(1)解本小题的关键是令h(x)=2x
2-3(1+a)x+6a,根据Δ
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,然后根据a的值分类讨论,求出h(x)>0的解集,从而可确定D.
(2)先求出f′(x)=6x
2-6(1+a)x+6a=6(x-1)(x-a),然后再根据(1)中a在不同取值下对应的D,确定f(x)的极值.
解:(1)x∈D?x>0且2x
2-3(1+a)x+6a>0.
令h(x)=2x
2-3(1+a)x+6a,Δ
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.
①当
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<a<1时,Δ<0,所以?x∈R,h(x)>0,所以B=R.于是D=A∩B=A=(0,+∞).
②当a=
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时,Δ=0,此时方程h(x)=0有唯一解,x
1=x
2=
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=
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=1,
所以B=(-∞,1)∪(1,+∞).于是D=A∩B=(0,1)∪(1,+∞).
③当a<
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时,Δ>0,此时方程h(x)=0有两个不同的解x
1=
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,x
2=
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.
因为x
1<x
2且x
2>0,所以B=(-∞,x
1)∪(x
2,+∞).
又因为x
1>0?a>0,所以
i)当0<a<
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时,D=A∩B=(0,x
1)∪(x
2,+∞);
ii)当a≤0时,D=(x
2,+∞).
(2)f′(x)=6x
2-6(1+a)x+6a=6(x-1)(x-a).
当a<1时,f(x)在R上的单调性如下表:
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①当
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<a<1时,D=(0,+∞).由表可得,x=a为f(x)在D内的极大值点,x=1为f(x)在D内的极小值点.
②当a=
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时,D=(0,1)∪(1,+∞).由表可得,x=
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为f(x)在D内的极大值点.
③当0<a<
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时,D=(0,x
1)∪(x
2,+∞).
因为x
1=
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=
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≥
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[3+3a-(3-5a)]=2a>a且x
1<
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<1,
x
2=
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=
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>
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=1,
所以a∈D,1∉D.
由表可得,x=a为f(x)在D内的极大值点.
④当a≤0时,D=(x
2,+∞)且x
2>1.
由表可得,f(x)在D内单调递增.因此f(x)在D内没有极值点.