Question

已知函数f（x）=4sinxcos（x-π/3）-√3

Answer 1

f（x）=4sinxcos（x-π/3）-√3
=2×2sinxcos（x-π/3）-√3
=2[sin(x+x-π/3)+sin(x-x+π/3)]-√3
=2sin(2x-π/3)+2sinπ/3-√3
=2sin(2x-π/3)
故函数的周期T=π，令2x-π/3=kπ，k属于Z，
故函数的零点为x=kπ/2+π/6,k属于Z
2x属于【π/24，3π/4】
则2x属于【π/12，3π/2】
则2x-π/3属于【-π/4，7π/6】
故函数最大值为y=2，最小值y=-根2.

Answer 2

f(x)=2sin^2(π/4-x)-√3cos2x
=1-cos(π/2-2x)-√3cos2x
=1-sin2x-√3cos2x
=1-2(1/2sin2x
√3/2cos2x)
=1-2(sin2xcosπ/3
sinπ/3cos2x)
=1-2sin(2x
π/3)
故最小正周期t=π
当2kπ
π/2≤2x
π/3≤2kπ
3π/2时，函数单调递减
即函数的单调递减区间为：kπ
π/12≤x≤kπ
7π/12
当0＜x＜π/2时
π/3＜2x
π/3＜4π/3
故当2x
π/3=π/2时
即x=π/12
f(x)min=f(π/12)=-1