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Solution for F(X) = X √ 1 − X , X < 1 . - CBSE (Science) Class 12 - Mathematics

Question

f(x) = $x\sqrt{1 - x} , x\frac{<}{}1$ .

Solution

$\text { Given }: f\left( x \right) = x\sqrt{1 - x}$

$\Rightarrow f'\left( x \right) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}$

$\text { For the local maxima or minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} = 0$

$\Rightarrow \sqrt{1 - x} = \frac{x}{2\sqrt{1 - x}}$

$\Rightarrow 2 - 2x = x$

$\Rightarrow 3x = 2$

$\Rightarrow x = \frac{2}{3}$

$\text { Thus, x } = \frac{2}{3} \text { is the possible point of local maxima or local minima }.$

$\text { Now },$

$f''\left( x \right) = \frac{- 1}{\sqrt{1 - x}} - \frac{1}{2}\left( \frac{\sqrt{1 - x} + \frac{x}{2\sqrt{1 - x}}}{\left( 1 - x \right)} \right) = \frac{- 1}{\sqrt{1 - x}} - \frac{1}{2}\left[ \frac{2 - x}{\left( 1 - x \right)\sqrt{1 - x}} \right]$

$\text { At }x = \frac{2}{3}:$

$f''\left( \frac{2}{3} \right) = \frac{- 1}{\sqrt{1 - \frac{2}{3}}} - \frac{1}{2}\left[ \frac{2 - \frac{2}{3}}{\left( 1 - \frac{2}{3} \right)\sqrt{1 - \frac{2}{3}}} \right] = - \sqrt{3} - \frac{\frac{4}{3}}{\frac{1}{3 \times \sqrt{3}}} = - \sqrt{3} - 4\sqrt{3} < 0$

$\text { So,} x = \frac{2}{3}\text { is the point of local maximum }.$

$\text { The local maximum value is given by }$

$f\left( \frac{2}{3} \right) = \frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3\sqrt{3}}$

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Solution F(X) = X √ 1 − X , X < 1 . Concept: Graph of Maxima and Minima.
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