Q42 / 120 · VITEEE 2017 · Mathematics
A value of $\mathrm{c}$ for which conclusion of Mean Value Theorem holds for the function $f(x)=\log _{e} x$ on the interval $[1,3]$ is
- A$\log _{3} \mathrm{e}$
- B$\log _{e} 3$
- Correct answerC$2 \log _{3} \mathrm{e}$
- D$\frac{1}{2} \log _{3} \mathrm{e}$
Solution
Using Lagrange's Mean Value Theorem
Let $\mathrm{f}(\mathrm{x})$ be a function defined on $[\mathrm{a}, \mathrm{b}]$
then, $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
$\mathrm{c} \in[\mathrm{a}, \mathrm{b}]$
$\therefore \quad$ Given $f(x)=\log _{e} x \quad \therefore f^{\prime}(x)=\frac{1}{x}$
$\therefore \quad$ equation (i) become
$$
\begin{aligned}
& \frac{1}{c}=\frac{f(3)-f(1)}{3-1} \\
\Rightarrow & \frac{1}{c}=\frac{\log _{e} 3-\log _{e} 1}{2}=\frac{\log _{\mathrm{e}} 3}{2} \Rightarrow c=\frac{2}{\log _{\mathrm{e}} 3} \\
\Rightarrow & c=2 \log _{3} e
\end{aligned}
$$
Hence, the correct answer is option: 'C'.