Zadanie 6, Paper 1, maj 2026, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2026, TZ1, Paper 1 (zadania wzorowane), pytanie 6. Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Polecenie
A circular filter of radius r cm has a concentric coated disk of radius 4 cm, with r>4. A wedge of angle π/4 is removed from the entire filter, including its coating. In the remaining filter the uncoated area is 5/2 times the coated area. Find r in the form for integer k.
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Odpowiedź z klucza
The remaining angle is 7π/4. Its uncoated area is and coated area . Their ratio gives (r2 − 16)/16 = 5/2, so r²=56 and cm. Equivalently the coated part is 2/7 of the total remaining area.
Schemat punktowania
M1: express the uncoated area as the difference of two sectors, permitting an attempted common angle. M1, dependent on the first: impose the ratio 5/2. A1: correct equation with consistent sector angles, or the simplified ratio (r²−16)/16=5/2. A1: √56. Alternative: M1 identify coating as 2/7 of the total, dependent M1 express that area proportion with a common angle, A1 correct equation, A1 √56.
Komentarz. Paper 1: no calculator. Give exact results and complete reasoning. The AA HL formula booklet is available. Equivalent valid methods are accepted unless a particular method is requested.