Question

定义在[-2,2]上的偶函数f(x) 当x≥0时，f(x)单调递减，且f(1-m)<f(m)成立，求

定义在[-2,2]上的偶函数f(x) 当x≥0时，f(x)单调递减，且f(1-m)<f(m)成立，求m取值范围

Answer 1

由f(1-m)<f(m)，根据定义域要求
-2<=1-m<=2, -1<=m<3
-2<=m<=2
:. -1<=m<=2
偶函数f(x)=f(-x)=f(|x|)
:. f(1-m)<f(m)变为 f(|1-m|)<f(|m|)
因为|1-m|， |m|同处单调减区间，
则有 |1-m|>|m|
平方得：1-2m>0, :. m<1/2
又 -1<=m<=2
:. -1<=m<1/2

Answer 2

由f(1-m)<f(m)，根据定义域要求
-2<=1-m<=2,
-1<=m<3
-2<=m<=2
:.
-1<=m<=2
偶函数f(x)=f(-x)=f(|x|)
:.
f(1-m)<f(m)变为
f(|1-m|)<f(|m|)
因为|1-m|，
|m|同处单调减区间，
则有
|1-m|>|m|
平方得:1-2m>0,
:.
m<1/2
又
-1<=m<=2
:.
-1<=m<1/2