函數(shù)f(x)是定義在R上的奇函數(shù)且在[0,+∞)上是增函數(shù)→易知f(x)在(-∞,+∞)上是增函數(shù)
那么f(4m-2mx)>f(4-2x^2)→4m-2mx>4-2x^2→x^2+mx+(2m-2)>0
設(shè)g(x)=x^2+mx+(2m-2),其對稱軸是x=m/2;
①當m/2≤0時,m≤0;使g(0)=0+(2m-2)>0→m>1,則不成立:{m≤0}∩{m>1}=空集
②當m/2≥1時,m≥2;使g(1)=1+m+(2m-2)>0→m>1/3解該不等式得m≥2且m>1/3→m≥2
③當0<(m/2)<1時,00→m∈{-1-√3,-1+√3}
?、佗冖鄣牟⒓?得∈{-1-√3,-1+√3}∪{m≥2}
^^^老總,給我加分啊