解答:
解:(Ⅰ) f'(x)=ae
x(x+2),g'(x)=2x+b----------------------(1分)
由題意,兩函數(shù)在x=0處有相同的切線.
∴f'(0)=2a,g'(0)=b,
∴2a=b,f(0)=a=g(0)=2,
∴a=2,b=4,
∴f(x)=2e
x(x+1),g(x)=x
2+4x+2.----------------------(3分)
(Ⅱ)f'(x)=2e
x(x+2),由f'(x)>0得x>-2,由f'(x)<0得x<-2,
∴f(x)在(-2,+∞)單調(diào)遞增,在(-∞,-2)單調(diào)遞減.----------------------(4分)
∵t>-3,∴t+1>-2
①當-3<t<-2時,f(x)在[t,-2]單調(diào)遞減,[-2,t+1]單調(diào)遞增,
∴
f(x)min=f(-2)=-2e-2.----------------------(5分)
②當t≥-2時,f(x)在[t,t+1]單調(diào)遞增,∴
f(x)min=f(t)=2et(t+1);
∴
f(x)=&2et(t+1) (t≥-2)----------------------(6分)
(Ⅲ)由題意F(x)=4e
x(x+1)-x
2-4x
求導得F'(x)=4e
x(x+1)+4e
x-2x-4=2(x+2)(2e
x-1),----------------------(8分)
由F'(x)>0得x>-ln2或x<-2,由F'(x)<0得-2<x<-ln2
∴F(x)在(-∞,-2),(-ln2,+∞)上單調(diào)遞增,在(-2,-ln2)上單調(diào)遞減----------(10分)
∴
F(x)極小值=F(-ln2)=2+2ln2-(ln2)2=2+ln2(2-ln2)>0----------------------(11分)
∵F(-4)=4e
-4×(-4+1)-16+16=-12e
-4<0----------------------(12分)
故函數(shù)F(x)=2f(x)-g(x)+2只有一個零點.----------------------(13分)