Question

已知函数f(x)=2sinxcosx-2根号3cos^2x+根号3

已知函数f(x)=2sinxcosx-2根号3cos^2x+根号3（1）求f(x单调区间)

Answer 1

f(x)=2sinxcosx-2根号3cos^2x+根号3
=sin2x-√3(1+cos2x)+√3
=sin2x-√3cos2x
=2[(1/2)sin2x-(√3/2)cos2x]
=2sin(2x-π/3)
所以递增区间为2x-π/3∈[2kπ-π/2, 2kπ+π/2]
即x∈[kπ-π/12, kπ+5π/12]
递减区间为2x-π/3∈[2kπ+π/2, 2kπ+3π/2]
即x∈[kπ+5π/12, kπ+11π/6]

Answer 2

已知函数f(x)=2sinxcosx-2(√3)cos²x+√3；求f(x)的单调区间)
解：f(x)=sin2x-(√3)(1+cos2x)+√3=sin2x-(√3)cos2x=2[(1/2)sin2x-(√3/2)cos2x]
=2[sin2xcos(π/3)-cos2xsin(π/3)]=2sin(2x-π/3)
①由-π/2+2kπ≦2x-π/3≦π/2+2kπ，得单增区间为[-π/12+kπ，5π/12+kπ] (k∈Z)
②由π/2+2kπ≦2x-π/3≦3π/2+2kπ，得单减区间为[π/12+kπ，11π/12+kπ] (k∈Z)

Answer 3

f(x)=2sinxcosx-2√3cos^2x+√3
＝sin2x-√3cos2x
=2sin(2x-π/3）
因此周期是2π/2＝π
2kπ-π/2≤2x-π/3≤2kπ+π/2时单增，
即2kπ-π/6≤2x≤2kπ+5π/6
kπ-π/12≤x≤kπ+5π/12时单增
kπ+5π/12≤x≤kπ+11π/12时单减

Answer 4

-2根号3cos^2x+根号3=-根号3（2cos^2x-1)=-根号3cos2x

f(x)可以化为f(x)=sin2x-根号3cox2x=2(0.5sin2x-0.5根号3cos2x)=2sin(2x-60°）

根据正弦函数单调性 在区间[-15°，75°]上，函数是单调递增的，在区间[75°，165°]上是单调递减的

Answer 5

解：
f(x)=5sinxcosx-5√3cos^2x+(5√3)/2
=(5sin2x)/2-5√3(1+cos2x)/2+(5√3)/2
=5[sin2x*(1/2)-cos2x*(√3/2)]
=5[sin2xcos(π/3)-cos2xsin(π/3)]
=5sin(2x-π/3)
对称轴
2x-π/3=kπ+π/2
2x=kπ+5π/6
x=kπ+5π/12
所以，在y轴右侧最靠近y轴的对称轴为x=5π/12
所以
将函数y=f（x）向左平移ψ（0<ψ<π/2）个单位长度，所的图像关于y轴对称
则
ψ=5π/12