解(1)∵f(x)=2sin
2x+2
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sinxcosx-1=
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sin2x-cos2x=2(sin2xcos
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-cos2xsin
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),
∴f(x)=2sin(2x-
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).
∴函數(shù)f(x)的圖象可由y=sinx的圖象按如下方式變換得到:
①將函數(shù)的y=sinx圖象向右平移
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個(gè)單位,得到函數(shù)y=sin(x-
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)的圖象;
②將函數(shù)y=sin(x-
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)的圖象上所有點(diǎn)的橫坐標(biāo)縮短到原來(lái)的
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倍(縱坐標(biāo)不變),
得到函數(shù)y=sin(2x-
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)的圖象;
③將函數(shù)y=sin(2x-
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)的圖象上所有點(diǎn)的縱坐標(biāo)伸長(zhǎng)到原來(lái)的2倍(橫坐標(biāo)不變),
得到函數(shù)f(x)=2sin(2x-
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)的圖象.
(2)由(1)知,f(x)=2sin(2x-
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),x∈R
∴g(x)=
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|f(x+
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)|+
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|f(x+
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)|=
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|2sin2x|+
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|2sin(2x+π)|=2|sin2x|.
又對(duì)任意x∈R,有g(shù)(-x)=2|sin(-2x)|=2|sin2x|=g(x),
∴函數(shù)g(x)是偶函數(shù).
∵函數(shù)y=2sin2x的最小正周期是π,
∴結(jié)合函數(shù)圖象可知,函數(shù)g(x)=2|sin2x|的最小正周期是T=
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.
(3)先求函數(shù)g(x)在一個(gè)周期[0,
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]內(nèi)的單調(diào)區(qū)間和函數(shù)值的取值范圍.
當(dāng)x∈[0,
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]時(shí),2x∈[0,π],此時(shí)g(x)=2sin2x.
易知,此時(shí)g(x)的單調(diào)增區(qū)間是[0,
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],單調(diào)減區(qū)間是[
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,
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];
函數(shù)的取值范圍是g(x)∈[0,2].
因此,由周期函數(shù)的性質(zhì),可知函數(shù)g(x)=2|sin2x|的單調(diào)增區(qū)間是[
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kπ,
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+
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kπ];
單調(diào)減區(qū)間是[
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+
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kπ,
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kπ],其中k∈Z.函數(shù)的g(x)值域是[0,2].
分析:(1)根據(jù)三角函數(shù)的二倍角公式進(jìn)行降次,再用輔助角公式合并,得f(x)=2sin(2x-
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).再用函數(shù)y=Asin(ωx+φ)的圖象變換的公式,可得到函數(shù)由曲線y=sinx的圖象經(jīng)變換的過(guò)程.
(2)根據(jù)(1)得到的表示式代入化簡(jiǎn),得g(x)=2|sin2x|.因此不難由正弦函數(shù)的奇偶性,證出g(x)是偶函數(shù),再結(jié)合正弦曲線的形狀,可得g(x)的最小正周期.
(3)注意到函數(shù)g(x)的最小正周期是
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,只需研究g(x)在區(qū)間[0,
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]上的單調(diào)性和最值,再結(jié)合函數(shù)的周期性,即可得到函數(shù)g(x)的單調(diào)區(qū)間和值域.
點(diǎn)評(píng):本題以一個(gè)特殊三角函數(shù)式為例,叫我們求函數(shù)的單調(diào)區(qū)間與值域,著重考查了三角恒等變換、三角函數(shù)的單調(diào)性、奇偶性和周期性等知識(shí)點(diǎn),屬于基礎(chǔ)題.