JEE Mains · Maths · STD 12 - 5. continuity and differentiation
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 \(\frac{\sqrt{3}}{7}\)
- B \(\frac{1}{12}\)
- C \(-\frac{1}{24}\)
- D \(-\frac{1}{12}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{12}\)
Step-by-step Solution
Detailed explanation
\(\frac{9+\alpha}{21}=\frac{16+\alpha}{28} \Rightarrow \alpha=12\) 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, \(c=2 \sqrt{3}\)…
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