解:(1)若數(shù)列{a
n}在某個(gè)區(qū)間上是遞增數(shù)列,
則a
n+1-a
n>0,
即a
n+1-a
n=f(a
n)-a
n=-2a
n2+2a
n-a
n=-2a
n2+a
n>0,
∴a
n∈(0,
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)(2分)
又當(dāng)a
n∈(0,
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),n≥1時(shí),
a
n+1=f(a
n)=-2a
n2+2a
n=-2a
n(a
n-1)
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,
所以對(duì)一切n∈N
*,均有a
n∈(0,
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),
且a
n+1-a
n>0,(3分)
所以數(shù)列{a
n}在區(qū)間(0,
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)上是遞增數(shù)列.…(4分)
(2)由(1)知a
n∈(0,
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),
從而
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);
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,
即
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;
令b
n=
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,
則有b
n+1=2b
n2且b
n∈(0,
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);
從而有l(wèi)gb
n+1=2lgb
n+lg2,(7分)
可得lgb
n+1+lg2=2(lgb
n+lg2),
所以數(shù)列{lgb
n+lg2}是lgb
1+lg2=lg
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為首項(xiàng),公比為2的等比數(shù)列. (8分)
(3)由(2)得lgb
n+lg2=lg
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,
即lgb
n=lg
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,
所以 b
n=
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,
所以
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,
所以log
3(
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,(10分)
所以,log
3(
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)=nlog
32+
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2-1.(11分)
即2
n+nlog
32-12
n+(log
32)n-1>(-1)
n-12λ+nlog
32-1nlog
32-1,
所以,2
n-1>(-1)
n-1λ恒成立
當(dāng)n為奇數(shù)時(shí),即λ<2
n-1恒成立,
當(dāng)且僅當(dāng)n=1時(shí),2
n-1有最小值1為.
∴λ<1
當(dāng)n為偶數(shù)時(shí),即λ>-2
n-1恒成立,
當(dāng)且僅當(dāng)n=2時(shí),有最大值-2為.
∴λ>-2(13)
所以,對(duì)任意n∈N
*,有-2<λ<1.
又λ非零整數(shù),
∴λ=-1(14分)
分析:(1)若數(shù)列{a
n}在某個(gè)區(qū)間上是遞增數(shù)列,則a
n+1-a
n>0,即a
n+1-a
n=f(a
n)-a
n=-2a
n2+2a
n-a
n=-2a
n2+a
n>0?a
n∈(0,

).所以對(duì)一切n∈N
*,均有a
n∈(0,

)且a
n+1-a
n>0,所以數(shù)列{a
n}在區(qū)間(0,

)上是遞增數(shù)列.
(2)由a
n∈(0,
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),知
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),所以
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.令b
n=
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,則有l(wèi)gb
n+1=2lgb
n+lg2,所以lgb
n+1+lg2=2(lgb
n+lg2),故數(shù)列{lgb
n+lg2}是lgb
1+lg2=lg
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為首項(xiàng),公比為2的等比數(shù)列.
(3)由(2)得b
n=
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,所以log
3(
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).故log
3(
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)=nlog
32+
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2-1,所以2
n-1>(-1)
n-1λ恒成立.由此能求出λ的值.
點(diǎn)評(píng):本題首先考查等差數(shù)列、等比數(shù)列的基本量、通項(xiàng),結(jié)合含兩個(gè)變量的不等式的處理問題,用兩邊夾的方法確定整數(shù)參數(shù).第Ⅲ小題對(duì)數(shù)學(xué)思維的要求比較高,要求學(xué)生理解“存在”、“恒成立”,以及運(yùn)用一般與特殊的關(guān)系進(jìn)行否定,本題有一定的探索性.綜合性強(qiáng),難度大,易出錯(cuò).