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VITEEE 2017 Paper

Section: Mathematics

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Question : 82 of 125

Marks: +1, -0

A value of c for which conclusion of Mean Value Theorem holds for the function f (x) =

\log_e

x on the interval [1, 3] is

$\log_3$ e
$\log_e$ 3
2 $\log_3$ e
$\frac{1}{3}$ $\log_3$ e

Solution:

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}

... (i)

c ∊ [a , b]

∴ Given f (x) =

\log_c

c

∴ f ' (x) =

\frac{1}{x}

∴ equation (i) become

\frac{1}{c}

=

\frac{f(3)-f(1)}{3-1}

⇒

\frac{1}{c}

=

\frac{\log_e 3 - \log_e 1}{2}

=

\frac{\log_e 3}{2}

⇒ c =

\frac{2}{\log_e 3}

⇒ c =

2\log_3 e

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