Let $\mathrm{A}=\{\mathrm{x}\in \mathrm{Z}:0\le \mathrm{x}\le 12\}$. Show that $R=\left\{\right(\mathrm{a},\mathrm{b}):\mathrm{a},\mathrm{b}\in \mathrm{A},|\mathrm{a}-\mathrm{b}|$ 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 $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ defined by $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{x}}^2-1},\mathrm{\forall }\mathrm{x}\in \mathrm{R}$ is neither one-one nor onto. Also, if $\mathrm{g}:\mathrm{R}\to \mathrm{R}$ is defined as $\mathrm{g}\left(\mathrm{x}\right)=2\mathrm{x}-1$, find fog(x).

Given: $\mathrm{R}=\left\{\right(\mathrm{a},\mathrm{b}):\mathrm{a},\mathrm{b}\in \mathrm{A},|\mathrm{a}-\mathrm{b}|$ is divisible by 4} 
For Reflexivity:
For any $\mathrm{a}\in \mathrm{A}$
$|\mathrm{a}-\mathrm{a}|=0$ which is divisible by 4
$(\mathrm{a},\mathrm{a})\in \mathrm{R}$
So, R is reflexive
For Symmetry:
Let $(\mathrm{a},\mathrm{b})\in \mathrm{R}$
$\Rightarrow |\mathrm{a}-\mathrm{b}|$ is divisible by 4
$\Rightarrow |\mathrm{b}-\mathrm{a}|$ is divisible by 4 $[\because |\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}\left|\right]$
$\Rightarrow (\mathrm{b},\mathrm{a})\in \mathrm{R}$
So, R is symmetric.
For Transitive:
Let $(\mathrm{a},\mathrm{b})\in \mathrm{R}\text{ and }(\mathrm{b},\mathrm{c})\in \mathrm{R}$
$\Rightarrow |\mathrm{a}-\mathrm{b}|$is divisible by 4
$\begin{array}{l}\Rightarrow |\mathrm{a}-\mathrm{b}|=4\mathrm{k}\\ \therefore \mathrm{a}-\mathrm{b}=\pm 4\mathrm{k},\mathrm{k}\in \mathrm{Z}\dots \text{ (i) }\end{array}$
Also, $|\mathrm{b}-\mathrm{c}|$is divisible by 4
$\begin{array}{l}\Rightarrow |\mathrm{b}-\mathrm{c}|=4\mathrm{m}\\ \therefore \mathrm{b}-\mathrm{c}=\pm 4\mathrm{m},\mathrm{m}\in \mathrm{Z}\end{array}$
Adding equations (i) and (ii)
a – b + b – c = ±4 (k + m)
⇒ a – c = ±4 (k + m)
|a – c | is divisible by 4,
⇒ (a, c) ϵ R
So, R is symmetric.
Therefore, R is reflexive, symmetric and transitive.
Now, Let x be an element of R such that (x, 1) ϵ R
Then | x – 1| is divisible by 4
x – 1 = 0, 4, 8, 12,
⇒ -x = 1, 5, 9 (∵ x ≤ 12)
∴ Set of all elements of A which are related to 1 are {1, 5, 9}.
Equivalence class of 2 i.e.
[2] = {(a, 2): a ϵ A, |a – 2| is divisible by 4}
⇒ | a – 2| = 4k(k is whole number, k ≤ 3) ⇒ a = 2, 6, 10
Therefore, set of all elements of A which are related to 1 are {1, 5, 9} and equivalence class [2] is {2, 6, 10}.

OR

Given: $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{x}}^2-1},\mathrm{\forall }\mathrm{x}\in \mathrm{R}$
For one-one, $\mathrm{f}\left(\mathrm{x}\right)=\ddot{\mathrm{f}}\left(\mathrm{y}\right)$
$\begin{array}{l}\frac{\mathrm{x}}{{\mathrm{x}}^2+1}=\frac{\mathrm{y}}{{\mathrm{y}}^2+1}\\ \Rightarrow {\mathrm{xy}}^2+\mathrm{x}={\mathrm{yx}}^2+\mathrm{y}\\ \Rightarrow {\mathrm{xy}}^2-{\mathrm{yx}}^2=\mathrm{y}-\mathrm{x}\\ \Rightarrow \mathrm{xy}(\mathrm{y}-\mathrm{x})=\mathrm{y}-\mathrm{x}\\ \Rightarrow \mathrm{xy}=1\\ \Rightarrow \mathrm{x}=\frac{1}{\mathrm{y}}\end{array}$
Since $\mathrm{x}\ne \mathrm{y}$, therefore, f(x) is not one-one.
For onto, $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y}$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{x}}{{\mathrm{x}}^2+1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\mathrm{y}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}={\mathrm{yx}}^2+\mathrm{y}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}^2\mathrm{y}+\mathrm{y}-\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\end{array}$
x cannot be expressed in terms of y
$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)$ is not onto.
As $\mathrm{g}\left(\mathrm{x}\right)=2\mathrm{x}-1$
$\begin{array}{r}\therefore \mathrm{f}\circ \mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left[\mathrm{g}\right(\mathrm{x}\left)\right]=\mathrm{f}(2\mathrm{x}-1)\\ =\frac{2\mathrm{x}-1}{(2\mathrm{x}-1{)}^2+1}=\frac{2\mathrm{x}-1}{4{\mathrm{x}}^2-4\mathrm{x}+2}\end{array}$
Therefore, the function $\mathrm{f}:\mathrm{R}\to \mathrm{R}$  defined by $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{x}}^2-1},\mathrm{\forall }\mathrm{x}\in \mathrm{R}$ is neither one-one nor onto has proved and for if $\mathrm{g}:\mathrm{R}\to \mathrm{R}$ is defined as $\mathrm{g}\left(\mathrm{x}\right)=2\mathrm{x}-1$then the $\mathrm{f}\circ \mathrm{g}\left(\mathrm{x}\right)=\frac{2\mathrm{x}-1}{4{\mathrm{x}}^2-4\mathrm{x}+2}$.