Cambridge IGCSE Mathematics - Additional 0606 — 2016 Oct/Nov Paper 1 · Variant 2

This document consists of 16 printed pages. DC (ST/SG) 126424 © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 3 1 0 9 5 9 4 6 7 9 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 2016 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/16 © UCLES 2016 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/16 © UCLES 2016 [Turn over 1 (a) Sets , A and B are such that n() , ( ) , ( ) A B A B 26 7 3 n n + + = = = l and ( ) B 15 n = . Using a Venn diagram, or otherwise, find (i) ( ) A n , [1] (ii) ( ) A B n , , [1] (iii) ( ) A B n , l. [1] (b) It is given that  { : } x x 0 30 1 1 = , P = {multiples of 5}, Q = {multiples of 6} and R = {multiples of 2}. Use set notation to complete the following statements. (i) Q … R, [1] (ii) P Q + = … [1]

4 0606/12/O/N/16 © UCLES 2016 2 Given that ( ) p qr p q r p q r a b c 5 3 2 3 1 2 1 2 3 = - - , find the value of each of the integers a, b and c. [3] 3 By using the substitution log y x 3 = , or otherwise, find the values of x for which ( ) log log log x x 3 9 0 3 2 3 5 3 + - = . [6]

5 0606/12/O/N/16 © UCLES 2016 [Turn over 4 (i) Find the first 3 terms in the expansion of x x 2 3 1 2 5 - J L KK N P OO , in descending powers of x. [3] (ii) Hence find the coefficient of x7 in the expansion of x x x 3 1 2 3 1 3 2 5 + - J L KK J L KK N P OO N P OO . [2]

6 0606/12/O/N/16 © UCLES 2016 5 (i) Find the equation of the normal to the curve ( ) ln y x 2 1 3 2 = + at the point P where x 3 1 =- . [4] The normal to the curve at the point P intersects the y-axis at the point Q. The curve ( ) ln y x 2 1 3 2 = + intersects the y-axis at the point R . (ii) Find the area of the triangle PQR. [3]

7 0606/12/O/N/16 © UCLES 2016 [Turn over 6 (a) Matrices X, Y and Z are such that X 2 4 6 3 1 5 = - J L K KK N P O OO , Y 1 1 0 = - ^ h and Z 0 5 1 3 = - J L KK N P OO . Write down all the matrix products which are possible using any two of these matrices. Do not evaluate these products. [2] (b) Matrices A, B and C are such that A 1 2 4 7 = - J L KK N P OO , B 4 10 2 4 = - J L KK N P OO and AC B = . (i) Find A 1 - . [2] (ii) Hence find C. [3]

8 0606/12/O/N/16 © UCLES 2016 7 y x O A B C cos y x 2 6 r = - J L KK N P OO The diagram shows part of the graph of cos y x 2 6 r = - J L KK N P OO. The graph intersects the y-axis at the point A, has a maximum point at B and intersects the x-axis at the point C. (i) Find the coordinates of A. [1] (ii) Find the coordinates of B. [2]

9 0606/12/O/N/16 © UCLES 2016 [Turn over (iii) Find the coordinates of C. [2] (iv) Find cos x x 2 6 d r - J L KK N P OO y . [1] (v) Hence find the area of the shaded region. [2]

10 0606/12/O/N/16 © UCLES 2016 8 i rad 12 cm B C A D O The diagram shows a sector AOB of the circle, centre O, radius 12 cm, together with points C and D such that ABCD is a rectangle. The angle AOB is i radians and the perimeter of the sector AOB is 47 cm. (i) Show that .1 92 i = radians correct to 2 decimal places. [2] (ii) Find the length of CD. [2]

11 0606/12/O/N/16 © UCLES 2016 [Turn over (iii) Given that the total area of the shape is 425 cm2, find the length of AD. [5]

12 0606/12/O/N/16 © UCLES 2016 9 Do not use a calculator in this question. The polynomial p(x) is ax x bx 4 18 3 2 - + + . It is given that p(x) and ( )x pl are both divisible by x 2 3 - . (i) Show that a 4 = and find the value of b. [4] (ii) Using the values of a and b from part (i), factorise p(x) completely. [2]

13 0606/12/O/N/16 © UCLES 2016 [Turn over (iii) Hence find the values of x for which p(x) x 2 = + . [3]

14 0606/12/O/N/16 © UCLES 2016 10 (a) v ms–1 U 0 20 30 U 2 t s The diagram shows part of the velocity-time graph for a particle, moving at v ms 1 - in a straight line, t s after passing through a fixed point. The particle travels at U ms 1 - for 20 s and then decelerates uniformly for 10 s to a velocity of U 2 ms 1 - . In this 30 s interval, the particle travels 165 m. (i) Find the value of U. [3] (ii) Find the acceleration of the particle between t = 20 and t = 30. [2]

15 0606/12/O/N/16 © UCLES 2016 [Turn over (b) A particle P travels in a straight line such that, t s after passing through a fixed point O, its velocity, v ms 1 - , is given by v t 8 3 2 = 4 e - J L KK N P OO . (i) Find the speed of P at O. [1] (ii) Find the value of t for which P is instantaneously at rest. [2] (iii) Find the acceleration of P when t = 1. [4] Question 11 is printed on the next page.

16 0606/12/O/N/16 © UCLES 2016 11 The variables x and y are such that when lny is plotted against x, a straight line graph is obtained. This line passes through the points , . , . ln ln x y x y 4 0 20 12 0 08 and = = = = . (i) Given that y Abx = , find the value of A and of b. [5] (ii) Find the value of y when x 6 = . [2] (iii) Find the value of x when . y 1 1 = . [2] 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.

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 6 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 2016 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 October/November 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 12 © UCLES 2016 Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied www without wrong working Question Answer Marks Part Marks 1 (a) (i) 10 B1 (ii) 22 B1 (iii) 4 B1 (b) (i) ⊂ Q R B1 (ii) ∩ = ∅ P Q , or { } B1 2 1, 3, 1 = = − = − a b c B3 B1 for each 3 2 3 5 2 0 + − = y y 1, 2 3 = = − y y 1 2 3 3 , 3− = = x x 1 1.44, 9 = = x x B1, B1 M1 M1 A1, A1 B1 for 5y or 3 5log x , B1 for –2 for correct attempt at the solution of their quadratic equation for dealing with one base 3 logarithm correctly A1 for each 4 (i) 10 7 4 80 80 32 3 9 − + x x x B3 B1 for each term, powers of x must be simplified (ii) Coefficients needed: ( ) 80 3 1 32 3 their their   × − + ×     = –48 M1 A1 for dealing with 2 terms Allow A1 for 7 48 − x

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 12 © UCLES 2016 Question Answer Marks Part Marks 5 (i) ( ) d 3 d 2 3 2 = + y x x When 1 d 3 , 0, 3 d 2 = − = = y x y x Equation of normal: 2 1 3 3   = − +     y x B1 B1 M1 A1 for correct derivative of log function for 0 = y M1 for attempt at a gradient of a perpendicular from differentiation and the equation of the normal (ii) 2 0, 9 Q  −     or ( ) 0,0.22 or better 1 0, ln 2 2       R or ( ) 0, 0.35 or better Area of PQR 1 1 2 1 ln 2 2 2 9 3   = + ×     = 0.0948 B1 ft B1 B1 Follow through on their c from part (i) Allow 0.095 6 (a) YX, XZ B2 B2 for both with no extras B1 for 1 correct with or without extras B1 for both correct with extras B0 for anything else (b) (i) 7 1 1 4 2 18     −   B1, B1 B1 for 1 18 , B1 for 7 1 4 2     −   (ii) C = A-1B 7 1 4 2 1 4 2 10 4 18 −    =    −    1 1 2 0 −   =     M1 A1, A1 for pre-multiplication A1 for any correct pair of elements, but must be from correct matrices

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 12 © UCLES 2016 Question Answer Marks Part Marks 7 (i) ( ) 0, 3 or ( ) 0,1.73 or better B1 (ii) π , 2 6       or ( ) 0.524, 2 or better B1, B1 B1 for each (iii) π cos 0 6   − =     x 2π 3 = x oe or 2.09 or better M1 A1 for correct attempt to solve trigonometric equation (iv) π 2sin 6   −     x ( +c) B1 (v) Area = 2π 3 0 π 2sin 6     −         x = 2 +1 = 3 M1 A1 for correct use of their limits, in radians, into π sin 6   −     k x . 8 (i) 47 24 12θ − = 23 12 θ = , so 1.917 θ = or better 1.92 θ = to 2dp M1 A1 for complete correct method to get θ = must have evidence of working to more than 2 dp, allow if 1.916 seen (truncated) (ii) 2 sin 2 12 θ = CD CD = awrt 19.6 or 19.7 M1 A1 for a complete method, may use cosine rule to get CD (iii) Area of sector = awrt 138 Area of triangle AOB = awrt 67 or 68 Area of segment = awrt 70 or 71 AD × AB + segment area = 425 leading to AD = awrt 18.1 or 18.0 Alternative method: Area of sector = awrt 138 Difference in length between BC (or AD) and OM where M is the midpoint of CD = 6.88, allow awrt 6.9 Remaining area consists of two trapezia each of width 9.85 and each of area 143.4 ( ) 1 2 6.88 9.85 143.4 2 − × = BC oe leading to AD = awrt 18.1 or 18.0 B1 M1 M1 M1 A1 B1 M1 M1 M1 A1 for sector area, allow unsimplified for a correct attempt at area for segment area (their sector area – their triangle area) for complete method to find AD Allow A1 for 18 for sector area for attempt to find difference between parallel sides for area of one trapezium ( ) 1 2 6.88 9.85 2 BC their their − × oe for attempt to find either BC or AD

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 12 © UCLES 2016 Question Answer Marks Part Marks 9 (i) ( ) 3 27 9 3 p : 4 18 0 2 8 4 2     − × + + =         a b ( ) 3 9 3 p 3 8 0 2 4 2       ′ = × − × + =             a b leading to 9 4 24 0 + + = a b oe and 27 4 48 0 + − = a b oe leading to 4, 15 = = − a b M1 M1 M1 A1 for attempt at 3 p 2       for differentiation and attempt at 3 p 2   ′    for solution of simultaneous equations, to get either a or b for both (ii) ( )( ) 2 2 2 3 + − x x oe M1, A1 M1 for attempt at long division or factorisation (iii) ( )( ) 2 2 2 3 2 + − = + x x x 2 0, 2 + = = − x x ( ) 2 2 3 1 − = x leading to 1, 2 = = x x B1 M1 A1 Must be using ( ) 2 + x correctly using part (ii) to get 2 = − x for solution of the quadratic equation 10 (a) (i) 1 20 10 165 2 2   + + =     U U U leading to 6 = U M1 DM1 A1 for realising that area under the graph is needed and attempt to find an area for equating their area to 165 and attempt to solve (ii) Gradient of line: 0.3 − M1, A1 M1 for use of the gradient, must be negative (b) (i) 27 B1 (ii) 2 8ln 4 = t 3.33 = t or better M1 A1 for a correct attempt to solve 2 8 e 4 = t (iii) acceleration = 2 2 2 8 8 2 3 e e 4 8     −     t t t When 1 = t , a = 6.98 M1, A1 M1, A1 M1 for a correct attempt to differentiate using the chain rule M1for use of 1 = t in their acceleration

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 12 © UCLES 2016 Question Answer Marks Part Marks 11 (i) ln ln ln = + y A x b Gradient: 0.12 ln 8 = − b , = 0.015 − 0.985 = b Intercept: ln 0.26 = A 1.30 = A Alternative 1 ln ln ln = + y A x b 0.2 4ln ln = + b A 0.08 12ln ln = + b A 1.30 = A and 0.985 = b Alternative 2 4 1.22 = Ab 12 1.08 = Ab 1.30 = A and 0.985 = b B1 M1 A1 DM1 A1 B1 M1 DM1 A1, A1 B1 B1 M1 A1, A1 may be implied, if equation not seen specifically, by correct values for A and b for use of gradient to obtain ln b Allow A1 for 0.015 e− for use of one of the given points correctly Allow A1 for 0.26 e or 1.3 for one correct equation for attempt to obtain either lnA or lnb from simultaneous equations Allow A1 for b = 0.015 e− and a = 0.26 e or 1.3 for correct attempt to obtain b or A, must already have B2 Allow A1 for b = 0.015 e− and a = 0.26 e or 1.3 (ii) When 6, ln 0.17 = = x y 1.19 = y M1 A1 for ln ln 6 ln y their A their b = + or ( ) 6 y their A their b = × allow awrt 1.18 to 1.20 (iii) When 1.1, ln 0.095 = = y y 11 = x M1 A1 for ln1.1 ln ln their A x their b = + or ( ) 1.1 x theirA theirb = × allow 10.5 to 11.5