Zadanie 11c, Paper 2, listopad 2025, TZ3, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2025, TZ3, Paper 2 (zadania wzorowane), pytanie 11(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
The usable lifetime X, in days, of a prototype coating is modelled by density f(x) = a(x − 2)3 + b(x − 2) for 2≤x≤7 and f(x)=0 otherwise. The real constants a,b are nonzero, and f(7)=0.
Polecenie
Show that the median lifetime is smaller than the mode.
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The coefficients from normalization and f(7)=0 are a=−4/625,b=4/25. Put u=x−2. Differentiating 4u(25−u²)/625 gives 4(25−3u²)/625, so the mode is 2+5/√3=4.886751346…. The cumulative probability there is 2u²/25−u⁴/625=5/9>.5. Since the density is positive inside the support, the median is smaller than the mode. Alternatively solving F(m)=.5 gives median 2+5√(1−1/√2)=4.705980501….
Schemat punktowania
M1: locate the density maximum. A1:mode4.886751… days. M1: evaluate the cumulative probability at the mode or solve the median equation. A1:F(mode)=5/9, or correct median equation/value. R1: conclude median <mode from 5/9>.5 or the correctly identified median 4.70598…<4.88675…. In the median-solving route, award the final A1 for a correct numerical median clearly identified as the median; a further written comparison is not required because the target comparison is given. A correct unidentified numerical value alone does not earn that last mark.
Komentarz. Paper 2: calculator and AA HL formula booklet allowed. Give exact answers or three significant figures unless directed otherwise. Retain unrounded values during calculation.