Zadanie 11d, Paper 1, maj 2025, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2025, TZ2, Paper 1 (zadania wzorowane), pytanie 11(d). 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 local approximation to a cube root is constructed by comparing F(x) = (1 + 6x)1/3 with R(x) = (1 + px)(1 + qx)−1, where p, q are rational numbers.
Polecenie
Using p = 4, q = 2, the rational expression approximates F(x) near zero, with ∣x∣ < 1/6.
(i) Choose a suitable x and use this approximation to estimate as a fraction of integers.
(ii) Explain why the same approximation is less accurate for .
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Odpowiedź z klucza
(i) Set 1 + 6x = 7/6, giving x = 1/36. Then R(1/36) = (1 + 1/9)/(1 + 1/18) = 20/19. (ii) The second input requires x = 1/12, which is farther from the expansion centre 0 than 1/36. The neglected terms are larger here, so the local approximation is less accurate. Indeed for positive x the rational approximation underestimates the cube root and its error increases; no error-bound calculation is required.
Schemat punktowania
(i) M1: set up the input relation. A1: x=1/36. M1: substitute into the rational expression (or the corresponding binomial approximation). A1: 20/19 for the rational version; accept the correctly evaluated polynomial version with working. (ii) A1: x=1/12. R1: compare its distance from zero with 1/36 to explain the poorer local approximation.
Komentarz. Paper 1: no calculator. Show mathematical reasoning. Equivalent exact forms and valid alternative methods are accepted unless a method is specified.