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ExamsJEE AdvancedMaths

Let m1 and m2 be the slopes of two adjacent sides of a square of side length a such that a² + 11a + 3(m1² + m2²) = 220. One vertex of the square is at the point (10(cos(alpha) - sin(alpha)), 10(sin(alpha) + cos(alpha))), where alpha is in (0, pi/2), and one diagonal of the square has equation (cos(alpha) - sin(alpha))x + (sin(alpha) + cos(alpha))y = 10. Find the value of 72(sin⁴(alpha) + cos⁴(alpha)) + a² - 3a + 13.

  1. 119
  2. 128
  3. 145
  4. 155

Correct answer: 128

Solution

The perpendicular distance from the given vertex to the diagonal equals a/sqrt(2) = 5*sqrt(2), giving a = 10. Substituting into the constraint gives m1²+m2² = 10/3, which forces sin²(2*alpha) = 3/4, so sin⁴(alpha)+cos⁴(alpha) = 5/8. The final expression equals 72*(5/8) + 100 - 30 + 13 = 128.

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