 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/10e/c/ii/solution

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

No. At

a = 0

, let

f (x) = 3 x, g (y) = y^{3} .

g

f (0) = 0

, and

(g \circ f) (x) = x,

so the composite has derivative one at zero. But

\frac{f ( h ) - f ( 0 )}{h} = ∣ h ∣^{- 2/3}

is unbounded, so

f

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