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JEE Main202224 Jun 2022Evening ShiftMathematicsDifferentiationActual

If y = tan - 1 sec x 3 - tan x 3 , π 2 < x 3 < 3 π 2 , then

Options

  1. Ax y ' ' + 2 y ' = 0
  2. Bx 2 y ' ' - 6 y + 3 π 2 = 0
  3. Cx 2 y ' ' - 6 y + 3 π = 0
  4. Dx y ' ' - 4 y ' = 0

Correct answer

B. x 2 y ' ' - 6 y + 3 π 2 = 0

Step-by-step solution

Given, y = tan - 1 sec x 3 - tan x 3 = tan - 1 1 - sin x 3 cos x 3 = tan - 1 1 - cos π 2 - x 3 sin π 2 - x 3 = tan - 1 tan π 4 - x 3 2 Since π 4 - x 3 2 ∈ - π 2 , 0 as π 2 < x 3 < 3 π 2 So, y = π 4 - x 3 2 Now differentiating we get, y ' = - 3 x 2 2 , y ' ' = - 3 x Now putting the value of x in term of y ' ' in 4 y = π - 2 x 3 We get, 4 y = π - 2 x 2 - y ' ' 3 12 y = 3 π + 2 x 2 y ' ' x 2 y ' ' - 6 y + 3 π

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