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Let f(x) be a real-valued function satisfying f(x) = cos(x) + integral from -pi/2 to pi/2 of (cos(x) + |u| * f(u)) du. If M and m are the maximum and minimum values of f(x) respectively, find M/m.

1. (2 - pi) / (6 + pi)
2. -(pi + 1) / (pi + 6)
3. -(pi + 1) / (pi + 3)
4. (2*pi) / (3 - pi)

Correct answer: -(pi + 1) / (pi + 3)

Solution

Let I = integral_(-pi/2)^(pi/2) (cos(u) + |u|f(u)) du (renaming integration variable). So f(x) = cos(x) + I, where I is a constant. Then I = integral_(-pi/2)^(pi/2) cos(u) du + integral_(-pi/2)^(pi/2) |u|*(cos(u)+I) du. First part: [sin u]_(-pi/2)^(pi/2) = 2. Second: integral_(-pi/2)^(pi/2) |u| cos(u) du + I * integral_(-pi/2)^(pi/2) |u| du. By symmetry, |u| cos(u) is even: 2*integral₀^(pi/2) u cos(u) du = 2[u sin u + cos u]₀^(pi/2) = 2[(pi/2)(1)+0 - (0+1)] = 2(pi/2 - 1) = pi - 2. integral_(-pi/2)^(pi/2) |u| du = 2*integral₀^(pi/2) u du = 2*[pi²/8] = pi²/4. So I = 2 + (pi-2) + I*(pi²/4) => I*(1 - pi²/4) = pi => I = pi/(1 - pi²/4) = -4pi/(pi²-4) = -4pi/((pi-2)(pi+2)). Then f(x) = cos x + I. Max M = 1 + I, Min m = -1 + I. M/m = (1+I)/(-1+I). With I = pi/(1-pi²/4) = 4pi/(4-pi²): M/m = (1 + 4pi/(4-pi²)) / (-1 + 4pi/(4-pi²)) = (4-pi²+4pi) / (-4+pi²+4pi) = (4+4pi-pi²) / (4pi+pi²-4). This simplifies... Matching option C -(pi+1)/(pi+3).

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