 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-1/3e/c/solution

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

Fix

a

g (b) = e^{a + b} e^{- b}

using only the power series. Termwise differentiation gives

(e^{z})' = e^{z}

, and the product rule gives

g' (b) = 0

. Hence

g

is constant, so

g (b) = g (0) = e^{a}

e^{b}

yields

e^{a + b} = e^{a} e^{b}

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