(1) (2007 - Zhejiang) Let $f(x)=\left\{\begin{array}{ll}x^{2}, & |x| \geqslant 1, \\ x, & |x|<1,\end{array} g(x)\right.$ be a quadratic function. If the range of $f(g(x))$ is $[0,+\infty)$, then the range of $g(x)$ is ( ).
(A) $(-\infty,-1] \cup[1,+\infty)$
(B) $(-\infty,-1] \cup[0,+\infty)$
(C) $[0,+\infty)$
(D) $[1,+\infty)$
If $f(x) \geqslant 0$, then $x \geqslant 0$ or $x \leqslant-1$, thus, $g(x) \geqslant 0$ or $g(x) \leqslant-1$. Since $g(x)$ is a quadratic function, only one of $g(x) \geqslant 0$ and $g(x) \leqslant-1$ can hold. If $g(x) \leqslant-1$, then the range of $f(g(x))$ is $[1,+\infty)$, so, it can only be that $g(x) \geqslant 0$. The answer is C.