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JEE Main201912 Apr 2019Morning ShiftMathematicsDifferentiationActual

If e y + x y = e , the ordered pair d y d x , d 2 y d x 2 at x = 0 is equal to

Options

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

Correct answer

B. - 1 e , 1 e 2

Step-by-step solution

e y + x y = e differentiate w. r. t. x e y d y d x + x d y d x + y = 0 ⇒ d y d x x + e y = - y ,     d y d x ( 0,1 ) = - 1 e Again differentiate w. r. t. x e y . d 2 y d x 2 + d y d x . e y . d y d x + x . d 2 y d x 2 + d y d x + d y d x = 0 ⇒ x + e y d 2 y d x 2 + d y d x 2 . e y + 2 d y d x = 0 Now, at 0 , 1 e d 2 y d x 2 + 1 e 2 e + 2 - 1 e = 0 ∴   d 2 y d x 2 = 1 e 2 Hence, ordered pair d y d x , d 2 y d x 2 = - 1 e , 1 e 2

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