5. Continuity and Differentiation
easy

મધ્યકમાન પ્રમેય મુજબ $f(b) - f(a) = $ $(b - a)f'({x_1});$  $a < {x_1} < b$ જો $f(x) = {1 \over x}$, તો ${x_1} = $

A

$\sqrt {ab} $

B

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

C

${{2ab} \over {a + b}}$

D

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

Solution

(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} $.

Standard 12
Mathematics

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