Cambridge IGCSE Mathematics - International 0607 — 2021 Oct/Nov Paper 6 · Variant 1

This document has 12 pages. Cambridge IGCSE™ DC (LK/SG) 212584/2 © UCLES 2021 [Turn over CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 Investigation and Modelling (Extended) October/November 2021 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 7) and part B (Questions 8 to 12). ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You should use a graphic display calculator where appropriate. ● You may use tracing paper. ● You must show all necessary working clearly, including sketches, to gain full marks for correct methods. ● In this paper you will be awarded marks for providing full reasons, examples and steps in your working to communicate your mathematics clearly and precisely. INFORMATION ● The total mark for this paper is 60. ● The number of marks for each question or part question is shown in brackets [ ]. * 1 3 2 8 8 3 9 6 1 0 *

2 0607/61/O/N/21 © UCLES 2021 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 7) PYTHAGOREAN SETS OF FOUR (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at finding the integer lengths of the sides of a cuboid that has an integer length for its diagonal. 1 S R Q P 3 5 6 NOT TO SCALE The diagram shows a cuboid with sides of length 3, 5 and 6. (a) Using Pythagoras’ Theorem in triangle PQR gives PR 3 5 2 2 2 = + . Find the value of PR2. … [1] (b) Using Pythagoras’ Theorem in triangle PRS gives PS PR 6 2 2 2 = + . Find the value of PS2. … [2]

3 0607/61/O/N/21 © UCLES 2021 [Turn over 2 S R Q P a b c d NOT TO SCALE The diagram shows a cuboid with sides of integer length a, b and c. Its diagonal, PS, has integer length d. (a) Use Pythagoras’ Theorem in triangle PQR to write down an expression for PR2 in terms of a and b. … [1] (b) Use your answer to part (a), and Pythagoras’ Theorem in triangle PRS, to show that d a b c 2 2 2 2 = + + . [1] 3 A cuboid has sides of length a, b and c, where a, b and c are integers and a b c G G . If the length of the diagonal, d, is also an integer then (a, b, c, d) is a Pythagorean set of four. Use d a b c 2 2 2 2 = + + to show that a cuboid with sides of length 4, 17 and 28 gives a Pythagorean set of four. Complete the Pythagorean set of four. (4, 17, 28, … ) [3]

4 0607/61/O/N/21 © UCLES 2021 4 (a) In a Pythagorean set of four (a, b, c, d ) d a b c 2 2 2 2 = + + . When d a c = + , show that ac b 2 2 = . [2] (b) Explain why b must be even. … … … [2]

5 0607/61/O/N/21 © UCLES 2021 [Turn over 5 Here is a method for finding Pythagorean sets of four using Question 4: • Choose any even integer for b. • Calculate ac using Question 4(a). • Find all the possible pairs of integers for a and c, where a c 1 . Use this method to find all the Pythagorean sets of four when you choose b 8 = . … [7]

6 0607/61/O/N/21 © UCLES 2021 6 (a) Which one of these two sets is a Pythagorean set of four? (18, 24, 72, 78) or (18, 24, 72, 84)? Show how you decide. (18, 24, 72, … ) [2] (b) (ka, kb, kc, kd ) is a Pythagorean set of four, where k is a positive integer greater than 1. Use algebra to show that (a, b, c, d ) must also be a Pythagorean set of four. [2] (c) (a, b, c, d ) is a basic Pythagorean set of four if the numbers a, b, c and d have no common factor greater than 1. Find the basic Pythagorean set of four for your answer to part (a). … [2]

7 0607/61/O/N/21 © UCLES 2021 [Turn over 7 The method in Question 5 to find Pythagorean sets of four is: • Choose any even integer for b. • Calculate ac using Question 4(a). • Find all the possible pairs of integers for a and c, where a c 1 . Use this method to find two basic Pythagorean sets of four where the smallest integer, a, is 2. (… , … , … , …) and (… , … , … , …) [5]

8 0607/61/O/N/21 © UCLES 2021 B MODELLING (QUESTIONS 8 to 12) REFLECTING A LASER BEAM (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at models for the height of the image of a reflected laser beam on a vertical wall. In this task all the measurements are in metres. The diagram shows, by a dashed line, the side view of the path of a laser beam. The laser beam • starts at source L • travels to a point R on horizontal ground AB • reflects at the point R so that angle LRA = angle NRB • travels to N, its image, on a vertical wall. Wall L A R B N h 4 Ground 6 15 NOT TO SCALE The height of L above the horizontal ground is LA 4 = . The height of N above the horizontal ground is h NB = . AR 6 = and RB 15 = . 8 Complete the statements to show that h 10 = . Triangle LRA is similar to triangle NRB. h 4 15 = … # = = 10 h … … … [2]

9 0607/61/O/N/21 © UCLES 2021 [Turn over 9 The laser source, L, can move towards or away from the wall. It now moves x metres to the right so that AR becomes x 6- metres. The point R does not move. The other given measurements remain the same. When the laser beam reflects at R, triangle LRA and triangle NRB will always be similar. L A R B N h 4 15 NOT TO SCALE ( )x 6- (a) Use the method in Question 8 to find a model for h in terms of x. … [2] (b) Sketch the graph of h against x for x 6 6 1 1 - . h x – 6 0 6 [2] (c) (i) Write down the equation of the vertical asymptote. … [2] (ii) Give a reason why there is a vertical asymptote. Refer to the path of the laser beam. … … [1]

10 0607/61/O/N/21 © UCLES 2021 10 The laser source, L, now stays fixed. At the start AR 6 = and RB 15 = . The point R then moves x metres towards B along the ground. The dashed line shows the path of the laser beam. L A R B h x 4 6 (a) Show that h x x 6 60 4 = + - . [2] (b) Sketch the graph of h x x 6 60 4 = + - for x 6 15 1 G - . h x – 6 0 15 [3] (c) When the point R has moved x metres towards B, the height, h, is 6. Find the value of x. … [2] NOT TO SCALE

11 0607/61/O/N/21 © UCLES 2021 [Turn over 11 L A R B x 4 6 h 10 y At the start, when AR 6 = , the height of the image is 10. After the point R moves x metres, the height of the image is h x x 6 60 4 = + - . y is the change in the height of the image, so y h 10 = - . (a) Show that a model for y is y x x 6 14 = + . [3] (b) (i) When the point R moves one metre to the left, away from B, x 1 =- . Use the model in part (a) to find the change in height of the image. … [2] (ii) The point R moves an additional one metre to the left, away from B. (a) Write down the value of x. … [1] (b) Find the additional change in height. … [3] Question 12 is printed on the next page. NOT TO SCALE

12 0607/61/O/N/21 © UCLES 2021 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 Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. 12 L A R B N 4 a h b (a) Find h in terms of a and b. … [2] (b) The point R moves x metres to the right, towards B, along the horizontal ground. y is the change in h. Find a model for y in terms of a, b and x. Do not simplify your answer. … [3] NOT TO SCALE

This document consists of 7 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) October/November 2021 MARK SCHEME Maximum Mark: 60 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 2021 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 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.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 3 of 7 Maths-Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5 Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. 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

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION PYTHAGOREAN SETS OF FOUR 1(a) 34 isw 1 1(b) 34 + 36 C1 FT their 34 70 1 FT their 34 2(a) a2 + b2 1 2(b) 2 2 2 d PR c = + leading to 2 2 2 2 d a b c = + + 1 3 2 2 2 4 17 28 + + 2 B1 for 42 + 172 + 282 [= 1089] B1 for 1089 their (4, 17, 28, 33) 1 4(a) (a + c)2 = a2 + 2ac + c2 soi oe 1 b2 = 2ac 1 4(b) Since ac is an integer so 2 2 b is an integer oe So b2 is even and then b is even 2 B1 for ac is an integer so 2 2 b is an integer oe B1 for so b2 is even, then b is even OR If b is odd, then b2 is odd so 2 2 b is not an integer thn ac is not an integer (2) OR B1 for b is odd then b2 is odd so 2 2 b is not an integer B1 for then ac is not an integer 5 2 8 2 C1 ac = 32 1 1, 32 2, 16 4, 8 C1 Substitution of one of their pairs into a + c or a2 + 82 + c2 oe C1 (1, 8, 32, 33) (2, 8, 16, 18) (4, 8, 8, 12) 3 B1 for each If 0 scored SC1 for d = 33, 18 or 12 following correct substitution

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 7 Question Answer Marks Partial Marks 6(a) 182 + 242 + 722 M1 (18, 24, 72, 78) A1 6(b) (ka)2 + (kb)2 + (kc)2 = (kd)2 1 k2a2 + k2b2 + k2c2 = k2d2 leading correctly to a2 + b2 + c2 = d2 1 6(c) Divide by 6 or common factor = 6 oe C1 (3, 4, 12, 13) 1 FT (18, 24, 72, 84) giving (3, 4, 12, 14) 7 Even b identified C1 Corresponding value for c with correct working shown C1 Substitution into d = 2 + c oe or into d = 22 + (their b)2 +(their c)2 oe C1 Two from (2, 6, 9, 11) (2, 10, 25, 27) (2, 14, 49, 51) (2, 18, 81, 83) etc. 2 B1 for one correct or B1 for two from: (2, 4, 4 ,6) (2, 8, 16, 18) (2, 12, 36, 38) (2, 16, 64, 66) etc. B MODELLING REFLECTING A LASER BEAM 8 15 4 6 = h 15 4 10 6 × = = h 2 B1 for each statement 9(a) 15 4 6 = − h x C1 60 6 = − h x oe 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 7 Question Answer Marks Partial Marks 9(b) Correct sketch 1 Curve must not touch the x-axis or go beyond x = 6 Always increasing. Start close to x = –6 and cross h-axis Intercept marked as 10 C1 9(c)(i) x = 6 2 B1 for x = k 9(c)(ii) The laser beam reflects vertically upwards so does not reach the wall. oe 1 10(a) AR = 6 + x and BR = 15 – x OR seen in two equal ratios M1 e.g. 15 4 6 − = + h x x 4(15 ) 60 4 6 6 − −   = =   + +   x x h x x A1 10(b) Correct sketch 1 for correct shape Curve always decreasing. Cross h axis h-intercept marked as 10. C1 x-intercept at 15. C1 10(c) 60 4 6 6 − = + x x oe or h = 6 drawn on the sketch in 10(b) C1 2.4 1 −6 6 10 20 30 40 x H 15 10 20 30 40 x H

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 7 Question Answer Marks Partial Marks 11(a) [y =] 60 4 10 6 − − + x x oe B1 10(6 ) 60 4 6 6 + − − + + x x x x 1 10(6 + x) seen correctly in expression or equation 10(6 ) 60 4 6 + − + + x x x [ ] 60 10 60 4 6 + − + + x x x [= 14 6 + x x ] 1 11(b)(i) 14 ( 1) 6 1 × − − C1 [–] 2.8 1 11(b)(ii)(a) –2 1 11(b)(ii)(b) 14 ( 2) 6 2 × − − or 60 4 ( 2) 6 2 −× − − C1 FT their (a) unless –1 [–]7 or 17 1 FT their correct calculation [–]4.2 final answer 1 12(a) Equal ratios using similar triangles e.g. 4 = h b a or relating to the pattern in previous questions C1 [ 4 ] = b h a 1 12(b) [y =] 4 4( ) − − + b b x a a x oe isw 3 FT their 12(a) B2 for 4( ) − + b x a x oe isw or B1 for a + x and b – x seen