Zadanie 11b, Paper 1, listopad 2022, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2022, Paper 1 (zadania wzorowane), pytanie 11(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
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The set D={1,3,4,5,6,8} supplies symbols for ordered three-symbol codes. Also consider P(x) = x3 − ax2 + bx − c, where a,b,c are distinct members of D.
Polecenie
Suppose x2 − 3x + 2 is a factor of P. (i) Express b in terms of a. (ii) Hence show that a=5,b=8,c=4 is the only permitted assignment. (iii) Factor P completely into linear factors. (iv) Sketch y=P(x), labelling all intercepts.
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Odpowiedź z klucza
(i) Write P = (x2 − 3x + 2)(x − d) = x3 − (d + 3)x2 + (3d + 2)x − 2d. Hence a=d+3,b=3d+2,c=2d and b=3a−7,c=2a−6. (ii) For a=1,3,4,5,6,8 the corresponding b is −4,2,5,8,11,17. Only a=4,5 survive the b-condition, but a=4 gives c=2∉D. Therefore a=5,b=8,c=4, all distinct. (iii)P = (x − 1)(x − 2)2. (iv) A positive leading cubic crosses at (1,0), has a local maximum between 1 and 2, touches at the local minimum (2,0), and crosses the y-axis at (0,−4).
Schemat punktowania
(i) M1:introduce a linear remaining factor. A1:correct expansion. M1:compare coefficients. A1:a=d+3,b=3d+2. M1:eliminate d. A1:b=3a−7. Equivalent factor-theorem equations at 1,2, polynomial division, or Vieta reasoning earn corresponding six marks. (ii) R1:restrict to a=4,5 using b∈D. R1:exclude a=4 using c=2 and verify the surviving distinct values. (iii) A1:(x−1)(x−2)². (iv) A1:positive cubic orientation and y-intercept −4. A1:crossing at 1 with a local maximum between 1 and 2. A1:tangency/local minimum at 2. In (ii), alternatively award R1 for c=2a−6 and R1 for a complete argument from that relation establishing the unique permitted assignment; the order of filtering b and c is not compulsory.
Komentarz. Paper 1: no calculator; AA HL formula booklet allowed. Give exact values and show working.