Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 6A / b

Solution

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}) .

There are no discussions about this article yet.

There are currently no matching articles.