Zadanie 2f, Paper 3, listopad 2024, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2024, Paper 3 (zadania wzorowane), pytanie 2(f). 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
Write F(x) = (x4 + ux2 + 1)(x4 + vx2 + 1), where u,v are integers with u<v. Determine u,v by forming and solving an appropriate system of equations.
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Expansion gives x8 + (u + v)x6 + (uv + 2)x4 + (u + v)x2 + 1. Comparing coefficients gives u+v=1 and uv+2=−4, so uv=−6. Substitute v=1−u to obtain u²−u−6=0, with roots −2 and 3. The condition u<v selects u=−2,v=3.
Schemat punktowania
M1: expand the proposed factors. A1: correct expanded coefficients. M1: compare with F. A1: equations u+v=1 and uv=−6. M1: solve this system. A1: u=−2,v=3 in the required order. Working backwards from calculator roots of F, exact or decimal, earns no marks without attempting to form and solve an appropriate system. Division route: M1 attempt division by x⁴+ux²+1; A1 correct quotient x⁴+(1−u)x²+u²−u−5 or correct remainder; M1 equate quotient to x⁴+vx²+1 or remainder to zero; A1 equations u²−u−6=0 and v=1−u; M1 solve; A1 ordered u=−2,v=3. Evaluation route: M1 consider F(1), A1 (u+2)(v+2)=0; M1 consider F(2), A1 (4u+17)(4v+17)=261; M1 solve the system; A1 ordered values. Listing both ordered assignments loses the final A1.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Give exact forms unless an approximation is requested.