A) \[-2\sqrt{2}\]
B) \[2\sqrt{2}\]
C) 3
D) -3
Correct Answer: B
Solution :
\[F(x)={{x}^{2}}+\frac{1}{{{x}^{2}}},\,\,g(x)=x-\frac{1}{x}\] \[h(x)=\frac{F(x)}{g(x)}\] \[h(x)=\frac{{{x}^{2}}+\frac{1}{{{x}^{2}}}}{x-\frac{1}{x}}\] \[f(\alpha )=\frac{{{\alpha }^{2}}+2}{\alpha }\] \[f(\alpha )=\alpha +\frac{2}{\alpha }\] \[F(\alpha )={{\left( \sqrt{\alpha }-\sqrt{\frac{2}{\alpha }} \right)}^{2}}+2\sqrt{2}\] Minimum value will be \[2\sqrt{2}\] At \[\sqrt{\alpha }-\sqrt{\frac{2}{\alpha }}=0\] \[\alpha =\sqrt{2}\]which is possible as \[x-\frac{1}{x}\in R\]
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