Question

已知函数f(x)=根号3sinxcosx-sin²x+1/2 1 求f（x）的最小正周期值 2 求f（x）的单调递增区间

3 求f（x）在[0,π/2]上的最值及取最值为x的值

Answer 1

解：
(1)
f(x)=√3sinxcosx-sin²x+½
=½[√3·(2sinxcosx)+(1-2sin²x)]
=½[√3sin(2x)+cos(2x)]
=(√3/2)sin(2x)+½cos(2x)
=sin(2x)cos(π/6)+cos(2x)sin(π/6)
=sin(2x+π/6)
最小正周期T=2π/2=π
(2)
2kπ-π/2≤2x+π/6≤2kπ+π/2，(k∈Z)时，函数单调递增
此时，kπ-π/3≤x≤kπ+π/6，(k∈Z)
函数的单调递增区间为[kπ-π/3，kπ+π/6]，(k∈Z)
(3)
x∈[0，π/2]，π/6≤2x+π/6≤7π/6
sin(7π/6)≤sin(2x+π/6)≤sin(π/2)
-½≤sin(2x+π/6)≤1
2x+π/6=π/2时，即x=π/6时，函数取得最大值，f(x)max=1
2x+π/6=7π/6时，即x=π/2时，函数取得最小值，f(x)min=-½