Cambridge IGCSE Mathematics - International 0607 — 2019 May/June Paper 6 · Variant 2

This document consists of 12 printed pages. DC (LEG/CB) 170704/3 © UCLES 2019 [Turn over * 6 7 0 7 9 0 5 9 9 6 * CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) May/June 2019 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 6) and B (Questions 7 to 9). 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 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 Assessment International Education Cambridge International General Certificate of Secondary Education

2 0607/62/M/J/19 © UCLES 2019 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 6) SQUARE ROOTS WITHIN SQUARE ROOTS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation looks at sequences of terms with square roots. You can form a sequence by using square roots within square roots. 6 , 6 6 + , 6 6 6 + + , 6 6 6 6 + + + , ... You can calculate the first four terms of this sequence as follows. 6 = 2.4494.…. 6 6 + = . ... 6 2 4494 + = . ... 8 4494 = 2.9068.…. 6 6 6 + + = . ... 6 2 9068 + = . ... 8 9068 = 2.9844.…. 6 6 6 6 + + + = . ... 6 2 9844 + = . ... 8 9844 = 2.9974….. 1 (a) Calculate the next term of the sequence, writing the decimal as far as the 4th decimal place. … (b) As the sequence continues, the terms get closer and closer to an integer. This integer is the integer limit of the sequence. Write down the integer limit of this sequence. …

3 0607/62/M/J/19 © UCLES 2019 [Turn over 2 Here is a similar sequence of square roots. 30 , 30 30 + , 30 30 30 + + , 30 30 30 30 + + + , ... (a) Calculate the first three terms, writing each decimal as far as the 4th decimal place. … , … , … (b) Write down the integer limit of this sequence. …

4 0607/62/M/J/19 © UCLES 2019 3 (a) Complete this table for similar sequences. First term Integer limit 2 6 12 20 30 42 7 (b) (i) Use part (a) to find the first term of the sequence that has an integer limit of 8. … (ii) Calculate the 4th term of the sequence in part (b)(i). Give your answer correct to 4 decimal places. …

5 0607/62/M/J/19 © UCLES 2019 [Turn over 4 The general sequence is k , k k + , k k k + + , k k k k + + + , ... So, when x is one of the terms, the next term is k x + . The sequence reaches its limit, x, when one term equals the next term. That is, when x k x = + . (a) Make k the subject of the equation x k x = + . … (b) Use part (a) to check that the sequence below has a limit, x, equal to 7. 42 , 42 42 + , 42 42 42 + + , 42 42 42 42 + + + , ... 5 You will find the following information useful in this question. For the equation ax bx c 0 2 + + = , x a b b ac 2 4 2 ! = - - The limit of 3 , 3 3 + , 3 3 3 + + , 3 3 3 3 + + + , ... is not an integer. Find the exact value of this limit. …

6 0607/62/M/J/19 © UCLES 2019 6 Here is another type of sequence. 5 , 5 13 + , 5 13 5 + + , 5 13 5 13 + + + , 5 13 5 13 5 + + + + , ... (a) The limit of this sequence is an integer, x. By calculating some terms of this sequence, find x. … (b) The general form of this sequence is a , a b + , a b a + + , a b a b + + + , ... where a and b are positive integers and a b ! . The sequence has a limit, x, when x a b x = + + . Show that this equation can be written as x a b x 2 - = + .

7 0607/62/M/J/19 © UCLES 2019 [Turn over (c) (i) Use your answer to part (a) to show that the equation x a b x 2 + - = is correct for this sequence. 5 , 5 13 + , 5 13 5 + + , 5 13 5 13 + + + , 5 13 5 13 5 + + + + , ... (ii) There are other pairs of positive integers, a and b, where a b ! , such that the sequence a , a b + , a b a + + , a b a b + + + , ... has the same limit as in part (a). Find all these pairs of positive integers.

8 0607/62/M/J/19 © UCLES 2019 B MODELLING (QUESTIONS 7 to 9) MAKING CONES (20 marks) You are advised to spend no more than 45 minutes on this part. You may find some of these formulas useful. Circumference, C, of a circle of radius r. C r 2r = Volume, V, of cone of radius r, height h. V r h 3 1 2 r = Curved surface area, A, of cone of radius r, sloping edge l. A rl r = This task looks at modelling the volume of a cone, which is made from a circular disc. In this part lengths are given in centimetres and angles in degrees. Raj makes a cone from a circular disc of radius 18. He cuts out part of the circular disc, leaving the shaded sector with sector angle x. He joins the two radii of 18 together. The cone that he makes has base radius r and height h. 18 18 NOT TO SCALE 18 x r h 7 (a) Show that the arc length of the shaded sector is x 10 r . (b) This arc length forms the circumference of the base of the cone. Find r in terms of x. …

9 0607/62/M/J/19 © UCLES 2019 [Turn over (c) Show that h x 324 400 2 = - . (d) (i) Show that a model for the volume, V, of this cone, is . V x x 0 0026 324 400 2 2 = - . (ii) Use part (i) to find the volume of the cone if Raj uses a semicircle of radius 18. …

10 0607/62/M/J/19 © UCLES 2019 (e) Sketch the graph of the model for 0 < x < 360. 0 360 V x (f) For the cone with maximum volume, find (i) the angle, x, … (ii) the maximum volume, V, … (iii) the curved surface area. …

11 0607/62/M/J/19 © UCLES 2019 [Turn over 8 Raj decides that he can make two cones from his circular disc. 18 NOT TO SCALE 18 x y One comes from the sector with sector angle x. The other comes from the remaining sector with sector angle y. (a) Write down a formula for y in terms of x. … (b) The total volume of the two cones is . . V x x y y 0 0026 324 400 0 0026 324 400 2 2 2 2 = - + - . Use your answer to part (a) to write down, in terms of x, a model for the total volume of the two cones. … (c) Sketch the graph of this model. 0 2500 2700 V x (d) Find the two angles, x, that give the maximum total volume for the two cones. … and … Question 9 is printed on the next page.

12 0607/62/M/J/19 © UCLES 2019 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. 9 Instead of the circular disc of radius 18, Raj now uses a disc of radius 36 to make one cone with the maximum volume. Find (a) the volume of this cone, … (b) the angle that gives this volume. …

This document consists of 7 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) May/June 2019 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 May/June 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 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 May/June 2019 © UCLES 2019 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 Question Answer Marks Partial Marks A INVESTIGATION SQUARE ROOTS WITHIN SQUARE ROOTS 1(a) 2.9995[…] 1 C opportunity 1(b) 3 1 2(a) 5.4772[…] 5.9562[…] 5.9963[…] 2 FT their 5.4772[…] FT their 5.9562[…] B1 for two correct including FT If 0 scored, SC1 for answer 5.477, 5.956, 5.996 2(b) 6 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 7 Question Answer Marks Partial Marks 3(a) 1st term Limit 2 2 6 3 12 4 20 5 30 6 42 7 1 3(b)(i) 56 1 C opportunity 3(b)(ii) 7.9999 1 FT their exact square root seen in part (i) with answer rounded correctly to 4 decimal places. 4(a) k = x2 – x or k = x(x – 1) final answer 1 4(b) 42 = 72 – 7 = 49 – 7 oe or 42 7 + = 49 = 7 or 42 = 7(7 – 1) = 7 × 6 or x2 – x = 42 with working leading to x = 7 1 5 1 13 2 + oe as final answer 2 M1 for substitution of a = 1, b = –1, c = –3 soi If 0 scored SC1 for x2 + x = 3 leading to 1 13 2 −+ oe as final answer C opportunity 6(a) 3 2 B1 for calculating a correct 2nd, 3rd, 4th or 5th term 6(b) x2 = a + + b x 1 6(c)(i) 32 – 5 = 4 = 13 3 + oe 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 7 Question Answer Marks Partial Marks 6(c)(ii) [a =] 1 [b =] 61 [a =] 2 [b =] 46 [a =] 3 [b =] 33 [a =] 4 [b =] 22 [a =] 7 [b =] 1 3 B2 for three or four correct pairs and no incorrect or five correct and one or more incorrect or B1 for one or two correct pairs and no incorrect or three or four correct pairs and one or more incorrect. C opportunity Communication: seen in three of the following questions 1 1(a) 6 6 6 6 6 + + + + or 6 2.9974[...] + 3(b)(i) 8 × 7 [=] 56 or 42 + 14 [=] 56 or 72 + 7 [=] 56 or 82 – 8 [=] 56 or, from the table, all first differences 4, 6, 8, 10, 12 or multiplications 1 × 2, 2 × 3, 3 × 4, 4 × 5, 5 × 6, 6 × 7 or 12 + 1, 22 + 2, 32 + 3, 42 + 4, 52 + 5, 62 + 6 or 22 – 2, 32 – 3, 42 – 4 , 52 – 5, 62 – 6, 72 – 7 3(b)(ii) 56 56 56 56 + + + or 7.4833[…] or 7.9676[…] or 7.9979[…] or 7.998[0] used for the next term 5 their(x2 – x) = 3 oe 6(c)(ii) Substitution of a or b in 9 – a = 3 + b , or a correct rearrangement, with working leading to other variable or a statement about finding square number for b + 3. oe or b = a2 – 18a + 78

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 7 Question Answer Marks Partial Marks B MODELLING MAKING CONES 7(a) 2 18 360 x × × π× oe 1 7(b) [r = ] 20 x oe 1 C opportunity 7(c) h2 = 182 – r2 or h = 2 2 18 −r M1 C opportunity Replacing r2 by 2 20       x their and 182 by 324 M1 7(d)(i) 1 3 π ÷ 400 oe = 0.00261 to 0.00262 1 7(d)(ii) 1310 or 1313 to 1313.2 1 C opportunity 7(e) Correct sketch 2 B1 for any sketch with a single maximum to the right of x = 240 or correct sketch, except that the maximum is between x = 180 and x = 240. C opportunity 7(f)(i) 294 1 7(f)(ii) 2330 or 2334 to 2335 1 7(f)(iii) awrt 831 2 FT their part (i) for 0 < x < 360. M1 for 2 18 360 x ×π× or 18 20 x π× × OR 828[. …] if volume formula used to find r 2 M1 for π × r × 18 where 14.6 ⩽ r ⩽ 14.7 8(a) y = 360 – x 1 90 180 270 360 1000 2000 3000 x y

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 7 Question Answer Marks Partial Marks 8(b) 2 2 0.0026 324 400 x V x = − + 2 2 (360 ) 0.0026(360 ) 324 400 x x − − − oe isw 1 FT substituting their y from part (a), only if y an expression in x with at least two terms. 8(c) Correct sketch 1 C opportunity 8(d) 116.6 to 117 243 to 243.4 2 B1 for each 9(a) 17957 to 18900 1 FT 8×their 7(f)(ii) C opportunity 9(b) 294 1 FT their positive 7(f)(i) less than 360 Communication: seen in three of the following questions 1 7(b) 10 xπ = 2πr 7(c) r2 + h2 = 182 or uses the word “Pythagoras” or a2 + b2 = c2 oe 7(d)(ii) Both substitutions of 180 seen 7(e) Appropriate scale indicated for V if approximately correct shape 8(c) Scale for x-axis e.g. 360 shown and gaps to 0 and soi 360 on V = 2500. 9(a) Factor of 2 or 8 seen, or implied from a written calculation. or 300 π 2 2 1296 100 −x x oe 0 90 180 270 360 2500 2600 2700 x V