Question

已知函数f(x)=√3(sin²x-cos²x)-2sinxcosx

求:(1)求f(x)的最小正周期 (2)设x∈[-π/3,π/3] , 求f(x)的知遇和单调递增区间

Answer 1

解f(x)=√3(sin^2x-cos^2x)-2sinxcosx
=-√3cos2x-sin2x
=-2（√3/2cos2x+1/2sin2x）
=-2sin（2x+π/3)
知1函数的T=2π/2=π，
2由x∈[-π/3,π/3]

知2x∈[-2π/3,2π/3]

即2x+π/3∈[-π/3,π]
即sin（2x+π/3）∈[-√3/2,1]
即-2sin（2x+π/3）∈[-2，√3]
即y∈[-2，√3]
故函数的值域[-2，√3]
由2x+π/3∈[-π/3,π]
知当2x+π/3∈[-π/3,π/2]
即当2x∈[-2π/3,π/6]
即x∈[-π/3,π/12] 时，y=√3(sin²x-cos²x)-2sinxcosx是增函数
故函数的增区间为[-π/3,π/12] 。

Answer 2

f(x)=√3(sin^2x-cos^2x)-2sinxcosx
=-√3cos2x-sin2x
=-2sin(2x+π/3)
1.求最小正周期T=π
2.设x∈[-π/3,π/3]，求函数的值域和单调区间
-π/2<2x+π/3<π/2递增
-5π/6<2x<π/6
-5π/12<x<π/12
π/2<2x+π/3<3π/2递减
π/12<x<7π/12
所以在-π/3<x<π/12递增π/12<x<π/3递减
函数的值域[-2,√3]
求采纳为满意回答。

Answer 3

f(x)=√3(sin²x-cos²x)-2sinxcosx
=-√3cos2x-sin2x
=-2(√3/2cos2x-1/2sin2x)
=-2cos(2x+π/6)

=2cos[π+(2x-5π/6)]
=2cos(2x-5π/6)]
T=2π/2=π

x∈[-π/3,π/3]

2x-5π/6∈[-3π/2,-π/6]
值域为[ -√3，0]

Answer 4

f(x)=√3(sin²x-cos²x)-2sinxcosx=-√3cos2x-sin2x=-2sin(2x+π/3)

f(x)最小正周期 π

f(x)在[-2,2]

单增 2kπ+π/2≤2x+π/3≤ 2kπ+3π/2
kπ+π/12≤x≤ kπ+7π/12