設(shè)函數(shù)f(x)=x|x-1|+m,g(x)=lnx.
(1)當(dāng)m>1時,求函數(shù)y=f(x)在[0,m]上的最大值;
(2)記函數(shù)p(x)=f(x)-g(x),若函數(shù)p(x)有零點,求m的取值范圍.
【答案】
分析:(1)化簡函數(shù)f(x)的解析式,分別在[0,1]和(1,m]上求函數(shù)的最大值.
(2)函數(shù)有零點即對應(yīng)方程有解,得到m的解析式m=h(x),通過導(dǎo)數(shù)符號確定h(x)=lnx-x|x-1|的單調(diào)性,由h(x)的單調(diào)性確定h(x)的取值范圍,即得m的取值范圍.
解答:解:(1)當(dāng)x∈[0,1]時,f(x)=x(1-x)+m=
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∴當(dāng)
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時,
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當(dāng)x∈(1,m]時,f(x)=x(x-1)+m=
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∵函數(shù)y=f(x)在(1,m]上單調(diào)遞增,∴f(x)
max=f(m)=m
2由
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得:
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又m>1
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.
∴當(dāng)
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時,f(x)
max=m
2;
當(dāng)
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時,
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.
(2)函數(shù)p(x)有零點即方程f(x)-g(x)=x|x-1|-lnx+m=0有解,
即m=lnx-x|x-1|有解
令h(x)=lnx-x|x-1|,當(dāng)x∈(0,1]時,h(x)=x
2-x+lnx
∵
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∴函數(shù)h(x)在(0,1]上是增函數(shù),∴h(x)≤h(1)=0
當(dāng)x∈(1,+∞)時,h(x)=-x
2+x+lnx.
∵
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=
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<0
∴函數(shù)h(x)在(1,+∞)上是減函數(shù),∴h(x)<h(1)=0
∴方程m=lnx-x|x-1|有解時,m≤0,
即函數(shù)p(x)有零點時m≤0
點評:本題考查用分類討論的方法求函數(shù)最大值,利用導(dǎo)數(shù)求函數(shù)值域,及化歸與轉(zhuǎn)化的思想方法.