∴P(1,1),
∵y=xn+1,∴y′=(n+1)xn,当x=1时,y′=n+1,即切线的斜率为:n+1,
故y=xn+1在(1,1)处的切线方程为y-1=(n+1)(x-1),
令y=0可得x=
| n |
| n+1 |
即该切线与x轴的交点的横坐标为xn=
| n |
| n+1 |
所以log2013x1+log2013x2+…+log2013x2012
=log2013
| 1 |
| 2 |
| 2 |
| 3 |
| 3 |
| 4 |
| 2012 |
| 2013 |
| 1 |
| 2013 |
故选B.
| n |
| n+1 |
| n |
| n+1 |
| 1 |
| 2 |
| 2 |
| 3 |
| 3 |
| 4 |
| 2012 |
| 2013 |
| 1 |
| 2013 |