Question

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

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

Answer 1

Answer: (D)

$f(x - y) = f(x) f (y) - f (a - x) f (a + y)$
Put $x = 0 = y$.
$\therefore \:\: f(0) = f(0) f(0) -f(a) f(a) = (f(0))^2 - (f (a))^2 $
$\Rightarrow 1= 1-(f(a))^2 \Rightarrow f(a) = 0$
Now $f(2a-x) = f(a - (x - a)) $
$ = f(a)f(x - a)-f(a- a) f(a + x - a) $
$0\cdot f(x-a)-f(0) f(x) $
$= 0 - 1. f(x) = - f(x) $.