函数y=x2(x>0)的图象在点(ak,ak2)处的切线与x轴交点的横坐标为ak+1,k为正整数,a1=16,则a1+a3+a5=( ) A.18 B.21 C.24 D.30
问题描述:
函数y=x2(x>0)的图象在点(ak,ak2)处的切线与x轴交点的横坐标为ak+1,k为正整数,a1=16,则a1+a3+a5=( )
A. 18
B. 21
C. 24
D. 30
答
依题意,y′=2x,
∴函数y=x2(x>0)的图象在点(ak,ak2)处的切线方程为y-ak2=2ak(x-ak)
令y=0,可得x=
ak,即ak+1=1 2
ak,1 2
∴数列{an}为等比数列an=16×(
)n-11 2
∴a1+a3+a5=16+4+1=21
故选B