MCQ
If $c$ is a point at which Rolle's theorem holds for the function, $f(\mathrm{x})=\log _{\mathrm{e}}\left(\frac{\mathrm{x}^{2}+\alpha}{7 \mathrm{x}}\right)$ in the interval $[3,4],$ where $\alpha \in \mathrm{R},$ then $f^{\prime \prime}(\mathrm{c})$ is equal to
  • A
    $\frac{\sqrt{3}}{7}$
  • $\frac{1}{12}$
  • C
    $-\frac{1}{24}$
  • D
    $-\frac{1}{12}$

Answer

Correct option: B.
$\frac{1}{12}$
b
$\frac{9+\alpha}{21}=\frac{16+\alpha}{28} \Rightarrow \alpha=12$

Also, $f^{\prime}(\mathrm{x})=\frac{7 \mathrm{x}}{\mathrm{x}^{2}+12} \times \frac{\mathrm{x}^{2}-12}{7 \mathrm{x}^{2}}=\frac{\mathrm{x}^{2}-12}{\mathrm{x}\left(\mathrm{x}^{2}+12\right)}$

Hence, $c=2 \sqrt{3}$

Now, $f^{\prime \prime}(c)=\frac{1}{12}$

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