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AIEEE 2002

JEE Main 2002 Mathematics question on Trigonometry

The number of integral values of kk for which the equation 7cos⁡x+5sin⁡x=2k+17\cos x + 5\sin x = 2k + 1 has a solution, is
  1. A.
    4
  2. B.
    8
  3. C.
    10
  4. D.
    12
Show answer and solution

Correct option: B

Solution

The equation 7cos x + 5sin x = 2k + 1 has a solution only if |2k + 1| ≤ √(7² + 5²) = √74 ≈ 8.60.

So -8.60 ≤ 2k + 1 ≤ 8.60, giving -9.60 ≤ 2k ≤ 7.60, i.e. -4.8 ≤ k ≤ 3.8.

The integral values of k are -4, -3, -2, -1, 0, 1, 2, 3 — that's 8 values.

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