已知函数 f(x)=ex(ax−1)−x+2f\left( x\right) = {\mathrm{e}}^{x}\left( {{ax} - 1}\right) - x + 2f(x)=ex(ax−1)−x+2 ，若存在唯一的整数 x0{x}_{0}x0 ，使 f(x0)<0f\left( {x}_{0}\right) < 0f(x0)<0 ，则 aaa 的取值范围是（ ）A. [14+12e4,13e3+13)\left\lbrack {\frac{1}{4} + \frac{1}{2{\mathrm{e}}^{4}},\frac{1}{3{\mathrm{e}}^{3}} + \frac{1}{3}}\right)[41+2e41,3e31+31) B. (3e−1,2e2−12]\left( {3\mathrm{e} - 1,2{\mathrm{e}}^{2} - \frac{1}{2}}\right\rbrack(3e−1,2e2−21]C. [12,1−1e)∪(3e−1,2e2−12]\left\lbrack {\frac{1}{2},1 - \frac{1}{\mathrm{e}}}\right) \cup \left( {3\mathrm{e} - 1,2{\mathrm{e}}^{2} - \frac{1}{2}}\right\rbrack[21,1−e1)∪(3e−1,2e2−21] D. [12,1−1e)∪[14+12e4,13e3+13)\left\lbrack {\frac{1}{2},1 - \frac{1}{\mathrm{e}}}\right) \cup \left\lbrack {\frac{1}{4} + \frac{1}{2{\mathrm{e}}^{4}},\frac{1}{3{\mathrm{e}}^{3}} + \frac{1}{3}}\right)[21,1−e1)∪[41+2e41,3e31+31)