Question

A value of $c$ for which the conclusion of mean value theorem holds for the function $f\left(x\right)={\log }_e\left(x\right)$ on the interval $\left[1,3\right]$ is

• A. $2{\log }_3\left(e\right)$
• B. $\frac{1}{2}{\log }_e\left(3\right)$
• C. ${\log }_3\left(e\right)$
• D. ${\log }_e\left(3\right)$

Answer 1

The correct option is A

$2{\log }_3\left(e\right)$

Explanation for correct answer:

Finding the value.

Given, $f\left(x\right)={\log }_e\left(x\right)$ on the interval $\left[1,3\right]$.

By using, Lagrange's mean value theorem $f\text{'}\left(c\right)=\frac{\left(f\left(b\right)-f\left(a\right)\right)}{\left(b-a\right)}$.

$\Rightarrow$$\frac{d}{dx}\left({\log }_e\left(x\right)\right)=\frac{\left(f\left(3\right)-f\left(1\right)\right)}{\left(3-1\right)}$

$\Rightarrow$ $\frac{1}{x}=\frac{\left({\log }_e\left(3\right)-{\log }_e\left(1\right)\right)}{2}$

$\Rightarrow$ $\frac{1}{c}=\frac{\left({\log }_e\left(3\right)\right)}{2}\left(\because {\log }_e\left(1\right)=0\right)$

$\Rightarrow$ $c=\frac{2}{\left({\log }_e\left(3\right)\right)}$

$\Rightarrow$ $c=2\left({\log }_3\left(e\right)\right)\left(\because {\log }_e\left(x\right)=\frac{1}{{\log }_x\left(e\right)}\right)$

Hence, option $\left(A\right)$ is correct.