Zadanie 2h, Paper 3, listopad 2024, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2024, Paper 3 (zadania wzorowane), pytanie 2(h). 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 every monic palindromic polynomial of odd degree has −1 as a root.
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Odpowiedź z klucza
For odd degree n, evaluate p(x) = xnp(1/x) at x=−1. This gives p(−1) = (−1)np(−1) = −p(−1), since n is odd. Consequently 2p(−1)=0 and p(−1)=0, which proves that −1 is a root. Alternatively, coefficients of paired powers cancel at −1.
Schemat punktowania
M1: use the reciprocal identity at −1 or evaluate paired terms. A1: correct substituted identity. A1: use odd n to obtain p(−1)=−p(−1). R1: deduce p(−1)=0 and hence the root. A coefficient-pair cancellation proof earns the corresponding marks. Product-of-roots proof: M1 for considering Vieta product; A1 for product (−1)^n=−1; A1 for pairing non-self-reciprocal roots with their reciprocals, counting multiplicities; R1 because each pair has product 1 and the remaining self-reciprocal roots are ±1, so product −1 forces at least one −1.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Give exact forms unless an approximation is requested.