Check whether a function $\displaystyle \mathrm{f}: \mathbb{R} \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]$ defined as $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+\mathrm{x}^{2}}$ is one-one and onto or not.

Let $\displaystyle \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right) \Rightarrow \frac{\mathrm{x}_{1}}{1+\mathrm{x}_{1}{ }^{2}}=\frac{\mathrm{x}_{2}}{1+\mathrm{x}_{2}{ }^{2}}$ $\displaystyle \Rightarrow \mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{x}_{2}{ }^{2}=\mathrm{x}_{2}+\mathrm{x}_{1}{ }^{2} \mathrm{x}_{2}$ $\displaystyle \Rightarrow\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)\left(1-\mathrm{x}_{1} \mathrm{x}_{2}\right)=0$ for $\displaystyle \mathrm{x}_{1}=2, \mathrm{x}_{2}=\frac{1}{2}$ wehave $\displaystyle \left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)\left(1-\mathrm{x}_{1} \mathrm{x}_{2}\right)=0$ but $\displaystyle \mathrm{x}_{1} \neq \mathrm{x}_{2}$ $\displaystyle \Rightarrow \mathrm{f}$ is not one-one. Let $\displaystyle \mathrm{x} \in R$ such that $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{y} \Rightarrow \mathrm{y}=\frac{\mathrm{x}}{1+\mathrm{x}^{2}}$ $\displaystyle \mathrm{x}^{2} \mathrm{y}-\mathrm{x}+\mathrm{y}=0$ $\displaystyle \Rightarrow \mathrm{x}=\frac{1 \pm \sqrt{1-4 \mathrm{y}^{2}}}{2 \mathrm{y}}(\mathrm{y} \neq 0)$. [For $\displaystyle \mathrm{y}=0 \in\left[-\frac{1}{2}, \frac{1}{2}\right]$, we have $\displaystyle 0 \in R$ such that $\displaystyle \mathrm{f}(0)=0$ ] $\displaystyle \mathrm{x} \neq 0, \mathrm{x} \in R \Rightarrow 1-4 \mathrm{y}^{2} \geq 0, \mathrm{y} \neq 0$