Cambridge IGCSE Mathematics - Additional 0606 — 2004 Oct/Nov Paper 1 · Variant 1

This document consists of 5 printed pages and 3 blank pages. MCS UCJ135 S69945/1 © UCLES 2004 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/01 Paper 1 October/November 2004 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Graph paper Mathematical tables READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Write your answers on the separate Answer Booklet/Paper provided. 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. 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. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. www.XtremePapers.com

© UCLES 2004 0606/1/O/N/04 2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x = . Binomial Theorem (a + b)n = an + ( ) an – 1b + ( ) an – 2b2 + … + ( ) an – rbr + … + bn, where n is a positive integer and ( ) = . 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1. sec2 A = 1 + tan2 A. cosec2 A = 1 + cot2 A. Formulae for ∆ABC = = . a2 = b2 + c2 – 2bc cos A. ∆= bc sin A. c –––– sin C b –––– sin B a –––– sin A n! –––––––– (n – r)! r! n r n r n 2 n 1 –b ± √b2 – 4ac ––––––––––––– 2a 2 1–

1 The position vectors of points A, B and C, relative to an origin O, are i ! 9j, 5i 0 3j and k(i ! 3j) respectively, where k is a constant. Given that C lies on the line AB, find the value of k. [4]
2 A youth club has facilities for members to play pool, darts and table-tennis. Every member plays at least one of the three games. P, D and T represent the sets of members who play pool, darts and table-tennis respectively. Express each of the following in set language and illustrate each by means of a Venn diagram. (i) The set of members who only play pool. [2] (ii) The set of members who play exactly 2 games, neither of which is darts. [2]
3 Without using a calculator, solve, for x and y, the simultaneous equations . [5] The diagram shows a sector COD of a circle, centre O, in which angle COD # radians. The points A and B lie on OD and OC respectively, and AB is an arc of a circle, centre O, of radius 7 cm. Given that the area of the shaded region ABCD is 48 cm
2, find the perimeter of this shaded region. [6]
5 Given that the expansion of (a ! x)(1 0 2x)
n in ascending powers of x is 3 0 41x ! bx
2 ! …, find the values of the constants a, n and b. [6]
6 The function f is defined, for 0 ` x ` π, by f
(x) # 5 ! 3 cos 4x. Find (i) the amplitude and the period of f, [2] (ii) the coordinates of the maximum and minimum points of the curve y # f
(x). [4]
7 (a) Find the number of different arrangements of the 9 letters of the word SINGAPORE in which S does not occur as the first letter. [2] (b) 3 students are selected to form a chess team from a group of 5 girls and 3 boys. Find the number of possible teams that can be selected in which there are more girls than boys. [4] 4 3 )
y−1 = 81
1 9 3
4x × ( 8
x ÷ 2
y = 64, © UCLES 2004 0606/1/O/N/04 [Turn over 3 D C B 7 cm O A rad 4 3
4

8 The function f is defined, for x â
, by (i) Find f
01 in terms of x and explain what this implies about the symmetry of the graph of y # f(x). [3] The function g is defined, for x â
, by (ii) Find the values of x for which f(x) # g
01(x). [3] (iii) State the value of x for which gf(x) # 02. [1]
9 (a) Solve, for 0° ≤x ≤360°, the equation sin
2 x # 3 cos
2 x ! 4 sin x. [4] (b) Solve, for 0 ` y ` 4, the equation cot 2y # 0.25, giving your answers in radians correct to 2 decimal places. [4] 10 A curve has the equation y # x
3 ln x, where x p 0. (i) Find an expression for . [2] Hence (ii) calculate the value of ln x at the stationary point of the curve, [2] (iii) find the approximate increase in y as x increases from e to e ! p, where p is small, [2] (iv) find . [3] 11 The line 4y # 3x ! 1 intersects the curve xy # 28x 0 27y at the point P(1, 1) and at the point Q. The perpendicular bisector of PQ intersects the line y # 4x at the point R. Calculate the area of triangle PQR. [9]

x
2 ln x
d
x d
y d
x g : x
√

x −3 2 . f : x
√

3x + 11 x −3 , x
=C
3. © UCLES 2004 0606/1/O/N/04 4

12 Answer only one of the following two alternatives. EITHER (a) At the beginning of 1960, the number of animals of a certain species was estimated at 20 000. This number decreased so that, after a period of n years, the population was 20 000e
00.05n. Estimate (i) the population at the beginning of 1970, [1] (ii) the year in which the population would be expected to have first decreased to 2000. [3] (b) Solve the equation 3
x ! 1 0 2 # 8 " 3
x 0 1. [6] OR A curve has the equation . (i) Show that the exact value of the y-coordinate of the stationary point of the curve is . [4] (ii) Determine whether the stationary point is a maximum or a minimum. [2] (iii) Calculate the area enclosed by the curve, the x-axis and the lines x # 0 and x # 1. [4] 2√3 y = e
1 2
x + 3e
−
1 2
x © UCLES 2004 0606/1/O/N/04 5

BLANK PAGE 0606/1/O/N/04 8 Every reasonable effort has been made to trace all copyright holders. The publishers would be pleased to hear from anyone whose rights we have unwittingly infringed. University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the November 2004 question paper 0606 ADDITIONAL MATHEMATICS 0606/01 Paper 1, maximum raw mark 80 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2004 question papers for most IGCSE and GCE Advanced Level syllabuses. www.XtremePapers.com

Grade thresholds taken for Syllabus 0606 (Additional Mathematics) in the November 2004 examination. minimum mark required for grade: maximum mark available A C E Component 1 80 64 30 20 Grade A* does not exist at the level of an individual component.

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 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.

The following abbreviations may be used in a mark scheme or used on the scripts: AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) Penalties MR -1 A penalty of MR -1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. OW -1,2 This is deducted from A or B marks when essential working is omitted. PA -1 This is deducted from A or B marks in the case of premature approximation. S -1 Occasionally used for persistent slackness – usually discussed at a meeting. EX -1 Applied to A or B marks when extra solutions are offered to a particular equation. Again, this is usually discussed at the meeting.

November 2004 INTERNATIONAL GCSE MARK SCHEME MAXIMUM MARK: 80 SYLLABUS/COMPONENT: 0606/01 ADDITIONAL MATHEMATICS (Paper 1)

Page 1 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 1 © University of Cambridge International Examinations 2005 1 OA = i + 9j OB = 5i – 3j OC = k(i + 3j) AB or BA = 4i – 12j AC or CA = (k – 1)i + (3k – 9)j CB or BC = (5 – k)i – (3k + 3)j M1 A1 For | one relevant vector |. Must be “–” Any 2 of these correct unsimplified. Ratio of i to j is the same → k = 2 M1 A1 Using ratio idea (not DM). Co. If use of AB = a–b and correct answer obtained, allow full marks. [or m = –3 = (3k + 3) ÷(k – 5) = (3k – 9) ÷ (k – 1) → M1 A1 → k = 2 M1 A1] [4] [or (1, 9) → (5, –3) y = –3x + 12 M1 A1 subs (k, 3k) or solve with y = 3x M1A1] Drawing M2 A2 2 (i) T D P ′ ∩ ′ ∩ or ( )′ ∪T D P I or T D ′ ∩ ′ or ( )′ ∪T D B1 B1 Co.co (ii) T D P ∩ ′ ∩ B1 B1 [4] Co.co 3 Change to powers of 2. 23x ÷ 2y = 26 → 3x – y = 6 M1 A1 Needs to try all terms as powers of 2 and have ±. Co Change to powers of 3. 34x × 3-2y+2 = 34 → 4x – 2y + 2 = 4 M1 A1 Needs to try all terms as powers of 3 and have ±. Co with bracket sorted. → x = 5 and y = 9 A1 [5] Co 4 Let OR = r ½r2 (4/3) – ½72 (4/3) = 48 M1 A1 Any use of ½r2θ for M mark. A1 unsimplified, but complete. → r = 11 A1 Co. If x + 7 for r allow A1 for x = 4. Perimeter = 11 × (4/3) + 7 × (4/3) + 2 (11 – 7) M1 A1√ M1 for any use of s = rθ. A1√ unsimplified and √ on his r or x only. → 32 A1 [6] Co – allow anything rounding to 32.0

Page 2 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 1 © University of Cambridge International Examinations 2005 5 (a + x) (1 – 2nx + …) = 3 – 41x + bx2 term in x3 = nC2 (± 2x)2 B1 Wherever it comes. → a = 3 B1 Co – anywhere 1 – 2an = –41 → n = 7 M1 A1 Must use 2 terms. Coeff of x2 is 3 × 84 – 1 × 14 M1 Must use sum of 2 products. → 238 A1 [6] Co. 6 f(x) = 5 + 3cos4x (i) a = 3, period = ½π B1 B1 Co. allow 90° for period. (ii) max/min x = π/4 or 2π/4 or 3π/4 → max of 8 → min of 2 B1 B1 When “8” is used as stationary value. When “2” is used as stationary value. (π/4, 2) (2π/4, 8) (3π/4, 2) B2, 1√ [6] √ for 5 ± his “a”. [B0 if degrees here] Ignore inclusion of max/min at 0 or π. 7 (a) 8 × 8! or 9 8 × 9! or 9! – 8! → 322 560 M1 A1 [2] Must be nCr – knows what to do. Ans only is ok for 2 marks. (b) 2G, 1B 5C2 × 3C1 = 10 × 3 = 30 3G, 0B 5C3 = 10 M1 A1 B1 Needs to be a product of nCr’s. Co. Anywhere. total = sum of these = 40 A1 [4] Co. 8 (i) y = (3x + 11)/(x – 3) Makes x the subject. f -1(x) = (3x + 11)/(x – 3) f and f -1 are the same functions. M1 A1 Good algebra in making x the subject. → Graph has y = x as line of symmetry. B1 [3] Co accept any mention of y = x. (ii) g(x) = ½(x – 3) g -1(x) = 2x + 3 → 2x + 3 = (3x + 11)/(x – 3) → 2x2 – 6x – 20 = 0 → x = –2 or 5 B1 M1 A1 [3] Anywhere. Algebra must lead to quadratic. Co. (iii) gf(x) = –2 → f(x) = g -1 (–2) → x = –2 B1 [1] However obtained.

Page 3 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 1 © University of Cambridge International Examinations 2005 9 (a) s2 = 3c2 + 4s Use of s2 + c2 = 1 → 4s2 – 4s – 3 = 0 → s = –½ or 3/2 → x = 210° and 330° M1 DM1 A1 A1√ [4] Used to eliminate cos completely. Scheme for quadratic. Co. √ for 2nd value from incorrect sine. (A1√ not given for extra values in the range, but could be given if soln of quadratic led to 2 values of sine<1.) (b) cot = 1/tan used tan 2y = 4 2y = 1.326 → y = 0.66 or 2y = π + 1.326 or 2π + 1.326 M1 A1 Use of cot = 1/tan even if “2” removed incorrectly. Not for tan and 2y split. Co (must be radians) – not for 0.67. → y = 2.23 → y = 3.80 or 3.81 A1√ A1√ [4] For (i) + ½π For (i) + π or (ii) + ½π [S–1 for extra values in the range] [sc All answers in degrees B1.] 10 y = x3lnx (i) dy/dx = 3x2lnx + x3 (1/x) = 3x2lnx + x2 M1 A1 [2] M1 correct “uv”. A1 ok unsimplified. (ii) dy/dx = 0 lnx = –⅓ M1 A1 [2] Not DM – setting his dy/dx to 0 + attempt to solve. (ii) δy = dy/dx × δx = (e2 + 3e2)p = 4e2p or 29.6p M1 A1 [2] Use of small increases. Allow for use of dy/dt. ∆x = p essential for M mark. Alg expression with “p” ok for M1. (iii) d/dx (x3lnx) = x2 + 3x2 lnx M1 A1 ∫ is reverse of diff used. A1 needs ⅓x3 Integrating → x3lnx = ⅓x3 + ∫3x2lnxdx ∫x2lnxdx = ⅓(x3lnx – ⅓x3) A1 [3] co Integration by parts ok. M1 A1 A1. 11 4y = 3x + 1 and xy = 28x – 27y Sim equations. → x2 – 10x + 9 = 0 or y2 – 8y + 7 = 0 → (9, 7) [(1, 1) was given] M1 DM1 A1 Complete elimination of x or y Soln of quadratic (by scheme) Co for (9, 7) P(1, 1), Q(9, 7) → gradient of PQ = ¾ Gradient of perp bisector is –4/3 M (mid-point of PQ) = (5, 4) M1 M1 Use of m1m2 = –1 with his PQ Use of (½(x1 + x2), ½(y1 + y2)) Eqn of perp bis. y−4 = – 4/3(x – 5) 3y + 4x = 32 meets y = 4x at R(2, 8) Area of ∆PQ = ½ × PQ × MR = ½ × 10 × 5 (or matrix method) → 25 A1 M1 M1 A1 [9] Co – unsimplified ok. Simulataneous eqns. Must be with a perp Any correct method. Co.

Page 4 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 1 © University of Cambridge International Examinations 2005 12 (a) (i) EITHER N = 20 000e -0.05n n = 10, N = 12 130 or more places. B1 [1] Co (ii) 2000 = 20 000e -0.05n e -0.05n = 0.1 Take logs n = 45.1 → 2006 M1 M1 A1 [3] Isolating exponential – or taking logs to get 3 terms. Taking logs. Co. needs 2006, not 2005. (b) Put y = 3x 3x + 1 = 3y or 3x – 1 = ⅓y 3y – 2 = 8y/3 → y = 6 M1 A1 A1 A1 Used. For each expression. Co. 3x = 6, xlg3 = lg6 → x = 1.63 [or ÷ by 3x – 1 M1 A1 3x – 1 = 2 M1 A1] [or ÷ by 3x + 1 M1 A1 3x + 1 = 18 M1 A1] M1 A1 [6] Taking logs for his 3x. co. 12 OR y = e 2 x + 3e 2 x − (i) dy/dx = 2 2 e 2 3 e 2 1 x x − − = 0 when ex = 3 y = √3 + 3 ÷ √3 = 2√3 B1 B1 M1 A1 [4] Anywhere – Setting his dy/dx to 0 and reasonable attempt at making ex the subject. Co. Decimal check – A0. (ii) d2y/dx2 = 0 e 4 3 e 4 1 2 2 > + −x x , MIN M1 A1 [2] M1 Reasonable attempt by any method. A1 Correct deduction but needs second differential correct. (iii) ∫         + − 2 2 e 3 e x x dx = 2 2 2 e 6 e x x − − [ ] at 1 – [ ] at 0 = 4 + 2√e - 6√e = 3.66 M1 A1 M1 A1 [4] Knowing to integrate for area + any attempt with exponentials. A1 co. DM0 if [ ] at 0 is ignored. Co. DM1 for quadratic equation. Formula Equation must be set to 0. Formula must be correct and correctly used, but allow for numerical and algebraic errors. Brackets Equation must be set to 0. Must be an attempt to get two linear brackets. Each bracket must then be equated to 0 and solved.