As $x\to-\infty$, the term $-1/\phi$ dominates the implicit equation, so
$$
\phi(x)\sim\frac1{A-x}.
$$
As $x\to+\infty$, put $\delta=1-\phi$. Then
$$
x-A=\frac12\log\frac2\delta-1+o(1),
$$
and hence
$$
\phi(x)\sim1-e^{-2(x-b)},
\qquad
\boxed{b=A-1+\frac12\log2}.
$$
The profile rises monotonically from $0$ to $1$, with an algebraic left tail and an exponential right tail.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-313/1/iv/solution