Friday 24 June 2016 Morning

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1 Oxford Cambridge and RSA Friday June 6 Morning A GCE MATHEMATICS (MEI) 75/A Applications of Advanced Mathematics (C) Paper A QUESTION PAPER *68656* Candidates answer on the Printed Answer Book. OCR supplied materials: Printed Answer Book 75/A MEI Examination Formulae and Tables (MF) Other materials required: Scientific or graphical calculator Duration: hour minutes INSTRUCTIONS TO CANDIDATES These instructions are the same on the Printed Answer Book and the Question Paper. The Question Paper will be found inside the Printed Answer Book. Write your name, centre number and candidate number in the spaces provided on the Printed Answer Book. Please write clearly and in capital letters. Write your answer to each question in the space provided in the Printed Answer Book. If additional space is required, you should use the lined page(s) at the end of the Printed Answer Book. The question number(s) must be clearly shown. Use black ink. HB pencil may be used for graphs and diagrams only. Read each question carefully. Make sure you know what you have to do before starting your answer. Answer all the questions. Do not write in the bar codes. You are permitted to use a scientific or graphical calculator in this paper. Final answers should be given to a degree of accuracy appropriate to the context. INFORMATION FOR CANDIDATES This information is the same on the Printed Answer Book and the Question Paper. The number of marks is given in brackets [ ] at the end of each question or part question on the Question Paper. 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. The total number of marks for this paper is 7. The Printed Answer Book consists of 6 pages. The Question Paper consists of 8 pages. Any blank pages are indicated. This paper will be followed by Paper B: Comprehension. INSTRUCTION TO EXAMS OFFICER / INVIGILATOR Do not send this Question Paper for marking; it should be retained in the centre or recycled. Please contact OCR Copyright should you wish to re-use this document. OCR 6 [T//65] DC (NF/FD) 76/ OCR is an exempt Charity Turn over

2 Section A (6 marks) Express cosi- sin i in the form R cos^i+ ah, where R and a Hence show that the equation cosi- sin i = has no solution. [6] q x Given that c+ m = - x+ x + f, find p and q, and state the set of values of x for which the expansion p is valid. [7] r. Fig. shows the curve y x = and the line y =. y O x Fig. The finite region enclosed by the curve and the line is rotated through 8 about the y-axis. Find the exact volume of revolution generated. [] Solve the equation sini = + cos i for G i G 8. [5] 5 In Fig. 5, triangles ABC, ACD and ADE are all right-angled, and angles BAC, CAD and DAE are all i. AB = x and AE = x. E D x C θ θ θ A x B Fig. 5 (i) Show that sec i =. [] (ii) Hence show the ratio of lengths ED to CB is. [] OCR 6 75/A Jun6

3 6 P is a general point on the curve with parametric equations x = t, y =. This is shown in Fig. 6. The t tangent at P intersects the x- and y-axes at the points Q and R respectively. y R P (t, ) t O Q x Fig. 6 Show that the area of the triangle OQR, where O is the origin, is independent of t. [7] OCR 6 75/A Jun6 Turn over

4 Section B (6 marks) 7 Fig. 7 shows a cuboid OABCDEFG with coordinates as shown. The point P has coordinates (,, ). z G (,,5) D (,,5) F (,,5) E (,,5) y C (,,) O P (,,) B (,,) A (,,) x Fig. 7 (i) Find the length of the diagonal AG. [] (ii) Show that the vector n = 5i- j+ k is normal to the plane DPF. Hence find the cartesian equation of this plane. [6] The diagonal AG intersects the plane DPF at Q. (iii) Write down a vector equation of the line AG. Hence find the coordinates of the point Q, and the ratio AQ : QG. [6] (iv) Find the acute angle between the line AG and the plane DPF. [] OCR 6 75/A Jun6

5 5 8 (i) Show that + x + - x = ^ + xh^ -xh. [] In a chemical reaction, the time t minutes taken for a mass x mg of a substance to be produced is modelled by the equation (ii) Show that when t =, x =. + x t = lnc m. - x [] (iii) Show that the rate of change of x is proportional to the product of ( + x) and ( x), and find the constant of proportionality. [] ^ -t - e h (iv) Show that x = -t. + e Hence determine the long-term mass of the substance predicted by this model. [] In another chemical reaction, the mass x mg at time t minutes is modelled by the differential equation dx - = k^+ xh^- xh e t, dt where k is a positive constant, and x = when t =. x (v) Show by integration that, for this reaction, lnc m = k^ e - x h. [5] (vi) Given that the long-term mass of this substance is.85 mg, find the value of k. [] END OF QUESTION PAPER OCR 6 75/A Jun6

6 75A Mark Scheme June 6 Question Answer Marks Guidance cos sin R cos cos sin sin Rcos, Rsin M A Correct pairs. Condone sign errors for the M mark (so accept Rsin ) R R B Or. or better, not unless chosen tan.9 M A ft their pairs (condone sign errors but division must be the correct way round), A for.9 or better (accept.5), with no errors seen in method for angle Maximum value of cos sin is B Or equivalent convincing numerical statement that no solutions exist e.g.. Maybe embedded in an attempt at a solution. Do not accept general statements e.g. doesn t work must be clear why no solutions exist dependent on first B SC: If candidates state that cos,sin tan this could score MABMAB (so max /6) [6] Note that those candidates who state R and tan with no (wrong) working seen could go on to score full marks 7

7 75A Mark Scheme June 6 Question Answer Marks Guidance q x ( ) x q q x q p p! p M* One of q x p or q( q ) x! p q p qq ( ) A A Allow x s on both sides of equations (if correct) p q (soi), for example, scores M A p p( p) qq ( ) q p or Mdep* Eliminating p (or q ) from simultaneous equations (not involving x) involving p q both variables oe if MAA awarded followed by either p or q correct (www) this implies this M mark p A p www (or q ) q Aft q (or p ) for second value, ft their p or q (e.g. the negative of their p or q) provided first marks awarded and only a single computational error in the method so must be a correct method for solving their equation in p or q (ignore mention of p and/or q ) x Valid for x A or x www, allow x but not say, x SC If M M awarded and no wrong working seen then B for p = and q = -, B for - < x < (oe) so max marks [7] Guidance for solving quadratics on this paper: use of correct quadratic equation formula (if formula is quoted correctly then only one sign slip is permitted, if the formula is quoted incorrectly M, if not quoted at all substitution must be completely correct to earn the M) or factorising (giving their x term and one other term when factors multiplied out) or completing the square (must get to the square root stage involving and arithmetical errors may be condoned provided that perfect square term was correct) 8

8 75A Mark Scheme June 6 Question Answer Marks Guidance V x dy M M for ( ) (d ) k x y with correct limits and k or, allow correct limits seen or implied later, if formula not stated then must substitute for their correctly to imply this formula condone lack of for the M mark and dy throughout (condone incorrect use of dx too) V y dy A Correct (or with a /) limits may be seen or implied through later working y y 6 B A [] y or y (but only if k ), condone Exact mark final answer (so no isw if correct answer is subsequently halved) but if exact value seen and is then followed by then isw y.5.5 (oe) x 9

9 75A Mark Scheme June 6 Question Answer Marks Guidance sin cos cos M* Use of correct double angle formulae: sin sin cos and any one of cos cos sin or sin or cos cos sin cos A Correct equation in solvable form e.g. sincos (oe) or 5sin 6sin or 5cos cos but not sin cos cos tan Mdep* Use of sin tan on their sin cos cos or correct method for solving quadratic in eithersin or cos (See guidance in question for solving quadratics) 6.6 A www (6.6 or better) 9 B Not from incorrect working Ignore additional solutions outside the range. If any additional solutions given inside the range 8 and full marks would have been awarded then remove last mark (so /5) Both answers in radians:.6 (or better) and / scores B [5] Answers with no working scores B B(so max /5)

10 75A Mark Scheme June 6 Question Answer Marks Guidance 5 (i) AC xsec B Accept any equivalent forme.g. ACcos x. If AC not seen then there must be a diagram as evidence of correct sides - xsec with no AC is B AD sec and AE sec x x B Accept x xsec (as AE = x) or any equivalent form. Otherwise there must be a corresponding diagram as evidence of correct sides. Accept cos x/ AC AC / AD AD / xfor the first two marks xsec x This line (oe) must be seen before the x s cancelled sec * B NB AG dependent on all previous marks [] 5 (ii) OR AD xcos B Same principles as above for each corresponding mark AC xcos and AB xcos B or x xcos (as AB = x) xcos x sec * B Must see xcos x(oe) before given answer ED xsin B oe e.g. ED x AD or ED ADtan with AD correctly expressed in terms of x and (or using 7.5 or better) - see (i) for alternatives for AD. Allow ED =.x (or better) but B if ED = missing CB xtan B oe e.g. CB AC x or CB ACsin with AC correctly expressed in terms of x and (or using 7.5 or better) - see (i) for alternatives for AC. Allow CB =.77x (or better) but B if CB = missing ED xsin cos CB x tan B www must come from exact working (so not using oe) - accept ED ED or sec (oe) (as from (i): sec ) CB sec CB / * B NB AG dependent on all previous marks in (ii) must be one step of intermediate working from cos to given answer []

11 75A Mark Scheme June 6 Question Answer Marks Guidance dy d y / dt / t M* M for their (d y/ d t ) / theird x/ dt in terms of t with at least one term correct dx d x / dt t 6 A A cao (oe) allow unsimplified even if subsequently cancelled incorrectly i.e. can isw with any non-zero gradient expressed as a function of t - or t any equivalent form (e.g. y = mx + c) but must have used the correct point, t t - if using y = mx + c must explicitly have c = before M can be awarded Mdep* y f ( t) x t y x t t t When x =, y t Aft Must be a function of t When y =, x = t Aft Must be a function of t A oe need not be simplified So area of triangle = t 8 (which is t independent of t) OR (for the first two marks) dy y x x dx x dy d x ( t) A [7] No ft on this mark an answer of 8 (www) with no additional comment is sufficient to award this mark M* Attempt to eliminate t and correctly differentiates their Cartesian equation A

12 75A Mark Scheme June 6 Question Answer Marks Guidance 7 (i) B AG 5 Condone etc. if recovered = 5 5 B Correct answer implies both marks accept 7. or better [] 7 (ii) DP = i + j 5k or PD= i j + 5k B One correct direction vector in plane DPF (oe e.g. expressed as a column vector) DF= i + j (or PF = j + 5k) B Any other correct direction vector in plane DPF n DP = 5 + ( 5) = B Scalar product with a direction vector in the plane (including evaluation and = ) (OR M forms vector cross product with at least two correct terms in solution) n DF = 5 = (or n P F 5 ) B Scalar product with a second direction vector in the plane (including evaluation and = ) (following OR above, A all correct ie a multiple of 5i j k ) (NB finding only one direction vector and its scalar product is B B B B) M 5x y z c oe (accept any non-cartesian form for M only) rn an 5x y + z = A MA for 5i j k or 5x y + z SC: if states if 5i j k is normal then of the form5x y z c and substitutes one coordinate gets MA, then substitutes the other two coordinates A (not A, A, A). Then states so 5i j k is normal and states the correct equation of the plane this can get B provided that there is a clear argument ie MAAB. Without a clear argument this is B [6] SC: if finds two relevant direction vectors B B and then finds equation of the plane from vector form, r a b c gets B. Eliminating parameters B cao. If then states so 5i j k is normal can get B, and then a valid reason why 5i j k is normal scores the final B mark (each B mark is dependent on the previous one)

13 75A Mark Scheme June 6 Question Answer Marks Guidance 7 (iii) x r = i + B Need r (or another single letter) = or y for first B z r j 5k i j5k - accept column.+ ( i + j + 5k) B NB answer is not unique e.g. vector form and condone row vectors (non-vector form scores B only) 5( ) () + (5) = M Substituting their line in their plane equation from (ii) (condone a slip if intention clear) their line and plane must be of the correct form (e.g. the line must be of the form r a tb ) =, =. A www cao NB is not unique as depends on choice of line seen in this part Q is (.,., ) A www - condone answer given as a position vector 7 (iv) AQ : QG = : A oe www [6] M* Selecting 5i j k and their direction vector from (iii) Angle between i j 5k and (5i j k) is where 5) () (5 cos Mdep* Correct formula (including cosine), with correct substitution, using their direction 5 6 vector from (iii) and the correct normal vector - condone either a single numerical slip or a single sign slip (but not one of each) if intention is clear. So it must be clear where the slip comes from e.g. if the magnitude of one vector is stated incorrectly with no working then this is M 56. or. A www cao (accept 56 or better, or better,.6 radians or better,.978 radians or better) angle between line and plane =. A www cao (accept or better,.59 radians or better) []

14 75A Mark Scheme June 6 Question Answer Marks Guidance 8 (i) x x * B NB AG must be at least one intermediate step before given answer correct x x ( x)( x) ( x)( x) application of partial fractions is fine [] 8 (ii) x x ln x x B or e or ln( x) ln( x) x x x B If only this line seen then award BB 8 (iii) x t ln ln( x) ln( x) x dt dx x x [] B M SC: Allow B only for verifying that when x =, t = Correct differentiation of their t OR for first two marks - If no subtraction law of logs seen e.g. dt ( x)() ( x)( ) d x x award B for correct first bracket ( x) x (reciprocal expression) and B for second correct bracket (quotient/chain rule)(oe) if additional constant(s) added (e.g. t = k ln( )) then award B only for a constant times a fully correct derivative dt dx ( x )( x ) d x ( x)( x) dt A dt dx and d x must be correctly attributed to the correct expression for this mark dt k A Explicitly stating (that the constant of proportionality is) possible to score AAin this part See next page for an alternative solution [] - therefore it is 5

15 75A Mark Scheme June 6 Question Answer Marks Guidance OR dx dx k( x)( x) k dt d t ( x)( x) B Allow omission of dx and/or dt ln( x) ln( x) kt( c) M Any non-zero constant, - further guidance in (v) for this and the next mark A www oe (condone absence of c) x ) ln( x ) kt ( c ) x, t c x ln kt k A www must include c and show that c =. Must explicitly state that k x 8 (iv) t x e x B t ( x)e x t t e xe x t t x( e ) e M Multiplying out, collecting x terms (condone sign slips and numerical errors (eg loss of a ) only but M if e t incorrectly replaced with e t ) and factorising their x terms correctly t t t t (e ) (e )e ( e ) A x t t t t * e ( e )e e ( x ) (as t ) B [] OR (for first three marks) t x e B x t ( x)e x t t e xe x t x( e ) e t ( e ) x t * e t M A www NB AG as AG must be an indication of how previous line leads to the required result (eg stating or showing multiplying by e t ) Multiplying out, collecting x terms (condone sign slips as above) and factorising their x terms correctly www NB AG 6

16 75A Mark Scheme June 6 Question Answer Marks Guidance 8 (v) Separating variables - condone sign slips and issues with placement of k but M t dx k e dt xx M* for ( x)( x)d x... or equivalent algebraic error in separating variables unless recovered. If no subsequent work integral signs needed, but allow omission of dx and/or dt but must be correctly placed if present t ln( x) ln( x) e ( c) A Any non-zero constants,, - this line must be seen and cannot be implied by later working (as this is an AG) condone absence of c or if a constant present condone the use of k for their constant. Do not condone invisible brackets e.g. ln + x unless recovered before subtraction law of logs applied all of these points apply to the next A mark too t ln( x) ln( x) ke ( c) A www oe When t, x c k Mdep* Substituting x, t into each term in an attempt to find their c (must get c = ) if they integrate and use k as their constant they must use x, t to find this single k term only x t ln k( e )* x A www NB AG must have obtained all previous marks in this part [5] OR (for first marks) final MA as above t dx k e dt xx M* Separating variables. If no subsequent work integral signs needed, but allow omission of dx or dt, but must be correctly placed if present x t t dx k e dt ln ke ( c) x x A Must see / x on lhs please note that one A mark cannot be awarded 8 (vi) as t, x.85 ln.85/.5 = k M Sets e t to and substitutes x =.85 condone substitution of a large value of t only if it leads to the correct value of k k =.8 A k.5ln 77 / or.8 or better [] 7

17 OCR Report to Centres June 6 75 Applications of Advanced Mathematics (C) General Comments: The performance of candidates on Paper A was similar and comparable to recent papers and the standard of work in the majority of cases was very high. This paper was accessible to all candidates but there were sufficient questions for the more able candidates to show their skills. Paper B, the comprehension, was well understood and most candidates scored good marks here. Candidates made similar errors as in previous years and these included: Sign and basic algebraic errors (Question ) Failure to include a constant of integration (Question 8(v)) Poor anti-logging and rules of logarithms (Questions 8(iv) and 8(v)) Failure to give clear descriptions in the comprehension paper (Questions and 6) Inappropriate accuracy, for example in Question, candidates either gave insufficient accuracy (answers to the nearest integer) or they gave too much accuracy (answers to or more decimal places) candidates are reminded to give answers to decimal place for questions involving trigonometry Failure to give exact answers when required (Questions and 5(ii)) Failure to give sufficient detail when verifying given results (Questions 5(i), 5(ii), 6, 7(ii), 8(i), 8(ii), 8(iii), 8(iv) and 8(v)). Some candidates assume that showing that a vector is perpendicular to one vector in the plane is sufficient to show that it is a normal vector. Quite a number of candidates failed to attempt some parts but there did not appear to be a shortage of time for either Paper. Centres are again reminded that as Papers A and B are marked separately any supplementary sheets used must be attached to the appropriate paper. Furthermore, centres are requested that Papers A and B are not attached to each other and they must be sent separately for marking. Comments on Individual Questions: Paper A Question The majority of candidates correctly worked out the values of R and although some candidates lost the first method mark by not including R in the expanded trigonometric statements Rcos and Rsin. Some failed to give in radians and a small minority stated R as rather than the correct. Candidates were less successfully in showing that cossin had no solutions with many simply stating that.9 arccos does not work or gives a math error. Many candidates failed to explain or give an equivalent mathematical statement that the maximum value of cos sin is which is less than and so did not score the final mark in this question.

18 OCR Report to Centres June 6 Question q x The binomial expansion of was done extremely well by the vast majority of candidates p with the most common error being the failure to correctly deal with the x term with many giving the qq coefficient (of this term) as ( ) qq ( ) rather than the correct. It was surprising how few p p candidates could go on to form the correct pair of simultaneous equations and fewer still who could solve this pair of equations accurately and successfully. Those candidates who correctly found the value of p usually went on to state the set of values of x for which the expansion was valid. Question The vast majority of candidates considered the correct integral (with correct limits and including the factor of ) for the volume of revolution generated by rotating the given curve about the y -axis and most went on to integrate y correctly. A number of candidates, however, misread the question and instead tried to calculate the volume or revolution generated by rotating the curve about the x -axis. It was the mention of a rotation of 8 that seemed to concern many candidates and a considerable number divided the correct answer of 6 by. Question The majority of candidates correctly replacedsin with sincos and cos with one of cos sin or sin or cos, although a minority of candidates made the costly mistake of forgetting the in the latter two identities. While a majority of candidates correctly obtained sin cos cos many then cancelled cos from both sides of the equation instead of cos sin cos and so lost the solution to the equationcos. factorising to obtain Most candidates correctly simplified sincos to tan and obtained the correct answer of 6.6 although a small minority gave an answer in radians or additional answers both inside and outside of the given range. A number of candidates, after applying suitable double angle identities, squared their equation leading to either a quadratic equation in either sin or cos. These attempts usually contained sign and/or algebraic errors or even, when the correct disguised quadratic was obtained, and solved, additional incorrect solutions (coming from the earlier squaring of the single angle equation) were given. Question 5 This question provided a certain amount of discrimination between candidates with some producing clear, concise arguments for why sec and why the ratio of the lengths ED to CB was : while a significant number left both parts of this question blank or scored no marks. The majority of candidates, however, scored at least one mark in (i) for starting that AC xsec (or equivalent) or that AD xcos but many failed to find corresponding expressions for either AD and AE or AC and AB in terms of x and one of sec or cos. Examiners noted that many candidates did not make it clear which expression corresponded to which side of the three triangles given in the question making it almost impossible for examiners to award any marks. 5

19 OCR Report to Centres June 6 In part (ii) many candidates scored at least two marks for stating that ED xsin and CB xtan although many then substituted in the angle from part (i) and tried to derive the exact value of using approximate values for these two lengths. Candidates who correctly found that ED cos usually went on to obtain the correct ratio although many did not show sufficient steps CB of working to explain how they obtained the given answer. Question 6 Candidates found this unstructured question on parametric equations quite demanding with the modal mark scored being out of a possible 7. Most candidates scored these two marks by dy correctly obtaining although the vast majority failed to make any further progress worthy dx t of merit with many trying to argue that the area of triangle OQR is independent of t without any further working of a mathematical nature. Of those that failed to obtain the correct derivative a number incorrectly stated that d y ln t dt or implied that d y d y d x. All that was required from dx dt dt those candidates who had obtained the correct gradient function in terms of t was to write down the equation of the tangenty x t t t, substitute x and y to obtain R, t and Q t, respectively and hence calculate the area of the triangle as 8 t which is t clearly independent of t. A number of candidates began by finding the Cartesian equation of the dy curve and correctly obtaining but they then incorrectly used this gradient function in their dx x equation of the tangent. Question 7 Nearly all candidates correctly obtained the length of the diagonal AG in part (i) with only a small minority stating only the direction vector AG. In part (ii) a number of candidates seemed to think that just showing one direction vector was normal to the plane was sufficient and some candidates showed all three. It was unfortunate that so many lost marks by not showing the evaluation of the scalar product(s) even though this was a show that question and so examiners had to be convinced that the candidates were indeed showing the required results and not simply stating them. The vast majority found the correct Cartesian equation of the plane as 5x y z. Most candidates correctly found the direction part of the vector equation of the line AG in part (ii) but very few candidates stated a correct vector equation which needed to begin with either x y... or r... The rest of this part was answered well with many correctly obtaining the value z of the parameter and hence the coordinates of the point Q. The ratio of AQ : QG was often found correctly although a number of candidates stated this ratio as either : or : 5. Candidates attempts to find the acute angle between the line AG and the plane DPF in part (iv) were varied with the vast majority considering the correct direction vectors of i j5k and 5i j k although a number attempted to find this angle using one of the direction vectors in 6

20 OCR Report to Centres June 6 the plane (rather than the normal to the plane). While many correctly obtained an angle of either 56. or. few failed to realise that these were the acute and obtuse angles between the line and the normal and so many did not subtract the relevant right-angle to obtain the correct answer of.. Question 8 Nearly all candidates correctly showed the required result in part (i) although a few attempted to use partial fractions with varying degrees of success. In part (ii) many candidates incorrectly verified that when x, t rather than the required result of showing that whent, x. Those that did begin by setting t usually went on to score both marks in this part. Part (iii) proved to be quite discriminating with many candidates unable to show that the rate of change of x was proportional to the given product. The most common method seen was to write t as ln( x) ln( x) and then to differentiate this expression with respect to x and obtain dt and then use part (i) to show that d t d x ( x )( x ), and dx x x d x ( x)( x) dt hence the constant of proportionality is clearly. The most common error was a failure to differentiate t correctly with many retaining the negative between the two terms. A number of candidates attempted instead to derive the given result by starting with the differential equation dx k( x)( x) and attempting to solve this using the method of separation of variables. While dt a number were successful in obtaining the required constant many failed to deal with dx correctly or forgot to include the required constant of integration. ( x)( x) t x t x Part (iv) was answered well with many correctly starting by either writing e or e, x x the latter of these two usually lead to the correct given answer while the former lead to t e x with the vast majority of candidates being unable to explain clearly why this would t e lead to the given result. As this was a show that question there needed to be a clear indication of t ( e ) how this result would leads to x t. Finally in this part many candidates correctly stated e that the long-term mass of the substance was mg. Part (v) was answered with varying degrees of success with the vast majority correctly separating dx t the variables to obtain k e dt ( x)( x) - however, from this point it was all too clear that a number of candidates did not, as requested, show by integration the given result, but simply wrote down the given answer (or an answer only a single step away from the given answer) without clearly showing how either side of the given equation was obtained. In many cases candidates failed to include a constant of integration that needed to be found using the given initial conditions. 7

21 OCR Report to Centres June 6 Part (vi) was answered extremely well with many candidates obtaining the correct answer of.8 which was achieved by setting e t equal to zero and substituting.85 for x. The most common error x seen by examiners was to set ln equal to.85 and solve for k with e t equal to zero. x Paper B Question Nearly all candidates correctly stated the hub height of the turbine although a number incorrectly found this height as 59.5 m (which came from using the diameter of the blade rather than the radius). Question While the majority of candidates correctly explained how the figure of m was obtained with the most common method being via the calculation which lead to a value of.88 many 6.7 did not give sufficient detail or tried to justify the figure of without showing any calculation at all. Question Parts (i) and (ii) were nearly always correct although in part (iii) many candidates failed to use the value of 7. to show that when the photomontage was printed on A paper, the height of the wind turbine was consistent with the angle of elevation found in part (ii). Question The responses to this question were mixed with many candidates failing to find the height above A of the lowest visible point as 99.5 m with many incorrectly using a height of 9.5 m. Furthermore, many candidates failed to read the question carefully and stated that the distance AC was m rather the horizontal distance from A to C. Question 5 7 Nearly all candidates correctly stated that arctan although many assumed incorrectly 8 that triangle TQB was right-angled. Of those that did realise that triangle TQB was not right-angled many took the slightly more long-winded approach of using the cosine rule to find instead of 9 8 realising that arctan arctan. Finally, many candidates did not realise that if they 8 8 are required to show that two answers agree to significant figures then they must quote both value correct to significant figures and so in this case examiners needed to see as a minimum of 5. and 5. followed by 5.. 8

22 GCE Mathematics (MEI) Max Mark a b c d e u 75 C MEI Introduction to advanced mathematics (AS) Raw C MEI Concepts for advanced mathematics (AS) Raw (C) MEI Methods for Advanced Mathematics with Coursework: Written Paper Raw (C) MEI Methods for Advanced Mathematics with Coursework: Coursework Raw (C) MEI Methods for Advanced Mathematics with 8 Coursework: Carried Forward Coursework Mark Raw C MEI Applications of advanced mathematics (A) Raw FP MEI Further concepts for advanced mathematics (AS) Raw FP MEI Further methods for advanced mathematics (A) Raw FP MEI Further applications of advanced mathematics (A) Raw (DE) MEI Differential Equations with Coursework: Written Paper Raw (DE) MEI Differential Equations with Coursework: Coursework Raw (DE) MEI Differential Equations with Coursework: Carried 8 Forward Coursework Mark Raw M MEI Mechanics (AS) Raw M MEI Mechanics (A) Raw M MEI Mechanics (A) Raw M MEI Mechanics (A) Raw S MEI Statistics (AS) Raw S MEI Statistics (A) Raw S MEI Statistics (A) Raw S MEI Statistics (A) Raw D MEI Decision mathematics (AS) Raw D MEI Decision mathematics (A) Raw DC MEI Decision mathematics computation (A) Raw (NM) MEI Numerical Methods with Coursework: Written Paper Raw (NM) MEI Numerical Methods with Coursework: Coursework Raw (NM) MEI Numerical Methods with Coursework: Carried 8 Forward Coursework Mark Raw NC MEI Numerical computation (A) Raw FPT - Further pure mathematics with technology (A) Raw Published: 7 August 6 Version.

23 GCE Statistics (MEI) Max Mark a b c d e u G Statistics MEI (Z) Raw UMS G Statistics MEI (Z) Raw UMS G Statistics MEI (Z) Raw UMS GCE Quantitative Methods (MEI) Max Mark a b c d e u G Introduction to Quantitative Methods MEI Raw G Introduction to Quantitative Methods MEI Raw G5 Statistics MEI Raw UMS G6 Decision MEI Raw UMS Level Certificate and FSMQ raw mark grade boundaries June 6 series For more information about results and grade calculations, see Level Certificate Mathematics for Engineering H86 Mathematics for Engineering H86 Mathematics for Engineering Max Mark a* a b c d e u This unit has no entries in June 6 Level Certificate Mathematical Techniques and Applications for Engineers Max Mark a* a b c d e u H865 Component Level Certificate Mathematics - Quantitative Reasoning (MEI) (GQ Reform) Raw Max Mark a b c d e u H866 Introduction to quantitative reasoning Raw H866 Critical maths Raw Overall Level Certificate Mathematics - Quantitive Problem Solving (MEI) (GQ Reform) Max Mark a b c d e u H867 Introduction to quantitative reasoning Raw H867 Statistical problem solving Raw Overall Advanced Free Standing Mathematics Qualification (FSMQ) Max Mark a b c d e u 699 Additional Mathematics Raw Intermediate Free Standing Mathematics Qualification (FSMQ) Max Mark a b c d e u 6989 Foundations of Advanced Mathematics (MEI) Raw Published: 7 August 6 Version.