A ral valued functin `f (x)` satisfies the functional equation `f (x-y)=f(x)f(y)- f(a-x) f(a+y)` where 'a' is a given constant and `f (0) =1, f(2a-x)` is equal to :

Options

A. `-f(x)`
B. `f(x)`
C. `f(a)+ f(a-x)`
D. `f(-x)`

Answer

To solve the problem, we need to find the value of \( f(2a - x) \) given the functional equation:

\[ f(x - y) = f(x)f(y) - f(a - x)f(a + y) \]

with the condition \( f(0) = 1 \).

### Step 1: Substitute \( x = 0 \) and \( y = 0 \)

We start by substituting \( x = 0 \) and \( y = 0 \) into the functional equation:

\[ f(0 - 0) = f(0)f(0) - f(a - 0)f(a + 0) \]

This simplifies to:

\[ f(0) = f(0)^2 - f(a)f(a) \]

Given that \( f(0) = 1 \), we have:

\[ 1 = 1^2 - f(a)^2 \]

This simplifies to:

\[ 1 = 1 - f(a)^2 \]

Thus, we find:

\[ f(a)^2 = 0 \implies f(a) = 0 \]

### Step 2: Substitute \( x = a \) and \( y = x \)

Next, we substitute \( x = a \) and \( y = x \) into the original functional equation:

\[ f(a - x) = f(a)f(x) - f(a - a)f(a + x) \]

This simplifies to:

\[ f(a - x) = 0 \cdot f(x) - f(0)f(a + x) \]

Since \( f(0) = 1 \), we have:

\[ f(a - x) = -f(a + x) \]

### Step 3: Analyze the symmetry

From the equation \( f(a - x) = -f(a + x) \), we can see that the function \( f \) is symmetric about the line \( x = a \). This means:

\[ f(2a - x) = -f(x) \]

### Step 4: Substitute \( x = 2a - x \)

To find \( f(2a - x) \), we can use the symmetry we just derived:

\[ f(2a - x) = -f(x) \]

### Conclusion

Thus, we conclude that:

\[ f(2a - x) = -f(x) \]