Let $\mathrm{a},\mathrm{b} \in \mathrm{R}, \left(\mathrm{a}\ne 0\right).$ If the function defined as

$\text{⨍ }\left(\mathrm{x}\right)\text{=}\left\{\begin{array}{ccc}\frac{2{\mathrm{x}}^2}{\mathrm{a}}& ,& 0\le \mathrm{x}<1\\ \mathrm{a}& ,& 1\le \mathrm{x}<\sqrt{2}\\ \frac{2{\mathrm{b}}^2-4\mathrm{b}}{{\mathrm{x}}^3}& ,& \sqrt{2}\text{≤x<∞}\end{array}\right.$ 

continuous in the interval $[0,\infty ),$ then an ordered pair $\left(\mathrm{a},\mathrm{b}\right)$ is :

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{c}\frac{2{\mathrm{x}}^2}{\mathrm{a}}\mathrm{ }:0\le \mathrm{x}<1\\ \mathrm{a}\mathrm{ }:1\le \mathrm{x}<\sqrt{2}\\ \frac{2{\mathrm{b}}^2-4\mathrm{b}}{{\mathrm{x}}^3}:\sqrt{2}\le \mathrm{x}<\mathrm{\infty }\end{array}\right.$

Continuous in  $[0, \infty ]$

$\underset{\mathrm{x}\to 1^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(1\right)$

$$\begin{gathered} \underset{\mathrm{x}\to 1^{-}}{\lim }\frac{2{\mathrm{x}}^2}{\mathrm{a}}=\mathrm{a}⟹\frac{2}{\mathrm{a}}=\mathrm{a} \\ \therefore {\mathrm{a}}^2=2\Rightarrow a=\pm \sqrt{2} \end{gathered}$$

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

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

$$\begin{gathered} \underset{\mathrm{x}\to {\sqrt{2}}^{\mathrm{ }}}{\lim }\frac{2{\mathrm{b}}^2-4b}{x^3}=\sqrt{2} \\ \Rightarrow \frac{2{\mathrm{b}}^2-4\mathrm{b}}{2\sqrt{2}}=\sqrt{2},\Rightarrow 2{\mathrm{b}}^2-4\mathrm{b}=4,\Rightarrow \Rightarrow {\mathrm{b}}^2-2\mathrm{b}-2=0 \\ \frac{2{\mathrm{b}}^2-4\mathrm{b}}{2\sqrt{2}}=-\sqrt{2}\Rightarrow 2{\mathrm{b}}^2-4\mathrm{b}=-4\Rightarrow \Rightarrow {\mathrm{b}}^2-2\mathrm{b}+2=0 \end{gathered}$$

, 

$$\begin{gathered} a=\sqrt{2 }\Rightarrow \mathrm{b}=\frac{2\pm \sqrt{4+8}}{2}=1\pm \sqrt{3}\mathrm{ }, \\ \mathrm{a}=-\sqrt{2}\Rightarrow \mathrm{b} \mathrm{is} \mathrm{imaginary} \end{gathered}$$

$\therefore \left(\mathrm{a},\mathrm{b}\right)=(\sqrt{2},\mathrm{ }1-\sqrt{3})$

Option (3) is correct