1. The range of the function $y=\sqrt{x-4}+\sqrt{15-3 x}$ is ( ).
(A) $[1,2]$.
(B) $(0,2)$.
(C) $(0, \sqrt{3})$.
(D) None of the above.
1. (A).

Obviously, the domain of the function is $4 \leqslant x \leqslant 5$. Then we can set $x$ $=4+\sin ^{2} \alpha \quad\left(0 \leqslant \alpha \leqslant \frac{\pi}{2}\right)$. Thus,
$$
\begin{aligned}
y= & \sqrt{\sin ^{2} \alpha}+\sqrt{3 \cos ^{2} \alpha} \\
& =\sin \alpha+\sqrt{3} \cos \alpha \\
& =2 \sin \left(\alpha+\frac{\pi}{3}\right) .
\end{aligned}
$$

From $\frac{\pi}{3} \leqslant \alpha+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$, we get $\quad 1 \leqslant 2 \sin \left(\alpha+\frac{\pi}{3}\right)$ $\leqslant 2$, which means $1 \leqslant y \leqslant 2$.