If $f(x) = \log(\sin x)$,$x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$,then the value of $c$ by applying Lagrange's Mean Value Theorem $(LMVT)$ is:

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

Explore More

Similar Questions

$A$ value of $c$ for which the conclusion of the Mean Value Theorem holds for the function $f(x) = \log_{e}x$ on the interval $[1, 3]$ is

If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$,where $a_1 \neq 0$ and $n \geq 2$,has a positive root $x = \alpha$,then the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a positive root which is:

Difficult
View Solution

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $[-1, 1]$ because

Consider the function $f(x)=2x^3-3x^2-x+1$ and the intervals $I_1=[-1,0]$,$I_2=[0,1]$,$I_3=[1,2]$,$I_4=[-2,-1]$. Then,

The constant $c$ of Rolle's theorem for the function $f(x)=(x-1)^3(x-2)^5$ in the interval $[1, 2]$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo