SetTheory
Question 1 
7 
If ‘H” is a subgroup of finite group (G,*) then O(H) is the divisor of O(G).
Given that the order of group is 35. Its divisors are 1,5,7,35.
It is asked that the size of largest possible subgroup other than G itself will be 7.
Question 2 
41  
42  
43  
44 
For any group ‘G’ with order ‘n’, every subgroup ‘H’ has order ‘k’ such that ‘n’ is divisible by ‘k’.
Solution:
Given order n = 84
Then the order of subgroups = {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}
As per the proper subgroup definition, it should be “42”.
Question 3 
Consider the set X = {a,b,c,d e} under the partial ordering

R = {(a,a),(a,b),(a,c),(a,d),(a,e),(b,b),(b,c),(b,e),(c,c),(c,e),(d,d),(d,e),(e,e)}.
The Hasse diagram of the partial order (X,R) is shown below.
The minimum number of ordered pairs that need to be added to R to make (X,R) a lattice is _________.
0  
1  
2  
3 
As per the definition of lattice, each pair should have GLB, LUB.
The given ‘R’ has GLB, LUB for each and every pair.
So, no need to add extra pair.
Thus no. of required pair such that Hasse diagram to become lattice is “0”.
Question 4 
A function f:N^{+} → N^{+}, deﬁned on the set of positive integers N^{+}, satisﬁes the following properties:

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

P: R is reﬂexive
Q: Ris transitive
Which one of the following statements is TRUE?
Both P and Q are true.  
P is true and Q is false.  
P is false and Q is true.  
Both P and Q are false. 
a≤c ∨ b≤d
Let a≤a ∨ b≤b is true for all a,b ∈ N
So there exists (a,a) ∀ a∈N.
It is Reflexive relation.
Consider an example
c = (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f)
This does not hold for any (a>e) or (b>f)
eg:
(2,2)R(1,2) as 2≤2
(1,2)R(1,1) as 1≤1
but (2,2) R (1,1) is False
So, Not transitive.
Question 6 
Consider a set U of 23 different compounds in a Chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U. Consider the following statements:

I. Each compound in U\S reacts with an odd number of compounds.
II. At least one compound in U\S reacts with an odd number of compounds.
III. Each compound in U\S reacts with an even number of compounds.
Which one of the above statements is ALWAYS TRUE?
Only I  
Only II  
Only III  
None 
U = 23
∃S ∋ (S⊂U)
Each component in ‘S’ reacts with exactly ‘3’ compounds of U,
If a component ‘a’ reacts with ‘b’, then it is obvious that ‘b’ also reacts with ‘a’.
It’s a kind of symmetric relation.>br> If we connect the react able compounds, it will be an undirected graph.
The sum of degree of vertices = 9 × 3 = 27
But, in the graph of ‘23’ vertices the sum of degree of vertices should be even because
(d_{i} = degree of vertex i.e., = no. of edges)
But ‘27’ is not an even number.
To make it an even, one odd number should be added.
So, there exists atleast one compound in U/S reacts with an odd number of compounds.
Question 7 
pr = 0  
pr = 1  
0 < pr ≤ 1/5  
1/5 < pr < 1 
Let A be the event that an element (x,y,z)∈ L^{3} satisfies x ∨(y∧z) = (x∨y) ∧ (x∨z) Since q∨(r∧s) = q∨p = q
and (q∨r)∧(q∨s)=t∧t=t q∨(r∧s)≠(q∨r)∧(q∨s)
Therefore, (x,y,z) = (q,r,s),(q,s,r),(r,q,s),(r,s,q),(s,r,q),(s,q,r)
i.e., 3! = 6 elements will not satisfy distributive law and all other (x,y,z) choices satisfy distributive law
n(A) = 1256 = 119
∴ required probability is 119/125
⇒ 1/5
Question 8 
Both {f} and {g} are functionally complete  
Only {f} is functionally complete
 
Only {g} is functionally complete
 
Neither {f} nor {g} is functionally complete 
f(X,X,X)=X'XX'+XX'+X'X'
=0+0+X'
=X'
Similarly, f(Y,Y,Y)=Y' and f(X,Z,Z)=Z'
f(Y',Y',Z')=(Y')'Y'Z'+Y'(Y')'+(Y')'(Z')'
=YY'Z'+Y'Y+YZ
=0+0+YZ
=YZ
We have derived NOT and AND. So f(X,Y,Z) is functionally complete.
g(X,Y,Z)=X'YZ+X'YZ'+XY
g(X,X,X)=X'XX+X'XZ'+XX
=0+0+X
=X
Similarly, g(Y,Y,Y)=Y and g(Z,Z,Z)=Z
NOT is not derived. Hence, g is not functionally complete.
Question 9 
R is symmetric and reflexive but not transitive  
R is reflexive but not symmetric and not transitive  
R is transitive but not reflexive and not symmetric  
R is symmetric but not reflexive and not transitive 
In aRb, 'a' and 'b' are distinct. So it can never be reflexive.
Symmetric:
In aRb, if 'a' and 'b' have common divisor other than 1, then bRa, i.e., 'b' and 'a' also will have common divisor other than 1. So, yes symmetric.
Transitive:
Take (3, 6) and (6, 2) elements of R. For transitivity (3, 2) must be the element of R, but 3 and 2 don't have a common divisor. So not transitive.
Question 10 
2046  
2047  
2048  
2049 
Question 11 
Suppose U is the power set of the set S = {1, 2, 3, 4, 5, 6}. For any T ∈ U, let T denote the number of element in T and T' denote the complement of T. For any T, R ∈ U, let T \ R be the set of all elements in T which are not in R. Which one of the following is true?
∀X ∈ U (X = X')  
∃X ∈ U ∃Y ∈ U (X = 5, Y = 5 and X ∩ Y = ∅)  
∀X ∈ U ∀Y ∈ U (X = 2, Y = 3 and X \ Y = ∅)  
∀X ∈ U ∀Y ∈ U (X \ Y = Y' \ X') 
(A) False. Consider X = {1,2}. Therefore, X' = {3,4,5,6}, X = 2 and X' = 4.
(B) False. Because for any two possible subsets of S with 5 elements should have atleast 4 elements in commonc. Hence X∩Y cannot be null.
(C) False. Consider X = {1,4}, Y= {1,2,3} then X\Y = {4} which is not null.
(D) True. Take any possible cases.
Question 12 
Both reflexive and symmetric  
Reflexive but not symmetric  
Not reflexive but symmetric  
Neither reflexive nor symmetric 
∴(p,q) R (p,q)
⇒ R is not reflexive.
Let (p,q) R (r,s) then ps = qr
⇒ rq = sp
⇒ (r,s) R (p,q)
⇒ R is symmetric.
Question 13 
S1: There is a subset of S that is larger than every other subset. S2: There is a subset of S that is smaller than every other subset.Which one of the following is CORRECT?
Both S1 and S2 are true  
S1 is true and S2 is false  
S2 is true and S1 is false  
Neither S1 nor S2 is true 
U⊂S, V⊂S
Let U = {1, 2, 3}
V = {2, 3, 4}
Symmetric difference:
(U – V) ∪ (V – U) = {1} ∪ {4} = {1, 4}
The minimum element in the symmetric difference is 1 and 1∈U.
S1: Let S = Universal set = {1, 2, … 2014}
This universal set is larger than every other subset.
S2: Null set is smaller than every other set.
Let U = { }, V = {1}
Symmetric difference = ({ } – {1}) ∪ ({1} – { }) = { } ∪ {1} = {1}
So, U < V because { } ∈ U.
Question 14 
For any subsets A and B of X, f(A ∪ B) = f(A)+f(B)  
For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B)  
For any subsets A and B of X, f(A ∩ B) = min{ f(A),f(B)}  
For any subsets S and T of Y, f^{ 1} (S ∩ T) = f^{ 1} (S) ∩ f^{ 1} (T) 
We need to consider subsets of 'x', which are A & B (A, B can have common elements are exclusive).
Similarly S, T are subsets of 'y'.
To be a function, each element should be mapped with only one element.
(a) f(A∪B)=f(A)+f(B)
{a,b,c}∪{c,d,e} = {a,b,c} + {c,d,e}
{a,b,c,d,e} = 3+3
5 = 6 FALSE
(d) To get inverse, the function should be oneone & onto.
The above diagram fulfills it. So we can proceed with inverse.
f^{1} (S∩T )=f^{1} (S)∩f^{1} (T)
f^{1} (c)=f^{1} ({a,b,c})∩f^{1} ({c,d,e})
2={1,2,3}∩{2,4,5}
2=2 TRUE
Question 15 
5  
6  
7  
8 
So, 15 is divided by {1, 3, 5, 15}.
As minimum is 4 and total is 15, we eliminate 1,3,15.
Answer is 5.
Question 16 
2  
3  
4  
5 
For eg: a two dimensional vector space have x, y axis. For dimensional vector space, it have x, y, z axis.
In the same manner, 6 dimensional vector space has x, y, z, p, q, r (assume).
Any subspace of it, with 4 dimensional subspace consists any 4 of the above. Then their intersection will be atmost 2.
[{x,y,z,p} ∩ {r,q,p,z}] = #2
V_{1} ∩ V_{2} = V_{1} + V_{2}  V_{1} ∪ V_{2} = 4+4+(6) = 2
Question 17 
P, Q and R are true  
Only Q and R are true  
Only P and Q are true  
Only R is true 
So f(i)should be resulting only {0, 1, …2014}
So, every element in range has a result value to domain. This is onto. (Option R is correct)
We have ‘2015’ elements in domain.
So atleast one element can have f(i) = i,
so option ‘Q’ is also True.
∴ Q, R are correct.
Question 18 
In any planar embedding, the number of faces is at least n/2+ 2  
In any planar embedding, the number of faces is less than n/2+ 2  
There is a planar embedding in which the number of faces is less than n/2+ 2  
There is a planar embedding in which the number of faces is at most n/(δ+1) 
v – e + f = 2 →①
Point ① degree of each vertex is minimum ‘3’.
3×n≥2e
e≤3n/2
From ①
n3n/2+f=2⇒
Question 19 
Commutative but not associative  
Both commutative and associative  
Associative but not commutative  
Neither commutative nor associative 
A binary relation on a set S is called cumulative if a*b = b*a ∀ x,y∈S.
Associative property:
A binary relation on set is called associative if (a*b)*c = a*(b*c) ∀ x,y∈S.
Given x⊕y = x^{2} + y^{2} (1)
Replace x, y in (1)
y⊕x = y^{2} + x^{2} which is same as (1), so this is cumulative
(x⊕y)⊕z = (x^{2} + y^{2}) ⊕ z
= (x^{2} + y^{2}) + z^{2}
= x^{2} + y^{2} + z^{2} + 2x^{2}y^{2} (2)
x⊕(y ⊕ z) = x ⊕ (y^{2} + z^{2})
= x^{2} + (y^{2} + z^{2})^{2}
= x^{2} + y^{2} + z^{2} + 2y^{2z2  (3) (2) & (3) are not same so this is not associative. }
Question 20 
Let f be a function from a set A to a set B, g a function from B to C, and h a function from A to C, such that h(a) = g(f(a)) for all a ∈ A. Which of the following statements is always true for all such functions f and g?
g is onto ⇒ h is onto  
h is onto ⇒ f is onto  
h is onto ⇒ g is onto  
h is onto ⇒ f and g are onto 
If h: A→C is a onto function, the composition must be onto, but the first function in the composition need to be onto.
So, B→C is must be onto.
Question 21 
n  
n + 1  
2^{(n1)} + 1  
n! 
Let A = {a, b, c}, here n = 3
Now, P(A) = {Ø, {a}, {b}, {c}, {a,b}, {b,c}, {{a}, {a,b,c}}
Now C will be contain Ø (empty set) and {a,b,c} (set itself) as Ø is the subset of every set. And every other subset is the subset of {a,b,c}.
Now taking the subset of cardinality, we an take any 1 of {a}, {b}, {c} as none of the set is subset of other.
Let's take {2}
→ Now taking the sets of cardinality 2 {a,b}, {b,c}
→ {b} ⊂ {a,b} and {b,c} but we can't take both as none of the 2 is subset of the other.
→ So let's take {c,a}.
So, C = {Ø, {b}, {b,c}, {a,b,c}}
→ So, if we observe carefully, we can see that we can select only 1 set from the subsets of each cardinality 1 to n
i.e., total n subsets + Ø = n + 1 subsets of A can be there in C.
→ So, even though we can have different combinations of subsets in C but maximum cardinality of C will be n+1 only.
Question 22 
p(p  1)  
pq  
(p^{2}  1)(q  1)  
p(p  1)(q  1) 
→ No. of multiples of p in n = pq [n = p⋅p⋅q]
→ No. of multiples of q in n = p^{2} [n = p^{2}q]
→ Prime factorization of n contains only p & q.
→ gcd(m,n) is to be multiple of p and (or) 1.
→ So, no. of possible m such that gcd(m,n) is 1 will be
n  number of multiples of either p (or) q
= n  p^{2}  pq + p
= p^{2}q  p^{2}  pq + p
= p(pq  p  q + 1)
= p(p  1)(q  1)
Question 23 
2  
4  
8  
None of the above 
P(S) = {φ, {{φ}}, {1}, {{2, 3}}, {{φ}, 1}, {1, {2, 3}}, {{φ}, 1, {2, 3}}}
In P(S) it contains 8 elements.
Question 24 
2^{n1} 
And so, no. of subsets with odd cardinality is half of total no. of subsets = 2^{n} /n = 2^{n1}
Question 25 
at least one of the set S_{i} is a finite set  
not more than one of the set S_{i} can be finite  
at least one of the sets S_{i} is an infinite set  
None of the above 
Question 26 
2^{2n}  
2^{n2}  
(2^{n})^{2}  
(2^{2})^{n} 
Question 27 
disjoint 