Zadanie 2f, Paper 3, maj 2024, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2024, TZ2, Paper 3 (zadania wzorowane), pytanie 2(f). 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 random quadratic is generated as q(x) = ax2 + bx + c. Choose a and c independently and uniformly from {1,2,3,4}, and choose b independently and uniformly from {1,2,3,4,5,6,7,8}. Ordered coefficient choices are distinct and equally likely. Parts (a)–(f) concern this discrete generator; the separate continuous model is specified explicitly in (g) and (h).
Polecenie
Find the probability p that a generated quadratic has two distinct x-intercepts. Show how you count the possibilities.
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Odpowiedź z klucza
Group by ac. For products 1,2,3,4,6,8,9,12 the ordered-pair counts are 1,2,2,3,2,2,1,2. The respective numbers of b satisfying b²>4ac are 6,6,5,4,4,3,2,2. The resulting triple counts are 6,12,10,12,8,6,2,4, totaling 60. Thus p=60/128=15/32=0.46875. Alternatively group by b=1,…,8: the favorable counts are 0,0,3,5,10,12,15,15, also totaling 60. A complement enumeration gives the same result after excluding the six repeated-root cases.
Schemat punktowania
M1: systematically vary ac or b and enforce a positive discriminant. For an ac table beyond products 1,2: A1 for one correct case among 3,4,6,8,9,12; A1 for two further cases; A1 for the remaining three. M1: sum disjoint counts/probabilities, with at least three groups included. A1: 15/32 or 0.469 with supporting count. For a by-b table, award the first accuracy mark for one correct nonzero case, the second for two further cases, and the third for the remaining three; equivalent complete complement counting is valid.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Retain exact intermediate expressions; give three significant figures for numerical probabilities.