Cambridge IGCSE Mathematics 0580 — 2010 Oct/Nov Paper 4 · Variant 1

8 (a) f(x) = 2x – 1 g(x) = x2 For Examiner's Work out Use (i) f(2), Answer(a)(i) [1] (ii) g( – 2), Answer(a)(ii) [1] (iii) ff(x) in its simplest form, Answer(a)(iii) ff(x) = [2] (iv) f –1(x), the inverse of f(x), Answer(a)(iv) f –1(x) = [2] (v) x when gf(x) = 4. Answer(a)(v) x = or x = [4] (b) y is inversely proportional to x and y = 8 when x = 2. Find, (i) an equation connecting y and x, Answer(b)(i) [2] 1 (ii) y when x = . 2 Answer(b)(ii) y = [1]

Mark scheme: 8 (a) (i) 3 1 (ii) 4 1 (iii) 4x – 3 final answer 2 M1 for 2(2x – 1) – 1 x + 1 y + 1 f ( x ) + 1 (iv) oe final answer 2 M1 for x = 2y – 1 or oe or oe 2 2 2 1 1 (v) – and 1 4 B1 for (2x – 1)2 soi 2 2 M2 for 2x – 1 = ± 2 M1 for 4x2 – 2x – 2x + 1 or M1 for 2x – 1 = 2 and M1 for (2x + 1)(2x – 3) or correct substitution in formula soi by (4 ± √64)/8 16 (b) (i) y = oe 2 Condone y = k/x and k = 16 stated x k M1 for y = oe x (ii) 32 1

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9 (a) The first five terms P1, P2, P3, P4 and P5 of a sequence are given below. For Examiner's 1 = 1 = P1 Use 1 + 2 = 3 = P2 1 + 2 + 3 = 6 = P3 1 + 2 + 3 + 4 = 10 = P4 1 + 2 + 3 + 4 + 5 = 15 = P5 (i) Write down the next term, P6, in the sequence 1, 3, 6, 10, 15… Answer(a)(i) [1] (ii) The formula for the nth term of this sequence is 1 Pn = n(n + 1). 2 Show this formula is true when n = 6. Answer (a)(ii) [1] (iii) Use the formula to find P50, the 50th term of this sequence. Answer(a)(iii) [1] (iv) Use your answer to part (iii) to find 3 + 6 + 9 + 12 +15 + ………… + 150. Answer(a)(iv) [1] (v) Find 1 + 2 + 3 + 4 + 5 +………… + 150. Answer(a)(v) [1] (vi) Use your answers to parts (iv) and (v) to find the sum of the numbers less than 150 which are not multiples of 3. Answer(a)(vi) [1] This question continues on the next page. (b) The first five terms, S1, S2, S3, S4 and S5 of a different sequence are given below. For Examiner's (1 × 1) = 1 = S1 Use (1 × 2) + (2 × 1) = 4 = S2 (1 × 3) + (2 × 2) + (3 × 1) = 10 = S3 (1 × 4) + (2 × 3) + (3 × 2) + (4 × 1) = 20 = S4 (1 × 5) + (2 × 4) + (3 × 3) + (4 × 2) + (5 × 1) = 35 = S5 (i) Work out the next term, S6, in the sequence 1, 4, 10, 20, 35… Answer(b)(i) [2] (ii) The formula for the nth term of this sequence is 1 Sn = n(n + 1)(n + 2). 6 Show this formula is true for n = 6. Answer(b)(ii) [1] (iii) Find (1 × 20) + (2 × 19) + (3 × 18) ………… + (20 × 1) . Answer(b)(iii) [1] (c) Show that S6 – S5 = P6, where P6 is your answer to part (a)(i). Answer(c) [1] 1 (d) Show by algebra that Sn – Sn – 1 = Pn . [Pn = n(n + 1)] 2 Answer(d) [3]

Mark scheme: 9 (a) (i) 21 1 (ii) P6 = ½ × 6 × 7 or better (= 21) 1 Allow 3(6 + 1) (iii) 1275 1 (iv) 3825 1ft ft for 3 × their (iii) (v) 11325 1 (vi) 7500 1ft ft their (v) – their (iv) provided > 0 (b) (i) 56 2 M1 for 1 × 6 + 2 × 5 + 3 × 4 + 4 × 3 + 5 × 2 + 6 × 1 1 (ii) S6 = × 6 × 7 × 8 or better (= 56) 1 6 (iii) 1540 1 (c) 56 – 35 = 21 1 1 1 (d) Correct algebraic proof with no errors 3 M1 for n(n + 1)(n + 2) – (n – 1)(n)(n + 1) oe 6 6 1 and M1 for n(n + 1)(3) oe 6

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