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Question

# Let A = {xϵZ:0≤x≤12 }. Show that R = {(a,b) : a, b ϵA,|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 the function f:R→R defined by f(x)=xx2+1∀ x ϵ R is neither one-one nor onto. Also, if g:R→R is defined as g(x) = 2x-1, find fog(x).

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Solution

## We have R={(a,b):a,bϵA|a−b| is divisible by 4} and A={xϵZ:0≤x≤12} i.e., A = {0,1,2,....,12} Reflexivity : For any aϵA we have |a-a| = 0 which is divisible by 4. That is , (a,a) ϵ R so, R is reflexive Symmetry: Let (a,b) ϵ R ∀ a,b ϵ A We have |a-b| is divisible by 4. As |a-b| = |b-a|. So |b-a| is also divisible by 4. that is,(b,a) ϵ R so, R is symmetric. Transitivity: Let a,b,c ϵ A such that (a, b ) ϵ R and (b,c) ϵ R then , |a-b| is divisible by 4 and |b-c| is divisible by 4. This implies that |a-b| = 4m and |b-c| = 4p where m, p ϵ Z. As (a−b)+(b−c)=±4(m+p)⇒|a−c|=4(m+p) That is |a-c| is also divisible by 4 so R is transitive. Since the relation R is reflexive, symmetric and transitive so, it is an equivalence relation. Also the set of all elements related to 1 in relation R is {1, 5, 9}. Now let (2, x) ϵ R so that |2-x| is divisible by 4 ∀ x ϵA That is , x = 2,6,10. Hence [2] = {2,6,10}. OR We have f:R→R defined by f(x)=xx2+1,∀ x ϵ R Let m, n ϵ R so that f(m)= f(n). That implies, mm2+1=nn2+1 ⇒mn2+m=nm2+n ⇒m−n=mn(m−n)→m−n=mn(m−n) ⇒(mn−1)(m−n)=0 Either m -n = 0 or mn-1 = 0 i.e., m = n or , m = 1n So, f(m) = f(n) does not necessarily imply m = n for all m, n ϵ R. Hence f(x) is not one-one. Let y = f(x) = xx2+1 ⇒yx2−x−y=0 ⇒=1±√1−4y22y Now for x to be real, we must have 1−4y2≥0 i.e., (1−2y)(1+2y)≥0 ie., yϵ[−12,12] That is for all x ϵ R (domain), we do not have y ϵ R (codomain) i.e., range ≠ codomain. Hence f(x) is not onto. Now fog(x) = f(g(x)) = f(2x-1) = 2x−1(2x−1)2+1=2x−14x2−4x+2=2x−12(2x2−2x+1)

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