解:(1)設
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=x的不動點為0和2
∴
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即
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即b、c滿足的關系式:b=1+
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且c≠0
(2)∵c=2∴b=2∴f(x)=
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(x≠1),
由已知可得2S
n=a
n-a
n2①,且a
n≠1.
當n≥2時,2S
n-1=a
n-1-a
n-12②,
①-②得(a
n+a
n-1)(a
n-a
n-1+1)=0,∴a
n=-a
n-1或a
n=-a
n-1=-1,
當n=1時,2a
1=a
1-a
12?a
1=-1,
若a
n=-a
n-1,則a
2=1與a
n≠1矛盾.∴a
n-a
n-1=-1,∴a
n=-n
∴要證待證不等式,只要證
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,
即證
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,
只要證nln(1+
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)<1<(n+1)ln(1+
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),即證
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<ln(1+
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)<
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.
考慮證不等式
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<ln(x+1)<x(x>0)**.
令g(x)=x-ln(1+x),h(x)=ln(x+1)-
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(x>0).
∴g'(x)=
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,h'(x)=
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,
∵x>0,∴g'(x)>0,h'(x)>0,∴g(x)、h(x)在(0,+∞)上都是增函數(shù),
∴g(x)>g(0)=0,h(x)>h(0)=0,∴x>0時,
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<ln(x+1)<x.
令x=
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則**式成立,∴
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,
(3)由(2)知b
n=
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,則T
n=
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在
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<ln(1+
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)<
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中,令n=1,2,3,…,2011,并將各式相加,
得
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<ln
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+ln
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+…+ln
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<1+
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.
即T
2012-1<ln2012<T
2011.
分析:(1)設
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=x的不動點為0和2,由此知
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推出b、c滿足的關系式.
(2)由c=2,知b=2,f(x)=
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(x≠1),2S
n=a
n-a
n2,且a
n≠1.所以a
n-a
n-1=-1,a
n=-n,要證待證不等式,只要證
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,利用分析法證明
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<ln(1+
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)<
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.考慮證不等式
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<ln(x+1)<x(x>0),由此入手利用函數(shù)的導數(shù)判斷函數(shù)的單調性,然后導出
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.
(3)由
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,利用(2)的結論,通過累加法證明所要證明的不等式T
2012-1<ln2012<T
2011即可.
點評:本題考查不等式的性質和應用,函數(shù)的導數(shù)判斷函數(shù)的單調性構造法的應用,分析法證明不等式的方法,解題時要認真審題,仔細解答,注意公式的合理運用.