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


Let $f: R-\{n\} \rightarrow R$ be a function defined by

$f(x)=\frac{x-m}{x-n}$, where $m \neq n .$ 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


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


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


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

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

$\Rightarrow x=y$

So, f is one-one.

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


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

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

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

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

$\Rightarrow x=\frac{n y-m}{y-1}$,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).






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