Cambridge IGCSE Mathematics - Additional 0606 — 2018 Oct/Nov Paper 1 · Variant 3

* 3 1 0 3 6 1 1 4 7 2 * This document consists of 15 printed pages and 1 blank page. DC (ST/JG) 153500/3 © UCLES 2018 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2018 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.

2 0606/13/O/N/18 © UCLES 2018 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/18 © UCLES 2018 [Turn over 1 (a) In the expansion of ( ) px 2 5 + the coefficient of x3 is equal to 25 8 - . Find the value of the constant p. [3] (b) Find the term independent of x in the expansion of x x 2 4 1 2 2 8 + e o . [3]

4 0606/13/O/N/18 © UCLES 2018 2 r cm O i rad P Q The diagram shows a sector POQ of a circle, centre O, radius r cm, where angle POQ i = radians. The perimeter of the sector is 20 cm. (i) Show that the area, A cm2, of the sector is given by A r r 10 2 = - . [3] It is given that r can vary and that A has a maximum value. (ii) Find the value of i for which A has a maximum value. [3]

5 0606/13/O/N/18 © UCLES 2018 [Turn over 3 Do not use a calculator in this question. In this question, all lengths are in centimetres. A triangle ABC is such that angle °, B AB 90 5 3 5 = = + and BC 5 3 5 = - . (i) Find, in its simplest surd form, the length of AC. [3] (ii) Find tanBCA, giving your answer in the form a b 3 + , where a and b are integers. [3]

6 0606/13/O/N/18 © UCLES 2018 4 In this question, the units of x are radians and the units of y are centimetres. It is given that ( ) cos y x 1 3 10 = + . (i) Find the value of x y d d when x 2 r = . [4] Given also that y is increasing at a rate of 6 cm s−1 when x 2 r = , (ii) find the corresponding rate of change of x. [2]

7 0606/13/O/N/18 © UCLES 2018 [Turn over 5 (i) Show that log log 4 2 9 3 = . [2] (ii) Hence solve log log x 4 3 9 3 + = . [3]

8 0606/13/O/N/18 © UCLES 2018 6 A particle P is moving in a straight line such that its displacement, s m, from a fixed point O at time t s, is given by s t 12 4 12 e . t 0 5 = + - - . (i) Find the value of t when P is instantaneously at rest. [3] (ii) Find an expression for the acceleration of P at time t s. [2] (iii) Find the value of s when the acceleration of P is 0.3 ms−2. [3] (iv) Explain why the acceleration of the particle will always be positive. [1]

9 0606/13/O/N/18 © UCLES 2018 [Turn over 7 A B D O C The diagram shows a quadrilateral OABC. The point D lies on OB such that OD DB 2 = and AD mAC = , where m is a scalar quantity. OA a = OB b = OC c = (i) Find AD in terms of m, a and c. [1] (ii) Find AD in terms of a and b. [2] (iii) Given that a b c 15 16 9 = - , find the value of m. [3]

10 0606/13/O/N/18 © UCLES 2018 8 ( ) sin x x 5 4 f = + for x 0 2 G G r radians ( )x x 3 g r = - for x R ! (i) Write down the range of ( )x f . [2] (ii) Find ( )x f 1 - and write down its range. [3] (iii) Solve ( )x 2 11 fg = . [4]

11 0606/13/O/N/18 © UCLES 2018 [Turn over 9 Find the equation of the normal to the curve ( ) ln y x x 3 1 2 2 = + at the point where x 2 = , giving your answer in the form y mx c = + , where m and c are correct to 2 decimal places. You must show all your working. [8]

12 0606/13/O/N/18 © UCLES 2018 10 x y = 12 + x – x2 y O B A y = x + 8 The diagram shows the curve y x x 12 2 = + - intersecting the line y x 8 = + at the points A and B. (i) Find the coordinates of the points A and B. [3] (ii) Find ( ) x x x 12 d 2 + - y . [2]

13 0606/13/O/N/18 © UCLES 2018 [Turn over (iii) Showing all your working, find the area of the shaded region. [4]

14 0606/13/O/N/18 © UCLES 2018 11 The polynomial ( )x ax x bx 17 8 p 3 2 = + + - is divisible by x 2 1 - and has a remainder of 35 - when divided by x 3 + . (i) By finding the value of each of the constants a and b, verify that a b = . [4] Using your values of a and b, (ii) find ( )x p in the form ) ( ) ( x x 2 1 q - , where ( )x q is a quadratic expression, [2]

15 0606/13/O/N/18 © UCLES 2018 (iii) factorise ( )x p completely, [1] (iv) solve sin sin sin a b 17 8 0 3 2 i i i + + - = for ° ° 0 180 1 1 i . [3]

16 0606/13/O/N/18 © UCLES 2018 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. BLANK PAGE

This document consists of 11 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2018 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 2018 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 2018 © UCLES 2018 Page 2 of 11 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 2018 © UCLES 2018 Page 3 of 11 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 2018 © UCLES 2018 Page 4 of 11 Question Answer Marks Guidance 1(a) ( ) 3 5 2 3 2 C px × × B1 3 8 40 25 p = − 3 8 1000 p = − M1 equating their coefficient of x3 to 8 25 − and finding p3 1 5 p = − or 0.2 p = − A1 1(b) ( ) 4 4 8 2 4 2 1 2 4 C x x   × ×    B1 1 70 16 256 × × M1 at least two of 1 70,16, 256 correct in an evaluation of a three-term product 35 8 , 4.375, 3 8 4 A1 cao 2(i) 20 2r r θ − = B1 Area = 2 1 20 2 2 r r r −       M1 use of their θ in terms of r in formula for sector area 2 10 A r r = − A1 simplification to get given answer Alternative 20 2 s r = − B1 1 (20 2 ) 2 r r = − M1 use of formula for sector area using their expression for s in terms of r 2 10 A r r = − A1 simplification to get given answer 2(ii) d 10 2 d A r r = − When d 0 d A r = , 5 r = M1 for d 10 d A kr r = − , equating to zero and solving for r ( ) 20 2 5 5 θ −× = M1 Dep substitution of their value of r to get θ 2 θ = A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 11 Question Answer Marks Guidance 3(i) ( ) ( ) 2 2 2 5 3 5 5 3 5 AC = + + − M1 correct use of Pythagoras or correct use of cosine rule with cos90 75 25 50 3 75 25 50 3 = + + + + − 200 = M1 correct expansion to 6 or 8 terms 10 2 AC = A1 from AC2 = 200 3(ii) 5 3 5 tan 5 3 5 BCA + = − oe B1 ( )( ) ( )( ) 5 3 5 5 3 5 5 3 5 5 3 5 + + = − + oe 100 50 3 50 + = oe M1 for rationalisation 2 3 = + A1 4(i) M1 for ( ) 9 10 1 cos3x + f(x) M1 for ( ) 9 sin3 1 cos3 k x x + ( ) 9 d 30sin3 1 cos3 d y x x x = − + A1 When π 2 x = , d 30 d y x = A1 4(ii) Use of d d d d d d y x y x t t × = with d 6 d y t = M1 d d 6 d d y x their x t × = d 1 or 0.2 d 5 x t = A1 FT from their answer from part (i)

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 6 of 11 Question Answer Marks Guidance 5(i) 3 9 3 log 4 log 4 log 9 = B1 change of base 3 1 log 4 2 = 2 3 1 log 2 2 = 3 or log 4 3 log 2 = B1 Dep must have B1 for change of base and full working Alternative A 9 9 log 4 2log 2 = B1 use of power rule 3 3 2log 2 log 9 = 3 3 2log 2 2log 3 = 3 log 2 = B1 Dep change of base and full working Alternative B 9 log 4 x = 9 4 x ⇒ = 9 4 x = 23 4 x ⇒ = B1 correct use of indices to reach 23 4 x = 3 3 2 log 2 x x ⇒ = ⇒ = 9 3 log 4 log 2 ∴ = B1 Dep full working Alternative C 10 9 10 log 4 log 4 log 9 = 10 10 2log 2 2log 3 = B1 change of base and use of power rule 3 log 2 = B1 Dep change of base and full working

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 11 Question Answer Marks Guidance 5(ii) 3 3 log 2 log 3 x + = 3 log 2 3 x = B1 for 3 log 2 3 x = 33 2x = B1 13.5, x = 27 2 x = B1 Alternative 3 3 3 log log 27 log 2 x = − B1 3 27 log 2 = B1 x = 13.5, 27 2 x = B1 6(i) 0.5 d 6e 4 d t s t − = − + M1 for 0.5 e 4 t k − + When d d s t = 0, 0.5 2 e 3 t − = 2 0.5 ln 3 t − = 2 2ln 3 t = − M1 Dep equating to zero and correct order of operations to solve for t 0.811 t = A1 6(ii) M1 for 0.5 e t k − 2 0.5 2 d 3e d t s t − = A1 6(iii) 0.5 3e 0.3 t − = 0.5 e 0.1 t − = ln0.1 0.5 t = − M1 correct order of operations and correct use of ln to solve 0.5 e 0.3 t k − = for t 0.5 4.605 12e 4 4.605 12 s − × = + × − M1 Dep use of t to obtain s 7.62 s = A1 6(iv) 0.5 e t − is always positive or 0.5 e t − can never be zero or negative B1 correct comment about 0.5 e t −

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 8 of 11 Question Answer Marks Guidance 7(i) ( ) AD m = − c a JJJG B1 7(ii) AD OD = −a JJJG JJJG B1 for OD JJJG 2 3 = b 2 3 = − b a B1 FT their OD JJJG if OD k = b JJJG 7(iii) ( ) m − c a 2 3 = − b a M1 equating parts (i) and (ii) ( ) 24 1 24 16 m m − + = a c b Comparing with 15 9 16 + = a c b M1 attempt to eliminate or compare like vectors using given condition 3 8 m = A1 8(i) ( ) 5 f 6 x - - or [ ] 5,6 oe B2 B1 for ( ) 5 f x p - - ( 5 p > ) or for ( ) f 6 q x - - ( 6 q < ) 8(ii) sin 5 4 y x = + M1 complete valid attempt to obtain the inverse with operations in correct order. ( ) 1 4sin 5 y x − = − A1 Range 0 2π y - - B1 8(iii) π 3 2 sin 5 ( 11) 4 x     −         + =       B1 for π 3 sin 5 4 x   −    + π 1 3 sin 4 2 x   −    = M1 for π 3 sin 4 x k   −    = 1 1 4sin 2 3 x π −  = +     oe M1 Dep for use of sin–1 and correct order of operations to obtain x. Allow one +/– or / × ÷ sign error π x = or 3.14 A1 π x = and no other solutions in range

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 9 of 11 Question Answer Marks Guidance 9 ( ) ( ) 2 2 d 6 ln 3 1 d 3 1 x x x x + = + B1 for 2 6 3 1 x x + ( ) 2 2 2 4 6 2 ln 3 1 d 3 1 d x x x x y x x x − + + = or ( ) ( ) 2 3 2 2 d 2 1 6 ln 3 1 d 3 1 y x x x x x x −     = + +         + M1 differentiation of a quotient or product ( ) 2 2 4 f( ) 2 ln 3 1 x x x x x − + or for ( ) 2 3 2 2 1 ln 3 1 f( ) x x x x     − + +         A1 When 2 x = , d 0.410 d y x = − A1 Gradient of perp = 2.436… M1 use of 1 m − with a gradient obtained by differentiation When 1 2, 0.641 or ln13 4 x y = = B1 Normal: ( ) 0.641 2.436 2 y x − = − M1 Dep 2.44 4.23 y x = − A1 10(i) 2 8 12 x x x + = + − 2 4 x = , 2 x = ± or 2 16 60 0 y y − + = y = 6 or y = 10 M1 correct method of solution to obtain x or y 2, 10 x y = = 2, 6 x y = − = A2 A1 for x = –2 and x = 2 or for y = 6 and y = 10 or for either point from a correctly solved equation. 10(ii) M1 for ( ) 2 3 12 x px qx c + + + ( ) 2 3 12 2 3 x x x c + − + A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 10 of 11 Question Answer Marks Guidance 10(iii) ( ) 2 2 3 2 1 12 6 10 4 2 3 2 x x x −     + − − + ×         B1 FT area of the trapezium unsimplified ( ) 1 6 10 4 2   + ×     or 2 2 2 ( 2) 8 2 8 ( 2) 2 2     − + × − + × −         (= 32) 2 3 2 3 2 2 ( 2) ( 2) 12 2 12 2 2 3 2 3     − − × + − − × −+ −         M1 correct use of correct limits for area under the curve using their integral of the form 2 3 12 x px qx + + 128 3 = oe A1 32 3 = oe A1 Alternative 2 2 212 8 d x x x x − + − − − ∫ 2 2 24 d x x − = − ∫ M1 subtraction of the two equations with intent to integrate the result 2 3 2 4 3 x x −   = −     A1 8 8 4 2 4 2 3 3     × − − × −+         M1 Dep for correct application of limits 32 3 = oe A1 11(i) 3 2 1 1 1 1 p 17 8 2 2 2 2 a b         = + + −                 M1 expression for 1 p 2       ( ) ( ) ( ) 3 2 p 3 3 17( 3) 3 8 a b − = − + − + − − M1 expression for ( ) p 3 − 17 8 0 8 4 2 a b + + − = 27 153 3 8 35 a b − + − −= − A1 both equations correct (allow equivalents and terms not collected but powers should be evaluated) Leading to 6 a b = = A1 from correct equations with evidence that both have been found correctly in order to verify that a = b

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 11 of 11 Question Answer Marks Guidance 11(ii) ( )( ) 2 2 1 3 10 8 x x x − + + B2 B1 for 2 3x and +8 from factorisation or for 2 3 10 ... x x + from long division 11(iii) ( )( )( ) 2 1 2 3 4 x x x − + + B1 cao 11(iv) 1 sin 2 θ = B1 o o 30 , 150 θ = B2 B1 for a first correct solution B1 for a second correct solution with no extras in range 0 180 θ - - and no solution arising from other factors.