【答案】
分析:(1)先求導(dǎo),求函數(shù)在已知區(qū)間上的極值,注意極值點(diǎn)是否在定義域內(nèi),進(jìn)行分類(lèi)討論,確定最值;(2)函數(shù)在區(qū)間上單調(diào)遞減,轉(zhuǎn)化為導(dǎo)函數(shù)小于等于0恒成立,再轉(zhuǎn)化為二次函數(shù)根的分布問(wèn)題.
解答:解:(1)當(dāng)n+3m
2=0時(shí),f(x)=x
2+mx-3m
2lnx.
則
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.
令f′(x)=0,得
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(舍),x=m.(3分)
①當(dāng)m>1時(shí),
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∴當(dāng)x=m時(shí),f
min(x)=2m
2-3m
2lnm.
令2m
2-3m
2lnm=0,得
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.(5分)
②當(dāng)0<m≤1時(shí),f′(x)≥0在x∈[1,+∞)上恒成立,
f(x)在x∈[1,+∞)上為增函數(shù),當(dāng)x=1時(shí),f
min(x)=1+m.
令m+1=0,得m=-1(舍).綜上所述,所求m為
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.(7分)
(2)∵對(duì)于任意的實(shí)數(shù)a∈[1,2],b-a=1,
f(x)在區(qū)間(a,b)上總是減函數(shù),則對(duì)于x∈(1,3),
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<0,
∴f′(x)≤0在區(qū)間[1,3]上恒成立.(9分)
設(shè)g(x)=2x
2+mx+n,∵x>0,
∴g(x)≤0在區(qū)間[1,3]上恒成立.
由g(x)二次項(xiàng)系數(shù)為正,得
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即
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亦即
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(12分)
∵(-n-2)
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=
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,
∴當(dāng)n<6時(shí),m≤
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,當(dāng)n≥6時(shí),m≤-n-2,(14分)
∴當(dāng)n<6時(shí),h(n)=
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,
當(dāng)n≥6時(shí),h(n)=-n-2,即
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(16分)
點(diǎn)評(píng):(1)利用導(dǎo)數(shù)求函數(shù)的最值問(wèn)題,體現(xiàn)了分類(lèi)討論的數(shù)學(xué)思想,是難點(diǎn);(2)題意的理解與轉(zhuǎn)化是難點(diǎn),在解答此題中用到了數(shù)形結(jié)合的數(shù)學(xué)思想.