Exams › JEE Advanced › Maths
Correct answer: (C) 6
Solve: -sqrt(3) <= 2*sqrt(2)*sin(theta) + 1 <= sqrt(3). Subtract 1: -1-sqrt(3) <= 2*sqrt(2)*sin(theta) <= sqrt(3)-1. Divide by 2*sqrt(2): (-1-sqrt(3))/(2*sqrt(2)) <= sin(theta) <= (sqrt(3)-1)/(2*sqrt(2)). Numerically: (-1-1.732)/2.828 <= sin(theta) <= (1.732-1)/2.828, i.e., -0.966 <= sin(theta) <= 0.259. In (pi/2, 3*pi/2): sin(theta) ranges from 1 down to -1. The constraint sin(theta) <= 0.259 in (pi/2, 3*pi/2) gives theta in [pi - arcsin(0.259), 3*pi/2]... wait, more carefully: sin(pi/2) = 1 > 0.259, so we need theta >= arccos or similar. In (pi/2, 3*pi/2): sin(theta) >= 0.259 occurs near theta in (pi/2, pi - arcsin(0.259)). The upper bound sin(theta) <= 0.259 is satisfied for theta in [pi - arcsin(0.259), 3*pi/2]. The lower bound sin(theta) >= -0.966: sin(theta) = -0.966 ~ sin(-75 deg) = sin(255 deg) = sin(17*pi/12). So solution interval: theta in [pi - arcsin(0.259), 17*pi/12] approx. arcsin(0.259) ~ pi/12 (15 deg). So pi - pi/12 = 11*pi/12. Solution: [11*pi/12, 17*pi/12]. m = 11, n = 17, |n - m| = 6.