$$
\begin{array}{l}
\text { 11. Let } A=\{x \mid-2 \leqslant x \leqslant a\}, \\
B=\{y \mid y=2 x+3, x \in A\}, \\
C=\left\{z \mid z=x^{2}, x \in A\right\} \text {, and } C \subseteq B, \text { then }
\end{array}
$$
$$
\begin{array}{l}
\text { 11. } B=\{y \mid y=2 x+3, x \in A\} \\
=[-1,2 a+3] \\
\text { When } a \geqslant 2 \text {, } C=\left\{2 \mid z=x^{2}, \quad x \in A\right\} \\
=\left[0, a^{2}\right], \| C=E \text { when } a^{2} \leqslant 2 a+3 \text {, } \\
\end{array}
$$
$$
2 \leqslant 0<3 \text {. } 0 \leqslant a<2 \text {, then } C=(0,4) \text {, }
$$

From $C \subseteq B$ we get $4 \leqslant 2 a+3, \frac{1}{2} \leqslant a<2$. When -2
$\subseteq a<0$, $C=\left[a^{2}, 4\right]$, from $C=B$ we get 4 $\leqslant 2 a+3$, which contradicts $-2 \leqslant a<0$.
In summary, the set of real numbers $a$ is $(2,3)$
$$
U\left[\frac{1}{2}, 2\right)=\left[\frac{1}{2}, 3\right]
$$