1. Given that $x$ is a real number. Then
$$
\sqrt{2016-x}+\sqrt{x-2000}
$$

the maximum value is $\qquad$
\begin{array}{l}-1.4 \sqrt{2} \text {. } \\ \text { Let } t=\sqrt{2016-x}+\sqrt{x-2000} \text {. } \\ \text { Then } t^{2}=16+2 \sqrt{(2016-x)(x-2000)} \\ \leqslant 16+2 \times 8=32 \\ \Rightarrow t \leqslant 4 \sqrt{2} \\ \Rightarrow t_{\text {max }}=4 \sqrt{2} \text {, at this point, } x=2008 \text {. } \\\end{array}