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 ensure the function \( f(x) = x^3 - 6x^2 + ax + b \) 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 \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

1. **Continuity and Differentiability**:
The function \( f(x) \) is a polynomial, which is continuous and differentiable everywhere, including the interval \([1, 3]\).

2. **Check \( f(1) \) and \( f(3) \)**:
We need to find \( f(1) \) and \( f(3) \) and set them equal to each other.

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

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

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

Simplifying this equation:
\[ a + b - 5 = 3a + b - 27 \]
\[ -5 + 27 = 3a - a \]
\[ 22 = 2a \]
\[ a = 11 \]

### Step 2: Determine the value of \( b \)

Since there are no additional conditions given for \( b \), it can take any real value. However, we also note that \( b \) cannot be equal to 6 based on the context provided in the video transcript.

### Conclusion

- The value of \( a \) is determined to be \( 11 \).
- The value of \( b \) can be any real number except \( 6 \).

### Final Answer

The correct statements based on the conditions of Rolle's theorem are:
- \( a = 11 \)
- \( b \) can be any real number except \( 6 \).