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

This document consists of 12 printed pages. DC (KN/FD) 129463/3 © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 6 5 9 5 4 0 4 1 3 2 * ADDITIONAL MATHEMATICS 0606/21 Paper 2 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.

2 0606/21/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 0606/21/M/J/17 © UCLES 2017 [Turn over 1 Find the equation of the curve which passes through the point (2, 17) and for which x y x 4 1 d d 3 = + . [4] 2 Do not use a calculator in this question. (a) Show that 24 27 15 9 30 # + can be written in the form a 2, where a is an integer. [3] (b) Solve the equation x x 3 1 2 3 + = - ^ ^ h h, giving your answer in the form b c 3 + , where b and c are integers. [3]

4 0606/21/M/J/17 © UCLES 2017 3 The variables x and y are such that ( ) ln y x 1 2 = + . (i) Find an expression for x y d d . [2] (ii) Hence, find the approximate change in y when x increases from 3 to h 3 + , where h is small. [2] 4 (a) Given that cos y x 7 10 3 = - , where the angle x is measured in degrees, state (i) the period of y, [1] (ii) the amplitude of y. [1] (b) 0 0° 45° 90° 135° 180° 225° 270° 315° 360° 2 4 6 8 10 12 y x Find the equation of the curve shown, in the form ( ) y a bx c g = + , where ( )x g is a trigonometric function and a, b and c are integers to be found. [4]

5 0606/21/M/J/17 © UCLES 2017 [Turn over 5 (i) Given that a is a constant, expand ( ) ax 2 4 + , in ascending powers of x, simplifying each term of your expansion. [2] Given also that the coefficient of x2 is equal to the coefficient of x3, (ii) show that a 3 = , [1] (iii) use your expansion to show that the value of 1.974 is 15.1 to 1 decimal place. [2]

6 0606/21/M/J/17 © UCLES 2017 6 Four cinemas, P, Q, R and S each sell adult, student and child tickets. The number of tickets sold by each cinema on one weekday were P: 90 adult, 10 student, 30 child Q: 45 student R: 25 adult, 15 child S: 10 adult, 100 child. (i) Given that 1 1 1 1 L = ^ h, construct a matrix, M, of the number of tickets sold, such that the matrix product LM can be found. [1] (ii) Find the matrix product LM. [1] (iii) State what information is represented by the matrix product LM. [1] An adult ticket costs $5, a student ticket costs $4 and a child ticket costs $3. (iv) Construct a matrix, N, of the ticket costs, such that the matrix product LMN can be found and state what information is represented by the matrix product LMN. [2]

7 0606/21/M/J/17 © UCLES 2017 [Turn over 7 (a) On each of the Venn diagrams below shade the region which represents the given set. ∩ ∩ ∩ X X P P' Q Q R R Y' Y [2] (b) In a group of students, each student studies at most two of art, music and design. No student studies both music and design. A denotes the set of students who study art, M denotes the set of students who study music, D denotes the set of students who study design. (i) Write the following using set notation. No student studies both music and design. [1] There are 100 students in the group. 39 students study art, 45 study music and 36 study design. 12 students study both art and music. 25 students study both art and design. (ii) Complete the Venn diagram below to represent this information and hence find the number of students in the group who do not study any of these subjects. A [3]

8 0606/21/M/J/17 © UCLES 2017 8 (a) A football club has 30 players. In how many different ways can a captain and a vice-captain be selected at random from these players? [1] (b) A team of 11 teachers is to be chosen from 2 mathematics teachers, 5 computing teachers and 9 science teachers. Find the number of different teams that can be chosen if (i) the team must have exactly 1 mathematics teacher, [2] (ii) the team must have exactly 1 mathematics teacher and at least 4 computing teachers. [4]

9 0606/21/M/J/17 © UCLES 2017 [Turn over 9 The curve x xy y y 3 4 3 0 2 2 + - + - = and the line ( ) y x 2 1 = - intersect at the points A and B. (i) Find the coordinates of A and of B. [5] (ii) Find the equation of the perpendicular bisector of the line AB, giving your answer in the form ax by c + = , where a, b and c are integers. [4]

10 0606/21/M/J/17 © UCLES 2017 10 The table shows values of the variables t and P. t 1 1.5 2 2.5 P 4.39 8.33 15.8 30.0 (i) Draw the graph of lnP against t on the grid below. [2] 0.5 1 2 3 ln P 0 1 1.5 2 2.5 t (ii) Use the graph to estimate the value of P when . t 2 2 = . [2] (iii) Find the gradient of the graph and state the coordinates of the point where the graph meets the vertical axis. [2] (iv) Using your answers to part (iii), show that P abt = , where a and b are constants to be found. [3] (v) Given that your equation in part (iv) is valid for values of t up to 10, find the smallest value of t, correct to 1 decimal place, for which P is at least 1000. [2]

11 0606/21/M/J/17 © UCLES 2017 [Turn over 11 (i) Prove that ( ) sin cot tan sec x x x x + = . [4] (ii) Hence solve the equation ( ) sin cot tan x x x 2 + = for x 0 360 G G c c. [4] Question 12 is printed on the next page.

12 0606/21/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. 12 A particle moves in a straight line so that, t seconds after passing a fixed point O, its displacement, s m, from O is given by cos s t t 1 3 5 = + - . (i) Find the distance between the particle’s first two positions of instantaneous rest. [7] (ii) Find the acceleration when t = π. [2]

® IGCSE is a registered trademark. This document consists of 7 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/21 Paper 2 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 of 7 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

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 3 of 7 Question Answer Marks Guidance 1 Integrates M1 must be clear attempt to integrate at least one term [ ] ( ) 4 y x x c = + + A1 Both terms correct 4 17 2 2 c = + + DM1 Substitution of x = 2, y = 17 to find c 4 1 y x x = + − cao A1 must have y = 2(a) 2 6 3 3 6 18 × = oe M1 method must be shown – simplifies and combines product 18 2 A1 If all over common denominator then consider the product for M1A1 9 2 oe soi leading to final answer of 27 2 B1 2(b) [ ] 6 3 2 3 x + = − M1 Expanding and making x subject – condone slips but must be of equivalent difficulty [ ] 6 3 2 3 2 3 2 3 x + + = × − + oe and multiplies out numerator and denominator M1 numerator at least 3 terms; 12 2 3 6 3 3 + + + 15 8 3 + A1 3(i) 2 2 1 x x + final answer B2 B1 for ( ) 2 1 1 ax b x × + + , a or b must be non-zero 3(ii) 2 2(3) (3) 1 y their h δ   = ×   +   or better M1 Substitutes ݔ= 3 into their d d y x and multiplies by h 6 10 h oe A1 4(a)(i) 36 B1 4(a)(ii) 7 B1 4(b) [ ] 5sin 4 7 y x = + B4 B1 for each of 5, 4 and 7 and B1 for sine Accept a = 5, b = 4, c = 7 for B3

0 © Qu 7 0606/21 © UCLES 2017 uestion 5(i) 16 5(ii) 2 5(iii) x 16 16 6(i) (M 6(ii) (L 6(iii) T c 6(iv) (N T co 7(a) 7(b)(i) n 6 32 24 ax + + 2 3 24 8 a a = an = −0.01 or a 6 + 32(3)(−0 6 − 0.96 + 0. ) 90 1 0 4 25 10    =    M )(1 1 = LM The total num cinemas oe ) 5 4 3   =    N he total inco omment e.g. ( ) n 0 M D ∩ = X C A 2 2 3 4 8 a x a x + nd solves to ax = −0.03 so 0.01) +24(9)( .0216 or 15.0 10 30 45 0 0 15 0 100       ) 90 0 1 1 25 10       mbers of each ome of all (4 total incom 0 or M D ∩ = Y Cambridge P Answer 3 4 4 x a x + given answe oi (−0.01)2 lead 06…isw 10 30 45 0 0 15 0 100   =    h type of tick ) cinemas or e from all tic = ∅ P IGCSE – M PUBLISHED Page 4 of 7 er ding to (125 55 = ket sold by al r other valid cket sales R Mark Schem D Ma ) 145 ll 4 Q R e arks B2 B1 fo or mi unsim SC1 16 + or all B1 or ve coeff must M1 A1 Must into t A1 a round B1 colum interc consi B1 Answ and m corre B1 B1 Calcu B1 Tota B2 B1 fo B1 No a M ∩ M Guid for at most 2 missing or for mplified form for 32 24 ax ax + l terms corre erifies that a f of 216 for b t be from cor t show clear their expansi and reach a v ds to 15.1 mns and/or r changed but istent wer must be must be cons ect M ulation not r al cost/value for each additional bra { } D ∩ = ∅ is May/June 201 ance terms incorr correct but m 2 3 8 x ax ax + + ect listed = 3 leads to both terms rrect terms in substitution ion for value which rows may be must be of correct or sistent with a equired of tickets etc ackets e.g. B0 17 ect 4x n (i) rder a c.

0 © Qu 7 8 8 0606/21 © UCLES 2017 uestion 7(b)(ii) 8(a)  8(b)(i) 2 20 8(b)(ii) ( or {2 10 9(i) Su − So va   30 2 870 P  = 2 14 1 10 C C × oe 002 ( 2 5 9 1 4 C C × × r ( 2 14 1 10 2 5 1 1 C C C C × × × 2002 – (10 + 092 ubstitution o 2 3 2 1 x x − + + olving their ariable (3x + 1 8 , 3 3  −  oe a M 33 C A e (2 × 1001) ) ( 9 2 6 1 C C + × 0 9 2 9 1 C C − × + × + 80 + 720)} of 2(1 y x = − 0 = oe ( 2 3x quadratic fou )( ) or 1 1 x + − and ( ) 1, 0 oe A 2 12 25 Cambridge P Answer ) 5 9 5 5 C C × 5 9 2 8 C C × + ) x ) 2 1 0 x − −= und from elim ( )( 3 1 x x + − e isw nfww D 5 11 17 IGCSE – M PUBLISHED Page 5 of 7 oe (840 + 25 2 5 9 1 3 C C C × × ) minating one ) 1 Mark Schem D Ma 52) ) 7 C e e arks B3 B1 co with B1 3 B1FT diagr requi FT o corre B1 M1 Cond A1 impli M3 M3 f M2 f but n produ M1 f produ A1 impli M1 Must subst error omis A1 Term M1 can b of x v A2 A1 f or A1 coord y-coo M Guid correct interse 12 and 25 co 3, 2, 11 corr T 17; must b ram and iden ired answer on 100− (sum ectly position done 14 4       fo ies M1 for fully corr for all necess not summed w ucts seen soi for one corre uct soi ies M3 t be attempt titution. Con r in substituti ssion of = 0 o ms collected be implied by values for each 1 for a corre dinates or a c ordinates May/June 201 ance ection of circ orrect rectly placed be on the Ve ntified as the m of their 5 ned values) or 14 10       rect method s sary products with no extra i ect three term at full ndone one sig ion. Condone or incorrect r y a correct p ct pair of x- correct pair o 17 cles enn e soi s a m gn e rhs air of

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 7 Question Answer Marks Guidance 9(ii) [ ] 1 2 m = cao B1 1 4 , 3 3       B1 FT 4 1 1 3 2 3 y their their x their   − = −     M1 or 1 2 y their x c = + and substitutes their midpoint and reaches c = … 6 3 7 y x − = A1 allow any equivalent form with integer coeffs/constant 10(i) t 1 1.5 2 2.5 lnP 1.48 2.12 2.76 3.4(0) M1 allow lnP values to 1 dp rounded or truncated (1.5, 2.1, 2.8, 3.4) single ruled line drawn within tolerance at least for t between 1 and 2.5 A1 All points within 1 square of line / must not pass through origin 10(ii) 3 etheir M1 18 to 22.2 A1 10(iii) (0, c) with 0.1 ⩽ c ⩽ 0.3 (0.2) B1 allow y = c condone c =… m in the range 1.25 ⩽ m ⩽ 1.34 (1.28) B1 10(iv) ( ) ln 1.28 0.2 P their t their = + M1 or ( ) ln ln ln P b t a = + ( 1.28) 0.2 e their t their P + = M1 or ln 1.28 b m their = = and ln 0.2 a c their = = 0.2 ( 1.28) e e their their t P = A1 or 1.10 ⩽ a ⩽1.35 3.49 ⩽ b ⩽ 3.82 10(v) 0.2 1.28 1000* their their t e e × or 1000* t their a their b × M1 A correct relationship e.g. ( ) 1.3 *ln 1000 0.2 t − where * is = or an inequality sign 5.3 A1 5.2 to 5.5 must be to 1dp

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 7 Question Answer Marks Guidance 11(i) 2 2 cos sin cos sin sin cos sin cos x x x x x x x x + + = oe B2 B1for either cos cot sin x x x = or sin tan cos x x x = used B1for correctly placing over a common denominator or for splitting into 3 correct terms not just for stating or working from both sides Valid use of Pythagorean identity e.g. 2 2 cos sin 1 x x + = B1 Simplification to secx (correct solution only) B1 not if working from both sides 11(ii) 1 cos 2 x = soi M1 60, 300 A1 Correct pair 1 cos 2 x = − soi M1 120, 240 A1 Correct pair 12(i) ( ) d 3 cos5 1 3 5sin5 d t t v t t − +   = = +     B2 B1 for either with no other terms or for both with 1 extra ( ) 3 5sin5 0 their t + = M1 Must be from an attempt to differentiate awrt 0.76 A1 0.7570187525 awrt 1.13 A1 1.12793684 substitutes their t values into s (4.07…, 3.58…) DM1 must be two values 0.48 to 0.49 [m] A1 Final A1 may imply earlier A1s 12(ii) 25cos 5t M1 Differentiating their v correctly providing at least 2 terms with one trig function −25 A1 Ignore +25 following −25