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प्रश्न
`f(x)=xsqrt(1-x), x<=1` .
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उत्तर
\[\text { Given }: f\left( x \right) = x\sqrt{1 - x}\]
\[ \Rightarrow f'\left( x \right) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} = 0\]
\[ \Rightarrow \sqrt{1 - x} = \frac{x}{2\sqrt{1 - x}}\]
\[ \Rightarrow 2 - 2x = x\]
\[ \Rightarrow 3x = 2\]
\[ \Rightarrow x = \frac{2}{3} \]
\[\text { Thus, x } = \frac{2}{3} \text { is the possible point of local maxima or local minima }. \]
\[\text { Now }, \]
\[ f''\left( x \right) = \frac{- 1}{\sqrt{1 - x}} - \frac{1}{2}\left( \frac{\sqrt{1 - x} + \frac{x}{2\sqrt{1 - x}}}{\left( 1 - x \right)} \right) = \frac{- 1}{\sqrt{1 - x}} - \frac{1}{2}\left[ \frac{2 - x}{\left( 1 - x \right)\sqrt{1 - x}} \right]\]
\[\text { At }x = \frac{2}{3}: \]
\[ f''\left( \frac{2}{3} \right) = \frac{- 1}{\sqrt{1 - \frac{2}{3}}} - \frac{1}{2}\left[ \frac{2 - \frac{2}{3}}{\left( 1 - \frac{2}{3} \right)\sqrt{1 - \frac{2}{3}}} \right] = - \sqrt{3} - \frac{\frac{4}{3}}{\frac{1}{3 \times \sqrt{3}}} = - \sqrt{3} - 4\sqrt{3} < 0\]
\[\text { So,} x = \frac{2}{3}\text { is the point of local maximum }. \]
\[\text { The local maximum value is given by }\]
\[f\left( \frac{2}{3} \right) = \frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3\sqrt{3}}\]
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