Let \(a, b \in(a \neq 0)\). If the function \(f\) is defined as
\(f(x)=\left\{\begin{array}{cc}\frac{2 x^2}{\mathrm{a}} & , 0 \leq x \lt 1 \ \mathrm{a} & , 1 \leq x \lt \sqrt{2} \ \frac{2 \mathrm{~b}^2-4 \mathrm{~b}}{x} ,\end{array}\right.\) \( \sqrt{2} \leq x \lt \infty\)
is continuous in the interval \([0, \infty)\), then an ordered pair \((\mathrm{a}, \mathrm{b})\) is

Function:
\(f(x)= \begin{cases}\frac{2 x^2}{a}, & 0 \leq x \lt 1 \\ a, & 1 \leq x \lt \sqrt{2} \\ \frac{2 b^2-4 b}{x}, & \sqrt{2} \leq x \lt \infty\end{cases}\)
Condition: Function \(f(x)\) must be continuous in the interval \([0, \infty)\). This implies:
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x), \text { and } \lim _{x \rightarrow \sqrt{2}^{-}} f(x)\) \(=\lim _{x \rightarrow \sqrt{2}^{+}} f(x) .\)
Step 1: Continuity at \(x=1\) :
\(\lim _{x \rightarrow 1^{-}} f(x)=\frac{2(1)^2}{a}=\frac{2}{a}, \quad \lim _{x \rightarrow 1^{+}} f(x)=a .\)
Equating these:
\(\frac{2}{a}=a \quad \Rightarrow \quad a^2=2 \quad \Rightarrow \quad a= \pm \sqrt{2}\)
Step 2: Continuity at \(x=\sqrt{2}\) :
\(\lim _{x \rightarrow \sqrt{2}^{-}} f(x)=a, \quad \lim _{x \rightarrow \sqrt{2}^{+}} f(x)=\frac{2 b^2-4 b}{\sqrt{2}}\)
Equating these:
\(a=\frac{2 b^2-4 b}{\sqrt{2}} \Rightarrow a \sqrt{2}=2 b^2-4 b .\)
Substitute \(a=\sqrt{2}\) :
\((\sqrt{2})(\sqrt{2})=2 b^2-4 b \quad \Rightarrow \quad 2=2 b^2-4 b\)
Simplify:
\(b^2-2 b-1=0\)
Solve for \(b\) :
\(b=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{3} .\)
Final Values: \(a=\sqrt{2}, b=1-\sqrt{3}\).
Answer: \((\sqrt{2}, 1-\sqrt{3})\), Option 3.