PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# F(X) = X + √ 1 − X , X ≤ 1 . - PUC Karnataka Science Class 12 - Mathematics

#### Question

f(x) = $x + \sqrt{1 - x}, x \leq 1$ .

#### Solution

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

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

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

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

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

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

$\Rightarrow 1 - x = \frac{1}{4}$

$\Rightarrow x = \frac{3}{4}$

$\text { Thus }, x = \frac{3}{4} \text { is the possible point of local maxima or local minima }.$

$\text { Now },$

$f''\left( x \right) = - \frac{\frac{1}{4\sqrt{1 - x}}}{4\left( 1 - x \right)}$

$\text { At }x = \frac{3}{4}:$

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

$\text { So,} x = \frac{3}{4} \text { is the point of local maximum }.$

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

$f\left( \frac{3}{4} \right) = \frac{3}{4} + \sqrt{1 - \frac{3}{4}} = \frac{5}{4}$

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