11. Solve the inequality $\sqrt{1-\frac{x^{2}}{4}}<a x+\frac{1}{2}$ for $x$.
11. Let $y=\sqrt{1-\frac{x^{2}}{4}}$, its graph is the upper half of the ellipse
$\frac{x^{2}}{4}+y^{2}=1$. Also, the graph of $y=a x+\frac{1}{2}$ is a line passing through the fixed point $B\left(0, \frac{1}{2}\right)$ with a slope of $a$, as shown in the figure.
$$
k_{A B}=\frac{1}{4}, k_{1 C}=-\frac{1}{4},
$$

Thus, the x-coordinates of the intersection points of the line with the upper half of the ellipse are the solutions to the equation
$\sqrt{1-\frac{x^{2}}{4}}=a x+\frac{1}{2}$: $x=\frac{-2 \pm \sqrt{16 a^{2}+3}}{4 a^{2}+1}$.
When $-\frac{1}{4}<a<\frac{1}{4}$, the solution to the original inequality is
$-2 \leqslant x<\frac{-2 a-\sqrt{16 a^{2}+3}}{4 a^{2}+1}$ or $\quad \frac{-2 a+\sqrt{16 a^{2}+3}}{4 a^{2}+1}<x \leqslant 2$.
When $a \geqslant \frac{1}{4}$, the solution to the original inequality is $\frac{-2 a+\sqrt{16 a^{2}+3}}{4 a^{2}+1}<x \leqslant 2$.
When $a \leqslant-\frac{1}{4}$, the solution to the inequality is $-2 \leqslant x<\frac{-2 a-\sqrt{16 a^{2}+3}}{4 a^{2}+1}$.