ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________

21. Find the absolute maximum and minimum values of

f(x, y) = 2 + 2x + 2y – x2 – y2

on triangular plate in the first quadrant, bounded by the lines x = 0, y = 0 and y = 9 – x.

a) – 4 b) – 2 c) 4 d) 2

21. Ans. (C)

f(0, 0) = 2
f(9, 0) = 2 + 18 – 81 = -61
f(0, 9) = 2 + 18 – 0 – 81 = -61
f = 2+2x+2y-x2-y2
Differentiating f partially w.r.to x
f
=2-2x …(1)
x
Differentiating f partially w.r.to y
f
=2-2y …(2)
y
f f
The stationary (Critical) points are given by =0 and =0
x y
 x=1, y=1
(1, 1) lies inside the triangular region
f(1, 1) = 2+2+2-1-1= 4
Absolute maximum value of f(x, y) is 4
22. Four the matrix

1 4 
A= 
2 3

the expression

A5 – 4A4 – 7A3 + 11A2 – A – 10I

is equivalent to

a) A2 + A + 5 I b) A + 5I c) A2 + 5I d) A2 + 2A + 6I

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ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________
22. (B)

1 4
A=  
 2 3

The characteristic equation of A is |A-I|=0

1-λ 4
=0
2 3-λ

 2 - 4 - 5 = 0

According to Cayley-Hamilton theorem A satisfies its own characteristic equation

 A2 – 4A – 5I = 0

Now

A5-4A4-7A3+11A2-A-10I

= A5-4A4-5A3-2A3+11A2-A-10I

= A3[A2-4A-5I]-2A3+11A2-A-10I

= -2A3+11A2-A-10I

[ A2-4A-5I=0]

= -2A3+8A2+10A+3A2-11A-10I

= -2A[A2-4A-5I+3A2-11A-10I

= 3A2-11A-10I = 3[4A+5I]-11A-10I

= 12A+15I-11A-10I = A+5I

23. The solution of the differential equation

(1 + y2)dx = (tan-1 y - x)dy

is
−1 −1
a) x = tan-1 y + 1 +ce− tan y
b) x = tan-1 y - 1 + ce− tan y

−1 1 −1
c) x = tan-1 y + 1 + ce− tan y
d) x = tan −1 y + 1 + ce − tan y
2

23. Ans. (B)

(1+y2) dx = (tan-1y-x) dy
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ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________
dx tan-1y-x
=
dy 1+y2

dx x tan-1y
+ =
dy 1+y2 1+y2

dx
+P(y)x=Q(y)
dy

1 tan-1y
where P= , Q=
1+y2 1+y2

Here we have taken x as dependent variable & y as independent

Integrating Factor (I.F) = e 

1
 1+y2 dy -1
=e =etan y

Complete solution is given as (Dependent variable) (I.F)

=  ( Q×I.F )dy+c

tan-1y tan-1y
xe tan y = 
-1
e dy+c
1+y2

Now to evaluate integral using integration by substitution

tan-1y = V (let)

1
dy=dV
1+y 2

xe tan y =  Ve VdV+c
-1

-1
xe tan y =Ve V -e V +c

xetan y = ( tan-1y ) etan y -etan y +c

-1
x=tan-1y-1+c e-tan y

24. The value of

10 [(1 − ax)(1 − bx 2 )(1 − cx 3 )(1 − dx 4 )]

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ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________
is

a) abcd (10!) b) abcd (9!) c) abcd (8!) d) abcd (7!)

24. Ans. (A)

 x 4 + y4  u u
25. If u = log e   , the value of x +y is
 x+y  x y

a) 6 b) 5 c) 4 d) 3

25. Ans. (D)

 x 4 +y4 
u=loge   =loge ( x 4 +y4 ) -loge ( x+y )
 x+y 

Partially differentiating with respect to x treating y as a constant

u
= 4 4 ( 4x 3 ) -
1 1
x x +y x+y

u 4x 4 x
x = 4 4-
x x +y x+y

Partially differentiating u w.r.to y treating x as a constant

u 1 1
= 4 4 4y3 -
y x +y x+y

u 4y4 y
y = 4 4-
y x +y x+y

u u 4x 4 x  4y4  y  x 4 +y4   x+y 

26. The general value of log (1 + i) + log(1 - i) is

a) log 2 – 4 nπi b) log 2 + 4 nπi c) log 2 + 2 nπi d) log 2 - 2 nπi

Ans. (C)

log(1+i) + log(1-i) = log(1+i) (1-i)

= log(1-i2) = log(1-(-1)) = log2

z=2

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ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________
|z| = 2, arg(z) = arg(2) = 0

General log of 2

= log2 = log|z| + i arg(z) + 2ni

 log2 (General) = log2 + i0 + 2ni

 log2 = log2+2ni

27. A bag contains 4 white and 2 blocks balls and another bag contains 3 of each colour. A bag is
selected at random and a ball is drawn at random from the bag chosen. The probability of the white
ball drawn is

1 1 5 7
a) b) c) d)
3 4 12 12

Ans (D)

We have two bags A and B (say)

A contains 4 white and 2 black balls.

B contains 3 white and 3 black balls.

W W
P ( W ) =P ( A ) P   +P ( B) P  
A  B

1 2 1 1 1 1 7
= × + × = + =
2 3 2 2 3 4 12

28. X is a continuous random variable with probability density function given by

f(x) = kx (0 ≤ x < 2)

= 2k (2≤ x < 4)

= -kx + 6k (4 ≤ x < 6)

The value of k will be

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ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________
2 1
a) b) c) 1 d) 8
3 8

Ans. (B)

f(x) is a pdf

  f ( x )dx=1,
-

kx dx+ 2kdx +  ( -kx+6k )dx=1

6 6
4
2 2 2
x x
k +2kx -k +6kx =1
2 0
2
2 4 4

k k
  4 +2k  2 -  20 +6k  2 =1
2 2

 2k+4k-10k+12k=1

 8k = 1

 k = 1/8

29. The first moment about origin of binomial distribution is

a) np b) npq c) n(1 - p) d) n(1 - p)q

Ans. (A)

The first moment about origin of random variable is equal to its mean. The mean of a random variable
following binomial distribution =  = np

30. For the regression equations

y = 0.516x + 33.73

and

x = 0.512y + 32.52

the means of x and y are nearly

a) 67.6 and 68.6 b) 68.6 and 68.6 c) 67.6 and 58.6 d) 68.6 and 58.6

Ans. (A)

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ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________
Regression equations are
y=0.516x+33.73
x=0.512y+32.52
Mean of x's and the mean y's lie on the two regression lines, we have
y=0.516x+33.73 …(1)
x=0.512y+32.52 …(2)
Now we have to solve these two simultaneous equations to find x and y
Putting value of x from (2) into (1)
y =0.516 0.512y+32.52 +33.73
 y =68.65
x=0.512 68.65 +32.52
= 67.66
98. The sum of all the natural numbers between 1 and 101 which are divisible by 5 is

a) 1000 b) 1050 c) 1500 d) 1500

Ans(B)
The numbers divisible by 5 are
5, 10, 15 …..100
Let n be the number of such numbers. Numbers are in A.P with common difference 5. nth term
an = a + (n-1)d
100 = 5+(n-1)5
n = 20
n
Sn =  a+ 
2
20
= 5+100 =1050
2
99. In a group of 100 people, 750 speak Hindi and 400 speak English. The number of only Hindi
speaking people is

a) 150 b) 350 c) 600 d) 750

Ans. (C)

Let A be the set of people who speak Hindi

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ESE Prelims Paper I 2020 Solutions (Engineering Mathematics)______________________________
Here each of 1000 people speak at least one of the two languages.
 n(AB)=1000
n(AB)=n(A)+n(B)-n(AB)
= 750+400-1000=150
n(only Hindi) = n(A B )
= n(A)-n(AB) = 750-150=600

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