题目

已知函数f(x)=(2sin(x+派/3)+sinx)cosx-根号3sin^2x,若存在
xo属于［0,5派／12］使得mf(x0)=4成立,求m的取值范围

提问时间：2020-08-03

答案

f(x)=[2sin(x+π/3)+sinx]cosx -√3sin²x
=[2sinxcos(π/3)+2cosxsin(π/3)+sinx]cosx -√3sin²x
=(2sinx+√3cosx)cosx -√3sin²x
=2sinxcosx+√3cos²x-√3sin²x
=sin(2x)+√3cos(2x)
=2[(1/2)sin(2x)+(√3/2)cos(2x)]
=2sin(2x+π/3)
x0∈[0,5π/12]
π/3≤2x+π/3≤7π/6
-1/2≤sin(2x+π/3)≤1
-1≤2sin(2x+π/3)≤2
mf(x0)=4≠0 m≠0
f(x0)=4/m
-1≤4/m≤2
4/m≥-1 4/m +1≥0 (m+4)/m≥0
m>0或m≤-4
4/m≤2 1- 2/m≥0 (m-2)/m≥0 m≥2或m