Cambridge IGCSE Mathematics - International 0607 — 2023 Oct/Nov Paper 6 · Variant 1

This document has 12 pages. [Turn over Cambridge IGCSE™ DC (EF/SG) 318329/3 © UCLES 2023 * 0 6 5 7 7 9 4 2 8 5 * CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 Investigation and Modelling (Extended) October/November 2023 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 5) and part B (Questions 6 to 10). ● 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 [ ].

2 0607/61/O/N/23 © UCLES 2023 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 5) F‑TYPE SEQUENCES (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation explores patterns in a special type of sequence of positive integers. In an F‑type sequence: • the first two terms are any two positive integers • after the first two terms, each term is equal to the sum of the previous two terms. 1 Here is a table of the first nine terms of an F‑type sequence. The first term F1 is 5. The second term F2 is 3. F1 F2 F3 F4 F5 F6 F7 F8 F9 5 3 8 11 19 30 49 79 128 In the table, F F 1 2 + = 5 + 3 = 8 = F3 F F 2 3 + = 3 + 8 = 11 = F4 F F 3 4 + = 8 + 11 = 19 = F5 and so on. (a) Calculate the 10th term. F10 = … [2] (b) (i) Complete the table. F2 = 3 F F 3 1 - = 3 F F 2 4 + = … F F 5 1 - = 14 F F F 2 4 6 + + = … F F 7 1 - = … F F F F 2 4 6 8 + + + = 123 F F 9 1 - = … [2]

3 0607/61/O/N/23 © UCLES 2023 [Turn over (ii) Complete this statement. F2 + F4 + F6 + F8 + F10 = F … - F … [1] (c) (i) Complete the table. F1 = 5 F F F 2 1 2 + - = 5 F F 1 3 + = … F F F 4 1 2 + - = 13 F F F 1 3 5 + + = … F F F 6 1 2 + - = … F F F F 1 3 5 7 + + + = 81 F F F 8 1 2 + - = … [2] (ii) Complete this statement. F1 + F3 + F5 + F7 + F9 = F … + F … - F … [1] (d) Use your statements in part (b)(ii) and part (c)(ii), and the definition of an F‑type sequence, to show that F F F F F F F F F F 1 2 3 4 5 6 7 8 9 10 + + + + + + + + + = F F 12 2 - . [2] (e) Use the statement in part (d) to complete this general statement. F1 + F2 + F3 + g + Fn = F … - F … [1]

4 0607/61/O/N/23 © UCLES 2023 2 In another F‑type sequence the first term is 3 and the second term is 1. (a) Complete the first five terms. 3, 1, … , … , … [1] (b) Is your statement in Question 1(e) correct for the sum of the first five terms in this sequence? … [3] 3 In another F‑type sequence the 2nd term is 3 and the 12th term is 652. (a) Use your answer to Question 1(e) to find the sum of the first 10 terms. … [2] (b) The sum of the first 12 terms of this sequence is 1704. Find the 10th term. … [3]

5 0607/61/O/N/23 © UCLES 2023 [Turn over 4 The Fibonacci sequence is a special F‑type sequence. The sequence starts 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … (a) Use this information and your answers to Question 1(c) to simplify this sum. f F F F F n 1 3 5 2 1 + + + + - … [1] (b) The 16th term in the Fibonacci sequence is 987. Find the 8 different terms in the Fibonacci sequence that add up to 987. … [2]

6 0607/61/O/N/23 © UCLES 2023 5 The first four terms of an F‑type sequence are a, b, c and d. (a) There is a relationship between c b 2 2 - and a simple combination of a and d. Investigate this relationship by making up at least three numerical examples of F‑type sequences. Write down this relationship. … [4] (b) The first term of the F‑type sequence is a and the second term is b. (i) Write c and d in terms of a and b, in their simplest form. c = … d = … [1] (ii) Use algebra to show that the relationship in part (a) is correct. [2]

7 0607/61/O/N/23 © UCLES 2023 [Turn over The modelling task starts on the next page.

8 0607/61/O/N/23 © UCLES 2023 B MODELLING (QUESTIONS 6 to 10) BIOLOGICAL AGE OF GOATS (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at the age, a, of a goat and its biological age, b, when compared to a human. A goat’s body ages more quickly than a human body. At birth, a goat’s age and its biological age are both 0. When a 0 = then b 0 = . 6 The life expectancy for a human is 73.5 years. The life expectancy for a goat is 10.5 years, which matches the biological life expectancy of 73.5 years for a human. When . a 10 5 = then . b 73 5 = . (a) Find a straight‑line model, in its simplest form, for b in terms of a. This is Model M. … [3] (b) Sketch the graph of your model. b a 0 [2] (c) A goat is 8 years old, so a 8 = . Find its biological age, b. … [1]

9 0607/61/O/N/23 © UCLES 2023 [Turn over 7 Goats age more quickly when young. A goat that is 2 years old has a biological age of 24 years. So, when a 2 = , b 24 = . (a) Find a straight‑line model for b in terms of a for a 0 2 G G . … [1] (b) After a goat reaches the age of 2 years, its biological age increases by 4 each year. (i) Find its biological age, b, when a 10 = . … [2] (ii) Find a straight‑line model for b in terms of a for a 2 H . Write the model in its simplest form. This is Model N. … [3] (c) Sketch the graphs of your straight‑line models in part (a) and part (b)(ii) on the axes on page 8. [2]

10 0607/61/O/N/23 © UCLES 2023 8 The most recent research gives this graph for a model of b in terms of a. b a 0 1 2 3 4 5 Goat’s age Biological age 6 7 8 9 10 10 20 30 40 50 60 70 (a) Use the graph to write down the biological age of a goat that is: • 2 years old … • 10 years old. … [1] (b) This model for the biological age is log b g a h = + where g and h are constants. (i) Use your answers to part (a) to write down two equations in g and h. … … [1] (ii) Use algebra to find g and h, correct to the nearest integer. Write down the model. This is Model P. … [3]

11 0607/61/O/N/23 © UCLES 2023 [Turn over (c) Find the age, correct to one decimal place, of a goat whose biological age is 70. … [3] 9 A goat lives until it is 18 years old, which is old for a goat. For each model calculate the biological age of the goat. Write down whether each model is valid or not valid for this goat. Model M in Question 6(a) … Model N in Question 7(b)(ii) … Model P in Question 8(b)(ii) … [4] Question 10 is printed on the next page.

12 0607/61/O/N/23 © UCLES 2023 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 Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. 10 Find the ages between which biological age from Model N 1 biological age from Model P 1 biological age from Model M. Between … and … [4]

This document consists of 7 printed pages. © UCLES 2023 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) October/November 2023 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 2023 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 2023 © UCLES 2023 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 2023 © UCLES 2023 Page 3 of 7 Mathematics-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, non-integer 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 or sign 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 A or B 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 2023 © UCLES 2023 Page 4 of 7 Question Answer Marks Partial Marks 1(a) 79 + 128 C1 207 1 1(b)(i) F2 = 3 F3 –F1 = 3 F2+F4 = 14 F5–F1 = 14 F2+F4+F6 = 44 F7–F1 = 44 F2 + F4 +F6 + F8 = 123 F9–F1 = 123 2 B1 for two correct cells 1(b)(ii) F11 – F1 1 1(c)(i) F1 = 5 F2 + F1–F2 = 5 F1 + F3 = 13 F4 + F1–F2 = 13 F1 + F3 + F5 = 32 F6 + F1–F2 = 32 F1 + F3 + F5 + F7 = 81 F8 + F1–F2 = 81 2 B1 for two correct cells 1(c)(ii) F10 + F1 – F2 1 1(d) F11 – F1 [ + ] F10 + F1 – F2 1 F10 + F11 – F2 = F12 – F2 or F10 + F11 = F12 leading to F12 – F2 1 Dep on first mark 1(e) Fn + 2 – F2 1 2(a) 4, 5, 9 1 2(b) 3 + 1 + 4 + 5 + 9 = 22 1 FT their 2(a) 14, 23 for the next two terms C1 Yes oe and 23 – 1 = 22 1 3(a) 652 – 3 C1 649 1 3(b) 1704 – their 649 or their 1707 – 652 OR their 649 + F11 + 652 = 1704 oe C1 FT F11 = their 1055 – 652 or F11 = 403 C1 FT 249 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 5 of 7 Question Answer Marks Partial Marks 4(a) F2n 1 4(b) 89, 144, 233, 377, 610, […] C1 1 2 5 13 34 89 233 610 1 5(a) Three correct new F-type sequences of at least 4 terms 2 B1 for one new F-type sequences Two c2 – b2 correctly evaluated 1 FT their sequences c2 – b2 = ad 1 5(b)(i) [c =] a + b [d =] a + 2b 1 5(b)(ii) (a + b)2 – b2 = a2 + 2ab + b2 – b2 = a2 + 2ab 1 a(a + 2b) = a2 + 2ab 1 6(a) 73.5 10.5 oe C1 b = 7a 2 B1 for 7 seen implied as gradient 6(b) Correct sketch 1 Straight ruled line through the origin Scales shown C1 Line passing through approximately (a, 7a) as shown by scales 6(c) 56 1 FT their 7 7(a) b = 12a 1 7(b)(i) (10 – 2)×4 + 24 C1 56 1 7(b)(ii) 4 is the gradient soi C1 Correctly substituting (2, 24) or (10, 56) into b = 4a + c oe 1 FT their 56 b = 4a + 16 oe 1 FT their 56

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 6 of 7 Question Answer Marks Partial Marks 7(c) Correct sketch 2 B1 for line from (0, 0) with gradient larger than for the graph from 6(b) or for line with gradient smaller than the graph from 6(b) starting at approx. (0, 16) or (2, 24) 8(a) 42 68 1 8(b)(i) 42 = g log 2 + h 68 = g log 10 + h 1 FT their 42 and their 68 8(b)(ii) 26 = 0.699 g C1 g = 37 1 h = 31 [b = 37 log a + 31] 1 8(c) 37 log a + 31 = 70 log a = 39 37 or 1.05[…] C2 C1 for each statement FT their g and their h for first statement only 11.2 or 11.3 1 9 126 88 77 3 B1 for each FT their models Not valid, valid, valid 1 FT their answers Do not award if no values seen

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 7 of 7 Question Answer Marks Partial Marks 10 Sketch showing curve (correct shape) and two straight lines with positive gradients, one from the origin and one from the b axis and identification of two correct intersection points of each line with the curve C2 C1 for one correct sketch or 7a = 37 log a + 31 and 4a + 16 = 37 log a + 31 seen 9.6[3] or 9 or 10 14.5 or 14 or 15 2 B1 for each If B0 scored SC1 for 67 or 67.3 to 67.4 and 74 or 73.9…