Question

From mean value theorem $f\left(b\right)-f\left(a\right)=(b-a)f^{\prime }\left(x_1\right);z<x_1<b$ 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. $\frac{a-b}{b+a}$

Answer 1

Answer: (A)

$\sqrt{ab}$

Given $f\left(x\right)=\frac{1}{x}$.

Now $f^{\prime }\left(x\right)=-\frac{1}{x^2}$.

Also given,

$f\left(b\right)-f\left(a\right)=(b-a)f^{\prime }\left(x_1\right)$

Or, $\frac{a-b}{ab}=(b-a).\frac{-1}{x_1^2}$

Or, $x_1^2=ab$

Or, $x_1=\sqrt{ab}$.