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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 ?

JEE · Math · previous-year question

  1. A.(2, –2)correct
  2. B.(2, –1)
  3. C.(–2, 2)
  4. D.(–2, 1)

Answer

A. (2, –2)

Explanation

Slope of the tangent to the curve y = x3 + ax – b at point (1, –5) m1 = $${\left. {{{dy} \over {dx}}} \right|_{\left( {1, - 5} \right)}}$$ = 3x2 + a = 3 + a Slope of the line –x + y + 4 = 0, m2 = 1 As line and tangent to the curve are perpendicular to each other, $$ \therefore $$ m1 $$ \times $$ m2 = -1 $$ \Rightarrow $$ (3 + a) $$ \times $$ 1 = -1 $$ \Rightarrow $$ a = - 4 $$ \therefore $$ Curve becomes y = x3 - 4x – b This curve goes through (1, –5) $$ \therefore $$ -5 = 1 - 4 - b $$ \Rightarrow $$ b = 2 So curve is y = x3 - 4x – 2 By checking each options you can see, (2, –2) lies on the curve.

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