解答:解:(Ⅰ) 當(dāng)a=-2,f(x)=-
+x-3lnx-30(x>0)∴
f′(x)=+1-=,
設(shè)f'(x)>0,即x
2-3x+2>0,
所以x<1,或x>2,
∴f(x)單調(diào)增區(qū)間是(0,1),(2,+∞);
(Ⅱ)∵F(x)=-2x
3+3(a+2)x
2+6x-6a-4a
2,當(dāng)x=1時(shí),函數(shù)F(x)有極值,
∴F'(x)=-6x
2+6(a+2)x+6,
且F'(1)=0,即a=-2,
∴F(x)=-2x
3+6x-4,
又F(x)=-2x
3+6x-4的圖象可由
F1(x)=-2x3+6x的圖象向下平移4個(gè)單位長度得到,而
F1(x)=-2x3+6x的圖象關(guān)于(0,0)對(duì)稱,
所以F(x)=-2x
3+6x-4的圖象的對(duì)稱中心坐標(biāo)為(0,-4);
(Ⅲ)假設(shè)存在a使g(x)在[a,-a]上為減函數(shù),
設(shè)h
1(x)=F(x)-6x
2+6(a-1)x•e
x,
h2(x)=e•f(x)=e•(+x+(a-1)lnx+15a),
h′1(x)=(-2x3+3(a-2)x2+12ax-4a2)•ex,
設(shè)m(x)=(-2x
3+3(a-2)x
2+12ax-4a
2),
當(dāng)g(x)在[a,-a]上為減函數(shù),則h
1(x)在[a,1]上為減函數(shù),h
2(x)在[1,-a]上為減函數(shù),且h
1(1)≥h
2(1).
由(Ⅰ)知當(dāng)a<-1時(shí),f(x)的單調(diào)減區(qū)間是(1,-a),
由h
1(1)≥h
2(1)得:4a
2+13a+3≤0,
解得:
-3≤a≤-,
當(dāng)h
1(x)在[a,1]上為減函數(shù)時(shí),對(duì)于?x∈[a,1],h'
1(x)≤0即m(x)≤0恒成立,
因?yàn)閙'(x)=-6(x+2)(x-a),
(1)當(dāng)a<-2時(shí),m(x)在[a,-2]上是增函數(shù),在(-∞,a],[-2,+∞)是減函數(shù),
所以m(x)在[a,1]上最大值為m(-2)=-4a
2-12a-8,
故m(-2)=-4a
2-12a-8≤0,
即a≤-2,或a≥-1,故a<-2;
(2)當(dāng)a>-2時(shí),m(x)在[-2,a]上是增函數(shù),在(-∞,-2],[a,+∞)是減函數(shù),
所以m(x)在[a,1]上最大值為m(a)=a
2(a+2),
故m(a)=a
2(a+2)≤0,則a≤-2與題設(shè)矛盾;
(3)當(dāng)a=-2時(shí),m(x)在[-2,1]上是減函數(shù),
所以m(x)在[a,1]上最大值為m(-2)=-4a
2-12a-8=0,
綜上所述,符合條件的a滿足[-3,-2].