Sets, Relations and Functions Test - 7

Since the domain is represented by the x- co ordinate of the ordered pair (x , y).Therefore, domain of the given relation is {1, 2}.

To make the relation an equivalence relation , the following ordered pairs are required (1,1),(2,2),(3,3)(2,1)(3,2)(1,3),(3,1)

Given, R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. 
(a) Since, (1, 1), (2, 2), (3, 3), (4, 4) ∉R. So, R is not reflexive. 
(b) Since, (1, 3) ∈R and (3, 1) ∈R but (1, 1) ∉R. So, R is not transitive. 
(c) Since, (2, 3) ∈R but (3,2) ∉R. So, B is not symmetric. 
(d) Since, (2, 4) ∈R and (2, 3) ∈R. So, R is not a function.

Since [x] ≤x, therefore , x –[x] ≥0. Also, x –[x] <1,∴0 ⩽x −[x]<1. Therefore ,Rf = [0,1).

A bijection from A to A is infact an arrangement of its n elements, taken all at a time, which can be done in    ways. Hence, the number of bijections from A to A is  

No. of elements in the set A = 4. Therefore , the no. of elements in  A  × A  = 4  × 4  = 16. As, the no. of relations in  A  × A = no. of subsets of A  × A  = 2¹⁶

The relation R = A x A is called Universal relation.

Consider any a ,b , c  ∈A .

1) Since both a and a must be either even or odd, so (a , a) ∈R  ⇒R is reflexive.

2) Let (a ,b) ∈R  ⇒ both a and b must be either even or odd, ⇒ both b and a must be either even or odd, ⇒ (b ,a) ∈R. Thus , (a ,b) ∈R  ⇒ (b ,a) ∈R  ⇒R is symmetric.

3) Let (a ,b) ∈R and (b ,c) ∈R  ⇒ both a and b must be either even or odd, also ,both b and c must be either even or odd, ⇒ all elements a, b and c must be either even or odd, ⇒ (a ,c) ∈R . Thus, (a ,b) ∈R  ⇒ (b,c) ∈R  ⇒ (a ,c) ∈R  ⇒R is transitive.

f(a+1)−f(a −1)
=(a+1)−(a+1)2 −{(a −1)−(a −1)2} = 2 −4a(a −1)2} = 2 −4a