Cambridge A Level Mathematics 9709 — 2002 Oct/Nov Paper 1 · Variant 1

CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level General Certificate of Education Advanced Level MATHEMATICS 9709/1 PAPER 1 Pure Mathematics 1 (P1) OCTOBER/NOVEMBER SESSION 2002 1 hour 45 minutes Additional materials: Answer paper Graph paper List of Formulae (MF9) TIME 1 hour 45 minutes INSTRUCTIONS TO CANDIDATES Write your name, Centre number and candidate number in the spaces provided on the answer paper/answer booklet. 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. INFORMATION FOR CANDIDATES 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 question paper consists of 5 printed pages and 3 blank pages. CIE 2002 [Turn over

2 1 Find the value of the term which is independent of x in the expansion of
x
3 x 4 . [3] 2 A geometric progression, for which the common ratio is positive, has a second term of 18 and a fourth term of 8. Find (i) the first term and the common ratio of the progression, [3] (ii) the sum to infinity of the progression. [2] 3 In the diagram, OPQ is a sector of a circle, centre O and radius r cm. Angle QOP  θ radians. The tangent to the circle at Q meets OP extended at R. (i) Show that the area, A cm2, of the shaded region is given by A  1 2r2(tan θ  θ). [2] (ii) In the case where θ  0.8 and r  15, evaluate the length of the perimeter of the shaded region. [4] 4 The gradient at any point (x, y) on a curve is (1
2x). The curve passes through the point (4, 11). Find (i) the equation of the curve, [4] (ii) the point at which the curve intersects the y-axis. [2] 5 (i) Show that the equation 3 tan θ  2 cos θ can be expressed as 2 sin2 θ
3 sin θ  2  0. [3] (ii) Hence solve the equation 3 tan θ  2 cos θ, for 0 Æ  θ  360 Æ. [3] 9709/1/O/N/02

3 6 In the diagram, triangle ABC is right-angled and D is the mid-point of BC. Angle DAC
30 Æ and angle BAD
x Æ. Denoting the length of AD by l, (i) express each of AC and BC exactly in terms of l, and show that AB
1 2l  7, [4] (ii) show that x
tan
1
2 3   30. [2] 7 Given that a
  2 2 1  , b
  2 6 3   and c
  p p p  1  , find (i) the angle between the directions of a and b, [4] (ii) the value of p for which b and c are perpendicular. [3] 8 A curve has equation y
x3  3x2  9x  k, where k is a constant. (i) Write down an expression for dy dx. [2] (ii) Find the x-coordinates of the two stationary points on the curve. [2] (iii) Hence find the two values of k for which the curve has a stationary point on the x-axis. [3] 9709/1/O/N/02 [Turn over

4 9 The diagram shows a rectangle ABCD, where A is (3, 2) and B is (1, 6). (i) Find the equation of BC. [4] Given that the equation of AC is y
x  1, find (ii) the coordinates of C, [2] (iii) the perimeter of the rectangle ABCD. [3] 10 The diagram shows the points A (1, 2) and B (4, 4) on the curve y
2  x. The line BC is the normal to the curve at B, and C lies on the x-axis. Lines AD and BE are perpendicular to the x-axis. (i) Find the equation of the normal BC. [4] (ii) Find the area of the shaded region. [4] 9709/1/O/N/02

5 11 (i) Express 2x2
8x  10 in the form a(x
b)2
c. [3] (ii) For the curve y  2x2
8x  10, state the least value of y and the corresponding value of x. [2] (iii) Find the set of values of x for which y  14. [3] Given that f : x  2x2
8x  10 for the domain x  k, (iv) find the least value of k for which f is one-one, [1] (v) express f
1(x) in terms of x in this case. [3] 9709/1/O/N/02

CAMBRIDGE INTERNATIONAL EXAMINATIONS CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level Advanced International Certificate of Education MARK SCHEME FOR the November 2002 question papers 9709 /1 9709 /2 9709 /3 8719 /3 9709 /4 9709 /5 8719 /5 9709 /6 0390 /6 9709 /7 8719 /7 9709 MATHEMATICS Paper 1 (Pure 1), maximum raw mark 75 Paper 2 (Pure 2), maximum raw mark 50 Paper 3 (Pure 3), maximum raw mark 75 Paper 4 (Mechanics 1), maximum raw mark 50 Paper 5 (Mechanics 2), maximum raw mark 50 Paper 6 (Probability and Statistics 1), maximum raw mark 50 Paper 7 (Probability and Statistics 2), maximum raw mark 50 These mark schemes are published as an aid to teachers and students, to indicate the requirements of the examination. They show the basis on which Examiners were initially instructed to award marks. They do 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 discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2002 question papers for most IGCSE, Advanced Subsidiary (AS) Level and Advanced Level syllabuses. UNIVERSITY of CAMBRIDGE Local Examinations Syndicate

Notes Mark Scheme Syllabus [ _ A Level Examinations — November 2002 9709 « 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 correetly 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. e 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 Y 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. e 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. e Fora fumerical 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.

Notes Mark Scheme Syllabus A Level Examinations — November 2002 9709 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 ofa 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 throughY“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.

CAMBRIDGE INTERNATIONAL EXAMINATIONS NOVEMBER 2002 GCE Advanced Subsidiary Level UNIVERSITY of CAMBRIDGE Local Examinations Syndicate

Page 1 | Mark Scheme Syllabus [| Paper [ AS Level Examinations — November 2002 9709 41 1 r=4-r m2 BI Guessing or attempt at r=2 Term is «Co x (3)? Bl For correct 4C2 x (3) ° for his r. = 34 Bax) Bl Correct only —isolated from expansion. 3 2. (i)ar=18 and ar?=8 Mi ‘Any 2 equations of type ar” Oars to give r=2/3 DMI Correct method on correct 2 equations. a=18+r = 27.0 Al 3 For his 18+r ae infinity = az(1- MIL Correct formula applied — even if r>1. (i) Sum to infinity = a+(1-1) A1V 2 | Follow through provided r<1. Answer = 81.0 (ignore r=+2/3) 3. RQ QR= stand BI Correct somewhere — in (ii) ok. iy ‘Area shaded = Yer?tane -Yer20 BI 2 | All correct — answer given, beware fortuitous. 5 R Gi) Are PQ= 15 x 0.8 = 12 Bl Anywhere (could be implied) OR=r+cos@ (21.53) M1 Must be correct with r and @ or Pythagoras. Perimeter = rtan@ + arc PQ + (r - r+cos6) MI Putting 4 things together — even if algebraic = 34.0 (33.9 ok) Al 4 | Correct only. rer 5 4. (i) y=(i+2x)*+G/2) + 2 GC) M1 ‘Attempt at Jn. Needs( +k Al Al for + 2 and k=3. use of (4,11) to find C C=2. MI Attempt to use (4,11) Al 4 | Correct only. (i) Ifx=0, y=7/3 M1 ALV | Use of x=0 providing there is some integration

Page 2 T Mark Scheme Syllabus AS Level Examinations — November 2002 Paper 41 9709 5. (i) 3tanO=2cos® —3sin@+cos0=2cos0 MI Use of t=s+c 3sin0=2cos?0 = 2(1-sin20) Mt Use of s?+c2=1 3s=2(1-s?). A 3 Everything correct — answer given. (ii) Soln of 2s?+3s-2=0 s=0.5 or -2 Mi Correct method of solution @= 30° or 150° ALAIN | Correct only, then V for 180 — first answer or consistent with his cosine-loses V mark if 3 | extra solutions. 6. (i) AC = lcos30 = 3/2 BL Correct only — not decimal BC = 2Isin30 =1 Bl Correct only AB = V(12+312/4) = BINT M1AI1 | Use of Pythagoras. Correct only. Answer 4 | given. Could be cosine rule. (ii) tan (x+30) = BC+AC = I + (1V3/2) ML Use of tangent in 90° triangle — tan=opp/adj. x = tan “(2/V3) - 30 B Al 2. | Xthe subject — beware fortuitous answers. » 40 46 Cc 4-12 +3 =-5 Mil Use of a;b;+a;by+a3b3 9 v49 cosO MIMI | Use of a.b.cos0 + Use of V(a,2+a;2+a;2) @ = 103.8° or 1.81 radians. Al 4{ Correct only (ii) Dot product = I1p+3 MI Use of aybytasbytasb; Dot product = 0 as P=-3/11 DMI | ~Oused Al 3 | correct only. 8. (i) dy/dx = 3x2+6x-9 B2,1 One off for each error including +k left. 2 if =0 when (x+3)(x-1 MI Use of dy/dx=0 “ x=-3 or n ¢ Mee Al Both values somewhere 2 Gi) Subbing the values into y=0. Mi Using y=0 at least once. — k=-27 ork=5. DMI Subbing his values for x into y=0 + soln. Al Both correct. 3

Page 3 I Mark Scheme Syllabus [ AS Level Examinations — November 2002 Paper_| 1 9709 9. (i) mof AB=-2 BL Correct only mof BC =-1(m)=% MI Used correctly . _ DMI Correct formula needed to be used. equation of BC y-6=V2(x-1)_ or 2y=x+11 AlV 4 | Av mark for any correct equation. (ii) Sim eqns y=x-1 and answer above ML Correct method Solution’ C (13,12) Al 2 | Correct only Gi) AB = 120 and BC = V180 MI Use of Pythagoras once - 20 ok perimeter = 2xV20 + 2«V180 DMI _| Use of 2a + 2b ~ with Pythagoras twice. = 35.8 or 35.7 or 16V5 or VI280 | Ay Correct only. 3 44 als > Aa) Dx 10. (Dy=2Vx. dy/dx =x M1 Realising the need to differentiate + use. _ =Y% Al Correct only Ifx=4, m= _ DMI m,m,=-1 numerical needed Perpendicular = - 2 Al 4 | correct only Eqn of y= -2x +12 or y-4=-2(x-4) (ii) Area P = J2Vx dx=2x'9/1.5 MI AIL _ | Knowing to integrate. Correct unsimplified. Evaluated from 1 to 4 DMI Correct use of 1 to 4 — not for 2 to 4. Answer = 32/3 — 4/3 = 28/3 Al 4 | Correct only. ie) -— TB 11 (i) 2x2+8x-10 = 2(x2)2 +¢ BIBI a=2 gets Bl, b=2 gets BI c=-18 Bl 3 | correct only (i) Least value = -18 when x=-2 BIVB W follow through for c and for —b: Calculus ok. (iii) 2x2+8x-10214 or 2(x+2)? -18014 MI setting the inequality to 0 x2+4x-12=0 or (x+2)?=16 Limit points 2 and ~6 Al correct only — irrespective of what they do x22 and x <-6 Al 3 | correct only (condone > or <) (iv) Smallest k is -2 BIO 1 | Follow through. (v) Makes x the subject and replaces x by y MI x the subject ~ reasonable attempt from completion of square. Mi x,y interchanged. f= AlV Correct form his answer to (i). 3