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The value of ‘c’ in Rolle’s theorem for $f (x) = \log (\frac{x^{2} + a b}{x (a + b)})$ in (a, b) where a > 0 is

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H.M. of a, b

$\frac{1}{a} + \frac{1}{b}$

A.M. of a, b

G.M. of a, b

answer is B.

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Detailed Solution

$f^{'} (c) = 0 \Rightarrow c^{2} = a b \Rightarrow c = \sqrt{a b}$

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