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JEE Main202126 Aug 2021Morning ShiftMathematicsDifferentiationActual

Let f ( x ) = cos 2 tan - 1 sin cot - 1 1 - x x , 0 < x < 1 . Then:

Options

  1. A( 1 - x ) 2 f ' ( x ) + 2 ( f ( x ) ) 2 = 0
  2. B( 1 + x ) 2 f ' ( x ) + 2 ( f ( x ) ) 2 = 0
  3. C( 1 - x ) 2 f ' ( x ) - 2 ( f ( x ) ) 2 = 0
  4. D( 1 + x ) 2 f ' ( x ) - 2 ( f ( x ) ) 2 = 0

Correct answer

A. ( 1 - x ) 2 f ' ( x ) + 2 ( f ( x ) ) 2 = 0

Step-by-step solution

Put x = sin 2 θ , 0 < x < 1 sin θ = x ⇒ f x = cos 2 tan - 1 sin cot - 1 1 - sin 2 θ sin 2 θ ⇒ f x = cos 2 tan - 1 ( sin θ ) ⇒ f x = cos 2 tan - 1 x = 1 - tan 2 tan - 1 x 1 + tan 2 tan - 1 x ⇒ f x = 1 - x 1 + x ⇒ f ' x = 1 + x - 1 - 1 - x · 1 1 + x 2 ⇒ f ' x = - 2 1 + x 2 Multiply, 1 - x 2 on both sides ⇒ 1 - x 2 f ' x = - 2 1 - x 2 1 + x 2 Now, 2 f x 2 = 2 1 - x 2 1 + x 2 ∴ 1 - x 2 f ' x + 2 f x 2 = 0 option 1 satisfied

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