If `f(x+y)=f(xy) AA x, y epsilon R, and f (2013) =2013`, then f(-2013) equals

Options

A. 2013
B. 0
C. `-2013`
D. none of these

Answer

To solve the problem where \( f(x+y) = f(xy) \) for all \( x, y \in \mathbb{R} \) and \( f(2013) = 2013 \), we need to find \( f(-2013) \).

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We start with the equation \( f(x+y) = f(xy) \). This means that the function \( f \) takes the same value for the sum of two numbers as it does for their product.

2. **Substituting Values**:
Let’s substitute \( x = 2013 \) and \( y = 0 \):
\[ f(2013 + 0) = f(2013 \cdot 0) \]
This simplifies to:
\[ f(2013) = f(0) \]
Since we know \( f(2013) = 2013 \), we conclude that:
\[ f(0) = 2013 \]

3. **Finding \( f(-2013) \)**:
Now, we substitute \( x = -2013 \) and \( y = -1 \):
\[ f(-2013 + (-1)) = f(-2013 \cdot -1) \]
This simplifies to:
\[ f(-2014) = f(2013) \]
Since \( f(2013) = 2013 \), we have:
\[ f(-2014) = 2013 \]

4. **Using \( f(-2013) \)**:
Next, we substitute \( x = -2012 \) and \( y = -1 \):
\[ f(-2012 + (-1)) = f(-2012 \cdot -1) \]
This simplifies to:
\[ f(-2013) = f(2012) \]

5. **Finding \( f(2012) \)**:
Now, we substitute \( x = 2013 \) and \( y = -1 \):
\[ f(2013 + (-1)) = f(2013 \cdot -1) \]
This simplifies to:
\[ f(2012) = f(-2013) \]

6. **Conclusion**:
From the previous steps, we have established that:
\[ f(-2013) = f(2012) \]
and since \( f(2012) = 2013 \), we conclude:
\[ f(-2013) = 2013 \]

### Final Answer:
Thus, \( f(-2013) = 2013 \).