Cambridge IGCSE Mathematics - International 0607 — 2018 Oct/Nov Paper 6 · Variant 3

* 7 7 5 7 8 6 1 6 3 5 * This document consists of 15 printed pages and 1 blank page. DC (CE/SW) 153620/4 © UCLES 2018 [Turn over CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 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 3) and B (Questions 4 to 6). 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 0607/63/O/N/18 © UCLES 2018 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 3) NEAREST NEIGHBOURS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation is about pairs of dots on rectangular grids. When two dots are next to each other, they are called nearest neighbours. Here are four equally spaced dots in a row. The dots marked with crosses are nearest neighbours. There are 3 pairs of nearest neighbours. 1 (a) Complete this table showing the number of pairs of nearest neighbours. Number of dots in row Number of pairs of nearest neighbours 2 1 3 4 3 5 6 n

3 0607/63/O/N/18 © UCLES 2018 [Turn over (b) With four dots in a row there are a total of 6 different arrangements of two crosses. In 3 of these arrangements the crosses mark dots that are nearest neighbours. Complete this table. Number of dots in row Number of pairs of nearest neighbours Total number of different arrangements of two crosses 2 1 1 3 4 3 6 5 6 15 n (c) (i) Multiply the number of dots in a row by the number of pairs of nearest neighbours. Write down a connection between your answer and the total number of different arrangements of two crosses. … … (ii) In a row of n dots, write down an expression, in terms of n, for the total number of different arrangements of two crosses. …

4 0607/63/O/N/18 © UCLES 2018 2 As well as pairs of nearest neighbours there are also pairs of 2nd nearest neighbours, 3rd nearest neighbours, 4th nearest neighbours and so on. Here are some examples. Nearest neighbours 2nd nearest neighbours 3rd nearest neighbours (a) Complete this table. Number of dots in row Number of pairs of Nearest neighbours 2nd nearest neighbours 3rd nearest neighbours 2 1 3 4 3 5 6 n (b) Find an expression for the number of pairs of kth nearest neighbours in a row of n dots. …

5 0607/63/O/N/18 © UCLES 2018 [Turn over (c) (i) You can find the total number of different arrangements of two crosses by adding the number of pairs of nearest neighbours, 2nd nearest neighbours, 3rd nearest neighbours and so on. Use this method to find the total number of different arrangements of two crosses in a row of 11 dots. … (ii) Use your expression in question 1(c)(ii) to show that your answer in question 2(c)(i) is correct.

6 0607/63/O/N/18 © UCLES 2018 3 Here is a 2 by 5 rectangle of dots. It has 2 rows of 5 dots. a b The distance between the dots in each column is a and the distance between each row is b. (a) When a = b, here are two pairs of crosses which are nearest neighbours. There are a total of 13 pairs of nearest neighbours for a 2 by 5 rectangle of dots. For a 2 by w rectangle of dots, where w H 2, find an expression for the number of pairs of nearest neighbours. …

7 0607/63/O/N/18 © UCLES 2018 [Turn over (b) When a ! b, the pairs of nearest neighbours are the dots that are closest to each other. Nearest neighbours Not nearest neighbours For a 2 by w rectangle of dots, where w H 2, find an expression for the number of pairs of nearest neighbours, (i) when a 2 b, … (ii) when a 1 b. …

8 0607/63/O/N/18 © UCLES 2018 (c) Here are two rectangles of dots. a a 3 by 5 rectangle of dots 6 by 4 rectangle of dots b b (i) For an h by w rectangle of dots, here are some expressions for the number of pairs of nearest neighbours, T. In all cases a 2 b and w H 2. h Number of pairs of nearest neighbours (T ) 3 2w 4 3w 5 4w 6 5w Find the formula for T in terms of h and w. … (ii) For a 2 b, there are 1000 pairs of nearest neighbours in a 101 by w rectangle of dots. Find the value of w. …

9 0607/63/O/N/18 © UCLES 2018 [Turn over (d) For an h by w rectangle of dots, find an expression for T when a = b and w H 2. …

10 0607/63/O/N/18 © UCLES 2018 THE MODELLING STARTS ON PAGE 11.

11 0607/63/O/N/18 © UCLES 2018 [Turn over B MODELLING (QUESTIONS 4 to 6) LONG JUMP (20 marks) You are advised to spend no more than 45 minutes on this part. This task is about modelling long jumping. Athletes run along a track and then jump as far as they can in one leap. The horizontal distance jumped is called their long jump distance. Run-up Take-off Flight through the air Landing At take-off, athletes jump up giving them a vertical speed. Their horizontal speed carries them forward in their flight through the air. The vertical speed and the horizontal speed give the take-off speed. 4 Here is a formula connecting the vertical speed, v m/s, and the maximum height, j metres, jumped by the athlete. v2 = 20j An athlete has a vertical speed of 3.5 m/s. Find the maximum height jumped by this athlete. …

12 0607/63/O/N/18 © UCLES 2018 5 This scatter diagram shows the horizontal speed and the long jump distance for each of 13 athletes in a recent World Championship. 6.5 6.5 7 7.5 8 8.5 9 7 7.5 8 8.5 9 9.5 Horizontal speed (m / s) Long jump distance (metres) d r The mean horizontal speed, r, is 8.5 m/s. The mean long jump distance, d, is 7.9 m. (a) (i) On the scatter diagram plot the mean point and draw, by eye, a line of best fit. (ii) Use your line of best fit to work out the equation that models the relationship between horizontal speed and long jump distance. …

13 0607/63/O/N/18 © UCLES 2018 [Turn over (iii) An athlete has a horizontal speed of 6.6 m/s. This gives a long jump distance of 4.45 m. Compare this long jump distance with the distance given by your equation in part (ii). … … (b) Long jump distance (d ) and horizontal speed (r) can also be modelled by the following quadratic equation. d = -0.46r 2 + 8.7r – 32.8 (i) The athlete in part (a)(iii) had a horizontal speed of 6.6 m/s and a long jump distance of 4.45 m. Which model gives a better approximation for this athlete’s long jump distance, the model you used in part (a)(iii) or this quadratic model? … … (ii) Sketch the graph of d = -0.46r 2 + 8.7r – 32.8 for 5 G r G 12. d r 12 5 (iii) Write down an inequality, in terms of r, for which the model is valid. …

14 0607/63/O/N/18 © UCLES 2018 6 When athletes jump they take-off at an angle a to the horizontal. The vertical speed and the horizontal speed give the take-off speed, T m/s. d a T A model for the long jump distance, d metres, is d = sin cos T a a 5 2 . (a) Sketch the graph of y = sin a cos a for 0° G a G 90°. 0° 90° a y (b) Show that when sin a cos a has its maximum value the greatest jump distance is d = T 10 2 . (c) An experienced long jumper has a take-off speed of 9.6 m/s. This gives a long jump distance of 7.89 m. Compare this long jump distance with the distance given by the formula in part (b). … …

15 0607/63/O/N/18 © UCLES 2018 (d) Take-off speed (T m/s), horizontal speed (h m/s) and vertical speed (v m/s) are related by the right-angled triangle below. T h a v (i) Use trigonometry to show that sin cos d T a a 5 2 = can be written as d vh 5 = . (ii) An athlete has a take-off speed of 10 m/s. This gives a greatest height of 1.25 m. Find the horizontal speed, h m/s, and the angle a. You may need to use the formula in question 4. …

16 0607/63/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 0607/63 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.

0607/63 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.

0607/63 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

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION NEAREST NEIGHBOURS 1(a) 2 B1 for 2,4,5 B1 for n – 1 1(b) 1 1(c)(i) ‘My answer is double the total’ or ‘the total is half my answer’ oe 1 1(c)(ii) ( ) 1 2 n n − oe 1 2(a) Nearest neighbours 2nd nearest neighbours 3rd nearest neighbours 2 1 3 2 1 4 3 2 1 5 4 3 2 6 5 4 3 7 n − 1 n − 2 n − 3 2 B1 for each correct column 2(b) n – k 1 2(c)(i) 55 2 M1 for 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 2(c)(ii) 11 10 2 × oe 1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 7 Question Answer Marks Partial Marks 3(a) 3w – 2 oe 2 M1 for at least 3 correct pairs of numerical results or sketches 3(b)(i) w 1 C opportunity 3(b)(ii) 2w – 2 oe 1 C opportunity 3(c)(i) T = (h – 1)w oe 1 3(c)(ii) 10 1 FT their (c)(i) if w an integer C opportunity 3(d) w(2h – 1) – h oe 2 B1 for 2 correct terms when expanded C opportunity Communication: seen in one of the following questions 1 3(b)(i) Two examples or an explanation 3(b)(ii) Sketch or 2 examples or an explanation 3(c)(ii) For correct working 3(d) Sketches (min 2) or at least 2 correct expressions or explanation

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 6 of 7 Question Answer Marks Partial Marks B MODELLING LONG JUMP 4 0.6125 isw 1 C opportunity 5(a)(i) Suitable ruled line within tolerance through (8.5, 7.9) 2 B1 for mean point plotted or line outside tolerance through (8.5, 7.9) or ruled line with positive gradient within tolerance but not passing through mean 5(a)(ii) d = (their m) r + (their c) oe 2 FT their ruled straight line with positive gradient B1 for correct gradient B1FT for correct intercept from their gradient C opportunity 5(a)(iii) Correct comparison of their d to 4.45 1 FT their (a)(ii) 5(b)(i) Jump length predicted by the quadratic model is 4.5(824 ...) 1 Compared with their calculated jump distance in 5(a)(iii), and correct decision 1 FT their (a)(iii) 5(b)(ii) Correct curve 1 n-shaped curve from r = 5.2 approx. C opportunity 5(b)(iii) 5.2 to 5.5 < r < 9.45… to 9.46… 2 B1 for each side of the inequality 6(a) Correct curve 1 n-shaped curve from (0,0) to (90, 0) C opportunity

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 7 Question Answer Marks Partial Marks 6(b) 2 0.5 [ ] 5 d T = × 1 6(c) Correct comparison using long jump distance given by the formula of 9.21 to 9.22 or difference of 1.32 to 1.33 1 6(d)(i) At least 3 of the following four seen in working: sin = v a T oe cos = h a T oe 2 sin cos = hv a a T oe 2 2 5 T vh d = × T oe 2 M1 for 2 of the expressions seen in working 6(d)(ii) [h = ] 8.66 [a = ] 30 3 B1 for [v = ]5 B1 for 8.66 or 30 C opportunities Communication: seen in three of the following questions 1 4 Correct units 4 Working shown 5(a)(ii) Gradient drawn on grid (lines or triangle) or calculation shown 5(b)(ii) Appropriate scale on d-axis or co-ordinates of maximum 6(a) Appropriate scale on y-axis or co-ordinates of maximum 6(d)(ii) Showing working – substitution 6(d)(ii) Showing working – Pythagoras 6(d)(ii) Showing working – trigonometry