11. Let $f(x)$ and $g(x)$ be odd and even functions defined on $R$, respectively, and $f(x) + g(x) = 2^x$. If for $x \in [1,2]$, the inequality $a f(x) + g(2x) \geq 0$ always holds, then the range of the real number $a$ is $\qquad$
Solution: $\left[-\frac{17}{6},+\infty\right]$.
Since $f(x)+g(x)=2^{x}$.
Therefore, $f(-x)+g(-x)=2^{-x}$, which means $-f(x)+g(x)=2^{-x}$.
(1)
From (1) and (2), we get $g(x)=\frac{2^{x}-2^{-x}}{2}, h(x)=\frac{2^{x}+2^{-x}}{2}$.
From $a f(x)+g(2 x) \geqslant 0$, we get $a\left(2^{x}-2^{-x}\right)+2^{2 x}+2^{-2 x} \geqslant 0$.
Let $t=2^{x}-2^{-x}$, then for $x \in[1,2]$, we have $t \in\left[\frac{3}{2}, \frac{15}{4}\right]$, and $2^{2 x}+2^{-2 x}=t^{2}+2$,
So from (3), we get $-a \leqslant t+\frac{2}{t}$ for $t \in\left[\frac{3}{2}, \frac{15}{4}\right]$ to always hold.
Since the function $t+\frac{2}{t}$ is monotonically increasing on $\left[\frac{3}{2}, \frac{15}{4}\right]$, when $t=\frac{3}{2}$, $\left(t+\frac{2}{t}\right)_{\min }=\frac{17}{6}$.
Therefore, $-a \leqslant \frac{17}{6}$, which means $a \geqslant-\frac{17}{6}$.