Cambridge IGCSE Mathematics - Additional 0606 — 2019 Oct/Nov Paper 1 · Variant 3

This document consists of 15 printed pages and 1 blank page. DC (SC/FC) 172297/1 © UCLES 2019 [Turn over * 9 4 2 8 0 1 3 3 1 3 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 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

2 0606/13/O/N/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

3 0606/13/O/N/19 © UCLES 2019 [Turn over 1 In a group of 145 students, the numbers studying mathematics, physics and chemistry are given below. All students study at least one of the three subjects. x students study all 3 subjects 24 students study both mathematics and chemistry 23 students study both physics and chemistry 28 students study both mathematics and physics 50 students study chemistry 75 students study physics 80 students study mathematics (i) Using the Venn diagram, find the value of x. [4] Mathematics Physics Chemistry (ii) Find the number of students who study mathematics only. [1]

4 0606/13/O/N/19 © UCLES 2019 2 (i) On the axes below, sketch the graph of cos y x 5 4 3 = - for ° °. x 90 90 G G - y x 0 2 - 2 - 4 - 6 - 8 - 10 4 6 45° -45° -90° 90° [4] (ii) Write down the amplitude of y. [1] (iii) Write down the period of y. [1]

5 0606/13/O/N/19 © UCLES 2019 [Turn over 3 (i) Differentiate y x 3 1 2 3 1 = - - ` j with respect to x. [2] (ii) Find the approximate change in y as x increases from 3 to ,p 3 + where p is small. [1] (iii) Find the equation of the normal to the curve y x 3 1 2 3 1 = - - ` j at the point where . x 3 = [3]

6 0606/13/O/N/19 © UCLES 2019 4 It is given that . A 5 4 2 1 = - e o (i) Find A-1. [2] (ii) Hence find, in radians, the acute angles x and y such that , . tan tan tan tan x y x y 5 2 12 4 7 + = - = [5]

7 0606/13/O/N/19 © UCLES 2019 [Turn over 5 (i) Differentiate ( ) ( ) ln x x 3 3 2 2 + + with respect to x. [3] (ii) Hence find ( ) . ln x x x 3 d 2 + y [2]

8 0606/13/O/N/19 © UCLES 2019 6 x 1 1.5 2 2.5 3 y 6 14.3 48 228 1536 The table shows values of the variables x and y such that , y Abx2 = where A and b are constants. (i) Draw a straight line graph to show that . y Abx2 = [4]

9 0606/13/O/N/19 © UCLES 2019 [Turn over (ii) Use your graph to find the value of A and of b. [4] (iii) Estimate the value of x when . y 100 = [2]

10 0606/13/O/N/19 © UCLES 2019 7 (a) A 5-digit code is to be chosen from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. Each digit may be used once only in any 5-digit code. Find the number of different 5-digit codes that may be chosen if (i) there are no restrictions, [1] (ii) the code is divisible by 5, [1] (iii) the code is even and greater than 70 000. [3] (b) A team of 6 people is to be chosen from 8 men and 6 women. Find the number of different teams that may be chosen if (i) there are no restrictions, [1] (ii) there are no women in the team, [1] (iii) there are a husband and wife who must not be separated. [3]

11 0606/13/O/N/19 © UCLES 2019 [Turn over 8 (a) Given that log x p a = and , log y q a = find, in terms of p and q, (i) , log axy a 2 [2] (ii) log ay x a 3 e o, [2] (iii) . log log a a x y + [1] (b) Using the substitution , m 3x = or otherwise, solve . 3 3 4 0 x x 1 2 - + = + [3]

12 0606/13/O/N/19 © UCLES 2019 9 P Q R O S i rad 2r cm r cm The diagram shows a sector OPQ of the circle centre O, radius 3r cm. The points S and R lie on OP and OQ respectively such that ORS is a sector of the circle centre O, radius 2r cm. The angle POQ i = radians. The perimeter of the shaded region PQRS is 100 cm. (i) Find i in terms of r. [2] (ii) Hence show that the area, A cm2, of the shaded region PQRS is given by . A r r 50 2 = - [2]

13 0606/13/O/N/19 © UCLES 2019 [Turn over (iii) Given that r can vary and that A has a maximum value, find this value of A. [2] (iv) Given that A is increasing at the rate of 3 cm2 s-1 when r = 10, find the corresponding rate of change of r. [3] (v) Find the corresponding rate of change of i when r = 10. [3]

14 0606/13/O/N/19 © UCLES 2019 10 (a) v ms-1 t s 0 - 20 20 5 10 The velocity-time graph for a particle P is shown by the two straight lines in the diagram. (i) Find the deceleration of P for . t 5 10 G G [2] (ii) Write down the value of t when the speed of P is zero. [1] (iii) Find the distance P has travelled for . t 0 10 G G [2]

15 0606/13/O/N/19 © UCLES 2019 (b) A particle Q has a displacement of x m from a fixed point O, t s after leaving O. The velocity, v ms-1, of Q at time t s is given by . v 6 1 e t2 = + (i) Find an expression for x in terms of t. [3] (ii) Find the value of t when the acceleration of Q is 24 ms-2. [3]

16 0606/13/O/N/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. BLANK PAGE

This document consists of 9 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 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 October/November 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 2 of 9 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 3 of 9 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 4 of 9 Question Answer Marks Guidance 1(i) M1 for a Venn diagram showing at least 4 correct ‘parts’ in terms of x A1 for all 7 ‘parts’ correct in terms of x on a Venn diagram or in working. May be implied by a correct equation. 80 24 23 3 145 + + + − + + = x x x 50 28 28 24 145 + + + − + + = x x x 75 28 24 3 145 + + + − + + = x x x ( ) 50 80 75 23 28 24 145 + + − + + + = x or equivalents M1 for forming an equation in x using sum of ‘parts’ = 145 or ( ) 50 80 75 23 28 24 145 + + − + + + = x Equations must be seen 15 = x A1 from correct working only 1(ii) 43 B1ft for their x plus 28 2(i) B4 B1 for a maximum at ( ) 0, 2 B1 for minimums at 8 = − y and no other minimums B1 for starting at ( ) o 90 , 2 − and finishing at ( ) o 90 , 2 B1 for a fully correct curve with correct shape, particularly at end points, that has earned all three previous B marks. 2(ii) 5 B1 2(iii) o 90 B1 3(i) ( ) 4 2 3 d 3 1 d − = − y kx x x M1 ( ) 4 2 3 d 1 6 3 1 d 3 − = − × − y x x x A1 3(ii) When d 3 3, d 8 = = − y x x 3 or 0.217 8 − − p p B1 FT on their d d y x of the form ( ) 4 2 3 3 1 − − kx x

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 5 of 9 Question Answer Marks Guidance 3(iii) When 3, = x 1 2 = y B1 for 1 2 = y Normal: ( ) 1 8 3 2 3 − = − y x M1 Dep on M1in part(i). An equation of the normal using their normal gradient, 3 and their y A1 allow unsimplified 4(i) 1 2 1 4 5 13 − −   −   −   oe B2 B1 for 1 13 − B1 for 1 2 4 5 − −     −   4(ii) 1 2 12 1 4 5 7 13       −    26 1 13 13   =     M1 for pre-multiplication by their inverse from (i) M1 for correct method for matrix multiplication 2 1  =   A1 1.11 = x B1 π 4 = y or 0.785 B1 5(i) ( ) ( ) 2 d ln 3 d + x x = ( ) 2 2 3 + x x B1 ( )( ) ( ) 2 2 2 d 2 3 2 ln 3 d 3 = + + + + y x x x x x x M1 for product rule A1 FT their ( ) 2 2 3 + x x 5(ii) ( ) ( ) ( ) 2 2 2 3 ln 3 2 2 ln 3 d + + = + + ∫ x x x x x x M1 for using their result from (i) for ( ) 2 2 ln 3 + + x kx x ( ) ( ) ( ) ( ) 2 2 2 2 ln 3 d 1 3 ln 3 2 2 + = + + − + ∫x x x x x x c A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 6 of 9 Question Answer Marks Guidance 6(i) 2 ln ln ln = + y A x b or 2 lg lg lg = + y A x b B1 May be implied by a table of values for 2 x and ln y or lg y or axes labelled ln y or lg y and 2 x M1 for attempt to plot either ln y or lg y against 2 x using an evenly spaced scale on each axis. A2 A2 All points on a correct line (for 1 ⩽ x2 ⩽ 9) with axes correctly labelled A1 One point not on the correct line or a correct line with axes not correctly labelled. A0 Two or more points not on the correct line or one point not on the line and axes incorrect 6(ii) Gradient = lnb or lgb ln 0.7 orlg 0.3 ≈ ≈ b b leading to M1 for a complete method using the gradient of their straight-line graph of lgy or lny against x2 to obtain b 2 = b (allow 1.6 – 2.4) A1 from correct working Intercept = ln A or lg A ln 1.1 ≈ A lg 0.5 ≈ A leading to M1 for a complete method using intercept of their straight-line graph of lgy or lny against x2 to find A 3 = A (allow 2.5 – 3.6) A1 from correct working 6(iii) ( ) 2 100 3 2 = x or 2 ln100 1.1 0.7 = + their their x or 2 lg100 0.5 0.3 = + their their x or reading from lgy = 2 to obtain 2 x or from lny = 4.6 to obtain 2 x M1 for a valid method to find 2 x Substitution methods should be using values of A and b in range leading to 2.25 = x (allow 2.0 – 2.7) A1 for an answer in range from correct working 7(a)(i) 15 120 B1 7(a)(ii) 1680 B1 7(a)(iii) Method 1 Total = 2310 B3 B1 1st digit is 7 or 9 1680 or 210 8 × B1 1st digit is 8 630 or 210 3 × 7(a)(iii) Method 2 Total = 2310 B3 B1 for 5th digit is 2,4 or 6 1890 or 210 9 × B1 for 5th digit is 8 420 or 210 2 ×

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 7 of 9 Question Answer Marks Guidance 7(b)(i) 3003 B1 7(b)(ii) 28 B1 7(b)(iii) Total 1419 B3 B1 Including husband and wife 495 B1 Excluding husband and wife 924 8(a)(i) log 2log log + + a a a a y x M1 for 2 log log log + + a a a a x y and 2 log 2log = a a y y 1 2 + + q p A1 8(a)(ii) 3log log log − − a a a x y a M1 for 3 log (log log ) − + a a a x a y and 3 log 3log = a a x x 3 1 − − p q or ( ) 3 1 − + p q A1 8(a)(iii) 1 1 + p q B1 8(b) 2 3 4 0 − + = m m 4 3 = m , ( ) 1 − 4 3 lg lg3 = x , 4 3 ln ln3 = x or 3 4 lg 3 M1 for obtaining a quadratic in m or 3x M1 Dep for attempt to solve quadratic and deal with 3x correctly 0.262 = x only A1 9(i) 100 2 2 3 θ θ = + + r r r M1 for addition of 2r and two arc lengths with at least one correct arc length 100 2 5 θ − = r r or 20 2 5 − r oe A1 9(ii) 2 2 1 1 9 4 2 2 θ θ − r r M1 for subtraction of two sector areas with at least one sector area correct. 2 5 100 2 2 5 −       r r r A1 Must expand and simplify to obtain given answer 2 50 − r r 9(iii) d 50 2 d = − A r r 0 50 2 = −r leading to 25 = r M1 for differentiation and equating to zero and obtaining r or for using completing the square ( ) 2 2 25 25 − − + r Max when 625 = A A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 8 of 9 Question Answer Marks Guidance 9(iv) When d 10, 30 d = = A r r B1 d 3 d 30 = r t M1 for d 3 d 30 = r t their where their 30 has been obtained from an evaluation of d d A r at r =10 d 0.1 d = r t or 1 10 A1 9(v) 2 d 20 d θ = − r r oe B1 d 1 d 5 θ = − r oe d 1 1 d 10 5 θ = × − t oe M1 for d d d d θ × r their their t r with both evaluated at r =10 d 1 or 0.02 d 50 θ = − − t A1 10(a)(i) 20 20 5 −− ± M1 for finding the gradient of the relevant part 8 A1 10(a)(ii) 7.5 B1 10(a)(iii) ( ) 1 1 5 7.5 20 2.5 20 2 2   + + × ×     or 1 1 20 5+ 2.5 20 2.5 20 2 2     × × × + × ×         oe M1 for a correct expression for total area using their 7.5 150 A1 10(b)(i) 2 3e = + + t x t c M1 for 2e + t k t Condone omission of c 0 0 3e 0 = + + c When 0, 0 = = t x so 3 = − c M1 Dep for substitution to find c 2 3e 3 = + − t x t A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 9 of 9 Question Answer Marks Guidance 10(b)(ii) 2 d 12e d = t v t so 2 12e 24 = t M1 for 2e t k equated to 24 2 ln2 = t M1 Dep for correct order of operations to obtain 2t 1 ln 2 2 = t , ln 2 or 0.347 A1