题目
| x | ┅┅ | 0 | 1 | 2 | 3 | ┅┅ |
| y | ┅┅ | -3 | -1.999 | -1.001 | 0 | ┅┅ |
| 3 |
| 2 |
| f(a+1) |
| a-1 |
| f(a) |
| a |
A.
|
B.
|
||||||||
C.
|
D.不确定 |
答案
∴
| f(a+1) |
| a-1 |
| f(a) |
| a |
| a-2 |
| a-1 |
| a-3 |
| a |
| a(a-2)-(a-3)(a-1) |
| a(a-1) |
| 2a-3 |
| a(a-1) |
∵1<a<
| 3 |
| 2 |
∴2a-3<0,a(a-1)>0即
| 2a-3 |
| a(a-1) |
∴
| f(a+1) |
| a-1 |
| f(a) |
| a |
故选A.
| x | ┅┅ | 0 | 1 | 2 | 3 | ┅┅ |
| y | ┅┅ | -3 | -1.999 | -1.001 | 0 | ┅┅ |
| 3 |
| 2 |
| f(a+1) |
| a-1 |
| f(a) |
| a |
A.
|
B.
|
||||||||
C.
|
D.不确定 |
| f(a+1) |
| a-1 |
| f(a) |
| a |
| a-2 |
| a-1 |
| a-3 |
| a |
| a(a-2)-(a-3)(a-1) |
| a(a-1) |
| 2a-3 |
| a(a-1) |
| 3 |
| 2 |
| 2a-3 |
| a(a-1) |
| f(a+1) |
| a-1 |
| f(a) |
| a |