Cambridge IGCSE Mathematics - Additional 0606 — 2019 Oct/Nov Paper 1 · Variant 2

This document consists of 16 printed pages. DC (LK/TP) 172296/2 © UCLES 2019 [Turn over * 7 2 1 8 1 4 9 8 0 6 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 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 0606/12/O/N/19 © UCLES 2019 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 0606/12/O/N/19 © UCLES 2019 [Turn over 1 (i) On the axes below, sketch the graph of cos y x 2 3 1 = - for ° ° x 90 90 G G - . y x 0 1 -1 -2 -3 -4 2 3 4 30° -30° -60° -90° 60° 90° [3] (ii) Write down the amplitude of cos x 2 3 1 - . [1] (iii) Write down the period of cos x 2 3 1 - . [1]

4 0606/12/O/N/19 © UCLES 2019 2 When lgy2 is plotted against x, a straight line is obtained passing through the points (5, 12) and (3, 20). Find y in terms of x, giving your answer in the form y 10ax b = + , where a and b are integers. [5]

5 0606/12/O/N/19 © UCLES 2019 [Turn over 3 The first three terms in the expansion of ( ) x x 1 7 1 2 14 4 - - b l can be written as ax bx 1 2 + + . Find the value of each of the constants a and b. [6]

6 0606/12/O/N/19 © UCLES 2019 4 (i) On the axes below, sketch the graph of y x x 2 9 5 2 = - - showing the coordinates of the points where the graph meets the axes. [4] y x O (ii) Find the values of k for which x x k 2 9 5 2 - - = has exactly 2 solutions. [3]

7 0606/12/O/N/19 © UCLES 2019 [Turn over 5 (a) It is given that f : x x 7 for x 0 H , g : x x 5 7 + for x 0 H . Identify each of the following functions with one of f -1, g-1, fg, gf, f 2, g2. (i) x 5 + [1] (ii) x 5 - [1] (iii) x2 [1] (iv) x 10 + [1] (b) It is given that ( ) h x a x b 2 = + where a and b are constants. (i) Why is x 2 2 G G - not a suitable domain for h(x)? [1] (ii) Given that ( ) h 1 4 = and ( ) h 1 16 = l , find the value of a and of b. [2]

8 0606/12/O/N/19 © UCLES 2019 6 (a) Write p p qr r qp 1 2 - 3 a k in the form p q r a b c, where a, b and c are constants. [3] (b) Solve log log x 2 7 3 x 7 + = . [4]

9 0606/12/O/N/19 © UCLES 2019 [Turn over 7 It is given that ( ) e y x 1 5 x2 = + + ^ h . (i) Find d d x y. [3] (ii) Find the approximate change in y as x increases from 0.5 to . p 0 5+ , where p is small. [2] (iii) Given that y is increasing at a rate of 2 units per second when . x 0 5 = , find the corresponding rate of change in x. [2]

10 0606/12/O/N/19 © UCLES 2019 8 (a) Five teams took part in a competition in which each team played each of the other 4 teams. The following table represents the results after all the matches had been played. Team Won Drawn Lost A 2 1 1 B 1 3 0 C 1 1 2 D 0 1 3 E 3 0 1 Points in the competition were awarded to the teams as follows 4 for each match won, 2 for each match drawn, 0 for each match lost. (i) Write down two matrices whose product under matrix multiplication will give the total number of points awarded to each team. [2] (ii) Evaluate the matrix product from part (i) and hence state which team was awarded the most points. [2]

11 0606/12/O/N/19 © UCLES 2019 [Turn over (b) It is given that A 1 2 1 4 = - c m and B 5 1 0 2 = - c m. (i) Find A-1. [2] (ii) Hence find the matrix C such that AC = B. [3]

12 0606/12/O/N/19 © UCLES 2019 9 A solid circular cylinder has a base radius of r cm and a height of h cm. The cylinder has a volume of cm 1200 3 r and a total surface area of S cm2. (i) Show that S r r 2 2400 2 r r = + . [3]

13 0606/12/O/N/19 © UCLES 2019 [Turn over (ii) Given that h and r can vary, find the stationary value of S and determine its nature. [5]

14 0606/12/O/N/19 © UCLES 2019 10 18 cm 18 cm O A B C 10 cm The diagram shows a circle centre O, radius 10 cm. The points A, B and C lie on the circumference of the circle such that AB = BC = 18 cm. (i) Show that angle . AOB 2 24 = radians correct to 2 decimal places. [3] (ii) Find the perimeter of the shaded region. [5]

15 0606/12/O/N/19 © UCLES 2019 [Turn over Continuation of working space for Question 10(ii). (iii) Find the area of the shaded region. [3] Question 11 is printed on the next page.

16 0606/12/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 A curve is such that ( ) d d x y x 2 3 1 2 2 3 2 = - - . Given that the curve has a gradient of 6 at the point (3, 11), find the equation of the curve. [8]

This document consists of 10 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 1 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.

0606/12 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.

0606/12 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 4 of 10 Question Answer Marks Guidance 1(i) B3 B1 for y intercept ( ) 0,1 , must have a graph B1 for starting and finishing at ( ) 90, 1 ± − B1 for all correct, must be attempt at a curve passing through ( ) 30, 1 ± − and ( ) 60, 3 ± − 1(ii) 2 B1 1(iii) o 120 or 2π 3 B1 2 2 lg = + y mx c B1 May be implied by subsequent work Gradient = ( ) 4 − = m B1 32 = c B1 2 2 10 + = c mx their their y M1 Dep on first B1 Use of 2 lg 2lg = y y and 2 2 10 + c mx their their Or use of ( ) 2 10 + = their c their mx y and 2 2 10 + c mx their their 16 2 10 − = x y A1 3 14 2 13 1 1 2 7 7   − = − +     x x x B2 All terms correct or B1 for 2 correct terms ( ) 4 2 1 2 1 8 24 − = − + x x x « B2 First three terms correct or B1 for one incorrect term Product = 2 293 1 10 7 − + x x M1 For attempt to multiply out to obtain ( ) 2 1 10 − + x mx , 16 ≠ m 293 10, 7 = − = a b A1 For both, need to identify a and b 4(i) B4 B1 for shape, with max in first quadrant B1 for ( ) 0.5,0 − and ( ) 5,0 B1 for ( ) 0,5 B1 all correct, with cusps and correct curvature for 0.5 < x and 5 > x

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 5 of 10 Question Answer Marks Guidance 4(ii) 0 = k B1 Not from incorrect work Stationary point when 121 8 = ± y or 15.125 ± M1 For attempt to find y-coordinate of stationary point, must be a complete method i.e. Use of calculus Use of discriminant, Use of completing the square Use of symmetry Allow if seen in part (i), but must be used in (ii) 121 8 > k A1 cao 5a(i) fg B1 5a(ii) 1 g− B1 5a(iii) 1 f − B1 5a(iv) 2 g B1 5(b)(i) Undefined at 0 = x oe B1 5(b)(ii) 4 = + a b ( ) 3 h′ = p x x and attempt at ( ) h 1 ′ M1 For attempt at ( ) h 1 and differentiation to obtain ( ) h 1 ′ , must have the form ( ) 3 h′ = p x x oe 8 = − b 12 = a A1 For both 6(a) 5 7 7 3 3 2 − p q r B3 B1 for each term or for each of 7 2 = a , 5 3 = b , 7 3 = − c

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 6 of 10 Question Answer Marks Guidance 6(b) Either 7 7 2 log 3 log + = x x M1 For change of base. ( ) 2 7 7 log 3log 2 0 − + = x x 7 7 log 1, log 2 = = x x M1 Dep for forming a 3 term quadratic equation in 7 log x and a correct attempt to solve 7, 49 = = x x M1 Dep on both previous M marks for dealing with a base 7 logarithm correctly A1 For both Or 1 2log 7 3 log 7 + = x x M1 For change of base ( ) 2 2 log 7 3log 7 1 0 − + = x x log 7 1, log 7 0.5 = = x x M1 Dep for forming a 3 term quadratic equation in log 7 x and a correct attempt to solve 7, 49 = = x x M1 Dep on both previous M marks for dealing with a base x logarithm correctly A1 For both Or lg lg7 2 3 or lg1000 lg7 lg + = x x M1 For change of base ( ) ( ) ( ) 2 2 lg 3lg7 lg 2 lg7 0 − + = x x lg 2lg7 = x lg lg7 = x M1 Dep for forming a 3 term quadratic equation in lg x and a correct attempt to solve 7, 49 = = x x M1 Dep on both previous M marks for dealing with a base 10 logarithm correctly A1 For both, must be exact 7(i) ( ) ( ) 2 2 d e 1 2 e 5 d = + + + x x y x x x B1 For 2 2 ex x M1 For attempt at differentiating a product or expanding brackets and differentiating a product A1 For all other terms, apart from 2 2 ex x , correct

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 7 of 10 Question Answer Marks Guidance 7(ii) When d 0.5, 9.35 d = = y x x M1 For attempt to find their d d y x when 0.5 = x and multiplication by p Approximate change = 9.35p A1 7(iii) d d d d d d × = y x y x t t d 9.346 2 d × = x t M1 For use of correct rates of change equation using their d d y x when 0.5 = x and d 2 d = y t d 0.214 d = x t A1 FT on 2 9.346 their Must be correct to at least 3 sf 8(a)(i) Either 2 1 1 4 1 3 0 2 1 1 2 0 0 1 3 3 0 1                  or 2 1 1 3 4 1 1 2 0 1 3 0                 B2 For correct matrices in correct order or B1 if one correct matrix and a slip in one element of the other matrix Or ( ) 2 1 1 0 3 4 2 0 1 3 1 1 0 1 0 2 3 1           or ( ) 2 1 1 0 3 4 2 1 3 1 1 0       B2 For correct matrices in correct order or B1 if one correct matrix and a slip in one element of the other matrix 8(a)(ii) 10 10 6 2 12                 or ( ) 10 10 6 2 12 Team E M1 For matrix multiplication of their (i), with at least 2 elements correct, must be in correct form, may be unsimplified A1 All correct and identifying team E 8(b)(i) 4 1 1 2 1 6     −   B2 B1 for 1 6 and B1 for 4 1 2 1     −  

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 8 of 10 Question Answer Marks Guidance 8(b)(ii) C = A-1B M1 For pre-multiplication by their inverse from (i) C 4 1 5 0 1 2 1 1 2 6    =    − −    M1 Dep for matrix multiplication, using their inverse from (i), at least 2 elements correct 21 2 1 9 2 6 −   =   − −   oe A1 9(i) 2 π 1200π = r h B1 2 1200 = h r or 1200π π = rh r and substitution into their S B1 Must have attempt to use in an equation for S 2 2 1200 2π 2π   = + ×     S r r r leading to given answer B1 9(ii) 2 d 2400π 4π d = − S r r r M1 Must obtain the form 2 + B Ar r When d 0 d = S r , 3 600, 8.43 = r M1 Dep for equating to zero and attempt to solve to obtain r = … A1 For correct r min 1340 = S or 1341 A1 Either 2 2 3 d 4800π 4π d = + S r r 2 2 d 0 d > S r so minimum B1 For a correct method to reach a correct conclusion If r is not calculated, then must state that r > 0 Or Consideration of gradient e.g. r 8.43 < 8.43 >8.43 d d S r − 0 + Minimum point B1 Must be making a correct and convincing argument with sufficient detail

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 9 of 10 Question Answer Marks Guidance 10(i) Either 2 2 2 18 10 10 200cos = + − AOB M1 Attempt to use cosine rule cos 0.62 = − AOB A1 Allow unsimplified 2.2395 = AOB or greater accuracy, so 2.24 (to 2 dp) or 2.239... = AOB so 2.24 (to 2 dp) 2.240 = AOB so 2.24 (to 2 dp) A1 Must justify 2 dp 10(i) Or 9 sin 2 10 = AOB or 9 tan 2 19 = AOB or 19 cos 2 10 = AOB M1 Attempt at trig using a right angled triangle 2 = AOB awrt 1.12 A1 2.2395 = AOB or greater accuracy, so 2.24 (to 2 dp) or 2.239... = AOB so 2.24 (to 2 dp) 2.240 = AOB so 2.24 (to 2 dp) A1 Must justify 2 dp 10(ii) ( ) 2π 2 2.2395 = − AOC or ( ) or π 2.2395 2 = − AOC ABC oe M1 For attempt to find angle AOC or ABC ( ) 2π 2 = − AOC their AOB ( ) π = − ABC their AOB oe 1.804 = AOC or 1.803 A1 Condone 1.8 or 1.80 Arc length = 18.04 or 18.03 M1 For attempt at arc length using 10 ×their AOC 20sin 2 = AOC AC or 36sin 2 ABC or 2 2 10 10 200cos + − AOC or 2 2 18 18 648cos + − ABC 15.69 = or 15.7 M1 For attempt at AC using their AOC, or ABC but 2π 2.24 or 3 ≠ AOC Perimeter = 33.7 A1 Allow awrt 33.7

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 10 of 10 Question Answer Marks Guidance 10(iii) Area of sector = 50 1.804 × = 90.2 or 90.15 M1 For attempt at sector area 2 1 10 2 × ×their AOC AOC must be in radians Area of triangle = 50sin1.804 = 48.6 or 48.66 M1 For attempt at area of triangle 2 1 10 sin 2 × × their AOC AOC must be in radians Shaded area = 41.6 or 41.5 A1 Lack of accuracy is penalised here 11 ( ) 1 3 d 2 3 1 d = − + y x c x M1 For ( ) 1 3 d 3 1 d   = −     y a x x , condone omission of + c A1 All correct, condone omission of c 6 4 = + c M1 Dep for attempt to find c ( ) 1 3 d 2 3 1 2 d   = − +     y x x A1 All correct, may be implied by 2 = c ( ) 4 3 1 3 1 2 2 = − + + y x x d M1 For attempt to integrate their d d y x to obtain the form ( ) ( ) 4 3 3 1 = − + + y b x mx d A1 All correct, condone omission of d 11 14 = + d M1 Dep for attempt to find d, a second arbitrary constant, having used an arbitrary constant for d d y x ( ) 4 3 1 3 1 2 3 2 = − + − y x x A1