11. (13 points) Given 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 sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and satisfy $2 b \cos A \leqslant 2 c-\sqrt{3} a$, find the range of $f(B)$.
(1) Notice,
$$
\begin{array}{l}
f(x)=\sqrt{3} \sin \frac{x}{2} \cdot \cos \frac{x}{2}-\cos ^{2} \frac{x}{2}+1 \\
=\frac{\sqrt{3}}{2} \sin x-\frac{1+\cos x}{2}+1 \\
=\frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x+\frac{1}{2} \\
=\sin \left(x-\frac{\pi}{6}\right)+\frac{1}{2} .
\end{array}
$$

Given $f(x)=\frac{11}{10}$, then $\sin \left(x-\frac{\pi}{6}\right)=\frac{3}{5}$.
Also, $x \in\left[0, \frac{\pi}{2}\right]$, then $x-\frac{\pi}{6} \in\left[-\frac{\pi}{6}, \frac{\pi}{3}\right]$.
Thus, $\cos \left(x-\frac{\pi}{6}\right)=\frac{4}{5}$.
Therefore, $\cos x=\cos \left(\left(x-\frac{\pi}{6}\right)+\frac{\pi}{6}\right)$
$$
\begin{array}{l}
=\cos \left(x-\frac{\pi}{6}\right) \cdot \cos \frac{\pi}{6}-\sin \left(x-\frac{\pi}{6}\right) \cdot \sin \frac{\pi}{6} \\
=\frac{4 \sqrt{3}}{10}-\frac{3}{10} .
\end{array}
$$

(2) Given $2 b \cos A \leqslant 2 c-\sqrt{3} a$,
$$
\Rightarrow 2 \sin B \cdot \cos A \leqslant 2 \sin C-\sqrt{3} \sin A \\
\Rightarrow 2 \sin B \cdot \cos A \leqslant 2 \sin (A+B)-\sqrt{3} \sin A \\
\Rightarrow 2 \sin B \cdot \cos A \\
\leqslant 2(\sin A \cdot \cos B+\sin B \cdot \cos A)-\sqrt{3} \sin A \\
\Rightarrow 2 \sin A \cdot \cos B \geqslant \sqrt{3} \sin A \\
\Rightarrow \cos B \geqslant \frac{\sqrt{3}}{2} \Rightarrow \angle B \in\left(0, \frac{\pi}{6}\right] . \\
\text { Hence } \sin \left(B-\frac{\pi}{6}\right) \in\left(-\frac{1}{2}, 0\right] \\
\Rightarrow f(B)=\sin \left(B-\frac{\pi}{6}\right)+\frac{1}{2} \in\left(0, \frac{1}{2}\right] .
$$