Let a, b ∈ R, (a ≠ 0) . If the function f , defined as f(x) = \...

Here, $f\left(x\right)$ is continuous at $x = 1$ $⇒ \lim_{x \to {1}^{-}}f\left(x\right) = \lim_{x \to {1}^{+}}f\left(x\right)$
$⇒ \dfrac{2}{a} = a ⇒ a = \pm \sqrt{2}$

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

$⇒ \lim_{x \to {\sqrt{2}}^{-}}f\left(x\right) = \lim_{x \to {\sqrt{2}}^{+}}f\left(x\right)$

$⇒ a = \dfrac{2{b}^{2} - 4b}{2\sqrt{2}}$

Put $a = \sqrt{2}$

$⇒ 2 = {b}^{2} - 2b ⇒ {b}^{2} - 2b - 2 = 0$

$⇒ b = \dfrac{2 \pm \sqrt{4 + 4.2}}{2} = 1 \pm \sqrt{3}$

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