Let A = R − {3} and B = R − {1}.

Solution:

$A=\mathbf{R}-\{3\}, B=\mathbf{R}-\{1\}$

$f: \mathrm{A} \rightarrow \mathrm{B}$ is defined as $f(x)=\left(\frac{x-2}{x-3}\right)$.

Let $x, y \in$ A such that $f(x)=f(y)$.

$\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}$

$\Rightarrow(x-2)(y-3)=(y-2)(x-3)$

$\Rightarrow x y-3 x-2 y+6=x y-3 y-2 x+6$

$\Rightarrow-3 x-2 y=-3 y-2 x$

$\Rightarrow 3 x-2 x=3 y-2 y$

$\Rightarrow x=y$

∴ f is one-one.

Let $y \in B=\mathbf{R}-\{1\} .$ Then, $y \neq 1$.

The function f is onto if there exists x ∈A such that f(x) = y.

Now,

$f(x)=y$

$\Rightarrow \frac{x-2}{x-3}=y$

$\Rightarrow x-2=x y-3 y$

$\Rightarrow x(1-y)=-3 y+2$

$\Rightarrow x=\frac{2-3 y}{1-y} \in \mathrm{A} \quad[y \neq 1]$

Thus, for any $y \in \mathrm{B}$, there exists $\frac{2-3 y}{1-y} \in \mathrm{A}$ such that

$f\left(\frac{2-3 y}{1-y}\right)=\frac{\left(\frac{2-3 y}{1-y}\right)-2}{\left(\frac{2-3 y}{1-y}\right)-3}=\frac{2-3 y-2+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}=y .$

∴ f is onto.

Hence, function f is one-one and onto.