From mean value theorem f(b) - f(a) = (b - a)f'({x₁}); a < {x₁} < b if f

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5. Continuity and Differentiation

easy

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} = $

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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