Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 / 1 / iv

Table of contents
- Solution iv

Solution

x \to - \infty

, the term

- 1/ ϕ

dominates the implicit equation, so

ϕ (x) \sim \frac{1}{A - x} .

x \to + \infty

, put

δ = 1 - ϕ

. Then

x - A = \frac{1}{2} lo g \frac{2}{δ} - 1 + o (1),

and hence

ϕ (x) \sim 1 - e^{- 2 (x - b)}, b = A - 1 + \frac{1}{2} lo g 2 .

The profile rises monotonically from

0

to

1

, with an algebraic left tail and an exponential right tail.

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