EngineeringMathematics
Question 2 
Let G be an arbitrary group. Consider the following relations on G:
R1: ∀a,b ∈ G, aR1b if and only if ∃g ∈ G such that a = g1bg R2: ∀a,b ∈ G, aR2b if and only if a = b1Which of the above is/are equivalence relation/relations?
R_{2} only  
R_{1} and R_{2}  
Neither R_{1} and R_{2}  
R_{1} only 
Consider Statement R_{1}:
Reflexive:
aR_{1}a
⇒ a = g^{1}ag
Left multiply both sides by g
⇒ ga = gg^{1}ag
Right multiply both sides by g^{1}
⇒ gag^{1} = gg^{1}agg^{1}
⇒ gag^{1} = a [∴ The relation is reflexive]
Symmetric:
If aR_{1}b, then ∃g ∈ G such that gag^{1} = b then a = g^{1}bg, which is Correct.
⇒ So, given relation is symmetric.
Transitive:
The given relation is Transitive.
So, the given relation R_{1} is equivalence.
R_{2}:
The given relation is not reflexive. So, which is not equivalence relation. Such that a ≠ a^{1}. So, only R_{1} is true.
Question 3 
Let X be a square matrix. Consider the following two statements on X.
I. X is invertible. II. Determinant of X is nonzero.Which one of the following is TRUE?
I implies II; II does not imply I.  
II implies I; I does not imply II.  
I and II are equivalent statements.  
I does not imply II; II does not imply I. 
That means we can also say that determinant of X is nonzero.
Question 4 
Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to
n!  
1  
(n1)!  
The total number of hamiltonian cycles in a complete graph are
(n1)!/2, where n is number of vertices.
Question 5 
Let U = {1,2,...,n}. Let A = {(x,X)x ∈ X, X ⊆ U}. Consider the following two statements on A.
Which of the above statements is/are TRUE?
Only II  
Only I  
Neither I nor II  
Both I and II 
and given A = {(x, X), x∈X and X⊆U}
Possible sets for U = {Φ, {1}, {2}, {1, 2}}
if x=1 then no. of possible sets = 2
x=2 then no. of possible sets = 2
⇒ No. of possible sets for A = (no. of sets at x=1) + (no. of sets at x=2) = 2 + 2 = 4
Consider statement (i) & (ii) and put n=2
Statement (i) is true
Statement (i) and (ii) both are true. Answer: (C)
Question 6 
Suppose Y is distributed uniformly in the open interval (1,6). The probability that the polynomial 3x^{2} + 6xY + 3Y + 6 has only real roots is (rounded off to 1 decimal place) _____.
0.3  
0.9  
0.1  
0.8 
3x^{2} + 6xY + 3Y + 6
= 3x^{2} + (6Y)x + (3Y + 6)
whch is in the form: ax^{2} + bx + c
For real roots: b^{2}  4ac ≥ 0
⇒ (6Y)^{2}  4(3)(3Y + 6) ≥ 0
⇒ 36Y^{2}  36Y  72 ≥ 0
⇒ Y^{2}  Y  2 ≥ 0
⇒ (Y+1)(Y2) ≥ 0
Y = 1 (or) 2
The given interval is (1,6).
So, we need to consider the range (2,6).
The probability = (1/(61)) * (62) = 1/5 * 4 = 0.8
Question 7 
 ∀x[(∀z zx ⇒ ((z = x) ∨ (z = 1))) ⇒ ∃w (w > x) ∧ (∀z zw ⇒ ((w = z) ∨ (z = 1)))]
Here 'ab' denotes that 'a divides b', where a and b are integers. Consider the following sets:
S1. {1, 2, 3, ..., 100} S2. Set of all positive integers S3. Set of all integersWhich of the above sets satisfy φ?
S1 and S3  
S1, S2 and S3  
S2 and S3  
S1 and S2 
Question 9 
3  
4  
5  
6 
→ Correction in Explanation:
⇒ (1  λ)(2  λ)  2 = 0
⇒ λ^{2}  3λ=0
λ = 0, 3
So maximum is 3.
Question 10 
Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equiprobable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is __________.
0.021  
0.022  
0.023  
0.024 
⇾ A person wins who gets lower number compared to other person.
⇾ There could be “tie”, if they get same number.
Favorable cases = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Probability (tie) = 6/36 (when two dice are thrown, sample space = 6 × 6 = 36)
= 1/6
“Find the probability that one of them wins in the third attempt"
⇾ Which means, first & second time it should be tie and third time it should not be tie
⇾ P (tie) * P (tie) * P (not tie)
⇒ 1/6* 1/6 * (1  1/6)
⇒ (5/36×6) = 0.138/6 = 0.023
Question 11 
1  
2  
3  
4 
Question 12 
41  
42  
43  
44 
For any group ‘G’ with order ‘n’, every subgroup ‘H’ has order ‘k’ such that ‘n’ is divisible by ‘k’.
Solution:
Given order n = 84
Then the order of subgroups = {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}
As per the proper subgroup definition, it should be “42”.
Question 13 
3/(1x)^{2}  
3x/(1x)^{2}  
2x/(1x)^{2}  
3x/(1x)^{2} 
Question 15 
Assume that multiplying a matrix G_{1} of dimension p×q with another matrix G_{2} of dimension q×r requires pqr scalar multiplications. Computing the product of n matrices G_{1}G_{2}G_{3}…G_{n} can be done by parenthesizing in different ways. Define G_{i}G_{i+1} as an explicitly computed pair for a given parenthesization if they are directly multiplied. For example, in the matrix multiplication chain G_{1}G_{2}G_{3}G_{4}G_{5}G_{6} using parenthesization(G_{1}(G_{2}G_{3}))(G_{4}(G_{5}G_{6})), G_{2}G_{3} and G_{5}G_{6} are the only explicitly computed pairs.
Consider a matrix multiplication chain F_{1}F_{2}F_{3}F_{4}F_{5}, where matrices F_{1}, F_{2}, F_{3}, F_{4} and F_{5} are of dimensions 2×25, 25×3, 3×16, 16×1 and 1×1000, respectively. In the parenthesization of F_{1}F_{2}F_{3}F_{4}F_{5} that minimizes the total number of scalar multiplications, the explicitly computed pairs is/ are
F_{1}F_{2} and F_{3}F_{4} only
 
F_{2}F_{3} only
 
F_{3}F_{4} only  
F_{1}F_{2} and F_{4}F_{5} only

→ Optimal Parenthesization is:
((F_{1}(F_{2}(F_{3} F_{4})))F_{5})
→ But according to the problem statement we are only considering F_{3}, F_{4} explicitly computed pairs.
Question 16 
Consider the firstorder logic sentence
where ψ(s,t,u,v,w,x,y) is a quantifierfree firstorder logic formula using only predicate symbols, and possibly equality, but no function symbols. Suppose φ has a model with a universe containing 7 elements.
Which one of the following statements is necessarily true?
There exists at least one model of φ with universe of size less than or equal to 3.
 
There exists no model of φ with universe of size less than or equal to 3.
 
There exists no model of φ with universe of size greater than 7.  
Every model of φ has a universe of size equal to 7.

"∃" there exists quantifier decides whether a sentence belong to the model or not.
i.e., ~∃ will make it not belong to the model. (1) We have ‘7’ elements in the universe, So max. size of universe in a model = ‘7’
(2) There are three '∃' quantifiers, which makes that a model have atleast “3” elements. So, min. size of universe in model = ‘7’.
(A) is False because: (2)
(B) is true
(C) is false because of (1)
(D) is false, because these all models with size {3 to 7} not only ‘7’.
Question 17 
Consider Guwahati (G) and Delhi (D) whose temperatures can be classified as high (H), medium (M) and low (L). Let P(H_{G}) denote the probability that Guwahati has high temperature. Similarly, P(M_{G}) and P(L_{G}) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(H_{D}), P(M_{D}) and P(L_{D}) for Delhi.
The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.
Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (H_{G}) then the probability of Delhi also having a high temperature (H_{D}) is 0.40; i.e., P(H_{D} ∣ H_{G}) = 0.40. Similarly, the next two entries are P(M_{D} ∣ H_{G}) = 0.48 and P(L_{D} ∣ H_{G}) = 0.12. Similarly for the other rows.
If it is known that P(H_{G}) = 0.2, P(M_{G}) = 0.5, and P(L_{G}) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is _______ .
0.60  
0.61  
0.62  
0.63 
The first entry denotes that if Guwahati has high temperature (H_{G} ) then the probability that Delhi also having a high temperature (H_{D} ) is 0.40.
P (H_{D} / H_{G} ) = 0.40
We need to find out the probability that Guwahati has high temperature.
Given that Delhi has high temperature (P(H_{G} / H_{D} )).
P (H_{D} / H_{G} ) = P(H_{G} ∩ H_{D} ) / P(H_{D} )
= 0.2×0.4 / 0.2×0.4+0.5×0.1+0.3×0.01
= 0.60
Question 18 
Let N be the set of natural numbers. Consider the following sets,

P: Set of Rational numbers (positive and negative)
Q: Set of functions from {0, 1} to N
R: Set of functions from N to {0, 1}
S: Set of finite subsets of N
Which of the above sets are countable?
Q and S only  
P and S only  
P and R only  
P, Q and S only

Set of functions from {0,1} to N is countable as it has one to one correspondence to N.
Set of functions from N to {0,1} is uncountable, as it has one to one correspondence to set of real numbers between (0 and 1).
Set of finite subsets of N is countable.
Question 19 
Consider the following statements.

(I) P does not have an inverse
(II) P has a repeated eigenvalue
(III) P cannot be diagonalized
Which one of the following options is correct?
Only I and III are necessarily true
 
Only II is necessarily true  
Only I and II are necessarily true  
Only II and III are necessarily true

Though the multiple of a vector represents same vector, and each eigen vector has distinct eigen value, we can conclude that ‘p’ has repeated eigen value.
If the unique eigen value corresponds to an eigen vector e, but the repeated eigen value corresponds to an entire plane, then the matrix can be diagonalized, using ‘e’ together with any two vectors that lie in plane.
But, if all eigen values are repeated, then the matrix cannot be diagonalized unless it is already diagonal.
So (III) holds correct.
A diagonal matrix can have inverse, So (I) is false.
Then (II) and (III) are necessarily True.
Question 20 
Let G be a graph with 100! vertices, with each vertex labeled by a distinct permutation of the numbers 1, 2, …, 100. There is an edge between vertices u and v if and only if the label of u can be obtained by swapping two adjacent numbers in the label of v. Let y denote the degree of a vertex in G, and z denote the number of connected components in G.
Then, y + 10z = ___________.
109  
110  
111  
112 
There exists edge between two vertices iff label of ‘u’ is obtained by swapping two adjacent numbers in label of ‘v’.
Example:
12 & 21, 23 & 34
The sets of the swapping numbers be (1, 2) (2, 3) (3, 4) … (99).
The no. of such sets are 99 i.e., no. of edges = 99.
As this is regular, each vertex has ‘99’ edges correspond to it.
So degree of each vertex = 99 = y.
As the vertices are connected together, the number of components formed = 1 = z
y + 102 = 99 + 10(1) = 109
Question 21 
Let T be a binary search tree with 15 nodes. The minimum and maximum possible heights of T are:
Note: The height of a tree with a single node is 0.4 and 15 respectively  
3 and 14 respectively  
4 and 14 respectively  
3 and 15 respectively 
The height of a tree with single node is 0.
Minimum possible height is when it is a complete binary tree.
Maximum possible height is when it is a skewed tree left/right.
So the minimum and maximum possible heights of T are: 3 and 14 respectively.
Question 22 

I. p ⇒ q
II. q ⇒ p
III. (¬q) ∨ p
IV. (¬p) ∨ q
I only  
I and IV only  
II only  
II and III only 
Construct Truth tables:
~p ⇒ ~q
II, III are equivalent to (~p) ⇒ (~q)
Method 2:
(I) p⇒q ≡ ~p∨q
(II) q⇒p ≡ ~q∨p
(III) (~q) ∨ p ≡ ~q∨p
(IV) (~p) ∨ p ≡ ~p∨q
Also, from question:
(~p) ⇒ (~q)
≡ p∨~q
So, (II) & (III) are equivalent to the statement given in question.
Question 23 
Consider the firstorder logic sentence F: ∀x(∃yR(x,y)). Assuming nonempty logical domains, which of the sentences below are implied by F?

I. ∃y(∃xR(x,y))
II. ∃y(∀xR(x,y))
III. ∀y(∃xR(x,y))
IV. ¬∃x(∀y¬R(x,y))
IV only  
I and IV only  
II only  
II and III only 
F: ∀x(∃yR(x,y)) (given)
: For all girls there exist a boyfriend
(x for girls and y for boys)
I: ∃y(∃xR(x,y))
: There exist some boys who have girlfriends.
(Subset of statement F, so True)
II: ∃y(∀xR(x,y))
: There exists some boys for which all the girls are girlfriend. (False)
III: ∀y(∃xR(x,y))
: For all boys exists a girlfriend. (False)
IV: ~∃x(∀y~R(x,y))
= ∀x(~∀y~R(x,y))
= ∀x(∃yR(x,y)) (∵ ~∀y=∃y, ~∃x=∀x)
(True)
Question 24 
a unique solution at x = J_{n} where J_{n} denotes a ndimensional vector of all 1  
no solution  
infinitely many solutions  
finitely many solutions 
AX = B
As given that
and c_{1}&c_{n} ≠ 0
means c_{0}a_{0} + c_{1}a_{1} + ...c_{n}a_{n} = 0, represents that a_{0}, a_{1}... a_{n} are linearly dependent.
So rank of 'A' = 0, (so if ‘B’ rank is = 0 infinite solution, ‘B’ rank>0 no solution) ⇾(1)
Another condition given here is, 'Σa_{i} = b',
so for c_{1}c_{2}...c_{n} = {1,1,...1} set, it is having value 'b',
so there exists a solution if AX = b →(2)
From (1)&(2), we can conclude that AX = B has infinitely many solutions.
Question 25 
Let X be a Gaussian random variable with mean 0 and variance σ^{2}. Let Y = max(X, 0) where max(a,b) is the maximum of a and b. The median of Y is __________.
0  
1  
2  
3 
Median is a point, where the probability of getting less than median is 1/2 and probability of getting greater than median is 1/2.
From the given details, we can simply conclude that, median is 0. (0 lies exactly between positive and negative values)
Question 26 
is 0  
is 1  
is 1  
does not exist 
If "x=1" is substituted we get 0/0 form, so apply LHospital rule
Substitute x=1
⇒ (7(1)^{6}10(1)^{4})/(3(1)^{2}6(1)) = (710)/(36) = (3)/(3) = 1
Question 27 
Let p, q and r be prepositions and the expression (p → q) → r be a contradiction. Then, the expression (r → p) → q is
a tautology  
a contradiction  
always TRUE when p is FALSE  
always TRUE when q is TRUE 
So r = F and (p→q) = T.
We have to evaluate the expression
(r→p)→q
Since r = F, (r→p) = T (As F→p, is always true)
The final expression is T→q and this is true when q is true, hence option D.
Question 28 
Let u and v be two vectors in R^{2} whose Euclidean norms satisfy u=2v. What is the value of α such that w = u + αv bisects the angle between u and v?
2  
1/2  
1  
1/2 
Let u, v be vectors in R^{2}, increasing at a point, with an angle θ.
A vector bisecting the angle should split θ into θ/2, θ/2
Means ‘w’ should have the same angle with u, v and it should be half of the angle between u and v.
Assume that the angle between u, v be 2θ (thus angle between u,w = θ and v,w = θ)
Cosθ = (u∙w)/(∥u∥ ∥w∥) ⇾(1)
Cosθ = (v∙w)/(∥v∥ ∥w∥) ⇾(2)
(1)/(2) ⇒ 1/1 = ((u∙w)/(∥u∥ ∥w∥))/((v∙w)/(∥v∥ ∥w∥)) ⇒ 1 = ((u∙w)/(∥u∥))/((v∙w)/(∥v∥))
⇒ (u∙w)/(v∙w) = (∥u∥)/(∥v∥) which is given that ∥u∥ = 2 ∥v∥
⇒ (u∙w)/(v∙w) = (2∥v∥)/(∥v∥) = 2 ⇾(3)
Given ∥u∥ = 2∥v∥
u∙v = ∥u∥ ∥v∥Cosθ
=2∙∥v∥^{2} Cosθ
w = u+αv
(u∙w)/(v∙w) = 2
(u∙(u+αv))/(v∙(u+αv)) = 2
(u∙u+α∙u∙v)/(u∙v+α∙v∙v) = 2a∙a = ∥a∥^{2}
4∥v∥^{2}+α∙2∙∥v∥^{2} Cosθ = 2(2∥v∥^{2} Cosθ+α∙∥v∥^{2})
4+2αCosθ = 2(2Cosθ+α)
4+2αCosθ = 4Cosθ+2α ⇒ Cosθ(uv)+2α4 = 0
42α = Cosθ(42α)
(42α)(Cosθ1) = 0
42α = 0
Question 29 
Both (I) and (II)  
(I) only  
(II) only  
Neither (I) nor (II) 
be a real valued, rank = 2 matrix.
a^{2}+b^{2}+c^{2}+d^{2} = 50
Square values are of order 0, 1, 4, 9, 16, 25, 36, …
So consider (0, 0, 5, 5) then Sum of this square = 0+0+25+25=50
To get rank 2, the 2×2 matrix can be
The eigen values are,
AλI = 0 (The characteristic equation)
λ(λ)25 = 0
λ^{2}25 = 0
So, the eigen values are within [5, 5], Statement I is correct.
The eigen values with largest magnitude must be strictly greater than 5: False.
So, only Statement I is correct.
Question 30 
The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is _________.
271  
272  
273  
274 
Let A = number divisible by 3
B = numbers divisible by 5
C = number divisible by 7
We need to find “The number of integers between 1 and 500 that are divisible by 3 or 5 or 7" i.e., A∪B∪C
We know,
A∪B∪C = A+B+CA∩BA∩CB∩C+A∩B
A = number of integers divisible by 3
[500/3 = 166.6 ≈ 166 = 166]
B = 100
[500/5 = 100]
C = 71
[500/7 = 71.42]
A∩B = number of integers divisible by both 3 and 5 we need to compute with LCM (15)
i.e.,⌊500/15⌋ ≈ 33
A∩B = 33
A∩C = 500/LCM(3,7) 500/21 = 23.8 ≈ 28
B∩C = 500/LCM(5,3) = 500/35 = 14.48 ≈ 14
A∩B∩C = 500/LCM(3,5,7) = 500/163 = 4.76 ≈ 4
A∪B∪C = A+B+CA∩BA∩CB∩C+A∩B∩C
= 166+100+71332814+4
= 271
Question 32 
Let p, q, r denote the statements “It is raining”, “It is cold”, and “It is pleasant”, respectively. Then the statement “It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold” is represented by
(¬p ∧ r) ∧ (¬r → (p ∧ q))  
(¬p ∧ r) ∧ ((p ∧ q) → ¬r)  
(¬p ∧ r) ∨ ((p ∧ q) → ¬r)  
(¬p ∧ r) ∨ (r → (p ∧ q)) 
q: It is cold
r: It is pleasant
“If it is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold.”
We can divide the statement into two parts with “Conjunction”.
i.e., ¬r→(p∧q) ⇾(2)
From (1) & (2), the given statement can be represented as
Question 33 
Nβ (1  β)  
Nβ  
N (1  β)  
Not expressible in terms of N and β alone 
Given g_{y} (z) = (1  β + βz)^{N} ⇾ it is a binomial distribution like (x+y)^{n}
Expectation (i.e., mean) of a binomial distribution will be np.
The polynomial function ,
given
Mean of Binomial distribution of b(x_{j},n,p)=
The probability Mass function,
Given:
Question 34 
Consider the set X = {a,b,c,d e} under the partial ordering

R = {(a,a),(a,b),(a,c),(a,d),(a,e),(b,b),(b,c),(b,e),(c,c),(c,e),(d,d),(d,e),(e,e)}.
The Hasse diagram of the partial order (X,R) is shown below.
The minimum number of ordered pairs that need to be added to R to make (X,R) a lattice is _________.
0  
1  
2  
3 
As per the definition of lattice, each pair should have GLB, LUB.
The given ‘R’ has GLB, LUB for each and every pair.
So, no need to add extra pair.
Thus no. of required pair such that Hasse diagram to become lattice is “0”.
Question 35 
2  
3  
4  
5 
R_{2}→R_{2}+R_{1}
The number of nonzero rows of P + Q = 2,
So Rank = 2
Note: “Rank” is the number of independent vectors.
Method1:
Each vector is a row in matrix.
Echelon form of a matrix has no. of zeroes increasing each rows.
The total nonzero rows left give the rank.
Method2:
Find determinant of matrix, for 3×5, if determinant is ‘0’, the max rank can be 2.
If determinant of any n×n is nonzero, then rows proceed with (n1)×(n1).
Question 36 
G is an undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is ___________.
16  
17  
18  
19 
Degree of each vertex ≥ 3
v = 2E
The relation between max and min degree of graph are
m ≤ 2E / v ≤ M
Given minimum degree = 3
So, 3 ≤2 E / v
3v ≤ 2E
3(n) ≤ 2(25)
n ≤ 50/3
n ≤ 16.6
(n = 16)
Question 37 
Probability that ‘P’ applies for the job given that Q applies for the job = P(p/q) = 1/2 ⇾ (2)
Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job = P(p/q) = 1/3 ⇾ (3)
Bayes Theorem:
(P(A/B) = (P(B/A)∙P(A))/P(B) ; P(A/B) = P(A∩B)/P(B))
⇒ P(p/q) = (P(q/p)∙P(p))/p(q)
⇒ 1/2 = (1/3×1/4)/p(q)
p(q) = 1/12×2 = 1/(6) (P(q) = 1/6) ⇾ (4)
From Bayes,
P(p/q) = (P(p∩q))/(P(q))
1/2 = P(p∩q)/(1⁄6)
(p(p∩q) = 1/12)
We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job = P(p'/q')
From Bayes theorem,
P(p'/q') = (P(p'∩q'))/P(q') ⇾ (5)
We know,
p(A∩B) = P(A) + P(B)  P(A∪B)
also (P(A'∩B') = 1  P(A∪B))
P(p'∩q') = 1  P(p∪q)
= 1  (P(p) + P(q)  P(p∩q))
= 1  (P(p) + P(q)  P(p) ∙ P(q))
= 1  (1/4 + 1/6  1/12)
= 1  (10/24  2/24)
= 1  (8/24)
= 2/3
(P(p'∩q') = 2/3) ⇾ (6)
Substitute in (5),
P(p'⁄q') = (2⁄3)/(1P(q)) = (2⁄3)/(11/6) = (2⁄3)/(5⁄6) = 4/5
(P(p'/q') = 4/5)
Question 39 
If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)^{2}] equals _________.
54  
55  
56  
57 
Mean = Variance
E(X) = E(X^{2})  (E(X))^{2} = 5
E(X^{2}) = 5 + (E(X))^{2} = 5 + 25 = 30
So, E[(X + 2)^{2}] = E[X^{2} + 4 + 4X]
= E(X^{2}) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
Question 40 
If the characteristic polynomial of a 3 × 3 matrix M over R (the set of real numbers) is λ^{3}  4λ^{2} + aλ + 30, a ∈ ℝ, and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is ________.
5  
6  
7  
8 
λ^{3}  4λ^{2} + aλ + 30 = 0 ⇾ (1)
One eigen value is ‘2’, so substitute it
2^{3}  4(2)^{2} + a(2) + 30 = 0
8  16 + 2a + 30 = 0
2a = 22
a = 11
Substitute in (1),
λ^{3}  4λ^{2}  11 + 30 = 0
So, (1) can be written as
(λ  2)(λ^{2}  2λ  15) = 0
(λ  2)(λ^{2}  5λ + 3λ  15) = 0
(λ  2)(λ  3)(λ  5) = 0
λ = 2, 3, 5
Max λ=5
Question 41 
Let p,q,r,s represent the following propositions.

p: x ∈ {8,9,10,11,12}
q: x is a composite number
r: x is a perfect square
s: x is a prime number
The integer x≥2 which satisﬁes ¬((p ⇒ q) ∧ (¬r ∨ ¬s)) is _________.
11  
12  
13  
14 
~((p→q) ∧ (~r ∨ ~S))
⇒ first simplify the given statement by converging them to ∧, ∨
⇒ [~(p→q) ∨ (~(~r ∨ ~s)]
Demorgan’s law:
⇒ [~(~p ∨ q) ∨ (r ∧ s)]
∵ p→q ≡ ~p ∨ q
⇒ [(p ∧ ~q) ∨ (r ∧ s)]
p ∧ ~q is {8,9,10,11,12} ∧ {not a composite number} i.e. {11}
r ∧ s is {perfect square} ∧ {prime} i.e. no answer
So, the one and only answer is 11.
Question 42 
Let a_{n} be the number of nbit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for a_{n}?
a_{n} = a_{(n1)} + 2a_{(n2)}  
a_{n} = a_{(n1)} + a_{(n2)}  
a_{n} = 2a_{(n1)} + a_{(n2)}  
a_{n} = 2a_{(n1)} + 2a_{(n2)} 
If n=1, we have {0,1}
# Occurrences = 2
If n=2, we have {00,01,10}
# Occurrences = 3
If n=3, we have {000,001,010,100,101}
# Occurrences = 5
It is evident that a_{3} = a_{1} + a_{2}
Similarly, a_{n} = a_{n1} + a_{n2}
Question 44 
A probability density function on the interval [a,1] is given by 1/x^{2} and outside this interval the value of the function is zero. The value of a is _________.
0.7  
0.6  
0.5  
0.8 
or
where (a, b) is internal and f(x) is probability density function.
Given,
f(x) = 1/x^{2} , a≤x≤1
The area under curve,
 1 + 1/a = 1
1/a = 2
a = 0.5
Question 45 
Two eigenvalues of a 3 × 3 real matrix P are (2 + √1) and 3. The determinant of P is __________.
18  
15  
17  
16 
So, For the given 3×3 matrix there would be 3 eigen values.
Given eigen values are : 2+i and 3.
So the third eigen value should be 2i.
As per the theorems, the determinant of the matrix is the product of the eigen values.
So the determinant is (2+i)*(2i)*3 = 15.
Question 46 
The coefﬁcient of x^{12} in (x^{3} + x^{4} + x^{5} + x^{6} + ...)^{3} is _________.
10  
11  
12  
13 
⇒ [x^{3}(1 + x + x^{2} + x^{3} + ...)]^{3}
= x^{9}(1 + x + x^{2} + x^{3} + ...)^{3}
First Reduction:
As x^{9} is out of the series, we need to find the coefficient of x^{3} in (1 + x + x^{2} + ⋯)^{3}
Here, m=3, k=3, the coefficient
= ^{5}C_{3} = 5!/2!3! = 10
Question 47 
Consider the recurrence relation a_{1} = 8, a_{n} = 6n^{2} + 2n + a_{n1}. Let a_{99} = K × 10^{4}. The value of K is ___________.
198  
199  
200  
201 
Replace a_{(n1)}
⇒ a_{n} = 6n^{2} + 2n + 6(n1)^{2} + 2(n1) + 6(n2)^{2} + 2(n2) + ⋯ a_{1}
Given that a_{1} = 8, replace it
⇒ a_{n} = 6n^{2} + 2n + 6(n1)^{2} + 2(n1) + 6(n2)^{2} + 2(n2) + ⋯8
= 6n^{2} + 2n + 6(n1)^{2} + 2(n1) + 6(n2)^{2} + 2(n2) + ⋯ + 6(1)^{2} + 2(1)
= 6(n^{2} + (n1)^{2} + (n2)^{2} + ⋯ + 2^{2} + 1^{2}) + 2(n + (n1) + ⋯1)
Sum of n^{2} = (n(n+1)(2n+1))/6
Sum of n = (n(n+1))/2
= 6 × (n(n+1)(2n+1))/6 + 2×(n(n+1))/2
= n(n+1)[1+2n+1]
= n(n+1)[2n+2]
= 2n(n+1)^{2}
Given a_{99} = k×10^{4}
a_{99} = 2(99)(100)^{2} = 198 × 10^{4}
∴k = 198
Question 48 
A function f:N^{+} → N^{+}, deﬁned on the set of positive integers N^{+}, satisﬁes the following properties:

f(n) = f(n/2) if n is even
f(n) = f(n+5) if n is odd
Let R = {i∃j: f(j)=i} be the set of distinct values that f takes. The maximum possible size of R is __________.
2  
3  
4  
5 
f(n)= f(n+5) if n is odd
We can observe that
and f(5) = f(10) = f(15) = f(20)
Observe that f(11) = f(8)
f(12) = f(6) = f(3)
f(13) = f(9) = f(14) = f(7) = f(12) = f(6) = f(3)
f(14) = f(9) = f(12) = f(6) = f(3)
f(16) = f(8) = f(4) = f(2) = f(1) [repeating]
So, we can conclude that
‘R’ can have size only ‘two’ [one: multiple of 5’s, other: other than 5 multiples]
Question 49 
Consider the following experiment.

Step1. Flip a fair coin twice.
Step2. If the outcomes are (TAILS, HEADS) then output Y and stop.
Step3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop.
Step4. If the outcomes are (TAILS, TAILS), then go to Step 1.
The probability that the output of the experiment is Y is (up to two decimal places) ________.
0.33  
0.34  
0.35  
0.36 
Stop conditions:
If outcome = TH then Stop [output 4]  (1)
else
outcome = HH/ HT then Stop [output N]  (2)
We get ‘y’ when we have (1) i.e., ‘TH’ is output.
(1) can be preceded by ‘TT’ also, as ‘TT’ will reset (1) again
Probability of getting y = TH + (TT)(TH) + (TT)(TT)(TH) + …
= 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 + ...
= (1/4)/(11/4)
= 1/3
= 0.33
Question 50 
Consider the following expressions:

(i) false
(ii) Q
(iii) true
(iv) P ∨ Q
(v) ¬Q ∨ P
The number of expressions given above that are logically implied by P ∧ (P ⇒ Q) is _________.
4  
5  
6  
7 
(P ∧ (P → Q))→ expression is a tautology. So we have to find
How many tautological formulas are there for the given inputs.
(P ∧ (P → Q)) → True is always tautology
(P ∧ (P → Q)) → False is not a tautology
(P ∧ (P → Q)) → Q is a tautology
(P ∧ (P → Q)) → ¬Q ∨ P is a tautology
(P ∧ (P → Q)) → P ∨ Q is a tautology
So there are 4 expressions logically implied by (P ∧ (P → Q))
Question 51 
Let f(x) be a polynomial and g(x) = f'(x) be its derivative. If the degree of (f(x) + f(x)) is 10, then the degree of (g(x)  g(x)) is __________.
9  
10  
11  
12 
It is given that f(x) + f(x) degree is 10.
It means f(x) is a polynomial of degree 10.
Then obviously the degree of g(x) which is f’(x) will be 9.
Question 52 
The minimum number of colours that is sufﬁcient to vertexcolour any planar graph is ________.
4  
5  
6  
7 
Here it is asked about the sufficient number of colors, so with the worst case of 4 colors we can color any planar graph.
Question 53 
Consider the systems, each consisting of m linear equations in n variables.
 I. If m < n, then all such systems have a solution
II. If m > n, then none of these systems has a solution
III. If m = n, then there exists a system which has a solution
Which one of the following is CORRECT?
I, II and III are true  
Only II and III are true  
Only III is true  
None of them is true 
If R(A) ≠ R(AB)
then there will be no solution.
ii) False, because if R(A) = R(AB),
then there will be solution possible.
iii) True, if R(A) = R(AB),
then there exists a solution.
Question 54 
Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. The probability that an LED bulb chosen uniformly at random lasts more than 100 hours is _________.
0.55  
0.56  
0.57  
0.58 
The bulbs of Type 1, Type 2 are same in number.
So, the probability to choose a type is 1/2.
The probability to choose quadrant ‘A’ in diagram is
P(last more than 100 hours/ type1) = 1/2 × 0.7
P(last more than 100 hours/ type2) = 1/2 × 0.4
Total probability = 1/2 × 0.7 + 1/2 × 0.4 = 0.55
Question 55 
Suppose that the eigenvalues of matrix A are 1, 2, 4. The determinant of (A^{1})^{T} is _________.
0.125  
0.126  
0.127  
0.128 
Given that eigen values are 1, 2, 4.
So, its determinant is 1*2*4 = 8
The determinant of (A^{1})^{T} = 1/ A^{T} = 1/A = 1/8 = 0.125
Question 56 
A binary relation R on ℕ × ℕ is deﬁned as follows: (a,b)R(c,d) if a≤c or b≤d. Consider the following propositions:

P: R is reﬂexive
Q: Ris transitive
Which one of the following statements is TRUE?
Both P and Q are true.  
P is true and Q is false.  
P is false and Q is true.  
Both P and Q are false. 
a≤c ∨ b≤d
Let a≤a ∨ b≤b is true for all a,b ∈ N
So there exists (a,a) ∀ a∈N.
It is Reflexive relation.
Consider an example
c = (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f)
This does not hold for any (a>e) or (b>f)
eg:
(2,2)R(1,2) as 2≤2
(1,2)R(1,1) as 1≤1
but (2,2) R (1,1) is False
So, Not transitive.
Question 57 
Which one of the following wellformed formulae in predicate calculus is NOT valid?
(∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x ¬p(x) ∨ ∀x q(x))  
(∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))  
∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x))  
∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x)) 
But in option (D), we can generate T → F.
Hence, not valid.
Question 58 
Consider a set U of 23 different compounds in a Chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U. Consider the following statements:

I. Each compound in U\S reacts with an odd number of compounds.
II. At least one compound in U\S reacts with an odd number of compounds.
III. Each compound in U\S reacts with an even number of compounds.
Which one of the above statements is ALWAYS TRUE?
Only I  
Only II  
Only III  
None 
U = 23
∃S ∋ (S⊂U)
Each component in ‘S’ reacts with exactly ‘3’ compounds of U,
If a component ‘a’ reacts with ‘b’, then it is obvious that ‘b’ also reacts with ‘a’.
It’s a kind of symmetric relation.>br> If we connect the react able compounds, it will be an undirected graph.
The sum of degree of vertices = 9 × 3 = 27
But, in the graph of ‘23’ vertices the sum of degree of vertices should be even because
(d_{i} = degree of vertex i.e., = no. of edges)
But ‘27’ is not an even number.
To make it an even, one odd number should be added.
So, there exists atleast one compound in U/S reacts with an odd number of compounds.
Question 59 
The value of the expression 13^{99}(mod 17), in the range 0 to 16, is ________.
4  
5  
6  
7 
a^{(p1)} ≡ 1 mod p (p is prime)
From given question,
p = 17
a^{(171)} ≡ 1 mod 17
a^{16} ≡ 1 mod 17
13^{16} ≡ 1 mod 17
Given:
13^{99} mod 17
13^{3} mod 17
2197 mod 17
4
Question 60 
5  
6  
7  
8 
l_{11} = 2 (1)
l_{11}u_{12} = 2
u_{12} = 2/2
u_{12} = 1  (2)
l_{21} = 4  (3)
l_{21}u_{12}+l_{22} = 9
l_{22} = 9  l_{21}u_{12} = 9  4 × 1 = 5
Question 61 
h(x)/g(x)  
1/x  
g(x)/h(x)
 
x/(1x)^{2}

Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1h(x)=1x/x1=1/x1
h(g(x))=g(x)/g(x)1=1x/x
⇒ g(h(x))/h(g(x))=x/(x1)(1x)=(x/x1)/1x=h(x)/g(x)
Question 62 
pr = 0  
pr = 1  
0 < pr ≤ 1/5  
1/5 < pr < 1 
Let A be the event that an element (x,y,z)∈ L^{3} satisfies x ∨(y∧z) = (x∨y) ∧ (x∨z) Since q∨(r∧s) = q∨p = q
and (q∨r)∧(q∨s)=t∧t=t q∨(r∧s)≠(q∨r)∧(q∨s)
Therefore, (x,y,z) = (q,r,s),(q,s,r),(r,q,s),(r,s,q),(s,r,q),(s,q,r)
i.e., 3! = 6 elements will not satisfy distributive law and all other (x,y,z) choices satisfy distributive law
n(A) = 1256 = 119
∴ required probability is 119/125
⇒ 1/5
Question 63 
a=6,b=4  
a=4,b=6
 
a=3,b=5
 
a=5,b=3 
By properties,
⇒ 6=1+a and 7=a4b
⇒ a=5 ⇒ 7=54b
⇒ b=3
Question 64 
1  
0  
1  
2 
Then the difference between the d(u) and d(v) is not more than '1'.
In the option 'D' the difference is given as '2' it is not possible in the undirected graph.
Question 66 
24  
25  
26  
27 
V+R=E+2 (1) where V, E, R are respectively number of vertices, edges and faces (regions)
Given V=10 (2) and number of edges on each face is three
∴3R=2E⇒R=2/3E (3)
Substituting (2), (3) in (1), we get
10+2/3E=E+2⇒E/3=8⇒E=24
Question 67 
Both {f} and {g} are functionally complete  
Only {f} is functionally complete
 
Only {g} is functionally complete
 
Neither {f} nor {g} is functionally complete 
f(X,X,X)=X'XX'+XX'+X'X'
=0+0+X'
=X'
Similarly, f(Y,Y,Y)=Y' and f(X,Z,Z)=Z'
f(Y',Y',Z')=(Y')'Y'Z'+Y'(Y')'+(Y')'(Z')'
=YY'Z'+Y'Y+YZ
=0+0+YZ
=YZ
We have derived NOT and AND. So f(X,Y,Z) is functionally complete.
g(X,Y,Z)=X'YZ+X'YZ'+XY
g(X,X,X)=X'XX+X'XZ'+XX
=0+0+X
=X
Similarly, g(Y,Y,Y)=Y and g(Z,Z,Z)=Z
NOT is not derived. Hence, g is not functionally complete.
Question 68 
0.99  
1.00  
2.00  
3.00 
=21/1(2)+32/2(3)+43/3(4)+…+10099/99(100)
=1/11/2+1/21/3+1/3…+1/981/99+1/991/100
=11/100
=99/100
=0.99
Question 69 
R is symmetric and reflexive but not transitive  
R is reflexive but not symmetric and not transitive  
R is transitive but not reflexive and not symmetric  
R is symmetric but not reflexive and not transitive 
In aRb, 'a' and 'b' are distinct. So it can never be reflexive.
Symmetric:
In aRb, if 'a' and 'b' have common divisor other than 1, then bRa, i.e., 'b' and 'a' also will have common divisor other than 1. So, yes symmetric.
Transitive:
Take (3, 6) and (6, 2) elements of R. For transitivity (3, 2) must be the element of R, but 3 and 2 don't have a common divisor. So not transitive.
Question 70 
If a person is known to corrupt, he is kind
 
If a person is not known to be corrupt, he is not kind
 
If a person is kind, he is not known to be corrupt
 
If a person is not kind, he is not known to be corrupt

q: candidate will be elected
r: candidate is kind
then S1=p→~q
=q→~p (conrapositive rule)
and S2:r→q⇒r→~p (transitive rule)
i.e., If a person is kind, he is not known to be corrupt ∴ Option is C
Question 71 
6  
7  
8  
9 
⇒ λ^{2}5λ6=0⇒(λ6)(λ+1)=0⇒λ=6,1
∴ Larger eigen value is 6
Question 72 
2046  
2047  
2048  
2049 
Question 73 
36  
37  
38  
39 
=2^{2}+3×5^{2}×7 (i.e., product of primes)
Then the number of divisions of 2100 is
(2+1)∙(1+1)∙(2+1)∙(1+1) i.e., (3)(2)(3)(2) i.e., 36
Question 74 
A tree has no bridges  
A bridge cannot be part of a simple cycle  
Every edge of a clique with size 3 is a bridge (A clique is any complete sub graph of a graph)  
A graph with bridges cannot have a cycle

∴ (A) is false
Since, every edge in a complete graph kn(n≥3) is not a bridge ⇒
(C) is false
Let us consider the following graph G:
This graph has a bridge i.e., edge ‘e’ and a cycle of length ‘3’
∴ (D) is false
Since, in a cycle every edge is not a bridge
∴ (B) is true
Question 75 
200KBand 300 KB
 
200KBand 250 KB  
250KBand 300 KB  
300KBand 400 KB 
Since Best fit algorithm is used. So, process of size,
357KB will occupy 400KB
210KB will occupy 250KB
468KB will occupy 500KB
491KB will occupy 600KB
So, partitions 200KB and 300KB are NOT alloted to any process.
Question 76 
36  
37  
38  
39 
m = 4, n = 3 ⇒ number of onto function is
Question 77 
0  
1  
2  
3 
Method 2: Determinant is unaltered by the operations (i) and (ii)
∴ Determinant of the resultant matrix = Determinant of the given matrix
(Since C_{1},C_{3} are proportional i.e., C_{3}=15C_{1})
Question 78 
∀x ∃y R(x,y)↔ ∃y ∀x R(x,y)
 
(∀x [∃y R(x,y)→S(x,y)])→ ∀x∃y S(x,y)
 
[∀x ∃y (P(x,y)→R(x,y)]↔[∀x ∃y ( ¬ P(x,y)∨R(x,y)]  
∀x ∀y P(x,y)→ ∀x ∀y P(y,x)

[∀x ∃y (P(x,y)→R(x,y)]↔[∀x ∃y ( ¬ P(x,y)∨R(x,y)] is a tautology.
Question 79 
A multiple of 4  
Even  
Odd  
Congruent to 0 mod 4, or, 1 mod 4

Question 80 
x_{b} – (f_{b}–f(x_{a}))f_{b} /(x_{b}–x_{a})  
x_{a} – (f_{a}–f(x_{a}))f_{a} /(x_{b}–x_{a})  
x_{b} – (x_{b}–x_{a})f_{b} /(f_{b}–f(x_{a}))  
x_{a} – (x_{b}–x_{a}) f_{a} /(f_{b}–f(x_{a})) 
The first two iterations of the secant method. The red curve shows the function f and the blue lines are the secants. For this particular case, the secant method will not converge.
Question 81 
II only  
III only  
II and III only  
I, II and III 
∴ f is not bounced in [1, 1] and hence f is not continuous in [1, 1].
∴ Statement II & III are true.
Question 82 
0.95  
0.96  
0.97  
0.98 
Number of functions from X to Y is 20^{2} i.e., 400 and number of oneone functions from X to Y is ^{20}P_{2} i.e., 20×19 = 380
∴ Probability of a function f being oneone is 380/400 i.e., 0.95
Question 83 
The result is head  
The result is tail
 
If the person is of Type 2, then the result is tail  
If the person is of Type 1, then the result is tail 
Case 1:
The person who speaks truth. This definitely implies that result of toss is Head.
Case 2:
The person who lies. In this the reality will be the negation of the statement.
The negation of (x⇔y) is exactly one of x or y holds. "The result of the toss is head if and only if I am telling the truth". So here two possibilities are there,
→ It is head and lie spoken.
→ It is not head and truth spoken.
Clearly, the second one cannot speaks the truth. So finally it is head.
Hence, option (A).
Question 84 
Suppose U is the power set of the set S = {1, 2, 3, 4, 5, 6}. For any T ∈ U, let T denote the number of element in T and T' denote the complement of T. For any T, R ∈ U, let T \ R be the set of all elements in T which are not in R. Which one of the following is true?
∀X ∈ U (X = X')  
∃X ∈ U ∃Y ∈ U (X = 5, Y = 5 and X ∩ Y = ∅)  
∀X ∈ U ∀Y ∈ U (X = 2, Y = 3 and X \ Y = ∅)  
∀X ∈ U ∀Y ∈ U (X \ Y = Y' \ X') 
(A) False. Consider X = {1,2}. Therefore, X' = {3,4,5,6}, X = 2 and X' = 4.
(B) False. Because for any two possible subsets of S with 5 elements should have atleast 4 elements in commonc. Hence X∩Y cannot be null.
(C) False. Consider X = {1,4}, Y= {1,2,3} then X\Y = {4} which is not null.
(D) True. Take any possible cases.
Question 85 
15  
16  
17  
18 
1 1 1 1
1 1 1 2
1 1 1 3
1 1 2 2
1 1 2 3
1 1 3 3
1 2 2 2
1 2 2 3
1 2 3 3
1 2 3 3
1 3 3 3
2 2 2 2
2 2 2 3
2 2 3 3
2 3 3 3
3 3 3 3
Hence, total 15 4digit no. are possible.
Question 86 
{α(4,2,1)  α≠0, α∈R}  
{α(4,2,1)  α≠0, α∈R}  
{α(2,0,1)  α≠0, α∈R}  
{α(2,0,1)  α≠0, α∈R} 
AX = λX ⇒ (A  I)X = 0
⇒ y+2z = 0 and x+2y = 0
⇒ y = 2z and x/(2) = y
∴ x/(2) = y = 2z ⇒ x/(4) = y/2 = z/1 = α(say)
∴ Eigen vectors are {α(4,2,1  α≠0, α∈R}
Question 87 
E, 201  
F, 201  
E, E20  
2, 01F 
ac No. of cache lines in cache is 2^{12} bytes which needs 12 bits. So next lower 12 bits are line indexing bits.
And the remaining top 4 bits are tag bits (out of 20). So answer is (A).
Question 88 
af(x) + bf(1/x) = 1/x  25  (1)
Put x = 1/x,
af(1/x) + bf(x) = x  25  (2)
Multiply equation (1) with 'a' and Multiply equation (2) with 'b', then
abf(1/x) + a^{2} = a/x  25a  (3)
abf(1/x) + b^{2} = bk  25b  (4)
Subtract (3)  (4), we get
(a^{2}  b^{2}) f(x) = a/x 25a  bx + 25b
f(x) = 1/(a^{2}  b^{2}) (a/x  25a  bx +25b)
Now from equation,
Hence option (A) is the answer.
Question 89 
309.33  
309.34  
309.35  
309.36 
∵v = velocity
= 2/3[(0+0)+4(10+25+32+11+2)+2(18+29+20+5)]
= 309.33 km
(Here length of each of the subinterval is h = 2)
Question 90 
pq+r = 0 or p = q = r  
p+qr = 0 or p = q = r
 
p+q+r = 0 or p = q = r  
pq+r = 0 or p = q = r 
Question 91 
Both reflexive and symmetric  
Reflexive but not symmetric  
Not reflexive but symmetric  
Neither reflexive nor symmetric 
∴(p,q) R (p,q)
⇒ R is not reflexive.
Let (p,q) R (r,s) then ps = qr
⇒ rq = sp
⇒ (r,s) R (p,q)
⇒ R is symmetric.
Question 92 
G_{1}=(V,E_{1}) where E_{1}={(u,v)(u,v)∉E}  
G_{2}=(V,E_{2} )where E_{2}={(u,v)│(u,v)∈E}  
G_{3}=(V,E_{3}) where E_{3}={(u,v)there is a path of length≤2 from u to v in E}  
G_{4}=(V_{4},E) where V_{4} is the set of vertices in G which are not isolated 
→ It strongly connected.
(A) G_{1}=(V,E_{1}) where E_{1}={(u,v)(u,v)∉E}
If (u, v) does not belong to the edge set ‘E’, then it indicates there are no edges. So, it is not connected.
(B) G_{2}=(V,E_{2} )where E_{2}={(u,v)│(u,v)∈E}
Given that ‘G’ is directed graph, i.e., it has path from each vertex to every other vertex.
Though direction is changed from (u, v) to (v, u), it is still connected component same as ‘G’.
(C) G_{3}=(V,E_{3}) where E_{3}={(u,v)there is a path of length≤2 from u to v in E}
This can also be true.
eg:
Both from each vertex to other vertex is also exists. So it is also strongly connected graph.
(D) G_{4}=(V_{4},E) where V_{4} is the set of vertices in G which are not isolated.
If ‘G’ has same ‘x’ no. of isolated vertices, one strongly connected component
then no. of SCC = x + 1
G_{4} contain only ‘1’ component, which is not same as G.
Question 93 
0  
1  
2  
3 
The finite dimensional spectral theorem says that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix.
Question 94 
I only  
II only  
Both I and II  
Neither I nor II 
Rolle’s theorem states that for any continuous, differentiable function that has two equal values at two distinct points, the function must have a point on the function where the first derivative is zero. The technical way to state this is: if f is continuous and differentiable on a closed interval [a,b] and if f(a) = f(b), then f has a minimum of one value c in the open interval [a, b] so that f'(c) = 0.
We can observe that, sin, cos are continuous, but, Tan is not continuous at π/2. As the mentioned interval does not contain π/2, we can conclude that it is continuous.
As per Rolls theorem both statement 1 and statement 2 are true.
Question 95 
2  
3  
4  
5 
f ’(x) = x(Sinx)’ + Sin(x)(x’)
= xCosx + Sinx ①
f ’’(x) = x (Cosx)’ + Cos (x)’+ Cos x
= x Sinx + 2Cosx ②
Given: f ’’(x) + f(x) + t Cosx = 0
Replace ① & ②,
xSinx + 2Cosx + xSinx + tCosx = 0
2Cosx + tCosx = 0
t = 2
Question 96 
There exists a y in the interval (0,1) such that f(y)=f(y+1)  
For every y in the interval (0,1),f(y)=f(2y)  
The maximum value of the function in the interval (0,2) is 1  
There exists a y in the interval (0, 1) such that f(y)=f(2y) 
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1]
g(0) = 2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0,1]. Therefore at some point g would be 0 in (0,1).
g=0 ⇒ f(y) = f(y+1) for some y in (0,1).
Apply similar logic to option D, Let g(y) = f(y) + f(2  y)
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1] (sum of two continuous functions is continuous)
g(0) = 2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0, 1]. Therefore at some point g would be 0 in (0, 1).
There exists y in the interval (0, 1) such that:
g=0 ⇒ f(y) = f(2 – y)
Both A, D are answers.
Question 97 
10  
11  
12  
13 
To get ‘22’ as Sum of four outcomes
x_{1} + x_{2} + x_{3} + x_{4} = 22
The maximum Sum = 6+6+6+6 = 24 which is near to 22
So, keeping three 6’s, 6+6+6+x = 22
x = 4 combination① = 6 6 6 4
Keeping two 6’s, 6+6+x_{1}+x_{2} = 22
x_{1}+x_{2} = 10 possible x’s (5, 5) only
combination② = 6 6 5 5
No. of permutation with 6664 = 4!/ 3! = 4
“ “ “ 6655 = 4!/ 2!2! = 6
Total = 4+6 = 10 ways out of 6×6×6×6 = 1296
Pnb (22) = 10/1296 ⇒ x = 10
Question 98 
89  
90  
91  
92 
Single two: 211111111 ⇒ 9!/8!1! = 9 pennants
Two twos: 22111111 ⇒ 8!/6!2! = 28
Three twos: 2221111 ⇒ 7!/3!4! = 35
Four twos: 222211 ⇒ 6!/4!2! = 15
Five twos: 22222 ⇒ 1
Total = 89 pennants.
Question 99 
16  
17  
18  
19 
{0,1}^{4}={0,1}×{0,1}×{0,1}×{0,1}=16
S=2^{16}
N=2^{S}
loglogN=loglog2^{S}=log S =log2^{16}=16
Question 100 
506  
507  
508  
509 
If we observe the graph, it looks like a 12 by 12 grid. Each corner vertex has a degree of 3 and we have 4 corner vertices. 40 external vertices of degree 5 and remaining 100 vertices of degree 8.
From Handshaking theorem, sum of the degrees of the vertices is equal to the 2*number of edges in the graph.
⇒ (4*3) + (40*5) + (100*8) = 2*E
⇒ 1012=2*E
⇒ E=506
Question 101 
(1, 1, 1, 1, 1, 1)  
(2, 2, 2, 2, 2, 2)  
(3, 3, 3, 1, 0, 0)  
(3, 2, 1, 1, 1, 0) 
A) (1, 1, 1, 1, 1, 1)
Yes, it is a graph.
We will see that option (C) is not graphic.
Question 102 
((p↔q)∧r)∨(p∧q∧∼r)  
(∼(p↔q )∧r)∨(p∧q∧∼r)  
((p→q)∧r)∨(p∧q∧∼r)  
(∼(p↔q)∧r)∧(p∧q∧∼r) 
Method2: directly check with one of {TTF, TFT, FTT} options.
As there are two T’s in each option, replace them and check with the third value.
Eg: Place p=q= T
(∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(T↔T)∧r)∨(T∧T∧∼r)
=(∼(T)∧r)∨(T∧∼r)
=(F∧r)∨(T∧∼r)
=(F)∨(∼r)
=∼r
This is true for r=F.
Similarly with p=r=T and q=F.
q=r=T and p=F
Option B is the answer.
Question 103 
11.90  
11.91  
11.92  
11.93 
Out of which ‘4’ are chosen at random without replaced and created.
If at least three of them are working then system is deemed functional
i.e., there should be only ‘one’ nonworking system in set of ‘4’.
It is possible with combination
W – W – W – N,
W – W – N – W,
W – N – W – W,
N – W – W – W.
For W – W – W – N, the probability = (choosing working out of 10) × (choosing working out of 9) × (choosing working out of 8) × (choosing nonworking out of 7)
=4/10×3/9×2/8×1/7
where 4/10 ⇒ 4 working out of 10
3/9 ⇒ 3 working are remaining out of ‘9’ as ‘1’ is already taken
For ‘4’ Sum combinations
Total probability =4×[4/10×3/9×2/8×1/7]=600/5040
We need 100p ⇒100×600/5040=11.90
Question 104 
3.9  
4.0  
4.1  
4.2 
There are ‘9’ words in this sentence.
No. of characters in each word
The (3)
quick (5)
brown (5)
fox (3)
jumps (5)
over (4)
the (3)
lazy (4)
dog (3)
Each word has equal probability.
So expected length = 3×1/9+5×1/9+5×1/9+3×1/9+5×1/9+ 4×1/9+3×1/9+4×1/9+3×1/9
=35/9
=3.9
Question 105 
36  
37  
38  
39 
Total no. of edges = 6×6 = 36
Question 107 
f(0)f(4) < 0  
f(0)f(4) > 0  
f(0) + f(4) > 0  
f(0) + f(4) < 0 
Polynomial will be
f(x) = (x1)(x2)(x3)
f(0) = 1 × 2 × 3 = 6
f(4) = 3 × 2 × 1 = 6
f(0)∙f(4) =  36
f(0) + f(4) = 6  6 = 0
Option (A) is correct.
Question 108 
Only I  
Only II  
Both I and II  
Neither I nor II 
Question 109 
6  
7  
8  
9 
AX = λX
x_{1} + x_{5} = λx_{1}  (1)
x_{1} + x_{5} = λx_{5}  (2)
(1) + (2) ⇒ 2(x_{1} + x_{5}) = λ(x_{1} + x_{5}) ⇒ λ_{1} = 2
x_{2} + x_{3} + x_{4} = λ∙x_{2}  (4)
x_{2} + x_{3} + x_{4} = λ∙x_{3}  (5)
x_{2} + x_{3} + x_{4} = λ∙x_{4}  (6)
(4)+(5)+(6) = 3(x_{2} + x_{3} + x_{4}) = λ(x_{2} + x_{3} + x_{4} ) ⇒ λ_{2} =3
Product = λ_{1} × λ_{2} = 2×3 = 6
Question 110 
0.259 to 0.261  
0.260 to 0.262  
0.261 to 0.263  
0.262 to 0.264 
n(A)=50, n(B)=33, n(C)=20
n(A∩B)=16, n(B∩C)=6, n(A∩C)=10
n(A∩B∩C)=3
P(A∪B∪C)=P(A)+P(B)+P(C)P(A∩B)P(B∩C) P(A∩C)+P(A∩B∩C)=74/100
∴ Required probability is P(A∩B∩C)=1P(A∪B∪C)=0.26
Question 111 
0.26  
0.27  
8  
0.29 
= 2' × 19' × 53'
Now number of distinct integral factors of 2014 will be,
(1+1)×(1+1)×(1+1) = 2×2×2 = 8
Question 112 
Both S1 and S2 are true  
S1 is true and S2 is false  
S2 is true and S1 is false  
Neither S1 nor S2 is true 
U⊂S, V⊂S
Let U = {1, 2, 3}
V = {2, 3, 4}
Symmetric difference:
(U – V) ∪ (V – U) = {1} ∪ {4} = {1, 4}
The minimum element in the symmetric difference is 1 and 1∈U.
S1: Let S = Universal set = {1, 2, … 2014}
This universal set is larger than every other subset.
S2: Null set is smaller than every other set.
Let U = { }, V = {1}
Symmetric difference = ({ } – {1}) ∪ ({1} – { }) = { } ∪ {1} = {1}
So, U < V because { } ∈ U.
Question 113 
5  
6  
7  
8 
In a cycle of n vertices, each vertex is connected to other two vertices. So each vertex degree is 2.
When we complement it, each vertex will be connected to remaining n3 vertices ( one is self and two other vertices in actual graph).
As per given question,
n3 =2
n=5
Cycle of 5 vertices is
Complement of the above graph1 is
Graph1 and Graph2 are complement each other.
So, the value of n is 5.
Question 114 
((a ⟶ b) ∧ (b ⟶ c)) ⟶ (a ⟶ c)  
(a ⟷ c) ⟶ (∽ b ⟶ (a ∧ c))  
(a ∧ b ∧ c) ⟶ (c ∨ a)  
a ⟶ (b ⟶ a) 
((a → b) ∧ (b → c)) → (a → c)
If (a → b) is false with a = T, b = F,
then (F ∧ (b → c)) → (a → c)
F → (a → c)
which is True for any (a → c)
This is tautology.
B:
(a ⟷ c) ⟶ (∽b ⟶ (a ∧ c))
For (a ⟷ c) be True and
∽b → (a ∧ c) should be False
Let a = c = F
(F → F) → (∽b (F ∩ F))
T → (∽b → F)
This is False for b = F
So, this is not True.
C:
(a ∧ b ∧ c) ⟶ (c ∨ a)
(c ∨ a) is False only for a = c = F
if (c ∨ a) is False
(F ∧ b ∧ F) → F
F → F which is Tautology
True always.
D:
a ⟶ (b ⟶ a)
a ⟶ (~b ∨ a)
(~a ∨ a) ∨ ~b = T ∨ ~b = T which is tautology
Question 115 
Only L is TRUE.  
Only M is TRUE.  
Only N is TRUE.  
L, M and N are TRUE. 
So, given statement can be sub divided such that we can utilize the negation of this atomic statements.
Suppose, X is Good mobile and Y is cheap then
P: (Good(x) → ~cheap(x)) → (~good(x) ∨ ~cheap(x))
Q: cheap(x) → ¬good(x) ⟺ ((¬cheap(x) ∨ good(x)) ⟺ ¬good(x) ∨ ¬cheap(x))
All these are contra positive.
All L, M, N are true.
Question 116 
For any subsets A and B of X, f(A ∪ B) = f(A)+f(B)  
For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B)  
For any subsets A and B of X, f(A ∩ B) = min{ f(A),f(B)}  
For any subsets S and T of Y, f^{ 1} (S ∩ T) = f^{ 1} (S) ∩ f^{ 1} (T) 
We need to consider subsets of 'x', which are A & B (A, B can have common elements are exclusive).
Similarly S, T are subsets of 'y'.
To be a function, each element should be mapped with only one element.
(a) f(A∪B)=f(A)+f(B)
{a,b,c}∪{c,d,e} = {a,b,c} + {c,d,e}
{a,b,c,d,e} = 3+3
5 = 6 FALSE
(d) To get inverse, the function should be oneone & onto.
The above diagram fulfills it. So we can proceed with inverse.
f^{1} (S∩T )=f^{1} (S)∩f^{1} (T)
f^{1} (c)=f^{1} ({a,b,c})∩f^{1} ({c,d,e})
2={1,2,3}∩{2,4,5}
2=2 TRUE
Question 117 
5  
6  
7  
8 
So, 15 is divided by {1, 3, 5, 15}.
As minimum is 4 and total is 15, we eliminate 1,3,15.
Answer is 5.
Question 118 
If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative.  
If the trace of the matrix is positive, all its eigenvalues are positive.  
If the determinant of the matrix is positive, all its eigenvalues are positive.  
If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. 
• The product of the n eigenvalues of A is the same as the determinant of A. •
A: Yes, for sum to be negative there should be atleast one negative number.
B: There can be one small negative number and remaining positive, where sum is positive.
C: Product of two negative numbers is positive. So, there no need of all positive eigen values.
D: There is no need for all eigen values to be positive, as product of two negative numbers is positive.
Question 119 
2  
3  
4  
5 
For eg: a two dimensional vector space have x, y axis. For dimensional vector space, it have x, y, z axis.
In the same manner, 6 dimensional vector space has x, y, z, p, q, r (assume).
Any subspace of it, with 4 dimensional subspace consists any 4 of the above. Then their intersection will be atmost 2.
[{x,y,z,p} ∩ {r,q,p,z}] = #2
V_{1} ∩ V_{2} = V_{1} + V_{2}  V_{1} ∪ V_{2} = 4+4+(6) = 2
Question 120 
4  
5  
6  
7 
We have xSinx,
We can observe that it is positive from 0 to π and negative in π to 2π.
To get positive value from π to 2π we put ‘‘ sign in the (π, 2π)
Question 121 
I only  
II only  
Both I and II  
Neither I nor II 
Question 123 
0.25  
0.26  
0.27  
0.28 
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2≥√(P(A)∙P(B))
1/2≥√(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4≥P(A)∙P(B)
P(A)∙P(B)≤1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
Question 124 
4  
5  
6  
7 
a*a^{1} = e
1. x*x=e So x^{1} is x ⇒ x is element of Group
2. y*y=e So y^{1} = y ⇒ y is element of Group
4. (y*x)*(y*x)=x*y*y*x=x*x*e=e So (y*x)^{1}=(y*x)
In ③, ④
x*y,y*x has same inverse, there should be unique inverse for each element.
x*y=y*x (even with cumulative law, we can conclude)
So {x, y, e, x*y} are element of Group.
Question 125 
P, Q and R are true  
Only Q and R are true  
Only P and Q are true  
Only R is true 
So f(i)should be resulting only {0, 1, …2014}
So, every element in range has a result value to domain. This is onto. (Option R is correct)
We have ‘2015’ elements in domain.
So atleast one element can have f(i) = i,
so option ‘Q’ is also True.
∴ Q, R are correct.
Question 126 
⌊n/k⌋  
⌈n/k⌉  
n–k  
nk+1 
Option 1, 2 will give answer 1. (i.e. one edge among them),
Option 3: nk = 0 edges.
Option 4: nk+1 : 1edge, which is false.
Question 127 
In any planar embedding, the number of faces is at least n/2+ 2  
In any planar embedding, the number of faces is less than n/2+ 2  
There is a planar embedding in which the number of faces is less than n/2+ 2  
There is a planar embedding in which the number of faces is at most n/(δ+1) 
v – e + f = 2 →①
Point ① degree of each vertex is minimum ‘3’.
3×n≥2e
e≤3n/2
From ①
n3n/2+f=2⇒
Question 128 
∀d (Rainy(d) ∧∼Cold(d))  
∀d (∼Rainy(d) → Cold(d))  
∃d (∼Rainy(d) → Cold(d))  
∃d (Rainy(d) ∧∼Cold(d)) 
= ∼[∀rainy days are cold]
= ∼[∀ days (rainy days ⇒ cold days]
= ∃ days[∼(cold days ∨ ∼rainy days)]
= ∃ days[rainy days ∧ ∼cold days]
Question 129 
Commutative but not associative  
Both commutative and associative  
Associative but not commutative  
Neither commutative nor associative 
A binary relation on a set S is called cumulative if a*b = b*a ∀ x,y∈S.
Associative property:
A binary relation on set is called associative if (a*b)*c = a*(b*c) ∀ x,y∈S.
Given x⊕y = x^{2} + y^{2} (1)
Replace x, y in (1)
y⊕x = y^{2} + x^{2} which is same as (1), so this is cumulative
(x⊕y)⊕z = (x^{2} + y^{2}) ⊕ z
= (x^{2} + y^{2}) + z^{2}
= x^{2} + y^{2} + z^{2} + 2x^{2}y^{2} (2)
x⊕(y ⊕ z) = x ⊕ (y^{2} + z^{2})
= x^{2} + (y^{2} + z^{2})^{2}
= x^{2} + y^{2} + z^{2} + 2y^{2z2  (3) (2) & (3) are not same so this is not associative. }
Question 130 
8/(2e^{3})  
9/(2e^{3})  
17/(2e^{3})  
26/(2e^{3}) 
P(x:λ)=(e^{λ} λ^{x})/x! for x = 0,1,2….
‘λ’ is the average number (mean)
Given that mean = λ = 3
The probability of observing fewer than three cars is
P(zero car) + P(one car) + P(two cars)
=(e^{3} 3^{0})/0!+(e^{3} 3^{1})/1!+(e^{3} 3^{2})/2!
=e^{3}+e^{3}∙3+(e^{3})∙9)/2
=(17e^{3<})/2
=17/(2e^{3} )
Question 131 
Try to derive options from the given matrix.
Observe that col 2 + col 3 will reuse x(x+1) term
C_{2} → C_{1} + C_{2}
Question 132 
At x = 3, f(x) = 2
LHL (3), f(3^{} )=(x+3 )/3=(3+3)/3=2
RHL (3), f(3^{+} )=x1=31=2
∴ f(x) is continuous.
Option B:
At x = 3, f(x) = 4
For x ≠ 3, f(x)=f(3^{+} )=f(3^{} )=8^{x}=83=5
This is not continuous.
Option C:
At x ≤ 3, f(x) = x+3 = 3+3 = 6
At RHL(3), f(3^{+} )=4
This is not continuous.
Option D:
f(x) at x=3 is not defined.
There is a break at x=3, so this is not continuous.
Question 133 
8.983  
9.003  
9.017  
9.045 
Question 134 
∃x(F(x)∧¬P(x))  
∃x(¬F(x)∧P(x))  
∃x(¬F(x)∧¬P(x))  
¬∃x(F(x)∧P(x)) 
P(x) = x is perfect
The meaning of ∃x(P(x)∧F(x)) is atleast one person who is my friend and perfect.
The negation of ∃x(P(x)∧F(x)) is “This is not the case that atlease one person who is my friend and perfect”.
So ~∃x(P(x)∧F(x)) is none of my friends are perfect.
Question 135 
0  
1  
2  
3 
In the above code total number of spills to memory is 1.
Question 136 
Both I_{1} and I_{2} are correct inferences  
I_{1} is correct but I_{2} is not a correct inference  
I_{1} is not correct but I_{2} is a correct inference  
Both I_{1} and I_{2} are not correct inferences 
The cricket match was played.
Let p = it rains
q = playing cricket/ match played
If (it rains) then (the match will not be played)
p ⇒ (∼q)
Inference: There was no rain. (i.e., p = F)
So for any F ⇒ (∼q) is true.
So this inference is valid.
I_{2}: If it rains then the cricket match will not be played.
It did not rain.
p ⇒ (∼q)
Inference: The cricket match was played.
q = T
p ⇒ (∼q)
p ⇒ (∼T)
p ⇒ F
This is false for p = T, so this is not true.
Question 137 
One, at π/2  
One, at 3π/2  
Two, at π/2 and 3π/2  
Two, at π/4 and 3π/2 
f’(x) = cos x
[just consider the given interval (π/4, 7π/4)]
f'(x) = 0 at π/2, 3π/2
To get local minima f ’’(x) > 0, f ’’(x) =  sin x
f ’’(x) at π/2, 3π/2
f ’’(x) = 1< 0 local maxima
f ’’ (3π/2) = 1 > 0 this is local minima
In the interval [π/4, π/2] the f(x) is increasing, so f(x) at π/4 is also a local minima.
So there are two local minima for f(x) at π/4, 3π/2.
Question 138 
1024 and 1024  
1024√2 and 1024√2  
4√2 and 4√2  
512√2 and 512√2 
The 2×2 matrix =
Cayley Hamilton theorem:
If matrix A has ‘λ’ as eigen value, A^{n} has eigen value as λ^{n}.
Eigen value of
AλI = 0
(1λ)(1+λ)1=0
(1λ^{2} )1=0
1=1λ^{2}
λ^{2}=2
λ=±√2
A^{19} has (√2)^{19}=2^{9}×√2 (or) (√2)^{19}=512√2
=512√2
Question 139 
∃x (real(x) ∨ rational(x))  
∀x (real(x) → rational(x))  
∃x (real(x) ∧ rational(x))  
∃x (rational(x) → real(x)) 
∃x (real(x) ∧ rational(x))
(A) ∃x(real(x) ∨ rational(x))
means There exists some number, which are either real or rational.
(B) ∀x (real(x)→rational(x))
If a number is real then it is rational.
(D) ∃x (rational(x)→real(x))
There exists a number such that if it is rational then it is real.
Question 140 
3  
4  
5  
6 
ve+f=2
Given 10 vertices & 15 edges
1015+f=2
f=2+1510
f=7
There will be an unbounded face always. So, number of faces = 6.
Question 141 
0 and 0.5  
0 and 1  
0.5 and 1  
0.25 and 0.75 
The sum of probabilities at x=1, x=1 itself is 0.5+0.5 =1. It is evident that, there is no probability for any other values.
The F(x=1) is 0.5 as per given probabilities and
F(x=1) = sum of F(x=1) +F(x=0)=...f(X=1) = 0.5 +0.5 =1
Question 142 
(A) 3 cycle graph not in original one.
(B) Correct 5 cycles & max degree is 4.
(C) Original graph doesn’t have a degree of 3.
(D) 4 cycles not in original one.
Question 143 
1  
3  
5  
7 
Question 144 
10/21  
5/12  
2/3  
1/6 
The value on second time can be {1, 2, 3, 4, 5, 6}
So the Sum can be
We have Sample space = 36
The no. of events where (Sum = atleast 6) = {6, 7, 6, 7, 8, 6, 7, 8, 9}
So the probability atleast ‘6’ while getting {1, 2, 3} in first time = 9/36 → ①
If we get ‘6’ in the first time itself, then we do not go for rolling die again.
So, its probability = 1/6
Total probability = 1/6 + 9/36 = 1/6 + 1/4 = 10/24 = 5/12
Question 145 
2^{n}  
2^{n}1  
2^{n}2  
2(2^{n}– 2) 
Onto function is possible if m ≥ n. So, no. of onto functions possible is,
n^{m}  ^{n}C_{1} (n1)^{m} + ^{n}C_{2} (n2)^{m} + .......
Here in Question,
m = n, n = 2
So, the final answer will be,
= 2^{n}  ^{2}C_{1} (21)^{n} + ^{2}C_{2} (22)^{n}
= 2^{n}  2 × 1 + 0
= 2^{n}  2
Question 146 
15  
30  
45  
360 
It is asked to find the distinct cycle of length 4. As it is complete graph, if we chose any two vertices, there will be an edge.
So, to get a cycle of length 4 (means selecting the 4 edges which can form a cycle) we can select any four vertices.
The number of such selection of 4 vertices from 6 vertices is ^{6}C_{4} => 15.
From each set of 4 vertices, suppose a set {a, b, c, d} we can have cycles like
abcd
abdc
acbd
acdb
adbc
adcb (Total 6, which is equal to number of cyclic permutations (n1)! )
As they are labelled you can observe, abcd and adcb are same, in different directions.
So, we get only three combinations from the above 6.
So, total number of distinct cycles of length 4 will be 15*3 = 45.
If it is asked about just number of cycles then 15*6 = 90
Question 147 
R = 0  
R < 0  
R ≥ 0  
R > 0 
So the answer will be R≥0.
Question 148 
1, 4, 3  
3, 7, 3  
7, 3, 2
 
1, 2, 3 
Question 149 
P(x) being true means that x is a prime number  
P(x) being true means that x is a number other than 1
 
P(x) is always true irrespective of the value of x
 
P(x) being true means that x has exactly two factors other than 1 and x 
This is the definition of prime nos.
Question 150 
0  
2  
–i  
i 
Question 151 
Index position of mode of X in X is the same as the index position of mode of Y in Y.  
Index position of median of X in X is the same as the index position of median of Y in Y.
 
μ_{y} = aμ_{x} + b  
σ_{y} = aσ_{x} + b 
(σ_{y})^{2} is variance so,
y_{i} = a * x_{i} + b
(σ_{y})^{2} = a^{2 }(σ_{x})^{2}
⇒ σ_{y} = a σ_{x}
Hence option (D) is incorrect.
Question 152 
1/5  
4/25  
1/4  
2/5 
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
^{5}C_{1} × ^{4}C_{1} = 20
So probability = 4/20 = 1/5
Question 153 
S = 2T  
S = T  1  
S = T  
S = T + 1 
i_{d}= no. of vertices of degree ‘d’ in ‘G’
Eg:
No. of vertices with degree ‘2’ = 3
ξ(G')=3×2='6' i.e., sum of degrees
By Handshaking Theorem,
The sum of degrees would be equal to twice the no. of edges
V=2E
It is given that ξ(G)=ξ(S) then
Sum of degrees of vertices in G is equal to sum of degrees of vertices in S
i.e., 2*(no. of edges in G)=2*no. of edges in S no. of edges in G=no. of edges in S
Eg:
ξ(G)=(2×2)+(2×3)=4+6=10
ξ(S)=2×5=10
You can observe that, though no. of vertices are different, but still no. of edges are same.
Question 154 
3.575  
3.676  
3.667  
3.607

Question 155 
2^{10}
 
2^{15}  
2^{20}  
2^{25} 
Definition of Reflexive relation:
A relation ‘R’ is reflexive if it contains xRx ∀ x∈A
A relation with all diagonal elements, it can contain any combination of nondiagonal elements.
Eg:
A={1, 2, 3}
So for a relation to be reflexive, it should contain all diagonal elements. In addition to them, we can have possible combination of (n^{2}n)nondiagonal elements (i.e., 2^{n2n})
Ex:
{(1,1)(2,2)(3,3)}  ‘0’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)}  ‘1’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)(1,3)} “
___________ “
___________ “
{(1,1)(2,2)(3,3)(1,2)(1,3)(2,1)(2,3)(3,1)(3,2)} (n^{2}n) diagonal elements
____________________
Total: 2^{n2n}
For the given question n = 5.
The number of reflexive relations =2^{(255)}=2^{20}
Question 156 
A group
 
A ring  
An integral domain  
A field 
1) closure
2) Associativity
3) Have Identity element
4) Invertible
Over ‘*’ operation the S = {1, ω, ω^{2}} satisfies the above properties.
The identity element is ‘1’ and inverse of 1 is 1, inverse of ‘w’ is 'w^{2}' and inverse of 'w^{2}' is 'w'.
Question 158 
pq + (1  p)(1  q)
 
(1  q)p
 
(1  p)q
 
pq 
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of tetsing process giving incorrect result × Probability that computer is not faulty
= p × q + (1  p) (1  q)
Question 159 
1/625  
4/625  
12/625
 
16/625 
We can write 10^{99} as 10^{96}×10^{3}
So, (10^{99})/(10^{96}) to be a whole number, [10^{96}×10^{3}/10^{96}]➝ (1)
We can observe that every divisor of 10^{3} is a multiple of 10^{96}
So number of divisor of 10^{3} to be found first
⇒ 10^{3}=(5×2)^{3}=2^{3}×5^{3}
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 10^{99} are 10^{99}=2^{99}×5^{99}=100×100=10000
Probability that divisor of 10^{99} is a multiple of 10^{96} is
⇒16/10,000
Question 160 
I and II
 
III and IV  
IV only  
II and IV 
⇾ Arrange the degree of vertices in descending order
eg. d_{1},d_{2}, d_{3}...d_{n}
⇾ Discard d_{1}, subtrack ‘1’ from the next 'd_{1}'degrees
eg:
⇒ 1 1 0 1
⇾ We should not get any negative value if its negative, this is not valid sequence
⇾ Repeat it till we get ‘0’ sequence
I. 7, 6, 5, 4, 4, 3, 2, 1
➡️5, 4, 3, 3, 2, 1, 0
➡️3, 2, 2, 1, 0, 0
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
II. 6, 6, 6, 6, 3, 3, 2, 2
➡️5, 5, 5, 2, 2, 1, 2
put them in descending order
➡️5, 5, 5, 2, 2, 2, 1
➡️4, 4, 1, 1, 1, 1
➡️3, 0, 0, 0, 1 (descending order)
➡️3, 1, 0, 0, 0
➡️0, 1, 1, 0
[This is not valid]
III. 7, 6, 6, 4, 4, 3, 2, 2
➡️5, 5, 3, 3, 2, 1, 1
➡️4, 2, 2, 1, 0, 1
➡️4, 2, 2, 1, 1, 0 (descending order)
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
IV. 8, 7, 7, 6, 4, 2, 1, 1
There is a degree ‘8’, but there are only ‘8’ vertices.
A vertex cannot have edge to itself in a simple graph. This is not valid sequence.
Question 161 
x=4, y=10  
x=5, y=8  
x=3, y=9  
x=4, y=10 
Trace = {Sum of diagonal elements of matrix}
Here given that eigen values are 4, 8
Sum = 8 + 4 = 12
Trace = 2 + y
⇒ 2 + y = 12
y = 10
Determinant = 2y  3x
Product of eigen values = 8 × 4 = 32
2y  3x = 32
(y = 10)
20  3x = 32
12 = 3x
x = 4
∴ x = 4, y = 10
Question 162 
Everyone can fool some person at some time  
No one can fool everyone all the time  
Everyone cannot fool some person all the time
 
No one can fool some person at some time 
For better understanding propagate negation sign outward by applying Demorgan's law.
∀x∃y∃t(¬F(x, y, t)) ≡ ¬∃x∀y∀t(F(x,y,t))
Now converting ¬∃x∀y∀t(F(x,y,t)) to English is simple.
¬∃x∀y∀t(F(x,y,t)) ⇒ There does not exist a person who can fool everyone all the time.
Which means "No one can fool everyone all the time".
Hence, Option (B) is correct.
Question 163 
No two vertices have the same degree.
 
At least two vertices have the same degree.  
At least three vertices have the same degree.  
All vertices have the same degree.

If all vertices have different degrees, then the degree sequence will be {1,2,3,....n1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n1) vertices). Degree sequence has only (n1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) consider a triangle, all vertices has same degree, so it is false
C) consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.
Question 164 
2  
3  
n1  
n 
Eg: Consider a square, which has 4 edges. It can be represented as bipartite ,with chromatic number 2.
Question 165 
Commutativity
 
Associativity  
Existence of inverse for every element
 
Existence of identity

So, commutativity is not required.
Question 166 
R is symmetric but NOT antisymmetric
 
R is NOT symmetric but antisymmetric
 
R is both symmetric and antisymmetric
 
R is neither symmetric nor antisymmetric

Antisymmetric Relation: A relation R on a set A is called antisymmetric if (a,b)€ R and (b,a) € R then a = b is called antisymmetric.
In the given relation R, for (x,y) there is no (y,x). So, this is not Symmetric. (x,z) is in R also (z,x) is in R, but as per antisymmetric relation, x should be equal to z, where this fails.
So, R is neither Symmetric nor Antisymmetric.
Question 167 
0.453
 
0.468
 
0.485  
0.492

P(e) = Probability of getting even no. face.
It is given that,
P(0) = 0.9 P(e)  (I)
Also we know that,
P(0) + P(e) = 1  (II)
Solving equation (I) and (II) we get,
P(e) = 0.52
Also even no. can be 2 or 4 or 6.
And given in question that P(2) = P(4) = P(6).
So, 3 × P(2) = 0.52
P(2) = 0.175
So, P(2) = P(4) = P(6) = 0.175
Also in question it is given that,
P(e/>3) = 0.75
P(even no. greater than 3)/ P(no. greater than 3) = 0.75
P(4,6)/P(>3) = 0.75
(0.175×2)/P(>3) = 0.75
P(>3) = 0.35/0.75 = 0.467
Question 168 
a, b are generators
 
b, c are generators
 
c, d are generators
 
d, a are generators

We can observe that, a is an identity element. ( a *x = x ). An identity element cannot be a generator, as it cannot produce any other element ( always a*a*... = a).
Also, b*b =a, so it also cannot produce all other elements ( always b*b*... =a , where a is identify element).
c,d are able to produce other elements like { c*c =b, c*(c*c) = c*b= d, c*(c*(c*c))) = c*(c*b)= c*d=a. }. Similar with d.
Question 169 
∀x(P(x) → (G(x) ∧ S(x)))
 
∀x((G(x) ∧ S(x)) → P(x))  
∃x((G(x) ∧ S(x)) → P(x)
 
∀x((G(x) ∨ S(x)) → P(x))

(A) for all ornaments, if it is precious then they should be gold and silver.
But, given statement does not says that, “ only gold and silver are precious “ . So this is wrong.
(B) For all ornaments, which contains gold and silver are precious.
Which is only the shaded region in the venn diagrams. But, it misses p,r regions. So, this is wrong option.
C) Some ornaments, which are gold and silver are precious. It is false, because all gold or silver ornaments are precious.
D) For all ornaments, Any ornament which is gold or silver is precious. Which is true.
Question 170 
¬Q□¬P
 
P□¬Q
 
¬P□Q
 
¬P□¬Q

P∨Q=P□️Q
So, option B is correct.
Question 172 
I and III  
I and IV
 
II and III
 
II and IV

II ) ¬∃x(P(x))= ∀x(~P(x))
III) ¬∃x(¬P(x)) = ∀x(P(x))
Question 174 
Q^{c} ∪ R^{c}
 
P ∪ Q^{c} ∪ R^{c}
 
P^{c} ∪ Q^{c} ∪ R^{c}
 
U 
It can be written as the p.q.r + p'.q.r +q'+r'
=> (p+p').q.r + q' +r'
=> q.r +(q'+r')
=> q.r + q'+r' = 1 i.e U
Question 175 
0  
either 0 or 1  
one of 0, 1 or 1  
any real number

When a5=0, then rank(A)=rank[AB]<3,
So infinite number of solutions.
But, it is given that the given system has unique solution i.e., rank(A)=rank[AB]=3 will be retain only if a5≠0.
Question 176 
1000e  
1000
 
100e  
100

Question 177 
square of R
 
reciprocal of R  
square root of R
 
logarithm of R

Question 178 
The graph is connected  
The graph is Eulerian
 
The graph has a vertexcover of size at most 3n/4
 
The graph has an independent set of size at least n/3

(A) Consider the following disconnected graph which is planar.
So false.
(B) A graph is Eulerian if all vertices have even degree but a planar graph can have vertices with odd degree.
So false.
(D) Consider K_{4} graph. It has independent set size 1 which is less than 4/3.
So false.
Hence, option (C) is correct.
Question 179 
P = Q  k
 
P = Q + k  
P = Q
 
P = Q +2 k

P=1+3+5+7+...+(2k1)
=(21)+(41)+(61)+(81)+...+(2k1)
=(2+4+6+8+...+2k)+(1+1+1+k times)
=Q(1+1+...+k times)
=Qk
Question 180 
0  
1  
2  
3 
f’(x) = 12x^{3} + 48x^{2} + 48x = 0
12x(x^{2}  4x + 4) = 0
x=0; (x2)^{2} = 0
x=2
f’’(x) = 36x^{2}  96x + 48
f ”(0) = 48
f ”(2) = 36(4)  96(2) + 48
= 144  192 + 48
= 0
At x=2, we can’t apply the second derivative test.
f’(1) = 12; f’(3) = 36, on either side of 2 there is no sign change then this is neither minimum or maximum.
Finally, we have only one Extremum i.e., x=0.
Question 181 
0.24
 
0.36  
0.4  
0.6

(i) She study Mathematics on Tuesday and computer science on wednesday.
⇒ 0.6×0.4
⇒ 0.24
(ii) She study computer science on Tuesday and computer science on wednesday.
⇒ 0.4×0.4
⇒ 0.16
→ The probability that she study computer science on wednesday is
0.24+0.16=0.40
Question 182 
one  
two  
three  
four 
Answer: We have only one matrix with eigen value 1.
Question 183 
3  
2  
√2  
1 
We can compare their values using standard normal distributions.
The above equation satisfies when σ_{y} will be equal to 3.
Question 184 
(∀x fsa(x)) ⇒ (∃y pda(y) ∧ equivalent(x,y))
 
∼∀y(∃x fsa(x) ⇒ pda(y) ∧ equivalent(x,y))  
∀x ∃y(fsa(x) ∧ pda(y) ∧ equivalent(x,y))
 
∀x ∃y(fsa(y)∧ pda(x) ∧ equivalent(x,y)) 
Option A:
If everything is a FSA. Then there exists an equivalent PDA for everything.
Option B:
Not for the case Y, if there exists a FSA then it can have equivalent PDA.
Option C:
Everything is a PDA and consists equivalent PDA.
Option D:
Everything is a PDA and has exist an equivalent FSA. In option A we are getting the equivalent of a and b.
So answer is option A.
Question 185 
Only I and II
 
Only I, II and III  
Only I, II and IV
 
All of I, II, III and IV 
II. ∼(∼P∧Q)⇒(P∨∼Q)≡I (✔️)
III. (P×Q)∨(P×∼Q)∨(∼P×∼Q)
P∧(Q∨∼Q)∨(∼P∧∼Q)
P∨(∼P×∼Q)
(P∨∼P)×(P∨∼Q)
(P∨∼Q)≡I=II (✔️)
IV. (P×Q)∨(P∧∼Q)∨(∼P×Q)
P×(Q∨∼Q)∨(∼P∧Q)
P∨(∼P×Q)
(P∨∼P)×(P∨Q)
(P∨Q)≠I (❌)
So I≡II≡III (✔️)
Question 186 
11/12  
10/12  
9/12  
8/12 
P(A') = 1/3; P(A) = 2/3
P(B') = 1/3; P(B) = 2/3
P(A ∪ B) = P(A) +P(B)  P(A ∩ B)
= 2/3 + 2/3  1/2
= 4+43/ 6
= 5/6
= 10/12
Question 187 
2  
3  
4  
5 
→ Chromatic number of a graph is the smallest number of colours needed to colour the vertices so that no two adjacent vertices share the same colour.
Question 188 
5  
4  
3  
2 
(2, 5, 8) is the maximal independent set for a chain of 9 nodes. If we add any start node to the set then it will not be MIS.
Independent set:
A set of vertices is called independent set such that no two vertices in the set are adjacent.
Question 189 
[β→(∃x,α(x))]→[∀x,β→α(x)]  
[∃x,β→α(x)]→[β→(∀x,α(x))]  
[(∃x,α(x))→β]→[∀x,α(x)→β]  
[(∀x,α(x))→β]→[∀x,α(x)→β]

L.H.S. : If there is an x such that α(x) is true, then β is true.
R.H.S. : For all x, if α(x) true, then β is true.
Here, the given LHS and RHS are to be same as β is a formula which can be independent of x (if β is true for one x, it is true for every x, and viceversa).
Here, LHS = RHS
So, Option C is valid.
Question 190 
[∃ x, α → (∀y, β → (∃u, ∀ v, y))]  
[∃ x, α → (∀y, β → (∃u, ∀ v, ¬y))]  
[∀ x, ¬α → (∃y, ¬β → (∀u, ∃ v, ¬y))]  
[∃ x, α ʌ (∀y, β ʌ (∃u, ∀ v, ¬y))] 
Question 191 
Question 192 
27  
28  
29  
30 
Question 193 
Question 194 
only S1  
S1 and S3  
S2 and S3  
S1 and S2 
As real + real = real and real * real = real
S2: It is closed as rational + rational = rational and rational * rational = rational
S3: It is not closed.
⇒ (0.3+0.4i) + (0.7+0.6i) = 1+i
Both (0.3+0.4i) & (0.7+0.6i) are complex numbers follows S3 but addition of them doesn't follows.
⇒ 1+i, (a^{2}+b^{2}) <= 1
⇒ 1+1 is not less than or equal to 1.
S4: {ia  a ia real}
In this there is no multiplicative identity exists (i.e., 1).
Question 195 
(i) and (iv) only  
(ii) and (iii) only  
(iii) only  
(i), (ii) and (iv) only 
(ii) and (iii), having more than one lub or glb for some pairs due to which they are not lattice.
Question 196 
Regular  
Complete  
Hamiltonian  
Euler 
→ In Euler graph all degrees must be even for all nodes. And number of odd degree vertices should be even.
→ So, degree of this new node will be even and as a new edge is formed between this new node and all other nodes of odd degree hence here is not a single node exists with degree odd so this is Euler graph.
Question 197 
S3 and S2  
S1 and S4  
S1 and S3  
S1, S2 and S3 
If determinant of some square matrix is zero then matrix do not have any inverse.
Hence, from the above definitions we can conclude that S1, S2, S3 are true and S4 is false.
Question 198 
2.417  
2.419  
2.423  
2.425 
Question 200 
P is true and Q is false.
 
P is false and Q is true.
 
Both P and Q are true.
 
Both P and Q are false. 
→ f(x) is continuous for all real values of x
For every value of x, there is corresponding value of f(x).
For x is positive, f(x) is also positive
x is negative, f(x) is positive.
So, f(x) is continuous for all real values of x.
→ f(x) is not differentiable for all real values of x. For x<0, derivative is negative
x>0, derivative is positive.
Here, left derivative and right derivatives are not equal.
Question 201 
n and n  
n^{2} and n  
n^{2} and 0  
n and 1 
→ Reflexive
→ Symmetric
→ Transitive
Let a set S be,
S = {1, 2, 3}
Now, the smallest relation which is equivalence relation is,
S×S = {(1,1), (2,2), (3,3)}
= 3
= n (for set of n elements)
And, the largest relation which is equivalence relation is,
S×S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
= 9
= 3^{2}
= n^{2} (for set of n elements)
Question 202 
9 edges and 5 vertices  
9 edges and 6 vertices  
10 edges and 5 vertices
 
10 edges and 6 vertices

if n ≥ 3 then e ≤ 3n6 (for planarity)
where n = no. of vertices
e = no. of edges
Now lets check the options.
A) e=9, n=5
9 ≤ 3(5)  6
9 ≤ 15  6
9 ≤ 9
Yes, it is planar.
B) e=9, n=6
9 ≤ 3(6)  6
9 ≤ 18  6 9 ≤ 12 Yes, it is planar.
iii) e=10, n=5
10 ≤ 3(5)  6
10 ≤ 15  6
10 ≤ 9 No, it is not planar.
So, option C is nonplanar graph.
iv) e=10, n=6
10 ≤ 3(6)  6
10 ≤ 18  6
10 ≤ 12
Yes, it is planar.
Question 203 
2  
3  
4  
5 
4 = 2^{2}
So, prime no. is 2 and power of 2 is 2. So exponent value 2 is considered now.
Now the no. of ways we can divide 2 into sets will be the answer.
So division can be done as,
{1,1}, {0,2}
in two ways. Hence, answer is 2.
Question 204 
¬∀x (Graph (x) ⇒ Connected (x))  
¬∃x (Graph (x) ∧ ¬Connected (x))  
¬∀x (¬Graph (x) ∨ Connected (x))  
∀x (Graph (x) ⇒ ¬Connected (x))

Given expression is
¬∀x(¬Graph(x) ∨ Connected(x)
which can be rewritten as,
¬∀x(Graph(x) ⇒ Connected(x)
which is equivalent to option (A)
(∵ ¬p∨q ≡ p→q)
So, option (A) and (C) cannot be the answer.
Coming to option (B), the given expression is,
∃x (Graph (x) ∧ ¬Connected (x))
"There exist some graph which is not connected", which is equivalent in saying that "Not every graph is connected".
Coming to option (D),
For all x graph is not connected, which is not correct.
Hence, option (D) is the answer.
Question 205 
Any kregular graph where k is an even number.
 
A complete graph on 90 vertices.
 
The complement of a cycle on 25 vertices.
 
None of the above.

→ all vertices in the graph have an "even degree".
→ And the graph must be corrected.
Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.
Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24  2 = 22.
There is a theorem which says "G be a graph with n vertices and if every vertex has a degree of atleast n1/2 then G is connected."
So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.
Question 206 
1/2  
1/10  
9!/20!  
None of these

→ Total no. of possible even number = 10
→ Here we are not considering odd number.
→ The probability that 2 appears at an earlier position than any other even number is =1/10
Question 207 
5  
7  
2  
1 
(AλI)^{2}I=0 [a^{2}b^{2}=(a+b)(ab)]
(AλI+I)(AλII)=0
(A(λI)I)(A(λ+I)I=0
Let us assume λ1=k & λ +1=k
λ =k+1 λ =k1
⇓ ⇓
for k=5; λ=4 λ =6
k=2; λ=1 λ =3
k=1; λ=2 λ = 0
k=4; λ=5 λ = 3
So; λ=4,1,2,5,6,3,0,3
Check with the option
Option C = 2
Question 208 
And, neither π_{2} refines π_{3}, nor π_{3} refines π_{2}.
Here, only π_{1} refined by every set, so it has to be at the top.
Finally, option C satisfies all the property.
Question 209 
{[1,1,0]^{T}, [1,0,1]^{T}} is a basis for the subspace X.  
{[1,1,0]^{T}, [1,0,1]^{T}} is a linearly independent set, but it does not span X and therefore is not a basis of X.
 
X is not a subspace of R^{3}  
None of the above

Question 210 
1.5
 
√2
 
1.6  
1.4 
Equation based on NewtonRapson is
x_{n+1}=x_{n}f(x_{n})/f'(x_{n})⟶ (II)
Equate I and II
x_{n}f(x_{n})/f'(x_{n})=x_{n}/2+9/8x_{n}
x_{n}f(x_{n})/f'(x_{n})=x_{n}x_{n}/2+9/8x_{n}
x_{n}f(x_{n})/f'(x_{n})=x_{n}(4x^{n2}9)/8x_{n}
So, f(x)=4x^{n2}9
4x^{2}9=0
4x^{2}=9
x^{2}=9/4
x=±3/2
x=±1.5
Question 211 
2^{20}
 
2^{10}  
None of the above 
So now we have 10 u's and 10 r's, i.e.,
uuuuuuuuuurrrrrrrrrr
So, finally the no. of arrangements of above sequences is,
Question 212 
2^{9}  
2^{19}  
So, no. of paths possible if line segment from (4,4) to (5,4) is taken is,
= paths possible from (0,0) to (4,4) * paths possible from (5,4) to (10,10)
= {uuuurrrr} * {uuuuuurrrrr}
Hence, the final answer is
Question 213 
7/8  
1/2  
7/16  
5/32 
The probability of obtaining heads
= (1/2)(5/8) + (1/2)(1/4)
= (5/16) + (1/8)
= 7/16
Question 215 
5  
8  
12  
16 
a^{p1 mod p = 1 Here, p = 17 So, p1 = 16 is the answer.}
Question 216 
∀x(P(x) ⇒ Q(x)) ⇒ (∀xP(x) ⇒ ∀xQ(x))  
∃x(P(x) ∨ Q(x)) ⇒ (∃xP(x) ⇒ ∃xQ(x))  
∃x(P(x) ∧ Q(x)) (∃xP(x) ∧ ∃xQ(x))  
∀x∃y P(x, y) ⇒ ∃y∀x P(x, y)

RHS = if P(x) holds for all x, then Q(x) holds for all x
LHS ⇒ RHS (✔)
RHS ⇒ LHS (️❌)
Question 217 
(iv) only  
(iii) and (iv) only  
(ii), (iii) and (iv) only  
(i), (ii), (iii) and (iv) 
Question 218 
(i) and (iii)  
(ii) and (iv)  
(i) and (iv)  
(iii) and (iv) 
The given condition can be satisfied by
(iii) (145, 265) → 5 ≤ 5, 4 < 6 and 1< 2
(iv) (0, 153) → 0 < 3
Question 219 
5  
6  
7  
10 
Probability of collision for each entry = 1/20
After inserting X values then probability becomes 1/2
i.e., (1/20)X = 1/2
X = b
Question 220 
In a multiuser operating system on an average, 20 requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three or five requests are made in 45 minutes is given by :
6.9 × 10^{6} × e^{20}  
1.02 × 10^{6} × e^{20}  
6.9 × 10^{3} × e^{20}  
1.02 × 10^{3} × e^{20} 
So, λ=15
Question 221 
Then,
x_{1} = c ⋅ (x_{0})^{2}  2 = 1 ⋅ (1)^{2}  2 = 1
x_{2} = c ⋅ (x_{1})^{2}  2 = 1 ⋅ (1)^{2}  2 = 1
So, the value converges to 1, which is equal to
Exactly, only (B) is answer. As all the term of x converges to 1.
Question 222 
(i) 0nly  
(i) and (ii) only  
(i), (ii) and (iii) only  
(i), (ii), (iii) and (iv) 
From the recurrence we should have c(x_{n})^{2}  x_{n}  2 < 0
For all the above values of c we have above equation as negative.
Question 223 
P_{a} and P_{b} are adjacent to each other with respect to their xcoordinate  
Either P_{a} or P_{b} has the largest or the smallest ycoordinate among all the points  
The difference between xcoordinates P_{a} and P_{b} is minimum  
None of the above 
Question 224 
Θ(n)  
Θ(nlogn)  
Θ(nlog^{2}n)  
Θ(n^{2}) 
For gradient to be maximum x_{2}x_{1} should be minimum. So, sort the points (in Θ(n logn) time) according to n coordinate and find the minimum difference between them (in Θ(n) time).
∴ Complexity = Θ(n logn + n) = Θ(n logn)
Question 225 
2^{n2}  
2^{n3} × 3  
2^{n1} 
(k(_{c}))_{2} 2^{nk}
∴ We need to find 'k' value such that, the value will be maximum.[k should be an integer].
If you differentiate (k(_{c}))_{2} 2^{nk} w.r.t. k and equal to 0.
You will get k = 2/(log_{e})2 which is not an integer.
So you can see it like
∴ The maximum degree 3⋅2^{n3} at k=3 or k=4.
Question 226 
n  
n+2  
2^{n/2}  
2^{n} / n 
While other nodes are connected so that total number of connected components is (n+1)+1
(here we are adding 1 because it is connected corresponding remaining vertices)
= n+2
Question 227 
Neither a Partial Order nor an Equivalence Relation
 
A Partial Order but not a Total Order
 
A Total Order
 
An Equivalence Relation

i) Symmetric
ii) Reflexive
iii) Transitive
If a relation is partial order relation then it must be
i) Reflexive
ii) Antisymmetric
iii) Transitive
If a relation is total order relation then it must be
i) Reflexive
ii) Symmetric
iii) Transitive
iv) Comparability
Given ordered pairs are (x,y)R(u,v) if (x
Here <, > are using while using these symbol between (x,y) and (y,v) then they are not satisfy the reflexive relation. If they uses (x<=u) and (y>=u) then reflexive relation can satisfies.
So, given relation cannot be a Equivalence. Total order relation or partial order relation.
Question 228 
It is not closed
 
2 does not have an inverse
 
3 does not have an inverse
 
8 does not have an inverse

Option A:
It is not closed under multiplication. After multiplication modulo (10) we get ‘0’. The ‘0’ is not present in the set.
(2*5)%10 ⇒ 10%10 = 0
Option B:
2 does not have an inverse such as
(2*x)%10 ≠ 1
Option C:
3 have an inverse such that
(3*7)%10 = 1
Option D:
8 does not have an inverse such that
(8*x)%10 ≠ 1
Question 229 
1  
n  
n+1  
2^{n} 
= no. of subsets with size less than or equal to 1
= n+1, because in question it is given that the two vertices are connected if and only if the corresponding sets intersect in exactly two elements.
Question 230 
Z^{2}^{xy}  
Z×2^{xy}  
Z^{2}^{x+y}  
2^{xyz} 
A set ‘P’ consists of m elements and ‘Q’ consists of n elements then total number of function from P to Q is m^{n}.
⇒ E be the no. of subsets of W = 2^{w} = 2^{xxy} = 2^{xy}
No. of function from Z to E is = (2xy)^{z} = (2xy)^{z} = 2^{xyz}
Question 231 
3  
4  
6  
9 
p(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} where a_{i}≠0
This can be written as
p(x) = a_{0} +x( a_{1} + a_{2}x + a_{3}x^{2})=a_{0}+(a_{1}+(a_{2}+a_{3}x)x)x
Total no. of multiplications required is 3
i.e., a_{3}x = K.....(i)
(a_{2}+K)x = M..... (ii)
(a_{1}+M)x=N...... (iii)
Question 232 
P1 is a tautology, but not P2
 
P2 is a tautology, but not P1
 
P1 and P2 are both tautologies  
Both P1 and P2 are not tautologies

Both P1 and P2 are not Tautologies.
Question 233 
(~A B)  
~(A ~B)  
~(~A ~B)  
~(~ A B) 
Question 234 
1/n  
2  
√n  
n 
The given probability P_{i} is for selection of each item independently with probability 1/2.
Now, Probability for x_{1} to be smallest in S = 1/2
Now, Probability for x_{2} to be smallest in S = Probability of x_{1} not being in S × Probability of x_{2} being in S
= 1/2 × 1/2
Similarly, Probability x_{i} to be smallest = (1/2)^{i}
Now the Expected value is
Question 235 
(nA ∪ B) A B  
(A^{2}+B^{2})n^{2}
 
n!(A∩B/A∪B)  
Two arbitrary subsets A⊆N and B⊆N.
Out of n! permutations π from N to N, to satisfy
min(π(A)) = min (π(B))
*) π(S) is the set of integers obtained by applying permutation π to each element of S.
If min(π(A)) =min (π(B)), say y = π(x) is the common minimum.
Since the permutation π is a 1to1 mapping of N,
x ∈ A∩B
∴ A∩B cannot be empty.
⇒ y = π(x)
= π(A∩B) is the minimum of π(A∪B) is the minimum of π(A) and π(B) are to be same.
You can think like
*) If the minimum of π(A) and π(B) are same [min π(A)] = min [π(B)]
then min(π(A∩B)) = min(π(A∪B))
∴ Total number is given by n! A∩B/A∪B
*) Finally
Considering all possible permutations, the fraction of them that meet this condition π(A∩B) / π(A∪B)
[The probability of single permutation].
Ex: N = {1, 2, 3, 4} A = {1, 3} B = {1, 2, 4}
Since π is one to one mapping
π(A∩B) = A∩B
∴ π(A) = {1, 2}
π(B) = {1, 4, 3}
π(A∩B) = {1}
π(A∪B) = {1, 2, 3, 4}
4! × 1/4 = 6
Question 236 
Determinant of F is zero  
There are an infinite number of solutions to Fx=b  
There is an x≠0 such that Fx=0  
F must have two identical rows 
Fu = Fv
Fu  Fv = 0
F(u  v) = 0
Given u ≠ v
F = 0 (i.e., Singular matrix, so determinant is zero)
Option A is true.
⇾ Fx = b; where F is singular
It can have no solution (or) infinitely many solutions.
Option B is true.
⇾ x ≠ 0 such that Fx = 0 is True because F is singular matrix (“stated by singular matrix rules). Option C is true.
⇾ F can two identical columns and rows.
Option D is false.
Question 237 
∀x [(tiger(x) ∧ lion(x)) → {(hungry(x) ∨ threatened(x))
→ attacks(x)}]  
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x))
∧ attacks(x)}]  
∀x [(tiger(x) ∨ lion(x)) → {(attacks(x) → (hungry (x)) ∨ threatened (x))}]  
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]

Here we have two cases.
i) If Tiger is hungry (or) threaten that will attack.
ii) If Lion is hungry (or) threaten that will attack.
If Tiger is hungry (or) threaten then both lion and tiger will not attack only Tiger will attack and viceversa.
Then answer is
∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
Note: Don’t confuse with the statement Tiger and Lion.
Question 238 
3m  
3n  
2m+1  
2n+1 
n = ^{m}C_{3}
Which subsets contains element i then size is
= ^{(m1)}C_{2}
Because 1 element is already known
Question 239 
X ⊂ Y  
X ⊃ Y
 
X = Y  
X  Y ≠ ∅ and Y  X ≠ ∅

E = {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)}
F = {(1,1), (2,2), (3,3)}
G = {(1,3), (2,1), (2,3), (3,1)}
X = (E∩F)  (F∩G)
= {(1,1), (2,2), (3,3)  ∅}
= {(1,1), (2,2), (3,3)} (✔️)
Y = (E  (E∩G)  (E  F))
= (E  {(1,3), (2,3), (3,1)}  {(1,2), (1,3), (2,3), (3,1)})
= {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)}  {(1,3), (2,2), (2,1)}  (1,2), (1,3), (2,3), (3,1)}
= {(1,1), (1,2), (2,2), (3,3)}  {(1,2), (1,3), (2,3), (3,1)}
= {(1,1), (2,2), (3,3)} (✔️)
X = Y
X = (E∩F)  (F∩G) = {2,5}  {5} = {2}
Y = (E  (E∩G)  (E  F))
= {(1,2,4,5)  (4,5)  (1,4)}
= {(1,2)  (1,4)}
= {2}
X = Y
Question 240 
No. of elements selected = n
Probability of getting head = ½
Probability of n heads out of 2n coin tosses is
2nC_{n}*(1/2)^{n}*(1/2)^{n}=2nC_{n}/4^{n}
Question 241 
In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?
0.4  
0.6  
0.8  
0.9 
0.6 = (0.5×0.4) + (0.5×P(rain at temp>25)
0.6 = (2) + (0.5×P(rain at temp>25)
P(rain at temp>25) = 0.8
Question 242 
1 and 2 only  
2 and 3 only  
1 and 3 only  
None of these 
y = 3
Here, identity element is 3.
(2) f(x, y) = max(x, y) = x = max(y, x)
⇒ y = 1
Here, identity element = 1
(3) f(x, y) = x^{n}y is not equal to f(y, x) = y^{n}x
So, no identity element.
Question 243 
f (x_{1}, x_{2}, …, x_{n}) = x_{1}’f(x_{1}, x_{2}, …, x_{n}) + x_{1}f(x_{1}, x_{2}, …, x_{n})  
f (x_{1}, x_{2}, …, x_{n}) = x_{2}f(x_{1}, x_{2}, …, x_{n}) + x_{2}’f(x_{1}, x_{2}, …, x_{n})  
f (x_{1}, x_{2}, …, x_{n}) = x_{n}’f(x_{1}, x_{2}, …, 0) + x_{n}f(x_{1}, x_{2}, …,1)  
f (x_{1}, x_{2}, …, x_{n}) = f(0, x_{2}, …, x_{n}) + f(1, x_{2}, .., x_{n}) 
LHS: f(x_{1}) = 0 where x_{1} = 0
LHS: f(x_{1}) = 1 when x1 = 1
RHS: f(0) + f(1) = 0 + 1 = always 1
Question 244 
satisfiable and valid  
satisfiable and so is its negation  
unsatisfiable but its negation is valid  
satisfiable but its negation is unsatisfiable 
Question 245 
When a coin is tossed, the probability of getting a Head is p, 0<p<1. Let N be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is
1/p  
1/(1p)  
1/p^{2}  
1/(1p^{2}) 
Multiply both sides with (1  p) and subtract,
E  (1  p) E = 1 × p + (1  p) p + (1  p) (1  p) p + ......
E  (1  p) E = p/(1  (1  p))
(1  1 + p) E = 1
pE = 1
E = 1/p
Question 246 
1 only  
2 only  
Neither 1 nor 2  
Both 1 and 2 
P = {1, 2, 4, 5}
Q = {2, 3, 5, 6}
R = {4, 5, 6, 7}
(1) P Δ (Q ∩ R) = (P Δ Q) ∩ (P Δ R)
P Δ ({2,3,5,6} ∩ {4,5,6,7}) = ({1,2,4,5} Δ {2,3,5,6} ∩ {1,2,4,5} Δ {4,5,6,7})
P Δ {5,6} = ({1,2,3,4,5,6}  {2,5}) ∩ ({1,2,4,5,6,7}  {4,5})
({1,2,4,5} Δ {5,6}) = {1,3,4,6} ∩ {1,2,6,7}
{1,2,4,5,6}  {5} = {1,6}
{1,2,4,6} ≠ {1,6}
Statement1 is False.
(2) P ∩ (Q ∩ R) = (P ∩ Q) Δ (P Δ R)
LHS:
{1,2,4,5} ∩ {5,6} = {5}
RHS:
({1,2,4,5} ∩ {2,3,5,6}) Δ ({1,2,4,5} Δ {4,5,6,7})
{2,5} Δ ({1,2,4,5,6,7}  {4,5})
{2,5} Δ {1,2,6,7}
{1,2,5,6,7}  {2}
{1,5,6,7}
LHS ≠ RHS
Statement  2 is also wrong.
Question 247 
28  
33  
37  
44 
A = set of numbers divisible by 2
B = set of numbers divisible by 3
C = set of numbers divisible by 5
n(A∪B∪C) = n(A) + n(B) + n(C)  n(A∩B)  n(B∩C)  n(C∩A) + n(A∩B∩C)
= ⌊123/2⌋ + ⌊123/3⌋ + ⌊123/5⌋  ⌊123/6⌋  ⌊123/15⌋  ⌊123/10⌋ + ⌊123/30⌋
= 61 + 41 + 24  20 12  8 + 4
= 90
Total no. that are not divisible
= n  n(A∪B∪C)
= 123  90
= 33
Question 248 
Consider the undirected graph G defined as follows. The vertices of G are bit strings of length n. We have an edge between vertex u and vertex v if and only if u and v differ in exactly one bit position (in other words, v can be obtained from u by flipping a single bit). The ratio of the chromatic number of G to the diameter of G is
1/(2^{n1})  
1/n  
2/n  
3/n 
That will give us a bipartite graph, with chromatic number = 2
Also from the same we can conclude that we need for a 'n' bit string, to traverse no more than (n1) edges or 'n' vertices to get a path between two arbitary points. So the ratio is (2/n).
Question 249 
a, a, a  
0, a, 2a  
a, 2a, 2a 
Question 250 
1R, 2S, 3P, 4Q  
1S, 2R, 3Q, 4P  
1S, 2Q, 3R, 4P  
1S, 2P, 3Q, 4R 
Question 251 
None of the above 
Question 252 
0.5  
0.75  
1.5  
2.0 
Question 253 
it will converge  
it will diverse  
it will neither converge nor diverse  
It is not applicable 
1 + 1/2 <= 9
and 3 + 1 <= 10
Question 254 
X = Y
 
X ⊂ Y  
Y ⊂ X
 
None of these

B = {1, 3, 4, 5}
C = {2, 4, 5, 6}
X = (A  B)  C
X = {2, 6}  {2, 4, 5, 6}
= ∅
Y = (A  C)  (B  C)
= {1, 3}  { 1, 3}
= ∅
X = Y
X = (A  B)  C
= (1, 5)  (5, 7, 4, 3)
= (1)
Y = (A  C)  (B  C)
= (1, 4)  (2, 4)
= (1)
X = Y
Question 255 
6  
8  
9  
13 
F = E  V + 2 [From Euler's formula i.e., F + V  E = 2]
F = 19  13 +2
F = 8
Question 256 
12  
8  
Less than 8
 
More than 12

Edges = 100
Minimum cover of vertex G is = 8
Maximum Independent set of G = No. of vertices  Minimum cover of vertex G
= 20  8
= 12
Question 257 
f(b  a)
 
f(b)  f(a)  
Then the probablity be area of the corresponding curve i.e.,
Question 258 
3 and 13
 
2 and 11  
4 and 13
 
8 and 14

Inverse of 4 = m; Inverse of 7 = n
(4×m)%15=1; (7*n)%15=1
Option A: m=3 n=13
12%15≠1 (✖️) 91%15=1 (✔️)
Option B: m=2 n=11
8%15≠1 (✖️) 11%15≠1 (✖️)
Option C: m=4 n=13
16%15=1(✔️) 91%15=1 (✔️)
Option D: m=8 n=14
120%15≠1(✖️) 98%15≠1(✖️)
Question 259 
1/2^{n}
 
1  1/n
 
1/n!
 
1(1/2^{n})

Hence Probability = (2^{n}  1) /2^{n} = 1  1/2^{n}
Question 260 
4/19
 
5/19
 
2/9  
19/30 
Q → 3 red, 1 blue
The probability of selecting a red ball is
(1/3)(2/5) + (2/3)(3/4)
2/15 + 1/2 = 19/30
The probability of selecting a red ball from P
(1/3) * (2/5) = 2/15
→ The colour of ball is selected is to be red and that is taken from the box P.
⇒ Probability of selecting a red ball from P/Probability of selecting a red ball
⇒ (2/15)/(19/30)
⇒ 4/19
Question 261 
1 and 1
 
1 and 6  
2 and 5  
4 and 1

A = (2  λ)(5  λ)  (4) = 0
10  7λ+ λ^{2}  4= 0
λ^{2}  7λ + 6 = 0
λ^{2}  6λ  λ + 6 = 0
(λ  6) 1(λ  6) = 0
λ = 1 (or) 6
Question 262 
i
 
i+1
 
2i  
2^{i} 
Put g(i) = i+1
S = 1 + 2x + 3x^{2} + 4x^{3} + .....
Sx = 1x + 2x^{2} + 3x^{3} + 4x^{4} + ......
S  Sx = 1 + x + x^{2} + x^{3} + .....
[Sum of infinite series in GP with ratio < 1 is a/1r]
S  Sx = 1/(1x)
S(1x) = 1/(1x)
S = 1/(1x)^{2}
Question 263 
G_{1}  
G_{2}  
G_{3}  
G_{4} 
which is planar
G_{3} can also be drawn as
which is planar
G_{4} can also be drawn as
which is planar
But G_{1} cannot be drawn as planar graph.
Hence, option (A) is the answer.
Question 264 
no solution
 
a unique solution
 
more than one but a finite number of solutions
 
an infinite number of solutions

2(2  20) +1(3 + 5) + 3(12  2)
= 44 + 8 + 30
= 6 ≠ 0
→ A ≠ 0, we have Unique Solution.
Question 265 
a group
 
a monoid but not a group
 
a semi group but not a monoid
 
neither a group nor a semi group

The algebraic structure is a group because the given matrix can have inverse and the inverse is nonsingular.
Question 266 
4  
6  
16  
24 
That means a=0,1,2 ⇒ 3
b = dmod5
That means b=0,1,2,3,4 ⇒ 4
→ Total no. of order pairs = 3 * 5 = 15
→ Ordered pair (c,d) has 1 combination.
Then total no. of combinations = 15+1 = 16
Question 267 
f and g should both be onto functions
 
f should be onto but g need not be onto
 
g should be onto but f need not be onto
 
both f and g need not be onto

f: B→C and g: A→B are two functions.
h = f∘g = f(g(x))
→ If his onto function, that means for every value in C, there must be value in A.
→ Here, we are mapping C to A using B, that means for every value in C there is a value in B then f is onto function.
→ But g may (or) may not be the onto function i.e., so values in B which may doesn't map with A.
Question 268 
R∪S, R∩S are both equivalence relations.  
R∪S is an equivalence relation.
 
R∩S is an equivalence relation.
 
Neither R∪S nor R∩S is an equivalence relation.

Let (a,b) present in R and (b,c) present in S and (a,c) is not present in either of them. Then R∪S will contain (a,b) and (b,c) but not (a,c) and hence not transitive.
And equivalence relation must satisfy 3 property:
(i) Reflexive
(ii) Symmetric
(iii) Transitive
But as we have seen that for R∪S, Transitivity is not satisfied.
Question 269 
∀(x) [teacher(x) → ∃ (y) [student(y) → likes (y, x)]]  
∀(x) [teacher(x) → ∃ (y) [student(y) ∧ likes (y, x)]]
 
∃(y) ∀(x) [teacher(x) → [student(y) ∧ likes (y, x)]]
 
∀(x) [teacher(x) ∧ ∃ (y)[student(y) → likes (y, x)]]

Option B: If x is a teacher, then there exists some y, who is a student and like x. (✔️)
Option C: There exists a student who likes all teachers.
Option D: If x is a teacher and then there exists some y, if y is a student then y likes x.
Question 270 
X ≡ Y  
X → Y
 
Y → X
 
¬Y → X

⇒ ∼(P∨Q) ∨ R
⇒ (∼P∧∼Q) ∨ R
⇒ (∼P∨R) × (∼Q∨R)
⇒ (P→R) ∧ (Q→R)
Option B: X→Y
[(P→R) × (Q→R)] → [(P→R) ∨ (Q→R)]
∼[(P→R) × (Q→R) ∨ (P→R) ∨ (Q→R)]
[∼(P→R) ∨ ∼(Q→R)] ∨ [(P→R) ∨ (Q→R)]
[∼(P→R) ∨ (P→R)] ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] ∨ [∼(Q→R) ∨ (Q→R)]
T ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] V T
T (✔️)
Question 271 
A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is
1/36  
1/6  
1/4  
1/3 
Question 272 
is always > (1/3)  
is always < (1/3)  
is always = (1/3)  
may be greater or lesser than (1/3) 
Question 273 
1  
0  
1  
2 
determinant = product of diagonal element [upper triangular matrix]
= 1 * 1 * 1 * 1
= 1
Question 274 
Let f be a function from a set A to a set B, g a function from B to C, and h a function from A to C, such that h(a) = g(f(a)) for all a ∈ A. Which of the following statements is always true for all such functions f and g?
g is onto ⇒ h is onto  
h is onto ⇒ f is onto  
h is onto ⇒ g is onto  
h is onto ⇒ f and g are onto 
If h: A→C is a onto function, the composition must be onto, but the first function in the composition need to be onto.
So, B→C is must be onto.
Question 275 
n  
n + 1  
2^{(n1)} + 1  
n! 
Let A = {a, b, c}, here n = 3
Now, P(A) = {Ø, {a}, {b}, {c}, {a,b}, {b,c}, {{a}, {a,b,c}}
Now C will be contain Ø (empty set) and {a,b,c} (set itself) as Ø is the subset of every set. And every other subset is the subset of {a,b,c}.
Now taking the subset of cardinality, we an take any 1 of {a}, {b}, {c} as none of the set is subset of other.
Let's take {2}
→ Now taking the sets of cardinality 2 {a,b}, {b,c}
→ {b} ⊂ {a,b} and {b,c} but we can't take both as none of the 2 is subset of the other.
→ So let's take {c,a}.
So, C = {Ø, {b}, {b,c}, {a,b,c}}
→ So, if we observe carefully, we can see that we can select only 1 set from the subsets of each cardinality 1 to n
i.e., total n subsets + Ø = n + 1 subsets of A can be there in C.
→ So, even though we can have different combinations of subsets in C but maximum cardinality of C will be n+1 only.
Question 276 
3  
4  
5  
6 
Question 277 
p(p  1)  
pq  
(p^{2}  1)(q  1)  
p(p  1)(q  1) 
→ No. of multiples of p in n = pq [n = p⋅p⋅q]
→ No. of multiples of q in n = p^{2} [n = p^{2}q]
→ Prime factorization of n contains only p & q.
→ gcd(m,n) is to be multiple of p and (or) 1.
→ So, no. of possible m such that gcd(m,n) is 1 will be
n  number of multiples of either p (or) q
= n  p^{2}  pq + p
= p^{2}q  p^{2}  pq + p
= p(pq  p  q + 1)
= p(p  1)(q  1)
Question 278 
1  
1  
0  
π 
In the limits are be π to π, one is odd and another is even product of even and odd is odd function and integrating function from the same negative value to positive value gives 0.
Question 279 
((∀x(P(x)∨Q(x))))⟹((∀xP(x))∨(∀xQ(x)))  
(∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))  
(∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))  
(∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))

RHS: (∀xP(x)) and (∀xQ(x)) both becomes False for assumed values which implies F→F and result will be True.
∴ LHS = RHS
Question 280 
(∃x) (boy(x) → (∀y) (girl(y) ∧ taller(x,y)))
 
(∃x) (boy(x) ∧ (∀y) (girl(y) ∧ taller(x,y)))
 
(∃x) (boy(x) → (∀y) (girl(y) → taller(x,y)))
 
(∃x) (boy(x) ∧ (∀y) (girl(y) → taller(x,y)))

'∧' → predicts statements are always true, no matter the value of x.
'→' → predicts there is no need of left predicate to be true always, but whenever it becomes true, then right predicate must be true.
Option D:
There exists a some boys who are taller than of all girls y.
Question 281 
{(x, y)y > x and x, y ∈ {0, 1, 2, ... }}  
{(x, y)y ≥ x and x, y ∈ {0, 1, 2, ... }}
 
{(x, y)y < x and x, y ∈ {0, 1, 2, ... }}
 
{(x, y)y ≤ x and x, y ∈ {0, 1, 2, ... }}

Answer is option B.
{(x, y)y ≥ x and x, y ∈ {0, 1, 2, ... }}
Question 282 
3/8  
1/2  
5/8  
3/4 
Then total number of possibilities = 2^{4} = 16
No. of possibilities getting 2 heads and 2 tails is
HHTT, HTHT, TTHH, THTH, THHT, HTTH = 6
Probability of getting 2 heads and 2 tails is
= No. of possibilities/Total no. of possibilities = 6/16 = 3/8
Question 283 
power (2,n)
 
power (2,n^{2})
 
power (2, (n^{2} + n)/2)
 
power (2, (n^{2}  n)/2)

A [i] [j] = A [j] [i]
So, we have only two choices, they are either upper triangular elements (or) lower triangular elements.
No. of such elements are
n + (n1) + (n2) + ... + 1
n(n+1)/2
We have two choices, thus we have
2^{(n(n+1)/2)} = 2^{((n2+n)/2) choices i.e., Power (2, (n2+n)/2). }
Question 284 
D^{1}C^{1}A^{1}
 
CDA
 
ADC
 
Does not necessarily exist

ABCD = I
Pre multiply A^{1} on both sides
A^{1}ABCD = A^{1}⋅I
BCD = A^{1}
Pre multiply B^{1} on both sides
B^{1}BCD = B^{1}A^{1}
CD = B^{1}A^{1}
Post multiply A on both sides
CDA = B^{1}A^{1}⋅A
∴ CDA = B^{1}(I)
∴ CDA = B^{1}
Question 285 
satisfiable but not valid
 
valid
 
a contradiction
 
None of the above

(P→(Q∨R)) → (P∨Q)→R
If P=T; Q=T; R=T
(P→(T∨T)) → ((T∨T)→R)
(P→T) → (T→R)
(T→T) → (T→T)
T→T
T(Satisfiable)
Question 286 
infinitely many
 
two distinct solutions
 
unique
 
none

rank = r(A) = r(AB) = 2
rank = total no. of variables
Hence, unique so