Find all possible values of the following expression: $\frac{1}{x^2-2x+3}$

$\left[\frac{1}{2},1\right)$ $\left[\frac{1}{2},0\right)$ $\left(0,\frac{1}{2}\right]$ $\left(0,\frac{1}{3}\right]$

$\left(0,\frac{1}{2}\right]$

$\frac{1}{x^2-2x+3}=\frac{1}{(x-1)^2+2}$ Now we know that $(x-1)^2≥0\,∀\,x∈R$. Thus, $(x-1)^2+2≥2\,∀\,x∈R$ or $2≤(x-1)^2+2<∞$ or $\frac{1}{2}≥\frac{1}{(x-1)^2+2}>0$ or $\frac{1}{x^2-2x+3}∈\left(0,\frac{1}{2}\right]$