Cambridge IGCSE Mathematics - Additional 0606 — 2019 May/June Paper 1 · Variant 1

This document consists of 15 printed pages and 1 blank page. DC (CE/CB) 165273/2 © UCLES 2019 [Turn over * 2 0 4 9 1 1 2 3 0 9 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 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 0606/11/M/J/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 0606/11/M/J/19 © UCLES 2019 [Turn over 1 (a) On the Venn diagrams below, shade the region indicated. A B C A B C ( ) A B C + , ( ) ' A B C , + [2] (b) On the Venn diagram below, draw sets P, Q and R such that P R 1 , Q R 1 and P Q + Q = .  [2]

4 0606/11/M/J/19 © UCLES 2019 2 (i) Write down the amplitude of sin x 4 3 1 - . [1] (ii) Write down the period of sin x 4 3 1 - . [1] (iii) On the axes below, sketch the graph of sin y x 4 3 1 = - for ° ° ° x 90 90 G G - . 6 4 2 –90 –60 –30 0 30 60 90 –2 –4 –6 y x [3]

5 0606/11/M/J/19 © UCLES 2019 [Turn over 3 The polynomial ( ) ( )( ) x x x k 2 1 12 p = - + - , where k is a constant. (i) Write down the value of ( )k p - . [1] When p(x) is divided by x + 3 the remainder is 23. (ii) Find the value of k. [2] (iii) Using your value of k, show that the equation ( )x 25 p =- has no real solutions. [3]

6 0606/11/M/J/19 © UCLES 2019 4 (i) The first 3 terms, in ascending powers of x, in the expansion of bx 2 8 + ` j can be written as a + 256x + cx2. Find the value of each of the constants a, b and c. [4] (ii) Using the values found in part (i), find the term independent of x in the expansion of bx x x 2 2 3 2 8 + - ` b j l . [3]

7 0606/11/M/J/19 © UCLES 2019 [Turn over 5 A particle P is moving with a velocity of 20 ms-1 in the same direction as 4 3 e o. (i) Find the velocity vector of P. [2] At time t = 0 s, P has position vector 2 1 e o relative to a fixed point O. (ii) Write down the position vector of P after t s. [2] A particle Q has position vector 18 17 e o relative to O at time t = 0 s and has a velocity vector 12 8 e o ms-1. (iii) Given that P and Q collide, find the value of t when they collide and the position vector of the point of collision. [3]

8 0606/11/M/J/19 © UCLES 2019 6 y x P O Q y x x 3 2 1 2 = - + y x 2 5 = + The diagram shows the curve y x x 3 2 1 2 = - + and the straight line y x 2 5 = + intersecting at the points P and Q. Showing all your working, find the area of the shaded region. [8]

9 0606/11/M/J/19 © UCLES 2019 [Turn over 7 (a) Solve log log x x 12 3 9 + = . [3] (b) Solve ( ) ( log log y 3 10 2 2 1 4 2 4 - = ) y 1 - + . [5]

10 0606/11/M/J/19 © UCLES 2019 8 It is given that ( )x 5 1 f ex = - for x R d . (i) Write down the range of f. [1] (ii) Find f -1 and state its domain. [3] It is given also that ( )x x 4 g 2 = + for x R d . (iii) Find the value of fg(1). [2]

11 0606/11/M/J/19 © UCLES 2019 [Turn over (iv) Find the exact solutions of ( )x 40 g2 = . [3]

12 0606/11/M/J/19 © UCLES 2019 9 In this question all lengths are in centimetres. A closed cylinder has base radius r, height h and volume V. It is given that the total surface area of the cylinder is 600r and that V, r and h can vary. (i) Show that V r r 300 3 r r = - . [3] (ii) Find the stationary value of V and determine its nature. [5]

13 0606/11/M/J/19 © UCLES 2019 [Turn over 10 When lg y is plotted against x2 a straight line graph is obtained which passes through the points (2, 4) and (6, 16). (i) Show that y 10A Bx2 = + , where A and B are constants. [4] (ii) Find y when x 3 1 = . [2] (iii) Find the positive value of x when y = 2. [3]

14 0606/11/M/J/19 © UCLES 2019 11 It is given that y x x 1 2 3 2 2 1 = + - ` `j j . (i) Show that x y x Px Qx 2 3 1 d d 2 1 2 = - + + ` j , where P and Q are integers. [5]

15 0606/11/M/J/19 © UCLES 2019 (ii) Hence find the equation of the normal to the curve y x x 1 2 3 2 2 1 = + - ` `j j at the point where x = 2, giving your answer in the form ax + by + c = 0, where a, b and c are integers. [4]

16 0606/11/M/J/19 © UCLES 2019 BLANK PAGE 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.

This document consists of 12 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 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 May/June 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 2 of 12 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 3 of 12 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

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 12 Question Answer Marks Guidance 1(a) E B1 E B1 1(b) E R P Q B2 B1 for ⊂ P R and ⊂ Q R B1 for ∩ = ∅ P Q 2(i) 4 B1 2(ii) o 120 or 2π 3 B1 2(iii) B3 B1 for a complete curve starting at ( ) o 90 , 3 − and finishing at ( ) o 90 , 5 − B1 for 5 3 y −≤ ≤for a complete curve Minimum point(s) at y = –5 Maximum point(s) at y = 3 DepB1 for a fully correct sine curve satisfying both the above and passing through ( ) o 60 , 1 − − ,( ) o 0 , 1 − and( ) o 60 , 1 − 3(i) –12 B1 3(ii) ( )( ) 2 3 1 3 12 23 k × −− − − = oe or ( ) ( )( ) 2 2 3 2 1 3 12 23 − + − − − − = k k M1 2 = − k A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 12 Question Answer Marks Guidance 3(iii) ( )( ) 2 1 2 12 25 − − − = − x x 2 2 5 15 0 − + = x x M1 expansion and simplification to a 3 term quadratic equation equated to zero, using their k. Discriminant: ( ) 25 4 2 15 − × × = –95 M1 using discriminant for their three term quadratic equation which is < 0 so no real solutions A1 cao for correct discriminant and correct conclusion 4(i) 256 = a B1 [ ] 7 8 2 256 bx x × × = oe or ( ) 2 6 2 8 7 2 2 bx cx × × ×   =   oe M1 1 4 = b oe, 112 = c A2 A1 for each 4(ii) ( ) 2 2 2 9 256 256 112 4 12   + + − +     x x x x B1 for 2 2 9 4 12   − +     x x Terms independent of x are ( ) ( ) ( ) 256 12 112 9 × − + × = 3072 1008 − + M1 adding and selecting ( ) ( ) ( ) 256 12 112 9 × − + × their their their their = –2064 A1 5(i) 2 2 3 1 20 4 3 4 v  = ×   + oe M1 finding and using the magnitude of 3 4    12 16   =     v A1 5(ii) 1 12 2 16    = +       p t r M1 correct use of position vector and their velocity vector A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 12 Question Answer Marks Guidance 5(iii) 17 8 1 12 18 12 2 16        + = +               t t Leading to 17 8 1 12 + = + t t or 18 12 2 16 + = + t t M1 equating position vectors of both particles at time t and solve either equation for t 4 = t A1 Position vector of collision 49 66       A1 6 Method 1 2 3 2 1 2 5 − + = + x x x leading to M1 equating the equations of the line and the curve and rearranging to obtain a three term quadratic equated to zero 2 3 4 4 0 − − = x x A1 2 3 = − x and 2 = x A1 ( ) ( ) 2 2 2 3 2 5 3 2 1 d − + − − + ∫ x x x x M1 subtraction (either way round) ( ) 2 2 2 3 4 4 3 d − + − ∫ x x x M1 integration to 2 3 + + Ax Bx Cx 2 2 3 2 3 4 2 −   + −   x x x A1 for 2 3 4 2 + − x x x oe ( ) 8 8 8 8 8 8 3 9 27   + − −− + +     40 8 27 = −− M1 Dep on preceding M1 correct use of limits 256 27 = or 9.48 or 13 27 9 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 12 Question Answer Marks Guidance 6 Method 2 2 3 2 1 2 5 − + = + x x x leading to M1 equating the line and the curve and rearranging to obtain a three term quadratic equated to zero 2 3 4 4 0 − − = x x A1 2 3 = − x and 2 = x A1 Area of trapezium = 1 11 8 9 2 3 3   + ×     B1 area of the trapezium, allow unsimplified Area under curve = 2 2 2 3 3 2 1 d − − + ∫ x x x M1 integration to 2 3 + + Ax Bx Cx 2 3 2 2 3 x x x −   = − +   A1 for 3 2 − + x x x = ( ) 8 4 2 8 4 2 27 9 3     − + −− − −         38 6 27 −− M1 DepM1 for correct use of limits. Shaded Area = 152 200 9 27 − 256 27 = or 9.48 or 13 27 9 A1 7(a) Method 1 3 3 3 log log 12 log 9 + = x x B1 change to base 3 logarithm 3 3log 12 2 = x 83 = x or 3 24 3 M1 simplification and dealing with base 3 logarithms to obtain a power of 3 6561 = x A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 12 Question Answer Marks Guidance 7(a) Method 2 9 9 9 log log 12 log 3 + = x x B1 change to base 9 9 3log 12 = x 4 9 = x or 3 12 9 M1 simplification and dealing with base 9 logarithms to obtain a power of 9 6561 = x A1 7(b) Method 1 ( ) ( ) 2 2 4 4 1 log 3 10 log 1 2 − = − + y y B1 use of power rule ( ) 2 4 2 3 10 1 log 2 1 − = − y y B1 DepB1 for use of division rule ( ) 2 2 3 10 2 1 − = − y y B1 for 4 1 log 2 2 = 2 4 12 0 + − = y y M1 Dep on first two B marks simplification to a three term quadratic. 2 = y only A1 7(b) Method 2 ( ) ( ) 2 2 4 4 1 log 3 10 log 1 2 − = − + y y B1 use of power rule ( ) ( ) 2 2 4 4 4 log 3 10 log 1 log 2 − = − + y y B1 for 4 log 2 ( ) 2 2 3 10 2 1 − = − y y B1 Dep on first B1 use of the multiplication rule 2 4 12 0 + − = y y M1 Dep on first and third B marks. simplification to a 3 term quadratic 2 = y only A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 9 of 12 Question Answer Marks Guidance 8(i) f 1 > − B1 or f ( ) 1 > − x , 1 > − y ,( ) 1, −∞, { } : 1 > − y y 8(ii) 1 e 5 + = y x oe M1 a complete valid method to obtain the inverse function 1 ln 5 +   =     x y or 1 1 f ( ) ln 5 − +   =     x x oe A1 Domain 1 > − x or ( ) 1, −∞ B1 FT their (i) or correct 8(iii) ( ) g 1 5 = so ( ) ( ) fg 1 f 5 = M1 evaluation using correct order of operations 5 5e 1 741 −= A1 awrt 741 or 5 5e 1 − 8(iv) ( ) ( ) 2 2 2 g 4 4 = + + x x M1 correct use of g2 4 2 8 16 4 40 + + + = x x ( ) 2 2 4 36 + = x or 4 2 8 20 0 + − = x x ( )( ) 2 2 10 2 0 + − = x x M1 DepM1 for forming and solving a quadratic in 2 x 2 = ± x only A1 9(i) Method 1 2 600π 2π 2π = + r rh B1 2 600π 2π 2π − = r h r M1 making h subject from a two term expression for SA. 2 π = V r h 2 2 600π-2π π 2π   =     r V r r 2 300 V r r r   = π −     3 300π π = − V r r A1 correct substitution and manipulation to obtain given answer

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 10 of 12 Question Answer Marks Guidance 9(i) Method 2 2 600π 2π 2π = + r rh B1 3 2 600πr 2π 2π = + r r h M1 multiplying both sides by r 3 2 600πr-2π π 2 = r r h 2 π = V r h 3 300π π = − V r r A1 correct manipulation to obtain 2 πr h 9(ii) 2 d 300π 3π d = − V r r M1 differentiation of given formula to A+Br2 When 2 d 300π 3π 0 d = − = V r r M1 equating to zero and attempt to solve 10 = r A1 2000π = V or 6280 or 6283 A1 2 2 d 6π d = − V r r , 2 2 d 0 d < V r so maximum B1 cao for 2 2 d 6π d = − V r r , 2 2 d 60π d = − V r or other correct method leading to maximum 10(i) Method 1 2 lg = + y A Bx B1 statement soi 16 6 = + A B 4 2 = + A B M1 one correct equation leading to 2 = − A and 3 = B A2 A1 for each 10(i) Method 2 2 lg = + y A Bx B1 statement soi Gradient = B 3 = B B1 16 6 = + A B or 4 2 = + A B M1 a correct equation 2 = − A A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 11 of 12 Question Answer Marks Guidance 10(i) Method 3 2 lg 4 3( 2) − = − y x or 2 lg 16 3( 6) − = − y x OR ( ) ( ) 4 3 2 or 16 3 6 c c = + = + M1 correct equation or for correct method for finding constant. 2 lg = + y A Bx B1 statement soi by their A and B Hence 2 3 2 10 − = x y B = 3 B1 A = –2 A1 10(ii) 2 1 2 3 3 10 y   −+     = M1 correct use of their A and B 0.1 = y oe A1 10(iii) 2 3 2 2 10 − = x M1 correct use of their A and B 2 lg2 3 2 = − x lg2 2 3 x + = M1 complete correct method to solve for x 0.876 = x A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 12 of 12 Question Answer Marks Guidance 11(i) ( )( ) ( ) 1 1 2 2 2 d 1 2 3 2 2 3 d − = + − + − y x x x x x M1 differentiation of a product B1 for ( ) ( ) 1 1 2 2 d 1 2 3 2 2 3 d 2 − − = × − x x x oe A1 all else correct i.e. ( ) ( ) 1 2 2 d 1 f ( ) 2 2 3 d = + + − y x x x x x ( ) ( ) ( ) 1 2 2 2 3 1 2 2 3 − = − + + − x x x x M1 correctly taking out a factor of ( ) 1 2 2 3 − − x or correctly using ( ) 1 2 2 3 − x as denominator ( ) 2 1 2 5 6 1 2 3 − + = − x x x A1 11(ii) When 2 x = , 5 y = B1 d 9 d y x = , so gradient of normal = 1 9 − M1 substitution to obtain gradient and correct method for gradient of normal Equation of normal ( ) 1 5 2 9 y x − = − − M1 DepM1 for equation of normal 9 47 0 x y + − = or 9 47 0 x y −− + = A1 Must be in this form