Engineering Mathematics With Solutions

CHAPTER 1 ENGINEERING MA MATHEMA THEMATICS TICS

YEAR 2012 MCQ 1.1

MCQ 1.2

MCQ 1.3

ONE MARK

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

6 @

6 @

(C) 1/4

(D) 2/3

If x = − 1, then the value of x x is (A) e − π/2

(B) e π/2

(C) x

(D) 1 1

2 . If C is a counter clockwise path in the z -plane z + 1 z + 3 such that z + 1 = 1, the value of 1 f (z) dz is 2π j C (A) − 2 (B) − 1 Given f (z ) =

−

#

(C) 1 MCQ 1.4

(D) 2

With initial condition x (1) = 0.5, the solution of the differential equation t dx + x = t , is dt

(A) x = t − 1 2

(B) x = t 2 − 1 2

2

(C) x = t 2

(D) x = t 2

YEAR 2012

MCQ 1.5

Given that

TWO MARKS

A

=

>

H

−5 −3

2

0

and I =

(A) 15A + 12I (C) 17A + 15I MCQ 1.6

> H

1 0 , the value of A3 is 0 1 (B) 19A + 30I (D) 17A + 21I

The maximum value of f (x) = x 3 − 9x 2 + 24x + 5 in the interval [1, 6] is (A) 21 (B) 25 (C) 41

(D) 46

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MCQ 1.7

ENGINE ERING MATHEMATICS

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

MCQ 1.8

(D) 3/4

The direction of vector A is radially outward from the origin, with A = kr n . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is (A) − 2 (B) 2 (C) 1

MCQ 1.9

CHAP 1

(D) 0

Consider the differential equation d 2 y (t ) dy (t ) dy +2 + y (t ) = δ (t ) with y (t ) t = − 2 and 2 =0 dt

dt

−

dt t = 0

=0

−

dy

The numerical value of is dt t = 0 (A) − 2

(B) − 1

(C) 0

(D) 1

+

YEAR 2011 MCQ 1.10

ONE MARK

Roots of the algebraic equation x 3 + x2 + x + 1 = 0 are (A) (+ 1, + j, − j ) (B) (+ 1 , − 1 , + 1) (C) (0 , 0, 0)

MCQ 1.11

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

MCQ 1.12

(D) (− 1, + j, − j )

dy = e −3x is dx

(A) − 1 e−3x + K 3

(B) − 1 e3x + K 3

(C) − 1 e−3x + K 3

(D) − 3e−x + K

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

The plot of the complex number Y = 1 is Z



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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

MCQ 1.8

(D) 3/4

The direction of vector A is radially outward from the origin, with A = kr n . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is (A) − 2 (B) 2 (C) 1

MCQ 1.9

CHAP 1

(D) 0

Consider the differential equation d 2 y (t ) dy (t ) dy +2 + y (t ) = δ (t ) with y (t ) t = − 2 and 2 =0 dt

dt

−

dt t = 0

=0

−

dy

The numerical value of is dt t = 0 (A) − 2

(B) − 1

(C) 0

(D) 1

+

YEAR 2011 MCQ 1.10

ONE MARK

Roots of the algebraic equation x 3 + x2 + x + 1 = 0 are (A) (+ 1, + j, − j ) (B) (+ 1 , − 1 , + 1) (C) (0 , 0, 0)

MCQ 1.11

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

MCQ 1.12

(D) (− 1, + j, − j )

dy = e −3x is dx

(A) − 1 e−3x + K 3

(B) − 1 e3x + K 3

(C) − 1 e−3x + K 3

(D) − 3e−x + K

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

The plot of the complex number Y = 1 is Z



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CHAP 1

ENGINEE RING MATHEMATICS

YEAR 2011 MCQ 1.13

TWO MARKS

Solution of the variables x 1 and x 2 for the following equations is to be obtained by employing the Newton-Raphson iterative method. Equation (1) 10x 2 sin x 1 − 0.8 = 0 Equation (2) 10x 22 − 10x2 cos x 1 − 0.6 = 0 Assuming the initial values are x 1 = 0.0 and x 2 = 1.0 , the jacobian matrix is (A)

(C) MCQ 1.14

PAGE 3

> >

H H

10 − 0.8 0 − 0.6

(B)

0 − 0.8 10 − 0.6

(D)

> H > H 10 0 0 10

10 0 10 − 10

The function f (x) = 2x − x2 − x 3 + 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 b etween 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) a 3 (C) a 2

(B) a 2 (D) a 3



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MCQ 1.16


> H

2 1 is decomposed into a product of a lower 4 −1 triangular matrix [L] and an upper triangular matrix [U ]. The properly decomposed [L] and [U ] matrices respectively are 1 0 1 1 2 0 1 1 (A) and (B) and 4 −1 0 −2 4 −1 0 1 The matrix [A] =

(C) MCQ 1.17

CHAP 1

> H > H > H > H 1 0 2 1 and 4 1 0 −1

(D)

> H > H > H > H 2 0 1 1.5 and 4 −3 0 1

c

m

The two vectors [1,1,1] and [1, a, a 2] where a = − 1 + j 3 , are 2 2 (A) Orthonormal (B) Orthogonal (C) Parallel

(D) Collinear

YEAR 2010 MCQ 1.18

ONE MARK

The value of the quantity P , where P = # xex dx , is equal to 0 (A) 0 (B) 1 1

(C) e MCQ 1.19

(D) 1/e

Divergence of the three-dimensional radial vector field r is (A) 3 (B) 1 /r t (C) ti + tj + k

(D) 3 (ti + tj + kt)

YEAR 2010 MCQ 1.20

MCQ 1.21

MCQ 1.22

TWO MARKS

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

At t = 0 , the function f (t ) = sin t has t (A) a minimum

(B) a discontinuity

(C) a point of inflection

(D) a maximum

J1 1 0N K O An eigenvector of P = K0 2 2O is K0 0 3O (A) − 1 1 1 T L P

(B) 1 2 1 T

8 8

B B

(C) 1 − 1 2

T

8 B 8 B

(D) 2 1 − 1 T



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CHAP 1

MCQ 1.23

MCQ 1.24


PAGE 5

2 dx 8x 0 with initial conditions For the differential equation d x + = 2 +6

dt dt dx x (0) = 1 and = 0 , the solution is dt t = 0

(A) x (t) = 2e− 6t − e− 2 t

(B) x (t) = 2e− 2t − e− 4 t

(C) x (t) =− e− 6t + 2e− 4 t

(D) x (t) = e− 2t + 2e− 4 t

For the set of equations, x1 + 2x2 + x 3 + 4x 4 = 2 and 3x1 + 6x2 + 3x 3 + 12x 4 = 6 . The following statement is true. (A) Only the trivial solution x1 = x2 = x 3 = x 4 = 0 exists (B) There are no solutions (C) A unique non-trivial solution exists (D) Multiple non-trivial solutions exist YEAR 2009

MCQ 1.25

ONE MARK

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 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 y = 1

(A) #

#

y = 1

(C) #

y = 0

x = y

f (x, y ) dxdy

x = y 2

y = 0

MCQ 1.27

TWO MARKS

#

x = 1

x = 0

f (x, y ) dxdy

#

(D) #

x

y = x 2 y=

y = 0

x = 1

x = y 2

f (x, y ) dxdy

#

x = y

x = 0

f (x, y ) dxdy

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

MCQ 1.28

y = 1

(B) #

(D) 16

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 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 1

(D) Will always have an equal number of extrema and zero crossings MCQ 1.29

Let x 2 − 117 = 0 . The iterative steps for the solution using Newton-Raphon’s method is given by (A) xk + 1 = 1 x k + 117 (B) xk + 1 = x k − 117 2 x k x k

b

l

(C) xk + 1 = x k − x k 117 MCQ 1.30

b

(D) xk + 1 = xk − 1 x k + 117 2 x k

F (x, y) = (x2 + xy) at x + (y 2 + xy ) at y . 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 MCQ 1.31

MCQ 1.32

ONE MARKS

X is a uniformly distributed random variable that takes values between 0 and 1. The value of E {X 3} will be

(A) 0

(B) 1/8

(C) 1/4

(D) 1/2

The characteristic equation of a (3 # 3) matrix P is defined as a (λ) = λI − P = λ3 + λ2 + 2λ + 1 = 0 If I denotes identity matrix, then the inverse of matrix P will be (A) (P2 + P + 2I ) (B) (P2 + P + I ) (C) − (P2 + P + I )

MCQ 1.33

l

(D) − (P2 + P + 2I )

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 T

(C) QQ will be invertible (D) QT Q will be invertible YEAR 2008 MCQ 1.34

TWO MARKS

Consider function f (x) = (x 2 − 4) 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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CHAP 1

MCQ 1.35


Equation e x − 1 = 0 is required to be solved using Newton’s method with an initial guess x 0 =− 1. Then, after one step of Newton’s method, estimate x 1 of the solution will be given by (A) 0.71828 (B) 0.36784 (C) 0.20587

MCQ 1.36

(D) 0.00000

A is m # n full rank matrix with m > n and I is identity matrix. Let matrix A' = (AT A) - 1A T , Then, which one of the following statement is FALSE ? (A) AA' A = A (B) (AA') 2 (C) A'A = I

MCQ 1.37

(D) AA' A = A'

A differential equation dx/dt = e- 2t u (t) , 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

MCQ 1.38

PAGE 7

(D) 0.0198

Let P be a 2 # 2 real orthogonal matrix and x is a real vector [x1, x2] T with length x = (x12 + x22) 1/2 . Then, which one of the following statements is correct ? (A) P x # x where at least one vector satisfies P x < x (B) P x # x for all vector x (C) P x $ x where at least one vector satisfies P x > x (D) No relationship can be established between x and P x YEAR 2007

MCQ 1.39

8

ONE MARK

B

T

x n is an n-tuple nonzero vector. The n # n matrix x = x1 x2 T V = xx (A) has rank zero (B) has rank 1 g

(C) is orthogonal

(D) has rank n

YEAR 2007 MCQ 1.40

TWO MARKS

1 - x The differential equation dx dt = τ is discretised using Euler’s numerical integration method with a time step 3 T > 0 . What is the maximum permissible value of 3 T to ensure stability of the solution of the corresponding discrete time equation ? (A) 1 (B) τ /2

(C)

τ

(D) 2τ



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MCQ 1.41


The value of

where C is the contour # (1 dz + z ) 2

C

(A) 2πi

(B)

(C) tan - 1 z MCQ 1.42

z − i /2 = 1 is

π

(D) πi tan - 1 z

The integral 1 2π (A) sin t cos t

# sin (t − 2π

0

τ ) cos τd τ equals

(B) 0

(C) (1/2) cos t MCQ 1.43

CHAP 1

(D) (1/2) sin t

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 < x, x > < x, y > det < y, x > < y, y > (A) is zero when x and y are linearly independent

G

=

(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 , T T where b = 0 1 0 and x = x1 x2 x 3 are three dimensional vectors. The 3 # 3 matrix M of this operations satisfies

8 B

Rx V S 1W L (x) = M Sx 2 W SSx 3 WW T X

8

B

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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CHAP 1


PAGE 9

Statement for Linked Answer Question 46 and 47. Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix −3 2 A= −2 0 A satisfies the relation (A) A + 3I + 2A - 1 = 0 (B) A2 + 2A + 2I = 0

= G

MCQ 1.46

MCQ 1.47

(C) (A + I) (A + 2I )

(D) exp (A) = 0

A9 equals (A) 511A + 510I

(B) 309A + 104I

(C) 154A + 155I

(D) exp (9A)

YEAR 2006 MCQ 1.48

The expression V = to

TWO MARKS

#

H

πR

0

2

(1 − h/H) 2 dh for the volume of a cone is equal

# R (1 − h/H) dr (C) # 2 rH (1 − r/R) dh (A)

R

π

2

2

0

H

π

0

MCQ 1.49

R

π

2

2

0

R

π

2

0

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

MCQ 1.50

# R (1 − h/H) dh (D) # 2 rH `1 − r j dr R (B)

2 /17

(D) 0

Two fair dice are rolled and the sum r of the numbers turned up is considered (A) Pr (r > 6 ) = 1 6 (B) Pr (r /3 is an integer) = 5 6 (C) Pr (r = 8 ; r /4 is an integer) = 5 9 (D) Pr (r = 6 ; r /5 is an integer) = 1 18

Statement for Linked Answer Question 51 and 52.

R− 10 VT R− 2VT R 2 VT S W S W S W W W S S P = 1 , Q = − 5 , R = S− 7 W are three vectors. SS 3 WW SS 9 WW SS 12 WW T X T X T X



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PAGE 10

MCQ 1.51

MCQ 1.52


CHAP 1

An orthogonal set of vectors having a span that contains P,Q,Ris R− 6 V R 4 V R− 4 V R 5 V R 8 V S W S W S W S W S W (A) S− 3 W S− 2 W (B) S 2 W S 7 W S 2 W SS− 6 WW SS 3 WW SS 4 WW SS− 11WW SS− 3WW RT 6 VX RT− 3VX R 3 V TR 4 VX RT1 V RX5 VT X S W S W S W S W S W S W (C) S 7 W S 2 W S 9 W (D) S 3 W S31W S 3 W SS− 1WW SS− 2WW SS− 4 WW SS11WW SS 3 WW SS 4 WW X T Xvector T X is linearly dependent upon T the X T solution X T X to the previous The Tfollowing problem R8 V R −2 V S W S W (A) S 9 W (B) S− 17 W SS 3 WW SS 30 WW RT4 VX TR 13 V X S W S W (C) S 4 W (D) S 2 W SS 5 WW SS− 3 WW

T X

T

X

YEAR 2005 MCQ 1.53

ONE MARK

In the matrix equation P x = q , which of the following is a necessary condition for the existence of at least on solution for the unknown vector x (A) Augmented matrix [P q ] 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 3

1

-3

dx , then S has the value

(A) − 1 3 (C) 1 2 MCQ 1.56

(B) 1 4 (D) 1

The solution of the first order DE x' (t) =− 3x (t), x (0) = x 0 is (A) x (t ) = x0 e - 3t (B) x (t) = x0 e - 3 (C) x (t) = x0 e - 1/3

(D) x (t) = x0 e - 1



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CHAP 1


PAGE 11

YEAR 2005

MCQ 1.57

MCQ 1.58

TWO MARKS

R3 − 2 2 V S W For the matrix p = S0 − 2 1 W, one of the eigen values is equal to − 2 SS0 0 1 WW T is anXeigen vector ? Which of the following R3 V R− 3V S W S W (A) S− 2 W (B) S 2 W SS 1 WW SS− 1WW RT 1 VX TR2 V X S W S W (C) S− 2 W (D) S 5 W SS 3 WW SS 0 WW T RX T X V S1 0 − 1W If R = S2 1 − 1W, then top row of R - 1 is SS2 3 2 WW X (A) 5 T6 4 (B) 5 − 3 1

8 B 8 B

8 8

(C) 2 0 − 1

B

(D) 2 − 1 1/2

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) 1 (B) 1 8 2 (C) 3 (D) 3 8 4

MCQ 1.60

For the function f (x) = x2 e - x , the maximum occurs when x is equal to (A) 2 (B) 1 (C) 0

(D) − 1 2

MCQ 1.61

MCQ 1.62

2 y For the scalar field u = x + , magnitude of the gradient at the point (1, 2 3 3) is 13 9 (A) (B) 9 2 (C) 5 (D) 9 2

For the equation x'' (t) + 3x' (t) + 2x (t) = 5,the solution x (t ) approaches which of the following values as t ? (A) 0 (B) 5 2 "

3

(C) 5

(D) 10 ***********



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PAGE 12


CHAP 1

SOLUTION

SOL 1.1

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

1 P [max (x, y )] <

&

0

&

0

2 Since X and Y are independent. 1 =P X<1 PY< 1 P [max (x, y )] < 2 2 2 1 = shaded area = 3 P X < 2 4 Similarly for Y : P Y < 1 = 3 2 4 So P [max (x, y )] < 1 = 3 # 3 = 9 2 4 4 16

b b

l l

&

b

lb

l

0

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 max X,Y < 1/2 : The points for max X, Y < 1/2 will be inside the region of square AEFG . GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia

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CHAP 1


PAGE 13

Area of AEFG & 6 @ 2 0 = Area of square ABCD

1 P max X, Y <

So,

4

3 3 # 2 2 = 9 = 2#2 16 SOL 1.2

Option (A) is correct. x=

− 1 = i = cos π + i sin π

2

2

π

x = e i 2

So,

^ h ^e h = e π

x x = e i 2 SOL 1.3

x

i π i

&

2

−

π

2

Option (C) is correct. 1 − 2 f (z ) = z + 1 z + 3 1 f (z) dz = sum of the residues of the poles which lie inside the given 2π j C closed region. C & z + 1 = 1 Only pole z =− 1 inside the circle, so residue at z =− 1 is. (z + 1) (− z + 1) 2 − z + 1 f (z ) = = lim = =1 2 (z + 1) (z + 3) z − 1 (z + 1) (z + 3) 1 So f (z) dz = 1 2π j C

#

"

#

SOL 1.4

Option (D) is correct. t dx + x = t dt dx x 1 + = dt t dx Px Q (General form) + = dt IF = e # = e

Integrating factor, Solution has the form

Pdt

t

= e ln t = t

# ^Q IF hdt + C t = # (1) (t) dt + C

x # IF = x#

# 1dt

#

2 xt = t + C

2 Taking the initial condition x (1) = 0.5 0.5 = 1 + C 2



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PAGE 14


CHAP 1

C = 0 2 xt = t

So, SOL 1.5

2

&

x = t

2

Option (B) is correct. Characteristic equation. A − λI

=0

−5 − λ −3 =0 2 −λ

5λ + λ2 + 6 = 0 2

λ

+ 5λ + 6 = 0

Since characteristic equation satisfies its own matrix, so 2 2 A + 5A + 6 = 0 & A =− 5A − 6I Multiplying with A 3 2 A + 5A + 6A = 0 3 A + 5 (− 5A − 6I ) + 6A = 0 3 A = 19A + 30I SOL 1.6

Option (B) is correct. f (x ) = x3 − 9x2 + 24x + 5 df (x ) = 3x2 − 18x + 24 = 0 dx

&

df (x ) = x2 − 6x + 8 = 0 dx

x = 4 , x = 2 d 2 f (x ) = 6x − 18 dx 2 d 2 f (x ) = 12 − 18 =− 6 < 0 dx 2 So at x = 2, f (x ) will be maximum

For x = 2, f (x )

max

= (2) 3 − 9 (2) 2 + 24 (2) + 5 = 8 − 36 + 48 + 5 = 25

SOL 1.7

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 ,........... 1 1 3 + 1 5 + ..... Probability P = + 2 2 2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia

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CHAP 1


P =

SOL 1.8

1 2

1−1 4

=

2 3

Option (A) is correct. Divergence of A in spherical coordinates is given as 1 2 (r 2 A ) = 1 2 (kr n + 2) d:A = r

r 2 2r = k 2 (n + 2) r n + 1 r

r 2 2r

= k (n + 2) r n − 1 = 0

(given)

n + 2 = 0 n =− 2 SOL 1.9

Option (D) is correct. d 2 y (t ) 2dy (t ) + + y (t ) = δ (t ) dt dt 2

By taking Laplace transform with initial conditions + 2 [sy (s) − y (0)] + Y (s ) = 1 ;s Y (s) − sy ( 0) − dy dt E 6s Y (s) + 2s − 0@ + 26sY (s) + 2@ + Y (s) = 1 2

t = 0

&

2

Y (s) [s2 + 2s + 1] = 1 − 2s − 4 2s 3 Y (s ) = 2− − s + 2s + 1

We know If,

y (t ) dy (t ) dt

then,

L

Y (s )

L

sY (s) − y ( 0)

(− 2s − 3) s +2 (s2 + 2s + 1) 2s2 − 3s + 2s2 + 4s + 2 =− (s2 + 2s + 1) 2 1 1 sY (s) − y ( 0) = s + 2 = s + 2 + (s + 1) (s + 1) (s + 1) 2 1 = 1 + s + 1 (s + 1) 2 By taking inverse Laplace transform dy (t ) = e−t u (t) + te−t u (t ) So,

sY (s) − y ( 0) =

dt

At t = 0+ ,

dy = e 0 + 0 = 1 dt t = 0 +



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PAGE 16

SOL 1.10


Option (D) is correct. x 3 + x 2 + x + 1 = 0 x 2 (x + 1) + (x − 1) = 0

(x + 1) (x 2 + 1) = 0 x + 1 = 0 & x =− 1 x 2 + 1 = 0 & x =− j, j x =− 1, − j, j

or and

SOL 1.11

Option (A) is correct. dy = e −3x dx dy = e−3x dx

by integrating, we get 1 y =− e−3x + K , where K is constant. 3

SOL 1.12

Option (D) is correct.

Z is Z = 0 where

Thus

θ is

around 45c or so.

Z = Z 45c where Z < 1 Y =

1 =

Z

Y > 1 [a

1 1 − 45c = Z Z 45c Z < 1]

So Y will be out of unity circle. SOL 1.13

Option (B) is correct. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 1


PAGE 17

f 1 = 10x2 sin x 1 − 0.8 f 2 = 10x 22 − 10x 2 cos x 1 − 0.6

Jacobian matrix is given by

R2 f S 1 S2x 1 J = S2 f 2 S2x 1 T

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

SOL 1.14

SOL 1.15

V

2 f 1 W 2x 2 W 2 f 2 W = 2x 2 W

> H 10 0 0 10

>

H

10x2 cos x 1 10 sin x 1 10x2 sin x 1 20x2 − 10 cos x 1

X

Option (C) is correct. f (x ) = 2x − x 2 + 3 f l (x ) = 2 − 2x = 0 x = 1 f m (x ) =− 2 f m (x ) is negative for x = 1, so the function has a maxima at x = 1. Option (A) is correct. Let a signal p (x ) is uniformly distributed between limits − a to + a .

Variance

σp

=

# x p (x) dx = # x a

a

2

−a

2

:

−a

3 = 1 x 2a 3

a

: D

1 dx 2a

3 2 = 2a = a

6 3 It means square value is equal to its variance −a

2

2 = σp = a p rms

3

p rms = a

3

SOL 1.16

Option (D) is correct. We know that matrix A is equal to product of lower triangular matrix L and upper triangular matrix U .

6 @6 @

A = L U

only option (D) satisfies the above relation. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 18

SOL 1.17


CHAP 1

Option (B) is correct. Let the given two vectors are X 1 = [1 , 1 , 1] 2 X 2 = [1, a, a ] Dot product of the vectors

R1V S W T X 1 $ X 2 = X1 X 2 = 1 1 1 S a W = 1 + a + a 2 SSa 2WW T X 1 3 a =− + j = 1 −2π/3

8 B

Where

2

2

so, 1 + a + a 2 = 0 X1, X 2 are orthogonal Note: We can see that X1, X 2 are not orthonormal as their magnitude is ! 1 SOL 1.18

Option (B) is correct. P = = =

# xe dx 1

x

0

6 @ 6xe @ #

1 x # e x dx 0 − # 1 d (x) # ex dx dx 0 dx

x 1

− 0

1

0

:

D

6@

x 1

(1) ex dx = (e 1 − 0) − e

0

= e1 − [e1 − e 0] = 1 SOL 1.19

Option (A) is correct. t Radial vector r = xti + ytj + z k

Divergence = 4$ r

m _

t : xt = 2 ti + 2 tj + 2 k i + ytj + z kt 2x 2y 2z

c

= SOL 1.20

2x 2y 2z + + = 2x 2y 2z

i

1+1+1 = 3

Option (C) is correct. No of white balls = 4 , no of red balls = 3 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 1


PAGE 19

If first removed ball is white then remaining no of balls = 6 (3 white, 3 red) we have 6 balls, one ball can be choose in 6 C 1 ways, since there are three red balls so probability that the second ball is red is 6 C 3 1 P = 3 1 = = 6 2 C 1 SOL 1.21

Option (D) is correct. Function f (t )= sin t = sin ct has a maxima at t = 0 as shown below t

SOL 1.22

Option (B) is correct. Let eigen vector

8

B

X = x1 x2 x 3 T Eigen vector corresponding to λ1 = 1 A − λ1 I X = 0

8

B

R0 1 0V Rx 1V R0V S W S W SW S0 1 2W Sx 2W = S0W SS0 0 2WW SSx 3WW SS0WW T X T x X = T0 X 2

x 3 = 0 (not given in the option) Eigen vector corresponding to λ2 = 2 x 2 + 2x 3 = 0

8

&

B

A − λ2 I X = 0

R− 1 1 0V Rx 1V R0V S W S W SW S 0 0 2W Sx 2W = S0W SS 0 0 1WW SSx 3WW SS0WW T X T X TX GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 1

− x1 + x 2 = 0

2x 3 = 0 & x 3 = 0 (not given in options.) Eigen vector corresponding to λ3 = 3 A − λ3 I X = 0

8

B

R− 2 1 0V Rx 1V R0V S W S W SW S 0 − 1 2W Sx 2W = S0W SS 0 0 0WW SSx 3WW SS0WW T X T X TX − 2x + x = 0 1

2

− x2 + 2x 3 = 0

Put x1 = 1, x 2 = 2 and x 3 = 1 So Eigen vector

Rx 1V R1V S W SW X = Sx 2W = S2W = 1 2 1 T SSx 3WW SS1WW T X TX

8 B

SOL 1.23

Option (B) is correct. d2x 6 dx 8x 0 + + = dt dt 2

Taking Laplace transform (with initial condition) on both sides s2 X (s) − sx (0) − x' (0) + 6 [sX (s ) − x (0 )] + 8X (s ) = 0 s2 X (s) − s (1) − 0 + 6 [sX (s) − 1] + 8X (s ) = 0 X (s) [s2 + 6s + 8] − s − 6 = 0 (s + 6) X (s ) = 2 (s + 6s + 8) By partial fraction 2 − 1 X (s ) = s + 2 s + 4 Taking inverse Laplace transform x (t ) = (2e− 2t − e − 4t ) SOL 1.24

Option (C) is correct. Set of equations .....(1) x1 + 2x 2 + x 3 + 4x 4 = 2 3x1 + 6x2 + 3x 3 + 12x 4 = .....(2) 6 or 3 (x1 + 2x2 + x 3 + 4x 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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CHAP 1

SOL 1.25


PAGE 21

Option (C) is correct. Let the matrix is

A =

> H a b c d

Trace of a square matrix is sum of its diagonal entries Trace A = a + d =− 2 Determinent ad − bc =− 35 Eigenvalue A − λI = 0 a − λ b =0 c d − λ

(a − λ) (d − λ) − bc = 0 λ2 − (a + d) λ + (ad − bc) = 0 λ2 − (− 2) λ + (− 35) = 0 λ2 + 2λ − 35 = 0 (λ − 5) (λ + 7) = 0 λ1, λ2 = 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 = y 2 to x2 = y & x = So volume under f (x,y ) V = SOL 1.27

y = 1

# # y = 0

y

x = y

x = y 2

f (x, y ) dx dy

Option (B) is correct. No of events of at least two people in the room being born on same date = n C 2

three people in the room being born on same date = n C 3 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 22


CHAP 1

Similarly four for people = n C 4 Probability of the event, 0.5 $

SOL 1.28

SOL 1.29

C 2 $ n C 3 $ n C 4 gn C n & N = 7 N

n

Option ( ) is correct. Assume a Cubic polynomial with real Coefficients P (x ) = a 0 x3 + a1 x3 + a 2 x + a 3 P' (x ) = 3a 0 x2 + 2a1 x + a2 P'' (x ) = 6a 0 x + 2a 1 P''' (x ) = 6a 0 Piv (x ) = 0

a 0, a 1, a 2, a 3 are real

Option (D) is correct. An iterative sequence in Newton-Raphson’s method is obtained by following expression f (x k ) x k + 1 = x k − f' (x k ) f (x ) = x 2 − 117 f' (x ) = 2x

So

f (x k ) = x k 2 − 117 f' (x k ) = 2x k = 2 # 117

So

SOL 1.30

x k + 1 = x k −

x k 2 − 117 1 117 = xk − x k + 2x k 2 x k

:

D

Option (D) is correct. Equation of straight line 0 2 y − 2 = − (x − 0) 2−0 y − 2 =− x

F $ dl = [(x2 + xy) at x + (y2 + xy) at y][dxat x + dyat y + dzat z] = (x2 + xy) dx + (y2 + xy) dy Limit of x : 0 to 2 Limit of y : 2 to 0

# F $ dl = # (x 2

2

0

Line

+ xy) dx +

# (y 0

2

2

+ xy) dy

y − 2 =− x dy =− dx

So

# F $ dl = #

2

0

=

[x2 + x (2 − x)] dx +

# y

0 2

2

+ (2 − y) y dy

# 2xdx + # 2y dy 2

0

0

2



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CHAP 1


2 2 y 2 x =2 +2

0

0

2

:2D ;2E

SOL 1.31

PAGE 23

= 4−4 = 0

Option (C) is correct. X is uniformly distributed between 0 and 1 So probability density function 1, 0 < x < 1 fX (X ) = 0, otherwise 1 1 So, E {X 3} = # X3 fX (X) dx = # X3 (1) dx

)

0

=

X 4

0

1

:4D

0

=

1 4

SOL 1.32

Option (D) is correct. According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation. Characteristic equation a (λ) = λI − P = λ3 + λ2 + 2λ + 1 = 0 Matrix P satisfies above equation P3 + P2 + 2P + I = 0 I =− (P3 + P2 + 2P ) Multiply both sides by P − 1 P − 1 =− (P2 + P + 2I )

SOL 1.33

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

SOL 1.34

Option (B) is correct. Given function f (x ) = (x 2 − 4) 2 f' (x ) = 2 (x2 − 4) 2x To obtain minima and maxima f' (x ) = 0 4x (x 2 − 4) = 0 x = 0, x 2 − 4 = 0 & x = ! 2 So, x = 0, + 2, − 2 GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 24


CHAP 1

f'' (x ) = 4x (2x) + 4 (x 2 − 4) = 12x 2 − 16 x = 0, f '' (0) = 12 (0) 2 − 16 =− 16 < 0 x =+ 2, f '' (2) = 12 (2) 2 − 16 = 32 > 0 x =− 2, f '' (− 2) = 12 (− 2) 2 − 16 = 32 > 0 So f (x ) has only two minima

For

SOL 1.35

(Maxima) (Minima) (Minima)

Option (A) is correct. An iterative sequence in Newton-Raphson method can obtain by following expression f (x n ) x n + 1 = x n − f' (x n ) We have to calculate x 1 , so n = 0 f (x 0) , Given x 0 =− 1 x 1 = x 0 − f' (x 0) f (x 0) = ex − 1 = e − 1 − 1 =− 0.63212 0

f' (x 0) = ex = e − 1 = 0.36787 0

So,

x 1 =− 1 −

(− 0.63212) (0.36787)

=− 1 + 1.71832 = 0.71832 SOL 1.36

Option (D) is correct. A' = (AT A) − 1 AT = A− 1 (AT ) − 1 AT = A− 1 I Put A' = A− 1 I in all option. option (A)

AA'A = A AA− 1 A = A A = A

option (B)

option (C)

(AA') 2 (AA− 1 I ) 2 (I ) 2 A'A

(true)

= I = I = I

(true)

= I

A− 1 IA = I I = I

option (D)

(true)

AA'A = A' AA− 1 IA = A = Y A'

SOL 1.37

Option (C) is correct. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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CHAP 1


PAGE 25

dx e− 2t u (t ) = dt x =

1

#

# e

e− 2t u (t) dt =

− 2t

dt

0 1

#

= f (t) dt , 0

t = .01 s

From trapezoid rule

#

6

t 0

@

f (t) dt = h f (0) + f (.01)

t0 + nh

2 e0 + e − .02 , h = .01 # 0 1 f (t) dt = .01 2

6

SOL 1.38

@

= .0099

Option (B) is correct. P is an orthogonal matrix So cos θ Let assume P = sin θ cos θ P X = sin θ cos θ = sin θ

> > >

− sin θ x1 x 2 T cos θ

H

x1 cos θ − x 2 sin θ − sin θ x 1 = cos θ x 2 x1 sin θ + x 2 cos θ

x 12 + x 22

=

SOL 1.39

H H8 B H> H >

(x1 cos θ − x 2 sin θ) 2 + (x1 sin θ + x 2 cos θ) 2

P X = P X

PPT = I − sin θ cos θ

= X

Option (D) is correct.

8

B

x = x1 x2 g x n T V = xxT

Rx V S 1W Sx 2W =S W ShW Sx nW T X

So rank of V is n . SOL 1.40

Option ( ) is correct.

SOL 1.41

Option (A) is correct. Given

Rx V S 1W Sx 2W ShW S W Sx n W T X

dz = # # 1 dz (z + i) (z − i) + z 2

C

C



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z − i = 1

Contour

2

P(0, 1) lies inside the circle z − i = 1 and P (0, 1) does not lie. 2 So by Cauchy’s integral formula 1 (z − i ) # 1 dz 2 = 2πi lim z i (z + i) (z − i) + z C = 2πi lim 1 = 2πi # 1 = π 2i z i z + i "

"

SOL 1.42


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 (1 ) P (5 )P (6 ) 1 1 1 1 = # # = 8 128 4 4 In option (A) P (3 , 4 , 5) = P (3) P (4) P (5) 1 1 1 1 = # # = 8 8 8 512 In option (B) P (1 , 2 , 5) = P (1) P (2) P (5) 1 1 1 1 = # # = 8 8 256 4

SOL 1.44

Option (D) is correct. x$x x$y det $ y x y $ y = (x : x) (y : y) − (x : y) (y : x)

>

H

= 0 only when x or y is zero

SOL 1.45


SOL 1.46

Option (C) is correct. For characteristic equation 2 −3 − λ =0 0−λ −1

>

H

or (− 3 − λ) (− λ) + 2 = 0 (λ + 1) (λ + 2) = 0 According to Cayley-Hamiliton theorem GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 27

(A + I) (A + 2I) = 0 SOL 1.47

Option (A) is correct. According to Cayley-Hamiliton theorem (A + I) (A + 2I ) = 0 or A2 + 3A + 2I = 0 or A2 =− (3A + 2I ) or A 4 = (3A + 2I) 2 = (9A2 + 12A + 4I ) 15A − 14I = 9 (− 3 A − 2I) + 12A + 4I =− 8 2 2 A = (− 15A − 14I) = 225A + 420A + 196 255A − 254I = 225 (− 3A − 2I) + 420A + 196I =− 9 2 A =− 255A − 254A =− 255 (− 3A − 2I) − 254A = 511A + 510I

SOL 1.48

Option (D) is correct. Volume of the cone V =

#

H

0

b

l

h 2 dh πR 2 1 − H

Solving the above integral 1 V = πR2H 3 Solve all integrals given in option only for option (D) # 0 R 2πrH 1 − Rr 2 dr = 13 πR2H

a

k

SOL 1.49


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 5 Probability of coming out 8 = 9

SOL 1.51


SOL 1.52

Option ( ) is correct. GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


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PAGE 28

SOL 1.53


Option (A) is correct. Matrix equation P X = q has a unique solution if ρ (P ) = ρ (r ) Where ρ (P ) " rank of matrix P ρ (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 (P) probability (Q ) probability (P + Q ) probability (Q ) = #1 probability (P ) so

SOL 1.55

probability (P + Q ) # probability (P )

Option (C) is correct. S =

#

3

1

−2 x− 3 dx = x −2

: D

SOL 1.56

Option (A) is correct. We have xo(t ) =− 3x (t ) or xo(t) + 3x (t) = 0 A.E. D + 3 = 0 Thus solution is x (t ) = C1 e − 3t From x (0) = x 0 we get C 1 = x 0 Thus x (t ) = x 0 e − 3t

SOL 1.57

Option (D) is correct. For eigen value λ =− 2 R3 − (− 2) V 2 −2 S W 1 − 2 − (− 2) S 0 W SS 0 0 1 − (− 2)WW R5 − 2 2VX T S W S0 0 1W SS0 0 1WW

T

SOL 1.58

3

1

=

1 2

Rx 1V R0V S W SW Sx 2W = S0W SSx 3WW SS0WW RTx 1VX RT0VX S W SW Sx 2W = S0W SSx 3WW SS0WW X T X TX

5x1 − 2x2 + x 3 = 0

Option (B) is correct. C 11 = 2 − (− 3) = 5



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CHAP 1


PAGE 29

C 21 =− (0 − (− 3)) =− 3 C 31 = (− (− 1)) = 1

R = (1) C11 + 2C 21 + 2C 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.

SOL 1.60

Option (A) is correct. We have f (x ) = x2e − x or f' (x ) = 2xe− x − x2 e− x = xe− x (2 − x ) f'' (x ) = (x2 − 4x + 2) e − x Now for maxima and minima, f' (x ) = 0 xe− x (2 − x ) = 0 or x = 0, 2 at x = 0 f '' (0) = 1 (+ ve) at x = 2 f '' (2) =− 2e − 2 (− ve) Now f '' (0) = 1 and f'' (2) =− 2e − 2 < 0 . Thus x = 2 is point of maxima

SOL 1.61

Option (C) is correct. 4 u = t i2

c

2x

2 2 2 u = ti u + tj u = xti + 2y 2x 2y

+ tj

m

At (1, 3) magnitude is SOL 1.62

4 u

=

x2 +

b 23 l

2

y =

2 y tj 3 1+4 =

Option (B) is correct. d2x 3dx 2x (t ) 5 + = 2 + dt

dt

Taking Laplace transform on both sides of above equation. s2 X (s) + 3sX (s) + 2X (s ) = 5 s

X (s ) =

From final value theorem

5 s (s + 3s + 2) 2

5 5 = 2 s (s + 3s + 2)

lim x (t ) = lim X (s ) = lim s

t " 3

s " 0

s " 0

2

***********



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Engineering Mathematics With Solutions

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