Zadanie 2f, Paper 3, maj 2026, TZ3, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2026, TZ3, Paper 3 (zadania wzorowane), pytanie 2(f). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
A scanning head moves on a straight guide, with displacement measured from its central position O. One test run has metres for 0 ≤ t ≤ 2π/3 seconds. The general model is , where t≥0, A,n>0 and −π≤β≤π. Write v=ds/dt and a=d²s/dt². Restoring acceleration is modelled by a=−n²s. For an observation taken at a uniformly random phase, the displacement random variable S has density for −A<s<A and zero otherwise. This density concerns random observation times, not a change to the deterministic motion law.
Polecenie
(i) Find E(S), with justification. (ii) Interpret this value in the scanning-head model.
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Odpowiedź z klucza
(i) E(S)=0. The density is even on the symmetric support (−A,A), so s f_S(s) is odd and its integral is zero; equivalently paired opposite displacements have equal probability. (ii) The mean observed position is the central position O. This does not mean the head remains at O or spends all its time there.
Schemat punktowania
(i) A1: E(S)=0. R1: valid density symmetry or odd-integrand justification; do not award this reason mark with an incorrect expectation. (ii) A1: O is the mean/central position of the motion. An interpretation already supplied in (i) may earn this mark. As explicitly accepted in the source, average velocity zero over complete cycles is also an acceptable contextual interpretation in (ii); merely naming harmonic motion is not enough.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Supply complete proofs and sufficient reasoning; numerical evidence alone does not establish a general identity.