For  $f\left(x\right)=\ln \left(\frac{x^2+\alpha }{7x}\right)$  Roll's theorem is applicable on $\text{ [3, 4]}$ with $c\in (3,4)$ such that $f^{\mathrm{\prime }}\left(c\right)=0\text{. }$ The value of $f^{\prime \prime }\left(c\right)$ is

Solution:

Roll's theorem is applicable to  $f\left(x\right)=\ln \left(\frac{x^2+\alpha }{7x}\right)$ on $[3,4]\text{. }$

 $\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}f\left(3\right)=f\left(4\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\ln \left(\frac{9+\alpha }{21}\right)=\ln \left(\frac{16+\alpha }{28}\right)\end{array}$

$\Rightarrow \frac{9+\alpha }{21}=\frac{16+\alpha }{28}\Rightarrow 36+4\alpha =48+3\alpha \Rightarrow \alpha =12$

Now,  $f\left(x\right)=\ln \left(\frac{x^2+\alpha }{7x}\right)$

$\begin{array}{l}\Rightarrow f\left(x\right)=\ln \left(x^2+12\right)-\ln 7x\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{2x}{x^2+12}-\frac{1}{x}=\frac{x^2-12}{x\left(x^2+12\right)}\\ \therefore f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow x^2-12=0\Rightarrow x=\pm \sqrt{12}\\ \therefore c=\sqrt{12}\end{array}$

Now,  $f^{\mathrm{\prime }}\left(x\right)=\frac{x^2-12}{x\left(x^2+12\right)}$

$\begin{array}{l}x\left(x^2+12\right)\\ \Rightarrow f^{\prime \prime }\left(x\right)=\frac{2x^2\left(x^2+12\right)-\left(x^2-12\right)\left(x^2+12+2x^2\right)}{x^2{\left(x^2+12\right)}^2}\\ \Rightarrow f^{\prime \prime }\left(x\right)=\frac{-x^4+48x^2+144}{x^2{\left(x^2+12\right)}^2}\\ \Rightarrow f^{\prime \prime }\left(\sqrt{12}\right)=\frac{1}{12}\end{array}$