Let `f:RtoR` be a function given by `f(x+y)=f(x)f(y)` for all x,y` in` R .If `f'(0)=2` then `f(x) ` is equal to`

Options

A. `A e^(x)`
B. `A e^(2x)`
C. 2x
D. none of these

Explanation

To solve the problem, we start with the functional equation given by \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \) and the condition that \( f'(0) = 2 \).

### Step-by-Step Solution:

1. **Evaluate \( f(0) \)**:
- Set \( x = 0 \) and \( y = 0 \) in the functional equation:
\[ f(0 + 0) = f(0)f(0) \implies f(0) = f(0)^2. \]
- This implies \( f(0)(f(0) - 1) = 0 \). Hence, \( f(0) = 0 \) or \( f(0) = 1 \).

2. **Determine which case to consider**:
- If \( f(0) = 0 \), then for any \( x \):
\[ f(x) = f(x + 0) = f(x)f(0) = 0, \]
which means \( f(x) = 0 \) for all \( x \). However, this contradicts \( f'(0) = 2 \).
- Therefore, we must have \( f(0) = 1 \).

3. **Differentiate the functional equation**:
- Differentiate both sides of \( f(x+y) = f(x)f(y) \) with respect to \( y \):
\[ \frac{d}{dy}f(x+y) = f'(x+y) = f(x)f'(y). \]
- Set \( y = 0 \):
\[ f'(x) = f(x)f'(0). \]
- Since \( f'(0) = 2 \), we have:
\[ f'(x) = 2f(x). \]

4. **Separate variables and integrate**:
- Rearranging gives:
\[ \frac{f'(x)}{f(x)} = 2. \]
- Integrate both sides:
\[ \int \frac{f'(x)}{f(x)} \, dx = \int 2 \, dx. \]
- This results in:
\[ \ln |f(x)| = 2x + C, \]
where \( C \) is a constant.

5. **Exponentiate to solve for \( f(x) \)**:
- Exponentiating both sides gives:
\[ f(x) = e^{2x + C} = e^{2x} e^C. \]
- Let \( A = e^C \), then:
\[ f(x) = A e^{2x}. \]

6. **Determine the constant \( A \)**:
- Since \( f(0) = 1 \):
\[ f(0) = A e^{2 \cdot 0} = A = 1. \]
- Therefore, we conclude:
\[ f(x) = e^{2x}. \]

### Final Answer:
Thus, the function \( f(x) \) is given by:
\[ f(x) = e^{2x}. \]