【答案】
分析:(1)當(dāng)a=-2時(shí)
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故函數(shù) 在(1,+∞)上是增函數(shù).
(2)
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,當(dāng)x∈[1,e],2x
2+a∈[a+2,a+2e
2].若a≥-2,f'(x)在[1,e]上非負(fù),故函數(shù)f(x)在[1,e]上是增函數(shù).
若-2e
2<a<-2,當(dāng)
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時(shí)f'(x)=0,當(dāng)
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時(shí),f'(x)<0,此時(shí)f(x)是減函數(shù); 當(dāng)
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時(shí),f'(x)>0,此時(shí)f(x)是增函數(shù).
所以此時(shí)有最值.若a≤-2e
2,f'(x)在[1,e]上非正,所以[f(x)]
min=f(e)=a+e
2.
(3)由題意可化簡(jiǎn)得
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(x∈[1,e]),令
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(x∈[1,e]),利用導(dǎo)數(shù)判斷其單調(diào)性求出最小值為g(1)=-1.
解答:解:(1)當(dāng)a=-2時(shí),f(x)=x
2-2lnx,當(dāng)x∈(1,+∞),
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,
(2)
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,當(dāng)x∈[1,e],2x
2+a∈[a+2,a+2e
2].
若a≥-2,f'(x)在[1,e]上非負(fù)(僅當(dāng)a=-2,x=1時(shí),f'(x)=0),故函數(shù)f(x)在[1,e]上是增函數(shù),此時(shí)[f(x)]
min=f(1)=1.
若-2e
2<a<-2,當(dāng)
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時(shí),f'(x)=0;
當(dāng)
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時(shí),f'(x)<0,此時(shí)f(x)是減函數(shù);
當(dāng)
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時(shí),f'(x)>0,此時(shí)f(x)是增函數(shù).
故[f(x)]
min=
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=
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.
若a≤-2e
2,f'(x)在[1,e]上非正(僅當(dāng)a=-2e
2,x=e時(shí),f'(x)=0),
故函數(shù)f(x)在[1,e]上是減函數(shù),此時(shí)[f(x)]
min=f(e)=a+e
2.
綜上可知,當(dāng)a≥-2時(shí),f(x)的最小值為1,相應(yīng)的x值為1;當(dāng)-2e
2<a<-2時(shí),f(x)
的最小值為
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,相應(yīng)的x值為
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;當(dāng)a≤-2e
2時(shí),f(x)的最小值為a+e
2,
相應(yīng)的x值為e.
(3)不等式f(x)≤(a+2)x,可化為a(x-lnx)≥x
2-2x.
∵x∈[1,e],∴l(xiāng)nx≤1≤x且等號(hào)不能同時(shí)取,所以lnx<x,即x-lnx>0,
因而
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(x∈[1,e])
令
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(x∈[1,e]),又
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,
當(dāng)x∈[1,e]時(shí),x-1≥0,lnx≤1,x+2-2lnx>0,
從而g'(x)≥0(僅當(dāng)x=1時(shí)取等號(hào)),所以g(x)在[1,e]上為增函數(shù),
故g(x)的最小值為g(1)=-1,所以a的取值范圍是[-1,+∞).
點(diǎn)評(píng):本題主要考查利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)及研究單調(diào)性與函數(shù)的最值,還考查求參數(shù)的范圍,解決此類問(wèn)題的關(guān)鍵是分離參數(shù)后轉(zhuǎn)化為恒成立問(wèn)題,即求新函數(shù)的最值問(wèn)題,是近年高考考查的熱點(diǎn).