Question

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

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

Answer 1

The correct option is D

$\frac{1}{12}$

Explanation for the correct options:

Finding the value of function:

$f\left(3\right)=f\left(4\right)$

$ln\left(\frac{9+\mathrm{\alpha }}{21}\right)=ln\left(\frac{16+\mathrm{\alpha }}{28}\right)$

$\Rightarrow$$9+\alpha /21=16+\alpha /28$

$\Rightarrow$$36+4\alpha =48+3\alpha$

$\Rightarrow$$\alpha =12$

Now,

$$\begin{gathered} f\left(x\right)=\ln \left(\frac{(x^2+12)}{7x}\right) \\ f\text{'}\left(x\right)=\left[\frac{7x}{(x^2+12)}\right]\times \frac{(7x\times 2x-(x^2+12)\times 7)}{{\left(7x\right)}^2} \end{gathered}$$

$f\text{'}\left(x\right)=\frac{(x^2–12)}{\left(x\right(x^2+12\left)\right)}$

$$\begin{gathered} f\text{'}\left(c\right)=0 \\ c=2\sqrt{3} \end{gathered}$$

$f\text{'}\text{'}\left(x\right)=\frac{(-x^4+48x^2+144)}{x^2{(x^2+12)}^2}$

Therefore,

$f\text{'}\text{'}\left(c\right)=\frac{1}{12}$

Hence, the correct option is (D)