Zadanie 2e, Paper 3, maj 2022, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2022, TZ2, Paper 3 (zadania wzorowane), pytanie 2(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Polecenie
Consider Fc(x) = x3 + cx2 − 6x + 1, where c is real. You may use the sufficient condition: a real cubic x³+px²+qx+r, r≠0, has a nonreal root if q²<3pr. A student claims that q²≥3pr guarantees all roots are real. (i) By varying the positive integer c, find the smallest c that disproves this claim. (ii) Explain why F_c has at least one real root for every real c.
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Odpowiedź z klucza
(i) A GDC polynomial-root table for c=1,…,8 shows three real roots in each case. At c=9, roots are approximately −9.633595563 and.3167977814±.05867340475i. Yet q²=36≥27=3pr, so c=9 is the smallest positive-integer counterexample. To certify the search, the reciprocal cubics z³−6z²+cz+1 have discriminants 761,733,729,725,697,621,473,229 for c1…8 and −135 for c9. A numerical root table is sufficient; the discriminant formula is not required. (ii) The real cubic has opposite unbounded signs at ±∞ and is continuous, so it crosses the x-axis; alternatively, nonreal roots occur in conjugate pairs.
Schemat punktowania
(i) M1:systematic GDC root/graph/table search. A1:c=9 satisfying the reversed inequality while retaining a nonreal pair. (ii) R1:odd-degree sign behavior and continuity, or conjugate pairing.
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.