Let the function f satisfies f(x).f ′ (−x)=f(−x).f ′ (x) for all x and f(0)=3 The value of `f(x).f(-x)` for all x is

Options

A. 4
B. 9
C. 12
D. 16

Answer

To solve the problem, we need to analyze the given condition and find the value of \( f(x) \cdot f(-x) \) for all \( x \).

### Step-by-Step Solution:

1. **Given Condition**: We start with the equation:
\[ f(x) \cdot f'(-x) = f(-x) \cdot f'(x) \]
for all \( x \), and we know that \( f(0) = 3 \).

2. **Rearranging the Equation**: We can rearrange the given condition:
\[ \frac{f'(-x)}{f(-x)} = \frac{f'(x)}{f(x)} \]
This implies that the ratio of the derivatives of \( f \) at \( -x \) and \( x \) is constant.

3. **Integrating Both Sides**: Integrating both sides with respect to \( x \):
\[ \int \frac{f'(-x)}{f(-x)} \, dx = \int \frac{f'(x)}{f(x)} \, dx \]
This gives us:
\[ \log |f(-x)| = -\log |f(x)| + C \]
where \( C \) is a constant of integration.

4. **Exponentiating Both Sides**: Exponentiating both sides, we have:
\[ |f(-x)| = \frac{K}{|f(x)|} \]
where \( K = e^C \) is a positive constant.

5. **Multiplying Both Sides**: Now, multiplying \( f(x) \) and \( f(-x) \):
\[ f(x) \cdot f(-x) = K \]
This means that the product \( f(x) \cdot f(-x) \) is a constant for all \( x \).

6. **Finding the Constant**: To find the value of \( K \), we substitute \( x = 0 \):
\[ f(0) \cdot f(-0) = K \]
Since \( f(0) = 3 \), we have:
\[ 3 \cdot 3 = K \implies K = 9 \]

7. **Conclusion**: Therefore, we conclude that:
\[ f(x) \cdot f(-x) = 9 \]
for all \( x \).

### Final Answer:
The value of \( f(x) \cdot f(-x) \) for all \( x \) is \( \boxed{9} \).