3. The solution set of the inequality $|x-3|+|x-a|<1$ is $\varnothing$, then the range of the real number $a$ is
A. $(0,1)$
B. $(1,+\infty)$
C. $(-\infty, 2]$
D. $(-\infty, 2] \cup[4,+\infty)$
3. D Hint: $|x-3|+|x-a| \geqslant|a-3|$, so the solution set of $|a-3| \geqslant 1$ is the range of $a$ that satisfies the condition.