Engineering Mathematics Test - 3 - PDF Flipbook
Engineering Mathematics Test - 3
378 Views
12 Downloads
PDF 3,176,275 Bytes
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
Data Loading...