解:(1)①∵F(x)=f(x)-g(x)=x
2+ax+c-lnx-c=x
2+ax-lnx,
∴
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,
∵函數(shù)F(x)在[1,2]上是減函數(shù),
∴當(dāng)x∈[1,2]時,
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恒成立,即
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恒成立,
又
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,
∴a∈
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;
②由G(x)=F(x)-x
2=x
2+ax-lnx-x
2=ax-lnx,
由
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<0,得x<
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,若
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,則G(x)在(0,e]上為減函數(shù),此時G(x)
min=ae-1,
由ae-1=3,得a=
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(舍);若
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,則函數(shù)G(x)在(0,
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)上為減函數(shù),在(
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,e)上為增函數(shù),
所以
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,由
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,得a=e
2
所以存在a=e
2,使函數(shù)G(x)的最小值是3.
(2)令
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,
則
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,
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,
∴
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∴g(x)=0,
在(x
1,x
2)內(nèi)必有一個實根.即?x
0∈(x
1,x
2),使
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成立.
分析:(1)①求出函數(shù)F(x),由導(dǎo)函數(shù)在[1,2]上小于等于0恒成立,采用分離變量求a的范圍;②求出函數(shù)G(x),求其導(dǎo)函數(shù),分最小值在不在給定的區(qū)間兩種情況討論a的存在性;
(2)構(gòu)造函數(shù)
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,判出g(x
1)•g(x
2)<0,則說明?x
0∈(x
1,x
2),使
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成立.
點評:本題(1)考查了利用函數(shù)的導(dǎo)函數(shù)判斷函數(shù)的單調(diào)性以及函數(shù)零點的判斷,考查了運用導(dǎo)函數(shù)的符號判斷函數(shù)單調(diào)性的方法,同時考查了分類討論的數(shù)學(xué)思想;
(2)考查了函數(shù)零點的判定,判斷函數(shù)在某區(qū)間內(nèi)有零點,只要滿足區(qū)間兩端點處的函數(shù)值的乘積小于0即可.