Question

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

• A. $\sqrt{ab}$
• B. $\frac{[a+b]}{2}$
• C. $\left[\frac{2ab}{(a+b)}\right]$
• D. $\frac{[b–a]}{[b+a]}$

Answer 1

The correct option is A

$\sqrt{ab}$

Explanation for the correct option:

Step 1. Find the value of $x$

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

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

Differentiate equation (1) with respect to $x$

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

Also given, $f\left(b\right)-f\left(a\right)=(b-a)f\text{'}\left(x\right)$ …(2)

Step 2. Put the values of $f\left(a\right),f\left(b\right)$ and $f\text{'}\left(x_1\right)$ in equation (2), we get

$\Rightarrow$$\frac{1}{b}-\frac{1}{a}=(b-a)\times \frac{-1}{{x_1}^2}$

$\Rightarrow$ $\frac{a-b}{ab}=\left(b-a\right)\times \frac{-1}{{x_1}^2}$

$\Rightarrow$ ${x_1}^2=ab$

$\Rightarrow$ $x_1=\sqrt{\mathbf{a}\mathbf{b}}$

Hence, Option ‘A’ is Correct.