sweetms
幼苗
共回答了16个问题采纳率:100% 举报
设m表示整数.
①当x=2m时,
[
x+1
2 ] =[m+0.5]=m, [
x
2 ] =[m]=m.
∴此时恒有y=0.
②当x=2m+1时,
[
x+1
2 ] =[m+1]=m+1, [
x
2 ] =[m+0.5]=m.
∴此时恒有y=1.
③当2m<x<2m+1时,
2m+1<x+1<2m+2
∴m<
x
2 <m+0.5
m+0.5<
x+1
2 <m+1
∴ [
x
2 ] =m, [
x+1
2 ] =m
∴此时恒有y=0
④当2m+1<x<2m+2时,
2m+2<x+1<2m+3
∴m+0.5<
x
2 <m+1
m+1<
x+1
2 <m+1.5
∴此时 [
x
2 ] =m, [
x+1
2 ] =m+1
∴此时恒有y=1.
综上可知,y∈{0,1}.
故答案为{0,1}.
1年前
2