【答案】
分析:(1)先求函數(shù)f(x)的導(dǎo)數(shù),f′(x),再對k進(jìn)行奇偶數(shù)討論:1°當(dāng)k 為奇數(shù)時,f′(x)=
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;2°當(dāng)k 為偶數(shù)時,f′(x)=
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;最后綜合即可;
(2)當(dāng)k 為偶數(shù)時,由(1)知f′(x)=
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,由條件得{a
n 2+1}是一個公比為2的等比數(shù)列,從而得到a
n2=2
n-1,最后利用反證法進(jìn)行證明即可;
(3)當(dāng)k為奇數(shù)時,f′(x)=2(x+
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),欲證原不等式成立,即證:(x+
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)
n-(x
n+
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)≥2
n-2,由二項(xiàng)式定理得,即證:C
n1x
n-2+C
n2x
n-4+…+C
n 2-n x
2-n≥2
n-2,
設(shè)Sn=C
n1x
n-2+C
n2x
n-4+…+C
n 2-nx
2-n,利用倒序相加法即可證得.
解答:解:(1)函數(shù)f(x)的定義域?yàn)椋?,+∞),又f′(x)=2x-2(-1)
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=
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,
1°當(dāng)k 為奇數(shù)時,f′(x)=
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,∵x∈(0,+∞),∴f′(x)>0恒成立;
2°當(dāng)k 為偶數(shù)時,f′(x)=
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,∵∴x+1>0,f′(x)>0得x>1,即f(x)的單調(diào)增區(qū)間為(1,+∞),
綜上所述,當(dāng)k 為奇數(shù)時,f(x)的單調(diào)增區(qū)間為(0,+∞),當(dāng)k 為偶數(shù)時,即f(x)的單調(diào)增區(qū)間為(1,+∞),
(2)當(dāng)k 為偶數(shù)時,由(1)知f′(x)=
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,∴f′(an)=
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,
由條件得:2(a
n2-1)=a
n+12-3,故有:a
n+12+1=2(a
n2+1),
∴{a
n 2+1}是一個公比為2的等比數(shù)列,∴a
n2=2
n-1,
假設(shè)數(shù)列{a
n2}中的存在三項(xiàng)a
r2,s
2,a
t2,能構(gòu)成等差數(shù)列
不妨設(shè)r<s<t,則2a
s 2=a
r 2+a
t 2,
即2(2
s-1)=2
r-1+2
t-1,∴2
s-r+1=1+2
t-r,
又s-r+1>0,t-r>0,∴2
s-r+1為偶數(shù),1+2
t-r為奇數(shù),故假設(shè)不成立,
因此,數(shù)列{a
n2}中的任意三項(xiàng)不能構(gòu)成等差數(shù)列;
(3)當(dāng)k為奇數(shù)時,f′(x)=2(x+
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),即證:(x+
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)
n-(x
n+

)≥2
n-2,
由二項(xiàng)式定理得,即證:C
n1x
n-2+C
n2x
n-4+…+C
n 2-n x
2-n≥2
n-2,
設(shè)Sn=C
n1x
n-2+C
n2x
n-4+…+C
n 2-nx
2-n,
Sn=C
n2-nx
2-n+…+C
n2x
n-4+C
n1x
n-2,
兩式相加得:
2Sn=C
n1(x
n-2+x
2-n)+C
n2(x
n-4+x
4-n)+…+C
nn-1(x
n-2+x
2-n)≥2(C
n1+C
n2+…+C
n2-n)=2(2
n-1),
∴Sn≥2
n-2,
即原不等式成立.
點(diǎn)評:本小題主要考查等差關(guān)系的確定、利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、用數(shù)學(xué)歸納法證明不等式等基礎(chǔ)知識,考查運(yùn)算求解能力,考查化歸與轉(zhuǎn)化思想.屬于基礎(chǔ)題.