Question

If $c$ is a point at which Rolle's theorem holds for the function, $f\left(x\right)={\log }_e\left(\frac{x^2+\alpha }{7x}\right)$ in the interval $[3,4]$, where $\alpha \in R$, then $f^{\prime \prime }\left(c\right)$ is equal to:

• A. $\frac{1}{12}$
• B. $-\frac{1}{12}$
• C. $-\frac{1}{24}$
• D. $\frac{\sqrt{3}}{7}$

Answer 1

The correct option is A $\frac{1}{12}$

Rolle's theorem is applicable on $f\left(x\right)$ in $[3,4]$
$\Rightarrow f\left(3\right)=f\left(4\right)$
$\Rightarrow \ln \left(\frac{9+\alpha }{21}\right)=\ln \left(\frac{16+\alpha }{28}\right)$
$\Rightarrow \frac{9+\alpha }{21}=\frac{16+\alpha }{28}$
$\Rightarrow 36+4\alpha =48+3\alpha \Rightarrow \alpha =12$

Now, $f\left(x\right)=\ln \left(\frac{x^2+12}{7x}\right)$ $\Rightarrow f^{\prime }\left(x\right)=\frac{7x}{x^2+12}\times (\frac{1}{7}-\frac{12}{7x^2})$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x^2-12}{x(x^2+12)}$

$f^{\prime }\left(c\right)=0\Rightarrow c=2\sqrt{3}$

$f^{\prime \prime }\left(x\right)=\frac{-x^4+48x^2+144}{x^2(x^2+12{)}^2}$
$\therefore f^{\prime \prime }\left(c\right)=f^{\prime \prime }\left(2\sqrt{3}\right)=\frac{1}{12}$