Cambridge IGCSE Mathematics - Additional 0606 — 2019 May/June Paper 1 · Variant 3

0606/13/M/J/19 · 80 marks · ≈90 min

The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.

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Question paper16 pages

Cambridge IGCSE Mathematics - Additional 0606 2019 May/June Paper 1 · Variant 3 question paper, page 1 of 16
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Mark scheme8 pages

Answers below. Sit the paper first if you are practising.

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Paper as text

Question paper, page 1

This document consists of 16 printed pages. DC (KN/TP) 165276/2 © UCLES 2019 [Turn over * 6 9 3 8 3 3 3 3 6 1 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2019 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic 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. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80. Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education

Question paper, page 2

2 0606/13/M/J/19 © UCLES 2019 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x a b b ac 2 4 2 ! = - - Binomial Theorem (a + b)n = an + ( n 1)an–1 b + ( n 2)an–2 b2 + … + ( n r)an–r br + … + bn, where n is a positive integer and ( n r) = n! (n – r)!r! 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A ∆ = 1 2 bc sin A

Question paper, page 3

3 0606/13/M/J/19 © UCLES 2019 [Turn over 1 Describe, using set notation, the relationship between the sets shown in each of the Venn diagrams below. A X Y Z B … … [3] 2 Given that p q r qr p q r p 1 a b c 2 2 3 1 = - ` j , find the value of each of the constants a, b and c. [3]

Question paper, page 4

4 0606/13/M/J/19 © UCLES 2019 3 Show that the line y mx 4 = + will touch or intersect the curve y x x m 3 2 = + + for all values of m. [4]

Question paper, page 5

5 0606/13/M/J/19 © UCLES 2019 [Turn over 4 It is given that ( ) ln y x x 1 2 5 3 = - + for x 1 2 . (i) Find the value of x y d d when x 2 = . You must show all your working. [4] (ii) Find the approximate change in y as x increases from 2 to p 2+ , where p is small. [1]

Question paper, page 6

6 0606/13/M/J/19 © UCLES 2019 5 (i) On the axes below, sketch the graph of y x x 3 14 5 2 = - - , showing the coordinates of the points where the graph meets the coordinate axes. y x 0 –1 6 [4] (ii) Find the exact value of k such that x x k 3 14 5 2 - - = has 3 solutions only. [3]

Question paper, page 7

7 0606/13/M/J/19 © UCLES 2019 [Turn over 6 (a) (i) Show that sec tan cosec cos i i i i - = . [3] (ii) Solve sec cosec tan 2 2 2 2 3 i i i - = for 0 180 G G i c c. [3] (b) Solve r sin 2 3 1 2 z+ = b l for r 0 2 1 1 z radians. [4]

Question paper, page 8

8 0606/13/M/J/19 © UCLES 2019 7 Do not use a calculator in this question. In this question, all lengths are in centimetres. A B C 2 5 1 - 2 5 2+ The diagram shows the triangle ABC such that AB 2 5 1 = - , BC 2 5 = + and angle ABC 90 = c. (i) Find the exact length of AC. [3]

Question paper, page 9

9 0606/13/M/J/19 © UCLES 2019 [Turn over (ii) Find tanACB, giving your answer in the form p q r + , where p, q and r are integers. [3] (iii) Hence find sec ACB 2 , giving your answer in the form s t u + where s, t and u are integers. [2]

Question paper, page 10

10 0606/13/M/J/19 © UCLES 2019 8 x f e x 3 7 | for x R d x x 2 1 g 2 7 | + for x 0 H (i) Write down the range of g. [1] (ii) Show that ( ) ln 62 5 f g 1 = - . [3] (iii) Solve ( ) ( ) x x 6 f g = l l l , giving your answer in the form lna, where a is an integer. [3]

Question paper, page 11

11 0606/13/M/J/19 © UCLES 2019 [Turn over (iv) On the axes below, sketch the graph of g y = and the graph of g y 1 = - , showing the points where the graphs meet the coordinate axes. y x O [3]

Question paper, page 12

12 0606/13/M/J/19 © UCLES 2019 9 (a) Jack has won 7 trophies for sport and wants to arrange them on a shelf. He has 2 trophies for cricket, 4 trophies for football and 1 trophy for swimming. Find the number of different arrangements if (i) there are no restrictions, [1] (ii) the football trophies are to be kept together, [3] (iii) the football trophies are to be kept together and the cricket trophies are to be kept together. [3]

Question paper, page 13

13 0606/13/M/J/19 © UCLES 2019 [Turn over (b) A team of 8 players is to be chosen from 6 girls and 8 boys. Find the number of different ways the team may be chosen if (i) there are no restrictions, [1] (ii) all the girls are in the team, [1] (iii) at least 1 girl is in the team. [2]

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14 0606/13/M/J/19 © UCLES 2019 10 A curve is such that x y x 2 3 d d 2 2 2 1 = + - ` j . The curve has a gradient of 5 at the point where x 3 = and passes through the point , 2 1 3 1 - e o. (i) Find the equation of the curve. [7]

Question paper, page 15

15 0606/13/M/J/19 © UCLES 2019 [Turn over (ii) Find the equation of the normal to the curve at the point where x 3 = , giving your answer in the form ax by c 0 + + = , where a, b and c are integers. [4] Question 11 is printed on the next page.

Question paper, page 16

16 0606/13/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. 11 A pilot wishes to fly his plane from a point A to a point B. The bearing of B from A is 050°. A wind is blowing from the north at a speed of 120 km h–1. The plane can fly at 600 km h–1 in still air. (i) Find the bearing on which the pilot must fly his plane in order to reach B. [4] (ii) Given that the distance from A to B is 2500 km, find the time taken to fly from A to B. [4]

Mark scheme, page 1

This document consists of 8 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2019 MARK SCHEME Maximum Mark: 80 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.

Mark scheme, page 2

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 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.

Mark scheme, page 3

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 3 of 8 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

Mark scheme, page 4

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 8 Question Answer Marks Guidance 1 ∩ = ∅ A B B1 ( ) ⊂ ∩ Z X Y B2 B1 for identifying ∩ X Y 2 3 2 = a B1 7 3 = b B1 3 = c B1 3 ( ) 2 3 4 0 + − + − = x m x m M1 For equating line and curve and attempting to obtain a quadratic equation equated to zero Discriminant:( ) ( ) 2 3 4 4 − − − m m M1 Dep For use of 2 4 − b ac , could be implied by use of quadratic formula ( ) 2 5 − m A1 Always positive or zero for any m, so line and curve will always touch or intersect A1 For a suitable comment/conclusion 4(i) B1 For ( ) 3 3 6 2 5 + x x M1 For attempt to differentiate a quotient ( )( ) ( ) ( ) 2 3 3 2 6 1 ln 2 5 2 5 d d 1 − − + + = − x x x x y x x A1 For all other terms correct When 2, x = d 24 ln21 d 21 y x = − or 8 ln 21, 7 − or –1.90 A1 4(ii) 1.90 − p oe B1

Mark scheme, page 5

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 8 Question Answer Marks Guidance 5(i) B1 For shape with maximum in 1st quadrant B1 For 1, 0 3   −     and ( ) 5, 0 B1 For ( ) 0, 5 B1 All correct with cusps and correct shape for 1 3 < − x and 5 > x 5(ii) M1 For attempt to find maximum point Maximum point when 7 3 = x A1 For 7 3 = x 64 3 = y so 64 3 = k A1 6(i) 1 sin sin cos cos θ θ θ θ − × oe M1 For dealing with sec, tan and cosec in terms of sin and cos 2 1 sin cos θ θ − M1 For simplification and use of identity 2 cos cos θ θ A1 For simplification to AG 6(ii) 3 cos2 2 θ = M1 For use of part (i) and attempt to solve to get as far as 2 ... θ = o o 2 30 , 330 θ = M1 For dealing with double angle correctly, may be implied by one correct solution o o 15 , 165 θ = A1 For both 6(iii) π 1 sin 3 2 φ   + = ±     π π 3π 5π 7π 9π , , , , 3 4 4 4 4 4 φ + = M1 For correct attempt to solve, may be implied by π π 3 4 φ + = M1 Dep For dealing with compound angle correctly 5π 11π 17π 23π , , , 12 12 12 12 φ = A2 A1 for one correct pair, A1 for a second correct pair with no extra solutions in the range.

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0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 8 Question Answer Marks Guidance 7(i) ( ) ( ) 2 2 2 2 5 1 2 5 = − + + AC M1 For use of Pythagoras’ theorem and attempt to expand brackets 20 4 5 1 4 4 5 5 = − + + + + A1 For correct unsimplified, must be convinced of non-calculator use 30 = AC A1 7(ii) 2 5 1 2 5 tan 2 5 2 5 − − = × + − ACB M1 For attempt at tan ACB and rationalisation 4 5 2 10 5 4 5 − − + = − oe M1 Dep For seeing at least 3 terms in the numerator 12 5 5 = − A1 7(iii) 2 2 sec tan 1 = + ACB ACB 144 120 5 125 1 = − + + M1 For use of identity using their (ii) 270 120 5 = − A1 8(i) g 1 ≥ B1 Must be using correct notation 8(ii) ( ) g 62 125 = B1 ( ) 1 1 f ln 3 − = x x B1 1ln125 ln5 3 = B1 For correct order and manipulation to obtain the given answer, need to see 1ln125 3 8(iii) 3 3e 24 = x M1 For dealing with derivatives correctly 1ln8 3 = x A1 ln 2 = x A1 8(iv) B3 B1 for correct g with intercept B1 for = y x and/or implication of symmetry B1 for correct 1 g− with intercept 9(a)(i) 7! 5040 = B1

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0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 8 Question Answer Marks Guidance 9(a)(ii) Treating the 4 trophies as 1 unit so there are 4! ways B1 Maybe implied by a correct answer There are also 4! ways of arranging the football trophies amongst themselves B1 Total = 4! 4! 576 × = B1 9(a)(iii) Treating the 4 football trophies as 1 unit and the 2 cricket trophies as 1 unit so there are 3! ways B1 Maybe implied by a correct answer There are also 4! ways of arranging the football trophies amongst themselves and 2 ways of arranging the cricket trophies B1 Maybe implied by a correct answer Total = 3! 4! 2 288 × × = B1 9(b)(i) 3003 B1 9(b)(ii) 28 B1 9(b)(iii) 3003 – 1 M1 For their (i) – 1 3002 A1 FT 10(i) M1 Attempt to integrate to obtain ( ) 1 2 2 3 + k x ( ) ( ) 1 2 d 2 3 d = + + y x c x A1 All correct, condone omission of +c 5 3 = + c M1 Dep For attempt at c ( ) 1 2 d 2 3 2 d = + + y x x M1 For a further attempt to integrate ( ) ( ) 3 2 1 2 3 2 3 = + + + y x x d A1 All correct, condone omission of +d 1 8 1 3 3 − = + + d M1 For attempt at d ( ) 3 2 1 2 3 2 4 3 = + + − y x x A1 Must have y =

Mark scheme, page 8

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 8 Question Answer Marks Guidance 10(ii) When 3, 11 = = x y M1 For attempt to find y using their (i) M1 Dep For attempt at normal Normal: ( ) 1 11 3 5 − = − − y x A1 Al1 correct unsimplified 5 58 0 + − = x y A1 For correct form 11(i) B1 For correct triangle, may be implied by subsequent work 120 600 sin sin130 α = M1 For use of the correct sine rule o 8.81 α = A1 Allow greater accuracy Bearing o 041.2 or o 041 A1 Allow greater accuracy 11(ii) 600 120 sin41.19 sin130 sinα = = rv M1 For use of sine rule using their α or cosine rule 515.8 = rv awrt 516 A1 Time taken = 2500 515.8 M1 For attempt to find time using their rv , not 600, 720 or 480 4.85 = or 4.84 A1

What you needed in this session

Cambridge’s own grade thresholds for 2019 May/June, Paper 1 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.

A64/80
B48/80
C31/80
D25/80
E19/80