Mathematics | VITEEE 2017 | Discussion Page

A value of c for which conclusion of Mean Value Theorem holds for the function f (x) = log_ex on the interval [1, 3] is

$\log_{3}{e}$

$\log_{e}{3}$

$2\log_{3}{e}$

$\frac{1}{2}\log_{3}{e}$

Option C

Using Lagrange's Mean Value Theorem Let f(x) be a function defined on [a, b]

then f'(c) = $\frac{f(b) - f(a)}{b - a}$ ...(1)

$c\epsilon[a, b]$

$\therefore Given f(x) =\log_{e}{x}$ $\therefore f(x) =\frac{1}{x}$

equation (1) become

$\frac{1}{c}=\frac{f(3) - f(1)}{3 - 1}$

$\Rightarrow\frac{1}{c}=\frac{\log_{e}{3} - \log_{e}{1}}{2}$ = $\frac{\log_{e}{3}}{2}$

c = $2\log_{3}{e}$

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