If `f(x)` is a twice differentiable function such that `f(0)=f(1)=f(2)=0`. Then

Options

A. `F(x)=0` has exactly 3 roots
B. `f'(x)=` for atleast 3 real values of x
C. `f''(x)=0` for atleast 2 real value of x
D. `f''(x)=0` for atleast 1 real value of x

Answer

To solve the problem, we will use Rolle's Theorem, which states that if a function is continuous on a closed interval and differentiable on the open interval, and if the function takes the same value at the endpoints of the interval, then there exists at least one point in the interval where the derivative of the function is zero.

### Step-by-Step Solution:

1. **Identify the given conditions**: We know that \( f(0) = f(1) = f(2) = 0 \). This means that the function \( f(x) \) has the same value (which is 0) at three different points: 0, 1, and 2.

2. **Apply Rolle's Theorem on the interval [0, 1]**:
- Since \( f(0) = f(1) \), and \( f(x) \) is continuous and differentiable (as it is twice differentiable), we can apply Rolle's Theorem.
- According to Rolle's Theorem, there exists at least one point \( c_1 \) in the interval \( (0, 1) \) such that \( f'(c_1) = 0 \).

3. **Apply Rolle's Theorem on the interval [1, 2]**:
- Similarly, since \( f(1) = f(2) \), we can apply Rolle's Theorem again on the interval [1, 2].
- This gives us at least one point \( c_2 \) in the interval \( (1, 2) \) such that \( f'(c_2) = 0 \).

4. **Conclusion about the derivative**:
- From the two applications of Rolle's Theorem, we have found at least two points \( c_1 \) and \( c_2 \) where \( f'(x) = 0 \).
- Therefore, \( f'(x) \) has at least two roots in the interval \( (0, 2) \).

5. **Further analysis**:
- Since \( f'(x) \) is continuous (as \( f(x) \) is twice differentiable), and we have established that \( f'(x) = 0 \) at least twice, we can conclude that there are at least two points where the derivative is zero.

### Final Answer:
Thus, we conclude that \( f'(x) = 0 \) has at least 2 real roots in the interval \( (0, 2) \).