已知f(x)=2x可以表示成一個(gè)奇函數(shù)g(x)與一個(gè)偶函數(shù)h(x)之和,若關(guān)于x的不等式ag(x)+h(2x)≥0對于x∈[1,2]恒成立,則實(shí)數(shù)a的最小值是________.
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分析:由題意可得g(x)+h(x)=2
x①,g(-x)+h(-x)=-g(x)+h(x)=2
-x②從而可得h(x)=
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,g(x)=
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而ag(x)+h(2x)≥0對于x∈[1,2]恒成立即a
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對于x∈[1,2]恒成立即
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對于x∈[1,2]恒成立,只要求出函數(shù)
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的最大值即可
解答:f(x)=2
x可以表示成一個(gè)奇函數(shù)g(x)與一個(gè)偶函數(shù)h(x)之和
∴g(x)+h(x)=2
x①,g(-x)+h(-x)=-g(x)+h(x)=2
-x②
①②聯(lián)立可得,h(x)=
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,g(x)=
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ag(x)+h(2x)≥0對于x∈[1,2]恒成立
a
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對于x∈[1,2]恒成立
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對于x∈[1,2]恒成立
t=2
x-2
-x,x∈[1,2],t∈
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則t
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在t∈
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單調(diào)遞增,
t=
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時(shí),則t
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=
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a
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故答案為:
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點(diǎn)評:本題主要考查了奇偶函數(shù)的定義的應(yīng)用,函數(shù)的恒成立的問題,常會轉(zhuǎn)化為求函數(shù)的最值問題,體現(xiàn)了轉(zhuǎn)化思想的應(yīng)用.