【答案】
分析:(Ⅰ)先求出導(dǎo)數(shù),對a分類討論即可得出;
(Ⅱ)利用導(dǎo)數(shù)的幾何意義求出切線的斜率,進(jìn)而得到切線的方程g(x)=0,構(gòu)造函數(shù)h(x)=g(x)-f(x),利用導(dǎo)數(shù)證明h(x)的最小值≥0即可.
解答:解:(Ⅰ)易知,函數(shù)f(x)的定義域為(0,+∞),
∵
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,①當(dāng)a≤0時,f
′(x)≥0,∴函數(shù)f(x)單調(diào)遞增,因此函數(shù)在(0,+∞)上無最大值,不符合題意,應(yīng)舍去;
②當(dāng)a>0時,
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,令f
′(x)=0,則
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.
當(dāng)
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時,f
′(x)>0,函數(shù)f(x)單調(diào)遞增;當(dāng)
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時,f
′(x)<0,函數(shù)f(x)單調(diào)遞減.
∴當(dāng)
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時,函數(shù)f(x)取得極大值,也即最大值.
∴
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=1,即
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,解得
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.
(Ⅱ)設(shè)P(x
,lnx
-ax
)是曲線f(x)=lnx-ax的圖象上的任意一點,則過點P的切線的斜率為
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,
∴切線為
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,化為y=g(x)=
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,
令h(x)=g(x)-f(x)=
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-(lnx-ax),
∴h
′(x)=
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=
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,令h
′(x)=0,解得x=x
.
當(dāng)0<x<x
時,h
′(x)<0,函數(shù)h(x)單調(diào)遞減;當(dāng)x>x
時,h
′(x)>0,函數(shù)h(x)單調(diào)遞增.
因此當(dāng)x=x
時,函數(shù)h(x)取得最小值,∴h(x)≥h(x
)=
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=0,
∴g(x)≥f(x),函數(shù)f(x)=1nx-ax的圖象除切點外恒在直線l的下方.
點評:熟練掌握利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性極值等性質(zhì)、分類討論的思想方法是解題的關(guān)鍵.