1. Given the inequality about $x$
$$
\sqrt{x}+\sqrt{2-x} \geqslant k
$$

has real solutions. Then the range of the real number $k$ is ( ).
(A) $(0,2]$
(B) $(-\infty, 0]$
(C) $(-\infty, 0)$
(D) $(-\infty, 2]$
-.1. D.
Let $y=\sqrt{x}+\sqrt{2-x}(0 \leqslant x \leqslant 2)$. Then $y^{2}=x+(2-x)+2 \sqrt{x(2-x)} \leqslant 4$. Therefore, $0<y \leqslant 2$, and when $x=1$, the equality holds. Hence, the range of real number $k$ is $(-\infty, 2]$.