1 |
3 |
1 |
2 |
∴f′(x)=ax2+bx+c,则f′(1)=−
a |
2 |
即3a+2b+2c=0
∵3a>2c>2b
∴a>0且b>0,故选项D正确
∵3a>2c>2b,2b=-3a-2c
∴3a>2c>-3a-2c即−
3 |
4 |
c |
a |
3 |
2 |
∵3a>2c>2b,2c=-3a-2b
∴3a>-3a-2b>2b,即−3<
b |
a |
3 |
4 |
∵3a>2c>2b,3a=-2b-2c
∴-2b-2c>2c>2b,即−
1 |
2 |
c |
b |
故选A.
1 |
3 |
1 |
2 |
a |
2 |
1 |
4 |
c |
a |
3 |
2 |
b |
a |
3 |
4 |
1 |
2 |
c |
b |
1 |
3 |
1 |
2 |
a |
2 |
3 |
4 |
c |
a |
3 |
2 |
b |
a |
3 |
4 |
1 |
2 |
c |
b |