Zadanie 1a, Paper 3, maj 2024, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2024, TZ2, Paper 3 (zadania wzorowane), pytanie 1(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 seeks pairs of differentiable functions f,g satisfying (fg)′ = 2f′g′. The ordinary product rule still gives (fg)′ = fg′ + gf′. For general derivations assume f>0 and 2g′ − g ≠ 0 on the interval under consideration. Each concrete pair is specified in its own part.
Polecenie
Take f(x) = (5 − x)−3 and g(x) = (x + 1)3 on x<5. (i) Find f′(x). (ii) Show that 2f′(x)g′(x) = 18(x + 1)2/(5 − x)4. (iii) Show that f(x)g′(x) + g(x)f′(x) has the same value.
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Odpowiedź z klucza
(i) Chain rule gives f′ = 3(5 − x)−4. (ii) g′ = 3(x + 1)2, so 2f′g′ = 2 ⋅ 3(5 − x)−4 ⋅ 3(x + 1)2 = 18(x + 1)2/(5 − x)4. (iii) The ordinary product-rule sum is 3(x + 1)2/(5 − x)3 + 3(x + 1)3/(5 − x)4. Put both terms over (5−x)⁴ and factor 3(x+1)²: the remaining bracket is (5−x)+(x+1)=6. The result is the same expression as in (ii).
Schemat punktowania
(i) M1: chain or quotient differentiation. A1: 3/(5−x)^4. (ii) A1: g′=3(x+1)². A1: correctly multiply, including the factor 2, to establish the target. (iii) M1: substitute into the product-rule sum. A1: both correct terms. M1: combine with a common denominator or factor. A1: simplify to the stated common value. Follow-through can earn at most the first three marks of (iii) from earlier derivative errors; direct differentiation of fg is also valid.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Retain exact intermediate expressions; give three significant figures for numerical probabilities.