4・33 The range of $x$ that satisfies the inequality $|x+1|+|x|<2$ is
(A) $-\frac{3}{2}<x<-1$.
(B) $-\frac{3}{2}<x<0$.
(C) $-\frac{3}{2}<x<\frac{1}{2}$.
(D) $0<x<\frac{1}{2}$.
(E) $-\frac{1}{2}<x<\frac{3}{2}$.
(China Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1986)
[Solution] If $x \geqslant 0$, then $x+1+x<2$, which means $2 x<1$,
$$
\therefore \quad 0 \leqslant x<\frac{1}{2} \text {. }
$$

If $-1<x<0$, then $x+1-x<2$, which means $1<2$, the inequality always holds.
If $x \leqslant-1$, then $-(x+1)-x<2$, which means $-2 x<3$,
$$
\therefore \quad-\frac{3}{2}<x \leqslant-1 \text {. }
$$

In summary,
$$
-\frac{3}{2}<x<\frac{1}{2} \text {. }
$$

Therefore, the correct choice is $(C)$.