解:(1)∵g(x)=f(x)-acosx+2(b>0)=b•cos2x+3,將函數(shù)y=g(x)的圖象左移
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個單位得函數(shù)y=bcos2(x+
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)+3=
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的圖象,
故h(x)=
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(b>0).…1′
故函數(shù)y=h(x)的周期為π,由2kπ-π
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2kπ,k∈z,可得
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≤x≤
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,故單調(diào)增區(qū)間為
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,(k∈Z).…6′
(2)因?yàn)閎≤0,對任意x恒有f(x)≥0成立,則2bcos
2x+acosx+1-b≥0
令t=cosx∈[-1,1],g(t)=2bt
2+at+1-b(t∈[-1,1]),則有g(shù)(t)≥0.…7′
當(dāng)b=0時,g(t)=at+1有g(shù)(1)≥0且g(-1)≥0,即-1≤a≤1,(a+b)
max=1;…9′
當(dāng)b<0時,g(t)=2bt
2+at+1-b(t∈[-1,1])有:
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,
即
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,即-1≤a+b≤2b+1<1,…11′
綜上可得:(a+b)
max=1.…12′
分析:(1)先求出 g(x)的解析式,根據(jù)函數(shù)y=Asin(ωx+∅)的圖象變換規(guī)律求得函數(shù)h(x)的解析式,從而求出函數(shù)y=h(x)的周期與單調(diào)增區(qū)間.
(2)令t=cosx∈[-1,1],g(t)=2bt
2+at+1-b(t∈[-1,1]),則g(t)≥0,再分當(dāng)b=0、和當(dāng)b<0兩種情況,分別求出a+b的最大值,從而得出結(jié)論.
點(diǎn)評:本題主要考查余弦函數(shù)的增區(qū)間,函數(shù)y=Asin(ωx+∅)的圖象變換規(guī)律,求三角函數(shù)的最值,屬于中檔題.