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Relation & Functions

Theory and Formula

Question


Q1. If two sets A and B are having 21 element in common, then the number of elements common to each of the sets A x B and B x A is .

(a) 2

(b) (21)2

(c) 21

(d) 14

Solution. b
Note : If m elements are common in A and B . then m2 elements will be common in (A x B) and (B x A) .
Solution

Q2. If S1 = {1,2,3,4,-------,20}, S2 = {x,y,z,a}, S3 = {x,y,w,b} . The number of elements of (S1 x S2)U(S1 x S3) is

(a) 120

(b) 100

(c) 80

(d) 160

Solution.a
n(S1 x S2) = 80 , n(S1 x S3) = 80, but two elements are common in S2 and S3 , So (S1 x S2) and (S1 x S3) has 40 common elements .

= 80 + 80 - 40

= 120
Solution

Q3. If n(A) = 5 , n(B) = 7 and n(A ∩ B) = 2 , find n[(A x B)∩ (B x A)] =

(a) 4

(b) 3

(c)2

(d) 1

Solution. a
n(A x B) = 35 , n(B x A) = 35 , 2 elements are common so 22 elements are common in (A x B) and (B x A) ,
that is why n[ (A x B)∩ (B x A)] = 4
Solution

Q4. If A has m and B has n elements then total number of relations from A to B will be

(a) 2

(b) 2mn

(c) mn

(d) m + n

Solution.b
2mn
Solution

Q5. A Relation R is defined in the set of integers I as follows (x,y)∑ R if x2 + y2 = 4 , find its Domain .

(a) {-1,0,1}

(b) {0,1}

(c) {-2,0}

(d) {-2,0,2}

Solution.d
clearly x2 + y2 = 4

x = √(4 - y2)

R = {(0,2)(0,-2)(2,0)(-2,0)}

So Domain is = {-2,0,2}
Solution