定义在[-2,2]上的偶函数f(x)在区间[0,2]上单调递减，若f(1-m)<f(m),实数m的取值范围

定义在[-2,2]上的偶函数f(x)在区间[0,2]上单调递减，若f(1-m)<f(m),实数m的取值范围

【解】：
偶函数f(x)在区间[0,2]上单调递减
则
在区间[-2，0]上单调递增
若f(1-m)<f(m)
当0<=1-m<=2,===>-1<=m<=1
0<=m<=2
f(1-m)<f(m)
1-m>m
m<1/2

0<=m<1/2

当-2<=1-m<=0,===>1<=m<=3
-2<=m<=0
无解

偶函数
f(1-m)=f(m-1)
f(m-1)<f(m)
当0<=m-1<=2,===>1<=m<=3
0<=m<=2
f(m-1)<f(m)
m-1>m
无解

当-2<=m-1<=0,===>==>-1<=m<=1
-2<=m<=0
f(m-1)<f(m)
m-1<m
m取任何实数
-1<=m<=0

偶函数
f(m)=f(-m)
f(1-m)<f(-m)
当0<=1-m<=2,==>-1<=m<=1
0<=-m<=2
f(1-m)<f(-m)
1-m>-m
m取任何实数
-1<=m<=0

当-2<=1-m<=0,==>1<=m<=3
-2<=-m<=0
f(1-m)<f(-m)
1-m<-m
无解

综上：0<=m<1/2 或-1<=