JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Differentiation

The point $(1,-5)$ lies on $y=x^3+ax-b$, so $-5=1+a-b$, i.e. $a-b=-6$. The slope there is $y^{\prime }=3x^2+a=3+a$. The line $-x+y+4=0$ has slope $1$, and for the tangent to be perpendicular to it we need $(3+a)(1)=-1$, so $a=-4$ and then $b=2$. The curve is $y=x^3-4x-2$, and testing the offered points, $(2,-2)$ works since $8-8-2=-2$.