【解】
1.(√3/2)sinx+1/2cosx
=cosπ/6·sinx+sinπ/6·cosx
=sin(π/6+x)
∴最小正周期Tmin=|2π/ω|=|2π/1|=2π
(ω为x前系数)
2.
∵sin(π/6+x)在[-π/2+2kπ,π/2+2kπ]上单调递增(k为整数)
∴-π/2+2kπ≤π/6+x≤π/2+2kπ
=>-2π/3+2kπ≤x≤π/3+2kπ(k∈Z)
即x的单调递增区间为:
[-π2/3+2kπ,π/3+2kπ]
3.函数最大值为1,此时x=π/2+2kπ
即x的取值集合为{x|π/2+2kπ|k∈Z}