$\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ where $\mathrm{A}=\mathrm{R}-\{3\}, \mathrm{B}=\mathrm{R}-\{1\}$ f is defined by $\mathrm{f}(\mathrm{x})$
$$
=\frac{x-2}{x-3}
$$
(a) $f\left(x_1\right)=\frac{x_1-2}{x_1-3}, f\left(x_2\right)=\frac{x_2-2}{x_2-3}$
$$
f\left(x_1\right)=f\left(x_2\right)=\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}
$$
or $\left(x_1-2\right)\left(x_2-3\right)=\left(x_2-2\right)\left(x_1-3\right)$
or $x_1 x_2-3 x_1-2 x_2+6=x_1 x_2-2 x_1-3 x_2+6$
i.e. $-3 x_1-2 x_2=-2 x_1-3 x_2$
or $-\mathrm{x}_1=-\mathrm{x}_2 \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \quad \therefore$ fis one-one
(b) Let $y=\frac{x-2}{x-3}, x y-3 y=x-2$
or $x(y-1)=3 y-2 \Rightarrow x=\frac{3 y-2}{y-1}$
For every value of $y$ except $y=1$, there is a pre-image.
$\mathrm{x}=\frac{3 \mathrm{y}-2}{\mathrm{y}-1} \Rightarrow \mathrm{f}$ is onto. Thus, $\mathrm{f}$ is one-one and onto.