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由f(x)是奇函数，得f(0) = 0;则f(2cos^2θ-4)+f(4m-2mcosθ)＞0；f(2cos^2θ-4)>-f(4m-2mcosθ) = f(-4m+2mcosθ)由于f(x)是增函数,故有2cos^2θ-4>-4m+2mcosθ化简cos^2θ-mcosθ-2m+2>0θ∈[0, π/2],则cosθ∈[0,1],令y =cos^2θ-mcosθ-2m+2则有三种情况：1. 对称轴在(0,0)左侧，m/20即1-m-2m+2>0 m1 m>2所以 m>23. 对称轴在(0,1)之间... |
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