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JEE Main202329 Jan 2023Evening ShiftMathematicsTrigonometric EquationsActual

The set of all values of λ for which the equation cos 2 2 x - 2 sin 4 x - 2 cos 2 x = λ

Options

  1. A- 2 , - 1
  2. B- 2 , - 3 2
  3. C- 1 , - 1 2
  4. D- 3 2 , - 1

Correct answer

D. - 3 2 , - 1

Step-by-step solution

Given: λ = cos 2 2 x - 2 sin 4 x - 2 cos 2 x ⇒ λ = 2 cos 2 x - 1 2 - 2 1 - cos 2 x 2 - 2 cos 2 x ⇒ λ = 4 cos 4 x - 4 cos 2 x + 1 - 2 1 - 2 cos 2 x + cos 4 x - 2 cos 2 x ⇒ λ = 2 cos 4 x - 2 cos 2 x + 1 - 2 ⇒ λ = 2 cos 4 x - 2 cos 2 x - 1 ⇒ λ = 2 cos 4 x - cos 2 x - 1 2 ⇒ λ = 2 cos 2 x - 1 2 2 - 3 4 So, λ max = 2 1 4 - 3 4 = - 1 (max Value) λ min = 2 0 - 3 4 = - 3 2 (Minimum Value) So, range of λ is - 3 2 , - 1

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