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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?

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Solution

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$$y=x^2+ax-b$$

$$(1, -5)$$ 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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