Cambridge IGCSE Mathematics - International 0607 — 2021 Oct/Nov Paper 4 · Variant 2

This document has 20 pages. Any blank pages are indicated. Cambridge IGCSE™ DC (LK/FC) 212427/1 © UCLES 2021 [Turn over CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/42 Paper 4 (Extended) October/November 2021 2 hours 15 minutes You must answer on the question paper. You will need: Geometrical instruments INSTRUCTIONS ● Answer all questions. ● 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 and you will be given marks for correct methods, including sketches, even if your answer is incorrect. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. ● For r, use your calculator value. INFORMATION ● The total mark for this paper is 120. ● The number of marks for each question or part question is shown in brackets [ ]. * 1 0 5 9 6 7 0 8 6 1 *

2 0607/42/O/N/21 © UCLES 2021 Formula List For the equation ax bx c 0 2 + + = x a b b ac 2 4 2 ! = - - Curved surface area, A, of cylinder of radius r, height h. r A rh 2 = Curved surface area, A, of cone of radius r, sloping edge l. r A rl = Curved surface area, A, of sphere of radius r. r A r 4 2 = Volume, V, of pyramid, base area A, height h. V Ah 3 1 = Volume, V, of cylinder of radius r, height h. r V r h 2 = Volume, V, of cone of radius r, height h. r V r h 3 1 2 = Volume, V, of sphere of radius r. r V r 3 4 3 = sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin bc A 2 1 Area = A C B c b a

3 0607/42/O/N/21 © UCLES 2021 [Turn over Answer all the questions. 1 (a) Stella and Tomas share $200 in the ratio 11 : 14. (i) Show that Stella receives $88. [1] (ii) Stella invests her $88 at a rate of 1.5% per year simple interest. Calculate the amount of interest Stella has at the end of 6 years. $ … [2] (b) Urs buys some clothes in a sale. (i) He buys a jacket for $22. The original price of the jacket was $25. Calculate the percentage reduction in the price of the jacket. … % [3] (ii) Urs buys a shirt for $13.50 . This is the price after a reduction of 10% of the original price. Calculate the original price of the shirt. $ … [2]

4 0607/42/O/N/21 © UCLES 2021 2 ( ) f sin x x = for ° ° x 0 360 G G ( ) ( ) g sin x x 2 = for ° ° x 0 360 G G 90 180 270 360 - 1 1 0 y x (a) On the diagram, sketch the graph of ( ) f y x = . [2] (b) Write down the coordinates of the local minimum point on the graph of ( ) f y x = . ( … , … ) [1] (c) Write down the period and amplitude of the graph of ( ) f y x = . Period = … Amplitude = … [2] (d) On the same diagram, sketch the graph of ( ) g y x = . [2] (e) Write down the range of (i) ( ) f x , … [1] (ii) ( ) g x . … [1] (f) On the diagram, shade the regions where ( ) sin sin x x 2 H . [1]

5 0607/42/O/N/21 © UCLES 2021 [Turn over 3 (a) The number of members in a social media group increases exponentially at a rate of 5% per month. At the start of the first month there are 882 members. (i) Calculate the number of members at the end of 10 months. Give your answer correct to the nearest integer. … [3] (ii) Calculate the number of complete months from the start until the group has 2000 members. … [4] (b) The mass of a radioactive substance decreases exponentially at a rate of r % per month. At the end of 10 months, its mass has decreased from 500 g to 242 g. Find the value of r. r = … [3]

6 0607/42/O/N/21 © UCLES 2021 4 The mass of each of 200 potatoes is measured. The cumulative frequency curve shows the results. 100 0 20 40 60 80 100 120 140 160 180 200 120 140 160 180 200 Mass (g) Cumulative frequency

7 0607/42/O/N/21 © UCLES 2021 [Turn over (a) (i) Write down the mass of the heaviest potato. … g [1] (ii) Find the median. … g [1] (iii) Find the interquartile range. … g [2] (iv) Find the number of potatoes with a mass greater than 170 g. … [2] (b) This frequency table also shows information about the masses of the 200 potatoes. Mass (m g) m 100 140 1 G m 1 0 14 4 6 1 G m 1 1 46 62 1 G m 1 1 0 62 9 1 G Frequency 50 10 90 50 Calculate an estimate of the mean mass. … g [2]

8 0607/42/O/N/21 © UCLES 2021 5 (a) P T - 10 - 10 - 8 - 6 - 4 - 2 2 4 6 8 10 - 8 - 6 - 4 - 2 0 2 4 6 x y (i) Reflect shape T in the y-axis. [1] (ii) Translate shape T by the vector 10 5 - e o. [2] (iii) Rotate shape T through 90° clockwise about the point (2, 0). [2] (iv) Enlarge shape T with scale factor -2 and centre (0, 0). [2] (v) Describe fully the single transformation that maps shape T onto shape P. … … [3]

9 0607/42/O/N/21 © UCLES 2021 [Turn over (b) ( ) f x x 2 = (i) The graph of ( ) f y x = is mapped onto the graph of ( ) g y x = by a translation with vector 0 2 e o. Find ( ) g x in terms of x. ( ) g x = … [1] (ii) The graph of ( ) f y x = is mapped onto the graph of ( ) h y x = by a stretch with factor 2 and the x-axis invariant. Find ( ) h x in terms of x. ( ) h x = … [1]

10 0607/42/O/N/21 © UCLES 2021 6 (a) (i) Work out 3 5 2 1 2 - - - e e o o. f p [2] (ii) A is the point (3, 5) and C is the point (4, 3). Find the column vector that maps the point A onto the point C. f p [2] (iii) D is the point (1, 3) and the vector from D to E is 3 2 e o. Find the coordinates of E. ( … , … ) [1] (iv) Find the magnitude of the vector 3 4 - - e o. … [2]

11 0607/42/O/N/21 © UCLES 2021 [Turn over (b) (i) P is the point (–1, 6) and Q is the point (3, 4). Find the equation of the perpendicular bisector of the line PQ. … [5] (ii) Find the coordinates of the point where the perpendicular bisector in part(b)(i) crosses the x-axis. ( … , … ) [2]

12 0607/42/O/N/21 © UCLES 2021 7 (a) The cost of a newspaper is $p. The cost of a magazine is $m. The total cost of 3 newspapers and 5 magazines is $13.30 . The total cost of 1 newspaper and 7 magazines is $15.90 . Find the value of p and the value of m. p = … m = … [5]

13 0607/42/O/N/21 © UCLES 2021 [Turn over (b) x x 1 + NOT TO SCALE x 2 1 – x 2 1 – The area of the rectangle is equal to the area of the square. Find the value of x. x = … [7]

14 0607/42/O/N/21 © UCLES 2021 8 (a) ( ) f x x 3 2 = - ( ) g x x 5 1 = - ( ) , h x x x 1 1 1 ! = + - (i) Find (a) f(3), … [1] (b) h(f(3)). … [1] (ii) Find f(g(x)) in its simplest form. … [2] (iii) Solve ( ) ( ) f g x x = . x = … [2] (iv) Find ( ) g x 1 - . ( ) g x 1 = - … [2]

15 0607/42/O/N/21 © UCLES 2021 [Turn over (v) Simplify ( ) ( ) h h x x 2 1 + + . Give your answer as a single fraction, in terms of x, in its simplest form. … [4] (b) ( ) j x 5x = (i) Find the value of x when ( ) j x 5 5 1 = . x = … [1] (ii) Find ( ) j x 1 - . ( ) j x 1 = - … [2]

16 0607/42/O/N/21 © UCLES 2021 9 (a) Complete the table for each sequence. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 7 5 3 1 B 16 25 36 49 C 2 1 1 2 4 [9] (b) y x 1 ? and z y3 ? . When x 36 = , y 2 = and z 24 = . Find z in terms of x. z = … [4]

17 0607/42/O/N/21 © UCLES 2021 [Turn over 10 Fast trains and slow trains travel from City A to City B. 40% of the trains from City A to City B are fast trains. The probability that a fast train arrives in City B on time is 0.9 . The probability that a slow train arrives in City B on time is 0.95 . Manuela goes to the station in City A and takes the next train to City B. (a) Complete the tree diagram. Train Fast Slow 0.4 Arrives On time Not on time … … … … … On time Not on time [3] (b) Find the probability that Manuela arrives in City B on time. … [3]

18 0607/42/O/N/21 © UCLES 2021 11 5 cm 8 cm 20 cm A Q P C B D 100° NOT TO SCALE The diagram shows a solid triangular prism of length 20 cm. The cross-section of the prism is triangle BCP and three faces are rectangles. BC = 8 cm, CP = 5 cm and angle ADQ = angle BCP = 100°. (a) Calculate the total surface area of the prism. … cm2 [7]

19 0607/42/O/N/21 © UCLES 2021 (b) (i) On the diagram of the prism, draw two straight lines and mark angle PAC. [1] (ii) Angle APC = 73.45°. Calculate angle PAC. Angle PAC = … [4]

20 0607/42/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. BLANK PAGE

This document consists of 8 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/42 Paper 4 (Extended) October/November 2021 MARK SCHEME Maximum Mark: 120 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.

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 2 of 8 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.

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 3 of 8 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

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 4 of 8 Question Answer Marks Partial Marks 1(a)(i) 11 200 11 14 × + or 11 200 25 × oe M1 1(a)(ii) 7.92 2 M1 for 88 1.5 6 100 × × oe 1(b)(i) 12 3 M2 for 25 22 25 − oe or 22 100 25 × oe or M1 for 22 25 1(b)(ii) 15 2 M1 for100 10 [...] 13.50 100 − × = oe 2(a) Correct sketch 2 B1 for sine curve but different period and/or different amplitude 2(b) (270, –1) 1 2(c) 360 1 2 B1 for each 2(d) Correct sketch 2 B1 for (sine)2 curve but different period and/or different amplitude 2(e)(i) –1 ⩽ f(x) ⩽ 1 1 2(e)(ii) 0 ⩽ g(x) ⩽ 1 1 2(f) Correct shading 1 3(a)(i) 1437 3 B2 for 1440 or 1436.6 to 1436.7 or M1 for 882 × 10 5 1 100   +     oe 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1 350 300 250 200 150 100 50 0 1 0.5 0 -0.5 -1

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 8 Question Answer Marks Partial Marks 3(a)(ii) 17 nfww 4 B3 for 16.8 or 16.78... OR or M3 5 2000 log 1 log 100 882     + =         n oe or correct trials reaching 16 and 17 or good sketch indicating value between 16 and 17 or M2 for 5 2000 1 100 882   + =     n oe or at least 3 correct trials, n > 10 or suitable graph or M1 for 5 882 1 2000 100   × + =     n oe soi by at least 2 trials with n > 10 3(b) 7 or 6.999... nfww 3 M2 for 10 242 500 or M1 for 500 × (...)10 = 242 4(a)(i) 190 1 4(a)(ii) 156 1 4(a)(iii) 22 2 B1 for LQ = 140 or UQ = 162 4(a)(iv) 28 2 B1 for 172 seen 4(b) 150.45 2 M1 for at least 3 mid-values soi 5(a)(i) Image at (–2, 1), (–5, 1), (–5, 2) 1 5(a)(ii) Image at (–8, 6), (–5, 6) (–5, 7) 2 B1 for translation 10 −       k or 5    k 5(a)(iii) Image at (3, 0), (3, –3), (4, –3) 2 B1 for rotation 90˚ anticlockwise about (2, 0) or rotation 90˚ clockwise but incorrect centre. 5(a)(iv) Image at (–4, –2), (–10 –2), (–10, –4) 2 B1 for enlargement factor –2 correct orientation but incorrect position 5(a)(v) Stretch [factor] 3 x-axis oe invariant 3 B1 for each 5(b)(i) x2 + 2 1 5(b)(ii) 2x2 1

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 8 Question Answer Marks Partial Marks 6(a)(i) 5 9    2 B1 for 5    k or 9    k or M1 for 3 2 5 4 −    −    −    or 3 2 5 4   +     6(a)(ii) 1 2     −   2 B1 for 1    k or 2 k     −   If 0 scored SC1 for answer 1 2 −       6(a)(iii) (4, 5) 1 6(a)(iv) 5 2 M1 for 2 2 ( 3) ( 4) − + − oe 6(b)(i) 2 3 = + y x oe 5 B4 for answer 2x + 3 OR M1 for grad = 6 4 1 3 − −− oe or better M1 for perp grad = 1 gradient − their B1 for (1, 5) M1 for substituting their (1, 5) into y = (their m)x + c oe 6(b)(ii) (–1.5, 0) 2 FT their (b)(i) M1 for y = 0 substituted in their equation in (b)(i) 7(a) [p =] [0].85 [m =] 2.15 5 B1 for 3p + 5m = 13.30 B1 for p + 7m = 15.90 M1 for correctly eliminating one variable from their equations in terms of p and m or correct sketch B1 for p = [0].85 B1 for m = 2.15 If M0 scored then SC1 for values satisfying one of their equations

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 8 Question Answer Marks Partial Marks 7(b) 1.43 or 1.434... only 7 B6 for 1.43 or 1.434... and 0.232 or 0.2324... OR M1 for 2 (2 1) ( 1) − = + x x x B1 for 2 4 2 2 1 − − + x x x oe B1 for 2 + x x M1 for correct three term quadratic from their areas M1 for 2 ( 5) ( 5) 4(3)(1) 2(3) −− ± − − for their quadratic or sketch leading to roots of their quadratic 8(a)(i)(a) 7 1 8(a)(i)(b) 1 8 oe 1 FT their 7 8(a)(ii) 15x – 5 or 5(3x – 1) final answer 2 M1 for 3(5x – 1) – 2 8(a)(iii) 1 2 − oe 2 M1 for 3x – 2 = 5x – 1 8(a)(iv) 1 5 + x oe final answer 2 M1 for x = 5y – 1 or y + 1 = 5x or 1 5 5 = − y x 8(a)(v) 3 5 ( 1)( 2) + + + x x x or 2 3 5 3 2 + + + x x x final answer 4 M1 for 2 1 1 1 1 + + + + x x FT their fractions provided there are two fractions of the form + a b c to be collected and the denominators are different M1 for 2(x + 2) + x + 1 or better or M1 for denominator (x + 1)(x + 2) 8(b)(i) 3 2 − oe 1 8(b)(ii) 5 log x or log log5 x 2 M1 for 5 = y x or log log5 = y x or 5 log = y x

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 8 Question Answer Marks Partial Marks 9(a) Sequence A –1 9 – 2n oe final answer 3 B1 for –1 B1 for k – 2n or 9 – cn, 0 ≠ c or correct answer spoiled Sequence B 64 ( ) 2 3 + n oe final answer 3 B1 for 64 B1 for ( ) 2 + n k , k any integer or M1 for second differences = 2 Sequence C 8 2 2 − n oe final answer 3 B1 for 8 B1 for 2 + n k , k any integer 9(b) 3 2 5184 x oe 4 B3 for z = 3 2 k x or for 12 = y x and 3 3 = z y or B2 for 12 = y x or 3 3 = z y or M1 for = k y x or 3 = z ky 10(a) Correct tree diagram 0.6 oe 0.9 and 0.1 oe 0.95 and 0.05 oe 3 B1 for each 10(b) 0.93 oe 3 M2 for 0.4 × 0.9 + 0.6 × 0.95 or M1 for 0.4 × 0.9 or 0.6 × 0.95 11(a) 502 or 502.1 to 502.3 nfww 7 M2 for 2 2 8 5 2 8 5 cos100 + − × × × or M1 for 2 2 2 8 5 cos100 2 8 5 + − = × × BP M1 for [2 ×] 1 8 5 sin100 2 × × × M1 for 20 × 8 M1 for 20 × 5 M1 for their BP × 20 11(b(i) PA and CA drawn and arc at A 1 11(b)(ii) 12.9 or 12.85 to 12.86 4 M3 for 2 2 5sin73.45 sin[... ] 20 8 = + OR M1 for 2 2 20 8 + M1 for 5 sin73.45 sin = their AC PAC oe