 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/15e/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Setting

a (t_{0}) = 1

gives

t_{0} = \frac{2 ( β + γ - β )}{γ} = \frac{2}{β + γ + β} .

At the present epoch,

H_{0} = \overset{a}{˙} (t_{0}) = β + γ,

t_{0} H_{0} = \frac{2 β + γ}{β + γ + β} .

For

β ≫ γ

, this tends to

1

t_{0} \sim H_{0}^{- 1}

. For

γ ≫ β

, it tends to

2

t_{0} \sim 2 H_{0}^{- 1}

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