已知函數(shù)f(x)=2x+1定義在R上.
(1)若f(x)可以表示為一個偶函數(shù)g(x)與一個奇函數(shù)h(x)之和,設h(x)=t,p(t)=g(2x)+2mh(x)+m2-m-1(m∈R),求出p(t)的解析式;
(2)若p(t)≥m2-m-1對于x∈[1,2]恒成立,求m的取值范圍;
(3)若方程p(p(t))=0無實根,求m的取值范圍.
【答案】
分析:(1)利用f(x)=g(x)+h(x)和f(-x)=g(-x)+h(-x)求出g(x)和h(x)的表達式,再求出p(t)關于t的表達式即可.
(2)先有x∈[1,2]找出t的范圍,在把所求問題轉(zhuǎn)化為求p(t)在[
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,
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]的最小值.讓大于等于m
2-m-1即可.
(3)轉(zhuǎn)化為關于p(t)的一元二次方程,利用判別式的取值,再分別討論即可.
解答:解:(1)假設f(x)=g(x)+h(x)①,其中g(x)偶函數(shù),h(x)為奇函數(shù),
則有f(-x)=g(-x)+h(-x),即f(-x)=g(x)-h(x)②,
由①②解得
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,
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.
∵f(x)定義在R上,∴g(x),h(x)都定義在R上.
∵
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,
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.
∴g(x)是偶函數(shù),h(x)是奇函數(shù),∵f(x)=2
x+1,
∴
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,
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.
由
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,則t∈R,
平方得
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,∴
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,
∴p(t)=t
2+2mt+m
2-m+1.
(2)∵t=h(x)關于x∈[1,2]單調(diào)遞增,∴
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.
∴p(t)=t
2+2mt+m
2-m+1≥m
2-m-1對于
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恒成立,
∴
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對于
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恒成立,
令
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,則
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,
∵
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,∴
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,故
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在
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上單調(diào)遞減,
∴
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,∴
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為m的取值范圍.
(3)由(1)得p(p(t))=[p(t)]
2+2mp(t)+m
2-m+1,
若p(p(t))=0無實根,即[p(t)]
2+2mp(t)+m
2-m+1①無實根,
方程①的判別式△=4m
2-4(m
2-m+1)=4(m-1).
1°當方程①的判別式△<0,即m<1時,方程①無實根.
2°當方程①的判別式△≥0,即m≥1時,
方程①有兩個實根
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,
即
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②,
只要方程②無實根,故其判別式
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,
即得
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③,且
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④,
∵m≥1,③恒成立,由④解得m<2,∴③④同時成立得1≤m<2.
綜上,m的取值范圍為m<2.
點評:本題是在考查指數(shù)函數(shù)的基礎上對函數(shù)的恒成立問題,函數(shù)奇偶性以及一元二次方程根的判斷的綜合考查,是一道綜合性很強的難題.