Cambridge IGCSE Mathematics - Additional 0606 — 2005 May/June Paper 1 · Variant 1

This document consists of 5 printed pages and 3 blank pages. SP (NF/KN) S87030 © UCLES 2005 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/01 Paper 1 May/June 2005 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

2 0606/01/M/J/05 © UCLES 2005 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, . Binomial Theorem (a + b)n = an + an–1 b + an–2 b2 + … + an–r br + … + 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. 1 2 c sin C b sin B a sin A n! (n – r)!r! ) n r( ) n r( ) n 2( ) n 1( x b b ac a = −± − 2 4 2

3 0606/01/M/J/05 © UCLES 2005 [Turn over 1 Given that A =
 , find (A2)–1. [4] 2 A student has a collection of 9 CDs, of which 4 are by the Beatles, 3 are by Abba and 2 are by the Rolling Stones. She selects 4 of the CDs from her collection. Calculate the number of ways in which she can make her selection if (i) her selection must contain her favourite Beatles CD, [2] (ii) her selection must contain 2 CDs by one group and 2 CDs by another. [3] 3 Given that θ is acute and that sinθ = , express, without using a calculator, in the form a + , where a and b are integers. [5] 4 The position vectors of points A and B relative to an origin O are –3i – j and i + 2j respectively. The point C lies on AB and is such that AC →= AB →. Find the position vector of C and show that it is a unit vector. [6] 5 The function f is defined, for 0°
x
180°, by f(x) = A + 5cos Bx, where A and B are constants. (i) Given that the maximum value of f is 3, state the value of A. [1] (ii) State the amplitude of f. [1] (iii) Given that the period of f is 120°, state the value of B. [1] (iv) Sketch the graph of f. [3] 6 Given that each of the following functions is defined for the domain –2
x
3, find the range of (i) f : x  2 – 3x, [1] (ii) g : x  2 – 3x , [2] (iii) h : x  2 –  3x . [2] State which of the functions f, g and h has an inverse. [2] 3–5 b sinθ ––––––––– cosθ – sinθ 1 3 2 1 –1 1

4 0606/01/M/J/05 © UCLES 2005 7 (a) Variables l and t are related by the equation l = l0(1 + α)t where l0 and α are constants. Given that l0 = 0.64 and α = 2.5 × 10–3, find the value of t for which l = 0.66. [3] (b) Solve the equation 1 + lg(8 – x) = lg(3x + 2). [4] 8 The table above shows experimental values of the variables x and y which are related by an equation of the form y = kxn, where k and n are constants. (i) Using graph paper, draw the graph of lg y against lgx. [3] (ii) Use your graph to estimate the value of k and of n. [4] 9 (i) Determine the set of values of k for which the equation x2 + 2x + k = 3kx – 1 has no real roots. [5] (ii) Hence state, giving a reason, what can be deduced about the curve y = (x + 1)2 and the line y = 3x – 1. [2] 10 The remainder when 2x3 + 2x2 – 13x + 12 is divided by x + a is three times the remainder when it is divided by x – a. (i) Show that 2a3 + a2 – 13a + 6 = 0. [3] (ii) Solve this equation completely. [5] 11 A particle travels in a straight line so that, t seconds after passing a fixed point A on the line, its acceleration, a ms–2, is given by a = –2 – 2t. It comes to rest at a point B when t = 4. (i) Find the velocity of the particle at A. [4] (ii) Find the distance AB. [3] (iii) Sketch the velocity-time graph for the motion from A to B. [1] x 10 100 1000 10 000 y 1900 250 31 4

5 0606/01/M/J/05 © UCLES 2005 12 Answer only one of the following two alternatives. EITHER The diagram, which is not drawn to scale, shows part of the graph of y = 8 – e2x, crossing the y-axis at A. The tangent to the curve at A crosses the x-axis at B. Find the area of the shaded region bounded by the curve, the tangent and the x-axis. [10] OR A piece of wire, of length 2m, is divided into two pieces. One piece is bent to form a square of side xm and the other is bent to form a circle of radius rm. (i) Express r in terms of x and show that the total area, Am2, of the two shapes is given by A = . [4] Given that x can vary, find (ii) the stationary value of A, [4] (iii) the nature of this stationary value. [2] (π + 4)x2 – 4x + 1 ––––––––––––––– π y x O A B

8 0606/01/M/J/05 BLANK PAGE 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 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 June 2005 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 June 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Grade thresholds taken for Syllabus 0606 (Additional Mathematics) in the June 2005 examination. minimum mark required for grade: maximum mark available A C E Component 1 80 59 32 21 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.

JUNE 2005 IGCSE MARK SCHEME MAXIMUM MARK: 80 SYLLABUS/COMPONENT: 0606/01 ADDITIONAL MATHEMATICS (Paper 1)

Page 1 Mark Scheme Syllabus Paper IGCSE – JUNE 2005 0606 1 © University of Cambridge International Examinations 2005 1       − =       − = 0 3 3 3 1 1 1 2 2 2 A (A²)–1 =       − 3 3 3 0 9 1 or A–1 first B1 B1 followed by squaring B1√B1√ B2,1 B1√,B1√ [4] One off for each error. B1√ for 1÷9. B1√ for rest. √ from his attempt at A². If       1 1 1 4 used, could get last 2 marks. 2 9 CDs →4 Beatles, 3 Abba, 2 Rolling (i) 8C3 = (8×7×6)÷(3×2×1) = 56 (ii) 2B 2A 4C2×3C2 = 18 2B 2R 4C2×1 = 6 2A 2R 3C2×1 = 3 → Total of 27 M1 A1 [2] M1 M1 A1 [3] 2 if correct without working 9C3 M0. 4×8C3 gets M1 A0 One correct product with nCrs 3 products added – even if nPr CAO 3 θ θ 2 sin 1 cos − = = 3 2 c s s − = 3 1 3 2 3 1 − = 1 2 1 − × top and bottom by )1 2 ( + → 1 + 2 M1 A1 M1 M1 A1 [5] Use of s² + c² = 1 to obtain cos as a surd – or correctly from 90o triangle. Correct algebra – getting rid of √3 Correct technique used to rationalise the denominator. This form ok. No need for a =, b = . (decimals get no credit) 4 OA =       − − 1 3 , OB =       2 1 , AB =       3 4 AC = 5 3 AB =         5 9 5 12 OC =OA + AC =       − − 1 3 +         5 9 5 12 =        − 5 4 5 3 OC = ( ) 25 16 25 9 + = 1 M1 A1 M1 A1 M1 A1 [6] Use of b – a or a – b – not for a + b CAO – not for negative of this. Could be implied by correct OC . Any correct method ok. CAO Correct method on his OC. Answer was given.

Page 2 Mark Scheme Syllabus Paper IGCSE – JUNE 2005 0606 1 © University of Cambridge International Examinations 2005 5 f(x) = A + 5cosBx (i) A = −2 (ii) Amplitude = 5 (iii) B = 3 (iv) Range 3 to –7 B1 B1 B1 B1 B2,1 [6] CAO CAO CAO −3 to 7 implied somewhere – table ok – even if no graph Needs 1½ oscillations – over-rides rest. √ on 3 and –7 Start at max – finishes at second min. Curves – but be tolerant 6 (i) −7 ≤ f(x) ≤ 8 (i) 0 ≤ g(x) ≤ 8 (ii) −7 ≤ h(x) ≤ 2 f yes g no h no B1 B1 B1 B1 B1 B2,1 [7] CAO Allow < for ≤ CAO As above CAO As above Loses one for each wrong decision. (answer f on its own – allow B2) 7 (a) t l l ) 1 ( 0 α + = Subs and divides 1.031 = 1.0025t t = lg 1.031÷ lg 1.0025 = 12.3 (b) 1 = log 10 LHS = lg 10(8 − x) 80 − 10x = 3x +2 → x = 6 M1 M1 A1 [3] B1 M1 M1 A1 [4] Sub + division before taking logs. (or lgl = lgl0 + tlg(1+α) + use) Taking logs. CAO to 3 sf or more. Anywhere in the question. Putting any 2 logs together Complete elimination of 3 logs CAO 8 lgx 1 2 3 4 lgy 3.28 2.40 1.49 0.60 (i) Knows what to do. Pts within ½ square. (ii) Gradient = ±n n = −0.88 to − 0.92 log k = y-intercept k = 14 000 to 16 000 M1 A2,1 [3] B1 A1 B1 A1 [4] For part (ii) – use of sim eqns is ok if points used are on line, not from table. Knows what to do. Accuracy within ½ square. B1 even if just stated without graph. B1 even if just stated without graph.

Page 3 Mark Scheme Syllabus Paper IGCSE – JUNE 2005 0606 1 © University of Cambridge International Examinations 2005 9 (i) x² + 2x + k = 3kx − 1 → x² + (2 – 3k)x + (k + 1) = 0 Uses b² − 4ac =, > or < 0 → 9k² − 16k End-points of 0 and 16/9 Use of b² − 4ac < 0 Solution set 0 < k < 16/9 (ii) Same case with k = 1 No intersection since k inside the range Special case. Solves simultaneous. eqns → √−7. B1 M1 A1 DM1 M1 A1 [5] B1 B1√ [2] Any use of b² − 4ac This quadratic only. Solution of this quadratic →2 values Definite recognition of − ve. CAO NB No intersection on its own without k = 1 gets no credit. 10 (i) x = –a → 12 13 2 2 2 3 + + + − a a a x = a → 12 13 2 2 2 3 + − + a a a 12 13 2 2 2 3 + + + − a a a = 3( 12 13 2 2 2 3 + − + a a a ) 2a³ + a² – 13a + 6 = 0 (ii) Tries a = 2 : fits ok. ( or −3, ½) ÷ (x – 2) → 2a² + 5a – 3 Solution → a = –3 and ½ If factors left as final answer, loses the last 2 marks. M1 M1 A1 [3] M1A1 M1 M1 A1 [5] For either of these – ignore simple algebraic and numeric slips Allow M1 if 3 wrong side. Answer given. Tries a search for first value Must be (x – ) for M. CAO for A mark. CAO for both. T & I : M1 A1 for first value, A1 for second value, A2 for third. 11 a = −2 − 2t (i) v = −2t − t² (+ c) v = 0 when t = 4 → c = 24 if t = 0, v = 24 ms–1 (ii) s = −t² −t³/3…+…(24t)… Put t = 4 → 3 2 58 m (iii) M1 A1 DM1 A1 [4] M1A1√ A1 [3] B1 [1] T Attempt at ∫. Ignore omission of c Attempt at c. CAO Attempt at ∫. “24t” not needed. CAO Curve necessary.

Page 4 Mark Scheme Syllabus Paper IGCSE – JUNE 2005 0606 1 © University of Cambridge International Examinations 2005 12 EITHER y = 8 − e–2x Tangent crosses y-axis at (3½, 0) y = 0, x = ½ln8 or 1.04 Area of triangle = ½×3.5×7 = 12.25 ∫ curve = [8x − ½e2x] From 0 to his “x” [4ln8 − 4] − [0 − 0.5] 12.25 − (4ln8 − 3.5) = 7.43 M1 A1 M1 A1 [4] B1 M1 M1 A1 DM1 A1 [6] For differential. CAO for gradient of −2. Any method ok providing calculus used. Numeric gradient for M1. Anywhere in the question. Even if no integration later. Attempt at ∫. CAO DM0 if value at 0 assumed to be 0. CAO 12 OR (i) Perimeter of square + circumference = 2 m → 4x + 2π r = 2 → r = π x 2 1− → A = x² + 2 2 1      − π π x → π π 1 4 ) 4 ( 2 + − + = x x A (ii) ) 4 8 2 ( 1 dx dA − + = x x π π = 0 when x = 8 2 4 + π = 0.28 m A = 0.14 (iii) ) 8 2 ( 1 dx d 2 2 + = π π A +ve → MIN M1 A1 M1 A1 [4] M1 A1 DM1 A1 [4] M1 A1 [2] Allow for π d or π r and for 2x or 4x CAO – in any form Needs π r² and l² (both) CAO – answer given Attempt at diff. A0 if π missing, but can then gain rest of marks. Sets his differential to 0. CAO – 2 sig figures sufficient. Any valid method ok. Needs correct algebraic 2 2 dx d A for A mark. DM1 for quadratic equation. Equation must be set to 0 if using formula or factors. Formula Factors Must be correct Must attempt to put quadratic into 2 factors – ignore arithmetic and algebraic slips. Each factor then equated to 0. x = 0, y = 7 dy/dx = −2e2x At = x = 0, m = −2