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A value of $C$ for which the conclusion of Mean Value Theorem holds for the function $f(x)=$ $\log _{\mathrm{e}} x$ on the interval $[1,3]$ is
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The correct answer is:
$2 \log _3 e$
$2 \log _3 e$
Using mean value theorem
$\begin{aligned}
& f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\
& \Rightarrow \frac{1}{c}=\frac{\log 3-\log 1}{2} \\
& \Rightarrow c=\frac{2}{\log _e 3}=2 \log _3 e.
\end{aligned}$
$\begin{aligned}
& f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\
& \Rightarrow \frac{1}{c}=\frac{\log 3-\log 1}{2} \\
& \Rightarrow c=\frac{2}{\log _e 3}=2 \log _3 e.
\end{aligned}$
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