高一数学题：定义在[-2,2]上的偶函数g(x),当x≥0时，g(x)单调递减，若g(1-m)＜g(m)成立，求m的取值范围.

g(x)定义在[-2,2]上的偶函数,
g(x)=g(-x)
当x≥0时，g(x)单调递减
则x<0
g(x)单调递增
g(1-m)<g(m)
等价于
|1-m|<|m|
(1-m)^2<m^2
(1-m+m)(1-m-m)<0
1-2m<0
m>1/2

又
-2<=1-m<=2
-2<=m<=2

-3<=-m<=1
-1<=m<=3

所以
-1<=m<=2

综上,
1/2<m<=2