XII maths

Biyani's Think Tank

Concept based notes

Maths

Class -XII

Poonam Fatehpuria

Varsha

Information Technology Biyani Girls College, Jaipur

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Published by :

Think Tanks

Biyani Group of Colleges

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Biyani Shikshan Samiti

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First Edition : 2009

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While every effort is taken to avoid errors or omissions in this Publication, any mistake or omission that may have crept in is not intentional. It may be taken note of that neither the publisher nor the author will be responsible for any damage or loss of any kind arising to anyone in any manner on account of such errors and omissions.

Preface

I

am glad to present this book, especially designed to serve the needs of the students. The book has been written keeping in mind the general weakness in understanding the fundamental concepts of the topics. The book is self-explanatory and adopts the “Teach Yourself” style. It is based on question-answer pattern. The language of book is quite easy and understandable based on scientific approach.

Any further improvement in the contents of the book by making corrections, omission and inclusion is keen to be achieved based on suggestions from the readers for which the author shall be obliged.

I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyani, Director (Acad.) Biyani Group of Colleges, who are the backbones and main concept provider and also have been constant source of motivation throughout this Endeavour. They played an active role in coordinating the various stages of this Endeavour and spearheaded the publishing work.

I look forward to receiving valuable suggestions from professors of various educational institutions, other faculty members and students for improvement of the quality of the book. The reader may feel free to send in their comments and suggestions to the under mentioned address.

Unit 1

Chapter – 1

Relations and functions

Q.1 Let A be set of first ten natural numbers. If R be a relation on A defined byxRy x + 2y = 10 then

i. Express R and R-1 as set of ordered pairs ii. Find domain of R and R-1

iii. Find range of R and R-1

Ans. Here A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } Relation R is defined as xRy x + 2y = 10 10 - x y = 2 Now for x = 1, y = 10 - 1= 9 A 2 2

Hence 1 is not related to any element of A.

Similarly use can observe that 3, 5, 7, 9 and 10 are also not related to any element of A.

Again use observe that

When

x = 2,y =

10 - 2

= 4 A

2R4

When

x = 4, y =

10 - 4

= 3 A

4R3

2

When

x = 6, y =

10- 6

= 2 A

6

R2

2

When

x = 8, y =

10- 8

= 1 A

8

R1

2

Hence i. R = {(2,4), (4,3), (6,2), (8,1) } and

R

-1 = (4,2),(3,4),(2,6),(1,8) ii. Domain of R = {2, 4, 6, 8} Domain of

R

-1= 4, 3, 2,1 iii. Range of R = {4, 3, 2, 1} Range of

R

-1 = 2,4,6,8

Q.2 Prove that the relation R on the set N x N defined by (a, b) R (c, d) a+d =b+ c for all (a, b), (c, d) N x N is an equivalence relation.

Ans. To prove that the given relation is an equivalence relation we have relation to show that it is reflexive, symmetric and transitive.

1) Reflexive – Let (a, b) be an arbitary element of N x N. Then,

( a, b) N x N a+ b = b+ a [by commutativity of addition on N]

(a, b) R (a, b)

Thus (a, b) R (a, b) for all (a, b) N x N

Hence the given relation R is reflexive relation on N x N. 2) Symmetric – Let (a. b), (c, d) N x N, Such that (a, b) R (c, d)

Since (a, b) R (c, d) a + d = b + c

c + b = d + a [by commutativity of addition on N]

So R is symmetric relation on N x N 3) Transitive – Let (a, b), (c, d) and (e, f) N x N.

Such that (a, b) R (c, d) and (c, d) R (e, f)

Since (a, b) R (c, d) a + d = b + c ………(1) And (c, d) R (e, f) c + f = d + e ………(2) Adding equation (1) & (2), we get

a + d + c + f = b + c + d + e

a + f = b+ e (a, b) R (e, f)

Thus (a, b) R (c, d) and (c, d) R (e, f) (a, b) R (e, f) for all (a, b), (c d), (e, f) N x N.

So, R is transitive relation on N x N

Hence R being reflexive, symmetric and transitivite is an equivalence relation on N x N.

Hence proved.

Q. 3 On the set N of natural numbers a relation R is defined as

a R b a2-4ab+3b2=0

v

(a, b N) . Prove that R is reflexive but not symmetric not transitivity.

Ans. Given set is N = {1, 2, 3,……..} Relation defined on N is a R b a2-4ab+3b2=0

v

a, b N. 1) Reflexivity – Let a N 2 2 a - 4a.a + 3a 2 2 2 = a - 4a + 3a 2 2 = 4a - 4a = 0 (a, a) R

v

a N. R is reflexive

2) Symmetry - Let a, b N such that (a, b) R 2 2 2 2 4 3 0 4 3 0 ( , )

( , )

ab b b ba a b a R

a b

R

a

Hence (a, b) R but (b, a) R R is not symmetric relation

Ex. (3, 1) R because 32 4 3 1 3(1)2 = 9-12+3 =12-12 = 0 But (1, 3) R because = (1)2 4(1)(3) 3(3)2 1 12 27 0 3) Transitivity – Let a, b, c N such that

(a, b) R and (b, c) R

( , )

a b

R

a

2

4

ab

3

b

2

0

and (b, c) R b2-4bc+3c2

Then it is not necessary true that a2-4ac+3c2=0

Ex. – (9, 3) because

9

2 4(9)(3) 3(3)2 = 81 – 108 + 27 = 0 and (3, 1) R because 32 4(3)(1) 3(1)2 = 9-12+3 = 0 but (9, 1) R because

9

2 4(9)(1) 3(1)2 = 81-36+3 0

R is not transitivity. Hence from (1), (2) & (3) it is clear that R is reflexive but not symmetric and transitivity.

Q. 4 Let A = {1,2,3} then give examples of relations which are – 1) Reflexive, symmetric and transitive

2) Symmetric and transitive but not reflexive 3) Reflexive and transitive but not symmetric 4) Reflexive and symmetric but not transitive

2)

R

₂ {(1,1), (2, 2)} is symmetric and transitive but not reflexive

3)

R

₃ {(1,1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)} is reflexive and transitive but not symmetric

4)

R

₄ {(1,1), (1, 2), (2,1), (2, 2), (2, 3), (3, 2), (3, 3)} is reflexive and symmetric but not transitive

Q. 5 If a, b, {1, 2, 3, 4}, then check whether the following is function or not

f

= {(a, b):b=a+1} also find its range.

Ans. Here f = { (1, 2), (2, 3), (3, 4)}. Here we observe that an element 4 of the given set is not related to any element of the given set. So f is not a function.

Q. 6 If 3 1 x x

(x) =

f then find

f f f x

[ { ( )}]

Ans.

f

(x) = 3 1 x x Now

f

f x ( ) 3 ( ) 1 f x f x 3 3 1 3 1 1 x x x x = 3 3 3 2 6 3 3 1 2 2 1 x x x x x x x x Again – f [f {f(x)}] = f 3 1 x x = 3 3 1 3 1 1 x x x x 3 3 3 4 3 1 4 x x x x x x [ { ( )}] f f f x x

Q. 7 Prove that the function f: R R, f(x) = cos x is many one function? Ans. Given f: R R such that f (x) = cos x

many one function: -

Let a, b R such that f (a) = f (b)

cos

cos

2

,

a

b

a

n

b n

I

f is many one function

Into function – Let y R (Co – domain) If it is possible let f (x) =y 1

cos

cos

y

x

y

x

x

will exist if

1

y

1

When

y

R

[ 1,1]

then pre – image of y does not exist in R (Domain) Hence f is not on to function

f is in to function Hence f is many – one in to function.

Q. 8 If f: R R, f (x) = 2x – 3 and g: R R, g(x) = 3

2 x

then prove that fog = gof =

R I

Ans. Given functions

f R

:

R

, f (x) = 2x-3

3

:

, ( )

2

x

g R

R g x

:

gof R

R

, so (gof) (x) = g [f(x)] = g (2x-3) =

(2

3) 3

2

2

2

x

x

x

gof (x) = x ……….. (1) Again

fog (x) = f [g(x)] = x + 3 x + 3 = 2 - 3 2 2 = x + 3

= f

- 3 = x fog (x) = x ……… (2) and R I : R R such that R

I

(

x

) =

x

,

v

x R ………….(3)

from (1), (2) & (3) we get,

fog = gof =

I

_R Hence proved

Q. 9 Let f: R R be defined by f (x) = 3x – 7. Show that f is invertible and hence find

f

1

Ans. A function f is invertible if f is a bijection 1) Injectivity – Let x, y R then

f(x) = f (y)

3

x

7 3

y

7

x

y

Thus f (x) = f(y) x=y for all x, y R. So, f is an injection 2) Surjectivity – Let y be an arbitary element of R, then f (x) = y

3

7

7

3

x

y

y

x

Clearly 7 3 y R for all y R

Thus for all y R, there exists

7

3

R

y

7 3 7 3 7 3 ( )

( )

y y f x y

f x

f

f is surjection. Hence

f R

:

R

is bijection. Consequently it is invertible Let f(x) = y 1 7 3 7 ( ) 3

3

7

y y f y

x

y

x

Therefore,

f

1: R R is given by 1 7 ( ) 3 x x

f

Chapter – 2

Binary Operation

Q. 1. Discuss the commutativity and associativity of the binary operation on R defined by a * b = 4 ab _{for all a, b R.} Ans. Commutativity – a * b = 4 ab _{and b*a =} 4 ba

We know that multiplication on R is commutative

4 4 ab ba _{for all a, b R} * * a b b a for all a, b R So ‘*’ is commutative on R. Associatively – Let a, b, c R then (a * b) * c 4

4

16

ab c

abc

_...(1) and a * (b * c) = a * 4 4

4

16

bc bc

a

abc

………..(2)

From (1) & (2), we observe that a * (b*c) = (a*b) *c Hence ‘*’ is associative.

Q. 2. Let ‘*’ be a binary operation on set Q – {1} defined by a*b = a+ b – a b , a, b Q - {1}

Find the identity element with respect to * on Q. Also prove that every element of Q – {1} is invertible.

Ans. Let the identity element e exist in Q – {1} w.r.t * on Q - {1}, then a *e = a = e*a for all a Q - {1}

a*e = a for all a Q - {1} a+ e – ae =a

e (1-a) = 0 e = 0

Thus o, is the identity element for * on Q – {1}. Let a be an arbitary element of Q – {1} and

let b be inverse of a, then.

a * b = 0 = b *a [0 is identity element] a * b = 0 a + b – ab = 0 b (1 – a) = - a b = 1 a a {1} 1 0 a Q a Since

a

Q {1}, therefore b= {1} 1 a Q a

Thus every element of Q – {1} is invertible and the inverse of an element a is

1 a a

Q. 3 Let * be an associative binary operation on a set S and a be an invertible element of

S then

(

a

1) 1 a

1 1 1 1 * * *

*

e a a a a e a a

a a

a is inverse of

a

1 a =

(a

1) 1 Hence proved

Chapter – 3

Inverse trigonometric functions

Q. 1 Find the principal values of the followings: - 1) cos-1 -1 2 (2) -1 -sec 2 3) cosec-1 1 (4) -1 3

-1

cot

Ans. (1) Let

_cos

-1

-1

₌

2

cos = -1/2 -1 π 3 π cosθ = cos π -3 2π cosθ = cos 3 2π θ = 3 -1 cos = 2

cosθ = -cos

2π

3

Hence principal value of cos-1 -1 is 2π

(2) Let -1 -2 =

sec

-1 2 secθ = - π secθ = -sec 4 π secθ = sec π -4 3π secθ = sec 4 3π 4 3π sec - 2 = 4

θ =

Hence principal value of sec-1(-√2) is 3π

4 3) Let -1 (+1) = θ cosec -1 (1) = π/2

cosecθ = +1

cosecθ = cosecπ/2

θ = π/2

cosec

Hence principal value of

cosec

-1(1) is 2

4) Let -1 -1 = θ 3

cot

1 3 cot = -π cotθ = -cot 3 cotθ = cot π - π/3 2π cotθ = cot 3 2π θ = 3 -1 2π -1 cot = 3 3

Hence principal value of cot 1 1 2 3 3 is

.

Q. 2 Prove that

tan

1

cos

1

tan

1

1

cos

1

2

4

2

4

2

a

a

b

b

b

a

Ans. Let 1 cos

1

2

a b

cos 2

a b LHS = tan 4 4

tan

2 2 2 π 4 π tanθ 4 = = + tanθ + tanθ - tanθ 1 + tanθ 1 - tanθ π tan = 1 1 - tanθ 1 + tanθ 4 + (1 - tanθ) = 1 - tanθ 1 + tanθ 1 + tan θ + 2 tanθ = π tan + 4 π 1 - tan 4 (1 + tanθ)

tan

1 + tan

2 + 1 + tan θ - 2 tan 2 2 2 2 2 2 = θ θ θ θ θ) 1 + tan θ = 2 1 - tan θ = = 2b = RHS a

(1 - tan

2

1 - tan

1 + tan

2

cos

=

tanA + tanB

tan(A + B) =

1- tanAtanB

tanA - tanB

tan(A - B)

1+ tanAtanB

Hence proved. Q. 3 If sin 1 5 sin 1 12 90

x x then find the value of x

Ans. sin 1 5 sin 1 12 90

-1 • -1 -1 -1 -1 -1 • -1 -1 2 2 5 12 sin = 90 - sin x x 5 12 sin = cos x x sin x + cos x = 90 25 12 cos 1 - = cos x x 25 12 1 - = x x

Squaring both sides, we get

2 2 2 2 2 2 2 25 144 1 - = x x x - 25 144 = x x x - 25 = 144 x = 169 x = ±13

Since x = - 13 does not satisfy given equation . So x = 13 is correct solution. Q. 4 Prove that

2

1 1

2 1 cos tan {sin(cot )}

2

x x

x

1 1 2 2 2 1 1 2 2 2 2 2 2 1 sin(cot ( )) sin sin

1 1

1

1 1

cos tan cos cos

1 2 1 1 2 2 x x x x x x x x x x Hence proved Q. 5 If cos 1 x cos 1 y

a b then prove that

2 2 2 2 2 2 cos sin x xy y a ab b

Ans. Given cos 1 x cos 1 y

a b 2 2 -1 2 2 -1 -1 -1 2 2 2 2 2 2 2 _{2 2} 2 2 2 2 2 2 x y x y = cos - 1- 1- = α a b a b

cos x + cos y = cos xy + 1- x 1- y

xy x y - 1- 1- = cosα ab a b xy x y - cosα = 1- 1-ab a b x y a b 2 2 2 2 2 2 2 2 2 2xy x y x y - cosα + cos = 1- - + ab a b a b 2 2 2 2 2 2 2 2 2 2 x 2xy y - cosα + = 1-cos α ab a b x 2xy y - cosα + = sin α ab a b

Hence proved

Q. 6 Solve the following equation –

-1 -1 -1

2

1 1 2

tan + tan = tan

1+ 2x 4x +1 x

Ans. Given tan-1 1 + tan-1 1 = tan-1 2₂

1+ 2x 4x +1 x -1 -2 2 -1 -1 -1 2 1 1 + ₂ 1 + 2x 1 + 4x tan = tan 1 1 x 1 -1 + 2x 1 + 4x x + y tan x + tan y = tan

1 - xy 4x + 1 + 1 + 2x 2 = 1 + 2x 4x + 1 -1 x 6x + 2 1+ 4x + 8x + 2x - 12 2 2 = x 2 3x + 1 2 2 2 3 2 2 3 2 2 2 2 = x 4x + 3x 3x + x = 8x + 6x 3x - 7x - 6x = 0 x 3x - 7x - 6 = 0 x 3x - 9x + 2x - 6 = 0 x 3x x - 3 + 2 x - 3 = 0 x 3x + 2 x - 3 = 0 -2 x = 0, 3,

Practice Problems –

1) Find principal values of the followings i)sin 1 1

2 ii)

1

tan 3

2) Solve the following equation

-1 x -1 x -1 -1

sec - sec = sec b - sec a

a b

3) If tan 1 3 tan 1 2 4

x x then find value of x.

4) Solve the following - tan 1 x 1 tan ( ) tan1 x 1 x 1 tan 1 3x .

UNIT – II

Chapter – 1

Matrices

Q. 1 If A =

4 - 2 1 3 5 7 9 6 21 15 18 - 25

Write the order of A and find the

elements

a and a

24 34

.

Ans.

Since the given matrix has 3 rows and 4 columns. Therefore it’s

order is 3×4.

24

a

is the element of second row and fourth column. So

a

₂₄

= 6 .

Similarly

34

a

is the element of third row and fourth column. So

a

₃₄

= -25

Q. 2 Construct a 2 × 3 matrix whose element a

ij

are given by:

a

ij

= 2i – j.

Ans. A 2 × 3 matrix will be of the from

[a

ij

] =

a a a 13 11 12 a a a 21 22 23

So

a 11 =

2×1–1 = 1

a 12

= 2×1–2 = 0

a = 2 ×1 - 3 = -1 13 a = 2 × 2 - 1 = 4 - 1 = 3 21 a = 2 × 2 - 2 = 4 - 2 = 2 22 a = 2 × 2 - 3 = 1 23

The required matrix is

[a

ij

] =

1 0 - 1 3 2 1

Q. 3 If the matrix

a b c d e f

is equal to

5 - 1 1 0 3 4

then find a, b, c, d, e, f

.

Ans. Given

a b c d e f

=

5 - 1 1 0 3 4

So by the definition of equality of two

matrices

a = 5, b = -1, c =1, d = 0, e = 3, f = 4.

Q. 4 Find the transpose of the matrix

1 3 A = 2 6 5 - 3

and verify that

T T

A = A

Ans. From the definition of transpose of a matrix,

T

A

is obtained by

interchanging the rows and

Columns of the matrix A. Thus

T 1 2 5

A =

3 6 - 3

Further taking the transpose of the matrix

T

T T _{= = A}

( )

1 32 6 5 - 3

A

Hence

T T A = A

Q. 5 Express the matrix

3 2 3 A = 4 5 3 2 4 5

as the sum of symmetric

and skew– symmetric matrix

Ans. We have

3 2 3 A = 4 5 3 2 4 5 3 4 2 T A = 2 5 4 3 3 5

So

3 2 3 3 4 2 6 6 5 T A + A = 4 5 3 + 2 5 4 = 6 10 7 2 4 5 3 3 5 5 7 10

And

3 2 3 3 4 2 0 -2 1 T A - A = 4 5 3 - 2 5 4 = 2 0 -1 2 4 5 3 3 5 -1 1 0

Let

3 3 5/2 1 _T P = A + A = 3 5 7/2 2 5/2 7/2 5

And

0 -1 1/2 1 _T Q = A - A = 1 0 1/2 2 1 - 1/2 0 2

Then

3 3 5/2 T P = 3 5 7/2 = P 5/2 7/2 5

P is symmetric matrix.

Now

0 1 -1/2 T Q = -1 0 1/2 = - Q 1/2 1/2 0

Q is skew-symmetric matrix.

Also

3 3 5/2 0 -1 1/2 P + Q = 3 5 7/2 + 1 0 1/2 5/2 7/2 5 -1/2 1/2 0 3 2 3 = 4 5 3 2 4 5

= A

Since we have expressed A as a sum of P and R. Hence A

can be expressed as a sum of

a

symmetric

and

a

skew–

symmetric matrix.

Q. 6 Given the matrix –

2 1 1 9 7 -1 2 -4 3

A = 3 -1 0 , B = 3 5 4 and C = 1 -1 0

0 2 4 2 1 6 9 4 5

Verify that (A+B) +C = A+ (B+C) i.e addition of matrices is

associative.

(A+B) =

2 1 1 9 7 -1 11 8 0 3 -1 0 + 3 5 4 = 6 4 4 0 2 4 2 1 6 2 3 10

Now (A+B) +C =

11 8 0 2 -4 3 13 4 3 6 4 4 + 1 -1 0 = 7 3 4 2 3 10 9 4 5 11 7 15

RHS

B+C =

9 7 -1 2 -4 3 11 3 2 3 5 4 + 1 -1 0 = 4 4 4 2 1 6 9 4 5 11 5 11

Now A+ (B+C) =

2 1 1 11 3 2 13 4 3 3 -1 0 + 4 4 4 = 7 3 4 0 2 4 11 5 11 11 7 15

Here LHS = RHS.

Hence verified.

Q. 7 If

1 2 A = 3 -4 5 6

and B =

4 5 6

7 -8 2

find AB and show that

AB BA .

Ans. Here we observe that the order of matrix A is 3×2 and B is 2×3.

Hence AB and BA both

are defined and are of orders 3×3 and 2×2 respectively.

So,

1 2 4 5 6 AB = 3 - 4 7 - 8 2 5 6 1x4 + 2x7 1x5 + 2x(-8) 1x6 + 2x2 = 3x4 + (-4) + (+7) 3x5 + (-4)x(-8) 3x6 + (-4)x2 5x4 + 6x7 5x5 - 6x8 5x6 + 6x2

18 -11 10 AB = -16 47 10 62 -23 42

.……….(1)

And BA =

1 2 4 5 6 × 3 - 4 7 - 8 2 5 6

=

4×1 + 5× 3 + 6× 5 4× 2 + 5×(-4) + 6× 6 7×1 + (-8)× 3 + 2× 5 7× 2 + (-8)×(-4) + 2× 6

BA =

49 24 -7 58

………..(2)

From (1) & (2) we observe that

AB BA

.

Q. 8 If

2 3 -4 1 0 6 -2 1 5

A =

and B =

5 1 2 6 -1 4 5 3 -4

then find 2A - 3B.

Ans. 2A = 2

4 6 -8 = 2 0 12 -4 2 10 2 3 -4 1 0 6 -2 1 5

And 3B = 3

15 3 6 = 18 -3 12 15 9 -12 5 1 2 6 -1 4 5 3 -4

So 2A – 3B =

15 3 6 -11 3 -14 18 -3 12 = -16 3 0 15 9 -12 -19 -7 22 4 6 -8 2 0 12 -4 2 10 -11 3 -14 2A - 3B = -16 3 0 -19 -7 22

Q. 9 Find AB if

0 2 0 3

A =

and

4 6 0 0

B =

Ans.

0 2 4 6 0 3 0 0

AB =

0× 4 + 2× 0 0× 6 + 2× 0 = 0 0 0× 4 + 3× 0 0× 6 + 3× 0 0 0

=

Hence the product of two non

– zero matrices can be a zero

matrix. Thus if the product of

two matrices is a zero matrix, it is not necessary that one of the

matrix is zero matrix.

Q. 10 Let

x

2 - 5x + 6,

f(x) =

find f (A) if A =

2 0 1 2 1 3 1 -1 0

Ans.

Here

f

(A)

=

A - 5A + 6I2 3

………..(1)

So

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

A

 4 + 0 + 1 0 + 0 + (-1) 2 + 0 + 0 4 + 2 + 3 0 + 1 - 3 2 + 3 + 0 2 - 2 + 0 0 - 1 + 0 1 - 3 + 0

=

2 5 -1 2 9 -2 5 0 -1 -2

A

-5A = - 5

2 0 1 -10 0 -5 2 1 3 = -10 -5 -15 1 -1 0 -5 5 0

And

3 1 0 0 6 0 0 6I = 6 0 1 0 = 0 6 0 0 0 1 0 0 6

5 -1 2 -10 0 -5 6 0 0 f(A) = 9 -2 5 + -10 -5 -15 + 0 6 0 0 -1 -2 -5 5 0 0 0 6 1 -1 -3 f(A) = -1 -1 -10 -5 4 4

Practice Problem –

1) For the matrix A, B and C given by

0 0 1 0 5 7 -1 3 5 A = 0 1 0 , B = 0 0 6 , C = 1 -3 -5 1 1 0 0 0 0 -1 3 5

Show that

2 (i)A = I 4 2 (ii)B = 0 (iii)C = C

2) If

5 -1 , B = 2 1 ,C = 1 3 6 7 3 4 -1 4

A =

Verify –

_{(i) (A )}T T_{= A} T T T T T T T T T T _T (ii)(A + B) = A + B (iii)(AC) = C A (iv)(AB) = B A v 3A = 3A

3) Express A =

6 1 -5 -2 -5 4 -3 3 -1

as a sum of symmetric and skew

–

4) If A =

2 3 4 5

, verify that

2 - 7A - 2I = 0

A

.

5) Find a, b, c, d so that

3a + b c + 3 b + 4 d + 2a 7 -2 = 8 5

6) Construct a 2×2 matrix A= [aij] whose elements are given

by

2 i + 2j aij = 2 .

7) If A = [x, y, z], B=

a h g h b f g f c

and C=

x y z

Chapter – 2

Determinant

Q. 1 Find determinant of A =

-1 6 -2 2 1 1 4 1 -3

Ans. |A| = (-1)

1 1 - 62 1 - 22 1 1 -3 4 -3 4 1

= -1 (-3-1) -6 (-6-4) -2 (2-4) = 4+60+4

|A| = 68

Q. 2 Find determinant of A =

1 2 -1 3 2 1 -2 3 3 1 2 1 1 -1 0 2

Ans. |A| =

1 -2 3 2 -2 3 2 1 3 2 1 -2 1 2 1 - 2 3 2 1 - 1 3 1 1 - 3 3 1 2 -1 0 2 1 0 2 1 -1 2 1 -1 0

1

= 1{1(4-0) +2(2+1) +3(0+2) {-2{2(4-0) +2(6-1)

+3(0-2)}-1{2(2+1)-1(6-1)

+3(-3-1)}-3{2(0+2)-1(0-2)-2(-3-1)}

= 1{4+6+6}-2{8+10-6}-1{6-5-12}-3{4+2+8}

= 1{16}-2{12}-1{-11}-3{14}

= -39

A =

1 0 2 1 2 3 2 4 6

Ans. A matrix is singular if

A

=0

1 0 2

A = 1 2 3 = 1[12 - 12]- 0[6 - 6] + 2[4 - 4] = 0 2 4 6

A = 0

Hence A is singular matrix.

Q. 4 Find the mines and cofactors of elements of the

determinant

2 -3 5

6 0

4

1 5 -7

0 4 M₁₁= = 0 - 20 = -20 5 -7 6 4 M₁₂ = = -42 - 4 = -46 1 -7 6 0 M₁₃ = = 30 - 0 = 30 1 5 -3 5 M₂₁= = 21 - 25 = -4 5 -7 2 5 M₂₂= = -14 - 5 = -19 1 -7 2 -3 M₂₃= = 10 + 3 = 13 1 5 -3 5 M₃₁= = -12 - 0 = -12 0 4 2 5 M = = 8 - 30 = -22 32 _{6 4} 2 -3 M = = 0 + 18 = 18 33 _{6 0}

Co- factors –

1+1 2 A = -1 .M = (-1) × (-20) = -20 11 11 1+2 3 A = -1 .M = -1 × (-46) = 46 12 12 1+3 ₄ A = -1 .M = (-1) × 30 = 30 13 13 2+1 3 A = -1 .M = (-1) × (-4) = 4 21 21 2+2 4 A = (-1) .M = (-1) × (-19) = -19 22 22 2+3 5 A = (-1) .M = (-1) × (13) = -13 23 23 3+1 4 A = (-1) .M = (-1) × (-12) = -12 31 31 A 3 3+2 5 = (-1) .M = (-1) × (-22) = 22 2 32 3+3 6 A = (-1) .M = (-1) ×18 = 18 33 33

Q. 5 If W is one of the imaginary cubs root of unity, find the

value of

2 2 2 1 w w w w 1 w 1 w

=



Ans. Given

2 2 2 1 w w w w 1 w 1 w

=



C C + C + C 1 1 2 3

gives

2 2 2 2 2 1 + w + w w w w + w + 1 w 1 w + w + 1 1 w

=



Since

1 + w + w = 02

So

2 2 0 w w = 0 w 1 0 1 w



Now finding its determinant & expending along first column

= 0 w -1 - 0 w -w3 2 2 + 0 w-w4 = 0



Q. 6 Prove that

2 2 2 a b ax + by b c bx + cy = (b - ac) ax + 2bxy + cy ax + by bx + cy 0

Ans. Let A =

a b ax + by b c bx + cy ax + by bx + cy 0

Appling

c

₃

c - xc - yc

_{3 1 2}

, we get

A =

a b ax + by - ax - by b c bx + cy - bx - cy ax + by bx + cy 0 - x(ax + by) - y(bx + cy)

A =

2 2

a b 0

b c 0

ax + by bx + cy -(ax + 2bxy + cy )

Now expending along

c

3

we get

2 2 2

b bx + cy - c(ax + by) - 0[a(bx + cy) - a(ax + by)] + 2bxy + cy )[ac - b ]

A = 0

-(ax

b

(a

(

2 2 2 2 2 2 + 2bxy + cy )(ac - b ) ac)(ax + 2bxy + cy )

A = - x

A =

Hence proved

Q. 7 Find the area of triangle with vertices at the points (3, 8), (-4,

2) and (5,-1).

Ans. Let A (3, 8), B (-4, 2), C (5,-1) are three given vertices of

triangle. So the area of ΔABC is

given by –

1

2

3 8 1 -4 2 1 5 -1 1



1 3 2 + 1 - 8{-4 - 5} + 1{4 - 10} 2 1 3 × 3 + 8 × 9 + 1× (-6) 2 1 9 + 72 - 6 2 75 2

=

=

=

=









Q. 8 If the points (a, 0), (0, b) and (1, 1) are collinear, Prove that a

+ b = ab

Ans. The given points are collinear so

a 0 1 0 b 1 = 0 1 1 1 a[b - 1] + 1[0 - b] = 0 ab - a - b = 0 ab = a + b

Hence proved.

Q. 9 using determinant find the equation of the line joining the

points (1, 2) and (3, 6).

Ans. Let P (x, y) be a point on line AB i.e. the points A (1, 2), B (3, 6)

and P(x, y) are collinear so

ABP = 0



1 2 1 1 3 6 1 2 x y 1

= 0

0 1 2 1 3 6 1 x y 1 1[6 - y]- 2[3 - x] + 1[3y - 6x] = 0 6 - y - 6 + 2x + 3y - 6x = 0 -4x + 2y = 0 y = 2x

Which

is the required equation of line.

Q. 10 Find the adjoint of the matrix

1 3 3 A = 1 4 3 1 3 4

1+1 11 1+2 12 1+3 13 2+1 21 2+2 22 2+3 23 3+1 31 3+2 32 3+3 33 c = (-1) × 7 = 7 c = (-1) × (1) = -1 c = (-1) × (-1) = -1 c = (-1) × (3) = -3 c = (-1) × (1) = 1 c = (-1) × (0) = 0 c = (-1) × (-3) = -3 c = (-1) × (0) = 0 c = (-1) × (1) = 1 T 7 -1 -1 adjA = -3 1 0 -3 0 0 7 -3 -3 adjA = -1 1 0 -1 0 1

Q. 11 Find the inverse of matrix

A = 2 -1 3 4 .

Ans.

A = 8 + 3 = 11 0

So, A is non – singular matrix and therefore it is invertible. Now

finding co-factors

1+1 11 1+2 12 2+1 21 2+2 22 c = (-1) × 4 = 4 c = (-1) × 3 = -3 c = (-1) × (-1) = 1 c = (-1) × 2 = 2 T 4 -3 adj A = 1 2 4 1 adjA = -3 2

Hence

-1 1 A = A .

adj(A)

=

1 4 1 = 4/11 1/11 -3 2 -3/11 2/11 11 4/11 1/11 Α -3/11 2/11

Q. 12 Find the inverse of

1 3 3 A = 1 4 3 1 3 4

and verify that

-1 3

A A = I

Ans. From question No. 10 we find that

adj A =

7 -3 -3 -1 1 0 -1 0 1 A = 1[16 - 9]- 3[4 - 3]+ 3[4 - 3] = 7 - 3 - 3 = 1 0

So A is invertible

Hence

-1 -1 1 A = adjA A 7 -3 -3 7 -3 -3 1 A = -1 1 0 = -1 1 0 1 -1 0 1 -1 0 1

Now

-1 -1 -1 -1 3 7 -3 -3 1 3 3 A A = -1 1 0 1 4 3 -1 0 1 1 3 4 7 - 3 - 3 21 - 12 - 9 21 - 9 - 12 A A = -1 + 1 + 0 -3 + 4 + 0 -3 + 3 + 0 -1 + 0 + 1 -3 + 0 + 3 -3 + 0 + 4 1 0 0 A A = 0 1 0 0 0 1 A A = I

Hence verified

Q. 13 Solve the following system of equations by using cramer’s

rule.

x + 2y = 3 4x + 8y = 12

Ans. We have

1 2 1 2 D = = 8 - 8 = 0 4 8 3 2 D = = 24 - 24 = 0 12 8 1 3 D = = 12 - 12 = 0 4 12

Since D,

D1

and

D2

all are equal to zero so the given system of

equations has infinitely

many solutions.

Let y=k then from equations x+2y=3

x+2k=3

x= 3-2k

Hence, x=3-2k, y= k is the solution of the given system of

equations, where k is arbitrary

real number.

Q. 14 solve the following system of equations by cramer’s rule

x – 2y = 4

- 3x +5y = -7

Ans. We have

1 2 1 -2 D = = 5 - 6 = -1 0 -3 5 4 -2 D = = 20 - 14 = 6 -7 5 1 4 D = = -7 + 12 = 5 -3 -7

So, by Cramer’s rule, we have

1 2 D 6 x = = = -6 D -1 D 5 y = = = -5 D -1

x= -6, y= -5 is required solution.

Q. 15 Solve the following system of equations

2x+3y+4z=0

x+ y+ z = 0

2x-y+3z=0

Ans. We have

2 3 4 D = 1 1 1 2 -1 3 = 2(3 + 1) - 3(3 - 2) + 4(-1 - 2) = 8 - 3 - 12 D = -7 0

So, the given system of equations has only the trivial solutions

i.e x=0, y=0, z=0

Q. 16 Solve the following homogeneous system of equations

x+y-2z=0...(1)

2x+y-3z=0………...(2)

5x+4y-9z=0……….(3)

Ans. We have,

1 1 -2 D = 2 1 -3 5 4 -9 = 1(-9 + 12) - 1(-18 + 15) - 2(8 - 5) = 3 + 3 - 6 D = 0

So, the system of equations has infinitely many solutions.

Consider eq. (1) & (2). Put z=k in equations(1) and (2), we get

–

x+y =2k

2x+y=3k

Solving these equations by cramer’s rule

1 2 1 2 1 1 D = = 1 - 2 = -1 2 1 2k 1 D = = 2k - 3k = -k 3k 1 1 2k D = = 3k - 4k = -k 2 3k D -k x = = = k D -1 D -y = = D k -1 = k

Q. 17 Use matrix method to solve the following system of

equations –

5x – 7y =2

7x – 5y =3

Ans. The given system of equations can be written as

5 -7 x 2 = 7 -5 y 3

Or A x = B, where

5 -7 x 2 A = , X , B 7 -5 y 3

So, the solution is given by

-1

X = A B

. So the find

A-1

we have to

find co factors

1+1 11 1+2 12 2+1 21 2+2 22 T -1 -1 C = (-1) (-5) = -5 C = (-1) (7) = -7 C = (-1) (-7) = 7 C = (-1) (5) = 5 -5 -7 -5 7 adj A = = 7 5 -7 5 A = -25 + 49 = 24 -5 7 1 1 A = adj(A) = -7 5 A 24 X = A B

-5 7 2 1 X = -7 5 3 24 -10 + 21 11 1 1 X = = -14 + 15 1 24 24 11/24 X = 1/24

Hence

x = 11 24

and y=

1 24

Q. 18 Show that the following system of equations is consistent

–

2x – y + 3z = 5

3x + 2y – z = 7

4x + 5y – 5z =9

Ans. The given of equation can be written as

–

2 -1 3 x 5 3 2 -1 y = 7 4 5 -5 z 9

A X =B

Where

2 -1 3 x 5 A 3 2 -1 , X y , B = 7 4 5 -5 z 9

Now

2 -1 3 A 3 2 -1 4 5 -5 = 2[-10 + 5] + 1[-15 + 4] + 3[15 - 8] = -10 - 11 + 21 = 0

So A is singular. So the given system of equation is either

inconsistent on consistent with

infinitely many solutions according as (adj A) B

0 or (adj A)

B= 0 respectively.

1+1 11 1+2 12 1+3 13 2+1 21 2+2 22 2+3 23 3+1 31 3+2 32 3+3 33 c = (-1) (-10 + 5) = -5 c = (-1) (-15 + 4) = 11 c = (-1) (15 - 8) = 7 c = (-1) (5 - 15) = 10 c = (-1) (-10 - 12) = -22 c = (-1) × (10 + 4) = -14 c = (-1) (1 - 6) = -5 c = (-1) (-2 - 9) = 11 c = (-1) (4 + 3) = 7 -5 11 7 adj A = 10 -5 11 7 -5 10 -5 22 -14 = 11 -22 11 -5 11 7 -7 -14 7 -5 10 5 adj A) (B) = 11 -22 7 7 -14 9 -25 + 70 - 45 0 = 55 - 154 + 99 = 0 35 - 98 + 63 0 (adj A) (B) = 0

Thus AX=B has infinitely many solutions and the given system

of equation is consistent.

Unit – III

Chapter – 1

Continuity and Differentiability

1. Check the continuity of the function f(x) at the origin :

( )

;

0

1;

0

x

f x

x

x

x

Ans. We have to show that the given function is continuous at x= 0,

so

LHL

lim

( )

lim

(0

)

0

f x

0

f

h

x

h

lim

(

)

0

lim

lim

1

(

)

0

1

f

h

h

h

_h

h

h

h

h

o

LHL

RHL lim

( )

lim

(0

)

0

0

f x

h

f

h

x

lim

( )

0

lim

lim

1

( )

0

1

f h

h

h

_h

h

h

h

h

o

RHL

Now f(0) = 1

Since LHL ≠ RHL, so the function f(x) is not continuous at the

origin.

2.

Test the continuity of the function at x= 0

sin

cos ,

0

( )

2,

0

x

x

when

x

f x

x

when

x

Ans. LHL

lim

( )

lim

(0

)

0

f x

0

f

h

x

h

lim

(

)

0

f

h

h

sin(

)

lim

cos(

)

(

)

0

h

h

h

h

sin( )

lim

lim cos( )

0

0

h

h

h

h

h

1 1

2

2

LHL

RHL

lim

( )

lim

(0

)

0

0

f x

h

f

h

x

( )

lim

cos( )

( )

0

( )

lim

lim

( )

0

0

1 1 2

2

Sin h

h

h

h

Sin h

Cos h

h

h

h

RHL

And f(0) = 2

Since f(0) = LHL = RHL

So the given function f(x) is continuous.

3.

Find the values of a and b for which the following

function is continuous at x = 1.

2

1

( )

1

5

2

1

x a

when

x

f x

b

when

x

x

when

x

(1)

(1 0)

(1)

lim

(1

)

lim 5(1

) 2

0

lim 5 5

2

0

lim 3 5

0

3

f

f

f

f

h

h

o

h

h

b

h

h

b

h

h

b

And f(1+0) = f(1-0)

lim

(1

)

lim

(1

)

0

0

lim 2(1

)

lim 5(1

) 2

0

2

3

1

f

h

f

h

h

h

h

a

h

h

o

h

a

a

Hence a = 1, b = 3

4.

Show that the function

1

1

,

0

( )

1

1

0

,

0

x

e

x

f x

x

e

x

is

discontinuous at x =0

Ans. LHL

lim

( )

lim

(0

)

0

f x

0

f

h

x

h

1 1 lim ( ) lim 1 0 0 ₁ 1 1 ₁ lim 1 0 1 ₁ 1 1 _{1 0 1} lim 1 1 0 1 0 1 ₁ h e f h h h h _e h e h h e h e h h e

RHL

lim

( )

0

f x

x

lim

(0

)

lim

( )

0

0

1

1

1

1

1

1 0

lim

lim

1

1

1

1 0

0

0

1

1

1

f

h

f h

h

h

h

h

e

e

h

h

h

h

e

e

Since LHL ≠ RHL

So that function f(x) is discontinuous at x = 0.

5.

Show that f(x) = (x) is not differentiate at x = 0.

Ans. LHD

lim

( )

(0)

0

0

f x

f

x

x

(0

)

(0)

lim

0

0

0

f

h

f

h

h

(

)

(0)

lim

(

)

0

f

h

f

h

h

0

lim

lim

(

)

0

0

h

h

h

h

h

h

-1

RHD

( )

(0)

lim

0

0

f x

f

x

x

(0

)

(0)

lim

0

( )

(0)

lim

0

0

lim

lim

1

0

0

f

h

f

h

h

f h

f

h

h

h

_h

h

h

h

h

6. Check the differentiate of the following function at x=

2

1 sin

0

₂

( )

₂

2

₂

₂

x

when

x

f x

x

when

x

Ans. When x=

2

2

( )

₂

2

₂

₂

2

2

f x

f

f

LHD

lim

2

2

0

2

2

f

h

f

h

h

2

2

lim

0

2

f

h

f

h

2

h

1 sin

2

2

lim

0

h

h

h

1 cos

2

2

lim

0

h

h

2

2sin

1 cos( )

2

lim

lim

0

0

h

h

h

h

h

h

2

sin

₂

₁

₂

lim

.

. 1 .0 0

2

2

0

2

h

h

h

h

RHD

lim

2

2

0

2

f

h

f

h

2

h

2

lim

0

h

2

h

2

2

2

2

lim

lim

0

0

0

h

h

h

h

h

h

Since LHD = RHD, so the given of f(x) is differentiable at

2

x

7.

If f(2) = 4 and '(2) 1

f

, then find

lim

(2) 2 ( )

2

2

xf

f x

x

x

Ans. We have

lim

( ) 2 ( )

2

2

xf x

f x

x

x

(2) 2 ( )

lim

2

2

xf

f x

x

x

(2) 2 (2) 2 (2) 2 ( )

lim

2

2

xf

f

f

f x

x

x

2

(2) 2

2

lim

(

2)

2

x

f

f x

f

x

x

2

(2)

( )

(2)

lim

2 lim

(

2)

(

2)

2

2

x

f

f x

f

x

x

x

x

'

(2) 2

(2)

( )

(2)

'(2)

lim

2

2

4 2 1

4 2

2

f

f

f x

f

f

x

x

8.

Differentiate the function f(x) = e

sin x

by first principle.

Ans.

( )

lim

(

)

( )

0

d

f x h

f x

f x

sin(

)

sin

lim

x h

x

d

e

e

f x

dx

h

o

h

sin(

) sin

₁

sin

_lim

0

x h

x

e

x

e

h

h

sin(

) sin

₁

_sin(

_{) sin}

sin

_lim

sin(

) sin

0

x h

x

e

x h

x

x

e

x h

x

h

h

sin(

) sin

1

sin(

) sin

sin

_lim

_lim

sin(

) sin

0

0

x h

x

e

x h

x

x

e

x h

x

h

h

h

2sin

cos

1

₂

₂

sin

_lim

_lim

0

0

₂

2

h

h

y

_x

e

x

e

y

h

y

h

Where y= sin (x + h) – sin x and when h → 0 ,y→ 0

sin

1

2

sin

( )

lim

lim

lim cos

₂

0

0

0

2

sin

( )

(1) (1) cos

sin

_cos

h

y

d

_x

e

_h

f x

e

x

h

dx

y

y

h

h

d

_x

f x

e

x

dx

x

e

x

Hence

d

e

sin

x

e

sin

x

cos

x

dx

.

9.

Differentiate the following function w.r.t. x

2

sin

1

f x

x

Ans. Let

y

sin

x

2

1

putting

u

x

2 1

, we get y= sin u and

2 1

u

x

cos

dy

u

du

and

2

dy

x

du

Now

dy

dy

du

dx

du

dx

cos( ) 2

2

2 cos

1

u

x

x

x

Hence

d

sin

x

2

1

2 cos

x

x

2

1

dx

10.

Differentiate

logsin x w. r .t x

2

Ans. Let

y

logsin

x , putting

2

u

sin

x and

2

v

x

2

log ,

sin

y

u u

v

and

v

x

2

1

,

cos

dy

du

v

du

u dv

and

2

dv

x

dx

Now

dy

dy

du dv

dx

du

dv

dx

1

cos

2

dy

v

x

dx

u

1

cos

2

sin

dy

v

x

dx

v

cot

2

dy

v

x

dx

2

2 cot

dy

x

x

dx

Hence

d

log sin

x

2

2 cot

x

x

2

dx

11.

Differentiate

tan

1

1

1

1

1

x

x

y

x

x

w. r. t. x

Ans. Put

x

cos

cos

1

x

So

tan

1

1 cos

1 cos

1 cos

1 cos

2 cos 2 sin 2 2 1 tan 2 cos 2 sin 2 2 y 2 2 2 cos 2sin 2 2 1 tan 2 2 2 cos 2sin 2 2 y 1 tan 2 1 tan 1 tan 2 y 1 tan tan 2 4 4 2 1 ₁ cos 4 2 y y y x

Differentiating w. r. t. x, we get

1

1

1

1

0

2

₁

2

2

₁

2

dy

dx

_x

_x

1

1

2

₁

2

dy

dx

_x

12. Differentiate

sec 1 1 , 0 1 2 ₂ 2 1 x x

1

1

sec

2

2cos

1

1

1

2

1

1

cos

2cos

1

sec

cos

1

cos

cos 2

2

1

2cos

2

2

1

y

y

x

x

y

y

y

x

dy

dx

_x

13. Differentiate

y

x

3 sin

x

w. r. t. x.

Ans.

y

x

3 sin

x

3

_sin

_sin

3

dy

d

d

x

x

x

x

dx

dx

dx

3

_cos

_sin

₃

2

dy

x

x

x

x

dx

3

_cos

₃

2

_sin

dy

x

x

x

x

dx

14.

Differentiate y = x sin x log x w. r. t. x.

Ans. y = x sin x log x

sin

log

log

sin

sin log

dy

d

d

d

x

x

x

x

x

x

x

x

x

dx

dx

dx

dx

dy

x

dx

1

sin x

x

log

cos

sin log

1

sin

log

cos

sin

log

x

x

x

x

x

dy

x

x

x

x

x

x

dx

15. Differentiate

1 sin

x

e

x

Ans. Let y =

e

x