Cambridge IGCSE Mathematics 0580 — 2012 May/June Paper 4 · Variant 3

1 For 9 f(x) = 1 – 2x g(x) = , x ≠ 0 h(x) = x3 + 1 Examiner's x Use (a) Find the value of (i) gf(2), Answer(a)(i) [2] (ii) h(–2). Answer(a)(ii) [1] (b) Find fg(x). Write your answer as a single fraction. Answer(b) fg(x) = [2] (c) Find h –1(x) , the inverse of h(x). Answer(c) h –1(x) = [2] (d) Write down which of these sketches shows the graph of each of y = f(x), y = g(x) and y = h(x). For Examiner's Use y y y 0 x 0 x 0 x Graph A Graph B Graph C y y y 0 x 0 x 0 x Graph D Graph E Graph F Answer(d) y = f(x) Graph y = g(x) Graph y = h(x) Graph [3] (e) k(x) = x5 O=3 Solve the equation k –1 (x) = 2. Answer(e) x = [2]

Mark scheme: 9 (a) (i) 2 B1 for f(2) = –3 soi − 1 3 oe (ii) –7 1 x − 2 2 (b) 2 M1 for 1− seen x final answer www x (c) y – 1 = x3 or x = y3 + 1 x = 3 y − 1 or x − 1 = y 3 M1 i.e. two correct steps For M1, accept a correct reverse flowchart 3 x − 1 oe final answer www2 A1 After 0 scored allow SC1 for 3 x − 1 seen then spoilt (d) A, F, D 3 B1 each 5 x + 3 = 2 (Variable can be (e) 29 2 M1 for x = k(2) or y in second method)

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10 (a) Rice costs $x per kilogram. For Potatoes cost $(x + 1) per kilogram. Examiner's The total cost of 12 kg of rice and 7 kg of potatoes is $31.70 . Use Find the cost of 1 kg of rice. Answer(a) $ [3] (b) The cost of a small bottle of juice is $y. The cost of a large bottle of juice is $(y + 1). When Catriona spends $36 on small bottles only, she receives 25 more bottles than when she spends $36 on large bottles only. (i) Show that 25y2 + 25y O 36 = 0 . Answer(b)(i) [3] (ii) Factorise 25y2 + 25y O 36 . Answer(b)(ii) [2] (iii) Solve the equation 25y2 + 25y O 36 = 0 . Answer(b)(iii) y = or y = [1] (iv) Find the total cost of 1 small bottle of juice and 1 large bottle of juice. Answer(b)(iv) $ [1]

Mark scheme: 10 (a) 1.3[0] 3 M2 for (31.7[0] – 7) ÷ (12 + 7) or better Or M1 for 12x + 7(x + 1) = 31.7[0] or better or 31.7[0] – 7 or better) (b) (i) 36 36 36 36 = 25 oe M2 SC1 for oe or oe seen y −y + 1 y y + 1 36 ( y + )1 − 36 y = 25 y ( y + )1 oe Accept both all over y(y + 1) 2 Must see at least one of these lines before E mark 36 y + 36 − 36 y = 25 y + 25 y oe 25 y 2 + 25 y − 36 = 0 E1 Final line reached without any errors or omissions (ii) (5 y + 9 )(5 y − 4 ) 2 Accept (25y – 20)(y + 1.8) oe SC1 for (5 y + m )(5 y + n ) where mn = − 36 or m + n = 5 (iii) –1.8 oe, 0.8 oe 1ft ft only SC1 from (b)(ii) (iv) 2.6[0] 1ft ft 2 × positive root from (b)(iii) +1 Dep on pos and neg root in (b)(iii) IGCSE – May/June 2012 0580 43

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11 For Examiner's Use Diagram 1 Diagram 2 Diagram 3 The diagrams show a sequence of dots and circles. Each diagram has one dot at the centre and 8 dots on each circle. The radius of the first circle is 1 unit. The radius of each new circle is 1 unit greater than the radius of the previous circle. (a) Complete the table for diagrams 4 and 5. Diagram 1 2 3 4 5 Number of dots 9 17 25 Area of the largest circle π 4π 9π Total length of the circumferences of the circles 2π 6π 12π [4] (b) (i) Write down, in terms of n, the number of dots in diagram n. Answer(b)(i) [2] (ii) Find n, when the number of dots in diagram n is 1097. Answer(b)(ii) n = [2] (c) Write down, in terms of n and π, the area of the largest circle in (i) diagram n, Answer(c)(i) [1] (ii) diagram 3n. Answer(c)(ii) [1] (d) Find, in terms of n and π, the total length of the circumferences of the circles in diagram n. Answer(d) [2]

Mark scheme: 11 (a) 33, 41 1 16 ,π 25 π 1 2 B1 each 20 ,π30 π (b) (i) 8 n + 1 oe final answer 2 e.g. 9 + 8(n –1), condone n = 8n + 1 SC1 for 8 n + k (ii) 137 www2 2 M1 for their (b)(i) = 1097 (c) (i) n 2 π oe final answer 1 (ii) 9 n 2 π oe final answer 1 Allow (3n ) 2 π (d) n ( n + )1 π oe final answer 2 SC1 for a quadratic expression e.g. n ( n + )1 , n2 + 5, n2 + n π

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