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Q.

By mean value theorem, $f (b) - f (a) = (b - a) . f^{1} (x_{1}); a < x_{1} < b$ if $f (x) = \frac{1}{x}$ then $x_{1} =$

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a

$\frac{a + b}{2}$

b

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

c

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

d

$\sqrt{a b}$

answer is A.

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Detailed Solution

$f^{1} (x_{1}) = \frac{f (b) - f (a)}{b - a}$

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