Question:
If $c$ is a point at which Rolle's theorem holds for the function, $f(x)=\log _{e}\left(\frac{x^{2}+a}{7 x}\right)$ in the interval $[3,4]$, where $\alpha \in R$, then $f^{\prime \prime}(c)$ is equal to:
Correct Option: , 2
Solution:
Since, Rolle's theorem is applicable
$\therefore f(a)=f(b)$
$f(3)=f(4) \Rightarrow \alpha=12$
$f^{\prime}(x)=\frac{x^{2}-12}{x\left(x^{2}+12\right)}$
As $f^{\prime}(c)=0$ (by Rolle's theorem)
$x=\pm \sqrt{12}, \quad \therefore \quad c=\sqrt{12}, \therefore \quad f^{\prime \prime}(c)=\frac{1}{12}$