Zadanie 2h, Paper 3, maj 2022, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2022, TZ2, Paper 3 (zadania wzorowane), pytanie 2(h). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Polecenie
Consider x4 + 8x3 − 15x2 − 9x − 2 = 0, which has exactly one integer root. You may use two sufficient criteria: a quartic x⁴+px³+qx²+rx+s has a nonreal root if r²<2qs; a cubic x³+px²+qx+r, r≠0, has one if q²<3pr. (i) State what the quartic criterion tells you here. (ii) State the integer root. (iii) Factor into a linear factor and a cubic factor, then use the cubic criterion to prove that the quartic has a nonreal root.
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Odpowiedź z klucza
(i) r²=81 and 2qs=2(−15)(−2)=60. Since81≥60, the sufficient condition is not satisfied and gives no conclusion about reality of all roots. (ii) x=2. (iii) Division gives (x−2)(x³+10x²+5x+1). For the cubic, q²=25<3pr=30, so it has a nonreal root, which is also a root of the quartic.
Schemat punktowania
(i) R1:no conclusion from 81≥60; no follow-through for a false converse. (ii) A1:2. (iii) M1:linear-times-cubic factorization attempt. A1:correct linear factor. A1:correct cubic factor written as a product; if factors are not written as a product award at most one of these accuracy marks. R1:25<30 proves a nonreal cubic and hence quartic root.
Komentarz. Paper 3: graphic display calculator and AA HL formula booklet allowed. Show proof and algebra where requested. Give exact values when requested, otherwise three significant figures. Credit valid alternative methods and appropriate follow-through.