Question

已知函数f（x)=2sin^2x+2根号3sinxcosx+1，求f(x)的最小正周期，单调区间，最值

最值是在【0，π/2】上的最值

Answer 1

f(x)=2(1-cos2x)/2+√3(sin2x)+1
=√3sin2x-cos2x+2
=2sin2xcosπ/6-2cos2xsinπ/6+2
=2(sin2xcosπ/6-cos2xsinπ/6)+2
=2sin(2x-π/6)+2
所以T=2π/2=π

增函数时2kπ-π/2<2x-π/6<2kπ+π/2
所以增区间(kπ-π/6,kπ+π/3)
同理，减区间(kπ+π/3,kπ+5π/6)

sin(2x-π/6)最值是±1
所以最大=4，最小=0