Show that the function

Solution:

To prove: function is neither one-one nor onto

Given: $f: R \rightarrow R: f(x)=x^{2}$

Solution: We have,

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

For, $f\left(x_{1}\right)=f\left(x_{2}\right)$

$\Rightarrow \mathrm{x}_{1}^{2}=\mathrm{x}_{2}^{2}$

$\Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$ or, $\mathrm{x}_{1}=-\mathrm{x}_{2}$

Since $x_{1}$ doesn't has unique image

$\therefore f(x)$ is not one-one

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

Let $f(x)=y$ such that $y \in R$

$\Rightarrow y=x^{2}$

$\Rightarrow x=\sqrt{y}$

If $y=-1$, as $y \in R$

Then $x$ will be undefined as we cannot place the negative value under the square root Hence $f(x)$ is not onto

Hence Proved