Question

Let $f\left(x\right)$ be a real -valued function such that $f\left(0\right)=\frac{1}{2}$ and $f(x+y)=f\left(x\right)f(a-y)+f\left(y\right)f(a-x)\forall x,y\in R$, then for some real $a$

• A. $f\left(x\right)$ is a periodic function
• B. $f\left(x\right)$ is a constant function
• C. $f\left(x\right)=\frac{1}{2}$
• D. $f\left(x\right)=\frac{\cos x}{2}$

Answer 1

Answer: (A), (B), (C)

$f\left(x\right)$ is a periodic function
$f\left(x\right)$ is a constant function
$f\left(x\right)=\frac{1}{2}$

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

Put $x=y=0$, we get $f\left(a\right)=\frac{1}{2}$ ...(i)

Let $y=0$

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

$\Rightarrow f\left(x\right)=\frac{1}{2}f\left(x\right)+\frac{1}{2}f(a-x)$

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

Put $y=a-x$ in Eq. (i),

$f\left(a\right)=\left(f\right(x){)}^2+(f(a-x){)}^2$

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

$f\left(x\right)=\pm \frac{1}{2}\left[f\right(x)\ne -\frac{1}{2}]$

Hence, $f\left(x\right)=\frac{1}{2}$