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Question
If the tangent to the curve, y=x3+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?
If the tangent to the curve, y=x3+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
The correct option is B (2,−2)
y=x2+ax−b
(1,−5) lies on the curve
⇒−5=1+a−b⇒a−b=−6 .(i)
Also, y′=3x2+a
y′(1,−5)=3+a (slope of tangent)
∵ this tangent is ⊥ to −x+y+4=0
⇒(3+a)(1)=−1
⇒a=−4 .(ii)
By (i) and (ii): a=−4, b=2
∴y=x3−4x−2
(2,−2) lies on this curve.
y=x2+ax−b
(1,−5) lies on the curve
⇒−5=1+a−b⇒a−b=−6 .(i)
Also, y′=3x2+a
y′(1,−5)=3+a (slope of tangent)
∵ this tangent is ⊥ to −x+y+4=0
⇒(3+a)(1)=−1
⇒a=−4 .(ii)
By (i) and (ii): a=−4, b=2
∴y=x3−4x−2
(2,−2) lies on this curve.
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