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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