 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/12/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Let

D = 1 + x^{2} + y^{2}

. The gradient has components

f_{x} = [D - 2 x (x + y)] / D^{2}

and

f_{y} = [D - 2 y (x + y)] / D^{2}

(x - y) (x + y) = 0

. The case

x + y = 0

cannot make either numerator zero, so

x = y = t

. The remaining equation is

1 + 2 t^{2} = 4 t^{2}

(x, y) = (1/ 2, 1/ 2), f = 1/ 2; (x, y) = (- 1/ 2, - 1/ 2), f = - 1/ 2 .

These are the only stationary points; the denominator is positive everywhere.

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