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A function y = f(x) satisfies f'(x) sin x + f(x) cos x = 1, with f(x) remaining bounded as x -> 0. Let I = integral from 0 to pi/2 of f(x) dx. Which statement is correct?
- pi/2 < I < pi²/4
- pi/4 < I < pi²/2
- f(pi/4) < f(pi/3)
- f(pi/4) > f(pi/3)
Correct answer: f(pi/4) < f(pi/3)
Solution
Solving gives f(x) sin x = x (C = 0 by boundedness), so f(x) = x/sin x, which is increasing on (0, pi/2); hence f(pi/4) < f(pi/3).
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