11. Given the vectors $\boldsymbol{m}=\left(\cos \frac{x}{2},-1\right), \boldsymbol{n}=\left(\sqrt{3} \sin \frac{x}{2}, \cos ^{2} \frac{x}{2}\right)$, let the function $f(x)=\boldsymbol{m} \cdot \boldsymbol{n}+1$.
(1) If $x \in\left[0, \frac{\pi}{2}\right], f(x)=\frac{11}{10}$, find the value of $\cos x$;
(2) In $\triangle A B C$, the angles $A, B, C$ have opposite sides $a, b, c$ respectively, and satisfy $2 b \cos A \leqslant 2 c-\sqrt{3} a$, find the range of $f(B)$.
(1) $f(x)=\sqrt{3} \sin \frac{x}{2} \cos \frac{x}{2}-\cos ^{2} \frac{x}{2}+1=\frac{\sqrt{3}}{2} \sin x-\frac{1}{2}(1+\cos x)+1$
$=\sin \left(x-\frac{\pi}{6}\right)+\frac{1}{2}$. Therefore, $f(x)=\frac{11}{10} \Rightarrow \sin \left(x-\frac{\pi}{6}\right)=\frac{3}{5}$, since $x-\frac{\pi}{6} \in\left[-\frac{\pi}{6}, \frac{\pi}{3}\right]$
$\Rightarrow \cos \left(x-\frac{\pi}{6}\right)=\frac{4}{5}$, so $\cos x=\cos \left(x-\frac{\pi}{6}+\frac{\pi}{6}\right)=\frac{4}{5} \cdot \frac{\sqrt{3}}{2}-\frac{3}{5} \cdot \frac{1}{2}=\frac{4 \sqrt{3}-3}{10}$.
(2) $2 \sin B \cos A \leqslant 2 \sin C-\sqrt{3} \sin A \Rightarrow 2 \cos B \sin A-\sqrt{3} \sin A \geqslant 0$
$\Rightarrow \cos B \geqslant \frac{\sqrt{3}}{2} \Rightarrow B \in\left(0, \frac{\pi}{6}\right] \Rightarrow B-\frac{\pi}{6} \in\left(-\frac{\pi}{6}, 0\right]$, so $f(B)=\sin \left(B-\frac{\pi}{6}\right)+\frac{1}{2} \in\left(0, \frac{1}{2}\right]$.