 52

Solution ): Put 2 x + 3 = l (6 x - 4 + m = 6lx - 4l + m 1 4 13

6 l = 2 l = , - 4 l + m = 3 ie m = 3 + = 3 3 3


2x + 3 1 6x - 4 13 dx 1 13 dx 2dx = dx + = log( 3 x - 4 x + 5) +

4 5 3 3 3 9 2 2 2 3 x - 4 x + 5 3x - 4x + 5 3x - 4 x + 5 - + 2x x 3 3

1 13 dx 1 13 dx 2 2= log( 3 x - 4 x + 5) + = log( 3 x - 4 x + 5) + 3 9 3 92 22 4 5- 2 - 2 11x - + x +3 9 3 3 3

- 2 x 1 13 3 1 13 - 3x 23 2 - 1 2 - 1 = log(3 x - 4x + 5) + × tan = log(3 x - 4 x + 5 ) + tan + c 3 9 311 11 3 11 11

3

5 x - 7(2) Evaluate dx 23x - x - 2

Solution ): Put 5 x - 7 = l (3 - 2 x + m = - 2 lx + 3 l + m 5 15 1

- 2 l = 5 l = - & 3l + m = - 7 m = - 7 - 3 l = - 7 + = 2 2 2

5x - 7 5 3 - 2 x 1 dx dx = - dx +

2 2 2 2 2 3 x - x - 2 3 x - x - 2 - 2 - ( x - 3 x )

5 1 dx 1 dx 2 2= - · 2 3 x - x - 2 + = - 5 3 x - x - 2 +

2 2 22 2 23 9 1 3- -- - x + - x 22 4 2 2

3- x 1 1 22 - 1 2 -1= - 5 3 x - x - 2 + × sin 3 = - 5 3x - x - 2 + sin (2 x - ) + c 12 2 2

Type IV

dx a cos x + b sin x + c

x 1 x dt 2to evalute put tan = t then differentiatingw.r.t. x sec =2 2 2 dx

2 dt 2 dt 2dt x x 1 t 1 - t 2 2 dx = = = & cos x = cos - sin = - =2 2ie

x x 2 22 2 2 2 1 + t 1 + t 1 + t 1 + t sec 1 + tan2 2

2 2

x x t 1 2 t & sin x = sin cos = × =2 2

2 2 1 + t 2+ t + t 2 21 1

x 2dt 1 - t 2 t 2tan = t , dx = , cos x = & sin x =when

2 2 2 21 + t 1 + t 1 + t


2 dt dx 2 dt 2 dt 2(1 + t )= = =

a cos x + b sin x + c (1 - t ) 2 t a (1 - t ) + 2 bt + c (1+ t ) (c - a ) t + 2 bt + a + c 2 2 2 2a + b + c

2 2(1 + t ) (1 + t )

which is Type I and hence can be evaluated.

Examples

dx (1 ) Evaluate

2cos x - 3 sin x + 5 x 2 dt 1 - t 2 t 2

Solution : Put tan = t , then dx = , cos x = & sin x = 2 2 2 21 + t 1 + t 1 + t

2 dt 2 dt 2dt 2 dt 2 1 + t = = = =G.I.

7 32 2 2 22 (1 - t ) 3 × 2 t 2(1 - t ) - 6 t + 5 (1 + t ) 3t - 6 t + 7 t - 2 t +2 - + 53 2 21 + t 1 + t

2 dt 2 dt 2 1 t - 1 1 3 = = = × tan = tan (t - 1 )1 1 - -

7 2 23 3 3 2322( t - 1 ) - 1 +2 ( t - 1 ) +23 3 3 3

dx 1 3 x --1 = tan tan 1 + c 2cos x - 3 sin x + 5 2 23

dx ( 52) Evaluate3- cos x

x dt - t 22 1 Solution : Put tan = t , then dx = & cos x =

2 1 + t 1 + t 2 2

dt 2 dx ( + t ) 2 dt 2dt 2 dt 1 dt 21 = = = = =

23 - 5 cos x 8 4 2 2 2 2 25(1 - t ) 3(1 + t ) - 5 (1 - t ) 8 t - 2 12t -3 - t -2821 + t 2

1 x 1t - tan -1 1 1 2 2 2= × log = log

1 1 x 14 4 2 × t + tan + 2 2 2 2

x 2 tan - 1 dx 1 2= log + c

x 3 tan- 5 cos x 4 2 + 1 2

dx ( 23 ) Evaluate

3+ sin x x 2 dt 2 t

Solution : Put tan = t , then dx = and sin x =2 + t + t 2 2 1 1


2 dt dx 2 dt 2 dt 2 dt 1 + t 2 = = = =

4 t 4 3 + 2 sin x 3 32 23(1 + t ) + 4 t 2 4+23 + 1 + t + t t - + 11 + t 3 2 3 9

2 t +2 dt 2 1 2 3t + 2 3 = = × tan = tan- 1 - 1 3 3 5 5 5 5 222 5 +t + 3 3 3 3

x +3tan 2 dx 2 2= tan + c -1 3 + 2 sin x 5 5

Typ e V

a cos x + b sin x dx

c sin x + e cos x + = +Solution :to evaluate put a cos x b sin x l (Denominator) m (derivative of denominator)

ie a cos x + b sin x = l ( c sin x + e cos x ) + m (c cos x - e sin x )where l & m are constantsto be foundout by equating the co - efficients of sin x & cos x separately .ie fromthe equations lc - me = b & le + mc = a

a cos x + b sin x c sin x + e cos x c cos x - e sin x then dx = l dx + m dx = lx + m log( c sin x + e cos x ) + c

c sin x + e cos x c sin x + e cos x c sin x + e cos x

Examples

3cos x - 2 sin x Evaluate dx

4 sin x + cos x Solution : Put 3 cos x - 2sin x = l (4 sin x + cos x ) + m (4 cos x - sin x )

4l - m = - 2 (1) (2) l + 4 m = 3

(1 ) × 4 16l - 4 m = - 8 (2)×1 l + 4m = 3

5 adding 17l = - 5 l = -

17

20 - 20 + 34 14from (1), m = 4 l + 2 = - + 2 = =

17 17 17

3 cos x - 2 sin x 5 4 sin x + cos x 14 4 cos x - sin x 5 14dx = - dx + = - 1 · dx + log( 4 sin x + cos x )

4 sin x + cos x 17 4 sin x + cos x 17 4 sin x + cos x 17 17

5 14= - x + log(4 sin x + cos x ) + c

17 17


Type VI

'x f (x )e dx where f (x ) = f (x ) + f (x )

ò 'x x x Solution : f ( x ) e dx = f (x ) e dx + f (x ) e dx (1)

x Consider f (x ) e dx

put u = f (x ), v ' = e , u ' = f (x ), v = e x x

'x x x f (x ) e dx = f (x ) e - f (x )e dx substitutingthis in (1),we have

ò ' 'x x x x x f (x )e dx = f (x )e - f (x )e dx + f (x ) e dx = f ( x ) e + c

Examples

xe x (1 )) Evaluate dx

2(1 + x

ò òx x x x x x xe (1 + x - 1) e (1 + x )e e e e Solution : dx = dx = dx - dx = dx - dx (1)(1 + x ) (1 + x ) (1 + x ) (1+ x ) 1+ x (1 + x )2 2 2 2 2

x e Consider dx (1+ x )

-1 1' 'x x put u = , v = e , u = , v = e 1 + x ( + )21 x

x x x x x e e - e e e dx = - dx = + dx (1 + x ) (1+ x ) (1+ x )2 2 (1 + x ) (1 + x )

substituting in (1)

2 x x x x xe e e e e dx = + dx - dx = + c (1 + x ) 1 + x 2 2 2(1 + x ) (1 + x ) (1 + x )

x - sin x ( 12) Evaluate dx

- cos x x x

2 sin cos òx - sin x x sin x x 2 2 Solution : dx = dx - dx = dx - dx

x x 1 - cos x 1 - cos x 1 - cos x 2 22 sin 2 sin2 2

ò1 x x 2 (1)= x cosec dx - cot dx

2 2 2

1Consider x cosec x dx 2

2

1 x x ' '2put u = x , v = cosec , u = 1, v = - cot

2 2 2


1 x x x 2x cosec dx = - x cot + cot dx

2 2 2 2

substituting in (1)

x - sin x x x x x dx = - x cot + cot dx - cot dx = - x cot + c

1 - cos x 2 2 2 2

Other examples

sin x cos x (1 ) Evaluate dx

41 + sin x 2Solution : Put sin 2x = t , then sin x cos x dx = dt

1dt

sin x cos x 1 1 2 - 1 - 1 2dx = = tan t = tan (sin x ) + c 2 24 21 + sin x 1 + t

2x + 1 ( 22) Evaluate dx 2(x + 1 )(x + )

2x +1 A Bx + C Solution : Let = +x + 12 2(x + 1 )(x + 2 ) x + 2

multiplying throughou t by ( 2 x +1)(x + )2 2 2then x +1 = A ( 1x + 2 ) + (Bx + C )(x + )

2 put x = - 1 , 2 = A (1 + 2 ) + 0 A =3

4 1 put x = 0, 1 = 2 A + C C = 1 - = -3 3

2Equating co - efficient of x on bothsides 2 1A + B = 1 B = 1 - A = 1 - =3 3

2 1 1 x -x + 12 3 3 3= +( x + 1)(x + 2 ) x + 1 x + 2 2 2

x +1 2 dx 1 x - 1 2 dx 1 2x 1 dx 2 dx = + dx = + dx -

3 x +1 3 3 x + 1 6 3 (x +1 )( x + 2) x + 2 x + 2 x + 22 2 2 2

2 1 1 x 2 - 1= log( x + 1) + log( x + 2) - tan + c

3 6 3 2 2

Type VII

ò2 2 2 2 2 2 2 (1) a - x dx (2) a + x dx (3 ) x - a dx (4 ) Ax + Bx + C dx

( ) ( px + q Ax + Bx + C dx 25 )


2 2(1) To evaluate a - x dx put x = a sin , dx = a cos d

ò ò 2a a - x dx = a - a sin · a cos d = a cos · a cos d = a cos d = (1 + cos 2 ) d 2 2 2 2 2 2 22

a a sin 2 a a a a 2 2 2 2 2 22= + × = + sin cos = + · sin · 1 - sin

2 2 2 2 2 2 2 2 2 2 2 2a x a x x a x a x - 1 - 1 2 2= sin + · 1 - = sin + · a - x

2 a 2 a 2 a 22 2a a x a x 2

a - x dx = a - x + + c 2 2 2 2 - 12 sin 2 a

2 2( 2) To evaluate a + x dx put x = a sinh , then dx = a cosh d

ò 2 a a + x dx = a + a sinh · a cosh d = a cosh d = (1 + cosh 2 ) d 2 2 2 2 2 2 2 2

a a a a sinh 2 a a 2 2 2 2 2 2= 1 · d + cosh 2 d = + · = + ×sinh cosh

2 2 2 2 2 2 2

2 2 2 2 2 2 a a a x a x x a x x 2 - 1 - 1 2 2= + sinh 1 + sinh = sinh + · 1 + = sinh + a + x 2 2 2 a 2 a 2 a 2 2a x a x 2

a + x dx = a + x + + c 2 2 2 2 - 12 sinh 2 a

2 2(3 ) To evaluate x - a dx put x = a cosh , then dx = a sinh d 2a x - a dx = a cosh - a · a sinh d = a sinh · a sinh d = a sinh d = (cosh 2 - 1 ) d 2 2 2 2 2 2 2

2 a a a sinh 2 a a a x 2 2 2 2 2 2

- 1= cosh 2 d - 1 · d = · - = sinh cosh - cosh 2 2 2 2 2 2 2 a 2 2 2 2 2 2 a a x a x x a x x a x 2 - 1 - 1 2 2 - 1= cosh - 1 cosh - cosh = - 1 · - cosh = x - a - cosh2 2 a 2 a 2 2 2 2 a 2 a

x a x 2x - a dx = x - a - + c 2 2 2 2 - 1

2 cosh 2 a B C

and hence can be2 2( 4) To evaluate Ax + Bx + C dx = A x + x + dx . This will take the form (1), (2) or (3)A A

evaluated. 2(5 )) To evaluate ( px + q Ax + Bx + C dx

2put px + q = l ( 2 derivativeof Ax + Bx + C ) + m = l ( Ax + B ) + m where l & m are constants to be found out,

òthen, ) ( px + q ) Ax + Bx + C dx = l ( 2 Ax + B Ax + Bx + C dx + m Ax + Bx + C 2 2 2


l C 2 ) 32 2= ( Ax + Bx + C + m A x + x + dx 2

3 A A the second integral reduces to (1), (2) or (3) and hence can be evaluated.

Type VIII

dx 2( px + q ) Ax + Bx + C

1 - 1 1 - 1put px + q = then p dx = dt & x = q t p t 2 t

dt -dx - dt t 2= =A B (px + q ) Ax + Bx + C 2 1 A - 1 2 - 1 (1- tq ) + (t - qt ) + Ct 2 2 2q + q + C p p 2t t p t p 2

This integral reduces to any one of Type II and hence can be solved.

Type IX

e bx c dx and e bx c dx ax ax cos( + ) sin( + )

To evaluatewe haveto use integration by parts. Let C = e cos(bx + c ) dx & S = e sin(bx + c ) dx ax ax

Consider C = e cos(bx + c )dx ax

sin( bx + c )' 'a x ax put u = e , u = ae , v = cos( bx + c ), v =

b a x e sin( bx + c ) a

C = - e sin( bx + c ) dx ax b b

ax bC = e sin(bx + c ) - aS (1) ax aS + bC = e sin(bx + c )

ax Consider S = e sin(bx + c ) dx

- cos( bx + c )put u = e , u ' = ae , v ' = sin( bx + c ), v =a x ax

b ax - e cos( bx + c ) a

S = + e cos( bx + c ) dx a x b b

ax bS = - ea cos(bx + c ) + aC ie bS - aC = - e cos(bx + c ) (2) ax

2 a x (1 ) × a a S + abC = ae sin( bx + c )2 a x (2) × b b S - abC = - be cos( bx + c )

[]adding (a + b ) S = e a sin( bx + c ) - b cos( bx + c )2 2 ax


[]e a sin(bx + c ) - b cos(bx + c )ax S =

2 2 (a + b )

( ) × b abS + b C = be sin( bx + c )2 ax 12 a x (2 ) × a abS - a C = - ae cos( bx + c )

[](a + b ) C = e b sin( bx + c ) + a cos( bx + c )2 2 ax subtractin g

[]ax e a cos( bx + c ) + b sin(bx + c )C =

(a + b )2 2

Examples

ò1 1 1(1 ) e sin 3x cos 2 x dx = e [sin 5x + sin x ] dx = e sin 5x dx + e sin dx 2 x 2 x 2 x 2 x

2 2 2

1 (2sin 5 x - 5 cos 5x ) 1 (2 sin x - cos x )2x 2 x = e + e + c 2 29 2 5

1 1 1 1 e 1 (3cos 2x + 2 sin 2 x )3 x 3x 2 3x 3x 3x 3 x (2) e cos x dx = e (1 + cos 2 x ) dx = e dx + e cos 2x dx = × + e

2 2 2 2 3 2 9 + 4 3x 3x e e = + ( 23 cos 2 x + 2 sin x ) + c 6 26

Exercise

Integrate the following w.r.t. x

- 1- 1 s i n x sin x 1 e (1 ) ( 2) (3 )

cos (log )2 2 1 - x x x 2 1 - x 2sec x 3x - 2 x (4) (5 ) (6)

tanx (2 + tan x ) 2 2(x + 1) (x + 3 ) (x - 1 )(x + 4 )x - x - 2 4 x + 5 4 x + 13

( 7) (8 ) (9 )2 2 x - 1 x + 22 x + 2 2x - 6 x + 18

(1 + x ) 1 1 (10) e (11) (12)x 2 + cos x - sin x 3 + 4 cos x 2(2 + x )

2 sin x + 3 cos x 1 (1 + sin x ) x (13) (14) (15) e 4 sin + 5 cos (1 + cos )x x 4 cos x + 9 sin x x 2 2

1 2 2 (16) (17 ) 6 - 4 x - 2 x (18) (2 x - 5 ) x - 3 x + 2x (x + 1 )n

1 1 (19) (20) (21) e sin 4 x sin 2x 2x (x + 1 ) 2 x + 3x + 4 (x + 1) x - 1 2 2

2 x 3 x 3 4 x 3(22) e cos 3 x cos x (23) e cos x (24) e sin x


Definite Integrals

Let f (x ) be a function defined in the interval (a , b ) and f (x )dx = g (x ) + c [][] The value of the integral at x = b minus the value of the integral at x = a ie g (b ) + c - g (a ) + c

b ie g ( b ) - g ( a ) is defined as Definite integral and denoted as f (x ) dx

a ò b

ie If f ( x ) dx = g ( x )then f ( x ) dx = g (b ) - g ( a )a

b is called upper limit and a is called lower limit.

Examples

2(1) Evaluate ( x - 2 x + 3 ) dx 3 2

1

2 1+ ·x x 2 2 1 2 16 1 2 4 3 4 3 +2 Solution : ( x - 2x + 3) dx = - 2 × 3 x = - 2 × + 3 2 - - 3 = 4 = + 6 - + - 33 2

4 3 4 3 4 3 3 4 3 1 1

16 1 2 84 - 64 - 3 + 8 25= 7 - - + = =

3 4 3 12 12

- 1 2(sin x 1( 2 )) Evaluate dx

1 - x 0 2

1 Solution sin: Put x = t then dx = dt - 1

1 - x 2p -1 - 1 when x = 0 , t = sin 0 = 0 when x = 1, t = sin 1 =2

p p 3 p p (sin x ) t 1 - 1 2 3 2 31 2 2dx = t dt = = =

3 3 2 241 - x 0 2 00

dx 1(3 ) Evaluate dx

x + 2 x + 5- 1 2

1dx dx 1 x 1 1 1 1 1 - 1 1+ + +1 1 - 1 - 1 - 1Solution : dx = = tan = tan - tanx + 2 x + 5 (x +1 ) + 2 2 2 2 2 2 22- 1 2 - 1 2 - 1

1 1 1 p p = tan 1 - tan 0 = - 0 =- 1 - 18 2 2 2 4 dx p

( 4) Evaluate+ cos4 3 x 0

x 2dt 1 - t 2tan = t dx = , cos x =Solution : Put then

2 2 21 + t 1 + t p when x = 0 , t = tan0 = 0 when x = p , t = tan = 82


dt 28 p dx 2 dt 2 dt 1 t 8 8 8+ t 21 ()= = = = tan - 1

4 + 3 cos x 2 23(1 - t ) 4 (1 + t ) + 3 (1 - t ) 7 70 0 0 2 0 2 2t + 7 4 + 0 2(1 + t )

1 1 1 p p - 1 - 1= tan 8 - tan 0 = × =2 7 7 7 2 7

Properties of Definite Integrals

b b 1 . f ( x ) dx = f ( y ) dy

a a b a

2. f ( x ) dx = - f ( x ) dx a b b c b

3 ). f ( x ) dx = f ( x ) dx + f ( x dx where a < c < b a a c a a b b

4. also f ( x ) dx = f (a - x ) dx f ( x ) dx = f ( a + b - x ) dx 0 0 a a

a 2 f ( x ) d x a if f ( x ) is an even function05. f ( x ) dx =

- a if f ( x ) is an odd function 0

a 2 f ( x ) dx 2 a if f (2a - x ) = f ( x )06. f ( x ) dx =

0 if f ( 2a - x ) = - f ( x )0

Examples

p sin x n 2(1 ) Evaluate dx

cos x + sin x n n 0

p -sin x n

p p sin x n 2 a a 2 2 Solution : Let I = dx = dx using f (x ) dx = f ( a - x ) dx

p p cos x + sin x - -n n 0 0 0 0cos x + sin x n n 2 2

p cos x n 2= dx

n n sin x + cos x 0

]sin x cos x (sin x + cos x ) p p n p n p n n p p 2 2 2 2 22I = dx + dx = dx = 1· dx = x =

cos x + sin x sin x + cos x (cos x + sin x ) 2 n n n n n n 0 0 0 0 0 p I =4

x sin x p ( 2) Evaluate dx

1 + sin x 0

x sin x (p - x ) sin(p - x )p p a a Solution : Let I = dx = dx using f (x )dx = f (a - x ) dx

1 + sin x 1+ sin(p - x )0 0 0 0


p p ( - x ) sin x ie I = dx

1 + sin x 0

p p p p x sin x p ( - x ) sin x p x sin x + ( - x ) sin x p sin x p sin x p 2I = dx + dx = dx = dx =1 + sin x 1 + sin x 1 + sin x 1 + sin x 1 + sin x 0 0 0 0 0

òp sin x (1 - sin x ) p sin x - sin x pp 2p p p p = dx = dx = sec x tan x dx - tan x dx 2

(1 + sin x )( 1 - sin x ) cos x 0 0 2 0 0

+ [][ ][] p p p p p p p p p p p = sec x tan x - sec x dx 1 dx = sec x - tan x + x = sec - tan + - sec 0 - tan 0 + 020

0 0 0

[][][] = p - 1 + p - p 1 = p - 2 + p p I = ( 2 p - )2

p dx 3(3 ) Evaluate

p +1 tan x 6

p -cos x dx

p p 2cos x dx b b 3 3Solution : Let I = = using f ( x ) dx = f ( a - x ) dx p p - -p cos x + sin x p a a cos x + sin x 6 6 2 2

sin x dx p 3I = dx

sin x + cos x p 6

p p p p 3 cos x dx sin x dx cos x + sin x 3 3 3 2 I = + dx = dx = 1 · dx = x

p p p cos x + sin x sin x + cos x cos x + sin x p 6 6 66

p p p = - =

3 6 6 p

I =12

ExerciseEvaluate the following

sin x e t an x p p e (1 ) 2 dx (2 ) log x dx (3) 4 dx

1 + cos x cos x 0 2 1 0 2

dx 1 11

( ) ( ) x tan x dx ( ) xe dx - 1 - x 4 5 62

9 - x 2- 0 -1 12

dx x dx p p 82 4(7 ) (8 ) (9) log( 1 + tan ) d a cos x + b sin x ( x + 1)( x + 1 )2 2 20 2 0 2 0

f ( x ) x dx log( + x )1 p 2 a 12(10) dx (11 ) (12 ) dx f ( x ) + f ( a - x ) sin x + cos x 2 (1 + x )0 0 0 2

p 1 7 p 1 2 p p Answers : (1) , (2 ) 1, (3) e - 1, (4 ) log , (5) - , (6) - , (7 ) , (8) ,4 3 5 4 2 e 2 ab 4

()p p p ( 9) log 2 (10 ) a (11) log 2 + 1 (12 ) log 2

8 8 2 2


Reduction formulae

ò p I. 2n n To obtain the reduction formula for I = sin x dx and hence to evaluate sin x dx n

0 n - 1Solution : I = sin x · sin x dx n

u = sin x & v ' = sin x , u ' = (n - )sin x cos x & v = - cos x n - 1 n -2 put 1 n - 1 n - 2 2 n - 1 n - 2 2I = - sin x cos x + (n - 1 ) sin x cos x dx = - sin cos x + (n - 1) sin x (1 - sin x ) dx n

n -1 n - 2 n n -1= - sin x cos x + (n - 1 ) sin x dx - (n - 1 ) sin x dx = - sin cos x + (n - 1)I - (n - 1 )I n - 2 n I + (n - 1)I = - sin x cos x + (n - 1 )I n - 1n n n - 2

-ie (1+ n - 1)I = - sin x cos x + (n - 1 )I n 1n n - 2

n - 1- sin x cos x n - 1ie I = + I n n - 2n n the ultimate integral is I or I according as n is even or odd

0 1

If n is even I = 1 dx = x If n is odd I = sin x dx = - cos x 0 1

p 2 n If I = sin x dx then n

0 p

- sin x cos x n - 1 n - 1n - 1 2 I = + I = + I 0n n - 2 n - 2n n n

0

n - 1 n - 1 n - 3 n - 1 n - 3 n - 5I = I = × I = × × I n n - 2 n - 4 n - 6n n n - 2 n n - 2 n - 4

in general

n - 1 n - 3 1 p × ×L L × × if n is even. n n - 2 2 2

I =n

n - 1 n - 3 2 × ×L L × ×1 if n is odd.n n - 2 3

p p 5 3 1 5p 2 6Eg. (1) I = sin x dx = × × × =

6 6 4 2 2 320

4 2 8p ( 2) I = 2 sin x dx = × × 1 = 5

5 5 3 150

p II. 2n n To obtain the reduction formula for I = cos dx andto evaluate cos x dx n

0

Solution : using integratio n by parts as in I we obtain

n - 1 cos x sin x n - 1I = + I n n n n - p a a p p

andfurtherif I = cos x dx = cos x dx using f (x ) dx = f (a - x )dx 2 n 2 n n 2 0 0 0 0


p 2= sin x dx which is I n

0

6 4 2 16p 2 7Eg. (1) cos x dx = × × × 1 =

7 5 3 350

p p 7 5 3 1 35p 2 8(2 ) cos x dx = × × × × =

8 6 4 2 2 256 0

III. To obtain the reduction formula for I = tan x dx n n

òSolution : I = tan x · tan x dx = tan x · (sec x - 1 ) dx = tan x · sec x dx - tan x dx n - 2 2 n - 2 2 n - 2 2 n - 2 n

n -1tan x I = - I which isthe requiredformula. n n - 2n - 1

IV. To obtain the reduction formula for I = cot x dx n n

òSolution : I = cot x · cot x dx = cot x · (cosec x - 1 ) dx = cot x cosec x dx - cot x dx n - 2 2 n - 2 2 n - 2 2 n - 2n

n - 1- cot x I = - I n n - 2n - 1

n V. To obtain the reduction formula for I = sec x dx n n - 2 2Solution sec: I = sec x · x dx n

' 'n - 2 2 n - 3put u = sec x & v = sec x , u = (n - 2 ) sec x · sec x tanx & v = tan x I = sec tanx - (n - 2) sec x · tan x dx = sec x tanx - (n - 2) sec (sec x - 1)dx n - 2 n - 2 2 n - 2 n - 2 2n

ò= sec x tan x - (n - 2 sec) sec x dx + (n - 2) x dx n - 2 n n - 2

n - 2 I + (n - 2)I = sec x tanx + (n - 2 )I n n n - 2n - 2(1 + n - 2 )I = sec x tan x + (n - 2) I n n - 2

sec x tan x n - 2n - 2I = + I which isthe required reduction formula. n n - 2n - 1 n - 1

VI. To obtain the reduction formula for I = cosec x dx n n

Solution : I = cosec x ·cosec dx n - 2 2n

' 'n - 2 2 n - 3put u = cosec x & v = cosec x , u = - (n - 2 )cosec x · cosec x cot x & v = - cot x n - 2 n - 2 2 n - 2 n - 2 2I = - cosec x cot x - (n - 2 )cosec x · cot x dx = - cosec x cot x - (n - 2 ) cosec x (cosec x - 1 ) dx n

n - 2 n n - 2 n - 2 = - cosec x cot x - (n - 2) cosec x dx + (n - 2) cosec x dx = - cosec x cot x - (n - 2 )I + (n - 2 ) I n n - 2


- ie I + (n - 2)I = - cosec x cot x + (n - 2 )I n 2n n n - 2

ie (n - 1 )I = - cosec x cot x + (n - 2 ) I n - 2n n - 2

n - 2- cosec x cot x ( n - 2 )ie I = + I which isthe required reduction formula.n n - 2( n - 1 ) ( n - 1 )

p VII. I = sin x cos x dx 2 sin x cos x dx m n m n To obtain the reduction formula of and hence to evaluatem , n

0

m n - 1 Solution : I = sin x cos x · cos x dx m , n ' ' m n - 1 m - 1 n m +1 n - 2put u = sin x cos x & v = cos x , u = m sin x · cos x - (n - 1 ) sin x · cos x & v = sin x

()I = sin x cos x · sin x - m sin x cos x - (n - 1)sin cos x sin x dx m n - 1 m - 1 n m +1 n - 2m , n

m +1 n - 1 m n m + 2 n - 2 = sin x cos x - m sin x cos x dx + (n - 1 cos) sin x · x dx

m +1 n - 1 m 2 n - 2= sin x cos x - mI + (n - 1 cos) sin x (1 - cos x ) x dx m , n m +1 n - 1 m n - 2 m n = sin x cos x - mI + (n - 1 ) (sin x cos x - sin x cos x ) dx m , n m + 1 n - 1= sin 1 x cos x - mI + (n - 1) I - (n - ) I m , n m , m - 2 m , n

ie I + mI + ( n - 1) I = sin x cos x x + (n - 1) I m + 1 n - 1 m , n m , n m , n m , n - 2

sin x cos x (n - 1)m + 1 n - 1I = + I ie m , n m , n - 2(m + n ) m + n

p 2 m n which is the required reduction formula, if I = sin x cos x dx m , n

0

p n - 1sin x cos x n - n -m + 1 n - 1 2 1 1 I = I then I = + I = 0 + I m , n m , n - 2m , n m , n - 2 m , n - 2 m + n ( m + n ) m + n m + n

0

applying this reduction formula continuous ly, we have n - 1 n - 3 2 1

× × L L × × if n is odd & m odd or evenm + n m + n - 2 m + 3 m + 1n - 1 n - 3 1 m - 1 m - 3 2

I = × × L L × × × × L L × × 1 if n is even & m is odd m , n m + n m + n - 2 m + 2 m m - 2 3p n - 1 n - 3 1 m - 1 m - 3 1

× × L L × × × × L L × × if n is even & m is evenm + n m + n - 2 m + 2 m m - 2 2 2

Examples

4 2 1 1 p 2 5 5(1 ) I = sin x cos x dx = × × =

5 5, 10 8 6 600

4 2 1 8p 2 6 5( 2) I = sin x cos x dx = × × =

6 , 5 11 9 7 6930


3 1 6 4 2 48p 2(3 ) I = sin x cos x dx = × × × × × 1 = 7 4

7 4, 11 9 7 5 3 34650

p p 5 3 1 5 3 1 5p 2 6 6( 4) I = sin x cos x dx = × × × × × × =

6 , 6 12 10 8 6 4 2 2 2048 0

x 91(5 ) Evaluate dx

1 - x 0 2

p 2 when x = 0 , = 0 when x = 1, =Solution : put x = sin , dx = cos d

9 9 9 x sin · cos d sin · cos d 8 6 4 2 128p p p 19dx = = = sin d = × × × × 1 = 2 2 2

cos 9 7 5 3 3151 - x 1 - sin 0 2 0 2 0 0

x dx 32 a (6) Evaluate

ax - x 20 2

3 3x dx x dx 2 a 2 a Solution : = put x - a = a sin , dx = a cos d

ax - x a - ( x - a )0 2 0 2 2 2 p p

when x = 0 , sin = - 1 = - when x = 2 a , sin = 1 =2 2

( a + a sin ) · a cos d a (1 + sin ) a cos d 3 3 3p p p 3 3 2G.I. = = = a (1 + sin + 3sin + 3 sin ) d 2 2 2

a cos2 2 2 p a - a sin p p - - -2 2 2

ò ò+ + p p p p p p 3 3 2 3 2= a 1 d + sin d + 3 sin d 3 sin d = a + 0 + 0 6 sin d 2 2 2 2 2 2

p p p p p - - - - - 02 2 2 2 2

p p p p p × 1 + 3 5p 3 3 3= a - + 6 × = a = a 2 2 2 2 2 2

Exercise

Evaluate the following

()p p 1 36 5 7 4 2(1 ) sin 3 d (2 ) x sin x dx (3) x 1 - x dx 2

0 0 0

dx 8 1 2

() 56 2( 4) (5 ) x 1 - x dx (6) x 2 - x dx 27

1 + x 0 2 0 02

p 1 - cos p p (7 ) cot x dx (8 ) cosec x dx (9) sin d 2 4 2 5 2

1 + cos p p 04 6

7 3 7 x x x dx 1 1 a ()(10) dx (11) dx (12)

1 - x 1 + x a - x 0 4 0 4 2 0 2 2

()8 16p 3p 8 5p 5 p 3p - 8 11 3 3 8 2Answers : (1) , (2) , (3) , (4) , (5) , (6 ) , (7 ) , (8 ) + log 2 + 3 , (9) ,45 35 256 15 256 8 12 4 8 3

71 1 16a (10) , (11) , (12 )3 24 35


DIFFERENTIAL EQUATIONS

An equation which consists of one dependent variable and its derivatives with respect to one or more independent variables iscalled a 'Differential Equation'. A differential equation of one dependent and one independent variable is called 'OrdinaryDifferential Equation'. A differential equation having one dependent and more than one independent variable is called'Partial Differential Equation'.

Examples of ordinary differential equation

dy y 1. - = 0 2 . x dx - y dx = 0 dx x

2 d y 3. ax dx + by dy = 0 4 . = 0

2 dx dy 2 x + 3 y - 7 dy 2 x - 3y + 4

5. = 6 . =dx 3x - y + 4 dx 4 x - 6 y + 1

32 22 d y dy 7. a = 1 + 8 . y dx + x dy = 0

dx 2dx 2d y dy d y dy 2 2

2 9. - 4 + 3 y = 0 10. x - 2 x + 3 y = 0dx dx dx dx 2 2

Order and Degree of a differential equationOrder

The order of the highest derivative occurring in a differential equation is called 'Order' of the differential equation.

Degree

The highest degree of the highest order derivative occurring in a differential equation (after removing the radicals if any) iscalled 'Degree' of the differential equation.

In the examples given above (1), (2), (3), (5), (6) & (8) are of order one and degree one, (4), (9) & (10) are of order twoand degree one where as (7) is of order two and degree two after removing the radicals.

Formation of Differential EquationDifferential equations are formed by eliminating the arbitrary constants. Arbitrary constants are eliminated by differentiation.

Examples

( ) a x + y = a 2 2 21 Formthe differenti al equation by eliminatingthe constant from Solution : x + y = a 2 2 2

differenti ,ating w.r.t. x we have dy 2 x + 2 y = 0 x dx + y dy = 0 which isthe differenti al equation.dx

124 KSOU Differential Equations

(2) Formthe differenti al equation by eliminatin g 'm ' & ' c ' from y = mx + c Solution : y = mx + c

dy differenti ,atingw.r.t. x we have = m dx

again ,differenti ating w.r.t. x we have

2d y = 0 isthe required differenti al equation.

2dx (3) Obtain the differenti al equation by eliminatin g 'a ' & 'b ' from y = a cos 3 x + b sin 3 x Solution sin: y = a cos 3x + b 3x

differenti ,ating w.r.t. x we have

dy = - 3 cosa sin 3x + 3 b 3x dx

again ,differenti ating w.r.t. x we have

d y 2

= - 9a cos 3 x - 9b sin 3x = - 9 y dx 2

d y 2

ie + 9y = 0 which isthe required differenti al equation.dx 2

Note :- It can be seen from the above examples that the order of the differential equation depends on the number ofarbitrary constants in the equation. ie. if arbitrary constant is one then order is one and if the arbitrary constants aretwo then the order is two.

Solution of equations of first order and first degreeI. Variable Separable

solutionIf the given differenti al equation can bewritten as f ( x ) dx + g ( y ) dy = 0 then this type is called 'Variable separable' ,

is obtained by integration

òie f (x )dx + g (y ) dy = Constant

2 2Eg. 1. Solve y 1 - x dy + x 1 - y dx = 02 2Solution : divide throughou t by 1 - x 1 - y

y dy x dx equation becomes + = 0

1 - y 1 - x 2 2

y dy x dx integratin g + = Constant

2 21 - y 1 - x

- - y - - x = - C 2 2ie 1 1

ie 1 - y + 1 - x = C is the solution.2 2

x x 2 Eg. 2. Solve 3e tan y dx + (1- e )sec y dy = 0

Solution ),: dividingthroughout by tan y (1 - e we havex


x 23 e sec y dx + dy = 0

1 - e tan y x 3 e sec y x 2

integrating, dx + dy = Constanttan y x 1 - e

- 3 log(1 - e ) + log tan y = log C x tan y

ie log = log C x 3(1 - e )

tan y = C (1 - e ) isthe solution. x 3

dy Eg. 3. Solve xy = 1 + x + y + xy

dx dy

Solution : given equation is xy = (1 + x )( 1 + y )dx

ie xy dy = (1 + x )(1+ y ) dx dividethroughou t by x (1+ y )

y dy (1 + x ) dx then =

1 + y x 1 + y - 1 1 x +

ie dy = dx 1 + y x

1 1+ ie 1 - dy = 1 dx

1 + y x integrating, y - log(1 + y ) = log x + x + c isthe soluton

dy 2 2Eg. 4. Solve ( x - y ) = a

dx Solution : put x - y = u

dy du du dy then 1 - = ie 1 - =

dx dx dx dx - du

given equation becomes u 1 = a 2 2

dx du

2 2 2 ie - u = a - u dx

2 u du = dx ie

u - a 2 2

2 2 2u - a + a ie du = dx u - a 2 2

a 2 ie 1+ du = dx

2 2 u - a a u - a 2

integratin g u + log = x + c 2 a u + a


a x - y - a ie x - y + log = x + c

2 x - y + a a x - y - a

ie log = y + c isthe solution.2 x - y + a

dy Eg. 5. Solve = xe given y = 0when x = 0 y - x 2

dx dy y - x 2 Solution : given equation is = xe · e dx

2ie e dy = xe dx - y - x 1 2- y - x integratin g, - e = - e + c 2

1 1when x = 0 , y = 0 - 1 = - + c c = -

2 2

1 1 1 2 ie 2 e = e +2 - y - x solution is - e = - e -- y - x 2 2

II. Homogeneous Equation

( , )dy f x y Equation of the type = or f ( x , y ) dx + g ( x , y ) dy = 0where f ( x , y ) & g ( x , y ) are homogeneou s expression s

dx f ( x , y )in x & y of same degree is called a 'Homogeneou s Equation'.

To solve put y = vx dy dv

then = v + x or dy = v dx + x dv dx dx

by substituti ng this, the given equation reduces to variable separable form and hence can be solved.

dy 2 2Eg. (1) Solve 2 xy = 3 y + x dx

dy dv Solution : put y = vx then = v + x

dx dx dv +2 2 2 2given equation becomes 2 x v v x = 3 x v + x dx

2 dividethroughou t by x dv + 2 then 2 v v x = 3v + 1dx

dv 2 2ie 2v + 2 vx = 3v + 1dx

dv 2 v dv dx ie 2vx = v + 1 ie = 2

dx x v + 12

integrating,we have log( v + 1 ) = log x + log c 2


+ 2y ie log 1 = log x + log c

x 2

( y + x )2 2log = log x + log c ie

x 22 2 2 ie log ( y + x ) - log x = log x + log c

log ( y + x ) = 3 log x + log c = log cx 2 2 3 2 2 3solution is y + x = cx

- x + x x Eg. (2) Solve 1 e dx + e 1 dy = 0y y

y Solution : put x = vy then dx = v dy + y dv

v v equation becomes (1+ e )(v dy + y dv ) + e (1 - v ) dy = 0

ie v dy + y dv + ve dy + ye dv + e dy - ve dy = 0 v v v v ie (v + ve + e - e )dy + y (1 + e ) dv = 0 v v v v

v v ie (v + e ) dy + y (1 + e )dv = 0

v dividethroughou t by y (v + e )

dy (1 + e ) dv v + = 0

y v v + e v integrating, log y + log(v + e ) = log c

ie log y (v + e ) = log c v ie y (v + e ) = c v

+ x x y e = c y

y x ie x + ye = c isthe solutiony

Equations reducible to homogeneous equatins

dy a x + b y + c 1 1 1Give = or ( a x + b y + c ) dx + ( a x + b y + c ) dy = 0

1 1 1 2 2 2 dx a x + b y + c 2 2 2 a b

1 1 Case (i) If = put a x + b y = t then it reduces to homogeneou s equation and hence can be solved.1 1a b 2 2

a b 1 1Case (ii) If put x = X + h & y = Y + k a b

2 2 where h & k are constants to be found out such that a h + b k + c = 0 & a h + b k + c = 0

1 1 1 2 2 2

then given equationreducesto a X + b Y dY

= or ( a X + b Y ) dX + ( a X + b Y ) dY = 0which is homogeneou s and hence can be solved.1 11 1 2 2dX a X + b Y 2 2


dy x + y - 1 Eg. (1) Solve =

dx 2 x + 2 y + 3dy dt

Solution : put x + y = t then 1 + =dx dx

dt t - 1 given equation becomes - 1 =

dx 2 t + 3dt t - 1 t - 1 + 2 t + 3 3t + 2

= +1 = = dx 2t + 3 2 t + 3 2 t + 3

2 t + 3ie dt = dx

3 t + 2

2 t + 3integratin g dt = x + c

3 t + 2

put )2t + 3 = l (3 t + 2 + m = 3lt + 3l + m 2 4 5

3 l = 2 l = 2 l + m = 3 m = 3 - 2 l = 3 - =3 3 3

2 ò3 t + 2 5 dt 2 5 dt 2 5 2+x + c = dt + = 1 dt + = t + log t 23 3 t + 2 3 3 t + 2 3 9 3 9 3t +3

2 5 2 + x + c = (x + y ) + log x + y

3 9 3 1 2 5 2+

ie x + c = y + log x + y is the solution3 3 9 3

Eg. (2) Solve ( 32 x + y - 3 ) dy = ( x + 2 y - )dx Solution : put x = X + h , y = Y + k

choose h & k such that (1) 2h + k = 3

(2) h + 2 k = 3

(1 ) × 2 4 h + 2 k = 6 ( 2) ×1 h + 2k = 3

subtractin g 3h = 3 h = 1, k = 3 - 2 h = 3 - 2 = 1 The given equation reduces to ( 22 X + Y ) dY = ( X + Y ) dX put Y = vX then dY = v dX + X dv

( 2 2 X + vX )(v dX + X dv ) = (X + vX )dX dividethroughout by X

+ + + = +then ( 2 2 v )v dX 2 (2 v )X dv (1 v ) dX 2(2 v + v - 1 - 2 v ) dX + (2 + v )X dv = 0

2 ie (v - 1 ) dX + (v + 2)X dv = 0 dX (v + 2 ) dv

ie + = 0X 2v - 1


(v + 2 ) dv integratin g, log X + = Constant

2v - 1 ò1 2 v dv dv

ie log X + + 2 = Constant2 v - 1 v - 1 2 2 1 1 v - 1 2ie log X + log( v - 1) + 2 × log = log c 2 2 v + 1

Y - 1- Y 2 X ie 2 log 2X + log 1 + 2 log = log c

Y 2 X + 1X

2 2ie 2 log X + log(Y - X ) - 2 log X + 2log(Y - X ) - 2 log(Y + X ) = 2 log c ie log( Y + X ) + log( Y - X ) + 2 log( Y - X ) - 2 log( Y + X ) = 2 log c 3 log( 2Y - X ) - log(Y + X ) = log c

(Y - X )32ie log = log c

Y + X 3 2(Y - X ) = c (Y + X ) but X = x - 1, Y = y - 1

( y - x ) = c ( y + x - 2 ) isthe solution.3 2

III. Linear Equation (Leibnitz's Equation)

dy Equation of the type + Py = Q where P & Q are functions of x is called a 'Linear Equation'.

dx P d x

To find the solution multiply both sides of the given equation by e dy + P d x P d x

then Py e = Qe dx

dy P d x P d x P dx ie e + yPe = Qe

dx d P d x P d x

ie ye = Qe dx

P dx P dx ye = Qe dx + c isthe required solution.

dx P d y P dy Note &: - + Px = Q where P Q are function of y is also a linear equation and its solution is xe = Qe dy + c dy

dy Eg. (1) Solve the equation + y tan x = cos x

dx Solution : Comparingwith the standardequation

P = tan x & Q = cos x P dx = tanx dx = log sec x P dx e = e = sec x log sec x


P dx P dx Solution is ye = Qe dx + c

ie secy sec x = cos x · x dx + c = 1dx + c = x + c

2 - t an- 1 y Eg. (2) (1 + y ) dx + ( x - e ) dy = 0

2Solution : dividethroughou t by (1+ y ) dy - 1dx x - e - tan y

+ = 0dy 2(1 + y )

-1dx x e - t an y ie + =

dy + + 2 2y 1 y 1

Comparingwith standardequation -11 e - tan y

P = Q = 2 2y + 1 y + 1

P dy P dy P dy = tan y - 1 Solution is xe = Q e dy + c

e - t an- 1 y xe = e dy + c t an- 1 y t an- 1 y ie

( y + 1)2

dy t an- 1 y - 1ie xe = + c = tan y + c

y + 2 1

Solution is xe = tan y + c t an- 1 y - 1

Bernoulli's Equation

dy n The equation + Py = Qy where P & Q are functions of x is called Bernoulli' s Equation. dx

1 dy P n To find the solution divide by y then + = Q y dx y n n - 1

1 - n + 1put = z ie y = z n - 1 y

differentiatingw.r.t. x 1 dy 1 dz dy dz

ie = (n - )- n ( 1- n + ) y = y dx ( 1 - n + 1) dx n dx dx

1 dz equation becomes + Pz = Q

( 1- n + ) dx dz

ie + (- n + 1) Pz = (- n + 1) Q which is a linear equation and hence can be solved.dx

dx dz n Note : - + Px = Qx is also Bernoulli' s equation, whose solution is given by + ( 1- n + 1 ) Pz = ( - n + )Q dy dy where &P Q are functions of y


dy 2Eg. (1) Solve + y tan x = y sec x

dx Solution : dividethroughou t by y 2

1 dy 1then + tan x = sec x

dx y 2y 1 1 dy dz

put = z then - = y dx dx 2 y

dz equation becomes - + tan x · z = sec x

dx dz

ie - tan x · z = - sec x dx

which is linear where P = - tanx , Q = - sec x

òP dx = - tan x dx = log cos x

p d x e = e = cos x l og c os x

Solution cosis z cos x = - sec x · x dx + c = - 1 dx + c = - x + c

cos x Solution is = - x + c

y

dy tan y Eg. (2) Solve - = (1 + x ) e sec y x

dx (1 + x )dy sin y

x Solution :the given equation is cos y - = (1 + x ) e dx (1+ x )dy dz

put sin theny = z , cos y =dx dx

dz z x equation becomes - = (1 + x ) e which is a linear equation w here dx 1 + x

1P = - , Q = (1 + x ) e x

1 + x 1

P dx = - dx = - log( 1 + x )(1 + x )

1 1 e = e = e =P dx - log(1 log + x ) 1+ x 1 + x

1 1x x x Solution is z · = ( 11 + x ) e dx = e dx = e + c 1 + x ( + x )

y Solution sin is = e + c x

1 + x


IV. Exact Differential Equation

Let M dx + N dy = 0 bethe differenti al equation where M & N are functions of x & y , the equation is saidto 'Exact'if

M N = and to find the solution

y x Consider M dx (1)

wherethe integration is donew.r.t. x treating y as aconstant andtake N dy (2) herethe integration is donew.r.t. y omittingthe terms containing x in N then the soltuion is (1) + (2) is Constant.

3 2 2 Eg. (1) Solve (2 x - xy - 2 y + 3 )dx - (x y + 2 x ) dy = 0

Solution : Comparing with standardequation 3 2 2M = 2 x - xy - 2 y + 3 , N = - x y - 2 x

M N M N = - 2 xy - 2, = - 2xy - 2 = hence equation is exact

y x y x x x x x y 4 2 4 2 2

3 2 2M dx = (2x - xy - 2y + 3)dx = 2 · - y · - 2 xy + 3 x = - - 2 xy + 3 x (1)4 2 2 2

N dy x y 2x dy 0 dy 2= ( (omitting the terms which contain- - ) = · x )= Constant

x x y c 4 2 2 solution is - - 2xy + 3x =

2 2 24 2 2ie x - x y - 4 xy + 6 x = c

2 2 2 2 2 2Eg. (2) Solve (x + y - a )x dx + (x - y - b ) y dy = 0

Solution : Comparing with standardequation 3 2 2 2 3 2M = x + xy - a x , N = x y - y - b y

M N = 2 xy , = 2 xy

y x M N

= equation is exacty x

4 2 2 2 2x x y a x 3 2 2for solution, consider M dx = (x + xy - a x ) dx = + - (treating y as a constant)4 2 2

N dy = (x y - y - b y ) dy = (- y - b y ) dy omittingthe terms which contain x 2 3 2 3 2

y b y 4 2 2= - -

4 2 x x - y a x y b y c 4 2 2 2 2 4 2 2

solution is + - - - =4 2 2 4 2 4

4 2 2 2 2 4 2 2ie x + 2 x y - 2 a x - y - 2 b y = c


Exercise

Solve the following

dy dy (1 ) x (1 - y ) + y (1 + x ) = 0 ( 2) = e + 4x e 2 2 2 x - 3 y 2 - 3 y

dx dx dy x ( 2 log x + 1)

(3 ) = (4 ) cos( x + y ) dy = dx dx (sin y + y cos y )

dy + dy dy 2 - y (5 ) y - x = a y (6 ) ( x + 1) + 1 = e dx dx dx

dy y y 2 2 ( 7) x dy - y dx = x + y dx (8 ) = + sin

dx x x (9 ) x y dx - (x + y ) dy = 0 (10) (2x + 5 y + 1) dx - (5 x + 2 y - 1) dy = 02 3 3

dy 2 y - x - 4(11) ( 4x - 6 y - 1) dx + ( 3y - 2 x - 2) dy = 0 (12 ) =

dx y - 3 x + 3dy dy

2 3 (13) x log x + y = (log x ) (14 ) = x - 2xy if y = 2 when x = 1dx dx

dy dy (15 ) + y cos x = y sin 2 x (16 ) x + y = x y 3 3 6

dx dx 2 - 1 2 2 2(17) (1 + y )dx = (tan y - x ) dy (18) (2xy + y - tany ) dx + (x - x tan y + sec y ) dy = 0

2 x y - 3x + 1 2 2 + (19 ) dx + dy = 0 ( 20 ) y 1 cos y dy + ( x + log x - x sin y ) dy = 0

x 3 4y y

Answers

x 1 1 x + y (1) log - - = c (2 ) 3e - 2e + 8x = c (3 ) y sin y = x log x + c (4) y = tan + c (5 ) ( x + 1)( 2 - e ) = c 2 x 3 y 3 2 y

y x y 2

3 3x 2 -y 2 2 2 - 1 7(6) (x +1 )(1 - e ) = c (7 ) y + x + y = cx (8) y = 2 x tan (cx ) (9) = 3 log cy (10) (x + y ) = c x - y y 3

() 1

5 2Y + 5 + 21 X () 2 12 2 (11) (2 x - y ) + log( 7 - 8 x + 12 y ) = c (12) ( X - XY + Y ) = c where X = x - 2, Y = y - 3

4 2Y - 5 - 21 X 1 5+3 2 1- x 2 - 2 2 x 3 5 2 (13) y log x = (log x ) + c (14 ) 2 y - x + 1 = 4e (15) y = ce + 2 sin x + 1 (16 ) x y cx = 1s i n

3 2

- 1 - t an- 1 y 2 2 2 3(17 ) x = tan y - 1 + ce (18 ) x y + xy - x tan y + tan y = c (19 ) x - y = cy ( 20 ) y ( x + log x ) + x cos y = c

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Study Material for B.Tech Mathematics -Code -BTC 11 Sharada Vikas Trust (R