Question

A real valued function $f\left(x\right)$ satisfies the functional equation $f(x-y)=f\left(x\right)f\left(y\right)-f(a-x)f(a+y)$ where $a$ is a given constant and $f\left(0\right)=1,f(2a-x)$ is equal to

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

Answer 1

The correct option is D $-f\left(x\right)$

$f(x-y)=f\left(x\right)f\left(y\right)-f(a-x)f(a+y)$ ...(i)
It is given that $f\left(0\right)=1$
Substituting $x=y=0$ in (i)
$1=1-\left(f\right(a){)}^2$
Hence $f\left(a\right)=0$
Now $f(2a-x)$
$=f(a-(x-a\left)\right)$
$=f\left(a\right)f(x-a)-f(a-a)f(a+x-a)$
$=0-f\left(0\right)f\left(x\right)$
$=-f\left(x\right)$