Question

设a属于(0,π/2),函数f(x)的定义域为【0,1】且f(0)=0,f(1)=1,当x≥y时,有f((x+y)/2)

Answer 1

根据题中给定的条件：

令x=1,y=0,有f(1/2)=f(1)sina+(1-sina)f(0)=sina

令x=1/2,y=0,有f(1/4)=f(1/2)sina+(1-sina)f(0)=sin²a

令x=1,y=1/2,有f(3/4)=f(1)sina+(1-sina)f(1/2)=2sina-sin²a

令x=3/4,y=1/4,有f(1/2)=f(3/4)sina+(1-sina)f(1/4)=3sin²a-2sin³a

∴3sin²a-2sin³a=sina

考虑到a∈(0,π/2),那么sina∈(0,1),解得sina=1/2

(1) f(1/2)/f(1/4)=sina/sin²a=2

(2) a=π/6

(3) g(x)=sin(π/6-2x)

=sin(π/6)cos(2x)-cos(π/6)sin(2x)

=(1/2)(2cos²x-1)-(√3/2)(2sinx•cosx)

=cos²x-√3•sinx•cosx-1/2