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JEE Main20207 Jan 2020Morning ShiftMathematicsDifferentiationActual

Let x k + y k = a k , a , k > 0 and d y d x + y x 1 3 = 0 , then k is

Options

  1. A3 2
  2. B4 3
  3. C2 3
  4. D1 3

Correct answer

C. 2 3

Step-by-step solution

Differentiating the given equation with respect to x k . x k - 1 + k . y k - 1 d y d x = 0 ⇒ d y d x = - x y k - 1 ⇒ d y d x + x y k - 1 = 0 ⇒ k - 1 = - 1 3 ⇒ k = 1 - 1 3 = 2 3

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