 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2008-printed-variant/p/p3/c/solution

Write

q = A + B x

and

ℓ = lo g 3

x^{2} - q

1

and

x

gives

(ℓ 2 24) (A B) = (4 26/3) .

The Gram matrix is positive definite, with determinant

4 (ℓ - 1) > 0

, so the solution is unique. Solving yields

q (x) = - \frac{1}{3 ( lo g 3 - 1 )} + (\frac{13}{6} + \frac{1}{6 ( lo g 3 - 1 )}) x .

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