Cambridge IGCSE Mathematics - Additional 0606 — 2017 Oct/Nov Paper 1 · Variant 2

* 4 5 4 1 9 1 2 0 0 1 * This document consists of 16 printed pages. DC (KN/SG) 135362/4 © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 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 0606/12/O/N/17 © UCLES 2017 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/17 © UCLES 2017 [Turn over 1 (i) On the Venn diagram below, draw sets X and Y such that X Y 0 n + = ^ h . [1] (ii) On the Venn diagram below, draw sets A, B and C such that C A B , 1 l ^ h . [2]

4 0606/12/O/N/17 © UCLES 2017 2 The graph of ( ) sin y a bx c = + has an amplitude of 4, a period of r 3 and passes through the point r , 12 2 J L KK N P OO. Find the value of each of the constants a, b and c. [4]

5 0606/12/O/N/17 © UCLES 2017 [Turn over 3 (i) Find, in ascending powers of x, the first 3 terms in the expansion of x 2 4 2 5 - J L KK N P OO . [3] (ii) Hence find the term independent of x in the expansion of x x x 2 4 1 3 2 5 2 2 - - J L KK J L KK N P OO N P OO . [3]

6 0606/12/O/N/17 © UCLES 2017 4 Given that ln y x x 1 3 2 2 2 = + + ^ h , find the value of x y d d when x = 2, giving your answer as ln a b 14 + , where a and b are fractions in their simplest form. [6]

7 0606/12/O/N/17 © UCLES 2017 [Turn over 5 When lgy is plotted against x, a straight line is obtained which passes through the points (0.6, 0.3) and (1.1, 0.2). (i) Find lgy in terms of x. [4] (ii) Find y in terms of x, giving your answer in the form y A 10bx = ^ h, where A and b are constants. [3]

8 0606/12/O/N/17 © UCLES 2017 6 Functions f and g are defined, for x 0 2 , by ( ) ln x x f = , ( )x x 2 3 g 2 = + . (i) Write down the range of f. [1] (ii) Write down the range of g. [1] (iii) Find the exact value of ( ) 4 f g 1 - . [2] (iv) Find x g 1 - ^ h and state its domain. [3]

9 0606/12/O/N/17 © UCLES 2017 [Turn over 7 A polynomial x p^ h is ax x bx 8 5 3 2 + + + , where a and b are integers. It is given that x 2 1 - is a factor of x p^ h and that a remainder of –25 is obtained when x p^ h is divided by x + 2. (i) Find the value of a and of b. [5] (ii) Using your values of a and b, find the exact solutions of x 5 p = ^ h . [2]

10 0606/12/O/N/17 © UCLES 2017 8 A B 4 ms–1 120 m 50 m The diagram shows a river which is 120 m wide and is flowing at 4 ms–1. Points A and B are on opposite sides of the river such that B is 50 m downstream from A. A man needs to cross the river from A to B in a boat which can travel at 5 ms–1 in still water. (i) Show that the man must point his boat upstream at an angle of approximately 65° to the bank. [4]

11 0606/12/O/N/17 © UCLES 2017 [Turn over (ii) Find the time the man takes to cross the river from A to B. [6]

12 0606/12/O/N/17 © UCLES 2017 9 (a) 60 40 20 x m 0 10 20 30 t s The diagram shows the displacement-time graph of a particle P which moves in a straight line such that, t s after leaving a fixed point O, its displacement from O is x m. On the axes below, draw the velocity-time graph of P. 6 4 2 0 10 20 30 t s velocity ms–1 [3]

13 0606/12/O/N/17 © UCLES 2017 [Turn over (b) A particle Q moves in a straight line such that its velocity, v ms–1, t s after passing through a fixed point O, is given by e v t 3 2 3 t5 = + - , for t H 0. (i) Find the velocity of Q when t = 0. [1] (ii) Find the value of t when the acceleration of Q is zero. [3] (iii) Find the distance of Q from O when t = 0.5. [4]

14 0606/12/O/N/17 © UCLES 2017 10 6 . 2 c m 5 cm 5 cm A C D B E F The diagram shows an isosceles triangle ABC, where AB = AC = 5 cm. The arc BEC is part of the circle centre A and has length 6.2 cm. The point D is the midpoint of the line BC. The arc BFC is a semi-circle centre D. (i) Show that angle BAC is 1.24 radians. [1] (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [4]

15 0606/12/O/N/17 © UCLES 2017 [Turn over 11 (a) Solve cot 2 35 5 z + = c ^ h for 0 360 c G G z c. [4] Question 11(b) is printed on the next page.

16 0606/12/O/N/17 © UCLES 2017 16 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. (b) (i) Show that cot tan sec sin i i i i + = . [3] (ii) Hence solve cot tan sec 3 3 3 2 3 i i i + =- for r r 2 2 G G i - , giving your answers in terms of r. [4]

® IGCSE is a registered trademark. This document consists of 10 printed pages. © UCLES 2017 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 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 International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2017 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 2017 © UCLES 2017 Page 2 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 2017 © UCLES 2017 Page 3 of 10 Question Answer Marks Guidance 1(i) 1 1(ii) Either 2 B1 for C with no intersection with either A or B (allow if C is not represented by a circle) B1 for all correct, C must be represented by a circle Or 2 4 = a B1 6 = b B1 2 = − c M1, A1 M1 for use of π , 2 12       to obtain c, using their values of a and of b 3(i) 2 4 32 20 5 − + x x B3 B1 for each correct term 3(ii) ( ) 2 4 2 4 1 9 32 20 5 …   − + +     x x x x B1 2 1 x and 4 9 x Independent of x: 20 45 − + M1 attempt to deal with 2 terms independent of x, must be looking at terms in 2 x and 2 1 x and terms in 4 x and 4 1 x = 25 A1 FT their answers from (i) (their 20 1) − × + (their5 9 × ) X Y C A B C A B

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 4 of 10 Question Answer Marks Guidance 4 correct differentiation of ( ) 2 ln 3 2 + x B1 attempt to differentiate a quotient or a product M1 ( ) ( ) ( ) 2 2 2 2 2 6 1 2 ln 3 2 d 3 2 d 1   + − +   +   = + x x x x y x x x A1 all other terms correct. When 12 5 4ln14 d 14 2, d 25  −     = = y x x M1 M1dep for substitution and attempt to simplify 6 4 ln14 35 25 = − A2 A1 for each correct term, must be in simplest form 5(i) Either Gradient 0.2 = − B1 lg 0.2 = − + y x c B1 lg = + y mx c soi correct attempt to find c M1 must have previous B1 lg 0.42 0.2 = − y x or 21 lg 50 5 = −x y A1 line in either form, allow equivalent fractions Or 0.3 0.6 = + m c B1 0.2 1.1 = + m c B1 attempt to solve for both m and c M1 must have at least one of the previous B marks Leading to lg 0.42 0.2 = − y x or 21 lg 50 5 = −x y A1 line in either form, allow equivalent fractions

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 5 of 10 Question Answer Marks Guidance 5(ii) Either ( ) 0.42 0.2 10 − = x y M1 dealing with the index, using their answer to (i) ( ) 0.42 0.2 10 10− = x y ( ) 0.2 2.63 10− = x y A2 A1 for each Or ( ) 10 = bx y A leads to lg lg = + y A bx Compare this form with their equation from (i) M1 comparing their answer to (i) with lg lg = + y A bx may be implied by one correct term from correct work lg 0.42 = A so 2.63 = A A1 0.2 = − b A1 A1 for each 6(i) y∈\ oe B1 Must have correct notation i.e. no use of x 6(ii) 3 > y oe B1 Must have correct notation i.e. no use of x 6(iii) ( ) 1 f e − = x x or ( ) g 4 35 = B1 First B1 may be implied by correct answer or by use of 35 ( ) 1 35 f g 4 e − = B1 6(iv) 2 3 2 − = y x or 2 3 2 − = x y M1 valid attempt to obtain the inverse ( ) 1 3 g 2 − − = x x A1 correct form, must be ( ) 1 g− = x or = y Domain 3 > x B1 Must have correct notation 7(i) 1 p : 2 5 0 2 8 2 a b   + + + =     M1 substitution of 1 2 = x and equating to zero (allow unsimplified) ( ) p 2 : 8 32 2 5 25 a b − − + − + = − M1 substitution of 2 = − x and equating to –25 (allow unsimplified) leading to 4 56 0 + + = a b 4 31 0 + − = a b oe M1 M1dep for solution of simultaneous equations to obtain a and b 12, 17 = = − a b A2 A1 for each

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 6 of 10 Question Answer Marks Guidance 7(ii) 3 2 12 8 17 0 + − = x x x 0 = x B1 for 0 = x 1 55 3 6 = − ± x oe B1 8 8(i) o 67.4 ∠ = ABC B1 o 4 5 sin sin67.4 = BAC M1 attempt at the sine rule, using 4 and 5 (or e.g. use of cosine rule followed by sine rule on triangle shown) o 47.6 ∠ = BAC A1 may be implied by later work Angle required = o o o 180 47.6 67.4 − − o 65 = A1 Answer Given 8(ii) ( ) 2 2 2 o 5 4 2 5 4 cos65 = + − × × × V M1 attempt at the cosine rule or sine rule to obtain V – allow if seen in (i) 4.91 = V or 4 sin sin65 = V BAC A1 Distance to travel: o 120 sin67.4 M1 distance to travel – allow if seen in (i) 130 or 2 2 120 50 + A1 Time taken: 130 4.91 M1 M1dep for correct method to find the time, must have both of the previous M marks 26.5 A1 A 5 V 4 C B

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 7 of 10 Question Answer Marks Guidance Alternative method 120 cos25 = AC oe M1 correct attempt at AC 132.4 = A1 Allow 132 Speed for this distance = 5 M1A1 M1dep A1 for speed, it must be 5 exactly for A1, must have first M mark Time taken =132.4 5 M1 M1dep for a correct method to find the time, must have both of the previous M marks = 26.5 A1 9(a) B3 B1 for line joining ( ) 0,5 and ( ) 10,5 B1 for a line joining ( ) 10,0.5 and ( ) 30,0.5 B1 all correct with no solid line joining ( ) 10,5 to ( ) 10,0.5 9(b)(i) 3 B1 9(b)(ii) 5 d 3 15e d 2 − = − + t v t M1 attempt to differentiate, must be in the form 5 e− + t a b When 5 d 0, e 0.1 d − = = t v t M1 M1dep for equating to zero and attempt to solve, must be of the form 5 e , 0 − = > t a b b to obtain an equation in the form 5 − = t k where k is a logarithm or < 0 0.461 = t A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 8 of 10 Question Answer Marks Guidance 9(b)(iii) Either attempt to integrate, must be in the form 5 2 e− + t c dt M1 ( ) 5 2 3 3 e 5 4 − = − + + t s t c A1 When 0, 0 = = t s so 3 5 = c M1 M1dep for attempt to find c and substitute 0.5 = t 0.738 s = A1 Or attempt to integrate, must be in the form 5 2 e− + t c dt M1 0.5 5 2 0 3 3 e 5 4 −   − +     t t A1 correct use of limits M1 M1dep leading to 0.738 = s A1 10(i) 5 6.2 ∠ = BAC , 1.24 ∠ = BAC B1 10(ii) sin0.62 5 = BD , 2.905 = BD , 2.91 B1 valid method to find BD Arc BFC: ( ) π 9.13 × = BD M1 attempt to find arc length BFC, using their BD Perimeter: 9.13 6.2 15.3 + = A1 10(iii) Area: 2 2 2 1 π 2.91 2 1 1 5 1.24 5 sin1.24 2 2   × × −           × × − × ×             B3 B1 for area of semi circle (= 13.3) B1 for area of sector (= 15.5) B1 for area of triangle (= 11.8) 9.58⩽ Area ⩽ 9.62 B1 final answer

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 9 of 10 Question Answer Marks Guidance 11(a) ( ) o 2 tan 35 5 φ + = M1 dealing correctly with cot and an attempt at solution of ( ) tan 35 φ + = c , order must be correct, to obtain a value for 35 φ + o o o o 35 21.8 , 201.8 , 381.8 φ + = M1 M1dep for an attempt at a second solution in the range, ( o 180 + their first solution in the range oe) o o 166.8 , 346.8 φ = A2 A1 for each 11(b)(i) Either 1 cos cos sin sin cos θ θ θ θ θ + M1 expressing all terms in terms of sinθ and cosθ where necessary 2 2 1 sin cos cos cos sin θ θ θ θ θ   =   +   M1 dealing with the fractions correctly to get 2 2 sin cos sin cos θ θ θ θ + in denominator or as in left hand column ( ) sin 1 θ = A1 use of identity, together with a complete and correct solution, withhold A1 for incorrect use of brackets Or sec 1 tan tan θ θ θ + = 2 sec 1 tan tan θ θ θ + M1 dealing with fractions in the denominator correctly to get 2 1 tan tan θ θ + in the denominator, allow tanθ taken to the numerator = 2 sec tan sec θ θ θ M1 use of the identity to get 2 sec θ = tan sec θ θ = sin cos cos θ θ θ × =sinθ A1 expressing all terms in terms of sinθ and cosθ and simplification to the given answer, withhold A1 for incorrect use of brackets

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 10 of 10 Question Answer Marks Guidance 11(b)(ii) 3 sin3 2 θ = − M1 correct attempt to solve for θ , order must be correct, may be implied by one correct solution 2π π 4π 3 , , 3 3 3 θ = − − 2π π 4π , , 9 9 9 θ = − − A3 A1 for each