Zadanie 2b, Paper 3, listopad 2025, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2025, TZ1, Paper 3 (zadania wzorowane), pytanie 2(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
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A regional volunteer network had maintained 1000 active memberships before disruption reduced the count to 125. A recovery programme starts at t=0 years. At t=4 the network has250 active members; at t=8 an audit measures 380. Two models use the long-term level 1000: a logistic model x = L/(1 + Ce−kt), with L,C,k>0, and a Gompertz model dx/dt = axln(1000/x), a>0. Fit each model using the observations at t=0 and t=4; reserve the t=8 audit for comparison. In the adjusted model, active memberships also lapse at a continuous proportional rate 0.01 per year.
Polecenie
For the Gompertz model: (i) state dx/dt at x=1000; (ii) interpret it; (iii) for h(x) = ln(ln 1000 − ln x),0<x<1000, show h′(x) = −1/[xln(1000/x)]; (iv) hence separate variables to show ln x = ln 1000 − Ae−at with A>0; (v) find A as ln p for a positive integer p; (vi) show a=.101 to three significant figures; (vii) predict x at t=8 to the nearest whole member.
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Odpowiedź z klucza
(i) dx/dt=0 because ln1=0. (ii) 1000 is an equilibrium membership count, with no instantaneous net growth. (iii) The chain rule gives h′=[1/(ln1000−lnx)](−1/x)=−1/[x ln(1000/x)]. (iv) Separation gives −dx/[x ln(1000/x)]=−a dt. Integrating:h(x)=−at+D. Exponentiate:ln1000−lnx=eᴰe⁻ᵃᵗ=Ae⁻ᵃᵗ with A=eᴰ>0, proving the target. (v) At 0,ln125=ln1000−A, so A=ln8. (vi) At 4,ln250=ln1000−ln8·e⁻⁴ᵃ, hence e⁻⁴ᵃ=ln4/ln8=2/3. Thus a=(1/4)ln(3/2)=.1013662770…, which rounds to.101. (vii)x(8)=1000exp[−ln8·(2/3)²]=396.850263…→397 members. If the displayed three-significant-figure a=.101 is used consistently, the prediction is 395.77543691…→396 members, also acceptable.
Schemat punktowania
(i) A1:zero. (ii) A1:equilibrium/no change at 1000; a maximum-only assertion is insufficient and no follow-through from a nonzero (i). (iii) M1:chain rule. A1:correct derivative expression proving the target. (iv) M1:separate variables. A1:correct separated equation. A1 each:correct integrated left side and right side with arbitrary constant. A1:exponentiate and rearrange to the positive-A form. (v) A1:correct initial substitution. A1:A=ln8; unsupported decimal alone earns at most one. (vi) A1:correct t=4 equation. A1:exact a or at least four significant figures rounding to.101. (vii) A1:correct t=8 substitution. A1:correct unrounded result. A1:397 whole members, or 396 from the stated rounded rate. In (iii), the equivalent three-factor chain rule earns M1 when two of the three factors are correct; the final A1 still requires the correct derivative.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Show complete reasoning and reproducible numerical work. Give exact results unless a stated approximation is requested.