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Let f(x) = x³ + x² + 100x + 7*sin(x). How many real roots does the equation 1/(y - f(1)) + 2/(y - f(2)) + 3/(y - f(3)) = 0 have?

1. No real root
2. Exactly one real root
3. Exactly two real roots
4. More than two real roots

Correct answer: Exactly two real roots

Solution

f'(x) = 3x² + 2x + 100 + 7*cos(x). The quadratic 3x²+2x+100 has its minimum ~99.7 (discriminant < 0) and |7*cos(x)| <= 7, so f'(x) > 0 for all x: f is strictly increasing, giving f(1) < f(2) < f(3). The rational function g(y) has poles at these three values, equals 0 at y = +-infinity, and changes sign across each pole, so IVT guarantees exactly one root in (f(1), f(2)) and one in (f(2), f(3)) — two real roots in total.

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