# Engineering Mathematics Test - 3 - PDF Flipbook

Engineering Mathematics Test - 3

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GATE
EEE

Engineering
Mathematics

Test-03Solutions

ENGINEERING MATHEMATICS
1. For the following matrix �21 −31� the number of positive

characteristic roots is

a) One

b) Two

c) Four

d) Cannot be found

Answer: (d)

Solution: A = �12 −31�
Given

⇒ |A − λI| = 0

⇒ �1 − λ 3−−1λ�
2

⇒ λ2 − 4λ + 5 = 0

λ=2±i

2 1 and = �30 21� then L × M is
2. Given matrix L = �3 2�
5
4

81
a) �13 2�

22 5

65
b) � 9 8 �

12 13

18
c) �2 13�

5 22

1

62
d) �9 4�

05

Answer: (b)

Solution:

2 1 �03 12� = 6 5
= �3 2� �9 8�
5 12 13
4

3. Among the following, the pair of vectors orthogonal to each

other is

a) [3 4 7], [3 4 7]

b) [1 0 0], [1 1 0]

c) [1 0 2], [0 5 0]

d) [1 1 1], [-1 -1 -1]

Answer: (c)

Solution:
The two vectors 1× and 1 × are said to be orthogonal if
1× × 1 = 0 (or) YXT = 0

Let X = [1 0 2] and [0 5 0]

Then 0
XYT = [1 0 2]�5�

0

=0+0+0=0

∴X, Y are orthogonal to each other

2

13 2
4. If the determinant of the matrix �0 5 −6� is 26, then the
8
27

27 8
determinant of the matrix �0 5 −6� is

13 2

a) –26

b) 26

c) 0

d) 52

Answer: (a)

Solution:

By the properties of determinant of the matrices, if two rows are

interchanged in a determinant then the value of the determinant

does not change but sign will change (In this problem R1 and

R3 are interchanged.)

11 3
5. The minimum and maximum Eigen values of matrix �1 5 1�
1
31

are –2 and 6 respectively. What is the other Eigen value?

a) 5

b) 3

c) 1

d) –1

Answer: (b)

Solution:

Two Eigen values are given –2 & 6

3

Sum of Eigen values of A = tr(A)

λ + (−2) + 6 = 1 + 5 + 1

⇒λ=3

6. The number of linearly independent Eigen vectors of �02 21� is
a) 0

b) 1

c) 2

d) Infinite

Answer: (b)

Solution:

Eigen values of A are 2, 2

Consider (A − λI) = �2 − λ 1 λ�
0 −
2

For λ = 2, (A – 2I) = �00 10�

∴ The number of linearly independent eigen vectors of matrix A

= (number of variables) − ρ(A − λI)

=2−1=1

7. Area bounded by the curve y = x2 and the lines x = 4 and y = 0

is given by

a) 64

b) 64
3

c) 128
3

d) 128
4

Answer: (b)

4

Solution:

Area = ∫04 x2dx = 64
3

8. Limit of the function, nli→m∞ n is
√n2+n

a) 1/2

b) 0

c) ∞

d) 1

Answer: (d)

Solution:

l i→m∞ = l i→m∞ 1
√ 2+ �1+ 1

= 1 = 1
√1+0

9. The value of the following definite integral in ∫−ππ��22 Sin2x dx =
1+cosx

a) -2 log2

b) 2

c) 0

d) None

Answer: (d)

5

Solution:

f(x) = Sin2x , f(−x) = −f(x)
1+cosx

⇒ f(x) is odd function

∫−ππ��22 f(x)dx = 0

10. The following plot shows a function y which varies linearly

with x. The value of the integral I = ∫12 ydx

a) 1
b) 2.5
c) 4
d) 5
Answer: (b)
Solution:

I = ∫12 ydx

6

Equation of the straight line is x + y = 1
−1 1

⇒y=1+x

I = ∫12(1 + x)dx = 5 = 2.5
2

11. Consider the function f (x) = (x2 – 4)2 where x is a real number.

Then the function has

a) Only one minimum

b) Only two minima

c) Three minima

d) Three maxima

Answer: (b)

Solution:
f(x) = (x2 – 4)2 where x∈R

f '(x) = 2(x2 – 4)2x = 0

⇒ x = 0, 2, –2

f"(x) = 4(3x2 – 4)

f"(0) = –16 < 0 f(x) has a maximum at x = 0
ff′′′(′(−22))==3322>>00� ⇒ f(x)has a minima at x = ±2

12. Consider points P and Q in xy - plane with P = (1, 0) and Q =
(0, 1). The line integral 2 ∫PQ(x dx + y dy) along the semicircle
with the p line segment PQ as its diameter is

a) – 1

b) 0

c) 1

7

d) depends on the direction (clockwise (or) anti-clockwise) of

the semi-circle

Answer: (b)

Solution:

Applying Green’ Theorem,
∫ M dx + Ndy = ∬ �∂∂My − ∂∂Nx� dxdy

= 0 (since M = 2x and N = 2y)

13. 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� =
3ı̅ − 4ȷ̅ is

a) -4

b) -2

c) -1

d) 1

Answer: (b)

Solution:

f(x, y, z) = x2 + 2y2 + z, P = (1, 1, 2)

and a� = 3ı̅ − 4ȷ̅

∇f = 2xı̅ + 4yȷ̅ + k�

(∇f)P = 2ı̅ + 4ȷ̅ + k�

Directional derivative = (∇f)P. a�
|a�|

= 6−16
5

= − 10 = −2
5

8

14. For a scalar function f(x, y, z) = x2 + 3y2 + 2z2 the gradient at

the point P(1 , 2, –1) is
a) 2ı̅ + 6ȷ̅ + 4k�
b) 2ı̅ + 12ȷ̅ − 4k�
c) 2ı̅ + 12ȷ̅ + 4k�

d) √56

Answer: (b)

Solution:

f(x, y, z) = x2 + 3y2 + 2z2 P(1 , 2, –1)

grad f = ∇f = 2xı̅ + 6yȷ̅ + 4zk�
∇f]P = 2ı̅ + 12ȷ̅ − 4k�

15. The line integral of the vector function F� = 2x ı̂ + x2ȷ̂ along

the x - axis from x = 1 to x = 2 is

a) 0

b) 2.33

c) 3

d) 5.33

Answer: (c)

Solution:

F� = 2x ı̂ + x2ȷ̂

Along x-axis, y = 0 ⇒ dy = 0
∫c F�. dF� = ∫c 2xdx + x2dy
= ∫12 2xdx = 3

9

16. Divergence of the vector field x2z ̂ + xy ̂ + yz2 � at (1, –1, 1) is

a) 0

b) 3

c) 5

d) 6

Answer: (c)

Solution:
Let F� = x2zı̂ + xyȷ̂ + yz2k�

= F1ı̂ + F2ȷ̂ + F3k�

div F� = ∂F1 + ∂F2 + ∂F3 = 2xz + x − 2yz
∂x ∂y ∂z

(div F�) at (1, –1, 1) = 2 + 1 + 2 = 5

17. Divergence of the 3-dimensional radial vector field r̅ is

a) 3

b) 1
r

c) ı̂ + ȷ̂ +k�

d) 3(ı̂ + ȷ̂ +k�)

Answer: (a)

Solution:

r̅ = xı̅ + yȷ̅ + zk�

div r̅ = ∂F1 + ∂F2 + ∂F3
∂x ∂y ∂z

=1+1+1=3

10

18. The line integral ∫PP12(ydx + xdy) from P1(x1, y1) to P2(x2, y2)
along the semi-circle P1P2 shown in the figure is

y

P2(x2 y2)

P1(x1 y1)

x

a) x2y2 – x1y1
b) (y22 – y12) + (x22 – x12)
c) (x2 – x1) (y2 – y1)
d) (y2 – y1)2 + (x2 – x1)2

Answer: (a)
Solution:

∫PP12 ydx + xdy = ∫PP12 d(xy) = (xy)PP12
= x2y2 − x1y1

19. 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

Answer: (b)

11

20. If the standard deviation of the speed of vehicles in a highway

is 8.8 kmph and the mean speed of the vehicles is 33 kmph, the

coefficient of variation in speed is

a) 0.1517

b) 0.1867

c) 0.2666

d) 0.3646

Answer: (c)

Solution:

σ = 8.8, μ = 33

Coefficient of variation = σ = 8.8 = 0.266
μ 33

21. If X is a continuous random variable whose probability density

function is given by f(x) = �k(5x0−, 2x2), 0 ≤ x ≤ 2 then P(x >
Otherwise

1) is

a) 3/14

b) 4/5

c) 14/17

d) 17/28

Answer: (d)

Solution:

∫−αα f(x)dx = 1 ⇒ ∫02 k(5x − 2x2)dx = 1 ⇒ k = 3/14
P(X > 1) = ∫1α f(x)dx

= ∫12 3 (5x − 2x2)dx = 17
14 28

12

22. Which of the following is a solution of the differential equation

d2y + p dy + (q + 1)y = 0? Where p = 4, q = 3
dx2 dx

a) e-3x

b) xe-x

c) xe-2x

d) x2e-2x

Answer: (c)

Solution:
Given [D2 + 4D + (3 + 1)]y = 0

⇒ (D2 + 4D + 4)y = 0
⇒ (D + 2)2 = 0

D = −2, −2

∴ y = (C1 + C2x)e−2x = C1e−2x + C1xe−2x
Hence e−2x and xe−2x are independent solutions

23. The degree of the differential equation d2x + 2x3 = 0 is
dt2

a) 0

b) 1

c) 2

d) 3

Answer: (b)

Solution:

The degree of a differential equation is the exponent of highest

ordered derivative in the equation.

∴ The degree of given differential equation is 1.

13

24. The solution of dy = y2 with initial value y(0) = 1 is bounded
dx

in the interval is

a) –∞ ≤ x ≤ ∞

b) –∞ ≤ x ≤ 1

c) x < 1, x > 1

d) –2 ≤ x ≤ 2

Answer: (c)

Solution:

Given dy = y2 … . . (1) and y(0) = 1 ……. (2)
dx

⇒∫ 1 dy = ∫ dx + c
y2

⇒− 1 = x + c … … . (3)
y

Using (2), (3) becomes

−1 = 0 + c
1

⇒ c = −1

∴ Solution is − 1 = x − 1(or)y = 1
y 1−x

y = 1 is not defined at x = 1
1−x

∴ y = 1 is bounded in the internal x < 1, x > 1
1−x

25. Consider the differential equation dy = 1+ y2, which one of
dx

the following can be particular solution of this differential

equation?

a) y = tan (x + 3)

b) y = (tan x) + 3

14

c) x = tan (y + 3)

d) x = (tan y) + 3

Answer: (a)

Solution:

Given dy = 1 + y2 → (1)
dx

dy = dx ⇒ tan−1 y = x + c
1+y2

⇒ y = tan(x + c)

For c = 3, y = (x + 3) is a particular solution of (1)

26. It is given that y" + 2y1 = 0, y(0) = 0 & ), y(1) = 0. What is

y(0.5)?

a) 0

b) 0.37

c) 0.62

d) 1.13

Answer: (a)

Solution:
Given (D2 + 2D + 1)y = 0 → (1)

and y(0) = 0 → (2)

y(1) = 0 → (3)

The auxiliary equation is
D2 + 2D + 1 = 0 ⇒ (D + 1)2 = 0

⇒ D = −1, −1
y = (C1 + C2)e−x → (4)
By using (2), (4) becomes

15

0 = (C1 + 0)
C1 + C2 = 0 and C2 = −C1 = 0

∴ y = 0 and y (0.5) = 0

27. The Laplace Transform of a unit step function ua(t), defined as

ua(t), = 0 for t < a

= 1 for t > a, is

a) e-as/s

b) se-as

c) s – u(0)

d) se-as – 1

Answer: (a)

Solution:

{ ( )} = ∫0∞ − ( )

= −

28. If L {f(t)} = then the value of l →im∞ ( ) =
2+ 2

a) Cannot be determined

b) Zero

c) Unity

d) Infinite

Answer: (b)

Solution:

Lt→im∞ f(t) = Ls→im0 sF(s)
=0

16

29. The contour C in the adjoining figure is described by x2 + y2 =

16. Then the value of ∮C z2+8
0.5z−1.5j

a) –2πj

b) 2πj

c) 4πj

d) –4πj

Answer: (d)

Solution:

= ∫ 2+8 dz and C is x2 + y2 = 16 or | | = 4
12 −23

= 2 ∫ 2+8
−3

By Cauchy Integral Formula,

∴ = 2[2 (3 )] = −4

Where f(z) = z2 + 8

(Since z = 3 lies in the contour)

30. Let x2 – 117 = 0. The iterative steps for the solution using

Newton - Raphson's method is given by

a) xk+1 = 1 �xk + 1x1k7�
2

b) xk+1 = �xk − 1x1k7�

17

c) xk+1 = �xk − 1x1k7�

d) xk+1 = xk − 1 �xk + 1x1k7�
2

Answer: (a)

Solution:

Given f(x) = x2 – 117 = 0
⇒ f1(x) = 2x

Newton’s – Raphson’s formula is

xk+1 = xk − f(xk)
f1(xk)

xk+1 = xk − �xk2 −117�
2xk

= 2xk2 −x2k +117 = xk2 +117
2xk 2xk

∴ xk+1 = 1 �xk + 1x1k7�
2

18

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