Let f : R−{n}→R be a function defined by

Injectivity:
Let x and y be two elements in the domain R-{n}, such that

$f(x)=f(y)$

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

$\Rightarrow(x-m)(y-n)=(x-n)(y-m)$

$\Rightarrow x y-n x-m y+m n=x y-m x-n y+m n$

$\Rightarrow(m-n) x=(m-n) y$

$\Rightarrow x=y$

So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that

$f(x)=y$

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

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

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

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

$\Rightarrow x=\frac{n y-m}{y-1}$,which is not defined for $y=1$

So, $1 \in R$ (co domain) has no pre image in $R$ - $\{n\}$

$\Rightarrow f$ is not onto.

Thus, the answer is (b).