From mean value theorem f(b) - f(a) = (b - a)f'( x_1 ); a < x_1 <… - Vidyadip

MCQ

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} = $

✓
$\sqrt {ab} $
B
${{a + b} \over 2}$
C
${{2ab} \over {a + b}}$
D
${{b - a} \over {b + a}}$

✓

Answer

Correct option: A.

$\sqrt {ab} $

a
(a) $f'({x_1}) = \frac{{ - 1}}{{x_1^2}}$

$\therefore \frac{{ - 1}}{{x_1^2}} = \frac{{\frac{1}{b} - \frac{1}{a}}}{{b - a}} = - \frac{1}{{ab}} $

$\Rightarrow {x_1} = \sqrt {ab} $.

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