(i) $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $3-4 \mathrm{x}$, $\mathrm{f}\left(\mathrm{x}_1\right)=3-4 \mathrm{x}_1, \mathrm{f}\left(\mathrm{x}_2\right)=3-4 \mathrm{x}_2$
(a) $f\left(x_1\right)=f\left(x_2\right) \Rightarrow 3-4 x_1=3-4 x_2$
$\Rightarrow x_1=x_2$. This shows that $f$ is one-one
(b) $f(x)=y=3-4 x \quad \therefore x=\frac{3-y}{4}$
For every value of y belonging to its codomain. There is a pre-image in its domain $\Rightarrow$ fis onto. Hence, $f$ is one-one onto
(ii) $f: R \rightarrow R$ given by $f(x)=1+x^2$
(a) $\mathrm{f}(1)=1+1=2, \mathrm{f}(-1)=1+1=2$
$\therefore \mathrm{f}(-1)=\mathrm{f}(1)=2$ i.e. $-1$ and 1 have the same image 2 . $\Rightarrow \mathrm{f}$ is not one-one.
(b) No negative number belonging to its codomain has its pre-image in its domain
$\Rightarrow \mathrm{f}$ is not onto. Thus $\mathrm{f}$ is neither one-one nor onto.