GATE SOLVED PAPER - EE - mywbut.com chapter wise solved papers... · Q. 4 The curl of the gradient of the scalar field defined by Vxy yz zx=+23 422+2 is (A) ... is a continuous function

GATE SOLVED PAPER - EEENGINEERING MATHEMATICS

YEAR 2013 ONE MARK

Q. 1 Given a vector field F y xa yza x ax y z2 2= - -v v v v , the line integral F dl:v v# evaluated

along a segment on the x-axis from x 1= to x 2= is(A) .2 33- (B) 0

(C) .2 33 (D) 7

Q. 2 The equation xx

21

21

00

1

2

-- => > >H H H has

(A) no solution (B) only one solution xx

00

1

2=> >H H

(C) non-zero unique solution (D) multiple solutions

Q. 3 Square roots of i- , where i 1= - , are(A) i , i-

(B) cos sini4 4p p- + -d dn n, cos sini4

343p p+b bl l

(C) cos sini4 43p p+d bn l, cos sini4

34

p p+b dl n

(D) cos sini43

43p p+ -b bl l, cos sini4

343p p- +b bl l

Q. 4 The curl of the gradient of the scalar field defined by V x y y z z x2 3 42 2 2= + + is(A) 4 6 8xya yza zxax y z+ +v v v

(B) 4 6 8a a ax y z+ +v v v

(C) xy z a x yz a y zx a4 4 2 6 3 8x y z2 2 2+ + + + +v v v^ ^ ^h h h

(D) 0

Q. 5 A continuous random variable X has a probability density function f x e x= -^ h , x0 < < 3. Then P X 1>" , is

(A) 0.368 (B) 0.5

(C) 0.632 (D) 1.0

YEAR 2013 TWO MARKS

Q. 6 When the Newton-Raphson method is applied to solve the equation f x x x2 1 03= + - =^ h , the solution at the end of the first iteration with the initial value as .x 1 20 = is(A) .0 82- (B) .0 49

(C) .0 705 (D) .1 69

Q. 7 A function y x x5 102= + is defined over an open interval ,x 1 2= ^ h. Atleast at one point in this interval, /dy dx is exactly(A) 20 (B) 25

(C) 30 (D) 35

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GATE SOLVED PAPER - EE ENGINEERING MATHEMATICS

Q. 8 zz dz

44

2

2

+-# evaluated anticlockwise around the circle z i 2- = , where i 1= -

, is

(A) 4p- (B) 0

(C) 2 p+ (D) i2 2+

Q. 9 A Matrix has eigenvalues 1- and 2- . The corresponding eigenvectors are 11-> H

and 12-> H respectively. The matrix is

(A) 11

12- -> H (B)

12

24- -> H

(C) 1

002

--> H (D)

02

13- -> H

YEAR 2012 ONE MARK

Q. 10 Two independent random variables X and Y are uniformly distributed in the interval ,1 1-6 @. The probability that ,max X Y6 @ is less than /1 2 is(A) /3 4 (B) /9 16

(C) /1 4 (D) /2 3

Q. 11 If ,x 1= - then the value of xx is(A) e /2p- (B) e /2p

(C) x (D) 1

Q. 12 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 dz21

Cp # is

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

Q. 13 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=

YEAR 2012 TWO MARKS

Q. 14 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+

Q. 15 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

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Q. 16 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. 17 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. 18 Consider the differential equation

( ) ( )

( )dt

d y tdt

dy ty t22

2

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

tt

00

=- ==

=-

-

The numerical value of dtdy

t 0= +

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

YEAR 2011 ONE MARK

Q. 19 Roots of the algebraic equation x x x 1 03 2+ + + = are

(A) ( , , )j j1+ + - (B) ( , , )1 1 1+ - +(C) ( , , )0 0 0 (D) ( , , )j j1- + -

Q. 20 A point Z has been plotted in the complex plane, as shown in figure below.

The plot of the complex number Y Z1= is

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Q. 21 With K as a constant, the possible solution for the first order differential equation

dxdy e x3= - is

(A) e Kx31 3- +- (B) e Kx

31 3- +

(C) e Kx31 3- +- (D) e K3 x- +-

YEAR 2011 TWO MARKS

Q. 22 Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method.

Equation (1) .sinx x10 0 8 02 1- =Equation (2) .cosx x x10 10 0 6 02

22 1- - =

Assuming the initial values are .x 0 01 = and .x 1 02 = , the jacobian matrix is

(A) ..

100

0 80 6

--> H (B)

100

010> H

(C) ..

010

0 80 6

--> H (D)

1010

010-> H

Q. 23 The function ( )f x x x x2 32 3= - - + has(A) a maxima at x 1= and minimum at x 5=(B) a maxima at x 1= and minimum at x 5=-(C) only maxima at x 1= and

(D) only a minimum at x 5=

Q. 24 A zero mean random signal is uniformly distributed between limits a- and a+ and its mean square value is equal to its variance. Then the r.m.s value of the signal is

(A) a3

(B) a2

(C) a 2 (D) a 3

Q. 25 The matrix [ ]A24

11= -> H is decomposed into a product of a lower triangular

matrix [ ]L and an upper triangular matrix [ ]U . The properly decomposed [ ]L and [ ]U matrices respectively are

(A) 14

01-> H and

10

12-> H (B)

24

01-> H and

10

11> H

(C) 14

01> H and

20

11-> H (D)

24

03-> H and

.10

1 51> H

Q. 26 The two vectors [1,1,1] and [1, , ]a a2 where a j21

23= - +_ i, are

(A) Orthonormal (B) Orthogonal

(C) Parallel (D) Collinear

YEAR 2010 ONE MARK

Q. 27 The value of the quantity P , where P xe dxx

0

1= # , is equal to

(A) 0 (B) 1

(C) e (D) 1/e

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Q. 28 Divergence of the three-dimensional radial vector field r is(A) 3 (B) /r1

(C) i j k+ +t t t (D) 3( )i j k+ +t t t

YEAR 2010 TWO MARKS

Q. 29 A box contains 4 white balls and 3 red balls. In succession, two balls are randomly and removed form the box. Given that the first removed ball is white, the probability that the second removed ball is red is(A) 1/3 (B) 3/7

(C) 1/2 (D) 4/7

Q. 30 An eigenvector of P 100

120

023

=

J

L

KKK

N

P

OOO is

(A) 1 1 1 T-8 B (B) 1 2 1 T8 B

(C) 1 1 2 T-8 B (D) 2 1 1 T-8 B

Q. 31 For the differential equation dtd x

dtdx x6 8 02

2

+ + = with initial conditions ( )x 0 1=

and dtdx 0

t 0=

=, the solution is

(A) ( ) 2x t e et t6 2= -- - (B) ( ) 2x t e et t2 4= -- -

(C) ( ) 2x t e et t6 4=- +- - (D) ( ) 2x t e et t2 4= +- -

Q. 32 For the set of equations, x x x x2 4 21 2 3 4+ + + = and 3 6 3 12 6x x x x1 2 3 4+ + + = . The following statement is true.(A) Only the trivial solution 0x x x x1 2 3 4= = = = exists

(B) There are no solutions

(C) A unique non-trivial solution exists

(D) Multiple non-trivial solutions exist

Q. 33 At t 0= , the function ( ) sinf t tt= has

(A) a minimum (B) a discontinuity

(C) a point of inflection (D) a maximum

YEAR 2009 ONE MARK

Q. 34 The trace and determinant of a 2 2# matrix are known to be 2- and 35- respectively. Its eigen values are(A) 30- and 5- (B) 37- and 1-(C) 7- and 5 (D) 17.5 and 2-

YEAR 2009 TWO MARKS

Q. 35 ( , )f x y is a continuous function defined over ( , ) [ , ] [ , ]x y 0 1 0 1#! . Given the two constraints, x y> 2 and y x> 2, the volume under ( , )f x y is

(A) ( , )f x y dxdyx y

x y

y

y

0

1

2=

=

=

=## (B) ( , )f x y dxdy

x y

x

y x

y 11

22 =

=

=

=##

(C) ( , )f x y dxdyx

x

y

y

0

1

0

1

=

=

=

=## (D) ( , )f x y dxdy

x

x y

y

y x

00 =

=

=

=##

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Q. 36 Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5 ?(A) 20 (B) 7

(C) 15 (D) 16

Q. 37 A cubic polynomial with real coefficients(A) Can possibly have no extrema and no zero crossings

(B) May have up to three extrema and upto 2 zero crossings

(C) Cannot have more than two extrema and more than three zero crossings

(D) Will always have an equal number of extrema and zero crossings

Q. 38 Let x 117 02 - = . The iterative steps for the solution using Newton-Raphon’s method is given by

(A) x x x21 117

k kk

1 = ++ b l (B) x x x117

k kk

1 = -+

(C) x x x117k k

k1 = -+ (D) x x x x2

1 117k k k

k1 = - ++ b l

Q. 39 ( , ) ( ) ( )x y x xy y xyF a ax y2 2= + + +t t . It’s line integral over the straight line from

( , ) ( , )x y 0 2= to ( , ) ( , )x y 2 0= evaluates to(A) 8- (B) 4

(C) 8 (D) 0

YEAR 2008 ONE MARKS

Q. 40 X is a uniformly distributed random variable that takes values between 0 and 1. The value of { }E X3 will be(A) 0 (B) 1/8

(C) 1/4 (D) 1/2

Q. 41 The characteristic equation of a (3 3# ) matrix P is defined as ( )a I P 2 1 03 2l l l l l= - = + + + =If I denotes identity matrix, then the inverse of matrix P will be(A) ( )P P I22 + +

(B) ( )P P I2 + +

(C) ( )P P I2- + +

(D) ( )P P I22- + +

Q. 42 If the rank of a ( )5 6# matrix Q is 4, then which one of the following statement is correct ?(A) Q will have four linearly independent rows and four linearly independent

columns

(B) Q will have four linearly independent rows and five linearly independent columns

(C) QQT will be invertible

(D) Q QT will be invertible

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YEAR 2008 TWO MARKS

Q. 43 Consider function ( ) ( 4)f x x2 2= - where x is a real number. Then the function has(A) only one minimum (B) only tow minima

(C) three minima (D) three maxima

Q. 44 Equation e 1 0x - = is required to be solved using Newton’s method with an initial guess x 10 =- . Then, after one step of Newton’s method, estimate x1 of the solution will be given by(A) 0.71828 (B) 0.36784

(C) 0.20587 (D) 0.00000

Q. 45 A is m n# full rank matrix with m n> and I is identity matrix. Let matrix ' ( )A A A A1T T= - , Then, which one of the following statement is FALSE ?

(A) 'AA A A= (B) ( ')AA 2

(C) 'A A I= (D) ' 'AA A A=

Q. 46 A differential equation / ( )dx dt e u tt2= - , has to be solved using trapezoidal rule of integration with a step size .h 0 01= s. Function ( )u t indicates a unit step function. If ( )x 0 0=- , then value of x at .t 0 01= s will be given by(A) 0.00099 (B) 0.00495

(C) 0.0099 (D) 0.0198

Q. 47 Let P be a 2 2# real orthogonal matrix and x is a real vector [ ]x ,x1 2T with length

( )x x x /12

22 1 2= + . Then, which one of the following statements is correct ?

(A) Px x# where at least one vector satisfies Px x<

(B) Px x# for all vector x

(C) Px x$ where at least one vector satisfies Px x>

(D) No relationship can be established between x and Px

YEAR 2007 ONE MARK

Q. 48 x x xx n1 2T

g= 8 B is an n-tuple nonzero vector. The n n# matrixV xxT=(A) has rank zero (B) has rank 1

(C) is orthogonal (D) has rank n

YEAR 2007 TWO MARKS

Q. 49 The differential equation dtdx x1= t

- is discretised using Euler’s numerical integration method with a time step T 0>3 . What is the maximum permissible value of T3 to ensure stability of the solution of the corresponding discrete time equation ?(A) 1 (B) /2t(C) t (D) 2t

Q. 50 The value of ( )z

dz1

C2+

# where C is the contour 1z i2- = is

(A) i2p (B) p(C) tan z1- (D) tani z1p -

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Q. 51 The integral ( )sin cost d21

0

2

pt t t-

p# equals

(A) sin cost t (B) 0

(C) cos t21 (D) sin t2

1

Q. 52 A loaded dice has following probability distribution of occurrences

Dice Value 1 2 3 4 5 6

Probability 1/4 1/8 1/8 1/8 1/8 1/4

If three identical dice as the above are thrown, the probability of occurrence of values 1, 5 and 6 on the three dice is(A) same as that of occurrence of 3, 4, 5

(B) same as that of occurrence of 1, 2, 5

(C) 1/128

(D) 5/8

Q. 53 Let x and y be two vectors in a 3 dimensional space and x,y< > denote their dot product. Then the determinant

detx,xy,x

x,yy,y

< >< >

< >< >= G

(A) is zero when x and y are linearly independent

(B) is positive when x and y are linearly independent

(C) is non-zero for all non-zero x and y(D) is zero only when either x or y is zero

Q. 54 The linear operation ( )L x is defined by the cross product L(x) b x#= , where b 0 1 0 T= 8 B and x x xx 1 2 3

T= 8 B are three dimensional vectors. The 3 3# matrix M of this operations satisfies

( ) Mxxx

L x1

2

3

=

R

T

SSSS

V

X

WWWW

Then the eigenvalues of M are(A) , ,0 1 1+ - (B) , ,1 1 1-(C) , ,i i 1- (D) , ,i i 0-

Statement for Linked Answer Question 55 and 56Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix

A32

20=

--= G

Q. 55 A satisfies the relation(A) A I A3 2 01+ + =- (B) 2 2 0A A I2 + + =(C) ( )( )A I A I2+ + (D) ( ) 0exp A =

Q. 56 A9 equals(A) 511 510A I+ (B) 309 104A I+(C) 154 155A I+ (D) ( )exp A9

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YEAR 2006 TWO MARKS

Q. 57 The expression ( / )V R h H dh1H 2 2

0p= -# for the volume of a cone is equal to

(A) ( / )R h H dr1R 2 2

0p -# (B) ( / )R h H dh1

R 2 2

0p -#

(C) ( / )rH r R dh2 1H

0p -# (D) rH

Rr dr2 1

R 2

0p -` j#

Q. 58 A surface ( , )S x y x y2 5 3= + - is integrated once over a path consisting of the points that satisfy ( ) ( )x y1 2 1 2 2+ + - = . The integral evaluates to(A) 17 2 (B) 17 2

(C) /2 17 (D) 0

Q. 59 Two fair dice are rolled and the sum r of the numbers turned up is considered(A) ( 6)Pr r > 6

1=

(B) Pr ( /r 3 is an integer) 65=

(C) Pr ( /r r8 4;= is an integer) 95=

(D) ( 6 /Pr r r 5;= is an integer) 161=

Statement for Linked Answer Question 60 and 61

, ,P Q R1013

25

9

27

12

T T T

=-

=-- = -

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

are three vectors.

Q. 60 An orthogonal set of vectors having a span that contains P,Q,R is

(A) 636

42

3

---

-

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

(B) 4

24

5711

823

-

- -

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

(C) 671

322

394-

-

- -

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

(D) 4311

1313

534

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

Q. 61 The following vector is linearly dependent upon the solution to the previous problem

(A) 893

R

T

SSSS

V

X

WWWW

(B) 217

30

--

R

T

SSSS

V

X

WWWW

(C) 445

R

T

SSSS

V

X

WWWW

(D) 1323-

R

T

SSSS

V

X

WWWW

YEAR 2005 ONE MARK

Q. 62 In the matrix equation x qP = , which of the following is a necessary condition for the existence of at least on solution for the unknown vector x(A) Augmented matrix [ ]qP must have the same rank as matrix P

(B) Vector q must have only non-zero elements

(C) Matrix P must be singular

(D) Matrix P must be square

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Q. 63 If P and Q are two random events, then the following is TRUE(A) Independence of P and Q implies that probability ( )P Q 0+ =(B) Probability ( )P Q, $ Probability (P) + Probability (Q)

(C) If P and Q are mutually exclusive, then they must be independent

(D) Probability ( )P Q+ # Probability (P)

Q. 64 If S x dx3

1=

3 -# , then S has the value

(A) 31- (B)

41

(C) 21 (D) 1

Q. 65 The solution of the first order DE '( ) ( )x t x t3=- , (0)x x0= is(A) ( )x t x e t

03= -

(B) ( )x t x e03= -

(C) ( )x t x e /0

1 3= -

(D) ( )x t x e01= -

YEAR 2005 TWO MARKS

Q. 66 For the matrix p300

22

0

211

=--

R

T

SSSS

V

X

, one of the eigen values is equal to 2-

Which of the following is an eigen vector ?

(A) 32

1-

R

T

SSSS

V

X

WWWW

(B) 3

21

-

-

R

T

SSSS

V

X

WWWW

(C) 12

3-

R

T

SSSS

V

X

WWWW

(D) 250

R

T

SSSS

V

X

Q. 67 If R122

013

11

2=

--

R

T

SSSS

V

X

WWWW

, then top row of R 1- is

(A) 5 6 48 B (B) 5 3 1-8 B

(C) 2 0 1-8 B (D) /2 1 1 2-8 B

Q. 68 A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is(A) 1/6 (B) 1/2

(C) 3/6 (D) 3/4

Q. 69 For the function ( )f x x e x2= - , the maximum occurs when x is equal to(A) 2 (B) 1

(C) 0 (D) 1-

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Q. 70 For the scalar field u x y2 3

2 2

= + , magnitude of the gradient at the point (1, 3) is

(A) 913 (B)

29

(C) 5 (D) 29

Q. 71 For the equation '' ( ) ' ( ) ( )x t x t x t3 2 5+ + = ,the solution ( )x t approaches which of the following values as t " 3 ?

(A) 0 (B) 25

(C) 5 (D) 10

***********

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SOLUTION

Sol. 1 Option (B) is correct.Given, the field vector

Fv y xa yza x ax y z2 2= - -v v v

For the line segment along x -axis, we have

dlv dxax= v

So, F dl:v v y x dx2= ^ ^h h

Since, on x -axis y 0= so,

F dl:v v 0=

or, F dl:v v# 0=

Sol. 2 Option (D) is correct.Given the equations in matrix form as

xx

21

21

1

2

--> >H H

00= > H

So, it is a homogenous set of linear equation. It has either a trivial solution x x 01 2= =^ h or an infinite no. of solution. Since, for the matrix

A 21

21=

--> H

we have the determinant

A 0=Hence, it will have multiple solutions

Sol. 3 Option (B) is correct.We know that

i e /i 2= p (In phasor form)

or, i- e /i 2= p-

So, i- e e/ / /i i2 1 2 4! != =p p- -^ h

cos sini4 4! p p= -d dn n= G

cos sini4 4p p= -d dn n ; cos sini4 4

p p- +d dn n

i2

12

= - ; i12

- +

This is equivalent to the given option (B) only.

Sol. 4 Option (D) is correct.Given the scalar field

V x y y z z x2 3 42 2 2= + +Its gradient is given by

Vd xy z a x yz a y zx a4 4 2 6 3 8x y z2 2 2= + + + + +v v v^ _ ^h i h

So, the curl of the gradient is obtained as

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v#d d̂ h a

xy z

a

z yz

a

y zx4 4 2 6 3 8

x

x

y

y

z

z2 2 2

=+ + +22

22

22

v v v

8 8a y y a z z a x x6 6 4 4x y z= - - - + -v v v^ ^ ^h h h 0=Note : From the properties of curl, we know that curl of gradient of any scalar

field is always zero. So, there is no need to solve the curl and gradient.

Sol. 5 Option (A) is correct.Given, the PdF of random variable x as

f x^ h e x= - x0 < < 3

So, P x 1>^ h e dxx

1=

3-# e1x

1= -

3-

: D e 1= - .0 368=

Sol. 6 Option (C) is correct.Given, the equation

f x^ h x x2 1 03= + - =as initial condition is x0 .1 2=so, from N R- method we obtain

xn 1+ xf xf x

nn

n= -

l̂^

h

h

Here x0 .1 2= f x0^ h . 2 . 1 3.1281 2 1 23= + - =^ ^h h

Also, f xl̂ h x3 22= +So, f x0l̂ h . .3 1 2 2 6 322= + =^ h

Hence, 1st iterative value is

x1 xf xf x

00

0= -

l̂^

h

h . .

.1 2 6 323 128= -

.0 705=

Sol. 7 Option (B) is correct.Given the function

y 5 10f x x x2= = +^ h in the internal ,x 1 2= ^ h

Since, function y is continuous in the interval ,1 2^ h as well as its is

differentiable at each point so, from Lagranges mean value theorem there exist

at least a point where

f cl̂ h b af b f a

= --^ ^h h

Here, we have

a 1= , b 2=So, for x a 1= = , we obtain

y 5f a f 1 1 10 1 152= = = + =^ ^ ^ ^h h h h

and for x b 2= = y 5f b f 2 2 10 2 402= = = + =^ ^ ^ ^h h h h

Therefore,

f cl̂ h 2 140 15 25= -

- =

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Sol. 8 Option (A) is correct.Given the contour integral

zz dz

44

2

2

+-#

It has two poles given as

z i2!=Now, the contour is defined by circle z i 2- = which is shown in the figure

below

So, it can be observed that the given contour enclosed z i2= while z i2=- is

out of the contour. So, we obtain the residue at z i2= only as

residue z iz

24

z i

2

2= +

-=

i ii

i i2 22 4

48 2

2

= +-

= - =^ h

Hence, contour integral is given as

zz dz

44

2

2

+-# 2 ip= (sum of residues)

2 i i2p= ^ h

4p=-

Sol. 9 Option (D) is correct.We know that the characteristic equation is given by

A X6 6@ @ Xl= 6 @

where A6 @ is the matrix as l is the scalar which gives eigen values. Now, we

consider the matrix

A6 @ ac

bd= > H matrix2 2#^ h

For eigen value 1- as eigen vector is 11-> H, so, we have

ac

bd

11-> >H H 1

11=- -> H

or, a b- 1=- ....(1)

c d- 1= ....(2)

Similarly, for eigen value 2- with eigen vector 12-> H, we obtain

ac

bd

12-> >H H 2

12=- -> H

or, a b2- 2=- ....(3)

c d2- 4= ....(4)

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Solving Eqs. (1) and (3), we obtain

a ,b0 1= =and solving Eqs. (2) and (4), we obtain

c ,d2 3=- =Thus, the required matrix is

ac

bd> H

02

13= - -> H

Sol. 10 Option (B) is correct.Probability density function of uniformly distributed variables X and Y is shown as

[ ( , )]maxP x y 21<& 0

Since X and Y are independent.

[ ( , )]maxP x y 21<& 0 P X P Y2

121< <= b bl l

P X 21<b l shaded area 4

3= =

Similarly for Y : P Y 21<b l 4

3=

So [ ( , )]maxP x y 21<& 0 4

343

169

#= =

Alternate Method:

From the given data since random variables X and Y lies in the interval [ , ]1 1- as from the figure X , Y lies in the region of the square ABCD .Probability for , 1/2max X Y <6 @ : The points for , /max X Y 1 2<6 @ will be inside the region of square AEFG .

So, ,maxP X Y 21<6 @& 0 Area of square

Area ofABCD

AEFG4=

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2 223

23

169

#

#= =

Sol. 11 Option (A) is correct.

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

So, x ei 2=p

xx ei x2=p

^ h & ei i2p

^ h e 2=p-

Sol. 12 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 z

z1 3

1= + +- +

( )( )( )( )

limz z

z z1 3

1 122 1

z 1= + +

+ - + = ="-

So ( )j f z dz21

Cp # 1=

Sol. 13 Option (D) is correct.

t dtdx x+ t=

dtdx

tx+ 1=

dtdx Px+ Q= (General form)

Integrating factor, IF e e e tlntPdt tdt1

= = = =# #

Solution has the form x IF# Q IF dt C#= +^ h#

x t# ( )( )t dt C1= +#

xt t C22

= +

Taking the initial condition

( )x 1 .0 5=

0.5 C21= +

C 0=

So, xt t22

= x t2& =

Sol. 14 Option (B) is correct.Characteristic equation.

A Il- 0=

52

3ll

- - -- 0=

5 62l l+ + 0=

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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= +

Sol. 15 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= ( )

dxd f x

12 18 6 0<2

2

= - =-

So at ,x 2= ( )f x will be maximum

( )f xmax

( ) ( ) ( )2 9 2 24 2 53 2= - + + 8 36 48 5= - + + 25=

Sol. 16 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 1 3

241

21

=-

=

Sol. 17 Option (A) is correct.Divergence of A in spherical coordinates is given as

A:d ( )r r

r A1r2

2

22= ( )

r rkr1 n

22

22= + ( )

rk n r2 n2

1= + +

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

n 2+ 0= n 2=-


( ) ( )

( )dt

d y tdtdy t

y t2

2

2

+ + ( )td=

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 s

s2 1

2 32=+ +- -

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

then, ( )

dtdy t

( ) ( )sY s y 0L -

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

s s2 1

2 322=

+ +- - +

( )s ss s s s

2 12 3 2 4 2

2

2 2

=+ +

- - + + +

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

ss

s12

11

11

2 2 2=++ =

++ +

+

( )s s11

11

2= + ++

Taking inverse Laplace transform

( )

dtdy t

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

At t 0= +, dtdy

t 0= +

e 0 10= + =


x x x 13 2+ + + 0= ( ) ( )x x x1 12 + + - 0= ( )( )x x1 12+ + 0=or x 1+ 0 1x&= =-and 1x2 + 0 ,x j j&= =- x 1, ,j j=- -


Z is Z 0= where q is around 45c or so.

Thus Z Z 45c= where Z 1<

Y Z Z Z1

451 1 45

cc= = = -

Y [ ]Z1 1> <a

So Y will be out of unity circle.


dxdy e x3= -

dy e dxx3= -

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by integrating, we get

y e K31 x3=- +- , where K is constant.


f 1 .sinx x10 0 82 1= - f 2 10 10 0.6cosx x x2

22 1= - -

Jacobian matrix is given by

J cossin

sincos

xf

xf

xf

xf

x xx x

xx x

1010

1020 10

1

1

1

2

2

1

2

2

2 1

2 1

1

2 1

22

22

22

22= = -

R

T

SSSSS

>

V

X

WWWWW

H

For 0, 1x x1 2= = , J 100

010= > H


( )f x x x2 32= - + ( )f xl x2 2 0= - = x 1= ( )f xm 2=-

( )f xm is negative for x 1= , so the function has a maxima at x 1= .

Sol. 24 Option (A) is correct.Let a signal ( )p x is uniformly distributed between limits a- to a+ .

Variance ps ( )x p x dxa

a 2=-# x a dx2

1a

a 2 :=-#

ax

21

3 a

a3

=-

: D a a6

23

3 2

= =

It means square value is equal to its variance

p rms2 a

3p

2

s= =

prms a3

=


We know that matrix A is equal to product of lower triangular matrix L and

upper triangular matrix U .

A L U= 6 6@ @

only option (D) satisfies the above relation.


Let the given two vectors are

X1 [ , , ]1 1 1= X2 [ , , ]a a1 2=

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Dot product of the vectors

X X1 2$ 1aa

a aX X 1 1 11

T1 2

2

2= = = + +

R

T

SSSS

8

V

X

WWWW

B

Where a 1 /j21

23 2 3p=- + = -

so,

a a1 2+ + 0= ,X X1 2 are orthogonalNote: We can see that ,X X1 2 are not orthonormal as their magnitude is 1!


P xe dxx

0

1= # ( )dx

d x e dx dxx

0

1

0

1x e dxx= - :6 D@# ##

( )e dx1 x

0

1

0

1xex= -6 @ # ( 0)e1

0

1ex

= - -6 @

[ ]e e e1 1 0= - - 1=


Radial vector r x y zi j k= + +t t t

Divergence r4$=

x y z

x y zi j k i j k:22

22

22= + + + +t t t t t t

c _m i

xx

yy

zz

22

22

22= + + 1 1 1= + + 3=

Sol. 29 Option (C) is correct.No of white balls 4= , no of red balls 3=If first removed ball is white then remaining no of balls 6(3 ,3white red)=we have 6 balls, one ball can be choose in C6

1 ways, since there are three red balls so probability that the second ball is red is

P CC

31

61= 6

3= 21=

Sol. 30 Option (B) is correct.Let eigen vector

X x x x1 2 3T= 8 B

Eigen vector corresponding to 11l = A I X1l-8 B 0=

xxx

000

110

022

1

2

3

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

000

=

R

T

SSSS

V

X

WWWW

x2 0= x x2 02 3+ = x 03& = (not given in the option)

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Eigen vector corresponding to 22l = A I X2l-8 B 0=

xxx

100

100

021

1

2

3

-R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

000

=

R

T

SSSS

V

X

WWWW

x x1 2- + 0= 2 0x3 = x 03& = (not given in options.)Eigen vector corresponding to 33l = A I X3l-8 B 0=

xxx

200

11

0

020

1

2

3

--

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

000

=

R

T

SSSS

V

X

WWWW

x x2 1 2- + 0= x x22 3- + 0=

Put ,x x1 21 2= = and x 13 =So Eigen vector

X xxx

1

2

3

=

R

T

SSSS

V

X

WWWW

121

1 2 1 T= =

R

T

SSSS

8

V

X

WWWW

B


8dtd x

dtdx x62

2

+ + 0=

Taking Laplace transform (with initial condition) on both sides

( ) ( ) ' ( ) [ ( ) ( )] ( )s X s sx x sX s x X s0 0 6 0 82 - - + - + 0= ( ) ( ) [ ( ) ] ( )s X s s sX s X s1 0 6 1 82 - - + - + 0= ( ) [ ]X s s s s6 8 62 + + - - 0=

( )X s ( )

( )s s

s6 8

62=+ ++

By partial fraction

( )X s s s22

41= + - +

Taking inverse Laplace transform

( )x t ( )e e2 t t2 4= -- -

Sol. 32 Option (C) is correct.Set of equations

x x x x2 41 2 3 4+ + + 2= .....(1)

x x x x3 6 3 121 2 3 4+ + + 6= .....(2)

or ( )x x x x3 2 41 2 3 4+ + + 3 2#=Equation (2) is same as equation(1) except a constant multiplying factor of 3. So infinite (multiple) no. of non-trivial solution exists.


Function ( )f t sint

t= sinct= has a maxima at 0t = as shown below

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Let the matrix is A ac

bd= > H

Trace of a square matrix is sum of its diagonal entries

Trace A a d= + 2=-Determinent ad bc- 35=-Eigenvalue A Il- 0=

a

cb

dl

l-

- 0=

( )( )a d bcl l- - - 0= ( ) ( )a d ad bc2l l- + + - 0= ( ) ( )2 352l l- - + - 0= 2 352l l+ - 0= ( )( )5 7l l- + 0= ,1 2l l ,5 7= -

Sol. 35 Option (A) is correct.Given constraints x y> 2 and y x> 2

Limit of y : y 0= to y 1=Limit of x : x y2= to x y x y2 &= =So volume under ( , )f x y

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V ( , )f x y dx dyx y

x y

y

y

0

1

2=

=

=

=

=##

Sol. 36 Option (B) is correct.No of events of at least two people in the room being born on same date Cn

2=three people in the room being born on same date Cn

3=Similarly four for people Cn

4=

Probability of the event, 0.5N

C C C C N 7n n n n

n2 3 4 &$ $ g$ =

Sol. 37 Option ( ) is correct.Assume a Cubic polynomial with real Coefficients

( )P x a x a x a x a03

13

2 3= + + + , , ,a a a a0 1 2 3 are real

' ( )P x a x a x a3 202

1 2= + + '' ( )P x a x a6 20 1= + ''' ( )P x 6a0= ( )P xiv 0=

Sol. 38 Option (D) is correct.An iterative sequence in Newton-Raphson’s method is obtained by following expression

xk 1+ ' ( )( )

xf xf x

kk

k= -

( )f x x 1172= - ' ( )f x x2=So ( )f xk x 117k

2= - ' ( )f xk x2 2 117k #= =

So xk 1+ 117x xx

2kk

k2

= - - x x x21 117

k kk

= - +: D

Sol. 39 Option (D) is correct.Equation of straight line

y 2- ( )x2 00 2 0= -- -

y 2- x=-

F dl$ [( ) ( ) ] [ ]x xy y xy dx dy dza a a a ax y x y z2 2= + + + + +t t t t t

( ) ( )x xy dx y xy dy2 2= + + +Limit of x : 0 to 2Limit of y : 2 to 0

F dl$# ( ) ( )x xy dx y xy dy2 2

2

0

0

2= + + +##

Line y 2- x=- dy dx=-

So F dl$# [ ( )] ( )x x x dx y y y dy2 22

0

2 2

2

0= + - + + -# #

xdx y dy2 22

0

0

2= + ## x y2 2 2 2

2

0

2 2

2

0

= +: ;D E 4 4= - 0=

Sol. 40 Option (C) is correct.X is uniformly distributed between 0 and 1So probability density function

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( )f XX 1, 0 1

0,

x

otherwise

< <= )

So, { }E X3 ( )X f X dxX3

0

1= # ( )X dx13

0

1= #

X4

4

0

1

= : D 41=

Sol. 41 Option (D) is correct.According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation.Characteristic equation

( )a I Pl l= - 2 1 03 2l l l= + + + =Matrix P satisfies above equation

P P P I23 2+ + + 0= I ( )P P P23 2=- + +Multiply both sides by P 1-

P 1- ( 2 )P P I2=- + +

Sol. 42 Option (A) is correct.Rank of a matrix is no. of linearly independent rows and columns of the matrix.Here Rank ( )Q 4r =So Q will have 4 linearly independent rows and flour independent columns.

Sol. 43 Option (B) is correct.Given function

( )f x ( )x 42 2= - ' ( )f x ( )x x2 4 22= -To obtain minima and maxima

'( )f x 0= ( )x x4 42 - 0= ,x x0 4 02= - = x 2& !=So, x 0, 2, 2= + - '' ( )f x ( ) ( )x x x4 2 4 42= + - x12 162= -

For , '' ( ) ( )x f0 0 12 0 16 16 0<2= = - =- (Maxima) , '' ( ) ( )x f2 2 12 2 16 32 0>2=+ = - = (Minima) , '' ( ) ( )x f2 2 12 2 16 32 0>2=- - = - - = (Minima)So ( )f x has only two minima

Sol. 44 Option (A) is correct.An iterative sequence in Newton-Raphson method can obtain by following expression

xn 1+ ' ( )( )

xf xf x

nn

n= -

We have to calculate x1, so n 0=

x1 ' ( )( )

xf xf x

00

0= - , Given x 10 =-

( )f x0 1e e1x 10= - = -- .0 63212=- ' ( )f x0 e ex 10= = - .0 36787=

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So, x1 ( . )( . )

10 367870 63212=- - -

.1 1 71832=- + .0 71832=


'A ( )A A AT T1= - ( )A A AT T1 1= - - A I1= -

Put 'A A I1= - in all option.

option (A) 'AA A A= AA A1- A= A A= (true)

option (B) ( ')AA 2 I= ( )AA I1 2- I= ( )I 2 I= (true)

option (C) 'A A I= A IA1- I= I I= (true)

option (D) 'AA A 'A= AA IA1- 'A A= =Y (false)


dtdx ( )e u tt2= -

x ( )e u t dtt2= -# e dtt2

0

1

= -#

( )f t dt0

1

= # ,

t .01= sFrom trapezoid rule

( )f t dtt

t nh

0

0+# ( ) (. )h f f2 0 01= +6 @

( )f t dt0

1# . e e2

01 .0 02= + -6 @, .h 01=

.0099=

Sol. 47 Option (B) is correct.P is an orthogonal matrix So PP IT =

Let assume P cossin

sincos

qq

qq=

-> H

PX cossin

sincos x x T

1 2

qq

qq=

-> 8H B

cossin

sincos

xx

1

2

qq

qq=

-> >H H

cos sinsin cos

x xx x

1 2

1 2

q qq q=-+> H

PX ( ) ( )cos sin sin cosx x x x1 22

1 22q q q q= - + +

x x12

22= +

PX X=


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x x x xn1 2Tg= 8 B

V xxT=

xx

x

xx

xn n

1

2

1

2

h h=

R

T

SSSSS

R

T

SSSSS

V

X

WWWWW

V

X

WWWWW

So rank of V is n .

Sol. 49 Option ( ) is correct.


Given z

dz1C

2+#

( )( )z i z idz

C

= + -#

Contour z i2- 1=

P(0, 1) lies inside the circle 1z i2- = and ( , )P 0 1 does not lie.

So by Cauchy’s integral formula

z

dz1C

2+# 2 ( )

( )( )limi z i

z i z i1

z ip= - + -"

limi z i2 1z i

p= +" i i2 2

1#p= p=


Sol. 52 Option (C) is correct.Probability of occurrence of values 1,5 and 6 on the three dice is

( , , )P 1 5 6 ( ) ( ) ( )P P P1 5 6=

41

81

41

# #= 1281=

In option (A)

( , , )P 3 4 5 ( ) ( ) ( )P P P3 4 5= 81

81

81

# #= 5121=

In option (B)

( , , )P 1 2 5 ( ) ( ) ( )P P P1 2 5= 41

81

81

# #= 2561=


det x xy x

x yy y

$

$

$

$> H ( )( ) ( )( )x x y y x y y x: : : := -

0= only when x or y is zero


Sol. 55 Option (C) is correct.For characteristic equation

3

12

0l

l- -- -> H 0=

or ( )( )3 2l l- - - + 0= ( )( )1 2l l+ + 0=According to Cayley-Hamiliton theorem

( )( )A I A I2+ + 0=

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Sol. 56 Option (A) is correct.According to Cayley-Hamiliton theorem

( )( )A I A I2+ + 0=or A A I3 22 + + 0=or A2 ( )A I3 2=- +or A4 ( ) ( )A I A A I3 2 9 12 42 2= + = + + 9( 3 2 ) 12 4A I A I= - - + + 15 14A I=- - A8 ( )A I A A15 14 225 420 1962 2= - - = + + 225( 3 2 ) 420 196A I A I= - - + + 255 254A I=- - A9 255 254A A2=- - 255( 3 2 ) 254A I A=- - - - A I511 510= +

Sol. 57 Option (D) is correct.Volume of the cone

V R Hh dh1

H 2

0

2p= -b l#

Solving the above integral

V R H31 2p=

Solve all integrals given in option only for option (D)

rH Rr dr2 1

R

0

2p -a k# R H31 2p=


Sol. 59 Option (C) is correct.By throwing dice twice 6 6 36# = possibilities will occur. Out of these sample space consist of sum 4, 8 and 12 because /r 4 is an integer. This can occur in following way :(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2) and (6, 6)

Sample Space 9=Favourable space is coming out of 8 5=

Probability of coming out 8 95=



Sol. 62 Option (A) is correct.Matrix equation PX q= has a unique solution if

( )Pr ( )rr=Where ( )P "r rank of matrix P ( )r "r rank of augmented matrix [ ]P

r :P q= 8 B

Sol. 63 Option (D) is correct.for two random events conditional probability is given by

( )probability P Q+ ( ) ( )probability probabilityP Q=

( )Qprobability ( )

( )P

P Q1

probabilityprobability +

#=

so ( )P Qprobability + ( )Pprobability#

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S x dx3

1= 3 -# x

22

1= -

3-

: D 21=


We have ( )x to ( )x t3=-or ( ) 3 ( )x t x t+o 0=A.E. D 3+ 0=Thus solution is ( )x t C e t

13= -

From ( )x x0 0= we get C1 x0=Thus ( )x t x e t

03= -


For eigen value l 2=-

( )

( )( )

xxx

3 200

22 2

0

21

1 2

1

2

3

- - -- - -

- -

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

000

=

R

T

SSSS

V

X

WWWW

xxx

500

200

211

1

2

3

-R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

000

=

R

T

SSSS

V

X

WWWW

x x x5 21 2 3- + 0=


C11 ( )2 3 5= - - = C21 ( ( ))0 3 3=- - - =- C31 ( ( ))1 1= - - = R ( )C C C1 2 211 21 31= + + 5 6 2= - + 1=

Sol. 68 Option (B) is correct.If the toss produces head, then for exactly two head in three tosses three tosses there must produce one head in next two tosses. The probability of one head in two tosses will be 1/2.


We have ( )f x x e x2= -

or ' ( )f x xe x e2 x x2= -- -

( )xe x2x= --

'' ( )f x ( )x x e4 2 x2= - + -

Now for maxima and minima, ' ( )f x 0= ( )xe x2x -- 0=or x ,0 2=at x 0= '' ( )f 0 1( )ve= +at x 2= '' ( )f 2 2 ( )e ve2=- --

Now '' ( )f 0 1= and '' ( )f e2 2 0<2=- - . Thus x 2= is point of maxima


u4 x y

ui j22

22= +t t

c m xu

yui j

22

22= +t t x yi j3

2= +t t

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At (1, 3) magnitude is u4 x y322

2= + b l 1 4= + 5=


( )dtd x

dtdx x t3 22

2

+ + 5=

Taking Laplace transform on both sides of above equation.

( ) ( ) ( )s X s sX s X s3 22 + + s5=

( )X s ( )s s s3 2

52=+ +

From final value theorem

( )limx tt"3

( )limX ss 0

="

( )

lim ss s s3 2

5s 0 2=

+ +" 2

5=

***********

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GATE SOLVED PAPER - EE - mywbut.com chapter wise solved papers... · Q. 4 The curl of the gradient of the scalar field defined by Vxy yz zx=+23 422+2 is (A) ... is a continuous function