Zadanie 2e, Paper 3, listopad 2024, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2024, Paper 3 (zadania wzorowane), pytanie 2(e). 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 nonzero polynomial is palindromic when its coefficients read identically in ascending and descending powers. If its degree is n, this is equivalent to p(x) = xnp(1/x) for x≠0. In this investigation, polynomials have real coefficients and nonzero constant terms. One polynomial to be studied is F(x) = x8 + x6 − 4x4 + x2 + 1.
Polecenie
Prove that the product of two palindromic polynomials is palindromic.
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Let the nonzero polynomials p,q have degrees n,m, and put f=pq. Their identities are p(x) = xnp(1/x) and q(x) = xmq(1/x). Multiplying, f(x) = xn+mp(1/x)q(1/x) = xn+mf(1/x) for x≠0. Since f has degree n+m, this establishes its palindromic property.
Schemat punktowania
A1: both reciprocal identities with their degrees. M1: substitute them into the product. A1: combine powers and factors correctly. A1: identify the degree n+m and the required identity. An explicit coefficient argument proving the same property earns equivalent credit. The reverse-identity route f(1/x)=p(x)q(x)/x^(n+m) earns the same pattern. For coefficient multiplication, A1 for two general palindromic polynomials of degrees n,m with distinct coefficient symbols; M1 for multiplying them; A1 for correct corresponding leading terms; A1 for matching reversed trailing terms and general symmetric coefficient pairing.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Give exact forms unless an approximation is requested.