Question

已知函数f（x）sinxcosx-√3sin²x+√3/2（1）求函数f（x）的最小正周期

Answer 1

答：
f(x)=sinxcosx-√3sinx*sinx+√3/2
=(1/2)sin2x+(√3/2)cos2x
=sin(2x+π/3)
（1）f(x)的周期为kπ，所以
最小正周期
为π
（2）-π/2<=x<=π/2
-2π/3<=2x+π/3<=4π/3
所以f(x)在[-π/2,π/2]上的单调减区间为：-2π/3<=2x+π/3<=-π/2及
π/2<=2x+π/3<=4π/3
即单调减区间为[-π/2,-5π/12]及[π/12,π/2]
（3）0<=x<=π/4，π/3<=2x+π/3<=5π/6
f(x)min=sin(5π/6)=1/2，此时x=π/4；
f(x)max=sin(π/2)=1，此时x=π/12.
.

Answer 2

f(x)=sinxcosx-√3sin²x+√3/2

=sinxcosx+(√3/2)(1-2sin²x)

=(sin2x)/2+(√3/2)cos2x

=sin(2x+π/3)

f（x）最小正周期为2π/2=π

x∈[π
/12,π
/4]时，f(x)单调递减，f(x)∈[√3/2,1]

,f(x)+m²＞1恒成立

那么，当f(x)取最小值√3/2时同样成立

m²>1-√3/2