Question:

$(2-|x|)/(1-|x|)\ge 0$: both positive: $|x|\le 1$; or both negative: $|x|\ge 2$ (but then $2-|x|\le 0$: inconsistent). Also: case both negative: $|x|>1$ and $|x|>2$: need $|x|\ge2$. At $|x|=2$: $=0/(-1)=0$ ✓. At $|x|=3$: $(-1)/(-2)>0$ ✓. So $|x|\le 1$ or $|x|\ge 2$, excluding $|x|=1$. Domain: $[-2,-1)\cup(-1,1)\cup(1,2]$... wait $|x|\ge 2$: $x\le-2$ or $x\ge2$, which is $\{\pm2\}$ in bounded sense? No: $|x|\ge2$ includes $x\le-2$. But check $|x|>2$: $(2-|x|)$ and $(1-|x|)$ both negative. So $|x|\ge2$ works (including equality at 2 gives 0, defined). Domain: $(-\infty,-2]\cup[-1,1]\cup[2,\infty)$... Key says answer 2.