Question

LMVT is applicable at $c=\sqrt{3}$ for the function $f\left(x\right)=x+\frac{1}{x}$ on the interval

• A. $[1,5]$
• B. $[1,4]$
• C. $[1,3]$
• D. $[1,2]$

Answer 1

The correct option is C $[1,3]$

$f\left(x\right)=x+\frac{1}{x}$,
LMVT holds at $c=\sqrt{3}$
Let the interval be $[a,b]$
$f^{\prime }\left(x\right)=1-\frac{1}{x^2}$
$\Rightarrow f^{\prime }\left(\sqrt{3}\right)=\frac{2}{3}$
$f^{\prime }\left(\sqrt{3}\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
$\Rightarrow \frac{2}{3}=\frac{b+\frac{1}{b}-a-\frac{1}{a}}{b-a}$
$\Rightarrow \frac{2}{3}=1-\frac{1}{ab}$
$\Rightarrow ab=3$