3y ago

41 Views

2 Downloads

1.24 MB

31 Pages

Transcription

2014 MathematicsHigherFinalised Marking Instructions Scottish Qualifications Authority 2014The information in this publication may be reproduced to support SQA qualifications only on a non-commercialbasis. If it is to be used for any other purposes written permission must be obtained from SQA’s NQ Delivery:Exam Operations team.Where the publication includes materials from sources other than SQA (secondary copyright), this materialshould only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for anyother purpose it is the centre’s responsibility to obtain the necessary copyright clearance. SQA’s NQ Delivery:Exam Operations team may be able to direct you to the secondary sources.These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers whenmarking External Course Assessments. This publication must not be reproduced for commercial or tradepurposes.

General CommentsThese marking instructions are for use with the 2014 Higher Mathematics Examination.For each question the marking instructions are in two sections, namely Illustrative Scheme andGeneric Scheme. The Illustrative Scheme covers methods which are commonly seen throughout themarking. The Generic Scheme indicates the rationale for which each mark is awarded. In generalmarkers should use the Illustrative Scheme and only use the Generic Scheme where a candidate hasused a method not covered in the Illustrative Scheme.All markers should apply the following general marking principles throughout their marking:1 Marks must be assigned in accordance with these marking instructions. In principle, marks areawarded for what is correct, rather than deducted for what is wrong.2 Award one mark for each. There are no half marks.3 The mark awarded for each part of a question should be entered in the outer right hand margin,opposite the end of the working concerned. The marks should correspond to those on the questionpaper and these marking instructions. Only the mark, as a whole number, should be written.22/3Marks in this columnwhole numbers onlyDo not record marks onscripts in this manner.4 Where a candidate has not been awarded any marks for an attempt at a question, or part of aquestion, 0 should be written in the right hand margin against their answer. It should not be leftblank. If absolutely no attempt at a question, or part of a question, has been made, ie a completelyempty space, then NR should be written in the outer margin.5 IT IS ESSENTIAL that every page of a candidate’s script should be checked for working. Unlessblank, every page which is devoid of a marking symbol should have a tick placed in the bottom righthand margin.6 Where the solution to part of a question is fragmented and continues later in the script, the marksshould be recorded at the end of the solution. This should be indicated with a down arrow ( ), in themargin, at the earlier stages.7 Working subsequent to an error must be followed through, with possible full marks for thesubsequent working, provided that the level of difficulty involved is approximately similar. Where,subsequent to an error, the working for a follow through mark has been eased, the follow throughmark cannot be awarded.8 As indicated on the front of the question paper, full credit should only be given where the solutioncontains appropriate working. Throughout this paper, unless specifically mentioned in the markinginstructions, a correct answer with no working receives no credit.Page 2

9Marking SymbolsNo comments or words should be written on scripts. Please use the following symbols and thoseindicated on the welcome letter and from comment 6 on the previous page. A tick should be used where a piece of working is correct and gains a mark. Markers mustcheck through the whole of a response, ticking the work only where a mark is awarded.At the point where an error occurs, the error should be underlined and a cross used toindicate where a mark has not been awarded. If no mark is lost the error should only beunderlined, ie a cross is only used where a mark is not awarded.x A cross-tick should be used to indicate “correct” working where a mark is awardedas a result of follow through from an error.A double cross-tick should be used to indicate correct working which is irrelevant orinsufficient to score any marks. This should also be used for working which has beeneased.A tilde should be used to indicate a minor error which is not being penalised, eg bad form. This should be used where a candidate is given the benefit of the doubt. A roof should be used to show that something is missing, such as part of a solution or acrucial step in the working.These will help markers to maintain consistency in their marking and are essential for the laterstages of SQA procedures.The examples below illustrate the use of the marking symbols .Example 1Example 2y x 6xA(4, 4,0), B(2, 2,6), C(2, 2,0)32 dy 3 x 2 12dx3x2x 12 0x 2y 16 1 x 3 2 4 6 AB b a 6 6 6 AC 6 0 5Example 3x 1 2Example 4 3 sin x 5 cos xk sin x cos a cos x sin a 1k cos a 3, k sin a 5 2 Since the remainder is 0,must be a factor. Page 3

10 In general, as a consequence of an error perceived to be trivial, casual or insignificant, eg 6 6 12,candidates lose the opportunity of gaining a mark. But note example 4 in comment 9 and the secondexample in comment 11.11 Where a transcription error (paper to script or within script) occurs, the candidate should bepenalised, egThis is a transcription error andso the mark is not awarded. x Eased as no longer a solution ofa quadratic equation. Exceptionally this error is not treated as atranscription error as the candidate deals with theintended quadratic equation. The candidate hasbeen given the benefit of the doubt. 12 Cross markingWhere a question results in two pairs of solutions, this technique should be applied, but only ifindicated in the detailed marking instructions for the question.Example: Point of intersection of line with curveIllustrative Scheme: 5 x 2, x 4 6Cross marked: 5 x 2, y 5y 5, y 7 6x 4, y 7Markers should choose whichever method benefits the candidate, but not a combination of both.13 In final answers, numerical values should be simplified as far as possible.1543Examples:should be simplified to 45 or 1 41should be simplified to 43121150 364should be simplified to 50must be simplified to 8453should be simplified to415The square root of perfect squares upto and including 100 must be known.14 Commonly Observed Responses (COR) are shown in the marking instructions to help markcommon and/or non-routine solutions. CORs may also be used as a guide in marking similarnon-routine candidate responses.15 Unless specifically mentioned in the marking 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, eg angles in degrees rounded to nearest degree;Omission of units;Bad form;Repeated error within a question, but not between questions or papers.Page 4

16 In any ‘Show that . . .’ question, where the candidate has to arrive at a required result, the last markof that part is not available as a follow through from a previous error.17 All working should be carefully checked, even where a fundamental misunderstanding is apparentearly in the candidate’s response. Marks may still be available later in the question so referencemust be made continually to the marking instructions. All working must be checked: theappearance of the correct answer does not necessarily indicate that the candidate has gained all theavailable marks.18 Scored out working which has not been replaced should be marked where still legible. However,if the scored out working has been replaced, only the work which has not been scored out shouldbe marked.19 Where a candidate has made multiple attempts using the same strategy, mark all attempts andaward the lowest mark.Where a candidate has tried different strategies, apply the above ruling to attempts within eachstrategy and then award the highest resultant mark. For example:Strategy 1 attempt 1 is worth 3 marksStrategy 1 attempt 2 is worth 4 marksFrom the attempts using strategy 1, theresultant mark would be 3.Strategy 2 attempt 1 is worth 1 markStrategy 2 attempt 2 is worth 5 marksFrom the attempts using strategy 2, theresultant mark would be 1.In this case, award 3 marks.20 It is of great importance that the utmost care should be exercised in totalling the marks.A tried and tested procedure is as follows:Step 1Step 2Step 3Manually calculate the total from the candidate’s script.Check this total using the grid issued with these marking instructions.Electronically enter the marks and obtain a total, which should now be compared to themanual total.This procedure enables markers to identify and rectify any errors in data entry before submittingeach candidate’s marks.21 The candidate’s script for Paper 2 should be placed inside the script for Paper 1, and thecandidate’s total score (ie Paper 1 Section B Paper 2) written in the space provided on thefront cover of the script for Paper 1.Page 5

Paper 1 Section werCBADDACDBCCABDBCBCADABCD4565Page 6

Paper 1- Section BQuestionGeneric SchemeIllustrative Scheme21a 1ssknow to differentiate and one term correct 2ssthe other term correct and set derivative to 0 2dy 0dx 1 6 x. or . 3x 26 x 3x2 0 stated explicitly 3 4 3x 02 4y 04 3pdsolve 4pdevaluate 5pdjustify nature of stationary points 5 use 2nd derivative or nature table 6icinterpretation 6 min. at (0,0) and max. at (2, 4)coordinatesMaxMark6Notes:1. 2 is not available for statements such as ‘dy 0 ’ with no other working.dx2. Accept 3x2 6 x 0 for 2.03. For candidates using a nature table, the minimum response for is:25dy 0 dydxx values 0 and 2;or expression 6 x 3x 2 ; signs and zeroes; shape.dx4. For candidates who differentiate correctly but then solve0 dy 0 incorrectly, 4 may be awarded as adxfollow through mark. 5 and 6 are not available if a nature table has been used, but may be awardedwhere candidates have used the 2nd derivative.5. For candidates who differentiate incorrectly 3 and 4 may be awarded as follow through marks. 5and 6 are not available if a nature table has been used, but may be awarded where candidates haveused the 2nd derivative.6. At 6 stage accept min at x 0 and max at x 2 .7. Candidates who find the x-coordinates of the SPs correctly but correctly process only one of these todetermine its nature, gain 6 but not 5.Page 7

Commonly Observed Responses:Candidate Ad2y 6 6xdx 2d2yd2yat x 0, 0,atx 2, 0dx 2dx 2hence minimum SP at x 0, maximum SP at x 2 5 6 Candidate Bdy 6 x 3x 2 0dx3x(3 x) 0x 0, x 3 1 2 y 0, y 0 4 3xCase (i)d2y 6 6xdx 2d2yx 0 2 0 Minimum SPdxd2yx 3 2 0 Maximum SPdx21 7 5 6Case (ii)x 0 3 dy 0 ? ? dx 5 x 6 x? inconsistent. Different signs for 6x 3x2 or3x(3 x)bpd 7find intercepts3x 2 x3 0 and (3, 0) or x 3 ;(0, 0) [may appear in part a] 8ic 8sketchsketch2Notes:8. 7 accept 3x2 x3 0 and correctly annotated diagram with 0, 3 and no other intercepts marked on sketch.9. The minimum required for 8 is a cubic curve, consistent with the SPs found in part (a) andappropriate number of x intercepts appearing on their sketch. It must be possible to determine thecoordinates of the SPs from the sketch.Page 8

Commonly Observed Responses:The following are acceptable for 8(2, 4)43O2Do not accept the following for 8(2, 4)3O(2, 4)O3Page 9

Question22Generic SchemeIllustrative SchemeMaxMarka 1 ss know to use x 1 and obtain an equation 16 1 7 1 a 1 b 0 2 ss know to use x 2 and obtain an equation 26 2 7 2 a 2 b 72 3 pd process equations to find one value 3a 1 or b 2 4 pd find the other value 4b 2 or a 13232Alternative Method for 1 and 2 1 1667 61a 1a 1b a 1b a 1 071219ab382a 76a 38 2a b 76 72 2266Note:1. An incorrect value at 3 should be followed through for the possible award of 4. However, if theequations are such that no solution exists, then 3 and 4 are not available.Commonly Observed Responses:Candidate A 1 X616 2a13a 137613 repeated error 2 67a 12106 5a 10ba 13a b 13 0b 2a 20 2a b 20 72Solving to get a 35, b 22 3 4Leading to, in part (b), 6x3 7 x2 35x 22 ( x 1)(6x2 13x 22) 5 6 7Page 104

Question22b 5ssGeneric SchemeIllustrative Schemesubstitute for a and b and know todivide by x 1 5 6x3MaxMark 7 x 2 x 2 x 1 Stated or implied by 6 6pdobtain quadratic factor 6 7pdcomplete factorisation 7 x 1 6 x2 x 2 x 1 3x 2 2 x 1 3Notes:2. For candidates who substitute a 1 into the correct quotient from part (a), 5, 6 and 7 areavailable.3. Candidates who use incorrect values obtained in part (a) may gain 5, 6 and 74. Where the quadratic factor obtained is irreducible, candidates must clearly demonstrate thatb2 4ac 0 to gain 7.5. Do not penalise the inclusion of ‘ 0 ’ or for solving for x.6. Candidates who use values, ex nihilo, for a and b can gain 5, if division is correct, but 6 and 7 areonly available if ( x 1) is a factor of the resulting expression.Commonly Observed Responses:Candidate BCandidate C22a no solution22b a 4, b 5 ex nihilo22a no solution22b a 2, b 3 ex nihilo(6x3 7 x2 4x 5) ( x 1) 167 4 6 161 5 5 550(6x3 7 x2 2x 3) ( x 1) 1672 6 1611( x 1)(6x2 x 5) 6 ( x 1)( x 1)(6x 5)( x 1) and are not available76Candidate D22a no solution22b a 4, b 3 ex nihilo(6x3 7 x2 4x 3) ( x 1) 1674 6 1613 53 30( x 1)(6x2 x 3)b2 4ac 1 72 71 6 71 0 so does not factorise 7Page 117is not a factor 53 12

QuestionGeneric SchemeIllustrative Scheme23a 1sssubstitute 3x 5 1 2pd 2 3 4pdpdexpress in standardquadratic formfind x-coordinatesfind y-coordinates 3 4MaxMarkx 1 x 3y 2 y 4 3 44Notes:1. ‘ 0 ’ must appear at 1 or 2 for mark 2 to be awarded.12. If x ( y 5) is substituted at 1 then 10 y 2 20 y 80 0 is obtained at 2.33. Special Case: In cases where x 1 and x 3 do not appear as a result of 1 and 2, but aresubstituted into the equation of the line to obtain the y values, if the candidate then checks thatboth points lie on the circle, 3 4 marks are awarded. If, in addition, the candidate makes astatement to the effect that a line can only cut a circle in, at most, 2 points, then 4 4 marks areawarded. Otherwise, 0 4 marks.4. 3 and 4 are not available for any attempt to solve a quadratic equation of the form ax2 bx cCommonly Observed Responses:Candidate Ax2 (3x 5)2 2x 4(3x 5) 15 010x2 40x 40 0 1 2 Xx 2 and y 1 323 b 5 ss 76pdicstate centrecalculate gradientscommunicate result 4 5 ( 1,2) 6m 2 , m 7demonstrates121m1 m2 2 12 PT is perpendicular to QT[or other appropriate statement]Alternative Method ssstate centre 6pdcalculate vectors 6 7iccommunicate result 755( 1, 2) 2 4 eg and 4 2 2 4 . 2 4 4 2 0 4 2 PT is perpendicular to QT[or other appropriate statement]Page 123

Notes:4. Other valid strategies:a Converse of Pythagoras’ Theorem: 6 process lengths, PT QT 20, PQ 40 7 apply converse and communicate result clearly.b Cosine Rule: 6 process lengths, 7 apply cosine rule to obtain angle 90o and communicate result clearly.Commonly Observed Responses:Candidate BCandidate CT( 1,2) 5 T( 1,2) 5 1m , m 22 6 1m1 , m2 22 6 m1 m2 1 7 m1 m2 1 7 No link between required condition and gradientsfound.23 cRequired condition is linked to gradients found. 8 9 8 centre (2, 1)ss knows to find and states centrepd calculate radius 9 radius 10 10 ic state equation of circle3Alternative Method 8 ss 8x 2 y 2 2 gx 2 fy c 025 6 g 8 f c 05 2g 4 f c 05 2g 4 f c 0 9 pd find f or g or c 9f 1, or g 2, or c 5 10 ic state values of f , g and c 10 f 1, g 2, c 5substitute points into generalequation of circleNotes:( 10)2 must be simplified to gain 10For candidates who find P and Q correctly in part (a), award 8 if centre (2,1) appearswithout working.7. For the mid-point of PQ being (2,1), 8 is available unless subsequent working indicates thatthis is not the intended centre.8. 9 is only available as a result of PQ being a diameter, or using a valid strategy to find thecentre eg midpoint of PQ or point of intersection of the perpendicular bisectors of PT and TQ. 10is still available.9. Where an incorrect centre or an incorrect radius appear ex nihilo 10 is not available.5.6.Page 13

QuestionGeneric SchemeIllustrative SchemeMaxMark24Method 1 1ss take log9 of both sides of the equation 1 2pd apply laws of logarithms 2log9 y log9 k log9 a x 3pd apply laws of logarithms 3log9 y log9 k x log9 a4 pd find k 4log9 k 2, k 81 or k 92 81 5pd find a 511log9 a , a 3 or a 9 2 32 1ss know to use equation of the line 2pd write in exponential form 2y 92 3pd apply laws of indices 3y 9 2 92 4 5pd find kpd find a 4k 81a 31 5log9 y log9 ka xMethod 21log9 y x 221 x 21xNotes:1.Candidates who start with 3 log9 y log9 k x log9 a also gain 1 and 2 .2.In Method 1, base 9 must appear by 4 stage, for 1 to be awarded.3.For k 81 and a 3 with spurious or no working, 4 and 5 are not available.Page 145

Commonly Observed Responses:Candidate Alog y log ka x 1log y log k log a x 2log y log k x log a 3k 81 4No evidence of which base is being used.a 3 5Answers at both 4 and 5 are consistent with using base 9.See Note 2Candidate B : A combination of Method 1 and Method 2.1log9 y x 22log9 y log9 ka x log9 y x log9 a log9kM2M1 1 2 3equating gradients and intercepts1log9 a 21 4a 92 3log9 k 2 5k 9 812Candidate Cat (0, 2) log9 y 2 y 9 812at (6,5) log9 y 5 1Substitute into equation81 ka0 2 k 81 y 95 3substitute into equation95 ka6 95 81a6 a6 93 36 a 3 4Page 15 5

Paper 2QuestionGeneric SchemeIllustrative Scheme1a 1ssfind gradient of AB 1mAB 1 2pdfind perpendicular gradient 2mperp 1 stated or implied by 4 3pdfind midpoint of AB 3 4,1 stated or implied by 4 4pdobtain equation 4y 1 1 x 4 Notes:1. 4 is only available as a consequence of using a perpendicular gradient and a midpoint.2. The gradient must appear in simplified form at 4 stage for 4 to be awarded.Commonly Observed Responses:Candidate AmAB 1 1 Xmperp 1 2(4,1) 3 y 1 1( x 4) y x 3 4Leading to part (b)y x 3 5y 2x 6(3,0) 6 7 and 8 are not available as A T (3,0)Page 16MaxMark4

QuestionGeneric SchemeIllustrative Scheme1b 5ssknow to solve simultaneously 5y 2x 6y x 5 6pdsolve correctly for x and y 6x 1, y 4MaxMark2Commonly Observed Responses:Candidate BPart (a) y 1 1( x 4) 4 y x 3 errorPart (b) y 2x 6 and y x 3 5 x 3, y 0 1 6correct strategy used, pd mark not availablec 7ssknow and use m tan 7tan 2 8pdcalculate angle 8116 6 2accept 1170 or 2 03 radiansCommonly Observed Responses:Candidate CmAT Candidate D12base angle 26 6o 7 XmAT 2 angle 90 26 6 116 6o 8 Xangle tan 1(2) 63 4 7 XO 8Candidate E:Part (a)mAB 2 0 2 1 5 3 8 4Part (b) 1 Xmperp 4 2Midpoint of AB (4, 1) 3 y 1 4 x 1 y 4x 5 4y 4x 5 0y 2x 6 0 5 X y 2 x 6 6 Xy 4 x 51 2 x 1, x , y 725 is a strategy mark. The correct strategy is to solvethe given equation with the equation from part (a)simultaneously. 5 is not awarded as the givenequation has not been used.The equation obtained at stage 4, has beenrearranged incorrectly in part (b). The nex

2014 Mathematics Higher Finalised Marking Instructions . These marking instructions are for use with the 2014 Higher Mathematics Examination. For each question the marking instructions are in two sections, . 8 As indicated on the front of the question paper, full credit should only be given where the solution

Related Documents: