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# Solution for Find the Absolute Maximum and Minimum Values of the Function of Given by F ( X ) = Cos 2 X + Sin X , X ∈ [ 0 , π ] . - CBSE (Commerce) Class 12 - Mathematics

#### Question

Find the absolute maximum and minimum values of the function of given by $f(x) = \cos^2 x + \sin x, x \in [0, \pi]$ .

#### Solution

$\text { Given }: f\left( x \right) = \cos^2 x + \sin x$

$\Rightarrow f'\left( x \right) = 2 \cos x\left( - \sin x \right) + \cos x = - 2 \sin x \cos x + \cos x$

$\text { For a local maximum or a local minimum, we must have }$

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

$\Rightarrow - 2 \sin x \cos x + \cos x = 0$

$\Rightarrow \cos x \left( 2 \sin x - 1 \right) = 0$

$\Rightarrow \sin x = \frac{1}{2} or \cos x = 0$

$\Rightarrow x = \frac{\pi}{6} or \frac{\pi}{2} \left[ \because x \in \left( 0, \pi \right) \right]$

$\text { Thus, the critical points of f are } 0, \frac{\pi}{6}, \frac{\pi}{2} \text { and } \pi .$

$\text { Now },$

$f\left( 0 \right) = \cos^2 \left( 0 \right) + \sin \left( 0 \right) = 1$

$f\left( \frac{\pi}{6} \right) = \cos^2 \left( \frac{\pi}{6} \right) + \sin \left( \frac{\pi}{6} \right) = \frac{5}{4}$

$f\left( \frac{\pi}{2} \right) = \cos^2 \left( \frac{\pi}{2} \right) + \sin \left( \frac{\pi}{2} \right) = 1$

$f\left( \pi \right) = \cos^2 \left( \pi \right) + \sin \left( \pi \right) = 1$

$\text { Hence, the absolute maximum value when } x = \frac{\pi}{6}\text { is } \frac{5}{4} \text { and the absolute minimum value when }x = 0, \frac{\pi}{2}, \pi is 1 .$

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Solution Find the Absolute Maximum and Minimum Values of the Function of Given by F ( X ) = Cos 2 X + Sin X , X ∈ [ 0 , π ] . Concept: Graph of Maxima and Minima.
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