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If $f(x)=\log (\sin x), x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]$, then value of 'c'by applyng L.mv.T. is
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The correct answer is:
$\frac{\pi}{2}$
(B)
Value of ' ' by applying L.M.V.T. is to be found out
$\begin{aligned}
f(x) &=\log (\sin x) \text { on }\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right] \Rightarrow \operatorname{lot} a=\frac{\pi}{6}, b=\frac{5 \pi}{6} \\
f^{\prime}(x) &=\frac{\cos x}{\sin x}=\cot x \Rightarrow f^{\prime}(c)=\cot c \\
f^{\prime}(c) &=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{5 \pi}{6}\right)-f\left(\frac{\pi}{6}\right)}{5 \frac{\pi}{6}-\frac{\pi}{6}} \\
\cot c &=\frac{\log \left[\sin \left(\frac{5 \pi}{6}\right)\right]-\log \left[\sin \left(\frac{\pi}{6}\right)\right]}{\frac{2 \pi}{3}}=\frac{\log \frac{1}{2}-\log \left(\frac{1}{2}\right)}{2 \frac{\pi}{3}} \\
\cot c=0 & \Rightarrow c=\frac{\pi}{2}
\end{aligned}$
Value of ' ' by applying L.M.V.T. is to be found out
$\begin{aligned}
f(x) &=\log (\sin x) \text { on }\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right] \Rightarrow \operatorname{lot} a=\frac{\pi}{6}, b=\frac{5 \pi}{6} \\
f^{\prime}(x) &=\frac{\cos x}{\sin x}=\cot x \Rightarrow f^{\prime}(c)=\cot c \\
f^{\prime}(c) &=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{5 \pi}{6}\right)-f\left(\frac{\pi}{6}\right)}{5 \frac{\pi}{6}-\frac{\pi}{6}} \\
\cot c &=\frac{\log \left[\sin \left(\frac{5 \pi}{6}\right)\right]-\log \left[\sin \left(\frac{\pi}{6}\right)\right]}{\frac{2 \pi}{3}}=\frac{\log \frac{1}{2}-\log \left(\frac{1}{2}\right)}{2 \frac{\pi}{3}} \\
\cot c=0 & \Rightarrow c=\frac{\pi}{2}
\end{aligned}$
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