Cambridge IGCSE Mathematics - Additional 0606 — 2017 Feb/March Paper 1 · Variant 2

This document consists of 14 printed pages and 2 blank pages. DC (LK/SG) 129474/2 © UCLES 2017 [Turn over * 5 3 3 1 7 2 3 3 8 8 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 0606/12/F/M/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 0606/12/F/M/17 © UCLES 2017 [Turn over 1 (a) It is given that  = { : , } x x x 0 35 R 1 1 ! and sets A and B are such that A = {multiples of 5} and B = {multiples of 7}. (i) Find ( ) n A B + . [1] (ii) Find ( ) n A B , . [1] (b) It is given that sets X, Y and Z are such that X Y Y + = , X Z Z + = and Y Z + Q = . On the Venn diagram below, illustrate sets X, Y and Z. [3]

4 0606/12/F/M/17 © UCLES 2017 2 (i) On the axes below sketch, for ° ° x 0 360 G G , the graph of cos y x 1 3 2 = + . [3] 360° 270° 180° x y O 90° (ii) Write down the coordinates of the point where this graph first has a minimum value. [1] 3 The first three terms in the expansion of a x 4 5 + J L KK N P OO are bx cx 32 2 + + . Find the value of each of the constants a, b and c. [5]

5 0606/12/F/M/17 © UCLES 2017 [Turn over 4 (a) It is given that A 1 2 3 4 = - J L KK N P OO. (i) Find A 1 - . [2] (ii) Using your answer to part (i), find the matrix M such that AM = 1 4 5 2 - - J L KK N P OO. [3] (b) X is a 2 1 3 - - J L KK N P OO and Y is a 2 4 1 3 J L KK N P OO, where a is a constant. Given that det X = 4 det Y, find the value of a. [2]

6 0606/12/F/M/17 © UCLES 2017 5 (i) Show that cosec sin cot cos i i i i - = . [3] (ii) Hence solve the equation cosec sin cos 3 1 i i i - = , for 0 G i 2 G r radians. [4]

7 0606/12/F/M/17 © UCLES 2017 [Turn over 6 (a) The letters of the word THURSDAY are arranged in a straight line. Find the number of different arrangements of these letters if (i) there are no restrictions, [1] (ii) the arrangement must start with the letter T and end with the letter Y, [1] (iii) the second letter in the arrangement must be Y. [1] (b) 7 children have to be divided into two groups, one of 4 children and the other of 3 children. Given that there are 3 girls and 4 boys, find the number of different ways this can be done if (i) there are no restrictions, [1] (ii) all the boys are in one group, [1] (iii) one boy and one girl are twins and must be in the same group. [3]

8 0606/12/F/M/17 © UCLES 2017 7 (a) A vector v has a magnitude of 102 units and has the same direction as 8 15 - J L KK N P OO. Find v in the form a b J L KK N P OO, where a and b are integers. [2] (b) Vectors c 4 3 = J L KK N P OO and d p q p q 5 = - + J L KK N P OO are such that c d p 2 27 2 + = J L KK N P OO. Find the possible values of the constants p and q. [6]

9 0606/12/F/M/17 © UCLES 2017 [Turn over 8 A curve is such that d d sin x y x 4 2 2 2 = . The curve has a gradient of 5 at the point where x 2 r = . (i) Find an expression for the gradient of the curve at the point ( , ) x y . [4] The curve passes through the point P 12 r c , 2 1 - m. (ii) Find the equation of the curve. [4] (iii) Find the equation of the normal to the curve at the point P, giving your answer in the form y mx c = + , where m and c are constants correct to 3 decimal places. [3]

10 0606/12/F/M/17 © UCLES 2017 9 i rad A B C O The diagram shows a circle, centre O, radius 10 cm. Points A, B and C lie on the circumference of the circle such that AC BC = . The area of the minor sector AOB is cm 20 2 r and angle AOB is i radians. (i) Find the value of i in terms of r. [2] (ii) Find the perimeter of the shaded region. [4]

11 0606/12/F/M/17 © UCLES 2017 [Turn over (iii) Find the area of the shaded region. [3]

12 0606/12/F/M/17 © UCLES 2017 10 y O x A B y = (2x – 5) 3 2 y 3 3 = The diagram shows part of the curve ( ) y x 2 5 2 3 = - and the line y 3 3 = . The curve meets the x-axis at the point A and the line y 3 3 = at the point B. Find the area of the shaded region enclosed by the line AB and the curve, giving your answer in the form p 20 3, where p is an integer. You must show all your working. [8]

13 0606/12/F/M/17 © UCLES 2017 [Turn over Question 11 is printed on the next page.

14 0606/12/F/M/17 © UCLES 2017 11 It is given that e y A bx = , where A and b are constants. When lny is plotted against x a straight line graph is obtained which passes through the points (1.0, 0.7) and (2.5, 3.7). (i) Find the value of A and of b. [6] (ii) Find the value of y when x 2 = . [2]

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16 0606/12/F/M/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 7 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/12 Paper 12 March 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 March 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 March 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 www without wrong working Question Answer Marks Part Marks 1 (a) (i) 0 B1 (ii) 10 B1 (b) B1 B1 B1 either ∩ = X Y Y or ∩ = X Z Z ∩ = ∅ Y Z completely correct Venn diagram. Y Z X

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2017 © UCLES 2017 Page 3 of 7 Question Answer Marks Part Marks 2 (i) B1 B1 B1 2 complete cycles having a maximum at 4 = y and a minimum at 2 = − y completely correct curve (ii) ( ) o 90 , 2 − B1 3 2 5 4 3 5 10 4 4     + +         x x a a a 5 32 = a , so 2 = a ( ) 4 1 5 4 b their a = × × , leading to 20 = b ( ) 3 1 10 16 c their a = × × leading to 5 = c B1 M1 A1 M1 A1 correct attempt to obtain b 4 (a) (i) 4 3 1 2 1 10     −   B1 B1 for 1 determinant for matrix (ii) 4 3 1 5 1 2 1 4 2 10 − −    =    −    M 4 7 1 3 6 5 −   =     M oe M1 A2,1,0 pre-multiplication by the matrix from part (i) –1 each element error (b) ( ) 3 2 4 6 4 − + = − a a 2 3 = a M1 A1 correct use of a determinant

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2017 © UCLES 2017 Page 4 of 7 Question Answer Marks Part Marks 5 (i) LHS 1 sin sin θ θ = − 2 1 sin sin θ θ − = 2 cos sin θ θ = cot cos θ θ = M1 M1 A1 dealing with cosecθ and attempt at dealing with fractions correct use of identity completely correct proof (ii) 1 cot cos cos 3 θ θ θ = 3cot cos cos 0 θ θ θ − = ( ) cos 3cot 1 0 θ θ − = cos 0 θ = 1 cot 3 θ = , so tan 3 θ = π 3π , 2 2 θ = , 1.25, 4.39 θ = M1 M1 A1,A1 use of part (i), manipulation and factorisation dealing with cotθ and attempt to solve A1 for each pair of solutions (allow 1.57 and 4.71) 6 (a) (i) 40 320 B1 (ii) 720 B1 (iii) 5040 B1 (b) (i) 35 B1 (ii) 1 B1 (iii) Twins in team of 4 5 2 10 = C Twins in team of 3 = 5 Total = 15 www B1 B1 B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2017 © UCLES 2017 Page 5 of 7 Question Answer Marks Part Marks 7 (a) 8 102 15 17     −   48 90     −   M1 A1 attempt to obtain magnitude of 8 15     −   and use it (b) 2 2 2 4 10 2 3 27   − +  =     + +     p q p p q 2 2 2 4 − + = p q p 10 2 3 + + p q = 27 leading to 2 12 20 0 − + = p p 2, 2 = = p q 10, 38 = = − p q M1 M1 A1 M1 A1 A1 dealing with the scalar and with addition equating like vectors and simplifying both equations correct elimination of q and subsequent solution of quadratic 8 (i) ( ) d 2cos2 d = − + y x c x 5 2cosπ = − + c d 3 2cos2 d = − y x x M1 A1 M1 A1 integration to obtain the form cos2 a x correct, condone omission of c attempt to find c May be implied by a correct c (ii) ( ) 3 sin 2 = − + y x x c 1 π 1 2 4 2 − = − + c π 3 sin2 4 = − − y x x oe M1 A1 M1 A1 integration to obtain the form sin 2 a x correct, condone omission of c attempt to find c (iii) When π 12 = x , d 3 3 d = − y x Normal equation: 1 1 π 2 12 3 3   + = −   −  y x 0.789 0.294 = − − y x cao M1 A1FT A1 attempt to obtain perpendicular gradient and normal equation FT on their d d y x from (i). Allow unsimplified

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2017 © UCLES 2017 Page 6 of 7 Question Answer Marks Part Marks 9 (i) 2 1 10 20π 2 θ × × = 2π 5 θ = M1 A1 use of sector area to obtain θ (ii) Arc length 4π = AB ( ) 2 2 2 10 10 2 10 10 cos2θ = + − × × × BC or 10 4π π sin sin 5 10 = BC 19.02 = BC Perimeter = 50.6 B1FT M1 A1 A1 FT their θ valid attempt to obtain BC (iii) Area = Either 2 2 1 π 19.02 sin 2 5 1 2π 20π 10 sin 2 5   ×         + − ×         = 121.6 allow awrt 122 Or 1 4π 20π 2 10 10sin 2 5   + × ×     = 121.6 allow awrt 122 M1 M1 A1 M1,M1 A1 area of triangle ACB area of relevant segment M1 for area of triangle AOB or AOC M1 for a complete method

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2017 © UCLES 2017 Page 7 of 7 Question Answer Marks Part Marks 10 ( ) 3 2 2 5 3 3 − = x 4 = x At A 2.5 = x Either Area ( ) 3 4 2 2.5 1 3 3 3 2 5 d 2 2 x x = × × − − ∫ ( ) 4 2.5 2.5 9 3 1 2 5 4 5 x   = − −     ( ) 2.5 9 3 1 3 0 4 5   = − −     9 3 20 = Or line AB: 2 3 5 3 = − y x Area ( ) 3 4 2 2.52 3 5 3 2 5 d = − − − ∫ x x x ( ) 4 5 2 2 2.5 2 5 3 5 3 5   −   = − −       x x x 9 3 9 3 4 5 = − 9 3 20 = M1 A1 B1 M1 M1 A1 DM1 A1 M1 M1 A1 DM1 A1 attempt to find x-coordinate of B x-coordinate of B x-coordinate of A plan and attempt to find the area of the triangle. Allow unsimplified attempt at integration, must be in the form ( ) 2.5 2 5 − x correct integration attempt to use limits correctly equation of AB and attempt to integrate attempt at integration, must contain the form ( ) 2.5 2 5 − x correct integration attempt to use correct limits correctly 11 (i) ln ln = + y A bx 0.7 ln = + A b 3.7 ln 2.5 = + A b leading to 2 = b and ln 1.3 = − A , so 0.273 = A or 1.3 e− B1 M1 A1 A1 M1,A1 may be implied by later work use of either point correctly in above equation or equivalent one correct equation M1 for dealing with ln correctly to obtain A. (ii) ln 1.3 2 = − + y x ln 2.7 = y 14.9 = y M1 A1 valid attempt to find y. Must include correct substitution and dealing with ln correctly.