Zadanie 9, Paper 1, maj 2025, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2025, TZ2, Paper 1 (zadania wzorowane), pytanie 9. Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Polecenie
Prove by contradiction that for every real t > 0.
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Odpowiedź z klucza
Assume, for contradiction, that some t > 0 satisfies t + 9/t < 6. Multiplication by the positive t preserves the inequality, giving t2 + 9 < 6t. Thus t2 − 6t + 9 = (t − 3)2 < 0. A real square cannot be negative, a contradiction. Therefore the stated inequality holds for all positive t; equality occurs at t=3.
Schemat punktowania
M1: explicitly assume the strict negation. R1: use t>0 to justify multiplication without reversing the inequality. M1: form a quadratic inequality. A1: t²-6t+9<0. A1: write the left side as (t-3)² or identify its nonnegative minimum. R1: explain the contradiction and conclude. If the assumption incorrectly uses ≤, the first and final marks are unavailable; the other four may still be earned. The algebraic marks require a correct or weak negation at the assumption step. A calculus contradiction proof is also accepted: M1 assumption, M1 differentiation, A1 correct derivative, M1 locate the minimum, A1 minimum 6, R1 contradiction.
Komentarz. Paper 1: no calculator. Show mathematical reasoning. Equivalent exact forms and valid alternative methods are accepted unless a method is specified.