解答:解:(Ⅰ)若a=1,則f(x)=(x
2-2x+1)e
x,f′(x)=(x
2-1)e
x∴切線的斜率k=f′(2)=3e
2又切點(diǎn)的坐標(biāo)為(2,e
2),
∴切線方程為y-e
2=3e
2(x-2),即3e
2x-y-5e
2=0
(Ⅱ)由f′(x)=[ax
2+(a-1)x-a]e
x得h(x)=f(x)-f'(x)=[-2ax+(a+1)]e
x,h′(x)=(-2ax-a+1)e
x,(1)當(dāng)a=0時(shí),h′(x)=e
x>0對(duì)x∈[0,1]恒成立,所以h(x)在[0,1]上單調(diào)遞增,h(x)max=h(1)=e
(2)當(dāng)a∈(0,1]時(shí),由h′(x)=0,得x=
-≥0
①當(dāng)
-≥1時(shí),即a∈(0,
]時(shí),h′(x)≥0對(duì)x∈[0,1]恒成立,h(x)在[0,1]上單調(diào)遞增,h(x)max=h(1)=(1-a)e
②當(dāng)1>
->0時(shí),即a∈(
,1)時(shí),h(x)在[0,
-)上單調(diào)遞增,在(
-,1]上單調(diào)遞減,h(x)max=h(
-)=2a
e③當(dāng)
-=0時(shí),即a=1時(shí),h′(x)≤0對(duì)x∈[0,1]恒成立,h(x)在[0,1]上單調(diào)遞減,h(x)max=h(0)=a+1
綜上,當(dāng)a=0時(shí),h(x)max=e,當(dāng)a∈(0,
]時(shí),h(x)max=)=(1-a)e
當(dāng)a∈(
,1)時(shí),h(x)max=2a
e,當(dāng)a=1時(shí),h(x)max=a+1.
(Ⅲ)由(Ⅰ)知,問(wèn)題可轉(zhuǎn)換為判定方程(x-1)
2e
x=x,x>1的實(shí)根的個(gè)數(shù).設(shè)φ(x)=(x-1)
2e
x-x,則φ′(x)=(x
2-1)e
x-1,再設(shè)k(x)=(x
2-1)e
x-1,x>1,則k′(x)=e
x(x
2+2x-1)
x>1時(shí),k′(x)>0,k(x)在(1,+∞)上單調(diào)遞增,又k(1)=-1<0,k(2)=3e
2-1>0,所以在(1,2)上存在唯一x
0,使得k(x
0)=0即存在唯一x
0,使得φ′(x
0)=0.
從而φ(x)在(1,x
0)上單調(diào)遞減,在(x
0,+∞)上單調(diào)遞增,φ(x
0)<φ(1)=-1<0,又φ(2)=e
2-2>0故y=φ(x)的大致圖象如圖所示.
因此y=φ(x)在(1,+∞)上只能有一個(gè)零點(diǎn).即當(dāng)x>1時(shí),f(x)=x只有一個(gè)實(shí)根.