88 A. It is easy to know that the domain of the function is $[4,5]$, so we can set $x=4+\sin ^{2} x,\left(0 \leqslant \alpha \leqslant \frac{\pi}{2}\right)$,
thus
$$
y=\sqrt{\sin ^{2} x}+\sqrt{3 \cdot\left(1-\sin ^{2} x\right)}=2 \sin \left(\alpha+\frac{\pi}{3}\right) .
$$

Since
$$
\frac{\pi}{3} \leqslant \alpha+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}
$$

Therefore, $1 \leqslant y \leqslant 2$.
88 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. $[0,+\infty)$