Zadanie 1g, Paper 3, listopad 2024, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2024, Paper 3 (zadania wzorowane), pytanie 1(g). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
A digitization team begins year 1 with 4800 pages awaiting processing. During each year, incoming revisions increase the outstanding work by 12.5%, after which a fixed quota is processed at year-end. Treat page counts as continuous estimates until rounding is requested. With quota 400, T_n denotes the work at the start of year n, so T1 = 4800 and . With quota 750, use D_n instead. For either policy, a non-positive algebraic prediction means the queue has already cleared; actual work cannot be negative.
Polecenie
Consider the general recurrence un+1 = run − d, with r,d>0 and r≠1. Prove by induction that for every integer n≥1.
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For n=1 the right side is u1 − d(1 − 1)/(r − 1) = u1, so the base case holds. Assume the formula for n=k. Then uk+1 = r[u1rk−1 − d(rk−1 − 1)/(r − 1)] − d = u1rk − d[r(rk−1 − 1) + (r − 1)]/(r − 1) = u1rk − d(rk − 1)/(r − 1). This is the claimed formula for k+1. The verified base case and implication establish it for every n≥1.
Schemat punktowania
R1: verify the base case n=1; n=2 alone does not earn this mark. M1: explicitly assume truth at n=k, not merely write let n=k. M1: substitute this hypothesis into the recurrence. A1: correctly multiply the expression by r, including the quota term. A1: correctly combine the quota terms over r−1. A1: obtain the formula at k+1. Final R1: appropriate induction conclusion; at least four of the preceding six marks must have been awarded. Marks following the base-case and hypothesis marks are independent of those individual marks, as in the source.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Give exact forms unless an approximation is requested.