函数f(x)=4sinxsin^2(45°+2分之x)+cos2x,x∈[π/6,2π/3],都有|f(x)-m|<2成立,求m的取值范围

显示全部楼层 · 2012-2-2 09:56:16

∵f(x)=4sinxsin2(π/4+x/2)+cos(2x) =2sinx[2sin2(π/4+x/2)]+cos(2x) =2sinx[1-cos(2(π/4+x/2))]+cos(2x) =2sinx[1-cos(π/2+x)]+cos(2x) =2sinx(1+sinx)+1-2sin2x =2sinx+1 ∵x∈[π/6,2π/3],
f(π/6)=2sin(π/6)+1=2f(π/2)=2sin(π/2)+1=3 f(2π/3)=2sin(2π/3)+1=1+√3∴f(x)最大值是f(π/2)=3，最小值是f(π/6)=2。|f(x)-m|<2即...

千问 · 2012-2-2 09:56:16

f(x)=2 sinx[2sin^2(π/4+x/2)]+cos2x=2sinx[1-cos(π/2+x) ]+cos2x
=2sinx(1+sinx)+1-2sin^2(x)=2sinx+1∵x∈[π/6,2π/3]∴f(x)∈[2,3]∵|f(x)-m|<2∴1<m<4...