 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2/7c/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

The cups have equal temperatures up to the intervention at

t = 1

. For the later recrossing, subtract the formula for

T_{2}

from that for

T_{1}

and set the result to zero:

- e^{- a (t - 1)} + \frac{1}{a} (1 - e^{- a (t - 2)}) = 0.

z = e^{- a (t - 2)}

gives

z = (1 + a e^{- a})^{- 1}

. Hence the nontrivial meeting time is

t^{*} = 2 + \frac{1}{a} lo g (1 + a e^{- a}) .

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