Official solution
Slope of the tangent to the curve y = x3 + ax -- b at point (1, --5)
m1 = \[{\left. {{{dy} \over {dx}}} \right|_{\left( {1, - 5} \right)}}\] = 3x2 + a = 3 + a
Slope of the line --x + y + 4 = 0,
m2 = 1
As line and tangent to the curve are perpendicular to each other,
\[ \therefore \] m1 \[ \times \] m2 = -1
\[ \Rightarrow \] (3 + a) \[ \times \] 1 = -1
\[ \Rightarrow \] a = - 4
\[ \therefore \] Curve becomes y = x3 - 4x -- b
This curve goes through (1, --5)
\[ \therefore \] -5 = 1 - 4 - b
\[ \Rightarrow \] b = 2
So curve is y = x3 - 4x -- 2
By checking each options you can see,
(2, --2) lies on the curve.