Zadanie 1a
(i) Verify algebraically that C1(n) = n(n − 1) + 1. (ii) The count 421 occurs in this r=1 scheme. Determine its index 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 14 zadań z tego arkusza, razem 55 punktów, każde z kluczem i schematem punktowania.
(i) Verify algebraically that C1(n) = n(n − 1) + 1. (ii) The count 421 occurs in this r=1 scheme. Determine its index n.
(i) Show that C1(n) + C1(n + 1) = 2n2 + 2. (ii) Describe this result in words as a rearrangement of tokens into two equal square arrays and isolated tokens. (iii) For n=3 sketch such a rearrangement, distinguishing tokens from the two counts.
Show that 4C1(n) − 3 is the square of an odd integer for every positive integer n.
For r=17 the counts may be written C17(n) = 1 + 34 + 68 + ⋯ + 34(n − 1), with just 1 at n=1. Use an arithmetic-series sum to show that C17(n) = 17n(n − 1) + 1.
Using tables or another suitable method, find the smallest count greater than 1 that occurs in both the r=1 and r=17 schemes.
A weighted version of the counter scheme has , for positive integers r,n. Prove by mathematical induction that Dr(n) = n(n + 1)[2r(2n + 1) + 3]/6.
(i) Given the nonreal root 2+2i of f(z)=0, state the other nonreal root. (ii) Verify that the mean of these two roots is 2.
Show that y=24x−12 is tangent to y=f(x) at A=(2,36).
Sketch y=f(x) and its tangent at A, marking the tangent’s x-intercept.
Assume a≠r. (i) Show that g′(x) = 2(x − r)[(x − a)2 + b2] + 2(x − r)2(x − a). (ii) Hence, or otherwise, prove that the tangent at A meets the x-axis at T = ((a + r)/2, 0).
For a≠r, use g′(a) = 2(a − r)b2 to deduce that the two nonreal roots are .
The figure concerns g(x) = (x + 1)2[(x − a)2 + 4]. The curve touches the x-axis at R=(−1,0). Its tangent at A=(a,64) crosses the x-axis at T=(1,0). Points C₁,C₂ represent respectively the nonreal roots with positive and negative imaginary parts of g(z)=0, drawn on…
Let g(x) = (x − r)2[(x − a)2 + b2], b>0, with A=(a,g(a)) and R=(r,0). When a≠r the tangent at A has x-intercept T=((a+r)/2,0). Assume (a − r)2 > 2b2, so the quartic has two inflection points. (i) Show that their x-coordinates satisfy…
Let g(x) = (x − r)2[(x − a)2 + b2], b>0, with A=(a,g(a)) and R=(r,0). When a≠r the tangent at A has x-intercept T=((a+r)/2,0). (i) Sketch the curve for a=r=−1 and b=2. (ii) For general a=r, state the coordinates of A and of the curve’s minimum point.
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