 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-2/5a/b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

[(\frac{α}{2} - 1) T_{0} + β] e^{- k (t_{2} - t_{1})} = (α - 1) T_{0} .

Solving for the jump and eliminating

t_{1}

using

e^{k t_{1}} = 2 (α - 1) / (α - 2)

gives

β = (α - 1) T_{0} e^{k (t_{2} - t_{1})} - \frac{α - 2}{2} T_{0} = \frac{α - 2}{2} T_{0} (e^{k t_{2}} - 1) .

This is the unique required temperature increase, and it satisfies the stipulated lower bound. Indeed

t_{2} > t_{1}

implies

β - \frac{α T _{0}}{2} = (α - 1) T_{0} (e^{k (t_{2} - t_{1})} - 1) > 0.

Thus no additional restriction on the prescribed time is needed.

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