$4 \cdot 11$ If the non-empty set $A=\{x \mid 2 a+1 \leqslant x \leqslant 3 a-5\}, B=\{x \mid$ $3 \leqslant x \leqslant 22\}$, then the set of all $a$ that makes $A \subseteq A \cap B$ true is
(A) $\{a \mid 1 \leqslant a \leqslant 9\}$.
(B) $\{a \mid 6 \leqslant a \leqslant 9\}$.
(C) $\{a \mid a \leqslant 9\}$.
(D) $\emptyset$.
(China High School Mathematics League, 1998)
[Solution] According to the problem, we have $A \subseteq B$, which gives us
$$
\left\{\begin{array}{l}
2 a+1 \geqslant 3, \\
3 a-5 \leqslant 22, \\
3 a-5 \geqslant 2 a+1 .
\end{array}\right.
$$

Solving this, we get
$$
6 \leqslant a \leqslant 9 .
$$

Therefore, the correct choice is (B).