解:(1)由圖象可知A=2,T=π;
所以ω=
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所以f(x)=2sin(2x+
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);它的單調(diào)增區(qū)間為:[k
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,k
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]k∈Z
(2)f(x)=2sin(2x+
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)在區(qū)間
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上是單調(diào)減函數(shù),
在區(qū)間
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是單調(diào)增函數(shù),
x∈
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時,f(x)∈[-2,-1]
x∈
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時f(x)∈[-2,1]
當(dāng)-2<a≤-1時函數(shù)y=f(x)與y=a(a為常數(shù))的圖象的交點(diǎn)的個數(shù)為:2;
當(dāng)-2=a或-1<a≤1時函數(shù)y=f(x)與y=a(a為常數(shù))的圖象的交點(diǎn)的個數(shù)為:1;
當(dāng)1<a或a<-2時函數(shù)y=f(x)與y=a(a為常數(shù))的圖象的交點(diǎn)的個數(shù)為:0;
分析:(1)通過函數(shù)的圖象,求出A,T,轉(zhuǎn)化為ω,得到函數(shù)的解析式,直接求出單調(diào)增區(qū)間即可.
(2)當(dāng)
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時,求出函數(shù)的最值,以及函數(shù)的值域,利用單調(diào)性,說明函數(shù)y=f(x)與y=a(a為常數(shù))的圖象的交點(diǎn)的個數(shù).
點(diǎn)評:本題是基礎(chǔ)題,考查三角函數(shù)的基本知識,考查視圖能力,利用基本函數(shù)的基本性質(zhì),考查分析問題解決問題的能力.