2. Given $\sin \alpha=x, \cos \beta=y, \cos (\alpha+\beta)=-\frac{11}{14}$, and $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$. Then the functional relationship between $y$ and $x$ is ( ).
(A) $y=-\frac{11}{14} \sqrt{1-x^{2}}+\frac{5 \sqrt{3}}{14} x\left(\frac{11}{14}<x<1\right)$
(B) $y=-\frac{11}{14} \sqrt{1-x^{2}}+\frac{5 \sqrt{3}}{14} x(0<x<1)$
(C) $y=-\frac{11}{14} x-\frac{5 \sqrt{3}}{14} \sqrt{1-x^{2}}\left(-\frac{11}{14}<x<1\right)$
(D) $y=-\frac{11}{14} x-\frac{5 \sqrt{3}}{14} \sqrt{1-x^{2}}(0<x<1)$
2. (A).

From $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$ we know
$$
\cos \alpha=\sqrt{1-x^{2}}, \sin (\alpha+\beta)=\frac{5 \sqrt{3}}{14} .
$$

We have
$$
\begin{aligned}
y & =\cos \beta=\cos [(\alpha+\beta)-\alpha] \\
& =\cos (\alpha+\beta) \cdot \cos \alpha+\sin (\alpha+\beta) \cdot \sin \alpha \\
& =-\frac{11}{14} \sqrt{1-x^{2}}+\frac{5 \sqrt{3}}{14} x .
\end{aligned}
$$

Also, from $0x=\sin \alpha=-\cos \left(\alpha+\frac{\pi}{2}\right) \\
>-\cos (\alpha+\beta)=\frac{11}{14} .
\end{array}
$$

Thus, the domain of the function is $\frac{11}{14}<x<1$. The correct choice is (A).