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If the tangent to the curve, y=x^(3)+ax-...

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, 1)

B

(-2, 2)

C

(2, -1)

D

(2, -2)

Text Solution

Verified by Experts

The correct Answer is:
D
`y=x^(3)+ax-b rArr (dy)/(dx)=3x^(2)+a`
Now `(dy)/(dx)|_((1","-5)) =-1`
`rArr a+3=-1 rArr a=-4`
(1,-5) lies on the curve
`rArr -5=1-4-b , b=5-3=2`
So, curve is `y=x^(3)-4x-2`
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