A real valued function $$f(x)$$ 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

The functional equation is
$$f(x-y)=f(x)f(y)-f(a-x)f(a+y)\qquad (\,\star\,)$$
with the given condition $$f(0)=1$$. We have to find $$f(2a-x)$$ in terms of $$f(x)$$.

Step 1: Put $$x=a$$ in $$(\star)$$
Substituting $$x=a$$ gives
$$f(a-y)=f(a)f(y)-f(a-a)f(a+y).$$
Because $$f(0)=1$$, this becomes
$$f(a-y)=f(a)f(y)-f(a+y).$$
Re-arranging, we obtain the symmetric relation
$$f(a-y)+f(a+y)=f(a)f(y)\qquad (1).$$

Step 2: Find $$f(a)$$
Set $$y=0$$ in (1):
$$f(a-0)+f(a+0)=f(a)f(0)\;\Longrightarrow\;2f(a)=f(a)\times 1.$$
Hence $$f(a)=0.$$

Step 3: Use $$f(a)=0$$ in the symmetric relation
Putting $$f(a)=0$$ back into (1) gives
$$f(a-y)+f(a+y)=0\qquad\text{for all real }y.\qquad (2)$$

Step 4: Express $$f(2a-x)$$
Choose $$y=a-x$$ in (2). Then
$$a-y=x,\qquad a+y=2a-x.$$
Equation (2) becomes
$$f(x)+f(2a-x)=0\;\Longrightarrow\;f(2a-x)=-f(x).$$

Thus $$f(2a-x)=-f(x).$$

Option A which is: $$-f(x)$$