15.(2023•云南模拟)设函数f(x)=xex+axf(x)=xe^{x}+axf(x)=xex+ax,a>−1a>-1a>−1,若存在唯一整数x0x_{0}x0,使得f(x0)<0f(x_{0})<0f(x0)<0,则aaa的取值范围是 ___(−1e,−1e2](-\frac{1}{e},-\frac{1}{{e}^{2}}](−e1,−e21]___.