11. Given the function $f(x)=-x^{2}+x+m+2$, if the solution set of the inequality $f(x) \geqslant|x|$ contains exactly one integer, then the range of the real number $m$ is $\qquad$ .
$f(x) \geqslant|x| \Rightarrow 0 \leqslant-x^{2}+x-|x|+m+2 \leqslant-x^{2}+m+2 \Rightarrow m+2 \geqslant x^{2}$ can hold, then $m+2 \geqslant 0$ and integer 0 satisfies the condition. Substituting $x= \pm 1$ yields $\left\{\begin{array}{l}-1+1+m+2<1, \\ -1-1+m+2<1\end{array} \Rightarrow-2 \leqslant m<-1\right.$. Therefore, the range of $m$ is $[-2,-1)$.