Cambridge A Level Mathematics 9709 — 2004 Oct/Nov Paper 1 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/01 Paper 1 Pure Mathematics 1 (P1) October/November 2004 1 hour 45 minutes Additional materials: Answer Booklet/Paper Graph paper List of Formulae (MF9) 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. 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 75. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. This document consists of 3 printed pages and 1 blank page. © UCLES 2004 [Turn over

2 1 Find the coefficient of x in the expansion of
3x −2 x 5 . [4] 2 Find (i) the sum of the first ten terms of the geometric progression 81, 54, 36, . . . , [3] (ii) the sum of all the terms in the arithmetic progression 180, 175, 170, . . . , 25. [3] 3 In the diagram, AC is an arc of a circle, centre O and radius 6 cm. The line BC is perpendicular to OC and OAB is a straight line. Angle AOC = 1 3π radians. Find the area of the shaded region, giving your answer in terms of π and √3. [5] 4 (i) Sketch and label, on the same diagram, the graphs of y = 2 sin x and y = cos 2x, for the interval 0 ≤x ≤π. [4] (ii) Hence state the number of solutions of the equation 2 sin x = cos 2x in the interval 0 ≤x ≤π. [1] 5 The equation of a curve is y = x2 −4x + 7 and the equation of a line is y + 3x = 9. The curve and the line intersect at the points A and B. (i) The mid-point of AB is M. Show that the coordinates of M are 1 2, 71 2. [4] (ii) Find the coordinates of the point Q on the curve at which the tangent is parallel to the line y + 3x = 9. [3] (iii) Find the distance MQ. [1] 6 The function f : x →5 sin2x + 3 cos2x is defined for the domain 0 ≤x ≤π. (i) Express f(x) in the form a + b sin2x, stating the values of a and b. [2] (ii) Hence find the values of x for which f(x) = 7 sin x. [3] (iii) State the range of f. [2] 9709/01/O/N/04

3 7 A curve is such that dy dx = 6 √(4x −3) and P (3, 3) is a point on the curve. (i) Find the equation of the normal to the curve at P, giving your answer in the form ax + by = c. [3] (ii) Find the equation of the curve. [4] 8 The points A and B have position vectors i + 7j + 2k and −5i + 5j + 6k respectively, relative to an origin O. (i) Use a scalar product to calculate angle AOB, giving your answer in radians correct to 3 significant figures. [4] (ii) The point C is such that −−→ AB = 2−−→ BC. Find the unit vector in the direction of −−→ OC. [4] 9 The function f : x →2x −a, where a is a constant, is defined for all real x. (i) In the case where a = 3, solve the equation ff(x) = 11. [3] The function g : x →x2 −6x is defined for all real x. (ii) Find the value of a for which the equation f(x) = g(x) has exactly one real solution. [3] The function h : x →x2 −6x is defined for the domain x ≥3. (iii) Express x2 −6x in the form (x −p)2 −q, where p and q are constants. [2] (iv) Find an expression for h−1(x) and state the domain of h−1. [4] 10 A curve has equation y = x2 + 2 x. (i) Write down expressions for dy dx and d2y dx2 . [3] (ii) Find the coordinates of the stationary point on the curve and determine its nature. [4] (iii) Find the volume of the solid formed when the region enclosed by the curve, the x-axis and the lines x = 1 and x = 2 is rotated completely about the x-axis. [6] 9709/01/O/N/04

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary and Advanced Level MARK SCHEME for the November 2004 question papers 9709 MATHEMATICS 9709/01 – Paper 1, maximum raw mark 75 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.

Grade thresholds taken for Syllabus 9709 (Mathematics) in the November 2004 examination. minimum mark required for grade: maximum mark available A B E Component 1 75 65 58 33 The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.

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 marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of

10. The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) 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) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) 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. An MR-2 penalty may be applied in particular cases if agreed at the coordination meeting. PA -1 This is deducted from A or B marks in the case of premature approximation. The PA -1 penalty is usually discussed at the meeting.

November 2004 GCE A AND AS LEVEL MARK SCHEME MAXIMUM MARK: 75 SYLLABUS/COMPONENT: 9709/01 MATHEMATICS

Page 1 Mark Scheme Syllabus Paper GCE AS/A LEVEL EXAMINATIONS – NOVEMBER 2004 9709 1 © University of Cambridge International Examinations 2005 1 (3x–2/x)5 Required term has 5C2 or 5C3 = 10 Also has 33 and 22 → 1080 B1 B1 B1 B1 [4] Needs 10 or implied by answers. Can be implied or in the expansion Co. If all expansion given, gets ¾ unless the required term is isolated from the expansion – or ringed etc. 2 (i) 81,54,36 r = 54/81 or 36/54 S10 = 81 (1– ⅔10) ÷ (1– ⅔) → 239 (ii) n = (180 – 25) ÷ 5 + 1 = 32 Use of any Sn formula → 3280 B1 M1 A1 [3] B1 M1 A1 [3] Value of r – unsimplified – allow 0.66 Correct formula – power 10 and used Co. More than 3 s.f. ok, but needs 238.8 31 gets M0 Correct formula – not for n = 25,5,180 Co 3 tan 60 = BC ÷ 6 BC = 6 3 Area = ½ × 6 × “BC” – ½ × 62 × π/3 → 18 3 – 6π M1 A1 M1 M1 A1 [5] Use of tan = opp ÷ adj In this form somewhere with 3 Area of triangle as ½bh or ½absinC Area of Sector. Co. (Must be in this form, not decimals). No 3 , max 3 out of 5. 4 (i) (ii) → 2 points of intersection. B1 B1 B1 B1 [4] B1√ [1] Mark two graphs independently. Half a cycle – all above axis for 0 to π. 2 shown as the max with ½ cycle only. One whole cycle for 0 to π –1 to 1 shown with one cycle only. Providing 2 trig graphs used. (ignore other half if 0 to 2π used) 5 (i) x2 – 4x + 7 = 9 – 3x → x2 – x – 2 = 0 Solution of this x = 2 or –1 → (2, 3) and (–1, 12) Mid point is M (½, 7½) (ii) dy/dx = 2x – 4 Equate to m of line (–3) + solution → (½, 5¼) (iii) Distance = 2¼ M1 DM1 A1 A1 [4] B1 M1 A1 [3] B1√ [1] Complete elimination of y (or x) Correct solution of eqn = 0. All 4 values needed. Beware fortuitous ans. Answer given. Co Equates dy/dx to constant m, m ≠0. Must have calculus – not for perp m. Co For distance between “his” points.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL EXAMINATIONS – NOVEMBER 2004 9709 1 © University of Cambridge International Examinations 2005 6 (i) 5s2 + 3c2 = 5s2 + 3(1 – s2) → 3 + 2sin2x a = 3, b = 2 (ii) 3 + 2s2 = 7s Sets to 0 and solves. s = ½ or s = 3 Only values are π/6 and 5π/6 (iii) Minimum value = “a” = 3 Maximum value is “a + b” = 5 Range 3≤f(x) ≤ 5 M1 A1 [2] M1 A1A1√ [3] B1√B1√ [2] Use of s2 + c2 = 1 3 + 2sin2x gets both marks. Sets to 0 + correct method of soln. Co for one value. Other π = “1st” (If degrees, give A0,A1√ for 180 –) For his “a” and “a+b”. Condone <. Allow 3 and 5 on their own. 7 dy/dx = 6/√(4x – 3) P(3, 3) (i) x = 3, m = 2. Perpendicular m = –½ → y – 3 = –½ (x – 3) → x + 2y = 9 (ii) ∫ → 6(4x – 3)½ ÷ ½ ÷ 4 y = 3(4x – 3) + c Uses (3, 3) → c = –6 M1 M1 A1 [3] M1 A1 M1 A1 [4] Use of m1m2 = –1 even if algebraic Correct form of line eqn or y=mx + c Needs putting as x + 2y = 9 for A mark. (tangent gets 0/3). M1 for (4x – 3)k ÷ k. A1 for k = ½ and ÷ 4 Using (3, 3) to find c only after attempt at integration. Allow full marks once –6 obtained. 8 (i) (i + 7j + 2k).(–5i + 5j + 6k) → –5 + 35 + 12 = 42 42 =√54 √86 cosθ → angle AOB = 0.907 (ii) BC = ½ (b – a) = – 3i – j + 2k OC = OB + BC = –5i + 5j + 6k – 3i – j + 2k = –8i + 4j + 8k Unit Vector = (– 8i + 4j + 8k) ÷ 12 M1 M1 M1 A1 [4] M1 A1 M1A1√ [4] Use of → x1x2 + y1y2 + z1z2 Modulus used in dot product Everything linked correctly Accept if more accuracy given. Must be radians. Any combination of OA/AO OB/BO is ok for the three M1 marks. If AB used with OA/OB max M1 M1 Could be from OA + AC Correct only. Knowing to divide by length of vector. (leaving as √ is acceptable for both marks)

Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL EXAMINATIONS – NOVEMBER 2004 9709 1 © University of Cambridge International Examinations 2005 9 f : x → 2x – a (i) ff(x) = 11, 2(2x – 3) – 3 = 11 [or backwards 2x – 3 = 11, x = 7, 2x – 3 = 7 (M1), (M1)] → x = 5 (ii) 2x – a = x2 – 6x → x2 – 8x + a = 0 Use of b2 – 4ac = 0 → a = 16 (or inspection) (iii) x2 – 6x = (x – 3)2 – 9 → p = 3, q = 9 (iv) y = (x – 3)2 – 9 x = ± √(y + 9) + 3 y = h-1(x) = √(x + 9) + 3 Domain of h-1 = {x: x ≥ –9} M1 DM1 A1 [3] M1 M1 A1 [3] B1 B1 [2] M1 DM1A1 B1 [4] M1 for putting “x” as “2x – 3” Everything completed to give answer. (if –3 omitted → 4¼, allow M1 only) n.b. 2(2x – 3) = ff(x) gets M1 – not DM1 co Setting up a 3-term quadratic equation in x Using b2 –4ac on quadratic = 0 or ≥0) Co. Can be stated from the (–8x). Allow if (x – 3)2 – 9 without p or q stated Attempt to make x the subject, but only from completing square expression Replace y by x – sign lost for A. Special case “ans = √(y + 9) + 3” allow 2/3. Co. (allow ≥– 9 or y ≥– 9 etc.) 10 (i) dy/dx = 2x – 2/x2 d2y/dx2 = 2 + 4/x3 (ii) dy/dx = 0 2x – 2/x2 = 0 → x3 = 1 → x = 1, y = 3 If x = 1, d2y/dx2 > 0, Minimum (iii) Vol = π∫y2dx = π∫(x4 + 4/x2 + 4x) dx = π [ x5/5 – 4/x + 2x2 ] [ ]2 – [ ]1 = 71π/5 or 44.6 B1 B1 B1√ [3] M1 A1 M1A1√ [4] M1 3× A1 DM1A1 [6] For –2/x2 or for –2x-2 For “2x” and for “2” For +4/x3 or 4x-3 or for diff. his dy/dx as long as it is a negative power of x Putting his dy/dx = 0 and solving for x Co ( ± 1 gets M1A0 but can get next M1A1) Looking at sign of d2y/dx2 or other. √ for his x into his d2y/dx2 Attempt at squaring + integration Still gets M1 if (a+b)2 = a2+b2 For each term and π. Can get A1A1 for above error. Use of limits, “–” needed for M1. co. (no π – loses last A1 and one of first A marks) DM1 for quadratic. Quadratic must be set to 0. Factors. Attempt at two brackets. Each bracket set to 0 and solved. Formula. Correct formula. Correct use, but allow for numerical slips in b2 and –4ac.