## Engineering Mathematics | GATE Subject Wise

Question 1 |

Fill in the Blank Type Question |

Question 2 |

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

Fill in the Blank Type Question |

Question 3 |

Fill in the Blank Type Question |

Question 4 |

E ^{1/z} | |

In(z) | |

Cos (z) |

^{1/z}is NOT analytic at z = 0

(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 - z

^{2}+ z

^{4}/4! - z

^{6}/6! + z

^{8}/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 |

Fill in the Blank Type Question |

Question 7 |

Question 8 |

-2.5 | |

0 | |

15 | |

25 |

**If sum of all the rows or columns are same, then that**

sum = one eigen value of that matrix

sum = one eigen value of that matrix

Sum of all rows

**(15-x) * |Matrix| = 0**

15 is a factor .

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, =0-10=-10≠0

Rank = 2

Question 10 |

Consider the following statement about the linear dependence of the real valued functions y_{1}= 1, y_{2}= x and y _{3} =x^{2}, over the field of real numbers.

I. y_{1}, y_{2} and y_{3} are linearly independent on -1 ≤ x ≤ 0

II. y_{1} , y_{2} and y_{3} are linearly dependent on 0 ≤ x ≤1

III. y_{1} , y_{2} and y_{3} are linearly independent on 0 ≤x ≤1

IV. y_{1}, y_{2} and y_{3} 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 |

**e**for real x. From among the following, choose the Taylor series approximation of f(x) around x = 0, which included all powers of x less than or equal to 3.

^{x+x2}Question 13 |

Fill in the Blank Type Question |

Question 14 |

Fill in the Blank Type Question |

Question 15 |

Question 16 |

S1: has 4 linearly independent eigenvectors.

S2: has 4 distinct eigenvalues.

S3: is non-singular (invertible).

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 |

**U**and the output be

**Y**of a system, and the other parameters are real constants. Identify which among the following systems is not a linear system:

(with initial rest conditions) | |

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 ____________.

Fill in the Blank Type Question |

Question 19 |

Fill in the Blank Type Question |

Question 20 |

Fill in the Blank Type Question |

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 |

**C**given below is on the complex plane where

The value of the integral is __________.

Fill in the Blank Type Question |

Question 23 |

We need to use suitable substitution here,

Put,

After simplification we obtain the following relation,

Given that the curve passes through points,

, we can obtain the value of constant C.

So,

Question 24 |

Fill in the Blank Type Question |

Question 25 |

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 |

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.

has local maximum

For x = 1, local minimum value

Finding f(- 100) = -333433.33

f (100) = 333233.33

( x =100, -100 are end points of interval)

∴ Minimum occurs at x= - 100

Question 31 |

^{4}= I, (where I denotes the identity matrix) and M ≠ I, M

^{2}≠I and M

^{3}≠I. Then, for any natural number k, M

^{-1 }equals:

M ^{4k+1} | |

M ^{4k + 2} | |

M ^{4k+3} | |

M ^{4k} |

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 = x

_{o}, then it is also differentiable at x = x

_{o}.

Q: If f(x) is continuous at x = x

_{o}, then it may not be differentiable at x = x

_{o}.

R: If f(x) is differentiable at x = x

_{o}, then it is also continuous at x = x

_{o}

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 |

**->**Integration of an odd function is even in this logic A

**->**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-I : -**

**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 ?

Question 40 |

The residue of at z = 1 is 1/2 | |

(complex conjugate of z) is an analytical function |

Question 41 |

Fill in the Blank Type Question |

Question 42 |

(2-t)e ^{t} | |

(1+2t)e ^{-t} | |

(2+t)e ^{-t} | |

(1-2t)e ^{t} |

**Differential equation is**s

^{2}+ 2s + 1 = 0

roots are equal s

_{1}= s

_{2}= - 1

So, y = c

_{1}e

^{-t}+ c

_{2}t e

^{-t}

y(0) = C

_{1}= 1

yï = - C

_{1}e

^{-t}– C

_{2}t e

^{-t}+ C

_{2}e

^{-t}

yï(0) = - C

_{1}+ C

_{2}= 1

C

_{2}= 2

So solution is y = e

^{-t}+ 2te

^{-t}

y = (1 + 2t) e

^{-t}.

Question 43 |

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}(x

^{2}+x+1)

f(0) = 1

f(0.5) = 1.067

For positive values of x, function never goes negative

Question 44 |

Fill in the Blank Type Question |

Question 45 |

Fill in the Blank Type Question |

Question 46 |

Fill in the Blank Type Question |

^{3}+ b

_{1}|z|

^{3}

Given that f(z) is analytic.

which is possible only when b = 0

since |z|

^{3}is differentiable at the origin but not analytic.

2z

^{3}is analytic everywhere

∴ f(z) = 2z

^{3}+ b|z|

^{3}is analytic

only when b = 0

**Lets take an Example**Question 47 |

^{3}-3x

^{2}+ 1?

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 is a constant. When the vector magnitude |I| is at its minimum value, the angle 0 that I make with the x axis (in degrees, such that) is

Fill in the Blank Type Question |

The minimum magnitude will be 5

At '5' magnitude angle is 90 °.

Question 50 |

Fill in the Blank Type Question |

**Using Green's Theorem**

Question 51 |

The value of is

Fill in the Blank Type Question |

Question 52 |

*x*is unknown. If the eigenvalues of the matrix A are and the x is equal to

Product of eigen values = Det of A

Question 53 |

Fill in the Blank Type Question |

Residue = coefficient of =1

Question 54 |

Fill in the Blank Type Question |

P(Tail) = 0.7

Required probability = TTTTH

= (0.7)(0.7)(0.7)(0.7)(0.3)

=0.07203

Question 55 |

Fill in the Blank Type Question |

Question 56 |

with exact solution

**y(x) = x**For

^{2}+ e^{x}*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 ____

Fill in the Blank Type Question |

Question 57 |

With and

Question 58 |

^{T}A

^{–1}is

sec ^{2} x | |

cos 4x | |

1 | |

0 |

Question 59 |

^{2}+ y

^{2}with respect to y is equal to the partial derivative of 6y + 4x with respect to x, is

y = 2 | |

x = 2 | |

x + y = 4 | |

x – y = 0 |

Partial derivative of with respect to y

**=0+2Y**

= partial derivative of with respect to x

**=0+4**

So,

Question 60 |

_{0}, in the complex z-plane and if n is a non-zero integer, then

2pnj | |

0 | |

2pn |

Question 61 |

^{–t}sin (2πt) u (t) where u (t) is the unit step function. The area under g(t) is______

Fill in the Blank Type Question |

Question 62 |

Fill in the Blank Type Question |

Now multiply equ(i) with 1/2

Now subtrac eq(ii) from (i)

Question 63 |

^{3}– 5x

^{2}+ 6x – 8 = 0. Taking the initial guess as x = 5, is________.

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 is _____.

Fill in the Blank Type Question |

By using divergence theorem

Question 67 |

47/48 | |

1/4 | |

13/48 | |

35/48 |

**P(Ram) = 1/6, P(Ramesh = 1/8)**

**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 |

_{1}) = f(z

_{2}) for all z

_{1}≠ z

_{2}, a = 2, b = 4 and c = 5,

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 |

For the differential equation, we have

Question 72 |

(cM) ^{T} = c(M)^{T} | |

All other are properties of matrix.

Refer: www.khanacademy.org

Question 73 |

0.42 | |

0.52 | |

0.62 | |

0.67 |

E

_{1}= 1 children family

E

_{2}= 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 |

E

_{1}= 1 children family

E

_{2}= 2 children family

A = picking a child

Then by Baye’s theorem, Required probability is

Question 75 |

**|z| = 3.**The value of the integral

dz is

Question 76 |

^{2}= I, where I is the (4 × 4) identity matrix. The positive eigen value of A is __________.

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 |

Question 80 |

**{ 0 ≤ y ≤ x and 0 ≤ x ≤ 12 }**is _________.

644 | |

720 | |

748 | |

864 |

Question 81 |

Which is obtained by reversing the order of the columns of the identity matrix I

_{6}.

Let

**P= I**where

_{6}+ αJ_{6}**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 |

^{2}, respectively. The relation which always holds true is

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

Question 83 |

_{1}) X(t

_{2})]

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 | |

^{x}as x approaches infinity equal to any variable e.g. y, k. and take the natural logarithm of both sides.

Question 88 |

has two equal roots, then the values of a are

± 1 | |

0,0 | |

± j | |

±1/ 2 |

^{2}− 4AC = 0

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 |

**z = x + jy**is given by e

^{-y}cos(x). The imaginary part of f(z) is

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 |

**f**and

_{x}(x) =2e^{-2x}for x ≥ 0**f**

_{x}(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 Pd of input X is

Let x

_{T}be the threshold

Where x

_{1}and x

_{2}are two levels

Question 95 |

0 | |

1 | |

2 | |

-1 |

Question 96 |

B) is a first order linear equation and homogeneous

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 A ^{T} 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 = x

_{1}+ x

_{2}- x

_{3}

P{x

_{1}+ x

_{2}≤ x

_{3}} = P{x

_{1}+ x

_{2}- x

_{3}≤ 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 –x

_{3}is

Question 103 |

_{1}and Y

_{2}as shown in the figure.

The value of H(Y1) + H(Y2) in bits is _______.

2 | |

5 | |

8 | |

6 |

Question 104 |

2 | |

e |

Question 105 |

^{2}+ 3y

^{2}+ z

^{3}at the point (1,1,1) is

7 | |

9 | |

12 | |

5 |

Question 106 |

3 | |

5 | |

8 | |

4 |

Question 107 |

at x = 1 ________

0.44 | |

0.48 | |

0.54 | |

0.62 |

Question 108 |

0.22 | |

0.33 | |

0.44 | |

0.55 |

Question 109 |

^{o})

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 then Area,

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 (

*a*

_{1}) to (

*a*

_{0}) in the Routh matrix 1

^{st}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 |

_{0}for t > 0, we need to

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 and the forcing function to | |

change the initial condition to and the forcing function to |

*x*(

*t*) is

*y*(

*t*) so, we can define a function relating

*x*(

*t*) and

*y*(

*t*) as below

where

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

…(1)

Now, the solutions becomes

or,

So, Eq. (1) changes to

or, …(2)

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 – 2

*x*(

*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,

Residue of those poles

which are inside (C)

So the required integral is given

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

This residue is

Question 118 |

^{x}is

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 |

^{3}- 9x

^{2}+ 24x + 5 in the interval [1, 6] is

21 | |

25 | |

41 | |

46 |

f(x) = x

^{3}– 9x

^{2}+ 24x + 5 in the interval [1,6]

First find local maximum if any by putting

f'(x) = 0

i.e. f’(x) = 3x

^{2}– 18x + 24 = 0

i.e. x

^{2}– 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 |

^{3}is

15 A + 12 I | |

19A + 30 I | |

17A + 15 I | |

17 A + 21I |

_{(-5-λ)(-λ) + 6 = 0}

So,

(by Cayley Hamilton theorem)

⇒ A

^{2}= –5A – 6I

Multiplying by A on both sides, we have,

A

^{3}=-5A

^{2}-6A

⇒ A

^{3}= -5(-5A-6I)-6A

= 19A + 30I

Question 123 |

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 =

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).