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: Plan:
Rewrite $f$ as a difference of logs to differentiate more easily.

Step 2: Find a:
Equal values at the ends: $\frac{9+a}{3} = \frac{16+a}{4}$, so $36+4a = 48 + 3a$, giving $a = 12$.

Step 3: Derivatives:
$f(x) = \log(x^2+12) - \log x$. Then $f'(x) = \frac{2x}{x^2+12} - \frac1x$, and $f''(x) = \frac{2(x^2+12) - 4x^2}{(x^2+12)^2} + \frac{1}{x^2} = \frac{24-2x^2}{(x^2+12)^2}+\frac{1}{x^2}$.
$f'(c) = 0$ gives $2c^2 = c^2 + 12$, so $c^2 = 12$. Then $f''(c) = \frac{24-24}{576} + \frac{1}{12} = \frac{1}{12}$.

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