Cambridge IGCSE Mathematics - Additional 0606 — 2015 May/June Paper 2 · Variant 2

This document consists of 16 printed pages. DC (NH/SW) 91465/1 © UCLES 2015 [Turn over * 8 8 7 8 4 1 0 8 4 3 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 2015 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 0606/22/M/J/15 © UCLES 2015 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 0606/22/M/J/15 © UCLES 2015 [Turn over 1 The universal set contains all the integers from 0 to 12 inclusive. Given that A = {1, 2, 3, 8, 12}, B = {0, 2, 3, 4, 6} and C = {1, 2, 4, 6, 7, 9, 10}, (i) complete the Venn diagram, [3] A B C (ii) state the value of ( ) A B C n k k l l , [1] (iii) write down the elements of the set A B C k k l . [1]

4 0606/22/M/J/15 © UCLES 2015 2 The table shows the number of passengers in Economy class and in Business class on 3 flights from London to Paris. The table also shows the departure times for the 3 flights and the cost of a single ticket in each class. Departure time Number of passengers in Economy class Number of passengers in Business class 09 30 60 50 13 30 70 52 15 45 58 34 Single ticket price (£) 120 300 (i) Write down a matrix, P, for the numbers of passengers and a matrix, Q, of single ticket prices, such that the matrix product QP can be found. [2] (ii) Find the matrix product QP. [2] (iii) Given that 1 1 1 R = J L K KK N P O OO, explain what information is found by evaluating the matrix product QPR. [1]

5 0606/22/M/J/15 © UCLES 2015 [Turn over 3 Do not use a calculator in this question. A cm B C ( ) 6 3 5 + The diagram shows the right-angled triangle ABC, where 6 3 5 AB = + ^ hcm and angle B = 90°. The area of this triangle is 2 36 15 5 cm2 + J L KK N P OO . (i) Find the length of the side BC in the form 5 a b cm + ^ h , where a and b are integers. [3] (ii) Find AC 2 ^ h in the form 5 c d cm2 + ^ h , where c and d are integers. [2]

6 0606/22/M/J/15 © UCLES 2015 4 A river, which is 80 m wide, flows at 2 ms–1 between parallel, straight banks. A man wants to row his boat straight across the river and land on the other bank directly opposite his starting point. He is able to row his boat in still water at 3 ms–1. Find (i) the direction in which he must row his boat, [2] (ii) the time it takes him to cross the river. [3]

7 0606/22/M/J/15 © UCLES 2015 [Turn over 5 Solve the simultaneous equations 2 3 7 , 4. x y xy x y 2 2 + = + = [5] 6 (a) Solve 6 4 1 2 x = - . [2] (b) Solve 2 8 16 64 2 4 log log log log log y y y 2 a a a a a + + - = . [4]

8 0606/22/M/J/15 © UCLES 2015 7 In the expansion of 1 2x n + ^ h , the coefficient of x4 is ten times the coefficient of x2. Find the value of the positive integer, n. [6]

9 0606/22/M/J/15 © UCLES 2015 [Turn over 8 B A C O x y y = x + 10 y = x2– 6x + 10 The graph of 6 10 y x x 2 = - + cuts the y-axis at A. The graphs of 6 10 y x x 2 = - + and 10 y x = + cut one another at A and B. The line BC is perpendicular to the x-axis. Calculate the area of the shaded region enclosed by the curve and the line AB, showing all your working. [8]

10 0606/22/M/J/15 © UCLES 2015 9 Solutions by accurate drawing will not be accepted. O y = mx + 4 2 4 –1 Q R P (2, 10) x y The line 4 y mx = + meets the lines 2 x = and 1 x =- at the points P and Q respectively. The point R is such that QR is parallel to the y-axis and the gradient of RP is 1. The point P has coordinates (2, 10). (i) Find the value of m. [2] (ii) Find the y-coordinate of Q. [1] (iii) Find the coordinates of R. [2]

11 0606/22/M/J/15 © UCLES 2015 [Turn over (iv) Find the equation of the line through P, perpendicular to PQ, giving your answer in the form ax by c + = , where a, b and c are integers. [3] (v) Find the coordinates of the midpoint, M, of the line PQ. [2] (vi) Find the area of triangle QRM. [2]

12 0606/22/M/J/15 © UCLES 2015 10 (a) The function f is defined by : sin x x f 7 for 0° 360° x G G . On the axes below, sketch the graph of ( ) y x f = . [2] O 90° 180° 270° 360° x y (b) The functions g and hg are defined, for 1 x H , by 4 3 , . ln x x x x g hg = - = ^ ^ ^ h h h (i) Show that 4 3 x h ex = + ^ h . [2]

13 0606/22/M/J/15 © UCLES 2015 [Turn over (ii) O x 1 y = g(x) y The diagram shows the graph of y x g = ^ h. Given that g and h are inverse functions, sketch, on the same diagram, the graph of y x h = ^ h. Give the coordinates of any point where your graph meets the coordinate axes. [2] (iii) State the domain of h. [1] (iv) State the range of h. [1]

14 0606/22/M/J/15 © UCLES 2015 11 O A B C D Q P S R h 4 8 The diagram shows a cuboid of height h units inside a right pyramid OPQRS of height 8 units and with square base of side 4 units. The base of the cuboid sits on the square base PQRS of the pyramid. The points A, B, C and D are corners of the cuboid and lie on the edges OP, OQ, OR and OS, respectively, of the pyramid OPQRS. The pyramids OPQRS and OABCD are similar. (i) Find an expression for AD in terms of h and hence show that the volume V of the cuboid is given by 4 4 16 V h h h 3 2 = - + units3. [4]

15 0606/22/M/J/15 © UCLES 2015 [Turn over (ii) Given that h can vary, find the value of h for which V is a maximum. [4] Question 12 is printed on the next page.

16 0606/22/M/J/15 © UCLES 2015 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. 12 (i) Show that x = –2 is a root of the polynomial equation 15 26 11 6 0 x x x 3 2 + - - = . [1] (ii) Find the remainder when 15 26 11 6 x x x 3 2 + - - is divided by x – 3. [2] (iii) Find the value of p and of q such that 15 26 11 6 x x x 3 2 + - - is a factor of 15 37 6 x px x qx 4 2 3 + - + + . [4]

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the May/June 2015 series 0606 ADDITIONAL MATHEMATICS 0606/22 Paper 2, maximum raw mark 80 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 May/June 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

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Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 22 © Cambridge International Examinations 2015 3 (i)    −    − ×    +     + 5 3 6 5 3 6 5 3 6 5 15 36 oe 9 225 5 108 5 90 216 − − − + 5 2 1+ cao Alternative method: ( ) ( ) 5 6 3 15 6 5 15 36 b a b a + + + = + 6a + 15b = 36 3a + 6b =15 a = 1 and b = 2 M1 DM1 A1 M1 DM1 A1 or 5 2 5 2 5 2 5 5 12 −     − ×     +     + oe or 1 25 5 12 5 10 24 − − − + or 25 5 12 5 10 24 − −     + − allow a = 1 and b = 2 or 5 2 1+ (ii)     + + + + + =        + +    + = 20 5 4 1 45 5 36 36 5 2 1 5 3 6 2 2 2 their their AC 5 40 102 + cao M1 A1 correct or correct ft expansions, using Pythagoras with    + 5 3 6 and their BC ignore attempts to square root after correct answer seen 4 (i) ( ) 3 2 cos = x oe soi 48.189…° or 131.810…° or 0.8410… rad or 2.3(00…) rad oe isw with reference axis indicated by comment, e.g. “to the bank” or “upstream”, etc. or clearly marked on a diagram M1 A1 Alternatively ( ) 3 2 sin = y oe soi 41.810…° or 0.7297 ... or 0.73(0) rad oe isw with reference axis indicated by comment, e.g. “to the perpendicular with the bank”, etc. or clearly marked on a diagram If M0 then SC1 for an unsupported answer of 138.189…° or 2.4118… rad or 318.189...° or 5.5534... rad with reference axis indicated by comment, e.g. “on a bearing of” or “from North” or clearly marked on a diagram

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 22 © Cambridge International Examinations 2015 (ii) Speed    = − = 5 4 9 or 3sin 48.2 or 2tan 48.2 or 3cos41.8 or 8. 41 tan 2 or 2. 48 cos 3 2 2 3 2 2 2 × × − + oe or 2.236(0…) rot to 4 or more figs or 2.24 [m/s] soi time 5 80 their = oe 35.66 to 35.8 (seconds) oe B1 M1 A1 Or Distance ) 33 .( 107 2. 48 sin 80 K = = oe soi time 3 33 . 107 K their = ignore subsequent rounding or attempted conversion to, e.g. minutes but A0 if answer spoiled by continuation of method if no working, so B0 M0, then allow B3 for an answer 35.66 to 35.8 oe 5 Substitution of either 4 – x or 4 – y into equation of curve and brackets expanded 12x2 – 52x + 48 [= 0] or 12y2 – 44y + 32 [= 0] oe Solve their 3-term quadratic 3 4 = x and 3 isw 3 8 = y and 1 isw M1 A1 M1 A1 A1 condone one sign error or slip in either equation of curve or expansion of brackets; condone omission of = 0, BUT 4 – x or 4 – y must be correct dep on a valid substitution attempt or 3 8 3 4 = = y x not from wrong working or x = 3 y = 1 not from wrong working if no working, allow full marks for fully correct answer only. 6 (a) ( ) 1 2 log 6 log 4 x   − =     oe or 6 1 log 2 4 x  = −     oe 1.23 or 1.226(29…) rot to 4 or more figures isw M1 A1 or       = 4 36 log 6 log x oe or x log 6 – log 36 = log 1 – log 4 oe correct answer or 1.22 implies M1

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 22 © Cambridge International Examinations 2015 (b) Method 1 2 2 4 log 64 16 2 8 log =       × × y y y oe y = 2 B3 B1 or B2 if at most one error or omitted step or B1 if at most two errors or omitted steps not from wrong working Method 2 log 2 + 2 log y + 3 log 2 + 4 log 2 + log y – 6 log 2 – log y = 4 log 2 y = 2 B3,2,1,0 B1 LHS terms log 2 y2 = log 2 + 2 log y; log 8 = 3 log 2; log 16 y = 4 log 2 + log y; –log 64 y = –6 log 2 – log y; RHS term 2 log 4 = 4 log 2 not from wrong working 7 ( )( )( )( ) ( )( ) 1 2 2 1 10 1 2 3 4 2 3 2 1 2 4 × − = × × × − − − n n n n n n or better n2 – 5n – 24 [= 0] oe (n + 3) (n – 8) [= 0] n = 8 only M3 A1 M1 A1 condone omitting the factor of n and/or n – 1; must have dealt with factorials M2 if one slip/omission or M1 if two slips/omissions or B1 for ( )( ) [ ] 2 2 2 2 1 x n n − seen and B1 for ( )( )( )( ) [ ] 4 4 2 24 3 2 1 x n n n n − − − seen equivalent must be 3-terms, e.g. n2 – 5n = 24 or any valid method of solution for their 3-term quadratic A0 if −3 also given as a final solution, i.e. not discarded If zero scored, allow SC1 for n = 8 unsupported or without correct method

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 22 © Cambridge International Examinations 2015 8 Method 1 (Separate areas subtracted) [ ] 7 B C x x = = soi ( ) 3 2 2 6 6 10 d 10 3 2 x x x x x x   − + = − +   ∫ Correct or correct ft substitution of limits 0 and their 7 into     + − x x x their 10 2 6 3 2 3 ( ) 7 17 10 2 1 × + oe or ( ) ( ) ( ) 7 2 2 7 0 0 7 10 d 10 10 7 2 2 x x x x   + = + = +     ∫ oe       −3 112 2 189 their 6 343 or 1 57 6 or 57.2 to 3 sf or 57.16(6...) rot to 4 figs isw B1 M2 DM1 B2 M1 A1 or M1 for at least one term correct dep on at least M1 being earned; evidence of substitution must be seen in their integral which must be at least two terms; condone omission of lower limit; or M1 for 2 1 (their 10 + their 17) × their 7 oe or B1 for ( ) x x x x 10 2 d 10 2 + = + ∫ dep on a genuine attempt to integrate the equation of the curve; must be their area trapezium/under the line – their attempt at area under curve from full and correct working with no omitted steps Method 2 (Subtracting and using integration once) [ ] 7 B c x x = = soi ( ) ∫ + − x x x d 7 2     + − 2 7 3 2 3 x x oe or     −2 7 3 2 3 x x oe Correct or correct ft substitution of limits 0 and their 7 into their     + − 2 7 3 2 3 x x 6 343 or 1 57 6 or 57.2 to 3 sf or 57.16(6...) rot to 4 figs isw B1 B1 M3 M2 A1 condone omission of dx or M2 for ( ) ∫ + = + 2 3 d 2 3 2 qx px x qx px oe either with p = ±1 or q = ±7 or M1 for ( ) ∫ + = + 2 3 d 2 3 2 qx px x qx px with non-zero constants p and q, with p ≠ ±1 and q ≠ ±7 dep on a valid integration attempt; evidence of substitution must be seen; condone omission of lower limit; from full and correct working with no omitted steps

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 22 © Cambridge International Examinations 2015 9 (i) 10 = 2m + 4 soi m = 3 M1 A1 or [ ] 0 2 4 10 − − = m oe soi (ii) 1 B1 (iii) 1 1 2 10 = − − − R y oe soi (–1, 7) or x = −1 and y = 7 M1 A1 or y = x + 8 oe if y = 7 only stated, provided that x = −1 is soi in working allow both marks if M0 then B1 for y = 7 only with no working (iv) Use of m1 m2 = –1 with their m from (i) ( ) 2 3 1 10 −       − = − x their y 3y + x = 32 isw M1 A1 A1 may be implied by perpendicular gradient seen in equation or c x their +       −3 1 and c their +       − = 2 3 1 10 allow for correct equation with integer coefficients in any simplified form (v) 1 11 , 2 2 their       oe isw B1,B1ft ft their yQ or M1 for 2 1 10 1 , 2 2 − +       seen (vi) 4.5 oe cao B2 not from wrong working or M1 for any correct method with correct coordinates 10 (a) 90 180 270 360 1 y x O B2,1,0 correct sinusoidal/reflected sinusoidal shape, all above x-axis with intent to have all maximum points of equal height; 2 maximum points of intended equal height only over 0 to 360; all max points clearly at y = 1; cusp at 180

Page 8 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 22 © Cambridge International Examinations 2015 (b)(i) [ ] ln(4 3) e 3 hg( ) 4 x x − + = fully correct and completion to [ ] hg( )x = x M1 A1 Alternative method ln(4 3) y x = − and change of subject to x oe, fully correct and comment that 1 h( ) g ( ) x x − = oe (ii) O 1 1 y = g(x y = h(x) B2,1,0 correct shape; 1 marked on the y-axis or (0, 1) stated close by; curve with positive gradient in first quadrant only (iii) 0 x ≥ or [ ) 0, ∞ B1 not domain ≥ 0 (iv) 1 y ≥ or [ ) 1, ∞ B1 or ( ) h 1 x ≥ , h 1 ≥etc. 11 (i) 8 8 h − or 8 : 8 – h soi 4 8 8 × −h oe 2 4 8 8       × −h h oe expand and simplify to h h h 16 4 4 2 3 + − AG M1 A1 M1 A1 or h − 8 8 or 8 – h : 8 soi or h − ÷ 8 8 4 oe h must be in the numerator of the expression for this mark; (ii) 16 8 4 3 2 + −h h oe 0 16 8 4 3 2 =       + −h h their and attempt to solve 3 8 oe only B1 M1 A2 must be a 3-term quadratic; must be an attempt at a derivative or A1 for h = 3 8 and 8 allow 2.67 or 2.66(6...) rot to 4 or more figs for 3 8

Page 9 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 22 © Cambridge International Examinations 2015 12 (i) –120 + 104 + 22 – 6 = 0 or correct unsimplified form, e.g. 15(–2)3 + 26(–2)2 – 11(–2) – 6 = 0 or 15(–8) + 26(4) – 11(–2) – 6 = 0 B1 or correct synthetic division −2 15 26 −11 −6 −30 8 6 15 −4 −3 0 (ii) Substituting x = 3 into 15x3 + 26x2 – 11x – 6 600 M1 A1 or correct synthetic division 3 15 26 −11 −6 45 213 606 15 71 202 600 correct answer implies M1; must be explicitly identified as answer if using synthetic/long division methods by e.g. circling (iii) (x – 1)(15x3 + 26x2 – 11x – 6) soi Multiply out (x ± 1)(15x3 + 26x2 – 11x – 6) and compare coefficients of x3 or x to quartic p = 11 q = 5 B1 M1 A1 A1 by inspection or division; may be implied by e.g. (ax + b)(15x3 + 26x2 – 11x – 6) and a = 1, b = −1 seen in later work comparing coefficients or multiply out, e.g. (ax + b)(15x3 + 26x2 – 11x – 6) and compare coefficients of x3 or x to quartic correct p or q implies M1; correct p and q www implies B1 M1