11. Let $f(x)$ and $g(x)$ be odd and even functions defined on $\mathbf{R}$, respectively, and $f(x) + g(x) = 2^x$. If for $x \in [1,2]$, the inequality $a f(x) + g(2x) \geqslant 0$ always holds, then the range of the real number $a$ is $\qquad$.
$$
\begin{array}{l}
\text { 11. }\left[-\frac{17}{6},+\infty\right) . \\
\text { From } f(x)+g(x)=2^{x} \\
\Rightarrow f(-x)+g(-x)=2^{-x} \\
\Rightarrow-f(x)+g(x)=2^{-x} \text {. }
\end{array}
$$

From equations (1) and (2), we get
$$
g(x)=\frac{2^{x}+2^{-x}}{2}, f(x)=\frac{2^{x}-2^{-x}}{2} \text {. }
$$

From $a f(x)+g(2 x) \geqslant 0$
$$
\Rightarrow a\left(2^{x}-2^{-x}\right)+2^{2 x}+2^{-2 x} \geqslant 0 \text {. }
$$

Let $t=2^{x}-2^{-x}$, given $x \in[1,2]$, we have $t \in\left[\frac{3}{2}, \frac{15}{4}\right]$, and $2^{2 x}+2^{-2 x}=t^{2}+2$. Then for $t \in\left[\frac{3}{2}, \frac{15}{4}\right]$, from equation (3) we get $-a \leqslant t+\frac{2}{t}$.
Since the function $t+\frac{2}{t}$ is monotonically increasing in the interval $\left[\frac{3}{2}, \frac{15}{4}\right]$, when $t=\frac{3}{2}$, $\left(t+\frac{2}{t}\right)_{\min }=\frac{17}{6}$. Therefore, $-a \leqslant \frac{17}{6}$, which means $a \geqslant-\frac{17}{6}$.