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

• A. $f(2a-x)=f\left(x\right)$, $f\left(x\right)$ is symmetric about $x=a$
• B. $f(2a-x)=0=f\left(x\right)$, $f\left(x\right)$ is symmetric about $x=a$
• C. $f(2a-x)+f\left(x\right)=0$, $f\left(x\right)$ is not symmetric about $x=a$
• D. $f\left(x\right)=0$, $f\left(x\right)$ is symmetric about $x=a$

Answer 1

Answer: (C)

$f(2a-x)+f\left(x\right)=0$, $f\left(x\right)$ is not symmetric about $x=a$

Using the given relation

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

Using the given expression for $f(x-y)$, take $x=a$ and $y=x-a$

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

When $x=0,y=0\Rightarrow f\left(0\right)=\left(f\right(0){)}^2-(f\left(a\right){)}^2$

It is given that $f\left(0\right)=1$

$\Rightarrow f^2\left(a\right)=0\Rightarrow f\left(a\right)=0$

Substitute this in the expression,
$\Rightarrow f(2a-x)=0\left(f\right(x-a\left)\right)-1\left(f\right(x\left)\right)$

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

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

Take $x=a-x$

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

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

Hence, $f\left(x\right)$ is not symmetric about $x=a$.