Zadanie 2b, Paper 3, maj 2023, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2023, TZ2, Paper 3 (zadania wzorowane), pytanie 2(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
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Two arithmetic constraints on polynomials are considered. First, for L(x) = mx + c with m,c nonzero, let r be its root. Call L admissible when r,m,c in that order form an arithmetic progression. Second, let F(x) = ax3 + bx2 + cx with real a≠0 and real b,c. Its stationary abscissas r1, r2 are the real roots of F′(x) = 3ax2 + 2bx + c. Call F admissible when a, r1, b, r2, c in that order form an arithmetic progression. The labels r₁,r₂ refer to this order and need not be increasing.
Polecenie
(i) Show that r=−c/m. (ii) For admissible L, show that L(x) = mx + 2m2/(m − 1). (iii) State the further excluded value of m.
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Odpowiedź z klucza
(i) L(r)=0 gives mr+c=0, so r=−c/m since m≠0. (ii) The arithmetic condition is 2m=r+c. Substitute the root:2m=−c/m+c. Multiply by m to get 2m²=c(m−1). Thus c=2m²/(m−1), giving the stated L. (iii) m≠1: before division, m=1 would require 2=0. The initial m≠0 restriction still applies.
Schemat punktowania
(i) A1:derive from mr+c=0. (ii) M1:use the arithmetic midpoint or equal differences. A1:correct substitution of r. M1:clear the denominator and isolate c. A1:c=2m²/(m−1), establishing the target. (iii) A1:m≠1 in addition to m≠0. Equivalent forward proofs using a common difference or eliminating r first receive the same marks. Numerical verification of one function, or merely working backwards from the supplied result, does not establish the general claim.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Exact results require analytical working; graphs and numerical checks support but do not replace proofs.