Let f be two differentiable function satisfying `f(1)=1,f(2)=4, f(3)=9`, then

Options

A. `f"(x)=2, AA x epsilon (r)`
B. `f'(x)=5=f"(x)`, for some `x epsilon(1,3)`
C. there exists at least one `x epsilon (1,3)` such that `f"(x)=2`
D. None of the above

Answer

To solve the problem, we need to analyze the given conditions and derive the necessary information about the function \( f(x) \).

### Step-by-Step Solution:

1. **Given Information:**
We have the function \( f \) satisfying the following conditions:
\[ f(1) = 1, \quad f(2) = 4, \quad f(3) = 9 \]

2. **Define a New Function:**
We define a new function \( g(x) \) as follows:
\[ g(x) = f(x) - x^2 \]

3. **Evaluate \( g(x) \) at Given Points:**
Now we will evaluate \( g(x) \) at the points 1, 2, and 3:
- For \( x = 1 \):
\[ g(1) = f(1) - 1^2 = 1 - 1 = 0 \]
- For \( x = 2 \):
\[ g(2) = f(2) - 2^2 = 4 - 4 = 0 \]
- For \( x = 3 \):
\[ g(3) = f(3) - 3^2 = 9 - 9 = 0 \]

4. **Roots of \( g(x) \):**
We have found that \( g(1) = 0 \), \( g(2) = 0 \), and \( g(3) = 0 \). Therefore, \( g(x) \) has at least three roots: \( x = 1, 2, 3 \).

5. **Using Rolle's Theorem:**
Since \( g(x) \) is continuous and differentiable (as \( f(x) \) is differentiable), we can apply Rolle's Theorem. This theorem states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one \( c \) in \( (a, b) \) such that \( g'(c) = 0 \).

6. **Finding the Derivative:**
We compute the derivative of \( g(x) \):
\[ g'(x) = f'(x) - 2x \]

7. **Setting the Derivative to Zero:**
Since \( g(x) \) has at least three roots, by Rolle's Theorem, there must be at least two points \( c_1 \) and \( c_2 \) in the intervals \( (1, 2) \) and \( (2, 3) \) such that:
\[ g'(c_1) = 0 \quad \text{and} \quad g'(c_2) = 0 \]
This implies:
\[ f'(c_1) - 2c_1 = 0 \quad \text{and} \quad f'(c_2) - 2c_2 = 0 \]
Therefore:
\[ f'(c_1) = 2c_1 \quad \text{and} \quad f'(c_2) = 2c_2 \]

8. **Conclusion:**
Since \( c_1 \) and \( c_2 \) are in the intervals \( (1, 2) \) and \( (2, 3) \), we conclude that there exists at least one \( x \) in the interval \( [1, 3] \) such that \( f'(x) = 2 \).

Thus, we can say:
\[ \text{There exists at least one } x \in [1, 3] \text{ such that } f'(x) = 2. \]

### Final Answer:
The correct option is:
- There exists at least one \( x \) in \( [1, 3] \) such that \( f(x) = 2 \).