2.1.3 * Given that $\alpha, \beta$ are both acute angles, $\cos (\alpha+\beta)=-\frac{4}{5}$. If we set $\sin \beta=x, \cos \alpha=y$, then the functional relationship between $y$ and $x$ is ().
(A) $y=-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x(0<x<1)$
(B) $y=-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x\left(\frac{4}{5}<x<1\right)$
(C) $y=-\frac{4}{5} \sqrt{1-x^{2}}-\frac{3}{5} x\left(\frac{4}{5}<x<1\right)$
(D) $y=-\frac{4}{5} \sqrt{1-x^{2}}-\frac{3}{5} x(0<x<1)$
From the known conditions, we have $\sin (\alpha+\beta)=\frac{3}{5}, \cos \beta=\sqrt{1-x^{2}}$. Therefore, $y=\cos \alpha=$ $\cos [(\alpha+\beta)-\beta]=\cos (\alpha+\beta) \cdot \cos \beta+\sin (\alpha+\beta) \cdot \sin \beta=-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x$. The range of $x$ should satisfy the system of inequalities $\left\{\begin{array}{l}0<x<1, \\ 0<-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x<1\end{array}\right.$, solving which yields $\frac{4}{5}<x<1$. Therefore, the correct choice is B.