Question

The value of $c$ in Lagrange's theorem for the function $f(x) = \log (\sin \, x)$ in the interval $\left[ \frac{\pi}{6} , \frac{5 \pi}{6} \right]$ is

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{2}$
• C. $\frac{2\pi}{3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi}{2}$

Answer 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 $f'(x) = \frac{1}{\sin \,x} \cos \, x = \cot \, x$
Now, $f \left(\frac{\pi}{6} \right) = \log \left( \sin \left(\frac{\pi}{6} \right)\right) = \log \left(\frac{1}{2} \right)$
$f \left(\frac{5 \pi}{6} \right) = \log \, \sin \left(\frac{5 \pi}{6} \right) = \log \, \sin \left( \pi - \frac{\pi}{6} \right) = \log \left( \frac{1}{2} \right)$
Now, according to Lagrange's theorem, there exist a point $c \, \in \left( \frac{\pi}{6} , \frac{5\pi}{6} \right)$ such that
$f'\left(c\right) - \frac{f\left(\frac{\pi}{6} \right)-f\left(\frac{5\pi}{6}\right)}{\left(\frac{\pi }{6} - \frac{5\pi }{6}\right)}$
$\Rightarrow \:\:\: \cot \, c = 0 \:\:\: \Rightarrow \: c = \frac{\pi}{2}$