Cambridge A Level Mathematics 9709 — 2005 May/June 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) May/June 2005 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 4 printed pages. © UCLES 2005 [Turn over

2 1 A curve is such that dy dx = 2x2 −5. Given that the point (3, 8) lies on the curve, find the equation of the curve. [4] 2 Find the gradient of the curve y = 12 x2 −4x at the point where x = 3. [4] 3 (i) Show that the equation sin θ + cos θ = 2(sin θ −cos θ) can be expressed as tan θ = 3. [2] (ii) Hence solve the equation sin θ + cos θ = 2(sin θ −cos θ), for 0◦≤θ ≤360◦. [2] 4 (i) Find the first 3 terms in the expansion of (2 −x)6 in ascending powers of x. [3] (ii) Find the value of k for which there is no term in x2 in the expansion of (1 + kx)(2 −x)6. [2] 5 The diagram shows a rhombus ABCD. The points B and D have coordinates (2, 10) and (6, 2) respectively, and A lies on the x-axis. The mid-point of BD is M. Find, by calculation, the coordinates of each of M, A and C. [6] 6 A geometric progression has 6 terms. The first term is 192 and the common ratio is 1.5. An arithmetic progression has 21 terms and common difference 1.5. Given that the sum of all the terms in the geometric progression is equal to the sum of all the terms in the arithmetic progression, find the first term and the last term of the arithmetic progression. [6] © UCLES 2005 9709/01/M/J/05

3 7 A function f is defined by f : x →3 −2 sin x, for 0◦≤x ≤360◦. (i) Find the range of f. [2] (ii) Sketch the graph of y = f(x). [2] A function g is defined by g : x →3 −2 sin x, for 0◦≤x ≤A◦, where A is a constant. (iii) State the largest value of A for which g has an inverse. [1] (iv) When A has this value, obtain an expression, in terms of x, for g−1(x). [2] 8 In the diagram, ABC is a semicircle, centre O and radius 9 cm. The line BD is perpendicular to the diameter AC and angle AOB = 2.4 radians. (i) Show that BD = 6.08 cm, correct to 3 significant figures. [2] (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [3] 9 A curve has equation y = 4 √x. (i) The normal to the curve at the point (4, 2) meets the x-axis at P and the y-axis at Q. Find the length of PQ, correct to 3 significant figures. [6] (ii) Find the area of the region enclosed by the curve, the x-axis and the lines x = 1 and x = 4. [4] © UCLES 2005 9709/01/M/J/05 [Turn over

4 10 The equation of a curve is y = x2 −3x + 4. (i) Show that the whole of the curve lies above the x-axis. [3] (ii) Find the set of values of x for which x2 −3x + 4 is a decreasing function of x. [1] The equation of a line is y + 2x = k, where k is a constant. (iii) In the case where k = 6, find the coordinates of the points of intersection of the line and the curve. [3] (iv) Find the value of k for which the line is a tangent to the curve. [3] 11 Relative to an origin O, the position vectors of the points A and B are given by −−→ OA = 2i + 3j −k and −−→ OB = 4i −3j + 2k. (i) Use a scalar product to find angle AOB, correct to the nearest degree. [4] (ii) Find the unit vector in the direction of −−→ AB. [3] (iii) The point C is such that −−→ OC = 6j + pk, where p is a constant. Given that the lengths of −−→ AB and −−→ AC are equal, find the possible values of p. [4] Every reasonable effort has been made to trace all copyright holders where the publishers (i.e. UCLES) are aware that third-party material has been reproduced. The publishers would be pleased to hear from anyone whose rights they 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. © UCLES 2005 9709/01/M/J/05

CAMBRIDGE INTERNATIONA1 EXAMINATIONS GCE Advanced Subsidiary and Advanced Level MARK SCHEME for the June 2005 question paper 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 June 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Grade thresho1ds for Syllabus 9709 (Mathematics) in the June 2005 examination. minimum mark required for grade: maximum mark available A B E Component 1 75 60 53 30 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.

JUNE 2005 GCE A/AS LEVEL MARK SCHEME MAXIMUM MARK: 75 SYLLABUS/COMPONENT: 9709/01 MATHEMATICS

Page 3 Mark Scheme Sy11abus Paper GCE AS/A LEVEL – JUNE 2005 9709 1 © University of Cambridge International Examinations 2005 1 x x y 5 3 2 3 − = (+ c) (3,8) fits 5 5 3 2 3 + − = x x y M1 A1 DM1 A1 [4] Attempt at integration. CAO Uses (3,8) in an integrated expression. CAO 2 x x y 4 12 2 − = ) 4 2 ( ) 4 ( 12 2 2 − × − − = − x x x dx dy If x = 3, 3 8 − = dx dy B1 M1 A1√ A1 [4] −12(x2 − 4x)−2 correct. Use of chain rule. √ for B0 attempts. Quotient or product rule ok (M1A2,1) CAO Uncancelled ok. 3 (i) s + c = 2s − 2c → s = 3c → tanθ = 3 (ii) → θ = 71.6° or 251.6° M1 A1 [2] B1 B1√[2] Use of t = s/c + collection → tanθ = k. Algebra needed to reduce to this form. B1√ for 180 + ...as only soln in range. 4 (i) (2 − x)6 = 64 − 192x + 240x2 (ii) (1 + kx)(2 − x)6 coeff of x2 = 240 − 192k = 0 → k = 5/4 or 1.25 3 x B1 [3] M1 A1√ [2] One for each term. Allow 26. Must be considering sum of 2 terms. ft for his expansion. (allow M1 if looking for coeff of x). 5 M (4, 6) m of BD = -2 M of AC = ½ Eqn of AC ) 4 ( 2 1 6 − = − x y → x = − 8 when y = 0 A(−8, 0) → C = (16, 12) by vector move etc. B1 M1 M1 A1 M1 A1 [6] CAO Use of m1m2 = −1 Correct method leading to A - the equation may not be seen - y = 0 may be used with gradient. Any valid method - vectors, midpoint backwards, or solution of 2 sim eqns.

Page 3 Mark Scheme Sy11abus Paper GCE AS/A LEVEL – JUNE 2005 9709 1 © University of Cambridge International Examinations 2005 6 GP a = 192, r = 1.5, n = 6 AP a = a, d = 1.5, n = 21 S6 for GP = 192(1.56 − 1) ÷ 0.5 = 3990 S21 for AP = 2 21(2a + 20 x 1.5) Equate and solve → a = 175 21st term in AP = a + 20d = 205 (or from 3990 = 21(a + l)/2 M1 M1 DM1 A1 M1 A1 [6] Correct sum formula used. Correct sum formula used. Needs both M’s - soln of sim eqns. CAO Correct formula used. 7 f : x → 3 − 2sinx for 0°≤ x ≤ 360°. (i) Range 1 ≤ f(x) ≤ 5 (ii) g: x → 3 − 2sinx for 0o ≤ x ≤ Ao (iii) Maximum value of A = 90 or ½π (iv) y = 3 − 2 sinx g-1(x) = sin−1       − 2 3 x B2,1,0 [2] B2,1,0 [2] B1 [1] M1 A1 [2] Needs 1, 5, ≤. One off for each error. Must be exactly 1 full oscillation - this overrides the rest. Starts and ends at 3. Correct shape needed. Curves, not blatant lines. CAO Attempt to make x the subject and then to replace x by y. Needs sin-1 (). Everything correct inc sin−1. Allow these marks anywhere.

Page 3 Mark Scheme Sy11abus Paper GCE AS/A LEVEL – JUNE 2005 9709 1 © University of Cambridge International Examinations 2005 8 (i) BD = 9sin (π − 2.4) = 6.08 cm (ii) OD = 9cos (π − 2.4) or Pyth (6.64) Arc AB = 9 x 2.4 Perimeter = 21.6 + 6.08 + 9 + 6.64 → 43.3 cm (iii)Area of sector = ½ 92 2.4 Area of triangle = ½.6.08.6.64 → 117 cm2 M1 A1 [2] M1 M1 A1 [3] M1 M1 A1 [3] Any valid method for BD (ans given) Any valid method - not DM mark - this could come in part (iii). Correct use of s = rθ . CAO Correct use of ½r2θ . Use of ½ bh. CAO 9 x y 4 = (i) dy/dx = −2x−1.5 = −¼ m of normal = 4 Eqn of normal y − 2 = 4(x − 4) P (3.5, 0) and Q (0, −14) Length of PQ = √(3.52 + 142) = 14.4 (ii) Area = ∫ = − 4 1 5 . 0 5 . 0 ] 5 . 0 4 [ 4 x dx x = [8 x ] = 16 − 8 = 8 M1 A1 M1 M1 M1 A1 [6] M1 A1 DM1A1 [4] Reasonable attempt at differentiation with his power of x. CAO Use of m1m2 = −1 even if algebraic. Use of equation for a straight line + use of x = 0 and y = 0. Needs correct formula or method. CAO Attempt at integration. Correct unsimplified. Correct use of limits. CAO

Page 3 Mark Scheme Sy11abus Paper GCE AS/A LEVEL – JUNE 2005 9709 1 © University of Cambridge International Examinations 2005 10 y = x2 − 3x + 4 (i) dy/dx = 2x − 3 = 0 when x = 1.5, y = 1.75 This is a minimum point, 1.75 > 0 Curve lies above the x - axis. (ii) Decreasing function for x < 1.5. (iii) y = x2 − 3x + 4 with y + 2x = 6 Eliminate y to give x2 − x − 2 = 0 or eliminate x to give y2 − 10y +16 = 0 → (−1, 8) and (2, 2) (iv) x2 − 3x + 4 = k − 2x → x2 − x + 4 - k = 0 or 2x − 3 = −2 Use of b2 − 4ac = 0 or x = ½ → y = 23/4 k = 33/4 M1 A1 A1√ [3] A1√ [1] M1 DM1 A1 [3] M1 M1 A1 [3] Completing square or using calculus. Correct 1.75 from some method. Correct deduction for candidate’s +ve y. Correct deduction for candidate’s value of x. Allow ≤. Attempt at eqn in x or y and set to 0. Correct method of solution. All values. Equates and sets to 0 Uses b2 − 4ac on eqn = 0 CAO 11 OA = 2i + 3j − k OB = 4i − 3j + 2k (i) OA .OB = 8 − 9 − 2 = −3 OA .OB = √14 X √29 cos AOB → AOB = 99o (ii) AB = b − a = 2i − 6j + 3k Magnitude of AB = √49 = 7 → Unit vector = 7 1 (2i - 6j + 3k) (iii) AC = −2i + 3j +(p + 1)k 4 + 9 + (p + 1)2 = 49 → p = 5 or −7 M1 M1 M1 A1 [4] B1 M1 A1√ [3] B1 M1 A1√ A1 [4] Correct use of a1a2 + b1b2 + c1c2 Modulus. Correct use of abcos θ CAO CAO Use of Pythagoras + division. CAO (use of BA for AB has max 2/3). CAO - condone a − c here. Correct method for forming an equation CAO