Cambridge IGCSE Mathematics - Additional 0606 — 2018 May/June Paper 2 · Variant 2

This document consists of 15 printed pages and 1 blank page. DC (KN/SW) 145515/2 © UCLES 2018 [Turn over * 4 4 0 1 8 5 0 8 5 1 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 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. Cambridge International Examinations Cambridge International General Certificate of Secondary Education

2 0606/22/M/J/18 © UCLES 2018 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 0606/22/M/J/18 © UCLES 2018 [Turn over 1 (i) Show that cos cot sin cosec i i i i + = . [3] (ii) Hence solve cos cot sin 4 i i i + = for ° ° 0 90 G G i . [2]

4 0606/22/M/J/18 © UCLES 2018 2 (a) On the Venn diagram below, shade the region that represents A B + l. A B [1] (b) The universal set  and sets P, Q and R are such that ( ) P Q R , , Q = l , ( ) P Q R + + Q = l , n( ) Q R 8 + = , n( ) R P 8 + = , n( ) P Q 10 + = , n(P) = 21, n(Q) = 15, n() = 30. Complete the Venn diagram to show this information and state the value of n(R). P Q R n(R) = … [4]

5 0606/22/M/J/18 © UCLES 2018 [Turn over 3 It is given that x + 3 is a factor of the polynomial ( )x x ax x b 2 24 p 3 2 = + - + . The remainder when p (x) is divided by x − 2 is −15. Find the remainder when p (x) is divided by x + 1. [6]

6 0606/22/M/J/18 © UCLES 2018 4 Find the coordinates of the points where the line y x 2 3 6 - = intersects the curve x y 4 9 5 2 2 + = . [5]

7 0606/22/M/J/18 © UCLES 2018 [Turn over 5 (a) Four parts in a play are to be given to four of the girls chosen from the seven girls in a drama class. Find the number of different ways in which this can be done. [2] (b) Three singers are chosen at random from a group of 5 Chinese, 4 Indian and 2 British singers. Find the number of different ways in which this can be done if (i) no Chinese singer is chosen, [1] (ii) one singer of each nationality is chosen, [2] (iii) the three singers chosen are all of the same nationality. [2]

8 0606/22/M/J/18 © UCLES 2018 6 1.5 rad O C D E B A 5 cm In the diagram, ABC is an arc of the circle centre O, radius 5 cm, and angle AOC is 1.5 radians. AD and CE are diameters of the circle and DE is a straight line. (i) Find the total perimeter of the shaded regions. [3] (ii) Find the total area of the shaded regions. [3]

9 0606/22/M/J/18 © UCLES 2018 [Turn over 7 Vectors i and j are vectors parallel to the x-axis and y-axis respectively. Given that a = 2i + 3j, b = i − 5j and c = 3i + 11j, find (i) the exact value of a c + , [2] (ii) the value of the constant m such that a + mb is parallel to j, [2] (iii) the value of the constant n such that na − b = c. [2]

10 0606/22/M/J/18 © UCLES 2018 8 (a) A = 2 1 1 3 - - J L KK N P OO and B = 0 3 2 5 - - J L KK N P OO. Find (BA)−1. [4] (b) The matrix X is such that XC = D, where C = 2 0 5 10 3 4 - J L KK N P OO and D = 4 5 4 - ^ h. (i) State the order of the matrix C. [1] (ii) Find the matrix X. [2]

11 0606/22/M/J/18 © UCLES 2018 [Turn over 9 (i) Differentiate sin x x 4^ h with respect to x. [4] (ii) Hence find sin cos sin x x x x x x x 8 d 4 3 + + c ^ hm y . [3]

12 0606/22/M/J/18 © UCLES 2018 10 (a) (i) On the axes below, sketch the graph of ( )( ) y x x 3 5 = + - showing the coordinates of the points where the curve meets the x-axis. [2] O y x (ii) Write down a suitable domain for the function ( ) ( )( ) x x x 3 5 f = + - such that f has an inverse. [1] (b) The functions g and h are defined by ( )x x 3 1 g = - for x > 1, h( )x x 4 = for x ≠ 0. (i) Find hg (x). [1] (ii) Find (hg)–1(x). [2] (c) Given that p(a) = b and that the function p has an inverse, write down p–1 (b). [1]

13 0606/22/M/J/18 © UCLES 2018 [Turn over 11 (a) Find d x x 2 1 - 3y . [2] (b) (i) Find sin x x 4 d y . [2] (ii) Hence evaluate sin x x 4 d r r 8 4y . [2] (c) Show that x 3 e d ln x 3 0 8 = y . [5]

14 0606/22/M/J/18 © UCLES 2018 12 In this question all lengths are in centimetres. The volume of a cone of height h and base radius r is given by r V r h 3 1 2 = . It is known that r sin 12 4 6 2 = - , r cos 12 4 6 2 = + , r tan 12 2 3 = - . h r 6 rad r A water cup is in the shape of a cone with its axis vertical. The diagrams show the cup and its cross-section. The vertical angle of the cone is r 6 radians. The depth of water in the cup is h. The surface of the water is a circle of radius r. (i) Find an expression for r in terms of h and show that the volume of water in the cup is given by r( ) V h 3 7 4 3 3 = - . [4]

15 0606/22/M/J/18 © UCLES 2018 (ii) Water is poured into the cup at a rate of 30 cm3 s–1. Find, correct to 2 decimal places, the rate at which the depth of water is increasing when h = 5. [4]

16 0606/22/M/J/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

IGCSE™ is a registered trademark. This document consists of 10 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/22 Paper 1 May/June 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 May/June 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 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.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 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

0 © Q 0606/22 © UCLES 2018 Question 1(i) U U C 1(ii) s 1 1 i 2(a) 2(b) t Uses cotθ = 2 2 cos sin sin θ θ + Uses 2 cos θ + Completes to sinθ = 1 4 14.5° or 14.47[751…] sw their 12 A P 11 C Answer cos sin θ θ 2θ 2 sin 1 θ + = o 1 cos sinθ = ] rot to 4 or m R 8 4 1 2 5 [0] [0] Cambridge P ecθ more figures B Q [0] IGCSE – M PUBLISHED Page 4 of 10 Marks B3 M1 s A1 B1 B3 B1 Mark Schem D s B1 for us oe at som B1 for us B1 for co in a com fractions or for wr Maximu does not Not from B1 for 8 regions c B1 for 1 B1 for 4 maximu other va within th STRICT e Par sing cot = θ me stage se of 2 cos θ + ommon deno pound fractio riting 2 1 sin sinθ − um of 2 mark t complete to m wrong work correctly pla correct 1, 2, 5 correc correctly pla m of 2 mark alues such as he diagram T FT their V Ma rtial Marks cos sin θ θ oe or 2 sin 1 θ + = o ominator of s on or in two 2 1 as sin θ θ θ ks if not ful o cosecθ rking aced and all ctly placed aced ks if fully co s 30, 21 and/ Venn diagram ay/June 201 sin tan = cos θ oe sinθ oe eithe partial 2 sin sin θ θ − oe ly correct or the empty orrect but /or 15 prese m 18 s θ θ er r ent

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 10 Question Answer Marks Partial Marks 3 p(−3) = 0 or p(2) = −15 stated or implied M1 −54 + 9a + 72 + b = 0 or better A1 finds one correct equation; implies M1 16 + 4a − 48 + b = −15 or better A1 finds another correct equation; implies M1 Solves a pair of simultaneous equations in a and b M1 dep on first M1 condone one sign or arithmetic error in their solution; as far as finding one unknown a = −7, b = 45 A1 60 cao A1 4 Eliminates one of the unknowns M1 Simplifies to a correct 3-term quadratic: 2x2 + 4x – 16 [= 0] oe or 2y2 − 6y – 36 [= 0] oe A1 Factorises or solves ( )( ) 4 2 0 + − = x x oe or ( )( ) 3 6 0 + − = y y oe M1 FT their 3-term quadratic in x or y; (2, 6) and (−4, −3) oe A2 Not from wrong working A1 for either (2, 6) or (−4, −3) or A1 for x = 2 and x = −4 or y = 6 and y = −3 5(a) 7 4P or 7 × 6 × 5 × 4 oe M1 840 A1 5(b)(i) 20 B1 5(b)(ii) 5 4 2 1 1 1 × × C C C or 5 × 4 × 2 oe M1 40 A1 5(b)(iii) 5 4 3 3 + C C oe M1 14 A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 10 Question Answer Marks Partial Marks 6(i) (Arc length = ) 1.5 × 5 oe soi M1 implied by 7.5 (DE =) 10sin(0.75) oe soi M1 implied by awrt 6.82 34.3 or answer in range 34.31 to 34.32 A1 6(ii) (Area sector = ) 1 2 × 52 × 1.5 oe M1 implied by 18.75 (Area triangle =) 1 2 × 52 × sin(1.5) oe M1 implied by awrt 12.47 31.2 or answer in range 31.21 to 31.22 A1 7(i) |their(a + c)| = 2 2 (5 14 ) + their M1 221 A1 mark final answer 7(ii) [(2 + m)i + (3 – 5m)j therefore] their (2 + m ) = 0 M1 for attempting to form a + mb and equate the scalar of the i component to 0 m = −2 only A1 implies M1 7(iii) [(2n − 1)i + (3n + 5)j = 3i + 11j or n(2i + 3j) = (3i + 11j) + (i – 5j) oe leading to] 2n − 1 = 3 or 3n + 5 = 11 oe, soi M1 n = 2 only A1 implies M1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 10 Question Answer Marks Partial Marks 8(a) 2 6 1 12 −       B2 B1 for a 2 by 2 matrix with 2 or 3 correct elements 2 3 1 0 12 6 1   −       − −   −  their oe isw B2 FT their non-singular BA B1 FT for either 3 1 0 ( )      −  their or 12 6 1 2 ... −  ×   − −   their If their BA is singular, B0 then SC1 for 12 6 1 2 ... −  ×   − −   their OR Alternative method A−1B−1: B2 for A–1 = 3 1 1 1 2 5 −     − −  isw or B−1 = 5 2 1 3 0 6 −     −   isw or B1 for a multiplier of 1 5 − or for 3 1 1 2 −     −   or for a multipler of 1 6 or for 5 2 3 0 −     −   B2 FT for A−1 B−1 = 1 30 12 6 1 2 −   ×   − − −   their their or B1 FT for a 2 by 2 matrix with 2 or 3 correct elements Maximum of 3 marks if not fully correct 8(b)(i) 2 × 3 B1 8(b)(ii) 1 2 2   −     oe isw B2 B1 for each correct element; must be in a 1 by 2 matrix or M1 for a full method as far as finding values for the two elements

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 10 Question Answer Marks Partial Marks 9(i) ( ) ( ) 1 2 d 1 sin (sin ) cos d 2 − = x x x x oe B2 B1 for 1 2 1 (sin ) ... 2 −× x or for 1 2 1 (sin ) 2 − x or for 1 2 1 (...) cos 2 −× x or for 1 1 2 1 (sin ) cos 2  −     × their their x x ( ) ( ) ( ) 3 1 4 2 4 sin 1 sin cos oe 2 their x x x their x x −   +     M1 Applies correct form of product rule ( ) ( ) 1 3 4 2 1 4 sin sin cos 2 −   +     x x x x x oe isw A1 Not from wrong working 9(ii) ( ) ( ) ( ) 3 1 4 2 4 4 sin d 1 sin cos d 2 sin oe x x x x x x x x x −   + ×     = ∫ ∫ M1 or 4 3 cos d 2 4 sin d 2 sin   + +     ∫ ∫ x x x x x x x x oe FT their (i) [ ] 2 4 2 sin 2 + + x x x c A2 A1 for 4 d 2 sin + ∫x x x x 10(a)(i) y x -3 5 B2 B1 for correct shape B1 for roots marked on the graph or seen nearby provided graph drawn and one root is negative and one is positive 10(a)(ii) Any correct domain B1 10(b)(i) 4 3 1 − x B1 mark final answer

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 10 Question Answer Marks Partial Marks 10(b)(ii) Correct method for finding inverse function e.g. swopping variables and changing subject or vice versa; or indicates 1 1 1 (hg) ( ) g h ( ) − − − = x x and finds 1 1 g ( ) 3 − + = x x and 1 4 h ( ) − = x x M1 FT only if hg( ) of the form + a their x bx c where a, b and c are integers 1 1 4 (hg) ( ) 1 3 −     = +       x x oe isw or 1 4 (hg) ( ) 3 − +   =   x x x oe isw A1 FT 1 (hg) ( ) − − = a cx their x bx oe If M0 then SC1 for their hg(x) of the form = + a y b x oe leading to their (hg)-1(x) of the form = − a y x b isw 10(c) a cao B1 11(a) ( ) 4 3 2 1 [ ] 4 2 3 − + × x c oe isw B2 B1 for ( ) 1 1 3 2 1 1 1 3   +     − ×   +     x k where k ≠ 0 11(b)(i) cos4 [ ] + k x c where k < 0 or 1 4 = k M1 [ ] 1 cos4 4 − + x c A1 11(b)(ii) Sight of correct substitution of limits: 1 4π 1 4π cos cos 4 4 4 8   − −−     oe M1 FT their cos4 k x from (b)(i) dep on M1 awarded in (b)(i) 1 4 A1 does not imply M1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 10 of 10 Question Answer Marks Partial Marks 11(c) 3 3 e d e [ ] = + ∫ x x x k c M1 k any non-zero constant k = 3 A1 Sight of correct substitution of limits: ln8 0 3 e e − their k their k oe M1 dep on first M1 Shows how to deal with the power of the first term e.g. 1 3 ln8 ln8 3 = or ln8 ln2 3 = or ( ) 3 3 8 seen B1 6 – 3 = 3 A1 Not from wrong working 12(i) π tan12 = r h oe M1 (2 3) = − r h or π tan12 = r h oe A1 ( ) 2 2 1 [ ] π 2 3 3 = − × V h h oe M1 Correctly uses their expression for r in terms of h in formula for volume of a cone dependent on finding an expression connecting r and h ( ) 3 π 4 4 3 3 [ ] 3 − + = h V oe correctly leading to ( ) 3 π 7 4 3 [ ] 3 − = h V AG A1 12(ii) Correct derivative of V e.g. ( ) 2 3π 7 4 3 3 − h oe isw B1 d d d d d d = × h h V t V t soi B1 5 1 30 d d = ×      h V their h M1 if correct implies B1 B1; if incorrect, a correct FT statement implies the second B1 5.32 A1