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The value of ‘c’ in lagrange’s mean value theorem for $f (x) = \log x$ on [1, e] is

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$e - 2$

$\frac{e}{2}$

$1 - e$

$e - 1$

answer is B.

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Detailed Solution

$\begin{aligned} f (x) = \log x, [a, b] = [1, e] \\ f^{1} (x) = \frac{1}{x} \\ By L.M.V.T. \Rightarrow f^{1} (c) = \frac{f (b) - f (a)}{b - a} \\ \frac{1}{c} = \frac{1 - 0}{e - 1} \Rightarrow [c = e - 1] \end{aligned}$

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