Explanation: Given, $f(x)=2+x^2$For one-one, $f(x_1)=f(x_2)$$\Rightarrow 2+x_1^2=2+x_2^2$$\Rightarrow x_1^2=x_2^2$$\Rightarrow x_1=\pm x_2$$\Rightarrow x_1=x_2$$\text{ or } x_1=-x_2$Thus, $f(x)$ is not one-one.For ontoLet$\therefore f(x)=y \text{ such that }$$\Rightarrow x^2=y-2$$\Rightarrow x=\pm \sqrt{y-2}$ Put $y=-3$, we get$x=\pm \sqrt{-3-2}=\pm \sqrt{-5}$Which is not possible as root of negative is not a real number.Hence, $x$ is not real.So, $f(x)$ is not onto.