 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/11h/e/solution

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

Conversely, suppose a continuous

f : I \to {- 1, 1}

satisfies

f (- 1) = - 1

and

f (1) = 1

g = - f : I \to I

is continuous, so part (d) gives

x = g (x) = - f (x)

. If

f (x) = 1

x = - 1

, contradicting

f (- 1) = - 1

; if

f (x) = - 1

x = 1

, contradicting

f (1) = 1

. Hence no such

f

exists.

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