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# Solution for F(X) = X √ 32 − X 2 , − 5 < X < 5 . - CBSE (Commerce) Class 12 - Mathematics

#### Question

f(x) = $x\sqrt{32 - x^2}, - 5\frac{<}{}x\frac{<}{}5$ .

#### Solution

$\text { Given }: f\left( x \right) = x\sqrt{32 - x^2}$

$\Rightarrow f'\left( x \right) = \sqrt{32 - x^2} - \frac{x^2}{\sqrt{32 - x^2}}$

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

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

$\Rightarrow \sqrt{32 - x^2} - \frac{x^2}{\sqrt{32 - x^2}} = 0$

$\Rightarrow \sqrt{32 - x^2} = \frac{x}{\sqrt{32 - x^2}}$

$\Rightarrow 32 - x^2 = x^2$

$\Rightarrow x^2 = 16$

$\Rightarrow x = \pm 4$

$\text { Thus, x = 4 and x = - 4 are the possible points of local maxima or local minima }.$

$\text { Now,}$

$f''\left( x \right) = \frac{- x}{\sqrt{32 - x^2}} - \left( \frac{2x\sqrt{32 - x^2} + \frac{x^3}{\sqrt{32 - x^2}}}{32 - x^2} \right) = \frac{- x}{\sqrt{32 - x^2}} - \left( \frac{2x\left( 32 - x^2 \right) + x^3}{\left( 32 - x^2 \right)\sqrt{32 - x^2}} \right)$

$\text { At }x = 4:$

$f''\left( 4 \right) = \frac{- 4}{\sqrt{32 - 4^2}} - \left[ \frac{8\left( 32 - 4^2 \right) + 4^3}{\left( 32 - 4^2 \right)\sqrt{32 - 4^2}} \right] = - 1 - \frac{192}{64} = - 3 < 0$

$\text { So, x = 4 is the point of local maximum } .$

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

$f\left( 4 \right) = 4\sqrt{32 - 4^2} = 16$

$\text { At } x = - 4:$

$f''\left( - 4 \right) = \frac{4}{\sqrt{32 - 4^2}} + \left[ \frac{8\left( 32 - 4^2 \right) - 4^3}{\left( 32 - 4^2 \right)\sqrt{32 - 4^2}} \right] = 1 + 2 = 3 > 0$

$\text { So, x = - 4 is the point of local minimum } .$

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

$f\left( - 4 \right) = - 4\sqrt{32 - 4^2} = - 16$

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Solution F(X) = X √ 32 − X 2 , − 5 < X < 5 . Concept: Graph of Maxima and Minima.
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