Question

From mean value theorem, $f\left(b\right)-f\left(a\right)=(b-a)f^1\left(x_1\right);a<x_1<biff\left(x\right)=\frac{1}{x}then{x}_1=$

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

Answer 1

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

We have,

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

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

Given that,

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

$f\left(b\right)=\frac{1}{b}$

$f\left(a\right)=\frac{1}{a}$

Now,

$f^{\prime }\left(x_1\right)=\frac{\frac{1}{b}-\frac{1}{a}}{b-a}$

$=\frac{\frac{a-b}{ab}}{b-a}$

$f^{\prime }\left(x_1\right)=-\frac{1}{ab}$

Given that,

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

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

So,

$-\frac{1}{x^2}=-\frac{1}{ab}$

$x^2=ab$

$x=\sqrt{ab}$

Hence, this is the answer.