Q. 204.6( 10 Votes )

# Let A =</sp

Answer :

We have,

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

We observe the following properties of relation R.

Reflexivity : For any a A, we have

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

(a,a) R

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

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.,  = [1,5,9].

OR

f:R R,         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:R R

g(x) = 2x – 1 as g: R R

(fog)(x) = f(g(x)) = = = = Rate this question :

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