解:(1)當(dāng)a=1時(shí),H(x)=f(x)-g(x)=lnx+x-x
2,定義域?yàn)椋?,+∞),
H′(x)=
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+1-2x=-
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,
當(dāng)0<x<1時(shí),H′(x)>0,當(dāng)x>1時(shí),H′(x)<0,
所以函數(shù)H(x)的增區(qū)間為(0,1),減區(qū)間為(1,+∞);
(2)H(x)=lnx+x-ax
2,H′(x)=
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=
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,
因?yàn)镠(x)在定義域內(nèi)不單調(diào),則函數(shù)h(x)=1+x-2ax
2在(0,+∞)內(nèi)有零點(diǎn),且在零點(diǎn)兩側(cè)函數(shù)值異號(hào),
又h(0)=1>0,則有
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或
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,解得a>0.
故實(shí)數(shù)a的取值范圍為(0,+∞).
(3)設(shè)P(x
0,y
0),則lnx
0+x
0=
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①,f′(x
0)=g′(x
0),即
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,化簡(jiǎn)得x
0+1=
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②
聯(lián)立①②消a得,lnx
0+
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-
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=0,
令φ(x)=lnx+
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x-
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,易知φ(x)=lnx+
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x-
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在(0,+∞)上單調(diào)遞增,又φ(1)=0,
所以lnx+
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x-
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=0有唯一解1,即x
0=1,則y
0=f(1)=1,g(1)=a=1,
故P(1,1),a=1.
分析:(1)a=1時(shí)寫出H(x)的表達(dá)式,求導(dǎo)數(shù)H′(x),然后在定義域內(nèi)解不等式H′(x)>0,H′(x)<0即可;
(2)H(x)在定義域內(nèi)不單調(diào),則函數(shù)H′(x)在(0,+∞)內(nèi)有零點(diǎn),且在零點(diǎn)兩側(cè)函數(shù)值異號(hào),據(jù)此得一不等式組,解出即可;
(3)設(shè)P(x
0,y
0),則lnx
0+x
0=
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,f′(x
0)=g′(x
0),聯(lián)立消掉a可得關(guān)于x
0的方程,構(gòu)造函數(shù),根據(jù)函數(shù)單調(diào)性可求得唯一x
0值,進(jìn)而可求a值;
點(diǎn)評(píng):本題考查利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性及導(dǎo)數(shù)的幾何意義,考查學(xué)生靈活運(yùn)用所學(xué)知識(shí)分析問(wèn)題解決問(wèn)題的能力,屬難題.