A) \[\left( \sqrt{2},1-\sqrt{3} \right)\]
B) \[\left( \sqrt{2},1-\sqrt{3} \right)\]
C) \[\left( \sqrt{2},-1+\sqrt{3} \right)\]
D) \[\left( -\sqrt{2},1+\sqrt{3} \right)\]
Correct Answer: A
Solution :
\[f(x)=\left\{ \begin{matrix} 2{{x}^{2}} & 0\le x<1 \\ a & {} \\ a & 1\le x<\sqrt{2} \\ \frac{2{{b}^{2}}-4b}{{{x}^{3}}} & \sqrt{2}\le x<\infty \\ \end{matrix} \right.\] is continuous in \[[0,\infty )\] continuous at \[x=1\]and \[x=\sqrt{2}\] \[\frac{Lim}{x\to {{1}^{-}}}f(x)=\frac{Lim}{x\to {{1}^{+}}}f(x)=f(1)\] \[\Rightarrow \,\frac{2}{a}=a\Rightarrow {{a}^{2}}=2\] ------------ (a) and\[\frac{Lim}{x\to \sqrt{2}}f(x)=\frac{Lim}{x\to {{\sqrt{2}}^{+}}}f(x)=f(\sqrt{2})\] \[\Rightarrow a=\frac{2{{b}^{2}}-4b}{2\sqrt{2}}\] \[\Rightarrow {{b}^{2}}-2b=\sqrt{2}a\] If \[a=\sqrt{2}\]then \[{{b}^{2}}-2b-2=0\Rightarrow 1\pm \sqrt{3}\] If \[a=-\sqrt{2}\] then \[{{b}^{2}}-2b+2=0\Rightarrow b\] is imaginary which is not possible \[\Rightarrow \,(a,b)=\left( \sqrt{2},1+\sqrt{3} \right)\]or \[\left( \sqrt{2},1-\sqrt{3} \right)\]
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