EngineeringMathematics
Question 1 
1/5  
1/3  
1/4  
2/5 
Total number of seats = 5*40
= 40 seats
Total number of students= 30
Step2: Given constraint that, 6th seat in the fifth row is empty
When we are deleting 6th seat, 30 students have 39 choices of seats
Step3: Total number of choices = ^{39}C_{30}
Total ways to choose = ^{40}C_{30}
Step4: Final probability = ^{39}C_{30} /^{ 40}C_{30}
= 1/4
Question 2 
0 < x < π  
2nπ < x < (2n+1)π , for n in N  
Empty set  
None of the above 
→ So the domain of log(log sin(x)) is undefined which is empty Set.
Question 3 
x = x^{1}, for any x belonging to G  
x = x^{2}, for any x belonging to G  
(x * y )^{2} = x^{2} * y^{2}, for any x, y belonging to G  
G is of finite order 
(x * y )^{2} = x^{2} * y^{2}, for any x, y belonging to G
Question 4 
Linear  
Exponential  
Quadratic  
Cubic 
Question 5 
maximum of n and d  
n + d  
nd  
nd / 2 
d * n = 2 * E
∴ E = d*n/2
Question 6 
Connected  
Disconnected  
Euler  
A circuit 
Consider the graph with three nodes(n1,n2&n3) and has 2 edges.
n1>n2 and n2>n1 then the graph is disconnected.
Question 7 
1  
k  
k1  
k/2 
Question 8 
(∃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 9 
are
None Of the Above 
Question 10 
is of cardinality 2^{c}  
have the same cardinality as A  
forms a partition of A  
is of cardinality C^{2} 
Note: Ambiguous between answer with option D
Question 11 
2/9  
4/9  
2/3  
1/2 
Probability that chip came chosen from company Y = 4/9
Number of Defective chips from company X = 1
Number of Defective chips from company Y = 2
Probability that chip is defective from company X = 5/9 * 1/5
Probability that chip is defective from company Y = 4/9 * 2/4
Probability that chip is defective = 5/9 * 1/5 + 4/9 * 2/4
Given chip is defective, probability that it came from the company
Y = P(Defective Company Y)/ P(Defective)
= (4/9 * 2/4) / (5/9 * 1/5 + 4/9 * 2/4)
= 2/3
Question 12 
f_{1}(t) + f_{2}(t)  
∫^{t}_{0} f_{1}(x), f_{2}(x) dx  
∫^{t}_{0} f_{1}(x), f_{2}(tx) dx  
max (f_{1}(t), f_{2}(t)) 
∫^{t}_{0} f_{1}(x), f_{2}(tx) dx
Question 13 
diagonal matrix  
A  
A  
0 
→ If A is skew symmetric matrix then A^{T} = A
→ In terms of the entries of the matrix, if a_{ij} denotes the entry in the i^{th} row and j^{th} column, then the skewsymmetric condition is equivalent to
→ If A is skew symmetric matrix then a_{ji}=a_{ij}
Question 14 
P(A∩B) = P(A)P(B)  
P(A∪B) = P(A) + P(B)  
P(AB) = P(A ∩ B) + P(B)  
P(A∪B) <= P(A) + P(B) 
(B) Happens when A and B are mutually exclusive.
(C) Not happens.
(D) P(A∪B) ≤ P(A) + P(B) is true because P(A∪B) = P(A) + P(B)  P(A∩B).
Question 15 
2.222  
2.275  
2.279  
None of the above 
The NewtonRaphson iterative formula is
Since x_{3} and x_{4} are identical upto 3 places of decima,we take x_{4}=2.279 as the required root, correct to three places of the decimal.
Question 16 
{1, 3, 5, 6, 7, 8}  
{2, 4, 9}  
{2, 4}  
{1, 2, 3, 4, 5, 6, 7, 8, 9} 
Question 17 
((a → b) ∧ (b → c)) → (a → c)  
(a ↔ c) → ( ¬b → (a ∧ c))  
(a ∧ b ∧ c) → (c ∨ a)  
a → (b → a) 
Note: Above (A↔C) →( ¬B→(A∧C)) is not tautology.
Question 18 
Idempotent  
Symmetric  
Orthogonal  
Hermitian 
A square matrix A is an orthogonal matrix if its transpose is equal to its inverse. An orthogonal matrix A is necessarily invertible and unitary.
Conditions for an orthogonal matrix:
A^{T}= A^{1} and
A A^{T} = A^{T} A = I,
where I is an Identity matrix
Question 19 
a, c, d  
a, b, d  
b, c, d  
a, b, c 
The graph has (a, b, c) as a strongly connected component.
Question 20 
1/2  
√2  
± 1/2  
± 1/√2 
Determinant of 2nd Matrix = X2– X
Since the determinants are equal:
X2  1 = X2  1.
X2 + X  1 = 0
X = 1/2
Option (A) is correct
Question 21 
{n,1}  
{n, n}  
{n^{2}, 1}  
{1, n^{2}} 
S={1,2,3,4}
→ If a set is said to be equivalence, then the set must be
i) Reflexive
ii) Symmetric
iii) Transitive
i) Reflexive Relation: A relation ‘R’ on a set ‘A’ is said to be reflexive if (xRx)∀x∈A.
A = {1,2,3}
R = {(1,1), (2,2), (3,3)}
R = {(1,1), (2,2)} It is false.
R = {(1,1), (2,2), (3,3), (1,2)}
ii) Symmetric Relation: A relation on a set A is said to be symmetric if (xRy). Then (yRx)∀x,y∈A i.e., if ordered pair (x,y)∈R. Then (y,x)∈R ∀x,y∈A.
A={1,2,3}
R1={(1,2), (2,1)}
R2={(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}
Transitive Relation:
A relation ‘R’ on a set ‘A’ is said to be transitive if (xRy) and (yRz), then (xRz)∀(x,y,z∈A).
A={1,2,3}
R1={ }
R2={(1,1)}
R3={(1,2), (3,1)}
R4={(1,2), (2,1), (1,1)}
⇾ A={1,2,3,4}
Largest ordered set is
S×S={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)}
⇒ Total = 16 = 42 = n2
Smallest ordered set = {(1,1),(2,2),(3,3),(4,4)}
⇒ Total=4=n
Note: In question, they are clearly mentioned that Rank of an Equivalence relation is equal to the number of induced Equivalence classes. Since we have maximum number of ordered pairs(which are reflexive, symmetric and transitive ) in largest Equivalence relation, its rank is always 1.
Question 22 
Reflexive, not symmetric, transitive  
Not reflexive, not antisymmetric, transitive  
Reflexive, antisymmetric, transitive  
Not reflexive, not antisymmetric, not transitive 
A relation ‘R’ on a set ‘A’ is said to be reflexive if (xRx)∀x∈A.
Example: 4 D 8, 4 D 12, 4 D16, 4 D20…….(Here, D means divide)
8 D 16, 8 D 24……….
In this example, we didn’t get 4D4. So, it is not reflexive.
AntiSymmetric Relation:
For all x ∈ I, R(x,y) and R(y,x) then x=y is antisymmetric. We can easily make a violation as R(2,2) and R(2,2) are not antisymmetric.
It is violating. So, not antisymmetric relation.
Transitive relation:
A relation ‘R’ on a set ‘A’ is said to be transitive if (xRy) and (yRz), then (xRz)∀(x,y,z∈A).
Example: 4D8, 4D12, 4D16, 4D20…….(Here, D means divide)
8D16, 8D24……….
{ 4D8, 8D16, 1D16}. So, it is satisfied.
Question 23 
1  
1/21  
0  
0.5 
Step1: Bag contains 19 Red(R) and 19 Blue(B) balls.
BB (or) RR happen we are discarded.
If we get BR (or) RB then B is discarded and R is returned.
Step2: There are some conditions that,
→ If black balls will either come with black then both black balls are discarder.
→ If it will come with red then only black balls will be discarded.
→ Suppose 2 red balls will come together means we are discarding both red balls.
Step3: As per the above constraints, total 19 Red balls it means odd number.
→ Among 19 only 18 will be discarded.
Step3: Final content of bag at second last trail will be either R,B (or) R,R,R and finally in last
trail bag will left with one red ball in both the cases.
Question 24 
α = 6, β = 1/2  
α = 2, β = 1/2  
α = 2, β = 1/2  
α = 6, β =1/2 
Step1: x= 1 and x=2
f(x) = α log x + β x2 + x
f'(x)= α/x + 2βx + 1 = 0
Step2: for extreme points f'(x)=0
α/x + 2βx + 1=0
Step3: For x= 1 then we will get α+2β= 1 → (i)
For x= 2: then we will get α+8β= 2 → (ii)
from (i) and (ii) we can get the value of α=2 and β= 1/2
Question 25 
[1, 0]  
[1, 0)  
[∞, 0]  
[∞,1] 
Step1: f(x) = logx and given range is [∞ to +∞]
g(x) = sin(x) and given range is [1,1]
Step2: Given 2 variables are A and B
A= f(g(x))
= logg(x)
= logsin(x)
So, we will get A range is [∞ ,0]
Step3: B= g(f(x))
= sin(f(x))
= sin(logx)
So, we will get B range is [1, 1]
Step4: Common in both A and B is A∩B
A∩B = [1, 0]
Key point: Ranges [ 1 ≤ sin(x) ≤ 1 and ∞ ≤ logx ≤ ∞ ]
Question 26 
tautology  
contradiction  
contingency  
absurdity 
Question 27 
¬∃(x) [ T(x) ⋀ ¬P(x) ]  
¬∃(x) [ T(x) ⋁ ¬P(x) ]  
¬∃(x) [ ¬T(x) ⋀ ¬P(x) ]  
¬∃(x) [ T(x) ⋀ P(x) ] 
Hence it is correct .
Option B implies "It is not the case that some are trigonometric functions or they are not periodic”.
Option C implies "It is not the case that some of not trigonometric functions are not periodic”.
Option D implies "It is not the case that some trigonometric functions are periodic”.
Question 28 
3  
8  
4  
2 
Step1: Total number of elements in power set of given set with ‘n’ elements = 2^{n}
Example: {1,2}
{{∅},{1},{2},{1,2}} → Total 4 possibilities.
Step2: The given set in question contains 3 elements({1,2},{2,1,1}, {2,1,1,2} }. So, number of elements in power set of given set is 2^{3} =8.
Step3: The power set is {{∅},{X}} or {{∅},{1,2}}.
Question 29 
injective and surjective  
surjective but not injective  
injective but not surjective  
neither injective nor surjective 
Question 30 
2/5  
2  
3  
5/2 
Step1: It is clearly showing that two vectors are perpendicular. If two vectors are perpendicular
we are using dot product is zero. a.b =0
Step2: Calculating dot product is “i is multiplied with i” and “j is multiplied with coefficient of j”
Step3: we can write like this,
= (2i+λj+k).(i2j+3k)=0
= 22λ+3 =0
2λ = 5
λ=5/2
Question 31 
e  
e/2  
e^{2}  
2 e 
Handshaking theorem states that the sum of degrees of the vertices of a graph is twice the number of edges.
If G=(V,E) be a graph with E edges,then
Σ deg_{G}(V) = 2E
This theorem applies even if multiple edges and loops are present.
Since the given graph is undirected, every edge contributes as twice in sum of degrees. So the sum of degrees is 2e.
Question 32 
2^{x}  
2  
2  
1 
Poisson Formula. Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is μ.
Then, the Poisson probability is: P(x; μ) = (e^{μ}) (μ^{x}) / x!
where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828.
From the given function the “μ” value is 2 which is the mean.
Question 33 
(1000/√2π).e^{25}  
1000/√2π  
(1000/√2π ).e^{2.5}  
400/√2π 
Question 34 
y(0) = 1,
y(1) = 0,
y(2) = 1,
y(3) = 10 is
x3 +2 x2 + 1  
x3 + 3x2 + 1  
x3 + 1  
x3 – 2 x2 + 1 
x = 0: y(x) = 1
x = 1: y(x) = 0
x = 2: y(x) = 8  8 + 1 = 1
x = 3: y(x) = 27  18 + 1 = 10
Question 35 
Ellipse  
Hyperbola  
Circle  
Parabola 
x = a cos t
y = b sin t
where:
x,y are the coordinates of any point on the ellipse,
a, b are the radius on the x and y axes respectively,
t is the parameter, which ranges from 0 to 2π radians
Question 36 
3/2  
3/2  
0  
5/4 
To find the minimum value, calculate the derivative until at what point given function is minimum value.
Applying First order derivative to the function y is y’ = 2x – 3
Again applying derivative of y ’is y” = 2 (Since y” > 0, it has a minimum value)
So the minima value at that point is (2x – 3) = 0 and x = 3/2
Question 37 
Multigraph  
Non regular graph  
Regular graph  
Complete graph 
Question 38 
Path  
Walk  
Tree  
Circuit 
Let G be a graph and let there be exactly one path between every pair of vertices in G. So G is connected. Now G has no cycles, because if G contains a cycle, say between vertices u and v, then there are two distinct paths between u and v, which is a contradiction. Thus G is connected and is without cycles, therefore it is a tree.
A tree is a minimally connected graph i.e. removing a single edge will disconnect the graph. A tree with n vertices has n−1 edges and only one path exists between every pair of vertices.
Question 39 
n edges  
nk edges  
(n − k)(n − k + 1) edges  
(n − k)(n − k + 1)/2 edges 
So, G has maximum number of edges if each component is a complete graph.
Hence, the maximum possible number of edges in the graph G is:
And in every case, (n − k)(n − k + 1)/2 will be greater than or equal to the above expression.
So, at maximum, there can be (n − k)(n − k + 1)/2 edges in a simple graph with n vertices and k components.
Question 40 
3  
4  
6  
9 
=a_{0}+X(a_{1}+X(a_{2}+a_{3}X))
↓ ↓ ↓
③ ② ①
Here, minimum 3 multiplication needed
Question 41 
I  
A  
A  
I 
This leads to the equivalent characterization: a matrix Q is orthogonal if its transpose is equal to its inverse: A^{T}=A^{1}
Question 42 
becomes zero
 
remains unaltered  
becomes infinitive  
becomes negative of its original value 
The value of the determinant remains unchanged if both rows and columns are interchanged.
Question 43 
a null matrix  
A+B+C  
(A+B+C)’  
(A+B+C) 
Question 44 
E1  
E  
1 – E^{1}  
1 – E 
Question 45 
is called
Simpson rule  
Trapezoidal rule  
Romberg’s rule  
Gregory’s formula 
Question 46 
Newton’s backward formula  
Gauss forward formula  
Gauss backward formula  
Stirling’s formula 
Question 47 
2  
3  
4  
5 
The determinant of 2x2 matrix is (adbc), If the elements are a,b,c and d.
From the given matrix, a=3,b=3,c=x and d=5 then adbc =153x
153x=3 ⇒ 3x=12 ⇒ x=4
Question 48 
779  
679  
0  
256 
Question 49 
f (b − a)  
f (b) − f (a)  
Question 50 
k and n  
k1 and k+1  
k1 and n1  
k+1 and nk 
→ If a vertex is removed then it results that all the components are also be disconnected. So removal can create (n1) components
Question 51 
f (x_{0}) > 0  
f (x_{0}) f (x_{0})” > 0  
f (x_{0}) f (x_{0})” < 0  
f (x_{0})” > 0 
Question 52 
△(U_{k}V_{k}) = U_{k}△V_{k} + V_{k}△U_{k}  
△(U_{k}V_{k}) = U_{k+1}△V_{k} + V_{k+1}△U_{k}  
△(U_{k}V_{k}) = V_{k+1}△U_{k} + U_{k}△V_{k}  
△(U_{k}V_{k}) = U_{k+1}△V_{k} + V_{k}△U_{k} 
Question 53 
√3 + √7 = √10  
√3 + √7 ≤ √10  
√3 + √7 < √10  
√3 + √7 > √10 
=4.3778021186334678840290620951451
√10=3.1622776601683793319988935444327
So, √3 + √7 > √10 is true.
Question 54 
same degree  
even degree  
odd degree  
different degree 
Proof:
→ Let G(V, E) be an Euler graph. Thus G contains an Euler line Z, which is a closed walk.
→ Let this walk start and end at the vertex u ∈ V. Since each visit of Z to an intermediate vertex v of Z contributes two to the degree of v and since Z traverses each edge exactly once, d(v) is even for every such vertex.
→ Each intermediate visit to u contributes two to the degree of u, and also the initial and ﬁnal edges of Z contribute one each to the degree of u. So the degree d(u) of u is also even.
Question 55 
n^{2}  
n(n1)/2  
n1  
n(n+1)/2 
Question 56 
Skew symmetric  
Symmetric  
Orthogonal  
Idempotent 
So, BTAB = B'.A.B
Taking transpose of B’.A.B
(B'.A.B)' = B'.A'.(B')' = B'.A.B // (B')' = B
So, it is a symmetric matrix.
Question 57 
n x^{n1}  
n^{n} . n!  
nx^{n} !  
n! 
Question 58 
5/8  
1/8  
2/3  
3/8 
The combinations are (HHH,HHT,HTH,HTT,TTT,TTH,THT,THH)
The number of combinations with two heads and one tail is HHT,HTH,THH
The the probability is the number of combinations of the event/ total combinations of the event = 3/8
Question 59 
(v+1)−k  
(v+1)/2 −k  
v−k  
v+k 
→ Suppose, if each vertex is a component, then k=v, then there will not be any edges among them. So, vk= 0 edges.
Method2:
→ According to pigeonhole principle, every component have v/k vertices.
→ Every component there will be (v/k)1 edges.
→ Total k components and edges= k*((v/k)1)
= v–k
Question 60 
R ∩ S = (R ∪ S) − [(R − S) ∪ (S − R)]  
R ∪ S = (R ∩ S) − [(R − S) ∪ (S − R)]  
R ∩ S = (R ∪ S) − [(R − S) ∩ (S − R)]  
R ∩ S = (R ∪ S) ∪ (R − S) 
R∩S = (R ∪ S) − [(R − S) ∪ (S − R)]
Question 61 
Question 62 
n * (n1) / 2  
n * (n+1) / 2  
n^{2}  
n * (n+1) 
Question 63 
7  
8  
3  
4 
Given set = {{A, B},C}
Power set = { {{A, B}}, {C}, {{A, B}, C}}, ϕ}
So, total 4 elements are present in the power set.
Question 64 
0.5, 0.25  
0.25, 0.5  
0.5, 1  
1, 0.5 
→P(E) denote the probability of the occurrence of event E
→P(A) = 0.5 and P(B) = 1
→Conditional probability is a measure of the probability of an event (some particular situation occurring) given that another event has occurred.
→If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A  B), or P_{B}(A)
→P(A/B) = P(A ∩ B)/P(B)
→If two events A and B are independent, then the probability of both events is the product of the probabilities for each event: P(A ∩ B) = P(A)P(B)
→P(A/B) = P(A) * P(B) / P(B)
→P(A/B) = 0.5
→Similarly, P(B/A) = P(A) * P(B) / P(A) and P(B/A) = 1
Question 65 
15  
7  
11  
3 
Substitute all the value of x from 0 to 5
The value of f(x) is 3 when x=0
The value of f(x) is 9 when x=1
The value of f(x) is 11 when x=2
The value of f(x) is 9 when x=3
The value of f(x) is 3 when x=4
The value of f(x) is 7 when x=5
so the least value is 11
Second method:
→We can solve this one by using derivatives.
→Given function is f(x) = 2x^{2} 8x 3
→ f'(x)=4x8 (first derivative)
a f''(x)=4 (Second derivative)
Here we got constant value which means that it has minimal value at the point x=2
So we can find the minimum value by substituting value 2 in place of “x” in the given function.
Minimum value is 2⨉(2)^{2} 8⨉(2) 3= 11
Question 66 
1 degree  
1 radian  
90 degrees  
π radians 
It is not an SI unit, as the SI unit of angular measure is the radian, but it is mentioned in the SI brochure as an accepted unit.Because a full rotation equals 2π radians.
The radian (SI symbol rad) is the SI unit for measuring angles, and is the standard unit of angular measure used in many areas of mathematics. The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends;
Question 67 
32  
128  
160  
192 
→ Number of bit strings of length 8 that end with 00: 26 = 64.
→ Number of bit strings of length 8 that start with 1 and end with 00: 25 = 32.
→ Applying the subtraction rule, the number is 128+64−32 = 160
Question 68 
parabola  
hyperbola  
circle  
ellipse 
Question 69 
0  
1  
2  
3 
Question 70 
30  
25  
22.5  
27.5 
Mode = 3 Median – 2 Mean
15 = 3 Median  2(30)
Median = 25
Question 71 
x^{y}  
2^{ (x+y) }  
y^{x}  
y! / (yx)! 
Question 72 
1/6  
1/12  
1/144  
1/24 
Probability of both friends born in a same month is (1/12) * (1/12)
Suppose, they born in january = 1/12 * 1/12
Suppose, they born in february = 1/12 * 1/12
; ; Suppose, they born in December = 1/12 * 1/12
Probability that two friends are born in the same month is= 12*(1/12)*(1/12)
= 1/12
Question 73 
(0, 0, 0)  
(0, 1, 0)  
(0, 1, 0)  
(1, 1, 1) 
Question 74 
To guarantee correction of upto t errors, the minimum Hamming distance dmin in a block code must be
t+1  
t−2
 
2t−1  
2t+1 
Question 75 
Consider the following statements :

(a) False╞ True
(b) If α╞ (β ∧ γ) then α╞ β and α╞ γ.
Which of the following is correct with respect to the above statements ?
Both statement (a) and statement (b) are false.  
Statement (a) is true but statement (b) is false.  
Statement (a) is false but statement (b) is true.
 
Both statement (a) and statement (b) are true.

(b) α╞ (β ∧ γ) also write αV(β ∧ γ)
α╞ β and α╞ γ also write (αVβ)∧(αVγ)
Finally, we proved as LHS=RHS. αV(β ∧ γ) = (αVβ)∧(αVγ)
So, both the statements are correct.
Question 76 
Consider the following English sentence :
“Agra and Gwalior are both in India”.
A student has written a logical sentence for the above English sentence in FirstOrder Logic using predicate In(x, y), which means x is in y, as follows :
In(Agra, India) ⋁ In(Gwalior, India)
Which one of the following is correct with respect to the above logical sentence ?
It is syntactically valid but does not express the meaning of the English sentence.  
It is syntactically valid and expresses the meaning of the English sentence also.
 
It is syntactically invalid but expresses the meaning of the English sentence.  
It is syntactically invalid and does not express the meaning of the English sentence. 
• In(Gwalior, India) means Gwalior is in india.
• In(Agra, India) ⋁ In(Gwalior, India), in this statement “⋁” means “or” So the entire gives the meaning of Either Agra is in india or Gwalior is in india.
• The statement is not expressing the meaning of English sentence.
Question 77 
Consider the following two sentences :
 (a) The planning graph data structure can be used to give a better heuristic for a planning problem.
(b) Dropping negative effects from every action schema in a planning problem results in a relaxed problem.
Which of the following is correct with respect to the above sentences ?
Both sentence (a) and sentence (b) are false.  
Both sentence (a) and sentence (b) are true.
 
Sentence (a) is true but sentence (b) is false.
 
Sentence (a) is false but sentence (b) is true. 
• When these negative effects are dropped, then the number of actions increase and dropping all of the negative effects from the action schema results in a relaxed problem.
• A planning graph is a directed graph organized into levels: first a level S_{0} for the initial state, consisting of nodes representing each fluent that holds in S_{0}; then a level A_{0} consisting of nodes for each ground action that might be applicable in S_{0}; then alternating levels S_{i} followed by A_{i} ; until we reach a termination condition.
• As a tool for generating accurate heuristics, we can view the planning graph as a relaxed problem that is efficiently solvable.
Question 78 
A knowledge base contains just one sentence, ∃x AsHighAs (x, Everest). Consider the following two sentences obtained after applying existential instantiation.
 (a) AsHighAs (Everest, Everest)
(b) AsHighAs (Kilimanjaro, Everest)
Which of the following is correct with respect to the above sentences ?
Both sentence (a) and sentence (b) are sound conclusions.
 
Both sentence (a) and sentence (b) are unsound conclusions.  
Sentence (a) is sound but sentence (b) is unsound.
 
Sentence (a) is unsound but sentence (b) is sound.

• In the statement (a) AsHighAs (Everest, Everest), both are Everest then we can’t compare.
• The statement (b) AsHighAs (Kilimanjaro, Everest) means there kilimanjaro which is as highest as Everest So this valid statement.Because we are comparing Kilimanjaro with Everest.
Question 79 
Consider the set of all possible fivecard poker hands dealt fairly from a standard deck of fiftytwo cards. How many atomic events are there in the joint probability distribution ?
2,598,960
 
3,468,960  
3,958,590
 
2,645,590

Given random variables X, Y, …, that are defined on a probability space, the joint probability distribution for X, Y, … is a probability distribution that gives the probability that each of X, Y, … falls in any particular range or discrete set of values specified for that variable. In the case of only two random variables, this is called a bivariate distribution, but the concept generalizes to any number of random variables, giving a multivariate distribution.
Given data,
 Total cards = 52
 poker hand = 5 cards of all possibilities
Step1: This problem, we are simply finding combinations of ^{52}C_{5}
= C(n,r) = C(52,5)
= 52! / (5!(525)!)
= 2598960
Question 80 
Which of the following statements is false about convex minimization problem ?
If a local minimum exists, then it is a global minimum
 
The set of all global minima is convex set
 
The set of all global minima is concave set  
For each strictly convex function, if the function has a minimum, then the minimum is unique 
1. Every local minimum is a global minimum
2. The optimal set is convex
3. If the objective function is strictly convex, then the problem has at most one optimal point.
These results are used by the theory of convex minimization along with geometric notions from functional analysis (in Hilbert spaces) such as the Hilbert projection theorem, the separating hyperplane theorem, and Farkas' lemma.
Question 81 
The following LPP

Maximize z = 100x_{1} + 2x_{2} + 5x_{3}
Subject to 14x_{1} + x_{2} − 6x_{3} + 3x_{4} = 7
32x_{1} + x_{2} − 12x_{3} ≤ 10 3x_{1} − x_{2} − x_{3} ≤ 0 x_{1 }, x_{2} , x_{3} , x_{4} ≥ 0 has
Solution : x_{1} = 100, x_{2} = 0, x_{3} = 0
 
Unbounded solution  
No solution
 
Solution : x_{1} = 50, x_{2} = 70, x_{3} = 60

A linear programming problem is said to have unbounded solution if its solution can be made infinitely large without violating any of its constraints in the problem.
Question 82 
Digital data received from a sensor can fill up 0 to 32 buffers. Let the sample space be S = {0, 1, 2, .........., 32} where the sample j denote that j of the buffers are full and P(i) = 1/561 (33i).
Let A denote the event that the even number of buffers are full. Then p(A) is:0.515  
0.785  
0.758  
0.485 
• Probability of all even number of buffers are full is P(A).
• We are going find the values of P(0),P(2),P(4),P(6) …. P(16).
• P(0) is 1/562 (330)
• P(2) is 1/562 (332)
• …
• P(16) is 1/562 (3332)
• The probability of all even number of buffers are full is P(A) which is equal to P(0)+P(2)+P(4)+P(6)+ …. P(16).
• P(A) = 1/562(33+31+29+27+....+1)
• P(A) = 1562×289 = 0.51423
Question 83 
The equivalence of ¬∃ x Q(x) is :
∃x ¬Q(x)  
∀x ¬Q(x)  
¬∃ x ¬Q(x)  
∀x Q(x) 
Question 84 
Z  
Q  
R  
C 
Z number set:
• Z is the set of integers, ie. positive, negative or zero.
• Example: ..., 100, ..., 12, 11, 10, ..., 5, 4, 3, 2,  1, 0, 1, 2, 3, 4, 5, ... 10, 11, 12, ..., 100, ...
• The set N is included in the set Z (because all natural numbers are part of the relative integers).
• The set A consists of Positive, negative and zero numbers.
Question 85 
Match the following in ListI and ListII, for a function f :
ListI ListII (a) ∀x∀y(f(x)=f(y)⟶x=y) (i) Constant (b) ∀y∃x(f(x)=y) (ii) Injective (c) ∀xf(x)=k (iii) Surjective
(a)(i), (b)(ii), (c)(iii)
 
(a)(iii), (b)(ii), (c)(i)  
(a)(ii), (b)(i), (c)(iii)  
(a) (ii), (b)(iii), (c)(i)

x, x' ∈ X, f(x) = f(x') ⇒ x = x'
Or, equivalently (using logical transposition),
x, x' ∈ X, x ≠ x' f(x) ≠ f(x')
• The function is surjective (onto) if each element of the codomain is mapped to by at least one element of the domain. (That is, the image and the codomain of the function are equal.) A surjective function is a surjection. Notationally:
∀y ∈ Y, ∃x ∈ X such that y = f(x)
A constant function is a function whose (output) value is the same for every input value.
For example, the function y(x) = 4 is a constant function because the value y(x) is 4 regardless of the input value x.
Question 86 
Which of the relations on {0, 1, 2, 3} is an equivalence relation ?
{ (0, 0) (0, 2) (2, 0) (2, 2) (2, 3) (3, 2) (3, 3) }  
{ (0, 0) (1, 1) (2, 2) (3, 3) }
 
{ (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2) (2, 0) }  
{ (0, 0) (0, 2) (2, 3) (1, 1) (2, 2) }

1. Reflexivity: f(x) = f(x)
True, as given the same input, a function always produces the same output.
2. Symmetry: if f(x) = f(y) then f(y) = f(x)
True, by the definition of equality.
3. Transitivity: if f(x) = f(y) and f(y) = f(z) then f(x) = f(z)
True, by the definition of equality.
Option2: { (0,0), (1,1), (2,2), (3,3) }
Has all the properties, thus, is an equivalence relation.
Option1: { (0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3) }
Not reflexive: (1,1) is missing
Not transitive: (0,2) and (2,3) are in the relation, but not (0,3)
Option3: { (0,0), (0,1) (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3) }
Not symmetric: (1,2) is present, but not (2,1).
Not transitive: (2,0) and (0,1) are in the relation, but not (2,1).
Option4: Similarly, option4 also not TRUE
Question 87 
Which of the following is an equivalence relation on the set of all functions from Z to Z ?
{(f, g)  f(x)  g(x) = 1 ∀ x∈Z}
 
{(f, g)  f(0) = g(0) or f(1) = g(1)}  
{(f, g)  f(0) = g(1) and f(1) = g(0)}
 
{(f, g)  f(x)  g(x) = k for some k∈Z}

The relation "is equal to" is the canonical example of an equivalence relation, where for any objects a, b, and c:
a = a (reflexive property),
if a = b then b = a (symmetric property), and
if a = b and b = c then a = c (transitive property)
Question 88 
Which of the following statements is true ?
(Z, ≤) is not totally ordered
 
The set inclusion relation ⊆ is a partial ordering on the power set of a set S  
(Z, ≠) is a poset  
Option 2: This is TRUE, because the set inclusion relation ⊆ is a partial ordering on the power set of S.
Example: Suppose the set S = {a,b}, the maximum possibilities are {φ, a, b, ab}
Option 3: (z,≠)is false, because if violates reflexive relation.
Option 4: FALSE.
Question 89 
n^{3}  
2^{n(n1)/2}  
n½  
2^{(n1)/2} 
→ Each subset of these edges will be form a graph.
→ Number of possible undirected graphs is 2(^{n}C_{2}) 2^{(n(n1)/2)}
Question 90 
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 91 
0  
1  
15  
1 
So, For the given 3×3 matrix there would be 3 eigenvalues.
Given eigenvalues are : 2+i and 3.
So the third eigenvalue should be 2i.
As per the theorems, the determinant of the matrix is the product of the eigenvalues. So the determinant is (2+i)*(2i)*3 = 15.
Question 92 
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^{ (n2 n)}
For the given question n = 5.
The number of reflexive relations =2 ^{(255)} =2 ^{20}
Question 93 
9  
8  
7  
6 
Each is printed twice the no. of letters = 26×2 = 52
If Mala has k colours, she can have k pairs of same colours.
She also can have^{ k} C _{2} different pairs in which each pair is having different colours.
So total no. of pairs that can be coloured = k+ ^{ k } C _{2}
k+ ^{k} C _{2} ≥ 26
k+k(k1)/2 ≥ 26
k(k+1)/2 ≥ 26
k(k+1) ≥ 52
k(k+1) ≥ 7*8
k≥7
Question 94 
~p → q  
~p V q  
~p V ~q  
p → ~q 
Question 95 
Read ⋀ as AND, ⋁ as OR, ~ as NOT, → as one way implication and ⬌ as two way implication?
((x→ y)⋀ x)>y  
((~x→ y)⋀(~x⋀~y)) → x  
(x→ (x ⋁ y))  
((x ⋁ y) ⬌ (~x ⋁ ~y)) 
Question 96 
{a,b,ab,aa}  
{a,b,ba,bb}  
{a,b}  
{aa,ab,ba,bb} 
→ In option B and D, "ba" is presented which is generated by f. So, we can conclude based on this, option B and D are false.
→ "ab" can be generated by f but which is not present in the option C. So, we can conclude option C is wrong.
→ Option A satisfying all the favorable cases which is generated by f.
Question 97 
Consider the vocabulary with only four propositions A, B, C and D. How many models are there for the following sentence?
(⌐A∨⌐B∨⌐C∨⌐D)8  
7  
15  
16 
Question 98 
The number of substrings that can be formed from string given by
“a d e f b g h n m p” is
10  
45  
56  
55 
n*(n+1)/2 + 1
We have added 1 because it may include a NULL string also.
The number of substrings = 10*(11)/2 + 1
The number of substrings = 56
Question 99 
Match the List 1 and List 2 and choose the correct answer from the code given below
LIST 1 LIST 2 (a) Equivalence (i) p ⇒ q (b) Contrapositive (ii) p ⇒ q : q ⇒ p (c) Converse (iii) p ⇒ q : ∼q ⇒ ∼p (d) Implication (iv) p ⇔ qCode:
(a)(i), (b)(ii), (c)(iii), (d)(iv)
 
(a)(ii), (b)(i),(c)(iii), (d)(iv)  
(a)(iv), (b)(iii), (c)(ii), (d)(i)  
(a)(iii), (b)(iv), (c)(ii), (d)(i)

∼q⇒ ∼p
According to above table,
equivalence means p ⇔ q
Contrapositive means p⇒ q : ∼q⇒ ∼p
Converse means p ⇒ q : q ⇒ p
Implication means p ⇔ q
Question 100 
The Kcoloring of an undirected graph G = (V, E) is a function
C: V ➝ {0, 1, ......, K1} such that c(u)≠c(v) for every edge (u,v) ∈ E
Which of the following is not correct?
G has no cycles of odd length
 
G has cycle of odd length
 
G is 2colorable
 
G is bipartite

• A cycle of length n ≥ 3 is 2chromatic if n is even and 3chromatic if n is odd.
• A graph is bicolourable (2chromatic) if and only if it has no odd cycles.
• A nonempty graph G is bicolourable if and only if G is bipartite
Question 101 
If a graph (G) has no loops or parallel edges and if the number of vertices(n) in the graph is n≥3, then the graph G is Hamiltonian if
 (i) deg(v) ≥ n/3 for each vertex v
(ii) deg(v) + deg(w) ≥ n whenever v and w are not connected by an edge.
(iii) E(G) ≥ 1/3(n1)(n2) + 2
Choose the correct answer for the code given below:
Code:(i) and (iii) only
 
(ii) and (iii) only
 
(iii) only
 
(ii) only 
→ Dirac's theorem on Hamiltonian cycles, the statement that an nvertex graph in which each vertex has degree at least n/2 must have a Hamiltonian cycle.
→ Dirac's theorem on chordal graphs, the characterization of chordal graphs as graphs in which all minimal separators are cliques.
→ Dirac's theorem on cycles in kconnected graphs, the result that for every set of k vertices in a kvertexconnected graph there exists a cycle that passes through all the vertices in the set.
Question 102 
A survey has been conducted on methods of commuter travel. Each respondent was asked to check Bus, Train or Automobile as a major methods of travelling to work. More than one answer was permitted. The results reported were as follows :
Bus 30 people; Train 35 people; Automobile 100 people; Bus and Train 15 people; Bus and Automobile 15 people; Train and Automobile 20 people; and all the three methods 5 people. How many people complete the survey form ?
160  
120  
165  
115 
Question 103 
A full joint distribution for the Toothache, Cavity and Catch is given in the table below:
Which is the probability of Cavity, given evidence of Toothache ?
< 0.2, 0.8 >
 
< 0.6, 0.8 >
 
< 0.4, 0.8 >
 
< 0.6, 0.4 >

Question 104 
In mathematical logic, which of the following are statements ?
 (i) There will be snow in January
(ii) What is the time now ?
(iii) Today is Sunday
(iv) You must study Discrete Mathematics.
Choose the correct answer from the code given below :
Code :(iii) and (iv)  
(i) and (ii)
 
(i) and (iii)  
(ii) and (iv)

(a) a meaningful declarative sentence that is true or false, or
(b) the assertion that is made by a true or false declarative sentence.
From the above four statements, statement 2 and 4 wont give meaning like true or false answers and statements 1 and 3 will give either true or false answers.
Question 105 
Consider the statements below :
“There is a country that borders both India and Nepal.“
Which of the following represents the above sentence correctly ?
∃c Border(Country(c), India ∧ Nepal)
 
∃c Country(c) ∧ Border(c, India) ∧ Border(c, Nepal)
 
[∃c Country(c)] ⇒ [Border(c,India) ∧ Border(c, Nepal)]
 
∃c Country(c) ⇒ [ Border(c, India) ∧ Border(c, Nepal)]

→ Border(c, India) which represents border between c and india
→ Border(c, Nepal) which represents border between c and Nepal
→ “∧” represents both
Option2 represents the sentence “There is a country that borders both India and Nepal".
Question 106 
Use Dual Simplex Method to solve the following problem :
 Maximize z = 2x_{1}  3x_{2}
subject to:
x_{1} + x_{2} ≥ 2
2x_{1} + x_{2} ≤ 10
x_{1} + x_{2} ≤ 8
x_{1}, x_{2} ≥ 0
x_{1} = 6, x_{2} = 2 and z = 18  
x_{1} = 2, x_{2} = 6 and z = 22
 
x_{1} = 2, x_{2} = 0 and z = 4
 
x_{1} = 0, x_{2} = 2 and z = 6

→ Option 3 satisfies all conditions and maximize property in the given problem
Question 107 
A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls will be red and two will be green ?
1/35  
1/14
 
1/9  
3/7 
→ Green balls are 4
→ Total 10 balls in the box. We need to select 4 balls from 10 balls in which two red balls out of 6 red balls and two green balls from 4 green balls.
→ Total of number of ways for selecting 4 balls out of 10 balls is C(10,4)
→ Number of ways for selecting two red balls from 6 and two green balls from 4 is C(6,2)*C(4,2)
→ Probability of selecting two balls is number of ways for selecting four balls / total number of ways P = C(6,2)*C(4,2) / C(10,4)
= C(6,2) is 6x5/2 = 15
= C(4,2) is 4x3/2 = 6
= C(10,4) is (10x9x8x7) / (4x3x2x1) = 210
P = C(6,2)*C(4,2) / C(10,4)
= 15x6/210
= 3/7, So option 4 is correct
Question 108 
P: Number of odd degree vertices is even
Q: Sum of degrees of all vertices is even
P only  
Q only  
Both P and Q  
Neither P nor Q 
The sum of the degrees of all the vertices of a graph is an even number (exactly twice the number of edges).
In every graph, the number of vertices of odd degree must be even.
Question 109 
No bridges  
{d,e}  
{c,d}  
{c,d} and {c,f} 
Equivalently, an edge is a bridge if and only if it is not contained in any cycle.
A graph is said to be bridgeless or isthmusfree if it contains no bridges.
If we remove {d,e} edge then there is no way to reach e and the graph is disconnected.
The removal of edges {c,d} and {c,f} makes graph disconnect but this forms a cycle.
Question 110 
It has 7 vertices  
It is evenvalent(all vertices have even valence)  
It is not connected  
It does not have a Euler circuit 
Important Properties:
→ An undirected graph has an Eulerian cycle if and only if every vertex has even degree, and all of its vertices with nonzero degree belong to a single connected component.
→ An undirected graph can be decomposed into edgedisjoint cycles if and only if all of its vertices have even degree. So, a graph has an Eulerian cycle if and only if it can be decomposed into edgedisjoint cycles and its nonzerodegree vertices belong to a single connected component.
→ An undirected graph has an Eulerian trail if and only if exactly zero or two vertices have odd degree, and all of its vertices with nonzero degree belong to a single connected component.
→ A directed graph has an Eulerian cycle if and only if every vertex has equal in degree and out degree, and all of its vertices with nonzero degree belong to a single strongly connected component. Equivalently, a directed graph has an Eulerian cycle if and only if it can be decomposed into edgedisjoint directed cycles and all of its vertices with nonzero degree belong to a single strongly connected component.
→ A directed graph has an Eulerian trail if and only if at most one vertex has (outdegree) − (indegree) = 1, at most one vertex has (indegree) − (outdegree) = 1, every other vertex has equal indegree and outdegree, and all of its vertices with nonzero degree belong to a single connected component of the underlying undirected graph
Question 111 
ABCDCFDEFAEA  
AEDCBAF  
AEFDCBA  
AFCDEBA 
Here, Option A: A,F and E are repeated several times.
Option B: It is not a cycle. It means, not closed walk
Option C: It is closed walk and all vertex traversed. So this is final answer.
Option D: It’s not a closed walk.
Question 112 
e<=n  
e<=2n  
e<=en  
None of the option 
v − e + f = 2.
→ For a simple, connected, planar graph with v vertices and e edges and faces, the following simple conditions hold for v ≥ 3:
Theorem 1. e ≤ 3v − 6;
Theorem 2. If there are no cycles of length 3, then e ≤ 2v − 4.
Theorem 3. f ≤ 2v − 4.
Theorem (The handshaking theorem):
Let G be an undirected graph (or multigraph) with V vertices and N edges. Then
Exercise:
Suppose a simple graph has 15 edges, 3 vertices of degree 4, and all others of degree 3. How many vertices does the graph have?
3*4+(x3)*3=30
Question 113 
A simple graph which is isomorphic to hamiltonian graph  
A graph drawn in a plane in such a way that if the vertex set of graph can be partitioned into two nonempty disjoint subset X and Y in such a way that each edge of G has one end in X and one end in Y  
A graph drawn in a plane in such a way that any pair of edges meet only at their end vertices  
None of the option 
In other words, it can be drawn in such a way that no edges cross each other. Such a drawing is called a plane graph or planar embedding of the graph.
Question 114 
(p Vq) → q  
p V(q → p)  
pV(p → q)  
Both B) and C) 
Question 115 
F(x,y)=mx+by  
F(x,y)=mxy+b  
F(x,y)=mx+b  
F(x,y)=mxy+by 
F(x,y) = mx + b  y
For positive m at any given X,
If y is on the line, then F(x, y) = 0
If y is above the line, then F(x, y) < 0
If y is below the line, then F(x, y) > 0
Question 116 
1,2,3  
2,3,4  
2,4,5  
1,3,5 
→ In words, for any graph the sum of the degrees of the vertices equals twice the number of edges. Stated in a slightly different way, D_{v} =2e says that D_{v} is ALWAYS an even number
→ 1+2+3=6
Question 117 
A graph drawn in a plane in such a way that any pair of edges meet only at their end vertices  
A graph drawn in a plane in such a way that if the vertex set of graph can be partitioned into two nonempty disjoint subset X and Y in such a way that each edge of G has one end in X and one end in Y.  
A simple graph which is Isomorphism to Hamiltonian graph.  
A function between the vertex sets of two graphs that maps adjacent vertices to adjacent vertices 
→ In other words, it can be drawn in such a way that no edges cross each other. Such a drawing is called a plane graph or planar embedding of the graph.
→ A plane graph can be defined as a planar graph with a mapping from every node to a point on a plane, and from every edge to a plane curve on that plane, such that the extreme points of each curve are the points mapped from its end nodes, and all curves are disjoint except on their extreme points.
Question 118 
70  
280  
120  
45 
So, the total probability that all three edges of the above exists
= 1/2 × 1/2 × 1/2 (as they are independent events)
= 1/8
Total number of ways in which we can select ‘3’ such vertices among ‘16’ vertices = ^{16}C_{3} = 560
Total number of cycles of length ‘3’ out of 16 vertices = 560 × 1/8 = 70
Question 119 
1  
2  
3  
4 
p(0) = 5 ⇒ b = 5
p(1) = 4 ⇒ a+b = 4 ⇒ a = 1
p(2) = 9 ⇒ 40+b = 9 ⇒ 4+5 = 9, which is false.
So degree 1 is not possible.
Let's take p(x) = ax 2 + bx +c
p(0) = 5 ⇒ c = 5
p(1) = 4 ⇒ a+b+c = 4 ⇒ a+b = 1 (1)
p(2) = 9 ⇒ 4a+2b+c = 9 ⇒ 2a+b = 2 (2)
(2)  (1)
⇒ a = 3, b = 11 = 4
p(3) = 20 ⇒ 9a+3b+c = 20
⇒ 2712+5 = 20
⇒ 20 = 20, True
Hence, minimum degree it can have.
Question 120 
At least one of the set S_{ i }is a finite set  
not more than one of the sets S _{i} can be finite  
At least one of the sets S_{ i }is an infinite set  
not more than one of the sets S_{ i }can be infinite 
Question 121 
n1  
bn  
bn+1  
independent of "the number of nodes 
● A loop is said to be independent if it contains at least one branch which is not a part of any other independent loop.
● A network with ‘b’ branches, ‘n’ nodes and ‘L’ independent loops will satisfy the fundamental theorem of network topology.
b=L+n−1
L=b−n+1
Question 122 
a V b → b ⋀ c  
a ⋀ b → b V c  
a V b → (b → c)  
a → b → (b → c) 
Question 123 
Zero  
One  
n1  
n/2 
● A circuit in such a tree is impossible (unless you have some other data structure as well such as a linked list or you just program the tree so badly it ends up with circuits).
Question 124 
2chromatic  
(n/2) chromatic  
(n1) chromatic  
nChromatic 
Question 125 
2  
3  
4  
n2⌈(n/2)⌉+2 
Question 126 
P  
Q  
R  
True≅T 
Question 127 
( ⌐ A ∨ ⌐ B ∨ ⌐ C ∨ ⌐ D)
8  
7  
15  
16 
Question 128 
(a)(i), (b)(ii), (c)(iii), (d)(iv)  
(a)(ii), (b)(i),(c)(iii), (d)(iv)  
(a)(iv), (b)(iii), (c)(ii), (d)(i)  
(a)(iii), (b)(iv), (c)(ii), (d)(i) 
According to above table,
equivalence means p ⇔ q
Contrapositive means p⇒ q : ∼q⇒ ∼p
Converse means p⇒ q : q⇒ p
Implication means p ⇔ q
Question 129 
S 1 : A solution is a subtree that has a goal node at every leaf.
S 2 : OR nodes are analogous to the branching in a deterministic environment
S 3 : AND nodes are analogous to the branching in a nondeterministic environment.
Which one of the following is true referencing the above statements?
Choose the correct answer from the code given below:
S1 False, S2 True, S3 True  
S1 True, S2 True, S3 True  
S1 False, S2 True, S3 False  
S1 True, S2 True, S3 False 
● A solution in an ANDOR tree is a sub tree whose leaves are included in the goal set
Question 130 
G has no cycles of odd length  
G has cycle of odd length  
G is 2colorable  
G is bipartite 
● A cycle of length n ≥ 3 is 2chromatic if n is even and 3chromatic if n is odd.
● A graph is bicolourable (2chromatic) if and only if it has no odd cycles.
● A nonempty graph G is bicolourable if and only if G is bipartite
Question 131 
(i) deg(v) ≥n/3 for each vertex v
(ii) deg(v) + deg(w) ≥ n whenever v and w are not connected by an edge.
(iii) E (G) ≥ 1/3 (n − 1 )(n − 2 ) + 2
(i) and (iii) only  
(ii) and (iii) only  
(iii) only  
(ii) only 
→ Dirac's theorem on chordal graphs, the characterization of chordal graphs as graphs in which all minimal separators are cliques
→ Dirac's theorem on cycles in kconnected graphs, the result that for every set of k vertices in a kvertexconnected graph there exists a cycle that passes through all the vertices in the set
Question 132 
160  
120  
165  
115 
Let “B” denotes people travelled by Train.
Let “C” denotes people travelled by Automobile.
Let “(A⋂B)” denotes people travelled by Bus and Train.
Let “(A⋂C)” denotes people travelled by Bus and Automobile.
Let “(B⋂C)” denotes people travelled by Train and Automobile.
Let “(A⋂B⋂C)” denotes people travelled by Bus, Train and Automobile.
Number of people completed the survey = A+B+C(A⋂B)(A⋂C)(B⋂C)+(A⋂B⋂C)
= 30+35+100151520+5
= 120
Question 133 
Which is the probability of Cavity, given evidence of Toothache ?
< 0.2, 0.8 >  
< 0.6, 0.8 >  
< 0.4, 0.8 >  
< 0.6, 0.4 > 
P(Cavity/Toothache)=P(Cavity ⋀ Toothache) / P(toothache)
=(0.108 + 0.012) / (0.108 + 0.012 + 0.016 + 0.064)
=(0.12)/(0.2)
=0.6
P(~Cavity / Toothache) = (0.016 + 0.064) / 0.2
= 0.4
Question 134 
(i) There will be snow in January
(ii) What is the time now ?
(iii) Today is Sunday
(iv) You must study Discrete Mathematics.
(iii) and (iv)  
(i) and (ii)  
(i) and (iii)  
(ii) and (iv) 
(a) a meaningful declarative sentence that is true or false, or
(b) the assertion that is made by a true or false declarative sentence.
From the above four statements , statement 2 and 4 wont give meaning like true or false answers and statements 1 and 3 will give either true or false answers
Question 135 
“ There is a country that borders both India and Nepal. “
Which of the following represents the above sentence correctly ?
∃c Border(Country(c), India ∧ Nepal)  
∃c Country(c) ∧ Border(c, India) ∧ Border(c, Nepal)  
[∃c Country(c)] ⇒ [Border(c,India) ∧ Border(c, Nepal)]  
∃c Country(c) ⇒ [ Border(c, India) ∧ Border(c, Nepal)] 
→ Border(c, India) which represents border between c and india
→ Border(c, Nepal) which represents border between c and Nepal
→ “∧” represents both
Option2 represents the sentence “ There is a country that borders both India and Nepal.
Question 136 
Maximize z= 2x _{1} 3x_{ 2}
subject to:
x _{1} + x_{ 2} ≥ 2
2x _{1} + x_{ 2} ≤ 10
x _{1} + x_{ 2} ≤ 8
x _{1} , x_{ 2} ≥ 0
x _{1} =6, x_{ 2} =2 and z = 18  
x _{1} =2, x_{ 2} =6 and z = 22  
x _{1} =2, x_{ 2} =0 and z= 4  
x _{1} =0, x_{ 2} =2 and z= 6 
→ Option 3 satisfies all conditions and maximize property in the given problem
Question 137 
1/35  
1/14  
1/9  
3/7 
→ Green balls are 4
→ Total 10 balls in the box. We need to select 4 balls from 10 balls in which two red balls out of 6 red balls and two green balls from 4 green balls.
→ Total of number of ways for selecting 4 balls out of 10 balls is C(10,4)
→ Number of ways for selecting two red balls from 6 and two green balls from 4 is C(6,2)*C(4,2)
→ Probability of selecting two balls is number of ways for selecting four balls / total number of ways P= C(6,2)*C(4,2) / C(10,4) = C(6,2) is 6x5/2=15
= C(4,2) is 4x3/2=6
= C(10,4) is (10x9x8x7) / (4x3x2x1)=210
P=C(6,2)*C(4,2) / C(10,4)
=15x6/210=
= 3/7, So option 4 is correct
Question 138 
3  
4  
5  
6 
‘v’ is number of vertices and ‘e’ is number of edges
‘f’ is number of faces including bounded and unbounded
10  15 + f = 2
f = 7
There is always one unbounded face, so the number of bounded faces = 6
Question 139 
D(1)C(1)A(1)  
CDA  
ADC  
Does not necessarily exist 
ABCD = I
Multiply LHS, RHS by A^{ −1}
A ^{−1} ABCD = A^{ −1} I (position of A ^{ −1} on both sides should be left)
⇒ B CD = A ^{−1}
⇒ BCDD^{ −1} = A ^{−1} D ^{−1}
⇒ B C = A ^{−1} D ^{−1} ⇒BCC ^{−1} = A ^{−1}D ^{−1}C ^{−1}
⇒ B = A ^{−1} D ^{−1} C ^{−1}
⇒ B = A ^{−1} D ^{−1} C ^{−1}
Now, B ^{−1} = (A ^{−1} D ^{−1} C ^{−1} )^{−1}
B ^{−1} = C DA
Question 140 
One, at π/2  
One, at 3 π /2  
Two, at π /2 and 3 π /2  
Two, at π /4 and 3 π /2 
This is very obvious from the graph of f (x) = sin x
On a second look at the graph below, I believe x =π/4
is also a local minimum. This is because it is lesser than all other values within its locality.
Thus we have two local minima: x = π /4 , 3π
Question 141 
Commutativity  
Associativity  
Existence of inverse for every element  
Existence of identity 
1. CLOSURE: If a and b are in the group then a • b is also in the group.
2. ASSOCIATIVITY: If a, b and c are in the group then (a • b) • c = a • (b • c).
3. IDENTITY: There is an element e of the group such that for any element a of the group
a • e = e • a = a.
4. INVERSES: For any element a of the group there is an element a ^{1} such that ○ a • a ^{1} = e
and
○ a ^{1} • a = e
Question 142 
2 ^{n}  
2 ^{n 1}  
2 ^{n} 2  
2(2 ^{n} 2) 
n ^{m} – (2^{ n} – 2)
i.e., 2 ^{n} – (2^{ 2} – 2) = 2^{ n} – 2
If there are 'm' elements in set A, 'n' elements in set B then
The number of functions are : n ^{m} The number of injective or oneone functions are ^{n} P _{m}
The number of surjective functions are:
If m If m>n, then n! * ^{m} C_{ n}
Given that m=n, n=2
2! * ^{n} C_{ 2}
Question 143 
E  
2E  
V  
2V 
Suppose a graph has n vertices with degrees d _{1} , d _{2} , d_{ 3} , ..., d_{ n}.
Add together all degrees to get a new number
d _{1} + d _{2} + d_{ 3} + . .. + d_{ n} = D _{v} . Then D_{ v} = 2 e .
In words, for any graph the sum of the degrees of the vertices equals twice the number of edges.
Question 144 
Euler circuit  
Hamiltonian path  
Euler Path  
Hamiltonian Circuit 
● An Euler path is a path that uses every edge of a graph exactly once. An Euler circuit is a circuit that uses every edge of a graph exactly once.
● An Euler path starts and ends at different vertices.
● An Euler circuit starts and ends at the same vertex.
Question 145 
(a,b)(e,f)  
(a,b),(a,c)  
(c,d)(d,h)  
(a,b) 
● Equivalently, an edge is a bridge if and only if it is not contained in any cycle.
● The removal of edges (a,b) and (e,f) makes graph disconnected.
Question 146 
S1: The existence of an Euler circuit implies that an euler path exists.
S2: The existence of an Euler path implies that an Euler circuit exists.
S1 is true  
S2 is true  
S1 and S2 both are true  
S1 and S2 both are false 
An Euler path in G is a simple path containing every edge of G exactly once.
An Euler path starts and ends at different vertices.
An Euler circuit starts and ends at the same vertex.
Question 147 
5  
6  
7  
8 
Where V is the number of vertices, E is the number of edges, and R is the number of regions.
R=2V+E=28+13=7
Question 148 
One  
Two  
Zero  
Three 
● The empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero
● Common notations for the empty set include "{}", " ", and "∅".
Question 149 
{(1,a),(1,b),(2,a),(b,b)}  
{(1,1),(2,2),(a,a),(b,b)}  
{(1,a),(2,a),(1,b),(2,b)}  
{(1,1),(a,a),(2,a),(1,b)} 
● In the question, Set A consists of two elements and Set B consists of two elements. So the total ordered pairs are four.
● Each ordered pair consists of one element from SetA and another element from SetB.
Question 150 
8  
6  
7  
9 
Power set consists of {}, {0}, {1}, {2}, {0,1}, {0,2}, {1,2}, {0,1,2}
So power set consists of all unique items then cardinality is 8
Question 151 
6  
8  
9  
13 
● Where V is the number of vertices, E is the number of edges, and R is the number of regions.
● R=2V+E=213+19=8
● We can also term region as face
Question 152 
(P⋀Q)V(~P⋀Q)V(P⋀~Q) is equal to ~Q⋀~P  
(P⋀Q)V(~P⋀Q)V(P⋀~Q) is equal to QVP  
(P⋀Q)V(~P⋀Q)V(P⋀~Q) is equal to QV(P⋀~q)  
(P⋀Q)V(~P⋀Q)V(P⋀~Q) is equal to PV(Q⋀~p) 
The last two columns of the above table are different. So option A is false.
Question 153 
12  
7  
144  
264 
The number of bus lines between b and c =3
The number of possible combinations between a and c going through b are 4 * 3 = 12.
if you're talking about round trip, he has 12 possible ways to get there and 12 possible ways to get back, so the total possible ways is 12 * 12 = 144.
Question 154 
28  
48  
20  
None of the option 
So, we can construct 5*8 = 40 diagonals.
But we have constructed each diagonal twice, once from each of its ends. Thus there are 20 diagonals in a regular octagon.
Question 155 
Antisymmetric  
bisymmetric  
anti reflexive  
Asymmetric 
● R is antisymmetric if for all x,y A, if xRy and yRx, then x=y.
● R is a partial order relation if R is reflexive, antisymmetric and transitive.
Question 156 
B is a finite but not complemented lattice  
B is a finite, Complemented and distributive lattice  
B is a finite,distributive but not complemented lattice  
B is not distributive lattice 
(i)a.(b + c) =a.b + a.c
(ii) a+(b.c) = (a + b).(a + c)
Boolean algebra properites:
Question 157 
A= ∅  
B= ∅  
A != B  
A=B 
● AUB={1,2,3}
● A ∩ B ={1,2,3}
● If two sets consists of same elements then A U B = A ∩ B
Question 158 
Reflexive  
Transitive  
Symmetric  
Asymmetric 
The following conditions need to satisfy for each property.
a = a (reflexive property),
if a = b then b = a (symmetric property), and
if a = b and b = c then a = c (transitive property).
an asymmetric relation is a binary relation on a set X where:For all a and b in X, if a is related to b, then b is not related to a
The above relation is not reflexive because there is no ordered pairs (2,2) and (4,4)
The above relation is not Symmetric because if (1,2) present means the relation should consists of (2,1) but there is no such ordered pair in the relation.
Asymmetric property also invalid because of (1,3) and (3,1) ordered pairs.
Question 159 
4/52 X 4/52  
4/52 X 3/52  
4/52 X 3/51  
4/52 X 4/51 
E 2 : Last card being ace
Note that E 1 and E 2 are dependent events, i.e., probability of last card being ace if first is ace will be lesser than the probability of last card being ace if first card is not ace.
So, probability of first card being ace = 4/52
Probability of last card being ace given that first card is ace is,
P(E 2 / E 1 ) = 3/51
∴ P(E 1 and E 2 ) = P(E 1 ) ⋅ P(E 2 / E 1 ) = 4/52 * 3/51
Question 160 
aVb = a ⋀ b  
aV(bVc)=(aVb)Vc  
aV(b ⋀ c)=a  
aV(bVc)=b 
A distributive lattice is a lattice in which the operations of join and meet distribute over each other.
Question 161 
(2,6,24)  
(3,5,15)  
(2,9,16)  
(4,15,30) 
Question 162 
Let P, Q and R be three atomic prepositional assertions, and
 X : (P ∨ Q) → R
Y : (P → R) ∨ (Q → R)
Which one of the following is a tautology?
X → Y
 
Y → X  
X ≣ Y  
~Y → X

Question 163 
For what values of k, the points(k+1, 2k),(k, 22K) and (4k, 62k) are collinear?
0, 1
 
1, 1  
1, 1/2
 
1/2, 1/2

1/2[X1(Y2  Y3) + X2(Y3  Y1) + X3(Y1  Y2)] = 0
Here, X1 = k + 1, Y1 = 2k, X2 = k, Y2 = 2 – 2k, X3 = 4  k, Y3 = 6 – 2k
1/2[k + 1(2 – 2k  6 + 2k) + k(6 – 2k  2k)  4 k(2k  2 + 2k)] = 0
1/2[k + 1(4) + k(6  4k) 4  k(4k  2)] = 0
1/2[4k  4 + 6k  4k^{2}  16k + 8  4k^{2} + 2k)] = 0
1/2(8k^{2}  4k + 4) = 0
8k^{2}  4k + 4 = 0
8k^{2}  8k + 4k + 4 = 0
8K(k + 1) + 4(k + 1) = 0
(k + 1) (4  8k) = 0
k + 1 = 0
k = 1
4  8k = 0
k = 4/8
k = 1/2
So, the value of k is 1 and 1/2 .
Question 164 
If a connected graph G has planar embedding with 4 faces and 4 vertices, then what will be the number of edges in G?
7  
6  
4  
3 
For any(connected) planar graph with v vertices, e edges and faces, we have
V  E + F = 2
= 4  E + 4 =2
E = 4  2 + 4
E = 6
Question 165 
What is the area bounded by the parabola 2y = x^{2} and the line x = y  4?
18  
36  
72  
6 
and the line x = y – 4  (2)
Then y = x+4
Now, Substitute Y value in Equation(1)
x^{2} = 2 ( x + 4 )  (3)
Solving equation3 we get x = 4,  2.
Place Values of x in equation (1) and (2) we will get y = 8, 2.
Then the points of intersection are (8, 4), (2, –2).
After solving the integration, we will get 18.
Question 166 
What is the possible number of reflexive relations on a set of 5 elements?
2^{25}
 
2^{15}  
2^{10}
 
2^{20}

Step2: The possible number of reflexive relations on a set of 5 elements 2^{(n2n)} which is 2^{20} for n=5.
Question 167 
Which one of the following is most affected by the presence of outliers in sample data?
Variance
 
Mean
 
Median
 
Mode

For the sample data set:
1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4
We find the following mean, median, mode, and standard deviation:
Mean = 2.58
Median = 2.5
Mode = 2
Standard Deviation = 1.08
If we add an outlier to the data set:
1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 400
The new values of our statistics are:
Mean = 35.38
Median = 2.5
Mode = 2
Standard Deviation = 114.74
Note: Outliers often has a significant effect on your mean and standard deviation.
Question 168 
Consider the matrix A defined as follows:
What is the eigenvalue of 3A^{3 }+ 5A^{2 } 6A + 21, where I is an identity matrix?
4, 110, 10
 
1, 27, 8
 
1, 9, 4
 
4, 27, 9

Eigenvalues of A^{3} are 1, 27 and –8.
Eigenvalues of A^{2} are 1, 9 and 4.
Eigenvalues of A are 1, 3 and –2.
Eigenvalues of I are 1, 1 and 1.
∴ The eigenvalues of 3A^{3} + 5A^{2} – 6A + 2I.
First eigenvalue = 3(1) + 5(1) – 6(1) + 2(1) = 4
Second eigenvalue = 3(27) + 5(9) – 6(3) + 2(1) = 110
Third eigenvalue = 3(–8) + 5(4) – 6(–2) + 2(1) = 10
∴ The required eigenvalues are 4, 110, and 10.
Question 169 
If A and B are sets and AUB = A∩B, then which of the following is correct?
A=B  
A=∅
 
B=∅  
A⊂B

= x belongs to A intersection B
= x belongs to A and x Belongs to B
= x belongs to B
so A subset of B  (2)
Now we will let y belong to B which implies y belongs to A U B
= y belongs to A intersection B
= y belongs to A and y belongs to B
= y belongs to A
Therefore, B subset of A  (3) from (2) and (3) we get A=B.
Question 170 
Which of the following statements is/are correct?

(i) If the rank of the matrix of given vectors is equal to the number of vectors, then the vectors are linearly independent.
(ii) If the rank of the matrix of given vectors is less than the number of vectors, then the vectors are linearly dependent.
Both (i) and (ii)
 
Only (ii)
 
Only (i)  
Neither (i) nor (ii) 
→ You can think of an r x c matrix as a set of r row vectors, each having c elements; or you can think of it as a set of c column vectors, each having r elements.
→ The rank of a matrix is defined as (a) the maximum number of linearly independent column vectors in the matrix or (b) the maximum number of linearly independent row vectors in the matrix. Both definitions are equivalent.
For an r x c matrix,
1. If r is less than c, then the maximum rank of the matrix is r.
2. If r is greater than c, then the maximum rank of the matrix is c.
→ The rank of a matrix would be zero only if the matrix had no elements. If a matrix had even one element, its minimum rank would be one.
Question 171 
Every cut set of a connected euler graph has____
An odd number of edges
 
At least three edges
 
An even number of edges  
A single edge 
Question 172 
A fair coin is tossed 6 times. What is the probability that exactly two heads will occur?
11/32  
1/64  
15/64
 
63/64 
The probability can be calculated as:
n is a total number of experiments
k is an expected number of successes
p is the probability of a success
Step2: The first task can be simply solved by calculating the probability of 2 successes:
A "success" is "tossing heads in a single toss".
A "failure" is "tossing tails in a single toss".
Step3: The probability of a success is ½
The number of all experiments is n=6.
The number of successes is k=2
Question 173 
How many solutions does the equation x + y  z = 11 have, where x, y and z are non negative integers?
120  
78  
156  
130 
Here, n=3 and k=11, giving you
Question 174 
A graph G is dual if and only if G is a ___
Euler graph
 
Regular graph  
Complete graph  
Planar graph

→ The correspondence between edges of G and those of G* is as follows:
if e∈E(G) lies on the boundaries of faces X and Y, then the endpts of the dual edge e*∈(G*) are the vertices x and y that represent faces X and Y of G.
Question 175 
Which of the given options is the logical translation of the following statement, where F(x) and P(x) express the terms friend and perfect, respectively?
“None of my friends are perfect”?
∃x(~F(x)∧P(x))
 
∃x(F(x)∧ ~P(x))
 
~∃x(F(x)∧P(x))  
∃x(~F(x)∧~P(x))

P(x) represents x is perfect.
Given statement is “None of my friends are perfect”.
F(x)∧P(x) > X is both Friend and perfect.
~∃x > There is exist no one.
We can also write the above statement as follows
∀x[F(x) ⟹ ¬P(x)]
≡ ∀x[¬F(x) ∨ ¬P(x)]
≡∀x¬[F(x) ∧ P(x)]
≡¬∃x[F(x) ∧ P(x)]
Question 176 
The average rate of convergence of the bisection method is ____.
1/2  
3  
2  
1 
Question 177 
What is the chromatic number of an nvertex simple connected graph, which does NOT contain any oddlength cycle?
N  
N1  
2  
3 
→ A bipartite graph has the chromatic number 2.
Eg: Consider a square, which has 4 edges. It can be represented as bipartite ,with chromatic number 2.
Question 178 
Considering an undirected graph, which of the following statements is/are true?
 (i) Number of vertices of odd degree is always even
(ii) Sum of degrees of all the vertices is always even
Neither (i) nor (ii)
 
Both (i) and (ii)  
Only (ii)  
Only (i)

True: Sum of degrees of all the vertices is always even.
Question 179 
Select the option that best describes the relationship between the following graphs:
G and H are directed  
G and H are isomorphic
 
G and H are homomorphic  
G and H are not isomorphic 
1. V(G1) = V(G2)
2. E(G1) = E(G2)
3. Degree sequences of G1 and G2 are same.
4. If the vertices {V1, V2, ... Vk} form a cycle of length K in G1, then the vertices {f(V1), f(V2), … f(Vk)} should form a cycle of length K in G2.
→ Graph G and H are having same number of vertices and edges. So, it satisfied first rule.
→ u1 degree 2, u2 degree 3, u3 degree 3, u4 degree 2, u5 degree 3, u6 degree 3.
→ V1 degree 2, V2 degree 3, V3 degree 3, V4 degree 2, V5 degree 3, V6 degree 3.
→ But it violates the property of number of cycles. The number of cycles are not same.
Question 180 
Consider the following statements. Which one is/are correct?
 (i) The LU decomposition method fails if any of the diagonal elements of the matrix is zero.
(ii) The LU decomposition is guaranteed when the coefficient matrix is positive definite.
Both (i) and (ii)  
Only (i)
 
Neither (i) nor (ii)  
Only (ii) 
1. The LU decomposition method fails if any of the diagonal elements of the matrix is zero.
2. The LU decomposition is guaranteed when the coefficient matrix is positive definite.
Question 181 
Which of the following Hasse diagrams is are lattice(s)?
Only (a), (b) and (c)  
Only (a) and (b)
 
Only (b) and (c)
 
Only (a) and (c)

Question 182 
Choose the option that correctly matches each element of LIST1 with exactly one element of LIST2:
LIST1 LIST2 (i) NewtonRaphson method (a) Solving nonlinear equation (ii) Simpson’s rule (b) Solving ordinary differential equations (iii)RungeKutta method (c) Numerical integration (iv) Gauss elimination (d) Interpolation
(i)(b), (ii)(d), (iii)(a), (iv)(c)  
(i)(a), (ii)(b), (iii)(c), (iv)(d)
 
(i)(a), (ii)(c), (iii)(d), (iv)(b)
 
(i)(d), (ii)(c), (iii)(b), (iv)(a)

Simpson’s rule→ Numerical integration
RungeKutta method→ Solving ordinary differential equations
Gauss elimination→ Solving nonlinear equation
Question 183 
How many ways are there to arrange the nine letters of the word ALLAHABAD?
4560
 
4000
 
7560
 
7500

Step2: Total number of alphabets! / alphabets are repeating using formula (n!) / (r1! r2!..).
Step3: Repeating letters are 4A's and 2L's
= 9! / (4!2!)
= 7560
Question 184 
How many bit strings of length eight (either start with a 1 bit or end with the two bits 00) can be formed?
64  
255  
160  
128 
1. Number of bit strings of length eight that start with a 1 bit: 2^{7} = 128
2. Number of bit strings of length eight that end with bits 00: 2^{6} = 64
3. Number of bit strings of length eight that start with a 1 bit and end with bits 00: 2^{5} = 32
The number is 128+64+32 = 160
Question 185 
Which of the following statements are logically equivalent?
 (i) ∀x(P(x))
(ii) ~∃x(P(x))
(iii) ~∃x(~P(x))
(iv) ∃x(~P(x))
Only (ii) and (iv)  
Only (ii) and (iii)
 
Only (i) and (iv)
 
Only (i) and (ii) 
Question 186 
What is the solution of the recurrence relation a_{n} = 6a_{n1}  9a_{n2} with initial conditions a_{0}=1 and a_{1}=6?
a_{n} = 3^{n} + 3n^{2}
 
a_{n} = 3^{n} + n3^{n}
 
a_{n} = 3^{n} + 3n^{n}
 
a_{n} = n^{3} + n3^{n}

→ Suppose that the characteristic equation r^{k} −c_{1}r^{k}−1 −···−c_{k} = 0 has k distinct roots r_{1}, r_{2}, ..., r_{k}.
→ Then, a sequence {a_{n}} is a solution of the recurrence relation: a_{n} = c_{1}a_{n−1} + c_{2}a_{n−2} + ··· + c_{k}a_{n−k}
→ if and only if a_{n} = α_{1}r^{n}_{1} + α_{2}r^{n}_{2} + ··· + α_{k} r^{n}_{k} for n = 0,1,2,..., where α_{1}, α_{2}, ..., α_{k} are constants.
→ r^{2} − 6r + 9 = 0 has only 3 as a root.
→ So the format of the solution is a_{n} = α_{1}3^{n} + α_{2}n3^{n}.
→ Need to determine α_{1} and α_{2} from initial conditions:
a_{0} = 1 = α_{1}
a_{1} = 6 = α_{1}3 + α_{2}3
Solving these equations we get α_{1}=1 and α_{2}=1
Therefore, a_{n} = 3^{n} + n3^{n}.
Question 187 
Which of the following statements about the NewtonRaphson method is/are correct?
 (i) It is quadratic convergent
(ii) If f'(x) is zero, it fails
(iii) It is also used to obtain complex root
(i), (ii) and (iii)
 
only (i) and (iii)
 
only (i) and (ii)  
only (i)

Note: NewtonRaphson method is mainly used for Interpolation.
Question 188 
What is the minimum number of students in a class to be sure that three of them are born in the same month?
24  
12  
25  
13 
Question 189 
2  
3  
5  
4 
Question 190 
Cubic  
Quadratic  
Linear  
none 
Note:For the bisection you simply have that ε_{ i+1} /ε _{i} =1/2, so, by definition the order of convergence is 1 (linearly).
Question 191 
a _{r} =(3) ^{r} +(1)^{ r}  
2a _{r} =(2)^{ r} /3 (1)^{ r}  
a_{ r} =3^{ r+1} (1)^{ r}  
a _{r} =3(2)^{ r} (1)^{ r} 
For r =2, a _{2}= a_{ 21} +2a_{ 22} =a _{1} +2a_{ 0} =7+2*2=7+4=11
For r=3,a_{ 3}= a_{ 31} +2a_{ 32} =a_{ 2} +2a_{ 1} =11+2*7=11+14=25
From the above options , Substitute the r values 0,1,2,3 then option D gives the solution to recurrence relation.
Question 192 
(1.25,1.5)  
(1,1.25)  
(1,1.5)  
None of the options 
root= X^{4}X1.
Root lies Between 1 and 2,
After second iteration=?
Using bisection method.
f(1)=X^{4}X1
=111
= 1
f(2)=X^{4}X1
= 2^{4} 2 1
=13
Given constraint that “root lies between 1 and 2”
Iteration1: x1=(a+b)/2
=(1+2)/2
= 1.5
f(1.5) = 2.5625
Iteration2: x2=(a+b)/2
=(1+1.5)/2
=1.25
f(1.25)=0.19140625 >0
Root may lie in between (1, 1.25)
Algorithm  Bisection Scheme
Given a function f (x) continuous on an interval [a,b] and f (a) * f (b) < 0
Do
c=(a+b)/2
if f(a)*f(c)< 0 then b=c
else a=c
while (none of the convergence criteria C1, C2 or C3 is satisfied)
More info:
Bisection method is the simplest among all the numerical schemes to solve the transcendental equations. This scheme is based on the intermediate value theorem for continuous functions .
Consider a transcendental equation f(x)=0 which has a zero in the interval [a,b] and f(a)*f(b)<0. Bisection scheme computes the zero, say c, by repeatedly halving the interval [a,b]. That is, starting with
c = (a+b) / 2
The interval [a,b] is replaced either with [c,b] or with [a,c] depending on the sign of f (a) * f (c) . This process is continued until the zero is obtained. Since the zero is obtained numerically the value of c may not exactly match with all the decimal places of the analytical solution of f (x) = 0 in the interval [a,b]. Hence any one of the following mechanisms can be used to stop the bisection iterations :
C1. Fixing a priori the total number of bisection iterations N i.e., the length of the interval or the maximum error after N iterations in this case is less than  ba  / 2N.
C2. By testing the condition  c_{i}  c_{ i1} (where i are the iteration number) less than some tolerance limit, say epsilon, fixed a priori.
C3. By testing the condition  f (c_{i} )  less than some tolerance limit alpha again fixed a priori.
http://mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/bracketing%20methods/bisection/bisection.html
Question 193 
24  
27  
36  
38 
⇒162 =2×3×3×3×3 = 3^{3}×(2×3)
For (2×3) to be a perfect cube, it should be multiplied by (2^{2}×3^{2})
∴ Required number = y = 2^{2}×3^{2} = 36
Question 194 
(7!)^{2}  
(8!)^{2}  
7!8!  
15! 
→ Now place each girl between a pair of boys. So, total ways of seating arrangement of girls is 8!
→ Finally, 7!8! Possible ways are possible.
Question 195 
2  
1  
1/2  
1 
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/sup>) 4+2αCosθ=2(2Cosθ+α) 4+2αCosθ=4Cosθ+2α⇒Cosθ(uv)+2α4=0 42α=Cosθ(42α) (42α)(Cosθ1)=0 42α=0 }
Question 196 
15  
30  
56  
60 
8 vertices and distinct cycles length=5
Step1: To find number of distinct cycles
C(n,r)=C(8,5)
=8! / (5!(8−5)!)
= 56
Question 197 
"None of my friends are perfect"
~∃x(p(x) ⋀ q(x))  
∃x(~p(x) ⋀ q(x))  
∃x(~p(x) ⋀ ~q(x))  
∃x(p(x) ⋀ ~q(x)) 
P(x) ⇒ x is perfect
There doesn't exist any person who is my friend and perfect
Question 198 
269  
20  
271  
272 
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 199 
26  
24  
20  
22 
=(a*10+b)+(b*10+a)+(a*10+c)+(c*10+a)+(b*10+c)+(c*10+b)
=2(a+b+c)(10+1)
=22
Question 200 
M2 is non singular  
M2 is null Matrix  
M2 is the identity matrix  
M2 is transpose of M1 
The easiest way we can prove this is by looking at the determinant, since det(AB)=det(A)det(B)
and a matrix A is singular iff det(A)=0
Question 201 
n(n1)/2  
n^{n2}  
nx  
n 
→ Maximum number of edges are=n.
Question 202 
2^{15}  
2^{14}  
2^{13}  
2^{12} 
=2^{6(5)/2}
=2^{15}
Question 203 
20  
30  
40  
50 
Here as given, F=?,V=25 and E=17
→ F+2517=2
→ 40
Question 204 
5/32  
15/128  
35/128  
None of the options 
→ Past 6 tosses, What happen we don’t know. But according to 7th toss, we will find that head appeared for 3 times.
→ Any coin probability of a head=½ and probability of a tail=½
→ If we treat tossing of head as success,then this leads to a case of binomial distribution.
→ According to binomial distribution, the probability that in 6 trials we get 3 success is
^{6}C_{3}*(½)^{3} *(½)^{3}= 5/16 (3 success in 6 trials can happen in ^{6}C_{3} ways). → 7th toss, The probability of obtaining a head=1/2
→ In given question is not mention that what happens after 7 tosses.
→ Probability for that for 7th toss head appears exactly 4 times is =(5/16)*(1/2)
=5/32.
Question 205 
Maximum of n,d  
n+d  
nd  
nd/2 
d*n = 2*E
∴ E = (d*n)/2
Question 206 
Cubic  
Quadratic  
Linear  
none 
Note:For the bisection you simply have that ε _{i+1} /ε_{ i} =1/2, so, by definition the order of convergence is 1 (linearly).
Question 207 
6  
10  
12  
5,5 
A closed interval is an interval which includes all its limit points, and is denoted with square brackets. For example, [0,1] means greater than or equal to 0 and less than or equal to 1.
A halfopen interval includes only one of its endpoints, and is denoted by mixing the notations for open and closed intervals. (0,1] means greater than 0 and less than or equal to 1, while [0,1) means greater than or equal to 0 and less than 1.
Given function f(x)=2x^{ 2} 2x +6 and the given interval is [0,2]
According to the given interval , we need to check the function at the values 0,1,2.
f(0)=2x02x0+6=6
f(1)=2x1 ^{2} 2x1+6=22+6=6
f(2)=2x2^{ 2} 2x2+6=84+6=10
Question 208 
(x+2)/2  
2/(π2)  
π2  
π+2 
=[−( π/2) ^{2} cos( π/2 )+2 π/2 sin( π/2 )+2cos( π/2 )+C]  [−( 0) ^{2} cos( π/2 )+2(0)sin( π/2 )+2cos( 0 )+C]
=[0+ π+0+C2C]
=π2
Question 209 
a_{ r} =(3)_{ r} +(1)_{ r}  
2a _{r} =(2)^{ r} /3 (1)^{ r}  
a_{ r} =3^{ r+1} (1)^{ r}  
a _{r} =3(2)_{ r} (1)_{ r} 
● For r =2, a _{2}= a _{21} +2a_{ 22} =a _{1} +2a_{ 0} =7+2*2=7+4=11 ● For r=3,a 3= a 31 +2a 32 =a 2 +2a 1 =11+2*7=11+14=25
● From the above options ,Substitute the r values 0,1,2,3 then option D gives the solution to recurrence relation.
Question 210 
2  
3  
5  
4 
● Consider matrix order is 3, there are 3 rows and each row is multiplied by 2 means we need to multiply 8 to the existing determinant.
● The given existing determinant is 5 and each row multiplied by 2 means 8 *5 =40.
Question 211 
Hilbert curve order 3  
Hilbert curve order 2  
Hilbert curve of order 1  
Hilbert curve of order 4 
→ The path taken by a Hilbert Curve appears as a sequence  or a certain iteration  of up, down, left, and right.
Question 212 
If the mean of a poisson distribution is m, then standard deviation of the distribution is:
m^{2}  
m  
2*m  
√m 
Question 213 
The standard deviation of binomial distribution with n observations and probability of success p, probability of failure is q is:
√npq
 
Pq  
Np  
√pq 
The binomial distribution for a random variable X with parameters n and p represents the sum of n independent variables Z which may assume the values 0 or 1. If the probability that each Z variable assumes the value 1 is equal to p, then the mean of each variable is equal to l*p + 0*(lp) = p, and the variance is equal to p(lp).
By the addition properties for independent random variables, the mean and variance of the binomial distribution are equal to the sum of the means and variances of the n independent Z variables, so
These definitions are intuitively logical. Imagine, for example 8 flips of a coin. If the coin is fair, then p = 0.5. One would expect the mean number of heads to be half the flips, or np = 8*0.5 = 4.
The variance is equal to np(lp) = 8*0.5*0.5 = 2.
Question 214 
The normal curve is symmetrical about its:
Standard deviation  
Mean
 
Variance  
Probability

The probability density of the normal distribution is
Where
• μ is the mean or expectation of the distribution (and also its median and mode).
• σ is the standard deviation, and
• 2 is the variance
Question 215 
The density of uniform distribution over the interval ⍺ < a < b < ⍺ is given by:
f(x) = λe^{λx} , x>=0
 
f(x) = q^{k}p  
f(x) = 1/(ba), a  
f(x) = (⍺/c)x^{⍺1}

The values of f(x) at the two boundaries a and b are usually unimportant because they do not alter the values of the integrals of f(X) dX over any interval, not of X f(X) dX or any higher moment. Sometimes they are chosen to be zero, and sometimes chosen to be 1/(b – a). The latter is appropriate in the context of estimation by the method of maximum likelihood. In the context of Fourier analysis, one may lake the value of f(a) to be 12(b – a), since then the inverse transform of may integral transform of this uniform function will yield back the function itself, rather than a function which equal ‘almost everywhere’, i.e except on a set of points with zero measure. Also, it is consistent with the sign function which has no such ambiguity.
In terms of mean and variance σ^{2}, the probability density may be written as:
Question 216 
Exponential distribution is special case of ____ distribution.
Theta  
Alpha  
Beta  
Gamma 
Question 217 
0  
1  
1  
⍺ 
Question 218 
The mean, mode and median are connected by the empirical relationship:
Meanmode = 2(meanmedian)
 
Meanmode = 3(meanmedian)  
Meanmode = (meanmode)/2
 
Meanmode = (meanmode)/3

Mode = mean  3 [mean  median]
Mode = 3 median  2 mean
and Median = mode + ⅔ [meanmode]
Question 219 
If a random variable takes a finite set of values it is called:
Continuous variate  
Normal variate
 
Discrete variate
 
Exponential variate 
Examples:
number of students present
number of red marbles in a jar
number of heads when flipping three coins
students’ grade level
→ A discrete random variable X has a countable number of possible values.
Example:
Let X represent the sum of two dice.
→ A discrete random variable is one which may take on only a countable number of distinct values such as 0,1,2,3,4,........ Discrete random variables are usually (but not necessarily) counts. If a random variable can take only a finite number of distinct values, then it must be discrete.
Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's surgery, the number of defective light bulbs in a box of ten.
Question 220 
The root mean square deviation when measured from the mean is:
Greatest
 
Positive
 
Least
 
Negative

Question 221 
The values which divide the frequency into four equal parts are called:
Coefficient of variance
 
Range  
Dispersion  
Quartiles

Question 222 
How many 3 digit numbers are there with all different odd digits?
16  
48  
54  
60 
• But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)
• And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)
• 5 * 4 * 3 = 60
Question 223 
In how many ways can a committee of 4 people be chosen from a group of 12?
495  
595
 
395
 
295

C(n,r) = C(12,4)
= 12! / [(4!(12−4)!)]
= 495
Hence, a committee of 4 people be selected from a group of 12 people in 495 ways.
Question 224 
A straight line which cuts a curve on two points at an infinite distance from the origin and yet is not itself wholly at infinity is called:
Spiral
 
Asymptote
 
Parallel
 
Polar

Question 225 
If k parallel lines of a determinant Δ become identical when x=a, then ____ is a factor of Δ.
(xa)+(k+1)
 
(xa)/(k1)
 
(xa)*(k+1)
 
(xa)^{(k1)}

• In general, if k rows (or k columns) become identical (x=a) when a is substituted for x, then (xa)^{r1} is a factor of D.
Question 226 
A minimal subgraph G’ of G such that V(G’)=V(G) and G’ is connected is called:
A spanning tree
 
A connected graph
 
A directed graph
 
A biconnected component 
Question 227 
Rank of nonsingular square matrix of order r is:
r  
0  
r1  
1 
• Rank of singular matrix is less than “r”.
Question 228 
The newton’s Raphson iterative formula for finding 1/N is:
½(x_{n} + N/x_{n})  
x_{n}(1  Nx_{n})
 
½(x_{n}+1/Nx_{n})  
1/k((k1)x_{n} + N/x_{n}^{k1})

• The iterations x_{k+1} = x_{k} − ( f(x_{k})/ f ′(x_{k}) ) are called Newton’s iterations.
Question 229 
Portability is not a quality factor of:
Software coding
 
Software design
 
Software Process  
Software testing

This model classifies all software requirements into 11 software quality factors. The 11 factors are grouped into three categories – product operation, product revision, and product transition factors.
1. Product operation factors − Correctness, Reliability, Efficiency, Integrity, Usability.
2. Product revision factors − Maintainability, Flexibility, Testability.
3. Product transition factors − Portability, Reusability, Interoperability.
Question 230 
The number of strips required in simpson’s 3/8^{th} rule is a multiple of:
1  
2  
3  
6 
Area = 3h/ 8 [( a + 3 b + 3 c + d )]
Simpson's Second Rule:
Multipliers:
Question 231 
A declarative sentence which is either true(1) or false(0) is called:
Lattice
 
Tautology
 
Contradiction
 
Proposition

• Sentences that assert a fact that could either be true or false.
Question 232 
The points at which the function attains extreme values are called:
Turning points  
End points
 
Higher points  
Extreme points

Question 233 
If f(x) = ax^{2} + bx + c the f(x(b/2a)) is:
An even function for all a except a=0
 
An even function for all a
 
Neither even nor odd
 
An odd function for all a except a=0 
• A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(x) = f(x) for all x in the domain of f.
Question 234 
The sum of n terms of 1/(1*2) + 1/(2*3) + 1/(3*4) + ... is
(n+1)/n
 
n/(n+1)
 
n/(2n+1)
 
(2n+1)/n

where
1st term = 1/(1*2)
2nd term = 1/(2*3)
3rd term = 1/(3*4)
.
.
.
.
nth term = 1/(n*(n+1))
nth term = 1/(n*(n+1))
i.e. the kth term is of the form 1/(k*(k+1))
which can further be written as kth term = 1/k  1/(k+1)
So, sum upto n terms can be calculated as:
(1/1  1/1+1) + (1/2  1/2+1) + (1/3  1/3+1) + ......... + (1/n1  /1n) + (1/n  1/n+1)
= (1  1/2) + (1/2  1/3) + (1/3  1/4) + ......... + (1/n1  1/n) + (1/n  1/n+1)
= 1  1/n+1
= ((n+1)  1)/n+1
= n/n+1
Question 235 
How many integers are between 1 and 200 which are divisible by any one of the integers 2,3 and 5(Hint: use set operation)?
125  
145  
146  
136 
B) numbers divisible by 3: 200/3 = 66
C) numbers divisible by 5: 200/5 = 40
Counting twice:
AB) numbers divisible by 6: 200/6 = 33
AC) numbers divisible by 10: 200/10 = 20
BC) numbers divisible by 15: 200/15 = 13
Counting 3 times:
ABC) numbers divisible by 30: 200/30 = 6
Total of numbers = A + B + C  AB  AC  BC + ABC = 100 + 66 + 40  33  20 13 + 6 = 146
Question 236 
In algebra of logic, the conjunction of two tautologies is:
Contradiction
 
Tautology
 
Negation  
Disjunction

1. The negation of a contradiction is a tautology.
2. The disjunction of two contingencies can be a tautology.
3. The conjunction of two tautologies is a tautology.
Question 237 
Harmonic function  
Laplace equation
 
Wave equation
 
Homogeneous

Note that the wave equation can be factored as
Question 238 
The number of two digit numbers divisible by the product of digits is:
8  
14  
13  
5 
12=1*2=2 which is a factor of 12
15=1*5=5 which is a factor of 15
24=2*4=8 which is a factor of 24
36=3*6=12 which is a factor of 36
Question 239 
If a relation < from A={1,2,3,4} to B={1,3,5} i.e., (a,b)∈R if a < b, then R^{1} is:
{(1,3)(1,5)(2,3)(2,5)(3,5)(4,5)}
 
{(3,1)(5,1)(3,2)(5,2)(5,3)(5,4)}  
{(3,3)(3,5)(5,3)(5,5)}
 
{(3,3)(3,4)(4,5)} 
Question 240 
If a lattice (L,R) has a greatest and least element then it is said to be:
Sub Lattice  
Complemented Lattice
 
Bounded Lattice
 
Distributive Lattice

→ A bounded lattice is an algebraic structure of the form (L, ∨, ∧, 0, 1) such that (L, ∨, ∧) is a lattice, 0 (the lattice's bottom) is the identity element for the join operation ∨, and 1 (the lattice top) is the identity element for the meet operation ∧.
→ Let 'L' be a lattice w.r.t R if there exists an element I∈L such that (aRI)∀x∈L, then I is called Upper Bound of a Lattice L.
Similarly, if there exists an element O∈L such that (ORa)∀a∈L, then O is called Lower Bound of Lattice L.
Question 241 
The sum of the series: (1/2) + (1/3) + (1/4)  (1/6) + (1/8) + (1/9) + (1/16)  (1/12) + ... + α is:
(1/2)  log(√2)^{3}
 
1 + log(√2)^{3}
 
3/2  
1/2 + log(√2)^{3}

Question 242 
Find the boolean product A⊙B of the two matrices.
The boolean product truth table is
According to the truth table, we have to perform matrix multiplication.
Question 243 
720  
120  
60  
360 
BANANA= 6 letters
B is appearing 1 time
A is appearing 3 times
N is appearing 2 times
So, = 6! / (3! * 2!)
= 60
Question 244 
(I) P ↔ ¬ Q
(II) ¬ P ↔ Q
(III) ¬ P ↔ ¬ Q
(IV) Q → P
Only (I) and (II)  
Only (II) and (III)  
Only (III) and (IV)  
None of the above 
So, (I) and (Ii) are TRUE.
Question 245 
∃x ⌐H(x)  
∀x ⌐H(x)  
∀x H(x)  
⌐x H(x) 
Question 246 
F _{00} (n) = ((10 ∗ F _{00} (n – 1) + 100)/ F _{00} (n – 2)) for n ≥ 2 Then what shall be the set of values of the sequence F_{ 00} ?
(1, 110, 1200)  
(1, 110, 600, 1200)  
(1, 2, 55, 110, 600, 1200)  
(1, 55, 110, 600, 1200) 
Sequence F _{00} defined as
F _{00} (0) = 1,
F _{00} (1) = 1,
F _{00} (n) = ((10 ∗ F _{00} (n – 1) + 100)/ F _{00} (n – 2)) for n ≥ 2
Let n=2
F _{00} (2) = (10 * F _{00} (1) + 100) / F _{00} (2 – 2)
= (10 * 1 + 100) / 1
= (10 + 100) / 1
= 110
Let n=3
F _{00} (3) = (10 * F _{00} (2) + 100) / F _{00} (3 – 2)
= (10 * 110 + 100) / 1
= (1100 + 100) / 1
= 1200
Similarly, n=4
F _{00} (4) = (10 * F _{00} (3) + 1_{00}) / F _{00} (4 – 2)
= (12100) / 110
= 110
F _{00} (5) = (10 * F _{00} (4) + 100) / F _{00} (5 – 2)
= (10*110 + 100) / 1200
= 1
The sequence will be (1, 110, 1200,110, 1).
Question 247 
(x ^{2} + 1)4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ 4}  
(x ^{2} + 1)4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ 4}  
(x ^{2} + 1)4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ 4}  
(x ^{2} + 1)4 and [(x^{ 3} – 4x)^{ 2} + 1]^{ 4} 
f(x) = x ^{3} – 4x, g(x) = 1/(x^{ 2} + 1) and h(x) = x ^{4}
hog(x)=h(1/(x ^{2} + 1))
=h(1/(x ^{2} )+1) ^{4}
= 1/(x ^{2} +1) ^{4}
= (x ^{2} +1) ^{4}
hogof(x)= hog(x ^{3} 4x)
= h(1/(x ^{3} 4x)^{ 2} +1)
= h(1/(x ^{3} 4x)^{ 2} +1) 4
= h((x ^{3} 4x) ^{2} +1)^{ 4}
So, option D id is correct answer.
Question 248 
0 and 100 and –6 and 34
16 and 6  
17 and 6  
17 and 7  
16 and 7 
0 and 100 → Counting sequentially:
0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 Total=17
–6 and 34 → Counting sequentially: 6,0,6,12,18,24,30
Total=7
Method2: 0 and 100 → Maximum number is 100. Divide ⌊100/6⌋ = 16+1 =17
Here, +1 because of 0.
–6 and 34 → Maximum number is 34. Divide ⌊34/6⌋ = 5+1+1 =7
Here, +1 because of 0 and +1 for 6
Question 249 
deg(v) ≥n/2 for each vertex v.  
E(G) ≥1/2(n – 1) (n – 2) + 2  
deg (v) + deg(w) ≥ n whenever v and w are not connected by an edge  
All of the above 
Question 250 
P ∨ R  
P ∨ S  
Q ∨ R  
Q ∨ S  
None of These 
OptionB: Let P be TRUE and S be false then the conclusion PVS is TRUE. Now, if we make R as TRUE then the premises (P→ Q)∧(R→ S) will be false because (R→ S) will be false. Hence this option is not correct.
OptionC: Let Q be false, R be TRUE then conclusion QVR will be TRUE. Now if we make S as FALSE then Premises (P→ Q)∧(R→ S) will be FALSE because (R→ S) will be false. Hence, this option is not correct.
OptionD: Let Q be TRUE and S be FALSE then conclusion QVS will be TRUE. Now if we make R as TRUE then premises (P→ Q)∧(R→ S) will be FALSE because (R→ S) will be false.
Therefore None of the given options are correct.
Note: As per UGC NET key, given option D as correct answer.
Question 251 
A – B ≤ A ⊕ B ≤ A + B ≤ A ∪ B  
A ⊕ B ≤ A – B ≤ A ∪ B ≤ A + B  
A ⊕ B ≤ A + B ≤ A – B ≤ A ∪ B  
A – B ≤ A ⊕ B ≤ A ∪ B ≤ A + B 
Step2: A–B we can also write into AA ∩ B and equivalent Venn diagram is
Step3: A⊕B We can also write into A+B 2A∩B and equivalent Venn diagram is
Step4: A + B We can represented into A + B and equivalent Venn diagram is
Step5: A∪B We can also write into A+B A∩B and equivalent Venn diagram is
Step6: A – B ≤ A ⊕ B ≤ A ∪ B ≤ A + B is correct order.
Question 252 
1/64  
1/32  
1/8  
1⁄4 
Ex: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121...,
Step1: Here, string length is 10. If we consider first 5 numbers as random choices like 0 or 1.
Remaining 5 numbers are fixed. Total number of possibilities are 2^{ 10} . But we are only considering first 5 choices. The probability is 2 ^{5} .
Step2: The probability 2 ^{5} /2^{ 10}
= 1/2 ^{5}
= 1/32
Question 253 
Which of the following is correct?
G _{1} contains Euler circuit and G _{2} does not contain Euler circuit.  
G _{1} does not contain Euler circuit and G _{2} contains Euler circuit.  
Both G _{1} and G_{ 2} do not contain Euler circuit.  
Both G_{ 1} and G_{ 2} contain Euler circuit. 
Step1: G1 have odd number of vertices. So, it is not euler circuit.
Step2: G2 also have odd number of vertices. So, it not euler circuit.
Question 254 
10  
15  
25  
30 
Step2: The number of combinations for three equivalence classes are
2,2,1 chosen in ( ^{5} C_{ 2} *^{ 3} C _{2} *^{ 1} C_{ 1} )/2! = 15
3,1,1 chosen in( ^{5} C_{ 2} *^{ 3} C _{2} *^{ 1} C_{ 1} )/2! = 10
Step3: Total differential classes are 15+10
=25.
Question 255 
14, 14  
16, 14  
16, 4  
14, 4 
=n ^{(n2)}
= 4^{ 2}
=16
Step2: Given Bipartite graph K_{ 2,2} . To find number of spanning tree in a bipartite graph K _{m,n} having standard formula is m^{ (n1)} * n^{ (m1)} .
m=2 and n=2
= 2 ^{(21)} * 2^{ (21)}
= 2 * 2
= 4
Question 256 
R _{1} ∩ R_{ 2} is reflexive and R_{ 1} ∪ R_{ 2} is irreflexive.  
R _{1} ∩ R_{ 2} is irreflexive and R_{ 1} ∪ R_{ 2} is reflexive.  
Both R _{1} ∩ R_{ 2} and R_{ 1} ∪ R_{ 2} are reflexive.  
Both R _{1} ∩ R_{ 2} and R_{ 1} ∪ R_{ 2} are irreflexive. 
Ex: Let set A={0,1}
R _{1} ={(0,0),(1,1)} all diagonal elements we are considering for reflexive relation.
R _{2} ={(0,0),(1,1)} all diagonal elements we are considering for reflexive relation.
R_{ 1} ∩ R_{ 2} must have {(0,0),(1,1)} is reflexive.
R _{1} ∪ R_{ 2} must have {(0,0),(1,1)} is reflexive.
Question 257 
3/4  
2/3  
1/2  
1⁄3 
 3 cards in a box
 1 st card: Both sides of one card is black. The card having 2 sides. We can write it as BB.
 2 nd card: Both sides of one card is red. The card having 2 sides. We can write it as RR.
 3rd card: one black side and one red side. We can write it as BR.
Step1: The probability that the opposite side is the same colour as the one side we observed is 2⁄3 because total number of cards are 3
Question 258 
2  
4  
5  
6 
Definition: A clique in a simple undirected graph is a complete subgraph that is not contained in any larger complete subgraph.
Step1: b,c,e,f is complete graph.
Step2: ‘a’ is not connected to ‘e’ and ‘b’ is not connected to ‘d’. So, it is not complete graph.
Question 259 
(a) A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree.
(b) A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.
(c) A complete graph (K_{ n} ) has a Hamilton Circuit whenever n ≥ 3.
(d) A cycle over six vertices (C _{6} ) is not a bipartite graph but a complete graph over 3 vertices is bipartite.
(a) only  
(b) and (c)  
(c) only  
(d) only 
(b)TRUE: A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.
(c)TRUE: A complete graph (K _{n} ) has a Hamilton Circuit whenever n ≥ 3. (d) FALSE: A cycle over six vertices (C_{ 6} ) is not a bipartite graph but a complete graph over 3 vertices is bipartite.
Question 260 
(a) The set of negative integers is countable.
(b) The set of integers that are multiples of 7 is countable.
(c)The set of even integers is countable.
(d)The set of real numbers between 0 and 1⁄2 is countable.
(a) and (c)  
(b) and (d)  
(b) only  
(d) only 
Suppose negative integers set size is 10.
Ex: 1, 2, 3,.....,10 is countable
(b)TRUE: The set of integers that are multiples of 7 is countable.
Suppose set of integers size is 10.
Ex: 1*7, 2*7, 3*7, .....,10*7 is countable
(c)TRUE: The set of even integers is countable.
Suppose set of even integers size is 10.
Ex: 2,4,6,8,10,....,20
(d) FALSE: The set of real numbers between 0 and 1⁄2 is countable. We can’t count real numbers.
Ex: 0.1, 0.2, 0.3, .....,0.∞
Question 261 
Consider the graph given below: The two distinct sets of vertices, which make the graph bipartite are:
(v _{1} , v_{ 4} , v_{ 6} ); (v_{ 2} , v_{ }3 , v_{ 5} , v_{ 7} , v_{ 8} )  
(v _{1} , v_{ 7} , v_{ 8} ); (v_{ 2} , v_{ 3} , v_{ 5} , v_{ 6} )  
(v _{1} , v_{ 4} , v_{ 6} , v_{ 7} ); (v_{ 2} , v_{ 3} , v_{ 5} , v_{ 8} )  
(v _{1} , v_{ 4} , v_{ 6} , v_{ 7} , v_{ 8} ); (v_{ 2} , v_{ 3} , v_{ 5} ) 
→ The two sets U and V may be thought of as a coloring of the graph with two colors.
Option A: FALSE because V _{5} , V_{ 7} and V_{ 3} are adjacent. So, it not not bipartite graph.
OptionB FALSE because V _{5} , V_{ 6} and V_{ 2} are adjacent. So, it not not bipartite graph.
OptionC TRUE because it follows properties of bipartied and no two colours are adjacent.
OptionD FALSE because because V _{4} , V_{ 6} and V_{ 8} are adjacent. So, it not not bipartite graph.
Question 262 
(a)
(b)
(c)
(a) and (b)  
(b) and (c)  
(a) and (c)  
(a), (b) and (c) 
Question 263 
(a) “If Gora gets the job and works hard, then he will be promoted. If Gora gets promotion, then he will be happy. He will not be happy, therefore, either he will not get the job or he will not work hard”.
(b) “Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty”.
(c) If n is a real number such that n >1, then n^{ 2} >1. Suppose that n^{ 2} >1, then n >1.
(a) and (c)  
(b) and (c)  
(a), (b) and (c)  
(a) and (b) 
Question 264 
(i)(ii)(iii)(iv)  
(ii)(iii)(i)(iv)  
(iii)(ii)(iv)(i)  
(iv)(iii)(ii)(i) 
→ Trivial proof is a proof that the implication p → q is true based on the fact that q is true.
→ Direct proof is a proof that the implication p → q is true that proceeds by showing that q must be true when p is true.
→ Indirect proof is a proof that the implication p → q is true that proceeds by showing that p must be false when q is false.
Question 265 
(a)p ∨ ~(p ∧ q)
(b)(p ∧ ~q) ∨ ~(p ∧ q)
(c)p ∧ (q ∨ r)
Which of the above propositions are tautologies?
(a) and (c)  
(b) and (c)  
(a) and (b)  
only (a) 
Question 266 
(a)The distributive property
(b)The commutative property
(c)The symmetric property
(a) and (b)  
(b) and (c)  
(a) only  
(b) only 
Closure: For all a, b in A, the result of the operation a • b is also in A.
Associativity: For all a, b and c in A, the equation (a • b) • c = a • (b • c) holds.
Identity element: There exists an element e in A, such that for all elements a in A, the equation e • a = a • e = a holds.
Inverse element: For each a in A, there exists an element b in A such that a • b = b • a = e, where e is the identity element.
Commutativity: For all a, b in A, a • b = b • a.
A group in which the group operation is not commutative is called a "nonabelian group" or "noncommutative group".
Question 267 
4960  
2600  
23751  
8855 
 The equation x+y+z+u = 29
 Constraints are x ≥ 1, y ≥ 2, z ≥ 3 and u ≥ 0
 Possible solutions?
Step1: Combine all constraints (1+2+3+0) =6
Step2: Subtract all constraints values with total equation number 296 =23.
Possibility to get repetition of a numbers of x,y and z but no chance for ‘u’ because its value is 0.
Step3: So, Subtract repeated values into total equation value
= 293
= 26
Step4: Possible solutions= ^{26} C_{ 23}
= 2600
Question 268 
66  
330  
495  
99 
→ There is no specification about positive,negative digits and also repetition of digits.
→ We are assuming the positive digits which is greater than or equal to 0.
→ The starting digit of the string should not be zero. We will also consider the repetition of digits.
→ Example of such strings are 70000,61000,60100,60010..,
→ The possible number of strings are C((n+r1),(r1))
→ From the given data n=7,r=5 then We have to fine C(11,4) which is equal to 330
Question 269 
22 / 36  
12 / 36  
14 / 36  
6 / 36 
→ The possible ways are 1,2,3,4,5,6
→ The possible ways of one number divides another number are (1,1) (1,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (5,5) and (6,6).
→ The probability is the number of outcomes / total outcomes = 14/36
Question 270 
1001  
3876  
775  
200 
→ From the given question, n=15, r=5 We need to calculate C(14,4) and it’s value is 1001.
Question 271 
(a) deg (v) ≥ n / 2 for each vertex of G
(b) E(G) ≥ 1 / 2 (n  1) (n  2) + 2 edges
(c) deg (v) + deg (w) ≥ n for every n and v not connected by an edge.
(a) and (b)  
(b) and (c)  
(a) and (c)  
(a), (b) and (c) 
→ According to ore’s theorem,Let G be a (finite and simple) graph with n ≥ 3 vertices. We denote by deg v the degree of a vertex v in G, i.e. the number of incident edges in G to v. Then, Ore's theorem states that if deg v + deg w ≥ n for every pair of distinct non adjacent vertices v and w of G, then G is Hamiltonian.
→ A complete graph G of n vertices has n(n−1)/2 edges and a Hamiltonian cycle in G contains n edges. Therefore the number of edgedisjoint Hamiltonian cycles in G cannot exceed (n − 1)/2. When n is odd, we show there are (n − 1)/2 edgedisjoint Hamiltonian cycles.
So the statement(b) is false and statement a and c are true.
Question 272 
Convert this argument into logical notations using the variables c, b, r, p for propositions of computations, electric bills, out of money and the power respectively. (Where ¬ means NOT)
if (c Λ b)→r and ¬b→p, then (¬r Λ p)→¬c  
if (c ∨ b)→r and ¬b→¬p, then (r Λ p)→c  
if (c Λ b)→r and ¬p→b, then (¬r ∨ p)→¬c  
if (c ∨ b)→r and ¬b→¬p, then (¬r Λ p)→¬c 
“c” for my computations are correct
“b” for I pay the electric bill.
“r” for I will run out of money
“p” for the power is on.
(c Λ b) means my computations are correct and I pay the electric bill.
(¬r Λ p) means I don’t run out of money and the power is still on.
According to the statement , the option (A) is correct.
Question 273 
(a)(i), (b)(ii), (c)(iii), (d)(iv)  
(a)(ii), (b)(iii), (c)(i), (d)(iv)  
(a)(iii), (b)(ii), (c)(iv), (d)(i)  
(a)(iv), (b)(ii), (c)(iii), (d)(i) 
→ The equivalence relation translates verbally into "if and only if" and is symbolized by a doublelined, double arrow pointing to the left and right (↔). If A and B represent statements, then A ↔ B means "A if and only if B."
→ The exportation rule is a rule in logic which states that "if (P and Q), then R" is equivalent to "if P then (if Q then R)".
→ The exportation rule may be formally stated as: (P ∧ Q) → R is equivalent to P → (Q →R)
→ A mode of argumentation or a form of argument in which a proposition is disproven by following its implications logically to an absurd conclusion. Arguments that use universals such as, “always”, “never”, “everyone”, “nobody”, etc., are prone to being reduced to absurd conclusions. The fallacy is in the argument that could be reduced to absurdity  so in essence, reductio ad absurdum is a technique to expose the fallacy.
Question 274 
“ x ≥ 6, if x ^{2} ≥ 5 and its proof as:
If x ≥ 6, then x 2 = x.x ≥ 6.6 = 36 ≥ 25
Which of the following is correct w.r.to the given proposition and its proof?
(a)The proof shows the converse of what is to be proved.
(b)The proof starts by assuming what is to be shown.
(c)The proof is correct and there is nothing wrong.
(a) only  
(c) only  
(a) and (b)  
(b) only 
Question 275 
(a)(ii), (b)(iii), (c)(iv), (d)(i)  
(a)(iii), (b)(iv), (c)(ii), (d)(i)  
(a)(iv), (b)(i), (c)(iii), (d)(ii)  
(a)(iv), (b)(iii), (c)(ii), (d)(i) 
Mnemonic Table → Contains machine OP code
Segment Register Table → Uses array data structure
EQU → Assembler directive.
The EQU directive gives a symbolic name to a numeric constant, a registerrelative value or a PCrelative value.
Question 276 
Identity law  
De Morgan’ s law  
Idempotent law  
Complement law 
(i). (A V B)’ = A' ∧ B'
(ii). (A ∧ B)’ = A' V B'
→ Identity Law :
(i). 1 AND A = A
(ii). 0 OR A = A
→ Complement law:
(i). A AND A'=1
(ii). A OR A'=0
→ Idempotent law:
The idempotence in the context of elements of algebras that remain invariant when raised to a positive integer power, and literally means "(the quality of having) the same power", from idem + potence (same + power).
(i). A V A=A
(ii). A ∧ A=A
According to boolean algebra
Question 277 
g  
g+1  
g^{ 4}  
None of the above 
f(x)=x+1
g(x)=x+3
Constraint is f0 f0 f0 f
Step1: We can write into fo fo fo f is f(f(f(x+1)))
We can write into f(f(x+2)) and f(x+3).
Step2: Above constraint is equal to "x+4" because f(x+3)+1
Step3: We can also write into fog(x)=x+4 and gof(x)=x+4.
So, g+1 is appropriate answer.
Question 278 
(i) 1, 2, 3, 4, 5
(ii) 3, 4, 5, 6, 7
(iii) 1, 4, 5, 8, 6
(iv) 3, 4, 5, 6
then
(i) and (ii)  
(iii) and (iv)  
(iii) and (ii)  
(ii) and (iv) 
1. Sum of degrees of the vertices of a graph should be even.
2. Sum of degrees of the vertices of a graph is equal to twice the number of edges.
Statement(i) is violating property1.
= 1+2+3+4+5
= 15 is odd number.
Statement(ii) is violating property1.
= 3+4+5+6+7
= 25 is odd number.
Statement(iii) is violating property1.
= 1+4+5+8+6
= 24 is even number
Statement(iv) is violating property1
= 3+4+5+6
= 18 is even number
Question 279 
(i) Z _{23}
(ii) Z _{29}
(iii) Z _{31}
(iv) Z _{33}
Then
(i) only  
(i) and (ii) only  
(i), (ii) and (iii) only  
(i), (ii), (iii) and (iv) 
Question 280 
T is a tree  
T contains no cycles  
Every pairs of vertices in T is connected by exactly one path  
All of these 
Step1:
n= number of vertices
n1 = number of edges
Example: n=5 vertices and n1=4 edges
Step2: The above graph T won’t have cycle then we are calling as tree. Here, every pairs of vertices in T is connected by exactly one path.
Note: The above properties is nothing but minimum spanning tree properties.
Question 281 
True  
Multi  Valued  