5. Continuity and Differentiation
easy

मध्यमान प्रमेय $f(b) - f(a) = (b - a)f'({x_1});$   $a < {x_1} < b$ से यदि $f(x) = \frac{1}{x}$, तो${x_1} = $

A

$\sqrt {ab} $

B

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

C

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

D

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

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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.