分析:(Ⅰ)把a(bǔ)=-
代入函數(shù)f(x),再對(duì)其進(jìn)行求導(dǎo)利用導(dǎo)數(shù)研究函數(shù)f(x)的單調(diào)區(qū)間;
(Ⅱ)當(dāng)x∈[0,+∞)時(shí),不等式f(x)≤x恒成立,即ax
2+ln(x+1)-x≤0恒成立,只要求出ax
2+ln(x+1)-x的最小值即可,構(gòu)造新的函數(shù),利用導(dǎo)數(shù)研究其最值問(wèn)題;
(Ⅲ)由題設(shè)(Ⅱ)可知當(dāng)a=0時(shí),ln(x+1)≤x在[0,+∞)上恒成立,利用此不等式對(duì)所要證明的不等式進(jìn)行放縮,從而進(jìn)行證明;
解答:解:(Ⅰ)當(dāng)a=-
時(shí),f(x)=-
x2+ln(x+1)(x>-1),
f′(x)=-
x+
=-
(x>-1),
由f'(x)>0,解得-1<x<1,由f'(x)<0,解得x>1.
故函數(shù)f(x)的單調(diào)遞增區(qū)間為(-1,1),單調(diào)遞減區(qū)間為(1,+∞).
(Ⅱ)當(dāng)x∈[0,+∞)時(shí),不等式f(x)≤x恒成立,即ax
2+ln(x+1)-x≤0恒成立,
設(shè)g(x)=ax
2+ln(x+1)-x(x≥0),只需g(x)
max≤0即可.
由g′(x)=2ax+
-1=
,
(。┊(dāng)a=0時(shí),g′(x)=
,
當(dāng)x>0時(shí),g'(x)<0,函數(shù)g(x)在(0,+∞)上單調(diào)遞減,
故g(x)≤g(0)=0成立.
(ⅱ)當(dāng)a>0時(shí),由g′(x)=
=0,因x∈[0,+∞),所以x=
-1,
①若
-1<0,即a>
時(shí),在區(qū)間(0,+∞)上,g'(x)>0,
則函數(shù)g(x)在(0,+∞)上單調(diào)遞增,
g(x)在[0,+∞)上無(wú)最大值,此時(shí)不滿足條件;
②若
-1≥0,即0<a≤
時(shí),函數(shù)g(x)在(0,
-1)上單調(diào)遞減,在區(qū)間(
-1,+∞)上單調(diào)遞增,
同樣g(x)在[0,+∞)上無(wú)最大值,不滿足條件.
(ⅲ)當(dāng)a<0時(shí),g′(x)=
,
∵x∈[0,+∞),∴2ax+(2a-1)<0,
∴g'(x)≤0,故函數(shù)g(x)在[0,+∞)上單調(diào)遞減,
故g(x)≤g(0)=0成立.
綜上所述,實(shí)數(shù)a的取值范圍是(-∞,0].
(Ⅲ)據(jù)(Ⅱ)知當(dāng)a=0時(shí),ln(x+1)≤x在[0,+∞)上恒成立,
又
=2(
-
),
∵ln{(1+
)(1+
)(1+
)•…•[1+
]}
=ln(1+
)+ln(1+
)+ln(1+
)+…+ln[1+
]<
+
+
+…+
=2[(
-
)+(
-
)+(
-
)+…+(
-
)]
=2[(
-
)]<1,
∴(1+
)(1+
)(1+
)•…•[1+
]<e.