已知ω>0，函数f（x）=sin（ωx+π/4）在（π/2,π）上单调递减，则ω的取值范围是？
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w>0上，当x属于（π/2，π）时，wx+π/4属于（wπ/2+π/4，wπ+π/4）
函数f（x）=sin（ωx+π/4）在（π/2,π）上单调递减
故2kπ+π/2<=wπ/2+π/4 wπ+π/4<=2kπ+3π/2
得到4k+1/2<=w<=2k+5/4
故4k+1/2<=2k+5/4
k<=3/8
k=0
得到1/2<=w<=5/4
函数f（x）=sin（ωx+π/4）在（π/2,π）上单调递减
故2kπ+π/2<=wπ/2+π/4 wπ+π/4<=2kπ+3π/2
得到4k+1/2<=w<=2k+5/4
故4k+1/2<=2k+5/4
k<=3/8
k=0
得到1/2<=w<=5/4
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