【答案】
分析:①检验f(x+
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)=|sin(2x+π+
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)-
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|=|sin(2x+
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π)
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|≠f(x)可判断①
②y=sin(x-
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)=cosx在区在区间[π,
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π]上单调递增,可判断②
③根据正弦函数与余弦函数在对称轴处取得函数的最值,把x=
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代入到函数y=sin(2x+
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)=cos2x进行检验,可判断③
④由x∈(0,π)可得0<sinx≤1,结合函数y=sinx+
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的单调性可判断④
⑤、先设函数y=tan
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-cscx上任意一点M(x,y)关于点(π,0)对称的点N(x′,y′),
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,代入到y=tan
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-cscx中可求对称函数
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-csc(2π-x′),可判断⑤
解答:解:∵f(x+
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)=|sin(2x+π+
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)-
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|=|sin(2x+
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π)
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|≠f(x),而f(x+π)=|sin(2x+2π
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)-
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|=|sin(2x+
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)
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|=f(x),则函数的最小正周期是π,故①错误
②y=sin(x-
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)=cosx在区在区间[π,
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π]上单调递增,故②错误
③x=
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时,函数y=sin(2x+
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)=cos2x的值为0,不是最值点,不符合对称轴的性质,故③错误
④∵x∈(0,π)
∴0<sinx≤1
y=sinx+
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在sinx=1时取得最小值5
∴y的最小值不是4,故④错误
⑤设函数y=tan
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-cscx上任意一点M(x,y)关于点(π,0)对称的点N(x′,y′)
则
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,即
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代入到y=tan
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-cscx中可得
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-csc(2π-x′)
∴
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,即函数y=tan
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-cscx的图象关于点(π,0)对称,故⑤正确
故答案为:⑤
点评:本题主要考查了三角函数的周期的判断,三角函数的诱导公式及函数的单调性的应用,函数的单调性在函数的最值求解中的应用及三角函数的对称性的应用,属于函数知识的综合应用