Question

If Rolle's theorem is applicable for the function \(f(x) = log(\frac{x^2+a}{x})\) on \([3,4]\) with \(c\in (3,4)\) such that \(f^'(c) = 0\), then the value of \(f^{''}(c)\) is

• A. \(\frac{1}{48}\)
• B. \(\frac{1}{36}\)
• C. \(\frac{1}{24}\)
• D. \(\frac{1}{12}\)

Hint:

Use \(f(3)=f(4)\) to get \(a\), then \(f'(c)=0\) gives \(c^2=a\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Rolle's theorem needs \(f(3) = f(4)\). Write \(f(x) = \log\left(x + \frac{a}{x}\right)\).

Step 2: Find a:
\(3 + \frac{a}{3} = 4 + \frac{a}{4}\), so \(\frac{a}{3} - \frac{a}{4} = 1\), \(\frac{a}{12} = 1\), \(a = 12\).

Step 3: First derivative:
\[ f'(x) = \frac{1 - a/x^2}{x + a/x} = \frac{x^2 - a}{x(x^2+a)} = \frac{x^2-12}{x^3+12x} \]
\(f'(c) = 0\) gives \(c^2 = 12\), \(c = 2\sqrt3 \approx 3.46\), which lies in \((3,4)\).

Step 4: Second derivative:
Differentiate \(f' = \frac{N}{M}\), with \(N = x^2-12\) and \(M = x^3+12x\): \(f'' = \frac{N'M - NM'}{M^2}\).
At \(x = c\), \(N = 0\), so \(f''(c) = \frac{N'(c)}{M(c)} = \frac{2c}{c^3+12c} = \frac{2}{c^2+12} = \frac{2}{24} = \frac{1}{12}\).

Final Answer:
\(f''(c) = \frac{1}{12}\), option (D). \[ \boxed{\frac{1}{12}} \]