Cambridge IGCSE Mathematics - Additional 0606 — 2017 Oct/Nov Paper 1 · Variant 3

* 6 5 8 1 4 2 3 2 1 8 * This document consists of 15 printed pages and 1 blank page. DC (LK/FC) 135366/3 © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 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 0606/13/O/N/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 0606/13/O/N/17 © UCLES 2017 [Turn over 1 Given that sec y 2 2i = and tan x 5 i = - , express y in terms of x. [2] 2 A curve is such that its gradient at the point (x, y) is given by e 10 3 x 5 + . Given that the curve passes though the point (0, 9), find the equation of the curve. [4]

4 0606/13/O/N/17 © UCLES 2017 3 Find the set of values of k for which the equation kx x k 3 4 0 2 + - + = has no real roots. [4] 4 The graph of ( ) cos y a bx c = + has an amplitude of 3, a period of 4 r and passes through the point , 12 2 5 r c m. Find the value of each of the constants a, b and c. [4]

5 0606/13/O/N/17 © UCLES 2017 [Turn over 5 (i) Find ( ) d x x 7 10 5 3 - - y . [2] (ii) Given that (7 10) d x x 14 25 a 6 5 3 - = - y , find the exact value of a. [3]

6 0606/13/O/N/17 © UCLES 2017 6 When ln y is plotted against x2 a straight line is obtained which passes through the points (0.2, 2.4) and (0.8, 0.9). (i) Express ln y in the form px q 2 + , where p and q are constants. [3] (ii) Hence express y in terms of z, where e z x2 = . [3]

7 0606/13/O/N/17 © UCLES 2017 [Turn over 7 (i) Find, in ascending powers of x, the first 3 terms in the expansion of x 2 4 2 6 - c m . Give each term in its simplest form. [3] (ii) Hence find the coefficient of x2 in the expansion of x x x 2 4 1 2 6 2 - + c c m m . [4]

8 0606/13/O/N/17 © UCLES 2017 8 It is given that ( )( ) y x x 4 3 1 3 5 = - - . (i) Show that ( ) ( ) d d x y x Ax B 3 1 3 2 = - + , where A and B are integers to be found. [5] (ii) Hence find, in terms of h, where h is small, the approximate change in y when x increases from 3 to 3 + h. [3]

9 0606/13/O/N/17 © UCLES 2017 [Turn over 9 (a) A 6-digit number is to be formed using the digits 1, 3, 5, 6, 8, 9. Each of these digits may be used only once in any 6-digit number. Find how many different 6-digit numbers can be formed if (i) there are no restrictions, [1] (ii) the number formed is even, [1] (iii) the number formed is even and greater than 300 000. [3] (b) Ruby wants to have a party for her friends. She can only invite 8 of her 15 friends. (i) Find the number of different ways she can choose her friends for the party if there are no restrictions. [1] Two of her 15 friends are twins who cannot be separated. (ii) Find the number of different ways she can now choose her friends for the party. [3]

10 0606/13/O/N/17 © UCLES 2017 10 (a) Given that , and A B AB a b 4 1 2 5 3 4 13 18 8 4 = - = - = c c c m m m, find the value of a and of b. [4]

11 0606/13/O/N/17 © UCLES 2017 [Turn over (b) It is given that , X Y 3 4 5 1 1 4 2 0 = - - = - c c m m and XZ = Y. (i) Find X 1 - . [2] (ii) Hence find Z. [3]

12 0606/13/O/N/17 © UCLES 2017 11 A B D C O 14.8 cm 10 cm The diagram shows a circle, centre O, radius 10 cm. The points A, B, C and D lie on the circumference of the circle such that AB is parallel to DC. The length of the minor arc AB is 14.8 cm. The area of the minor sector ODC is 21.8 cm2. (i) Write down, in radians, angle AOB. [1] (ii) Find, in radians, angle DOC. [2]

13 0606/13/O/N/17 © UCLES 2017 [Turn over (iii) Find the perimeter of the shaded region. [4] (iv) Find the area of the shaded region. [3]

14 0606/13/O/N/17 © UCLES 2017 12 The line y x 2 1 = + intersects the curve xy y 14 2 = - at the points P and Q. The midpoint of the line PQ is the point M. (i) Show that the point , 10 8 23 - c m lies on the perpendicular bisector of PQ. [9]

15 0606/13/O/N/17 © UCLES 2017 The line PQ intersects the y-axis at the point R. The perpendicular bisector of PQ intersects the y-axis at the point S. (ii) Find the area of the triangle RSM. [3]

16 0606/13/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. BLANK PAGE

® 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 0606/13 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.

0606/13 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 3 of 10 Question Answer Marks Guidance 1 Using 2 2 tan 1 sec θ θ + = to obtain 2 2(tan 1) θ = + y or ( ) 2 2 5 sec 1 θ + = − x ( ) 2 5 1 2 + + = y x M1 use of correct identity 2 2(( 5) 1) = + + y x oe A1 2 5 d 10e 3 d = + x y x an attempt at integration in form ae5x+ bx M1 ( ) 5 10 e 3 5 = + + x y x c A1 condone omission of c attempt to find c using 0, 9 = = x y M1 M1dep 5 2e 3 7 = + + x y x A1 3 ( ) 9 4 4 < − k k 2 4 16 9 − − k k M1 use of the discriminant with correct values ( )( ) 2 9 2 1 − + k k M1 M1dep for solution of their quadratic to obtain critical values Critical values 9 1 , 2 2 − A1 1 9 , 2 2 < − > k k A1 4 a = 3 B1 b = 8 B1 5 3cos 8 2 12 c π   = × +     M1 substitution of 12 x π = and 5 2 = y to find c c = 4 A1 5(i) ( ) 2 5 5 7 10 14 − x B2 B1 for ( ) 2 5 7 10 − k x

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 4 of 10 Question Answer Marks Guidance 5(ii) ( ) 2 5 6 5 25 7 10 14 14   − =     a x ( ) ( ) 2 2 5 5 5 5 25 7 10 7 6 10 14 14 14 − − × − = a ( ) 2 5 7 10 4 5 − − = a M1 correct application of limits for ( ) 2 5 7 10 − k x 5 2 9 10 7 + = a M1 M1dep for evaluation of ( ) 2 5 7 6 10 × − and correct order of operations to find a, including dealing with power. 253 7 = a or 1 7 36 A1 6(i) Gradient = 2.4 0.9 0.2 0.8 − − (= –2.5) B1 2 5 ln 2 = − + y x c M1 straight line form and correct substitutions to find c 2 5 ln 2.9 2 = − + y x oe A1 Alternative method 2.4 (0.2) = + p q 0.9 (0.8) = + p q B1 Correct method of solution to find p and q from two correct equations M1 M1dep 2 5 ln 2.9 2 = − + y x A1 6(ii) 2 5 2.9 2 e   − +     = x y M1 dealing with ln 2 5 2.9 2 e e − = × x y M1 M1dep for dealing with the index 5 2 18.2 − = y z A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 5 of 10 Question Answer Marks Guidance 7(i) 2 4 64 48 15 − + x x B3 B1 for each correct term in final line of response 7(ii) ( ) 2 4 2 2 1 64 48 15 2 x x x x   − + + +     B1 B1for 2 2 1 2 + + x x oe at least two correctly obtained products leading to terms in x2 M1 Term in 2 x : 64 15 96 + − A1 FT for correct evaluation of their 64 + (2 × their–48) + their 15 17 = − A1 8(i) attempt to differentiate a product M1 ( ) ( ) ( ) 2 5 3 3 d 5 4 3 3 1 3 1 d 3   = − × × − + −     y x x x x A2 A1 for ( ) ( ) 2 3 5 ( ) 4 3 3 1 3   + − × × −     x x A1 for ( ) 5 3 ( ) 3 1 + − x ( ) ( ) ( ) ( ) 2 3 3 1 5 20 3 1 = − − + − x x x M1 use of ( ) ( ) ( ) 5 2 3 3 3 1 3 1 3 1 − = − − x x x ( ) ( ) 2 3 3 1 8 21 = − − x x A1 8(ii) When 3 = x , 2 3 d 8 3 d = × y x M1 ( ) 2 3 3 3 1 × − × k or ( ) 2 3 9 1 − × k or 4×k (where k is any number) 2 3 8 3 ∂= × × y h M1 M1dep for their ( ) ( ) 2 3 9 1 − × × k h 12 ∂= y h A1 9(a)(i) 720 B1 9(a)(ii) 240 B1 9(a)(iii) 4! 2 × × k or 240 4! 2 −× × k or correct equivalents with no extra terms added or subtracted B1 4 4! × × p or correct equivalents with no extra terms added or subtracted B1 192 B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 6 of 10 Question Answer Marks Guidance 9(b)(i) 6435 B1 9(b)(ii) With twins: 13 6 C or 1716 Without twins: 13 8 C or 1287 B2 B1 for 13 6 C or 1716 or 13 8 C or 1287 B1 for ( 13 6 C and 13 8 C ) or (1716 and 1287) with no multiples and no extra terms Total: 1716 + 1287 = 3003 B1 3003 from a correct method 10(a) matrix multiplication, must have at least 2 correct elements M1 AB = 13 8 2 5 3 4     − +   a b a b A1 2 5 18 − = a b 3 4 4 + = a b M1 formation and solution of simultaneous equations leading to 4, 2 = = − a b A1 Alternate scheme AB = 13 8 18 4       ABB–1 = 13 8 18 4       B–1 M1 Correct plan Correct inverse B1 A 4 1 13 8 4 3 1 18 4 5 2 23 − −      = =           a b M1 Correct order and method of multiplication with at least two correct elements leading to 4, 2 = = − a b A1 10(b)(i) 1 5 1 4 3 17   −     oe B2 B1 for 1 17 − B1 for 1 5 4 3       10(b)(ii) Z = 1 5 1 2 1 4 3 4 0 17 −    −       M1 pre-multiplication with two elements correct 19 2 1 8 8 17   = −     oe A2 A1 for four correct of 1 17 − , 19, 2, 8, 8

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 7 of 10 Question Answer Marks Guidance 11(i) 1.48 B1 11(ii) 2 1 10 21.8 2 θ × × = M1 correct use of sector area 0.436 θ = A1 11(iii) ( ) 2 1.48 0.436 2.18(4) 2 BOC π − − ∠ = = B1 2.18(4) or unsimplified 1 20sin 2   = ∠     BC BOC or 10 sin sin 2 π × = −       BOC BC BOC or (200 200cos = − BC BOC ( ) 17.7 5 = BC M2 M1 for a complete correct method to find BC using their angle BOC M1 for a correct plan using 14.8, their BC and 10 × their answer to (ii) Perimeter = ( ) 14.8 2 17.7(5) 4.36 + × + 54.7 = or 54.6 A1 awrt 54.7 or awrt 54.6

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 8 of 10 Question Answer Marks Guidance 11(iv) Area = 2 2 1 1 10 1.48 21.8 2 10 sin2.18(4) 2 2     × × + + ×         B2 B1 for 2 1 10 1.48 21.8 2   × × +     B1 for 2 1 2 10 sin 2.18(4) 2   ×     = 178 B1 awrt 178 from correct working Alternative method 1 Segment area = ( ) ( ) 2 1 10 2.18 sin 2.18 2 − B1 B1 for ( ) ( ) 2 1 2 10 2.18(4) sin 2.18(4 2 × − Area required = ( ) ( ) 2 1 100 2 10 2.18(4) sin 2.18(4) 2 π − × − B1 = 178 B1 awrt 178 from correct working Alternative method 2 Area of trapezium = ( )( ) ( ) 1 13.5 4.33 17.1 2 + B1 correct area of trapezium ABCD (allow unsimplified) Area of segments = ( ) ( ) 2 1 10 1.48 sin1.48 2 − + ( ) ( ) 2 1 10 0.436 sin0.436 2 − B1 correct area of both segments (allow unsimplified) = 178 B1 awrt 178 from correct working

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 9 of 10 Question Answer Marks Guidance 12(i) 2 2 5 12 0 + − = x x or 2 3 28 0 + − = y y M1 attempt to get in terms of one variable ( )( ) 2 3 4 0 − + = x x or ( )( ) 7 4 0 + − = y y M1 M1dep for solution of a three term quadratic leading to 4, 7 = − = − x y and 3 , 4 2 = = x y A2 A1 for each ‘pair’ Midpoint M 3 2 4 4 ( 7) , 2 2 −   + −     5 3 , 4 2     = − −         A1 correctly obtained midpoint Gradient of PQ = 2 B1 may be implied Perp gradient = 1 2 − M1 gradient 1 of their PQ − Perp bisector: 3 1 5 2 2 4   + = − +     y x M1 M1dep for equation of perp bisector using their perp gradient and their midpoint. (unsimplified) ( ) 1 17 23 10 2 8 8 = − − − = y or 23 1 17 10 8 2 8 = − − → = − x x A1 all correct so far and for verification using a correct equation

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 10 of 10 Question Answer Marks Guidance 12(ii) Area = 1 17 5 1 2 8 4   × + ×     M1 finding R, S and RS correct method for finding area M1 M1dep = 125 64 or 1.95 or 61 64 1 A1 Alternative method 1 Area = 1 125 125 2 4 8 × × M1 finding R, S, RM and MS correct method for finding area M1 M1dep = 125 64 or 1.95 or 61 64 1 A1 Alternative method 2 Area = 5 4 17 3 8 2 0 0 0 1 1 1 2 − − − M1 finding R and S to obtain their 5 4 17 3 8 2 0 0 0 1 1 1 2 − − − = 1 5 85 2 4 32 − − oe M1 M1dep for correct method of evaluation = 125 64 or 1.95 or 61 64 1 A1