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 $f(2a-x)=f(a-(x-a))$
$=f(a)f(x-a)-f(a-a)f(a+x-a)$
$0\cdot f(x-a)-f(0)f(x)$
$=0-1.f(x)=-f(x)$.