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2012 MathematicsHigherFinalised Marking Instructions Scottish Qualifications Authority 2012The 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 material shouldonly be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any otherpurpose it is the centre’s responsibility to obtain the necessary copyright clearance. SQA’s NQ Delivery: ExamOperations 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 trade purposes.

General CommentsThese marking instructions are for use with the 2012 Higher Mathematics Examination.For each question the marking instructions are in two sections, namely Illustrative Scheme and GenericScheme. The Illustrative Scheme covers methods which are commonly seen throughout the marking.The Generic Scheme indicates the rationale for which each mark is awarded. In general markers shoulduse the Illustrative Scheme and only use the Generic Scheme where a candidate has used a method notcovered 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 x . 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 column whole numbers onlyDo not record marks onscripts in this manner.4 Where a candidate has not been awarded any marks for a question, or part of a question, 0 shouldbe written in the right hand margin against their answer. It should not be left blank.5 Every page of a candidate’s script should be checked for working. Unless blank, every page which isdevoid of a marking symbol should have a tick placed in the bottom right hand 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.8As 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 and the symbolsindicated 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.RAt 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, i.e. a cross is only used where a mark is not awarded.x RA cross-tick should be used to indicate “correct” working where a mark is awardedas a result of follow through from an error.RA 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 been eased.A tilde should be used to indicate a minor error which is not being penalised, e.g. bad form.R 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.7These will help markers to maintain consistency in their marking and will assist the examiners in thelater stages of SQA procedures.The examples below illustrate the use of the marking symbols .Example 1ydydxx 6x3RA(4, 4,0), B(2, 2,6), C(2, 2,0)3 x 2 123 x 2 12xyExample 22xRx2 xx3 Rx10R27x4 7 16Rx5 RExample 3ABb aAC§6· 6 0 ¹4k sin x cos a cos x sin a R x13, k sin aR x2xx1(repeated error)Example 43 sin x 5 cos xk cos a§6· 6 6 ¹5Rx21 5 2 84 4 81 1 2 0R x1R x2Since the remainder is 0, x 4 must be a factor.R x3( x 2 x 2) R x4( x 4)( x 1)( x 2) R x5x4 or xPage 3 1 or x2R x6

10 In general, as a consequence of an error perceived to be trivial, casual or insignificant, e.g. 6 u 6 12,candidates lose the opportunity of gaining a mark. But note example 4 in comment 9 and comment 11.11 Where a transcription error (paper to script or within script) occurs, the candidate should bepenalised, e.g.This is a transcription error andso the mark is not awarded.x 2 5x 7Eased as no longer a solution ofa quadratic equation.9x 4x 4x 30x1x 2 5x 7Exceptionally this error is not treated as atranscription error as the candidate deals withthe intended quadratic equation. The candidatehas been given the benefit of the doubt.RxRx 4x 30( x 3)( x 1)0xRR9x 41 or 3R12 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: x5 x 2, x 4x6y5, yCross marked: x5 x 2, y 5 7x6x 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.4315must be simplified to 45 or 1 41must be simplified to 43Examples:112150 3644must be simplified to 50must be simplified to 853must be simplified to415The square root of perfect squares upto and including 100 must be known.14 Regularly occurring responses (ROR) are shown in the marking instructions to help mark commonand/or non-routine solutions. RORs may also be used as a guide in marking similar non-routinecandidate responses.15 Unless specifically mentioned in the marking instructions, the following should not be penalised:x Working subsequent to a correct answer;x Correct working in the wrong part of a question;x Legitimate variations in numerical answers, e.g. angles in degrees rounded to nearest degree;x Omission of units;x Bad form;x 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 formula, the last mark of thatpart 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 In the exceptional circumstance where you are in doubt whether a mark should or should not beawarded, consult your Team Leader (TL).19 Scored out working which has not been replaced should be marked where still legible. However, ifthe scored out working has been replaced, only the work which has not been scored out should bemarked.20 A valid approach, within Mathematical problem solving, is to try different strategies. Where thisoccurs, all working should be marked. The mark awarded to the candidate is from the highestscoring strategy. This is distinctly different from the candidate who gives two or more solutions to aquestion/part of a question, deliberately leaving all solutions, hoping to gain some benefit. All suchcontradictory responses should be marked and the lowest mark given.21 It is of great importance that the utmost care should be exercised in totalling the marks.The recommended procedure is as follows:Step 1Manually calculate the total from the candidate’s script.Step 2Check this total using the grid issued with these marking instructions.Step 3In EMC, 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.22 The candidate’s script for Paper 2 should be placed inside the script for Paper 1, and thecandidate’s total score (i.e. Paper 1 Section B Paper 2) written in the space provided on thefront cover of the script for Paper 1.23 In cases of difficulty, covered neither in detail nor in principle in these instructions, markers shouldcontact their TL in the first instance. A referral to the Principal Assessor (PA) should only be made inconsultation with the TL. Further details of PA Referrals can be found in The General MarkingInstructions.Page 5

Paper 1 Section 3D14A15D16C17D18B19B20AA5B6C4D5Page 6

Paper 1 Section B21(i) Show that ( x 4) is a factor of x3 5x2 2x 8 .(a)(ii) Factorise x3 5x2 2x 8 fully.(iii) Solve x3 5x2 2x 8 0.6Generic SchemeIllustrative Scheme21 (a)Method 1 : Using synthetic divisionx1ssknow to use xx2pdcomplete evaluation4x141 5x241 5284 4 8 1 201x3x4x5x62state conclusionx3icfind quadratic factorx4pdfactorise completelyx5( x 4)( x 2)( x 1)state solutionsx 1, 2, 4icic68' remainder is zero so ( x 4) is a factor'x2 x 2stated, or implied by x5stated explicitly in any orderMethod 2 : Using substitution and inspectionx1 know to use x 4x264 80 8 8x3( x 4) is a factorx4( x 4)( x 2 x 2)stated, or implied by x5x5( x 4)( x 2)( x 1)stated explicitly in any orderx 1, 2, 4606Notes1.x3 is only available as a consequence of the evidence for x1 and x2 .2. Communication at x3 must be consistent with working at x2 .i.e. candidate’s working must arrive legitimately at zero before x3 is awarded.If the remainder is not 0 then an appropriate statement would be '( x 4) is not a factor' .3. Accept any of the following for x3 :0 so ( x 4) is a factor ' ' f (4) ' since remainder is 0, it is a factor 'the 0 from table linked to word ' factor ' by e.g. ' so ' , ' hence ' ,' ? ', ' o ', ' ' .4. Do not accept any of the following for x3 : double underlining the zero or boxing in the zero, without a comment ' x 4 is a factor ' , ' ( x 4) is a factor ' , ' x 4 is a root ' , ' ( x 4) is a root ' the word ' factor ' only, with no link.5. To gain x6 , 4, 1, 2 must appear together in (a).6.( x 4)( x 2)( x 1) leading to 4, 0 , 2, 0 and 1, 0 only does not gain x6 .7.( x 2)( x 1) only, leading to x2, x 1 does not gain x6 as equation solved is not a cubic.8. Candidates who attempt to solve the cubic equation subsequent to xor no solutions, cannot gain x .6Page 7 1, 2, 4 and obtain different solutions,

The diagram shows the curve with equation y21 (b)x3 5x2 2x 8.The curve crosses the x-axis at P, Q and R.Determine the shaded area.6Generic SchemeIllustrative Scheme21 (b)x7identify xQ from working in (a)x72interpret appropriate limitsx80, 2know and start to integratex9integrate one term correctly (but see Note 10)x10 pdcomplete integrationx10x11 icsubstitute limitsx11x12 pdstate areax12x8x9icicss145322x 4 x 3 x 2 8 x or equivalent1(2)4432353 (2)3 2 2 8 u 2 0or 1023but not a decimal approximation6Notes9. Evidence for x7 and x8 may not appear until x11 stage.10. Where a candidate differentiates one or more terms at x9 , then x9 , x10 , x11 and x12 are not available.11. Candidates who substitute at x11 , without integrating at x9 , do not gain x9 , x10 , x11 and x12.12. For candidates who make an error in (a), x8 is only available if 0 is the lower limit and a positive integervalue is used for the upper limit.13. x11 is only available where both limits are numerical values.14. Candidates must show evidence that they have considered the lower limit 0 in their substitution at x11 stage.Regularly occurring responsesResponse 1Candidates who use Q throughoutCandidate AQ³( x 3 5x 2 2 x 8) dx0Qª1 4 5 3 2 2º« 4 x 3 x 2 x 8 x »¼ 01 4Q453 Q 3 Q 2 8Q 0 x7 x x8 x x9 R x10 R x11 x x12 xHowever, if Q is replaced by 2 at thisstage, and working continues, all 6marks may still be available .Response 2Dealing with negativesCandidate BQ( 1, 0)x x7 1³ (x0ª1 « 43 5x 2 2 x 8) dx9R 5x 3410R xº 1x x x2 8x»3¼01( 1)4461 1253 ( 1)3 ( 1)2 8( 1) 0cannot be negative sobut APage 8x x861126112R x12 x x126112R x11

The expression cos x 3 sin x can be written in the form k cos( x a) where k ! 0 and 0 d a 2S .22 (a)Calculate the values of k and a.4Generic SchemeIllustrative Scheme22 (a)x1ssuse compound angle formulax1k cos x cos a k sin x sin ax2iccompare coefficientsx2k cos a 1 and k sin ax3pdprocess kx32 (do not accept 4)x4pdprocess ax4S3stated explicitly3stated explicitlybut must be consistent with x24Notes1. Treat k cos x cos a sin x sin a as bad form only if the equations at the x2 stage both contain k.2.2cos x cos a 2sin x sin a or 2(cos x cos a sin x sin a) is acceptable for x1 and x3 .3. Accept k cos a 1 and k sin a 3 for x2 .4.x2 is not available for k cos x 1 and k sin x5.x4 is only available for a single value of a.3 , however, x4 is still available.6. Candidates who work in degrees and do not convert to radian measure in (a) do not gain x4 .7. Candidates may use any form of the wave equation for x1 , x2 and x3 , however, x4 is only available if thevalue of a is interpreted for the form k cos( x a).Regularly occurring responsesResponse 1 : Missing information in workingCandidate ACandidate B72 cos a 1R 2 sin a 3 R31πR3tan aa7cos a 17x1 xsin ax2 Rx Rtan ax Ra343 marks out of 4x1 x3x231π3x3x4xxxNot consistent with0 marks out of 4evidence at x2 .Response 2 : Correct expansion of k cos( x a) and possible errors for x2 and x4Candidate CCandidate Dk cos a 13Rxk sin a1tan a3so ak cos a2π63Candidate Ex x2k cos a 1k sin a 1x x4tan ak sin a1so a3πR x46tan a 3x x2 3 so a5π3R x4Response 3 : Labelling incorrect using cos(A B) cos Acos B sin Asin B from formula listCandidate FCandidate Gk cos A cos B k sin A sin Bk cos a 1k sin atan ax x1k cos A cos B k sin A sin Bk cos x 123R x3 so ak sin xπ3R x4Candidate Htan xx x2Page 9k cos A cos B k sin A sin Bk cos B 1k sin B33 so xx x1π3R x4tan B3 R x23 so BπR x43x x1

22 (b)cos x 3 sin x with the x and y axes, in the interval 0 d x d 2S Find the points of intersection of the graph of yGeneric Scheme3Illustrative Scheme22 (b)x5icinterpret y-interceptx51x6ssstrategy for finding rootsx6e.g. 2 cos x x7icstate both rootsx7S 7S,66S30 or3 sin xcos x3Notes8. Candidates should only be penalised once for leaving their answer in degrees in (a) and (b).9. If the expression used in (b) is not consistent with (a) then only x5 and x7 are available.10. Correct roots without working cannot gain x6 but will gain x7 .11. Candidates should only be penalised once for not simplifying 4 in (a) and (b).Regularly occurring responsesResponse 4 : Communication for x5Candidate I(1, 0) without working.Candidate Jcos 0 3 sin 0x x551R xso (1, 0).Response 5 : Follow through from a wrong value of aCandidate KCandidate Lπ6π3From (a) athen in (b) x7,4π3x R7onlyResponse 6 : Root or graphical approachCandidate Mπ2π6π3andand7π6 3π2xx6 π3x x4x660then in (b) x30 , 210 only x R77Candidate N(a) 60 x x4R x6R x7xFrom (a) aNote 10Candidate O(b)36027090moved 60 to leftRxWhen x 30q, 210q R x76cuts x-axis atπ 2π,6 3Response 7 : Circular argument not leading anywhereCandidate P122 cos x u 2 sin x u32cos x 3 sin xx6 R0x7 R0Candidate Q(b) 2 cos x y2 cos 0 π3π3x x65π60 so x2 cos π3S3is penalised as x S3obtained in (a).However x5 and x7 are still available asfollow through. See Note 9.Response 8 : Transcription error in (b)(a) correctx ,11π6R x71R x5Page 10R x6x x7

23 (a) Find the equation of "1, the perpendicular bisector of the line joining P(3, 3) to Q( 1, 9) .Generic Scheme4Illustrative Scheme23 (a)x1x2ssssfind midpoint of PQx1(1, 3)find gradient of PQx2 3313x3icinterpret perpendicular gradientxx4icstate equation of perp. bisectorx4y 31( x 1)34Notes1.x4 is only available if a midpoint and a perpendicular gradient are used.2. Candidates who use ymx c must obtain a numerical value for c before x4 is available.Regularly occurring responsesResponse 1 : Candidates who use wrong midpoint or no midpointCandidate Amidpoint M(2, 6)mMQx x 5 RR x2RR x315mAx15y ( 6)x 2 RCandidate B1mPQmAx x1 3 R13R x2Rusing R, y ( 2)7R x413x 1R x3xx x423 (b) Find the equation of "2 which is parallel to PQ and passes through R(1, 2).Generic SchemeP2Illustrative Scheme23 (b)x5icuse parallel gradientsx5 3x6icstate equation of linex6y ( 2)stated, or implied by x6 3( x 1)2Notes3.x6 is only available to candidates who use R and their gradient of PQ from (a).Regularly occurring responsesResponse 2 : Not using parallel gradient for equationCandidate Cy ( 2)13x 1xx5x6xxCandidate DCandidate EParallel so same gradientsmso m13y ( 2)x5x13x 1xx6 RIf mPQPage 11x5 R 3 Ry ( 2)13x 1x 3 only do not award x5x6x

Find the point of intersection of "1 and "2.23 (c)3Generic SchemeIllustrative Scheme23 (c)x7ssx7use valid approachx 3ye.g. 8 and 9 x 3 y13or 3x 1x or 3(3 y 8) yx8pdsolve for one variablex8e.g. xx9pdsolve for other variablex9e.g. y38311 2523Notesx 3y4. Neither 8 and 3x y 1nory 3x 1 and 3yx 8 are sufficient to gain x7 .5. x7 , x8 and x9 are not available to candidates who: Equate zeros Give answers only, without working Use R for equations in both (a) and (b) Use the same gradient for the lines in (a) and (b).23 (d)Hence find the shortest distance between PQ and "2.2Generic SchemeIllustrative Scheme23 (d)x10 ssidentify appropriate pointsx10 (1, 3) andx11 pdcalculate distancex1152accept12 ,10252or2 5Notes6. x10 and x11 are only available for considering the distance between the midpoint of PQ and the candidate’sanswer from (c) or for considering the perpendicular distance from P or Q to "2.7. At least one coordinate at x10 stage must be a fraction for x11 to be available.8. There should only be one calculation of a distance to gain x11 .Regularly occurring responsesResponse 3 : Following through from correct (a), (b) and (c)Candidate F(1, 3), (1, 2)dx x105 R x11Response 4 : Following through from correct (a), (b) and (c)Candidate G(1, 3),PR12 ,52R x105 , QR 125 , dIf reference was made to this being theperpendicular distance then x11 wouldbe available.2 511so 2 5 is shortest distance. x xPage 122

These marking instructions are for use with the 2012 Higher Mathematics Examination. For each question the marking instructions are in two sections, namely Illustrative Scheme and Generic Scheme. The Illustrative Scheme covers methods which are commonly seen throughout the marking. The Generic Scheme

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