parmenides51 wrote:
Determine all functions $f : R_+ \to R_+$ such that:
a) $f(xf(y))f(y) = f(x + y)$ for $x, y \ge 0$
b) $f(2) = 0$
c) $f(x) \ne 0$ for $0 \le x < 2$.
I suppose your $\mathbb R_+$ is $\mathbb R_{\ge 0}$ (and not, as usual on this forum, $\mathbb R_{>0}$). If so :
Let $P(x,y)$ be the assertion $f(xf(y))f(y)=f(x+y)$
$P(x,2)$ $\implies$ $f(x+2)=0$ and so $f(x)=0$ $\forall x\ge 2$
And so $f(x)=0$ $\iff$ $x\ge 2$
Let then $y<2$ (so that $f(y)\ne 0$) : $f(x+y)=0$ $\iff$ $f(xf(y))=0$
And so $x+y\ge 2$ $\iff$ $xf(y)\ge 2$
Which is $x\ge 2-y$ $\iff$ $x\ge\frac 2{f(y)}$
And so $f(y)=\frac 2{2-y}$
And solution $\boxed{f(x)=\frac 2{2-x}\quad\forall x\in[0,2)\text{ and }f(x)=0\quad\forall x\in[2,+\infty)}$, which indeed fits