f'(x)=e^x*[x^2+(a+2)x+2a]=e^x*(x+2)(x+a),a=-2时f'(x)=e^2*(x+2)^2>=0,f(x)是增函数；a∴a>=-2时f(x)在[a,+∞)上是增函数，f(x)<=e^a在[a,+∞)上有解，<==>f(a)<=e^a,<==>2a^2+a-1<=0,解得-1<=a<=1/2,已知a<0,故-1<=a<0.a<-2时f(x)<=e^a,<==>f(-2)=e^(-2)*(4-a)<=e^a,取对数得-2+ln(4-a)<=a,∴0综上，-1<=a<0,为所求。
