Cambridge IGCSE Mathematics - Additional 0606 — 2019 Oct/Nov Paper 2 · Variant 2

This document consists of 16 printed pages. DC (LEG/SG) 172298/2 © UCLES 2019 [Turn over * 5 1 9 2 8 8 5 9 0 1 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 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 0606/22/O/N/19 © UCLES 2019 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x b b ac a = − − 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 0606/22/O/N/19 © UCLES 2019 [Turn over 1 On each of the Venn diagrams below, shade the region indicated. A B A B A B + , + l l ^ ^ h h A B C A B C + , ^ h A B C A B C + + l [3]

4 0606/22/O/N/19 © UCLES 2019 2 Given that sin cos y x x 2 3 3 = + , show that d d d d sin x y x y y k x 3 3 2 2 + + = , where k is a constant to be determined. [5]

5 0606/22/O/N/19 © UCLES 2019 [Turn over 3 A 5-digit code is formed using the following characters. Letters a e i o u Numbers 1 2 3 4 5 6 Symbols @ ∗ # No character can be repeated in a code. Find the number of possible codes if (i) there are no restrictions, [2] (ii) the code starts with a symbol followed by two letters and then two numbers, [2] (iii) the first two characters are numbers, and no other numbers appear in the code. [2]

6 0606/22/O/N/19 © UCLES 2019 4 Find the values of k for which the line y kx 3 = + does not meet the curve y x x 5 12 2 = + + . [5]

7 0606/22/O/N/19 © UCLES 2019 [Turn over 5 At the point where x 1 = on the curve y x k 1 2 = + ^ h , the normal has a gradient of 3 1 . (i) Find the value of the constant k. [4] (ii) Using your value of k, find the equation of the tangent to the curve at x 2 = . [3]

8 0606/22/O/N/19 © UCLES 2019 6 (i) Show that sec tan tan sec sin x x x x x 1 1 2 + + + = . [5]

9 0606/22/O/N/19 © UCLES 2019 [Turn over (ii) Hence solve the equation sec tan tan sec sin x x x x x 1 1 1 3 + + + = + for x 0 180 c c G G . [4]

10 0606/22/O/N/19 © UCLES 2019 7 (a) The cubic equation x ax bx 40 0 3 2 + + - = has three positive integer roots. Two of the roots are 2 and 4. Find the other root and the value of each of the integers a and b. [4] (b) Do not use a calculator in this question. Solve the equation x x x 5 46 40 0 3 2 - = - - given that it has three integer roots, only one of which is positive. [4]

11 0606/22/O/N/19 © UCLES 2019 [Turn over 8 (i) A particle A travels with a speed of 6.5 ms–1 in the direction i j 5 12 - - . Find the velocity, vA , of A. [2] (ii) A particle B travels with velocity i j 12 9 = - vB . Find the speed, in ms–1, of B. [2] Particle A starts moving from the point with position vector i j 20 7 - . At the same time particle B starts moving from the point with position vector i j 1 67 1 - + . (iii) Find rA, the position vector of A after t seconds, and rB, the position vector of B after t seconds. [2] (iv) Find the time when the particles collide and the position vector of the point of collision. [3]

12 0606/22/O/N/19 © UCLES 2019 9 M P B (5,-1) A (-3,5) (r,s) The diagram shows the points A (–3, 5) and B (5, –1). The mid-point of AB is M and the line PM is perpendicular to AB. The point P has coordinates (r, s). (i) Find the equation of the line PM in the form y mx c = + , where m and c are exact constants. [5] (ii) Hence find an expression for s in terms of r. [1]

13 0606/22/O/N/19 © UCLES 2019 [Turn over (iii) Given that the length of PM is 10 units, find the value of r and of s. [5]

14 0606/22/O/N/19 © UCLES 2019 10 (i) Given that ln y x x 2 = , find d d x y. [3] (ii) Find the coordinates of the stationary point on the curve ln y x x 2 = . [3]

15 0606/22/O/N/19 © UCLES 2019 [Turn over (iii) Using your answer to part (i), find d ln x x x 3 y . [3] (iv) Hence evaluate 1 2 d ln x x x 3 y . [2] Question 11 is printed on the next page.

16 0606/22/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. 11 Do not use a calculator in this question. Solve the quadratic equation x x 5 3 3 5 3 0 2 - + + + = ^ ^ h h , giving your answers in the form a b 5 + , where a and b are constants. [6]

This document consists of 10 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/22 Paper 2 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.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 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.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 3 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

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 4 of 10 Question Answer Marks Guidance 1 A B B1 B1 B 1 2 d 6cos3 d y x x = B1 3sin3x − B1 2 2 d 18sin3 9cos3 d y x x x = − − B1 FT Correct derivative of their d d y x Insert and collect like terms M1 Must insert for y, their d d y x and 2 2 d d y x correctly resulting in 6 terms. k = –15 A1 Allow –15sin3x seen nfww 3(i) 14 5P or 14 × 13 × 12 × 11 × 10 M1 240 240 A1 cao 3(ii) 3 1P × 5 2P × 6 2P or 3 × (5 × 4) × (6 × 5) M1 Two of the three elements multiplied by … = 1800 A1 3(iii) 6 2P × 8 3P or (6 × 5) × (8 × 7 × 6) M1 One element multiplied by … Clear intention to multiply = 10 080 A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 5 of 10 Question Answer Marks Guidance 4 kx + 3 = x2 + 5x + 12 → x2 + (5 – k)x + 9(= 0) M1 Equate and attempt to simplify to all terms on one side. Use discriminant of their quadratic. M1 dep (5 – k)2 – 36 oe A1 Unsimplified k = –1 and 11 A1 Both boundary values –1 < k < 11 A1 Must be in terms of ݇. OR 2x + 5 ~ k M1 Connect gradients of line and curve ( ) 2 2 2 5 3 2 5 3 5 12 y x x x x x x = + + → + + = + + M1 Eliminate ݇ and y. x2 = 9 → x = ± 3 A1 k = 11 or k = –1 A1 –1 < k < 11 A1 5(i) ( ) 3 d 2 d 1 y k x x − = + B1 oe Unsimplified Gradient of normal = ( ) 3 1 2 x k + or Gradient of tangent = −3 M1 Gradient of normal = 1 gradientof tangent − 8 1 2 or 3 2 3 8 k k = = − M1 Equate gradient of normal to 1 3 at x = 1 or equate gradient of tangent to –3 at x =1 k = 12 A1 5(ii) d 8 2 d 9 y x x = → = − or their 2 27 k − B1 FT 4 3 y = or their 9 k B1 FT 4 3 8 8 28 or 2 9 9 9 y y x x − = − = − + − B1 isw

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 6 of 10 Question Answer Marks Guidance 6(i) sin 1 cos cos sin 1 cos cos 1 1 x x x x x x + + + M1 Use sin 1 tan and sec cos cos x x x x x = = throughout sin 1 cos 1 cos sin x x x x + + + M1 dep Multiply by cosݔ ( ) 2 2 sin 1 2cos cos 1 cos sin x x x x x + + + + M1 dep Add their fractions correctly and expand (1 + cosx)2 correctly ( ) ( ) 2 1 cos 1 cos sin x x x + + M1 dep Use sin2x + cos2x = 1 and take out a factor of 2. All correct AG A1 Do not award if brackets missing at any point or ݔ missing more than twice or ݔmisplaced. Do not credit mixed variables. OR ( ) ( ) 2 2 tan sec 1 tan sec 1 x x x x + + + M1 Add fractions ( ) 2 2sec 2sec tan sec 1 x x x x + = + M1 dep Expand brackets correctly and use 1 + tan2x = sec2x 2sec tan x x M1 dep Cancel secݔ+ 1 2 cos cos sin x x x × M1 dep Use sin 1 tan and sec cos cos x x x x x = = oe All correct AG A1 Do not award if brackets missing at any point or ݔ missing more than twice or ݔ misplaced. Do not credit mixed variables. 6(ii) 3sin2x + sinx – 2 = 0 oe B1 Obtain three term quadratic. (3sinx – 2)(sinx + 1) = 0 M1 Solve three term quadratic 41.8° awrt A1 138.2° awrt A1 Mark final answers This mark is not awarded if there are more solutions in the range.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 7 of 10 Question Answer Marks Guidance 7(a) 2 × 4 × p = 40 → p = 5 B1 May be obtained later. (x – 2)(x – 4)(x – p) = 0 M1 Factorise cubic a = –11 A1 Expand and identify b = 38 A1 OR 2 × 4 × p = 40 → p = 5 B1 May be obtained later. Obtain equations 4a + 2b = 32 16a + 4b = –24 and attempt to solve M1 a = –11 A1 b = 38 A1 7(b) Find x = –1 M1 Trial value/s and finds a root or shows that (x + 1) or (x + 4) or (x – 10) divides into x3 – 5x2 – 46x – 40. (x + 1)(x2 – 6x – 40) (= 0) or (x + 4)(x2 – 9x – 10)(= 0) or (x – 10)(x2 + 5x + 4)(= 0) A1 Factorise to give linear and quadratic factor (x + 1)(x + 4)(x – 10) (= 0) M1 Solve the quadratic to give 2 roots x = –1, –4, 10 A1 OR Uses factor theorem to find a root (–1)3 – 5(–12) – 46(–1) – 40 or –1 – 5 + 46 – 40 = 0 → x = –1 M1 This may be awarded for x = –4 or x = 10. Uses factor theorem to attempt to find further roots M1 At least two more trials. (–4)3 – 5(–4)2 – 46(–4) – 40 or – 64 – 80 + 184 – 40 = 0 → x = –4 A1 (10)3 – 5(10)2 – 46(10) – 40 or 1000 – 500 – 460 – 40 = 0 → x = 10 A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 8 of 10 Question Answer Marks Guidance 8(i) 2 2 5 12 13 + = M1 ( ) 5 1 6 or 5 12 2 2 = − − − − i j i j A v A1 8(ii) ( ) 2 12 12 9 B v = + − M1 Use Pythagoras 15 A1 Do not allow ± 15. Mark final answer. 8(iii) ( ) ( ) 20 2.5 7 6 or 20 2.5 7 6 t t t −     = +     − −     = − + −− i j A A r r B1 FT on their vA only if of the form k(–5i – 12j) where k ≠ 1 or 0. ( ) ( ) 67 12 11 9 or 67 12 11 9 t t t −     = +     −     = − + + − i j B B r r B1 8(iv) 20 – 2.5t = –67 + 12t or –7 – 6t = 11 – 9t M1 Equate ݔ or ݕ coordinates. Must have two terms in both coordinates. t = 6 A1 nfww Ignore other value of t. 5 43   =   −   r only or r = 5i – 43j A1 A0 if further value of ࢘ found. 9(i) Midpoint (1, 2) B1 May be seen on diagram Gradient of AB= 3 4 − B1 Gradient of PM 1 4 gradientof 3 their AB − = = M1 Use m1 × m2 = –1 Equation PM 2 4 1 3 y x − = − M1 dep Attempt to find equation of line with their midpoint and their gradient of PM. If y = mx + c used c must be found. 4 2 3 3 y x = + A1 9(ii) 4 2 3 3 s r = + B1 FT Insert (r, s) into their linear equation to

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 9 of 10 Question Answer Marks Guidance obtain ݏ =… 9(iii) (r – 1)2 + (s – 2)2 = 100 oe B1 FT Use Pythagoras with their (1, 2) Eliminate r or s M1 From one linear and one quadratic expression. Unsimplified 25r2 –50r – 875 = 0 oe or 25s2 –100s – 1500 = 0 oe A1 (5r +25)(5r –35) = 0 oe or (5s – 50)(5s + 30) = 0 oe M1 Solve three term quadratic Can be implied by correct solution. r = 7, s = 10 A1 Do not award if negative values of ݎand ݏ are also given nfww OR Equivalent method such as: 100 2 2 = + →       = b a b a MP and 4 3 b a = B1 Using distance =10 and gradient = 4 3 . Eliminate a or b M1 2 2 2 2 4 100 3 3 or 100 4 a a b b   + =      + =     A1 → a = (±)6 and b = (±)8 M1 Solve r = 7, s = 10 A1 10(i) Quotient rule or product rule M1 4 4 2 ln ln .2 or x x x x x x x x − − oe isw A2/1/0 Minus one each error. Allow unsimplified. 10(ii) x – 2xlnx = 0 M1 Set d 0 d y x = and attempt to solve. Must have two terms and obtain lnx = k only. aw 1.65 or e rt x = A1 y = 0.184 awrt or 1 2e A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 10 of 10 Question Answer Marks Guidance 10(iii) 2 3 3 ln 1 2ln d x x x x x x = − ⌠⌡ M1 Integrate their derivative from (i) which must have two terms. Condone omission of dݔ. 2 1 2x − A1 Find 3 1 dx x ⌠⌡ ( ) 3 2 2 ln 1 ln d 4 2 x x x C x x x = − − + ⌠⌡ A1 oe Rearrange and complete 10(iv) Insert limits and subtract correctly M1 dep Must be inserting into two terms in ݔ from (iii). Values explicitly seen if expression is incorrect. 3 ln 2 16 8 − or 0.101 awrt A1 11 ( )( ) 5 3 5 3 4 − + = − B1 Seen anywhere Attempt formula M1 ( ) 3 5 2 5 3 x −± = − A1 Multiply by their ( ) 5 3 + M1 Attempt must be seen with a further line of working. oe 5 3 x = + A1 oe Mark final answer ( ) 1 5 3 4 x − + = A1 oe Mark final answer