The function `f(x) =x^(3) - 6x^(2)+ax + b` satisfy the conditions of Rolle's theorem on [1,3] which of these are correct ?

Options

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

Answer

To solve the problem, we need to determine the values of \( a \) and \( b \) for the function \( f(x) = x^3 - 6x^2 + ax + b \) such that it satisfies the conditions of Rolle's theorem on the interval \([1, 3]\).

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

Given:
- The function \( f(x) \) is a polynomial, which is continuous and differentiable everywhere, including on the interval \([1, 3]\).
- We need to ensure that \( f(1) = f(3) \).

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

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

### Step 3: Set \( f(1) = f(3) \)
Since \( f(1) = f(3) \), we set the two expressions equal:
\[ a + b - 5 = 3a + b - 27 \]

### Step 4: Simplify the equation
Subtract \( b \) from both sides:
\[ a - 5 = 3a - 27 \]
Now, rearranging gives:
\[ -5 + 27 = 3a - a \]
\[ 22 = 2a \]
\[ a = 11 \]

### Step 5: Find the value of \( b \)
Now that we have \( a = 11 \), we can substitute it back into either expression for \( f(1) \) or \( f(3) \) to find \( b \). Using \( f(1) \):
\[ f(1) = 11 + b - 5 \]
\[ f(1) = 6 + b \]

Using \( f(3) \):
\[ f(3) = 3(11) + b - 27 \]
\[ f(3) = 33 + b - 27 = 6 + b \]

Since both expressions equal \( 6 + b \), we see that \( b \) can be any real number. Thus, \( b \) is not uniquely determined.

### Conclusion
The values we found are:
- \( a = 11 \)
- \( b \) can be any real number.

### Final Answer
The correct option is that \( a = 11 \) and \( b \) can be any real number.