Question

$I=\underset{\pi /4}{\overset{\pi /3}{\int }}\left(\frac{8\sin x-\sin 2x}{x}\right)dx$. Then

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

Answer 1

Answer: (C)

Consider
$f(x)=8\sin x-\sin 2x$
$f^{\prime }(x)=8\sin x-2\cos 2x$
$f^{\prime }(x)=-8\sin x+4\sin 2x$
$=-8\sin x(1-\cos x)$
$\therefore f^{\prime }(x)<0x\in \left(\frac{\pi }{4},\frac{\pi }{3}\right)$
$\therefore f^{\prime }(x)$ is $\downarrow$ function
$f^{\prime }\left(\frac{\pi }{3}\right)<f^{\prime }(x)<f^{\prime }\left(\frac{\pi }{4}\right)$
$5<f^{\prime }(x)<\frac{8}{\sqrt{2}}$
$5<f^{\prime }(x)<4\sqrt{2}$
$5x<f(x)<4\sqrt{2}x$
$5<\frac{f(x)}{x}<4\sqrt{2}$
$\underset{\pi /4}{\overset{\pi /3}{\int }}5<\int \frac{f(x)}{x}<\underset{\pi /4}{\overset{\pi /3}{\int }}4\sqrt{2}$
$\underset{\pi /4}{\overset{\pi /3}{\int }}5<\int \frac{8\sin x-\sin 2x}{x}<\underset{\pi /4}{\overset{\pi /3}{\int }}4\sqrt{2}$
$\frac{5\pi }{12}<I<\frac{\sqrt{2}\pi }{3}$