Zadanie 8b, Paper 1, maj 2022, TZ2, Matematyka AA IB SL
Wzorowane na: Matematyka AA IB SL — maj 2022, TZ2, Paper 1 (zadania wzorowane), pytanie 8(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
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Let f(x) = 4/(x − 4) + 2 for x≠4 and g(x)=x−2 for all real x. Their curves intersect at A=(2,0) and another point B.
Polecenie
A region lies above the x-axis, below g from A to B, and below f from B to the vertical line x=k, where k is an integer greater than 6. A circular hole of radius 1 centred at (5,1) is removed from the region; this disk lies entirely inside it. On integrating, the remaining area is collected as 4ln p + 20 − π, where p is a positive integer:20−π is specifically the contribution from the non-logarithmic terms, and all logarithmic terms are combined into 4ln p. Find k,p.
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The first portion is a triangle with base 4 and height 4, area 8. Removing the disk of area π leaves 8−π in this portion. The second is . Thus remaining area is 4ln((k−4)/2)+2k−4−π. The specified non-log contribution gives 2k−4−π=20−π, equivalently 2k−4=20, hence k=12. Then p=(12−4)/2=4.
Schemat punktowania
R1:split the region at B into intervals[2,6],[6,k]. A1:triangle-minus-disk area 8−π. A1:primitive 4ln(x−4). A1:primitive 2x. M1:substitute bounds 6,k. A1:correct evaluated second area. M1:combine the triangle-minus-disk with the integral area and compare with the stated form. M1:equate 2k−4=20. A1:k=12. A1:p=4. The first three accuracy marks are independent of the region-splitting R1.
Komentarz. Paper 1: no calculator. AA SL formula booklet allowed. Give exact answers. Credit valid alternative methods and follow-through with working unless inconsistent with a supplied result.