Question

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

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{3\pi }{4}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{\pi }{4}$
B $\frac{\pi }{2}$

Lagrange's mean value theorem states that if $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists some $c$ between $a$ and $b$ such that $f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Given $f\left(x\right)=log\left(sinx\right)$ and $[a,b]=[\frac{\pi }{6},\frac{5\pi }{6}]$

$⟹f^{{}^{\prime }}\left(x\right)=\frac{cosx}{sinx}=cotx$

Therefore, $f^{{}^{\prime }}\left(c\right)=\frac{log\left(sinb\right)-log\left(sina\right)}{b-a}$

$⟹cotc=\frac{log\left(0.5\right)-log\left(0.5\right)}{\frac{5\pi }{6}-\frac{\pi }{6}}$

$⟹cotc=0$

$⟹c=\frac{\pi }{2}$