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GATE SOLVED PAPERElectronics & Communication Engineering Mathematics

Copyright By NODIA & COMPANY

Information contained in this book has been obtained by authors, from sources believes to be reliable. However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineering or other professional services.

NODIA AND COMPANYB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039Ph : +91 - 141 - 2101150www.nodia.co.inemail : [email protected]

GATE SOLVED PAPER - ECENGINEERING MATHEMATICS

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2013 ONE MARK

Q. 1 The maximum value of q until which the approximation sin .q q holds to within 10% error is(A) 10c (B) 18c

(C) 50c (D) 90c

Q. 2 The minimum eigen value of the following matrix is352

5127

275

R

T

SSSS

V

X

WWWW

(A) 0 (B) 1

(C) 2 (D) 3

Q. 3 A polynomial ( )f x a x a x a x a x a4 4 3 3 2 2 1 0= + + + - with all coefficients positive has(A) no real roots

(B) no negative real root

(C) odd number of real roots

(D) at least one positive and one negative real root

2013 TWO MARKS

Q. 4 Let A be an m n# matrix and B an n m# matrix. It is given that determinant I ABm+ =^ h determinant I BAn+^ h, where Ik is the k k# identity matrix. Using

the above property, the determinant of the matrix given below is2111

1211

1121

1112

R

T

SSSSSS

V

X

WWWWWW

(A) 2 (B) 5

(C) 8 (D) 16

2012 ONE MARK

Q. 5 With initial condition ( ) .x 1 0 5= , the solution of the differential equation t dtdx x t+ = , is

(A) x t 21= - (B) x t 2

12= -

(C) x t22= (D) x t2=

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GATE SOLVED PAPER - EC ENGINEERING MATHEMATICS

Q. 6 Given ( )f z z z11

32= + - + .

If C is a counter clockwise path in the z -plane such that z 1 1+ = , the value of ( )j f z dz2

1Cp # is

(A) 2- (B) 1-(C) 1 (D) 2

Q. 7 If ,x 1= - then the value of xx is(A) e /2p- (B) e /2p

(C) x (D) 1

2012 TWO MARKS

Q. 8 Consider the differential equation

( ) ( )

( )dt

d y tdt

dy ty t22

2

+ + ( )td= with ( ) 2 0andy t dtdy

tt

00

=- == =- -The numerical value of dt

dyt 0= +

is(A) 2- (B) 1-(C) 0 (D) 1

Q. 9 The direction of vector A is radially outward from the origin, with krA n= . where r x y z2 2 2 2= + + and k is a constant. The value of n for which A 0:d = is(A) 2- (B) 2(C) 1 (D) 0

Q. 10 A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is(A) /1 3 (B) /1 2

(C) /2 3 (D) /3 4

Q. 11 The maximum value of ( )f x x x x9 24 53 2= - + + in the interval [ , ]1 6 is(A) 21 (B) 25

(C) 41 (D) 46

Q. 12 Given that

andA I52

30

10

01=

- - => >H H, the value of A3 is(A) 15 12A I+ (B) 19 30A I+(C) 17 15A I+ (D) 17 21A I+

2011 ONE MARK

Q. 13 Consider a closed surface S surrounding volume V . If rv is the position vector of a point inside S , with nt the unit normal on S , the value of the integral r n dS5

S$v t##

is

(A) V3 (B) V5

(C) V10 (D) V15


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Q. 14 The solution of the differential equation , (0)dxdy ky y c= = is

(A) x ce ky= - (B) x kecy=(C) y cekx= (D) y ce kx= -

Q. 15 The value of the integral ( )z z

z dz4 5

3 4

c2+ +- +# where c is the circle z 1= is given

by(A) 0 (B) 1/10

(C) 4/5 (D) 1

2011 TWO MARKS

Q. 16 A numerical solution of the equation ( )f x x 3 0+ - = can be obtained using Newton- Raphson method. If the starting value is x 2= for the iteration, the value of x that is to be used in the next step is(A) 0.306 (B) 0.739

(C) 1.694 (D) 2.306

Q. 17 The system of equations

4 6 20

4

x y z

x y y

x y z

6

ml

+ + =+ + =+ + =

has NO solution for values of l and given by(A) 6, 20ml = = (B) 6, 20ml = =Y(C) 6, 20ml = =Y (D) 6, 20ml = =Y

Q. 18 A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is(A) 2/36 (B) 2/6

(C) 5/12 (D) 1/2

2010 ONE MARKS

Q. 19 The eigen values of a skew-symmetric matrix are(A) always zero (B) always pure imaginary

(C) either zero or pure imaginary (D) always real

Q. 20 The trigonometric Fourier series for the waveform ( )f t shown below contains

(A) only cosine terms and zero values for the dc components

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(B) only cosine terms and a positive value for the dc components

(C) only cosine terms and a negative value for the dc components

(D) only sine terms and a negative value for the dc components

Q. 21 A function ( )n x satisfied the differential equation ( ) ( )

dxd n x

Ln x

022

2- =where L is a constant. The boundary conditions are : (0)n K= and ( )n 03 = . The solution to this equation is(A) ( ) ( / )expn x K x L= (B) ( ) ( / )expn x K x L= -(C) ( ) ( / )expn x K x L2= - (D) ( ) ( / )expn x K x L= -

2010 TWO MARKS

Q. 22 If e x /y x1= , then y has a(A) maximum at x e= (B) minimum at x e=(C) maximum at x e 1= - (D) minimum at x e 1= -

Q. 23 A fair coin is tossed independently four times. The probability of the event the number of time heads shown up is more than the number of times tail shown up(A) 1/16 (B) 1/3

(C) 1/4 (D) 5/16

Q. 24 If A xya x ax y2= +v t t , then A dlC

$v v# over the path shown in the figure is

(A) 0 (B) 3

2

(C) 1 (D) 2 3

Q. 25 The residues of a complex function

( )( )( )

x zz z z

z1 2

1 2= - --

at its poles are

(A) ,21

21- and 1 (B) ,2

121- and 1-

(C) , 121 and 2

3- (D) ,21 1- and 2

3

Q. 26 Consider differential equation ( )

( )dxdy x

y x x- = , with the initialcondition ( )y 0 0= . Using Eulers first order method with a step size of 0.1, the value of ( . )y 0 3 is(A) 0.01 (B) 0.031

(C) 0.0631 (D) 0.1


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Q. 27 Given ( )( )

f t Ls s k s

s4 3

3 113 2= + + -

+- ; E. If ( ) 1lim f tt ="3 , then the value of k is(A) 1 (B) 2

(C) 3 (D) 4

2009 ONE MARK

Q. 28 The order of the differential equation dtd y

dtdy y e t2

2 34+ + = -c m is

(A) 1 (B) 2

(C) 3 (D) 4

Q. 29 A fair coin is tossed 10 times. What is the probability that only the first two tosses will yield heads?

(A) 21 2c m (B) C 2110 2

2b l(C) 2

1 10c m (D) C 2110 210b l

Q. 30 If ( )f z c c z0 1 1= + - , then ( )zf z

dz1

unit circle

+# is given by(A) c2 1p (B) ( )c2 1 0p +(C) jc2 1p (D) ( )c2 1 0p +

2009 TWO MARKS

Q. 31 The Taylor series expansion of sinx

xp- at x p= is given by

(A) !

( )...

x1

3

2p+ - + (B) !

( )...

x1

3

2p- - - +

(C) !

( )...

x1

3

2p- - + (D) !

( )...

x1

3

2p- + - +

Q. 32 Match each differential equation in Group I to its family of solution curves from Group IIGroup I Group II

A. dxdy

xy= 1. Circles

B. dxdy

xy=- 2. Straight lines

C. dxdy

yx= 3. Hyperbolas

D. dxdy

yx=-

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

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Q. 33 The Eigen values of following matrix are

130

310

563

-- -R

T

SSSS

V

X

WWWW

(A) 3, 3 5 , 6j j+ - (B) 6 5 , 3 , 3j j j- + + -(C) 3 , 3 , 5j j j+ - + (D) 3, 1 3 , 1 3j j- + - -

2008 ONE MARKS

Q. 34 All the four entries of the 2 2# matrix Ppp

pp

11

21

12

22= = G are nonzero,

and one of its eigenvalue is zero. Which of the following statements is true?(A) p p p p 111 12 12 21- = (B) p p p p 111 22 12 21- =-(C) p p p p 011 22 12 21- = (D) p p p p 011 22 12 21+ =

Q. 35 The system of linear equations

x y4 2+ 7= x y2 + 6= has(A) a unique solution

(B) no solution

(C) an infinite number of solutions

(D) exactly two distinct solutions

Q. 36 The equation ( )sin z 10= has(A) no real or complex solution(B) exactly two distinct complex solutions(C) a unique solution(D) an infinite number of complex solutions

Q. 37 For real values of x , the minimum value of the function( ) ( ) ( )exp expf x x x= + - is

(A) 2 (B) 1

(C) 0.5 (D) 0

Q. 38 Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x 0= ?(A) ( )sin x3 (B) ( )sin x2

(C) ( )cos x3 (D) ( )cos x2

Q. 39 Which of the following is a solution to the differential equation( )

( )dt

dx tx t3 0+ = ?

(A) ( )x t e3 t= - (B) ( )x t e2 t3= -(C) ( )x t t2

3 2=- (D) ( )x t t3 2=

2008 TWO MARKS

Q. 40 The recursion relation to solve x e x= - using Newton - Raphson method is(A) x en x1 n=+ - (B) x x en n x1 n= -+ -

(C) (1 )x xe

e1n n x

x

1 n

n= + ++ --

(D) ( )

xx e

x e x1 1n

nx

nx

n1

2

n

n

= -- - -

+ --


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Q. 41 The residue of the function ( )f z ( ) ( )z z2 2

12 2= + - at z 2= is

(A) 321- (B)

161-

(C) 161 (D)

321

Q. 42 Consider the matrix P02

13= - -= G. The value of ep is

(A) e ee e

e ee e

2 32 2 5

2 1

2 1

1 2

2 1

--

--

- -- -

- -- -> H

(B) e ee e

e ee e2 423 2

1 1

1 2

2 1

1 2

+-

-+

- --

- -- -> H

(C) e ee e

e ee

52 6

34 6

2 1

2 1

1 2

2 1

--

-+

- -- -

- -- -> H

(D) e ee e

e ee e

22 2 2

1 2

1 2

1 2

1 2

-- +

-- +

- -- -

- -- -> H

Q. 43 In the Taylor series expansion of ( ) ( )exp sinx x+ about the point x p= , the coefficient of ( )x 2p- is(A) ( )exp p (B) . ( )exp0 5 p(C) ( )exp 1p + (D) ( )exp 1p -

Q. 44 The value of the integral of the function ( , )g x y x y4 103 4= + along the straight line segment from the point ( , )0 0 to the point ( , )1 2 in the x y- plane is(A) 33

(B) 35

(C) 40

(D) 56

Q. 45 Consider points P and Q in the x y- plane, with ( , )P 1 0= and ( , )Q 0 1= . The line integral 2 ( )xdx ydy

P

Q +# along the semicircle with the line segment PQ as its diameter(A) is 1-(B) is 0

(C) is 1

(D) depends on the direction (clockwise or anit-clockwise) of the semicircle

2007 ONE MARK

Q. 46 The following plot shows a function which varies linearly with x . The value of the

integral I ydx1

2= # is

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(A) 1.0 (B) 2.5

(C) 4.0 (D) 5.0

Q. 47 For x 1


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.in(A) ( )f t1 and ( )f t2 are orthogonal (B) ( )f t1 and ( )f t3 are orthogonal

(C) ( )f t2 and ( )f t3 are orthogonal D) ( )f t1 and ( )f t2 are orthonormal

Q. 54 If the semi-circular control D of radius 2 is as shown in the figure, then the value

of the integral ( )s

ds1

1

D2-# is

(A) jp (B) jp-(C) p- (D) p

Q. 55 It is given that , ...X X XM1 2 at M non-zero, orthogonal vectors. The dimension of the vector space spanned by the M2 vectors , ,... , , ,...X X X X X XM M1 2 1 2- - - is(A) M2 (B) M 1+(C) M

(D) dependent on the choice of , ,...X X XM1 2

Q. 56 Consider the function ( )f x x x 22= - - . The maximum value of ( )f x in the closed interval [ , ]4 4- is(A) 18 (B) 10

(C) 225- (D) indeterminateQ. 57 An examination consists of two papers, Paper 1 and Paper 2. The probability

of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is(A) 0.5 (B) 0.18

(C) 0.12 (D) 0.06

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2006 ONE MARK

Q. 58 The rank of the matrix 111

111

101

-R

T

SSSS

V

X

WWWW is

(A) 0 (B) 1

(C) 2 (D) 3

Q. 59 P4#4# , where P is a vector, is equal to(A) P P P2#4# 4- (B) ( )P P24 4 4#+(C) P P24 4#+ (D) ( )P P24 4$ 4-

Q. 60 ( )P ds4# $# # , where P is a vector, is equal to(A) P dl$# (B) P dl4 4 $# ##(C) P dl4 $## (D) Pdv4$# # #

Q. 61 A probability density function is of the form ( ) , ( , )p x Ke xx 3 3!= -a- . The value of K is(A) 0.5 (B) 1

(C) .0 5a (D) aQ. 62 A solution for the differential equation ( ) ( ) ( )x t x t t2 d+ =o with initial condition

(0 ) 0x =- is(A) ( )e u tt2- (B) ( )e u tt2

(C) ( )e u tt- (D) ( )e u tt

2006 TWO MARKS

Q. 63 The eigenvalue and the corresponding eigenvector of 2 2# matrix are given byEigenvalue Eigenvector

81l = v 11

1 = = G42l = v 1

12 = -= G

The matrix is

(A) 62

26= G (B) 46 64= G

(C) 24

42= G (D) 48 84= G

Q. 64 For the function of a complex variable lnW Z= (where, W u jv= + and Z x jy= +, the u = constant lines get mapped in Z-plane as(A) set of radial straight lines (B) set of concentric circles

(C) set of confocal hyperbolas (D) set of confocal ellipses

Q. 65 The value of the constant integral z 4

1

z j

2

2+- =

# dz is positive sense is(A) j

2p (B)

2p-

(C) j2p- (D)

2p


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Q. 66 The integral sin d30

q qp# is given by(A)

21 (B)

32

(C) 34 (D)

38

Q. 67 Three companies ,X Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below

Company % of Computer Supplied Probability of being supplied defective

X %60 .0 01

Y %30 .0 02

Z %10 .0 03

Given that a computer is defective, the probability that was supplied by Y is (A) 0.1 (B) 0.2

(C) 0.3 (D) 0.4

Q. 68 For the matrix 42

24= G the eigenvalue corresponding to the eigenvector 101101= G is

(A) 2

(B) 4

(C) 6

(D) 8

Q. 69 For the differential equation dxd y k y 02

22+ = the boundary conditions are

(i) y 0= for x 0= and (ii) y 0= for x a=The form of non-zero solutions of y (where m varies over all integers) are

(A) siny Aa

m xm

m

p=/ (B) cosy A am x

mm

p=/(C) y A xm a

m

m

= p/ (D) y A em am xm

= p-/

Q. 70 As x increased from 3- to 3, the function ( )f xe

e1 x

x= +(A) monotonically increases

(B) monotonically decreases

(C) increases to a maximum value and then decreases

(D) decreases to a minimum value and then increases

2005 ONE MARK

Q. 71 The following differential equation has

dtd y

dtdy y3 4 22

2 3 2+ + +c cm m x=(A) degree 2= , order 1= (B) degree 1= , order 2=(C) degree 4= , order 3= (D) degree 2= , order 3=

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Q. 72 A fair dice is rolled twice. The probability that an odd number will follow an even number is(A) 1/2 (B) 1/6

(C) 1/3 (D) 1/4

Q. 73 A solution of the following differential equation is given by dxd y

dxdy y5 6 02

2

- + =(A) y e ex x2 3= + - (B) y e ex x2 3= +(C) 3y e x x2 3= +- (D) y e ex x2 3= +- -

2005 TWO MARKS

Q. 74 In what range should ( )Re s remain so that the Laplace transform of the function e( )a t2 5+ + exits.(A) ( ) 2Re s a> + (B) ( )Re s a 7> +(C) ( )Re s 2< (D) ( )Re s a 5> +

Q. 75 The derivative of the symmetric function drawn in given figure will look like

Q. 76 Match the following and choose the correct combination:Group I Group 2E. Newton-Raphson method 1. Solving nonlinear equationsF. Runge-kutta method 2. Solving linear simultaneous equationsG. Simpsons Rule 3. Solving ordinary differential equationsH. Gauss elimination 4. Numerical integration 5. Interpolation 6. Calculation of Eigenvalues(A) E 6, F 1, G 5, H 3- - - - (B) E 1, F 6, G 4, H 3- - - -(C) E 1, F 3, G 4, H 2- - - - (D) E 5, F 3, G 4, H 1- - - -

Q. 77 Given the matrix 44

23

-= G, the eigenvector is(A)

32= G (B) 43= G

(C) 21-= G (D) 12-= G


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Q. 78 Let, .

A20

0 13=

-= G and A ab01 21

=- = G. Then ( )a b+ =(A) 7/20 (B) 3/20

(C) 19/60 (D) 11/20

Q. 79 The value of the integral expI x dx21

8

2

0p= -3 c m# is

(A) 1 (B) p(C) 2 (D) 2p

Q. 80 Given an orthogonal matrix

A

1110

1110

1101

1101

= -- -

R

T

SSSSS

V

X

WWWWW

AAT 1-6 @ is

(A) 000

0

00

00

0

000

41

41

21

21

R

T

SSSSSS

V

X

WWWWWW

(B) 000

0

00

00

0

000

21

21

21

21

R

T

SSSSSS

V

X

WWWWWW

(C)

1000

0100

0010

0001

R

T

SSSSS

V

X

WWWWW

(D) 000

0

00

00

0

000

41

41

41

41

R

T

SSSSSS

V

X

WWWWWW

***********

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SOLUTIONS

Sol. 1 Option (B) is correct.Here, as we know sinLim

0q

"q 0.

but for %10 error, we can check option (B) first,

q 18 .18180

0 314#c c cp= = =

sinq .sin18 0 309c= = % error .

. . % . %0 3090 314 0 309 100 0 49#= - =

Now, we check it for 50cq = q .50 50

1800 873#c c c

p= = = sinq .sin50 0 77c= = % error .

0.77 0.873 . %0 873 12 25=- =-

so, the error is more than %10 . Hence, for error less than 10%, 18cq = can have the approximation sinq . q

Sol. 2 Option (A) is correct.For, a given matrix A6 @ the eigen value is calculated as A Il- 0=where l gives the eigen values of matrix. Here, the minimum eigen value among the given options is l 0=We check the characteristic equation of matrix for this eigen value A Il- A= (for 0l = )

352

5127

275

=

3 5 260 49 25 14 35 24= - - - + -^ ^ ^h h h 33 55 22= - + 0=Hence, it satisfied the characteristic equation and so, the minimum eigen value is l 0=

Sol. 3 Option (D) is correct.Given, the polynomial

f x^ h a x a x a x a x a4 4 3 3 2 2 1 0= + + + -Since, all the coefficients are positive so, the roots of equation is given by f x^ h 0=It will have at least one pole in right hand plane as there will be least one sign change from a1^ h to a0^ h in the Routh matrix 1st column. Also, there will be a


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corresponding pole in left hand planei.e.; at least one positive root (in R.H.P)and at least one negative root (in L.H.P)Rest of the roots will be either on imaginary axis or in L.H.P

Sol. 4 Option (B) is correct.Consider the given matrix be

I ABm+ 2111

1211

1121

1112

=

R

T

SSSSSS

V

X

WWWWWW

where m 4= so, we obtain

AB

2111

1211

1121

1112

1000

0100

0010

0001

= -

R

T

SSSSSS

R

T

SSSSSS

V

X

WWWWWW

V

X

WWWWWW

1111

1111

1111

1111

=

R

T

SSSSSS

V

X

WWWWWW

1111

=

R

T

SSSSSS

V

X

WWWWWW

1 1 1 16 @

Hence, we get

A

1111

=

R

T

SSSSSS

V

X

WWWWWW

, B 1 1 1 1= 8 B

Therefore, BA 1 1 1 1= 8 B 1111

R

T

SSSSSS

V

X

WWWWWW

4=From the given property Det I ABm+^ h Det I BAm= +^ h

& Det

2111

1211

1121

1112

R

T

SSSSSS

V

X

WWWWWW

Det

1000

0100

0010

0001

4= +

R

T

SSSSSS

V

X

WWWWWW

Z

[

\

]]

]]

_

`

a

bb

bb

1 4= + 5=Note : Determinant of identity matrix is always 1.

Sol. 5 Option (D) is correct.

t dtdx x+ t=

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dtdx

tx+ 1=

dtdx Px+ Q= (General form)

Integrating factor, IF e e e tlntPdt t dt1= = = =# #

Solution has the form,

x IF# Q IF dt C#= +^ h# x t# ( )( )t dt C1= +# xt t C2

2= +Taking the initial condition,

( )x 1 .0 5= 0.5 C2

1= + & C 0=

So, xt t22= x t2& =

Sol. 6 Option (C) is correct.

( )f z z z11

32= + - +

( )j f z dz21

Cp # = sum of the residues of the poles which lie inside the given closed region.

C z 1 1& + =Only pole z 1=- inside the circle, so residue at z 1=- is. ( )f z

( )( )z zz1 3

1= + +- +

( )( )( )( )

limz zz z

1 31 1

22 1

z 1= + +

+ - + = ="-

So ( )j f z dz21

Cp # 1=Sol. 7 Option (A) is correct.

x i1= - = cos sini2 2p p= +

So, x ei 2= p

xx eix

2= p^ h & ei i2p^ h e 2= p-Sol. 8 Option (D) is correct.

( ) ( )

( )dt

d y tdtdy t

y t2

2

2

+ + ( )td=By taking Laplace transform with initial conditions

( ) ( ) [ ( ) ( )] ( )s Y s sy dtdy sy s y Y s0 2 0

t

2

0- - + - +

=; E 1=

& ( ) 2 ( ) ( )s Y s s sY s Y s2 0 22 + - + + +6 6@ @ 1= ( ) [ ]Y s s s2 12+ + s1 2 4= - - ( )Y s

s ss2 1

2 32= + +- -

We know that, If, ( )y t ( )Y sL


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then, ( )dt

dy t ( ) ( )sY s y 0L -

So, ( ) ( )sY s y 0- ( )( )s s

s s2 1

2 322= + +

- - +

( )s s

s s s s2 1

2 3 2 4 22

2 2= + +- - + + +

( ) ( )sY s y 0- ( ) ( ) ( )ss

ss

s12

11

11

2 2 2= ++ = +

+ + +

( )s s11

11

2= + + +Taking inverse Laplace transform

( )dt

dy t ( ) ( )e u t te u tt t= +- -

At t 0= +, dtdy

t 0= + e 0 10= + =

Sol. 9 Option (A) is correct.Divergence of A in spherical coordinates is given as

A:d ( )r r

r A1 r22

22= ( )

r rkr1 n2

2

22= +

( )rk n r2 n2

1= + +

( )k n r2 0n 1= + =- (given) n 2+ 0= & n 2=-

Sol. 10 Option (C) is correct.Probability of appearing a head is /1 2. If the number of required tosses is odd, we have following sequence of events.

,H ,TTH , ...........TTTTH

Probability P .....21

21

213 5= + + +b bl l

P 1 3

241

21

= - =

Sol. 11 Option (B) is correct.

( )f x x x x9 24 53 2= - + +

( )dxdf x

x x3 18 24 02= - + =

& ( )dxdf x

6 8 0x x2= - + = x 4= , x 2=

( )

dxd f x

2

2

x6 18= -

For ,x 2= ( )dx

d f x12 18 6 0

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Sol. 12 Option (B) is correct.Characteristic equation.

A Il- 0=

52

3ll

- - -- 0=

5 62l l+ + 0= 5 62l l+ + 0=Since characteristic equation satisfies its own matrix, so

5 6A A2+ + 0= 5 6A A I2& =- -Multiplying with A 5 6A A A3 2+ + 0= 5( 5 6 ) 6A A I A3+ - - + 0= A3 19 30A I= +


From Divergence theorem, we have

Adv4$v v### A n dss$= v t#

The position vector

rv u x u y u zx y z= + +t t t^ hHere, 5A r=v v, thus A4$ v u

xu

yu

zu x u y u zx y z x y z:2

222

22= + + + +t t t t t tc ^m h

5dxdx

dydy

dzdz= + +c m 3 5#= 15=

So, 5r n dss

$v t## 15dv V15= =###


We have dxdy ky=

Integrating ydy# k dx A= +#

or lny kx A= +Since ( )y 0 c= thus lnc A=So, we get, lny lnkx c= +or lny ln lne ckx= +or y cekx=

Sol. 15 Option (A) is correct.

C R Integrals is z z

z dz4 5

3 4

C2+ +- +# where C is circle z 1=

( )f z dzC# 0= if poles are outside C.

Now z z4 52+ + 0= ( )z 2 12+ + 0=Thus z ,1 2 j z2 1>,1 2&!=-So poles are outside the unit circle.


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We have ( )f x x x 3 0= + - = ( )f xl

x1

21= +

Substituting x0 2= we get ( )f x0l .1 35355= and ( ) .f x 2 2 3 0 4140 = + - =Newton Raphson Method

x1 ( )( )

xf xf x

00

0= -l

Substituting all values we have

x 1 ..2 1 3535

0 414= - .1 694=Sol. 17 Option (B) is correct.

Writing :A B we have

:::

111

144

16

620

l m

R

T

SSSS

V

X

WWWW

Apply R R R3 3 2" -

::: 20

110

140

16

6

620

l m- -

R

T

SSSS

V

X

WWWW

For equation to have solution, rank of A and :A B must be same. Thus for no solution; 6, 20!ml =

Sol. 18 Option (C) is correct.Total outcome are 36 out of which favorable outcomes are :(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6);(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15.

Thus ( )P E ..No of total outcomes

No of favourable outcomes3615

125= = =

Sol. 19 Option (C) is correct.Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in conjugate pairs.

Sol. 20 Option (C) is correct.For a function ( )x t trigonometric fourier series is

( )x t [ ]cos sinA A n t B n to n nn 1

w w= + +3=/

Where, Ao ( )T x t dt1

T0 0

= # T0" fundamental period

An ( )cosT x t n t dt2

T0 0

w= #

Bn ( )sinT x t n t dt2

T0 0

w= #For an even function ( ),x t B 0n =Since given function is even function so coefficient B 0n = , only cosine and constant

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terms are present in its fourier series representation.Constant term :

A0 ( )T x t dt1

/

/

T

T

4

3 4= -#

T Adt Adt1 2

/

/

/

/

T

T

T

T

4

4

4

3 4= + --: D# # T

TA AT1 2 2 2= -: D A2=-Constant term is negative.


Given differential equation

( ) ( )

dxd n x

Ln x

2

2

2- 0=Let ( )n x Ae x= lSo, A e

LAex x2

2l -ll

0=

L12

2l - L01& !l= =

Boundary condition, ( )n 03 = so take L1l =-

( )n x Ae Lx= -

( )n 0 Ae K A K0 &= = =So, ( )n x Ke ( / )x L= -


Given that ey x x1=

or lney lnx x1=

or y lnx x1=

Now dxdy lnx x x x

1 1x1

2= + - -^ h lnx x1

2 2= -For maxima and minima :

dxdy (1 ) 0ln

xx12= - =

lnx 1= " x e 1=Now

dxd y

2

2

lnx

xx x x

2 2 1 13 3 2=- - - -b bl l

lnx x

xx

2 2 12 3 3=- + -

dyd x

at x e2

2

1=

e e e2 2 1 0


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Av xya x ax y2= +t t dlv dxa dyax y= +t t A dl

C

:v v# ( ) ( )xya x a dxa dyax yC

x y2 := + +t t t t#

( )xydx x dyC

2= +#

xdx xdx dy dy3 34

31

/

/

/

/

2 3

1 3

1

3

3

1

1 3

2 3= + + +# # ##

[ ] [ ]21

34

31

23

31

34

34 3 1 3

1 1 3= - + - + - + -: :D D 1=

Sol. 25 Option (C) is correct.Given function

( )X z ( )( )z z z

z1 2

1 2= - --

Poles are located at 0, 1, 2andz z z= = =At Z 0= residues is R0 ( )z X z

Z 0:= = ( )( )0 1 0 2

1 2 0#= - -- 2

1=at z 1= , R1 ( 1) ( )Z X Z

Z 1:= - =

( )1 1 2

1 2 1 1#= -- =

At z 2= , R2 ( ) ( )z X z2z 2

:= - =

( )2 2 11 2 2

23#= -

- =-Sol. 26 Option (B) is correct.

Taking step size h 0.1= , ( )y 0 0=x y

dxdy x y= + y y hdx

dyi i1 = ++

0 0 0 . ( )y 0 0 1 0 01 = + =0.1 0 0.1 . ( . ) .y 0 0 1 0 1 0 012 = + =0.2 0.01 0.21 . . . .y 0 01 0 21 0 1 0 0313 #= + =0.3 0.031

From table, at . , ( . ) .x y x0 3 0 3 0 031= = =Sol. 27 Option (D) is correct.

Given that

( )f t ( )s s K s

s4 3

3 1L 1 3 2= + + -+- ; E

( )lim f tt"3

1=By final value theorem

( )lim f tt"3

( )lim sF s 1s 0

= ="

or ( )

( )lim

s s K ss s4 3

3 1s 0 3 2

:

+ + -+

" 1=

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or [ ( )]

( )lim

s s s Ks s4 33 1

s 0 2+ + -+

" 1=

K 31- 1=

or K 4=Sol. 28 Option (B) is correct.

The highest derivative terms present in DE is of 2nd order.

Sol. 29 Option (C) is correct.Number of elements in sample space is 210. Only one element

, , , , , , , , ,H H T T T T T T T T" , is event. Thus probability is 2110

Sol. 30 Option (C) is correct.We have

( )f z c c z0 1 1= + -

( )f z1 ( )

zf z

zc c z1 1 0 1 1= + = + + - ( )

zz c c1

20 1= + +

Since ( )f z1 has double pole at z 0= , the residue at z 0= is Res ( )f z z1 0= . ( )lim z f z

z 0

21=

" .

( )lim z

zz c c1

z 0

220 1= + +

" c m c1=Hence

( )f z dz1unit circle

# [ ( )]zf z

dz1

unit circle

= +# j2p= [Residue at z 0= ]

jc2 1p=Sol. 31 Option (D) is correct.

We have ( )f x sinx

xp= -

Substituting x p- y= ,we get ( )f y p+ ( )sin sin

yy

yyp= + =- ( )sin

yy1= -

! !

...yy y y1

3 5

3 5

= - - + -c mor ( )f y p+

! !...y y1

3 5

2 4

=- + - +Substituting x yp- = we get ( )f x

!( )

!( )

...x x

13 5

2 4p p=- + - - - +Sol. 32 Option (A) is correct.

(A) dxdy

xy=

or ydy# x

dx= #or logy log logx c= +or y cx= Straight LineThus option (A) and (C) may be correct.

(B) dxdy

xy=-


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or ydy# x

dx=- #or logy log logx c=- +or logy log log

xc1= +

or y xc= Hyperbola

Sol. 33 Option (D) is correct.Sum of the principal diagonal element of matrix is equal to the sum of Eigen values. Sum of the diagonal element is 1 1 3 1- - + = .In only option (D), the sum of Eigen values is 1.

Sol. 34 Option (C) is correct.The product of Eigen value is equal to the determinant of the matrix. Since one of the Eigen value is zero, the product of Eigen value is zero, thus determinant of the matrix is zero.

Thus p p p p11 22 12 21- 0=Sol. 35 Option (B) is correct.

The given system is

xy

42

21= =G G 76= = G

We have A 42

21= = G

and A 42

21 0= = Rank of matrix ( )A 2 , e 1>x and 0 1e<

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We have x e x= -or ( )f x x e x= - - '( )f x e1 x= + -The Newton-Raphson iterative formula is

xn 1+ '( )( )

xf xf x

nn

n= -Now ( )f xn x en xn= - - '( )f xn e1 xn= + -

Thus xn 1+ xe

x e1

n xn

x

n

n= - +-

-

-

( )ex e

11

xn

x

n

n= ++

-

-


Res ( )f z z a= ( )!( ) ( )

n dzd z a f z

11

n

nn

z a1

1= - ---

=6 @Here we have n 2= and a 2=Thus Res ( )f z z 2= ( )!

( )( ) ( )dz

d zz z2 1

1 22 2

1

z a

22 2= - - - + =; E

( )dz

dz 2

1

z a2= + =; E ( )z 2

2

z a3= +

-=

; E 64

2=- 321=-


eP ( )L sI A1 1= -- -6 @ L

ss00 0

213

11

= - - --

-e o= =G G L

ss2

13

11

= -+-

-e o= G L ( )( )

( )( )

( )( )

( )( )

s ss

s s

s s

s ss

1 1 23

1 22

1 21

1 2= - + +

+

+ +-

+ +

+ +f p> H

e ee e

e ee e

22 2 2

1 2

1 2

1 2

1 2= -- +-

- +- -

- -

- -

- -= GSol. 43 Option (B) is correct.

Taylor series is given as

( )f x ( )!

'( )!

( )"( ) ...f a x a f a

x af a

1 2

2

= + - + - +For x p= we haveThus ( )f x ( )

!'( )

!( )

"( )...f x fx

f x1 2

2

p p p p= + - + -

Now ( )f x sine xx= + '( )f x cose xx= + "( )f x sine xx= - "( )f p sine ep= - =p p


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Thus the coefficient of ( )x 2p- is !

"( )f2p

Sol. 44 Option (A) is correct.The equation of straight line from ( , )0 0 to ( , )1 2 is y x2= .Now ( , )g x y x y4 103 4= +or, ( , )g x x2 x x4 1603 4= +Now ( , )g x x2

0

1# ( )x x dx4 1603 40

1= +# [ ]x x32 334 5 01= + =


I ( )xdx ydy2P

Q= +# xdx ydy2 2

P

Q

P

Q= + ## xdx ydy2 2 0

0

1

1

0= + =##Sol. 46 Option (B) is correct.

The given plot is straight line whose equation is

x y1 1- + 1=

or y x 1= +Now I ydx

1

2= # ( )x dx11

2= +#

( )x21 2 2= +; E .29 24 2 5= - =


cothx sinhcosh

xx=

as 1, 1coshx x

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( )x e1 2 2= - - -6 @ Neglecting higher powers ( )x e3 2= - -


We have kdxd y2

2

2

y y2= -

or dxd y

ky

2

2

2- ky

22=-

A.E. Dk122- 0=

or D k1!=

C.F. C e C e1 2kx

kx= +-

P.I. D k

y y1

k2 1 2

22

22

= -- =c m

Thus solution is

y C e C e y1 2 2kx

kx= + +-

From ( )y y0 1= we get C C1 2+ y y1 2= -From ( )y y23 = we get that C1 must be zero.Thus C2 y y1 2= - y ( )y y e y1 2 2k

x= - +-Sol. 52 Option (B) is correct.

We have

( )f x x x x4 43 2= - + - '( )f x x x3 2 42= - +Taking x 20 = in Newton-Raphosn method x1 '( )

( )x

f xf x

00

0= - ( ) ( )

( )2

3 2 2 2 42 2 4 2 4

2

3 2

= - - +- + -

34=

Sol. 53 Option (C) is correct.For two orthogonal signal ( )f x and ( )g x

( ) ( )f x g x dx3

3

-

+# 0=i.e. common area between ( )f x and ( )g x is zero.

Sol. 54 Option (A) is correct.We know that

s

ds1

1

D2-# j2p= [sum of residues]

Singular points are at s 1!= but only s 1=+ lies inside the given contour, Thus Residue at s 1=+ is ( ) ( )lim s f s1

s 1-

" ( )lim s

s1

11

21

s 1 2= - - ="

s

ds1

1

D2-# j j2 2

1p p= =` j


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Sol. 55 Option (C) is correct.For two orthogonal vectors, we require two dimensions to define them and similarly for three orthogonal vector we require three dimensions to define them. M2 vectors are basically M orthogonal vector and we require M dimensions to

define them.

Sol. 56 Option (A) is correct.We have

( )f x x x 22= - + '( )f x x2 1 0= - = x

21" =

"( )f x 2=Since "( )f x 2 0>= , thus x

21= is minimum point. The maximum value in

closed interval ,4 4-6 @ will be at x 4=- or x 4=Now maximum value

[ ( 4), (4)]max f f= - ( , )max 18 10= 18=

Sol. 57 Option (C) is correct.Probability of failing in paper 1 is ( ) .P A 0 3=Possibility of failing in Paper 2 is ( ) .P B 0 2=Probability of failing in paper 1, when student has failed in paper 2 is 0.6P B

A =^ hWe know that

P BAb l ( )

( )P BP B+=

or ( )P A B+ ( )P B P BA= b l . . .0 6 0 2 0 12#= =

Sol. 58 Option (C) is correct.We have

A 111

111

101

110

110

100

+= - -R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW R R3 1-

Since one full row is zero, ( )A 3

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We know ( )p x dx3

3

-# 1=

Thus Ke dxx3

3 a--# 1=

or Ke dx Ke dxx x0

0 + 33

a a--

## 1=or

( )K e k ex x0

0a a+ -33a a

--6 6@ @ 1=

or K Ka a+ 1=or K

2a=


We have ( ) ( )x t x t2+o ( )s t=Taking Laplace transform both sides

( ) ( ) ( )sX s x X s0 2- + 1=or ( ) ( )sX s X s2+ 1= Since ( )x 0 0=- ( )X s

s 21= +

Now taking inverse Laplace transform we have

( )x t ( )e u tt2= -Sol. 63 Option (A) is correct.

Sum of the Eigen values must be equal to the sum of element of principal diagonal of matrix.

Only matrix 62

26= G satisfy this condition.


We have W ln z= u jv+ ( )ln x jy= +or eu jv+ x jy= +or e eu jv x jy= + ( )cos sine v j vu + x jy= +Now cosx e vu= and siny e vu=Thus x y2 2+ e u2= Equation of circle

Sol. 65 Option (D) is correct.We have

z

dz4

1

z j2

2+

- =

# ( )( )z i z i dz2 21

z j 2

= + -- =

#

( , )P 0 2 lies inside the circle z j 2- = and ( , )P 0 2- does not lie.Thus By cauchys integral formula

I ( )( )( )

limi z iz i z i

2 22 2

1z i2

p= - + -" i ii

2 22

2C

p p= + =#Sol. 66 Option (C) is correct.

I sin d30

q q= p#


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sin sin d4

3 30

q q q= -p ` j# sin sin sin3 3 4 3q q q= - cos s4

3123

0 0q w q= - =p p: :D D 43 43 121 121 34= + - + =8 8B B

Sol. 67 Option (D) is correct.Let d " defective and y " supply by Y

p dya k ( )

( )P d

P y d+= ( )P y d+ . . .0 3 0 02 0 006#= = ( )P d . . . . . . .0 6 0 1 0 3 0 02 0 1 0 03 0 015# # #= + + = P d

ya k .. .

0 0150 006 0 4= =


We have A 42

24= = G

Now [ ]A I Xl-6 @ 0=or

42

24

101101

ll

--= =G G 00= = G

or ( )( ) ( )101 4 2 101l- + 0=or l 6=


We have dxd y k y2

22+ 0=

or D y k y2 2+ 0=The AE is m k2 2+ 0=The solution of AE is m ik!=Thus sin cosy A kx B kx= +From x 0= , y 0= we get B 0= and ,x a y 0= = we get sinA ka 0=or sinka 0= k

am xp=

Thus y sinAa

m xm

m

p= ` j/Sol. 70 Option (A) is correct.

We have ( )f x e

e1 x

x= +For x " 3, the value of ( )f x monotonically increases.

Sol. 71 Option (B) is correct.Order is the highest derivative term present in the equation and degree is the power of highest derivative term.Order 2= , degree 1=

Sol. 72 Option (D) is correct.Probability of coming odd number is 2

1 and the probability of coming even number is 2

1 . Both the events are independent to each other, thus probability of coming

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odd number after an even number is 21

21

41# = .


We have dxd y

dxdy y5 62

2

- + 0=The . .A E is m m5 62- + 0= m ,3 2=The CF is yc C e C ex x1 3 2 2= +Since Q 0= , thus y C e C ex x1 3 2 2= +Thus only (B) may be correct.


We have ( )f t e( )a t2 5= + + .e e( )a t5 2= +Taking Laplace transform we get

( )F s ( )

es a 2

15= - +; E Thus ( ) ( )Re s a 2> +Sol. 75 Option (C) is correct.

For x 0> the slope of given curve is negative. Only (C) satisfy this condition.Sol. 76 Option (C) is correct.

Newton - Raphson " Method-Solving nonlinear eq.

Runge - kutta Method " Solving ordinary differential eq.

Simpsons Rule " Numerical Integration

Gauss elimination " Solving linear simultaneous eq.


We have A 44

23=

-= GCharacteristic equation is

A Il- 0=

or 4

42

3l

l-

- 0=or ( 4 )(3 ) 8l l- - - - 0=or 12 82l l- + + - 0=or 202l l+ - 0=or l ,5 4=- Eigen valuesEigen vector for 5l =- ( )A I Xil- 0=

( ) xx

1 54

28 4

1

2

- --= =G G 00= = G

xx

10

20

1

2= =G G 00= = G R R42 1-

x x21 2+ 0=Let x x2 11 2&- = =- ,Thus X

21= -= G Eigen vector


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Sol. 78 Option (A) is correct.We have

A .2

00 13=

-= G and A ab01 21

=- = GNow AA 1- I=

or . a

b20

0 13 0

21-= =G G 10 01= = G

or .a b

b10

2 0 13-= G 10 01= = G

or .a2 0 1- 0= and b3 1=Thus solving above we have b

31= and a

601=

Therefore a b+ 31

601

207= + =

Sol. 79 Option (A) is correct.Gaussian PDF is

( )f x e dx21 ( )x

2 2

2

p s= 33 -

-sm-# for x3 3# #-

and ( )f x dx3

3

-# 1=

Substituting 0m = and 2s = in above we get e dx

2 21 x

8

2

p 33 -

-# 1=

or 2 e dx2 21

0

x8

2

p3 -# 1=

or e dx21

0

x8

2

p3 -# 1=

Sol. 80 Option (C) is correct.From orthogonal matrix

[ ]AAT I=Since the inverse of I is I , thus

[ ]AAT 1- I I1= =-

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