Question

Let $f:R-\left\{n\right\}\to R$ be a function defined by

$f\left(x\right)=\frac{x-m}{x-n},\mathrm{where}m\ne n.$ Then,

(a) f is one-one onto
(b) f is one-one into
(c) f is many one onto
(d) f is many one into

Answer 1

Injectivity:
Let x and y be two elements in the domain R-{n}, such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{x-m}{x-n}=\frac{y-m}{y-n} \\ \Rightarrow \left(x-m\right)\left(y-n\right)=\left(x-n\right)\left(y-m\right) \\ \Rightarrow xy-nx-my+mn=xy-mx-ny+mn \\ \Rightarrow \left(m-n\right)x=\left(m-n\right)y \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that
$$\begin{gathered} f\left(x\right)=y \\ \Rightarrow \frac{x-m}{x-n}=y \\ \Rightarrow x-m=xy-ny \\ \Rightarrow ny-m=xy-x \\ \Rightarrow ny-m=x\left(y-1\right) \\ \Rightarrow x=\frac{ny-m}{y-1},\text{which is not defined for}y\text{=1} \\ \text{So, 1 ∈}R\left(\text{co domain}\right)\text{has no pre image in}R\text{-}\left\{n\right\} \end{gathered}$$
$\Rightarrow$f is not onto.
Thus, the answer is (b).