Question

已知曲线在点与相切,求ab.

Answer 1

y′=4ax³+3bx²+2x;

x=1;y′=4a+3b+2=11;

x=1;y=a+b+1+3=6;

4a+3b=9(1)

a+b=2(2)

(1)-(2)×3得：

4a-3a=9-6;

a=3;

b=-1;