 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/32e/b/i/solution

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

The graph of

F

is the symmetric tent with vertices

(- 1, 1 - μ), (0, 1), (1, 1 - μ) .

0 < μ < 1

, then

F \geq 0

and

F^{2} (x) = 1 - μ + μ^{2} ∣ x ∣,

F^{2}

is an upward-pointing V with its minimum at

(0, 1 - μ)

and equal endpoint values

1 - μ + μ^{2}

1 < μ < 2

, the zeros of

F

are

\pm 1/ μ

and

F^{2} (x) = {1 - μ + μ^{2} ∣ x ∣, 1 + μ - μ^{2} ∣ x ∣, ∣ x ∣ \leq 1/ μ, 1/ μ \leq ∣ x ∣ \leq 1.

Thus its five vertices are

(- 1, 1 + μ - μ^{2}), (- 1/ μ, 1), (0, 1 - μ), (1/ μ, 1), (1, 1 + μ - μ^{2}) .

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