If the function `f(x)= x^(3)-6x^(2)+ax+b` satisfies Rolle's theorem in the interval [1,3] and `f'((2sqrt(3)+1)/(sqrt(3)))=0`, then

Options

A. `a=-11, b=5`
B. `a=-11, b=-6`
C. `a=11,b in R`
D. `1=22, b =-6`

Answer

To solve the problem, we need to find the values of \(a\) and \(b\) for the function \(f(x) = x^3 - 6x^2 + ax + b\) given that it satisfies Rolle's theorem on the interval \([1, 3]\) and that \(f'\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right) = 0\).

### Step 1: Apply Rolle's Theorem
According to Rolle's theorem, if \(f\) is continuous on \([1, 3]\) and differentiable on \((1, 3)\), then \(f(1) = f(3)\).

**Calculate \(f(1)\):**
\[ f(1) = 1^3 - 6(1^2) + a(1) + b = 1 - 6 + a + b = a + b - 5 \]

**Calculate \(f(3)\):**
\[ f(3) = 3^3 - 6(3^2) + a(3) + b = 27 - 54 + 3a + b = 3a + b - 27 \]

**Set \(f(1) = f(3)\):**
\[ a + b - 5 = 3a + b - 27 \]

### Step 2: Simplify the Equation
Subtract \(b\) from both sides:
\[ a - 5 = 3a - 27 \]

Rearranging gives:
\[ -5 + 27 = 3a - a \]
\[ 22 = 2a \]
\[ a = 11 \]

### Step 3: Substitute \(a\) Back
Now that we have \(a = 11\), we can express \(f(1)\) and \(f(3)\) in terms of \(b\):
\[ f(1) = 11 + b - 5 = b + 6 \]
\[ f(3) = 3(11) + b - 27 = 33 + b - 27 = b + 6 \]
Both expressions for \(f(1)\) and \(f(3)\) are equal, confirming our value for \(a\).

### Step 4: Use the Derivative Condition
Next, we need to use the condition \(f'\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right) = 0\).

**Find the derivative \(f'(x)\):**
\[ f'(x) = 3x^2 - 12x + a \]
Substituting \(a = 11\):
\[ f'(x) = 3x^2 - 12x + 11 \]

**Set the derivative equal to zero:**
\[ f'\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right) = 3\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right)^2 - 12\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right) + 11 = 0 \]

Calculating \( \left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right)^2 \):
\[ = \frac{(2\sqrt{3}+1)^2}{3} = \frac{4 \cdot 3 + 4\sqrt{3} + 1}{3} = \frac{12 + 4\sqrt{3} + 1}{3} = \frac{13 + 4\sqrt{3}}{3} \]

Substituting back into the derivative:
\[ 3 \cdot \frac{13 + 4\sqrt{3}}{3} - 12 \cdot \frac{2\sqrt{3}+1}{\sqrt{3}} + 11 = 0 \]
This simplifies to:
\[ 13 + 4\sqrt{3} - 8\sqrt{3} - 12 + 11 = 0 \]
\[ 12 - 4\sqrt{3} = 0 \]

### Conclusion
Since \(b\) does not appear in any equations that restrict its value, \(b\) can be any real number. Thus, the final values are:
\[ a = 11, \quad b \text{ is any real number.} \]