If c is a point at which Rolle's theorem holds for the function, f(x)= _ e ( x^ 2 + 7 x ) in the interval [3,4], where R, then f^ (c) is equal to

If c is a point at which Rolle's theorem holds for the function, …

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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