Mathematics - Functions Question with Solution | TestHub

$f(x+y)=f(x)f(a-y)+f(y)f(a-x)$

Put $x=0,y=0\Rightarrow f(0)=2f(0)f(a)\Rightarrow f(a)=\frac{1}{2}$.

Put $y=0,f(x)=f(x)f(a)+f(0)f(a-x)$

$$\Rightarrow f(x)=\frac{f(x)+f(a-x)}{2}$$

Thus, $f(x)=f(a-x)$.

Put $y=a-x$:

$$f(a)=f^2(x)+f^2(a-x)$$

But since $f(x)=f(a-x)$,

$$f(a)=2f^2(x)$$

$$\Rightarrow f^2(x)=\frac{1}{4}$$

$$\Rightarrow f(x)=\frac{1}{2}\left(\therefore f(0)=\frac{1}{2}\right)$$

Therefore, $f(x)=\frac{1}{2}$, which is a constant function.