JEE Mains · Maths · STD 12 - 7.2 definite integral
\(I=\int \limits_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x\). Then
- A \(\frac{\pi}{2} < I < \frac{3 \pi}{4}\)
- B \(\frac{\pi}{5} < I < \frac{5 \pi}{12}\)
- C \(\frac{5 \pi}{12} < I < \frac{\sqrt{2}}{3} \pi\)
- D \(\frac{3 \pi}{4} < I < \pi\)
Answer & Solution
Correct Answer
(C) \(\frac{5 \pi}{12} < I < \frac{\sqrt{2}}{3} \pi\)
Step-by-step Solution
Detailed explanation
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 \(\downarrow\)…
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