 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-1/9f/c/solution

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

x^{α^{- 1}}

repeatedly gives

f (x) = f (x^{α^{- 1}}) = \dots = f (x^{α^{- n}}) .

Since

x^{α^{- n}} \to 1

, continuity at

1

gives

f (x) = f (1)

. Thus $f$ is constant.

Without continuity the claim is false. The function

f (1) = 1

and

f (x) = 0

for

x \neq = 1

satisfies

f (x) = f (x^{α})

because

x^{α} = 1

for positive

x

exactly when

x = 1

, but it is not constant.

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