Explanation: Given, $\mathit{f}\left(\mathit{x}\right)=2+{\mathit{x}}^2$
For one-one, $\mathit{f}\left({\mathit{x}}_1\right)=\mathit{f}\left({\mathit{x}}_2\right)$
$\Rightarrow 2+{\mathit{x}}_1^2=2+{\mathit{x}}_2^2$
$\Rightarrow {\mathit{x}}_1^2={\mathit{x}}_2{}^2$
$\Rightarrow {\mathit{x}}_1=\pm {\mathit{x}}_2$
$\Rightarrow {\mathit{x}}_1={\mathit{x}}_2$

Thus, $\mathit{f}\left(\mathit{x}\right)$ is not one-one.
For onto
Let

$\Rightarrow {\mathit{x}}^2=\mathit{y}-2$
$\Rightarrow \mathit{x}=\pm \sqrt{\mathit{y}-2}$

Put $\mathit{y}=-3$, we get
$\mathit{x}=\pm \sqrt{-3-2}=\pm \sqrt{-5}$
Which is not possible as root of negative is not a real number.
Hence, $\mathit{x}$ is not real.
So, $\mathit{f}\left(\mathit{x}\right)$ is not onto.