本试题主要是考查了函数单调性和函数最值的求解的综合运用。
(1)根据已知条件,对于参数a进行分类讨论,判定单调性得到结论。
(2)在第一问的基础上,进一步对于不同情况下的单调性分别研究得到最值。
选做题:(参加IB学习的学生必须做,不参加IB学习的学生原则上不要做)
题目:(本题满分值为10分)
解: (1) ∵f(x)=-
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ax
3+x
2+2
(a≠0),∴
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= -ax
2+2x.
①当a>0时,令
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>0,即-ax
2+2x>0,得0<x<
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.
∴f(x)在(-∞,0),
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上是减函数,在
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上是增函数. ………………4分
②当a<0时,令
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>0,即-ax
2+2x>0,得x>0,或x<
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.
∴f(x)在(-∞,
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),(0, +∞)上是增函数,在(
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,0)上是减函数.………………8分
(2)由(1)得:
①当0<
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<1,即a>2时,f(x)在(1,2)上是减函数,
∴f(x)
max=f(1)=3-
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. ……………10分
②当1≤
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≤2,即1≤a≤2时,f(x)在
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上是增函数,在
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上是减函数,
∴f(x)
max=f
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=
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. ………12分
③当
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>2时,即0<
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<1时,f(x)在(1,2)上是增函数,
∴f(x)
max=f(2)=
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. ……………14分
综上所述,当0<
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<1时,f(x)的最大值为3-
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,
当1≤
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≤2时,f(x)的最大值为
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,
当
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>2时,f(x)的最大值为
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. ………………15分