Here \(I_{ N }\) is onto. [Given]
But \(I_{ N }+I_{ N }\) is not onto 
[∵ We can find 3 in the co−domain N, where 3 is not a multiple of 2]

Here  \(f(x)=y=\frac { 1 }{ x } \Rightarrow x=\frac { 1 }{ y } .\)
\(\therefore \)   \(f^{ -1 }=\{ (y,x)\} =\left\{ \left( y,\frac { 1 }{ y } \right) \right\} \)
\(\Rightarrow \)  \(f^{ -1 }(x)=\frac { 1 }{ x } .\)
\(\therefore \)  \(fof^{ -1 }(x)=f\left( \frac { 1 }{ x } \right) =\frac { 1 }{ 1/x } =x.\)
Also     \(f^{ -1 }\quad of\quad (x)=f^{ -1 }\left( \frac { 1 }{ x } \right) =\frac { 1 }{ 1/x } =x.\)
Thus  \(fof^{ -1 }(x)=f^{ -1 }\quad of\quad (x)\)
Hence, \(fof^{ -1 }=f^{ -1 }\quad of.\)

(i) We have:
\(R=\{ (a,b):(a-b)\) is a multiple of 4},
Where \(a,b\in A=\{ x:x\in W,0\le x\le 12]\} .\) 
(I) For any \(a\in A\) , we have:
\((a-b)=0,\) which is multiple of 4
\(\Rightarrow \) \((a,a)\in R\forall a\in A\)
Thus R is reflexive.
(II) Let \(a,b\in A\)
Now \((a,b)\in R\)
\(\Rightarrow (a-b)\) is a multiple of 4
\(\Rightarrow (b-a)\) is a multiple of 4
\(\Rightarrow (b,a)\in R\) .
Thus R is symmetric.
(III) Let \(a,b,c\in A\)
Now \((a,b)\in R\) and \((b,c)\in R\)
\(\Rightarrow (a-b)\) is a multiple of 4 and \((b-c)\) is a multiple of 4
\(\Rightarrow (a-b)+(b-c)\) is a multiple of 4
\(\Rightarrow (a-c)\) is a multiple of 4
\(\Rightarrow (a,c)\in R\).
Thus R is transitive.
Hence, R is an equivalence relation.
(ii) Let x be an element of A such that \((x,2)\in R\).
Then (x-2) is a multiple of 4
\(\Rightarrow \) x-2 = 0, 4, 8, 12
\(\Rightarrow \) x = 2, 6,10.  \(\left[ \because \ 14\notin A \right] \)
Hence, the set of all elements, which are related to 2 = {2, 6,10}.