Cambridge IGCSE Mathematics - Additional 0606 — 2007 Oct/Nov Paper 1 · Variant 1

This document consists of 5 printed pages and 3 blank pages. SP (SC) T44477/2 © UCLES 2007 [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. * 1 4 5 2 3 2 2 2 0 0 * ADDITIONAL MATHEMATICS 0606/01 Paper 1 October/November 2007 2 hours Additional Materials: Answer Booklet/Paper Graph paper (1 sheet) Electronic calculator Mathematical tables www.XtremePapers.com

2 0606/01/O/N/07 © UCLES 2007 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/07 © UCLES 2007 [Turn over 1 Given that A = ( 2 –1 3 1), find the value of each of the constants m and n for which A2 + mA = nI, where I is the identity matrix. [4] 2 Show that 1 ––––––– 1 – cosθ – 1 ––––––– 1 + cosθ ≡ 2cosec θ cot θ. [4] 3 Given that p = 3 + 1 3 – 1 , express in its simplest surd form, (i) p, [3] (ii) p – 1–p. [2] 4 A badminton team of 4 men and 4 women is to be selected from 9 men and 6 women. (i) Find the total number of ways in which the team can be selected if there are no restrictions on the selection. [3] Two of the men are twins. (ii) Find the number of ways in which the team can be selected if exactly one of the twins is in the team. [3] 5 In this question, i is a unit vector due east, and j is a unit vector due north. A plane flies from P to Q where → PQ = (960i + 400j) km. A constant wind is blowing with velocity (–60i + 60j) km h–1. Given that the plane takes 4 hours to travel from P to Q, find (i) the velocity, in still air, of the plane, giving your answer in the form (ai + bj) km h–1, [4] (ii) the bearing, to the nearest degree, on which the plane must be directed. [2] 6 A curve is such that dy dx = 6 4x + 1 , and (6, 20) is a point on the curve. (i) Find the equation of the curve. [4] A line with gradient – 1 2 is a normal to the curve. (ii) Find the coordinates of the points at which this normal meets the coordinate axes. [4]

4 0606/01/O/N/07 © UCLES 2007 7 (i) Use the substitution u = 2x to solve the equation 22x = 2x + 2 + 5. [5] (ii) Solve the equation 2log93 + log5(7y – 3) = log28. [4] 8 (a) The remainder when the expression x3 – 11x2 + kx – 30 is divided by x – 1 is 4 times the remainder when this expression is divided by x – 2. Find the value of the constant k. [4] (b) Solve the equation x3 – 4x2 – 8x + 8 = 0, expressing non-integer solutions in the form a ± b, where a and b are integers. [5] 9 x 2 4 6 8 10 y 14.4 10.8 11.2 12.6 14.4 The table shows experimental values of two variables, x and y. (i) Using graph paper, plot xy against x2. [2] (ii) Use the graph of xy against x2 to express y in terms of x. [4] (iii) Find the value of y for which y = 83 ––x . [3] 10 A D B C 0.8 rad 10 cm 10 cm The diagram shows a sector ABC of the circle, centre A and radius 10 cm, in which angle BAC = 0.8 radians. The arc CD of a circle has centre B and the point D lies on AB. (i) Show that the length of the straight line BC is 7.79 cm, correct to 2 decimal places. [2] (ii) Find the perimeter of the shaded region. [4] (iii) Find the area of the shaded region. [4]

5 0606/01/O/N/07 © UCLES 2007 11 Answer only one of the following two alternatives. EITHER A curve has the equation y = xe2x. (i) Obtain expressions for dy dx and d2y dx2 . [5] (ii) Show that the y-coordinate of the stationary point of the curve is – 1 2e . [3] (iii) Determine the nature of this stationary point. [2] OR (i) Show that d dx ( ln x ––– x2 ) = 1 – 2 lnx ––––––– x3 . [3] (ii) Show that the y-coordinate of the stationary point of the curve y = ln x ––– x2 is 1–– 2e . [3] (iii) Use the result from part (i) to find ∫ ( ln x ––– x3 ) dx. [4]

8 0606/01/O/N/07 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 2007 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 2007 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 2007 0606 1 © UCLES 2007 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 2007 0606 1 © UCLES 2007 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 2007 0606 1 © UCLES 2007 1 A=       − 1 3 1 2 , A² =       − − 2 9 3 1       − − 2 9 3 1 + m       − 1 3 1 2 = n       1 0 0 1 1 + 2m = n and −3−m=0 → m = −3 and n = −5 M1 A1 M1 A1 [4] Reasonable attempt (needs 2 correct) All correct Identity matrix must be correct Equating their elements once co. 2 θ θ θ θ cot ec cos 2 cos 1 1 cos 1 1 ≡       + − − Manipulation of fractions (1−c)(1+c) = s² used θ θ 2 sin cos 2 ⇒ 2cosecθcotθ M1 B1 M1 (1−c)(1+c) in denominator +reasonable attempt at numerator (ignore signs) Knowledge of cot and cosec All correct A1 [4] When all correct a.g Beware fortuitous answers. 3 (i) 1 3 1 3 − + = p → 1 3 1 3 − + = p × 1 3 1 3 + + → 1 3 1 3 2 3 − + + = p = 2 + √3 (ii) either p p 1 − = 2 + √3 − 3 2 1 + or p p 1 − = p p 1 2 − → 2√3 M1 A1 A1 [3] M1 A1 [2] × top and bottom by √3 + 1 Denominator = 2 co Complete method. co. 4 (i) 4 men from 9 = 9C4 (126) 4 women from 6 = 6C4 (15) Multiply together → 1890 (ii) One twin included (7C3×6C4) To include other twin ×2 → 1050 B1 M1 A1 [3] M1 DM1 A1 [3] For either 9C4 or 6C4 Product of 2 nCrs. co For 2 nCrs. Two times his first answer. co 5 (i) Resultant vel = (960i +400j) ÷ 4 → (240i +100j) v (still air) = (240i +100j) − wind = 300i + 40j (ii) tanθ = 40 ÷ 300 ( → 7.6º) → Bearing of 082º (awrt 82º) M1 A1 M1 A1 [4] M1 A1 [2] Division of distance by time co (could be wind × 4) then ÷ 4 later Needs subtraction co Use of tan with their 2 components Not 960i +400j

Page 5 Mark Scheme Syllabus Paper IGCSE – October/November 2007 0606 1 © UCLES 2007 6 (i) 1 4 6 d d + = x x y y = 2 1 2 1 )1 4 ( 6 + x ÷ 4 ( +c) Uses (6,20) → c = 5 ( 5 1 4 3 + + = x y ) B1 B1 M1 A1 [4] For an expression involving( ) 1 2 4 1 x + For all correct Uses (6,20) in an integration involving ( ) 1 4 1 k x + , 1 2 k ≠− co (do not mark after +5) (ii) Perp to −½ = 2 M1 Use of m1m2=−1, with attempt to solve co on y-value, using x = 2 1 4 6 + x = 2 → x = 2, y = 14 A1√ Eqn → y – 14 = −½(x − 2) or 2y+x=30 → (0, 15) and (30, 0) M1 A1 [4] Correct method for line co 7 (i) 5 2 2 2 2 + = + x x x 2 2 = u² 2 2 + x = 4u Solution of quadratic u²=4u+5 2x = 5 → x = lg5 ÷ lg2 → x = 2.32 (ii) 2log93 + log5(7y − 3) = log28. 2×½ + ….. = 3 log5(7y − 3) = 2 (7y − 3) = 25 → y = 4 B1 B1 M1 M1 A1 [5] B1 B1 M1 A1 [4] co co Correct method of solution of quad=0 From 2x = k to x by correct method co – loses if more than one answer given. For ½ For RHS = 3 From log5 to 5p = k. co 8 (a) f(1) = 1−11+k-30 → k − 40 f(2) = 8−44+2k-30 → 2k − 66 f(1) = 4f(2) → k = 32 (b) 0 8 8 4 2 3 = + − − x x x Tries for a first solution → x = −2 Divides by (x − his first solution) → x² − 6x + 4 = 0 → x = 2 20 6 ± → 3 ± √5 M1 A1 M1 A1 [4] M1 A1 M1 DM1 A1 [5] Uses either x = 1 or 2, not −1 or −2. Both correct, unsimplified. Linked + solution – allow if 4 on LHS co Search shown for M,x = −2 gets M1A1. Correct method. Correct method for soln of quadratic Must be simplified.

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2007 0606 1 © UCLES 2007 9 (i) x 2 4 6 8 10 y 14.4 10.8 11.2 12.6 14.4 xy 29 43 67 101 144 x² 4 16 36 64 100 (ii) Gradient 1.2 (±0.1) ‘y’ intercept ( ) 2 ± M1 A1 [2] B1 B1 Knows what to do. Mark from graph – 5 points are in line. co co → x x y 24 2.1 + = M1 A1 [4] xy = (their grad)x + (their intercept) (iii) From graph xy = 83 → x² = 49 Valid method to obtain y y = 11.6 – 12.2 M1 M1 A1 [3] Reads on vertical axis at 83 Valid method to obtain y co 10 (i) BC = 2(10sin0.4) = 7.79 (ii) ∠ABC = ½(π−0.8) = 1.17 rads Arc CD = 7.79 × 1.17, Arc BC = 10×0.8 → P = sum of the arcs + BD (=7.79) → P = 24.9 M1 A1 [2] B1 M1 M1 A1 [4] Any correct method – cos rule ok. Anywhere in the question. Use of s=rθ in either arc. Overall plan – arc CD + arc BC + BD co. (iii) Area sector BDC = ½(7.79)²×1.17 M1 Use of A=½r²θ for sector BDC Area segment on BC =½.10²(0.8−sin0.8) B1 B1 for ( ) 2 0.5 10 0.8 B1 B1 for ( ) 2 0.5 10 sin 0.8 → Shaded area = 39.6 or 39.7 A1 [4] co 11 EITHER (i) x x y 2 e = d/dx(e2x) = 2e2x → dy/dx = e2x + 2x e2x → d²y/dx² = 2e2x + 2e2x + 4xe2x (ii) dy/dx = 0 when 1+2x = 0 → x = −½ → y = −½e-1 = e 2 1 − . (iii) If x = −½ → +ve result → Minimum (or gradient goes −,0,+ ) (or y value to left or right of (−½ )> 1 2e − ) B1 M1A1 M1A1 [5] M1 A1 A1 [3] M1 A1 [2] Anywhere – even if product not used Use of correct formula for “uv”. co Use of product formula again. co. Sets his dy/dx to 0 and tries to solve. co – ag – beware fortuitous results. Looks at sign. Correct deduction from correct x. (or by any other valid method)

Page 7 Mark Scheme Syllabus Paper IGCSE – October/November 2007 0606 1 © UCLES 2007 11 OR (i) d/dx(lnx) = 1/x       2 ln d d x x x = 4 ln 2 x x x x − = 3 ln 2 1 x x − B1 M1 A1 [3] Anywhere – even if quotient not used Use of correct quotient formula co (ii) dy/dx = 0 → lnx = ½ → x =√e M1 A1 Sets his dy/dx to 0 and tries to solve. → y = ln(√e)÷e = e 2 1 . A1 co – ag – beware fortuitous results. [3] (iii) 2 ln x x =∫       x x d 1 3 -∫ x x x d ln 2 3 ∫ x x x d ln 3 = ½ × [∫ −       2 3 ln d 1 x x x x ] → =       − − − 2 2 2 1 ln 2 x x x + c M1 B1 B1 A1 [4] Recognition that integration is the reverse of differentiation. B1 for ½. B1 for (x –2) ÷ (−2) All ok including +c. 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.