Cambridge IGCSE Mathematics - Additional 0606 — 2019 Feb/March Paper 1 · Variant 2

This document consists of 15 printed pages and 1 blank page. DC (JM/CB) 165270/2 © UCLES 2019 [Turn over * 6 7 1 5 0 4 6 2 1 2 * Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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.

2 0606/12/F/M/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 0606/12/F/M/19 © UCLES 2019 [Turn over 1 (a) Given that  { : }, x x 1 20 1 1 = A = {multiples of 3}, B = {multiples of 4}, find (i) n( ), A [1] (ii) n( ). A B + [1] (b) On the Venn diagram below, draw the sets P, Q and R such that P Q 1 and . Q R + Q =  [2] (c) Using set notation, describe the shaded areas shown in the Venn diagrams below. S T X Y Z … … [2]

4 0606/12/F/M/19 © UCLES 2019 2 On the axes below, sketch the graph of the curve y x x 2 5 3 2 = - - , stating the coordinates of any points where the curve meets the coordinate axes. x y 10 –10 0 –1 4 [4]

5 0606/12/F/M/19 © UCLES 2019 [Turn over 3 (i) Find the first 3 terms in the expansion, in ascending powers of x, of x 3 9 6 - f p . Give the terms in their simplest form. [3] (ii) Hence find the term independent of x in the expansion of x x x 3 9 2 6 2 - - f f p p . [3]

6 0606/12/F/M/19 © UCLES 2019 4 The polynomial ( )x x ax bx 2 49 p 3 2 = + + - , where a and b are constants. When ( )x pl is divided by x + 3 there is a remainder of -24 . (i) Show that a b 6 78 - = . [2] It is given that 2x - 1 is a factor of p (x). (ii) Find the value of a and of b. [4] (iii) Write p (x) in the form ( ) ( ) x x 2 1 Q - , where Q (x) is a quadratic factor. [2] (iv) Hence factorise p (x) completely. [1]

7 0606/12/F/M/19 © UCLES 2019 [Turn over 5 It is given that . log x p 4 = Giving your answer in its simplest form, find, in terms of p, (i) ( ) log x 16 4 , [2] (ii) log x 256 4 7 f p. [2] Using your answers to parts (i) and (ii), (iii) solve ( ) log log x x 16 256 5 4 4 7 - = f p , giving your answer correct to 2 decimal places. [3]

8 0606/12/F/M/19 © UCLES 2019 6 (a) Given that A 1 0 2 1 = - e o, B 1 2 3 4 5 1 = - f p and C 3 2 0 = - ` j, write down the matrix products which are possible. You do not need to evaluate your products. [2] (b) It is given that X 2 5 2 3 = - e o and Y 4 2 1 0 = e o. (i) Find X-1. [2] (ii) Hence find the matrix Z such that XZ = Y. [3]

9 0606/12/F/M/19 © UCLES 2019 [Turn over 7 Do not use a calculator in this question. All lengths in this question are in centimetres. A B C D 5 2 3 + 5 10 2 - 5 6 3 + The diagram shows the trapezium ABCD, where AB 2 3 5 = + , DC 6 3 5 = + , AD 10 2 5 = - and angle ADC = 90°. (i) Find the area of ABCD, giving your answer in the form a b 5 + , where a and b are integers. [3] (ii) Find cotBCD, giving your answer in the form c d 5 + , where c and d are fractions in their simplest form. [3]

10 0606/12/F/M/19 © UCLES 2019 8 (a) 0 5 10 15 20 v 2 4 6 8 10 t The diagram shows the velocity-time graph of a particle P moving in a straight line with velocity v ms-1 at time t seconds after leaving a fixed point. (i) Write down the value of the acceleration of P when t = 5. [1] (ii) Find the distance travelled by the particle P between t = 0 and t = 10. [2]

11 0606/12/F/M/19 © UCLES 2019 [Turn over (b) A particle Q moves such that its velocity, v ms-1, t seconds after leaving a fixed point, is given by sin v t 3 2 1 = - . (i) Find the speed of Q when t 7 12 r = . [2] (ii) Find the least value of t for which the acceleration of Q is zero. [3]

12 0606/12/F/M/19 © UCLES 2019 9 The area of a sector of a circle of radius r cm is 36 cm2. (i) Show that the perimeter, P cm, of the sector is such that P r r 2 72 = + . [3] (ii) Hence, given that r can vary, find the stationary value of P and determine its nature. [4]

13 0606/12/F/M/19 © UCLES 2019 [Turn over 10 A curve is such that when x = 0, both y = -5 and x y 10 d d = . Given that x y 4 3 d d e x 2 2 2 = + , find (i) the equation of the curve, [7] (ii) the equation of the normal to the curve at the point where x 4 1 = . [3]

14 0606/12/F/M/19 © UCLES 2019 11 (a) Solve sin cos tan x x x 2 1 = for ° x 0 180° G G . [3]

15 0606/12/F/M/19 © UCLES 2019 (b) (i) Show that sec cot sin cos i i i i - = . [3] (ii) Hence solve sec cot sin 3 3 3 2 1 i i i - = for 3 2 3 2 G G r r i - , where i is in radians. [4]

16 0606/12/F/M/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 8 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 12 March 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 March 2019 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 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2019 © UCLES 2019 Page 4 of 8 Question Answer Marks Partial Marks 1(a)(i) 6 B1 1(a)(ii) 1 B1 1(b) Q P R 2 B1 for P contained within Q B1 for Q and R separate 1(c) ′ ′ ∩ S T or ( )′ ∪ S T oe B1 ( ) ( ) ∩ ∪ ∩ X Y X Z or ( ) ∩ ∪ X Y Z oe B1 2 4 B1 for general shape with maximum point in 1st quadrant B1 for 1 ,0 2   −     and ( ) 3,0 soi B1 for ( ) 0,3 soi B1 dep on first B1, with cusps and correct shape for 1 2 < − x and 3 > x 3(i) 2 729 162 15 − + x x 3 B1 for 729 B1 for 162 − x B1 for 2 15x Mark final answer 3(ii) ( ) 2 2 2 4 729 162 15 4   − + − +     x x x x B1 for expansion of 2 2   −     x x Term independent of x = 2916 60 − + M1 for attempt to find independent term, must be considering 2 products using their answer to part (i) 2856 = − A1 4(i) ( ) 2 p 6 2 ′ = + + x x ax b B1 for ( ) 2 p 6 2 ′ = + + x x ax b ( ) p 3 54 6a b ′ − = − + , 24 = − leading to 6 78 − = a b B1 must be convinced of correct substitution and simplification AG

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2019 © UCLES 2019 Page 5 of 8 Question Answer Marks Partial Marks 4(ii) 1 2 p : 49 0 2 8 4 2   + + − =     a b M1 for attempt at 1 p 2       equated to 0 6 78 − = a b 2 195 + = a b oe M1 M Dep on previous M for attempt to solve both equations leading to 27 = a A1 84 = b A1 4(iii) ( )( ) 2 2 1 14 49 − + + x x x 2 M1 for factorisation by observation or by long division 4(iv) ( )( ) 2 2 1 7 − + x x B1 5(i) 4 4 log 16 log + p M1 for dealing with product correctly 2 + p A1 5(ii) 4 4 7log log 256 − x M1 for dealing with power and division correctly 7 4 − p A1 5(iii) 2 (7 4) 5 + − − = p p leading to 1 6 = p M1 for use of parts (i) and (ii) to obtain a value for p so 1 6 4 = x M1 for correct attempt to deal with 4 log in order to obtain x 1.26 = x A1 6(a) BA and CB 2 B1 for one correct product of 2 matrices B1 for a second correct product of 2 matrices, with no other incorrect products 6(b)(i) 3 2 1 5 2 16     −   oe 2 B1 for 1 16 soi B1 for 3 2 5 2     −  

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2019 © UCLES 2019 Page 6 of 8 Question Answer Marks Partial Marks 6(b)(ii) 1 1 − − = X XZ X Y 3 2 4 1 1 5 2 2 0 16    =    −    Z M1 for pre-multiplication by their inverse matrix attempt at matrix multiplication M1 M1 Dep on previous M mark, must have at least 2 correct elements 16 3 1 16 5 16   =   − −   Z oe A1 7(i) Area = ( )( ) 1 8 6 5 10 2 5 2 + − M1 for a correct method of finding the area of the trapezium 10 22 5 = + A2 A1 for 10 with sufficient working seen A1 for 22 5 with sufficient working seen 7(ii) 4 cot 10 2 5 θ = − B1 ( ) ( )( ) 4 10 2 5 10 2 5 10 2 5 + = − + M1 for attempt to rationalise an expression for cotθ , some evidence of expansion must be seen 1 5 2 10 = + A1 8(a)(i) 0 B1 8(a)(ii) Area under curve = ( ) ( ) ( ) 1 1 2 10 4 10 10 20 4 2 2 × + × + + × M1 for attempt to find the total area under the graph = 110 A1 8(b)(i) When 7 , 2.5 12 π = = − t v M1 for substitution of 7 12 π = t and correct attempt to evaluate Speed = 2.5 A1 must be positive 8(b)(ii) 6cos2 = a t M1 for differentiation to get acceleration, must be of the form cos2 m t When acceleration = 0, cos2 0 = t M1 M Dep on previous M mark for equating to zero and correct attempt to solve to get a solution in radians. π 4 = t or 0.785 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2019 © UCLES 2019 Page 7 of 8 Question Answer Marks Partial Marks 9(i) 2 1 36 2 θ = r 2 72 θ = r M1 for use of the area of the sector 2 θ = + P r r M1 for attempt to find P making use of the area 72 2 = + P r r A1 for attempt to simplify to obtain AG 9(ii) 2 d 72 2 d = − P r r M1 for attempt to differentiate to obtain the form 2 + b a r and equate to zero When d 0, 6 d = = P r r A1 P = 24 A1 2 2 3 d 144 d = P r r positive so minimum B1 FT on their positive r, for a correct method to determine the nature of the stationary point leading to a correct conclusion. If the second derivative is evaluated, it must be correct for their r. 10(i) ( ) 2 d 2e 3 d = + + x y x c x 2 M1 for attempt to integrate to obtain the form 2e + x m nx A1 all correct 8 = c M1 M1 Dep on previous M mark for attempt to get c ( ) 2 2 3 e 8 2 = + + + x x y x d 2 M1 for attempt to integrate again to obtain the form ( ) 2 2 + + x pe qx rx A1 all correct, FT on their 2e x k and their c 6 = − d M1 M1 Dep on previous M mark for attempt to get d 2 2 3 e 8 6 2 = + + − x x y x A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2019 © UCLES 2019 Page 8 of 8 Question Answer Marks Partial Marks 10(ii) When 1 4 = x , 2.26 = − y d 12.0 d = y x M1 for attempt to obtain both y and d d y x using their work from (i) 1 1 2.26 12 4   + = − −     y x 2 M1 Dep on previous M mark for attempt to obtain the equation of the normal A1 allow unsimplified, must be using correct accuracy or exact equivalents. 11(a) ( ) 2 2sin cos 1 0 − = x x M1 for obtaining in terms of sin and cos to obtain one solution correctly o o sin 0, 0 , 180 = = x x B1 for o o 0 , 180 = x and no other in the given range for the solution of this equation o o 1 cos , 45 , 135 2 = ± = x x A1 for o o 45 , 135 = x and no other in the given range for the solution of this equation 11(b)(i) 2 1 sin cos cos θ θ θ − M1 for dealing with cot and sec 2 cos cos θ θ M1 for correct use of identity cosθ A1 for all correct working to gain AG 11(b)(ii) 1 cos3 2 θ = 5π π or 9 9 θ = M1 for use of part (i) and attempt to solve correctly to obtain a positive angle, may be implied by one correct solution 5π π or 9 9 θ = − − M1 for use of part (i) and attempt to solve correctly to obtain a negative angle, may be implied by one correct solution π 5π , 9 9 θ = ± ± A2 A1 for one correct pair of solutions A1 for a second pair of solutions with no extra solutions within the range