If c is a point at which Rolle's theorem holds for function, f(mathrm{x})=log

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5. Continuity and Differentiation

hard

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

$\frac{\sqrt{3}}{7}$

$\frac{1}{12}$

$-\frac{1}{24}$

$-\frac{1}{12}$

(JEE MAIN-2020)

Solution

$\frac{9+\alpha}{21}=\frac{16+\alpha}{28} \Rightarrow \alpha=12$

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

Standard 12

Mathematics

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normal

(KVPY-2014)

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medium

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

Solution

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