 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-4/1c/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

f (a) = f (a^{'})

, applying

g

gives

(g \circ f) (a) = (g \circ f) (a^{'})

g \circ f

implies

a = a^{'}

, so

f

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