2. Let set $A=\left\{x|| 2 x-x^{2} \mid \leqslant x\right\}, B=\left\{x|| \frac{x}{1-x} \left\lvert\, \leqslant \frac{x}{1-x}\right.\right\}$, $C=\left\{x \mid a x^{2}+x+b<0\right\}$, if $(A \cup B) \cap C=\varnothing$, and $(A \cup B) \cup C=\mathrm{R}$, find the values of $a$ and $b$.
2. Since $A \cup B=\{x \mid 0 \leqslant x \leqslant 3\}, C=\overline{A \cup B}=\{x \mid x<0 \text{ or } x>3\}$, thus $a x^{2} +x+b=a(x-0)(x-3)<0 \quad(a<0)$. Therefore, $a=-\frac{1}{3}, b=0$.