Question

已知函数f(x)=2sin的平方(x+π\4)-根号3cos2x-1

1.当x属于R时，求函数f(x)的最小正周期。
2.当x属于[π\12,13π\24]时，求f(x)的最大值和最小值。

Answer 1

函数f(x)=2sin^2 (x+π/4)- √3cos2x-1
=-cos(2x+π/2)- √3cos2x
=sin2x- √3cos2x
=2sin(2x-π/3)
1.当x属于R时，函数f(x)的最小正周期T=2π/2=π。
2.当x属于[π/12,13π/24]时，求f(x)的最大值和最小值。
13π/24-π/12=11π/24<12π/24=π/2=T/2
f(π/12)=2sin(π/6-π/3)=2sin(-π/6)=-1
f(13π/24)=2sin(13π/12-π/3)=2sin(π/12)<2
f(5π/12)=2sin(5π/6-π/3)=2sin(π/2)=2
f max= f(5π/12)=2
f min= f(π/12)=-1