 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-ii/5/a/iii/solution

f (t) = ⎩ ⎨ ⎧ - 3 t, 6 t - 3, 3 - 3 t, 0 \leq t \leq 1/3, 1/3 \leq t \leq 2/3, 2/3 \leq t \leq 1.

The adjacent formulas agree at

1/3

and

2/3

, so the function is continuous, and

f (0) = f (1) = 0

. Any pair of points at distance

2/3

can be written

t, t + 2/3

with

0 \leq t \leq 1/3

. For every such

t

f (t + 2/3) - f (t) = (1 - 3 t) - (- 3 t) = 1.

Thus their values are never equal. The horizontal chord of reciprocal-integer length conclusion does not extend to this distance.

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