奇函数f(x)的定义域为R,且在[0,+∞)上是增函数,当0≤θ≤π/2时,是否存在实数m,使f(4m-2mcosθ)-f(2sin²θ+2)>f(0)对所有θ∈[0,π/2]求出所有适合条件的实数m.
问题描述:
奇函数f(x)的定义域为R,且在[0,+∞)上是增函数,当0≤θ≤π/2时,是否存在实数m,使f(4m-2mcosθ)-f(2sin²θ+2)>f(0)对所有θ∈[0,π/2]求出所有适合条件的实数m.
答
f(0) = -f(0)=> f(0) = 0f(4m-2mcosθ)-f(2sin²θ+2)>f(0)= 0=> f(4m-2mcosθ)> f(2sin²θ+2)0≤θ≤π/24m-2mcosθ > 0 and 2sin²θ+2 > 0=> 4m-2mcosθ > 2sin²θ+24m-2mcosθ > -...