Question

For the function $x+\left(\frac{1}{x}\right),x\in [1,3]$, the value of $\mathrm{c}$ for the mean value theorem is

• A. $1$
• B. $\sqrt{3}$
• C. $2$
• D. None of these

Answer 1

The correct option is B

$\sqrt{3}$

Finding $\mathrm{c}$ using mean value theorem :

Let $f\left(x\right)$be an function define in interval $[a,b]$ then M.V.T is given by $\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{f}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{-}\mathbf{}\mathbf{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{b}\mathbf{}\mathbf{-}\mathbf{}\mathbf{a}}$

$\begin{array}{rcl}Letf\left(x\right)& =& x+\left(\frac{1}{x}\right),x\in [1,3]\\ f\text{'}\left(x\right)& =& 1-\frac{1}{x^2}\\ f\text{'}\left(c\right)& =& 1-\frac{1}{c^2}(\because f\text{'}(c)=\frac{f\left(b\right)-f\left(a\right)}{b-a})\\ b& =& 3\&a=1\\ 1-\frac{1}{c^2}& =& \frac{(3+{\displaystyle \frac{1}{3}})-(1+{\displaystyle \frac{1}{1}})}{3-1}\\ 1-\frac{1}{c^2}& =& \frac{{\displaystyle \frac{4}{3}}}{2}\\ 1-\frac{1}{c^2}& =& \frac{2}{3}\\ 1-\frac{2}{3}& =& \frac{1}{c^2}\\ \frac{1}{3}& =& \frac{1}{c^2}\\ c& =& \sqrt{3}\end{array}$

Hence, option (B) is correct answer.