1. The maximum value of the real number $k$ for which the inequality
$$
\sqrt{x-3}+\sqrt{6-x} \geqslant k
$$

has solutions is ( ).
(A) $\sqrt{6}-\sqrt{3}$
(B) $\sqrt{3}$
(C) $\sqrt{6}+\sqrt{3}$
(D) $\sqrt{6}$
-.1.D.
Let $y=\sqrt{x-3}+\sqrt{6-x}, 3 \leqslant x \leqslant 6$, then $y^{2} \leqslant 2[(x-3)+(6-x)]=6$.
Therefore, $0<y \leqslant \sqrt{6}$. Hence, the maximum value of the real number $k$ is $\sqrt{6}$.