Question

Let \(I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x\). Then
(A) \(\frac{\sqrt{3}}{8} \leq \mathrm{I} \leq \frac{\sqrt{2}}{6}\)
(B) \(\frac{\sqrt{3}}{2 \pi} \leq \mathrm{I} \leq \frac{2 \sqrt{3}}{\pi}\)
(C) \(\frac{\sqrt{3}}{9} \leq \mathrm{I} \leq \frac{\sqrt{2}}{16}\)
(D) \(\pi \leq \mathrm{I} \leq \frac{4 \pi}{3}\)

Answer 1

Ans: (A)
Hint \(: f(x)=\frac{\sin x}{x}\) (decreasing function)
$$
\begin{aligned}
&\Rightarrow \mathrm{f}\left(\frac{\pi}{3}\right)<\mathrm{f}(\mathrm{x})<\mathrm{f}\left(\frac{\pi}{4}\right) \\
&\Rightarrow \frac{\sqrt{3}}{2} \times \frac{3}{\pi} \int_{\pi / 4}^{\frac{\pi}{3}} \mathrm{dx}<\mathrm{I}<\frac{1}{\sqrt{2}} \times \frac{4}{\pi} \times \int_{\pi / 4}^{\frac{\pi}{3}} \mathrm{dx} \\
&\Rightarrow \frac{\sqrt{3}}{8}<\mathrm{I}<\frac{\sqrt{2}}{6}
\end{aligned}
$$