(1)證明:設(shè)
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,
所以
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.…(1分)
當(dāng)x<0時,
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,當(dāng)x=0時,
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,當(dāng)x>0時,
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.
即函數(shù)φ
1(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,在x=0處取得唯一極小值,…(2分)
因?yàn)棣?sub>1(0)=0,所以對任意實(shí)數(shù)x均有 φ
1(x)≥φ
1(0)=0.
即f(x)-g
1(x)≥0,
所以f(x)≥g
1(x).…(3分)
(2)解:當(dāng)x>0時,f(x)>g
n(x).…(4分)
用數(shù)學(xué)歸納法證明如下:
①當(dāng)n=1時,由(1)知f(x)>g
1(x).
②假設(shè)當(dāng)n=k(k∈N
*)時,對任意x>0均有f(x)>g
k(x),…(5分)
令φ
k(x)=f(x)-g
k(x),φ
k+1(x)=f(x)-g
k+1(x),
因?yàn)閷θ我獾恼龑?shí)數(shù)x,
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,
由歸納假設(shè)知,
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.…(6分)
即φ
k+1(x)=f(x)-g
k+1(x)在(0,+∞)上為增函數(shù),亦即φ
k+1(x)>φ
k+1(0),
因?yàn)棣?sub>k+1(0)=0,所以φ
k+1(x)>0.
從而對任意x>0,有f(x)-g
k+1(x)>0.
即對任意x>0,有f(x)>g
k+1(x).
這就是說,當(dāng)n=k+1時,對任意x>0,也有f(x)>g
k+1(x).
由①、②知,當(dāng)x>0時,都有f(x)>g
n(x).…(8分)
(3)證明:先證對任意正整數(shù)n,g
n(1)<e.
由(2)知,當(dāng)x>0時,對任意正整數(shù)n,都有f(x)>g
n(x).
令x=1,得g
n(1)<f(1)=e.
所以g
n(1)<e.…(9分)
再證對任意正整數(shù)n,
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=
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.
要證明上式,只需證明對任意正整數(shù)n,不等式
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成立.
即要證明對任意正整數(shù)n,不等式
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(*)成立.…(10分)
以下分別用數(shù)學(xué)歸納法和基本不等式法證明不等式(*):
方法1(數(shù)學(xué)歸納法):
①當(dāng)n=1時,
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成立,所以不等式(*)成立.
②假設(shè)當(dāng)n=k(k∈N
*)時,不等式(*)成立,
即
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.…(11分)
則
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.
因?yàn)?img class='latex' src='http://thumb.zyjl.cn/pic5/latex/110672.png' />,…(12分)
所以
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.…(13分)
這說明當(dāng)n=k+1時,不等式(*)也成立.
由①、②知,對任意正整數(shù)n,不等式(*)都成立.
綜上可知,對任意正整數(shù)n,不等式
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成立.
…(14分)
方法2(基本不等式法):
因?yàn)?img class='latex' src='http://thumb.zyjl.cn/pic5/latex/110674.png' />,…(11分)
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,
…,
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,
將以上n個不等式相乘,得
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.…(13分)
所以對任意正整數(shù)n,不等式(*)都成立.
綜上可知,對任意正整數(shù)n,不等式
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成立.
…(14分)
分析:(1)設(shè)
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,可得函數(shù)φ
1(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,在x=0處取得唯一極小值,從而可得對任意實(shí)數(shù)x均有 φ
1(x)≥φ
1(0)=0,即可得到結(jié)論;
(2)當(dāng)x>0時,f(x)>g
n(x),用數(shù)學(xué)歸納法證明,第2步證明的關(guān)鍵是證明φ
k+1(x)=f(x)-g
k+1(x)在(0,+∞)上為增函數(shù);
(3)先證對任意正整數(shù)n,g
n(1)<e,再證對任意正整數(shù)n,
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=
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,利用分析法、再利用數(shù)學(xué)歸納法和基本不等式法可以證明結(jié)論.
點(diǎn)評:本小題主要考查函數(shù)、導(dǎo)數(shù)、不等式、數(shù)學(xué)歸納法、二項(xiàng)式定理等知識,考查數(shù)形結(jié)合、化歸與轉(zhuǎn)化、分類與討論的數(shù)學(xué)思想方法,以及運(yùn)算求解能力.