Let $f: R \rightarrow R$ be a function defined by

$f(x)=\frac{x-m}{x-n}$

For any $(\mathrm{x}, \mathrm{y}) \in \mathrm{R}$

Let $\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y})$

$\Rightarrow \frac{x-m}{x-n}=\frac{y-m}{y-n} \Rightarrow x=y$

$\therefore \mathrm{f}$ is one $-$ one

Let $\alpha \in \mathrm{R}$ such that $\mathrm{f}(\mathrm{x})=\alpha$

$$

\Rightarrow \alpha=\frac{x-m}{x-n} \Rightarrow(x-n) \alpha=x-m

$$

$\Rightarrow x \alpha-n \alpha=x-m$

$$

\Rightarrow x \alpha-x=n \alpha-m

$$

$\Rightarrow x(\alpha-1)=n \alpha-m$

$\Rightarrow \mathrm{x}=\frac{\mathrm{n} \alpha-\mathrm{m}}{\alpha-1} \cdot$ for $\alpha=1, \mathrm{x} \notin \mathrm{R}$

So, $\mathrm{f}$ is not onto.