engineering mathematics with solutions

CHAPTER 1ENGINEERING MATHEMATICS

GATE Previous Year Solved Paper By RK Kanodia & Ashish MuroliaPublished by: NODIA and COMPANY ISBN: 9788192276243Visit us at: www.nodia.co.in

YEAR 2012 ONE MARK

MCQ 1.1 Two independent random variables X and Y are uniformly distributed in the interval ,1 16 @. The probability that ,max X Y6 @ is less than /1 2 is(A) /3 4 (B) /9 16

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

MCQ 1.2 If ,x 1= then the value of xx is(A) e /2 (B) e /2

(C) x (D) 1

MCQ 1.3 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

C# is

(A) 2 (B) 1

(C) 1 (D) 2

MCQ 1.4 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

MCQ 1.5 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+

MCQ 1.6 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

ENGINEERING MATHEMATICS CHAP 1


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MCQ 1.7 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

MCQ 1.8 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

MCQ 1.9 Consider the differential equation

( ) ( )

( )dt

d y tdt

dy ty t22

2

+ + ( )t= 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

MCQ 1.10 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 +

MCQ 1.11 With K as a constant, the possible solution for the first order differential

equation dxdy e x3= is

(A) e K31 x3 + (B) e K3

1 x3 +

(C) e K31 x3 + (D) e K3 x +

MCQ 1.12 A point Z has been plotted in the complex plane, as shown in figure below.

The plot of the complex number Y Z1= is

CHAP 1 ENGINEERING MATHEMATICS PAGE 3


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YEAR 2011 TWO MARKS

MCQ 1.13 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 6 022 2 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

MCQ 1.14 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=

MCQ 1.15 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



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MCQ 1.16 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

MCQ 1.17 The two vectors [1,1,1] and [ , , ]a a1 2 where a j21

23= +c m, are

(A) Orthonormal (B) Orthogonal

(C) Parallel (D) Collinear

YEAR 2010 ONE MARK

MCQ 1.18 The value of the quantity P , where P xe dxx0

1= # , is equal to

(A) 0 (B) 1

(C) e (D) 1/e

MCQ 1.19 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

MCQ 1.20 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

MCQ 1.21 At t 0= , the function ( ) sinf t tt= has

(A) a minimum (B) a discontinuity

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

MCQ 1.22 An eigenvector of P 100

120

023

=

J

L

KKK

N

P

OOO is

(A) 1 1 1 T8 B (B) 1 2 1 T8 B(C) 1 1 2 T8 B (D) 2 1 1 T8 B



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MCQ 1.23 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= +

MCQ 1.24 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

YEAR 2009 ONE MARK

MCQ 1.25 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

MCQ 1.26 ( , )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 dxdyx y

x

y x

y 11

22 =

=

=

= ##

(C) ( , )f x y dxdyx

x

y

y

0

1

0

1

=

=

=

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

x y

y

y x

00 =

=

=

= ##

MCQ 1.27 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

MCQ 1.28 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



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

MCQ 1.29 Let x 117 02 = . The iterative steps for the solution using Newton-Raphons method is given by

(A) x x x21 117

k kk

1= ++ b l (B) x x x117k k k1= +(C) x x x117k k

k1= + (D) x x x x2

1 117k k k

k1= ++ b l

MCQ 1.30 ( , ) ( ) ( )x y x xy y xyF a ax y2 2= + + +t t . Its 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

MCQ 1.31 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

MCQ 1.32 The characteristic equation of a (3 3# ) matrix P is defined as ( )a I P 2 1 03 2 = = + + + =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 + +

MCQ 1.33 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

YEAR 2008 TWO MARKS

MCQ 1.34 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



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MCQ 1.35 Equation e 1 0x = is required to be solved using Newtons method with an initial guess x 10 = . Then, after one step of Newtons method, estimate x1 of the solution will be given by(A) 0.71828 (B) 0.36784

(C) 0.20587 (D) 0.00000

MCQ 1.36 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=

MCQ 1.37 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

MCQ 1.38 Let P be a 2 2# real orthogonal matrix and x is a real vector [ ]x ,x1 2 T 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

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

YEAR 2007 ONE MARK

MCQ 1.39 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

MCQ 1.40 The differential equation dtdx x1= - is discretised using Eulers 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) /2

(C) (D) 2



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MCQ 1.41 The value of ( )z

dz1

C2+

# where C is the contour /z i 2 1 = is(A) i2 (B)

(C) tan z1- (D) tani z1 -

MCQ 1.42 The integral ( )sin cost d21

0

2

# equals(A) sin cost t (B) 0

(C) ( / )cos t1 2 (D) (1/2)sin t

MCQ 1.43 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

MCQ 1.44 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

MCQ 1.45 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



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Statement for Linked Answer Question 46 and 47.Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix

A32

20=

= G

MCQ 1.46 A satisfies the relation(A) A I A3 2 01+ + =- (B) 2 2 0A A I2+ + =

(C) ( )( )A I A I2+ + (D) ( ) 0exp A =

MCQ 1.47 A9 equals(A) 511 510A I+ (B) 309 104A I+

(C) 154 155A I+ (D) ( )exp A9

YEAR 2006 TWO MARKS

MCQ 1.48 The expression ( / )V R h H dh1H 2 2

0= # for the volume of a cone is equal

to

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

0 # (B) ( / )R h H dh1R 2 2

0 #

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

0 # (D) rH R

r dr2 1R 2

0 ` j#

MCQ 1.49 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

MCQ 1.50 Two fair dice are rolled and the sum r of the numbers turned up is considered

(A) ( )Pr r 661> =

(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) 181=

Statement for Linked Answer Question 51 and 52.

, ,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.



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MCQ 1.51 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)

424

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

MCQ 1.52 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

MCQ 1.53 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

MCQ 1.54 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)

MCQ 1.55 If S x dx31

=3 -# , then S has the value

(A) 31 (B)

41

(C) 21 (D) 1

MCQ 1.56 The solution of the first order DE '( ) ( )x t x t3= , (0)x x0= is(A) ( )x t x e t0 3= - (B) ( )x t x e0 3= -

(C) ( )x t x e /0 1 3= - (D) ( )x t x e0 1= -



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YEAR 2005 TWO MARKS

MCQ 1.57 For the matrix p300

22

0

211

=

R

T

SSSS

V

X

WWWW, 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)

321

R

T

SSSS

V

X

WWWW

(C) 12

3

R

T

SSSS

V

X

WWWW (D)

250

R

T

SSSS

V

X

WWWW

MCQ 1.58 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 18 B(C) 2 0 18 B (D) /2 1 1 28 B

MCQ 1.59 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) 81 (B)

21

(C) 83 (D)

43

MCQ 1.60 For the function ( )f x x e x2= - , the maximum occurs when x is equal to(A) 2 (B) 1

(C) 0 (D) 1

MCQ 1.61 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

MCQ 1.62 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 1.1 Option (B) is correct.Probability density function of uniformly distributed variables X and Y is shown as

[ ( , )]maxP x y 21



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So, ,maxP X Y 21



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C 0=

So, xt t22

= x t2& =

SOL 1.5 Option (B) is correct.Characteristic equation.

A I 0=

52

3

0=

5 62 + + 0=

5 62 + + 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 1.6 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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P 21

1 41 3

2=

=

SOL 1.8 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 1.9 Option (D) is correct.

( ) ( )

( )dt

d y tdtdy t

y t2

2

2

+ + ( )t=

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 s

s2 1

2 32= + +

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 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 s1

11

12= + + +

By taking inverse Laplace transform

( )

dtdy t

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

At t 0= +, dtdy

t 0= + e 0 10= + =



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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=

SOL 1.11 Option (A) is correct.

dxdy e x3=

dy e dxx3=

by integrating, we get

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


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

Thus Z Z 45c= where Z 1



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f 1 .sinx x10 0 82 1= f 2 10 10 0.6cosx x x22 2 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

SOL 1.14 Option (C) is correct.

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

Variance p ( )x p x dxa

a 2=# x a dx21a

a 2:=

#

ax

21

3 a

a3

=

: D a a62 33 2

= =

It means square value is equal to its variance

p rms2 a3p2

= =

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.



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Let the given two vectors are

X1 [ , , ]1 1 1= X2 [ , , ]a a1 2=Dot product of the vectors

X X1 2$ 1aa

a aX X 1 1 11

T1 2

2

2= = = + +

R

T

SSSS

8V

X

WWWW

B

Where a 1 /j21

23 2 3= + =

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 dxx0

1= #

( )dxd x e dx dxx

0

1

0

1x e dxx= :6 D@# ## ( )e dx1 x

0

1

0

1xex= 6 @ # ( 0)e10

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 tc _m i

xx

yy

zz

22

22

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

SOL 1.20 Option (C) is correct.No of white balls 4= , no of red balls 3=



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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=


Function ( )f t sintt= sinct= has a maxima at 0t = as shown below

SOL 1.22 Option (B) is correct.Let eigen vector

X x x x1 2 3 T= 8 BEigen vector corresponding to 11 = A I X18 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)Eigen vector corresponding to 22 = A I X28 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



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x x1 2 + 0= 2 0x3 = x 03& = (not given in options.)Eigen vector corresponding to 33 = A I X38 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

8V

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 1.24 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.



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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 I 0=

a

cb

d

0=

( )( )a d bc 0= ( ) ( )a d ad bc2 + + 0= ( ) ( )2 352 + 0= 2 352 + 0= ( )( )5 7 + 0= ,1 2 ,5 7=

SOL 1.26 Option (A) is correct.Given constraints x y> 2 and y x> 2

h

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

V ( , )f x y dx dyx y

x y

y

y

0

1

2=

=

=

=

= ##

SOL 1.27 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=



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

( )P x a x a x a x a0 3 1 3 2 3= + + + , , ,a a a a0 1 2 3 are real ' ( )P x a x a x a3 20 2 1 2= + + '' ( )P x a x a6 20 1= + ''' ( )P x 6a0= ( )P xiv 0=

SOL 1.29 Option (D) is correct.An iterative sequence in Newton-Raphsons method is obtained by following expression

xk 1+ ' ( )( )

xf xf x

kk

k=

( )f x x 1172= ' ( )f x x2=So ( )f xk x 117k2= ' ( )f xk x2 2 117k #= =

So xk 1+ 117x x

x2k k

k2

= x x x21 117

k kk

= +: DSOL 1.30 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 22

0

0

2= + + +##

Line y 2 x= dy dx=

So F dl$# [ ( )] ( )x x x dx y y y dy2 220

2 2

2

0= + + + # #

xdx y dy2 22

0

0

2= + ##



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x y2 2 2 22

0

2 2

2

0

= +: ;D E 4 4= 0=SOL 1.31 Option (C) is correct.

X is uniformly distributed between 0 and 1So probability density function

( )f XX 1, 0 1

0,

x

otherwise

< 2= = = (Minima)So ( )f x has only two minima

SOL 1.35 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=

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)




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dtdx ( )e u tt2=

x ( )e u t dtt2= # e dtt20

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 dt

0

1# . e e201 .0 02= + 6 @, .h 01=

.0099=

SOL 1.38 Option (B) is correct.P is an orthogonal matrix So PP IT =

Let assume P cossin

sincos

=

> H PX

cossin

sincos x x

T1 2

=

> 8H B

cossin

sincos

xx

1

2

=

> >H H cos sinsin cosx xx x1 21 2 =+> H

PX ( ) ( )cos sin sin cosx x x x1 2 2 1 2 2 = + +

x x12 22= +

PX X=


x x x xn1 2 Tg= 8 B V xxT=

xx

x

xx

xn n

1

2

1

2

h h=

R

T

SSSSSS

R

T

SSSSSS

V

X

WWWWWW

V

X

WWWWWW

So rank of V is n .

SOL 1.40 Option ( ) is correct.


Given z

dz1C

2+# ( )( )z i z i

dz

C

= + #



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Contour z i2 1=

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

So by Cauchys integral formula

z

dz1C

2+# 2 ( )( )( )limi z i z i z i

1z i

= + "

limi z i21

z i= +" i i2 2

1#= =


SOL 1.43 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 1.46 Option (C) is correct.For characteristic equation

3

12

0

> H 0=

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



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

SOL 1.47 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 1.48 Option (D) is correct.Volume of the cone

V R Hh dh1

H 2

0

2= b l#

Solving the above integral

V R H31 2=

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

rH Rr dr2 1

R

0

2 a k# R H31 2=


SOL 1.50 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=





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SOL 1.53 Option (A) is correct.Matrix equation PX q= has a unique solution if ( )P ( )r=Where ( )P " rank of matrix P ( )r " rank of augmented matrix [ ]P r :P q= 8 B

SOL 1.54 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#


S x dx31

= 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 t1 3=

From ( )x x0 0= we get C1 x0=Thus ( )x t x e t0 3=


For eigen value 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= =



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

SOL 1.59 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

engineering mathematics with solutions

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