f(x)=2cos(x+兀/3)[sin(x+兀/3)-√3cos(x+兀/3)] 若对任意x0,兀/6],使得m[f(x)+√3]+2恒成立、求实数m的取值范围?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 19:22:42
![f(x)=2cos(x+兀/3)[sin(x+兀/3)-√3cos(x+兀/3)] 若对任意x0,兀/6],使得m[f(x)+√3]+2恒成立、求实数m的取值范围?](/uploads/image/z/7020673-25-3.jpg?t=f%28x%29%3D2cos%28x%2B%E5%85%80%2F3%29%5Bsin%28x%2B%E5%85%80%2F3%29-%E2%88%9A3cos%28x%2B%E5%85%80%2F3%29%5D+%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8Fx0%2C%E5%85%80%2F6%5D%2C%E4%BD%BF%E5%BE%97m%5Bf%28x%29%2B%E2%88%9A3%5D%2B2%E6%81%92%E6%88%90%E7%AB%8B%E3%80%81%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%3F)
f(x)=2cos(x+兀/3)[sin(x+兀/3)-√3cos(x+兀/3)] 若对任意x0,兀/6],使得m[f(x)+√3]+2恒成立、求实数m的取值范围?
f(x)=2cos(x+兀/3)[sin(x+兀/3)-√3cos(x+兀/3)] 若对任意x0,兀/6],使得m[f(x)+√3]+2恒成立、求实数m的取值范围?
f(x)=2cos(x+兀/3)[sin(x+兀/3)-√3cos(x+兀/3)] 若对任意x0,兀/6],使得m[f(x)+√3]+2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<=-2/m<=2,所以-2<=m<=-1
这个