Question

The value of $c$ in Rolle's theorem for $f\left(x\right)=\log \left(\frac{x^2+ab}{x(a+b)}\right)$

in $(a,b)$ where $a>0$ is

• A. A.M. of $a,b$
• B. G.M. of $a,b$
• C. H.M. of $a,b$
• D. $\frac{1}{a}+\frac{1}{b}$

Answer 1

The correct option is B G.M. of $a,b$

We have $f\left(x\right)=\log \left(\frac{x^2+ab}{x(a+b)}\right)$

$=\log (x^2+ab)-\log \left(x\right(a+b\left)\right)$

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

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

Now, $f^{\prime }\left(c\right)=0$ gives $c^2-ab=0$

$\Rightarrow c=\sqrt{ab}\in (a,b)$
Clearly, $c$ is G.M. of $a,b$