求解高中三角函数题
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答:
(1)
f(x)=√3sin2x+(1/2)f'(π/12)cos2x+f'(π/4)
求导:
f'(x)=2√3cos2x-2*(1/2)*f'(π/12)sin2x
=2√3cos2x-f'(π/12)sin2x
f'(π/12)=2√3cos(π/6)-f'(π/12)sin(π/6)=3-(1/2)f'(π/12)
f'(π/12)=2
f'(x)=2√3cos2x-2sin2x
f'(π/4)=0-2=-2
所以:
f(x)=√3sin2x+cos2x-2
=2sin(2x+π/6)-2
所以:f(x)的最小正周期T=π,最小值为-2-2=-4
(2)
|f(x)-m|<3对任意π/12<x<=π/3恒成立
-3<f(x)-m<3
-3<2sin(2x+π/6)-2-m<3
因为:
π/12<x<=π/3
π/6<2x<=2π/3
π/3<2x+π/6<=5π/6
所以:1/2<=sin(2x+π/6)<=1
所以:-3<1-2-m<=2sin(2x+π/6)-2-m<=2-2-m<3
所以:-3<-1-m<=2sin(2x+π/6)-2-m<=-m<3
解得:m<2并且m>-3
所以:-3<m<2
(1)
f(x)=√3sin2x+(1/2)f'(π/12)cos2x+f'(π/4)
求导:
f'(x)=2√3cos2x-2*(1/2)*f'(π/12)sin2x
=2√3cos2x-f'(π/12)sin2x
f'(π/12)=2√3cos(π/6)-f'(π/12)sin(π/6)=3-(1/2)f'(π/12)
f'(π/12)=2
f'(x)=2√3cos2x-2sin2x
f'(π/4)=0-2=-2
所以:
f(x)=√3sin2x+cos2x-2
=2sin(2x+π/6)-2
所以:f(x)的最小正周期T=π,最小值为-2-2=-4
(2)
|f(x)-m|<3对任意π/12<x<=π/3恒成立
-3<f(x)-m<3
-3<2sin(2x+π/6)-2-m<3
因为:
π/12<x<=π/3
π/6<2x<=2π/3
π/3<2x+π/6<=5π/6
所以:1/2<=sin(2x+π/6)<=1
所以:-3<1-2-m<=2sin(2x+π/6)-2-m<=2-2-m<3
所以:-3<-1-m<=2sin(2x+π/6)-2-m<=-m<3
解得:m<2并且m>-3
所以:-3<m<2
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