Engineering Mathematics | GATE Subject Wise
Question 1 |

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Question 2 |

Then Pr(Z > 2 |Z > 1), rounded off to two decimal places, is equal to __________.
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Question 3 |

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Question 4 |
E1/z | |
In(z) | |
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Cos (z) |
(b) ℓnz is NOT analytic in Domain D = {z / x ≤ 0 , y = 0}
(c) 1/(1-z) is not analytic at z=1
(d) cos (z) =1 - z2 + z4/4! - z6/6! + z8/8!
∴ cosz is analytic every where in the complex plane z
Question 5 |
for n = - 1 and n = + 1, respectively, are
Hyperbolas and Circles | |
Circles and Hyperbolas | |
Parabolas and Circles | |
Hyperbolas and Parabolas |

Question 6 |

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Question 7 |
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Question 8 |

-2.5 | |
0 | |
15 | |
25 |
sum = one eigen value of that matrix
Sum of all rows
(15-x) * |Matrix| = 0
15 is a factor .
Another Approach

Question 9 |

0 | |
1 | |
2 | |
3 |

= 5(0-12) – 10(6-6) +10(6-0)= -60-0+60=0
But a 2×2 minor,


Question 10 |
Consider the following statement about the linear dependence of the real valued functions y1= 1, y2= x and y 3 =x2, over the field of real numbers.
I. y1, y2 and y3 are linearly independent on -1 ≤ x ≤ 0
II. y1 , y2 and y3 are linearly dependent on 0 ≤ x ≤1
III. y1 , y2 and y3 are linearly independent on 0 ≤x ≤1
IV. y1, y2 and y3 are linearly dependent on -1 ≤ x ≤ 0
Which one among the following is correct?
Which one among the following is correct? | |
Both I and III are true | |
Both II and IV are true | |
Both III and IV are true |
Question 11 |
Fill in the Blank Type Question |
the number of outcomes =
{ (1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6) }
n(p) =6
therefore, probability = n(p) /n(s) = 6/216 =0.0277 =~0.028
Question 12 |
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Question 13 |

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Question 14 |

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Question 15 |








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Question 16 |

S1:

S2:

S3:

Which one among the following is TRUE?
S1 implies S2 | |
S1 implies S3 | |
S2 implies S1 | |
S3 implies S2 |
Eigen vectors corresponding to distinct eigen values are linearly independent.
So, “S2 implies S1”.
->We know that if a matrix A has'n' distinct eigen values then A has 'n' linearly independent eigenvectors
. Therefore"S2 implies S1"
Question 17 |
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Linear system must result zero output for zero input.
It must satisfy super position and homogenity principle
Which option c does not satisfy so Option (C) is Correct Answer.
Question 18 |


The coefficient (correct to two decimal places) is equal to ____________.
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Question 19 |




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Question 20 |






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Question 21 |

The initial conditions are and
The position (accurate to two decimal places) of the particle at
is _____________.
Fill in the Blank Type Question |
Given condition,
This can be solved easily in laplace domain,
By taking inverse Laplace transform we get y(t),
Now its value at
Question 22 |


The value of the integral is __________.
Fill in the Blank Type Question |

Question 23 |


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We need to use suitable substitution here,
Put,


After simplification we obtain the following relation,

Given that the curve passes through points,


So,

Question 24 |
Fill in the Blank Type Question |

Question 25 |

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Question 26 |


same and equal to 0.5 | |
same and equal to -0.5 | |
0.5 and – 0.5, respectively | |
- 0.5 and 0.5, respectively |


Question 27 |

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Question 28 |

0.3, –2.5, 0.5 | |
0.0, 3.0, 2.0 | |
0.3, 0.33, 0.5 | |
4.0, 3.0, 2.0 |

Question 29 |
Fill in the Blank Type Question |

Question 30 |


Fill in the Blank Type Question |


at x =1, f( x) has local minimum.

For x = 1, local minimum value

Finding f(- 100) = -333433.33
f (100) = 333233.33
(

∴ Minimum occurs at x= - 100
Question 31 |
M4k+1 | |
M4k + 2 | |
M4k+3 | |
M4k |

Question 32 |
Fill in the Blank Type Question |

Therefore Mean of random variables = First moment = 1
Question 33 |
P: If f(x) is continuous at x = xo, then it is also differentiable at x = xo.
Q: If f(x) is continuous at x = xo, then it may not be differentiable at x = xo.
R: If f(x) is differentiable at x = xo, then it is also continuous at x = xo
P is true, Q is false; R is false | |
P is false, Q is true, R is true | |
P is false, Q is true; R is false | |
P is true, Q is false, R is true |
Every differentiable function is continuous but converse need not be true
Question 34 |

The solutions have neither maxima nor minima anywhere except at the boundaries. | |
The solutions are not separable in the coordinates. | |
The solutions are not continuous. | |
The solutions are not dependent on the boundary conditions. |

Question 35 |


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-> B cannot be the answer as they are odd functions.
->C and D are even functions but the integration of a linear curve has to be parabolic in nature and it cannot be a constant function. Based on this Option C is correct.
Question 36 |


Fill in the Blank Type Question |

Method – II:-

(Changing into polar coordinates by


Another Approach

Question 37 |

The value of the integral is ______
Fill in the Blank Type Question |

Question 38 |

–1/2 | |
–1/3 | |
1/3 | |
1/2 |

Question 39 |
Suppose A and B are two independent events with probabilities P(A) ≠ 0 and P(B) ≠ 0.
Let and
be their complements.
Which one of the following statements is FALSE ?
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Question 40 |

The residue of ![]() | |
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Question 41 |


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Question 42 |

(2-t)et | |
(1+2t)e-t | |
(2+t)e-t | |
(1-2t)et |
roots are equal s1 = s2 = - 1
So, y = c1e-t + c2t e-t
y(0) = C1 = 1
yï = - C1e-t – C2t e-t + C2e-t
yï(0) = - C1 + C2 = 1
C2 = 2
So solution is y = e-t + 2te-t
y = (1 + 2t) e-t.
Question 43 |

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Putting f’(x) = 0, we get x = 0, or x = 1


At x = 0, f"(x) = 1 (so we have a minimum)
At x = 1,

(so we have a maximum), curve B. shows a single local minimum = 0 and a single local maximum at x = 1.
OR
f(1) = e-x (x2+x+1)
f(0) = 1
f(0.5) = 1.067
For positive values of x, function never goes negative
Question 44 |

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Question 45 |

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Question 46 |

Fill in the Blank Type Question |
Given that f(z) is analytic.
which is possible only when b = 0
since |z|3 is differentiable at the origin but not analytic.
2z3 is analytic everywhere
∴ f(z) = 2z3 + b|z|3 is analytic
only when b = 0
Lets take an Example

Question 47 |
f(x) increase monotonically | |
f(x) increases, then decreases and increases again | |
f(x) decreases, then increases and decreases again | |
f(x) increases and then decreases |
f (- 1) = − 3,
f (0) = 1
f (1) =− 1
f (2) =− 3
f (3) = 1
Correct Option is B
We can plot for various valve of x

f(x) increases, decreases and again increases.
Question 48 |
1 | |
2 | |
3 | |
4 or more |


Question 49 |


Where


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The minimum magnitude will be 5
At '5' magnitude angle is 90 °.

Question 50 |


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Question 51 |

The value of

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Question 52 |




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Product of eigen values = Det of A

Question 53 |

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Residue = coefficient of

Question 54 |
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P(Tail) = 0.7
Required probability = TTTTH
= (0.7)(0.7)(0.7)(0.7)(0.3)
=0.07203
Question 55 |

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Question 56 |

with exact solution y(x) = x2 + ex For x = 0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size h = 0.1 is ____
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Question 57 |

With


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Question 58 |

sec2 x | |
cos 4x | |
1 | |
0 |

Question 59 |
y = 2 | |
x = 2 | |
x + y = 4 | |
x – y = 0 |
Partial derivative of

= partial derivative of

So,

Question 60 |

2pnj | |
0 | |
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2pn |

Question 61 |
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Question 62 |

Fill in the Blank Type Question |


Now multiply equ(i) with 1/2

Now subtrac eq(ii) from (i)

Question 63 |
Fill in the Blank Type Question |

By Newton-Raphson method,

Question 64 |
Fill in the Blank Type Question |
Probability of getting

Probability of not getting

Now, the random variable X does represent number of throws required for getting 3. So,



Let

Equation (1) – (2),

Hence, the expected value of X is

Question 65 |

Given x(0) = 20 and x(1) = 10/e, where e = 2.718,
the value of x(2) is ________
Fill in the Blank Type Question |


This is a homogeneous equation. So, particular solution is zero. We get auxiliary equation as

Solving equations (1) and (2),

Taking t = 2

Question 66 |

S

Fill in the Blank Type Question |

By using divergence theorem

Question 67 |
47/48 | |
1/4 | |
13/48 | |
35/48 |
Probability that only one of them will be selected ,
is given by
P(only one) = P(Ram) × P(not Ramesh) + P(not Ram) × P(Ramesh)
= 1/6 × 7/8 + 5/6 × 1/8 = 12/48 = 1/4
Question 68 |

tan y – cot x = c | |
tan x – cot y = c | |
tan y + cot x = c | |
tan x + cot y = c |

Question 69 |

then d should be equal to _______________
150 | |
10 | |
50 | |
25 |
Given


Question 70 |
1.5 | |
4 | |
0.25 | |
1.6 |

Subtracting equation (2) from (1),


Hence,

Question 71 |

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For the differential equation, we have

Question 72 |
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(cM)T = c(M)T | |
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All other are properties of matrix.

Refer: www.khanacademy.org
Question 73 |
0.42 | |
0.52 | |
0.62 | |
0.67 |
E1 = 1 children family
E2 = 2 children family
A = picking a child
Then by Baye’s theorem, Required probability is

Question 74 |
0.42 | |
0.52 | |
0.62 | |
0.67 |
E1 = 1 children family
E2 = 2 children family
A = picking a child
Then by Baye’s theorem, Required probability is

Question 75 |

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Question 76 |
1 | |
2 | |
4 | |
8 |

Question 77 |
0.23 | |
0.27 | |
0.31 | |
0.33 |
P(x1>x2>x3),
P(x1>x3>x2),
P(x2>x1>x3),
P(x2>x3>x1),
P(x3>x2>x1),
P(x3>x1>x2).
All cases are equally likely and hence P(x1) being largest is P(x1>x2>x3) or P(x1>x3>x2).
Hence, P(x1 being largest)=2/6=1/3=0.33
Question 78 |
0 | |
0.2 | |
0.4 | |
0.5 |

Question 79 |
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Question 80 |
644 | |
720 | |
748 | |
864 |

Question 81 |

Which is obtained by reversing the order of the columns of the identity matrix I6.
Let P= I6 + αJ6 where α is a non-negative real number.
The value of α for which det(P) = 0 is ___________.
0 | |
1 | |
2 | |
4 |



Since, α is non-negative real number. Hence,
α = 1
Question 82 |

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i.e., variance cannot be negative .So ,we have

Question 83 |


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Question 84 |
1 | |
2 | |
3 | |
4 |

Question 85 |
100 | |
8 | |
1000 | |
200 |
Determinant of (A) * Determinant of (B)
5*40 = 200
so answer = 200
Question 86 |
40 | |
50 | |
60 | |
80 |

Question 87 |
ln2 | |
1.0 | |
e | |
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Question 88 |

has two equal roots, then the values of a are
± 1 | |
0,0 | |
± j | |
±1/ 2 |

Question 89 |

a unique solution | |
infinitely many solutions | |
no solution | |
exactly two solutions |



Clearly rank(A)=2 , rank(A/B) = 2 , number of unknowns = 3
so rank (A) = rank (A / B) =2
Since, rank (A) = rank (A / B) < number of unknowns
∴ Equations have infinitely many solutions.
Question 90 |
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Question 91 |
40 | |
45 | |
49 | |
52 |

Question 92 |
1 | |
2 | |
3 | |
4 |

Question 93 |
0.2 | |
0.4 | |
0.5 | |
0.8 |


We can use pasrevalis theorem

Question 94 |
fx(x) =2e-2x for x ≥ 0 and
fx(x) =0 for x < 0.
For outputs to be of equal probability, the quantizer threshold should be _____.
0.25 | |
0.35 | |
0.45 | |
0.50 |

One bit quantizer will give two levels. Both levels have probability of


Let xT be the threshold

Where x1 and x2 are two levels

Question 95 |
0 | |
1 | |
2 | |
-1 |

Question 96 |
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B)

C)The equation in option (C) is non-linear and non-homogenous.
D)The equation in option (D) is non-linear and homogenous.
Question 97 |

P1—M3, P2—M2, P3—M4, P4—M1 | |
P1—M3, P2—M1, P3—M4, P4—M2 | |
P1—M4, P2—M1, P3—M3, P4—M2 | |
P1—M2, P2—M1, P3—M3, P4—M4 |

Question 98 |
0.067 | |
0.073 | |
0.082 | |
0.091 |
So, The required probability,

Question 99 |


2 | |
4 | |
6 | |
8 |

Question 100 |
If A is upper triangular, the eigenvalues of A are the diagonal elements of it | |
If A is real symmetric, the eigenvalues of A are always real and positive | |
If A is real, the eigenvalues of A and AT are always the same | |
If all the principal minors of A are positive, all the eigenvalues of A are also positive |

Question 101 |
1 | |
3 | |
4 | |
2 |
with 2 tosses HT and TH are valid cases out of HH,HT,TH,TT
with 3 tosses HHT and TTH are valid cases out of HHH,HHT,HTH,HTT,THH,THT,TTH,TTT
And so on
X 2 3 4 5 ..........
Prob(X=i) 2/4 2/8 2/16 2/32 ........
E(x)= 2*2/4 + 3 *2/8 + 4*2/16 +...... (i)
E(x) /2= 2*2/8 + 3*2/16 +.........(ii)
(i)-(ii) is 2*2/4 + (3-2)2/8 + (4-3)2/16 +.....
E(x)/2 => 1+2/8+2/16+2/32+...
E(x)/2=>1+ 2(1/8 + 1/16 + 1/32 +...)
E(x)/2=>1 +2* (1/8 / (1 - 1/2))
E(x)/2=>1+1/2
E(x)/2=>3/2
so E(x)=>3
Question 102 |


0.12 | |
0.16 | |
0.20 | |
0.24 |

Let z = x1 + x2 - x3
P{x1 + x2 ≤ x3} = P{x1 + x2 - x3 ≤ o}
=P{z ≤ o}
Let us find probability density function of random variable z.
Since Z is summation of three random variable

Overall pdf of z is convolution of the pdf of



pdf of –x3 is


Question 103 |

The value of H(Y1) + H(Y2) in bits is _______.
2 | |
5 | |
8 | |
6 |

Question 104 |

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2 | |
e |

Question 105 |
7 | |
9 | |
12 | |
5 |

Question 106 |


3 | |
5 | |
8 | |
4 |

Question 107 |

0.44 | |
0.48 | |
0.54 | |
0.62 |

Question 108 |

0.22 | |
0.33 | |
0.44 | |
0.55 |

Question 109 |
12 | |
36 | |
30 | |
45 |
x (opposite side)
y (adjacent side)
z (hypotenuse) of a right angled triangle.
Given Z + y = K(constant) ......(1) and angle between them say


In order to have maximum area,



Question 110 |

If the output is 0, the probability that the input is also 0 equals ______________
0.4 | |
0.6 | |
0.7 | |
0.8 |

Question 111 |
10 ° | |
18 ° | |
50 ° | |
90 ° |

Now, we check it for θ= 50 °

From the above we can conclude that the error is more than 10%. Hence, for error less than 10%, till θ= 18 ° we have the approximation

Question 112 |

0 | |
1/3 | |
1 | |
3 |

Question 113 |

no real roots | |
no negative real root | |
odd number of real roots | |
at least one positive and one negative real root |

Since, all the coefficients are positive so, the roots of equation is given by

It will have at least one pole in right hand plane as there will be least one sign change from (a1) to (a0) in the Routh matrix 1st column. Also, there will be a corresponding pole in left hand plane
i.e., at least one positive root (in R.H.P.)
i.e., at least one negative root (in L.H.P.)
Rest of the roots will be either on imaginary axis or in L.H.P
Question 114 |



4/9 | |
½ | |
2/3 | |
5/9 |
Here, let Y=3V-2U, where Y is also a normal random variable with mean = 0 and variance = 32 (1/9) + 22 (1/4) = 2
So it will be symmetric about mean that is 0.
P(Y>=0)= 1/2 (by symmetry property)
Question 115 |




2 | |
5 | |
8 | |
16 |

Question 116 |
change the initial condition to –y(0) and the forcing function 2x(t) | |
change the initial condition to 2y(0) and the forcing function to –x(t) | |
change the initial condition to ![]() ![]() | |
change the initial condition to ![]() ![]() |

where P, Q, K are constant. Taking the Laplace transform both the sides, we get

Now, the solutions becomes

or,

So, Eq. (1) changes to

or,

Comparing Eq. (1) and (2), we conclude that

This makes the two equations to be same. Hence, we require to change the initial condition to -2y(0) and the forcing equation to – 2x(t)
Question 117 |


-2 | |
-1 | |
1 | |
2 |



Poles are at -1 and -3 i.e. (-1,0) and (-3,0).
From figure below of |z + 1| = 1, we see that (-1,0) is inside the circle and (- 3, 0) is outside the circle.

Residue theorem says,

which are inside (C)
So the required integral

by the residue of function at pole (-1, 0) (which is inside the circle).
This residue is

Question 118 |

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x | |
1 |

Question 119 |




2 | |
- 1 | |
0 | |
1 |

taking Laplace transform on both the sides we have








Question 120 |
1/3 | |
1/2 | |
2/3 | |
3/4 |





Question 121 |
21 | |
25 | |
41 | |
46 |
f(x) = x3 – 9x2 + 24x + 5 in the interval [1,6]
First find local maximum if any by putting
f'(x) = 0
i.e. f’(x) = 3x2 – 18x + 24 = 0
i.e. x2 – 6x + 8 = 0
x = 2,4
Now f”(x) = 6x – 18
f”(2) = 12 – 18 - -6 < 0
(So x = 2 is a point of local maximum) and
f”(4) = 24 – 18 = +6 > 0
(So x = 4 is a point of local minimum) Now tabulate the values of f at end point of interval and at local maximum point, to find absolute maximum in given range, as shown below:

Clearly the absolute maxima is at x = 6 and absolute maximum value is 41.
Question 122 |


15 A + 12 I | |
19A + 30 I | |
17A + 15 I | |
17 A + 21I |


(-5-λ)(-λ) + 6 = 0

So,

(by Cayley Hamilton theorem)
⇒ A2 = –5A – 6I
Multiplying by A on both sides, we have,
A3 =-5A2-6A
⇒ A3 = -5(-5A-6I)-6A
= 19A + 30I
Question 123 |

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As it has constant pdf that means it has uniform distribution. And for uniform distribution of pdf , we have formula
Variance= mean square value



Mean square value =

here, b=

therefore, variance =

Refer the Topic Wise Question for Statistics Engineering Mathematics
Question 124 |
(0, 0, 0) | |
(0, -1, 0) | |
(0, 1, 0) | |
(1, 1, 1) |

As per the above diagram the coordinates of the vertex which is diagonally opposite to the vertex whose coordinates are (1, 0, 1) = (0, -1, 0).