Engineering-Mathematics
Question 1 |

Let M be the adjacency matrix of G.
Define graph G2on the same set of vertices with adjacency matrix N, where

Which one of the following statements is true?
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |



Question 2 |

A | 3 |

Question 3 |
A | 0.37 |

Question 4 |
A | If a relation S is reflexive and circular, then S is an equivalence relation. |
B | If a relation S is transitive and circular, then S is an equivalence relation. |
C | If a relation S is circular and symmetric, then S is an equivalence relation. |
D | If a relation S is reflexive and symmetric, then S is an equivalence relation. |
Theorem: A relation R on a set A is an equivalence relation if and only if it is reflexive and circular.
For symmetry, assume that x, y ∈ A so that xRy, lets check for yRx.
Since R is reflexive and y ∈ A, we know that yRy. Since R is circular and xRy and yRy, we know that yRx. Thus R is symmetric.
For transitivity, assume that x, y, z ∈ A so that xRy and yRz. Check for xRz. Since R is circular and xRy and yRz, we know that zRx. Since we already proved that R is symmetric, zRx implies that xRz. Thus R is transitive.
Question 5 |
Which one of the following options is correct?
A | G is always cyclic, but H may not be cyclic. |
B | Both G and H are always cyclic. |
C | G may not be cyclic, but H is always cyclic. |
D | Both G and H may not be cyclic. |
If ‘G’ is a group with sides 6, its subgroups can have orders 1, 2, 3, 6.
(The subgroup order must divide the order of the group)
Given ‘H’ can be 1 to 6, but 4, 5 cannot divide ‘6’.
Then ‘H’ is not a subgroup.
G can be cyclic only if it is abelian. Thus G may or may not be cyclic.
The H can be cyclic only for the divisors of 6 and H cannot be cyclic for any non divisors of 6.
Question 6 |
S1: There exist random variables X and Y such that
EX-E(X)Y-E(Y)2>Var[X] Var[Y]
S2: For all random variables X and Y,
CovX,Y=E|X-E[X]| |Y-E[Y]|
Which one of the following choices is correct?
A | S1is false, but S2is true. |
B | Both S1and S2are true. |
C | S1is true, but S2is false. |
D | Both S1and S2 are false. |
For a dataset with single values, we have variance 0. EX-E(X)Y-E(Y)2>Var[X] Var[Y]
This leads to inequance of 0>0 which is incorrect.

Its not |x-E(x)|. Thus S2 is also incorrect.
Question 7 |
A | 11 |
v - e + f = 2
v is number of vertices
e is number of edges
f is number of faces including bounded and unbounded
8-e+5=2
=> 13-2 =e
The number of edges are =11
Question 8 |

If the received signal is H, the probability that the transmitted signal was H (rounded to 2 decimal places) is _______
A | 0.04 |
Bayes theorem:
Probability of event A happening given that event B has already happened is
P(A/B) = P(B/A)*P(A) / P(B)
Here, it is asked that P( H transmitted / H received).
S can send signal to H with 0.1 probability, S can send signal to L with 0.9 probability.
The complete diagram can be

Probability that H Transmitted (H_t) given that H received (H_r)is
P( H_t / H_r) = P( H_r/ H_t) * P(H_t) / P(H_r)
P(H-r) = probability that H received = P( H received from H)+ P(H received from L)
It can be observed from the graph that H can receive in two ways (S to H to H) and (S to L to H)
The P(H_r) = 0.1*0.3 + 0.9*0.8= 0.03+0.72 = 0.75
P(H_received given that H_transmitted) =0.3
P(H transmitted ) = 0.1 i.e.
P( H_t / H_r) = P( H_r/ H_t) * P(H_t) / P(H_r)
= 0.3*0.1 / 0.75 = 0.04
Question 9 |

The value of the above expression (rounded to 2 decimal places) is _______
A | 0.25 |

Question 10 |

Which one of the following choices is correct?
A | Both S1and S2 are tautologies. |
B | Neither S1and S2 are tautology. |
C | S1is not a tautology but S2is a tautology. |
D | S1is a tautology but S2is not a tautology. |
A tautology is a formula which is "always true" . That is, it is true for every assignment of truth values to its simple components.
Method 1:
S1: (~p ^ (p Vq)) →q
The implication is false only for T->F condition.
Let's consider q as false, then
(~p ^ (p Vq)) will be (~p ^ (p VF)) = (~p ^ (p)) =F.
It is always F->F which is true for implication. So there are no cases that return false, thus its always True i.e. its Tautology.
S2:
q->(~p (p Vq))
The false case for implication occurs at T->F case.
Let q=T then (~p (p Vq)) = (~p (p VT))= ~p. (It can be false for p=T).
So there is a case which yields T->F = F. Thus its not Valid or Not a Tautology.
Method 2:

Question 11 |
Let ({p,q{,*) be a semigroup where p * q = q. Show that:
(a) p * q = q * p and (b) q * q = q
A | Theory Explanation. |
Question 12 |
Show that proposition C is a logical consequence of the formula
A ∧ (A →(B ∨ C)) ∧ (B → ~A)
Using truth tables.
A | Theory Explanation. |
Question 13 |
The less-than relation, <, on reals is
A | a partial ordering since it is asymmetric and reflexive |
B | a partial ordering since it is antisymmetric and reflexive |
C | not a partial ordering because it is not asymmetric and not reflexive |
D | not a partial ordering because it is not antisymmetric and reflexive
|
E | none of the above |
1) not reflexive
2) irreflexive
3) not symmetric
4) Asymmetric
5) Antisymmetric
Question 14 |
Let A and B be sets with cardinalities m and n respectively. The number of one-one mappings (injections) from A to B, when m < n, is:
A | mn |
B | nPm |
C | mCn |
D | nCm |
E | mPn |
A = {a1, a2, ... am} and
B = {b1, b2, ... bn}
A one-one function 'f' assigns each element ai of A a distinct element, bj=f(ai) of Bi for a, there are n choices, for a2 there are n-1 choices, for am there are (n-(m-1)) choices.
i.e.,

Question 15 |
The proposition p ∧(~p ∨ q) is:
A | a tautology |
B | logically equivalent to p ∧ q |
C | logically equivalent to p ∨ q |
D | a contradiction |
E | none of the above |
(p ∧ ~p) ∨ (p ∧ q)
F ∨ (p ∧ q)
(p ∧ q)
Question 16 |
Let S be an infinite set and S1 ..., Sn be sets such that S1 ∪ S2 ∪ ... ∪ Sn = S. Then,
A | at least one of the set Si is a finite set |
B | not more than one of the set Si can be finite |
C | at least one of the sets Si is an infinite set |
D | not more than one of the sets Si can be infinite |
E | None of the above |
Question 17 |
Let A be a finite set of size n. The number of elements in the power set of A × A is:
A | 22n |
B | 2n2 |
C | (2n)2 |
D | (22)n |
E | None of the above |
A = {1,2}
|A| = {∅, {1}, {2}, {1,2}}
Cardinality of power set of A = 2n2
A × A = {1,2} × {1,2}
= {(1,1), (1,2), (2,1), (2,2)}
Cardinality of A × A = n2
Cardinality of power set of A × A = 2n2
Question 18 |
The Laplace transform of the periodic function f(t) described by the curve below, i.e.,

is _________

A | Out of syllabus. |
Question 19 |
Given and S the surface of a unit cube with one corner at the origin and edges parallel to the coordinate axes, the value of integral

A | Out of syllabus. |
Question 20 |
The differential equation yn + y = 0 is subjected to the boundary conditions.
y (0) = 0 y(λ) = 0
In order that the equation has non-trivial solution(s), the general value of λ is __________
A | Out of syllabus. |
Question 21 |
For X = 4.0, the value of I in the FORTRAN 77 statement
1 = -2**2 + 5.0*X/X*3 + 3/4 is _______
A | Out of syllabus. |
Question 22 |
The value of the double integral is
A | 1/3 |

Question 23 |
If the matrix A4, calculated by the use of Cayley-Hamilton theorem or otherwise, is _________
A | A4 = I |
(1-λ) (-1-λ) (i-λ) (-i-λ)
= (λ2-1) (λ2+1)
= λ4-1
Characteristic equation is λ4-1 = 0.
According to Cayley-Hamilton theorem, every matrix satisfies its own characteristic equation, so
A4 = I
Question 24 |
If the linear velocity is given by

The angular velocity at the point (1, 1, -1) is ________
A | Out of syllabus. |
Question 25 |
Given the differential equation, y′ = x − y with the initial condition y(0) = 0. The value of y(0.1) calculated numerically upto the third place of decimal by the second order Runga-Kutta method with step size h = 0.1 is ________
A | Out of syllabus. |
Question 26 |
The radius of convergence of the power series

A | Out of syllabus. |
Question 27 |
The function f(x,y) = x2y - 3xy + 2y + x has
A | no local extremum |
B | one local minimum but no local maximum |
C | one local maximum but no local minimum |
D | one local minimum and one local maximum |
Question 28 |

A | 1 |

Question 29 |
Which of the following improper integrals is (are) convergent?
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |
Question 30 |
The differential equation
d2y/dx2 + dy/dx + siny = 0 is:
A | linear |
B | non-linear |
C | homogeneous |
D | of degree two |
d2y/dx2 + dy/dx + siny = 0
In this DE, degree is 1 then this represent linear equation.
Question 31 |
Simpson’s rule for integration gives exact result when f(x) is a polynomial of degree
A | 1 |
B | 2 |
C | 3 |
D | 4 |
Question 32 |
Fourier series of the periodic function (period 2π) defined by

But putting x = π, we get the sum of the series.

A | π2/4 |
B | π2/6 |
C | π2/8 |
D | π2/12 |
Question 33 |
The eigen vector(s) of the matrix

is (are)
A | (0,0,α ) |
B | (α,0,0) |
C | (0,0,1) |
D | (0,α,0) |
E | Both B and D |
So the question as has
(A - λI)X = 0
AX = 0

What x1, x2, x3 are suitable?
Which means:
x1 times column 1 + x2 times column 2 + x3 times column 3 = zero vector
Since α is not equal to zero, so x3 must be necessarily zero to get zero vector.
Hence, only (B) and (D) satisfies.
Question 34 |
Every element a of some ring (R,+,0) satisfies the equation aoa = a.
Decide whether or not the ring is commutative.
A | Theory Explanation. |
Question 35 |
A 3-ary tree is a tree in which every internal node has exactly three children. Use induction to prove that the number of leaves in a 3-ary tree with n interval nodes is 2(n-1)+3.
A | Theory Explanation. |
Question 36 |
Let p and q be propositions. Using only the truth table decide whether p ⇔ q does not imply p → q is true or false.
A | True |
B | False |

So, "imply" is False making "does not imply" True.
Question 37 |
(a) Let * be a Boolean operation defined as
If C = A * B then evaluate and fill in the blanks:
(i) A * A = _______
(ii) C * A = _______
(b) Solve the following boolean equations for the values of A, B and C:
A | Theory Explanation. |
Question 38 |
Find the inverse of the matrix
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |
-λ3 + 2λ2 - 2 = 0
Using Cayley-Hamiltonian theorem
-A3 + 2A2 - 2I = 0
So, A-1 = 1/2 (2A - A2)
Solving we get,

Question 39 |
Match the following items

A | (i) - (b), (ii) - (c), (iii) - (d), (iv) - (a) |
Question 40 |
The Hasse diagrams of all the lattices with up to four elements are __________ (write all the relevant Hasse diagrams).
A | ![]() |
We can't draw lattice with 1 element.
For 2 element:

For 3 element:

For 4 element:

Question 41 |
Let A, B and C be independent events which occur with probabilities 0.8, 0.5 and 0.3 respectively. The probability of occurrence of at least one of the event is __________
A | 0.93 |
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)P(B) - P(B)P(C) - P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 - 0.4 - 0.5 - 0.24 + 0.12
= 0.93
Question 42 |
The number of subsets {1, 2, ... n} with odd cardinality is __________.
A | 2n-1 |
And so, no. of subsets with odd cardinality is half of total no. of subsets = 2n /n = 2n-1
Question 43 |
The number of edges in a regular graph of degree d and n vertices is _________.
A | d*n/2 |
d * n = 2 * |E|
∴ |E| = d*n/2
Question 44 |
The probability of an event B is P1. The probability that events A and B occur together is P2 while the probability that A and occur together is P3. The probability of the event A in terms of P1, P2 and P3 is __________.
A | P2 + P3 |
P3 = P(A) - P2
P(A) = P2 + P3
Question 45 |
On the set N of non-negative integers, the binary operation __________ is associative and non-commutative.
A | fog |
(fog)(x) = f(g(x))
It is associative, (fog)oh = fo(goh), but its usually not commutative. fog is usually not equal to gof.
Note that if fog exists then gof might not even exists.
Question 46 |
Amongst the properties {reflexivity, symmetry, anti-symmetry, transitivity} the relation R = {(x,y) ∈ N2 | x ≠ y } satisfies __________.
A | symmetry |
It is symmetric as if xRy then yRx.
It is not antisymmetric as xRy and yRx are possible and we can have x≠y.
It is not transitive as if xRy and yRz then xRz need not be true. This is violated when x=x.
So, symmetry is the answer.
Question 47 |
The number of substrings (of all lengths inclusive) that can be formed from a character string of length n is
A | n |
B | n2 |
C | n(n-1)/2 |
D | n(n+1)/2 |
n = 1
(n-1) = 2
(n-2) = 3
So, Total = n(n+1)/2
Question 48 |
The tank of matrix is:
A | 0 |
B | 1 |
C | 2 |
D | 3 |

Question 49 |
Some group (G,o) is known to be abelian. Then, which one of the following is true for G?
A | g = g-1 for every g ∈ G |
B | g = g2 for every g ∈ G |
C | (goh)2 = g2oh2 for every g,h ∈ G |
D | G is of finite order |
For Abelian group, commutative also holds
i.e., (aob) = (boa)
Consider option (C):
(goh)2 = (goh)o(gog)
= (hog)o(goh)
= (ho(gog)oh)
= ((hog2)oh)
= (g2oh)oh
= g2o(hoh)
= g2oh2 [True]
Question 50 |
In a compact single dimensional array representation for lower triangular matrices (i.e all the elements above the diagonal are zero) of size n × n, non-zero elements (i.e elements of the lower triangle) of each row are stored one after another, starting from the first row, the index of the (i, j)th element of the lower triangular matrix in this new representation is:
A | i + j |
B | i + j - 1 |
C | j + i(i-1)/2 |
D | i + j(j-1)/2 |
If we assume array index starting from 1 then, ith row contains i number of non-zero elements. Before ith row there are (i-1) rows, (1 to i-1) and in total these rows has 1+2+3......+(i-1) = i(i-1)/2 elements.
Now at ith row, the jth element will be at j position.
So the index of (i, j)th element of lower triangular matrix in this new representation is
j = i(i-1)/2