Cambridge A Level Mathematics 9709 — 2008 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 2008 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. 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. 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 75. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. This document consists of 4 printed pages. © UCLES 2008 [Turn over *4050671575*

2 1 Find the value of the coefficient of x2 in the expansion of
x 2 + 2 x 6 . [3] 2 Prove the identity 1 + sin x cos x + cos x 1 + sin x ≡ 2 cos x. [4] 3 The first term of an arithmetic progression is 6 and the fifth term is 12. The progression has n terms and the sum of all the terms is 90. Find the value of n. [4] 4 A O D N C M F P E B i j k 6 cm 8 cm 20 cm The diagram shows a semicircular prism with a horizontal rectangular base ABCD. The vertical ends AED and BFC are semicircles of radius 6 cm. The length of the prism is 20 cm. The mid-point of AD is the origin O, the mid-point of BC is M and the mid-point of DC is N. The points E and F are the highest points of the semicircular ends of the prism. The point P lies on EF such that EP = 8 cm. Unit vectors i, j and k are parallel to OD, OM and OE respectively. (i) Express each of the vectors −−→ PA and −−→ PN in terms of i, j and k. [3] (ii) Use a scalar product to calculate angle APN. [4] 5 The function f is such that f(x) = a −b cos x for 0◦≤x ≤360◦, where a and b are positive constants. The maximum value of f(x) is 10 and the minimum value is −2. (i) Find the values of a and b. [3] (ii) Solve the equation f(x) = 0. [3] (iii) Sketch the graph of y = f(x). [2] © UCLES 2008 9709/01/O/N/08

3 6 9 cm O Q P T 5 cm In the diagram, the circle has centre O and radius 5 cm. The points P and Q lie on the circle, and the arc length PQ is 9 cm. The tangents to the circle at P and Q meet at the point T. Calculate (i) angle POQ in radians, [2] (ii) the length of PT, [3] (iii) the area of the shaded region. [3] 7 x cm r cm A wire, 80 cm long, is cut into two pieces. One piece is bent to form a square of side x cm and the other piece is bent to form a circle of radius r cm (see diagram). The total area of the square and the circle is A cm2. (i) Show that A = (π + 4)x2 −160x + 1600 π . [4] (ii) Given that x and r can vary, find the value of x for which A has a stationary value. [4] 8 The equation of a curve is y = 5 −8 x. (i) Show that the equation of the normal to the curve at the point P (2, 1) is 2y + x = 4. [4] This normal meets the curve again at the point Q. (ii) Find the coordinates of Q. [3] (iii) Find the length of PQ. [2] © UCLES 2008 9709/01/O/N/08 [Turn over

4 9 O x y Q P 2 1 1 y x = (3 + 1)
The diagram shows the curve y = √(3x + 1) and the points P (0, 1) and Q (1, 2) on the curve. The shaded region is bounded by the curve, the y-axis and the line y = 2. (i) Find the area of the shaded region. [4] (ii) Find the volume obtained when the shaded region is rotated through 360◦about the x-axis. [4] Tangents are drawn to the curve at the points P and Q. (iii) Find the acute angle, in degrees correct to 1 decimal place, between the two tangents. [4] 10 The function f is defined by f : x →3x −2 for x ∈. (i) Sketch, in a single diagram, the graphs of y = f(x) and y = f −1(x), making clear the relationship between the two graphs. [2] The function g is defined by g : x →6x −x2 for x ∈. (ii) Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9. [5] The function h is defined by h : x →6x −x2 for x ≥3. (iii) Express 6x −x2 in the form a −(x −b)2, where a and b are positive constants. [2] (iv) Express h−1(x) in terms of x. [3] 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. © UCLES 2008 9709/01/O/N/08

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2008 question paper 9709 MATHEMATICS 9709/01 Paper 1, maximum raw mark 75 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.

Page 2 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9709 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 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.

Page 3 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9709 01 © UCLES 2008 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.

Page 4 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9709 01 © UCLES 2008 1 6 2 2     + x x Term in x2 15 2 2 2 4 ×         x x M1 A1 Correct term – needs powers 4 and 2 For × 15 Coeff = 4 15 or 3.75 A1 [3] Ignore inclusion of x2 2 x x x x x cos 2 sin 1 cos cos sin 1 ≡ + + + LHS ) sin 1( cos cos ) sin 1( 2 2 x x x x + + + M1 Reasonable algebra. Correct denominator and one term correct in numerator = ) sin 1( cos sin 2 2 x x x + + M1 A1 Use of sin2x + cos2x = 1 For 2 + 2sinx = x cos 2 A1 [4] Co – answer was given – check preceding line 3 1st term = a = 6 5th term = a + 4d = 12 → d = 1.5 B1 Correct value of d Sn = 2 n (12 + (n – 1)1.5) = 90 M1 Use of correct formula with his d → n2 + 7n – 120 = 0 DM1 Correct method for soln of quadratic → n = 8 A1 [4] Co (ignore inclusion of n = –15) 4 (i) PA = –6i – 8j – 6k B1 Co – column vectors ok PN = 6i + 2j – 6k B2, 1 One off for each error [3] (all incorrect sign – just one error) (ii) PA . PN = –36 – 16 + 36 = –16 M1 Use of x1x2 + y1y2 + z1z2 76 136 16 os c − = APN M1 M1 Modulus worked correctly for either one Division of "–16" by "product of moduli" → APN = 99° A1 Allow more accuracy [4]

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9709 01 © UCLES 2008 5 x a a – bcosx (i) a + b = 10 and a – b = –2 → a = 4 and b = 6 M1 A1 Al M1 for either correct. A1 both correct Co [3] (if a – b = 10, and a + b = –2, treat as MR –1, (i) a = 4, b = –6, (ii) 131.8, 228.2, (iii) Sketch is mirror image in y = 4) (ii) 4 – 6cosx = 0 → cosx = 2/3 → x = 48.2° or 311.8° M1 A1 A1√ [3] Makes cosx subject and uses inv cos. For 1st angle. √ for 360° – "his angle" (iii) B2,1 [2] Must be just one cycle Starts at – 2 and ends at –2 Max at 10. "V shapes " lose a mark. Parabolas lose 1 mark. 6 (i) Using s = rθ, 9 = 5θ → θ = 1.8 rad. (ii) Uses POT. Halves the angle Uses tangent in POT PT = 5 tan0.9 = 6.30 cm (not 6.31) (iii) area of sector = ½ × 52 × 1.8 (22.5) Area of POT = ½ × 5 × 6.30 (15.75) Shaded area = 2 triangles – sector → 9.00 (allow 8.95 to 9.05) Ml Al [2] Ml Ml Al [3] Ml Ml Al [3] Use of formula. co Realises the need to halve Use of tangent – even if angle not halved co Use of A = ½r2θ with 1.8 or 0.9. Use of ½bh and (2 triangles – sector) co 7 (i) 4x + 2πr = 80 A = x2 + πr2 → ( ) π π 1600 160 4 2 + − + = x x A B1 B1 M1 A1 [4] Connection of lengths Connection of areas Eliminates r. co but answer given. (ii) ( ) π π 160 4 2 d d − + = x x A = 0 when x = ) 4 ( 2 160 + π or 11.2 Ml Al DM1 Al [4] Attempt at differentiation. co Ignore omission of π. Sets to 0 and solves. co

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9709 01 © UCLES 2008 8 y = 5 – x 8 , P(2, l) (i) 2 8 d d x x y = m of tan = 2 m of normal = –½ Eqn of normal y – 1= – 2 1 (x – 2) → 2y + x = 4 B1 M1 M1 A1 [4] Correct differentiation Use of m1m2 = –1 Correct method for line Answer given (ii) Sim eqns 2y + x = 4, y = 5 – x 8 → x2 + 6x – 16 = 0 or y2 – 7y + 6 = 0 → (–8, 6) M1 DM1 A1 [3] Complete elimination of x or y Soln of quadratic. co (iii) Length = 125 5 10 2 2 = + → 11.2 (accept 125 or 5 5 etc) M1 Al [2] Correct use of Pythagoras For his points. 9 y = 1 3 + x (i) A = ∫x dy = ∫ − 2 1 2 3 1 y dy M1 Uses integration wrt y = 9 4 3 9 3 = −     y y (allow 0.44 to 0.45) Al DM1 A1 [4] Integration correct Use of limits 0 to 1. co [or 2 –∫ + x x d 1 3 = [2 – 3 )1 3 ( 2 3 2 3 × + x ] = 9 4 ] B1 B1 M1A1 B1 for everything but ÷3. B1 for ÷3. M1 for "2–" and use of limits 0 to 1. (ii) ( ) x x x y V d 1 3 d 2 ∫ ∫ + = = π π M1 M1 for correct formula used with y2 and integration wrt x. (does not need π) =       + x x 2 3 2 π from 0 to 1 A1 A1 integration correct, including π. Vol of cylinder = π × 22 × 1 = 4π B1 Or by integration of y2 = 4 → Subtraction → 1.5 π (4.71) A1 co [4]

Page 7 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9709 01 © UCLES 2008 (iii) ( ) 3 1 3 d d 2 1 2 1 × + = − x x y If x = 0, m = 2 3 . If x = 1, m = 4 3 B1 M1 B1 for everything but × 3. M1 for × 3. At x = 0, angle = 56.3° At x = 1, angle = 36.9° M1 Linking angle with tangent once → angle between = 19.4° A1 co Could use vectors, or tan(A – B) formula. Could also find tangents, point of intersection, 3 lengths and cosine rule. [4] 10 f : x a 3x – 2 (i) B1 B1 [2] Graph of y = 3x –2 Evidence of mirror image in y = x or graph of 1/3 (x + 2). Whichever way, there must be symmetry shown or quoted or implied by same intercepts. (ii) gf(x) = 6(3x – 2) – (3x – 2)2 M1 Must be gf, not fg = –9x2 + 30x –16 d/dx = –18x + 30 = 0 when x = 5/3 → Max of 9 A1 M1 DM1 A1 [5] Co Differentiates or completes square Sets to 0, solves and attempts to find y All ok – answer was given (gf(x) = 9 – (3x – 5)2 → Max 9) (iii) 6x – x2 = 9 – (x – 3)2 B1, B1 [2] Does not need a or b. (iv) y = 9 – (3 – x)2 y x − ± = − 9 3 → h–1(x) = 3 + √ (9 – x) M1 DM1 A1 [3] Order of operations in making x subject Interchanging x and y Allow if ± given (Special case → if correct with y instead of x, give 2 out of 3)