 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-12/4/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

For every real

r \geq 0

lo g (1 + r) = \int_{0}^{r} \frac{d t}{1 + t} \leq r .

Taking

r = ∣ w ∣

proves the required logarithm inequality, even without the lower bound

∣ w ∣ \geq 1/2

. Hence $C_{2} = 1$ works.

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