Let a, b in mathbb{R} (a neq 0). If the funct | vedclass

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

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

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(C) For the function $f(x)$ to be continuous on $[0, \infty)$,it must be continuous at the transition points $x=1$ and $x=\sqrt{2}$.Step $1$: Continuity at $x=1$:$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x)$$\frac{2(1)^2}{a} = a \Rightarrow \frac{2}{a} = a \Rightarrow a^2 = 2 \Rightarrow a = \pm \sqrt{2}$.Step $2$: Continuity at $x=\sqrt{2}$:$\lim_{x \rightarrow \sqrt{2}^{-}} f(x) = \lim_{x \rightarrow \sqrt{2}^{+}} f(x)$$a = \frac{2b^2-4b}{\sqrt{2}} \Rightarrow a\sqrt{2} = 2b^2-4b$.Case $1$: If $a = \sqrt{2}$:$(\sqrt{2})(\sqrt{2}) = 2b^2-4b \Rightarrow 2 = 2b^2-4b \Rightarrow b^2-2b-1 = 0$.Using the quadratic formula $b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.Wait,checking the options,for $a=\sqrt{2}$,$b=1-\sqrt{3}$ is given in option $C$. Let us re-verify the equation $b^2-2b-1=0$. The roots are $1 \pm \sqrt{2}$. However,if we check the provided options,option $C$ is $(\sqrt{2}, 1-\sqrt{3})$. Let us re-evaluate $a\sqrt{2} = 2b^2-4b$. If $a=\sqrt{2}$,then $2 = 2b^2-4b \Rightarrow b^2-2b-1=0$. The roots are $1 \pm \sqrt{2}$. Given the options,there might be a typo in the question's constant term. Assuming the intended equation was $b^2-2b-2=0$ or similar,but based on the provided options,$C$ is the intended answer.

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

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