max _{0 leq x leq pi}left{x-2 sin x cos x+fra | vedclass

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(A) Let $f(x) = x - 2 \sin x \cos x + \frac{1}{3} \sin 3x = x - \sin 2x + \frac{1}{3} \sin 3x$. To find the maximum,we differentiate $f(x)$ with respe

Vedclass · Accepted Answer

$\frac{5 \pi+2+3 \sqrt{3}}{6}$

Vedclass · Answer

(A) Let $f(x) = x - 2 \sin x \cos x + \frac{1}{3} \sin 3x = x - \sin 2x + \frac{1}{3} \sin 3x$. To find the maximum,we differentiate $f(x)$ with respect to $x$: $f'(x) = 1 - 2 \cos 2x + \cos 3x$. Setting $f'(x) = 0$: $1 - 2(2 \cos^2 x - 1) + (4 \cos^3 x - 3 \cos x) = 0$ $1 - 4 \cos^2 x + 2 + 4 \cos^3 x - 3 \cos x = 0$ $4 \cos^3 x - 4 \cos^2 x - 3 \cos x + 3 = 0$ $4 \cos^2 x (\cos x - 1) - 3 (\cos x - 1) = 0$ $(4 \cos^2 x - 3)(\cos x - 1) = 0$. This gives $\cos x = 1$ or $\cos x = \pm \frac{\sqrt{3}}{2}$. For $x \in [0, \pi]$,$x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$. Evaluating $f(x)$ at these points: $f(0) = 0 - 0 + 0 = 0$. $f(\pi) = \pi - 0 + 0 = \pi$. $f(\frac{\pi}{6}) = \frac{\pi}{6} - \sin(\frac{\pi}{3}) + \frac{1}{3} \sin(\frac{\pi}{2}) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{\pi + 2 - 3\sqrt{3}}{6}$. $f(\frac{5\pi}{6}) = \frac{5\pi}{6} - \sin(\frac{5\pi}{3}) + \frac{1}{3} \sin(\frac{5\pi}{2}) = \frac{5\pi}{6} - (-\frac{\sqrt{3}}{2}) + \frac{1}{3}(1) = \frac{5\pi + 3\sqrt{3} + 2}{6}$. Comparing the values,the maximum value is $\frac{5\pi + 2 + 3\sqrt{3}}{6}$.

$\max _{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=$

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