Engineering Mathematics

CHAPTER 1 ENGINEERING MATHEMATICS

YEAR 2012 MCQ 1.1

The area enclosed between the straight line y = x and the parabola y = x2 in the x -y plane is (B) 1/4 (A) 1/6 (C) 1/3

MCQ 1.2

ONE MARK

(D) 1/2

Consider the function f (x) = x in the interval − 1 # x # 1. At the point x = 0 , f (x) is (A) continuous and differentiable (B) non-continuous and differentiable (C) continuous and non-differentiable (D) neither continuous nor differentiable

MCQ 1.3

MCQ 1.4

x is lim b 1 − cos 2 l x"0 x (A) 1/4

(B) 1/2

(C) 1

(D) 2

At x = 0, the function f (x) = x3 + 1 has (A) a maximum value (B) a minimum value (C) a singularity

MCQ 1.5

(D) a point of inflection

For the spherical surface x2 + y2 + z2 = 1, the unit outward normal vector at the point c 1 , 1 , 0 m is given by 2 2 1 1 (B) 1 i − 1 j (A) i+ j 2 2 2 2 1 1 (C) k (D) i+ j+ 1 k 3 3 3

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ENGINEERING MATHEMATICS

CHAP 1

YEAR 2012 MCQ 1.6

MCQ 1.7

MCQ 1.8

(A) f (t) = sin t

1 is given by s (s + 1) (B) f (t) = e−t sin t

(C) f (t) = e−t

(D) f (t) = 1 − e−t

The inverse Laplace transform of the function F (s) =

5 3 For the matrix A = > , ONE of the normalized eigen vectors given as 1 3H J 1 N J 1N K O K 2O 2O O (A) K (B) K K −1 O KK 3 OO K O 2 L P L 2P J 3 N J 1 N K O K O 10 O 5O (C) K (D) K K −1 O K 2 O K O K O 10 L P L 5P A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is (A) 1/20 (B) 1/12 (C) 3/10

MCQ 1.9

TWO MARKS

(D) 1/2

Consider the differential equation x2 (d 2 y/dx 2) + x (dy/dx) − 4y = 0 with the boundary conditions of y (0) = 0 and y (1) = 1. The complete solution of the differential equation is (B) sin a πx k (A) x2 2 (C) ex sin a πx k 2

(D) e−x sin a πx k 2

MCQ 1.10

x + 2y + z = 4 2x + y + 2z = 5 x−y+z = 1 The system of algebraic equations given above has (A) a unique solution of x = 1, y = 1 and z = 1. (B) only the two solutions of (x = 1, y = 1, z = 1) and (x = 2, y = 1, z = 0) (C) infinite number of solutions (D) no feasible solution

YEAR 2011 MCQ 1.11

ONE MARK

A series expansion for the function sin θ is GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


(A) 1 − θ + θ − ... 2! 4!

(B) θ − θ + θ − ... 3! 5!

(C) 1 + θ + θ + θ + ... 2! 3! What is lim sin θ equal to ? θ"0 θ (A) θ

(D) θ + θ + θ + ... 3! 5!

(B) sin θ

(C) 0

(D) 1

2

4

2

MCQ 1.12

MCQ 1.13

3

3

5

5

(D) complex

The product of two complex numbers 1 + i and 2 − 5i is (A) 7 − 3i (B) 3 − 4i (C) − 3 − 4i

MCQ 1.15

3

Eigen values of a real symmetric matrix are always (A) positive (B) negative (C) real

MCQ 1.14

PAGE 3

(D) 7 + 3i

If f (x) is an even function and a is a positive real number, then equals (A) 0 (B) a

#−a f (x) dx a

(D) 2 # f (x) dx a

(C) 2a

0

YEAR 2011 MCQ 1.16

TWO MARKS

1 dx , when evaluated by using Simpson’s 1/3 rule on two x equal sub-intervals each of length 1, equals (A) 1.000 (B) 1.098 The integral

#1

3

(C) 1.111 MCQ 1.17

dy Consider the differential equation = (1 + y2) x . The general solution with dx constant c is 2 (B) y = tan2 a x + c k (A) y = tan x + tan c 2 2 (C) y = tan2 a x k + c 2

MCQ 1.18

(D) 1.120

2

(D) y = tan b x + c l 2

An unbiased coin is tossed five times. The outcome of each toss is either a head or a tail. The probability of getting at least one head is (B) 13 (A) 1 32 32 (C) 16 32

(D) 31 32



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MCQ 1.19


CHAP 1

Consider the following system of equations 2x1 + x2 + x 3 = 0 x2 − x 3 = 0 x1 + x 2 = 0 This system has (A) a unique solution (B) no solution (C) infinite number of solutions (D) five solutions

YEAR 2010 MCQ 1.20

ONE MARK

The parabolic arc y = x , 1 # x # 2 is revolved around the x -axis. The volume of the solid of revolution is (A) π/4 (B) π/2 (C) 3π/4 (D) 3π/2

MCQ 1.21

d 3f f d 2f = 0 , is a 3 + 2 dη2 dη (A) second order nonlinear ordinary differential equation

The Blasius equation,

(B) third order nonlinear ordinary differential equation (C) third order linear ordinary differential equation (D) mixed order nonlinear ordinary differential equation MCQ 1.22

The value of the integral (A) − π

#− 33 1 dx + x2

(C) π/2 MCQ 1.23

(B) − π/2 (D) π

The modulus of the complex number b 3 + 4i l is 1 − 2i (A) 5 (B) 5 (C) 1/ 5

MCQ 1.24

is

(D) 1/5

The function y = 2 − 3x (A) is continuous 6x ! R and differentiable 6x ! R (B) is continuous 6x ! R and differentiable 6x ! R except at x = 3/2 (C) is continuous 6x ! R and differentiable 6x ! R except at x = 2/3 (D) is continuous 6x ! R except x = 3 and differentiable 6x ! R GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 5

YEAR 2010 MCQ 1.25

TWO MARKS

2 2 H is One of the eigen vectors of the matrix A = > 1 3 2 (A) > H −1

2 (B) > H 1

4 (C) > H 1 MCQ 1.26

MCQ 1.27

1 (D) > H −1 The Laplace transform of a function f (t) is 2 1 . The function f (t) is s (s + 1) (A) t − 1 + e−t

(B) t + 1 + e−t

(C) − 1 + e−t

(D) 2t + et

A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is (A) 2/315 (B) 1/630 (C) 1/1260

MCQ 1.28

(D) 1/2520

Torque exerted on a flywheel over a cycle is listed in the table. Flywheel energy (in J per unit cycle) using Simpson’s rule is Angle (Degree)

0

60c

120c

180c

240c

300c

360c

Torque (N-m)

0

1066

− 323

0

323

− 355

0

(A) 542

(B) 993

(C) 1444

(D) 1986

YEAR 2009 MCQ 1.29

3/5 4/5 H, the transpose of the matrix is equal to the For a matrix 6M @ = > x 3/5 T −1 inverse of the matrix, 6M @ = 6M @ . The value of x is given by (B) − 3 (A) − 4 5 5 (C) 3 5

MCQ 1.30

ONE MARK

(D) 4 5

The divergence of the vector field 3xzi + 2xyj − yz2 k at a point (1, 1, 1) is equal to (A) 7 (B) 4 (C) 3

(D) 0



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MCQ 1.31


The inverse Laplace transform of 1/ (s2 + s) is (B) 1 − et (A) 1 + et (C) 1 − e−t

MCQ 1.32

CHAP 1

(D) 1 + e−t

If three coins are tossed simultaneously, the probability of getting at least one head is (A) 1/8 (B) 3/8 (C) 1/2

(D) 7/8

YEAR 2009 MCQ 1.33

An analytic function of a complex variable z = x + iy is expressed as f (z) = u (x, y) + iv (x, y) where i = − 1 . If u = xy , the expression for v should be (x + y) 2 x2 − y2 (B) (A) +k +k 2 2 (C)

MCQ 1.34

MCQ 1.35

MCQ 1.36

TWO MARKS

y2 − x2 +k 2

The solution of x

(D)

(x − y) 2 +k 2

dy + y = x 4 with the condition y (1) = 6 is 5 dx

4 (A) y = x + 1 5 x

4 (B) y = 4x + 4 5 5x

4 (C) y = x + 1 5

5 (D) y = x + 1 5

A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of (x + y) 2 on path AB traversed in a counterclockwise sense is

(A) π − 1 2

(B) π + 1 2

(C) π 2

(D) 1

The distance between the origin and the point nearest to it on the surface z2 = 1 + xy is GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


(A) 1 (C) MCQ 1.37

3 2

(B) (D) 2

3

The area enclosed between the curves y2 = 4x and x2 = 4y is (B) 8 (A) 16 3 (C) 32 3

MCQ 1.38

PAGE 7

(D) 16

The standard deviation of a uniformly distributed random variable between 0 and 1 is (B) 1 (A) 1 12 3 (C) 5 (D) 7 12 12 YEAR 2008

MCQ 1.39

ONE MARK

In the Taylor series expansion of ex about x = 2 , the coefficient of (x − 2) 4 is (A) 1/4 ! (B) 2 4 /4! (C) e2 /4!

MCQ 1.40

MCQ 1.41

MCQ 1.42

Given that xp + 3x = 0 , and x (0) = 1, xo(0) = 0 , what is x (1) ? (A) − 0.99 (B) − 0.16 (C) 0.16

(D) 0.99

1/3 The value of lim x − 2 x " 8 (x − 8) (A) 1 16

(B) 1 12

(C) 1 8

(D) 1 4

A coin is tossed 4 times. What is the probability of getting heads exactly 3 times ? (B) 3 (A) 1 8 4 (C) 1 2

MCQ 1.43

(D) e 4 /4!

R1 S The matrix S3 SS1 T other two eigen (A) p

(D) 3 4

2 4VW 0 6W has one eigen value equal to 3. The sum of the 1 pWW valueX is (B) p − 1

(C) p − 2

(D) p − 3



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MCQ 1.44


CHAP 1

The divergence of the vector field (x − y) i + (y − x) j + (x + y + z) k is (A) 0 (B) 1 (C) 2

(D) 3

YEAR 2008 MCQ 1.45

TWO MARKS

Consider the shaded triangular region P shown in the figure. What is ## xydxdy ? P

MCQ 1.46

(A) 1 6

(B) 2 9

(C) 7 16

(D) 1

The directional derivative of the scalar function f (x, y, z) = x2 + 2y2 + z at the point P = (1, 1, 2) in the direction of the vector a = 3i − 4j is

MCQ 1.47

(A) − 4

(B) − 2

(C) − 1

(D) 1

For what value of a, if any will the following system of equation in x, y and z have a solution ? 2x + 3y = 4 x+y+z = 4 3x + 2y − z = a

MCQ 1.48

MCQ 1.49

(A) Any real number

(B) 0

(C) 1

(D) There is no such value

Which of the following integrals is unbounded ? π/4

(A)

#0 tan xdx

(B)

#0 3 x2 1+ 1 dx

(C)

#0 3xe−x dx

(D)

#0

The integral

# f (z) dz

1

1 dx 1−x

evaluated around the unit circle on the complex plane



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CHAP 1

MCQ 1.50


for f (z) = cos z is z (A) 2πi

(B) 4πi

(C) − 2πi

(D) 0

The length of the curve y = 2 x3/2 between x = 0 and x = 1 is 3 (A) 0.27 (B) 0.67 (C) 1

MCQ 1.51

(D) 1.22

1 1 1 2 The eigen vector of the matrix > are written in the form > H and > H. H a b 0 2 What is a + b ? (A) 0 (B) 1 2 (C) 1

MCQ 1.52

Let f = yx . What is (A) 0 (C) 1

MCQ 1.53

PAGE 9

(D) 2 22 f at x = 2, y = 1 ? 2x2y (B) ln 2 (D) 1 ln 2

It is given that y m + 2yl + y = 0, y (0) = 0, y (1) = 0 . What is y (0.5) ? (A) 0 (B) 0.37 (C) 0.62

(D) 1.13

YEAR 2007 MCQ 1.54

The minimum value of function y = x2 in the interval [1, 5] is (A) 0 (B) 1 (C) 25

MCQ 1.55

(D) undefined

If a square matrix A is real and symmetric, then the eigen values (A) are always real (B) are always real and positive (C) are always real and non-negative pairs

MCQ 1.56

ONE MARK

(D) occur in complex conjugate

If ϕ (x, y) and ψ (x, y) are functions with continuous second derivatives, then ϕ (x, y) + iψ (x, y) can be expressed as an analytic function of x + iψ (i = − 1) , when 2ϕ 2ϕ 2ψ 2ϕ 2ψ 2ψ 2ϕ 2ψ (B) (A) =− , = =− , = 2x 2x 2y 2y 2y 2x 2x 2y (C)

22ϕ 22ϕ 22 ψ 22 ψ + = + =1 2x2 2y2 2x2 2y2

(D)

2ϕ 2ϕ 2ψ 2ψ + = + =0 2x 2y 2x 2y



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MCQ 1.57


The partial differential equation

CHAP 1

22 ϕ 22 ϕ 2ϕ 2ϕ + + + = 0 has 2x2 2y2 2x 2y

(A) degree 1 order 2

(B) degree 1 order 1

(C) degree 2 order 1

(D) degree 2 order 2

YEAR 2007 MCQ 1.58

If y = x + x + (A) 4 or 1

TWO MARKS

x + x + ......3 , then y ^2 h = (B) 4 only

(C) 1 only MCQ 1.59

The area of a triangle formed by the tips of vectors a, b and c is (A) 1 (a − b) : (a − c) 2

MCQ 1.60

(D) − 2 # x # 2

If F (s) is the Laplace transform of function f (t), then Laplace transform of

#0

MCQ 1.62

(B) 1 (a − b) # (a − c) 2

(C) 1 a # b # c (D) 1 (a # b) : c 2 2 dy The solution of = y2 with initial value y (0) = 1 bounded in the interval dx (A) − 3 # x # 3 (B) − 3 # x # 1 (C) x < 1, x > 1

MCQ 1.61

(D) undefined

t

f (τ) dτ is

(A) 1 F (s) s

(B) 1 F (s) − f (0) s

(C) sF (s) − f (0)

(D) # F (s) d s

A calculator has accuracy up to 8 digits after decimal place. The value of 2π

#0 sin xdx when evaluated using the calculator by trapezoidal method with 8 equal intervals, to 5 significant digits is (A) 0.00000 (B) 1.0000 (C) 0.00500 MCQ 1.63

(D) 0.00025

Let X and Y be two independent random variables. Which one of the relations between expectation (E), variance (Var) and covariance (Cov) given below is FALSE ? (A) E (XY ) = E (X ) E (Y ) (B) Cοv (X, Y ) = 0 (C) Var (X + Y ) = Var (X ) + Var (Y )

(D) E (X 2 Y 2) = (E (X )) 2 (E (Y )) 2



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CHAP 1


PAGE 11

2

ex − b1 + x + x l 2 lim = 3 x"0 x

MCQ 1.64

MCQ 1.65

(A) 0

(B) 1/6

(C) 1/3

(D) 1

2 1 H is The number of linearly independent eigen vectors of > 0 2 (A) 0 (B) 1 (C) 2

(D) infinite

YEAR 2006 MCQ 1.66

ONE MARK

Match the items in column I and II. Column I P. Gauss-Seidel method

Column II 1. Interpolation

Q. Forward Newton-Gauss method 2. Non-linear differential equations

MCQ 1.67

R. Runge-Kutta method

3. Numerical integration

S. Trapezoidal Rule

4. Linear algebraic equations

(A) P-1, Q-4, R-3, S-2

(B) P-1, Q-4, R-2, S-3

(C) P-1. Q-3, R-2, S-4

(D) P-4, Q-1, R-2, S-3

The solution of the differential equation (A) (1 + x) e+x (C) (1 − x) e+x

MCQ 1.68

2

2

2 dy + 2xy = e−x with y (0) = 1 is dx 2 (B) (1 + x) e−x

(D) (1 − x) e−x

2

Let x denote a real number. Find out the INCORRECT statement. (A) S = {x : x > 3} represents the set of all real numbers greater than 3 (B) S = {x : x2 < 0} represents the empty set. (C) S = {x : x ! A and x ! B} represents the union of set A and set B . (D) S = {x : a < x < b} represents the set of all real numbers between a and b, where a and b are real numbers.

MCQ 1.69

A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective ? (B) 1 (A) 1 5 25 (C) 20 99

(D) 19 495



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CHAP 1

YEAR 2006 MCQ 1.70

TWO MARKS

3 2 H are 5 and 1. What are the eigen Eigen values of a matrix S = > 2 3 values of the matrix S 2 = SS ? (A) 1 and 25 (B) 6 and 4 (C) 5 and 1

MCQ 1.71

(D) 2 and 10

Equation of the line normal to function f (x) = (x − 8) 2/3 + 1 at P (0, 5) is (A) y = 3x − 5 (B) y = 3x + 5 (C) 3y = x + 15

MCQ 1.72

Assuming i =

(D) 3y = x − 15 − 1 and t is a real number,

3 + i1 2 2

(A)

(B)

MCQ 1.74

π/3

eit dt is

3 − i1 2 2

(D) 1 + i c1 − 3 m 2 2

(C) 1 + i 3 2 2 MCQ 1.73

#0

2 If f (x) = 2x2 − 7x + 3 , then lim f (x) will be x"3 5x − 12x − 9

(A) − 1/3

(B) 5/18

(C) 0

(D) 2/5

Match the items in column I and II. Column I

Column II

P.

Singular matrix

1.

Determinant is not defined

Q.

Non-square matrix

2.

Determinant is always one

R.

Real symmetric

3.

Determinant is zero

S.

Orthogonal matrix

4.

Eigenvalues are always real

5.

Eigenvalues are not defined

(A) P-3, Q-1, R-4, S-2 (B) P-2, Q-3, R-4, S-1 (C) P-3, Q-2, R-5, S-4 (D) P-3, Q-4, R-2, S-1 MCQ 1.75

For

d 2y dy 2x 2 + 4 dx + 3y = 3e , the particular integral is dx

(A) 1 e2x 15

(B) 1 e2x 5

(C) 3e2x (D) C1 e−x + C2 e−3x GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1

MCQ 1.76

MCQ 1.77


PAGE 13

Multiplication of matrices E and F is G . matrices E and G are R1 0 0V Rcos θ − sin θ 0V S W S W E = S sin θ cos θ 0W and G = S0 1 0W SS0 0 1WW SS 0 0 1WW T X T X What is the matrix F ? Rcos θ − sin θ 0V R cos θ cos θ 0V W W S S (B) S− cos θ sin θ 0W (A) S sin θ cos θ 0W SS 0 SS 0 1WW 0 0 1WW TR VX RT cos θ sin θ 0XV W S S sin θ − cos θ 0W (C) S− sin θ cos θ 0W (D) Scos θ sin θ 0W W SS S S 0 0 0 1W 0 1WW X X T T Consider the continuous random variable with probability density function f (t) = 1 + t for − 1 # t # 0 = 1 − t for 0 # t # 1 The standard deviation of the random variable is (B) 1 (A) 1 3 6 (C) 1 3

(D) 1 6

YEAR 2005 MCQ 1.78

ONE MARK

Stokes theorem connects (A) a line integral and a surface integral (B) a surface integral and a volume integral (C) a line integral and a volume integral (D) gradient of a function and its surface integral

MCQ 1.79

A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is (A) 0.0036 (B) 0.1937 (C) 0.2234 (D) 0.3874

MCQ 1.80

#−a (sin6 x + sin7 x) dx a (A) 2 # sin6 x dx 0 a

is equal to

(C) 2 # (sin6 x + sin7 x) dx a

(B) 2 # sin7 x dx a

0

(D) zero

0



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MCQ 1.81


A is a 3 # 4 real matrix and Ax = b is an inconsistent system of equations. The highest possible rank of A is (A) 1 (B) 2 (C) 3

MCQ 1.82

CHAP 1

(D) 4

Changing the order of the integration in the double integral I = leads to I =

#r #p s

q

MCQ 1.84

f (x, y) dxdy What is q ?

(A) 4y

(B) 16 y2

(C) x

(D) 8

TWO MARKS

(D) 100 units

By a change of variable x (u, v) = uv, y (u, v) = v/u is double integral, the integrand f (x, y) changes to f (uv, v/u) φ (u, v). Then, φ (u, v) is (A) 2v/u (B) 2uv (C) v2

MCQ 1.86

f (x, y) dydx

Which one of the following is an eigen vector of the matrix R V S5 0 0 0W S0 5 0 0W S0 0 2 1W S W S0 0 3 1W R V R V S 1W S0W T X S− 2W S0W (A) S W (B) S W S 0W S1W S 0W S0W TR VX TR X V 1 S W S 1W S 0W S− 1W (C) S W (D) S W S 0W S 2W S− 2W S 1W T X T X With a 1 unit change in b, what is the change in x in the solution of the system of equations x + y = 2, 1.01x + 0.99y = b ? (A) zero (B) 2 units (C) 50 units

MCQ 1.85

2

4

YEAR 2005 MCQ 1.83

8

#0 #x

(D) 1

The right circular cone of largest volume that can be enclosed by a sphere of 1 m radius has a height of (A) 1/3 m (B) 2/3 m (C) 2 2 m 3

(D) 4/3 m



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CHAP 1

MCQ 1.87


2 ln (x) dy and y (1) = 0 , then what is y (e) ? + 2xy = x dx (A) e (B) 1

If x2

(C) 1/e MCQ 1.88

PAGE 15

(D) 1/e2

The line integral # V :dr of the vector V :(r) = 2xyzi + x2 zj + x2 yk from the origin to the point P (1, 1, 1) (A) is 1 (B) is zero (C) is – 1 (D) cannot be determined without specifying the path

MCQ 1.89

Starting from x 0 = 1, one step of Newton-Raphson method in solving the equation x3 + 3x − 7 = 0 gives the next value (x1) as (B) x1 = 1.406 (A) x1 = 0.5 (C) x1 = 1.5

MCQ 1.90

(D) x1 = 2

A single die is thrown twice. What is the probability that the sum is neither 8 nor 9 ? (A) 1/9 (B) 5/36 (C) 1/4

(D) 3/4

• Common Data For Q. 91 and 92

MCQ 1.91

The complete solution of the ordinary differential equation d 2y dy + p + qy = 0 is y = c1 e−x + c2 e−3x dx dx2 Then p and q are (B) p = 3, q = 4 (A) p = 3, q = 3 (C) p = 4, q = 3

MCQ 1.92

(D) p = 4, q = 4

Which of the following is a solution of the differential equation d 2y dy 2 + p dx + (q + 1) y = 0 dx −3x (A) e (B) xe−x (C) xe−2x

(D) x2 e−2x

YEAR 2004 MCQ 1.93

If x = a (θ + sin θ) and y = a (1 − cos θ), then

ONE MARK

dy will be equal to dx



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MCQ 1.94

MCQ 1.95


CHAP 1

(A) sin b θ l 2

(B) cos b θ l 2

(C) tan b θ l 2

(D) cot b θ l 2

The angle between two unit-magnitude coplanar vectors P (0.866, 0.500, 0) and Q (0.259, 0.966, 0) will be (A) 0c

(B) 30c

(C) 45c

(D) 60c

R1 S The sum of the eigen values of the matrix given below is S1 SS3 T (A) 5 (B) 7 (C) 9

TWO MARKS

From a pack of regular playing cards, two cards are drawn at random. What is the probability that both cards will be Kings, if first card in NOT replaced ? (B) 1 (A) 1 26 52 (C) 1 169

MCQ 1.97

(D) 1 221

0, for t < a A delayed unit step function is defined as U (t − a) = * Its Laplace 1 , for t $ a transform is −as (B) e (A) ae−as s as (C) e s

MCQ 1.98

3VW 1W 1WW X

(D) 18

YEAR 2004 MCQ 1.96

2 5 1

as (D) e a

The values of a function f (x) are tabulated below x

f (x)

0

1

1

2

2

1

3

10

Using Newton’s forward difference formula, the cubic polynomial that can be fitted to the above data, is GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 17

(A) 2x3 + 7x2 − 6x + 2

(B) 2x3 − 7x2 + 6x − 2

(C) x3 − 7x2 − 6x2 + 1

(D) 2x3 − 7x2 + 6x + 1

The volume of an object expressed in spherical co-ordinates is given by

MCQ 1.99

V=

MCQ 1.100

2π

#0 #0

π/3

#0

1

r2 sin φdrdφdθ

The value of the integral is (A) π 3

(B) π 6

(C) 2π 3

(D) π 4

For which value of x will the matrix given below become singular ? R 8 x 0V S W = S 4 0 2W SS12 6 0WW T X (A) 4 (B) 6 (C) 8

(D) 12

YEAR 2003 MCQ 1.101

MCQ 1.102

ONE MARK

2 lim sin x is equal to x x"0

(A) 0

(B) 3

(C) 1

(D) − 1

The accuracy of Simpson’s rule quadrature for a step size h is (A) O (h2) (B) O (h3) (C) O (h 4)

MCQ 1.103

(D) O (h5)

4 1 For the matrix > the eigen values are 1 4H (B) − 3 and − 5 (A) 3 and − 3 (C) 3 and 5

(D) 5 and 0

YEAR 2003 MCQ 1.104

TWO MARKS

Consider the system of simultaneous equations x + 2y + z = 6 2x + y + 2z = 6 x+y+z = 5 GATE Previous Year Solved Paper For Mechanical Engineering

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CHAP 1

This system has (A) unique solution (B) infinite number of solutions (C) no solution (D) exactly two solutions MCQ 1.105

The area enclosed between the parabola y = x2 and the straight line y = x is (A) 1/8 (B) 1/6 (C) 1/3

MCQ 1.106

(D) 1/2

The solution of the differential equation (A) y =

1 x+c

dy + y2 = 0 is dx 3 (B) y = − x + c 3

(C) cex linear MCQ 1.107

(D) unsolvable as equation is non-

The vector field is F = xi − yj (where i and j are unit vector) is (A) divergence free, but not irrotational (B) irrotational, but not divergence free (C) divergence free and irrotational (D) neither divergence free nor irrational

MCQ 1.108

Laplace transform of the function sin ωt is (B) 2 ω 2 (A) 2 s 2 s +ω s +ω (C) 2 s 2 (D) 2 ω 2 s −ω s −ω

MCQ 1.109

A box contains 5 black and 5 red balls. Two balls are randomly picked one after another form the box, without replacement. The probability for balls being red is (A) 1/90 (B) 1/2 (C) 19/90

(D) 2/9

YEAR 2002 MCQ 1.110

ONE MARK

Two dice are thrown. What is the probability that the sum of the numbers on the two dice is eight? (B) 5 (A) 5 36 18 (C) 1 (D) 1 3 4 GATE Previous Year Solved Paper For Mechanical Engineering



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CHAP 1


MCQ 1.111

Which of the following functions is not differentiable in the domain [− 1, 1] ? (B) f (x) = x − 1 (A) f (x) = x2 (C) f (x) = 2

MCQ 1.112

PAGE 19

(D) f (x) = maximum ( x, − x )

A regression model is used to express a variable Y as a function of another variable X .This implies that (A) there is a causal relationship between Y and X (B) a value of X may be used to estimate a value of Y (C) values of X exactly determine values of Y (D) there is no causal relationship between Y and X YEAR 2002

MCQ 1.113

TWO MARKS

The following set of equations has 3x + 2y + z = 4 x−y+z = 2

MCQ 1.114

− 2x + 2z = 5 (A) no solution

(B) a unique solution

(C) multiple solutions

(D) an inconsistency

The function f (x, y) = 2x2 + 2xy − y3 has (A) only one stationary point at (0, 0) (B) two stationary points at (0, 0) and b 1 , − 1 l 6 3 (C) two stationary points at (0, 0) and (1, − 1) (D) no stationary point

MCQ 1.115

Manish has to travel from A to D changing buses at stops B and C enroute. The maximum waiting time at either stop can be 8 min each but any time of waiting up to 8 min is equally, likely at both places. He can afford up to 13 min of total waiting time if he is to arrive at D on time. What is the probability that Manish will arrive late at D ? (B) 13 (A) 8 13 64 (C) 119 128

(D) 9 128

YEAR 2001 MCQ 1.116

ONE MARK

The divergence of vector i = xi + yj + zk is (A) i + j + k (B) 3 (C) 0 (D) 1 GATE Previous Year Solved Paper For Mechanical Engineering



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MCQ 1.117

MCQ 1.118

Consider the system of equations given below x+y = 2 2x + 2y = 5 This system has (A) one solution

(B) no solution

(C) infinite solutions

(D) four solutions

What is the derivative of f (x) = x at x = 0 ? (A) 1 (B) − 1 (C) 0

MCQ 1.119

CHAP 1

(D) Does not exist

The Gauss divergence theorem relates certain (A) surface integrals to volume integrals (B) surface integrals to line integrals (C) vector quantities to other vector quantities (D) line integrals to volume integrals

YEAR 2001

TWO MARKS 3

MCQ 1.120

MCQ 1.121

The minimum point of the function f (x) = b x l − x is at 3 (A) x = 1 (B) x =− 1 (C) x = 0 (D) x = 1 3 The rank of a 3 # 3 matrix C (= AB), found by multiplying a non-zero column matrix A of size 3 # 1 and a non-zero row matrix B of size 1 # 3 , is (A) 0 (B) 1 (C) 2

MCQ 1.122

(D) 3

An unbiased coin is tossed three times. The probability that the head turns up in exactly two cases is (B) 1 (A) 1 9 8 (C) 2 3

(D) 3 8

**********



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CHAP 1


PAGE 21

SOLUTION SOL 1.1

Option (A) is correct. For

y = x straight line and y = x 2 parabola, curve is as given. The shaded region is the area, which is bounded by the both curves (common area).

We solve given equation as follows to gett the intersection points : In y = x2 putting y = x we have x = x2 or x2 − x = 0 & x (x − 1) = 0 & x = 0, 1 Then from y = x , for x = 0 & y = 0 and x = 1 & y = 1 Curve y = x2 and y = x intersects at point (0, 0) and (1, 1) So, the area bounded by both the curves is y = x2

x=1

A=

# # dydx

x=0 3

y=x

x=1

=

y = x2

# dx # dy

x=0

y=x

x=1

=

# dx6y @

x2 x

x=0

2 1 = :x − x D = 1 − 1 =− 1 = 1 unit2 3 2 0 3 2 6 6

SOL 1.2

x=1

# (x

=

Area is never negative

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− x) dx

x=0

Option (C) is correct. Given f (x) = x (in − 1 # x # 1) For this function the plot is as given below.


2

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CHAP 1

At x = 0 , function is continuous but not differentiable because. For x > 0 and x < 0 f l (x) = 1 and f l (x) =− 1 lim f l(x) = 1 and lim f l(x) =− 1 x " 0+

x " 0−

R.H.S lim = 1 and L.H.S lim =− 1 Therefore it is not differentiable. SOL 1.3

Option (B) is correct. Let

y = lim x"0

(1 − cos x) x2

It forms : 0 D condition. Hence by L-Hospital rule 0 d

y = lim dx x"0

(1 − cos x) = lim sin x 2 d x " 0 2x dx (x )

Still these gives : 0 D condition, so again applying L-Hospital rule 0 y = lim x"0

SOL 1.4

(sin x) = lim cos x = cos 0 = 1 2 2 2 x"0 2 # dxd (x)

Option (D) is correct. We have f (x) = x3 + 1 f l(x) = 3x2 + 0 Putting f l(x) equal to zero f l(x) = 0 Now At x = 0,

SOL 1.5

d dx

3x2 + 0 = 0 & x = 0 f ll(x) = 6x fll(0) = 6 # 0 = 0

Hence x = 0 is the point of inflection.

Option (A) is correct. Given : x2 + y2 + z2 = 1 This is a equation of sphere with radius r = 1



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CHAP 1


PAGE 23

The unit normal vector at point c 1 , 1 , 0 m is OA 2 2 Hence OA = c 1 − 0 m i + c 1 − 0 m j + (0 − 0) k = 1 i + 1 j 2 2 2 2 SOL 1.6

Option (D) is correct. First using the partial fraction : A (s + 1) + Bs 1 =A+ B = s (s + 1) s s + 1 s (s + 1) (A + B) s 1 A + = s (s + 1) s (s + 1) s (s + 1) Comparing the coefficients both the sides, (A + B) = 0 and A = 1, B =− 1 1 So =1− 1 s (s + 1) s s + 1 F (s) =

F (t) = L−1 [F (s)] = L−1 ; 1 E = L−1 :1 − 1 D = L−1 :1D − L−1 : 1 D s s+1 s s+1 s (s + 1) = 1 − e−t SOL 1.7

Option (B) is correct. 5 A => 1 For finding eigen values, we write Given

3 3H the characteristic equation as

A − λI = 0 5−λ 3 =0 1 3−λ (5 − λ) (3 − λ) − 3 = 0 λ2 − 8λ + 12 = 0 & λ = 2, 6 Now from characteristic equation for eigen vector.

&

For λ = 2

&

6A − λI @"x , = 60@

5−2 3 X1 0 => H > H > H 1 3 − 2 X2 0 3 3 X1 0 >1 1H>X H = >0H 2 X1 + X 2 = 0

So

& X1 =− X2

1 eigen vector = * 4 −1

Magnitude of eigen vector =

(1) 2 + (1) 2 =

2



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CHAP 1

R 1 V W S 2W S Normalized eigen vector = S −1W W S 2 X T SOL 1.8

Option (D) is correct. Given : No. of Red balls = 4 No. of Black ball = 6 3 balls are selected randomly one after another, without replacement. 1 red and 2 black balls are will be selected as following Manners

Probability for these sequence

R B B

4 6 5=1 10 # 9 # 8 6

B R B

6 4 5=1 10 # 9 # 8 6

B B R

6 5 4=1 10 # 9 # 8 6

Hence Total probability of selecting 1 red and 2 black ball is P =1+1+1 = 3=1 6 6 6 6 2 SOL 1.9

Option (A) is correct. d2y dy We have x2 2 + x − 4y = 0 dx dx z Let x = e then z = log x dz = 1 x dx dy dy dy So, we get = b lb dz l = 1 x dz dx dz dx dy x = Dy dx Again

...(1)

where d = D dz

d 2y d dy d 1 dy − 1 dy + 1 d dy dz b l 2 = dx b dx l = dx b x dz l = x2 dz x dz dz dx dx dy d 2y d 2 y dy = −21 + 1 2 dz = 12 c 2 − m dz x dz x dz dx x dz

x2 d 2 y = (D2 − D) y = D (D − 1) y dx2 Now substitute in equation (i) [D (D − 1) + D − 4] y = 0 (D2 − 4) y = 0 & D = ! 2 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 25

y = C1 x2 + C2 x−2 y (0) = 0 , equation (ii) gives 0 = C1 # 0 + C 2 C2 = 0 And from y (1) = 1, equation (ii) gives 1 = C1 + C 2

So the required solution is From the given limits

...(ii)

C1 = 1 Substitute C1 & C2 in equation (ii), the required solution be y = x2 SOL 1.10

Option (C) is correct. For given equation matrix form is as follows R4V R1 2 1V W S W S A = S2 1 2W, B = S5W SS1WW SS1 − 1 1WW X T X The augmented matrix Tis R1 2 1 : 4V W S R2 " R2 − 2R1, R 3 " R 3 − R1 8A : BB = S2 1 2 : 5W SS1 − 1 1 : 1WW TR1 2 1 : X4V W S R 3 " R3 − R2 + S0 − 3 0 : − 3W SS0 − 3 0 : − 3WW RT1 2 1 : 4VX W S + S0 − 3 0 : − 3W R2 " R2 / − 3 SS0 0 0 : 0WW RT1 2 1 : 4V X S W + S0 1 0 : 1W SS0 0 0 : 0WW T X This gives rank of A, ρ (A) = 2 and Rank of 8A : BB = ρ 8A : BB = 2 Which is less than the number of unknowns (3) ρ 6A@ = ρ 8A : BB = 2 < 3 Hence, this gives infinite No. of solutions.

SOL 1.11

Option (B) is correct. 3 5 7 sin θ = θ − θ + θ − θ + ...... 3 5 7

SOL 1.12

Option (D) is correct. GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1

y = lim sin θ θ"0 θ

Let

d (sin θ) = lim dθ d = lim cos θ 1 θ"0 θ"0 ( θ ) dθ = cos 0 =1 1

SOL 1.13

Applying L-Hospital rule

Option (C) is correct Let a square matrix x y A => y xH We know that the characteristic equation for the eigen values is given by A − λI = 0 x−λ y =0 y x−λ (x − λ) 2 − y2 = 0 (x − λ) 2 = y2 x − λ =! y & λ = x ! y So, eigen values are real if matrix is real and symmetric.

SOL 1.14

Option (A) is correct. Let, z1 = (1 + i), z2 = (2 − 5i) z = z1 # z2 = (1 + i) (2 − 5i) = 2 − 5i + 2i − 5i2 = 2 − 3i + 5 = 7 − 3i

SOL 1.15

Option (D) is correct. For a function, whose limits bounded between − a to a and a is a positive real number. The solution is given by

#−a f (x) dx a

SOL 1.16

i 2 =− 1

2 # f (x) dx ;

f (x) is even

0

f (x) is odd

a

=*

0

;

Option (C) is correct. 1 dx x From this function we get a = 1, b = 3 and n = 3 − 1 = 2 Let,

So,

f (x) =

#1

3

h =b−a = 3−1 = 1 n 2

We make the table from the given function y = f (x) = 1 as follows : x GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


x

f (x) = y = 1 x

x=1

y1 = 1 = 1 1

x=2

y2 = 1 = 0.5 2

x=3

y 3 = 1 = 0.333 3

PAGE 27

Applying the Simpson’s 1/3 rd formula 3 #1 x1 dx = h3 6(y1 + y3) + 4y2@ = 13 6(1 + 0.333) + 4 # 0.5@ = 1 [1.333 + 2] = 3.333 = 1.111 3 3 SOL 1.17

SOL 1.18

Option (D) is correct. dy Given : = (1 + y2) x dx dy = xdx (1 + y2) Integrating both the sides, we get dy = # xdx # 1+ y2 2 tan−1 y = x + c & 2

2

y = tan b x + c l 2

Option (D) is correct. The probability of getting head p = 1 2 And the probability of getting tail q = 1 − 1 = 1 2 2 The probability of getting at least one head is 5 0 P (x $ 1) = 1 − 5C 0 (p) 5 (q) 0 = 1 − 1 # b 1 l b 1 l 2 2 = 1 − 15 = 31 32 2

SOL 1.19

Option (C) is correct. Given system of equations are, 2x1 + x2 + x 3 = 0 x2 − x 3 = 0 x1 + x 2 = 0 Adding the equation (i) and (ii) we have

...(i) ...(ii) ...(iii)



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PAGE 28


CHAP 1

2x1 + 2x2 = 0 ...(iv) x1 + x 2 = 0 We see that the equation (iii) and (iv) is same and they will meet at infinite points. Hence this system of equations have infinite number of solutions. SOL 1.20

Option (D) is correct. The volume of a solid generated by revolution about x -axis bounded by the function f (x) and limits between a to b is given by V = Given Therefore,

SOL 1.21

#a

y = V =

b

πy2 dx

x and a = 1, b = 2

#1

2

2 2 2 π ( x ) 2 dx = π # xdx = π :x D = π : 4 − 1 D = 3π 2 1 2 2 2 1

Option (B) is correct. d 3f f d 2f =0 + dη3 2 dη2 Order is determined by the order of the highest derivation present in it. So, It is third order equation but it is a nonlinear equation because in linear equation, the product of f with d 2 f/dη2 is not allow. Therefore, it is a third order non-linear ordinary differential equation. Given:

SOL 1.22

Option (D) is correct. Let

I =

#− 33 1 dx + x2

= 6tan−1 x @3 = [tan−1 (+ 3) − tan−1 (− 3)] −3 = π − a− π k = π 2 2 SOL 1.23

tan−1 (− θ) =− tan−1 (θ)

Option (B) is correct. z = 3 + 4i 1 − 2i Divide and multiply z by the conjugate of (1 − 2i) to convert it in the form of a + bi we have (3 + 4i) (1 + 2i) z = 3 + 4i # 1 + 2i = 1 − 2i 1 + 2i (1) 2 − (2i) 2 Let,

2 = 3 + 10i +2 8i = 3 + 10i − 8 1 − (− 4) 1 − 4i = − 5 + 10i =− 1 + 2i 5

z =

(− 1) 2 + (2) 2 =

5

a + ib =



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CHAP 1

SOL 1.24


PAGE 29

Option (C) is correct. Z if x < 2 ]2 − 3x 3 ] ] y = f (x) = [0 if x = 2 3 ] ]]− (2 − 3x) if x > 2 3 Checking the continuity\of the function. At x = 2 , Lf (x) = lim f b 2 − h l = lim 2 − 3 b 2 − h l 3 3 3 h"0 h"0

and

= lim 2 − 2 + 3h = 0 h"0 Rf (x) = lim f b 2 + h l = lim 3 b 2 + h l − 2 3 3 h"0 h"0 = lim 2 + 3h − 2 = 0 h"0

Since

L lim f (x) = R lim f (x) h"0

h"0

So, function is continuous 6 x ! R Now checking the differentiability : f ^ 23 − h h − f ^ 23 h 2 − 3 ^ 23 − h h − 0 Lf l (x) = lim = lim h"0 h"0 −h −h = lim 2 − 2 + 3h = lim 3h =− 3 h"0 h"0 −h −h and

Since SOL 1.25

f ^ 23 + h h − f ^ 23 h h"0 h

Rf l (x) = lim

3 ^ 23 + h h − 2 − 0 = lim = lim 2 + 3h − 2 = 3 h"0 h"0 h h Lf lb 2 l ! Rf lb 2 l, f (x) is not differentiable at x = 2 . 3 3 3

Option (A) is correct. 2 2 H A => 1 3 And λ1 and λ2 are the eigen values of the matrix A. The characteristic equation is written as Let,

A − λI = 0 2 2 1 0 > H − λ> H =0 1 3 0 1 2−λ 2 =0 1 3−λ

...(i)

(2 − λ) (3 − λ) − 2 = 0 λ2 − 5λ + 4 = 0 & λ = 1 & 4 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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PAGE 30


CHAP 1

Putting λ = 1 in equation (i), 0 2−1 2 x1 => H > H > H 0 1 3 − 1 x2

x1 where > H is eigen vector x2

2 x1 0 => H H > H 2 x2 0

1 >1

x1 + 2x2 = 0 or x1 + 2x2 = 0 Let x2 = K Then x1 + 2K = 0 & x1 =− 2K So, the eigen vector is − 2K −2 > K H or > 1H 2 Since option A> H is in the same ratio of x1 and x2 . Therefore option (A) −1 is an eigen vector. SOL 1.26

Option (A) is correct. f (t) is the inverse Laplace So,

f (t) = L − 1 ; 2 1 s (s + 1)E 1 = A + B2 + C s s+1 s s2 (s + 1)

As (1 + s) + B (s + 1) + Cs2 s2 (s + 1) s2 (A + C) + s (A + B) + B = s2 (s + 1) Compare the coefficients of s2, s and constant terms and we get A + C = 0 ; A + B = 0 and B = 1 Solving above equation, we get A =− 1, B = 1 and C = 1 Thus f (t) = L − 1 :− 1 + 12 + 1 D s s s+1 =

=− 1 + t + e−t = t − 1 + e−t

SOL 1.27

L − 1 : 1 D = e−at s+a

Option (C) is correct. The box contains : Number of washers = 2 Number of nuts = 3 Number of bolts = 4 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 31

Total objects = 2 + 3 + 4 = 9 First two washers are drawn from the box which contain 9 items. So the probability of drawing 2 washers is, 2 7!2! n P1 = 9C2 == 1 = Cn = 1 = 2 = 1 C2 9 # 8 # 7! 9 # 8 36 9! 7! 2! After this box contains only 7 objects and then 3 nuts drawn from it. So the probability of drawing 3 nuts from the remaining objects is, 3 4! 3! P2 = 7C 3 = 1 = = 1 C3 7! 7 # 6 # 5 # 4! 35 4! 3! After this box contain only 4 objects, probability of drawing 4 bolts from the box, 4 P3 = 4C 4 = 1 = 1 C4 1 Therefore the required probability is, P = P1 P2 P3 = 1 # 1 # 1 = 1 36 35 1260 SOL 1.28

Option (B) is correct. Given : h = 60c − 0 = 60c h = 60 # π = π = 1.047 radians 180 3 From the table, we have y 0 = 0 , y1 = 1066 , y2 =− 323 , y 3 = 0 , y 4 = 323 , y5 =− 355 and y6 = 0 From the Simpson’s 1/3rd rule the flywheel Energy is, E = h 6(y 0 + y6) + 4 (y1 + y 3 + y5) + 2 (y2 + y 4)@ 3 Substitute the values, we get E = 1.047 6(0 + 0) + 4 (1066 + 0 − 355) + 2 (− 323 + 323)@ 3 = 1.047 64 # 711 + 2 (0)@ = 993 Nm rad (Joules/cycle) 3

SOL 1.29

Option (A) is correct. Given : And

3 5

M => x

4 5 3 5

H

[M]T = [M] −1

Τ −1 We know that when 6A@ = 6A@ then it is called orthogonal matrix.

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PAGE 32


CHAP 1

6M @T 6M @ = I Substitute the values of M and M T , we get

>

3 5 4 5

x 3 5

H.>x 3 5

4 5 3 5

1 0

H = >0 1H

3 4 3 V b 5 # 5 l + 5 xW W = >1 0H 4 4 3 3 W 0 1 b 5 # 5 l + b 5 # 5 lW X 2 12 3 9 1 0 25 + 5 x 25 + x => >12 + 3 x H 0 1H 1 25 5 Comparing both sides a12 element, 12 + 3 x = 0 " x =− 12 5 =− 4 25 5 25 # 3 5 R 3 S b # 3 l + x2 5 S 5 S 4 # 3 + 3x Sb 5 5l 5 T

SOL 1.30

Option (C) is correct. Let, V = 3xzi + 2xyj − yz2 k We know divergence vector field of V is given by (4: V) So,

4: V = c 2 i + 2 j + 2 k m : ^3xzi + 2xyj − yz2 k h 2x 2y 2z

4: V = 3z + 2x − 2yz At point P (1, 1, 1) (4: V) P (1, 1, 1) = 3 # 1 + 2 # 1 − 2 # 1 # 1 = 3 SOL 1.31

Option (C) is correct. f (s) = L − 1 ; 2 1 E s +s 1 First, take the function 2 and break it by the partial fraction, s +s Solve by 1 = 1 1 1 = − * 1 =A+ B 4 s (s + 1) s (s + 1) s2 + s (s + 1) s s + 1 So, L − 1 c 2 1 m = L − 1 ;1 − 1 E = L − 1 :1D − L − 1 : 1 D = 1 − e−t s s+1 s (s + 1) s +s Let

SOL 1.32

Option (D) is correct. Total number of cases = 23 = 8 & Possible cases when coins are tossed simultaneously.



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CHAP 1


H H H T H T T T

H H T H T H T T

PAGE 33

H T H H T T H T

From these cases we can see that out of total 8 cases 7 cases contain at least one head. So, the probability of come at least one head is = 7 8 SOL 1.33

Option (C) is correct. Given : z = x + iy is a analytic function f (z) = u (x, y) + iv (x, y) u = xy Analytic function satisfies the Cauchy-Riemann equation. 2u = 2v and 2u =−2v 2x 2y 2y 2x So from equation (i), 2u = y 2x

&

..(i)

2v = y 2y

2u = x & 2v =− x 2y 2x Let v (x, y) be the conjugate function of u (x, y) dv = 2v dx + 2v dy = (− x) dx + (y) dy 2x 2y Integrating both the sides,

# dv

=− # xdx +

# ydy

2 y2 v =− x + + k = 1 (y2 − x2) + k 2 2 2

SOL 1.34

Option (A) is correct. dy Given x + y = x4 dx dy + 1 y = x3 dx b x l

...(i)

dy It is a single order differential equation. Compare this with + Py = Q dx and we get GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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PAGE 34


P =1 x

CHAP 1

Q = x3

Its solution will be y (I.F.) =

# Q (I.F.) dx + C

I.F. = e # Pdx = e Complete solution is given by, yx =

# x1 dx

= e log x = x

# x3 # xdx + C

e

=

# x4 dx + C

5 = x +C 5

...(ii)

and y (1) = 6 at x = 1 & y = 6 From equation (ii), 5 5 6 1 = 1+C & C = 6−1 = 1 5 5# 5 5 Then, from equation (ii), we get 5 4 yx = x + 1 & y = x + 1 5 5 x SOL 1.35

Option (B) is correct. The equation of circle with unit radius and centre at origin is given by, x2 + y2 = 1

Finding the integration of (x + y) 2 on path AB traversed in counter-clockwise sense So using the polar form Let: x = cos θ , y = sin θ , and r = 1 So put the value of x and y and limits in first quadrant between 0 to π/2 . Hence,

I =

#0

π/2

=

#0

π/2

=

#0

π/2

(cos θ + sin θ) 2 dθ (cos2 θ + sin2 θ + 2 sin θ cos θ) dθ (1 + sin 2θ) dθ

Integrating above equation, we get π/2 = :θ − cos 2θ D = ;a π − cos π k − b 0 − cos 0 lE 2 0 2 2 2 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 35

= b π + 1 l − b− 1 l = π + 1 2 2 2 2 SOL 1.36

Option (A) is correct. The given equation of surface is ...(i) z2 = 1 + xy Let P (x, y, z) be the nearest point on the surface (i), then distance from the origin is d =

(x − 0) 2 + (y − 0) 2 + (z − 0) 2

d 2 = x2 + y2 + z2 z2 = d 2 − x2 − y2 From equation (i) and (ii), we get

...(ii)

d 2 − x2 − y2 = 1 + xy d 2 = x2 + y2 + xy + 1 Let ...(iii) f (x, y) = d 2 = x2 + y2 + xy + 1 The f (x, y) be the maximum or minimum according to d 2 maximum or minimum. Differentiating equation (iii) w.r.t x and y respectively, we get 2f 2f = 2y + x = 2x + y or 2x 2y 2f 2f Applying maxima minima principle and putting and equal to zero, 2x 2y 2f 2f = 2x + y = 0 or = 2y + x = 0 2x 2y Solving these equations, we get x = 0 , y = 0 So, x = y = 0 is only one stationary point. 22 f Now p = 2 =2 2x 22 f q = =1 2x2y 22 f =2 2y2 or pr − q2 = 4 − 1 = 3 > 0 and r is positive. So, f (x, y) = d 2 is minimum at (0, 0). Hence minimum value of d 2 at (0, 0). r =

d 2 = x2 + y2 + xy + 1 = 1 d = 1 or f (x, y) = 1 So, the nearest point is &

z2 = 1 + xy = 1 + 0 z =! 1



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SOL 1.37


CHAP 1

Option (A) is correct. Given : y2 = 4x and x2 = 4y draw the curves from the given equations,

The shaded area shows the common area. Now finding the intersection points of the curves. y2 = 4x = 4 4y = 8 y Squaring both sides

x=

4y From second curve

y 4 = 8 # 8 # y & y (y3 − 64) = 0 y =4 & 0 Similarly put y = 0 in curve x2 = 4y x2 = 4 # 0 = 0 & x = 0 And Put y =4 x2 = 4 # 4 = 16 x = 4 So, x = 4, 0 Therefore the intersection points of the curves are (0, 0) and (4, 4). So the enclosed area is given by A=

#x

x2

(y1 − y2) dx

1

Put y1 and y2 from the equation of curves y2 = 4x and x2 = 4y 2 4 A = # b 4x − x l dx 4 0 4 x2 1 b 2 x − 4 l dx = 2 # x dx − 4 0 Integrating the equation, we get

=

#0

4

#0

4

x2 dx

3 4 4 A = 2 :2 x3/2D − 1 :x D 3 4 3 0 0 3 = 4 # 43/2 − 1 # 4 = 4 # 8 − 16 = 16 3 3 3 3 3 4



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CHAP 1

SOL 1.38


PAGE 37

Option (A) is correct. The cumulative distribution function Z0, x#a ] ]x − a , a x, x > b Mean

E (x) =

/ xf (x) = a +2 b b

x=a

Variance = x2 f (x) − x 2 = x2 f (x) − 6xf (x)@2 Substitute the value of f (x) Variance =

b

/ x=a

==

2

b

x2

/

1 dx − 1 ) x b − a dx 3 b−a x=a

b b 2 x3 − >) x 3 G 3 (b − a) a 2 (b − a) aH

2

3 3 (b2 − a2) 2 = b −a − 3 (b − a) 4 (b − a) 2

=

(b − a) (b2 + ab + a2) (b + a) 2 (b − a) 2 − 3 (b − a) 4 (b − a) 2

=

4 (b2 + ab + a2) + 3 (a + b) 2 (b − a) 2 = 12 12

Standard deviation = Given : b = 1, a = 0

Variance =

So, standard deviation = 1 − 0 = 12 SOL 1.39

(b − a) 2 (b − a) = 12 12

1 12

Option (C) is correct. Taylor’s series expansion of f (x) is given by, (x − a) (x − a) 2 (x − a) 3 f l (a) + f m (a) + f lll (a) + .... 1 2 3 f mm (a) Then from this expansion the coefficient of (x − a) 4 is 4 Given a =2 f (x) = ex f l (x) = ex f m (x) = ex f (x) = f (a) +



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CHAP 1

f n (x) = ex f mm (x) = ex 2 Hence, for a = 2 the coefficient of (x − a) 4 is e 4

SOL 1.40

Option (D) is correct. Given : xp + 3x = 0 and x (0) = 1 D= d dt

(D2 + 3) x = 0 The auxiliary Equation is written as m2 + 3 = 0 m =! 3 i = 0 ! Here the roots are imaginary m1 = 0 and m2 = Solution is given by

3i 3

x = em t (A cos m2 t + B sin m2 t) 1

= e0 [A cos 3 t + B sin 3 t] = [A cos 3 t + B sin 3 t] Given : x (0) = 1 at t = 0 , x = 1 Substituting in equation (i),

...(i)

1 = [A cos 3 (0) + B sin 3 (0)]= A + 0 A =1 Differentiateing equation (i) w.r.t. t , xo = 3 [− A sin 3 t + B cos 3 t] Given xo(0) = 0 at t = 0 , xo = 0 Substituting in equation (ii), we get

...(ii)

0 = 3 [− A sin 0 + B cos 0] B =0 Substituting A & B in equation (i) x = cos 3 t x (1) = cos 3 = 0.99 SOL 1.41

Option (B) is correct. Let

1/3 f (x) = lim x − 2 x " 8 (x − 8)

= lim

1 3

x"8

x−2/3 1

0 form 0 Applying L-Hospital rule

Substitute the limits, we get GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 39

f (x) = 1 (8) −2/3 = 1 (23) −2/3 = 1 = 1 3 3 4 # 3 12 SOL 1.42

Option (A) is correct. In a coin probability of getting Head p = 1 = No. of Possible cases 2 No. of Total cases Probability of getting tail q = 1−1 = 1 2 2 So the probability of getting Heads exactly three times, when coin is tossed 4 times is 3 1 P = 4C 3 (p) 3 (q) 1 = 4C 3 b 1 l b 1 l 2 2 = 4#1 #1 = 1 8 2 4

SOL 1.43

Option (C) is correct.

R1 2 4V W S Let, A = S3 0 6W SS1 1 pWW Let the eigen values of this matrix areT λ1, λ2 &Xλ3 Here one values is given so let λ1 = 3 We know that Sum of eigen values of matrix= Sum of the diagonal element of matrix A λ1 + λ2 + λ3 = 1 + 0 + p λ2 + λ3 = 1 + p − λ1 = 1 + p − 3 = p − 2

SOL 1.44

Option (D) is correct. We know that the divergence is defined as 4:V Let And So,

V = (x − y) i + (y − x) j + (x + y + z) k 4 = c 2 i + 2 j + 2 km 2x 2y 2z 4:V = c 2 i + 2 j + 2 k m : 6(x − y) i + (y − x) j + (x + y + z) k @ 2x 2y 2z = 2 (x − y) + 2 (y − x) + 2 (x + y + z) 2x 2y 2z = 1+1+1 = 3

SOL 1.45

Option (A) is correct. Given : GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1

The equation of line in intercept form is given by x +y =1 2 1

x +y =1 a b

x + 2y = 2 & x = 2 (1 − y) The limit of x is between 0 to x = 2 (1 − y) and y is 0 to 1, Now

##p xydxdy

=

y=1

2 (1 − y)

y=1

4 (1 − y) − 0E dy 2

#y = 0 #x = 0

=

#y = 0

=

#y = 0

y=1

y;

xydxdy =

y=1

#y = 0

x2 2 (1 − y) ydy :2D 0

2

2y (1 + y2 − 2y) dy =

y=1

#y = 0

2 (y + y3 − 2y2) dy

Again Integrating and substituting the limits, we get

##p xydxdy

SOL 1.46

y 2 y 4 2y 3 1 = 2; + − = 2 :1 + 1 − 2 − 0D 2 3 E0 2 4 3 4 = 2:6 + 3 − 8D = 2 = 1 12 12 6

Option (B) is correct. Direction derivative of a function f along a vector P is given by a =grad f : a a 2f 2f 2f where grad f = c i+ j+ k 2x 2y 2z m f (x, y, z) = x2 + 2y2 + z , a = 3i − 4j 3i − 4j a = grad (x2 + 2y2 + z) : (3) 2 + (− 4) 2 (3i − 4j) 6x − 16y = = (2xi + 4yj + k) : 5 25 At point P (1, 1, 2) the direction derivative is a = 6 # 1 − 16 # 1 =− 10 =− 2 5 5

SOL 1.47

Option (B) is correct. Given : 2x + 3y = 4 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 41

x+y+z = 4 x + 2y − z = a It is a set of non-homogenous equation, so the augmented matrix of this system is R V S2 3 0 : 4W 6A : B@ = S1 1 1 : 4W SS1 2 − 1 : aWW TR X V 4W S2 3 0 : S 4W R 3 " R 3 + R2 , R2 " 2R2 − R1 + 0 −1 2 : SS2 3 0 : 4 + aWW TR V X S2 3 0 : 4W + S0 − 1 2 : 4W R 3 " R 3 − R1 SS0 0 0 : aWW X T So, for a unique solution of the system of equations, it must have the condition ρ [A: B] = ρ [A] So, when putting a = 0 We get ρ [A: B] = ρ [A] SOL 1.48

Option (D) is correct. Here we check all the four options for unbounded condition. π/4 (A) #0 tan xdx = 8log sec x B0π/4 = 9log sec π4 − log sec 0 C = log 2 − log 1 = log 2 (B) (C) Let

#0 3 x2 1+ 1 dx #0 3xe−x dx I =

= 6tan−1 x @3 = tan−1 3 − tan−1 (0) = π − 0 = π 0 2 2

3

#0 xe

3 −x

#e

dx = x

0

= 6− xe @ + −x 3 0

dx −

# :dxd (x) # e

−x

dx D dx

0

3

#e

3

−x

−x

= 6− e−x (x + 1)@3 dx = 6− xe−x − e−x@3 0 0

0

=− [0 − 1] = 1 1 1 (D) #0 1 −1 x dx =− #0 x −1 1 dx =−6log (x − 1)@10 −6log 0 − log (− 1)@ Both log 0 and log (–1) undefined so it is unbounded.

SOL 1.49

Option (A) is correct. Let

I=

# f (z) dz

and f (z) = cos z z



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CHAP 1

cos z dz ...(i) z−0 Given that z = 1 for unit circle. From the Cauchy Integral formula ...(ii) # zf−(z)a dz = 2πi f (a) Compare equation (i) and (ii), we can say that, I =

Then

# cosz z dz = #

a = 0 and f (z) = cos z Or, f (a) = f (0) = cos 0 = 1 Now from equation (ii) we get # zf−(z)0 dz = 2πi # 1 = 2πi SOL 1.50

a=0

Option (D) is correct. y = 2 x3/2 3

Given

...(i)

#x

We know that the length of curve is given by

1

x2

)

dy 2 b dx l + 1 3 dx

...(ii)

Differentiate equation(i) w.r.t. x 3 dy = 2 # 3 x 2 − 1 = x1/2 = x 3 2 dx dy Substitute the limit x1 = 0 to x2 = 1 and in equation (ii), we get dx L =

#0

1

_ ( x ) 2 + 1 i dx =

#0

1

x + 1 dx

1

= :2 (x + 1) 3/2D = 1.22 3 0 SOL 1.51

Option (B) is correct. 1 2 λ1 and λ2 is the eigen values of the matrix. A => 0 2H For eigen values characteristic matrix is, Let

A − λI = 0 1 >0

2 1 − λ> H 2 0

0 =0 1H

(1 − λ) 2 =0 0 (2 − λ)

...(i)

(1 − λ) (2 − λ) = 0 & λ = 1 & 2 So, Eigen vector corresponding to the λ = 1 is, 0 2 1 >0 1H>a H = 0 2a + a = 0 & a = 0 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 43

Again for λ = 2 −1 2 1 > 0 0H>b H = 0 − 1 + 2b = 0 Then sum of

SOL 1.52

a &b & a + b = 0 + 1 = 1 2 2

Option (C) is correct. Given f (x, y) = yx First partially differentiate the function w.r.t. y 2f = xyx − 1 2y Again differentiate. it w.r.t. x 22 f = yx − 1 (1) + x ^yx − 1 log y h = yx − 1 ^x log y + 1h 2x2y At :

SOL 1.53

b=1 2

x = 2, y = 1 22 f = (1) 2 − 1 (2 log 1 + 1) = 1 (2 # 0 + 1) = 1 2x2y

Option (A) is correct. Given : y m + 2yl + y = 0 2 (D + 2D + 1) y = 0 The auxiliary equation is

where D = d/dx

m2 + 2m + 1 = 0 (m + 1) 2 = 0 , m =− 1, − 1 The roots of auxiliary equation are equal and hence the general solution of the given differential equation is, ..(i) y = (C1 + C2 x) em x = (C1 + C2 x) e−x Given y (0) = 0 at x = 0, & y = 0 Substitute in equation (i), we get 0 = (C1 + C2 # 0) e−0 0 = C1 # 1 & C1 = 0 Again y (1) = 0 , at x = 1 & y = 0 Substitute in equation (i), we get 0 = [C1 + C2 # (1)] e−1 = [C1 + C2] 1 e 1

C1 + C 2 = 0 & C 2 = 0 Substitute C1 and C2 in equation (i), we get y = (0 + 0x) e−x = 0 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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PAGE 44


And SOL 1.54

SOL 1.55

CHAP 1

y (0.5) = 0

Option (B) is correct. Given : y = x2 and interval [1, 5] At x=1 &y =1 And at x=5 y = (5) 2 = 25 Here the interval is bounded between 1 and 5 So, the minimum value at this interval is 1.

...(i)

Option (A) is correct Let square matrix x y A => y xH The characteristic equation for the eigen values is given by A − λI = 0 x−λ y =0 y x−λ (x − λ) 2 − y2 = 0 (x − λ) 2 = y2 x − λ =! y λ = x!y So, eigen values are real if matrix is real and symmetric.

SOL 1.56

Option (B) is correct. The Cauchy-Reimann equation, the necessary condition for a function f (z) to be analytic is 2ϕ 2ψ = 2y 2x 2ϕ 2ϕ 2ϕ 2ψ 2ψ 2ψ when , , , exist. =− 2x 2y 2x 2y 2y 2x

SOL 1.57

Option (A) is correct. 2 2 ϕ 2 2 ϕ 2ϕ 2ϕ Given : + 2 + + =0 2x 2y 2x 2 2y Order is determined by the order of the highest derivative present in it. Degree is determined by the degree of the highest order derivative present in it after the differential equation is cleared of radicals and fractions. So, degree = 1 and order = 2 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1

SOL 1.58


PAGE 45

Option (B) is correct. Given

y = x+

x+

y−x = x+ Squaring both the sides,

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

...(i)

x + x + ....3

(y − x) 2 = x + x + x + ......3 (y − x) 2 = y y2 + x2 − 2xy = y We have to find y (2), put x = 2 in equation (ii),

From equation (i) ...(ii)

y2 + 4 − 4y = y y2 − 5y + 4 = 0 (y − 4) (y − 1) = 0 y = 1, 4 From Equation (i) we see that For y (2) Therefore, SOL 1.59

y = 2+

2+

2 + 2 + .....3 > 2

y =4

Option (B) is correct.

Vector area of TABC , A = 1 BC # BA = 1 (c − b) # (a − b) 2 2 = 1 [c # a − c # b − b # a + b # b] 2 = 1 [c # a + b # c + a # b] 2 b # b = 0 and c # b =− (b # c) = 1 [(a − b) # (a − c)] 2 SOL 1.60

Option (C) is correct. dy dy Given : = y2 or 2 = dx dx y Integrating both the sides GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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PAGE 46


# dy y2

=

CHAP 1

# dx

−1 = x + C y Given y (0) = 1 at x=0 &y=1 Put in equation (i) for the value of C − 1 = 0 + C &C =− 1 1 From equation (i), −1 = x − 1 y y =− 1 x−1 For this value of y , x − 1 ! 0 or x ! 1 And x < 1 or x > 1 SOL 1.61

...(i)

Option (A) is correct. φ (t) =

Let

# f (t) dt and φ (0) = 0 then φl (t) = f (t) t

0

We know the formula of Laplace transforms of φl (t) is L 6φl (t)@ = sL 6φ ^ t h@ − φ (0) = sL 6φ (t)@ L 6φ (t)@ = 1 L 6φl (t)@ s Substitute the values of φ (t) and φl (t), we get t L ; f (t) dtE = 1 L 6f (t)@ s 0

φ (0) = 0

#

L;

or

# f (t) dtE = s1 F (s) t

0

SOL 1.62

Option (A) is correct. From the Trapezoidal Method b #a f (x) dx = h2 6f (x0) + 2f (x1) + 2f (x2) .....2f (xn − 1) + f (xn)@ Interval h = 2π − 0 = π 8 4 Find

2π

#0 sin xdx

...(i)

Here f (x) = sin x

Table for the interval of π/4 is as follows Angle θ

0

f (x) = sin x 0

π 4

π 2

0.707 1

3π 4

π

0.707 0

5π 4

3π 2

− 0.707 − 1

7π 4

− 0.707 0



2π

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CHAP 1


PAGE 47

Now from equation(i), 2π #0 sin xdx = π8 [0 + 2 (0.707 + 1 + 0.707 + 0 − 0.707 − 1 − 0.0707 + 0)] = π #0 = 0 8 SOL 1.63

Option (D) is correct. The X and Y be two independent random variables. So, E (XY) = E (X) E (Y) & covariance is defined as Cov (X, Y) = E (XY) − E (X) E (Y) = E (X) E (Y) − E (X) E (Y)

(i)

From eqn. (i)

=0 For two independent random variables Var (X + Y) = Var (X) + Var (Y) and E (X 2 Y 2) = E (X 2) E (Y 2) So, option (D) is incorrect. SOL 1.64

Option (B) is correct. 2

Let,

ex − b1 + x + x l 2 f (x) = lim 3 x"0 x x e − (1 + x) = lim x"0 3x2 x = lim e − 1 x " 0 6x

0 form 0 0 form 0 0 form 0

x 0 = lim e = e = 1 6 6 x"0 6

SOL 1.65

Option (B) is correct. 2 1 H A => 0 2 Let λ is the eigen value of the given matrix then characteristic matrix is 1 0 A − λI = 0 H = Identity matrix Here I = > 0 1 2−λ 1 =0 0 2−λ Let,

(2 − λ) 2 = 0 λ = 2, 2 So, only one eigen vector. GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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SOL 1.66


CHAP 1

Option (D) is correct. Column I P. Gauss-Seidel method

4. Linear algebraic equation

Q. Forward Newton-Gauss method

1. Interpolation

R. Runge-Kutta method

2. Non-linear differential equation

S. Trapezoidal Rule

3. Numerical integration

So, correct pairs are, P-4, Q-1, R-2, S-3 SOL 1.67

Option (B) is correct. 2 dy Given : + 2xy = e−x and y (0) = 1 dx It is the first order linear differential equation so its solution is y (I.F.) = So,

# Q (I.F.) dx + C

I. F . = e

# Pdx

=e

# 2xdx x2

= e2 # xdx = e2 # 2 = ex The complete solution is,

2

yex =

# e−x # ex dx + C

=

# dx + C = x + C

2

2

compare with dy + P (y) = Q dx

2

y = x +x2 c e Given y (0) = 1 At x =0 &y=1 Substitute in equation (i), we get 1 =C &C=1 1 2 Then y = x +x2 1 = (x + 1) e−x e

...(i)

SOL 1.68

Option (C) is correct. The incorrect statement is, S = {x : x ! A and x ! B} represents the union of setA and set B . The above symbol (!) denotes intersection of set A and set B . Therefore this statement is incorrect.

SOL 1.69

Option (D) is correct. Total number of items = 100 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 49

Number of defective items = 20 Number of Non-defective items = 80 Then the probability that both items are defective, when 2 items are selected at random is, 20! 20 # 19 80 20 C C 18 ! 2 ! 2 0 2 = P = 100 = = 19 100 # 99 495 C2 100! 2 98!2!

Alternate Method : Here two items are selected without replacement. Probability of first item being defective is P1 = 20 = 1 100 5 After drawing one defective item from box, there are 19 defective items in the 99 remaining items. Probability that second item is defective, P2 = 19 899 then probability that both are defective P = P1 # P2 = 1 # 19 = 19 5 99 495 SOL 1.70

Option (A) is correct. 3 2 H S => 2 3 Eigen values of this matrix is 5 and 1. We can say λ1 = 1 λ2 = 5 Then the eigen value of the matrix Given :

S 2 = S S is λ12 , λ22 Because. if λ1, λ2, λ3 .... are the eigen values of A, then eigen value of Am are λ1m, λm2 , λm3 .... Hence matrix S 2 has eigen values (1) 2 and (5) 2 & 1 and 25 SOL 1.71

Option (B) is correct. Given f (x) = (x − 8) 2/3 + 1 The equation of line normal to the function is (y − y1) = m2 (x − x1) Slope of tangent at point (0, 5) is m1 = f l (x) = :2 (x − 8) −1/3D 3 (0, 5) 1 m1 = f l (x) = 2 (− 8) −1/3 =− 2 (23) − 3 =− 1 3 3 3 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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...(i)

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PAGE 50


CHAP 1

We know the slope of two perpendicular curves is − 1. m1 m2 =− 1 m2 =− 1 = − 1 = 3 m1 − 1/3 The equation of line, from equation (i) is (y − 5) = 3 (x − 0) y = 3x + 5 SOL 1.72


it π/3 iπ/3 0 eit dt = :e D & e − e i 0 i i π = 1 6e 3 i − 1@ = 1 9cos π + i sin π − 1C i 3 3 i

f (x) =

#0

π/3

= 1 ;1 + i 3 − 1E = 1 ;− 1 + 3 iE i 2 2 i 2 2 = 1 # i ;− 1 + 3 iE =− i ;− 1 + 3 iE 2 2 i i 2 2

i2 =− 1

= i ; 1 − 3 iE = 1 i − 3 i 2 = 3 + 1 i 2 2 2 2 2 2 SOL 1.73

Option (B) is correct. Given

2 f (x) = 2x2 − 7x + 3 5x − 12x − 9

Then

2 lim f (x) = lim 2x2 − 7x + 3 x"3 x " 3 5x − 12x − 9

Applying L – Hospital rule = lim 4x − 7 x " 3 10x − 12 Substitute the limit, we get lim f (x) = 4 # 3 − 7 = 12 − 7 = 5 10 # 3 − 12 30 − 12 18 x"3 SOL 1.74

Option (A) is correct. (P) Singular Matrix " Determinant is zero A = 0 (Q) Non-square matrix " An m # n matrix for which m ! n , is called nonsquare matrix. Its determinant is not defined (R) Real Symmetric Matrix " Eigen values are always real. (S) Orthogonal Matrix " A square matrix A is said to be orthogonal if AAT = I Its determinant is always one.

SOL 1.75

Option (B) is correct. GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


Given :

PAGE 51

d 2y dy 2x 2 + 4 dx + 3y = 3e dx d =D dx

[D2 + 4D + 3] y = 3e2x The auxiliary Equation is, Then

SOL 1.76

m2 + 4m + 3 = 0 & m =− 1, − 3 C.F. = C1 e−x + C2 e−3x 2x 3e2x P.I. = 2 3e = D + 4D + 3 (D + 1) (D + 3) 2x 2x 3e2x = = 3e = e 5 (2 + 1) (2 + 3) 3 # 5

Option (C) is correct. Given EF = G Rcos θ − sin θ S cos θ S sin θ SS 0 0 T We know that the matrix

SOL 1.77

Put D = 2

where G = I = Identity matrix

R1 0VW S 0W # F = S0 SS0 1WW X T multiplication

0 1 0 of

0VW 0W 1WW X a matrix and its inverse be a identity

AA−1 = I So, we can say that F is the inverse matrix of E [adj.E] F = E −1 = E Rcos θ − (sin θ) 0VT R cos θ sin θ 0V W S W S adjE = S sin θ cos θ 0W = S− sin θ cos θ 0W SS 0 SS 0 1WW 0 0 1WW X X T T E = 6cos θ # (cos θ − 0)@ − 8^− sin θh # ^sin θ − 0hB + 0 = cos2 θ + sin2 θ = 1 R cos θ sin θ 0V W adj . E 6 @ S Hence, F = = S− sin θ cos θ 0W E SS 0 0 1WW X T Option (B) is correct. The probability density function is, 1+t for − 1 # t # 0 f (t) = ) 1−t for 0 # t # 1 For standard deviation first we have to find the mean and variance of the function. Mean (t ) = =

#−13t f (t) dt 0

=

0

#−1 t (1 + t) dt + #0

#−1 (t + t2) dt + #0

1

1

t (1 − t) dt

(t − t2) dt



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CHAP 1

2 3 0 2 3 1 = :t + t D + :t − t D = :− 1 + 1 D + :1 − 1 D = 0 2 3 −1 2 3 0 2 3 2 3

And

variance ^σ2h =

#− 33 (t − t ) 2 f (t) dt

t=0

0

=

#−1 t2 (1 + t) dt + #0

=

#−1 (t2 + t3) dt + #0

0

1

1

t2 (1 − t) dt

(t2 − t3) dt

3 4 0 3 4 1 = :t + t D + :t − t D 3 4 −1 3 4 0 =−:− 1 + 1 D + :1 − 1 − 0D = 1 + 1 = 1 3 4 3 4 12 12 6 Now, standard deviation (σ2) s = 1 = 1 6 6

SOL 1.78

Option (A) is correct. The Stokes theorem is,

#C F : dr

=

##S (4 # F) : ndS

=

##S (Curl F) : dS

Here we can see that the line integral # F : dr and surface integral C ## (Curl F) : ds is related to the stokes theorem. S

SOL 1.79

Option (B) is correct. Let, P = defective items Q = non-defective items 10% items are defective, then probability of defective items P = 0.1 Probability of non-defective item Q = 1 − 0.1 = 0.9 The Probability that exactly 2 of the chosen items are defective is = 10 C2 (P) 2 (Q) 8 = 10! (0.1) 2 (0.9) 8 8! 2! = 45 # (0.1) 2 # (0.9) 8 = 0.1937

SOL 1.80


f (x) = =

#−a (sin6 x + sin7 x) dx a

#−a sin6 xdx + #−a sin7 xdx a

a

We know that GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


#−a f (x) dx a

0 =* a 2 # f (x) 0

PAGE 53

when f (− x) =− f (x); odd function when f (− x) = f (x); even function

Now, here sin6 x is an even function and sin7 x is an odd function. Then, f (x) = 2 # sin6 xdx + 0 = 2 # sin6 xdx a

a

0

0

SOL 1.81

Option (C) is correct. We know, from the Echelon form the rank of any matrix is equal to the Number of non zero rows. Here order of matrix is 3 # 4 , then, we can say that the Highest possible rank of this matrix is 3.

SOL 1.82

Option (A) is correct. I =

Given

8

2

#0 #π/4 f (x, y) dydx

We can draw the graph from the limits of the integration, the limit of y is from y = x to y = 2 . For x the limit is x = 0 to x = 8 4

Here we change the order of the integration. The limit of x is 0 to 8 but we have to find the limits in the form of y then x = 0 to x = 4y and limit of y is 0 to 2 So

8

2

#0 #x/4 f (x, y) dydx

=

2

#0 #0

4y

f (x, y) dxdy =

#r #p s

q

f (x, y) dxdy

Comparing the limits and get r = 0 , s = 2 , p = 0 , q = 4y SOL 1.83

Option (A) is correct.

R S5 0 S0 5 Let, A =S S0 0 S0 0 T The characteristic equation for eigen values

V 0W 0W 1WW 1W X is given by, 0 0 2 3



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CHAP 1

A − λI = 0 5−λ 0 0 0 0 5−λ 0 0 A= =0 0 0 2−λ 1 0 0 3 1−λ Solving this, we get (5 − λ) (5 − λ) [(2 − λ) (1 − λ) − 3] = 0 (5 − λ) 2 [2 − 3λ + λ2 − 3] = 0 (5 − λ) 2 (λ2 − 3λ − 1) = 0 So, (5 − λ) 2 = 0 & λ = 5 , 5 and λ2 − 3λ − 1 = 0 − (− 3) ! 9 + 4 = 3 + 13 , 3 − 13 2 2 2 The eigen values are λ = 5 , 5, 3 + 13 , 3 − 13 2 2 R V Sx1W Sx2W Let X1 = S W Sx 3W Sx 4W T X be the eigen vector for the eigen value λ = 5 Then, (A − λI ) X1 = 0 λ =

(A − 5I ) X1 = 0 VR V 0 0 0WSx1W 0 0 0WSx2W =0 0 − 3 1WWSSx 3WW 0 3 − 4WSx 4W XT X or − 3x 3 + x 4 = 0 3x 3 − 4x 4 = 0 This implies that x 3 = 0 , x 4 = 0 Let x1 = k1 and x2 = k2 R V Sk1W Sk2W So, eigen vector, where k1 , k2 ε R X1 = S W S0W S0W T X Option (C) is correct. Given : ...(i) x+y = 2 ...(ii) 1.01x + 0.99y = b , db = 1 unit We have to find the change in x in the solution of the system. So reduce y R S0 S0 S0 S S0 T

SOL 1.84



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PAGE 55

From the equation (i) and (ii). Multiply equation (i) by 0.99 and subtract from equation (ii) 1.01x + 0.99y − (0.99x + 0.99y) = b − 1.98 1.01x − 0.99x = b − 1.98 0.02x = b − 1.98 Differentiating both the sides, we get 0.02dx = db dx = 1 = 50 unit 0.02 SOL 1.85

SOL 1.86

Option (A) is correct. Given, x (u, v) = uv dx = v , du And y (u, v) = v u 2y =− v2 2u u We know that, R2x S 2u φ (u, v) = S2y S S2u T v φ (u, v) = >− 2v u

dx = u dv

2y =1 2v u 2x VW 2v W 2y W 2v W X u 1 H = v # 1 − u # − v = v + v = 2v a u2 k u u u u u

Option (D) is correct.

Given : Radius of sphere r = 1 Let, Radius of cone = R Height of the cone = H GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


db = 1

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CHAP 1

Finding the relation between the volume and Height of the cone From ΔOBD , OB 2 = OD 2 + BD 2 1 = (H − 1) 2 + R2 = H 2 + 1 − 2H + R2 R2 + H 2 − 2H = 0 R2 = 2H − H 2 Volume of the cone, V = 1 πR 2 H 3 2 Substitute the value of R from equation (i), we get V = 1 π (2H − H 2) H = 1 π (2H 2 − H 3) 3 3 Differentiate V w.r.t to H dV = 1 π [4H − 3H 2] 3 dH

...(i)

d 2 V = 1 π [4 − 6H] 3 dH 2 For minimum and maximum value, using the principal of minima and maxima. Put dV = 0 dH 1 π [4H − 3H 2] = 0 3 H [4 − 3H] = 0 & H = 0 and H = 4 3 Again differentiate

d 2 V = 1 π 4 − 6 4 = 1 π [4 − 8] =− 4 π < 0 (Maxima) # 3D 3 3 : 3 dH 2 2 (Minima) And at H = 0 , d V2 = 1 π [4 − 0] = 4 π > 0 3 3 dH So, for the largest volume of cone, the value of H should be 4/3 At H = 4 , 3

SOL 1.87

Option (D) is correct. 2 ln (x) dy Given : x2 + 2xy = x dx 2 ln (x) dy 2y + = x dx x3 Comparing this equation with the differential equation

dy + P (y) = Q we dx

2 ln (x) have P = 2 and Q = x x3 The integrating factor is, 2

I.F.= e # Pdx = e # x dx e2 lnx = e lnx = x2 2



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PAGE 57

Complete solution is written as, y (I.F.) = y (x2) = Integrating the value Let,

# Q (I.F.) dx + C x 2 # 2 ln # x dx + C x3

# ln x # x1 dx

I =

= 2 # ln x # 1 dx + C x (II)

Separately

# ln x # x1 dx (I)

...(i)

(I)

...(ii)

(II)

= ln x # 1 dx − x = ln x ln x −

d (ln x) 1 # &dx # # x dx 0 dx

# x1 # ln xdx

From equation(ii)

1 444 2 444 3 I

2I = (ln x) 2 (ln x) 2 2 Substitute the value from equation (iii) in equation (i), 2 (ln x) 2 +C y (x2) = 2 I =

...(iii)

x2 y = (ln x) 2 + C Given y (1) = 0 , means at x = 1 &y = 0

...(iv)

or

then 0 = (ln 1) 2 + C & C = 0 So from equation (iv), we get Now at x = e , SOL 1.88

x2 y = (ln x) 2 (ln e) 2 y (e) = = 12 2 e e

Option (A) is correct. Potential function of v = x2 yz at P (1, 1, 1) is = 12 # 1 # 1 = 1 and at origin O (0, 0, 0) is 0. Thus the integral of vector function from origin to the point (1, 1, 1) is = 6x2 yz @ P − 6x2 yz @O

= 1−0 = 1 SOL 1.89

Option (C) is correct. Let, f (x) = x3 + 3x − 7 From the Newton Rapson’s method f (xn) xn + 1 = xn − f l (xn)

...(i)



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CHAP 1

We have to find the value of x1 , so put n = 0 in equation (i), f (x 0) x1 = x 0 − f l (x 0) = x3 + 3x − 7 = 13 + 3 # 1 − 7 = 1 + 3 − 7 =− 3 = 3x2 + 3 = 3 # (1) 2 + 3 = 6 (− 3) x1 = 1 − = 1 + 3 = 1 + 1 = 3 = 1.5 6 6 2 2

f (x) f (x 0) f l (x) f l (x 0) Then, SOL 1.90

x0 = 1

Option (D) is correct. We know a die has 6 faces and 6 numbers so the total number of ways = 6 # 6 = 36 And total ways in which sum is either 8 or 9 is 9, i.e. (2, 6), (3, 6) (3, 5) (4, 4) (4, 5) (5, 4) (5, 3) (6, 2) (6, 3) Total number of tosses when both the 8 or 9 numbers are not come = 36 − 9 = 27 Then probability of not coming sum 8 or 9 is, = 27 = 3 36 4

SOL 1.91

Option (C) is correct. d 2y dy Given : 2 + p dx + qy = 0 dx The solution of this equation is given by, y = c1 emx + c2 enx Here m & n are the roots of ordinary differential equation Given solution is, y = c1 e−x + c2 e−3x Comparing equation (i) and (ii), we get m =− 1 and n =− 3 Sum of roots, m + n =− p − 1 − 3 =− p & p = 4 and product of roots, mn = q (− 1) (− 3) = q & q = 3

SOL 1.92

...(i) ...(ii)

Option (C) is correct. d 2y dy Given : 2 + p dx + (q + 1) y = 0 dx [D2 + pD + (q + 1)] y = 0 From the previous question, put p = 4 and m = 3 [D2 + 4D + 4] y = 0 The auxilliary equation of equation (i) is written as GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 59

m2 + 4m + 4 = 0 & m =− 2, − 2 Here the roots of auxiliary equation are same then the solution is Let c1 = 0 y = (c1 + c2 x) emx = xe−2x e o c2 = 1 SOL 1.93

Option (C) is correct. Given : x = a (θ + sin θ), y = a (1 − cos θ) First differentiate x w.r.t. θ, dx = a [1 + cos θ] dθ And differentiate y w.r.t. θ dy = a [0 − (− sin θ)] = a sin θ dθ dy dy dθ = dy/dθ We know, = dx dθ # dx dx/dθ dy Substitute the values of and dx dθ dθ

SOL 1.94

2 sin θ cos θ dy θ 1 sin 2 2 = a sin θ # = = dx 2θ a [1 + cos θ] 1 + cos θ 2 cos 2 θ sin 2 = tan θ = cos θ + 1 = 2 cos2 θ 2 2 θ cos 2 Option (C) is correct. Given : P (0.866, 0.500, 0), so we can write P = 0.866i + 0.5j + 0k Q = (0.259, 0.966, 0), so we can write Q = 0.259i + 0.966j + 0k For the coplanar vectors P : Q = P Q cos θ cos θ =

P:Q P Q

P : Q = (0.866i + 0.5j + 0k) : (0.259i + 0.966j + 0k) = 0.866 # 0.259 + 0.5 # 0.966 So,

SOL 1.95

0.866 # 0.259 + 0.5 # 0.966 (0.866) 2 + (0.5) 2 + (0.259) 2 + (0.966) 2 0.70729 = 0.707 = 0.22429 + 0.483 = 0.99 # 1.001 0.99 # 1.001 θ = cos−1 (0.707) = 45c

cos θ =

Option (B) is correct. GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1

R1 2 3V W S Let A = S1 5 1W SS3 1 1WW X T We know that the sum of the eigen value of a matrix is equal to the sum of the diagonal elements of the matrix So, the sum of eigen values is, 1+5+1 = 7 SOL 1.96

Option (D) is correct. Given : Total number of cards = 52 and two cards are drawn at random. Number of kings in playing cards = 4 So the probability that both cards will be king is given by, 4 3 P = 52C1 # 51C1 = 4 # 3 = 1 52 51 221 C1 C1

SOL 1.97

n

Cr =

Option (B) is correct. Given :

0, U (t − a) = * 1,

for t < a for t $ a

From the definition of Laplace Transform L [F (t)] = L 6U (t − a)@ = =

#0 3e−st f (t) dt #0 3e−st U (t − a) dt #0

a −st

e

(0) +

#a 3e−st (1) dt = 0 + #a 3e−st dt

−as −as −st 3 L 6U (t − a)@ = :e D = 0 − :e D = e −s s −s a

SOL 1.98

Option (D) is correct. First we have to make the table from the given data

Take x 0 = 0 and h = 1 Then P = x − x0 = x h GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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n r n−r

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PAGE 61

From Newton’s forward Formula P (P − 1) 2 P (P − 1) (P − 2) 3 f (x) = f (x 0) + P Δf (0) + Δ f (0) + Δ f (0) 1 2 3 x (x − 1) 2 x (x − 1) (x − 2) 3 = f (0) + xΔf (0) + Δ f (0) + Δ f (0) 2 6 x (x − 1) x (x − 1) (x − 2) = 1 + x (1) + (− 2) + (12) 2 6 = 1 + x − x (x − 1) + 2x (x − 1) (x − 2) f (x) = 2x3 − 7x2 + 6x + 1 SOL 1.99

Option (A) is correct. Given :

V =

π/3

2π

#0 #0 #0

1

r2 sin φdrdφdθ

First integrating the term of r , we get V =

2π

#0 #0

π/3

r3 1 sin φdφdθ = :3D 0

2π

#0 #0

π/3

1 sin φdφdθ 3

Integrating the term of φ, we have

#0

2π

=− 1 3 =− 1 3

#0

V =1 3

π/3 6− cos φ@0 dθ

π 1 2π 1 9cos 3 − cos 0C dθ =− 3 #0 :2 − 1D dθ 2π 2π #0 b− 12 ldθ =− 13 # b− 12 l #0 dθ 2π

Now, integrating the term of θ, we have V = 1 6θ@ 20π = 1 [2π − 0] = π 6 6 3 SOL 1.100

Option (A) is correct.

R8 x S Let, A =S 4 0 SS12 6 T For singularity of the matrix A = 0 8 x 0 4 0 2 =0 12 6 0

0VW 2W 0WW X

8 [0 − 2 # 6] − x [0 − 24] + 0 [24 − 0] = 0 8 # (− 12) + 24x = 0 − 96 + 24x = 0 & x = 96 = 4 24 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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SOL 1.101


CHAP 1

Option (A) is correct 2 2 f (x) = lim sin x = lim sin x # x x x x x"0 x"0 2 = lim b sin x l # x x x"0

Let,

lim sin x = 1 x"0 x

= (1) 2 # 0 = 0 Alternative : 2

Let

f (x) = lim sin x x x"0 f (x) = lim 2 sin x cos x 1 x"0 = lim sin 2x = sin 0 = 0 1 1 x"0

0 : 0 formD Apply L-Hospital rule

SOL 1.102

Option (D) is correct. Accuracy of Simpson’s rule quadrature is O (h5)

SOL 1.103

Option (C) is correct. 4 1 A => 1 4H The characteristic equation for the eigen value is given by, Let,

A − λI = 0

1 0 I = Identity matrix > 0 1H

4 1 1 0 >1 4H − λ >0 1H = 0 4−λ 1

1 =0 4−λ

(4 − λ) (4 − λ) − 1 = 0 (4 − λ) 2 − 1 = 0 λ2 − 8λ + 15 = 0 Solving above equation, we get λ = 5, 3 SOL 1.104

Option (C) is correct. Given : x + 2y + z = 6 2x + y + 2z = 6 x+y+z = 5 Comparing to Ax = B ,we get GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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R1 2 1V R6V W S S W A = S2 1 2W, B = S6W SS1 1 1WW SS5WW X T X T Write the system of simultaneous equations in the form of Augmented matrix, R1 2 1 : 6V W S R2 " R2 − 2R1 and R 3 " 2R 3 − R2 6A: B@ = S2 1 2 : 6W SS1 1 1 : 5WW TR1 2 1 : X6V W S + S0 − 3 0 : − 6W R 3 " 3R 3 + R2 SS0 1 0 : 4WW RT1 2 1 : 6VX W S + S0 − 3 0 : − 6W SS0 0 0 : 6WW X T It is a echelon form of matrix. Since ρ 6A@ = 2 and ρ 5A: B? = 3 ρ [A] ! ρ [A: B ] So, the system has no solution and system is inconsistent. SOL 1.105

Option (B) is correct. Given : y = x2 and y = x . The shaded area shows the area, which is bounded by the both curves.

Solving given equation, we get the intersection points as, In y = x2 putting y = x we have x = x2 or x2 − x = 0 which gives x = 0, 1 Then from y = x we can see that curve y = x2 and y = x intersects at point (0, 0) and (1, 1). So, the area bounded by both the curves is y = x2

x=1

A=

# #

x=0

dydx =

y=x

x=1

#

x=0

dx

y = x2

#

y=x

x=1

dy =

# dx6y @

x2 x

x=0



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x=1

=

# (x

2

x=0

CHAP 1

3 2 1 − x) = :x − x D = 1 − 1 =− 1 = 1 unit2 3 2 0 3 2 6 6

Area is never negative SOL 1.106

Option (A) is correct. dy + y2 = 0 dx dy =− y2 dx dy − 2 = dx y Integrating both the sides, we have dy − # 2 = # dx y y−1 = x + c & y =

SOL 1.107

1 x+c

Option (C) is correct. Given : F = xi − yj First Check divergency, for divergence, Grade F = 4:F = ; 2 i + 2 j + 2 k E:6xi − yj @ = 1 − 1 = 0 2x 2y 2z So we can say that F is divergence free. Now checking the irrationalit;. For irritation the curl F = 0 Curl F = 4# F = ; 2 i + 2 j + 2 k E # [xi − yj] 2x 2y 2z R V j k W S i 2 2 2 W = i [0 − 0] − j [0 − 0] + k [0 − 0] = 0 =S S2x 2y 2z W S x −y 0 W T X So, vector field is irrotational. We can say that the vector field is divergence free and irrotational.

SOL 1.108

Option (B) is correct. Let f (t) = sin ωt From the definition of Laplace transformation L [F (t)] = =

#0 3e−st f (t) dt #0 3e−st b e

iωt

=

#0 3e−st sin ωtdt

− e−iωt dt l 2i



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PAGE 65

iωt −iωt 3 sin ωt = e − e = 1 # (e−st eiωt − e−st e−iωt) dt 2i 2i 0 3 = 1 # 6e(− s + iω) t − e− (s + iω) t@ dt 2i 0 Integrating above equation, we get − (s + iω) t (− s + iω) t sin ωt = 1 = e − e 2i − s + iω − (s + iω)G 0

3

− (s + iω) t (− s + iω) t = 1 =e +e 2i − s + iω (s + iω)G 0 Substitute the limits, we get −0 e0 sin ωt = 1 =0 + 0 − e + e 2i (− s + iω) s + iω oG =− 1 ; s + iω + iω − s E 2i (− s + iω) (s + iω) 2iω =− 1 # = −ω = 2ω 2 2i (iω) 2 − s 2 − ω2 − s 2 ω +s Alternative : From the definition of Laplace transformation 3

L [F (t)] =

#0 3e−st sin ωtdt

a =− s and eat a sin bt − b cos bt @ 26 e o a +b b=ω 3 −st Then, L [sin ωt] = ; 2e 2 ^− s sin ωt − ω cos ωt hE s +ω 0 −3 −0 e e =; 2 (− s sin 3 − ω cos 3)E − ; 2 (− s sin 0 − ω cos 0)E s + ω2 s + ω2 = 0 − 2 1 2 [0 − ω] =− 2 1 2 (− ω) s +ω s +ω L [sin ωt] = 2 ω 2 s +ω We know # eat sin btdt =

SOL 1.109

2

Option (D) is correct. Given : black balls = 5, Red balls = 5, Total balls=10 Here, two balls are picked from the box randomly one after the other without replacement. So the probability of both the balls are red is 5! 5! 5 # 3!2! 5 n C 0 ! 5 ! 0 # C2 P = = # = 1 # 10 = 10 = 2 n Cr = 10 45 45 9 C2 10! r n−r 3!2!

Alternate Method : Given :

Black balls = 5 ,



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CHAP 1

Red balls = 5 Total balls = 10 The probability of drawing a red bell, P1 = 5 = 1 10 2 If ball is not replaced, then box contains 9 balls. So, probability of drawing the next red ball from the box. P2 = 4 9 Hence, probability for both the balls being red is, P = P1 # P2 = 1 # 4 = 2 2 9 9 SOL 1.110

Option (A) is correct. We know that a dice has 6 faces and 6 numbers so the total number of cases (outcomes) = 6 # 6 = 36 And total ways in which sum of the numbers on the dices is eight, (2, 6) (3, 5) (4, 4) (5, 3) (6, 2) So, the probability that the sum of the numbers eight is, p = 5 36

SOL 1.111

Option (D) is correct. We have to draw the graph on x -y axis from the given functions.

−x f (x) = * 0 x

x #− 1 x=0 x$1

It clearly shows that f (x) is differential at x =− 1, x = 0 and x = 1, i.e. in the domain [− 1, 1]. So, (a), (b) and (c) are differential and f (x) is maximum at (x, − x). SOL 1.112

Option (B) is correct. If the scatter diagram indicates some relationship between two variables X GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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PAGE 67

and Y , then the dots of the scatter diagram will be concentrated round a curve. This curve is called the curve of regression. Regression analysis is used for estimating the unknown values of one variable corresponding to the known value of another variable. SOL 1.113

Option (B) is correct. Given : 3x + 2y + z = 4 x−y+z = 2 − 2x + 2z = 5 The Augmented matrix of the given system of equation is R 3 2 1 : 4V W S 6A : B@ = S 1 − 1 1 : 2W R 3 " R 3 + 2R2 , R2 " R2 − R1 SS− 2 0 2 : 5WW X T R 3 2 1 : 4V W S + S− 2 − 3 0 : − 2W SS 0 − 2 4 : 9WW X of unknown) Here ρ 6A : B@ = ρT6A@ = 3 = n (number Then the system of equation has a unique solution.

SOL 1.114

Option (B) is correct. Given : f (x, y) = 2x2 + 2xy − y3 Partially differentiate this function w.r.t x and y , 2f 2f = 2x − 3y2 = 4x + 2y , 2x 2y For the stationary point of the function, put 2f/2x and 2f/2y equal to zero. 2f ...(i) & 2x + y = 0 = 4x + 2y = 0 2x 2f ...(ii) and & 2x − 3y2 = 0 = 2x − 3y2 = 0 2y From equation (i), y =− 2x substitute in equation (ii), 2x − 3 (− 2x) 2 = 0 2x − 3 # 4x2 = 0 6x2 − x = 0 & x = 0 , 1 6 From equation (i), For x = 0 , y =− 2 # (0) = 0 and for x = 1 , y =− 2 # 1 =− 1 6 6 3 So, two stationary point at (0, 0) and b 1 , − 1 l 6 3 GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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SOL 1.115


CHAP 1

Option (B) is correct. Sample space = (1, 1), (1, 2) ... (1, 8) (2, 1), (2, 2) (3, 1), (3, 2) h h (8, 1), (8, 2)

f f h f

(2, 8) (3, 8) h (8, 8)

Total number of sample space = 8 # 8 = 64 Now, the favourable cases when Manish will arrive late at D = (6, 8), (8, 6)...(8, 8) Total number of favourable cases = 13 Probability = Total number of favourable cases Totol number of sample space = 13 64

So,

SOL 1.116

Option (B) is correct. Divergence is defined as d:r where r = xi + yj + zk and So,

d= 2 i+ 2 j+ 2 k 2x 2y 2z d:r = c 2 i + 2 j + 2 k m:(xi + yj + zk) 2x 2y 2z d:r = 1 + 1 + 1 = 3

SOL 1.117

Option (B) is correct. Given : x+y = 2 2x + 2y = 5 The Augmented matrix of the given system of equations is 1 1 : 2 6A : B@ = >2 2 : 5H Applying row operation, R2 " R2 − 2R1 1 1 : 2

6A : B@ = >0 0 : 1H ρ [A] = 1 ! ρ 6A : B@ = 2 So, the system has no solution. SOL 1.118

Option (D) is correct. Given : f (x) = x GATE Previous Year Solved Paper For Mechanical Engineering Published by: NODIA and COMPANY


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CHAP 1


PAGE 69

x if x > 0 f (x) = * 0 if x = 0 −x if x < 0 f (0 − h) − f (0) − (− h) Lf l (x) = lim = lim − 0 =− 1 h"0 h"0 −h −h f (0 + h) − f (0) Rf l (x) = lim = lim h − 0 = 1 h"0 h"0 h h Since Lfl (0) ! Rf l (0) So, derivative of f (x) at x = 0 does not exist. SOL 1.119

Option (A) is correct. The surface integral of the normal component of a vector function F taken around a closed surface S is equal to the integral of the divergence of F taken over the volume V enclosed by the surface S . Mathematically

## F:n dS S

=

### div Fdv V

So, Gauss divergence theorem relates surface integrals to volume integrals. SOL 1.120

Option (A) is correct. Given :

3 f (x) = x − x 3

f l (x) = x2 − 1 f m (x) = 2x Using the principle of maxima – minima and put f l (x) = 0 x2 − 1 = 0 & x = ! 1 Hence at x =− 1, (Maxima) f m (x) =− 2 < 0 at x = 1, (Minima) f m (x) = 2 > 0 So, f (x) is minimum at x = 1 SOL 1.121

Option (B) is correct.

Ra V S 1W Let A = Sb1W, B = 8a2 b2 c2B SSc WW 1 T X C = AB Ra V Ra a a b 1 2 S 1W S1 2 Let = Sb1W # 8a2 b2 c2B = Sb1 a2 b1 b2 SSc WW SSc a c b 1 2 1 1 2 X T matrix T all the The 3 # 3 minor of this is zero and zero. So the rank of this matrix is 1.

a1 c2VW b1 c2W c1 c2WW 2 # 2X minors are also



ISBN: 9788192276250

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PAGE 70


CHAP 1

ρ 6C @ = 1 SOL 1.122

Option (D) is correct. In a coin probability of getting head p = 1 and probability of getting tail, 2 1 1 q = 1− = 2 2 When unbiased coin is tossed three times, then total possibilities are H H H H H T H T H T H H H T T T T H T H T T T T From these cases, there are three cases, when head comes exactly two times. So, the probability of getting head exactly two times, when coin is tossed 3 times is, 2 P = 3C2 (p) 2 (q) 1 = 3 # b 1 l # 1 = 3 2 2 8

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ISBN: 9788192276250

Engineering Mathematics

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