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JEE Main20199 Apr 2019Morning ShiftMathematicsApplication of DerivativesActual

Let S be the set of all values of x for which the tangent to the curve y = f x = x 3 - x 2 - 2 x at ( x , y ) is parallel to the line segment joining the points ( 1 , f ( 1 ) ) and - 1 , f - 1 , then S is equal to

Options

  1. A  - 1 3 , - 1
  2. B- 1 3 , 1
  3. C1 3 , 1
  4. D1 3 , - 1

Correct answer

B. - 1 3 , 1

Step-by-step solution

Given y = f x = x 3 - x 2 - 2 x ⇒ f 1 = 1 - 1 - 2 = - 2 ⇒ f - 1 = - 1 - 1 + 2 = 0 The slope of a line joining the points x 1 ,   y 1   &   x 2 ,   y 2 is y 2 - y 1 x 2 - x 1 Thus, the slope of the line segment joining the points 1 ,   f 1   &   - 1 ,   f - 1 is m = f ( 1 ) - f - 1 1 + 1 = - 2 - 0 2 = - 1 Given, this line segment is parallel to the tangent of the curve y = f x , at x ,   y and we know that the slope of the tangent to y = f x is d y d x

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