【答案】
分析:先研究函數(shù)的性質(zhì),觀察知函數(shù)是個偶函數(shù),由于f′(x)=2x+sinx,在[0,
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]上f′(x)>0,可推斷出函數(shù)在y軸兩邊是左減右增,此類函數(shù)的特點是自變量離原點的位置越近,則函數(shù)值越小,欲使f(x
1)>f(x
2)恒成立,只需x
1,到原點的距離比x
2,到原點的距離大即可,由此可得出|x
1|>|x
2|,在所給三個條件中找符合條件的即可.
解答:解:函數(shù)f(x)為偶函數(shù),f′(x)=2x+sinx,
當(dāng)0<x≤
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時,0<sinx≤1,0<2x≤π,
∴f′(x)>0,函數(shù)f(x)在[0,
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]上為單調(diào)增函數(shù),
由偶函數(shù)性質(zhì)知函數(shù)在[-
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,0]上為減函數(shù).
當(dāng)x
12>x
22時,得|x
1|>|x
2|≥0,
∴f(|x
1|)>f(|x
2|),由函數(shù)f(x)在上[-
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,
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]為偶函數(shù)得f(x
1)>f(x
2),故②成立.
∵
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>-
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,而f(
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)=f(
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),
∴①不成立,同理可知③不成立.故答案是②.
故應(yīng)填②
點評:本題考查函數(shù)的性質(zhì)奇偶性與單調(diào)性,屬于利用性質(zhì)推導(dǎo)出自變量的大小的問題,本題的解題方法新穎,判斷靈活,方法巧妙.