解:(1)由題意:f(x)≥g(x)?x
2-ax≥lnx,(x>0)
分離參數(shù)α可得:a≤
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,(x>0)…(1分)
設(shè)
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,則
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=
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…(2分)
由于函數(shù)y=x
2,y=lnx在區(qū)間(0,+∞)上都是增函數(shù),所以
函數(shù)y=x
2+lnx-1在區(qū)間(0,+∞)上也是增函數(shù),顯然x=1時(shí),該函數(shù)值為0
所以當(dāng)x∈(0,1)時(shí),Φ
′(x)<0,當(dāng)x∈(1,+∞)時(shí),Φ
′(x)>0
所以函數(shù)Φ(x)在(0,1)上是減函數(shù),在(1,+∞)上是增函數(shù)
所以Φ(x)
min=Φ(1)=1,所以a≤Φ(x)
min=1即a∈(-∞,1)…(4分)
(2)由題意知道:h(x)=x
2-ax+lnx.則
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所以方程2x
2-ax+1=0,(x>0)有兩個(gè)不相等的實(shí)數(shù)根x
1,x
2,且
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,
又因?yàn)?img class='latex' src='http://thumb.zyjl.cn/pic5/latex/207264.png' />,所以
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,且
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…(6分)
而h(x
1)-h(x
2)=
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=
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=
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=
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═
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,(x
2>1)
設(shè)
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,則
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所以
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,即
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…(8分)
(3)
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所以
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=
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…(9分)
因?yàn)閍∈(1,2),所以
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所以當(dāng)
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時(shí),r(x)是增函數(shù),所以當(dāng)
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時(shí),
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,a∈(1,2)…(10分)
所以,要滿足題意就需要滿足下面的條件:
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,
若令
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,a∈(1,2),
即對(duì)任意a∈(1,2),
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>0恒成立
因?yàn)棣?sup>′(a)=
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=
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…(11分)
分類討論如下:
①若k=0,則
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,所以φ(a)在(1,2)遞減,
此時(shí)φ(a)<φ(1)=0不符合題意
②若k<0,則
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,所以φ(a)在區(qū)間(1,2)遞減,
此時(shí)φ(a)<φ(1)=0不符合題意.
③若k>0,則
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,那么當(dāng)
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時(shí),假設(shè)t為2與
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中較小的一個(gè)數(shù),即t={
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},
則φ(a)在區(qū)間(1,min{
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})上遞減,此時(shí)φ(a)<φ(1)=0不符合題意.
綜上可得
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解得
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,即實(shí)數(shù)k的取值范圍為
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…(14分)
分析:(1)由于f(x)≥g(x)恒成立,只需使x
2-ax≥lnx,(x>0)分離參數(shù)來(lái)解決,注意a≤F(x)即要a≤F(x)
min;a≥F(x)即要a≥F(x)
max;
(2)借助于極值點(diǎn)的范圍,利用函數(shù)的導(dǎo)數(shù)來(lái)處理;
(3)與(1)類似處理,注意分類討論.
點(diǎn)評(píng):本題考查導(dǎo)數(shù)的綜合應(yīng)用,屬于較難的題目,注意與不等式恒成立的有關(guān)的參數(shù)取值范圍問(wèn)題常用分離參數(shù)來(lái)解決.