Mathematics - Continuity - Differentiability Question with Solution | TestHub

Here, $f\left(x\right)$ is continuous at $x=\mathrm{1 }$  $\Rightarrow \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)$
$\Rightarrow \frac{2}{a}=a \Rightarrow a= \pm \sqrt{2}$

Since, $f\left(x\right)$ is continuous at $x=\sqrt{2}$

$\Rightarrow \underset{x\to {\sqrt{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to {\sqrt{2}}^{+}}{\lim }f\left(x\right)$

$\Rightarrow a=\frac{2b^2-4b}{2\sqrt{2}}$

Put $a= \sqrt{2}$

$\Rightarrow 2=b^2-2b \Rightarrow b^2-2b-2=0$

$\Rightarrow b=\frac{2\pm \sqrt{4+4.2}}{2}=1\pm \sqrt{3}$

Hence, $\left(a,b\right)\equiv \left(\sqrt{2},1\pm \sqrt{3}\right)$