9231_w02_er

SR9IN0201 1 FOREWORD ....................................................................................................................... 1 FURTHER MATHEMATICS................................................................................................ 2 GCE Advanced Level ...................................................................................................................................... 2 Paper 9231/01 Paper 1 ................................................................................................................................. 2 Paper 9231/02 Paper 2 ................................................................................................................................. 8 This booklet contains reports written by Examiners on the work of candidates in certain papers. Its contents are primarily for the information of the subject teachers concerned. www.XtremePapers.com 2 GCE Advanced Level Paper 9231/01 Paper 1 General comments The overall quality of work submitted in response to this Paper was good and provided clear evidence of a well-prepared candidature. Most candidates made a serious attempt at all the questions, submitted responses in order and set out their work in a clear way. The only opportunity for rubric infringement occurred in Question 11, but very few candidates wasted time by submitting responses to both options of that question. There were some misreads, especially in Question 10, and quite a lot of elementary arithmetic and algebraic errors. Such errors can lead to severe consequences for a question response, for not only do they necessarily lead to incorrect results, but also, they can in certain situations, drive the candidate into unworkable, or at least time consuming, strategies. In order, therefore, to avoid these unfortunate situations, which in extreme cases can seriously undermine the overall examination performance, it is essential that work is checked at each stage of its development. A feature of this new syllabus is that candidates must answer all questions if they are to obtain full credit. In this case, it is good to be able to record that with the exception of the vector product and linear spaces almost all candidates gave evidence of having an in depth knowledge of the entire syllabus. In particular, knowledge of complex numbers, tested in Question 7, was good. As in the case of previous A Level Further Mathematics Paper 1 examinations, the calculus topics, per se, were well understood and failures in questions, such as 4, 5, 6, and 8 which related to this material were usually due to non-calculus errors. Comments on specific questions Question 1 This short introductory question did not go as well as expected. It contained a somewhat unusual element and this seems to have baffled most of the candidature. Almost all candidates obtained a correct result for Sn = � � N n nu 1 , by application of the difference method. A few treated this series as the difference of two geometric series and so obtained a correct but unsimplified result for Sn and this, more often than not, led to difficulties in the final part of the question. The majority stated that the given series is convergent if x < 0, but made no mention of x = 0. Likewise, most stated, or implied, that S∞ = e x for x < 0, but very few identified S∞ = 0 when x = 0. Answers: SN = e x � e(N + 1)x ; infinite series is convergent for x < 0 ; S∞ = e x, for x < 0, = 0 for x = 0. 3 Question 2 In general this question was well answered. Almost all candidates began by substituting y 1 for x in the given polynomial equation, and so went on to obtain the correct polynomial equation in y. As often as not, however, the terms of this equation were not ordered in the standard way. For the rest, the majority validly obtained � 2 + � 2 + � 2 + � 2 = 1 – 2A and � � 2 + � � 2 + � � 2 + � � 2 = 4 + A, and so generally went on to obtain the correct value of A. Nevertheless there were some who, apparently, were unable to solve the equation 1 � 2A = 4 + A correctly. However, there were some who attempted to evaluate the second of these sums in terms of A by considering the coefficients of the x-equation. Although, in principle, this complicated strategy is feasible, hardly any candidates had the necessary algebraic expertise to argue successfully in this way. Answer: A = �1. Question 3 There were many essentially correct responses to this question, but only a minority of these were complete. Most responses showed the simplification of � = an+1 � an to a form which clearly indicates divisibility by 24. For the rest of the question, the main deficiencies were the general failure to state at the outset what the inductive hypothesis actually is, failure to verify that 24│� 0 (very often it was shown instead that 24│a1), failure to establish the key result, namely, that 24│� k � 24│ak + 1, and also failure to complete the induction argument in a satisfactory way. Of course, this was impossible if 24│� 0 had not been previously established. Answer: an+1 � an = 24[(12)(17) 2n + 9n]. Question 4 This turned out to be a successfully answered question. The majority of candidates produced a complete and correct response. (i) Not everyone integrated x 2 e x� correctly and moreover there were some errors in the application of the limits. (ii) Most responses showed correct working to establish the displayed reduction formula. Nevertheless, very few responses showed the optimal strategy, based on a consideration of D(x n + 1 2 e x� ). Instead it was common for candidates to begin by considering the integrand of In + 2 as (xn + l)(x 2 e x� ) and then to apply the integration by parts rule to this situation. Some, by a similar argument, obtained In in terms of In � 2 and then changed n to n + 2 so as to prove the required result. This is a satisfactory way to proceed, though it should be emphasised that if n is general then so is n + 2. (iii) Surprisingly, there were more errors here than in either of the preceding parts of this question. All that was required was to use I3 = I1 � e2 1 , I5 = 2I3 � e2 1 (common erroneous variants of these equations where I3 = 2I1 � e2 1 , I5 = 3I3 � e2 1 ) so as to express I5 in terms of Il and then to apply the result obtained in part (i). Answers: (i) I1 = 2 e1 1� � ; (iii) I5 = 1 � e2 5 . 4 Question 5 There were very few complete and correct responses to this question. (i) Only about half of all candidates could produce a satisfactory argument to show that lim 0�� y = 1. Moreover, few understood that in order to show that the line y = 1 is an asymptote of C it is also necessary to prove that r or x � + � as � � 0, yet only a small number of candidates produced such a proof. (ii) Again, many candidates were not able to produce a correct sketch of C. Here it was expected of candidates that the line y = 1 would appear, but this was frequently omitted. It was also common for the curve to be drawn starting from the pole. (iii) Almost all responses began with a correct integral representation of the area of the sector OPQ and went on to carry out the integration of �-�2 correctly. It was in the completion, which at the purely manipulative level involved only sub A Level mathematics, that some responses fell apart so that again there was evidence of lack of basic mathematical skills. (iv) This concluding section of the question was very well answered. Most candidates started (correctly) with something like �d dr = �� � 2 � s = ��� � � d 3/ 6/ 42 � �� � and so went on to obtain the required result. Answers: (i) y = r sin� = � �sin � 1 as � � 0, r = � 1 � + � as � � 0+ ; (iii) Area of sector OPQ = �2 3 . Question 6 Most responses showed a good understanding of implicit differentiation and, generally, the working was accurate. Nevertheless, few candidates produced a complete and correct response to this question. (i) Most responses got as far as exhibiting the preliminary result 3x2 + y2 + 2xy � � � � � � x y d d + y 2 � 3y2 � � � � � � x y d d = 0. From this the majority of candidates argued (essentially) that x y d d = 0 � 3x2 + y2 = 0 (*) which is impossible since x and y are necessarily real. This is an incorrect argument since in fact (*) does have the (unique) solution, x = y = 0. Thus to complete the argument it is necessary to say that as (0, 0) is not on the curve (this requires formal verification) then x y d d is not zero at any point on it. However, relatively few candidates argued in this way. (ii) The level of accuracy, both in the further differentiation and in the arithmetic, was generally impressive. The most persistent error was the writing of D(3y2 � � � � � � x y d d ) as 6y � � � � � � x y d d + 3y2 � � � � � � � � 2 2 d d x y and not as 6y 2 d d � � � � � � x y + 3y2 � � � � � � � � 2 2 d d x y . Answers: (ii) At (1, �1), x y d d = 5 4 , 125 198 d d 2 2 � x y . 5 Question 7 This question was generally well answered and the working was, for the most part, accurate. Strangely, it was in parts (i) and (ii), rather than in the later material, that the working sometimes ran into unnecessary complications. In part (i), for example, it was common to see � �� � �� �� �� �� ���� �� �� sinicos 1 sinicos sincos sinicos sinicossinicos sinicos sinicos 11 22 �� � � � � � �� � � � � z whereas in its place all that was required was � � � � � ����� ������ � sinicossincos 1 1 i z (by deMoivre’s theorem) = cos � � isin � and there were also similar complications in some responses to part (ii). For the rest, responses developed along correct lines in the way demanded by the question, and the working was generally accurate. Answers: a = 16 15 , b = 16 1 . Question 8 This question showed that almost all candidates had a good understanding of how to solve a linear second order differential equation. Nevertheless, there seemed to be a generally limited understanding of the form of the solution obtained, so that relatively few completely correct responses actually appeared. Almost all candidates obtained the AQE, solved it and formed the correct complementary function. Working for the obtaining of the particular integral was usually accurate so that a correct general solution for the given differential equation was a feature of most scripts. The majority of candidates understood, in principle, how to apply the given initial conditions and here again most responses led to a correct result for y in terms of t. A small minority of candidates, however, applied the given initial conditions to the complementary function and then added the particular integral in an attempt to find the required solution. Most candidates comprehended that the complementary function tends to zero as t tends to positive infinity and so concluded (correctly) that y � sin 3t when t is large and positive. However, from this point onwards, very few made any significant progress. Even the preliminary inequality sin 3t < �0.5 (*) was obtained by only a minority, even though it follows almost immediately from 108y + 109 < 9.5 x 108. The concluding argument would then follow immediately by observing that for 0 � 3t � 2�, (*) � 6 7� < 3t < 6 11� and that � �� 2 6 7 6 11 � = 3 1 . Nevertheless such reasoning appeared in few scripts even though it required no more than a knowledge of very basic A Level mathematics. Answer: y = 4e�2t � e�3t + sin 3t. Question 9 The responses to this question showed that the majority of candidates had a sound understanding of the application of scalar and vector products to problems involving 3-dimensional metric vectors. They also showed some understanding of the geometry specific to this question. Most arguments began with the use of vector products to determine vectors, n1 and n2, perpendicular to the planes �1 and � 2 and then went on evaluate n1 x n2 so as to obtain the vector equation of l. This strategy generated very few errors. 6 In contrast, a minority of candidates equated the components of the vector equations of �1 and � 2 and so expressed three of � 1, � 1, � 2, � 2 in terms of the fourth and again this will lead to the required result. However, this strategy inevitably generated more errors than the first, and thus showed the clear advantage that derives from a having a complete understanding of all the syllabus methodology. This question, and others like it, become much more formidable if no use can be made of the vector product. In the middle part of the question, almost all candidates obtained a vector not parallel to l (2i - j - 4k was the most popular) in the plane � 3. Also, they usually went on to obtain both a vector equation of � 3, often giving it in the form r.n = p, and a scalar equation of � 3, though, of course, lack of knowledge of the vector product made this exercise more difficult than it need have been. In the final part of this question, many candidates argued that as the three given linear equations represent � 1, � 2, � 3, respectively, and as also the line l is common to the three planes, then the given system, S, has an infinite number of solutions. This is correct as far as it goes, but it also requires that the first two equations be shown to represent � 1, � 2, and this vital part of the argument was omitted by some candidates. Of course, in the preceding part of the question, the third equation would have been shown to represent � 3. The alternative strategy, adopted by many other candidates, was to reduce the augmented matrix, A, to the echelon form and this simple operation was usually carried out accurately. Nevertheless, it was common for the concluding argument to be deficient in some way. Thus to say that as the rank of A is equal to 2, then S has an infinite number of solutions is incorrect, for S might have no solutions. Again, and similar, the fact that the echelon form contains a row of zeros does not of itself imply that S has an infinite number of solutions. That depends on the position of the zeros in the rows which are non-zero row vectors. Finally, the still weaker argument that det B = 0 (*), where B is the matrix of coefficients appertaining to S, is clearly false, for (*) is necessary for the conclusion but not sufficient. Answers: A vector equation of l is r = 2i + 4j + 6k + s(i + 2j - k); A vector equation of � 3 is r = 2i + 4j + 6k + s(i + 2j - k) + t(2i - j - 4k); A scalar equation of � 3 is 9x - 2y + 5z = 40. Question 10 This was the least successful question of the Paper, by quite a long way. Although some responses showed correct strategies for parts (i) and (ii), only a minority of all candidates made significant progress with part (iii). (i) A not uncommon error was the misreading of at least one of the elements of the matrix H and, of course, such an inaccuracy undermined much of the subsequent working. It must be emphasised therefore, that in situations, such as this, where there is a lot of numerical data, it is essential that the candidate carries out a thorough check of the copying of this information before becoming involved in the subsequent working. Moreover, there were also a number of arithmetic errors in attempts (not always complete) to obtain a valid echelon form and in some such cases the end product turned out to be a matrix of rank 4. This shows, yet again, the need for continuous checking. (ii) Most candidates understood that a basis for the null space of T could be obtained from three relevant linear equations. In this context, not everyone used the row echelon form they had obtained in part (i) from which the required basis can readily be obtained. Instead, a substantial minority used the given form of H and so embarked on a more extended strategy. Nevertheless the overall standard of working accuracy was very satisfactory so that most candidates obtained the required basis. (iii) Here there was a sharp dichotomy of the candidature into the majority who had no idea how to begin and the minority who knew a valid method and applied it in an accurate way. Such a method starts with a consideration of the general solution of the given vector equation, namely x = � � � � � � � � � � � � � � � � 2 1 3 1 + � � � � � � � � � � � � � � � � 1 1 1 0 7 From this it follows immediately that IxI2 = 1 + ( � � 3)2 + ( � � 1)2 + ( � � 2)2 and the rest is then a simple minimisation exercise which in the majority of effective responses was carried out by writing the above as a quadratic polynomial in � and then by applying the completion of the square technique. Thus the obtaining of another particular solution and then hoping for the best, a strategy employed by some candidates, is not a valid method. Among those who used calculus there were some who considered IxI rather than IxI2 and who thus became involved in unnecessary complication. Answers: (i) Dimension of range space of T is 3; (ii) A basis for the null space of T is � � � � � � � � � � � � � � � 1 1 1 0 ; (iii) Least possible value of IxI is 3 . Question 11 EITHER Although less popular than the alternative for Question 11, it nonetheless generated much good work. The level of numerical accuracy in responses to the later part of this question was impressive, especially in cases where candidates were involved in unnecessarily complicated strategies. For responses to part (i) something like the following was expected. Ge = � e � , (G + kI)e = Ge + kIe = � e + ke = ( � + k)e, and no comment would then be necessary. Similarly in part (ii) the following detail should appear in a complete response. G 2 e = G(Ge) = G( � e) = � (Ge) = � ( � e) = � 2e, and again no comment would be required. However, few candidates produced complete responses to both parts. For the rest of this question, most responses showed working leading to a correct characteristic equation for the matrix A from which the eigenvalues and a set of corresponding eigenvectors were obtained without error. It was also good to see that care was taken to ensure that eigenvalues and eigenvectors were paired off correctly. Many candidates failed to perceive that B = A – 8I and hence could not exploit the results of parts (i) and (ii) so as to obtain the eigenvalues and eigenvectors of B2 in a very simple way. Instead, they attempted to find these scalars and vectors for the matrix B and then went on to use (ii) to obtain the required results for B2. Finally it must be remarked that there was a small subset of candidates who evaluated B2 and then attempted to find its characteristic equation and so to work on to the final destination. Generally these attempts perished in the large amount of effort required. Answers: The eigenvalues of A are 0, 2, 3. Corresponding eigenvectors are . 1 0 1 , 1 1 3 , 1 1 1 � � � � � � � � � � �� � � � � � � � � � �� � � � � � � � � � � The above are also eigenvectors of B2. The corresponding eigenvalues of B2 are 64, 36, 25, respectively. Question 11 OR In the work for this more popular option of Question 11 there were many technical deficiencies. Very few completely correct responses appeared. (i) Usually candidates used an algebraic division process in order to determine P and Q. However, even this elementary process was not always applied correctly. The simple strategy of putting x = c in (x � c)(x + P) + Q ≡ (x � a)(x - b), so as to obtain Q and then considering the coefficient of x so as to obtain P, was attempted by very few. (ii) Most responses showed correct equations for the asymptotes and this was often the case