Cambridge IGCSE Mathematics - Additional 0606 — 2017 May/June Paper 1 · Variant 2

This document consists of 12 printed pages. DC (LK/CGW) 129467/2 © UCLES 2017 [Turn over * 0 1 4 7 3 1 1 6 5 7 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2017 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 International Examinations Cambridge International General Certificate of Secondary Education

2 0606/12/M/J/17 © UCLES 2017 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/M/J/17 © UCLES 2017 [Turn over 1 On each of the Venn diagrams below, shade the region which represents the given set. A B C (A + B) , C A B C (A + B ) + C A B C (A , B) + C [3] 2 It is given that ( ) y x x 1 5 4 2 2 1 = + + . Showing all your working, find the exact value of d d x y when x 3 = . [5]

4 0606/12/M/J/17 © UCLES 2017 3 Vectors i and j are unit vectors parallel to the x-axis and y-axis respectively. (a) The vector v has a magnitude of 3 5 units and has the same direction as i j 2 - . Find v giving your answer in the form i j a b + , where a and b are integers. [2] (b) The velocity vector w makes an angle of 30° with the positive x-axis and is such that w 2 = . Find w giving your answer in the form i j c d + , where c and d are integers. [2] 4 The first 3 terms in the expansion of x 3 6 n - J L KK N P OO are ax bx 81 2 + + . Find the value of each of the constants n, a and b. [5]

5 0606/12/M/J/17 © UCLES 2017 [Turn over 5 A particle P moves in a straight line, such that its displacement, x m, from a fixed point O, t s after passing O, is given by ( ) cos x t 4 3 4 = - . (i) Find the velocity of P at time t. [1] (ii) Hence write down the maximum speed of P. [1] (iii) Find the smallest value of t for which the acceleration of P is zero. [3] (iv) For the value of t found in part (iii), find the distance of P from O. [1]

6 0606/12/M/J/17 © UCLES 2017 6 (i) Show that cot tan cosec cos i i i i + = . [4] It is given that cot tan cosec d 2 2 2 4 3 a 0 i i i i + = y , where a 0 4 1 1 r. (ii) Using your answer to part (i) find the value of a, giving your answer in terms of r. [4]

7 0606/12/M/J/17 © UCLES 2017 [Turn over 7 It is given that ( ) y A 10bx = , where A and b are constants. The straight line graph obtained when lgy is plotted against x passes through the points ( . , . ) 0 5 2 2 and ( . , . ) 1 0 3 7 . (i) Find the value of A and of b. [5] Using your values of A and b, find (ii) the value of y when . x 0 6 = , [2] (iii) the value of x when y 600 = . [2]

8 0606/12/M/J/17 © UCLES 2017 8 (a) A 5-digit number is to be formed from the seven digits 1, 2, 3, 5, 6, 8 and 9. Each digit can only be used once in any 5-digit number. Find the number of different 5-digit numbers that can be formed if (i) there are no restrictions, [1] (ii) the number is divisible by 5, [1] (iii) the number is greater than 60 000, [1] (iv) the number is greater than 60 000 and even. [3] (b) Ranjit has 25 friends of whom 15 are boys and 10 are girls. Ranjit wishes to hold a birthday party but can only invite 7 friends. Find the number of different ways these 7 friends can be selected if (i) there are no restrictions, [1] (ii) only 2 of the 7 friends are boys, [1] (iii) the 25 friends include a boy and his sister who cannot be separated. [3]

9 0606/12/M/J/17 © UCLES 2017 [Turn over 9 (a) Given that A 3 1 4 1 2 5 = - J L K KK N P O OO, B 1 3 2 0 = - J L KK N P OO and C AB = , (i) state the order of A, [1] (ii) find C. [3] (b) The matrix X 5 4 12 7 = - - J L KK N P OO. (i) Find X 1 - . [2] (ii) Using X 1 - , find the coordinates of the point of intersection of the lines , . y x y x 12 5 26 7 4 52 = - = - [4]

10 0606/12/M/J/17 © UCLES 2017 10 4 cm A D O C B 8 cm The diagram shows a circle, centre O, radius 8 cm. The points A, B, C and D lie on the circumference of the circle such that AB is parallel to DC. The length of the arc AD is 4 cm and the length of the chord AB is 15 cm. (i) Find, in radians, angle AOD. [1] (ii) Hence show that angle . DOC 1 43 = radians, correct to 2 decimal places. [3]

11 0606/12/M/J/17 © UCLES 2017 [Turn over (iii) Find the perimeter of the shaded region. [3] (iv) Find the area of the shaded region. [4] Question 11 is printed on the next page.

12 0606/12/M/J/17 © UCLES 2017 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. 11 The curve ( ) f y x = passes through the point 7 , 2 1 2 J L KK N P OO and is such that ( ) f e x x 2 1 = - l . (i) Find the equation of the curve. [4] (ii) Find the value of x for which f ʺ(x) = 4, giving your answer in the form ln a b 2 + , where a and b are constants. [4]

® IGCSE is a registered trademark. This document consists of 11 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2017 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 3 of 11 Question Answer Marks Partial Marks 1 ( ) ∪ ∩ A B C ( ) ∩ ∪ A B C B3 B1 for each ( ) ′ ′ ∩ ∩ A B C 2 attempt at differentiating a quotient, must have minus sign and ( ) 2 1 + x in the denominator M1 for ( ) 1 2 2 5 4 − + x B1 for ( )( ) 1 2 2 1 10 5 4 2 − + x x DB1 ( ) ( )( ) ( ) ( ) 1 1 2 2 2 2 2 1 1 10 5 4 5 4 d 2 d 1 − + + − + = + x x x x y x x A1 all else correct When d 11 3, d 112 = = y x x A1 must be exact Alternative ( ) ( ) 1 1 2 2 5 4 1 − = + + y x x M1 attempt to differentiate a product for ( ) 1 2 2 5 4 − + x B1 for ( )( ) 1 2 2 1 10 5 4 2 − + x x DB1 ( ) ( ) ( ) ( ) ( ) 1 1 1 2 2 2 2 2 d 110 5 4 1 5 4 1 d 2 y x x x x x x − − − = + + + + − + A1 all else correct When d 11 3, d 112 = = y x x A1 A1 must be exact

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 4 of 11 Question Answer Marks Partial Marks 3(a) v ( ) 1 3 5 2 5 = × − i j M1 attempt to find the magnitude of ( ) 2 − i j and use 3 6 = − i j A1 for 3 6 − i j only 3(b) w o o 2cos30 2sin30 = + i j M1 attempt to use trigonometry correctly to obtain components 3 = + i j A1 4 ( ) 2 1 2 3 3 1 3 6 6 − −     − + −         n n n x x n n n 3 81 = n , so 4 = n B1 3 1 4 3 6 × × − = a M1 for 1 3 6 −  −     n x n , 1 13 6 −  −     n n x C or 1 3 1 6 −    −       n n x , with/without their n 18 = − a A1 using their n and equating to a to obtain 18 = − a 2 4 3 1 3 2 36 × × × = b M1 for ( ) 2 2 1 3 6 −  −     n x n n , 2 2 23 6 −      n n x C or 2 2 3 2 6 −          n n x , with/without their n 3 2 = b A1 using their n and equating to b to obtain 3 2 = b 5(i) 12sin3 = − v t B1 5(ii) 12 B1 FT on their (i) of the form sin3 k t , must be k 5(iii) 36cos3 = − a t B1 allow unsimplified 3 2 t π = , 1.57 or better B1 π 6 = t or 0.524 B1 5(iv) 4 cao B1 may be obtained from knowledge of cosine curve

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 11 Question Answer Marks Partial Marks 6(i) 1 1 cos sin sin sin cos θ θ θ θ θ × + M1 cos sin for cot , tan , sin cos 1 cosec sin θ θ θ θ θ θ θ θ = = = dealing with the fractions correctly M1 2 2 1 sin cos sin cos sin θ θ θ θ θ × + M1 use of identity cosθ = A1 correct simplification, with all correct Alternative 1 ( ) 2 cosec 1 1 tan tan θ θ θ + M1 dealing with fractions 2 tan cosec sec θ θ θ = M1 use of appropriate identity 2 sin 1 cos cos sin θ θ θ θ = × × M1 1 sin for cot , tan , tan cos 1 1 sec , cosec cos sin θ θ θ θ θ θ θ θ θ = = = = cosθ = A1 correct simplification, with all correct Alternative 2 ( ) 2 cosec 1 cot 1 cot θ θ θ + M1 dealing with fractions 2 cot cosec cosec θ θ θ = M1 use of appropriate identity cot cosec θ θ = cos sin sin θ θ θ = × M1 cos sin for cot , tan , sin cos 1 cosec sin θ θ θ θ θ θ θ θ = = = cosθ = A1 correct simplification, with all correct

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 11 Question Answer Marks Partial Marks 6(ii) 0 0 1 cos2 d sin 2 2 θ θ θ   =     ∫ a a B1 1 3 sin2 2 4 = a M1 use of [ ]0 3 sin2 4 θ = a k to obtain 3 sin2 4 = k a π 2 3 = a DM1 attempt to solve equation of the form 3 sin2 4 = k a , with 3 1 1 4 − k - - , must have a correct order of operations dealing with the double angle π 6 = a , 0.167π or better A1 7(i) lg lg = + y A bx B1 straight line form, may be implied by correct values of both A and b later Gradient = b, M1 equating gradient to b 3 = b A1 Use of substitution into one of the following 2.2 lg 0.5 = + A b 3.7 lg = + A b 0.5 158.489 10 = × b A 5011.872 10 = × b A or equivalent valid method leads to lg 0.7 = A M1 5, 5.01 = A or 0.7 10 A1 Alternative 1 lg lg = + y A bx B1 straight line form, may be implied by correct work later 2.2 lg 0.5 = + A b M1 one correct equation 3.7 lg = + A b A1 both equations correct attempt to solve 2 correct equations M1 leading to 3 = b and 5, 5.01 = A or 0.7 10 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 11 Question Answer Marks Partial Marks 7(i) Alternative 2 ( ) 10 = bx y A 0.5 158.489 10 = × b A M1 one correct equation 5011.872 10 = × b A A1 both correct 0.5 5011.872 10 158.489 = b M1 attempt to solve 2 correct equations leading to 3 = b A1 correct b Use of substitution leads to 5, 5.01 = A or 0.7 10 A1 correct A 7(ii) Substitute A and b correctly into either ( ) 0.6 10 = b y A , lg lg 0.6 = + y A b or lg lg 0.6lg10 = + b y A or using lg 1.8 0.7 = + y M1 correct statement using their A and b correctly in either equation or using lg 3 0.7 = + y x 316 = y , 315 or 2.5 10 A1 7(iii) Substitute A and b correctly into either ( ) 600 10 = bx A , lg600 lg = + A bx or lg600 lg lg10 = + b A x or using lg600 3 0.7 = + x M1 correct statement using their A and b correctly in either equation or using lg 3 0.7 = + y x 0.693 = x A1 8(a)(i) 2520 B1 8(a)(ii) 360 B1 8(a)(iii) 1080 B1 8(a)(iv) 6 or 8 to start with No of ways = 2 5 4 3 2 × × × × = 240 B1 9 to start with No of ways = 1 5 4 3 3 × × × × = 180 B1 Total number of ways = 420 DB1 Dependent on both previous B marks

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 11 Question Answer Marks Partial Marks 8(a)(iv) Alternative 1 All numbers > 6000 − all odd numbers > 6000 B1 plan and attempt to use, must be using 1080 1080 180 480 − − B1 for 180 and 480 Total number of ways = 420 DB1 Dependent on both previous B marks Alternative 2 Even numbers > 60000 : Odd numbers > 60000 7 : 11 B1 correct ratio Total number of ways = 7 1080 18 × B1 = 420 DB1 Dependent on both previous B marks 8(b)(i) 480700 B1 8(b)(ii) 26460 B1 8(b)(iii) With brother and sister 23 5 33649 = C B1 for 23 5 C or 23 5 × k k C C Without brother and sister 23 7 245157 = C B1 for 23 7 C or 23 7 × k k C C Total number of ways = 278806 B1 for 23 23 5 7 + C C and evaluation 9(a)(i) 3 2 × B1 9(a)(ii) correct attempt to multiply the 2 matrices M1 C 6 6 5 2 19 8 −     =     −   A2 –1 for each incorrect element 9(b)(i) 1 7 12 1 4 5 13 − −   =   −   X B2 B1 for correct use of determinant B1 for correct matrix 9(b)(ii) 7 12 26 1 4 5 52 13 x y −      =      −      B1 attempt to evaluate using inverse from (i) together with pre-multiplication to obtain a 2 1 × matrix M1 34, 12 = = x y A2 A1 for each 10(i) 0.5 B1 for 0.5 from correct work only

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 9 of 11 Question Answer Marks Partial Marks 10(ii) ( ) 2 2 2 15 8 8 2 8 8 cos = + − × × × AOB 2.43075 = AOB rads M1 use of cosine rule (or equivalent) to obtain angle AOB. ( ) 2 DOC AOB their AOD = − M1 use of angle AOD and symmetry 1.43 = DOC to 2 dp A1 Answer Given: need to have seen either 2.431 or better, or 1.431 or better in previous calculations Alternative 1 1 15 2 8 sin 2 +   = × ×     DOC M1 use of basic trigonometry use of 1 0.5 2 + DOC M1 may be implied 1.43 = DOC to 2 dp A1 Answer Given: need to have seen either 2.431 or better, or 1.431 or better or 1.215 or better in previous calculations Alternative 2 ( ) 2 2 2 15 8 8 2 8 8 cos = + − × × × AOB 2.43075 = AOB rads 8 ∠ × = AOB arc AB M1 use of cosine rule (or equivalent) to obtain angle AOB. arc 8 8 − = ∠ AB DOC M1 attempt at DOC, must be a complete method with AOB found 1.43 = DOC to 2 dp A1 Answer Given: need to have seen either 2.431 or better, or 1.431 or better or 1.215 or better in previous calculations Alternative 3 Equating 2 different forms for the area of triangle AOB 2 15 31 1 8 sin 4 2 = × AOB , 2.43075 = AOB rads M1 using both different forms of the area of triangle AOB ( ) 2 their = − DOC AOB AOD M1 use of angle AOD and symmetry 1.43 = DOC to 2 dp A1 Answer Given: need to have seen either 2.431 or better, or 1.431 or better in previous calculations

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 10 of 11 Question Answer Marks Partial Marks 10(iii) 1.43 2 sin 2 8  =     DC or ( ) 2 2 2 8 8 2 8 8 cos1.43 = + − × × × DC M1 use of cosine rule or basic trigomoetry to obtain DC 10.49 = DC A1 awrt 10.5, may be implied Perimeter = 10.49 4 4 15 + + + = 33.5 A1 awrt 33.5 10(iv) ( ) ( ) 2 2 1 1 8 2.43 sin2.43 8 1.431 sin1.431 2 2 × − − × − B1 area of one appropriate sector; allow unsimplified; may be implied by a correct segment area of one appropriate triangle, allow unsimplified B1 an appropriate segment, allow unsimplified B1 = 42.8 (allow awrt 42.8) B1 final answer Alternative 1 Area of a trapezium + 2 small segments B1 one appropriate small sector, allow unsimpified (could be doubled) Each small segment = ( ) 2 1 8 0.5 sin0.5 2 × − B1 an appropriate triangle, allow unsimplfied (could be doubled) Area of trapezium = ( ) ( ) 1 15 10.5 6.041 2.784 2 + × − B1 attempt at trapezium, must have a correct attempt at finding the distance between the parallel sides – allow unsimplified Total area = 42.8 (allow awrt 42.8) B1 final answer Alternative 2 Area of 2 small sectors + area of triangle ODC – the area of triangle OAB Area of a small sector = 2 1 1 8 2 2 × × B1 area of small sector, allow unsimplified, (could be doubled) Area of triangle ODC = 2 1 8 sin1.43 2 × × B1 area of triangle ODC, allow unsimplified Area of triangle OAB = 2 1 8 sin 2.43 2 × × B1 area of triangle OAB, allow unsimplified Total area = 42.8 (allow awrt 42.8) B1 final answer

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 11 of 11 Question Answer Marks Partial Marks 10(iv) Alternative 3 Area of rectangle + 2 small triangles + 2 small segments Each small segment = ( ) 2 1 8 0.5 sin0.5 2 × − B1 area of a small segment, allow unsimplified, could be doubled ( )( ) 15 10.49 1 6.041 2.784 2 2 − × − B1 area of a small triangle, allow unsimplified, could be doubled Area of rectangle = ( ) 10.49 6.041 2.784 × − B1 allow unsimplified, could be doubled Total area = 42.8 (allow awrt 42.8) B1 final answer Alternative 4 Sector AOB – sector AOD – sector COB – triangle DOC B1 area of one appropriate sector; allow unsimplified; may be implied by a correct segment 2 2 2 1 1 1 8 2.43 2 8 0.5 8 sin1.43 2 2 2       × × − × × − ×             Area = sector AOB – segment DC – triangle AOB B1 area of one appropriate triangle, allow unsimplified 2 2 1 1 8 2.43 (their segment) 8 sin 2.43 2 2     × × − − ×         B1 an appropriate segment, allow unsimplified Total area = 42.8 (allow awrt 42.8) B1 final answer 11(i) 2 1 e − x m where m is numeric constant M1 ( ) ( ) 2 1 1 f e 2 − = + x x c A1 condone omission of +c 7 1 2 2 = + c DM1 correct attempt to find arbitrary constant ( ) 2 1 1 f e 3 2 − = + x x A1 must be an equation 11(ii) 2 1 e − x k where k is a numeric constant M1 ( ) 2 1 f '' 2e − = x x A1 4 2 1 ln  −=     x k DM1 attempt to equate to 4 and use logarithms 1 ln 2 2 = + x A1