# SPECIMEN. XXXX June 2013 Morning/Afternoon A2 GCE MATHEMATICS (MEI) 4798 Further Pure Mathematics with Technology (FPT) Duration: Up to 2 hours

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2 1 This question concerns curves with parametric equations where k is positive and t takes all real values. 2 x = t k sin t, y = 1 cos t, (i) Investigate the curves for values of k where 0 < k < 1. Describe the common features of these curves and sketch, for 2π t 4π, a typical example. [5] (ii) Prove that, if 0 < k < 1, the curve is never parallel to the y-axis. [3] (iii) Sketch the curve for k = 1. Describe the main feature by which this curve differs from the curves where 0 < k < 1. [3] (iv) For the case k = 1, confirm the behaviour of the curve at the point where t = 0 by investigating the gradient as t 0. [5] (v) Sketch the curve for k = 2. Show that the width of each loop, measured parallel to the x-axis, is π. [6] 3 (vi) Form an equation to find the value of k where the loops will just touch each other. Find the value of k to 4 decimal places. [3] 2 (i) The function f is defined by f( ) = 3 (3 3i) 2 6i i. Solve the equation f( ) = 0 and plot the roots as points on an Argand diagram. Show that these points form an equilateral triangle. [6] (ii) Find f ( ) and show that the equation f ( ) = 0 has a repeated root. Plot this root on the Argand diagram drawn in part (i). [4] The function f has the following properties: ( ) the roots of the equation f( ) = 0 form an equilateral triangle when plotted in the Argand diagram; ( ) the equation f ( ) = 0 has a single repeated root. The rest of this question explores the relationship between ( ) and ( ) for other cubic functions. (iii) Let g( ) = 3 + b 2 + c + d. Find the relationship between b and c when there is a single repeated root of the equation g ( ) = 0. [3] (iv) (A) Obtain another cubic, with distinct roots, for which ( ) is true. Show that ( ) is also true for your cubic. (B) Obtain a cubic for which ( ) and ( ) are both false. [3] The three points representing 0, 1, 2 form an equilateral triangle on the Argand diagram. (v) Given that 2 = β 1, β, find the two possible values of β. [3] (vi) Use one of your values of β to write down a cubic with roots 0, 1 and 2. Show that ( ) is true for this cubic. [5] OCR 2012 SPECIMEN

3 3 3 (i) Create a program to find all the positive integer solutions to x 2 3y 2 = 1 with x 100, y 100. Write out your program in full and list the solutions it gives. [10] (ii) Show how the other solutions can be derived from the solution with the smallest x-value. Use each solution to give a rational approximation to 3. [5] (iii) Edit your program so that it will find solutions to x 2 ny 2 = 1, where n is a positive integer. Write out the lines of your program that you have changed. Use the edited program to find a rational approximation to 5 that is accurate to within 0.1%. [6] (iv) Explain why the edited program will not give any results if n is a square number. [2] OCR 2012 SPECIMEN

6 2 1 (i) 1 (ii) OCR 2012

7 3 1 (iii) 1 (iv) OCR 2012 Turn over

8 4 1 (v) 1 (vi) OCR 2012

9 5 2 (i) OCR 2012 Turn over

10 6 2 (ii) 2 (iii) OCR 2012

11 7 2(iv)(A) 2(iv)(B) 2 (v) OCR 2012 Turn over

12 8 2 (vi) OCR 2012

13 9 3 (i) OCR 2012 Turn over

14 10 3 (ii) OCR 2012

15 11 3 (iii) 3 (iv) OCR 2012

17 xxxx June 2013 GCE MATHEMATICS (MEI) 4798 Further Pure Mathematics with Technology (FPT) MARK SCHEME Duration: Up to 2 hours MAXIMUM MARK 72 Final Version Last updated: xx/xx/xxxx This document consists of 17 pages

20 4798 Mark Scheme Specimen 11. Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations Meaning in mark scheme E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working 4

21 4798 Mark Scheme Specimen 12. Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually 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, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. 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). Therefore M0 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E 5

23 4798 Mark Scheme Specimen If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 7

24 4798 Mark Scheme Specimen Question Answer Marks Guidance 1 (i) Periodic with period 2π B1 Maximum 2, minimum 0 B1 Both required for mark. Accept comment about coordinates of maxima/minima. Comment about symmetry B1 G2 Minima (zero gradient) clearly shown. Give B1 for general shape correct (three cycles). [5] 1 (ii) dy sint M1 = dx 1 kcost Denominator > 0 for all t. E1 Evidence of attempt to find d y dx. Accept denominator is never 0. OR dx 1 kcost dt = dx > 0 for all t. dt M1 E1 [3] Accept is never 0. 8

25 4798 Mark Scheme Specimen 1 (iii) G2 Periodic with three complete cycles. Cusps clearly shown. The curve has cusps. B1 [3] 1 (iv) sin t lim t k cos t sin t lim t 0 1 k cos t The point is defined at t = 0: x = 0, y =0. M1 M1 E1 Limit from one direction only scores M1M0A0 Therefore the curve has a cusp at (0,0) E1 [5] 9

26 4798 Mark Scheme Specimen 1 (v) G2 dy is infinite when 1 2cost = 0 dx π t = 3 π π x = 2sin 3 3 π = 3 3 π Hence width of loop is π = (vi) 1 1 arccos k sin arccos = π k k k = M1 M1 M1 [6] M1 Any correct value of t. π Substituting to find width of half the loop. 3 Taking positive value and doubling. For lhs correct but rhs = +π score M1A0A0 [3] 10

27 4798 Mark Scheme Specimen 2 (i) ( 3 2) 1 B2 z = i, i,1 2i B1 for 1 or 2 correct only. B2 Points marked correctly. B1 for two correct ( i ( 1 2 ) 2 2 i 2 2 = i i = 2 2 ( = i ( 1 2i) = M1 Evaluation of the distance between the points. All shown to be equal. Any other acceptable method should be awarded full marks. [6] 11

28 4798 Mark Scheme Specimen 2 (ii) 2 f'( z) = 3 z (6+ 6i) z 6i B1 2 f'( z) = 3( z (1 i)). Repeated root at z = 1 i. or 2 Discriminant =( (6 + 6i)) 4 3 6i ( ) = 0 or 2 csolve(3 z (6 + 6i) z 6i = 0, z) = 1 i. This root must be repeated as a quadratic will have two roots over the complex numbers. M1 Factorising f'( z ) = 0. Must show that the root is repeated for these marks by either factorising, finding the discriminant or stating that it must have two solutions over the complex numbers. 2 csolve(3 z (6 + 6i) z 6i = 0, z) = 1 i without further explanation scores M0A0. Point marked correctly. [4] 12

29 4798 Mark Scheme Specimen 2 (iii) 2 = 3z + g'( z) bz + c Discriminant: b Discriminant = c 2 b c = 3 M1 M1 Accept b=± 3c 2 Award full marks for integrating kz ( α ) and finding the same result by comparing coefficients. 2 (iv) Function that has distinct roots, repeated root to f '( z ) = 0. Showing that the roots are in an equilateral triangle. Function that has roots that are not in an equilateral triangle and distinct roots to f '( z ) = 0. [3] B1 B1 B1 [3] 13

30 4798 Mark Scheme Specimen 2 (v) π M1 β = 1 and arg( β ) =± 3 π β i ± i = ± or β = e 3 [3] 2 (vi) 1 3 M1 hz ( ) = zz ( z1) z + i z1 2 2 Evidence of attempt to find modulus and argument or suitable diagram. Either form acceptable. i π Or equivalent e 3 form 2 2 z1 z1 z z1 1 3(2 ) h'( z) = 3z 3z z+ i 2 2 (6 z z1(3 + 3 i)) = 12 So h'( z ) = 0 has a single repeated solution. 2 M1 Accept the alternative method of using the relationship between b and c from (iii) acceptable. [5] 14

31 4798 Mark Scheme Specimen 3 (i) Example program: Define program1(m)= Prgm Local i,j For i,1,m For j,1,m If i^2-3*j^2=1 Then Disp i,j EndIf EndFor EndFor EndPrgm x = 2, y = 1; x = 7, y = 4 ; x= 26, y = 15, x= 97, y= 56 M6 A4 [10] If some (or all) of the answers are incorrect allocate method marks as follows: M1 Appropriate structure program: nested loops or equivalent M1 Appropriate use of variables M1 Maximum value or 100 used. M1 Loop for x or equivalent M1 Loop for y or equivalent M1 Check (If) statement More efficient programs are possible. for each correct x, y pair Subtract (maximum) if any additional value(s) of x and/or y solutions are given (only penalise once). 15

32 4798 Mark Scheme Specimen 3 (ii) Using x = 2, y = 1. M1 Evidence of using x = 2, y = 1 to find at least 1 other 2 x solution. 2 + y2 3 = (2+ 3) At least one shown correctly. = So x= 7, y= 4 is a solution. x3 + y3 3 = (2+ 3) = So x= 26, y = 15 is a solution, x3 + y3 3 = (2+ 3) = So x= 97, y= 56 is a solution, 3 4 Other two shown to be solutions based on x = 2, y = 1. x If x² 3 y² = 1 then x² 3 y² 3 y ,,, M1 [5] Rearranging x² 3 y² = 1. Must be based on appropriate rearrangement. 16

33 4798 Mark Scheme Specimen 3 (iii) Change input to Define program1(m,n)= Change If i² 3j²=1 Then to If i² nj²=1 Then M1 Accept method based on changing to i² 5j²=1 if clear. Smallest solution is x= 9, y= 4 B1 9 4 is not accurate enough. Finding another solution (program or squares ) x= 161, y= is accurate to within 0.1% 72 3 (iv) 2 Writing n= d would give any solution as 2 2 x = ( dy) + 1 But there are no square numbers that are 1 apart. B1 M1 [6] E2 [2] Seen or implied 17

34 ASSESSMENT GRID Qualification type GCE Specification name Mathematics (MEI) Spec. No. Year 20XX Status choose Unit No Series Specimen Unit name Name Setting Task Max. no. of marks Further Pure Mathematics with Technology 72 Q. No. Spec. (and part) Ref Assessment Objectives (Defined below or on a separate tab) AO1 AO2 AO3 AO4 AO5 Demand of question ( e.g. high, medium, low, or target A C E Total (mark for question or part) Marks for each spec area (eg Calculus, c, Algorithms, A,...) C j T relevant additional attributes (e.g. synoptic content, extended writing, objective, SPAG, etc.) 1 (i) (ii) (iii) (iv) (v) (vi) (i) (ii) (iii) (iv) (v) (vi) (i) (ii) (iii) (iv) Auto totals Targets

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