Zadanie 2a, Paper 3, listopad 2023, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2023, Paper 3 (zadania wzorowane), pytanie 2(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
This investigation concerns intersecting curve families. In 2a, take parabolas y=kx², where k≠0; work locally where x,y≠0. In 2b–2d use Ea : x2/(a2 + 9) + y2/a2 = 1 for a>0 and Hb : x2/(9 − b2) − y2/b2 = 1 for 0<b<3. In 2e–2f a tangent of finite slope f is rotated anticlockwise through an angle α; where its new slope is finite, .
Polecenie
(i) Express the slope of y=kx² using only x,y. (ii) Hence show its orthogonal trajectories satisfy y′=−x/(2y). (iii) Solve this differential equation to show that these trajectories have equation x²+2y²=C, with constant C.
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Odpowiedź z klucza
(i) y′=2kx=2y/x. (ii) Perpendicular tangent slopes are negative reciprocals, giving y′=−x/(2y). (iii) Separate: y dy=−x dx/2. Integrate y²/2=−x²/4+K, hence x²+2y²=C. Nondegenerate real trajectories have C>0; the derivation applies on local arcs away from x,y=0.
Schemat punktowania
(i) A1:2y/x. (ii) A1:negative reciprocal −x/(2y), with no follow-through from an incorrect original slope. (iii) M1:separate variables. A1:correct integration with a constant yielding the target family. For (ii), a statement that the required gradient is the negative reciprocal of 2y/x, or a correct product-of-slopes equation, suffices for A1; the final displayed slope is the supplied target.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Exact claims and proofs require analytical working. Counts of intercepts mean distinct real points, unless repeated roots are explicitly requested.