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 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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