 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-317/3/iii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

4 lo g T = lo g (τ + 2/3) + constant

, gives

\frac{d lo g T}{d τ} = \frac{1}{4 ( τ + 2/3 )} = \frac{3}{8 + 12 τ} .

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