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)-
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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.
f(0) = -f(0)
=> f(0) = 0
f(4m-2mcosθ)-f(2sin²θ+2)>f(0)
= 0
=> f(4m-2mcosθ)> f(2sin²θ+2)
0≤θ≤π/2
4m-2mcosθ > 0 and 2sin²θ+2 > 0
=> 4m-2mcosθ > 2sin²θ+2
4m-2mcosθ > - cos²θ
cos²θ -2mcosθ + 4m > 0
△ = 4m^2-16m = 4m(m-4)≥ 0
m≤ 0 or m ≥ 4
=> f(0) = 0
f(4m-2mcosθ)-f(2sin²θ+2)>f(0)
= 0
=> f(4m-2mcosθ)> f(2sin²θ+2)
0≤θ≤π/2
4m-2mcosθ > 0 and 2sin²θ+2 > 0
=> 4m-2mcosθ > 2sin²θ+2
4m-2mcosθ > - cos²θ
cos²θ -2mcosθ + 4m > 0
△ = 4m^2-16m = 4m(m-4)≥ 0
m≤ 0 or m ≥ 4
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