Question

已知函数f(x)=根号3sinxcosx+cos²x 求（1）写出函数f(x)的最小正周期和单调递增区间

Answer 1

(1)f(x)=√3sinxcosx+cos²x

=√3/2sin2x+(cos2x+1)/2

=√3/2sin2x+1/2cos2x+1/2

=sin(2x+π/6)+1/2

f(x)的最小正周期为2π/2=π

2kπ-π/2≤2x+π/6≤2kπ+π/2 (k∈Z)

kπ-π/3≤x≤kπ+π/6

则f(x)的单调递增区间为[kπ-π/3,kπ+π/6] (k∈Z)

(2)2x+π/6=kπ+π/2 (k∈Z)

对称轴为x=kπ/2+π/6

因为0