I= _ / 4 ^ / 3 (8 x- 2 x x ) d x. Then - Vidyadip

MCQ

$I=\int \limits_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x$. Then

A
$\frac{\pi}{2} < I < \frac{3 \pi}{4}$
B
$\frac{\pi}{5} < I < \frac{5 \pi}{12}$
✓
$\frac{5 \pi}{12} < I < \frac{\sqrt{2}}{3} \pi$
D
$\frac{3 \pi}{4} < I < \pi$

✓

Answer

Correct option: C.

$\frac{5 \pi}{12} < I < \frac{\sqrt{2}}{3} \pi$

c
Consider

$f(x)=8 \sin x-\sin 2 x$

$f^{\prime}(x)=8 \sin x-2 \cos 2 x$

$f^{\prime \prime}(x)=-8 \sin x+4 \sin 2 x$

$=-8 \sin x(1-\cos x)$

$\therefore f^{\prime \prime}(x)<0 x \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$

$\therefore f ^{\prime}( x )$ is $\downarrow$ function

$f^{\prime}\left(\frac{\pi}{3}\right)$

$5 < f^{\prime}(x)<\frac{8}{\sqrt{2}}$

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