Q64 / 125 · BITSAT 2014 · Mathematics
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a function defined by
$f(x)=\frac{x-m}{x-n},$ where $m \neq n,$ then
- A$\mathrm{f}$ is one-one onto
- Correct answerB$\mathrm{f}$ is one-one into
- C$\mathrm{f}$ is many-one onto
- Dfis many-one into
Solution
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.
Hence, the correct answer is option: 'B'.