【答案】
分析:(Ⅰ)求函數(shù)f(x)=x-lnx的導(dǎo)數(shù),利用導(dǎo)數(shù)判斷在[e,e
2]上的單調(diào)性,便可求值域;
(Ⅱ)依題意就是求f(x)在[e,e
2]上的最大值,用a表示出函數(shù)最大值,再將恒成立轉(zhuǎn)化為函數(shù)最值問(wèn)題,結(jié)合導(dǎo)數(shù)法解決即可.
解答:解:(Ⅰ)當(dāng)a=-1時(shí),f(x)=x-lnx,
得
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,(2分)
令f'(x)>0,即
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,解得x>1,所以函數(shù)f(x)在(1,+∞)上為增函數(shù),
據(jù)此,函數(shù)f(x)在[e,e
2]上為增函數(shù),(4分)
而f(e)=e-1,f(e
2)=e
2-2,所以函數(shù)f(x)在[e,e
2]上的值域?yàn)閇e-1,e
2-2](6分)
(Ⅱ)由
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,令f'(x)=0,得
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,即x=-a,
當(dāng)x∈(0,-a)時(shí),f'(x)<0,函數(shù)f(x)在(0,-a)上單調(diào)遞減;
當(dāng)x∈(-a,+∞)時(shí),f'(x)>0,函數(shù)f(x)在(-a,+∞)上單調(diào)遞增;(7分)
若1≤-a≤e,即-e≤a≤-1,易得函數(shù)f(x)在[e,e
2]上為增函數(shù),
此時(shí),f(x)
max=f(e
2),要使f(x)≤e-1對(duì)x∈[e,e
2]恒成立,只需f(e
2)≤e-1即可,
所以有e
2+2a≤e-1,即
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而
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,即
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,所以此時(shí)無(wú)解.(8分)
若e<-a<e
2,即-e>a>-e
2,易知函數(shù)f(x)在[e,-a]上為減函數(shù),在[-a,e
2]上為增函數(shù),
要使f(x)≤e-1對(duì)x∈[e,e
2]恒成立,只需
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,即
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,
由
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和
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得
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.(10分)
若-a≥e
2,即a≤-e
2,易得函數(shù)f(x)在[e,e
2]上為減函數(shù),
此時(shí),f(x)
max=f(e),要使f(x)≤e-1對(duì)x∈[e,e
2]恒成立,只需f(e)≤e-1即可,
所以有e+a≤e-1,即a≤-1,又因?yàn)閍≤-e
2,所以a≤-e
2.(12分)
綜合上述,實(shí)數(shù)a的取值范圍是
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.(13分)
點(diǎn)評(píng):本題考查函數(shù)的導(dǎo)數(shù)研究函數(shù)的單調(diào)性,以及函數(shù)的導(dǎo)數(shù)在求函數(shù)最值的應(yīng)用,解題的關(guān)鍵是將恒成立問(wèn)題轉(zhuǎn)化為函數(shù)最值問(wèn)題解決,屬于中檔題.