Cambridge IGCSE Mathematics 0580 — 2011 Oct/Nov Paper 2 · Variant 1

11 Work out. 2  2 1  (a)    4 3      Answer(a)   [2]   −1  2 1  (b)    4 3      Answer(b)   [2]  

Mark scheme:   11 (a)   2 B1 two or three entries correct  20 13   1 12 − 12  1  a c   3 − 1 oe 2 B1 (b)     B1(k ) b d  2  − 2 1      − 4 2

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12 For Examiner's 50 Use 40 30mark test 20English 10 0 10 20 30 40 50 60 70 80 Mathematics test mark The scatter diagram shows the marks obtained in a Mathematics test and the marks obtained in an English test by 15 students. (a) Describe the correlation. Answer(a) [1] (b) The mean for the Mathematics test is 47.3 . The mean for the English test is 30.3 . Plot the mean point (47.3, 30.3) on the scatter diagram above. [1] (c) (i) Draw the line of best fit on the diagram above. [1] (ii) One student missed the English test. She received 45 marks in the Mathematics test. Use your line to estimate the mark she might have gained in the English test. Answer(c)(ii) [1]

Mark scheme: 12 (a) Negative 1 Ignore embellishments (b) Correct point 1 (c) (i) Accurate ruled line 1 (ii) English mark 1ft Follow through their (c)(i) 1 1

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13 For A C B D Examiner's Use a b O A and B have position vectors a and b relative to the origin O. C is the midpoint of AB and B is the midpoint of AD. Find, in terms of a and b, in their simplest form (a) the position vector of C, Answer(a) [2] (b) the vector . Answer(b) [2]

Mark scheme: 1 1 13 (a) a + b oe 2 M1 unsimplified or any correct route 2 2 1 e.g a + (b – a) or OA + AC 2 1 1 (b) –1 a + 1 b oe 2 M1 unsimplified or any correct route 2 2 1 1 e.g. CD = 1 AB or b – a + (b – a) 2 2

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l 14 T = 2π g (a) Find T when g = 9.8 and ℓ = 2. Answer(a) T = [2] (b) Make g the subject of the formula. Answer(b) g = [3]

Mark scheme: 14 (a) 2.84 2 M1 correct substitution of g and l seen 4 π 2 l (b) oe 3 M1 each correct move but third move marked on T 2 answer line

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15 A container ship travelled at 14 km/h for 8 hours and then slowed down to 9 km/h over a period of For 30 minutes. Examiner's Use It travelled at this speed for another 4 hours and then slowed to a stop over 30 minutes. The speed-time graph shows this voyage. 16 14 12 10 Speed 8 (km / h) 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Time (hours) (a) Calculate the total distance travelled by the ship. Answer(a) km [4] (b) Calculate the average speed of the ship for the whole voyage. Answer(b) km/h [1]

Mark scheme: 15 (a) 156 4 M1 intention to find area under graph B2 completely correct area statement or B1 two areas found correctly (or one trapezium area) (b) 12 1ft Their (a)/13

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16 For y Examiner's Use C (9,7) NOT TO SCALE A (1,3) x O B (3,0) The co-ordinates of A, B and C are shown on the diagram, which is not to scale. (a) Find the length of the line AB. Answer(a) AB = [3] (b) Find the equation of the line AC. Answer(b) [3]

Mark scheme: 16 (a) 3.61 3 M1 (3 – 1)2 + (0 – 3)2 oe M1 2 2 + 3 2 1 1 1 1 (b) y = x + 2 oe 3 B2 y = x + k or y = kx + 2 2 2 2 2 1 1 or B1 kx + 2 or x + k 2 2 1 If 0 scored B1 m = 2 1 B1 c = 2 clearly identified in working 2 IGCSE – October/November 2011 0580 21 1

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1 For 17 f(x) = (x ≠ O4) Examiner's x + 4 Use g(x) = x2 – 3x h(x) = x3 + 1 (a) Work out fg(1). Answer(a) [2] (b) Find hO1(x). Answer(b) h O1(x) = [2] (c) Solve the equation g(x) = O2. Answer(c) x = or x = [3] Question 18 is printed on the next page.

Mark scheme: 1 17 (a) 2 B1 f(–2) seen 2 (b) 3√(x – 1) or 3 x − 1 2 M1 x –1 = y3 or 3√(y – 1) (c) 1 2 3 M2 (x – 1)(x – 2) = 0 or M1 (x + a)(x + b) = 0 where ab = 2 or a + b = –3 If 0 scored give M1 for x2 – 3x + 2 = 0 1

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18 The first four terms of a sequence are For Examiner's T1 = 12 T2 = 12 + 22 T3 = 12 + 22 + 32 T4 = 12 + 22 + 32 + 42 . Use 1 (a) The nth term is given by Tn = n(n + 1)(2n + 1). 6 Work out the value of T23. Answer(a) T23 = [2] (b) A new sequence is formed as follows. U1 = T2 – T1 U2 = T3 – T2 U3 = T4 – T3 ……. (i) Find the values of U1 and U2. Answer(b)(i) U1 = and U2 = [2] (ii) Write down a formula for the nth term, Un . Answer(b)(ii) Un = [1] (c) The first four terms of another sequence are V1 = 22 V2 = 22 + 42 V3 = 22 + 42 + 62 V4 = 22 + 42 + 62 + 82. By comparing this sequence with the one in part (a), find a formula for the nth term, Vn . Answer(c) Vn = [2]

Mark scheme: 1 18 (a) 4324 cao 2 M1 × 23 × 24 × 47 or better 6 (b) (i) 4, 9 2 B1 either correct (ii) (n + 1)2 or n2 + 2n + 1 1 2 (c) n(n+ 1)(2n + 1) oe 2 M1 recognising Vn = 4Tn 3

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