 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/8h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

h_{i} = E_{i} T_{6}

. Then

h_{6} = 0

, and first-step analysis on the closed class gives

h_{1} h_{2} h_{5} = 1 + \frac{1}{4} h_{5}, = 1 + \frac{1}{4} h_{1} + \frac{1}{2} h_{5}, = 1 + \frac{1}{4} h_{1} + \frac{1}{2} h_{2} .

Solving these linear equations gives

h_{5} = \frac{20}{7}, h_{1} = \frac{12}{7} .

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