解:(1)由
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,所以當(dāng)a=b=1時(shí),
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則
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=-3(x
2-1).
在(0,1)內(nèi),
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,在(1,2)內(nèi),
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,
所以在(0,1)內(nèi),
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為增函數(shù),在(1,2)內(nèi)
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為減函數(shù).
則f
3(x)的極大值為f
3(1)=3,由f
3(0)=1,
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.
所以函數(shù)
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在[0,2]上的最大值為f
3(1)=3,最小值為f
3(2)=-1;
(2)因?yàn)閷?duì)任意x
1,x
2∈[-1,1],都有|f
3(x
1)-f
3(x
2)|≤1,
所以|f
3(1)-f
3(-1)|≤1,從而有|(-1+3a+b)-(1-3a+b)|=|6a-2|≤1,
所以
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.
又
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=-3(x
2-a),
在
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內(nèi)f
′3(x)0,
所以f
3(x)在
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內(nèi)為減函數(shù),
f
3(x)在
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內(nèi)為增函數(shù),
只需
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,則
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即
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,解得:
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.
所以a的取值范圍是
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.
(3)
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.
由f
4(x)在[-1,1]上的最大值為
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,則
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,
所以
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,即
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①
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,即
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②
①+②得,
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,又因?yàn)?img class='latex' src='http://thumb.zyjl.cn/pic5/latex/295931.png' />,所以
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,所以
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.
將
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代入①得:
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,
將
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代入②得:
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≤a≤0.
所以a=0.
綜上知a,b的值分別為0,
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.
分析:(1)把a(bǔ),b的值代入函數(shù)解析式求出
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,求導(dǎo)后利用導(dǎo)函數(shù)的零點(diǎn)將(0,2)分段,由單調(diào)性判出極值點(diǎn),求出極值,再求出端點(diǎn)值,則f
3(x)在[0,2]上的最大值和最小值可求;
(2)根據(jù)對(duì)任意x
1,x
2∈[-1,1],都有|f
3(x
1)-f
3(x
2)|≤1,說明當(dāng)x取兩個(gè)特殊值-1和1時(shí)|f
3(1)-f
3(-1)|≤1成立,由此求出a的初步范圍,然后把原函數(shù)f
3(x)求導(dǎo),得到導(dǎo)函數(shù)的兩個(gè)零點(diǎn)為
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,再求出函數(shù)f
3(x)在(-1,1)上的極大值和極小值,再由極大值和極小值差的絕對(duì)值小于等于1求出a的取值范圍,和由|f
3(1)-f
3(-1)|≤1求出的a的范圍取交集即可;
(3)由|f
4(x)|在[-1,1]上的最大值為
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,則x取-1和1時(shí)的函數(shù)值都在
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和
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之間,聯(lián)立解出b的范圍,再由x取0時(shí)的函數(shù)值也在
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和
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之間,得到b的范圍,兩者結(jié)合即可求出b的值,把b的值代入x取-1和1時(shí)的式子,即可得到a的值.
點(diǎn)評(píng):本題考查了利用導(dǎo)數(shù)研究函數(shù)的最值,考查了數(shù)學(xué)轉(zhuǎn)化思想方法,解答此題的關(guān)鍵是特值化思想的應(yīng)用,求具體參數(shù)的值時(shí)運(yùn)用了“兩邊夾”的思想方法,屬有一定難度題.