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

$\begin{array}{l}f\left(3\right)=f\left(4\right)\\ \ln \left(\frac{9+\alpha }{21}\right)=\ln \left(\frac{16+\alpha }{28}\right)\\ \Rightarrow 9+\alpha /21=16+\alpha /28\\ \Rightarrow 36+4\alpha =48+3\alpha \\ \Rightarrow \alpha =12\\ \text{ Now, }\\ f\left(x\right)=\ln \left(\frac{\left(x^2+12\right)}{7x}\right)\\ f^{\mathrm{\prime }}\left(x\right)=\left[\frac{7x}{\left(x^2+12\right)}\right]\times \frac{\left(7x\times 2x-\left(x^2+12\right)\times 7\right)}{(7x{)}^2}\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{\left(x^2-12\right)}{\left(x\left(x^2+12\right)\right)}\\ f^{\mathrm{\prime }}\left(c\right)=0\Rightarrow c=2\sqrt{3}\\ \\ \therefore f^{\prime \prime }\left(x\right)=\frac{\left(-x^4+48x^2+144\right)}{x^2{\left(x^2+12\right)}^2}\\ \Rightarrow f\text{'}\text{'}\left(c\right)=\frac{1}{12}\end{array}$