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# Solution for F(X) = 1+2 Sin X+3 Cos2x, 0 < X < 2 π 3 is (A) Minimum at X = π 2 (B) Maximum at X = Sin − 1 ( 1 √ 3 ) (C) Minimum at X = π 6 (D) Maximum at Sin − 1 ( 1 6 ) - CBSE (Science) Class 12 - Mathematics

#### Question

f(x) = 1+2 sin x+3 cos2x, $0\frac{<}{}x\frac{<}{}\frac{2\pi}{3}$ is

(a) Minimum at x =$\frac{\pi}{2}$

(b) Maximum at x = sin $- 1$ ( $\frac{1}{\sqrt{3}}$)

(c) Minimum at x = $\frac{\pi}{6}$

(d) Maximum at sin $- 1$ ($\frac{1}{6})$

#### Solution

$(a)\text { Minimum at } x = \frac{\pi}{2}$

$\text { Given }: f\left( x \right) = 1 + 2 \sin x + 3 \cos^2 x$

$\Rightarrow f'\left( x \right) = 2 \cos x - 6 \cos x \sin x$

$\Rightarrow f'\left( x \right) = 2 \cos x\left( 1 - 3 \sin x \right)$

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

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

$\Rightarrow 2 \cos x\left( 1 - 3 \sin x \right) = 0$

$\Rightarrow 2 \cos x = 0 or \left( 1 - 3 \sin x \right) = 0$

$\Rightarrow \cos x = 0 \ or \sin x = \frac{1}{3}$

$\Rightarrow x = \frac{\pi}{2} or x = \sin^{- 1} \left( \frac{1}{3} \right)$

$\text { Now,}$

$f''\left( x \right) = - 2 \sin x - 6 \cos 2x$

$\Rightarrow f''\left( \frac{\pi}{2} \right) = - 2 \sin \frac{\pi}{2} - 6 \cos \left( 2 \times \frac{\pi}{2} \right) = - 2 + 6 = 4 > 0$

$\text { So, x } = \frac{\pi}{2} \text { is a local minima }.$

$\text { Also },$

$f''\left( \sin^{- 1} \left( \frac{1}{3} \right) \right) = - 2 \sin \left( \sin^{- 1} \left( \frac{1}{3} \right) \right) - 6 \cos \left( \sin^{- 1} \left( \frac{1}{3} \right) \right) = \frac{- 2}{3} - 6 \times \frac{2\sqrt{2}}{3} = - \left( \frac{2}{3} + 4\sqrt{2} \right) < 0$

$\text { So,} x = \sin^{- 1} \left( \frac{1}{3} \right)\text { is a local maxima }.$

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Solution F(X) = 1+2 Sin X+3 Cos2x, 0 < X < 2 π 3 is (A) Minimum at X = π 2 (B) Maximum at X = Sin − 1 ( 1 √ 3 ) (C) Minimum at X = π 6 (D) Maximum at Sin − 1 ( 1 6 ) Concept: Graph of Maxima and Minima.
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