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JEE Main 2019 · 9 Apr · Shift I

JEE Main 2019 Mathematics question on Application of Derivatives

If the tangent to the curve, y=x3+ax−by = x^3 + ax - b at the point (1,−5)(1, -5) is perpendicular to the line, −x+y+4=0-x + y + 4 = 0, then which one of the following points lies on the curve?
  1. A.
    (−2,2)(-2, 2)
  2. B.
    (2,−2)(2, -2)
  3. C.
    (−2,1)(-2, 1)
  4. D.
    (2,−1)(2, -1)
Show answer and solution

Correct option: B

Solution

The tangent slope at (1,-5): dy/dx = 3x² + a, evaluated at x=1 gives 3+a. The line -x+y+4=0 has slope 1, so the perpendicular slope is -1. Thus 3+a = -1, giving a = -4.

Since (1,-5) lies on the curve: -5 = 1 + (-4)(1) - b → -5 = -3 - b → b = 2.

The curve is y = x³ - 4x - 2. Check (2,-2): 8 - 8 - 2 = -2 ✓

[JEE Main 2019 · 9 Apr · Shift I]

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