已知函数f(x)={-√3(tanx)²+2tanx+√3}/{(tanx)²+1},x属于R

题目

已知函数f(x)={-√3(tanx)²+2tanx+√3}/{(tanx)²+1},x属于R
(1)求函数f(x)的最小正周期以及一条对称轴
(2)若x属于【π/4,3π/4】,求f(x)的最小值和最大值
(3)若g(x)=f(x)-a 且x属于(-π/12,2π/3】,当g(x)=0的解有0个,1个和2个时,分别求出对应a的取值范围?

答案

已知函数f(x)=[-(√3)tan²x+2tanx+√3]/(tan²x+1),x属于R
(1)求函数f(x)的最小正周期以及一条对称轴
(2)若x属于【π/4,3π/4】,求f(x)的最小值和最大值
(3)若g(x)=f(x)-a 且x属于(-π/12,2π/3】,当g(x)=0的解有0个,1个和2个时,分别求出对
应a的取值范围
(1).f(x)=[-(√3)tan²x+2tanx+√3]/(tan²x+1)=[-(√3)tan²x+2tanx+√3]/(sec²x)
=-(√3)sin²x+2sinxcosx+(√3)cos²x=sin2x+(√3)(cos²x-sin²x)=sin2x+(√3)cos2x
=2[(1/2)sin2x+(√3/2)cos2x]=2[sin2xcos(π/3)+cos2xsin(π/3)]=2sin(2x+π/3)
故其最小正周期T=π；令2x+π/3=π/2,得2x=π/2-π/3=π/6,故x=π/12就是其一条对称轴.
(2).若x∈[π/4,3π/4],则当x=7π/12时f(x)获得最小值minf(x)=f(7π/12)=2sin(7π/6+π/3)
=2sin(3π/2)=-2；当x=π/4时f(x)获得最大值maxf(x)=f(π/4)=2sin(π/2+π/3)=2sin(5π/6)
=2sin(π-π/6)=2sin(π/6)=2×(1/2)=1.
(3).g(x)=2sin(2x+π/3)-a=0.①,(-π/12