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)

$f(x)=\log (\sin (x))$ 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 $f^{\prime }(x)=\frac{1}{\sin x}\cos x=\cot x$
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
$f^{\prime }\left(c\right)-\frac{f\left(\frac{\pi }{6}\right)-f\left(\frac{5\pi }{6}\right)}{\left(\frac{\pi }{6}-\frac{5\pi }{6}\right)}$
$\Rightarrow \cot c=0\Rightarrow c=\frac{\pi }{2}$