已知函数 f(x)=sinx^2+2sinxcosx+3cosx^2,x∈R .求:

2025-12-17 20:50:12
推荐回答(2个)
回答1:

f(x)=sin�0�5x+2sinxcosx+3cos�0�5x
=(1-cos2x)/2+sin2x+3(1+cos2x)/2
=2+cos2x+sin2x
=2+√2sin(2x+п/4)
当2x+п/4=2kп+п/2x=kп+п/8时,能取到最大值为2+√2
(2) 2kп-п/2≤2x+п/4≤2kп+п/2
2kп-3п/4≤2x≤2kп+п/4
kп-3п/8≤x≤kп+п/8
则函数的单调增区间为[kп-3п/8,kп+п/8]

回答2:

f(x)=sinx^2+2sinxcosx+3cosx^2
=1+sin2x+2cos^2x
=1+sin2x+1+cos2x
=2+根号2sin(2x+pi/4)
sin(2x+pi/4)=1时最大值=2+根号2
2x+pi/4=pi/2+2k*pi
x=pi/8+k*pi
sin(2x+pi/4)=-1时最小值=2-根号2
2x+pi/4=3pi/2+2k*pi
x=5pi/8+k*pi