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NDA 2 2025 Maths, question 29

Topic: Trigonometry

Passage · Q29 to Q30

Let cos⁡(2x+3y)=12\cos (2x+3y)=\frac{1}{2} and cos⁡(3x+2y)=32\cos (3x+2y)=\frac{\sqrt{3}}{2}, where −π<(2x+3y)<π-\pi<(2x+3y)<\pi and −π<(3x+2y)<π-\pi<(3x+2y)<\pi.

Q29

How many values does (x+y)(x+y) have?

  1. (a) Two
  2. (b) Three
  3. (c) FourCorrect answer
  4. (d) More than four

Solution.

Answer: (c) Four

  1. Let U=2x+3y and V=3x+2y.
  2. The angle restrictions give U=±π/3 and V=±π/6 independently.
  3. Since x+y=(U+V)/5, the values are ±π/10 and ±π/30: four distinct values.