A) \[\left( \frac{\sqrt{3}}{8},\frac{\sqrt{2}}{6} \right)\]
B) \[\left( \frac{\sqrt{2}}{2},\frac{\sqrt{3}}{2} \right)\]
C) \[\left( \frac{1}{2},\frac{\sqrt{2}}{2} \right)\]
D) None of these
Correct Answer: A
Solution :
| Let \[f\left( x \right)=\frac{\sin x}{x}\] |
| \[\therefore \]\[f'\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}\] |
| \[=\frac{(x-\tan x)\cos x}{{{x}^{2}}}<0\,\forall \,x\in \left[ \frac{\pi }{4},\frac{\pi }{3} \right]\] |
| \[\therefore \]\[f\left( x \right)=\frac{\sin x}{x}\] is decreases on the interval, \[\left[ \frac{\pi }{4},\frac{\pi }{3} \right]\] |
| \[\Rightarrow \] The least value of the function\[m=f\left( \frac{\pi }{3} \right)=\frac{\sin \left( \pi /3 \right)}{\left( \pi /3 \right)}=\frac{3\sqrt{3}}{2\pi }\] |
| and the greasiest value of the function\[M=f\left( \frac{\pi }{4} \right)=\frac{\sin \left( \pi /4 \right)}{\left( \pi /4 \right)}=\frac{2\sqrt{2}}{\pi }\] |
| Therefore\[\left( \frac{\pi }{3}-\frac{\pi }{4} \right)\frac{3\sqrt{3}}{2\pi }<\int\limits_{\pi /4}^{\pi /3}{\frac{\sin x}{x}}dx<\left( \frac{\pi }{3}-\frac{\pi }{4} \right)\frac{2\sqrt{2}}{\pi }\][Mean Value Theorem of Integral Calculus] |
| Hence\[\frac{\sqrt{3}}{8}<\int\limits_{\pi /4}^{\pi /3}{\frac{\sin x}{x}dx<\frac{\sqrt{2}}{6}}\] |
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