(i) f: R → R defined by f(x) = 3 − 4x.

(ii) f: R → R defined by f(x) = 1 + $x^2.$

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) \\ $.

$$ \begin{aligned} &\Rightarrow 3-4 x_{1}=3-4 x_{2} \\\\ &\Rightarrow-4 x_{1}=-4 x_{2} \\\\ &\Rightarrow x_{1}=x_{2} \\ \end{aligned} $$

$\therefore f$ 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 . \\ $

$\therefore f$ 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,

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

$\therefore f$ is not one-one.

Consider an element $-2$ in co-domain $\mathbf{R}$.

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.