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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

  1. A \(\frac{\sqrt{3}}{7}\)
  2. B \(\frac{1}{12}\)
  3. C \(-\frac{1}{24}\)
  4. D \(-\frac{1}{12}\)
Verified Solution

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, \(c=2 \sqrt{3}\)…