Let $A = R - \{ 3\} \;\;and\;\;B = R - \{ 1\} .$ Consider the function $f:A \to B$ defined by $f(x) = \left( {\cfrac{{x - 2}}{{x - 3}}} \right).$ Is f one-one and onto? Justify your answer.

$A = R - \{ 3\} ,\;\;B = R - \{ 1\} \;\;and\;\;f(x) = \cfrac{{x - 2}}{{x - 3}}$

Let ${x_1},\;{x_2}$ $\in A \Rightarrow \;\;f({x_1}) = \cfrac{{{x_1} - 2}}{{{x_1} - 3}}\;\;and\;\;f({x_2}) = \cfrac{{{x_2} - 2}}{{{x_2} - 3}}$

Injectivity

$f({x_1}) = f({x_2}) \Rightarrow \cfrac{{{x_1} - 2}}{{{x_1} - 3}} = \cfrac{{{x_2} - 2}}{{{x_2} - 3}}$
$\Rightarrow$ $({x_1} - 2)({x_2} - 3) = ({x_2} - 2)({x_1} - 3)$
$\Rightarrow$ ${x_1}{x_2} - 3{x_1} - 2{x_2} + 6 = {x_1}{x_2} - 2{x_1} - 3{x_2} + 6$

$\Rightarrow$ $- 3{x_1} - 2{x_2} = - 2{x_1} - 3{x_2} \Rightarrow {x_1} = {x_2}$

Therefore, f is a one-one function.

Surjectivity

$f:A \to B,$ let $y \in B$ (co-domain of f) be any element, then there exist $x \in A$ (domain of f) such that $f(x) = y$
$\Rightarrow$ $y = \cfrac{{x - 2}}{{x - 3}} \Rightarrow y(x - 3) = x - 2 \Rightarrow xy - 3y = x - 2$

$\Rightarrow$ $x(y - 1) = 3y - 2 \Rightarrow x = \cfrac{{3y - 2}}{{y - 1}}$

Therefore, $f\left( {\cfrac{{3y - 2}}{{y - 1}}} \right) = \cfrac{{\cfrac{{3y - 2}}{{y - 2}} - 2}}{{\cfrac{{3y - 2}}{{y - 1}} - 3}} = \cfrac{{3y - 2 - 2y + 2}}{{3y - 2 - 3y + 3}} = y$

Therefore, f is an onto function.

Hence, f is one-one and onto function.