Question

已知f(x)=ln(2－x)+ax.设曲线在（1,f(1)）处切线为l.若与圆（x+1）∧2+y∧2=1相切,求a...

Answer 1

f(1) = a

L 过（1,a)

f'(x) = - 1/(2-x) + a,切线斜率k= f'(1) = a - 1

L的方程为：y - a = (a-1)(x-1),y + (1-a)x - 1 = 0

圆心（-1,0）到L的距离为1,即 |a-1-1|/根号(1+(1-a)^2)=1

a = 1