Let f (x) be a real valued function such that  f0=12 and f(x+y)=f(x)f(a−y)+f(y)f(a−x) , ∀x,y∈R then for some real a,

f(x+y)=f(x)f(a−y)+f(y)f(a−x)----iPut x = y = 0, we get f(a)=12Let y = 0⇒f(x)=f(x)f(a)+f(0)⋅f(a−x)⇒f(x)=12f(x)+12f(a−x)⇒f(x)=f(a−x)Put y = a- x in Eq. (i), f(a)=(f(x))2+(f(a−x))2⇒(f(x))2=14⇒f(x)=±12Hence ,f(x)=12