题目内容：

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已知奇函数f(x)的定义域为R,且f(x)在【0,正无穷〕上是增函数,是否存在这样的实数m,使f(cos2x-3)+f(4m-2mcosx)>f(0)对所有x属于【0,pai/2】均成立?若成立,求出适合条件的实数免得值或取值范围?

优质解答

奇函数f(x)的定义域为R
所以f(0)=0
f(cos2x-3)+f(4m-2mcosx)>0
f(cos2x-3)>-f(4m-2mcosx)
f(cos2x-3)>f(-4m+2mcosx)
即cos2x-3>-4m+2mcosx
2(cosx)^2-2mcosx+4m-4>0
0≤cosx=t≤1
2t^2-2mt+4m-4>0
m>(t2-2)/(t-2)
又(t2-2)/(t-2)=4-[(2-t)+2/(2-t)]≤4-2√2
所以m>4-2√2