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1987

JEE Main 1987 Mathematics question on Trigonometry

The number of all possible triplets (a1,a2,a3)(a_1, a_2, a_3) such that a1+a2cos⁡(2x)+a3sin⁡2(x)=0a_1 + a_2 \cos(2x) + a_3 \sin^2(x) = 0, ∀\forall xx is
  1. A.
    0
  2. B.
    1
  3. C.
    3
  4. D.
    ∞\infty
Show answer and solution

Correct option: D

Solution

Rewrite the equation using cos(2x) = 1 - 2sin²(x), so:

a₁ + a₂(1 - 2sin²x) + a₃sin²x = 0

(a₁ + a₂) + (-2a₂ + a₃)sin²x = 0 for all x.

This requires two conditions: a₁ + a₂ = 0 and -2a₂ + a₃ = 0, i.e., a₂ = -a₁ and a₃ = 2a₂ = -2a₁. Since a₁ is a free parameter, there are infinitely many triplets of the form (t, -t, -2t) for any real t.

[1987]

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