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Let f: (0, pi) -> R be twice differentiable with lim(t->x) [(f(x)*sin t - f(t)*sin x)/(t - x)] = sin²(x) for all x in (0, pi). If f(pi/6) = -pi/12, which statement is TRUE?

1. f(pi/4) = pi/(4*sqrt(2))
2. f(x) < x⁴/6 - x² for all x in (0, pi)
3. There exists alpha in (0, pi) such that f'(alpha) = 0
4. f''(pi/2) + f(pi/2) = 0

Correct answer: f''(pi/2) + f(pi/2) = 0

Solution

The condition gives f(x) = -x*sin x; then f''(x) + f(x) = -2*cos x, but at x = pi/2, f''(pi/2) + f(pi/2) = -pi/2 - (-pi/2)... reduces to 0, so option D holds.

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