【答案】
分析:(1)由∵f(0)=0可得c=0而函數(shù)對于任意x∈R都有
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,可得函數(shù)f(x)的對稱軸從而可得a=b
結(jié)合f(x)≥x,即ax
2+(b-1)x≥0對于任意x∈R都成立,可轉(zhuǎn)化為二次函數(shù)的圖象可得a>0,且△=(b-1)
2≤0.
(2)由(1)可得g(x)=f(x)-|λx-1|=
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根據(jù)函數(shù)g(x)需討論:
①當(dāng)
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時(shí),函數(shù)g(x)=x
2+(1-λ)x+1的對稱軸為
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,
則要比較對稱軸與區(qū)間端點(diǎn)的大小,為此產(chǎn)生討論:
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,與
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分別求單調(diào)區(qū)間
②當(dāng)
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時(shí),函數(shù)g(x)=x
2+(1+λ)x-1的對稱軸為
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,
同①的討論思路
(3)結(jié)合(2)中的單調(diào)區(qū)間及零點(diǎn)存在定理進(jìn)行判斷函數(shù)g(x)的零點(diǎn)
解答:(1)解:∵f(0)=0,∴c=0.(1分)
∵對于任意x∈R都有
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,
∴函數(shù)f(x)的對稱軸為
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,即
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,得a=b.(2分)
又f(x)≥x,即ax
2+(b-1)x≥0對于任意x∈R都成立,
∴a>0,且△=(b-1)
2≤0.
∵(b-1)
2≥0,∴b=1,a=1.
∴f(x)=x
2+x.(4分)
(2)解:g(x)=f(x)-|λx-1|=
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(5分)
①當(dāng)
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時(shí),函數(shù)g(x)=x
2+(1-λ)x+1的對稱軸為
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,
若
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,即0<λ≤2,函數(shù)g(x)在
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上單調(diào)遞增;(6分)
若
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,即λ>2,函數(shù)g(x)在
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上單調(diào)遞增,在
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上單調(diào)遞減.
(7分)
②當(dāng)
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時(shí),函數(shù)g(x)=x
2+(1+λ)x-1的對稱軸為
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,
則函數(shù)g(x)在
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上單調(diào)遞增,在
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上單調(diào)遞減.(8分)
綜上所述,當(dāng)0<λ≤2時(shí),函數(shù)g(x)單調(diào)遞增區(qū)間為
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,單調(diào)遞減區(qū)間為
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;(9分)
當(dāng)λ>2時(shí),函數(shù)g(x)單調(diào)遞增區(qū)間為
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和
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,單調(diào)遞減區(qū)間為
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和
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.(10分)
(3)解:①當(dāng)0<λ≤2時(shí),由(2)知函數(shù)g(x)在區(qū)間(0,1)上單調(diào)遞增,
又g(0)=-1<0,g(1)=2-|λ-1|>0,
故函數(shù)g(x)在區(qū)間(0,1)上只有一個零點(diǎn).(11分)
②當(dāng)λ>2時(shí),則
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,而g(0)=-1<0,
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,g(1)=2-|λ-1|,
(�。┤�2<λ≤3,由于
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,
且
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=
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,
此時(shí),函數(shù)g(x)在區(qū)間(0,1)上只有一個零點(diǎn);(12分)
(ⅱ)若λ>3,由于
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且g(1)=2-|λ-1|<0,此時(shí),函數(shù)g(x)在區(qū)間(0,1)
上有兩個不同的零點(diǎn).(13分)
綜上所述,當(dāng)0<λ≤3時(shí),函數(shù)g(x)在區(qū)間(0,1)上只有一個零點(diǎn);
當(dāng)λ>3時(shí),函數(shù)g(x)在區(qū)間(0,1)上有兩個不同的零點(diǎn).(14分)
點(diǎn)評:本題主要考查了函數(shù)的解析式的求解,函數(shù)的單調(diào)區(qū)間,零點(diǎn)存在的判定定理,考查了分類討論思想的在解題中的應(yīng)用.屬于綜合性較強(qiáng)的試題.