已知函数f(x)=2sinx⼀4cosx⼀4+根3cosx⼀2(1)求函数最小正周期及最值

(2)令g(x)=f(x+π/3)判断其奇偶性说明理由
2025-02-25 01:46:48
推荐回答(1个)
回答1:

解:

f(x)=2sin(x/4)cos(x/4)+√3cos(x/2)

=sin(x/2)+√3cos(x/2)

=2sin(x/2 +π/3)

最小正周期是T=2π/(1/2)=4π

最大值是2,最小值是-2

g(x)

=f(x+π/3)

=2sin(x/2 +2π/3)

=2cos(x/2+π/6)

g(-x)=2cos(-x/2 +π/6)=2cos(x/2 -π/6)

g(x)=2cos(x/2 +π/6)

g(x)+g(-x)=2*[2cos(x/2)cos(π/6)]=2√3*cos(x/2)≠0

∴g(x)≠g(-x),且g(-x)≠-g(x)

所以g(x)是非奇非偶函数

谢谢