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 72.

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1 ADVANCED GCE UNIT 7/ MATHEMATICS (MEI Further Methods for Advanced Mathematics (FP THURSDAY 7 JUNE 7 Additional materials: Answer booklet (8 pages Graph paper MEI Examination Formulae and Tables (MF Morning Time: hour minutes INSTRUCTIONS TO CANDIDATES Write your name, centre number and candidate number in the spaces provided on the answer booklet. Answer all the questions in Section A and one question from Section B. You are permitted to use a graphical calculator in this paper. Final answers should be given to a degree of accuracy appropriate to the context. 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 7. ADVICE TO CANDIDATES Read each question carefully and make sure you know what you have to do before starting your answer. You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used. This document consists of printed pages. HN/ OCR 7 [H//] OCR is an exempt Charity [Turn over

2 Section A ( marks Answer all the questions (a A curve has polar equation r a( cos q, where a is a positive constant. (i Sketch the curve. [] (ii Find the area of the region enclosed by the section of the curve for which and the line [] q p. Û (b Use a trigonometric substitution to show that d x. [] Ù ı - x (c In this part of the question, f(x arccos (x. (i Find f (x. [] (ii Use a standard series to expand f (x, and hence find the series for f(x in ascending powers of x, up to the term in x. [] ( q p (a Use de Moivre s theorem to show that sin q sin q sin q sin q. [] (b (i Find the cube roots of j in the form re jq where r and p. [] p q These cube roots are represented by points A, B and C in the Argand diagram, with A in the first quadrant and ABC going anticlockwise. The midpoint of AB is M, and M represents the complex number w. (ii Draw an Argand diagram, showing the points A, B, C and M. [] (iii Find the modulus and argument of w. [] (iv Find w in the form a bj. [] OCR 7 7/ June 7

3 Let Ê ˆ M -. Á Ë - - (i Show that the characteristic equation for M is l l 8l. [] Ê ˆ You are given that - is an eigenvector of M corresponding to the eigenvalue. Á Ë (ii Find the other two eigenvalues of M, and corresponding eigenvectors. [8] (iii Write down a matrix P, and a diagonal matrix D, such that P M P D. [] (iv Use the Cayley-Hamilton theorem to find integers a and b such that M am bm. [] Section B (8 marks Answer one question Option : Hyperbolic functions Û (a Find Ù d, x giving your answer in an exact logarithmic form. [] ı 9x + (b (i Starting from the definitions of sinh x and cosh x in terms of exponentials, prove that sinh x sinh x cosh x. [] (ii Show that one of the stationary points on the curve has coordinates (ln, 9, and find the coordinates of the other two stationary points. [7] ln Û (iii Show that Ù ( cosh x - cosh x dx. [] ı -ln y cosh x cosh x [Question is printed overleaf.] OCR 7 7/ June 7 [Turn over

4 Option : Investigation of curves This question requires the use of a graphical calculator. The curve with equation y x kx k is to be investigated for different values of k. x k (i Use your graphical calculator to obtain rough sketches of the curve in the cases k, k. and k. [] (ii Show that the equation of the curve may be written as y x k k (k. x k Hence find the two values of k for which the curve is a straight line. [] (iii When the curve is not a straight line, it is a conic. (A Name the type of conic. [] (B Write down the equations of the asymptotes. [] (iv Draw a sketch to show the shape of the curve when k 8. This sketch should show where the curve crosses the axes and how it approaches its asymptotes. Indicate the points A and B on the curve where x and x k respectively. [] 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 (OCR 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. OCR 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. OCR 7 7/ June 7

5 Mark Scheme 7 June 7

6 7 Mark Scheme June 7 (a(i B Must include a sharp point at O and have infinite gradient at θ Give B for r increasing from zero for < θ <, or decreasing to zero for < θ < (ii Area is r dθ a ( cosθ d θ a a (b Put x sinθ ( cos θ + ( + cos θ dθ a sin sin θ θ + θ ( Integral is tanθ ( sin θ cosθ dθ 8cos θ (cosθ dθ sec θ dθ M A B BB ft B M A M A ag For integral of ( cosθ For a correct integral expression including limits (may be implied by later work Using cos θ ( + cos θ Integrating Accept.78a or x cosθ Limits not required For a + b cosθ and k cos θ sec θ dθ tanθ SR If x tanhu is used M for sinh( ln A for ( (max / 8 (c(i f ( x x B Give B for any non-zero real multiple of this (or for etc sin y (ii f ( x ( x ( + x + x f ( x C x x x f ( C x +... f ( x x x M A M A Binomial expansion ( terms, n Expansion of ( x correct (accept unsimplified form Integrating series for f ( x Must obtain a non-zero C not required x term 7

7 7 Mark Scheme June 7 OR by repeated differentiation ( Finding f ( x Evaluating ( f ( ( 88 f ( x x x +... M M A ft f ( x x x x +... A Must obtain a non-zero value ft from (c(i when B given 8

8 7 Mark Scheme June 7 (a (cosθ + jsinθ c + jc s c s jc s + cs + js Equating imaginary parts sin θ c ( s s c s + s s ( s s + s M M A M s s sinθ sin + s s + s θ + sin θ + s A ag (b(i + j 8, arg( + j r θ θ, BB B ft B ft M A Accept.8;., (Implies B for 8 One correct (Implies B for Adding or subtracting Accept θ + k, k,, (ii B Give B for two of B, C, M in the correct quadrants Give B ft for all four points in the correct quadrants (iii w arg w ( + 7 B ft B Accept.7 Accept.8 (iv w ( 7 8 arg( w w 8 j 7 ( cos + jsin M A ft A Obtaining either modulus or argument Both correct (ft Allow from arg w etc SR If B, C interchanged on diagram (ii B (iii B B for (iv MAA 9

9 7 Mark Scheme June 7 (i det( M λ I ( λ[( λ( λ ] [( λ + ] + [ ( λ] ( λ ( + λ + λ ( λ + ( + λ 8 + 9λ + λ λ λ + λ 8λ + λ λ Characteristic equation is λ λ 8λ (ii λ ( λ 8( λ + (iii (iv Other eigenvalues are 8, When λ 8, x + y + z 8x ( x + y z 8y x y z 8z y x and z ; eigenvector is When λ, x + y + z x x + y z y P M y x, z x ; eigenvector is D 8 M 8M M M + 8M M M (M M + 8M + 8M + 8M + 9M M A M A ag M A M M A M M A B ft M A M M A 8 Obtaining det( M λ I Any correct form Simplification Solving to obtain a non-zero value Two independent equations Obtaining a non-zero eigenvector ( x + y+ z 8x etc can earn MM Two independent equations Obtaining a non-zero eigenvector B if P is clearly singular Order must be consistent with P when B has been earned

10 7 Mark Scheme June 7 (a x dx arsinh 9x + arsinh 9 ln( + + ln M A A M A For arsinh or for any sinh substitution For x or for sinh x u For or for du OR M For ln( kx + k x +... [ Give M for ln( ax + bx +... ] ln( + 9x + x AA or ln( x + x + 9 ln 8 ln ln A (b(i sinh x cosh x ( e x ( e sinh x x e e x x ( e x + e x M A (e x e x (e For completion x + e x (e x x e (ii dy sinh x sinh x dx For stationary points, sinh x sinh x cosh x x, y 7 x ( ± ln( sinh x( cosh x sinh x or cosh x + 9 ln 9 y x ln, y 9 BB M A A ag A ag B 7 When exponential form used, give B for any terms correctly differentiated d y Solving to obtain a value of dx sinh x, cosh x or e x ( or x stated Correctly obtained Correctly obtained The last AA ag can be replaced by BB ag for a full verification (iii sinh x sinh 8 x ln ln 9 9 BB M A ag When exponential form used, give B for any terms correctly integrated Exact evaluation of sinh(ln and sinh( ln

11 7 Mark Scheme June 7 (i k B B Maximum on LH branch and minimum on RH branch Crossing axes correctly B B Two branches with positive gradient Crossing axes correctly k k. B B Maximum on LH branch and minimum on RH branch Crossing positive y-axis and minimum in first quadrant (ii ( x + k( x k + k + k y x + k k( k + x k + x + k Straight line when k( k + k, k M A (ag BB Working in either direction For completion (iii(a Hyperbola B (B x k y x k B B

12 7 Mark Scheme June 7 (iv B B B B B Asymptotes correctly drawn Curve approaching asymptotes correctly (both branches Intercept on y-axis, and not crossing the x-axis Points A and B marked, with minimum point between them Points A and B at the same height ( y

13 Report on the Units taken in June 7 7: Further Methods for Advanced Mathematics (FP General Comments On the whole, the candidates seemed to find this paper quite hard; and the full range of marks, from to 7, was obtained. There were many excellent scripts, with about % of candidates scoring marks or more, as well as some who were clearly not ready for an examination at this level; about % of candidates scored less than marks. Quite a few candidates appeared to be short of time, especially when they had used inefficient methods such as repeated differentiation in Q(c(ii or multiplying out brackets (instead of using the binomial theorem in Q(a. In Section A, the question on matrices (Q was much the best answered; and in Section B the overwhelming majority of candidates chose the hyperbolic functions option. Comments on Individual Questions This question, on polar coordinates and calculus, had an average mark of about (out of 8. (a(i The sketch was usually drawn well. Most showed the cusp at θ clearly, but some also had a sharp point at θ, instead of an infinite gradient. (a(ii Most candidates knew how to find the area, and the integration was very often carried out accurately. Sometimes the factors ½ and a were missing, and some candidates were unable to integrate cos θ. (b About half the candidates used the correct substitution x sinθ (or x cosθ to obtain dθ (or dθ, but many could not proceed beyond this cos θ sin θ point. Other candidates tried inappropriate substitutions such as x sinθ, x tanθ, u x, and achieved nothing useful. (c(i The differentiation of arccos x was often done correctly, although very common error. The denominator was sometimes given as x x. was a (c(ii This part was poorly answered. Few candidates realised that all they needed to do was to find the first three terms of the binomial expansion of ( x and then integrate; and those that did, often forgot about the constant term after integrating. There were very many attempts to differentiate arccos x five times, and hardly any of these were successful. This question, on complex numbers, had an average mark of about. (a (b(i For many candidates this seemed to be a familiar piece of work which was carried out confidently and accurately. Quite a few candidates confused this with the process of expressing powers in terms of multiple angles and considered ( z / z, without any success. The process of finding the cube roots of a complex number was quite well understood, and there were very many correct answers. The argument of one of the roots was sometimes given as 9, outside the required range, instead

14 Report on the Units taken in June 7 of ; but the most common error was to start by giving the argument of + j as instead of. (b(ii (b(iii (b(iv The Argand diagram was usually drawn correctly with ABC forming an equilateral triangle, although the midpoint M was sometimes placed on the imaginary axis instead of in the second quadrant. Relatively few candidates were able to use their Argand diagram to find the modulus and argument of w correctly. In particular, the modulus of w was often thought to be the same as that of the cube roots. Most candidates knew how to use the modulus and argument of w to find the modulus and argument of w, and hence evaluate w ; although errors in previous parts very often caused the final answer to be incorrect. This question, on matrices, was the best answered question, with half the candidates scoring marks or more. Candidates appeared to tackle it with confidence, and it was quite often the first question answered. (i (ii (iii (iv Almost all candidates knew how to find the characteristic equation, although there were many algebraic, and especially sign, errors in the working. The methods for finding eigenvalues and eigenvectors were very well known, and loss of marks was usually caused only by careless slips. However, some candidates did not appear to know that an eigenvector must be non-zero. Almost all candidates correctly found P as the matrix with the eigenvectors as its columns (but this earned no marks if the resulting matrix was obviously singular, for example having a column of zeros or two identical columns. The diagonal matrix D very often contained the eigenvalues instead of their squares. Most candidates knew the Cayley-Hamilton theorem, but many did not see how to obtain M. Some attempts began by cancelling a factor M from the cubic equation, which is invalid since M is a singular matrix. This question, on hyperbolic functions, was quite well answered, and the average mark was about. By far the most common reason for loss of marks was the failure to show sufficient working when the answer was given on the question paper. (a (b(i Most candidates realised that the integral involved arsinh, or went straight to the logarithmic form, although the factor was often omitted. This was usually adequately, if not always elegantly, demonstrated. (b(ii Most candidates differentiated correctly, although d (cosh x sinh x and d x d (cosh x sinh x were fairly common errors. However, it was rare to see full dx marks in this part. Having obtained cosh x it was not sufficient to write 9 so x ln, y since this point is given on the question paper. To earn the marks a candidate was expected to write (at least 9 x ln( + ( ln, y ( + (9 + 9 or an equivalent exact calculation using cosh x and cosh x (.

15 Report on the Units taken in June 7 Candidates frequently failed to find one or other of the remaining stationary points 9 (, 7 and ( ln,. (b(iii The integration was usually done correctly (more often than the differentiation in part (b(ii, but very many candidates lost marks for not showing sufficient detail in the evaluation. Since the answer is given, it is not sufficient to state, for example sinh(ln ; the calculation si nh(ln (9 should be shown. 9 9 This question, on the investigation of curves, was more popular than in the past; and in a few centres all the candidates answered this question. Even so it was only attempted by about % of the candidates. There were a few very good complete solutions, but most attempts were fragmentary; the average mark was about 9. (i In the cases k and k it was quite common for one branch of the curve to be missing, presumably caused by poor choice of scales for the axes on the calculator. (ii (iii (iv Several candidates were unable to do this simple algebra. Every type of conic was offered by at least one candidate, even though it is known to have asymptotes. Few sketches were drawn carefully enough to show clearly all the details requested.

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 72.

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 72. ADVANCED GCE UNIT / MATHEMATICS (MEI Further Methods for Advanced Mathematics (FP THURSDAY JUNE Additional materials: Answer booklet (8 pages Graph paper MEI Eamination Formulae and Tables (MF Morning