2017 Mathematics Paper 1 (Non-calculator) Higher. Finalised Marking Instructions

Transcription

1 National Qualifications Mathematics Paper (Non-calculator) Higher Finalised Marking Instructions Scottish Qualifications Authority 07 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is reproduced, SQA should be clearly acknowledged as the source. If it is to be used for any other purpose, written permission must be obtained from permissions@sqa.org.uk. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of eamination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Assessment team may be able to direct you to the secondary sources. These ing instructions have been prepared by eamination teams for use by SQA appointed ers when ing eternal course assessments. This publication must not be reproduced for commercial or trade purposes.

2 General ing principles for Higher Mathematics This information is provided to help you understand the general principles you must apply when ing candidate responses to questions in this Paper. These principles must be read in conjunction with the detailed ing instructions, which identify the key features required in candidate responses. For each question the ing instructions are generally in two sections, namely Illustrative Scheme and Generic Scheme. The illustrative scheme covers methods which are commonly seen throughout the ing. The generic scheme indicates the rationale for which each is awarded. In general, ers should use the illustrative scheme and only use the generic scheme where a candidate has used a method not covered in the illustrative scheme. (a) (b) (c) (d) (e) (f) (g) (h) (i) Marks for each candidate response must always be assigned in line with these general ing principles and the detailed ing instructions for this assessment. Marking should always be positive. This means that, for each candidate response, s are accumulated for the demonstration of relevant skills, knowledge and understanding: they are not deducted from a maimum on the basis of errors or omissions. If a specific candidate response does not seem to be covered by either the principles or detailed ing instructions, and you are uncertain how to assess it, you must seek guidance from your Team Leader. Credit must be assigned in accordance with the specific assessment guidelines. One is available for each. There are no half s. Working subsequent to an error must be followed through, with possible credit for the subsequent working, provided that the level of difficulty involved is approimately similar. Where, subsequent to an error, the working for a follow through has been eased, the follow through cannot be awarded. As indicated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Unless specifically mentioned in the ing instructions, a correct answer with no working receives no credit. Candidates may use any mathematically correct method to answer questions ecept in cases where a particular method is specified or ecluded. As a consequence of an error perceived to be trivial, casual or insignificant, eg 66 candidates lose the opportunity of gaining a. However, note the second eample in comment (j). page

3 (j) Where a transcription error (paper to script or within script) occurs, the candidate should normally lose the opportunity to be awarded the net process, eg This is a transcription error and so the is not awarded. Eased as no longer a solution of a quadratic equation so is not awarded. Eceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt and all s awarded ( )( ) 0 or (k) Horizontal/vertical ing Where a question results in two pairs of solutions, this technique should be applied, but only if indicated in the detailed ing instructions for the question. Eample: = = y = 5 y = 7 Horizontal: 5 = and = Vertical: 5 = and y = 5 6 y = 5 and y = 7 6 = and y = 7 Markers should choose whichever method benefits the candidate, but not a combination of both. (l) In final answers, unless specifically mentioned in the detailed ing instructions, numerical values should be simplified as far as possible, eg: 5 must be simplified to 5 or 5 must be simplified to must be simplified to 8* must be simplified to must be simplified to 5 *The square root of perfect squares up to and including 00 must be known. (m) Commonly Observed Responses (COR) are shown in the ing instructions to help common and/or non-routine solutions. CORs may also be used as a guide when ing similar non-routine candidate responses. page

4 (n) Unless specifically mentioned in the ing instructions, the following should not be penalised: Working subsequent to a correct answer Correct working in the wrong part of a question Legitimate variations in numerical answers/algebraic epressions, eg angles in degrees rounded to nearest degree Omission of units Bad form (bad form only becomes bad form if subsequent working is correct), eg ( )( ) written as ( ) 6 written as credit gains full Repeated error within a question, but not between questions or papers (o) (p) (q) (r) In any Show that question, where the candidate has to arrive at a required result, the last of that part is not available as a follow-through from a previous error unless specified in the detailed ing instructions. All working should be carefully checked, even where a fundamental misunderstanding is apparent early in the candidate's response. Marks may still be available later in the question so reference must be made continually to the ing instructions. The appearance of the correct answer does not necessarily indicate that the candidate has gained all the available s. Scored-out working which has not been replaced should be ed where still legible. However, if the scored out working has been replaced, only the work which has not been scored out should be ed. Where a candidate has made multiple attempts using the same strategy and not identified their final answer, all attempts and award the lowest. Where a candidate has tried different valid strategies, apply the above ruling to attempts within each strategy and then award the highest resultant. For eample: Strategy attempt is worth s. Strategy attempt is worth s. From the attempts using strategy, the resultant would be. Strategy attempt is worth. Strategy attempt is worth 5 s. From the attempts using strategy, the resultant would be. In this case, award s. page

5 Specific ing instructions for each question Question Generic scheme Illustrative scheme. (a) evaluate epression 0 Question Generic scheme Illustrative scheme. (b) interpret notation state epression for g f. For cos5 without working, award both and. g( 5 ) cos5. Candidates who interpret the composite function as either g f or g f do not gain any s. g f 0 award. However, 0 cos with no working does not gain any s. cos.. g f leading to cos and. Candidate A g f cos 5 0 cos 5 followed by incorrect simplification of the function award page 5

6 . state coordinates of centre (, ) find gradient of radius state perpendicular gradient determine equation of tangent y 5. Accept 6 for.. The perpendicular gradient must be simplified at or stage for to be available.. is only available as a consequence of trying to find and use a perpendicular gradient.. At, accept y5 0, y 5 or any other rearrangement of the equation where the constant terms have been simplified. Question Generic scheme Illustrative scheme. start to differentiate... complete differentiation.... is awarded for correct application of the chain rule. Candidate A dy d dy 6 d Working subsequent to a correct answer: General Marking Principle (n) Candidate B dy 6 d Incorrect answer with no working page 6

7 . Method use the discriminant Method ( k 5) apply condition and simplify determine the value of k Method communicate and epress in factorised form epand and compare 6 k 0 or 6 k k 9 Method equal roots k 5 leading to k 5 determine the value of k k 9. At the stage, treat k 5 as bad form only if the candidate treats k 5 as if it is bracketed in their net line of working. See Candidates A and B.. In Method if candidates use any condition other than discriminant 0 then is lost and is unavailable. Candidate A Candidate B k 5 /\/\/\/\/\/\ 6 k 0 k 9 k 5 k 0 k page 7

8 5. (a) evaluate scalar product Question Generic scheme Illustrative scheme 5. (b) calculate u 7 use scalar product 7 cos evaluate uw. 9 or 5. Candidates who treat negative signs with a lack of rigour and arrive at 7 gain.. Surds must be fully simplified for to be awarded. page 8

9 6. Method equate composite function to write h h h in terms of Method h h h 7 state inverse function Method write as y rearrange 7 and start to complete rearrangement h ( ) 7 or - h ( ) ( 7 ) y7 y 7 Method state inverse function Method interchange variables h ( ) 7 or h ( ) ( 7 ) y 7 Method complete rearrangement y 7 state inverse function h ( ) 7 or h ( ) ( 7 ). y 7 or y ( 7 ) does not gain.. At stage, accept h epressed in terms of any dummy variable eg. h ( ) 7 or h ( ) ( 7 ) with no working gains /. h ( y) y 7. page 9

10 Candidate A h 7 ^ 7 7 awarded for knowing to perform the inverse operations in reverse order 7 h ( ) 7 Candidate B BEWARE h( )... Candidate C h ( ) 7 With no working 0/ page 0

11 7. find midpoint of AB (,7 ) demonstrate the line is vertical m median undefined state equation. m median alone is not sufficient to gain. Candidates must use either vertical or 0 undefined. However is still available.. mmedian mmedian impossible mmedian infinite are not acceptable for However, if these are followed by either vertical or undefined then award, and is still available.. mmedian 0 undefined mmedian undefined are not acceptable for is not available as a consequence of using a numeric gradient; however, see notes 5 and 6. a,b, using the coordinates of A and B and 5. For candidates who find an incorrect midpoint find the median through C without any further errors award /. However, if a, then both and are available. 6. For candidates who find 5y (median through B) or y (median through A) award /. Candidate A,7 m 0 0 undefined Candidate B,7 m 0 0 y 7 Candidate C,7 m ^ 0 y Candidate D,7 Median passes through,7 and, Candidate E,7 Both coordinates have an value vertical line page

12 8. write in differentiable form differentiate t t evaluate derivative 50. Candidates who arrive at an epression containing more than one term at award 0/.. is only available for differentiating a term containing a negative power of t. Candidate A Candidate B Candidate C t t 5 t t t implied by Candidate D t t 00 Candidate E t t 5 Candidate F Bad form of chain rule t /\/\/\ t /\/\/\ 50 Candidate G t t /\/\/\ 5 page

13 9. (a) interpret information 8m 6 stated eplicitly state the value of m or in a rearranged form m or m 05. Stating m or simply writing with no other working gains only. Candidate A 8 6 u n Candidate B 8 m 6 un m page

14 9. (b) (i) communicate condition for limit to eist a limit eists as the recurrence relation is linear and. For accept: any of or or statements such as: lies between and or or 0 with no further comment; is a proper fraction. is not available for: or or statements such as: It is between and. or is a fraction.. Candidates who state m can only gain if it is eplicitly stated that m in part (a). 5. Do not accept a for. Candidate C Candidate D (a) m (b) m (a) (b) m page

15 9. (b) (ii) 6 know how to calculate limit or L= L+ 6 5 calculate limit 5 8 b 6. Do not accept L with no further working for. -a 7. and 5 are not available to candidates who conjecture that L = 8 following the calculation of further terms in the sequence. 8. For L = 8 with no working, award 0/. 9. For candidates who use a value of m appearing e nihilo or which is inconsistent with their answer in part (a) and 5 are not available. Candidate E no valid limit (a) m 6 (b) L L 5 page 5

16 0. (a) know to integrate between appropriate limits... 0 d Method use upper lower 0 integrate 5 substitute limits evaluate area 5 8 Method know to integrate between appropriate limits for both integrals... d and d integrate both functions and substitute limits into both functions 0 and 0 evaluation of both functions and 5 evidence of subtracting areas page 6

17 . is not available to candidates who omit d.. Treat the absence of brackets at stage as bad form only if the correct integral is obtained at. See Candidates A and B.. Where a candidate differentiates one or more terms at, then, and 5 are unavailable.. Accept unsimplified epressions at e.g.. 5. Do not penalise the inclusion of c. 6. Candidates who substitute limits without integrating do not gain, or is only available if there is evidence that the lower limit 0 has been considered. 8. Do not penalise errors in substitution of 0 at. Candidate A 0 d 5 Candidate B 0 d cannot be negative so 5 6 However,... so Area 6 5 Treating individual integrals as areas Candidate C - Method and Area is 8 5 Candidate D - Method and Area is 8 5 Candidate E - Method and Area cannot be negative Area is 8 5 page 7

18 0. (b) 6 use line quadratic Method 6 d 7 integrate 7 8 substitute limits and evaluate integral state fraction 9 6 use cubic - line Method ( ) d 6 7 integrate 7 8 substitute limits and evaluate integral state fraction 9 6 integrate line 6 Method d 0 7 substitute limits and evaluate integral evidence of subtracting integrals 8 0 or 0 9 state fraction 9 page 8

19 IMPORTANT: Notes prefied by *** may be subject to General Marking Principle (n). If a candidate has been penalised for the error in (a) then they must not be penalised a second time for the same error in (b). 9. *** 6 is not available to candidates who omit d. 0. In Methods and only, treat the absence of brackets at 6 stage as bad form only if the correct integral is obtained at 7.. Candidates who have an incorrect epression to integrate at the and 7 stage due solely to the absence of brackets lose, but are awarded 6.. Where a candidate differentiates one or more terms at 7, then 7, 8 and 9 are unavailable. *** In cases where Note has applied in part (a), 7 is lost but 8 and 9 are available.. In Methods and only, accept unsimplified epressions at 7 e.g.. Do not penalise the inclusion of c. 5. *** 8 in Methods and and 7 in method is only available if there is evidence that the lower limit 0 has been considered. 6. At the 9 stage, the fraction must be consistent with the answers at 5 and 8 for 9 to be awarded. 7. Do not penalise errors in substitution of 0 at 8 in Method & or 7 in Method. page 9

20 . Question Generic scheme Illustrative scheme determine the gradient of given line or of AB or a Method determine the other gradient find a a or 0 Method determine the gradient of given line stated or implied by equation of line and substitute solve for a a 5 7 y 7 0 Candidate A - using simultaneous equations mline y0 y 0 a a a 0 a 0 Candidate B a mab a a Candidate C Method y 7 y 0 y 5 0 y 0 y 0 No mention of a ^ page 0

21 . use laws of logs write in eponential form log 9 a a 9 solve for a 8. 6 must be simplified at or stage for to be awarded.. Accept log9 at.. may be implied by. Candidate A Candidate B Candidate C log a log a log 9 a a a a =9 a ^ a Candidate D log 6 log log a a 6 log a a 6 loga log a 8 a 8 page

22 . write in integrable form start to integrate 5 5 process coefficient of ( ) complete integration and simplify 5 c. For candidates who differentiate throughout, only is available.. For candidates who integrate the denominator without attempting to write in integrable form award 0/.. If candidates start to integrate individual terms within the bracket or attempt to epand a bracket no further s are available.. ' c' is required for. Candidate A 5 Candidate B 5 5 ^ 5 5 c 5 c 6 Candidate C Candidate D Differentiate in part: Differentiate in part: c c page

23 Question Generic Scheme Illustrative Scheme. (a) use compound angle formula k sin cos a k cos sin a stated eplicitly Mark compare coefficients process for k k cos a, k sin a stated eplicitly k process for a and epress in required form. Accept k sin cos a cos sin a sin( 0 ) for. Treat ksin cos a cos sin a as bad form only if the equations at the stage both contain k.. Do not penalise the omission of degree signs.. sin cos a cos sin a sin cos a cos sin a is acceptable for and. or. In the calculation of k =, do not penalise the appearance of. 5. Accept k cos a, ksin a for. 6. is not available for k cos, ksin, however, is still available. 7. is only available for a single value of k, k >0. 8. is not available to candidates who work with throughout parts (a) and (b) without simplifying at any stage. 9. is not available for a value of a given in radians. 0. Candidates may use any form of the wave equation for, and, however, is only available if the value of a is interpreted in the form ksin( a). Evidence for may only appear as a label on the graph in part (b). Responses with missing information in working: Candidate A ^ cosa sin a tan a, a0 sin( 0 ) Candidate B ksin cos a k cos sin a cosa sin a tan a a 0 Not consistent with equations at. sin( 0 ) page

24 Question Generic Scheme Illustrative Scheme Responses with the correct epansion of k sin - a but errors for either or. Candidate C Candidate D Candidate E Mark k cos a, k sin a k cos a, k sin a k cos a, k sin a tan a a 60 tan a a 60 sin( 60) tan a, a 0 sin( 0) Responses with the incorrect labelling; ksin Acos B k cos Asin B from formula list. Candidate F ksin AcosB kcosasin B kcos a ksin a Candidate G ksin AcosB kcosasin B kcos ksin Candidate H ksin AcosB kcosasin B k cosb ksin B tan a, a0 sin( 0) tan, 0 sin( 0) tan B, B 0 sin( 0) page

25 . (b) 5 roots identifiable from graph 5 0 and 0 6 coordinates of both turning points identifiable from graph 6 0, and 00, 7 y-intercept and value of y at 60 identifiable from graph 7. 5, 6 and 7 are only available for attempting to draw a sine graph with a period of 60.. Ignore any part of a graph drawn outwith Vertical ing is not applicable to 5 and Candidates sketch arrived at in (b) must be consistent with the equation obtained in (a), see also candidates I and J. 6. For any incorrect horizontal translation of the graph of the wave function arrived at in part(a) only 6 is available. Candidate I (a) sin 0 correct equation (b) Incorrect translation: sin 0 Sketch of Only 6 is available Candidate J (a) sin 0 incorrect equation (b) Sketch of sin 0 All s are available page 5

26 5. (a) state value of a 5 state value of b Question Generic scheme Illustrative Scheme Mark 5. (b) state value of integral 0. Candidates answer at (b) must be consistent with the value of b obtained in (a).. In parts (b) and (c), candidates who have 0 and -6 accompanied by working, the working must be checked to ensure that no errors have occurred prior to the correct answer appearing. Candidate A From (a) a b 5 hd Question Generic scheme Illustrative scheme 5. (c) state value of derivative -6 [END OF MARKING INSTRUCTIONS] page 6

27 National Qualifications Mathematics Paper Higher Finalised Marking Instructions Scottish Qualifications Authority 07 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is reproduced, SQA should be clearly acknowledged as the source. If it is to be used for any other purpose, written permission must be obtained from permissions@sqa.org.uk. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of eamination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Assessment team may be able to direct you to the secondary sources. These ing instructions have been prepared by eamination teams for use by SQA appointed ers when ing eternal course assessments. This publication must not be reproduced for commercial or trade purposes.

28 General ing principles for Higher Mathematics This information is provided to help you understand the general principles you must apply when ing candidate responses to questions in this Paper. These principles must be read in conjunction with the detailed ing instructions, which identify the key features required in candidate responses. For each question the ing instructions are generally in two sections, namely Illustrative Scheme and Generic Scheme. The illustrative scheme covers methods which are commonly seen throughout the ing. The generic scheme indicates the rationale for which each is awarded. In general, ers should use the illustrative scheme and only use the generic scheme where a candidate has used a method not covered in the illustrative scheme. (a) (b) (c) (d) (e) (f) (g) (h) (i) Marks for each candidate response must always be assigned in line with these general ing principles and the detailed ing instructions for this assessment. Marking should always be positive. This means that, for each candidate response, s are accumulated for the demonstration of relevant skills, knowledge and understanding: they are not deducted from a maimum on the basis of errors or omissions. If a specific candidate response does not seem to be covered by either the principles or detailed ing instructions, and you are uncertain how to assess it, you must seek guidance from your Team Leader. Credit must be assigned in accordance with the specific assessment guidelines. One is available for each. There are no half s. Working subsequent to an error must be followed through, with possible credit for the subsequent working, provided that the level of difficulty involved is approimately similar. Where, subsequent to an error, the working for a follow through has been eased, the follow through cannot be awarded. As indicated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Unless specifically mentioned in the ing instructions, a correct answer with no working receives no credit. Candidates may use any mathematically correct method to answer questions ecept in cases where a particular method is specified or ecluded. As a consequence of an error perceived to be trivial, casual or insignificant, eg 66 candidates lose the opportunity of gaining a. However, note the second eample in comment (j). page

29 (j) Where a transcription error (paper to script or within script) occurs, the candidate should normally lose the opportunity to be awarded the net process, eg This is a transcription error and so the is not awarded. Eased as no longer a solution of a quadratic equation so is not awarded. Eceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt and all s awarded ( )( ) 0 or (k) Horizontal/vertical ing Where a question results in two pairs of solutions, this technique should be applied, but only if indicated in the detailed ing instructions for the question. Eample: = = y = 5 y = 7 Horizontal: 5 = and = Vertical: 5 = and y = 5 6 y = 5 and y = 7 6 = and y = 7 Markers should choose whichever method benefits the candidate, but not a combination of both. (l) In final answers, unless specifically mentioned in the detailed ing instructions, numerical values should be simplified as far as possible, eg: 5 must be simplified to 5 or 5 must be simplified to must be simplified to 8* 5 must be simplified to must be simplified to 5 *The square root of perfect squares up to and including 00 must be known. (m) Commonly Observed Responses (COR) are shown in the ing instructions to help common and/or non-routine solutions. CORs may also be used as a guide when ing similar non-routine candidate responses. page

30 (n) Unless specifically mentioned in the ing instructions, the following should not be penalised: Working subsequent to a correct answer Correct working in the wrong part of a question Legitimate variations in numerical answers/algebraic epressions, eg angles in degrees rounded to nearest degree Omission of units Bad form (bad form only becomes bad form if subsequent working is correct), eg ( )( ) written as ( ) 6 written as credit gains full Repeated error within a question, but not between questions or papers (o) (p) (q) (r) In any Show that question, where the candidate has to arrive at a required result, the last of that part is not available as a follow-through from a previous error unless specified in the detailed ing instructions. All working should be carefully checked, even where a fundamental misunderstanding is apparent early in the candidate's response. Marks may still be available later in the question so reference must be made continually to the ing instructions. The appearance of the correct answer does not necessarily indicate that the candidate has gained all the available s. Scored-out working which has not been replaced should be ed where still legible. However, if the scored out working has been replaced, only the work which has not been scored out should be ed. Where a candidate has made multiple attempts using the same strategy and not identified their final answer, all attempts and award the lowest. Where a candidate has tried different valid strategies, apply the above ruling to attempts within each strategy and then award the highest resultant. For eample: Strategy attempt is worth s. Strategy attempt is worth s. From the attempts using strategy, the resultant would be. Strategy attempt is worth. Strategy attempt is worth 5 s. From the attempts using strategy, the resultant would be. In this case, award s. page

31 . (a) find mid-point of BC 6, calculate gradient of BC 6 use property of perpendicular lines determine equation of line in a simplified form y9. is only available as a consequence of using a perpendicular gradient and a midpoint.. The gradient of the perpendicular bisector must appear in simplified form at or stage for to be awarded.. At, accept y9 0, y 9 or any other rearrangement of the equation where the constant terms have been simplified. Question Generic scheme Illustrative scheme. (b) 5 use m tan 5 6 determine equation of AB 6 y. At 6, accept y 0, y or any other rearrangement of the equation where the constant terms have been simplified. Question Generic scheme Illustrative scheme. (c) 7 find or y coordinate 8 find remaining coordinate 7 8 or y 5 8 y 5 or 8 page 5

32 . (a) Method Method know to use in synthetic division complete division, interpret result and state conclusion Method know to substitute complete evaluation, interpret result and state conclusion Method start long division and find leading term in quotient Remainder 0 is a factor Method 5 0 is a factor Method ( -) 5 complete division, interpret result and state conclusion ( -) remainder = 0 ( -) is a factor page 6

33 . Communication at must be consistent with working at that stage i.e. a candidate s working must arrive legitimately at 0 before can be awarded.. Accept any of the following for : f 0 so ( ) is a factor since remainder = 0, it is a factor the 0 from any method linked to the word factor by e.g. so, hence,,,. Do not accept any of the following for : double underlining the zero or boing the zero without comment is a factor, is a root, is a root, is a factor, is a root is a root. the word factor only with no link Question Generic scheme Illustrative scheme. (b) state quadratic factor find remaining factors and 5 state solution 5,,. The appearance of 0 is not required for 5 to be awarded. 5. Candidates who identify a different initial factor and subsequent quadratic factor can gain all available s is only available as a result of a valid strategy at and. 7. Accept, 0,, 0,, 0 for 5. page 7

34 . substitute for y 5 or 5 epress in standard quadratic form factorise find coordinates find y coordinates 5 y 6 y 5. At the quadratic must lead to two distinct real roots for and 5 to be available.. is only available if 0 appears at or stage.. If a candidate arrives at an equation which is not a quadratic at stage, then, and 5 are not available. At do not penalise candidates who fail to etract the common factor or who have divided the quadratic equation by is available for substituting correctly into the quadratic formula. 6. and 5 may be ed either horizontally or vertically. 7. For candidates who identify both solutions by inspection, full s may be awarded provided they justify that their points lie on both the line and the circle. Candidates who identify both solutions, but justify only one gain out of 5. Candidate A y 6 y 9 5 Candidate B Candidates who substitute into the circle equation only Sub Sub y y 0 y y5 0 y 6 y 0 y y 5 0 y 6 or y y or y 5,6, 5 page 8

35 . (a) Method Method identify common factor ( 8... stated or implied by complete the square... process for c and write in required form Method Method epand completed square form a ab ab c equate coefficients a, ab, ab c 50. process for b and c and write in required form with no working gains and only; however, see Candidate G.. is only available for a calculation involving both multiplication and subtraction of integers. Candidate A Candidate B ^ further working is required Candidate C a ab ab c a, ab, b c 50 a, b, c Candidate D page 9

36 Candidate E Candidate F a b c a ab ab c a, ab, ab c 50 b, c Candidate G is awarded as all working relates to completed square form Check: a ab ab c a, ab, ab c 50 b, c Candidate H 50 is lost as no reference is made to completed square form 6 50 Award / Question Generic scheme Illustrative scheme. (b) differentiate two terms 5 complete differentiation is awarded for any two of the following three terms:,, 50 page 0

37 . (c) Method Method 6 link with (a) and identify sign of 7 communicate reason and 6 f ( ) always strictly increasing Method Method 6 identify minimum value of f 7 communicate reason. Do not penalise 0 or the omission of f 6 7 eg minimum value = or annotated sketch 0 f 0 always strictly increasing at 6 in Method. 5. Responses in part (c) must be consistent with working in parts (a) and (b) for 6 and 7 to be available. 6. Where erroneous working leads to a candidate considering a function which is not always strictly increasing, only 6 is available. 7. At 6 communication should be eplicitly in terms of the given function. Do not accept statements such as (something) 0, something squared 0. However, 7 is still available. Candidate I f ' 0 strictly increasing. Award out of Candidate J Since and is 0 for all then 66 0 for all. 50 Hence the curve is strictly increasing for all values of. 6 7 page

38 5. (a) identify pathway PR RQ stated or implied by state PQ. Award i j k i j k i j 5k Candidates who choose to work with column vectors and leave their answer in the form cannot gain. 5. is not available for simply adding or subtracting vectors within an invalid strategy.. Where candidates choose specific points consistent with the given vectors, only and are available. However, should the statement without loss of generality precede the selected points then s,, and are all available. Question Generic scheme Illustrative scheme 5. (b) interpret ratio or identify pathway and demonstrate result PR RQ or PQ QR leading to i jk 5. This is a show that question. Candidates who choose to work with column vectors must write their final answer in the required form to gain. does not gain. 6. Beware of candidates who fudge their working between and. page

39 Candidate A legitimate use of the section formula npq mpr PS = m n PQ PR PS = PS = PS = i jk Candidate B BEWARE - treating P as the origin QS = SR s= q r 9 s = 5 5 s = i jk page

40 5. (c) Method Method 5 evaluate PQ. PS 6 evaluate PQ 5 PQ. PS 6 PQ 50 7 evaluate PS 7 PS 8 8 use scalar product 8 cosqps = calculate angle Method 9 56 or 0795 radians Method 5 5 evaluate QS 5 QS 6 6 evaluate PQ 6 PQ 50 7 evaluate PS 7 PS 8 8 use cosine rule cosqps = calculate angle 9 56 or 0795 radians 7. For candidates who use PS not equal to i jk 5 is not available in Method or 7 in Method. 8. Do not penalise candidates who treat negative signs with a lack of rigour when calculating a magnitude. However, leading to 6 indicates an invalid method for calculating the magnitude. No can be awarded for any magnitude arrived at using an invalid method ab. is not available to candidates who simply state the formula cos θ = ab. PQ. PS However, cos = or cos = PQ PS Accept answers which round to 6 or 08radians.. Do not penalise the omission or incorrect use of units.. 9 is only available as a result of using a valid strategy.. 9 is only available for a single angle.. For a correct answer with no working award 0/5. is acceptable. Similarly for Method. 5 page

41 Candidate C Calculating wrong angle Candidate D Calculating wrong angle QPQS. 9 5 QP 50 6 QS cos ˆ 9 PQS = 50 8 PQS ˆ = strategy incomplete For candidates who continue, and use the angle found to evaluate the required angle, then all s are available. Candidate E From (a) PQ i j k PQ. PS 5 PQ 68 6 PS cos ˆ QPS = 68 8 QPS ˆ = PS. QP 5 QP 50 6 PS 8 7 cos = = 9 strategy incomplete For candidates who continue, and use the angle found to evaluate the required angle, then all s are available. Candidate F From (a)pq i j k PQ. PS 5 PQ 68 6 PS cos ˆ QPS = 68 8 QPS ˆ = 88 9 Candidate G From (b) PS i j+k PQ. PS 5 PQ 50 6 PS 6 7 cos ˆ QPS = QPS ˆ = 85 9 page 5

42 6. substitute appropriate double angle formula 5sin sin epress in standard quadratic form sin 5sin 6 0 factorise solve for sin ( sin )(sin ) 5 sin, sin 5 solve for 5 088, 9, sin. is not available for simply stating cos sin with no further working.. In the event of cos sin or cos being substituted for cos, cannot be awarded until the equation reduces to a quadratic in sin.. Substituting sin A or sin for cos at stage should be treated as bad form provided the equation is written in terms of at stage. Otherwise, is not available.. 0 must appear by stage for to be awarded. However, for candidates using the quadratic formula to solve the equation, 0 must appear at stage for to be awarded. 5. 5sin sin 6 0 does not gain unless is awarded. 6. sin 5 gains Candidates may epress the equation obtained at in the form s 5s 6 0 or In these cases, award for ( s )(s ) 0 or ( )( ) 0. However, is only available if sin appears eplicitly at this stage. 8. and 5 are only available as a consequence of solving a quadratic equation. 9., and 5 are not available for any attempt to solve a quadratic equation written in the form a b c is not available to candidates who work in degrees and do not convert their solutions into radian measure. 9. Accept answers which round to 0 85 and at 5 eg, Answers written as decimals should be rounded to no fewer than significant figures.. Do not penalise additional solutions at 5. 5 page 6

43 Candidate A Candidate B ( s )(s ) 0 s, s 088, 9 5 Candidate C 5sin sin sin 5sin 6 sin sin 5 6 sin 6, sin 5 6 no solution, sin 0 5, 89 5 sin 5sin 6 0 9sin 6 0 sin 070, 5 Candidate D 5sin sin sin 5sin 6 0 sin 5sin 6 sin sin 5 6 sin 6, sin 5 6 no solution, sin 0 5, 89 5 Candidate E - reading cos as cos 5sin cos 5sin sin sin 5sin 6 0 sin 5 7 sin 0886, sin 86 08, 05 5 page 7

44 7. (a) write in differentiable form stated or implied differentiate one term dy dy 6 or d d complete differentiation and equate to zero 0 or solve for. For candidates who do not differentiate a term involving a fractional inde, either or is available but not both.. is available only as a consequence of solving an equation involving a fractional power of.. For candidates who integrate one or other of the terms is unavailable. Candidate A differentiating incorrectly y 6 5 dy 6 d Candidate B integrating the second term y 6 5 dy 6 d page 8

45 7. (b) 5 evaluate y at stationary point consider value of y at end points 6 and 0 7 state greatest and least values 7 greatest 8, least 0 stated eplicitly. The only valid approach to finding the stationary point is via differentiation. A numerical approach can only gain is not available to candidates who do not consider both end points. 6. Vertical ing is not applicable to 6 and Ignore any nature table which may appear in a candidate s solution; however, the appearance of 8, at a nature table is sufficient for Greatest, 8 ; least, 90 does not gain and 7 are not available for evaluating y at a value of, obtained at stage, which lies outwith the interval For candidates who only evaluate the derivative, 5, 6 and 7 are not available. Question Generic scheme Illustrative scheme 8. (a) find epression for u find epression for u and epress in the correct form 5k 0 u k k leading to u 5k 0k 0 page 9

46 8. (b) interpret information epress inequality in standard quadratic form 5k 0k 0 5 5k 0k determine zeros of quadratic epression 5 5, 6 state range with justification 6 k 5 with eg sketch or table of signs. Candidates who work with an equation from the outset lose and. However, 5 and 6 are still available.. At 5 do not penalise candidates who fail to etract the common factor or who have divided the quadratic inequation by 5.. and 5 are only available to candidates who arrive at a quadratic epression at.. At 6 accept k and k 5 or k, k 5 together with the required justification. 5. For a trial and error approach award 0/. page 0

47 9. Method state linear equation introduce logs use laws of logs use laws of logs 5 state k and n Method log y log log y log log log ylog log log y log 5 k 8, n or y 8 5 Method state linear equation use laws of logs use laws of logs use laws of logs 5 state k and n log y log log ylog log y y Method 5 k 8, n or y 8 5 page

48 Question Generic Scheme Illustrative Scheme Method Method The equations at, and must be stated eplicitly. introduce logs to y k n log y log k n Mark use laws of logs log y nlog log k interpret intercept log k use laws of logs 5 interpret gradient Method interpret point on log graph k 8 5 n Method log and log y 0 5 convert from log to ep. form and y 0 interpret point and convert log 0, log y, y substitute into evaluate k n y k and n k k 8 5 substitute other point into n y k and evaluate n 5 0 n n 0 n 5. Markers must not pick and choose between methods. Identify the method which best matches the candidates approach.. Treat the omission of base as bad form at and in Method, at and for Method and Method, and at in Method.. m or gradient does not gain 5 in Method.. Accept 8 in lieu of throughout.,, 0 for. 5. In Method candidates may use 0for and followed by page

49 Candidate A Candidate B With no working. Method : With no working. Method : k 8 n 8 n 5 k 5 Award /5 Award 0/5 Candidate C Candidate D Method : Method : log k k 8 n Award /5 5 log y log log ylog y k, n 5 Award /5 Candidate E Method : y /\/\/\/\/\/\/\/\/\/\/\/\ log y log log ylog y ^ y 5 Award /5 page

50 0. (a) Method calculate m AB calculate m BC interpret result and state conclusion Method mab see Note 9 5 mbc 5 AB and BC are parallel (common direction), B is a common point, hence A, B and C are collinear. Method calculate an appropriate vector e.g. AB calculate a second vector e.g. BC and compare interpret result and state conclusion Method 9 AB see Note 5 BC 5 AB BC 5 AB and BC are parallel (common direction), B is a common point, hence A, B and C are collinear. calculate m AB Method find equation of line and substitute point communication Method mab 9 eg, y leading to 6 7 since C lies on line A, B and C are collinear. At and stage, candidates may calculate the gradients/vectors using any pair of points.. can only be awarded if a candidate has stated parallel, common point and collinear.. Candidates who state points A, B and C are parallel or m AB and m BC are parallel do not gain. page

51 Candidate A mab 9 5 mbc 5 ^ AB and BC are parallel, B is a common point, hence A, B and C are collinear. Candidate B AB 5 BC AB and BC are parallel, B is a common point, hence A, B and C are collinear. Candidate C 9 AB 5 BC 5 5 and 9 5 AB BC ignore working subsequent to correct statement at. AB and BC are parallel, B is a common point, hence A, B and C are collinear. page 5

52 0. (b) find radius determine an appropriate ratio 5 e.g. : or 5 (using B and C) 6 7 find centre state equation of circle 6 8, or :5 or y (using A and C) Where the correct centre appears without working 5 is lost, 6 is awarded and 7 is still available. Where an incorrect centre or radius from working then 7 is available. However, if an incorrect centre or an incorrect radius appears e nihilo 7 is not available. 5. Do not accept 6 0 for 7. Candidate D Radius 6 0 Interprets D as midpoint of BC 5 Centre D is95, 5 6 y Candidate F Radius 0 Interprets D as midpoint of AC 5 Centre D is5, 6 y Candidate E Radius 0 Interprets D as midpoint of AC 5 Centre D is5, 6 y Candidate G Radius 6 0 CD BD or simply 5 Centre D is, 6 y 60 7 page 6

53 . (a) Method substitute for sin simplify and factorise substitute for simplify cos and Method sin cos sin cos stated cos eplicitly as above or in a simplified form of the above sin cos sin sin sin leading to Method substitute for sin simplify and substitute for cos epand and simplify Method sin cos sin cos cos stated eplicitly as above or in a simplified form of the above sin sin sin sin sin sin leading to sin. is not available to candidates who simply quote sin sin cos without substituting into the epression given on the LHS. See Candidate B. In method where candidates attempt and in the same line of working may still be awarded if there is an error at.. is not available to candidates who work throughout with A in place of.. Treat multiple attempts which are not scored out as different strategies, and apply General Marking Principle (r). 5. On the appearance of LHS 0, the first available is lost; however, any further s are still available. Candidate A sin cos sin cos sin cos Candidate B sin LHS sin cos cos sin sin cos sin ^ cos sin sin sin In proving the identity, candidates must work with both sides independently ie in each line of working the LHS must be equivalent to the line above. sin sin cos sin cos sin sin cos cos cos sin page 7

54 . (b) know to differentiate sin d (sin ) d 5 start to differentiate 5 sin... 6 complete differentiation 6... cos [END OF MARKING INSTRUCTIONS] page 8