Cambridge IGCSE Mathematics - International 0607 — 2018 Oct/Nov Paper 6 · Variant 2

* 1 2 8 9 0 8 4 1 4 9 * This document consists of 14 printed pages and 2 blank pages. DC (NF/CGW) 153619/2 © UCLES 2018 [Turn over CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) October/November 2018 1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials: Graphics Calculator 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. Do not use staples, paper clips, glue or correction fluid. You may use an HB pencil for any diagrams or graphs. DO NOT WRITE IN ANY BARCODES. Answer both parts A (Questions 1 to 4) and B (Questions 5 to 8). You must show all relevant working to gain full marks for correct methods, including sketches. In this paper you will also be assessed on your ability to provide full reasons and to communicate your mathematics clearly and precisely. At the end of the examination, fasten all your work securely together. The total number of marks for this paper is 40. Cambridge International Examinations Cambridge International General Certificate of Secondary Education

2 0607/62/O/N/18 © UCLES 2018 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 4) RIGHT SPIRALS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation is about the lengths of spirals drawn on a square co-ordinate grid. A robot starts from (0, 0) and moves 1 unit to Corner 1 . It then turns right and moves 1 unit to Corner 2 . It then turns right and moves 2 units to Corner 3 . It then turns right and moves 2 units to Corner 4 . It then turns right and moves 3 units to Corner 5 . This forms a spiral, shown on the grid below. 4 0 6 4 2 −2 −4 −6 −6 −4 −2 2 4 6 x y 5 2 1 3 The robot continues to turn and move in the same way.

3 0607/62/O/N/18 © UCLES 2018 [Turn over 1 (a) By continuing the spiral, show that Corner 10 is at (3, 3). (b) The length of the spiral from (0, 0) to Corner 4 is 6 units. Find the length of the spiral from (0, 0) to Corner 10 . … (c) Use your spiral to complete this table. Corner number Lengths added Length from (0, 0) 1 1 1 2 1 + 1 2 3 1 + 1 + 2 4 4 6 5 6 1 + 1 + 2 + 2 + 3 + 3 12 7 1 + 1 + 2 + 2 + 3 + 3 + 4 16 8 20 9 10 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5

4 0607/62/O/N/18 © UCLES 2018 2 The table shows the length, L, of the spiral from (0, 0) to Corner k , where k is an even number. Term number (n) k Length (L) 1 2 2 2 4 6 3 6 12 4 8 20 n k n n 1 + ` j (a) Find a formula for n in terms of k. … (b) Use part (a) to show that the formula for the length, L, of the spiral from (0, 0) to Corner k is L k k 4 2 = + ` j. (c) Show that the formula gives the correct length of the spiral from (0, 0) to Corner 14 .

5 0607/62/O/N/18 © UCLES 2018 [Turn over 3 (a) When k is an even number, find an expression, in terms of k, for the length of the spiral (i) from Corner k - 1 to Corner k , … (ii) from Corner k to Corner k + 1 . … (b) (i) Using question 2(b), find a formula, in terms of k, for the length, L, of the spiral from (0, 0) to Corner k + 1 . … (ii) Use your formula to show that the length of the spiral from (0, 0) to Corner 7 is 16.

6 0607/62/O/N/18 © UCLES 2018 4 A corner on the spiral has co-ordinates (x, y). Horizontal lengths of the spiral are added to give the total horizontal length, H. x co-ordinate (x) Horizontal lengths Total horizontal length (H) 1 1 1 2 1 + 2 + 3 6 3 1 + 2 + 3 + 4 + 5 15 4 1 + 2 + 3 + 4 + 5 + 6 + 7 28 5 (a) Complete the table. (b) Find a formula, in terms of x, for the total horizontal length, H, of the spiral from (0, 0) to the corner with co-ordinates (x, y). … (c) Write down a formula, in terms of y, for the total vertical length, V, of the spiral from (0, 0) to the corner with co-ordinates (x, y). …

7 0607/62/O/N/18 © UCLES 2018 [Turn over (d) k is an even number. (i) Use your answers to part (b) and part (c) to show that a formula for the length, L, of the spiral from (0, 0) to Corner k with co-ordinates (x, y) is L x x 2 2 1 = - ` j. (ii) The spiral has length 1560 from (0, 0) to the corner with co-ordinates (x, y). Use the formula in part (i) to find the value of x. …

8 0607/62/O/N/18 © UCLES 2018 B MODELLING (QUESTIONS 5 to 8) OPEN BOXES (20 marks) You are advised to spend no more than 45 minutes on this part. This task looks at maximum volumes when open boxes are made using regular shaped pieces of metal. Jenny makes open triangular-based boxes from a piece of metal in the shape of an equilateral triangle. A O 30° 30 cm NOT TO SCALE She cuts equal sized pieces from each corner of the equilateral triangle. Each cut is at right angles to the side of the shape. 30° 30 cm cut = height NOT TO SCALE The sides are folded up to form the vertical sides of an open triangular box. height

9 0607/62/O/N/18 © UCLES 2018 [Turn over 5 (a) The length of one side of the metal equilateral triangle is 30 cm. Use trigonometry to show that OA is 17.32 cm, correct to 4 significant figures. (b) Here is an enlargement of one corner cut from the metal triangle. 30° 2 cm 2 cm NOT TO SCALE r cm (i) Jenny makes a cut of 2 cm at right angles to the side of the equilateral triangle. Show that r = 4.

10 0607/62/O/N/18 © UCLES 2018 (ii) 30° 30 cm NOT TO SCALE A O Using OA = 17.32 cm and r = 4, find the area of the shaded triangle. You should use this formula. Area = sin bc A 2 1 B A c a b C … (iii) Jenny cuts the corner shown in part (b) from each corner of the equilateral triangle. She folds the sides up to make an open box. Show that the volume of the box is approximately 461 cm3.

11 0607/62/O/N/18 © UCLES 2018 [Turn over 6 Jenny wants a model for the volume, V cm3, of the open box made from an equilateral triangle of side 30 cm. She makes a cut of length x cm at right angles to the side of the equilateral triangle. 30° x cm r cm 30° 30 cm NOT TO SCALE O 30° (a) (i) Find an expression for r, in terms of x. … (ii) Find an expression, in terms of x, for the area of the shaded isosceles triangle. … (iii) The height of the open box is x cm. Explain why the model for the volume, V cm3, of the open box is sin cos sin V x x 120 30 15 30 2 3 ° ° ° 2 = - e o . … … …

12 0607/62/O/N/18 © UCLES 2018 (iv) Sketch the graph of V against x on the axes below. V x 0 0 550 9 Volume (cm3) Height (cm) (b) For what values of x is the model valid? … (c) Find the possible heights of the open box when V = 400. …

13 0607/62/O/N/18 © UCLES 2018 [Turn over 7 The equilateral triangle now has side E cm. (a) Give a reason why the model in question 6(a)(iii) becomes sin cos sin V x E x 120 2 30 30 2 3 ° ° ° 2 = - e o . … … (b) Use this model to find the height that gives the greatest volume when E = 60. …

14 0607/62/O/N/18 © UCLES 2018 8 Jenny makes an open square-based box from a square piece of metal of side E cm. (a) Change the model in question 7(a) so that it gives the volume of this box. … (b) The model in part (a) simplifies to V x E x 2 2 = - ` j . Find the relationship between E and x which gives the maximum volume of the box. …

15 0607/62/O/N/18 © UCLES 2018 BLANK PAGE

16 0607/62/O/N/18 © UCLES 2018 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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. BLANK PAGE

This document consists of 7 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) October/November 2018 MARK SCHEME Maximum Mark: 40 Published 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 2 of 7 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 3 of 7 MARK SCHEME NOTES The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION RIGHT SPIRALS 1(a) Completely correct spiral to (3, 3) 1 1(b) 30 1 C opportunity 1(c) 1 + 1 + 2 + 2 1 + 1 + 2 + 2 + 3 9 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 25 (30) 2 B1 for four or five correct cells 2(a) n = 2 k oe 1 2(b) [ ] 1 2 2 k k L   = +     with no subsequent errors 1 Alternative methods: [ ] ( ) 2 2 2 4 n L n = + … n2 + n [ ] 2 2 2 4 4 2 2 2 k k k k L   = + … +     2(c) 14 (14 2) 56 4 + = 1 20 + 10 + 12 + 14 = 56 oe 1 3(a)(i) 2 k oe 1 C opportunity 3(a)(ii) 1 2 k + oe 1 C opportunity 3(b)(i) [ ] 1 1 1 2 2 2 k k k k L their +     = + + +         oe isw 1 FT their 3(a)(ii) Condone correct formula for k odd as final answer 3(b)(ii) Correct substitution in their (i) leading to 16 1 4(a) 1 + 2 + ... + 7 + 8 + 9 and 45 1 4(b) H = 2x2 − x oe 2 B1 for 2x2 seen C opportunity 4(c) V = 2y2 − y oe 1 FT their 4(b) 4(d)(i) y = x 1 2x2 − x + 2x2 − x oe or 2(2x2 – x) 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 7 Question Answer Marks Partial Marks 4(d)(ii) 20 1 C opportunity Communication: seen in two of the following questions 1 1(b) for showing working, 1 + 1 + ... + 5 3(a)(i) for 2 examples of lengths before k 3(a)(ii) for 2 examples of lengths after k 4(b) for at least three first differences seen or working with simultaneous equations 4(d)(ii) for appropriate working or using quadratic or sketch

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 6 of 7 Modelling Question Answer Marks Partial Marks B MODELLING OPEN BOXES 5(a) 15 cos30D oe 1 leading to 17.320 to 17.321 1 5(b)(i) 2 sin30D or 2 sin30 r = each leading to 4 1 5(b)(ii) 76.8 or 76.82 to 76.83 2 M1 for 17.32 – 4 soi 13.32 or B1 for 6.66 or 23.07 to 23.08 or 11.5… 5(b)(iii) [76.8 × 3 × 2 oe leading to] 460.8 to 460. 9... 1 6(a)(i) sin30 x D or 2x oe 1 6(a)(ii) 2 1 sin120 (17.32 ) 2 sin30 x − D D oe 1 6(a)(iii) [Area of cross-section =] 3 triangles oe 1 6(a)(iv) Correct curve 1 6(b) 0 < x < 8.66 1 6(c) 1.50 and 4.54 or 4.55 1 C opportunity 7(a) 15 becomes 2 E oe 1 7(b) 5.77 or 5.77 to 5.80 1 C opportunity

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 7 Question Answer Marks Partial Marks 8(a) 2 [ ] 2 sin90 ( ) 2cos45 sin 45 E x V x = − D D D oe 3 B1 for cos 45° or sin 45° correctly used B1 for sin 90° correctly used B1 for 2x oe correctly placed Max B2 if final answer incorrect If 0 scored SC2 for sketch with x, E and E - 2x correctly marked or SC1 for sketch with x and E correctly marked 8(b) 6 E x = oe 2 Accept x = (0.17 or 0.166...)E M1 for sketch of V against x for at least two values of E or table of values C opportunity Communication: seen in one of the following questions 1 6(c) for line on graph 7(b) for drawing graph indicating where maximum is 8(b) for suitable scale on sketch or working from table/pairs of values