Zadanie 1a
(i) Find T0(6). (ii) Write a simplified expression for T0(n).
Zadania wzorowane na pytaniach z tego arkusza: te same umiejętności, format i punktacja, inne dane i kontekst. Treści arkusza IB nie publikujemy.
W bazie 15 zadań z tego arkusza, razem 55 punktów, każde z kluczem i schematem punktowania.
(i) Find T0(6). (ii) Write a simplified expression for T0(n).
Use the arithmetic-sequence sum to show that T1(n) = n2 + 2n.
Find T2(4).
Rectangle k has width 1, spans k − 1 ≤ x ≤ k, and has height (2k + 1)2. A quadratic curve y=f(x) passes through their top-right corners. (i) State f(x). (ii) Compare areas to show . (iii) Use a similar area comparison to show…
Define un = (2n + 1)3 − (2n − 1)3. The pairs for n=1,2,3,4 are (26,26), (98,124), (a,b), (386,728). Find (i) a; (ii) b.
For un = (2n + 1)3 − (2n − 1)3, (i) suggest an expression for ; (ii) prove it using the definition of u.
The two bounds to verify are and . Again un = (2n + 1)3 − (2n − 1)3. (i) Express u_n as a quadratic in v = 2n + 1. (ii) Hence express in terms of T2(n), T1(n), T0(n). (iii) Hence show that…
First take a=6, b=0. Find all possible z when (i) z is real; (ii) z is purely imaginary; (iii) .
Sketch the locus r=6 on an Argand diagram, labelling every axis intercept.
Now let , −π/2 ≤ θ ≤ π/2. (i) State the maximum r. (ii) Find z at that maximum.
Use the coordinate identities to show that every point of , −π/2 ≤ θ ≤ π/2, satisfies (x − 3)2 + y2 = 9.
For (x − 3)2 + y2 = 9, (i) find dy/dx in terms of x and y; (ii) hence or otherwise find all x-values where the tangent gradient is undefined.
Sketch , −π/2 ≤ θ ≤ π/2, on an Argand diagram. Label all axis intercepts.
Finally take , 0 ≤ θ ≤ 2π. Derive wherever the denominator is nonzero.
For this part take , 0 ≤ θ ≤ 2π. The point A on this curve has negative imaginary part and a horizontal tangent. Using 2g, find its θ in [0, 2π].
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