Question

In each of the following cases, state whether the functions is one-one, onto or bijective. Justify answer.
(i) f$:$ R$\to$ R defined by $f\left(x\right)=3-4x$
(ii) f$:$ R $\to$ R defined by $f\left(x\right)=1+x^2$.

Answer 1

$\left(i\right)$ $f:$ R$\to$ R defined by $f\left(x\right)=3-4x$

Let $x_1,x_2\in R$ such that $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow$ $3-4x_1=3-4x_2$

$\Rightarrow$ $-4x_1=-4x_2$

$\Rightarrow$ $x_1=x_2$

$\therefore$ $f$ is one-one.

For any real number $\left(y\right)$ in $R,$ there $\frac{3-y}{4}$ in $R$ 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.

$\left(ii\right)$ $f:$ R $\to$ R defined by $f\left(x\right)=1+x^2$.

Let $x_1,x_2\in R$ 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 $x_1=x_2$

Fir instance,

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

$\therefore$ $f$ is not one-one.

Consider an element $-2$ in co-domain $R.$

It is seen that $f\left(x\right)=1+x^2$ is positive for all $x\in R$

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

$\therefore$ $f$ is nor onto.

Hence, $f$ is not bijective.