GATE by RK Kanodiamaths

Page 1

1. If A =-

-- l

é

ë

êêê

ù

û

úúú

0 1 2

1 0 3

2 2

is a singular matrix, then l is

(A) 0 (B) -2

(C) 2 (D) -1

2. If A and B are square matrices of order 4 4´ such

that A B= 5 and A B= a × , then a is

(A) 5 (B) 25

(C) 625 (D) None of these

3. If A and B are square matrices of the same order

such that AB A= and BA A= , then A and B are both

(A) Singular (B) Idempotent

(C) Involutory (D) None of these

4. The matrix, A =- -

-

é

ë

êêê

ù

û

úúú

5 8 0

3 5 0

1 2 1

is

(A) Idempotent (B) Involutory

(C) Singular (D) None of these

5. Every diagonal element of a skew–symmetric matrix

is

(A) 1 (B) 0

(C) Purely real (D) None of these

6. The matrix, A =- -

é

ëêê

ù

ûúú

1

2 2

2

1

2

i

iis

(A) Orthogonal (B) Idempotent

(C) Unitary (D) None of these

7. Every diagonal elements of a Hermitian matrix is

(A) Purely real (B) 0

(C) Purely imaginary (D) 1

8. Every diagonal element of a Skew–Hermitian matrix

is

(A) Purely real (B) 0

(C) Purely imaginary (D) 1

9. If A is Hermitian, then iA is

(A) Symmetric (B) Skew–symmetric

(C) Hermitian (D) Skew–Hermitian

10. If A is Skew–Hermitian, then iA is

(A) Symmetric (B) Skew–symmetric

(C) Hermitian (D) Skew–Hermitian.

11. If A =- - -

--

é

ë

êêê

ù

û

úúú

1 2 2

2 1 2

2 2 1

, then adj. A is equal to

(A) A (B) ct

(C) 3At (D) 3A

12. The inverse of the matrix-

-é

ëê

ù

ûú

1 2

3 5is

(A)5 2

3 1

é

ëê

ù

ûú (B)

5 3

2 1

é

ëê

ù

ûú

(C)- -- -

é

ëê

ù

ûú

5 2

3 1(D) None of these

CHAPTER

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525

LINEAR ALGEBRA

9.1

GATE EC BY RK Kanodia

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Page 2

13. Let A =é

ë

êêê

ù

û

úúú

1 0 0

5 2 0

3 1 2

, then A-1 is equal to

(A)1

4

4 0 0

10 2 0

1 1 2

-- -

é

ë

êêê

ù

û

úúú

(B)1

2

2 0 0

5 1 0

1 1 2

-- -

é

ë

êêê

ù

û

úúú

(C)

1 0 0

10 2 0

1 1 2

-- -

é

ë

êêê

ù

û

úúú

(D) None of these

14. If the rank of the matrix, A =-é

ë

êêê

ù

û

úúú

2 1 3

4 7

1 4 5

l is 2, then

the value of l is

(A) -13 (B) 13

(C) 3 (D) None of these

15. Let A and B be non–singular square matrices of the

same order. Consider the following statements.

(I) ( )AB A BT T T= (II) ( )AB B A

- - -=1 1 1

(III) adj adj adj( ) ( . )( . )AB A B= (IV) r = r r( )( ) ( )AB A B

(V) AB A B= ×

Which of the above statements are false ?

(A) I, III & IV (B) IV & V

(C) I & II (D) All the above

16. The rank of the matrix A =---

é

ë

êêê

ù

û

úúú

2 1 1

0 3 2

2 4 3

is

(A) 3 (B) 2

(C) 1 (D) None of these

17. The system of equations 3 0x y z- + = ,

15 6 5 0x y z- + = , l - + =x y z2 2 0 has a non–zero

solution, if l is

(A) 6 (B) -6

(C) 2 (D) -2

18. The system of equation x y z- + =2 0,

2 3 0x y z- + = , l + - =x y z 0 has the trivial solution as

the only solution, if l is

(A) l ¹ - 4

5(B) l = 4

3

(C) l ¹ 2 (D) None of these

19. The system equationsx y z+ + = 6, x y z+ + =2 3 10,

x y z+ + l =2 12 is inconsistent, if l is

(A) 3 (B) -3

(C) 0 (D) None of these.

20. The system of equations 5 3 7 4x y z+ + = ,

3 26 2 9x y z+ + = , 7 2 10 5x y z+ + = has

(A) a unique solution

(B) no solution

(C) an infinite number of solutions

(D) none of these

21. If A is an n–row square matrix of rank (n - 1), then

(A) adj A = 0 (B) adj A ¹ 0

(C) adj A = In (D) None of these

22. The system of equations x y z- + =4 7 14,

3 8 2 13x y z+ - = , 7 8 26 5x y z- + = has

(A) a unique solution

(B) no solution

(C) an infinite number of solution

(D) none of these

23. The eigen values of A =-

é

ëê

ù

ûú

3 4

9 5are

(A) ± 1 (B) 1, 1

(C) - -1 1, (D) None of these

24. The eigen values of A =-

- --

é

ë

êêê

ù

û

úúú

8 6 2

6 7 4

2 4 3

are

(A) 0, 3, -15 (B) 0 3 15, ,- -

(C) 0 3 15, , (D) 0 3 15, ,-

25. If the eigen values of a square matrix be 1 2, - and 3,

then the eigen values of the matrix 2A are

(A)1

21

3

2, ,- (B) 2 4 6, ,-

(C) 1 2 3, ,- (D) None of these.

26. If A is a non–singular matrix and the eigen values

of A are 2 3 3, , - then the eigen values of A-1 are

(A) 2 3 3, , - (B)1

2

1

3

1

3, ,

-

(C) 2 3 3A A A, , - (D) None of these

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Page 3

27. If -1 2 3, , are the eigen values of a square matrix A

then the eigen values of A2 are

(A) -1 2 3, , (B) 1, 4, 9

(C) 1, 2, 3 (D) None of these

28. If 2 4, - are the eigen values of a non–singular

matrix A and A = 4, then the eigen values of adj A are

(A) 1

21, - (B) 2 1, -

(C) 2 4, - (D) 8 16, -

29. If 2 and 4 are the eigen values of A then the

eigenvalues of AT are

(A) 1

2

1

4, (B) 2, 4

(C) 4, 16 (D) None of these

30. If 1 and 3 are the eigenvalues of a square matrix A

then A3 is equal to

(A) 13 2( )A I- (B) 13 12 2A I-

(C) 12 2( )A I- (D) None of these

31. If A is a square matrix of order 3 and A = 2 then

A A( )adj is equal to

(A)

2 0 0

0 2 0

0 0 2

é

ë

êêê

ù

û

úúú

(B)

1

2

1

2

1

2

0 0

0 0

0 0

é

ë

êêê

ù

û

úúú

(C)

1 0 0

0 1 0

0 0 1

é

ë

êêê

ù

û

úúú

(D) None of these

32. The sum of the eigenvalues of A =é

ë

êêê

ù

û

úúú

8 2 3

4 5 9

2 0 5

is

equal to

(A) 18 (B) 15

(C) 10 (D) None of these

33. If 1, 2 and 5 are the eigen values of the matrix A

then A is equal to

(A) 8 (B) 10

(C) 9 (D) None of these

34. If the product of matrices

A =é

ëê

ù

ûú

cos cos sin

cos sin sin

2

2

q q q

q q qand

B =f f f

f f f

é

ëê

ù

ûú

cos cos sin

cos sin sin

2

2

is a null matrix, then q and f differ by

(A) an odd multiple of p

(B) an even multiple of p

(C) an odd multiple of p2

(D) an even multiple p2

35. If A and B are two matrices such that A B+ and AB

are both defined, then A and B are

(A) both null matrices

(B) both identity matrices

(C) both square matrices of the same order

(D) None of these

36. If A =-é

ëê

ù

ûú

0

0

2

2

tan

tan

a

a

then ( )cos sin

sin cosI A- ×

-é

ëê

ù

ûú

a

a a

a2 is equal to

(A) I A+ (B) I A-

(C) I A+ 2 (D) I A- 2

37. If A =--

é

ëê

ù

ûú

3 4

1 1, then for every positive integer

n n, A is equal to

(A)1 2 4

1 2

++

é

ëê

ù

ûú

n n

n n(B)

1 2 4

1 2

+ --

é

ëê

ù

ûú

n n

n n

(C)1 2 4

1 2

-+

é

ëê

ù

ûú

n n

n n(D) None of these

38. If A a

a aa a

=-

é

ëê

ù

ûú

cos sin

sin cos, then consider the following

statements :

I. A A Aa b ab× = II. A A Aa b a b× = +( )

III. ( )cos sin

sin cosA a

a a

a an

n n

n n=

-

é

ëê

ù

ûú

IV. ( )cos sin

sin cosAa

a aa a

n n n

n n=

-é

ëê

ù

ûú

Which of the above statements are true ?

(A) I and II (B) I and IV

(C) II and III (D) II and IV

Chap 9.1

Page

527

Linear Algebra GATE EC BY RK Kanodia

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Page 4

39. If A is a 3-rowed square matrix such that A = 3,

then adj adj( )A is equal to :

(A) 3A (B) 9A

(C) 27A (D) none of these

40. If A is a 3-rowed square matrix, then adj adj( )A is

equal to

(A) A6

(B) A3

(C) A4

(D) A2

41. If A is a 3-rowed square matrix such that A = 2,

then adj adj( )A2 is equal to

(A) 24 (B) 28

(C) 216 (D) None of these

42. If A =é

ëê

ù

ûú

2 0x

x xand A

- =-

é

ëê

ù

ûú

1 1 0

1 2, then the value

of x is

(A) 1 (B) 2

(C)1

2(D) None of these

43. If A =é

ë

êêê

ù

û

úúú

1 2

2 1

1 1

then A-1 is

(A)

1 4

3 2

2 5

é

ë

êêê

ù

û

úúú

(B)

1 2

2 1

1 2

--

é

ë

êêê

ù

û

úúú

(C)

2 3

3 1

2 7

é

ë

êêê

ù

û

úúú

(D) Undefined

44. If A =-

-

é

ë

êêê

ù

û

úúú

2 1

1 0

3 4

and B =- -é

ëê

ù

ûú

1 2 5

3 4 0then AB is

(A)

- - -- -

é

ë

êêê

ù

û

úúú

1 8 10

1 2 5

9 22 15

(B)

0 0 10

1 2 5

0 21 15

-- - -

-

é

ë

êêê

ù

û

úúú

(C)

- - -- -

é

ë

êêê

ù

û

úúú

1 8 10

1 2 5

9 22 15

(D)

0 8 10

1 2 5

9 21 15

- -- -

é

ë

êêê

ù

û

úúú

45. If A =-

é

ëê

ù

ûú

1 2 0

3 1 4, then AA

T is

(A)1 3

1 4-é

ëê

ù

ûú (B)

1 0 1

1 2 3-é

ëê

ù

ûú

(C)2 1

1 26

é

ëê

ù

ûú (D) Undefined

46. The matrix, that has an inverse is

(A)3 1

6 2

é

ëê

ù

ûú (B)

5 2

2 1

é

ëê

ù

ûú

(C)6 2

9 3

é

ëê

ù

ûú (D)

8 2

4 1

é

ëê

ù

ûú

47. The skew symmetric matrix is

(A)

0 2 5

2 0 6

5 6 0

-

- -

é

ë

êêê

ù

û

úúú

(B)

1 5 2

6 3 1

2 4 0

é

ë

êêê

ù

û

úúú

(C)

0 1 3

1 0 5

3 5 0

é

ë

êêê

ù

û

úúú

(D)

0 3 3

2 0 2

1 1 0

é

ë

êêê

ù

û

úúú

48. If A =é

ëê

ù

ûú

1 1 0

1 0 1and B =

é

ë

êêê

ù

û

úúú

1

0

1

, the product of A and B

is

(A)1

0

é

ëê

ù

ûú (B)

1 0

0 1

é

ëê

ù

ûú

(C)1

2

é

ëê

ù

ûú (D)

1 0

0 2

é

ëê

ù

ûú

49. Matrix D is an orthogonal matrix D =é

ëê

ù

ûú

A B

C 0. The

value of B is

(A)1

2(B)

1

2

(C) 1 (D) 0

50. If A n n´ is a triangular matrix then det A is

(A) ( )-=

Õ 11

aiii

n

(B) aiii

n

=Õ

1

(C) ( )-=å 1

1

aiii

n

(D) aiii

n

=å

1

Page

528

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Page 5

51. If A =é

ëê

ù

ûú

t t

e tt

2 cos

sin, then

d

dt

Awill be

(A)t t

e tt

2 sin

sin

é

ëê

ù

ûú (B)

2t t

e tt

cos

sin

é

ëê

ù

ûú

(C)2t t

e tt

-é

ëê

ù

ûú

sin

cos(D) Undefined

52. If A RÎ ´n n , det A ¹ 0, then

(A) A is non singular and the rows and columns of A

are linearly independent.

(B) A is non singular and the rows A are linearly

dependent.

(C) A is non singular and the A has one zero rows.

(D) A is singular.

************

SOLUTIONS

1. (B) A is singular if A = 0

Þ-

-- l

é

ë

êêê

ù

û

úúú

=0 1 2

1 0 3

2 2

0

Þ - --

- l½½½ ½

½½+

-½½½ ½

½½+

- l½½½ ½

½½=( )1

1 2

22

1 2

0 30

0 3

20

Þ l - + =( ) ( )4 2 3 0 Þ l - + = Þ l = -4 6 0 2

2. (C) If k is a constant and A is a square matrix of

order n n´ then k knA A= .

A B A B B B= Þ = = =5 5 5 6254

Þ a = 625

3. (B) A is singular, if A = 0,

A is Idempotent, if A A2 =

A is Involutory, if A2 = I

Now, A AA A A A A A A2 = = = = =( ) ( )B B B

and B BB BA B B AB BA B2 = = = = =( ) ( )

Þ =A A2 and B B

2 = ,

Thus A & B both are Idempotent.

4. (B) Since, A2

5 8 0

3 5 0

1 2 1

5 8 0

3 5 0

1 2 1

=- -

-

é

ë

êêê

ù

û

úúú

- -

-

é

ë

êêê

ù

û

úúú

=é

ë

êêê

ù

û

úúú

1 0 0

0 1 0

0 0 1

= I, A I A2 = Þ is involutory.

5. (B) Let A = [ ]aij be a skew–symmetric matrix, then

A AT = - , Þ = -a aij ij ,

if i j= then a a a aii ii ii ii= - Þ = Þ =2 0 0

Thus diagonal elements are zero.

6. (C) A is orthogonal if AA IT =

A is unitary if AA IQ = , where A

Q is the conjugate

transpose of A i.e., A AQ T= ( ) .

Here,

AAQ

i

i

i

i=

- -

é

ë

êêêê

ù

û

úúúú

- -

é

ë

êêêê

ù

û

úúúú

=

1

2 2

2

1

2

1

2 2

2

1

2

1 0

0 12

é

ëê

ù

ûú = I

Thus A is unitary.

Chap 9.1

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Page 6

7. (A) A square matrix A is said to be Hermitian if

A AQ = . So a aij ji= . If i j= then a aii ii= i.e. conjugate of

an element is the element itself and aii is purely real.

8. (C) A square matrix A is said to be Skew-Hermitian

if A AQ = - . If A is Skew–Hermitian then A A

Q = -

Þ = -a aji ij ,

if i j= then a aii ii= - Þ + =a aii ii 0

it is only possible when aii is purely imaginary.

9. (D) A is Hermitian then A AQ =

Now, ( )i i i iQ Q QA A A A= = - = - , Þ = -( ) ( )i iQ

A A

Thus iA is Skew–Hermitian.

10. (C) A is Skew–Hermitian then A AQ = -

Now, ( ) ( )i i iQ QA A A A= = - - = then iA is Hermitian.

11. (C) If A = ´[ ]aij n n then det A = ´[ ]cij n n

T

Where cij is the cofactor of aij

Also c Mij

i j

ij= - +( )1 , where M ij is the minor of aij ,

obtained by leaving the row and the column

corresponding to aij and then take the determinant of

the remaining matrix.

Now, M11 = minor of a11 i.e. - =-

-½½½ ½

½½= -1

1 2

2 13

Similarly

M12 =2 2

2 1

-½½½ ½

½½= 6 ; M13 =

-½½½ ½

½½2 1

2 2= - 6

M21

2 2

2 1=

- --

½½½ ½

½½= - 6 ; M22

1 2

2 1=

- -½½½ ½

½½= 3 ;

M23

1 2

2 2=

- --

½½½ ½

½½= 6 ; M31

2 2

1 2=

- --

½½½ ½

½½= 6 ;

M32

1 2

2 2=

- --

½½½ ½

½½= 6 ; M33

1 2

2 1=

- -½½½ ½

½½= 3

C M11

1 1

111 3= - = -+( ) ; C M12

1 2

121 6= - = -+( ) ;

C M13

1 3

131 6= - = -+( ) ; C M21

2 1

211 6= - =+( ) ;

C M22

2 2

221 3= - =+( ) ; C M23

2 3

231 6= - = -+( ) ;

C M31

3 1

311 6= - =+( ) ; C M32

3 2

321 6= - = -+( ) ;

C M33

3 3

331 3= - =+( )

det A =é

ë

êêê

ù

û

úúú

C C C

C C C

C C C

T

11 12 13

21 22 23

31 32 33

=- - -

--

é

ë

êêê

ù

û

úúú

=- - -

--

é

ë

êêê

ù

û

ú3 6 6

6 3 6

6 6 3

3

1 2 2

2 1 2

2 2 1

T

úú

=

T

T3A

12. (A) Since AA

- =1 1adj A

Now, Here A =-

-= -

1 2

3 51

Also, adj A =- -- -

é

ëê

ù

ûú

5 3

2 1

T

Þ =- -- -

é

ëê

ù

ûúadj A

5 2

3 1

A- =

-1 1

1

- -- -

é

ëê

ù

ûú

5 2

3 1=

é

ëê

ù

ûú

5 2

3 1

13. (A) Since, AA

A- =1 1

adj

A = = ¹1 0 0

5 2 0

3 1 2

4 0,

adj A ==-

-é

ë

êêê

ù

û

úúú

=- -

é

ë

êêê

ù

û

4 10 10

0 2 1

0 0 2

4 0 0

10 2 0

1 1 2

T

úúú

A- =

- -

é

ë

êêê

ù

û

úúú

1 1

4

4 0 0

10 2 0

1 1 2

14. (B) A matrix A ( )m n´ is said to be of rank r if

(i) it has at least one non–zero minor of order r, and

(ii) all other minors of order greater than r, if any; are

zero. The rank of A is denoted by r( )A . Now, given that

r = ®( )A 2 minor of order greater than 2 i.e., 3 is zero.

Thus A =-

l½

½

½½

½

½

½½=

2 1 3

4 7

1 4 5

0

Þ - l + - l + - =2 35 4 1 20 3 16 7 0( ) ( ) ( ) ,

Þ - l + - l + =70 8 20 27 0,

Þ l = Þ l =9 117 13

15. (A) The correct statements are

( )AB B AT T T= , ( )AB B A

- - -=1 1 1,

adj adj adj( ) ( ) ( )AB B A=

r ¹ r r( ) ( ) ( )AB A B , A A BB = ×

Thus statements I, II, and IV are wrong.

16. (B) Since

A = - + + - + = - + =2 9 8 2 2 3 2 2 0( ) ( )

Þ r <( )A 3

Again, one minor of order 2 is2 1

0 36 0

½½½ ½

½½= ¹

Þ r =( )A 2

Page

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Page 7

Þ- l

- - - l½½½ ½

½½=

3 5

4 50

Þ - l - - l + =( )( )3 5 16 0 Þ - + l + l + =15 2 16 02

Þ l + l + =2 2 1 0 Þ ( )l + =1 02 Þ l = - -1 1,

Thus eigen values are - -1 1,

24. (C) Characteristic equation is A I- l = 0

Þ- l -

- - l -- - l

½

½

½½

½

½

½½

8 6 2

6 7 4

2 4 3

= 0

Þ l - l + l =2 218 45 0

Þ l l - l - =( )( )3 15 0 Þ l = 0 3 15, ,

25. (B) If eigen values of A are l1, l2, l3 then the eigen

values of kA are kl1, kl2 , kl3. So the eigen values of 2A

are 2 4, - and 6

26. (B) If l1 , l2 ,........, l n are the eigen values of a

non–singular matrix A, then A-1 has the eigen values

1

1l,

1

2l, ........,

1

l n

. Thus eigen values of A-1 are

1

2,

1

3,

-1

3.

27. (B) If l1, l2, ......, l n are the eigen values of a matrix

A, then A2 has the eigen values l1

2, l2

2, ........, l n

2 . So,

eigen values of A2 are 1, 4, 9.

28. (B) If l1, l2 ,...., l n are the eigen values of A then

the eigen values adj A areA

l1

,A

l2

,......,A

l n

; A ¹ 0. Thus

eigenvalues of adj A are4

2,

-4

4i.e. 2 and-1.

29. (B) Since, the eigenvalues of A and AT are square so

the eigenvalues of AT are 2 and 4.

30. (B) Since 1 and 3 are the eigenvalues of A so the

characteristic equation of A is

( ) ( )l - l - =1 3 0 Þ l - l + =2 4 3 0

Also, by Cayley–Hamilton theorem, every square

matrix satisfies its own characteristic equation so

A A I2

24 3 0- + =

Þ = -A A I2

24 3

Þ = -A A A3 24 3 = - -4 4 3 3( )A I A

Þ = -A A I3

213 12

31. (A) Since A A A I( )adj = 3

Þ =é

ë

êêê

ù

û

úúú

=é

ë

êêê

ù

û

úúú

A A( )adj 2

1 0 0

0 1 0

0 0 1

2 0 0

0 2 0

0 0 2

32. (A) Since the sum of the eigenvalues of an n–square

matrix is equal to the trace of the matrix (i.e. sum of the

diagonal elements)

so, required sum = + + =8 5 5 18

33. (B) Since the product of the eigenvalues is equal to

the determinant of the matrix so A = ´ ´ =1 2 5 10

34. (C)

AB =f - f f - f

f - fcos cos cos ( ) cos sin cos ( )

cos sin cos (

q q q qq q ) sin sin cos ( )q qf - f

é

ëê

ù

ûú = A

null matrix when cos ( )q - f = 0

This happens when ( )q - f is an odd multiple ofp2

.

35. (C) Since A B+ is defined, A and B are matrices of

the same type, say m n´ . Also, AB is defined. So, the

number of columns in A must be equal to the number of

rows in B i.e. n m= . Hence, A and B are square matrices

of the same order.

36. (A) Let tana2

= t, then, costan

tan

a

a

a=

-

+=

-+

12

12

12

2

2

2

t

t t

and sintan

tan

a

a

a=

+=

+

22

12

2

122

t

t

( )cos sin

sin cosI A- ×

-é

ëê

ù

ûú

a aa a

=-

é

ë

êêê

ù

û

úúú

´-é

ëê

ù

ûú

12

21

tan

tan

cos sin

sin cos

a

aa aa a

=-

é

ëê

ù

ûú ´

-+

-+

+-+

é

ë

êê

1

1

1

1

2

1

2

1

1

1

2

2 2

2

2

2

t

t

t

t

t

t

t

t

t

t

( )

( )

êê

ù

û

úúúú

=-é

ëê

ù

ûú =

-é

ë

êêê

ù

û

úúú

= +1

1

12

21

t

t

tan

tan

( )

a

aI A

Page

532

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Page 8

37. (B) A2 3 4

1 1

3 4

1 1

5 8

2 3=

--

é

ëê

ù

ûú

--

é

ëê

ù

ûú =

--

é

ëê

ù

ûú

=+ -

-é

ëê

ù

ûú

1 2 4

1 2

n n

n n, where n = 2.

38. (D) A Aa b

a aa a

b bb b

× =-

é

ëê

ù

ûú -é

ëê

ùcos sin

sin cos

cos sin

sin cos ûú

=+ +

- + +é

ëê

ù

ûú = +

cos ( ) sin ( )

sin ( ) cos ( )

a b a ba b a b a bA

Also, it is easy to prove by induction that

( )cos sin

sin cosA a

a aa a

n n n

n n=

-é

ëê

ù

ûú

39. (A) We know that adj adj( )A A A= ×-n 2

.

Here n = 3 and A = 3.

So, adj adj( ) ( )A A A= × =-3 33 2 .

40. (C) We have adj adj( )( )

A A=-n 1 2

Putting n = 3, we get adj adj( )A A=4.

41. (C) Let B A= adj adj( )2 .

Then, B is also a 3 3´ matrix.

adj adj adj adj{ ( )}A B B B2 3

3 1 2= = =

-

= = éëê

ùûú

= =-

adj adj( )( )

A A A2

22

3 12

16 162

2

[ ]K A A2 2

=

42. (C)2 0x

x x

é

ëê

ù

ûú

1 0

1 2-é

ëê

ù

ûú =

é

ëê

ù

ûú

1 0

0 1

Þé

ëê

ù

ûú =

é

ëê

ù

ûú

2 0

0 2

1 0

0 1

x

x, So, 2 1x = Þ x =

1

2.

43. (D) Inverse matrix is defined for square matrix only.

44. (C) AB =-

-

é

ë

êêê

ù

û

úúú

- -é

ëê

ù

ûú

2 1

1 0

3 4

1 2 5

3 4 0

=+ - - + - - + -( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )

( )(

2 1 1 3 2 2 1 4 2 5 1 0

1 1) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )

( )( ) ( )( )

+ - + - +- +

0 3 1 2 0 4 1 5 0 0

3 1 4 3 ( )( ) ( )( ) ( )( ) ( )( )- - + - - +

é

ë

êêê

ù

û

úúú3 2 4 4 3 5 4 0

=- - -

- --

é

ë

êêê

ù

û

úúú

1 8 10

1 2 5

9 22 15

45. (C) AAT =

-é

ëê

ù

ûú -é

ë

êêê

ù

û

úúú

1 2 0

3 1 4

1 3

2 1

0 4

=+ + + - +

+( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )

( )( ) (

1 1 2 2 0 0 1 3 2 1 0 4

3 1 - + + - - +é

ëê

ù

ûú1 2 4 0 3 3 1 1 4 4)( ) ( )( ) ( )( ) ( )( ) ( )( )

=é

ëê

ù

ûú

5 1

1 26

46. (B) if A is zero, A-1 does not exist and the matrix A

is said to be singular. Only (B) satisfy this condition.

A = = - =5 2

2 15 1 2 2 1( )( ) ( )( )

47. (A) A skew symmetric matrix A n n´ is a matrix with

A AT = - . The matrix of (A) satisfy this condition.

48. (C) AB =é

ëê

ù

ûú

é

ë

êêê

ù

û

úúú

1 1 0

1 0 1

1

0

1

=+ ++ +

é

ëê

ù

ûú =

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

1 1 1 0 0 1

1 1 0 0 1 1

1

2

é

ëê

ù

ûú

49. (C) For orthogonal matrix

det M = 1 And M M- =1 T, therefore Hence D D

- =1 T

D DT A C

B BC

B

C A=

é

ëê

ù

ûú = =

--

-é

ëê

ù

ûú

-

0

1 01

This implies BC

BC=

--

Þ = Þ = ±BB

B1

1

Hence B = 1

50. (B) From linear algebra for A n n´ triangular matrix

det A ==

Õ aiii

n

1

, The product of the diagonal entries of A

51. (C )d

dt

d t

dt

d t

dtd e

dt

d t

dt

tt

A=

é

ë

êêê

ù

û

úúú

=

( ) (cos )

( ) (sin )

2

2 -é

ëê

ù

ûú

sin

cos

t

e tt

52. (A) If det A ¹ 0, then A n n´ is non-singular, but if

A n n´ is non-singular, then no row can be expressed as a

linear combination of any other. Otherwise det A = 0

************

Chap 9.1

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Page 9

1. If f x x x x( ) = - + -3 26 11 6 is on [1, 3], then the point

c « ] , [1 3 such that f c¢ =( ) 0 is given by

(A) c = ±21

2(B) c = ±2

1

3

(C) c = ±21

2(D) None of these

2. Let f x x( ) sin= 2 , 02

£ £x p and f c¢ =( ) 0 for c « ] , [02

p .

Then, c is equal to

(A)p

4(B)

p

3

(C)p

6(D) None

3. Let f x x x ex

( ) ( )= +-

3 2 , - £ £3 0x . Let c « -] , [3 0 such

that f c¢ =( ) 0. Then, the value of c is

(A) 3 (B) -3

(C) -2 (D) -1

2

4. If Rolle’s theorem holds for f x x x kx( ) = - + +3 26 5 on

[1, 3] with c = +21

3, the value of k is

(A) -3 (B) 3

(C) 7 (D) 11

5. A point on the parabola y x= -( )3 2, where the

tangent is parallel to the chord joining A (3, 0) and B (4,

1) is

(A) (7, 1) (B)3

2

1

4,

æ

èç

ö

ø÷

(C)7

2

1

4,

æ

èç

ö

ø÷ (D) -æ

èç

ö

ø÷

1

2

1

2,

6. A point on the curve y x= - 2 on [2, 3], where the

tangent is parallel to the chord joining the end points of

the curve is

(A)9

4

1

2,

æ

èç

ö

ø÷ (B)

7

2

1

4,

æ

èç

ö

ø÷

(C)7

4

1

2,

æ

èç

ö

ø÷ (D)

9

2

1

4,

æ

èç

ö

ø÷

7. Let f x x x x( ) ( )( )= - -1 2 be defined in [ , ]0 1

2. Then, the

value of c of the mean value theorem is

(A) 0.16 (B) 0.20

(C) 0.24 (D) None

8. Let f x x( ) = -2 4 be defined in [2, 4]. Then, the value

of c of the mean value theorem is

(A) - 6 (B) 6

(C) 3 (D) 2 3

9. Let f x ex( ) = in [0, 1]. Then, the value of c of the

mean-value theorem is

(A) 0.5 (B) ( )e - 1

(C) log ( )e - 1 (D) None

10. At what point on the curve y x= -(cos )1 in ] ,0 2p[ ,

is the tangent parallel to x –axis ?

(A)p

21, -æ

èç

ö

ø÷ (B) ( , )p - 2

(C)2

3

3

2

p,

-æ

èç

ö

ø÷ (D) None of these

CHAPTER

9.2

DIFFERENTIAL CALCULUS

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Page 10

11. log sin ( )x h+ when expanded in Taylor’s series, is

equal to

(A) log sin cotx h x h x+ - +1

2

2 2cosec K

(B) log sin cotx h x h x+ + +1

2

2 2sec K

(C) log sin cotx h x h x- + +1

2

2 2cosec K

(D) None of these

12. sin x when expanded in powers of x -æ

èç

ö

ø÷

p

2is

(A) 12

2

2

3

2

4

2 3 2

+

-æ

èç

ö

ø÷

+

-æ

èç

ö

ø÷

+

-æ

èç

ö

ø÷

+

x x xp p p

! ! !K

(B) 12

2

2

4

2 2

-

-æ

èç

ö

ø÷

+

-æ

èç

ö

ø÷

-

x xp p

! !K

(C) x

x x

-æ

èç

ö

ø÷ +

-æ

èç

ö

ø÷

+

-æ

èç

ö

ø÷

+p

p p

2

2

3

2

5

2

3 5

! !K

(D) None of these

13. tanp

4+æ

èç

ö

ø÷x when expanded in Taylor’s series, gives

(A) 14

3

2 3+ + + +x x x K

(B) 1 2 28

3

2 3+ + + +x x x ...

(C) 12 4

2 4

+ + +x x

! !K

(D) None of these

14. If u exyz= , then¶

¶ ¶ ¶

3u

x y zis equal to

(A) e xyz x y zxyz[ ]1 3 2 2 2+ +

(B) e xyz x y zxyz [ ]1 3 3 3+ +

(C) e xyz x y zxyz [ ]1 3 2 2 2+ +

(D) e xyz x y zxyz [ ]1 3 3 3 3+ +

15. If z f x ay x ay= + + f -( ) ( ), then

(A)¶

¶=

¶

¶

2

2

22

2

z

xa

z

y(B)

¶

¶=

¶

¶

2

2

22

2

z

ya

z

x

(C)¶

¶= -

¶

¶

2

2 2

2

2

1z

y a

z

x(D)

¶

¶= -

¶

¶

2

2

22

2

z

xa

z

y

16. If ux y

x y=

+

+

æ

èçç

ö

ø÷÷

-tan 1 , then xu

xy

u

y

¶

¶+

¶

¶equals

(A) 2 2cos u (B)1

42sin u

(C)1

4tan u (D) 2 2tan u

17. If ux y x y xy

x xy y=

+ + -

- +-tan 1

3 3 2 2

2 2, then the value of

xu

xy

u

y

¶

¶+

¶

¶is

(A)1

22sin u (B) sin 2u

(C) sin u (D) 0

18. If uy

xx

y

x= fæ

èç

ö

ø÷ + yæ

èç

ö

ø÷, then the value of

xu

dxxy

u

dx dyy

u

y

22

2

22

2

22

¶+

¶+

¶

¶, is

(A) 0 (B) u

(C) 2u (D) -u

19. If z e y x tx

e= =sin , log and y t= 2, thendz

dtis given

by the expression

(A)e

ty t y

x

(sin cos )- 2 2 (B)e

ty t y

x

(sin cos )+ 2 2

(C)e

ty t y

x

(cos sin )+ 2 2 (D)e

ty t y

x

(cos sin )- 2 2

20. If z z u v u x xy y v a= = - - =( , ) , ,2 22 , then

(A) ( ) ( )x yz

xx y

z

y+

¶

¶= -

¶

¶(B) ( ) ( )x y

z

xx y

z

y-

¶

¶= +

¶

¶

(C) ( ) ( )x yz

xy x

z

y+

¶

¶= -

¶

¶(D) ( ) ( )y x

z

xx y

z

y-

¶

¶= +

¶

¶

21. If f x y y z( , ) , ( , )= f =0 0, then

(A)¶

¶×

¶f

¶=

¶

¶×

¶f

¶×

f

y z

f

x y

dz

dx(B)

¶

¶×

¶f

¶×

¶

¶=

¶

¶×

f

y z

f

x

f

x

dz

dx

(C)¶

¶×

¶f

¶× =

¶

¶×

¶f

¶

f

y z

dz

dx

f

x y(D) None of these

22. If z x y= +2 2 and x y axy a3 3 23 5+ + = , then at

x a y adz

dx= =, , is equal to

(A) 2a (B) 0

(C) 2 2a (D) a3

Chap 9.2

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Page 11

23. If x r y r= =cos , sinq q where r and q are the

functions of x, thendx

dtis equal to

(A) rdr

dtr

d

dtcos sinq q

q- (B) cos sinq q

qdr

dtr

d

dt-

(C) rdr

dt

d

dtcos sinq q

q+ (D) r

dr

dt

d

dtcos sinq q

q-

24. If r x y2 2 2= + , then¶

+¶

¶

2

2

2

2

r

dx

r

yis equal to

(A) rr

x

r

y

2

2 2

¶

¶æ

èç

ö

ø÷ +

¶

¶

æ

èçç

ö

ø÷÷

ìíï

îï

üýï

þï(B) 2 2

2 2

rr

x

r

y

¶

¶æ

èç

ö

ø÷ +

¶

¶

æ

èçç

ö

ø÷÷

ìíï

îï

üýï

þï

(C)1

2

2 2

r

r

x

r

y

¶

¶æ

èç

ö

ø÷ +

¶

¶

æ

èçç

ö

ø÷÷

ìíï

îï

üýï

þï(D) None of these

25. If x r y r= =cos , sinq q, then the value of¶

¶+

¶

¶

2

2

2

2

q q

x yis

(A) 0 (B) 1

(C)¶

¶

r

x(D)

¶

¶

r

y

26. If u x ym n= , then

(A) du mx y nx ym n m n= +- -1 1 (B) du mdx ndy= +

(C) udu mxdx nydy= + (D)du

um

dx

xn

dy

y= +

27. If y ax x3 2 33 0- + = , then the value ofd y

dx

2

2is equal

to

(A) -a x

y

2 2

5(B)

2 2 2

5

a x

y

(C) -2 2 4

5

a x

y(D) -

2 2 2

5

a x

y

28. zy

x= -tan 1 , then

(A) dzxdy ydx

x y=

-

+2 2(B) dz

xdy ydx

x y=

+

+2 2

(C) dzxdx ydy

x y=

-

+2 2(D) dz

xdx ydy

x y=

-

+2 2

29. If ux y

x y=

+

+log

2 2

, then xu

xy

u

y

¶

¶+

¶

¶is equal to

(A) 0 (B) 1

(C) u (D) eu

30. If u x yfy

x

n= æ

èç

ö

ø÷-1 , then x

u

xy

y

y x

¶

¶+

¶

¶ ¶

2

2

2

is equal to

(A) nu (B) n n u( )- 1

(C) ( )nu

x-

¶

¶1 (D) ( )n

u

y-

¶

¶1

31. Match the List–I with List–II.

List–I

(i) If ux y

x y=

+

2

then xu

xy

u

x y

¶

¶+

¶

¶ ¶

2

2

2

(ii) If ux y

x y

=-

+

1

2

1

2

1

4

1

4

then xu

xxy

u

x yy

u

y

22

2

22

2

22

¶

¶+

¶

¶ ¶+

¶

¶

(iii) If u x y= +1

2

1

2 then xu

xxy

u

x yy

u

y

22

2

22

2

22

¶

¶+

¶

¶ ¶+

¶

¶

(iv) If u fy

x= æ

èç

ö

ø÷ then x

u

xy

u

y

¶

¶+

¶

¶

List–II

(1) -3

16u (2)

¶

¶

u

x

(3) 0 (4) -1

4u

Correct match is—

(I) (II) (III) (IV)

(A) 1 2 3 4

(B) 2 1 4 3

(C) 2 1 3 4

(D) 1 2 4 3

32. If an error of 1% is made in measuring the major

and minor axes of an ellipse, then the percentage error

in the area is approximately equal to

(A) 1% (B) 2%

(C) p% (D) 4%

33. Consider the Assertion (A) and Reason (R) given

below:

Assertion (A): If u xyfy

x= æ

èç

ö

ø÷, then x

u

xy

u

yu

¶

¶+

¶

¶= 2

Reason (R): Given function u is homogeneous of

degree 2 in x and y.

Of these statements

(A) Both A and R are true and R is the correct

explanation of A

Page

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Page 12

(B) Both A and R are true and R is not a correct

explanation of A

(C) A is true but R is false

(D) A is false but R is true

34. If u x xy= log , where x y xy3 3 3 1+ + = , thendu

dxis

equal to

(A) ( log )12

2+ -

+

+

æ

èçç

ö

ø÷÷xy

x

y

x y

y x

(B) ( log )12

2+ -

+

+

æ

èçç

ö

ø÷÷xy

y

x

y x

x y

(C) ( log )12

2- -

+

+

æ

èçç

ö

ø÷÷xy

x

y

x y

y x

(D) ( log )12

2- -

+

+

æ

èçç

ö

ø÷÷xy

y

x

y x

x y

35. If z xyfy

x= æ

èç

ö

ø÷, then x

z

xy

z

y

¶

¶+

¶

¶is equal to

(A) z (B) 2z

(C) xz (D) yz

36. f x x x x( ) = - + +2 15 36 13 2 is increasing in the

interval

(A) ] 2, 3 [ (B) ] -¥, 3 [

(C) ] -¥, 2 [È ] 3, ¥ (D) None of these

37. f xx

x( )

( )=

+2 1is increasing in the interval

(A) ] -¥, - 1 [ È ] 1, ¥ [ (B) ] -1, 1 [

(C) ] -1, ¥ [ (D) None of these

38. f x x x( ) = -4 22 is decreasing in the interval

(A) ] -¥, -1 [ È ] 0, 1 [ (B) ] -1, 1 [

(C) ] -¥, -1 [ È ] 1, ¥ [ (D) None of these

39. f x x x( ) = + +9 73 6 is increasing for

(A) all positive real values of x

(B) all negative real values of x

(C) all non-zero real values of x

(D) None of these

40. If f x kx x x( ) = - + +3 29 9 3 is increasing in each

interval, then

(A) k < 3 (B) k £ 3

(C) k > 3 (D) k ³ 3

41. If a < 0, then f x e eax ax( ) = + - is decreasing for

(A) x > 0 (B) x < 0

(C) x > 1 (D) x < 1

42. f x x e x( ) = -2 is increasing in the interval

(A) ] -¥ ¥, [ (B) ] -2, 0 [

(C) ] 2, ¥ [ (D) ] 0, 2 [

43. The least value of a for which f x x ax( ) = + +2 1 is

increasing on ] 1, 2, [ is

(A) 2 (B) -2

(C) 1 (D) -1

44. The minimum distance from the point (4, 2) to the

parabola y x2 8= , is

(A) 2 (B) 2 2

(C) 2 (D) 3 2

45. The co-ordinates of the point on the parabola

y x x= + +2 7 2 which is closest to the straight line

y x= -3 3, are

(A) (-2, -8) (B) (2, -8)

(C) (-2, 0) (D) None of these

46. The shortest distance of the point (0, c), where

0 5£ <c , from the parabola y x= 2 is

(A) 4 1c + (B)4 1

2

c +

(C)4 1

2

c -(D) None of these

47. The maximum value of1

x

x

æ

èç

ö

ø÷ is

(A) e (B) e e-

1

(C)1

e

e

æ

èç

ö

ø÷ (D) None of these

48. The minimum value of xx

2 250+æ

èç

ö

ø÷ is

(A) 75 (B) 50

(C) 25 (D) 0

49. The maximum value of f x x x( ) ( cos ) sin= +1 is

(A) 3 (B) 3 3

(C) 4 (D)3 3

4

Chap 9.2

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50. The greatest value of

f xx

x

( )sin

sin

=+æ

èç

ö

ø÷

2

4

p

on the interval [ , ]02

p is

(A)1

2(B) 2

(C) 1 (D) - 2

51. If y a x bx x= + +log 2 has its extremum values at

x = -1 and x = 2, then

(A) a b= - =1

22, (B) a b= = -2 1,

(C) a b= = -21

2, (D) None of these

52. The co-ordinates of the point on the curve

4 5 202 2x y+ = that is farthest from the point (0, -2) are

(A) ( , )5 0 (B) ( , )6 0

(C) (0, 2) (D) None of these

53. For what value of x x02

£ £æ

èç

ö

ø÷

p, the function

yx

x=

+( tan )1has a maxima ?

(A) tan x (B) 0

(C) cot x (D) cos x

*************

SOLUTIONS

1. (B) A polynomial function is continuous as well as

differentiable. So, the given function is continuous and

differentiable.

f ( )1 0= and f ( )3 0= . So, f f( ) ( )1 3= .

By Rolle’s theorem Ec such that ¢ =f c( ) 0.

Now, f x x x¢ = - +( ) 3 12 112

Þ ¢ = - +f c c c( ) 3 12 112 .

Now, f c c c¢ = Þ - + =( ) 0 3 12 11 02

Þ = ±æ

èç

ö

ø÷c 2

1

3.

2. (A) Since the sine function is continuous at each

x R« , so f x x( ) sin= 2 is continuous in 02

,pé

ëêù

ûú.

Also, f x x¢ =( ) cos2 2 , which clearly exists for all

x « ] , [02

p.So, f x( ) is differentiable in x « ] , [0

2

p.

Also, f f( )02

0= æ

èç

ö

ø÷ =

p. By Rolle’s theorem, there exists

c « ] , [02

psuch that ¢ =f c( ) 0.

2 2 0cos c = Þ 22

c =p

Þ c =p

4.

3. (C) Since a polynomial function as well as an

exponential function is continuous and the product of

two continuous functions is continuous, so f x( ) is

continuous in [-3, 0].

f x x e e x x ex x

x x x

¢ = + × - + =+ -é

ëê

ù

ûú

- - -( ) ( ) ( )2 3

1

23

6

22 2 2 2

2

which clearly exists for all x « -] , [3 0 .

f x( ) is differentiable in ] -3, 0 [.

Also, f f( ) ( )- = =3 0 0.

By Rolle’s theorem c « -] 3, 0 [ such that f c¢ =( ) 0.

Now, f c¢ =( ) 0 Þ ec c

c- + -é

ëê

ù

ûú =2

26

20

c c+ - =6 02 i.e. c c2 6 0- - =

Þ ( ) ( )c c+ - =2 3 0 Þ c c= - =2 3, .

Hence, c = -2 « ] -3, 0 [ .

4. (D) f x c x k¢ = - +( ) 3 122

f c c c k¢ = Þ - + =( ) 0 3 12 02

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Page 14

fp

21

æ

èç

ö

ø÷ = , ¢æ

èç

ö

ø÷ =f

p

20, ¢¢æ

èç

ö

ø÷ = -f

p

21,

¢¢¢æ

èç

ö

ø÷ =f

p

20, ¢¢¢¢æ

èç

ö

ø÷ =f

p

21, ....

13. (B) Let f x x( ) tan= Then,

f x f xfx

fp p p p

4 4 4 2 4

2

+æ

èç

ö

ø÷ = æ

èç

ö

ø÷ + ¢æ

èç

ö

ø÷ + × ¢¢æ

èç

ö

ø÷

!+ ¢¢¢æ

èç

ö

ø÷+

xf

3

3 4!...

p

¢ =f x( ) sec2, ¢¢ =f x x x( ) tan2 2

sec ,

¢¢¢ = +f x x x x( ) tan2 44 2 2sec sec etc.

Now,

f f f fp p p p

41

42

44

4

æ

èç

ö

ø÷ = ¢æ

èç

ö

ø÷ = ¢¢æ

èç

ö

ø÷ = ¢¢¢æ

èç

ö

ø, , , ÷ = 16, ...

Thus tanp

41 2

24

616

2 3

+æ

èç

ö

ø÷ = + + × + × +x x

x xK

= + + + +1 2 28

3

2 3x x x K

14. (C) Here u exyz= Þ¶

¶= ×

u

xe yzxyz

¶

¶ ¶= + ×

2u

x yze yze xzxyz xyz = +e z xyzxyz ( )2

¶

¶ ¶ ¶= × + + + ×

321 2

u

x y ze xyz z xyz e xyxyz xyz( ) ( )

= + +e xyz x y zxyz( )1 3 2 2 2

15. (B) z f x ay x ay= + + f -( ) ( )

¶

¶= ¢ + + f¢ -

z

xf x ay x ay( ) ( )

¶= ¢¢ + + ¢¢f -

2

2

z

dxf x ay x ay( ) ( )....(1)

¶

¶= ¢ + - f¢ -

z

yaf x ay a x ay( ) ( )

¶

¶= ¢¢ + + ¢¢f -

2

2

2 2z

ya f x ay a x ay( ) ( )....(2)

Hence from (1) and (2), we get¶

¶=

¶

¶

2

2

22

2

z

ya

z

x

16. (B) ux y

x y=

+

+

æ

èçç

ö

ø÷÷

-tan 1

Þ =+

+=tan u

x y

x yf (say)

Which is a homogeneous equation of degree 1/2

By Euler’s theorem. xf

xy

f

yf

¶

¶+

¶

¶=

1

2

Þ¶

¶+

¶

¶=x

u

xy

u

yu

(tan ) (tan )tan

1

2

x uu

xy u

u

yusec sec tan2 2 1

2

¶

¶+

¶

¶=

Þ¶

¶+

¶

¶=x

u

xy

u

yu u

1

2sin cos =

1

42sin u

17. (A) Here tan ux y x y xy

x xy y=

+ + -

- +

3 3 2 2

2 2= f (say)

Which is homogeneous of degree 1

Thus xf

xy

f

yf

¶

¶+

¶

¶=

As above question number 16 xf

xy

u

yu

¶

¶+

¶

¶=

1

22sin

18. (A) Let vy

x= fæ

èç

ö

ø÷ and w x

y

x= Yæ

èç

ö

ø÷

Then u v w= +

Now v is homogeneous of degree zero and w is

homogeneous of degree one

Þ¶

¶+

¶

¶ ¶+

¶

¶=x

v

xxy

v

x yy

v

y

22

2

22

2

22 0....(1)

and xw

xxy

w

x yy

w

y

22

2

22

2

22 0

¶

¶+

¶

¶ ¶+

¶

¶= ....(2)

Adding (1) and (2), we get

xx

v w xyx y

v w yy

v w22

2

22

2

22 0

¶

¶+ +

¶

¶ ¶+ +

¶

¶+ =( ) ( ) ( )

Þ¶

¶+

¶

¶ ¶+

¶

¶=x

u

xxy

u

x yy

u

y

22

2

22

2

22 0

19. (B) z e yx= sin Þ¶

¶=

z

xe yx sin

And¶

¶=

z

ye yx cos , x te= log Þ =

dx

dt t

1

And y t= 2 Þ =dy

dtt2

dz

dt

z

x

dx

dt

z

y

dy

dt=

¶

¶× +

¶

¶×

= × + ×e yt

e y tx xsin cos1

2 = +e

ty t y

x

(sin cos )2 2

20. (C) Given that

z z u v u x xy y v a= = - - =( , ), ,2 22 ....(i)

¶

¶=

¶

¶×

¶

¶+

¶

¶×

¶

¶

z

x

z

u

u

x

z

v

v

x....(ii)

and¶

¶=

¶

¶×

¶

¶+

¶

¶×

¶

¶

z

y

z

u

u

y

z

v

v

y....(iii)

From (i),

¶

¶= -

¶

¶= - -

u

xx y

u

yx y2 2 2 2, ,

¶

¶=

v

x0,

¶

¶=

v

y0

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Page 15

Substituting these values in (ii) and (iii)

¶

¶=

¶

¶- +

¶

¶×

z

x

z

ux y

z

v( )2 2 0....(iv)

and¶

¶=

¶

¶× - - +

¶

¶×

z

y

z

ux y

z

v( )2 2 0....(v)

From (iv) and (v), we get

( ) ( )x yz

xy x

z

y+

¶

¶= -

¶

¶

21. (C) Given that f x y y z( , ) , ( , )= f =0 0

These are implicit functions

dy

dx

f

xf

y

dz

dy

y

z

= -

¶

¶¶

¶

= -

¶f

¶¶f

¶

,

dy

dx

dz

dy

f

xf

y

y

z

× =-

¶

¶¶

¶

æ

è

çççç

ö

ø

÷÷÷÷

´-

¶f

¶¶f

¶

æ

è

çççç

ö

ø

÷÷÷÷

or,¶

¶×

¶f

¶× =

¶

¶×

¶f

¶

f

y z

dz

dx

f

x y

22. (B) Given that z x y= +2 2

and x y axy a3 3 23 5+ + = ...(i)

dz

dx

z

x

z

y

dy

dx=

¶

¶+

¶

¶× ....(ii)

from (i),¶

¶=

+×

z

x x yx

1

22

2 2,

¶

¶=

+×

z

y x yy

1

22

2 2

and 3 3 3 3 1 02 2x ydy

dxax

dy

dxay+ + + =.

Þ = -+

+

æ

èçç

ö

ø÷÷

dy

dx

x ay

y ax

2

2

Substituting these value in (ii), we get

dz

dx

x

x y

y

x y

x ay

y ax=

++

+-

+

+

æ

èçç

ö

ø÷÷2 2 2 2

2

2

dz

dx

a

a a

a

a a

a aa

a a aa a

æ

èç

ö

ø÷ =

++

+-

+

+

æ

èçç

ö

ø÷÷ =

( , ) .2 2 2 2

2

20

23. (B) Given that x r= cos q, y r= sin q....(i)

dx

dt

x

r

dr

dt

x d

dt=

¶

¶× +

¶

¶×

q

q....(ii)

From (i),¶

¶=

x

rcos q,

¶

¶= -

xr

qqsin

Substituting these values in (ii), we get

dx

dt

dr

dtr

d

dt= - ×cos sinq q

q

24. (C) r x y2 2 2= + Þ¶

¶=

r

xx2 and

¶

¶=

r

yy2

and¶

¶=

2

22

r

xand

¶

¶=

2

22

r

yÞ

¶

¶+

¶

¶= + +

2

2

2

22 2 4

r

x

r

y

and¶

¶æ

èç

ö

ø÷ +

¶

¶

æ

èçç

ö

ø÷÷ = + =

r

x

r

yx y r

2 2

2 2 24 4 4

Þ¶

¶+

¶

¶=

2

2

2

2 2

1r

x

y

y r

¶

¶æ

èç

ö

ø÷ +

¶

¶

æ

èçç

ö

ø÷÷

ìíï

îï

üýï

þï

r

x

r

y

2 2

25. (A) x r y r= =cos , sinq q

Þ =tan qy

xÞ = æ

èç

ö

ø÷-q tan 1 y

x

Þ¶

¶=

+

-æ

èç

ö

ø÷ =

-

+

q

x y x

y

x

y

x y

1

1 2 2 2 2( )

and¶

¶=

-

+

2

2 2 2 2

2q

x

xy

x y( )

Similarly¶

¶=

+

2

2 2 2 2

2q

y

xy

x y( )and

¶

¶+

¶

¶=

2

2

2

20

q q

x y

26. (D) Given that u x ym n=

Taking logarithm of both sides, we get

log log logu m x n y= +

Differentiating with respect to x, we get

1 1 1

u

du

dxm

xn

y

dy

dx= × + × or,

du

um

dx

xn

dy

y= + ×

27. (D) Given that f x y y ax x( , ) = - + =3 2 33 0

f ax x f y f a xx y xx= - + = = - +6 3 3 6 62 2, , ,

f y fyy xy= =6 0,

d y

dx

f f f f f f f

f

xx y x y xy yy x

y

2

2

2 2

3

2= -

- +é

ëê

ù

ûú

( ) ( )

( )

= -- - + -é

ëê

ù

ûú

( ( ) ( )

( )

6 6 3 0 6 3 6

3

2 2 2 2

2 3

x a y y x ax

y

= - - - +2

45

3 3 2 2

yax ay a x( )

= - - + +2

45

3 3 2 2

ya a y a x[ ( ) ]

= - - +2

3 45

2 2 2

ya ax a x[ ( ) ] [ \ x y ax3 3 23 0+ - = ]

= -2 2 2

5

a x

y

28. (A) Given that zy

x= -tan 1 ....(i)

Chap 9.2

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dz

dx

z

x

z

y

dy

dx=

¶

¶+

¶

¶× ....(ii)

From (i)¶

¶=

+ æ

èç

ö

ø÷

×-æ

èç

ö

ø÷ =

-

+

z

x y

x

y

x

y

x y

1

1

2 2 2 2

¶

¶=

+ æ

èç

ö

ø÷

× æ

èç

ö

ø÷ =

+

z

y y

x

x

x

x y

1

1

12 2 2

Substituting these in (ii), we get

dz

dx

y

x y

x

x y

dy

dx=

-

++

+×

2 2 2 2, dz

xdy ydx

x y=

-

+2 2

29. (B) ux y

x y=

+

+log

2 2

, ex y

x y

u =+

+

2 2

= f (say)

f is a homogeneous function of degree one

xf

xy

f

yf

¶

¶+

¶

¶= Þ x

e

xy

e

ye

u uu¶

¶+

¶

¶=

or xeu

xye

u

yeu u u¶

¶+

¶

¶=

or, xu

xy

u

y

¶

¶+

¶

¶= 1

30. (C) Given that u x yfy

x

n= æ

èç

ö

ø÷-1 .

It is a homogeneous function of degree n

Euler’s theorem xu

xy

u

ynu

¶

¶+

¶

¶=

Differentiating partially w.r.t. x, we get

xu

x

u

xy

u

y x

n u

x

¶

¶+

¶

¶+

¶

¶ ¶=

¶

¶

2

2

2

Þ¶

¶+

¶

¶ ¶= -

¶

¶x

u

xy

u

y xn

u

x

2

2

2

1( )

31. (B) In (a) ux y

x y=

+

2

It is a homogeneous function of

degree 2.

xu

xy

u

x yn

u

x

u

x

¶

¶+

¶

¶ ¶= -

¶

¶=

¶

¶

2

2

2

1( ) (as in question 30)

In (b) ux y

x y=

-

+

1 2 1 2

1 4 1 4. It is a homogeneous function of

degree1

2

1

4

1

4-æ

èç

ö

ø÷ =

xu

xxy

u

x yy

u

yn n u2

2

2

22

2

22 1

¶

¶+

¶

¶ ¶+

¶

¶= -( )

= -æ

èç

ö

ø÷ = -

1

4

1

41

3

16u u

In (c) u x y= +1 2 1 2 It is a homogeneous function of

degree1

2.

xu

xxy

u

x dyy

u

yn n u2

2

2

22

2

22 1

¶

¶+

¶

¶+

¶

¶= -( )

= -æ

èç

ö

ø÷ = -

1

2

1

21

1

4u u

In (d)u fy

x= æ

èç

ö

ø÷ It is a homogeneous function of degree

zero.

xu

xy

u

yu

¶

¶+

¶

¶= =0 0.

Hence correct match is

a b c d

2 1 3 4

32. (B) Let 2a and 2b be the major and minor axes of the

ellipse

Area A ab= p

Þ = + +log log log logA a bp

Þ ¶ = ¶ + ¶ + ¶(log ) (log ) (log ) (log )A a bp

Þ¶

= +¶

+¶A

A

a

a

b

b0

Þ ¶ = ¶ + × ¶100 100 100

AA

aa

bb

But it is given that100

1a

a¶ = , and100

1b

b¶ =

1001 1 2

AA¶ = + =

Thus percentage error in A =2%

33. (A) Given that u xyfy

x= æ

èç

ö

ø÷. Since it is a homogeneous

function of degree 2.

By Euler’s theorem xu

xy

u

ynu

¶

¶+

¶

¶= (where n = 2)

Thus xu

xy

u

yu

¶

¶+

¶

¶= 2

34. (A) Given that u x xy= log ....(i)

x y xy3 3 3 1+ + = ....(ii)

we know that¶

¶=

¶

¶+

¶

¶

u

x

u

x

u

y

dy

dx....(ii)

From (i)¶

¶= × × +

u

xx

xyy xy

1log = +1 log xy

and¶

¶= × ×

u

yx

xyx

1=

x

y

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From (ii), we get

3 3 3 1 02 2x ydy

dxx

dy

dxy+ + + ×æ

èç

ö

ø÷ = Þ = -

+

+

æ

èçç

ö

ø÷÷

dy

dx

x y

y x

2

2

Substituting these in (A), we get

du

dxxy

x

y

x y

y x= + + -

+

+

æ

èçç

ö

ø÷÷

ìíî

üýþ

( log )12

2

35. (B) The given function is homogeneous of degree 2.

Euler’s theorem xz

xy

z

yz

¶

¶+

¶

¶= 2

36. (C) ¢ = - + = - -f x x x x x( ) ( )( )6 30 36 6 2 32

Clearly, ¢ >f x( ) 0 when x < 2 and also when x > 3.

f x( ) is increasing in ] -¥, 2 [ È ] 3, ¥ [.

37. (B) f xx x

x

x

x¢ =

+ -

+=

-

+( )

( )

( ) ( )

2 2

2 2

2

2 2

1 2

1

1

1

Clearly, ( )x2 21 0+ > for all x.

So, f x¢ >( ) 0 Þ - >( )1 02x

Þ ( ) ( )1 1 0- + >x x

This happens when - < <1 1x .

So, f x( ) is increasing in ] -1, 1 [.

38. (A) f x x x x x x¢ = - = - +( ) ( )( )4 4 4 1 13 .

Clearly, f x¢ <( ) 0 when x < - 1 and also when x > 1.

Sol. f x( ) is decreasing in ] -¥, -1 [ È ] 1, ¥ [.

39.(C) f x x x¢ = + >( ) 9 21 08 6 for all non-zero real values

of x.

40. (C) f x kx x kx x¢ = - + = - +( ) [ ]3 18 9 3 6 32 2

This is positive when k > 0 and 36 12 0- <k or k > 3.

41. (A) f x e e axax ax( ) ( ) cosh= + =- 2 .

¢ = <f x a ax( ) sinh2 0 When x > 0 because a < 0

42. (D) ¢ = - + = -- - -f x x e xe e x xx x x( ) ( )2 2 2 .

Clearly, ¢ >f x( ) 0 when x > 0 and x < 2.

43. (B) ¢ = +f x x a( ) ( )2

1 2 2 2 4< < Þ < <x x Þ + < + < +2 2 4a x a a

Þ + < ¢ < +( ) ( ) ( )2 4a f x a .

For f x( ) increasing, we have ¢ >f x( ) 0.

\2 0+ ³a or a ³ - 2. So, least value of a is -2.

44. (B) Let the point closest to (4, 2) be ( , )2 42t .

Now, D t t= - + -( ) ( )2 4 4 22 2 2 is minimum when

E t t= - + -( ) ( )2 4 4 22 2 2 is minimum.

Now, E t t= - +4 16 204

Þ = - = - + +dE

dtt t t t16 16 16 1 13 2( ) ( )

dE

dtt= Þ =0 1

d E

dtt

2

2

248= . So,d E

dtt

2

2

1

48 0é

ëê

ù

ûú = >

=( )

.

So, t = 1 is a point of minima.

Thus Minimum distance = - + - =( ) ( )2 4 4 2 2 22 2 .

45. (A) Let the required point be P x y( , ). Then,

perpendicular distance of P x y( , ) from y x- + =3 3 0 is

py x x x x

=- +

=+ + - +3 3

10

7 2 3 3

10

2

=+ +

=+ +x x x2 24 5

10

2 1

10

( )or p

x=

+ +( )2 1

10

2

So,dp

dx

x=

+2 2

10

( )and

d p

dx

2

2

2

10=

dp

dx= 0 Þ x = -2, Also,

d p

dxx

2

2

2

0æ

èçç

ö

ø÷÷ >

= -

.

So, x = -2 is a point of minima.

When x = -2, we get y = - + ´ - + = -( ) ( )2 7 2 2 82 .

The required point is ( , )- -2 8 .

46. (C) Let A c( , )0 be the given point and P x y( , ) be any

point on y x= 2.

D x y c= + -2 2( ) is shortest when E x y c= + -2 2( ) is

shortest.

Now,

E x y c y y c= + - = + -2 2 2( ) ( ) Þ E y y cy c= + - +2 22

dE

dyy c= + -2 1 2 and

d E

dy

2

22 0= > .

dE

dy= 0 Þ y c= -æ

èç

ö

ø÷

1

2

Thus E minimum, when y c= -æ

èç

ö

ø÷

1

2

Also, D c c c= -æ

èç

ö

ø÷ + - -æ

èç

ö

ø÷

1

2

1

2

2

. .. x y c2 1

2= = -æ

èç

ö

ø÷

é

ëê

ù

ûú

= - =-

cc1

4

4 1

2

Chap 9.2

Page

543

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Page 18

47. (B) Let yx

x

= æ

èç

ö

ø÷

1then, y x x= -

Þ = - +-dy

dxx xx ( log )1

d y

dxx x x

x

x x2

2

211

= + - ×- -( log )

dy

dx= 0 Þ 1 0+ =log x Þ x

e=

1

d y

dx ex

e

e2

21

11

10

é

ëê

ù

ûú = -æ

èç

ö

ø÷ <

=æ

èçç

ö

ø÷÷

- -

.

So, xe

=1

is a point of maxima. Maximum value = e e1 .

48. (A) ¢ = -f x xx

( ) 2250

2and ¢¢ = +æ

èç

ö

ø÷f x

x( ) 2

5003

¢ =f x( ) 0 Þ 2250

02

xx

- = Þ x = 5.

¢¢ = >f ( )5 6 0. So, x = 5 is a point of minima.

Thus minimum value = +æ

èç

ö

ø÷ =25

250

575.

49. (D) ¢ = - +f x x x( ) ( cos )(cos )2 1 1 and

¢¢ = - +f x x x( ) sin ( cos )1 4 .

¢ =f x( ) 0 Þ =cos x1

2or cos x = -1 Þ =x p 3 or

x = p.

¢¢æ

èç

ö

ø÷ =

-<f

p

3

3 3

20. So, x = p 3 is a point of maxima.

Maximum value = æ

èç

ö

ø÷ +æ

èç

ö

ø÷ =sin cos

p p

31

3

3 3

4.

50. (C) f xx x

x x( )

sin cos

sin cos=

+2

2

=+

=2 2 2 2

( )sec cosecx x z(say),

where z x x= +( )sec cosec .

dz

dxx x x x

x

xx= - = -sec cosectan cot

cos

sin(tan )

2

3 1 .

dz

dx= 0 Þ tan x = 1 Þ x =

p

4in 0

2,

pé

ëêù

ûú.

Sign ofdz

dxchanges from -ve to +ve when x passes

through the point p 4. So, z is minimum at x = p 4 and

therefore, f x( ) is maximum at x = p 4.

Maximum value =+

=2 2

4 41

[sec( ) ( )]p pcosec.

51. (C)dy

dx

a

xbx= + +2 1

dy

dx x

é

ëêù

ûú=

= -( )1

0 Þ - - + =a b2 1 0 Þ + =a b2 1....(i)

dy

dx x

é

ëêù

ûú=

=( )2

0 Þ + + =a

b2

4 1 0

Þ + = -a b8 2....(ii)

Solving (i) and (ii) we get b = -1

2and a = 2.

52. (C) The given curve isx y2 2

5 41+ = which is an

ellipse.

Let the required point be ( cos , sin )5 2f f . Then,

D = f - + f +( cos ) ( sin )5 0 2 22 2 is maximum

when z D= 2 is maximum

z = f + + f5 4 12 2cos ( sin )

Þf

= - f f + + f fdz

d10 8 1cos sin ( sin ) cos

dz

df= 0 Þ 2 4 0cos ( sin )f - f =

Þ f =cos 0 Þ f =p

2.

dz

df= - f + fsin cos2 8 Þ

f= - f - f

d z

d

2

22 2 8cos sin

when f =p

2,

d z

d

2

20

f< .

z is maximum when f =p

2. So, the required point is

52 2

cos , sinp pæ

èç

ö

ø÷ i.e. (0, 2).

53. (D) Let zx

x x

x

x=

+= +

1 1tan tan

Then,dz

dx xx= - +

12

2sec and

d z

dx xx x

2

2 3

222= + sec tan

dz

dx= 0 Þ - + =

10

2

2

xxsec Þ x x= cos .

d z

dxx x x

x x

2

2

3 22 2 0é

ëê

ù

ûú = + >

= cos

cos tansec .

Thus z has a minima and therefore y has a maxima at

x x= cos .

************

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Page 19

1.x

xdx

2 1+ò is equal to

(A)1

212log ( )x + (B) log ( )x2 1+

(C) tan -1

2

x(D) 2 1tan - x

2. If F aa

a( )log

,= >1

1 and F x a dx K( ) = +ò 2 is equal

to

(A)1

1log

( )a

a ax a- + (B)1

log( )

aa ax a-

(C)1

1log

( )a

a ax a+ + (D)1

1log

( )a

a ax a+ -

3.dx

x1 +ò sinis equal to

(A) - + +cot x x ccosec (B) cot x x c+ +cosec

(C) tan x x c- +sec (D) tan x x c+ +sec

4.( )3 1

2 2 32

x

x xdx

+

- +ò is equal to

(A)3

42 2 3

5

2

2 1

5

2 1log ( ) tanx xx

- + +-æ

èçç

ö

ø÷÷

-

(B)4

32 2 3 5

2 1

5

2 1log ( ) tanx xx

- + +-æ

èçç

ö

ø÷÷

-

(C)4

32 2 3

2

5

2 1

5

2 1log ( ) tanx xx

- + +-æ

èçç

ö

ø÷÷

-

(D)3

42 2 3

2

5

2 1

5

2 1log ( ) tanx xx

- + +-æ

èçç

ö

ø÷÷

-

5.dx

x1 3 2+ò sinis equal to

(A) 1

2

1tan (tan )- x (B) 2 1tan (tan )- x

(C) 1

2

1 2tan ( tan )- x (D) ( )2 1 1

2tan tan- x

6.2 3

3 4

sin cos

sin cos

x x

x xdx

+

+ò is equal to

(A)9

25

1

253 4x x x+ +log( sin cos )

(B)18

25

2

253 4x x x+ +log( sin cos )

(C)18

25

1

253 4x x x+ +log( sin cos )

(D) None of these

7. 3 8 3 2+ -ò x x dx is equal to

(A)3 4

3 33 8 3 2x

x x-

+ - --æ

èç

ö

ø÷-25

18 3

3 4

5

1sinx

(B)3 4

63 8 3 2x

x x-

+ - +-æ

èç

ö

ø÷-25 3

18

3 4

5

1sinx

(C)3 4

6 33 8 3 2x

x x-

+ - --æ

èç

ö

ø÷-25

18 3

3 4

5

1sinx

(D) None of these

8.dx

x x2 3 42 + +ò is equal to

(A)1

2

4 3

23

1sin - +x(B)

1

2

4 3

23

1sinh - +x

(C)1

2

4 3

23

1cosh - +x(D) None of these

CHAPTER

9.3

Page

545

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Page 20

9.2 3

12

x

x xdx

+

+ +ò is equal to

(A) 2 1 22 1

3

2 1x xx

+ + ++-sinh

(B) x xx2 11 2

2 1

3+ + +

+-sinh

(C) 2 12 1

3

2 1x xx

+ + ++-sinh

(D) 2 12 1

3

2 1x xx

+ + -+-sinh

10.dx

x x-ò 2

is equal to

(A) x x c- +2 (B) sin ( )- - +1 2 1x c

(C) log ( )2 1x c- + (D) tan ( )- - +1 2 1x c

11.1

1 1 2 2( )x x xdx

+ - -ò is equal to

(A) 22

1

1cosh -

+

æ

èçç

ö

ø÷÷x

(B)1

2

2

1

1cosh -

+

æ

èçç

ö

ø÷÷x

(C) -+

æ

èçç

ö

ø÷÷

-22

1

1coshx

(D) -+

æ

èçç

ö

ø÷÷

-1

2

2

1

1coshx

12.dx

x xsin cos+ò is equal to

(A)1

2 4log tan x +æ

èç

ö

ø÷

p(B)

1

2 2 6log tan

x+æ

èç

ö

ø÷

p

(C)1

2 2 8log tan

x+æ

èç

ö

ø÷

p(D)

1

2 4 4log tan

x+æ

èç

ö

ø÷

p

13.dx

x a x bsin( ) sin( )- -ò is equal to

(A) sin( ) log sin( )x a x b- -

(B) log sinx a

x b

-

-æ

èç

ö

ø÷

(C) sin( ) logsin( )

sin( )a b

x a

x b-

-

-

ìíî

üýþ

(D)1

sin( )log

sin( )

sin( )a b

x a

x b-

-

-

ìíî

üýþ

14.dx

ex -ò 1is equal to

(A) log ( )ex - 1 (B) log ( )1 - ex

(C) log ( )e x- - 1 (D) log ( )1 - ex

15.dx

x x x1 2 3+ + +ò is equal to

(A)1

2

1

1

2

2

1log( )

tanx

xx

+

++

é

ëê

ù

ûú

-

(B)1

4

1

12

2

2

1log( )

tanx

xx

+

++

é

ëê

ù

ûú

-

(C)1

2

1

12

2

2

1log( )

tanx

xx

+

+-

é

ëê

ù

ûú

-

(D) None of these

16.sin

sin

x

xdx

1 -ò is equal to

(A) - + + +x x x ksec tan (B) - + +x x xsec tan

(C) - + -x x xsec tan (D) - - -x x xsec tan

17. e f x f x dxx { ( ) ( )}+ ¢ò is equal to

(A) e f xx ¢( ) (B) e f xx ( )

(C) e f xx + ( ) (D) None of these

18. The value of ex

xdxx 1

1

+

+

æ

èçç

ö

ø÷÷ò

sin

cosis

(A) ex

cx tan2

+ (B) ex

cx cot2

+

(C) e x cx tan + (D) e x cx cot +

19.x

xdx

3

2 1+ò is equal to

(A) x x c2 2 1+ + +log ( )

(B) log ( )x x c2 21+ - +

(C)1

2

1

212 2x x c- + +log ( )

(D)1

2

1

212 2x x c+ + +log( )

20. sin -ò 1 x dx is equal to

(A) x x x csin - + - +1 21 (B) x x x csin - - - +1 21

(C) x x x csin - + + +1 21 (D) x x x csin - - - +1 21

21.sin cos

sin

x x

xdx

+

+ò1 2

is equal to

(A) sin x (B) x

(C) cos x (D) tan x

22. The value of 5 30

1

x dx-ò is

(A) -1/2 (B) 13/10

(C) 1/2 (D) 23/10

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Page 21

39. ( )x y dydxx

x

2 2

0

1

+òò is equal to

(A)7

60(B)

3

35

(C)4

49(D) None of these

40. The value of dy dx

x

0

1

0

12+

òò is

(A)p

42 1log ( )+ (B)

p

42 1log ( )-

(C)p

22 1log ( )+ (D) None of these

41. If A is the region bounded by the parabolas y x2 4=

and x y2 4= , then ydxdyA

òò is equal to

(A)48

5(B)

36

5

(C)32

5(D) None of these

42. The area of the region bounded by the curves

x y a2 2 2+ = and x y a+ = in the first quadrant is given

by

(A) dxdya x

a xa

-

-

òò2 2

0

(B) dxdy

a xa

00

2 2-

òò

(C) dxdya

a x

a y

0

2 2

òò-

-

(D) None of these

43. The area bounded by the curves y x y x= = -2 , ,

x = 1 and x = 4 is given by

(A) 25 (B)33

2

(C)47

4(D)

101

6

44. The area bounded by the curves y x2 9= , x y- + =2 0

is given by

(A) 1 (B)1

2

(C)3

2(D)

5

4

45. The area of the cardioid r a= +( cos )1 q is given by

(A) 20

1

0rdrd

r

a

qq

q

p

=

+

= òò( cos )

(B) 20

1p qqò ò =

+

r a

a

rdrd( cos )

(C) 20

1

0

2

rdrdr

a

qqp

=

+

òò( cos )

(D) 20

1

0

4

rdrdr

a

qqp

=

+

òò( cos )

46. The area bounded by the curve r = q qcos and the

lines q = 0 and qp

=2

is given by

(A)p p

4 161

2

-æ

èçç

ö

ø÷÷ (B)

p p

16 61

2

-æ

èçç

ö

ø÷÷

(C)p p

16 161

2

-æ

èçç

ö

ø÷÷ (D) None of these

47. The area of the lemniscate r a2 2 2= cos q is given by

(A) 40 0

24p

qq

ò ò rdrda cos

(B) 20

2

0

2

rdrda

qqp cos

òò

(C) 40

2

0

2

rdrda

qqp cos

òò (D) 20

2

0rdrd

a

qqp cos

òò

48. The area of the region bounded by the curve

y x x( )2 2 3+ = and 4 2y x= is given by

(A)0

1

0

2 4

ò ò =y

x

dxdy (B)0

1

0

2 4

ò ò =y

x

dydx

(C)0

2

42

3 2 2

ò ò =

+

y x

x x

dydx( )

(D)y y x

x x

dxdy= =ò ò

+

0

1

42

3 2 2( )

49. The volume of the cylinder x y a2 2 2+ = bounded

below by z = 0 and bounded above by z h= is given by

(A) pah (B) pa h2

(C)1

3

3pa h (D) None of these

50. e dxdydzx y z+ +òòò 0

1

0

1

0

1

is equal to

(A) ( )e - 1 3 (B)3

21( )e -

(C) ( )e - 1 2 (D) None of these

51.- -

+

ò ò ò + +1

1

0

z

x z

x z

x y z dy dx dz( ) is equal to

(A) 4 (B) -4

(C) 0 (D) None of these

*************

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Page 22

SOLUTIONS

1. (A)x

xdx

2 1+ò

Put x t2 1+ = Þ 2 xdx dt=x

xdx

tdt

2 1

1

2

1

+= ×ò ò

=1

2log t = +

1

212log ( )x

2. (A) F x a dx Kx( ) = +ò = +a

aK

x

log

Þ = +F aa

aK

a

( )log

Ka

a

a

a

a

a a

= - =-1 1

log log log

F xa

a

a

a

x a

( )log log

= +-1

= - +1

1log

[ ]a

a ax a

3. (C)dx

x1 +ò sin

=+æ

èç

ö

ø÷ +

òdx

x x x xsin cos sin cos2 2

2 22

2 2

=

+æ

èç

ö

ø÷

òdx

x xcos sin

2 2

2=

+æ

èç

ö

ø÷

òsec

2

2

2

12

x

xdx

tan

Put 12

+ =tanx

t

Þ =sec2

22

xdx dt Þ = - +ò

2 22

dt

tdt

tK

=-

++

2

12

tanx

K =-

++

22

2 2

cos

cos sin

x

x xK

=-

+´

-

-+

22

2 2

2 2

2 2

cos

cos sin

cos sin

cos sin

x

x x

x x

x xK

=- +

-+

22

22 2

2 2

2

2 2

cos sin cos

cos sin

x x x

x xK

=- + +

+( cos ) sin

cos

1 x x

xk = - - +tan x x Ksec 1

= - +tan x x csec

4. (A) Let Ix

x xdx=

+

- +ò3 1

2 2 32

Let 3 1 4 2x p x q+ = - +( ) Þ p q= =3

4

5

2,

Ix

x xdx=

-

- +ò3

4

4 2

2 2 32+

- +ò5

2 2 2 32

dx

x x

= - +3

42 2 32log ( )x x +

-æ

èç

ö

ø÷ +

æ

èçç

ö

ø÷÷

ò5

4 1

2

5

2

2 2

dx

x

= - + +æ

èçç

ö

ø÷÷

--3

42 2 3

5

4

1

5

2

1

2

5

2

2 1log ( ) tanx xx

5. (C) Let Idx

x=

+ò 1 3 2sin

=+ò

cosec

cosec

2

2 3

x dx

x=

+ +òcosec

2

21 3

x dx

x( cot )

Put cot x t x dx dt= Þ - =cosec2

Idt

t

t x=

-

+= = æ

èç

ö

ø÷ò - -

4

1

2 2

1

2 22

1 1cot cotcot

= -1

221tan ( tan )x

6. (C) Let Ix x

x xdx=

+

+ò2 3

3 4

sin cos

sin cos

Let ( sin cos ) ( cos sin )2 3 3 4x x p x x+ = -

+ +q x x( sin cos )3 4

p =1

25, q =

18

25

Ix x

x xdx

x x=

-

++

+ò

1

25

3 4

3 4

18

25

3 4

3

cos sin

sin cos

sin cos

sin x xdx

+ò 4 cos

= + +1

253 4

18

25log ( sin cos )x x x

7. (B) 3 8 3 2+ -ò x x dx = æ

èç

ö

ø÷ - -æ

èç

ö

ø÷ò3

5

3

4

3

2 2

x dx

= -æ

èç

ö

ø÷

æ

èç

ö

ø÷ - -æ

èç

ö

ø÷ + æ

èç

ö

ø÷ -3

1

2

4

3

5

3

4

3

5

3

2 2 2

1x x sinx -æ

è

çççç

ö

ø

÷÷÷÷

ì

íï

îï

ü

ýï

þï

4

35

3

=-

+ - +--3 4

63 8 3

25 3

18

3 4

5

2 1xx x

xsin

8. (B)dx

x x2 3 42 + +ò =

+æ

èç

ö

ø÷ +

æ

èçç

ö

ø÷÷

ò1

23

4

23

4

2 2

dx

x

Chap 9.3

Page

549

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Page 23

=+

æ

èçç

ö

ø÷÷

-1

2

3

4

23

4

1sinhx

=+-1

2

4 3

23

1sinhx

9. (B)2 3

12

x

x xdx

+

+ +ò

=+

+ ++

+ +ò ò

2 1

1

2

12 2

x

x xdx

dx

x x

=+

+ ++

+æ

èç

ö

ø÷ +

æ

èçç

ö

ø÷÷

ò ò2 1

12

1

2

3

2

2 2 2

x

x xdx

dx

x

=+ +

++

-( )sinh

x xx2 1 2

11

1

2

2

1

2

3

2

= + + ++-2 1 2

2 1

3

2 1x xx

sinh

10. (B)dx

x xI

1 -=ò

Put x dx d= Þ =sin sin cos2 2q q q q

I d d=-

=ò ò2

1

22

sin cos

sin sin

sin cos

sin cos

q q

q qq

q q

q qq

I d c= = +ò2 2q q = +-2 1sin x c

I x c= - +-sin ( )1 2 1

11. (D) Let Ix x x

dx=+ - -

ò1

1 1 2 2( )

Put xt

+ =11

Þ dxt

dt= -12

I tdt

t t t

dt

t=

-

- -æ

èç

ö

ø÷ - -æ

èç

ö

ø÷

= --

ò ò1

11 2

11

11

2 1

2

2 2

= -

-æ

èçç

ö

ø÷÷

ò1

21

2

2

2

dt

t

= - -1

2

1

1 2

cosht

= -+

æ

èçç

ö

ø÷÷

-1

2

2

1

1coshx

12. (C)dx

x xsin cos+ò

=+

ò1

2

4 4

dx

x xsin cos cos sinp p

=+æ

èç

ö

ø÷

ò1

2

4

dx

xsinp

= +æ

èç

ö

ø÷ò

1

2 4cosec x dx

p

= - +æ

èç

ö

ø÷

é

ëê

ù

ûú

1

2

1

2 4log cot x

p= +æ

èç

ö

ø÷

1

2 2 8log tan

x p

13. (D)dx

x a x bsin( ) sin( )- -ò

=-

-

- -ò1

sin( )

sin( )

sin( ) sin( )a b

a b dx

x a x b

=-

- - -

- -ò1

sin( )

sin [( ) ( )]

sin( ) sin( )a b

x b x a

x a x bdx

=-

1

sin ( )a b

´- - - - -

- -

sin( ) cos( ) cos( ) sin( )

sin( ) sin( )

x b x a x b x a

x a x bdxò

=-

- - -ò1

sin( )[cot( ) cot( )]

a bx a x b dx

=-

- - -1

sin ( )[log sin ( ) log sin ( )]

a bx a x b dx

=-

-

-

ìíî

üýþ

1

sin ( )log

sin( )

sin( )a b

x a

x b

14. (D) Let Idx

e

e dx

ex

x

x=

-=

-ò ò-

-1 1

Put 1 - =-e tx Þ e dx dtx- =

Idt

tt e x= = = -ò -log log ( )1

15. (B) Let Idx

x x x=

+ + +ò 1 2 3

=+ +ò

dx

x x( ) ( )1 1 2

Let1

1 1 1 12 2( )( )+ +=

++

+

+x x

A

x

Bx C

x

1 1 12= + + + +A x Bx C x( ) ( )( )

Comparing the coefficients of x x2 , and constant terms,

A B+ = 0, B C+ = 0, C A+ = 1

Solving these equations, we get

A =1

2, B C= - =

1

2

1

2,

Ix

dxx

xdx=

+-

-

+ò ò1

2

1

1

1

2

1

12

= + - + + -1

21

1

21

1

2

2 1log ( ) log ( ) tanx x x

=+

++

é

ëê

ù

ûú

-1

4

1

12

2

2

1log( )

tanx

xx

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550

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Page 24

16. (B) Let Ix

xdx=

-òsin

sin1

=- -

-ò1 1

1

( sin )

sin

x

xdx

=-

-ò ò1

1 sin xdx dx =

+

--ò

1

1 2

sin

sin

x

xdx x

=+

-ò1

2

sin

cos

x

xdx x = + -ò ( tan )sec sec

2 x x x dx x

= + -tan x x xsec

17. (B) Let I e f x f x dxx= + ¢ò { ( ) ( )}

= + ¢ò òe f x dx e f x dxx x( ) ( )

= - ¢ + ¢ = ×ò ò{ ( ) ( ) } ( ) ( )f x e f x e dx e f x dx f x ex x x x

18. (A) Let I ex

xdxx=

+

+

æ

èçç

ö

ø÷÷ò

1

1

sin

cos

=+æ

è

çççç

ö

ø

÷÷÷÷

ò e

x x

xdxx

1 22 2

22

2

sin cos

cos

= +ò ò1

2 2 2

2ex

dx ex

dxx xsec tan

= × - ×ìíî

üýþ

+ò ò1

22

22

2 2e

xe

xdx e

xdxx x xtan tan tan

= +ex

cx tan2

19. (C) Ix

xdx=

+ò3

2 1=

×

+òx x

xdx

2

2 1

=+ -

+òx x

xdx

( )2

2

1 1

1= -

+ò òxdxx

xdx

2 1

= - + +1

2

1

212 2x x clog ( )

20. (A) Let I x dx= -ò sin 1 = × ×-ò sin 1 1x dx

= × --

×- òsin 1

2

1

1x x

xx dx

= --

- òx xx

xdxsin 1

21

In second part put 1 2 2- =x t

xdx tdt= - = +- òx x dtsin 1

= +-x x tsin 1 = + - +-x x x csin 1 21

21.sin cos

sin

x x

xdx

+

+ò1 2

=+

+ +ò

sin cos

(sin cos ) sin cos

x x

x x x xdx

2 2 2

=+

+ò

sin cos

(cos cos )

x x

x xdx

2

=+

+= =ò ò

sin cos

sin cos

x x

x xdx dx x

22. (D) 5 3 5 3 5 30

3 5

0

3 5

3 5

1

x dx x dx x dx- = - - + -ò ò ò

= - +æ

èç

ö

ø÷ + -

æ

èçç

ö

ø÷÷

5

23

5

232

0

3 5 2

3 5

1

x xx

x

= - +æ

èç

ö

ø÷ + -æ

èç

ö

ø÷ - -æ

èç

ö

ø÷

é

ëê

ù

ûú

9

10

9

5

5

23

9

10

9

5

= + - +æ

èç

ö

ø÷ =

9

10

1

2

9

10

13

10

23. (B)dx

e ex x+ -ò0

1

=+ò

e dx

e

x

x2

0

1

1

Put e tx = Þ e dx dtx = =+

=ò -dt

tt

e

e

2

1

1

11

[tan ]

= -- -tan tan1 1 1e = --tan 1

4e

p

24. (D) x x dx x x dxc c

( ) ( )10

2

0

- = -ò ò

= -æ

èç

ö

ø÷

1

2

1

3

2 3

0

x x

c

= -1

63 22c c( )

x x dxc

( )1 00

- =ò Þ1

63 2 02c c( )- =

Þ =c3

2

25. (D) Put x x t2 + = Þ ( )2 1x dx dt+ =

2 12 2 2

20

1

0

2

1 2

0

2x

x xdx

dt

tt

+

+= = =ò ò ( )

26. (A) x xdx4 5sinp

p

ò

Since, f x x x( ) ( ) sin ( )- = - -4 5 = -x x4 5sin

f x( ) is odd function thus

x x dx4 5 0sin =òp

p

27. (A) cos (cos )2

0

2

0

21

22 1x dx x dx

p p

ò ò= +

Chap 9.3

Page

551

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Page 25

= +æ

èç

ö

ø÷

1

2

1

22

0

2

sin x x

p

= - + -æ

èç

ö

ø÷

é

ëê

ù

ûú

1

2

1

20

20(sin sin )p

p

= - - +é

ëêù

ûú=

1

2

1

20 0 0

2 4( )

p p

Aliter 1. cos2

0

2

x dx

p

ò =

æ

èç

ö

ø÷

æ

èç

ö

ø÷

æ

èç

ö

ø÷

G G

G

3

2

1

2

24

2

=

1

22

p=

p

4

Aliter 2. Use Walli’s Rule cos2

0

2

1

2 2 4x

pp p

ò = × =

28. (B) Let I a x dxa

= -ò 2 2

0

Put x a dx a d= Þ =sin cosq q q when x = 0, q = 0,

when x a= , qp

=2

I a a a d= -ò 2 2 2

0

2

sin cosq q qp

= = × ×òa d a2 2

0

2

2

1

2 2cos q q

pp

(By Walli’s Formula)

=pa2

4

Aliter: a x dxa

2 2

0

-ò

= - +é

ëêù

ûú-1

2

1

2

2 2 2 1

0

x a x ax

a

a

sin = +é

ëê

ù

ûú0

4

2pa=

pa2

4

29. (D) Let I x dx= ò log (tan )0

2p

....(1)

I x dx= -æ

èç

ö

ø÷ò log tan

pp

20

2

I x= ò log (cot )0

2p

....(2)

Adding (1) and (2), we get

20

2

I x x dx= +ò [log (tan ) log (cot )]

p

= ×ò log (tan cot )x x dx0

2p

= =ò log 1 00

2

dx

p

Þ I = 0

30. (D) Let It

dt= -æ

èç

ö

ø÷ò2

2 40

1

sinp p

....(i)

= - -æ

èç

ö

ø÷ò2

21

40

1

sin ( )p p

t dt = -æ

èç

ö

ø÷ò2

4 20

1

sinp p

t dt

= - -æ

èç

ö

ø÷ = -ò2

2 41

0

1

sinp p

t dt

2 0 0I I= Þ =

31. (C) Let If x

f x f a xdx

a

=+ -ò

( )

( ) ( )20

2

....(1)

If a x

f a x f xdx

a

=-

- +ò( )

( ) ( )

2

20

2

....(2)

Adding (1) and (2), we get

22

20

2

If x f a x

f x f a xdx

a

=+ -

+ -ò( ) ( )

( ) ( )= × = =ò1 2

0

2

0

2dx x aa

a[ ]

Þ =I a

32. (C) Let Ie

xxdx

x

=-

-

ò1

20

1 2

1

Put 1 2- =x t

Þ-

- =1

2 12

2xx dx dt( )

when x t= =0 1, , when x t= =1 0,

I e dt e e e et t= - = - = - - = -ò1

0

1

0 0 1 1[ ] [ ]

33. (B) Let Idx

x x=

- +ò 1 2

0

1

=

-æ

èç

ö

ø÷ +

æ

èçç

ö

ø÷÷

òdx

x1

2

3

2

2 2

0

1

=-

é

ë

êêêê

ù

û

úúúú

-1

3

2

1

2

3

2

1

0

1

tanx

= - -æ

èçç

ö

ø÷÷

é

ëê

ù

ûú

- -2

3

1

3

1

3

1 1tan tan = +æ

èç

ö

ø÷

2

3 6 6

p p

= =2

3 3

2 3

9

p p

34. (B) Let Ix

xdx=

-ò1

1

=-

+-ò ò

x

xdx

x

xdx

1

0

0

1

= - + ×-ò ò1 11

0

0

1

dx dx = - +-[ ] [ ]x x1

0

0

1

= - - - + - =[ ( )] [ ]0 1 1 0 0

35. (C) | |sin xdx0

100p

ò | |= ò1000

sin xdxp

[ . .. sin x is periodic with period p]

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552

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Page 26

= = -ò100 1000

0sin ( cos )x dx xp

p

= - +100 0( cos cos )p = +100 1 1( ) = 200.

36. (C) Let I x nx dx f x dxm= =ò òcos sin ( )0 0

p p

Where f x x xm n( ) cos sin=

f x x xm n( ) cos ( ) sin ( )p p p- = - -

= -( cos ) (sin )x xm n

= -cos sinm nx x, if m is odd

I x x dxm n= =ò cos sin0

0p

, if m is odd

37. (A) Let I xF x dx= ò (sin )0

p

....(1)

= - -ò ( ) [sin ( )]x F x dxp pp

0

I x F x dx= -ò ( ) (sin )pp

0

....(2)

Adding (1) and (2), we get

20

I F x dx= ò pp

(sin )

Þ I F x dx= ò1

20

pp

(sin )

38. (B) Let Ie x x

dxx

= +æ

èç

ö

ø÷ò 2 2

22

2

0

2

sec tan

p

= +ò ò1

2 2 2

2

0 0

2 2

ex

dx ex

dxx xsec

p p

tan = +I I1 2

Here, I ex

dxx

1

2

0

1

2 2

2

= ò sec

p

= ×é

ëêù

ûú- ×ò

1

22

2

1

22

20 0

2 2

ex

ex

dxx xtan tan

p p

= -æ

èç

ö

ø÷ - òe e

xdxxp

p

p2

2

40

20

tan tan

= -e Ip 2

2 , I I e1 2

2+ = p

I I I e= + =1 2

2p

39. (B) ( )x y dy dxx

x

2 2

0

1

+òò = +é

ëêù

ûúò x y y dx

x

x

2 3

0

11

3

= + - -é

ëêù

ûúò x x x x dx5 2 3 21

3

1

3

3 3

0

1

= + -é

ëêù

ûú=

2

7

2

15

1

3

3

35

7 2 5 2 4

0

1

x x x

40. (D) dydx

x

0

1

0

12+

òò =0

1

0

1 2

ò +[ ]y dx

x

= +ò 1 2

0

1

x dx

= + + + +1

21 12 2

0

1[ log( )]x x x x

= + +1

22 1 2[ log ( )]

41. (A) Let I ydxdyA

= òò ,

Solving the given equations y x2 4= and x y2 4= , we get

x x= =0 4, . The region of integration A is given by

A ydydxx

x

= òò2 4

2

0

4

=é

ëê

ù

ûúò

ydx

x

x2

2

0

4

2 2 4

= -æ

èçç

ö

ø÷÷ò

1

24

10

4

0

4

xx

dx = -é

ëê

ù

ûú =x

x25

0

4

160

48

5

42. (A) The curves are

x y a2 2 2+ = ... ....(i)

x y a+ = ... ....(ii)

The curves (i) and (ii) intersect at A (a, 0) and B (0,a)

The required area A dydxy a x

a x

x

a

== -

-

=òò

2 2

0

43. (D) The given equations of the curves are

y x= 2 i.e., y x2 4= ....(i) y x= - ....(ii)

If a figure is drawn then from fig. the required area is

A dydxx

x

=-òò

2

1

4

= -ò [ ]y x

x2

1

4

= +ò [ ]21

4

x x dx

= +æ

èç

ö

ø÷ - +æ

èç

ö

ø÷

32

38

4

3

1

2=

101

6

44. (B) The equations of the given curves are

y x2 9= ....(i) x y- + =2 0....(ii)

The curves (i) and (ii) intersect at

A(1, 3) and B(4, 6)

If a figure is drawn then from fig. the required area is

A dydxx

x

=+òò

2

3

1

4

= +ò [ ]y dxx

x

2

3

1

4

= - +ò [ ( )]3 21

4

x x dx = - -é

ëêù

ûú2

1

223 2 2

1

4

x x x

= - - - - -æ

èç

ö

ø÷( )16 8 8 2

1

22 =

1

2

Chap 9.3

Page

553

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Page 27

45. (A) The equation of the cardioid is

r a= +( cos )1 q ....(i)

If a figure is drawn then from fig. the required area is

Required area A rdrdr

a

==

+

=òò2

0

1

0

qq

q

p ( cos )

46. (C) The equation of the given curve is

r = q qcos ....(i)

The required area

A rdrdr

===òò q

q q

q

p

00

2 cos

= é

ëêù

ûúò

1

2

2

0

2

r do

q qp

qcos

= ò1

2

2 2

0

2

q q qp

cos d = +ò1

41 22

0

2

q q qp

( cos )d

= +ò ò1

4

1

422

0

2

2

0

2

q q q q qp p

d dcos

= é

ëêù

ûú+ æ

èç

ö

ø÷ - ò

1

4

1

3

1

4

2

22

2

2

3

0

2

0 0

22 2

q qq

qq

p p psin sin

dqé

ëêê

ù

ûúú

= + -é

ëê

ù

ûúò

pq q q

p3

0

2

96

1

42sin d

= - -æ

èç

ö

ø÷ - -æ

èç

ö

ø÷

é

ëêê

ùò

pq

q qq

p p3

0 0

2

96

1

4

2

2

2

2

2

cos cosd

ûúú

= +-

-æ

èç

ö

ø÷ - ò

p pq q

p3

0

2

96

1

4 40

1

82cos d

= - - æ

èç

ö

ø÷

p pq

p3

096 16

1

8

1

22

2

sin = -æ

èçç

ö

ø÷÷

p p

16 161

2

47. (A) The curve is r a2 2 2= cos q

If a figure is drawn then from fig. the required area is

A rdrdr

a

===òò4

0

2

0

4

qq

q

p cos

= é

ëêù

ûúò4

1

2

2

0

2

0

4

r d

a cos qp

q

= ò2 22

0

4

a dcos q qp

= é

ëêù

ûú=2

2

2

2

0

2

4

a asin q

p

48. (C) The equations of given curves are

y x x( )2 2 3+ = ....(i) and 4 2y x= ....(ii)

The curve (i) and (ii) intersect at A (2, 1).

If a figure is drawn then from fig. the required area is

The required area A dxdyy x

x x

x

==

+

=òò

2

2

4

3 2

0

2 ( )

49. (B) The equation of the cylinder is x y a2 2 2+ =

The equation of surface CDE is z h= .

If a figure is drawn then from fig. the required area is

Thus the equation volume is V zdxdyA

= ò4

=-

òò400

2 2

hdydx

a xa

= -ò4 0

0

2 2

h y dxa x

a

[ ] = -ò4 2 2

0

h a x dxa

Let x a= sin q, Þ dx a d= cos q q,

Volume V h a a a d= - ×ò4 2 2 2

0

2

sin cosq q qp

= ò4 2 2

0

2

ha dcos q qp

= × × =41

2 2

2 2ha a hp

p .

50. (A) e dxdydzx y z+ +òòò0

1

0

1

0

1

= + +òò [ ]e dydzx y z

0

1

0

1

0

1

= -+ + +òò [ ]e e dydzy z y z1

0

1

0

1

= -+ + +ò [ ]e e dzy z y z1

0

1

0

1

= - - -+ + +ò [( ) ( )]e e e e dzz z z z2 1 1

0

1

= - ++ +ò ( )e e e dzz z z2 1

0

1

2 = - ++ +[ ]e e ez z z2 1

0

12

= - + - - +( ) ( )e e e e e3 2 22 2 1

= - + - = -e e e e3 2 33 3 1 1( )

51. (C) ( )x y z dydxdzx z

x zz

+ +-

+

-òòò

01

1

=+ +é

ëê

ù

ûú

-

+

-òò

( )x y zdxdz

x y

x zz 2

01

1

2

=+

- æ

èç

ö

ø÷

é

ëêê

ù

ûúú

òò-

( )2 2

2

2

2

2 2

01

1x z x

dxdzz

= + -é

ëê

ù

ûúòò

-

2 2 2

0

3

1

1

(( ) )x z x dx dz =+

-é

ëê

ù

ûú

-ò2

3 3

3 3

01

1( )x z x

dz

z

= - --ò

2

32 3 3

1

1

[( ) ]z z z dz = =é

ëê

ù

ûú

- -ò

2

36 4

4

3

1

1 4

1

1

z dzz

= -æ

èç

ö

ø÷ =4

1

4

1

40

********

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Page 28

10. The integration of f z x ixy( ) = +2 from A(1, 1) to B(2,

4) along the straight line AB joining the two points is

(A)-

+29

311i (B)

29

311- i

(C)23

56+ i (D)

23

56- i

11.e

zdz

z

c

2

41( )?

+=ò where c is the circle of z = 3

(A)4

9

3pie- (B)

4

9

3pie

(C)4

3

1pie- (D)

8

3

2pie-

12.1 2

1 2

-

- -=ò

z

z z zdz

c( )( )

? where c is the circle z = 15.

(A) 2 6+ i p (B) 4 3+ i p

(C) 1 + ip (D) i3p

13. ( ) ?z z dzc

- =ò 2 where c is the upper half of the circle

z = 1

(A)-2

3(B)

2

3

(C)3

2(D)

-3

2

14.cos

?pz

zdz

c-

=ò 1where c is the circle z = 3

(A) i2 p (B) - i2 p

(C) i6 2p (D) - i6 2p

15.sin

( )( )?

pz

z zdz

c

2

2 1- -=ò where c is the circle z = 3

(A) i6p (B) i2p

(C) i4p (D) 0

16. The value of1

2 12p

p

i

z

zdz

c

cos

-ò around a rectangle with

vertices at 2 ± i , - ±2 i is

(A) 6 (B) i e2

(C) 8 (D) 0

Statement for Q. 17–18:

f zz z

z zdz

c

( )( )

0

2

0

3 7 1=

+ +

-ò , where c is the circle

x y2 2 4+ = .

17. The value of f ( )3 is

(A) 6 (B) 4i

(C) -4i (D) 0

18. The value of ¢ -f i( )1 is

(A) 7 2( )p + i (B) 6 2( )+ ip

(C) 2 5 13p ( )+ i (D) 0

Statement for 19–21:

Expand the given function in Taylor’s series.

19. f zz

z( ) =

-

+

1

1about the points z = 0

(A) 1 2 2 3+ + +( ......)z z z (B) - - - +1 2 2 3( ......)z z z

(C) - + - +1 2 2 3( ......)z z z (D) None of the above

20. f zz

( ) =+

1

1about z = 1

(A)-

- - + -é

ëêù

ûú1

21

1

21

1

21

2

2( ) ( ) .......z z

(B)1

21

1

21

1

21

2

2- - + -é

ëêù

ûú( ) ( ) .......z z

(C)1

21

1

21

1

21

2

2+ - + -é

ëêù

ûú( ) ( ) .......z z

(D) None of the above

21. f z z( ) sin= about z =p

4

(A)1

21

4

1

2 4

2

+ -æ

èç

ö

ø÷ - -æ

èç

ö

ø÷ -

é

ëêê

ù

ûúú

z zp p

!.......

(B)1

21

4

1

2 4

2

+ -æ

èç

ö

ø÷ + -æ

èç

ö

ø÷ +

é

ëêê

ù

ûúú

z zp p

!.......

(C)1

21

4

1

2 4

2

- -æ

èç

ö

ø÷ - -æ

èç

ö

ø÷ -

é

ëêê

ù

ûúú

z zp p

!.......

(D) None of the above

22. If z + <1 1, then z-2 is equal to

(A) 1 1 1 1

1

+ + + -

=

¥

å ( )( )n z n

n

(B) 1 1 1 1

1

+ + + +

=

¥

å ( )( )n z n

n

(C) 1 11

+ +=

¥

å n z n

n

( )

(D) 1 1 11

+ + +=

¥

å ( )( )n z n

n

Chap 9.5

Page

565

Complex Variables GATE EC BY RK Kanodia

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Page 29

Statement for Q. 23–25.

Expand the function1

1 2( )( )z z- -in Laurent’s

series for the condition given in question.

23. 1 2< <z

(A)1 2 3

2 3z z z+ + + .......

(B) K K- - - - - - - -- - -z z z z z z3 2 1 2 31

2

1

4

1

8

1

18

(C)1 3 7

2 2 4z z z+ + ...........

(D) None of the above

24. z > 2

(A)6 13 20

2 3z z z+ + + ........ (B)

1 8 132 3z z z

+ + + .........

(C)1 3 7

2 3 4z z z+ + + ......... (D)

2 3 42 3 4z z z

- + - ........

25. z < 1

(A) 1 37

2

15

4

2 2++

+z z z .....

(B)1

2

3

4

7

8

15

16

2 3+ + +z z z ...

(C)1

4

3

4 8 16

2 3

+ + +z z

.......

(D) None of the above

26. If z - <1 1 , the Laurent’s series for1

1 2z z z( )( )- -is

(A) - - --

--

-( )( )

!

( )

!...........z

z z1

1

2

1

5

3 5

(B) - - --

--

--( )( )

!

( )

!.........z

z z1

1

2

1

5

13 5

(C) - - - - - - -( ) ( ) ( ) ..........z z z1 1 13 5

(D) - - - - - - - - --( ) ( ) ( ) ( ) .........z z z z1 1 1 11 3 5

27. The Laurent’s series of1

1z ez( )-for z < 2 is

(A)1 1

2

1

126

1

7202

2

z zz z+ + + + + ..........

(B)1 1

2

1

12

1

7202

2

z zz- + - + ..........

(C)1 1

12

1

634

1

720

2 2

zz z+ + + + ..........

(D) None of the above

28. The Laurent’s series of f zz

z z( )

( )( )=

+ +2 21 4is,

where z < 1

(A)1

4

5

16

21

64

3 5z z z- + ..........

(B)1

2

1

4

5

16

21

64

2 4 6+ + +z z z ..........

(C)1

2

3

4

15

8

3 5z z z- + ..........

(D)1

2

1

2

3

4

15

8

2 4 6+ + +z z z ..........

29. The residue of the function1

4

- e

z

Zz

at its pole is

(A)4

3(B)

-4

3

(C)-2

3(D)

2

3

30. The residue of zz

cos1

at z = 0 is

(A)1

2(B)

-1

2

(C)1

3(D)

-1

3

31.1 2

1 2

-

- -=ò

z

z z zdz

c( )( )

? where c is z = 15.

(A) - i3p (B) i3 p

(C) 2 (D) -2

32.z z

z

dzc

cos?

-æ

èç

ö

ø÷

=ò p

2

where c is z - =1 1

(A) 6 p (B) - 6p

(C) i2p (D) None of the above

33. z e dzz

c

2

1

ò = ? where c is z = 1

(A) i3p (B) - i3p

C)ip

3(D) None of the above

34.dq

q

p

20

2

+=ò cos

?

(A)-2

2

p(B)

2

3

p

(C) 2 2p (D) -2 3p

Page

566

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Page 30

35.x

x a x bdx

2

2 2 2 2( )( )?

+ +=

-¥

¥

ò

(A)p ab

a b+(B)

p ( )a b

ab

+

(C)p

a b+(D) p ( )a b+

36.dx

x1 6

0+

=¥

ò ?

(A)p

6(B)

p

2

(C)2

3

p(D)

p

3

***************

SOLUTIONS

1. (C) Since, f z u ivx i y i

x yz( )

( ) ( );= + =

+ - -

+¹

3 3

2 2

1 10

Þ =-

+=

+

+u

x y

x yv

x y

x y

3 3

2 2

3 3

2 2;

Cauchy Riemann equations are

¶

¶=

¶

¶

u

x

v

yand

¶

¶= -

¶

¶

u

y

v

x

By differentiation the value of¶

¶

¶

¶

¶

¶

¶

¶

u

x

y

y

v

x

v

y, , , at( , )0 0

we get0

0, so we apply first principle method.

At the origin,

¶

¶=

+ -= =

® ®

u

x

u h u

h

h h

hh hlim

( , ) ( , )lim

0 0

3 20 0 0 01

¶

¶=

+ -=

-= -

® ®

u

v

u k u

k

k k

kh klim

( , ) ( , )lim

0 0

3 20 0 0 01

¶

¶=

+ -= =

® ®

v

x

v h v

h

h h

hh hlim

( , ) ( , )lim

0 0

3 20 0 0 01

¶

¶=

+= =

® ®

v

y

v k v

k

k k

kk klim

( , ), ( , )lim

0 0

3 20 0 0 01

Thus, we see that¶

¶=

¶

¶

u

x

v

yand

¶

¶= -

¶

¶

u

y

v

x

Hence, Cauchy-Riemann equations are satisfied at

z = 0.

Again, ¢ =-

®f

f z f

zz( ) lim

( ) ( )0

0

0

=- + +

+ +

é

ëê

ù

ûú®

lim( ) ( )

( ) ( )z

x y i x y

x y x iy0

3 3 3 3

2 2

1

Now let z ® 0 along y x= , then

fx y i x y

x y x iyz¢ =

- + +

+ +

é

ëê

ù

ûú®

( ) lim( ) ( )

( ) ( )0

1

0

3 3 3 3

2 2=

+=

+2

2 1

1

2

i

i

i

( )

Again let z ® 0 along y = 0, then

fx i x

x xi

x¢ =

+é

ëê

ù

ûú = +

®( ) lim

( )

( )0

11

0

3 3

2

So we see that ¢f ( )0 is not unique. Hence ¢f ( )0 does not

exist.

2. (A) Since, ¢ = =®

f zdf

dz

f

zz( ) lim

D

D

D0

or ¢ =+

+®f z

u i v

x i yz( ) lim

D

D D

D D0....(1)

Now, the derivative ¢f z( ) exits of the limit in equation

(1) is unique i.e. it does not depends on the path along

which Dz ® 0.

Chap 9.5

Page

567

Complex Variables GATE EC BY RK Kanodia

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