(a) Show that a function f: R → R defined by f ( x )=(2 x)/(1+ x^2) is neither one-one nor onto. Further, find set A so that the given function f: R…

$$\begin{aligned} & \text { Let } \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right) \text { for some } \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R} \\ & \text { Then } \frac{2 \mathrm{x}_{1}}{1+\mathrm{x}_{1}^{2}}=\frac{2 \mathrm{x}_{2}}{1+\mathrm{x}_{2}^{2}} \\ & \Rightarrow \mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{x}_{2}^{2}=\mathrm{x}_{2}+\mathrm{x}_{1}^{2} \mathrm{x}_{2} \\ & \Rightarrow\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)-\mathrm{x}_{1} \mathrm{x}_{2}\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)=0 \\ & \Rightarrow\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)\left(1-\mathrm{x}_{1} \mathrm{x}_{2}\right)=0 \\ & \Rightarrow \mathrm{x}_{1}-\mathrm{x}_{2}=0 \text { or } 1-\mathrm{x}_{1} \mathrm{x}_{2}=0 \\ & \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2} \text { or } \mathrm{x}_{1} \mathrm{x}_{2}=1 \text {, so if } \mathrm{x}_{1} \mathrm{x}_{2}=1, \mathrm{x}_{1} \neq \mathrm{x}_{2} \end{aligned} $$ Hence f is not one -one Let $\displaystyle \mathrm{y}=\mathrm{f}(\mathrm{x})$ where $\displaystyle \mathrm{x} \in \mathrm{R}$ Then $\displaystyle \mathrm{y}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}$. Here, for $\displaystyle \mathrm{x}=0, \mathrm{y}=0$ If $\displaystyle \mathrm{y} \neq 0$, then $\displaystyle \mathrm{y}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}$ $$\begin{aligned} & \Rightarrow \mathrm{y} \mathrm{x}^{2}-2 \mathrm{x}+\mathrm{y}=0 \\ & \Rightarrow \mathrm{x}=\frac{2 \pm \sqrt{4-4 \mathrm{y}^{2}}}{2 \mathrm{y}} \end{aligned} $$ For x to be real, $\displaystyle 4-4 \mathrm{y}^{2} \geq 0$ $$\begin{aligned} & \Rightarrow \mathrm{y}^{2} \leq 1 \\ & \Rightarrow-1 \leq \mathrm{y} \leq 1 \end{aligned} $$ Hence, range $\displaystyle =[-1,1] \neq$ codomain Hence, f is not onto. For the given function to become onto, $\displaystyle A=[-1,1]$ Let $\displaystyle (\mathrm{a}, \mathrm{b}) \in \mathrm{N} \times \mathrm{N}$ We have $$\mathrm{a}-\mathrm{a}=\mathrm{b}-\mathrm{b} $$ This implies that $\displaystyle (\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \forall(\mathrm{a}, \mathrm{b}) \in \mathrm{N} \times \mathrm{N}$ Hence R is reflexive Let (a, b) R (c, d) for some ( $\displaystyle \mathrm{a}, \mathrm{b}$ ), ( $\displaystyle \mathrm{c}, \mathrm{d}) \in \mathrm{N} \times \mathrm{N}$ Then $\displaystyle \mathrm{a}-\mathrm{c}=\mathrm{b}-\mathrm{d}$ $$\Rightarrow \mathrm{c}-\mathrm{a}=\mathrm{d}-\mathrm{b} $$ $\displaystyle \Longrightarrow(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})$ Hence, $\displaystyle \mathrm{R}$ is symmetric. Let (a, b) R (c, d), (c, d) R (e, f) for some ( $\displaystyle \mathrm{a}, \mathrm{b}$ ), ( $\displaystyle \mathrm{c}, \mathrm{d}$ ), ( $\displaystyle \mathrm{e}, \mathrm{f}) \in \mathrm{N} \times \mathrm{N}$ Then $\displaystyle \mathrm{a}-\mathrm{c}=\mathrm{b}-\mathrm{d}, \mathrm{c}-\mathrm{e}=\mathrm{d}-\mathrm{f}$ $\displaystyle \Rightarrow \mathrm{a}-\mathrm{c}+\mathrm{c}-\mathrm{e}=\mathrm{b}-\mathrm{d}+\mathrm{d}-\mathrm{f}$ $\displaystyle \Rightarrow \mathrm{a}-\mathrm{e}=\mathrm{b}-\mathrm{f}$ ⇒ $\displaystyle (\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{e}, \mathrm{f})$ Hence, R is transitive Thus, R is an equivalence relation.