Given that $$f(x)=\sqrt{\log(2x-x^2)}$$ then find the domain of $f(x)$
My solution:
$\log(2x-x^2)\geq0$
$\implies 2x-x^2\geq1 \implies x\leq1$
Also,
$2x-x^2\gt0$
$\implies x\gt0$ ,$\ x\lt2$
On finding the intersection,
$x\in(0,1]$
I can't figure out where I went wrong here, The given solution is ${\{1\}}$
Notice that:
$$2x - x^2 \ge 1 \iff x^2-2x+1 \le 0 \iff (x-1)^2\le 0 \iff x=1$$
Besides, the second inequality $2x - x^2 > 0$ is implied by the first.