In each of the following cases,

Solution:

(i) $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=3-4 x$.\

Let $x_{1}, x_{2} \in \mathbf{R}$ such that $f\left(x_{1}\right)=f\left(x_{2}\right)$.

$\Rightarrow 3-4 x_{1}=3-4 x_{2}$

$\Rightarrow-4 x_{1}=-4 x_{2}$

$\Rightarrow x_{1}=x_{2}$

∴ f is one-one.

For any real number $(y)$ in $\mathbf{R}$, there exists $\frac{3-y}{4}$ in $\mathbf{R}$ such that

$f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)=y$

∴f is onto.

Hence, f is bijective.

(ii) $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined as

$f(x)=1+x^{2}$

Let $x_{1}, x_{2} \in \mathbf{R}$ such that $f\left(x_{1}\right)=f\left(x_{2}\right)$.

$\Rightarrow 1+x_{1}^{2}=1+x_{2}^{2}$

$\Rightarrow x_{1}^{2}=x_{2}^{2}$

$\Rightarrow x_{1}=\pm x_{2}$

$\therefore f\left(x_{1}\right)=f\left(x_{2}\right)$ does not imply that $x_{1}=x_{2}$.

For instance,

$f(1)=f(-1)=2$

∴ f is not one-one.

Consider an element −2 in co-domain R.

It is seen that $f(x)=1+x^{2}$ is positive for all $x \in \mathbf{R}$.

Thus, there does not exist any $x$ in domain $\mathbf{R}$ such that $f(x)=-2$.

$\therefore f$ is not onto.

Hence, $f$ is neither one-one nor onto.