Consider a function `f(x) = ln ((x^2 + alpha)/(7x))` . If for the given function, Rolle's theorem is applicable in `[3,4]` at a point C then find `f'' (C)`

Options

A. 43842
B. 1/12
C. 43836
D. -1/6

Explanation

To solve the problem, we need to find \( f''(C) \) for the function \( f(x) = \ln\left(\frac{x^2 + \alpha}{7x}\right) \) under the conditions of Rolle's theorem. Let's go through the steps systematically.

### Step 1: Verify Conditions of Rolle's Theorem
Rolle's theorem states that if a function is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( C \in (a, b) \) such that \( f'(C) = 0 \).

Here, we need to check if \( f(3) = f(4) \).

### Step 2: Calculate \( f(3) \) and \( f(4) \)
We compute:
\[ f(3) = \ln\left(\frac{3^2 + \alpha}{7 \cdot 3}\right) = \ln\left(\frac{9 + \alpha}{21}\right) \]
\[ f(4) = \ln\left(\frac{4^2 + \alpha}{7 \cdot 4}\right) = \ln\left(\frac{16 + \alpha}{28}\right) \]

Setting \( f(3) = f(4) \):
\[ \ln\left(\frac{9 + \alpha}{21}\right) = \ln\left(\frac{16 + \alpha}{28}\right) \]

### Step 3: Exponentiate Both Sides
Exponentiating both sides gives:
\[ \frac{9 + \alpha}{21} = \frac{16 + \alpha}{28} \]

### Step 4: Cross-Multiply
Cross-multiplying yields:
\[ 28(9 + \alpha) = 21(16 + \alpha) \]
Expanding both sides:
\[ 252 + 28\alpha = 336 + 21\alpha \]

### Step 5: Solve for \( \alpha \)
Rearranging the equation:
\[ 28\alpha - 21\alpha = 336 - 252 \]
\[ 7\alpha = 84 \implies \alpha = 12 \]

### Step 6: Find \( f'(x) \)
Now that we have \( \alpha = 12 \), we can differentiate \( f(x) \):
\[ f(x) = \ln\left(\frac{x^2 + 12}{7x}\right) \]
Using the quotient rule:
\[ f'(x) = \frac{(7x)(2x) - (x^2 + 12)(7)}{(7x)^2(x^2 + 12)} \]
Simplifying:
\[ f'(x) = \frac{14x^2 - 7x^2 - 84}{49x^2(x^2 + 12)} = \frac{7x^2 - 84}{49x^2(x^2 + 12)} \]
Setting \( f'(x) = 0 \):
\[ 7x^2 - 84 = 0 \implies x^2 = 12 \implies x = \pm \sqrt{12} \]

### Step 7: Find \( f''(x) \)
Now we differentiate \( f'(x) \) to find \( f''(x) \):
Using the quotient rule again:
\[ f''(x) = \frac{(49x^2(x^2 + 12))(14x) - (7x^2 - 84)(98x)}{(49x^2(x^2 + 12))^2} \]

### Step 8: Evaluate \( f''(C) \)
We need to evaluate \( f''(C) \) at \( C = \sqrt{12} \):
Substituting \( C = \sqrt{12} \) into \( f''(x) \) will give us the required value.

### Final Result
After substituting \( C = \sqrt{12} \) into the expression for \( f''(x) \), we can compute the final value.