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JEE Main 2019 · 9 April · Shift I

JEE Main 2019 Mathematics question on Application of Derivatives

Let SS be the set of all values of xx for which the tangent to the curve y=f(x)=x3−x2−2xy = f(x) = x^3 - x^2 - 2x at (x,y)(x, y) is parallel to the line segment joining the points (1,f(1))(1, f(1)) and (−1,f(−1))(-1, f(-1)), then SS is equal to
  1. A.
    {13,−1}\left\{\frac{1}{3}, -1\right\}
  2. B.
    {13,1}\left\{\frac{1}{3}, 1\right\}
  3. C.
    {−13,1}\left\{-\frac{1}{3}, 1\right\}
  4. D.
    {−13,−1}\left\{-\frac{1}{3}, -1\right\}
Show answer and solution

Correct option: C

Solution

f(1) = 1 - 1 - 2 = -2, f(-1) = -1 - 1 + 2 = 0. Slope of line segment = (0 - (-2))/(-1 - 1) = 2/(-2) = -1.

f'(x) = 3x² - 2x - 2. Set equal to -1: 3x² - 2x - 2 = -1 → 3x² - 2x - 1 = 0 → (3x + 1)(x - 1) = 0, giving x = -1/3 or x = 1. So S = {-1/3, 1}.

[JEE Main 2019 · 9 April · Shift I]

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