Given that A = {x ∊ Z ; 0 ≤ x ≤ 12} = {0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12} R = {(a , b) : a,b ∊ Z , |a - b| is divisible by 4} For each a ∊ A ⇒ |a - a| = 0 ⇒ Zero is divisible by 4 ⇒ Given function is reflexive ... (i) If (a , b) ∊ R then (b , a) ∊ R |a - b| = |b - a| Given function is symmetric ... (ii) For transitive If (a , b) ∊ R ⇒ |a - b| is divisible by 4. ⇒ a - b = ± 4m (b , c) ∊ R ⇒ |b - c| is divisible by 4. ⇒ b - c = ± 4n ⇒ |a - c| = |± 4m ± 4n| which is divisible by 4. ⇒ (a , c) ∊ R Function is transitive......(iii) The relation is equivalence. from (i),(ii),(iii) Set of elements relates to 1 is {(1 , 1) , (1 , 5) , (1 , 9) , (5 , 1) , (9 , 1)} Let (y , 2) ∊ R , y ∊ A |y - 2| = 4c where c is whole number c ≤ 3 ⇒ y = 2 , 6 , 10 ⇒ Equivalence class [2] is {2,6,10} OR Given y = $\frac{x}{x^2+1}$ ⇒ $yx^2$ - x + y = 0 a = y , b = - 1 , c = y ⇒ x = $\frac{(-1)\pm\sqrt{1-4y^2}}{2y}$ For each value of y we will get two distinct values. Function is many one not one one As 1 - $4y^2$ ≥ 0 ⇒ $\frac{-1}{2}$ ≤ y ≤ $\frac{1}{2}$ Function is not onto. f o g (x) = f (2x - 1) = $\frac{2x-1}{(2x-1)^2+1}$ = $\frac{2x-1}{2(2x^2-2x+1)}$