作业帮已知函数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.
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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.](/uploads/image/z/1535082-42-2.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%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E5%B1%9E%E4%BA%8E%5B0%2C%CF%80%2F6%5D%2C%E4%BD%BF%E5%BE%97m%5Bf%28x%29%2B%E2%88%9A3%5D%2B2%3D0%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
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.
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.
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.
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
∵f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
∴f(x)=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)＝sin[2(x+π/3)]-√3cos[2(x+π/3)]－√3
＝2sin[2(x+π/3)－π/3]－√3＝2sin﹙2x－π/3﹚－√3
∵m[f(x)+√3]+2=0 ...
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∵f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
∴f(x)=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)＝sin[2(x+π/3)]-√3cos[2(x+π/3)]－√3
＝2sin[2(x+π/3)－π/3]－√3＝2sin﹙2x－π/3﹚－√3
∵m[f(x)+√3]+2=0 ∴2msin﹙2x－π/3﹚＝﹣2
∴sin﹙2x－π/3﹚＝﹣1/m
∵x∈[0,π/6] ∴2x－π/3∈[﹣π/3，0] ∴sin﹙2x－π/3﹚∈[﹣√3/2，0]
∵对任意x属于[0,π/6],使得m[f(x)+√3]+2=0恒成立
∴﹣1/m∈[﹣√3/2，0]
∴1/m∈[0，√3/2]
∴m∈[2√3/3，﹢∞﹚
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