Question:
If \(c\) is a point at which Rolle's theorem holds for the function,
\(f(x)=\log_e\left(\frac{x^2+\alpha}{7x}\right)\)
in the interval \([3,4]\), where \(\alpha\in R\), then \(f''(c)\) is equal to
If \(c\) is a point at which Rolle's theorem holds for the function,
\(f(x)=\log_e\left(\frac{x^2+\alpha}{7x}\right)\)
in the interval \([3,4]\), where \(\alpha\in R\), then \(f''(c)\) is equal to
\(f(x)=\log_e\left(\frac{x^2+\alpha}{7x}\right)\)
in the interval \([3,4]\), where \(\alpha\in R\), then \(f''(c)\) is equal to
Show Hint
Rolle's theorem needs \(f(3)=f(4)\) to find \(\alpha\); then solve \(f'(c)=0\) and compute \(f''(c)\).
Updated On: Nov 5, 2026
- \(\frac{1}{12}\)
- \(\frac{\sqrt3}{7}\)
- \(-\frac{1}{12}\)
- \(-\frac{1}{24}\)
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The Correct Option is A
Solution and Explanation
\(f(3)=f(4)\Rightarrow\frac{9+\alpha}{21}=\frac{16+\alpha}{28}\Rightarrow 28(9+\alpha)=21(16+\alpha)\Rightarrow 252+28\alpha=336+21\alpha\Rightarrow\alpha=12\).
\(f(x)=\ln(x^2+12)-\ln7-\ln x\), so \(f'(x)=\frac{2x}{x^2+12}-\frac1x\).
\(f'(c)=0\Rightarrow 2c^2=c^2+12\Rightarrow c^2=12\), \(c=2\sqrt3\approx3.46\in(3,4)\).
\(f''(x)=\frac{2(x^2+12)-4x^2}{(x^2+12)^2}+\frac{1}{x^2}=\frac{24-2x^2}{(x^2+12)^2}+\frac1{x^2}\).
At \(x^2=12\): \(\frac{24-24}{576}+\frac1{12}=\frac1{12}\).
Answer: (1) \(\frac{1}{12}\)
\(f(x)=\ln(x^2+12)-\ln7-\ln x\), so \(f'(x)=\frac{2x}{x^2+12}-\frac1x\).
\(f'(c)=0\Rightarrow 2c^2=c^2+12\Rightarrow c^2=12\), \(c=2\sqrt3\approx3.46\in(3,4)\).
\(f''(x)=\frac{2(x^2+12)-4x^2}{(x^2+12)^2}+\frac{1}{x^2}=\frac{24-2x^2}{(x^2+12)^2}+\frac1{x^2}\).
At \(x^2=12\): \(\frac{24-24}{576}+\frac1{12}=\frac1{12}\).
Answer: (1) \(\frac{1}{12}\)
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