Cambridge IGCSE Mathematics - Additional 0606 — 2013 Oct/Nov Paper 2 · Variant 1

This document consists of 18 printed pages and 2 blank pages. DC (LEG/SW) 67890/2 © UCLES 2013 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 0 9 1 7 8 8 6 3 7 4 * ADDITIONAL MATHEMATICS 0606/21 Paper 2 October/November 2013 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 a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, 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. www.theallpapers.com

2 0606/21/O/N/13 © UCLES 2013 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 www.theallpapers.com

3 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 1 Find the set of values of x for which x x 6 5 2 1 - . [3] www.theallpapers.com

4 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 2 Do not use a calculator in this question. Express 5 4 5 1 2 2 - - ^ h in the form p q 5 + , where p and q are integers. [4] www.theallpapers.com

5 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 3 (i) Given that y x 4 1 5 8 = - J L KK N P OO , find d d x y. [2] (ii) Hence find the approximate change in y as x increases from 12 to p 12 + , where p is small. [2] www.theallpapers.com

6 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 4 Given that log X 5 p = and log Y 2 p = , find (i) log X p 2, [1] (ii) log X 1 p , [1] (iii) log p XY . [2] www.theallpapers.com

7 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 5 Solve the simultaneous equations , 256 4 1024 y x = 3 9 243. x y 2 # = [5] www.theallpapers.com

8 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 6 (a) (i) Find the coefficient of x3 in the expansion of x 1 2 6 - ^ h . [2] (ii) Find the coefficient of x3 in the expansion of x 1 2 + e o x 1 2 6 - ^ h . [3] (b) Expand x x 2 1 4 + e o in a series of powers of x with integer coefficients. [3] www.theallpapers.com

9 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 7 x cm The diagram shows a box in the shape of a cuboid with a square cross-section of side x cm. The volume of the box is 3500 cm3. Four pieces of tape are fastened round the box as shown. The pieces of tape are parallel to the edges of the box. (i) Given that the total length of the four pieces of tape is L cm, show that L x x 14 7000 2 = + . [3] (ii) Given that x can vary, find the stationary value of L and determine the nature of this stationary value. [5] www.theallpapers.com

10 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 8 The table shows experimental values of two variables x and y. x 2 4 6 8 y 9.6 38.4 105 232 It is known that x and y are related by the equation y ax bx 3 = + , where a and b are constants. (i) A straight line graph is to be drawn for this information with x y on the vertical axis. State the variable which must be plotted on the horizontal axis. [1] (ii) Draw this straight line graph on the grid below. [2] O 10 20 30 y x www.theallpapers.com

11 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Use your graph to estimate the value of a and of b. [3] (iv) Estimate the value of x for which y x 2 25 = . [2] www.theallpapers.com

12 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 9 40 m 70 m 1.8 m s–1 P Q The diagram shows a river with parallel banks. The river is 40 m wide and is flowing with a speed of 1.8 ms–1. A canoe travels in a straight line from a point P on one bank to a point Q on the opposite bank 70 m downstream from P. Given that the canoe takes 12 s to travel from P to Q, calculate the speed of the canoe in still water and the angle to the bank that the canoe was steered. [8] www.theallpapers.com

13 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use www.theallpapers.com

14 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 10 1.4 rad O A B 12 cm The diagram shows a circle with centre O and a chord AB. The radius of the circle is 12 cm and angle AOB is 1.4 radians. (i) Find the perimeter of the shaded region. [5] www.theallpapers.com

15 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Find the area of the shaded region. [4] www.theallpapers.com

16 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 11 O Q P (9, e3) y x y e x 3 = The diagram shows part of the curve e y x 3 = . The tangent to the curve at , e P 9 3 ^ h meets the x-axis at Q. (i) Find the coordinates of Q. [4] www.theallpapers.com

17 0606/21/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Find the area of the shaded region bounded by the curve, the coordinate axes and the tangent to the curve at P. [6] www.theallpapers.com

18 0606/21/O/N/13 © UCLES 2013 For Examiner’s Use 12 (a) Solve the equation cosec cos x x 2 7 0 + = for 0° 360° x G G . [4] (b) Solve the equation sin y 7 2 1 5 - = ^ h for y 0 5 G G radians. [5] www.theallpapers.com

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20 0606/21/O/N/13 © UCLES 2013 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. University of 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 www.theallpapers.com

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2013 series 0606 ADDITIONAL MATHEMATICS 0606/21 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 October/November 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.XtremePapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 21 © Cambridge International Examinations 2013 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Accuracy 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 generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2, 1, 0 means that the candidate can earn anything from 0 to 2.

Page 3 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 21 © Cambridge International Examinations 2013 1 ( )( )1 6 − + x x Critical values –6 and 1 1 6 < < − x M1 A1 A1 [3] Attempt to solve a three term quadratic Allow 6 − > x AND 1 < x but not OR or a comma. Mark final answer. 2 ( ) 4 5 16 80 2 5 4 2 + − = − Multiply top and bottom by 1 5 + 1 5 17 + OR ( ) 4 5 16 80 2 5 4 2 + − = − ( )( ) ) ( 5 5 5 1 5 p q q p q p − + − = + − Leading to 16 , 84 5 − = − = − p q q p p = 17 q = 1 M1 M1 A1 A1 [4] M1 M1 A1 A1 Attempt to expand, allow one error, must be in the form 5 b a + . Must be attempt to expand top and bottom. Allow A1 for c 4 5 68 + Must get to a pair of simultaneous equations for this mark 3 (i) (ii) 7 5 4 1 d d       − = x k k y k = 2 Use x x y y ∂ × = ∂ d d with x = 12 and p x = ∂ –256p M1 A1 [2] M1 A1 [2] on k needs both M marks only for –128kp and must be evaluated 4 (i) (ii) (iii) 10 –5 7 log log log = + = Y X XY p p p 7 1 B1 [1] B1 [1] B1 B1 [2] Not 5 1 log − p Or XY p p XY log 1 log = Do not allow just 7 log log = + Y X p p on XY p log 1

Page 4 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 21 © Cambridge International Examinations 2013 5 5 4 = −y x oe 5 2 2 = + y x oe Solve their linear simultaneous equations 3 = x or 5.0 − = y OR from log 01 .3 408 .2 602 .0 = − y x 386 .2 954 .0 954 .0 = + y x OR from ln 931 .6 545 .5 386 .1 = − y x 493 .5 197 .2 197 .2 = + y x Final M1A1A1 follows as before B1 B1 M1 A1,A1 [5] B1 B1 B1 B1 Each in two variables and not quadratic as far as x = … or y = … 6 (a) (i) (ii) (b) –8 or 20 ( ) 3 160 x − isw ( ) 2 60 x (i) 2 1 + (their 60) ( ) 3 130 x − 2 2 1 8 24 32 16 x x x x + + + + oe B1 B1 [2] B1 M1 A1 [3] B3,2,1,0 [3] 40 ± implies 20 2× ± or +160 hence B1 OK if seen in expansion Can be implied Terms must be evaluated (allow 24x0) B2 for 4 terms correct. B1 for 2 or 3 terms correct. ISW once expansion is seen. 7 (i) (ii) 2 3500 x l = l x x L 2 2 4 3 + + × = Substitute for l and correctly reach 2 7000 14 x x L + = 3 14000 14 d d x x L − = Equate x L d d to 0 and solve 10 = x 210 = L 4 2 2 42000 d d x x y = and minimum stated B1 B1 DB1ag [3] M1A1 DM1 A1 B1 [5] allow 2 lx = 3500 RHS 3 terms e.g.       + + 2 3500 2 2 12 x x x or better Dependent on both previous B marks M1 either power reduced by one A1 both terms correct Must get = n x Both values Or use of gradient either side of turning point.

Page 5 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 21 © Cambridge International Examinations 2013 8 (i) (ii) (iii) (iv) 2 x Plot x y against 2 x with linear scales 2 x 4 16 36 64 x y 4.8 9.6 17.5 29 Finds gradient (0.4) 02 .0 4.0 ± = a 4.0 2.3 ± = b Read 5. 12 = x y or substitute in formula 4.8 B1 [1] B1 B1 [2] M1 A1 B1 [3] M1 A1 [2] Implied by axes or values in a table. May be seen in (ii) Must be linear scales At least 3 correct points plotted and no incorrect points Line must be ruled and through at least 2 correct points Condone use of correct values from table/graph to find gradient and /or equation. Values read from graph must be correct. Obtaining ( ) 24 to 22 2 = x from graph As far as 2 x = +ve constant 4.7 to 4.9 ignore –4.8 or 0 9 Method A Takes components 40 sin 12 = α v ( ) 70 8.1 cos 12 = + α v 4. 48 cos 12 = α v Solve for v or α 6. 39 = α 23 .5 = v M1 A1 A1 M1A1 DM1 A1 A1 [8] Allow 0.691 radians Method B 6. 21 12 8.1 = × = x 4. 48 6. 21 70 = − = y ( ) 56 . 3942 4. 48 40 2 2 2 = + = D 8. 62 = D 12 D V = 23 .5 = V 4. 48 40 tan = α ° = 6. 39 α B1 B1 M1 A1 DM1 A1 M1 A1 [8] 5.23 or better Allow 0.691 radians

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 21 © Cambridge International Examinations 2013 Method C ( ) 6. 80 70 40 2 2 = + = z ( ) 72 .6 12 70 40 2 2 = + = v ( ) 74 . 29 7 4 tan = → = δ δ oe 74 . 29 cos 72 .6 8.1 2 72 .6 8.1 2 2 2 × × − + = V 23 .5 = V 23 . 5 74 . 29 sin 8. 1 sin = β ( ) ( ) 2 8.9 or 3 8.9 = β 6. 39 74 . 29 = + = β α B1 B1 B1 M1 A1 M1 A1 A1 [8] Or ( ) 4 7 90 tan = −δ Allow 0.172 radians Allow 0.691 radians Method D ( ) 6. 80 70 40 2 2 = + = z 6. 21 12 8.1 = × = x ( ) 74 . 29 7 4 tan = → = δ δ oe 74 . 29 cos 6. 80 .6. 21 .2 6. 80 6. 21 2 2 2 − + = D ( ) 23 .5 12 / 8. 62 = = V 8. 62 74 . 29 sin 6. 21 sin = β ( ) 3 8.9 = β or ( ) 2 8.9 6. 39 74 . 29 = + = β α B1 B1 B1 M1 A1 M1 A1 A1 [8] This method has extra steps so note at this point the M mark is for an equation in D but the A mark is for a value of V. Allow 0.172 radians Allow 0.691 radians

Page 7 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 21 © Cambridge International Examinations 2013 10 (i) (ii) 4.1 cos 12 12 2 12 12 2 2 2 × × × − + = AB 15.4 to 15.5 ( ) 88 .4 4.1 2 = − = π θ Use ( ) 6. 58 = = θ r s 74.1 (Sector) ( )( ) 352 4.1 2 12 2 1 2 = − × × π or 4.1 12 2 1 12 2 2 × × − × π (Triangle) ( ) 71 or 9. 70 4.1 sin 12 12 2 1 = × × × = Area of major sector + Area of triangle 422 or 423 M1 A1 B1 M1 A1 [5] M1 M1 M1 A1 [4] 7.0 sin 12 2× = AB May be implied May be implied 9.4 12 × or better oe May be implied . May be implied 11 (i) (ii) x x y 3 1 e 3 1 d d = 3e 3 1 = m ( ) 9 e 3 1 e 3 3 − = − x y At Q y = 0, x = 6 Area triangle 3e 5.1 or 30.1 ∫ = x x x 3 1 3 1 e 3 d e oe Uses limits of 0 and 9 in integrated function. 3 e 3 3 − or 57.3 Area under curve subtract area of triangle 3 e 5.1 3 −or 27.1 B1 M1 DM1 A1 [4] B1 B1 M1 A1 M1 A1 [6] For insertion of x = 9 into their x y d d . 6.7 or better if correct. Using their evaluated m to find eqn 2. 40 7.6 − = x y or better if correct. Accept value that rounds to 6.0 to 2sf ± must see both values inserted if incorrect answer Condone 27.2 if obtained from 57.3 – 30.1.

Page 8 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 21 © Cambridge International Examinations 2013 12 (a) (b) x x sin 1 cosec = inserted into equation 7 2 tan − = x 164.1 344.1 (2y – 1) = 0.79..or 2.34… Find y using radians 0.898 (or 0.9 or 0.90) 1.67, 4.04 and 4.81(45) B1 DB1 B1 B1 [4] B1 M1 A1 A1 A1 [5] One correct value. on ( )1. 164 180 + Must come from tanx = Condone164 and 344 Deduct 1 mark for extras in range Allow 0.8 , 2.3 or 45.6° Add 1 then divide by 2 on a correct angle One correct value Another correct value Final two values Deduct 1 mark for extras in range