已知函数f(x)=asinxcosx-根号3acos^2+根号3/2a(a>0),(1)当a=1时,写单调递减区间

题目

已知函数f(x)=asinxcosx-根号3acos^2+根号3/2a(a>0),(1)当a=1时,写单调递减区间
(2)当x∈[0,π/2]时,f(x)的最小值是-根号3,求在区间[-兀,兀]上,使函数f(x)取得最值时自变量x的和

答案

1、f(x)=sinxcosx-√3(cosx)^2+√3/2
=(1/2)sin2x-(√3/2)cos2x
=sin(2x-π/3)
递减区间为2kπ+π/2≤2x-π/3≤2kπ+3π/2
kπ+5π/12≤x≤kπ+11π/12
2、f(x)=asin(2x-π/3)
最小值是-根号3,说明a=√3 sin(2x-π/3)=-1
x∈[-兀,兀] -5π/3≤2x-π/3≤7π/3,故可取三种：
2x-π/3=-3π/2时 x=-7π/12
2x-π/3=-π/2时 x=-π/12
2x-π/3=+3π/2时 x=11π/12
它们的和=-7π/12-π/12+11π/12=π/4