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JEE Main 2019 · 12 Apr · Shift II

JEE Main 2019 Mathematics question on Trigonometry

Let SS be the set of all α∈R\alpha \in R such that the equation, cos⁡2x+αsin⁡x=2α−7\cos 2x + \alpha \sin x = 2\alpha - 7 has a solution. Then, SS is equal to
  1. A.
    RR
  2. B.
    [1,4][1, 4]
  3. C.
    [3,7][3, 7]
  4. D.
    [2,6][2, 6]
Show answer and solution

Correct option: D

Solution

Substitute cos2x = 1 - 2sin²x to get 1 - 2sin²x + α sinx = 2α - 7, which rearranges to 2sin²x - α sinx + (2α - 8) = 0.

Let t = sinx, so t ∈ [-1, 1]. The equation becomes 2t² - αt + (2α - 8) = 0, giving α(t - 2) = 2t² - 8 = 2(t-2)(t+2), so α = 2(t+2) (since t ≠ 2).

Thus α = 2t + 4, and since t ∈ [-1, 1], α ranges from 2(-1)+4 = 2 to 2(1)+4 = 6. Hence S = [2, 6].

[JEE Main 2019 · 12 Apr · Shift II]

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