The value of $\mathrm{c}$ for the function $\mathrm{f}(x)=\log x$ on $[1, \mathrm{e}]$ if LMVT can be…
The value of $\mathrm{c}$ for the function $\mathrm{f}(x)=\log x$ on $[1, \mathrm{e}]$ if LMVT can be applied, is
- $e-2$
- $e+1$
- $e-1$
- 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,
$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{e})-\mathrm{f}(1)}{\mathrm{e}-1} \\
& \Rightarrow \frac{1}{\mathrm{c}}=\frac{\log \mathrm{e}-\log 1}{\mathrm{e}-1} \\
& \Rightarrow \frac{1}{\mathrm{c}}=\frac{1}{\mathrm{e}-1} \\
& \Rightarrow \mathrm{c}=\mathrm{e}-1
\end{aligned}$
Asked in: MHT CET 2023 (10 May Shift 2)
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