Let the function f : R−{−b}→R−{1} be defined by

Solution:

(c) f is both one-one and onto

Injectivity:

Let $x$ and $y$ be two elements in the domain $R-\{-b\}$, such that

$f(x)=f(y)$

$\Rightarrow \frac{x+a}{x+b}=\frac{y+a}{y+b}$

$\Rightarrow(x+a)(y+b)=(x+b)(y+a)$

$\Rightarrow x y+b x+a y+a b=x y+a x+b y+a b$

$\Rightarrow b x+a y=a x+b y$

$\Rightarrow(a-b) x=(a-b) y$

$\Rightarrow x=y$

So, f is one-one.

Surjectivity:

Let $y$ be an element in the co-domain of $f_{z}$ i.e. $R\{1\}$, such that $f(x)=y$

$f(x)=y$

$\Rightarrow \frac{x+a}{x+b}=y$

$\Rightarrow x+a=y x+y b$

$\Rightarrow x-y x=y b-a$

$\Rightarrow x(1-y)=y b-a$

$\Rightarrow x=\frac{y b-a}{1-y} \in R-\{-b\}$

So, $f$ is onto.