Question

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

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

Answer 1

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

Given that :
$f(x-y)=f\left(x\right)f\left(y\right)-f(a-x)f(a+y)\forall x,y\in R$
Putting $y=0$, we get
$f\left(x\right)=f\left(x\right)-f(a-x)f\left(a\right)$
$\Rightarrow f(a-x)f\left(a\right)=0$
$\Rightarrow f\left(a\right)=0$ $\left(\text{As}f\right(a-x)\ne 0)$

Now, $f(2a-x)=f(a-(x-a\left)\right)$
$=f\left(a\right)f(a-x)-f(a-a)f(a+x-a)$
$=0-f\left(x\right)$
Therefore, $f(2a-x)=-f\left(x\right)$