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The value of $c$ in Lagrange's mean value theorem for the function $f(x)=\log (\sin x)$ in the interval $\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]$ is
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The correct answer is:
$\frac{\pi}{2}$
$f(x)=\log (\sin (x))$ is continuous in $\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]$ and differentiable in $\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ as its derivative
$$
f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x
$$
$\operatorname{Now}, f\left(\frac{\pi}{6}\right)=\log \left(\sin \left(\frac{\pi}{6}\right)\right)=\log \left(\frac{1}{2}\right)$
$$
\int\left(\frac{5 \pi}{6}\right)=\log \sin \left(\frac{5 \pi}{6}\right)=\log \sin \left(\pi \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)
$$
Now, according to Lagrange's mean value theorem, there exists a point $c \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ such that
$$
f^{\prime}(c)=\frac{f\left(\frac{\pi}{6}\right)-f\left(\frac{5 \pi}{6}\right)}{\left(\frac{\pi}{6}-\frac{5 \pi}{6}\right)} \Rightarrow \cot c=0 \Rightarrow c=\frac{\pi}{2}
$$
$$
f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x
$$
$\operatorname{Now}, f\left(\frac{\pi}{6}\right)=\log \left(\sin \left(\frac{\pi}{6}\right)\right)=\log \left(\frac{1}{2}\right)$
$$
\int\left(\frac{5 \pi}{6}\right)=\log \sin \left(\frac{5 \pi}{6}\right)=\log \sin \left(\pi \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)
$$
Now, according to Lagrange's mean value theorem, there exists a point $c \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ such that
$$
f^{\prime}(c)=\frac{f\left(\frac{\pi}{6}\right)-f\left(\frac{5 \pi}{6}\right)}{\left(\frac{\pi}{6}-\frac{5 \pi}{6}\right)} \Rightarrow \cot c=0 \Rightarrow c=\frac{\pi}{2}
$$
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