x2−x+n |
x2+x+1 |
n−1 |
2 |
则y(x2+x+1)=x2-x+n,
整理得:(y-1)x2+(y+1)x+y-n=0,
△=(y+1)2-4(y-1)(y-n)≥0,
解得:
3+2n−2
| ||
3 |
3+2n+2
| ||
3 |
∴f(x)的最小值为an=
3+2n−2
| ||
3 |
最大值为bn=
3+2n+2
| ||
3 |
∴cn=(1-an)(1-bn)=-
4 |
3 |
∴数列{cn}是常数数列
故选A.
x2−x+n |
x2+x+1 |
n−1 |
2 |
x2−x+n |
x2+x+1 |
n−1 |
2 |
3+2n−2
| ||
3 |
3+2n+2
| ||
3 |
3+2n−2
| ||
3 |
3+2n+2
| ||
3 |
4 |
3 |