If the tangent to the curve, y=x^{3}+a x-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? A \quad(-2,2) B (2,-2) C \quad(2,-1) D \quad(-2,1)
If the tangent to the curve, y=x^{3}+a x-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? A \quad(-2,2) B (2,-2) C \quad(2,-1) D \quad(-2,1)

Correct option is
B (2,-2)
y=x^{2}+a x-b
(1,-5) lies on the curve
\Rightarrow-5=1+\mathrm{a}-\mathrm{b} \Rightarrow \mathrm{a}-\mathrm{b}=-6.(i)
\mathrm{AlsO}, y^{\prime}=3 x^{2}+a
y_{(1,-5)}^{\prime}=3+\mathrm{a} (slope of tangent)
\because this tangent is \perp to -x+y+4=0

    \[\begin{array}{l} \Rightarrow(3+\mathrm{a})(1)=-1 \\ \Rightarrow \mathrm{a}=-4 \text {.(ii) } \end{array}\]

By (i) and (ii): a=-4, b=2
\therefore y=x^{3}-4 x-2
(2,-2) lies on this curve.