(C) Given function $f(x) = \log x$ on the interval $[1, e]$.First,we find the derivative of the function:$f'(x) = \frac{1}{x}$.According to Lagrange's

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(C) Given function $f(x) = \log x$ on the interval $[1, e]$.First,we find the derivative of the function:$f'(x) = \frac{1}{x}$.According to Lagrange's Mean Value Theorem $(LMVT)$,there exists at least one point $c \in (1, e)$ such that:$f'(c) = \frac{f(e) - f(1)}{e - 1}$.Substituting the values:$f(e) = \log e = 1$ and $f(1) = \log 1 = 0$.So,$\frac{1}{c} = \frac{1 - 0}{e - 1}$.$\frac{1}{c} = \frac{1}{e - 1}$.Therefore,$c = e - 1$.

Class 12 Mathematics Continuity and Differentiability Rolle’s theorem, Lagrange's mean value theorem

Medium

The value of $c$ for the function $f(x) = \log x$ on $[1, e]$ if Lagrange's Mean Value Theorem $(LMVT)$ is applied,is

MHT CET 2023 All MHT CET

A
$e-2$
B
$e+1$
C
$e-1$
D
$e$

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Continuity and Differentiability

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Rolle’s theorem, Lagrange's mean value theorem

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