- 一道数学题已知函数f(x)的最小正周期为π,且在(π/4,π/2
- 已知f(x)的最小正周期为π,且在(π/4,π/2)上单调递增,则f(x)的解析式为( )
A.f(x)=3tanx/(1+tanx^2)
B.f(x)=2sinx/(1-cosx)
C.f(x)=sin2x^2-cos2x^2
D.f(x)=sin(x+π/4)sin(x-π/4)
- 淘汰C,f(x)=sin2x^2-cos2x^2 ,最小正周期为π/2;
f(x)=3tanx/(1+tanx^2) ,在(π/4,π/2)上单调递减;
f(x)=2sinx/(1-cosx)=2sinx/(2sin(x/2)*2sin(x/2))
=1/tan(x/2),在(π/4,π/2)上单调递减;
选D