Cambridge IGCSE Mathematics - Additional 0606 — 2020 Feb/March Paper 1 · Variant 2

DC (NF/CB) 196735/3 R © UCLES 2020 [Turn over This document has 16 pages. Blank pages are indicated. * 6 5 6 2 3 4 6 4 7 4 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 2020 2 hours You must answer on the question paper. No additional materials are needed. 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 calculator where appropriate. ● You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ● 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. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ]. Cambridge IGCSE™

2 0606/12/F/M/20 © UCLES 2020 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 a n a b n a b n r a b b 1 2 n n n n n r r n 1 2 2 + = + + + + + + - - - J L KK J L KK J L KK N P OO N P OO N P OO where n is a positive integer and ( )! ! ! n r n r r n = - J L KK N P OO Arithmetic series u a n d S n a l n a n d 1 2 1 2 1 2 1 n n − − = + = + = + ^ ^ ^ h h h # - Geometric series u ar S r a r r S r a r 1 1 1 1 1 n n n n 1 1 ! − − − = = = 3 − ^ ^ ^ h h h 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC sin sin sin A a B b C c = = a2 = b2 + c2 – 2bc cos A ∆ = 2 1 bc sin A

3 0606/12/F/M/20 © UCLES 2020 [Turn over 1 (a) On the axes below sketch the graph of ( )( )( ) y x x x 3 2 4 1 =- - - + , showing the coordinates of the points where the curve intersects the coordinate axes. [3] O y x (b) Hence find the values of x for which ( )( )( ) x x x 3 2 4 1 0 2 - - - + . [2]

4 0606/12/F/M/20 © UCLES 2020 2 Find the values of k for which the line y kx 3 = + is a tangent to the curve y x x k 2 4 1 2 = + + - . [5]

5 0606/12/F/M/20 © UCLES 2020 [Turn over 3 The first 3 terms in the expansion of ( ) ax 3 5 - , in ascending powers of x, can be written in the form b x cx 81 2 - + . Find the value of each of a, b and c. [5]

6 0606/12/F/M/20 © UCLES 2020 4 The tangent to the curve ln y x x 3 4 6 2 3 = - - ` j , at the point where , x 2 = meets the y-axis at the point P. Find the exact coordinates of P. [6]

7 0606/12/F/M/20 © UCLES 2020 [Turn over 5 DO NOT USE A CALCULATOR IN THIS QUESTION. In this question all lengths are in centimetres. A B C D 2 4 3 + 5 3 - The diagram shows the isosceles triangle ABC, where AB AC = and BC 2 4 3 = + . The height, AD, of the triangle is 5 3 - . (a) Find the area of the triangle ABC, giving your answer in the form a b 3 + , where a and b are integers. [2] (b) Find , tan ABC giving your answer in the form c d 3 + , where c and d are integers. [3] (c) Find sec ABC 2 , giving your answer in the form e f 3 + , where e and f are integers. [2]

8 0606/12/F/M/20 © UCLES 2020 6 Solutions by accurate drawing will not be accepted. The points A and B have coordinates ( , ) 2 4 - and ( , ) 6 10 respectively. (a) Find the equation of the perpendicular bisector of the line AB, giving your answer in the form ax by c 0 + + = , where a, b and c are integers. [4] The point C has coordinates ( , ) p 5 and lies on the perpendicular bisector of AB. (b) Find the value of p. [1] It is given that the line AB bisects the line CD. (c) Find the coordinates of D. [2]

9 0606/12/F/M/20 © UCLES 2020 [Turn over 7 ( )x ax x bx 3 12 p 3 2 = + + - has a factor of x 2 1 + . When ( )x p is divided by x 3 - the remainder is 105. (a) Find the value of a and of b. [5] (b) Using your values of a and b, write ( )x p as a product of x 2 1 + and a quadratic factor. [2] (c) Hence solve ( )x 0 p = . [2]

10 0606/12/F/M/20 © UCLES 2020 8 In this question all distances are in km. A ship P sails from a point A, which has position vector 0 0 e o, with a speed of 52 kmh-1 in the direction of 5 12 - e o. (a) Find the velocity vector of the ship. [1] (b) Write down the position vector of P at a time t hours after leaving A. [1] At the same time that ship P sails from A, a ship Q sails from a point B, which has position vector 12 8 e o , with velocity vector 25 45 - e o kmh-1. (c) Write down the position vector of Q at a time t hours after leaving B. [1] (d) Using your answers to parts (b) and (c), find the displacement vector PQ at time t hours. [1]

11 0606/12/F/M/20 © UCLES 2020 [Turn over (e) Hence show that PQ t t 34 168 208 2 = - + . [2] (f) Find the value of t when P and Q are first 2 km apart. [2]

12 0606/12/F/M/20 © UCLES 2020 9 (a) (i) Find how many different 4-digit numbers can be formed using the digits 2, 3, 5, 7, 8 and 9, if each digit may be used only once in any number. [1] (ii) How many of the numbers found in part (i) are divisible by 5? [1] (iii) How many of the numbers found in part (i) are odd and greater than 7000? [4]

13 0606/12/F/M/20 © UCLES 2020 [Turn over (b) The number of combinations of n items taken 3 at a time is 92n. Find the value of the constant n. [4]

14 0606/12/F/M/20 © UCLES 2020 10 (a) Solve ° tan 45 2 1 a+ =- ` j for 0 360 ° ° G G a . [3] (b) (i) Show that sin sin sec a 1 1 1 1 2 i i i - - + = , where a is a constant to be found. [3]

15 0606/12/F/M/20 © UCLES 2020 [Turn over (ii) Hence solve sin sin 3 1 1 3 1 1 8 z z - - + =- for 3 3 G G r r z - radians. [5] Question 11 is on the next page.

16 0606/12/F/M/20 © UCLES 2020 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 Given that . ln x x x x 2 3 2 3 1 3 1 2 4 d a 1 + + - - = J L KK N P OO y and that a 1 2 , find the value of a. [7]

This document consists of 10 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 12 March 2020 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 March 2020 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 2 of 10 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 3 of 10 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 4 of 10 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 Guidance 1(a) 3 B1 For correct shape with minimum point in the fourth quadrant and the maximum point in the first quadrant. Ends of the curve must be in the 2nd and 4th quadrants B1 for correct x- intercepts ( ) ( ) ( ) 1,0 , 2,0 , 4,0 − B1 for correct y-intercept ( ) 0, 24 − 1(b) 1 < − x B1 2 4 < < x B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 5 of 10 Question Answer Marks Guidance 2 2 2 4 1 3 + + −= + x x k kx ( ) ( ) 2 2 4 4 0 + − + − = x k x k 2 M1 for attempt to equate the line and curve and simplify to a 3 term quadratic equation = 0 A1 for a correct equation, allow equivalent form ( ) ( ) 2 4 4 2 4 − = × × − k k M1 Use of discriminant in any form 2 16 48 0 − + = k k 12 = k , 4 = k Do not isw 2 Dep M1 on previous M mark, for attempt to solve a quadratic equation in k A1 for both Alternative 1 2 2 4 1 3 + + −= + x x k kx ( ) ( ) 2 2 4 4 0 + − + − = x k x k (2 M1 for attempt to equate the line and curve and simplify A1 for a correct equation, allow equivalent form 4 4 = + k x ( ) ( ) 2 4 4 2 4 4 0 4 4 − −     + − + − =         k k k k M1 Equating gradients and substitution to obtain a quadratic equation in terms of k 2 16 48 0 − + = k k 12 = k and 4 = k Do not isw 2) Dep M1 on previous M mark, for attempt to solve a quadratic equation in k A1 for both Alternative 2 2 2 4 1 3 + + −= + x x k kx ( ) ( ) 2 2 4 4 0 + − + − = x k x k (2 M1 for attempt to equate the line and curve and simplify A1 for a correct equation, allow equivalent form 4 4 = + k x 2 2 4 0 − = x x 0, 2 = x M1 Equating gradients and substitution to obtain a quadratic equation in terms of x and solution of this equation to obtain 2 x values 4 4 = + k x 12 = k and 4 = k Do not isw 2) Dep M1 on previous M mark, for substitution of their x values to obtain k values A1 for both

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 6 of 10 Question Answer Marks Guidance 3 243 = b B1 Must be evaluated ( ) 5 4 1 3 81 × × − = − C a M1 Allow equivalent with no negative signs, allow sign error 1 5 = a oe A1 ( ) 2 5 3 2 3 × × − C a M1 Allow with their 2 a 54 5 = c or 10.8 oe A1 Must be from correct working 4 2 2 d 6 d 2 3 4 = − − y x x x x 2 M1 for attempt to differentiate, must have at least one term correct A1 All correct When d 1 2, d 2 = = − y x x B1 When 4 2, ln8 3 = = − x y , or exact equivalent B1 Allow 8 ln8 6 − Equation of tangent ( ) 4 1 ln8 2 3 2   − − = − −     y x oe M1 Dep on first M mark, allow unsimplified, allow use of decimals 1 0, ln8 3   −     , or exact equivalent A1 Allow 1 0, ln8 3 = = − x y 5(a) ( )( ) 1 5 3 2 4 3 2 − + ( ) 1 10 2 3 20 3 12 2 − + − M1 Need to see ( ) 1 10 18 3 12 2 − − or ( ) 5 9 3 6 − − minimum for M1 9 3 1 − A1 5(b) 5 3 1 2 3 tan 1 2 3 1 2 3 − − = × + − ABC 5 3 10 3 6 1 12 − − + = − M1 Attempt at trig ratio and attempt to rationalise. Need to see 5 11 3 6 − + in the numerator as a minimum for M1 Allow one error only 3 1 = − 2 A1 for 3 , A1 for 1 −

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 7 of 10 Question Answer Marks Guidance 5(c) 2 2 sec tan 1 = + ABC ABC ( ) 2 3 1 1 = − + oe M1 Allow use of correct identity with their (b) 5 2 3 = − A1 Alternative ( ) ( ) 2 2 2 2 5 3 1 2 3 sec 1 2 3   − + +   =   +     ABC leads to 41 6 3 13 4 3 − + leads to 533 72 242 3 121 + − (M1 For a complete method using triangle ABD, with sufficient detail in the expansions and rationalisation 5 2 3 = − A1) 6(a) Midpoint = ( ) 2,7 B1 Gradient of AB = 6 8 oe B1 Perp bisector: ( ) 4 7 2 3 − = − − y x M1 Must be using a perp gradient and a mid-point 4 3 29 0 + − = x y A1 Allow in any order but must be equated to zero. 6(b) 3 B1 FT on their (a) 6(c) Displacement vector 3 4 CM −   =      M1 Allow equivalent vectors or other methods. May be implied by one correct coordinate. ( ) 1,11 − A1 Allow 1, 11 = − = x y

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 8 of 10 Question Answer Marks Guidance 7(a) 1 3 p : 12 0 2 8 4 2   − − + − − =     a b ( ) p 3 : 27 27 3 12 105 + + − = a b M1 For attempt at an equation using either 1 p 2   −     or ( ) p 3 4 90 + = − a b A1 Allow equivalent with constants collected 9 30 + = a b A1 Allow equivalent with constants collected 6, 24 = = − a b 2 M1 for attempt to solve their equations, dep on first M mark A1 for both 7(b) ( )( ) 2 2 1 3 12 + − x x 2 B1 for 2 3x B1 for 12 − and no extra term in x 7(c) 1 2 = − x B1 2 = ± x B1 Dep on both B marks in part (b) 8(a) 20 48 −       B1 8(b) 20 48 t −       B1 Follow through on their (a) 8(c) 12 25 8 45 t −     +         oe B1 8(d) 12 5 8 3 −     +     −     t oe B1 8(e) ( ) ( ) 2 2 2 12 5 8 3 = − + −  PQ t t M1 Attempt to find modulus of their (d) which must contain terms in t 2 2 144 120 25 64 48 9 = − + + − +  PQ t t t t 2 34 168 208 = − + PQ t t A1 Must see correct expansion leading to given answer.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 9 of 10 Question Answer Marks Guidance 8(f) 2 34 168 204 0 − + = t t M1 For dealing with square root correctly and attempt to solve a 3 term quadratic equation 2.15 only A1 9(a)(i) 360 B1 9(a)(ii) 60 B1 FT on their (b)(i) divided by 6 9(a)(iii) A complete plan for dealing with odd numbers and numbers greater than 7000, see below M1 Must be considering each case Starts with 8 and ends with odd = 48 B1 Starts with 7 or 9 and ends with odd = 72 B1 120 A1 Alternative Their answer to (a)(i) –odd numbers starting with 2–odd number starting with 3 or 5–all even numbers (M1 Must be considering each case All even numbers =120 Odd and starting with 2 = 48 Odd and starting with 3 or 5 = 72 2 B1 for 1 correct 120 A1) 9(b) ( ) ! 92 3 !3! = − n n n B1 ( )( ) 1 2 552 − − = n n n n M1 Attempt to simplify factorials ( ) 2 3 550 0 − − = n n n ( )( ) 25 22 0 − + = n n n M1 Dep on previous M mark for expansion and simplification to a cubic or quadratic in n and attempt to solve 25 = n A1 For 25 = n only 10(a) o o o 45 144.7 , 324.7 α + = o o 99.7 , 279.7 α = 3 M1 for attempt to solve using a correct order of operations, may be implied by one correct solution A1 for 1 correct solution A1 for a second correct solution and no extras

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2020 © UCLES 2020 Page 10 of 10 Question Answer Marks Guidance 10(b)(i) ( ) ( ) 2 sin 1 sin 1 sin 1 θ θ θ + − − − M1 For dealing with fractions 2 2 cos θ − M1 For simplification of numerator and use of the correct identity 2 2sec θ − 2 = − a A1 Must see previous line for A1 10(b)(ii) 2 2sec 3 8 φ − = − oe sec3 2 φ = ± M1 For making use of (i) and attempt to simplify in terms of 3φ 1 cos3 2 φ = ± A1 2π π π 2π 3 , , , 3 3 3 3 φ = − − 2π π π 2π , , , 9 9 9 9 φ = − − or 0.349 ± , 0.698 ± , 3 Dep M1 for attempt to solve, may be implied by one correct solution A1 for each pair of correct solutions 11 ( ) ( ) 1 ln 2 3 ln 3 1 ln  + + − −    a x x x 2 B1 for 1 term correct B1 all correct ( ) ( ) ( ) ( ) ln 2 3 ln 3 1 ln ln5 ln2 + + − − − + a a a M1 Correct substitution of limits, dep on first B1, ignore equality Must have 3 terms involving x ( )( ) 2 3 3 1 ln ln 2.4 10 + − = a a a M1 For use of both addition and subtraction rules, ignore equality Or for use of addition rule on each side of an equation. 2 6 17 3 0 − − = a a A1 3 = a 2 M1 for solution of their quadratic A1 for 3 = a only