Question

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your 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

(i)

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

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

Hence f is one-one

Also the range of f is R = codomain of f,

since it is first order polynomial

Therefore f is both one-one and onto function

(ii)

$f:R\to R$ is defined as $f\left(x\right)=1+x^2$

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 $x_1=x_2$.
For 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 $f\left(x\right)=1+x_2^2$ 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 not onto.

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