 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-4/14c/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

1 = (B / g) e^{- (μ + r) s_{*} / g}

, hence

r = (g / s_{*}) lo g (B / g) - μ

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