【答案】
分析:(I)先求函數(shù)f(x)在區(qū)間(1,2)上遞減的充要條件,
f(x)在區(qū)間(1,2)上遞減?f'(x)=3x
2-4ax+a
2≤0在區(qū)間(1,2)上恒成立,處理二次不等式恒成立問(wèn)題可用實(shí)根分布求解.
(II)x∈[0,|a|+1]時(shí),f(x)<2a
2恒成立?f(x)
max<2a
2,x∈[0,|a|+1],問(wèn)題轉(zhuǎn)化為求函數(shù)的最值問(wèn)題.
解答:解:(I)∵f(x)在區(qū)間(1,2)上遞減,
∴其導(dǎo)函數(shù)f'(x)=3x
2-4ax+a
2≤0在區(qū)間(1,2)上恒成立.
∴
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故a≤3是函數(shù)f(x)在區(qū)間(1,2)上遞減的必要而不充分的條件
解法二:f'(x)=3x
2-4ax+a
2=(3x-a)(x-a)≤0在區(qū)間(1,2)上恒成立,
∴a只能大于0,∴
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,∴
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∴2≤a≤3⇒a≤3
故a≤3是函數(shù)f(x)在區(qū)間(1,2)上遞減的必要而不充分的條件
(II)∵f(x)=x(x-a)
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當(dāng)a>0時(shí),函數(shù)y=f(x)在(
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)上遞增,
在
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上遞減,在
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上遞增,
故有
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當(dāng)a<0時(shí),函數(shù)y=f(x)在
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上遞增,
∴只要f(1-a)<2a
2⇒4a
3-6a
2+5a-1>0
令g(a)=4a
3-6a
2+5a-1,
則
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所以g(a)在(-∞,0)上遞增,
又g(0)=-1<0∴f(1-a)<2a
2不能恒成立
故所求的a的取值范圍為
點(diǎn)評(píng):本題考查已知函數(shù)的單調(diào)區(qū)間求參數(shù)范圍問(wèn)題和不等式恒成立問(wèn)題,體現(xiàn)分類(lèi)討論和化歸思想.