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Let f: R -> R and g: R -> R be defined by f(x) = ln(x² + 1) - e^(-x) + 1 and g(x) = (1 - 2e^(2x)) / e^x. For which of the following ranges of alpha does the inequality f(g(((alpha-1)²)/3)) > f(g(alpha - 5/3)) hold?

1. (2, 3)
2. (-2, -1)
3. (1, 2)
4. (-1, 1)

Correct answer: (1, 2)

Solution

f is strictly increasing (f'(x) > 0). g(x) = e^(-x) - 2e^x is strictly decreasing (g'(x) = -e^(-x) - 2e^x < 0). Since both are strictly monotone, f(g(A)) > f(g(B)) iff g(A) > g(B) iff A < B. So the inequality becomes (alpha-1)²/3 < alpha - 5/3. Solving: (alpha-1)² < 3alpha - 5 => alpha² - 2alpha + 1 < 3alpha - 5 => alpha² - 5alpha + 6 < 0 => (alpha-2)(alpha-3) < 0 => 2 < alpha < 3. Answer: (2, 3).

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