2. Let the real number $a$ be such that the inequality
$$
|2 x-a|+|3 x-2 a| \geqslant a^{2}
$$

holds for all real numbers $x$. Then the set of all $a$ that satisfy this condition is ( ).
(A) $\left[-\frac{1}{3}, \frac{1}{3}\right]$
(B) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(C) $\left[-\frac{1}{4}, \frac{1}{3}\right]$
(D) $[-3,3]$
2.A.

For $k \in \mathbf{R}$, let $x=\frac{1}{2} k a$. Then the original inequality becomes
$$
|a| \cdot|k-1|+\frac{3}{2}|a| \cdot\left|k-\frac{4}{3}\right| \geqslant|a|^{2} \text {. }
$$

From this, it is easy to see that the original inequality is equivalent to
$$
|a| \leqslant|k-1|+\frac{3}{2}\left|k-\frac{4}{3}\right|
$$

for any $k \in \mathbf{R}$. Since
$$
|k-1|+\frac{3}{2}\left|k-\frac{4}{3}\right|=\left\{\begin{array}{ll}
\frac{5}{2} k-3, & k \geqslant \frac{4}{3} \\
1-\frac{1}{2} k, & 1 \leqslant k<\frac{4}{3} \\
3-\frac{5}{2} k, & k<1
\end{array}\right.
$$

Therefore, $\min _{k \in \mathbf{R}}\left\{|k-1|+\frac{3}{2}\left|k-\frac{4}{3}\right|\right\}=\frac{1}{3}$. 
Thus, the above inequality is equivalent to $|a| \leqslant \frac{1}{3}$.