以下以不等式考慮
\(\displaystyle A: a-1\leq x \leq (a^2-a), x\in \mathbb{R}\)
\(\displaystyle B: 2-a\leq x \leq (5-a), x\in \mathbb{R}\)
考慮\(A\cap B =\emptyset\)
畫圖可知: 可能為\(\displaystyle 5-a\leq a-1\)或是\(\displaystyle a^2-a\leq 2-a\)
得\(\displaystyle a\geq 3\)或是 \(\displaystyle -\sqrt{2}\leq x \leq \sqrt{2}\)
將上式不等式個別取補集得\(\displaystyle a<3\) 或是\(\displaystyle a>\sqrt{2} \ or \ a<-\sqrt{2}\)
亦可合併為\(\displaystyle \sqrt{2}<a<3 \ or \ a<-\sqrt{2}\)
