9231_s10_er

Cambridge International Advanced Level 9231 Further Mathematics June 2010 Principal Examiner Report for Teachers © UCLES 2010 FURTHER MATHEMATICS Paper 9231/11 Paper 11 General comments The scripts for this paper were of a generally good quality. There were a number of outstanding scripts and once again there were very few poor scripts. Work was well presented by the vast majority of candidates. Solutions were set out in a clear and logical order. The standard of numerical accuracy was good. Algebraic manipulation, where required, was sound. This year there was a marked improvement at vector work, which was good to see. There was no evidence to suggest that candidates had any difficulty completing the paper in the time allowed. A very high proportion of scripts had substantial attempts at all eleven questions. Once again there were few misreads and few rubric infringements. The Examiners felt candidates had a sound knowledge of most topics on the syllabus. As well as the vector work, already mentioned, candidates performed particularly well on matrices, roots of equations and whichever alternative of the final question that they chose, each of which was answered by a large number of candidates. Comments on specific questions Question 1 Many candidates scored full marks on this question. Only the weakest candidates found difficulty dealing with cube roots and differentiating, either implicitly or explicitly after rearranging the equation. Those differentiating implicitly found difficulty with the third term, or left 29 on the right-hand side. Those differentiating explicitly sometimes failed to employ the chain rule correctly. Answers: 3, 27 4 . Question 2 In part (i), most sketches scored at least 1 mark for being of approximately the correct shape and location for 0 Y θ Y 2π. Marks were lost for not showing the initial line to be a tangent at the pole, or for not showing asymptotic approach to the circle ar = . In part (ii), those who expanded 2)e1( θ−− invariably went on to correctly obtain the printed answer. Question 3 Almost all candidates correctly obtained the length of C, since the derivatives of the parametric equations were given and the expression for t s d d was a basic integral. A substantial number of candidates were unable to determine y and were thus unable to make further progress. Many of those who found y, successfully completed the question. Answer: 15 2128 π . 1 www.XtremePapers.com Cambridge International Advanced Level 9231 Further Mathematics June 2010 Principal Examiner Report for Teachers © UCLES 2010 Question 4 The initial mark in this question was a method mark for applying the method of differences to ∑ =               −−      + N n nn 1 66 2 1 2 1 . Those not realising this were unable to score any of the first 4 marks. Candidates were not required to simplify their expression for NS , although some did, which wasted valuable time. There were a considerable number of completely correct answers, although some who obtained NS correctly wrote at considerable length in part (i) to no avail. Those who wrote NS in a form such as that below quickly realised that part (i) required 1 6 1 × and part (ii) 0 6 1 × . Answers: ( ) ( ) 384 1 1 32 1 1 24 5 2 1 6 1 22 6 −+−+−      + NNNNN ; (i) 6 1 ; (ii) 0. Question 5 Many candidates proved the reduction formula correctly, using integration by parts. There were a disappointing number of candidates who tried, in vain, to integrate ( )nxln . A considerable number of candidates had difficulty with the notation in the proof by induction. Consequently they were unable to formulate a correct inductive hypothesis and make progress. Some candidates did not realise that they needed to show that 4 1 e 4 1 2 1 +=I for the base case, thus producing a good deal of unnecessary work. There were, pleasingly, a substantial number of correct proofs, however. Question 6 There were many completely correct answers to this question. Only the weaker candidates had difficulty in obtaining the initial result and both values in part (i). Some candidates did not obtain the correct values in part (ii) because they relied on mis-remembered formulae. Some candidates, realising that they had gone wrong in finding 6S , began again with a different approach and obtained 8S correctly. Mostly, candidates employed the method in the mark scheme. Answers: (i) −2; (ii) 1, −6. Question 7 The responses to this question were most encouraging. Nearly all candidates were able to show in part (i) that the lines 1l and 2l intersected, with very few forgetting to check the values obtained for λ and µ in the third equation. In part (ii) most candidates were able to find the normal vector to the plane containing 1l and 2l using the relevant vector product. The usual method, then, was to find the equation of the plane containing 1l and 2l , before using the distance of a point from a plane formula. Others considered the distance between parallel planes having the normal vector that they had obtained. Some, unnecessarily long, solutions found the point where the perpendicular from P met the plane containing 1l and 2l , before using Pythagoras to obtain the result. In part (iii) few used the method of finding the magnitude of the cross product between a vector joining P to any point of 1l and the unit vector in the direction of 1l . The preferred approach was to call Q the foot of the perpendicular from P to 1l and obtain the value,       − 21 29 , of the parameter at Q, then use Pythagoras to determine the required distance. Where candidates employed methods that required calculating the value of a parameter, the rather awkward fractions involved inevitably led to arithmetical errors for some. Answers: (ii) 38 7 or 1.14; (iii) 21 125 or 2.44. 2 Cambridge International Advanced Level 9231 Further Mathematics June 2010 Principal Examiner Report for Teachers © UCLES 2010 Question 8 Almost all candidates were able to use the result Ae = λe to obtain the required eigenvalue and the required eigenvector when the eigenvalue was 4. Many could then write down the required matrices P and D and state, correctly, that Q = P –1 . Errors rather frequently occurred in calculating P –1 , whether by row operations or co-factors and determinant. These methods occurred with approximately equal frequency. Answers: 3,           − − 4 4 1 ,           −− −−− 411 424 111 ,           102400 02430 001 . Question 9 Once again, candidates demonstrated rather limited knowledge of complex numbers. Part (i) was successfully done by many, possibly more from memory than reasoning. The chief problem in part (ii) was being able to express 5z as 3 i2 e32 π − . Often the argument was incorrect. Those who accomplished this step mostly obtained the fifth roots correctly. Thus, many only obtained 1 mark for their Argand diagram as the roots were not in the correct place. Many, of the relatively few who attempted part (iii), made the fatal mistake of summing only 4 terms of the GP, rather than 5. Those who summed 5 usually realised the connection with part (ii) and were able to obtain the printed result. Part (iv) was seldom and not very successfully attempted. It might be helpful for candidates to think of w−2 as ‘the distance of w from 2 in the Argand diagram’. Answers: (i) 5 i2 e πk − , k = 0, 1, 2, 3, 4; (ii)       +− 5 i2 15 i2 2e ππ k , k = 0, 1, 2; (iv) 15 i2 e2 π − . Question 10 This question was somewhat different to recent linear algebra questions. There were few complete answers and a good deal of muddled thinking. Those attempting the first part by row operations seldom gave a complete set of values of a for which the system of equations had a unique solution. This was surprising, as intelligent reading of the next two parts of the question would give a good clue as to what was expected. Formal set notation was not required here. Those who said that the system of equations has no unique solution when the appropriate determinant was zero fared much better. A sizeable number of candidates thought that because the appropriate determinant was zero, when a took the value 18, that the system would have no solution. The distinction between ‘no solution’ and ‘no unique solution’ was clearly not appreciated. Those who used row operations usually found a contradiction such as 0z = –5 and deduced that no solution was possible. In the case when a took the value 18, only a small number obtained the two equations 24 =+ yx and 4 1 =z and hardly any of those gave the parametric solution, although some said there were infinite solutions. Some, who ignored any reference to infinite solutions, did manage to find the particular solution for which 1=++ zyx . Answers: all values of a except a = 8 and a = 18; 3 1 =x , 12 5 =y , 4 1 =z . 3 Cambridge International Advanced Level 9231 Further Mathematics June 2010 Principal Examiner Report for Teachers © UCLES 2010 Question 11 EITHER The initial request in this question was much more easily dealt with by expressing x y d d and 2 2 d yd x as x z x d d 3 2 and 2 2 2 2 d d 3 d d 6 x z z x z z +      respectively, rather than trying to express x z d d and 2 2 d d x z in terms of x and y. The majority realised this and had little difficulty in establishing the first result. The remainder of the question was well done by a large number of those doing this alternative. The only significant difficulty was in finding the coefficient of x2sin in the particular solution. Most knew how to do it, but only a few could do it accurately. This did not prevent success on the final request, where nearly all said that 0e →−x as ∞→x and obtained the printed result. Answer: ( ){ }3 1 2sin2cose xxxz x +−= − . Question 11 OR Nearly all candidates attempting this alternative were able to find the equation of the horizontal and vertical asymptote. Likewise, almost all were able to find the only point of intersection of C with the asymptotes. Most candidates were able to accurately find the only stationary point, although some slipped up in calculating the y-coordinate. Finding the set of values for which the gradient of C was negative caused problems. This was largely due to candidates looking at the equation they had solved for the stationary points, rather than at the expression for the gradient function itself. This led them to give       <<− 1 3 1 : xx (set notation was not required). Nevertheless, they invariably drew the correct graph. Answers: (i) 1=x , 1=y ; (ii) ( 3 1 , 1); (iii)(a)       −− 8 1 , 3 1 , (b) 3 1 − 1. 4 Cambridge International Advanced Level 9231 Further Mathematics June 2010 Principal Examiner Report for Teachers © UCLES 2010 FURTHER MATHEMATICS Paper 9231/12 Paper 12 General comments The scripts for this paper were of a generally good quality. There were a number of outstanding scripts and once again there were very few poor scripts. Work was well presented by the vast majority of candidates. Solutions were set out in a clear and logical order. The standard of numerical accuracy was good. Algebraic manipulation, where required, was sound. This year there was a marked improvement at vector work, which was good to see. There was no evidence to suggest that candidates had any difficulty completing the paper in the time allowed. A very high proportion of scripts had substantial attempts at all eleven questions. Once again there were few misreads and few rubric infringements. The Examiners felt candidates had a sound knowledge of most topics on the syllabus. As well as the vector work, already mentioned, candidates performed particularly well on matrices, roots of equations and whichever alternative of the final question that they chose, each of which was answered by a large number of candidates. Comments on specific questions Question 1 Many candidates scored full marks on this question. Only the weakest candidates found difficulty dealing with cube roots and differentiating, either implicitly or explicitly after rearranging the equation. Those differentiating implicitly found difficulty with the third term, or left 29 on the right-hand side. Those differentiating explicitly sometimes failed to employ the chain rule correctly. Answers: 3, 27 4 . Question 2 In part (i), most sketches scored at least 1 mark for being of approximately the correct shape and location for 0 Y θ Y 2π. Marks were lost for not showing the initial line to be a tangent at the pole, or for not showing asymptotic approach to the circle ar = . In part (ii), those who expanded 2)e1( θ−− invariably went on to correctly obtain the printed answer. Question 3 Almost all candidates correctly obtained the length of C, since the derivatives of the parametric equations were given and the expression for t s d d was a basic integral. A substantial number of candidates were unable to determine y and were thus unable to make further progress. Many of those who found y, successfully completed the question. Answer: 15 2128 π . 5 Cambridge International Advanced Level 9231 Further Mathematics June 2010 Principal Examiner Report for Teachers © UCLES 2010 Question 4 The initial mark in this question was a method mark for applying the method of differences to ∑ =               −−      + N n nn 1 66 2 1 2 1 . Those not realising this were unable to score any of the first 4 marks. Candidates were not required to simplify their expression for NS , although some did, which wasted valuable time. There were a considerable number of completely correct answers, although some who obtained NS correctly wrote at considerable length in part (i) to no avail. Those who wrote NS in a form such as that below quickly realised that part (i) required 1 6 1 × and part (ii) 0 6 1 × . Answers: ( ) ( ) 384 1 1 32 1 1 24 5 2 1 6 1 22 6 −+−+−      + NNNNN ; (i) 6 1 ; (ii) 0. Question 5 Many candidates proved the reduction formula correctly, using integration by parts. There were a disappointing number of candidates who tried, in vain, to integrate ( )nxln . A considerable number of candidates had difficulty with the notation in the proof by induction. Consequently they were unable to formulate a correct inductive hypothesis and make progress. Some candidates did not realise that they needed to show that 4 1 e 4 1 2 1 +=I for the base case, thus producing a good deal of unnecessary work. There were, pleasingly, a substantial number of correct proofs, however. Question 6 There were many completely correct answers to this question. Only the weaker candidates had difficulty in obtaining the initial result and both values in part (i). Some candidates did not obtain the correct values in part (ii) because they relied on mis-remembered formulae. Some candidates, realising that they had gone wrong in finding 6S , began again with a different approach and obtained 8S correctly. Mostly, candidates employed the method in the mark scheme. Answers: (i) −2; (ii) 1, −6. Question 7 The responses to this question were most encouraging. Nearly all candidates were able to show in part (i) that the lines 1l and 2l intersected, with very few forgetting to check the values obtained for λ and µ in the third equation. In part (ii) most candidates were able to find the normal vector to the plane containing 1l and 2l using the relevant vector product. The usual method, then, was to find the equation of the plane containing 1l and 2l , before using the distance of a point from a plane formula. Others considered the distance between parallel planes having the normal vector that they had obtained. Some, unnecessarily long, solutions found the point where the perpendicular from P met the plane containing 1l and 2l , before using Pythagoras to obtain the result. In part (iii) few used the method of finding the magnitude of the cross product between a vector joining P to any point of 1l and the unit vector in the direction of 1l . The preferred approach was to call Q the foot of the perpendicular from P to 1l and obtain the value,       − 21 29 , of the parameter at Q, then use Pythagoras to determine the required distance. Where candidates employed methods that required calculating the value of a parameter, the rather awkward fractions involved inevitably led to arithmetical errors for some. Answers: (ii) 38 7 or 1.14; (iii) 21 125 or 2.44. 6 Cambridge International Advanced Level 9231 Further Mathematics June 2010 Principal Examiner Report for Teachers © UCLES 2010 Question 8 Almost all candidates were able to use the result Ae = λe to obtain the required eigenvalue and the required eigenvector when the eigenvalue was 4. Many could then write down the required matrices P and D and state, correctly, that Q = P –1 . Errors rather frequently occurred in calculating P –1 , whether by row operations or co-factors and determinant. These methods occurred with approximately equal frequency. Answers: 3,           − − 4 4 1 ,           −− −−− 411 424 111 ,           102400 02430 001 . Question 9 Once again, candidates demonstrated rather limited knowledge of complex numbers. Part (i) was successfully done by many, possibly more from memory than reasoning. The chief problem in part (ii) was being able to express 5z as 3 i2 e32 π − . Often the argument was incorrect. Those who accomplished this step mostly obtained the fifth roots correctly. Thus, many only obtained 1 mark for their Argand diagram as the roots were not in the correct place. Many, of the relatively few who attempted part (iii), made the fatal mistake of summing only 4 terms of the GP, rather than 5. Those who summed 5 usually realised the connection with part (ii) and were able to obtain the printed result. Part (iv) was seldom and not very successfully attempted. It might be helpful for candidates to think of w−2 as ‘the distance of w from 2 in the Argand diagram’. Answers: (i) 5 i2 e πk − , k = 0, 1, 2, 3, 4; (ii)       +− 5 i2 15 i2 2e ππ k , k = 0, 1, 2; (iv) 15 i2 e2 π − . Question 10 This question was somewhat different to recent linear algebra questions. There were few complete answers and a good deal of muddled thinking. Those attempting the first part by row operations seldom gave a complete set of values of a for which the system of equations had a unique solution. This was surprising, as intelligent reading of the next two parts of the question would give a good clue as to what was expected. Formal set notation was not required here. Those who said that the system of equations has no unique solution when the appropriate determinant was zero fared much better. A sizeable number of candidates thought that because the appropriate determinant was zero, when a took the value 18, that the system would have no solution. The distinction between ‘no solution’ and ‘no unique sol