k |
k+1 |
-1≤-
k |
k+1 |
解得k>0 x在(0,1]g(x)=x-kx-k=(1-k)x-k 令g(x)=0 x=
k |
k+1 |
0<
k |
k+1 |
1 |
2 |
2−k |
k+1 |
1<
2−k |
k+1 |
1 |
2 |
x在(2,3]g(x)=x-2-kx-k=(1-k)x-2-k 令g(x)=0 x=
k+2 |
1−k |
2<
k+2 |
1−k |
1 |
4 |
1 |
4 |
故答案为:(0,
1 |
4 |
k |
k+1 |
k |
k+1 |
k |
k+1 |
k |
k+1 |
1 |
2 |
2−k |
k+1 |
2−k |
k+1 |
1 |
2 |
k+2 |
1−k |
k+2 |
1−k |
1 |
4 |
1 |
4 |
1 |
4 |