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 any constant and $f(0)=1$, then $f(2a-x)$ is equal to

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

Answer 1

Answer: (B)

Given $f\left(x-y\right)=f\left(x\right)f\left(y\right)-f\left(a-x\right)f\left(a+y\right)$
Let $x=0=y$
$\therefore f\left(0\right)={\left(f\left(0\right)\right)}^2-{\left(f\left(a\right)\right)}^2$
$\Rightarrow 1=1-{\left(f\left(a\right)\right)}^2$
$\Rightarrow f\left(a\right)=0$
$\therefore f\left(2a-x\right)=f\left(a-\left(x-a\right)\right)$
$=f\left(a\right)f\left(x-a\right)-f\left(a+x-a\right)f\left(0\right)$
$=0-f\left(x\right)\left(1\right)=-f\left(x\right)\left(\because f\left(a\right)=0,f\left(0\right)=1\right)$