11. Given the function $f(x)=-x^{2}+x+m+2$. If the solution set of the inequality $f(x) \geqslant|x|$ contains exactly 1 integer, then the range of the real number $m$ is $\qquad$ .
11. $[-2,-1)$.

Notice,
$$
\begin{array}{l}
f(x) \geqslant|x| \Leftrightarrow 2-|x| \geqslant x^{2}-x-m . \\
\text { Let } g(x)=2-|x|, h(x)=x^{2}-x-m .
\end{array}
$$

Plot the graphs of the two functions in the same Cartesian coordinate system, as shown in Figure 3, and we know the integer solution is $x=0$.
$$
\text { Therefore }\left\{\begin{array}{l}
f(0) \geqslant 0-0-m, \\
f(1)<1-1-m
\end{array} \Rightarrow-2 \leqslant m<-1\right. \text {. }
$$