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Question: From mean value theorem \(f(b) - f(a) = (b - a)f^{'}(x_{1}),a < x_{1} < b\) if \(f(x) = \frac{1}{x}\...

From mean value theorem f(b)−f(a)=(b−a)f′(x1),a<x1<bf(b) - f(a) = (b - a)f^{'}(x_{1}),a < x_{1} < b if f(x)=1xf(x) = \frac{1}{x} then x1x_{1}

A

ab\sqrt{ab}

B

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

C

a+b2\frac{a + b}{2}

D

b−ab+a\frac{b - a}{b + a}

Answer

ab\sqrt{ab}

Explanation

Solution

f′(x1)=−1x12f^{'}(x_{1}) = \frac{- 1}{x_{1}^{2}}, ∴\therefore −1x12=1b−1ab−a=−1ab⇒x1=ab\frac{- 1}{x_{1}^{2}} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a} = - \frac{1}{ab} \Rightarrow x_{1} = \sqrt{ab}.