Q51 / 150 · MHT CET 2023 (10 May Shift 2) · Mathematics

The value of $\mathrm{c}$ for the function $\mathrm{f}(x)=\log x$ on $[1, \mathrm{e}]$ if LMVT can be applied, is

  • A
    $e-2$
  • B
    $e+1$
  • Correct answer
    C
    $e-1$
  • D
    e

Solution

$\begin{aligned} & \mathrm{f}(x)=\log x \\ & \Rightarrow \mathrm{f}^{\prime}(x)=\frac{1}{x} \end{aligned}$ By Lagrange's Mean value theorem, $\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{e})-\mathrm{f}(1)}{\mathrm{e}-1} \\ & \Rightarrow \frac{1}{\mathrm{c}}=\frac{\log \mathrm{e}-\log 1}{\mathrm{e}-1} \\ & \Rightarrow \frac{1}{\mathrm{c}}=\frac{1}{\mathrm{e}-1} \\ & \Rightarrow \mathrm{c}=\mathrm{e}-1 \end{aligned}$

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

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