Cambridge IGCSE Mathematics - Additional 0606 — 2008 Oct/Nov Paper 1 · Variant 1

This document consists of 5 printed pages and 3 blank pages. SPA (CM/DT) T56700 © UCLES 2008 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education 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. 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. 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. * 9 6 0 6 9 4 1 3 0 0 * ADDITIONAL MATHEMATICS 0606/01 Paper 1 October/November 2008 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Mathematical tables www.XtremePapers.com

2 0606/01/O/N/08 © UCLES 2008 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 0606/01/O/N/08 © UCLES 2008 [Turn over 1 B A C
(i) Copy the Venn diagram above and shade the region that represents A ∪ (B ∩ C). [1] (ii) Copy the Venn diagram above and shade the region that represents A ∩ (B ∪ C). [1] (iii) Copy the Venn diagram above and shade the region that represents (A ∪ B ∪ C)
. [1] 2 Find the set of values of x for which (2x + 1)2 > 8x + 9. [4] 3 Prove that sin A 1 + cos A + 1 + cos A sin A ≡ 2 cosec A. [4] 4 A function f is such that f(x) = ax3 + bx2 + 3x + 4. When f(x) is divided by x – 1, the remainder is 3. When f(x) is divided by 2x + 1, the remainder is 6. Find the value of a and of b. [5] 5 Given that a = 5i – 12j and that b = pi + j, find (i) the unit vector in the direction of a, [2] (ii) the values of the constants p and q such that qa + b = 19i – 23j. [3] 6 (i) Solve the equation 2t = 9 + 5 t . [3] (ii) Hence, or otherwise, solve the equation 2x 1 2 = 9 + 5x– 1 2. [3] 7 (i) Express 4x2 – 12x + 3 in the form (ax + b)2 + c, where a, b and c are constants and a > 0. [3] (ii) Hence, or otherwise, find the coordinates of the stationary point of the curve y = 4x2 – 12x + 3. [2] (iii) Given that f(x) = 4x2 – 12x + 3, write down the range of f. [1]

4 0606/01/O/N/08 © UCLES 2008 8 A curve is such that d2y dx2 = 4e–2x. Given that dy dx = 3 when x = 0 and that the curve passes through the point (2, e–4), find the equation of the curve. [6] 9 (i) Find, in ascending powers of x, the first 3 terms in the expansion of (2 – 3x)5. [3] The first 3 terms in the expansion of (a + bx)(2 – 3x)5 in ascending powers of x are 64 – 192x + cx2. (ii) Find the value of a, of b and of c. [5] 10 (a) Functions f and g are defined, for x 
, by f(x) = 3 – x, g(x) = x x + 2 , where x ≠ –2. (i) Find fg(x). [2] (ii) Hence find the value of x for which fg(x) = 10. [2] (b) A function h is defined, for x 
, by h(x) = 4 + ln x, where x > 1. (i) Find the range of h. [1] (ii) Find the value of h–1(9). [2] (iii) On the same axes, sketch the graphs of y = h(x) and y = h–1(x). [3] 11 Solve the equation (i) tan 2x – 3 cot 2x = 0, for 0° < x < 180°, [4] (ii) cosec y = 1 – 2cot2 y, for 0°  y  360°, [5] (iii) sec(z + π 2) = –2, for 0 < z < π radians. [3]

5 0606/01/O/N/08 © UCLES 2008 12 Answer only one of the following two alternatives. EITHER A curve has equation y = x2 x + 1 . (i) Find the coordinates of the stationary points of the curve. [5] The normal to the curve at the point where x = 1 meets the x-axis at M. The tangent to the curve at the point where x = –2 meets the y-axis at N. (ii) Find the area of the triangle MNO, where O is the origin. [6] OR A curve has equation y = ex – 2 – 2x + 6. (i) Find the coordinates of the stationary point of the curve and determine the nature of the stationary point. [6] The area of the region enclosed by the curve, the positive x-axis, the positive y-axis and the line x = 3 is k + e – e–2. (ii) Find the value of k. [5]

8 0606/01/O/N/08 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 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.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2008 question paper 0606 ADDITIONAL MATHEMATICS 0606/01 Paper 1, 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. 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 discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 01 © UCLES 2008 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 2008 0606 01 © UCLES 2008 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) 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.

Page 4 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 01 © UCLES 2008 1 (i) correct diagram (ii) correct diagram (iii) correct diagram B1 B1 B1 [3] 2 (2x + 1)2 > 8x + 9 4x2 – 4x – 8 > 0 x2 – x – 2 > 0 (x + 1)(x – 2) > 0 Leads to critical values x = –1,2 x < –1 and x > 2 M1 DM1 A1 √A1 [4] M1 for simplification to 3 term quadratic DM1 for factorisation A1 for critical values Follow through on their critical values. 3 LHS = A A A A A sin ) cos 1( cos 2 cos 1 sin 2 2 + + + + = A A A sin ) cos 1( cos 2 2 + + = A sin 2 leading to 2cos ecA M1 A1 M1 A1 [4] M1 for attempt to deal with fractions and attempt to obtain numerator A1 correct M1 for use of sin2 A + cos2 A = 1 4 Substitution of x = 1 leading to a + b + 4 = 0 Substitution of 2 1 − = x leading to –a +2b – 28 = 0 Leading to a = –12, b = 8 M1 M1 A1 M1 A1 [5] M1 for substitution of x = 1 and equated to 3 M1 for substitution of x = 2 1 − and equated to 6 A1 for both correct M1 for solution A1 for both 5 (i) a = 13 1 (5i – 12j) (ii) q(5i –12j) + pi + j = 19i – 23j 5q + p = 19 –12q + 1 = –23 Leading to q = 2, p = 9 M1, A1 [2] M1 M1 A1 [3] M1 for a valid attempt to obtain magnitude. M1 for equating like vectors M1 for solution of (simultaneous) equations A1 for both 6 (i) 2t 2 – 9t – 5 = 0 (2t + 1)(t – 5) = 0 2 1 = t , t = 5 (ii) 5 0.5, 2 1 − = x x = 0.25, 25 M1 DM1 A1 [3] M1 A1,A1 [3] M1 for attempting to form a quadratic in t DM1 for attempt to solve a 3 term quadratic A1 for both M1 for realising that x0.5 is equivalent to t (or valid attempt at solution)

Page 5 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 01 © UCLES 2008 7 (i) y = 4x2 – 12x + 3 y = (2x – 3)2 – 6 (ii)     −6 , 2 3 (iii) f ≥ –6 B1 B1 B1 [3] √B1, √B1 [2] √B1 [1] B1 for 2 (part of linear factor) B1 for –3 (part of linear factor) B1 for –6 Follow through on their a, b and c Allow calculus method. Follow through on their c 8 ) ( e 2 d d 2 c x y x + − = − When 5 ,0 ,3 d d = ∴ = = 1c x x y 5 e 2 d d 2 + − = −x x y y = e–2x + 5x(+c2) When x = 2, y = e–4 ∴c2 = –10 y = e–2x +5x – 10 B1 M1 A1 B1 M1 √A1 [6] B1 for –2e–2x M1 for attempt to find c1 B1 for –2e–2x M1 for attempt to find c2 √ –2 times their c1 9 (i) 25 + 5C124(–3x) + 5C223(–3x)2 32 – 240x + 720x2 (ii) 32a = 64, a = 2 32b – 240a = –192, b = 9 –240b + 720a = c c = –720 B1 B1 B1 [3] B1 M1 A1 M1 A1 [5] B1 for 32 or 25 B1 for –240 B1 for 720. B1 for a = 2 M1 for equation in a and b equated to ±192 A1 for b = 9 M1 for equation in a and b equated to c A1 for c = –720 10 (a) (i) fg(x) = f       + 2 x x = 3 – 2 + x x (ii) 3 – 2 + x x = 10 leading to x = –1.75 (b) (i) h(x) > 4 (ii) h–1(x) = ex – 4 h–1(9) = e5 (≈ 148) or 4 + lnx = 9, leading to 5e x = (iii) correct graphs M1 A1 [2] DM1 A1 [2] B1 [1] M1 A1 [2] B1 B1 B1 [3] M1 for order DM1 for dealing with fractions sensibly M1 for attempting to obtain inverse function B1 for each curve B1 for idea of symmetry

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 01 © UCLES 2008 11 (i) tan22x = 3 tan2x = (±) 3 2x = 60°, 120°, 240°, 300° x = 30°, 60°, 120°, 150° (ii) 2cosec2 y + cosec y –3 = 0 (2cosec y + 3)(cosec y – 1) = 0 cosec y = – 2 3 , 1 sin y = – 3 2 , 1 y = 221.8°, 318.2°, y = 90° (iii) cos       + 2 π z = – 2 1 3 4 , 3 2 2 π π π = + z 6 6 5 , π π = z , allow 0.52, 2.62 rads M1 DM1 A1, A1 [4] M1, A1 M1 A1, A1 [5] M1 A1,A1 [3] M1 for an equation in tan22x M1 for attempt to solve using 2x correctly A1 for any pair M1 for correct use of identity or other valid method A1 for a correct quadratic M1 for solution of quadratic and attempt to solve correctly A1 for 221.8°, 318.2°, A1 for 90° M1 for dealing with sec and order of operations A1 for each 12 EITHER (i) 2 2 )1 ( 2 )1 ( d d + − + = x x x x x y 2)1 ( ) 2 ( + + = x x x 0 d d = x y , x = 0, –2 y = 0, –4 (ii) gradient of normal = 3 4 − normal y = 3 4 − x + 6 11 , leads to M (1.375,0) N (0, –4) Area = 2.75 M1 A1 DM1 A1,A1 [5] M1 A1 √ B1 B1 M1 √ A1 [6] M1 for attempt to differentiate a quotient A1 correct allow unsimplified DM1 for equating to zero and an attempt to solve A1 for each pair (could be x = 0 and x = –2) M1 for attempt to obtain gradient of the normal A1 for a correct (unsimplified) normal equation Follow through on their normal B1 for N M1 for attempt to get area of triangle Ft on their M and N (must be on axes)

Page 7 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 01 © UCLES 2008 12 OR (i) 2 2 e d d − − = x x y 2 2 e 0, d d = − = x x y x = 2 + ln2 (2.69) y = 4 – 2ln2 (2.61) 2 2 2 e − = x dx y d , always +ve ∴min (ii) [ ] ∫ + = + − − − − 3 0 3 0 2 2 2 6 e d ) 6 2 e( x x x x x x = (e – 9 + 18) – (e–2) = e – e–2 + 9 k = 9 B1 B1 M1 A1 A1 B1 [6] M1, A1 M1 A1 B1 [5] B1 for 2 ex− B1 for –2 only M1 for equating to zero and attempt to solve A1 for x A1 for y B1 for conclusion from a valid method M1 for attempt to integrate M1 for correctly applying limits A1 for e – e–2 B1 for k