Q67 / 150 · MHT CET 2024 (02 May Shift 2) · Mathematics

The value of C for which Mean value Theorem holds for the function $\mathrm{f}(x)=\log _{\mathrm{e}} x$ on the interval $[1,3]$ is

  • A
    $\log _3 \mathrm{e}$
  • B
    $\log _{\mathrm{e}} 3$
  • C
    $\quad \frac{1}{2} \log _{\mathrm{e}} 3$
  • Correct answer
    D
    $2 \log _3 \mathrm{e}$

Solution

$\begin{aligned} \mathrm{f}(x) & =\log _{\mathrm{e}} x \\ \mathrm{f}(1) & =\log _{\mathrm{e}} 1=0 \\ \mathrm{f}(3) & =\log _{\mathrm{e}} 3 \text { and } \mathrm{f}^{\prime}(x)=\frac{1}{x} \end{aligned}$
By Lagrange's mean value theorem, $\begin{aligned} & f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\ & \Rightarrow \frac{1}{c}=\frac{\log _e 3-0}{2} \Rightarrow c=\frac{2}{\log _e 3} \Rightarrow \mathrm{c}=2 \log _3 e \end{aligned}$

Hence, the correct answer is option: 'D'.

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