Engineering Mathematics Test - 5 - PDF Flipbook
Engineering Mathematics Test - 5
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GATE
EEE
Engineering
Mathematics
Test-05Solutions
ENGINEERING MATHEMATICS
1. If A is a real square matrix then AAT is
a) Unsymmetric
b) Always symmetric
c) Skew symmetric
d) Sometimes symmetric
Answer: (b)
Solution:
Given that A is real square matrix
Consider, (AAT)T = (AT)TAT
(∵ (AB)T = BTAT)
= AAT(∵ (AT)T = A)
∵ A is symmetric matrix
2. The eigen values of the matrix A = �01 10� are
a) 1, 1
b) –1, –1
c) j, –j
d) 1, –1
Answer: (d)
Solution:
Given A = �01 10� ⇒ |A − λI| = 0
⇒ �0 − λ 1 λ� ⇒ λ2 − 0. λ + (−1) = 0
1 −
0
λ2 − 1 = 0 ⇒ λ = ±1
1
50 2
3. If A = �0 3 0� then A-1 =
1
20
−2
10 0�
a) � 0 1/3 5
−2 0
502
b) �0 −1/3 0�
201
1/5 0 1/2
c) � 0 1/3 0 �
1/2 0 1
1/5 0 −1/2
d) � 0 1/3 0 �
−1/2 0
1
Answer: (a)
Solution:
502
Given A = �0 3 0�
201
3 0 −6
⇒ |A| = 3 and adj(A) = � 0 1 0 �
−6 0 15
∴ A−1 = adj(A)
|A|
= 1 3 0 −6 1 0 −2
3 �0 1 0 �=�0 1/3 0�
−6 0 15 −2 5
0
2
2 −1 0 0
4. The eigen values of the matrix �00 3 0 00� are
0 −2
0 0 −1 4
a) 2, –2, 1, –1
b) 2, 3, –2, 4
c) 2, 3, 1, 4
d) None
Answer: (b)
Solution:
If λ is an eigen value(s) of matrix A4×4 then the eigen value(s)
are given by |A − λI| = 0.
2−λ −1 0 0
3−λ 0
i.e. � 0 −2 − λ 0 �=0
0 0 −1 0
4−λ
0 0
3−λ 0 0
⇒ (2 − λ) � 0 −2 − λ 0 � = 0
0 −1 4 − λ
⇒ (2 − λ)(3 − λ)[−(2 + λ)(4 − λ)] = 0
⇒ (2 − λ)(3 − λ)(2 + λ)(4 − λ) = 0
∴The eigen values are given by
λ = 2, −2, 3, 4
5. The system of equations x + y + z = 6, x + 4y + 6z, x + 4y + λz
= μ has no solution for values of λ and μ given by
a) λ = 6, μ = 20
b) λ = 6, μ ≠ 20
c) λ ≠ 6, μ = 20
3
d) λ ≠ 6, μ ≠ 20
Answer: (b)
Solution:
Given AX = B
Consider, 1 1 16
[A|B] = �1 4 6� 20�
1 4 λμ
11 1 6
~ �0 3 5 � 14 �
0 3 λ−1 μ−6
11 1 6
~ �0 3 5 � 14 �
0 0 λ − 6 μ − 20
For no solution, λ = 6 and μ ≠ 20
6. One pair of eigen vectors corresponding to the two eigen values
of the matrix �10 −01� is
a) �−1j� , �−j1�
b) �01� , �−01�
c) �1j � , �01�
d) �1j � , �1j �
Answer: (d)
4
Solution:
Given A = �10 −01�
Eigen values:
∴Eigen value of A are λ =i, – i
Eigen vector:
�0 − λ 0−−1λ� �xx12� = �00� … … . . (1)
1
Case - (i) put λ = i in (1)
i. e. �−1i −−1i � �xx12� = �00�
⇒ –ix1 – x2 = 0
∴ X = �xx12� = �−1i� (or) �1i �
Case - (ii) put λ = –i in (1)
i. e. �1i −i1� �xx12� = �00�
⇒ –ix1 – x2 = 0
∴ X = �xx12� = �1i � (or) �−i1�
Hence the eigen vectors of a corresponding to eigen values
λ = i & − i are X1 = �−1i� &X2 = �1i �
Or
X1 = �1i � &X2 = �−i1�
5
7. If y = x + �x + �x + √x + ⋯ … … . . ∞ then y(2) =
a) 4 or 1
b) 4 only
c) 1 only
d) Undefined
Answer: (b)
Solution:
y = x + �x + �x + √x + ⋯ … … . . ∞
⇒ (y − x)2 = y ⇒ y2 − 2xy + x2 = y
at x = 2, y2 − 4y + 4 = y ⇒ y2 − 5y + 4 = 0
⇒ y = 1 or 4
but x = 2
8. The value of ∫0∞ ∫0∞ e−x2e−y2 dx dy is
a) √
2
b) √
c)
d)
4
Answer: (d)
Solution:
∫0α ∫0α e−x2e−y2dxdy = ∫0α ∫0α e−(x2+y2)dxdy
Put x = rcosθ, y = rsinθ, |J| = r
∫0π�2 ∫0α e−r2 rdrdθ = π
4
6
9. A parabolic cable is held between two supports at the same
level. The horizontal span between the supports is L. The sag at
the mid-span is h. The equation of the parabola is y = 4h 22,
where x is the horizontal coordinate and y is the vertical
coordinate with the origin at the centre of the cable. The
expression for the total length of the cable is
a) ∫0L �1 + 64 h2x2 dx
L4
b) 2 ∫0L/2 �1 + 64 h3x2 dx
L4
c) ∫0L/2 �1 + 64 h2x2 dx
L4
d) 2 ∫0L/2 �1 + 64 h2x2 dx
L4
Answer: (d)
Solution:
Total length = 2 ∫0L/2 �1 + �ddyx�2 dx
= 2 ∫0L/2 �1 + 64 h2x2 dx
L4
10. The area enclosed between the curves y2 = 4x and x2 = 4y is
a) 16/3
b) 8
c) 32/3
d) 16
Answer: (a)
7
Solution:
Area = ∫04 �2√x − x42� dx = 16
3
11. At t = 0, the function f(t) = has
a) a minimum
b) a discontinuity
c) a point of inflection
d) a maximum
Answer: (b)
Solution:
f(t) =sitnt
l i→m0 ( ) = 1 but f(0) does not exists
12. The magnitude of the gradient for the function f(x, y, z) = x2 +
3y2 + z3 at the point (1, 1, 1) is _______.
Answer: 7
Solution:
∇f = ∂f ı̅ + ∂f ȷ̅ + ∂f k�
∂x dy ∂z
= 2xı̅ + 6yȷ̅ + 3z2k�
(∇f)at(1,1,1) = 2ı̅ + 6ȷ̅ + 3k�
∴ �(∇f)at(1,1,1)� = √4 + 36 + 9 = √49 = 7
8
13. The line integral of function F = yzi, in the counterclockwise
direction, along the circle x2 + y2 = 1 at z = 1 is
a) –2π
b) –π
c) π
d) 2π
Answer: (b)
Solution:
∫C F�. dr̅ = ∫C yzdx = ∫C ydx (∵ z = 1)
= ∫C(ydx − 0dy) = ∬s −dxdy
(from Green’s theorem)
= −π
14. The integral ∮C(ydx − xdy) is evaluated along the circle x2 + y2
= 1 traversed in counter clockwise direction. The integral is
4
equal to
a) 0
b) − π
4
c) − π
2
d) π
4
Answer: (c)
Solution:
Applying Green’s theorem,
∮C(ydx − xdy) = ∬R(−1 − 1)dxdy
9
= −2 (Area of the given circle)
= −2 �π4� = − π
2
15. Directional derivative of ϕ = 2xz – y2 at the point (1, 3, 2)
becomes maximum in the direction of
a) 4i + 2j – 3k
b) 4i – 6j + 2k
c) 2i – 6j + 2k
d) 4i – 6j – 2k
Answer: (b)
Solution:
∇∅ = ∂∅ ı̅ + ∂∅ ȷ̅ + ∂∅ k�
∂x ∂y ∂z
= 2zı̅ − 2yȷ̅ + 2xk�
∴ Required direction vector = (∇∅) at (1, 3, 2) = (4ı̅ − 6ȷ̅ + 2k�)
16. Let be an arbitrary smooth real valued scalar function and �V⃗
be an arbitrary smooth vector valued function in a three
dimensional space. Which one of the following is an identity?
a) Curl (ϕV�⃗) = ∇(ϕDiv�V⃗)
b) Div�V⃗ = 0
c) Div Curl �V⃗ = 0
d) Div (ϕ�V⃗) = (ϕDivV�⃗)
Answer: (c)
Solution:
Option (c) is True (from vector identities)
10
17. The life of a bulb (in hours) is a random variable with an
exponential distribution f(t) = αe−αt,0 ≤ t ≤ ∞. The probability
that its value lies b/w 100 and 200 hours is
a) e–100α – e–200α
b) e–100 – e–200
c) e–100α + e–200α
d) e–200α – e–100α
Answer: a
Solution:
P (100 < X < 200)
= ∫120000 ( ) = ∫120000 − = −100 − −200
18. For a random variable x(−∞ < x < ∞) following normal
distribution, the mean is µ=100 If the probability is P = α for
x ≥110. Then the probability of x lying between 90 and 110 i.e.,
P (90 ≤ x ≤ 110) and equal to
a) 1 – 2α
b) 1 – α
c) 1 – α/2
d) 2α
Answer: (a)
Solution:
11
P(x ≥ 110) = α ⇒ P(x ≤ 90) = α
⇒ P(90 ≤ x ≤ 110) = 1 – 2α
19. 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
d) 1/2520
Answer: (c)
Solution:
Required probability = 2 × 1 × 3 × 2 × 1 × 4 × 3 × 2 × 1
9 8 7 6 5 5 3 2 1
= 1
1260
20. t is estimated that the average number of events during a year
is three. What is the probability of occurrence of not more than
two events over a two-year duration? Assume that the number of
events follow a poison distribution
a) 0.052
b) 0.062
c) 0.072
d) 0.082
Answer: (b)
12
Solution:
In a Poisson distribution mean λ = 3 per year
= 6 per two years
Required probability
= P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= e−λ �1 + λ + λ22� = 0.0619
21. The probability that a given positive integer lying between 1
and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is _____.
Answer: 0.26
Solution:
Number of integers in the set which are divisible by 2 or 3 or 5
= n(2) + n(3) + n(5) − n(2⋀3) − n(3⋀5) −
n(5⋀2) + n(2⋀3⋀5)
= 50 + 33 + 20 – 16 – 10 – 6 + 3
= 74
∴ The number of integers between 1 and 100, which are not
divisible by 2 or 3 or 5 = 100 – 74 = 26
∴ Required probability = 26 = 0.26
100
22. If C is a constant, then the solution of dy = 1 + y2
dx
a) y = sin (x + c)
b) y = cos (x + c)
c) y = tan (x + c)
d) y = ex + c
Answer: (c)
13
Solution:
Given dy = 1 + y2
dx
⇒ ∫ 1 dy = ∫ dx + c
1+y2
⇒ tan−1(y) = x + c
∴ y = tan(x + c)
23. The solution to the ordinary differential equation d2y + dy −
dx2 dx
6y = 0
a) y = C1e3x + C2e-2x
b) y = C1e3x + C2e2x
c) y = C1e-3x + C2e2x
d) y = C1e-3x + C2e-2x
Answer: (c)
Solution:
Given d2y + dy − 6y = 0
dx2 dx
⇒ (D2 + D – 6)y = 0
⇒ f(D) = 0 where f(D) = D2 + D – 6
The auxiliary equation is f(D) = 0
⇒ D2 + D – 6 = 0
⇒ D = 2, -3
∴ y = C1e2x + C2e−3x (or) y = C1e−3x + C2e2x
24. The solution of differential equation dy = Ky, y(0) = C is
dx
a) x = C eky
b) x = keky
14
c) y = ekx C
d) y = Ce-kx
Answer: (c)
Solution:
Given dy = Ky ……….. (1)
dx
and y(0) = c ……….. (2)
⇒ ∫ 1 dy = K ∫ dx + c1
y
⇒ log y = Kx + c1 ⇒ y = eKx + c1
⇒ y = eKx c2
Where c2 = ec1
Using (2), (3) becomes
c = c2
∴ y = ekxc
25. Consider the differential equation dy = (1 + y2)x. The general
dx
solution with constant ‘C’ is
a) y = tan �x22� + C
b) y = tan2 �2x + C�
c) y = tan2 �2x� + C
d) y = tan �x22 + C�
Answer: (d)
15
Solution:
Given dy = (1 + y2)x ………… (1)
dx
⇒ ∫ dy = ∫ x dx + C ⇒ tan−1(y) = x2 + C
1+y2 2
∴ y = tan �x22 + C�
26. The Laplace transform of the function f(t) = k, 0 < t < c
= k, c < t < ∞, is
a) (k/s) e-sc
b) (k/s) esc
c) ke-sc
d) (k/s) (1 – e-sc)
Answer: (d)
Solution:
{ ( )} = ∫0 − . + 0 = �1− − �
27. The inverse Laplace transform of 1/(s2 + 2s) is
a) (1 – e-2t)
b) (1 + e+2t)/2
c) (1 – e+2t)/2
d) (1 – e-2t)/2
Answer: (d)
Solution:
−1 � 2+12 � = −1 � ( 1+2)� = −1 �21 �1 − +12��
= 1 [1 − −2 ]
2
16
28. If F(s) is the Laplace transform of the function f(t) then
Laplace transform of ∫01 f(x)dx is
a) 1 F(s)
s
b) 1 F(s) − f(0)
s
c) s F(s) − f(0)
d) ∫ F(s)ds
Answer: (a)
Solution:
�∫0 ( ) � = ( )
29. For an analytic function f(x + iy) = u(x, y) + iv(x, y), u is given
by u = 3x2 – 3y2. The expression for v, considering k is to be
constant is
a) 3y2 – 3x2 + k
b) 6x – 6y + k
c) 6y – 6x + k
d) 6xy + k
Answer: (d)
Solution:
Given u = 3x2 – 3y2
for f(x + iy) = u(x, y) + iv (x, y)
dv = vs dx + vs dy
= −uydx + uxdy �∵ ux = vy & vx = −uy�
dv = – (0 – 6y) dx + 6x dy which is exact differential equation,
Integrating v = 6xy + k
17
30. The integral ∫xx12 x2dx with x2 > x1> 0 is evaluated analytically
as well as numerically using a single application of the
trapezoidal rule. If I is the exact value of the integral obtained
analytically and J is the approximate value obtained using the
trapezoidal rule, which of the following statements is correct
about their relationship?
a) J > I
b) J < I
c) J = I
d) Insufficient data to determine the relationship
Answer: (a)
Solution:
In trapezoidal rule the integral f(x) is approximated with a
polynomial of degree 0 or 1 (a straight line). The straight line
segments lie above the parabola y = x2 as shown in the figure by
using trapezoidal rule.
Hence J > 1
Correct option: (c)
18
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