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(C) For the function $f(x)$ to be continuous in $[0, \infty)$,it must be continuous at the transition points $x=1$ and $x=\sqrt{2}$.$1$. Continuity at

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$(\sqrt{2}, 1 - \sqrt{3})$

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(C) For the function $f(x)$ to be continuous in $[0, \infty)$,it must be continuous at the transition points $x=1$ and $x=\sqrt{2}$.$1$. Continuity at $x=1$:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$$\frac{2(1)^2}{a} = a \implies \frac{2}{a} = a \implies a^2 = 2 \implies a = \pm \sqrt{2}$.$2$. Continuity at $x=\sqrt{2}$:$\lim_{x 	o \sqrt{2}^-} f(x) = \lim_{x 	o \sqrt{2}^+} f(x) = f(\sqrt{2})$$a = \frac{2b^2 - 4b}{(\sqrt{2})^3} = \frac{2b^2 - 4b}{2\sqrt{2}} = \frac{b^2 - 2b}{\sqrt{2}}$.Case $1$: If $a = \sqrt{2}$,then $\sqrt{2} = \frac{b^2 - 2b}{\sqrt{2}} \implies b^2 - 2b = 2 \implies b^2 - 2b - 2 = 0$.Using the quadratic formula $b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4+8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.So,$(a, b) = (\sqrt{2}, 1 \pm \sqrt{3})$.Comparing with the given options,the pair $(\sqrt{2}, 1 - \sqrt{3})$ matches option $C$.

Let $a, b \in R, (a \ne 0)$. If the function $f$ defined as
$f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \le x < 1 \\ a, & 1 \le x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3}, & \sqrt{2} \le x < \infty \end{cases}$
is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

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Let $a, b \in R, (a \ne 0)$. If the function $f$ defined as$f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \le x < 1 \\ a, & 1 \le x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3}, & \sqrt{2} \le x < \infty \end{cases}$is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is