From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$ $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $
From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$ $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $
- A
$\sqrt {ab} $
- B
${{a + b} \over 2}$
- C
${{2ab} \over {a + b}}$
- D
${{b - a} \over {b + a}}$
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