If the tangent to the curve, y=x^3+ax-b at the point (1, -5) is perpendicular to the line, -x+y+4=0, then which one of the following points lies on the curve?

Correct option is B (2,-2)
y=x^3+ax-b
(1,-5) lies on the curve ⇒ -5=1+a-b ⇒ a-b=-6 (i)
Also, y'=3x^2+a
y'(1,-5)=3+a (slope of tangent)
This tangent is ⊥ to -x+y+4=0 ⇒ (3+a)(1)=-1 ⇒ a=-4 (ii)
By (i) and (ii): a=-4, b=2
∴y=x^3-4x-2
(2,-2) lies on this curve.