Cambridge IGCSE Mathematics 0580 — 2013 Oct/Nov Paper 2 · Variant 3
0580/23/O/N/13 · 3 questions · 20 marks · ≈23 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper12 pages
Mark scheme4 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q17 · (p, q) is the image of the point (x, y) under this combined transformation
17 (p, q) is the image of the point (x, y) under this combined transformation. For Examiner′s Use p -1 0 x 3 = + 2 q y f p f 0 1 p f p f p (a) Draw the image of the triangle under the combined transformation. y 4 3 2 1 x 0 –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 [3] -1 0 (b) Describe fully the single transformation represented by . f 0 1 p Answer (b) ................................................................................................................................ [2] _____________________________________________________________________________________
Mark scheme: 17 (a) triangle at (0, 3) (2, 3) and (2, 4) 3 B1 for each correct vertex If 0 scored then M1 for correct reflection in the y axis or correct translation of their first stage 3 right 2 up (b) reflection in y axis 2 B1 for reflection B1 for y axis or x = 0
Q18 · A gardener measured the lengths of 50 green beans from his garden
18 A gardener measured the lengths of 50 green beans from his garden. For Examiner′s The results have been used to draw this cumulative frequency diagram. Use 50 40 30 Cumulative frequency 20 10 0 5 10 15 20 25 30 35 Length of green bean (cm) Work out (a) the median, Answer(a) ......................................... cm [1] (b) the number of green beans that are longer than 26 cm, Answer(b) ............................................... [2] (c) the inter-quartile range, Answer(c) ......................................... cm [2] (d) the probability that a green bean chosen at random is more than 14 cm long. Answer(d) ............................................... [2] _____________________________________________________________________________________ Question 19 is printed on the next page.
Mark scheme: 18 (a) 19–19.1 1 (b) 3 2 M1 for 47 seen (c) 4.9 to 5.7 2 B1 for [UQ] 21.7 to 22.2 and [LQ] 16.5 to 16.8 45 2 B1 for 45 seen or (d) oe 50 5 SC1 for isw 50
Q19 · F(x) = 2x + 3 g(x) = x2 For Examiner′s Use (a) Find fg(6)
19 f(x) = 2x + 3 g(x) = x2 For Examiner′s Use (a) Find fg(6). Answer(a) ............................................... [2] (b) Solve the equation gf(x) = 100. Answer(b) x = .................... or x = .................... [3] (c) Find f –1(x). Answer(c) f –1(x) = ............................................... [2] (d) Find ff –1(5). Answer(d) ............................................... [1]
Mark scheme: 19 (a) 75 2 B1 for [g(6) =] 36 (b) 3.5 –6.5 3 M1 for (2x + 3)2 = 100 M1 for 2x + 3 = [±]10 If 0 scored, SC1 for one correct value as answer x − 3 3 y (c) oe final answer 2 M1 for x = 2y + 3 or y – 3 = 2x or = x + 2 2 2 or better (d) 5 1
What was in this paper
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