11. For the system of inequalities about $x$ $\left\{\begin{array}{l}x^{2}-x-2>0 \\ 2 x^{2}+(2 k+5) x+5 k<0\end{array}\right.$, the set of integer solutions is $\{-2\}$. Find the range of real numbers for $k$.
11. From $x^{2}-x-2>0$, we get $x>2$ or $x<-1$. Suppose the solution set of $(2 x+5)(x+k)<0$ is $A$, and $-2$ $\in A$, then substituting into the above inequality yields $k<2, A=\left(-\frac{5}{2},-k\right)$, thus $-\frac{5}{2}<-k \leqslant 3$, which gives $-3 \leqslant k<\frac{5}{2}$. In summary, $-3 \leqslant k<2$.