Cambridge IGCSE Mathematics (with coursework) 0581 — 2009 Oct/Nov Paper 2 · Variant 2

This document consists of 12 printed pages. IB09 11_0581_22/FP © UCLES 2009 [Turn over *5286560077* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MATHEMATICS 0581/22 Paper 2 (Extended) October/November 2009 1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials: Electronic calculator Geometrical instruments Mathematical tables (optional) Tracing paper (optional) READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. If working is needed for any question it must be shown below that question. Electronic calculators should be used. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For π , use either your calculator value or 3.142. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 70. www.XtremePapers.com

2 © UCLES 2009 0581/22/O/N/09 For Examiner's Use 1 For the diagram above write down (a) the order of rotational symmetry, Answer(a) [1] (b) the number of lines of symmetry. Answer(b) [1] 2 Write down the next two prime numbers after 31. Answer and [2] 3 Use your calculator to find the value of o 2 o 2 o o _ (cos30 ) (sin30 ) 2(sin120 )(cos120 ) . Answer [2] 4 Simplify 5 8 x 3 2 ÷ 1 2 x 5 2 _ . Answer [2]

3 © UCLES 2009 0581/22/O/N/09 [Turn over For Examiner's Use 5 In 1977 the population of China was 9.5 x 108. In 2007 the population of China was 1.322 x 109. Calculate the population in 2007 as a percentage of the population in 1977. Answer %[2] 6 A = 0 1 4 12 − − B = 3 4 0 4 − Calculate the value of 5 |A| + |B|, where |A| and |B| are the determinants of A and B. Answer [2] 7 Shade the region required in each Venn Diagram. A C B B' ∩ (A ∩ C ) A C B B' ∩ (A ∪ C ) [2]

4 © UCLES 2009 0581/22/O/N/09 For Examiner's Use 8 Find the length of the line joining the points A(−2, 10) and B(−8, 2). Answer AB = [2] 9 Solve the simultaneous equations 6x + 18y = 57, 2x – 3y = −8. Answer x = y = [3] 10 The braking distance, d, of a car is directly proportional to the square of its speed, v. When d = 2, v = 5. Find d when v = 40. Answer d = [3]

5 © UCLES 2009 0581/22/O/N/09 [Turn over For Examiner's Use 11 20 cm 20 cm 12 cm 12 cm 12 cm 12 cm 12 cm NOT TO SCALE 12 cm 20 cm 20 cm 20 cm 20 cm Each of the lengths 20 cm and 12 cm is measured correct to the nearest centimetre. Calculate the upper bound for the perimeter of the shape. Answer cm [3] 12 Simplify 16 – 4(3x – 2)2. Answer [3]

6 © UCLES 2009 0581/22/O/N/09 For Examiner's Use 13 Solve the inequality 6(2 − 3x) − 4(1 − 2x)=Y= 0. Answer [3] 14 Zainab borrows $249 from a bank to pay for a new bed. The bank charges compound interest at 1.7 % per month. Calculate how much interest she owes at the end of 3 months. Give your answer correct to 2 decimal places. Answer $ [3] 15 M O R P Q r p O is the origin and OPQR is a parallelogram whose diagonals intersect at M. The vector OP is represented by p and the vector is represented by r. (a) Write down a single vector which is represented by (i) p + r, Answer(a)(i) [1] (ii) 2 1 p – 2 1 r. Answer(a)(ii) [1] (b) On the diagram, mark with a cross (x) and label with the letter S the point with position vector 2 1 p + 4 3 r. [2]

7 © UCLES 2009 0581/22/O/N/09 [Turn over For Examiner's Use 16 85° North North B A C 3 km 3 km NOT TO SCALE A, B and C are three places in a desert. Tom leaves A at 06 40 and takes 30 minutes to walk directly to B, a distance of 3 kilometres. He then takes an hour to walk directly from B to C, also a distance of 3 kilometres. (a) At what time did Tom arrive at C? Answer (a) [1] (b) Calculate his average speed for the whole journey. Answer (b) km/h [2] (c) The bearing of C from A is 085°. Find the bearing of A from C. Answer (c) [1] 17 (a) In 2007, a tourist changed 4000 Chinese Yuan into pounds (£) when the exchange rate was £1 = 15.2978 Chinese Yuan. Calculate the amount he received, giving your answer correct to 2 decimal places. Answer(a) £ [2] (b) In 2006, the exchange rate was £1 = 15.9128 Chinese Yuan. Calculate the percentage decrease in the number of Chinese Yuan for each £1 from 2006 to 2007. Answer(b) % [2]

8 © UCLES 2009 0581/22/O/N/09 For Examiner's Use 18 y = mx + c y = 2x + 8 y x 9 units A B NOT TO SCALE The line y = mx + c is parallel to the line y = 2x + 8. The distance AB is 9 units. Find the value of m and the value of c. Answer m = and c = [4]

9 © UCLES 2009 0581/22/O/N/09 [Turn over For Examiner's Use 19 D E F A O B C 68° NOT TO SCALE Points A, B and C lie on a circle, centre O, with diameter AB. BD, OCE and AF are parallel lines. Angle CBD = 68°. Calculate (a) angle BOC, Answer(a) Angle BOC = [2] (b) angle ACE. Answer(b) Angle ACE = [2]

10 © UCLES 2009 0581/22/O/N/09 For Examiner's Use 20 The number of hours that a group of 80 students spent using a computer in a week was recorded. The results are shown by the cumulative frequency curve. 80 60 40 20 0 10 20 30 40 50 60 70 Cumulative frequency Number of hours Use the cumulative frequency curve to find (a) the median, Answer(a) h [1] (b) the upper quartile, Answer(b) h [1] (c) the interquartile range, Answer(c) h [1] (d) the number of students who spent more than 50 hours using a computer in a week. Answer(d) [2]

11 © UCLES 2009 0581/22/O/N/09 [Turn over For Examiner's Use 21 40 30 20 10 0 10 20 30 40 50 60 Speed (m / s) Time (seconds) car truck The graph shows the speed of a truck and a car over 60 seconds. (a) Calculate the acceleration of the car over the first 45 seconds. Answer(a) m/s2 [2] (b) Calculate the distance travelled by the car while it was travelling faster than the truck. Answer(b) m [3] Question 22 is printed on the next page.

12 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2009 0581/22/O/N/09 For Examiner's Use 22 f(x) = 4x + 1 g(x) = x3 + 1 h(x) = 3 1 2 + x (a) Find the value of gf(0). Answer(a) [2] (b) Find fg(x). Simplify your answer. Answer(b) [2] (c) Find h -1(x). Answer(c) [2]

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2009 question paper for the guidance of teachers 0581 MATHEMATICS 0581/22 Paper 22 (Extended), maximum raw mark 70 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2009 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Page 2 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 22 © UCLES 2009 Qu Answers Mark Part Marks 1 (a) 6 (b) 0 1 1 2 37, 41 2 B1, B1 independent 3 –0.577 or 3 3 − or 3 1 − 2 B1 numerator 0.5 oe or B1 denominator –0.866……or 2 3 − 4 1.25 x4 (or 1 4 1 x4) 2 B1 1.25 B1 x4 5 139 2 M1 1.322 × 109 / 9.5 × 108 (× 100) 6 8 2 M1 |A| = 0 × 12 – 1 × –4 or better or |B| = 3 × –4 – 0 × 4 or better det symbol can be implied by the working 7 2 B1, B1 8 10 www 2 M1 (–2 – –8)2 + (10 – 2)2 or better 9 x = 0.5 y = 3 www 3 M1 consistent × and – for y or consistent × and + for x A1 one correct provided M1 scored 10 128 3 M1 d = kv2 A1 k = 2/25 (= 0.08) or M1 v2 = kd A1 k = 12.5 11 198 cao 3 M1 12.5 and 20.5 seen M1 6 × sum of their two upper bounds 12 –36x2 + 48x or 12x(4 – 3x) oe or other partly factorised versions 3 M1 squaring to “9x2 –12x + 4” algebraic M1 multiplying by –4 terms M1 adding 16 only 13 x [ 0.8 or x [ 4/5 cao 3 B1 12 – 18x B1 –4 + 8x these terms may be reversed if moved to the other side of the inequality allow >= 14 $12.92 3 M1 249 × r3 r can be anything dep M1 r = 1.017 and subtracting 249 SC2 261.92 on answer line

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 22 © UCLES 2009 15 (a) (i) OQ (ii) RM or MP (b) × S 1 1 2 Allow ½RP B1, B1 correct position wrt each direction of the vector ± 1 mm 16 (a) (0)810 or 8:10 etc. (b) 4 (c) 265 1 2 1 M1 (3 + 3)/(1 + 0.5) 17 (a) 261.48 cao (b) (±)3.86(48…) or 3.865 2 2 M1 4000 / 15.2978 M1 (15.9128 – 15.2978)/15.9128 (×100) oe or (“261.48” – 4000/15.9128) / “261.48” 18 m = 2 c = –10 4 B1 B(5, 0) or A(–4, 0) seen or used B1 m = 2 M1 substituting (5,0) into y = 2x + c or 0 5 0 − −c = 2 19 (a) 44 (b) 158 2 2 M1 OCB = 68 20 (a) 38 (b) 45 to 46 (c) 15 to 16 (d) 10 or 11 1 1 1 2 SC1 70 on answer line M1 speed/time 21 (a) 0.8 or 4/5 cao (b) 960 www 2 3 M1 30 × (12 + 36)/2 M1 10 × (12 + 36)/2 M1 12 × 40 M1 ½ × 40 × 24

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 22 © UCLES 2009 22 (a) 2 (b) 4x3 + 5 (c) 2 )1 3 ( − x 2 2 2 M1 f(0) = 1 M1 4(x3 + 1) + 1 M1 rearranging y = (2x + 1)/3 to make x the subject and interchanging x and y. Allow any one error in the working 70