试题
| f(2x+1) |
| 4x-1 |
| 3 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
解：∵f（x）是定义域为（-∞，0）∪（0，+∞）的奇函数，在区间（0，+∞）上单调递增，| f(2x+1) |
| 4x-1 |
| { | f(2x+1)＞0 4x-1＜0 |
| { | f(2x+1)＜0 4x-1＞0 |
| { | 2x+1＞2或-2＜2x+1＜0 4x-1＜0 |
| { | 2x+1＜-2或0＜2x+1＜2 4x-1＞0 |
| 3 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
