Zadanie 1a
(i) Show that the normal to H at x=t has slope −t2/9. (ii) Hence show that its equation is y = −(t2/9)x + t3/9 − 9/t.
Matematyka AA IB HL — listopad 2025, TZ1, Paper 3 (zadania wzorowane)
Zadania wzorowane na pytaniach z tego arkusza: te same umiejętności, format i punktacja, inne dane i kontekst. Treści arkusza IB nie publikujemy.
W bazie 10 zadań z tego arkusza, razem 55 punktów, każde z kluczem i schematem punktowania.
Zadanie 1b
(i) Eliminate t to show that every such normal has m<0 and (−m)c2 = 9(m2 − 1)2. (ii) State the set of possible intercepts c.
Zadanie 1c
For which intercepts c is there (i) exactly one distinct normal to H through (0,c); (ii) exactly two distinct normals through (0,c)?
Zadanie 1d
On one pair of axes sketch (i) H; (ii) its two distinct normals passing through (0,1). Mark each point of normal contact. Exact contact coordinates and line equations are not required.
Zadanie 1e
The normal to F at A is y = −t2x + kt3 − k/t. (i) Show that its intersections with F satisfy x2 − k(t − t−3)x − k2/t2 = 0. (ii) Hence, using a sum/product of roots or another valid method, find B.
Zadanie 1f
Prove that angle BCA is a right angle.
Zadanie 2a
For the logistic model: (i) interpret choosing L=1000; (ii) find C; (iii) find k; (iv) predict the membership count at t=8, to the nearest whole member.
Zadanie 2b
For the Gompertz model: (i) state dx/dt at x=1000; (ii) interpret it; (iii) for h(x) = ln(ln 1000 − ln x),0<x<1000, show h′(x) = −1/[xln(1000/x)]; (iv) hence separate variables to show ln x = ln 1000 − Ae−at with A>0; (v) find A as ln p for a positive…
Zadanie 2c
Compare the two predictions with the t=8 audit.
Zadanie 2d
The adjusted model is dx/dt = .101xln(1000/x) − .01x. (i) Starting at x₀=250 when t=4, use Euler’s method with step 0.4 years to estimate the count at t=8, to the nearest whole member. (ii) Comment on this prediction.