Question

设函数f(x)=2cos^2x+√3sin2x+m（1）求f(x)的最小正周期和在[0,∏]上的单调递增区间

Answer 1

f(x)=2cos^2x+√3sin2x+m

=cos2x+1+√3sin2x+m

=2(1/2cos2x+√3/2sin2x)+m+1

=2sin(2x+π/6)+m+1

所以

(1)f(x)的最小正周期T=2π/2=π；

(2)当x∈[0,∏]时,2x+π/6∈[π/6,13π/6],

sin(2x+π/6)的单调递增区间为2x+π/6∈[π/6,π/2]U[3π/2,13π/6]

即x∈[0,π/6]U[2π/3,π].