(A)4x-y-3=0 (B)x+4y-5=0
(C)4x-y+3=0 (D)x+4y+3=0
A
解析:y=x4,则y′=4x3,∵y=x4的一切线与x+4y-8=0垂直则切线斜率为4.设切点为(x0,y0)则4x30=y0=4则x30=1,x0=1.y0=14=1。故切点坐标为(1,1)设切线方程为y=4x+b,则代入切点坐标求得切线方程为4x-y-3=0.