113 Given the function $f(x)=\sqrt{x+2}+k$, and there exist $a, b(a<b)$ such that the range of $f(x)$ on $[a, b]$ is $[a, b]$, find the range of the real number $k$.
13 From the problem, we know that the domain of $f(x)$ is $x \geqslant -2$. By the monotonicity of $f(x)$, the range of $f(x)$ on $[a, b]$ being $[a, b]$ is equivalent to the equation $f(x)=\sqrt{x+2}+k=x$ having two distinct real roots, i.e.,
$$
x^{2}-(2 k+1) x+k^{2}-2=0 \quad (x \geqslant -2)
$$

having two distinct real roots. Therefore,
$$
(2 k+1)^{2}-4\left(k^{2}-2\right)>0,
$$

solving this gives $k>-\frac{9}{4}$. From $f(x)=\sqrt{x+2}+k=x$, we know $x \geqslant k$, i.e.,
$$
\frac{2 k+1-\sqrt{(2 k+1)^{2}-4\left(k^{2}-2\right)}}{2} \geqslant k,
$$

solving this gives $k \leqslant -2$, and when $k \in\left(-\frac{9}{4},-2\right]$, 
$$
\frac{2 k+1-\sqrt{(2 k+1)^{2}-4\left(k^{2}-2\right)}}{2} \geqslant -2
$$

always holds. Therefore, $k \in\left(-\frac{9}{4},-2\right]$.