A) A local maxima
B) A local minima
C) Neither a local maxima nor a local minima
D) None of these
Correct Answer: A
Solution :
Let \[f(x)=x\sqrt{1-{{x}^{2}}}\] Þ \[f'(x)=\frac{1-2{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}=0\Rightarrow x=\pm \frac{1}{\sqrt{2}}\] But as \[x>0\], we have \[x=\frac{1}{\sqrt{2}}\] Now, again \[{f}''(x)=\frac{\sqrt{1-{{x}^{2}}}(-4x)-(1-2{{x}^{2}})\frac{-x}{\sqrt{1-{{x}^{2}}}}}{(1-{{x}^{2}})}\] \[=\frac{2{{x}^{3}}-3x}{{{(1-{{x}^{2}})}^{3/2}}}\] \[\Rightarrow {f}''\left( \frac{1}{\sqrt{2}} \right)=-ve\]. Then \[f(x)\] is maximum at \[x=\frac{1}{\sqrt{2}}\].
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