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(C) Given the functional equation: $f(x - y) = f(x)f(y) - f(a - x)f(a + y)$.Step $1$: Find $f(a)$.Substitute $x = 0$ and $y = 0$ into the equation:$f(

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$-f(x)$

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(C) Given the functional equation: $f(x - y) = f(x)f(y) - f(a - x)f(a + y)$.Step $1$: Find $f(a)$.Substitute $x = 0$ and $y = 0$ into the equation:$f(0 - 0) = f(0)f(0) - f(a - 0)f(a + 0)$$f(0) = (f(0))^2 - (f(a))^2$Since $f(0) = 1$,we have:$1 = 1^2 - (f(a))^2$$(f(a))^2 = 0 \Rightarrow f(a) = 0$.Step $2$: Find $f(2a - x)$.Substitute $y = a$ in the original equation:$f(x - a) = f(x)f(a) - f(a - x)f(a + a)$Since $f(a) = 0$,this simplifies to:$f(x - a) = -f(a - x)f(2a) \dots (i)$Alternatively,substitute $x = a$ and $y = x - a$:$f(a - (x - a)) = f(a)f(x - a) - f(a - a)f(a + x - a)$$f(2a - x) = 0 \cdot f(x - a) - f(0)f(x)$Since $f(0) = 1$,we get:$f(2a - x) = -f(x)$.

$A$ 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 $f(0) = 1$. Then $f(2a - x)$ is equal to:

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