Question

# 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?

A
$$(-2, 2)$$
B
$$(2, -2)$$
C
$$(2, -1)$$
D
$$(-2, 1)$$
Solution
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#### Correct option is B. $$(2, -2)$$$$y=x^2+ax-b$$$$(1, -5)$$ lies on the curve$$\Rightarrow -5=1+a-b\Rightarrow a-b=-6$$ .(i)Also, $$y'=3x^2+a$$$$y'_{(1, -5)}=3+a$$ (slope of tangent)$$\because$$ this tangent is $$\perp$$ to $$-x+y+4=0$$$$\Rightarrow (3+a)(1)=-1$$$$\Rightarrow a=-4$$ .(ii)By (i) and (ii): $$a=-4$$, $$b=2$$$$\therefore y=x^3-4x-2$$$$(2, -2)$$ lies on this curve.

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