Question

From mean value theorem $f\left(b\right)-f\left(a\right)=(b-a)f^{\prime }\left(x_1\right);a<x_1<b$ if $f\left(x\right)=\frac{1}{x}$, then $x_1$

• A. $\sqrt{ab}$
• B. $\frac{a+b}{2}$
• C. $\frac{2ab}{a+b}$
• D. $\frac{b-a}{a+b}$

Answer 1

The correct option is A $\sqrt{ab}$

$f^{\prime }\left(x_1\right)=\frac{-1}{x_1^2}\therefore \frac{-1}{x_1^2}=\frac{\frac{1}{b}-\frac{1}{a}}{b-a}=-\frac{1}{ab}\Rightarrow x_1=\sqrt{ab}$