已知函数f(x)=2sin(2x-π/3)+1,,x属于[π/4,π/2],
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答:
f(x)=2sin(2x-π/3)+1
π/4<=x<=π/2
所以:π/2<=2x<=π
所以:π/6<=2x-π/3<=2π/3
所以:1/2<=sin(2x-π/3)<=1
所以:2*(1/2)+1<=f(x)<=2*1+1
所以:2<=f(x)<=3
|f(x)-m|<2成立
则有:-2<f(x)-m<2
所以:m-2<f(x)<m+2包含2<=f(x)<=3
所以:
m-2<=2
m+2>=3
所以:
1<=m<=4
f(x)=2sin(2x-π/3)+1
π/4<=x<=π/2
所以:π/2<=2x<=π
所以:π/6<=2x-π/3<=2π/3
所以:1/2<=sin(2x-π/3)<=1
所以:2*(1/2)+1<=f(x)<=2*1+1
所以:2<=f(x)<=3
|f(x)-m|<2成立
则有:-2<f(x)-m<2
所以:m-2<f(x)<m+2包含2<=f(x)<=3
所以:
m-2<=2
m+2>=3
所以:
1<=m<=4
更多追问追答
追问
不能等于吧??
追答
嗯,这个忽略了:
答:
f(x)=2sin(2x-π/3)+1
π/43
所以:
1<m<4
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