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JEE Main202331 Jan 2023Morning ShiftMathematicsDifferentiationActual

Let y = f x = sin 3 π 3 cos π 3 2 - 4 x 3 + 5 x 2 + 1 3 2 . Then, at x = 1 ,

Options

  1. A2 y ' + 3 π 2 y = 0
  2. B2 y ' + 3 π 2 y = 0
  3. C2 y ' - 3 π 2 y = 0
  4. Dy ' + 3 π 2 y = 0

Correct answer

B. 2 y ' + 3 π 2 y = 0

Step-by-step solution

The given equation can be written as y = sin 3 π 3 cos g x where, g x = π 3 2 - 4 x 3 + 5 x 2 + 1 3 / 2 ⇒ g ' x = π 2 2 - 4 x 3 + 5 x 2 + 1 1 / 2 - 12 x 2 + 10 x ⇒ g ' 1 = π 2 2 2 - 2 = - π And, g 1 = 2 π 3 = π - π 3 Now, y ' = 3 sin 2 π 3 cos g x × cos π 3 cos g x × π 3 - sin g x g ' x y ' 1 = 3 sin 2 - π 6 · cos π 6 · π 3 - sin 2 π 3 g ' 1 y ' 1 = 3 4 · 3 2 · π 3 - 3 2 - &#

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