If `f(0)=f(1)=f(2)=0` and function f(x) is twice differentiable in (0, 2) and continuous in [0, 2], then which of the following is/are definitely true ?

Options

A. (a) `f''(c)=0, AA c in (0,2)`
B. (b) `f'(c)=0, "for atleast two "c in (0, 2)`
C. (c) `f'(c)=0, "for exactly one "c in (-0,2)`
D. (d) `f''(c)=0,` for atleast one `c in (0,2)`

Answer

To solve the given problem, we will use the Mean Value Theorem (MVT) and analyze the implications of the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We know that \( f(0) = f(1) = f(2) = 0 \). The function \( f(x) \) is twice differentiable in the interval \( (0, 2) \) and continuous in the closed interval \( [0, 2] \).

2. **Applying the Mean Value Theorem (MVT)**:
According to the MVT, if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point \( c \) in the open interval such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

3. **Finding \( c_1 \) in \( (0, 1) \)**:
Let’s apply MVT on the interval \( [0, 1] \):
\[ f'(c_1) = \frac{f(1) - f(0)}{1 - 0} = \frac{0 - 0}{1} = 0 \]
Thus, there exists a point \( c_1 \in (0, 1) \) such that \( f'(c_1) = 0 \).

4. **Finding \( c_2 \) in \( (1, 2) \)**:
Now, apply MVT on the interval \( [1, 2] \):
\[ f'(c_2) = \frac{f(2) - f(1)}{2 - 1} = \frac{0 - 0}{1} = 0 \]
Thus, there exists a point \( c_2 \in (1, 2) \) such that \( f'(c_2) = 0 \).

5. **Analyzing the First Derivative**:
From the above two results, we have:
- \( f'(c_1) = 0 \) for some \( c_1 \in (0, 1) \)
- \( f'(c_2) = 0 \) for some \( c_2 \in (1, 2) \)

6. **Applying MVT Again for the Second Derivative**:
Now, we apply MVT to \( f' \) on the interval \( [c_1, c_2] \):
\[ f''(c) = \frac{f'(c_2) - f'(c_1)}{c_2 - c_1} = \frac{0 - 0}{c_2 - c_1} = 0 \]
Thus, there exists a point \( c \in (c_1, c_2) \) such that \( f''(c) = 0 \).

### Conclusion:
From the above analysis, we can conclude:
- There are points \( c_1 \) and \( c_2 \) where the first derivative is zero.
- There is a point \( c \) where the second derivative is zero.

### Options Analysis:
1. **Option A**: \( f''(c) = 0 \) for some \( c \in [0, 2] \) - This is true since we found \( f''(c) = 0 \) for some \( c \) in the interval.
2. **Option B**: \( f'(c) = 0 \) for some \( c \in [0, 2] \) - This is also true since we found \( f'(c_1) = 0 \) and \( f'(c_2) = 0 \).
3. **Option C**: \( f'(c) = 0 \) for all \( c \in [0, 2] \) - This is false, as we only found specific points where the derivative is zero.
4. **Option D**: \( f''(c) = 0 \) for at least one \( c \in [0, 2] \) - This is true since we found \( f''(c) = 0 \) for some \( c \).

### Final Answer:
The definitely true statements are:
- Option A: True
- Option B: True
- Option D: True
- Option C: False