 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/6a/b/solution

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

Differentiate the map:

g^{'} (x) = r + k e^{- λ x} (1 - λ x) .

At the positive equilibrium,

k e^{- λ x_{*}} = 1 - r

, so

g^{'} (x_{*}) = 1 - (1 - r) λ x_{*} = 1 - (1 - r) lo g (\frac{k}{1 - r}) .

∣ g^{'} (x_{*}) ∣ < 1

. Existence already makes the logarithm positive, so the upper inequality is automatic. The lower inequality gives

(1 - r) lo g (\frac{k}{1 - r}) < 2.

1 - r < k < (1 - r) exp (\frac{2}{1 - r}) .

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