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 $f\left(0\right)=1$, then $f(2a-x)$ equals

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

Answer 1

Answer: (B)

$-f\left(x\right)$

$f(x-y)=f\left(x\right)f\left(y\right)-f(a-x)f(a+y)......\left(i\right)$

Substitute $x=a$, and $y=x-a$

$\Rightarrow f(a-(x-a\left)\right)=f\left(a\right)f(x-a)-f(a-a)f(a+x-a)$

$\Rightarrow f(2a-x)=f\left(a\right).f(x-a)-f\left(0\right).f\left(x\right)=f\left(a\right).f(x-a)-f\left(x\right)......\left(ii\right)$

Now put $x=y=0$ in $\left(i\right)$,

$\Rightarrow f\left(0\right)=\left\{f\right(0){\}}^2-\{f\left(a\right){\}}^2\Rightarrow f\left(a\right)=0$ since $f\left(0\right)=1$ (given)

$\therefore f(2a-x)=-f\left(x\right)$