1. Given the functions
$$
f(x)=x^{2}-2 x, \quad g(x)=\left\{\begin{array}{ll}
x-2, & x \geqslant 1, \\
-x, & x<1 .
\end{array}\right.
$$

Then the solution set of the inequality $f(x) \leqslant 3 g(x)$ is $\qquad$ (Supplied by An Zhenping)
1. $[-1,0] \cup[2,3]$.
(Method 1) Since $g(x)=|x-1|-1$, the inequality $f(x) \leqslant 3 g(x)$ is equivalent to
$$
x^{2}-2 x \leqslant 3|x-1|-3,
$$

which is equivalent to
$$
(x-1)^{2} \leqslant 3|x-1|-2,
$$

that is,
$$
|x-1|^{2}-3|x-1|+2 \leqslant 0,
$$

thus $1 \leqslant|x-1| \leqslant 2$, solving this yields $-1 \leqslant x \leqslant 0$ or $2 \leqslant x \leqslant 3$.
(Method 2) The original inequality is equivalent to
$$
\left\{\begin{array}{l}
x^{2}-2 x \leqslant 3(x-2), \\
x \geqslant 1,
\end{array} \text { or } \quad \left\{\begin{array}{l}
x^{2}-2 x \leqslant-3 x, \\
x<1 .
\end{array}\right.\right.
$$

Solving this yields $-1 \leqslant x \leqslant 0$ or $2 \leqslant x \leqslant 3$.