The value of C for which Mean value Theorem holds for the function $\mathrm{f}(x)=\log _{\mathrm{e}} x$ on…
- $\log _3 \mathrm{e}$
- $\log _{\mathrm{e}} 3$
- $\quad \frac{1}{2} \log _{\mathrm{e}} 3$
- $2 \log _3 \mathrm{e}$
Solution
By Lagrange's mean value theorem, $\begin{aligned} & f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\ & \Rightarrow \frac{1}{c}=\frac{\log _e 3-0}{2} \Rightarrow c=\frac{2}{\log _e 3} \Rightarrow \mathrm{c}=2 \log _3 e \end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 2)