1. Given the function $f(x)=k-\sqrt{x+2}$, there exist real numbers $a, b (a<b)$, such that the range of $f(x)$ on the interval $[a, b]$ is $[a, b]$. Then the range of values for $k$ is $\qquad$ .
$-1 .\left(-\frac{5}{4},-1\right]$.
From the problem, we have
$$
\begin{array}{l}
k-\sqrt{a+2}=b, k-\sqrt{b+2}=a . \\
\text { Then } \sqrt{b+2}+\sqrt{a+2}=1 \\
\Rightarrow k=b+1-\sqrt{b+2} . \\
\text { Let } \sqrt{b+2}=t(0 \leqslant t \leqslant 1) . \\
\text { Hence } k=t^{2}-t-1 \in\left(-\frac{5}{4},-1\right] .
\end{array}
$$