Question 1 
Commutative but not associative  
Both commutative and associative  
Associative but not commutative  
Neither commutative nor associative 
Discuss it
Question 2 
A group  
A ring  
An integral domain  
A field 
Discuss it
Question 3 
Commutativity  
Associativity  
Existence of inverse for every element  
Existence of identity 
Discuss it
Question 4 
R is symmetric but NOT antisymmetric  
R is NOT symmetric but antisymmetric  
R is both symmetric and antisymmetric  
R is neither symmetric nor antisymmetric 
Discuss it
Question 5 
* 
a 
b 
c 
d 
a 
a 
b 
c 
d 
b 
b 
a 
d 
c 
c 
c 
d 
b 
a 
d 
d 
c 
a 
b 
a, b are generators  
b, c are generators  
c, d are generators  
d, a are generators 
Discuss it
Check for all: a^{1} = a , a^{2} = a * a = a a^{3} = a^{2} * a = a * a = a a is not the generator since we are not able to express other members of the group in powers of a Check for c  c^{1} = c c^{2} = c * c = b c^{3} = c^{2} * c = b * c = d c^{4} = c^{2} * c^{2} = b * b = a We are able to generate all the members of the group from c , Hence c is the generator Similarly check for d
Question 6 
Q^{c} U R^{c}  
P U Q^{c} U R^{c}  
P^{c} U Q^{c} U R^{c}  
U 
Discuss it
Question 7 
n and n  
n^{2} and n  
n^{2} and 0  
n and 1 
Discuss it
Consider an example set, S = (1,2,3) Equivalence property follows, reflexive, symmetric and transitive Largest ordered set are s x s = { (1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3) } which are 9 which equal to 3^2 = n^2 Smallest ordered set are { (1,1) (2,2) ( 3,3)} which are 3 and equals to n. number of elements.
Question 8 
2  
3  
4  
5 
Discuss it
2 can be written as 2 power 2. Number of partitioning of 2 = no. of non isomorphic abelian groups 2 can be partitioned as {(2),(1,1)}
Question 10 
A  
B  
C  
D 
Discuss it
Question 11 
z^{2xy}  
z x 2 ^{xy}  
z^{2x + y}  
2^{xyz} 
Discuss it
So number of functions from Z to E =
Question 12 
It is not closed  
2 does not have an inverse  
3 does not have an inverse  
8 does not have an inverse 
Discuss it
Question 13 
Neither a Partial Order nor an Equivalence Relation  
A Partial Order but not a Total Order  
A Total Order  
An Equivalence Relation 
Discuss it
Solution:
An equivalence relation on a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are 1. Reflexive: a R a for all a Є R, 2. Symmetric: a R b implies that b R a for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R.
An partial order relation on a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are
1. Reflexive: a R a for all a Є R, 2. AntiSymmetric: a R b and b R a implies that for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R.
An total order relation a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are
1. Reflexive: a R a for all a Є R, 2. Symmetric: a R b implies that b R a for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R. 4. Comparability : either a R b or b R a for all a,b Є R.
As given in question, a relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v , reflexive property is not satisfied here , because there is > or < relationship between (x ,y) pair set and (u,v) pair set . Other way , if there would have been x <= u and y>= v (or x=u and y=v) kind of relation amongs elements of sets then reflexive property could have been satisfied. Since reflexive property in not satisfied here , so given realtion can not be equivalence ,partial order or total order relation.So ,Answer (A) is true .
This solution is contributed by Nirmal Bharadwaj.
Question 14 
12  
13  
15  
16 
Discuss it
The given mapping S is defined by f:{0,1}^4 > {0,1} . So, number of functions from S will be 2^16. Now N is defined by f : S> {0,1}. So Number of functions from S to {0,1} will be 2^S. Hence log_{2}log_{2}N = log_{2}S = 16
Question 15 
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 
Discuss it
S1 is true because NULL set is smaller than every other set.
S2 is true because the UNIVERSAL set {1, 2, ..., 2014} is larger than every other set.
Thus, both S1 and S2 are true.
Please comment below if you find anything wrong in the above post.
Question 16 
A  
B  
C  
D 
Discuss it
Let x = {a, b, c} and y = {1, 2} A Function f maps each element of x to 1 in y. f(a)=1 , f(b)=1 , f(c) =1 A = {a, b} B = {b, c}  A ]  f(A u B)  = f({a, b, c}) = 3  f(A)+f(B) = 2 + 2 = 4 , LHS != RHS.  B ] f(A ∩ B) = f({b}) = { 1 } f(A) ∩ f(B) = {1, 1} ∩ {1, 1} = {1, 1} LHS != RHS  C ] f(A ∩ B) = f({b}) = { 1 } = 1 min{f(A),f(B)} = min(2,2) = 2 LHS != RHS  D ] In a function a value can be mapped only to one value.
Question 17 
3  
5  
7  
9 
Discuss it
Question 18 
1  
2  
3  
4 
Discuss it
Question 19 
x ∗ x = y ∗ y = x ∗ y ∗ x ∗ y = y ∗ x ∗ y ∗ x = ewhere e is the identity element. The maximum number of elements in such a group is __________.
2  
3  
4  
5 
Discuss it
x * x = e, x is its own inverse y * y = e, y is its own inverse (x*y) * (x* y) = e, x*y is its own inverse (y*x) * (y*x) = e, y*x is its own inverse also x*x*e = e*e can be rewritten as follows x*y*y*x = e*y*y*e = e, (Since y *y = e) (x*y) * (y*x) = e shows that (x *y) and (y *x) are each other’s inverse and we already know that (x*y) and (y*x) are inverse of its own. As per (G,*) to be group any element should have only one inverse element (unique) This implies x*y = y*x (is one element) So the elements of such group are 4 which are {x, y, e, x*y}.See following definition of group from wikipedia. A group is a set, G, together with an operation • (called the group law of G) that combines any two elements a and b to form another element, denoted a • b or ab. To qualify as a group, the set and operation, (G, •), must satisfy four requirements known as the group axioms:[5] Closure For all a, b in G, the result of the operation, a • b, is also in G.b[›] Associativity For all a, b and c in G, (a • b) • c = a • (b • c). Identity element There exists an element e in G, such that for every element a in G, the equation e • a = a • e = a holds. Such an element is unique (see below), and thus one speaks of the identity element. Inverse element For each a in G, there exists an element b in G such that a • b = b • a = e, where e is the identity element. Source: https://en.wikipedia.org/wiki/Group_%28mathematics%29
Question 20 
P. For each such function it must be the case that for every i, f(i) = i. Q. For each such function it must be the case that for some i, f(i) = i. R. Each such function must be onto.Which one of the following is CORRECT?
P, Q and R are true  
Only Q and R are true  
Only P and Q are true  
Only R is true 
Discuss it
Question 21 
X ⊂ Y  
X ⊃ Y  
X = Y  
X  Y ≠ φ and Y  X ≠ φ 
Discuss it
Question 22 
(n  A ∪ B) A B  
(A^{2}+B^{2})n^{2}  
n! A∩B / A∪B  
A∩B^{2}nCA∪B 
Discuss it
Question 23 
3m  
3n  
2m + 1  
2n + 1 
Discuss it
Question 24 
X = Y  
X ⊂ Y  
Y ⊂ X  
none of these 
Discuss it
Question 25 
not a lattice  
a lattice but not a distributive lattice  
a distributive lattice but not a Boolean algebra  
a Boolean algebra 
Discuss it
It is a lattice but not a distributive lattice. Table for Join Operation of above Hesse diagram V a b c d e ________________ a a a a a a b a b a a b c a a c a c d a a a d d e a b c d e Table for Meet Operation of above Hesse diagram ^ a b c d e _______________ a a b c d e b b b e e e c c e c e e d d e e d e e e e e e e Therefore for any two element p, q in the lattice (A,<=) p <= p V q ; p^q <= p This satisfies for all element (a,b,c,d,e). which has 'a' as unique least upper bound and 'e' as unique greatest lower bound. The given lattice doesn't obey distributive law, so it is not distributive lattice, Note that for b,c,d we have distributive law b^(cVd) = (b^c) V (b^d). From the diagram / tables given above we can verify as follows, (i) L.H.S. = b ^ (c V d) = b ^ a = b (ii) R.H.S. = (b^c) V (b^d) = e v e = e b != e which contradict the distributive law. Hence it is not distributive lattice. so, option (B) is correct.
Question 26 
3 and 13  
2 and 11  
4 and 13  
8 and 14 
Discuss it
For a number 'n', (n) x (n') = I, where
n' = inverse of 'n' I = Identity element Now, the identity element for multiplication is 1. So, we need to find two numbers 'm' and 'n' such that (4 x m) % 15 = 1 and (7 x n) % 15 = 1,
where 'm' is the inverse of 4 and 'n' is the inverse of 7, and both 'm' and 'n' belong to the given set. Thus, from the given set, it can be easily identified by putting values that m = 4 and n = 13. So, C is the correct choice.
Please comment below if you find anything wrong in the above post.
Question 27 
R ∪ S, R ∩ S are both equivalence relations  
R ∪ S is an equivalence relation  
R ∩ S is an equivalence relation  
Neither R ∪ S nor R ∩ S is an equivalence relation 
Discuss it
Question 28 
f and g should both be onto functions.  
f should be onto but g need not be onto  
g should be onto but f need not be onto  
both f and g need not be onto 
Discuss it
Given that, f: B → C and g: A → B and h = f o g. Note that the sign o represents composition. h is basically f(g(x)). So h is a function from set A to set C. It is also given that h is an onto function which means for every value in C there is a value in A.We map from C to A using B. So for every value in C, there must be a value in B. It means f must be onto. But g may or may not be onto as there may be some values in B which don't map to A. Example :
Let us consider following sets A : {a1, a2, a3} B : {b1, b2} C : {c1} And following function values f(b1) = c1 g(a1) = b1, g(a2) = b1, g(a3) = b1 Values of h() would be, h(a1) = c1, h(a2) = c1, h(a3) = c1 Here h is onto, therefore f is onto, but g is onto as b2 is not mapped to any value in A.Given that, f: B → C and g: A → B and h = f o g.
Question 29 
4  
6  
16  
24 
Discuss it
a = c mod 3 (given) Thus, ‘a’ can be any one of these values : 0, 1, 2
b = d mod 5 (given) Thus, ‘b’ can be any one of these values : 0, 1, 2, 3, 4
Thus, ordered pair for (a, b) are : (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4)
Therefore, ordered pair (a, b) has 15 combinations and ordered pair (c, d) has 1 combination. Total combinations = 15 + 1 = 16
Hence, option (C) is correct.
Please comment below if you find anything wrong in the above post.
Question 30 
S = {(x, y)  y = x+1 and x, y ∈ {0, 1, 2, ...}}The reflexive transitive closure of S is
{(x, y)  y > x and x, y ∈ {0, 1, 2, ... }}  
{(x, y)  y ≥ x and x, y ∈ {0, 1, 2, ... }}  
{(x, y)  y < x and x, y ∈ {0, 1, 2, ... }}  
{(x, y)  y ≤ x and x, y ∈ {0, 1, 2, ... }} 
Discuss it
Reflexive closure of a relation R on set S is the smallest reflexive relation which contains R.
If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}.
Now transitive closure is defined as smallest transitive relation which contains S.
We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now.
Thus, option (B) matches the final set S.
Please comment below if you find anything wrong in the above post.
Question 31 
*  e  a  b  c 
e  e  a  b  c 
a  a  b  c  e 
b  
c 
c a e b
 
c b a e  
c b e a  
c e a b 
Discuss it
Question 32 
S = {{1, 2}, {1, 2, 3}, {1, 3, 5}, (1, 2, 4), (1, 2, 3, 4, 5}}is necessary and sufficient to make S a complete lattice under the partial order defined by set containment ?
{1}  
{1}, {2, 3}  
{1}, {1, 3}  
{1}, {1, 3}, (1, 2, 3, 4}, {1, 2, 3, 5) 
Discuss it
 A partially ordered set L is called a complete lattice if every subset M of L has a least upper bound called as supremum and a greatest lower bound called as infimum.
 We are given a set containment relation.
 So, supremum element is union of all the subset and infimum element is intersection of all the subset.
 Set S is not complete lattice because although it has a supremum for every subset, but some subsets have no infimum. We take subset {{1,3,5},{1,2,4}}.Intersection of these sets is {1}, which is not present in S. So we have to add set {1} in S to make it a complete lattice
Question 33 
A  
B  
C  
D 
Discuss it
Let there be total number of edges of a graph to be formed be e and total number of vertices be n.
So the total number of ways a graph can be formed having exactly e edges and v vertices is given by the total number of ways we can select e edges among v edges
i. e.
C (v, e)
Now, here e = (n^{2} – 3n)/2 and v = n(n1)/2 (maximum number of edges in a simple graph).
So to select a edge we can do that in C (v, (n^{2} – 3n)/2) ways.
Since minimum no. of edges to be selected for a graph are (n^{2} – 3n)/2.
So total number of graphs possible will be :
C(v, e) + C(v, e+1) + C(v, e+2) +…………...+ C(v, v).
C(v, ve) + C(v, v(e+1)) + C(v, v(e+2)) +…………...+ C(v, vv).
Since v  e = n(n1)/2  (n^{2} – 3n)/2 = n
Therefore
C(v, n) + C(v, n1) + C(v, n2) +…………...+ C(v, 0).
Solving this we will get
This solution is contributed by Namita Singh.
Question 34 
does not form a group  
forms a noncommutative group  
does not have a right identity element  
forms a group if the empty string is removed from ∑* 
Discuss it
Question 35 
Let P : S → {True, False} be a predicate defined on S. Suppose that P(a) = True, P(b) = False and P(x) ⇒ P(y) for all x, y ∈ S satisfying x ≤ y, where ⇒ stands for logical implication.Which of the following statements CANNOT be true ?
P(x) = True for all x ∈ S such that x ≠ b  
P(x) = False for all x ∈ S such that x ≠ a and x ≠ c  
P(x) = False for all x ∈ S such that b ≤ x and x ≠ c  
P(x) = False for all x ∈ S such that a ≤ x and b ≤ x 
Discuss it
'a' and 'b' are given as minimal elements. No other element in S is of lower order than either a or b. 'c' is given as maximum element. So, c is of higher order than any other element in S.
P(a) = True means all elements 'x' which have an edge from element 'a' have to be true. Since there is an edge from 'a', we have to satisfy formula P(a) => P(x), which can only be done by setting P(x) = True.
Elements which have an edge from b can be anything because formula P(b) => P(x) is satisfied as P(b) = False.
(A) This statement is true because making all elements true trivially satisfy formula P(x) => P(y).
(B) This statement is true if all elements are connected from b then all elements can be false.
(C) This statement is true because b<=x ensures x!=a and for all other elements P(x) can be false without violating the given implication.
(D) This statement is false. Since, P(a) = true , for all 'x' such that a<=x, P(x) must be true. We do have at least one such 'x', which is 'c' as it is the maximum element.
Thus, option (D) is the answer.
Please comment below if you find anything wrong in the above post.
Question 36 
Define g : 2^{A} → 2^{B} as : g(C) = {f(x)  x ∈ C}, for all subsets C of A. Define h : 2^{B} → 2^{A} as : h(D) = {x  x ∈ A, f(x) ∈ D}, for all subsets D of B.Which of the following statements is always true ?
g(h(D)) ⊆ D  
g(h(D)) ⊇ D  
g(h(D)) ∩ D = ф  
g(h(D)) ∩ (B  D) ≠ ф 
Discuss it
Question 37 
2^{7} 3^{7} 5^{7}  
2^{8} 3^{8} 5^{8}  
2^{9} 3^{9} 5^{9}  
2^{10} 5^{10} 7^{10} 
Discuss it
Since, the answer is a product of three prime numbers 2, 3 and 5. So, we have three nonempty sequence of strings : "a", "a" and "a".
f(s) = 2^{x} for some x Since, 7 and 9 are not multiple of 2. So, options (A) and (C) are eliminated.
f(a) = 2^{3} = 8 h = 2^{8} 3^{8}5^{8}
Thus, option (B) is correct.
Please comment below if you find anything wrong in the above post.
Question 38 
The set of all rational negative numbers forms a group under multiplication.  
The set of all nonsingular matrices forms a group under multiplication.  
The set of all matrices forms a group under multiplication.  
Both (2) and (3) are true. 
Discuss it
Question 39 
Neither reflexive nor symmetric  
Symmetric and reflexive  
Transitive and reflexive  
Transitive and symmetric 
Discuss it
 Reflexive : A relation is reflexive if every element of set is paired with itself. Here none of the element of A is paired with themselves, so S is not reflexive.
 Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
 Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive. Thus, option (D) is correct. Please comment below if you find anything wrong in the above post.
Question 40 
R1(a,b) iff (a+b) is even over the set of integers R2(a,b) iff (a+b) is odd over the set of integers R3(a,b) iff a.b > 0 over the set of nonzero rational numbers R4(a,b) iff a  b <= 2 over the set of natural numbersWhich of the following statements is correct?
R1 and R2 are equivalence relations, R3 and R4 are not  
R1 and R3 are equivalence relations, R2 and R4 are not  
R1 and R4 are equivalence relations, R2 and R3 are not  
R1, R2, R3 and R4 are all equivalence relations 
Discuss it
 R1(a,b)
 Reflexive : Yes, because (a+a) is even.
 Symmetrix : Yes, (a+b) is even ⟹ (b+a) is even.
 Transitive : Yes, because (a+b) is even and (b+c) is even ⟹ (a+c) is even.
 R2(a,b)
 Reflexive : No, because (a+a) is even.
 R3(a,b)
 Reflexive : Yes, because a.a > 0.
 Symmetrix : Yes, a.b > 0 ⟹ b.a > 0.
 Transitive : Yes, because a.b > 0 and b.c > 0 ⟹ a.c > 0.
 R4(a,b)
 Reflexive : Yes, because aa ≤ 2.
 Symmetrix : Yes, ab ≤ 2 ⟹ ba ≤ 2.
 Transitive : No, because ab ≤ 2 and bc ≤ 2 ⇏ (ac) is even.
Question 41 
S1: There exists infinite sets A, B, C such that A ∩ (B ∪ C) is finite. S2: There exists two irrational numbers x and y such that (x+y) is rational.Which of the following is true about S1 and S2?
Only S1 is correct  
Only S2 is correct  
Both S1 and S2 are correct  
None of S1 and S2 is correct 
Discuss it
Question 42 
R is not an equivalence relation  
R is an equivalence relation having 1 equivalence class  
R is an equivalence relation having 2 equivalence classes  
R is an equivalence relation having 3 equivalence classes 
Discuss it
Question 43 
a  
b  
c  
d 
Discuss it
Question 44 
Both commutative and associative  
Commutative but not associative  
Not commutative but associative  
Neither commutative nor associative 
Discuss it
Question 45 
Pr = 0  
Pr = 1  
0 < Pr ≤ 1/5  
1/5 < Pr < 1 
Discuss it
Number of triplets in L^{3} = Number of ways in which we can choose 3 elements from 5 with repetition = 5 * 5 * 5 = 125. Now, when we take x = t, then the given condition for L is satisfied for any y and z. Here, y and z can be taken in 5 * 5 = 25 ways. Take x = r, y = p, z = p. Here also, the given condition is satisfied. So, pr > 25 / 125 > 1/5. For x = q, y = r, z = s, the given condition is not satisfied as q ⋁ (r ⋀ s) = q ⋁ p = q, while (q ⋁ r) ⋀ (q ⋁ s) = t ⋀ t = t. So, pr ≠ 1. Hence D choice.
Question 46 
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 
Discuss it
Question 47 
1024  
1023  
2048  
2043 
Discuss it
Question 48 
a = (1, 5, null), b = (1, null, 7), c = (3, null, 9), d = (4, 7, null), e = (1, 5, 7), f = (3, 7, null), g = (4, null, 9).Which one of the following statements is correct?
R contains a, b, e, f, g but not c, d  
R contains a, b, c, d, e, f, g  
R contains e, f, g but not a, b  
R contains e but not f, g 
Discuss it
Below is R1 A  B  1  5 3  7 Below is R2 A  C  1  7 4  9 Full outer join of above two is A  B  C  1  5  7 3  7  NULL 4  NULL  9So the full outer join contains e = (1, 5, 7), f = (3, 7, null), g = (4, null, 9).
Question 49 
36  
64  
81  
72 
Discuss it
Question 50 
0.95  
0.80  
0.75  
0.70 
Discuss it
Question 51 
Both reflexive and symmetric  
Reflexive but not symmetric  
Not reflexive but symmetric  
Neither reflexive nor symmetric 
Discuss it
((p, q), (r, s)) ∈ R if and only if p–s = q–r (p, q) is not related to (p, q) as pq is not same as qp. The relation is symmetric because if p–s = q–r, then sq = sp.
Question 52 
 An element x in A is related to an element y in B (under R_{1}) if x + y is divisible by 3.
 An element x in B is related to an element y in C (under R_{2}) if x + y is even but not divisible by 3.
R1R2 = {(1, 2), (1, 4), (3, 3), (5, 4), (7, 3)}  
R1R2 = {(1, 2), (1, 3), (3, 2), (5, 2), (7, 3)}  
R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}  
R1R2 = {(3, 2), (3, 4), (5, 1), (5, 3), (7, 1)} 
Discuss it
R1 is a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} . Under R1, an element x in A is related to an element y in B if x + y is divisible by 3.
Thus, R1 = {(1, 2), (1, 8), (3, 6), (5, 4), (7, 2), (7, 8)}
R2 is a relation from B = {2, 4, 6, 8} to C = {1, 2, 3, 4} Under R2, an element y in B is related to an element z in C if y + z is even but not divisible by 3.
Thus, R2 = {(2, 2), (4, 4), (6, 2), (6, 4), (8, 2)}
Thus, R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
Thus, option (C) is correct.
Please comment below if you find anything wrong in the above post.
Question 53 
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 
Discuss it
Question 54 
n  
n + 1  
2n1 + 1  
n! 
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Question 55 
1  
2  
3  
4 
Discuss it
Let us assume: f(1) = x. Then, f(2) = f(2/2) = f(1) = x f(3) = f(3+5) = f(8) = f(8/2) = f(4/2) = f(2/1) = f(1) = x Similarly, f(4) = x f(5) = f(5+5) = f(10/2) = f(5) = y.So it will have two values. All multiples of 5 will have value y and others will have value x. It will have 2 different values.
Question 56 
(a, b) R (c, d) if a <= c or b <= d.Consider the following propositions:
P: R is reflexive Q: R is transitiveWhich 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. 
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Question 57 
 f (x, y) = x + y  3
 f (x, y) = max(x, y)
 f (x, y) = x^{y}
I and II only  
II and III only  
I and III only  
None of these 
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Question 58 
f (x1, x2, ..., xn) = x1'f(x1, x2, ..., xn) + x1f(x1, x2, ..., xn)  
f (x1, x2, ..., xn) = x2f(x1, x2, …, xn) + x2'f(x1, x2, …,xn)  
f (x1, x2, ..., xn) = xn'f(x1, x2, …, 0) + xnf(x1, x2, …,1)  
f (x1, x2, ..., xn) = f(0, x2, …, xn) + f(1, x2, .., xn) 
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Question 59 
satisfiable and valid  
satisfiable and so is its negation  
unsatisfiable but its negation is valid  
satisfiable but its negation is unsatisfiable 
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Question 60 
I only  
II only  
Neither I nor II  
Both I and II 
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Question 61 
28  
33  
37  
44 
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Question 62 
(i) { (a, p) (b, q) (c, r) }
(ii) { (a, p) (b, q) (c, s) (d, r) }
(iii) { (a, p) (b, s) (b, r) (c, q) }
(i) and (ii) only
 
(ii) and (iii) only
 
(i) and (iii) only  
None of these

Discuss it
(iii) is not a function since b (∈A) is associated with two different elements of B.
Question 63 
( A, + ) is a semi group but not monoid
 
( A, + ) is a monoid but not group
 
( A, + ) is a group
 
( A, + ) is not a semi group

Discuss it
A: ( A,+ ) to be semi group, it has to satisfy Closure property & Associative property
Closure: Given, a + b = ab => 1 + 2 = 3
2 + 3 = 5
So what ever value we take for a,b their ab value belongs to A so it satisfies closure.
Associative: I n order to satisfy associative property it needs to hold
( a + b) + c = a + ( b + c )
Checking:
( a + b) + c = a + ( b + c )
ab + c = a + bc
abc = abc
So for all values for a,b,c associative property satisfies
Monoid:
a + e = e + a = a, ∀ a ∈ A
ae = a
e = 1
Identity element is 1, so A is monoid.
Group:
a + b = b + a = e
It doesn’t satisfy the property because for all values of a,b it is not equal to e. So it not a group.