Mathematics - Functions Question with Solution | TestHub

Given $f(x+y)+f(x-y)=2f(x)f(y)$ and $f(0)=1,f(1)=0$. Substitute $x=0$: $f(y)+f(-y)=2f(0)f(y)=2(1)f(y)=2f(y)$. This implies $f(-y)=f(y)$, so $f$ is an even function. Substitute $y=x$: $f(2x)+f(0)=2f(x)f(x)⟹f(2x)+1=2(f(x){)}^2$. Using $f(1)=0$, we find $f(2)$: $f(2\cdot 1)+1=2(f(1){)}^2⟹f(2)+1=2(0{)}^2⟹f(2)=-1$. Now, to find $f(3)$, substitute $x=2,y=1$ into the original equation: $f(2+1)+f(2-1)=2f(2)f(1)$. This gives $f(3)+f(1)=2f(2)f(1)$, so $f(3)+0=2(-1)(0)⟹f(3)=0$. Therefore, $f(2)+f(3)=-1+0=-1$.