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

This document consists of 16 printed pages. DC (NH/FC) 135359/3 © UCLES 2017 [Turn over * 1 4 0 4 3 0 1 9 6 4 * ADDITIONAL MATHEMATICS 0606/11 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. Cambridge International Examinations Cambridge International General Certificate of Secondary Education

2 0606/11/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 0606/11/O/N/17 © UCLES 2017 [Turn over 1 Express in set notation the shaded regions shown in the Venn diagrams below. (i) A B … [1] (ii) A B C … [1] (iii) A B … [1]

4 0606/11/O/N/17 © UCLES 2017 2 The polynomial p(x) is ax bx x 13 4 3 2 + - + , where a and b are integers. Given that x 2 1 - is a factor of ( )x p and also a factor of ( )x pl , (i) find the value of a and of b. [5] Using your values of a and b, (ii) find the remainder when ( )x p is divided by x 1 + . [2]

5 0606/11/O/N/17 © UCLES 2017 [Turn over 3 (a) Given that T l g 2 2 1 2 1 r = - , express l in terms of T, g and r. [2] (b) By using the substitution y x 3 1 = , or otherwise, solve x x 4 3 0 3 2 3 1 - + = . [4]

6 0606/11/O/N/17 © UCLES 2017 4 When lgy is plotted against x2 a straight line is obtained which passes through the points (4, 3) and (12, 7). (i) Find the gradient of the line. [1] (ii) Use your answer to part (i) to express lgy in terms of x. [2] (iii) Hence express y in terms of x, giving your answer in the form y A 10bx2 = ^ h where A and b are constants. [3]

7 0606/11/O/N/17 © UCLES 2017 [Turn over 5 y y = 4e2x + 16e–2x x 1 O A B The diagram shows part of the graph of y 4 16 e e x x 2 2 = + - meeting the y-axis at the point A and the line x = 1 at the point B. (i) Find the coordinates of A. [1] (ii) Find the y-coordinate of B. [1] (iii) Find x 4 16 e e d x x 2 2 + - ^ h y . [2] (iv) Hence find the area of the shaded region enclosed by the curve and the line AB. You must show all your working. [4]

8 0606/11/O/N/17 © UCLES 2017 6 (a) Functions f and g are such that, for x R d , x x 3 f 2 = + ^ h , x x 4 1 g = - ^ h . (i) State the range of f. [1] (ii) Solve ( )x 4 fg = . [3]

9 0606/11/O/N/17 © UCLES 2017 [Turn over (b) A function h is such that x x x 4 2 1 h = - + ^ h for x R d , x 4 ! . (i) Find x h 1 - ^ h and state its range. [4] (ii) Find x h2^ h, giving your answer in its simplest form. [3]

10 0606/11/O/N/17 © UCLES 2017 7 (i) Write ln x x 2 1 2 1 - + J L KK N P OO as the difference of two logarithms. [1] A curve has equation ln y x x x 2 1 2 1 4 = - + + J L KK N P OO for x 2 1 2 . (ii) Using your answer to part (i) show that x y x x b a 4 1 d d 2 2 = - + , where a and b are integers. [4]

11 0606/11/O/N/17 © UCLES 2017 [Turn over (iii) Hence find the x-coordinate of the stationary point on the curve. [2] (iv) Determine the nature of this stationary point. [2]

12 0606/11/O/N/17 © UCLES 2017 8 (a) 10 people are to be chosen, to receive concert tickets, from a group of 8 men and 6 women. (i) Find the number of different ways the 10 people can be chosen if 6 of them are men and 4 of them are women. [2] The group of 8 men and 6 women contains a man and his wife. (ii) Find the number of different ways the 10 people can be chosen if both the man and his wife are chosen or neither of them is chosen. [3]

13 0606/11/O/N/17 © UCLES 2017 [Turn over (b) Freddie has forgotten the 6-digit code that he uses to lock his briefcase. He knows that he did not repeat any digit and that he did not start his code with a zero. (i) Find the number of different 6-digit numbers he could have chosen. [1] Freddie also remembers that his 6-digit code is divisible by 5. (ii) Find the number of different 6-digit numbers he could have chosen. [3] Freddie decides to choose a new 6-digit code for his briefcase once he has opened it. He plans to have the 6-digit number divisible by 2 and greater than 600 000, again with no repetitions of digits. (iii) Find the number of different 6-digit numbers he can choose. [3]

14 0606/11/O/N/17 © UCLES 2017 9 P A Q B 10 cm 24 cm The diagram shows a circle, centre A, radius 10 cm, intersecting a circle, centre B, radius 24 cm. The two circles intersect at the points P and Q. The radii AP and AQ are tangents to the circle with centre B. The radii BP and BQ are tangents to the circle with centre A. (i) Show that angle PAQ is 2.35 radians, correct to 3 significant figures. [2] (ii) Find angle PBQ in radians. [1] (iii) Find the perimeter of the shaded region. [3]

15 0606/11/O/N/17 © UCLES 2017 [Turn over (iv) Find the area of the shaded region. [4] Question 10 is printed on the next page.

16 0606/11/O/N/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. 10 (a) Solve cosec sin x x 3 2 4 2 0 - = for ° x 0 180° G G . [4] (b) Solve tan y 3 4 3 r - = J L KK N P OO for y 0 2 G G r radians, giving your answers in terms of r. [4]

® IGCSE is a registered trademark. This document consists of 7 printed pages. © UCLES 2017 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/11 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 3 of 7 Question Answer Marks Guidance 1(i) ′∩ A B B1 1(ii) ∩ ∩ A B C B1 1(iii) ∪ A B B1 2(i) 1 13 p 4 2 8 4 2  = + − +     a b M1 attempt at 1 2       p ( ) 2 p 3 2 13 ′ = + − x ax bx 1 3 p 13 2 4   ′ = + −     a b M1 attempt at 1 2   ′    p leading to 2 20 + = a b and 3 4 52 0 + − = a b A1 at least one correct equation solution of simultaneous equations DM1 12, 4 = = a b A1 for both 2(ii) ( ) 1 12 4 13 4 − = − + + + p M1 9 A1 FT on their integer values of a and b 3(a) 1 1 2 2 2π = Tg l 2 2 4π = T g l B1 multiplication/dealing with power of 1 2 or squaring 2 2 4π = T g l or 2 1 2 2π           Tg B1 for either 3(b) 2 4 3 0 − + = y y leading to 1, 3 = = y y M1 reduction to quadratic equation and attempt to solve 1 3 1, x = 1 3 3 = x DM1 attempt to solve 1 3 = x k (positive k) 1, 27 = = x x A2 A1 for each

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 4 of 7 Question Answer Marks Guidance 4(i) 1 2 B1 4(ii) 2 lg = + y mx c 2 1 lg 1 2 = + y x B2 –1 for each error 4(iii) 2 1 2 10   +       = x y B1 dealing with lg on their (ii) 2 2 10 10     =     x y B2 B1 for each, dependent on first B1 5(i) (0, 20) B1 5(ii) 31.7 B1 5(iii) ( ) 2 2 2e 8e − − + x x c B2 B1 for each correct term 5(iv) Area of trapezium = ( ) 1 20 31.7 2 + 25.86 = or 25.85 B1 ( ) ( ) 1 2 2 2 2 0 2e 8e 2e 8e 6 − −   − = − −−   x x M1 substitution of both limits, must have come from integration of the form 2 2 e e− + x x a b . 19.7 A1 Required area = 6.15, 6.16, 6.17 A1 6(a)(i) f 3 . B1 must be using a correct notation 6(a)(ii) ( ) 2 4 1 3 4 − + = x M1 correct order solution of resulting quadratic equation DM1 1 0, 2 = = x x A1 both required

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 5 of 7 Question Answer Marks Guidance 6(b)(i) 4 2 1 − = + xy y x M1 ‘multiplying out’ ( ) 2 4 1 − = + x y y 4 1 2 + = − y x y M1 collecting together like terms ( ) 1 4 1 h 2 − + = − x x x A1 correct answer with correct notation Range 1 h 4 −≠ B1 must be using a correct notation 6(b)(ii) ( ) 2 2 1 h h 4 +   =   −   x x x = 2 1 2 1 4 2 1 4 4 +  +   −   +  −   −   x x x x M1 dealing with h2 correctly dealing with fractions within fractions M1 5 2 17 2 − = − x x oe A1 7(i) ( ) ( ) ln 2 1 ln 2 1 + − − x x B1 7(ii) attempt to differentiate M1 d 2 2 4 d 2 1 2 1 = − + + − y x x x A1 all correct attempt to obtain in required form DM1 2 2 16 8 4 1 − = − x x A1 A1 all correct 7(iii) When d 0 d = y x , 2 16 8 0 −= x M1 setting d 0 d = y x and attempt to solve 1 2 = x only A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 6 of 7 Question Answer Marks Guidance 7(iv) ( ) ( ) ( ) 2 2 2 2 2 2 32 4 1 8 16 8 d d 4 1 − − − = − x x x x y x x M1 attempt at second derivative and conclusion or equivalent method When 1 2 = x 2 2 d d y x is + ve, so minimum A1 8(a)(i) 8 6 6 4 C C × B1 either 8 6 C or 6 4 C 420 B1 8(a)(ii) 12 12 8 10 + C C B2 B1 for each = 561 B1 Alternate scheme: 1001 ( ) 12 9 2 − × C B1 B1 = 561 B1 8(b)(i) 136 080 B1 8(b)(ii) No of ways ending with 0 - 15 120 B1 No of ways ending with 5 - 13 440 B1 Total 28 560 B1 8(b)(iii) Starting with 6 or 8 - 13 440 B1 Starting with 7 or 9 - 16 800 B1 Total = 30 240 B1 9(i) tan 2.4 2  =     PAQ M1 valid method 2.352(01…) = PAQ 2.35 = PAQ correct to 3 sf A1 must see greater than 3 sf then rounding 9(ii) 0.790 = PBQ or 0.792 B1 9(iii) ( ) ( ) 2.352 10 0.790 24 × + × M1,A1 M1 for correct attempt at an arc length A1 for one correct arc length = awrt 42.5 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 7 of 7 Question Answer Marks Guidance 9(iv) 2 2 1 1 24 0.790 24 sin0.790 2 2       × × − × ×             B1,B1 B1 for a correct sector area allow, unsimplified B1 for a correct area of a triangle, allow unsimplified 2 2 1 1 10 2.352 10 sin 2.352 2 2       + × × − × ×             B1 correct plan, dependent on both previous B marks 22.94 82.1 = + = 105 B1 10(a) 2 3 sin 2 4 = x B1 dealing correctly with cosec 3 sin2 2 = ± x 2 60, 120, 240, 300 = x M1 correct method of solution including dealing with 2x correctly, may be implied by one correct solution. 30, 60, 120, 150 = x A2 A1 for each correct pair 10(b) π 1 tan 4 3   − =     y M1 dealing with order of operations to obtain a first solution π π 7π , 4 6 6 − = y M1 M1 for attempt to obtain a second solution 5π 17π , 12 12 = y A2 A1 for each