Let A =</sp

We have,

R = , where a,b A = {x= {0,1,2,...,12}.

We observe the following properties of relation R.

Reflexivity : For any aA, we have

|a – a| = 0, which is a multiple of 4..

(a,a) R

Thus, (a,a) R for all aA.

So, R is reflexive.

Symmetry: Let (a,b) R. Then,

(a,b) R

|a – b| is a multiple of 4.

|a – b| = 4 for some λ N

|b – a| = 4 for some λ N

(b – a) R

So, R is symmetric.

Transitivity: Let (a,b) R and (b,c) R

|a – b| is a multiple of 4 and |b – c| is a multiple of 4

|a – b| = 4 and |b – c| = 4 for some λ, N

|a – b| = and b – c =

|a – c| =

|a – c| is a multiple of 4

|a – c|is a multiple of 4

(a,c) R

Thus, (a,b) R and (b,c) R

Hence (a,c) R

So R is Transitive.

Hence R is an equivalence relation.

Let x be an element of A such that (x,1) R. Then,

| x – 1|is a multiple of 4

|x – 1| = 0,4,8,12

|x| = 1,5,9 {13 }

Hence the set of all elements of A which are related to 1 is {1,5,9} i.e., [1] = [1,5,9].

OR

f:RR,

f(x) is not one – one.

Also, y =

if x is real

– 4AC 0

1 – 4 0

(1 – 2y)(1 + 2y) 0

(2y – 1)(2y + 1) 0

Codomain R

But range

Function is not onto.

as f:RR

g(x) = 2x – 1 as g: RR

(fog)(x) = f(g(x)) =

=

=

=

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