搜狗做题网已知函数f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
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已知函数f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
对任意x属于[0,π/6],使得m[f(x)+√3]+2=0恒成立,求实数m的取值范围.
对任意x属于[0,π/6],使得m[f(x)+√3]+2=0恒成立,求实数m的取值范围.
![已知函数f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]](/uploads/image/z/14911016-32-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2cos%28x%2B%CF%80%2F3%29%5Bsin%28x%2B%CF%80%2F3%29-%E2%88%9A3cos%28x%2B%CF%80%2F3%29%5D)
f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
=4cos(x+π/3)[1/2sin(x+π/3)-√3/2cos(x+π/3)]
=4cos(x+π/3)[sin(x+π/3-π/3)]
=4[cosxcos(π/3)-sinxsin(π/3)]*sinx
=2(cosx-√3sinx)sinx
=2cosxsinx-2√3sin²x
=sin2x+√3cos2x-√3
=2sin(2x+π/3)-√3,在[0,π/6]上,f(x)∈[1-√3,2-√3]
m[f(x)+√3]+2=0恒成立,即[f(x)+√3]=-2/m恒成立,所以1
=4cos(x+π/3)[1/2sin(x+π/3)-√3/2cos(x+π/3)]
=4cos(x+π/3)[sin(x+π/3-π/3)]
=4[cosxcos(π/3)-sinxsin(π/3)]*sinx
=2(cosx-√3sinx)sinx
=2cosxsinx-2√3sin²x
=sin2x+√3cos2x-√3
=2sin(2x+π/3)-√3,在[0,π/6]上,f(x)∈[1-√3,2-√3]
m[f(x)+√3]+2=0恒成立,即[f(x)+√3]=-2/m恒成立,所以1
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