CBSE Class 12 Math 2018 Solved Paper

Question : 24

Total: 29

Let A = {x ∊ Z:0 ≤ x ≤ 12). Show that
R = {(a,b): a,b ∊ A, |a - b| is divisible by 4} is an equivalence relation.
Find the set of all elements related to 1. Also write the equivalence class {2}
OR
Show that function f : R → R defined by f(x) =

x2+1

, for all x ∊ R is neither one-one nor onto. Also, if g : R → R is defined as g(x) = 2x - 1 , find f o g (x)

Solution:

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 =

x2+1

⇒ yx2 - x + y = 0
a = y , b = - 1 , c = y ⇒ x =

(−1)±√1−4y2

For each value of y we will get two distinct values. Function is many one not one one
As 1 - 4y2 ≥ 0 ⇒

−1

≤ y ≤

Function is not onto.
f o g (x) = f (2x - 1) =

2x−1

(2x−1)2+1

2x−1

2(2x2−2x+1)

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