 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/25i/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Again set

h = G

and fix

θ

. For

0 < α < C^{- 1/2}

f (r) = h (r, θ) - α sin (C r) .

Since

h^{''} = - K h

and

K \leq C

f^{''} + C f = (C - K) h \geq 0, f (0) = 0, f^{'} (0) = 1 - α C > 0.

For

0 < r < π / C

, differentiation gives

\frac{d}{d r} (f^{'} sin (C r) - C f cos (C r)) = (f^{''} + C f) sin (C r) \geq 0.

The expression in parentheses vanishes at zero, and hence

(\frac{f}{s i n ( C r )})^{'} \geq 0.

Its limit at zero is

(1 - α C) / C > 0

, so

f > 0

. Letting

α ↑ C^{- 1/2}

h (r, θ) \geq \frac{s i n ( C r )}{C} (0 \leq r < π / C) .

Area B (p, ε) \geq \frac{2 π}{C} (1 - cos (C ε)) .

Now

\frac{2 ( 1 - c o s x )}{x ^{2}} ⟶ 1

x \to 0

. Given

0 < γ < 1

, choose

x_{0} \in (0, π)

such that

2 (1 - cos x) / x^{2} \geq γ

for

0 < x \leq x_{0}

, and set

ε_{0} = x_{0} / C

. Then every geodesic polar coordinate ball of radius

ε \leq ε_{0}

satisfies

Area B (p, ε) \geq γπ ε^{2} .

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