 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/36d/c/solution

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

f

tends to

- \infty

both as

r \to 0^{+}

and as

r \to \infty

. Its only positive stationary point is

r_{0} = (\frac{m}{H ^{2}})^{1/3},

and it is a maximum with

f (r_{0}) = 1 - 3 (m H)^{2/3} .

Therefore

f

has two positive simple roots exactly when

m H < \frac{1}{3 3} = \frac{1}{27} .

r_{c} = 1/ (3 H)

f (r_{c}) = \frac{2}{3} - 23 m H > 0.

Since

f

is positive precisely between its two positive roots, this proves

r_{-} < \frac{1}{3 H} < r_{+} .

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