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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
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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