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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