2. If the solution set of the inequality
$|x-a|<|x|+|x+1|$ is all real numbers, then the range of values for $a$ is $\qquad$
2. $-1<a<0$.

Since the inequality $|x-a|<|x|+|x+1|$ means that the graph of $y=|x-a|$ must be below the graph of $y=|x|+|x+1|$, it is necessary and sufficient that $-1<a<0$.