2010 Mathematics. Higher. Finalised Marking Instructions

Transcription

1 00 Mathemats Higher Finalised Marking Instructions Scottish Qualifations Authority 00 The information in this publation may be reproduced to support SQA qualifations only on a noncommercial basis. If it is to be used for any other purposes written permiion must be obtained from the Eternal Print Team, Centre Serves, Dalkeith. Where the publation includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of eamination or aement. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the neceary copyright clearance. SQA s Eternal Print Team, Centre Serves, at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Eamination Teams for use by SQA Appointed Markers when marking Eternal Course Aements. This publation must not be reproduced for commercial or trade purposes.

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3 Detailed Marking Instructions : Higher Mathemats 00 Vs 00 Mathemats Higher Marking Instructions Eam date: May 00 CONTENTS General Marking Principles Marking Instructions for each Question Strtly Confidential These instructions are strtly confidential and, in common with the scripts you will view and mark, they must never form the subject of remark of any kind, ecept to Scottish Qualifations Authority staff. Page Marking The utmost care must be taken when entering Item level marks into Appointees Online. It is of partular importance that you enter a zero (0) when the candidate has attempted a question but has not gained a mark and a dash (-) when the candidate has not attempted a question. Page

4 Part One : General Marking Principles for Mathemats Higher This information is provided to help you understand the general principles you must apply when marking candidate responses to questions in this Paper. These principles must be read in conjunction with the specif Marking Instructions for each question. For each question the marking instructions are split into two sections, namely and. The indates what each mark is being awarded for. The cover methods whh you will commonly see throughout your marking. In general you should use the for marking and revert to the only where a candidate has used a method not covered in the or you are unsure of whether you should award a mark or not. All markers should apply the following general marking principles throughout their marking:. Marks for each candidate response must always be aigned in line with these general marking principles and the specif Marking Instructions for the relevant question. If a specif candidate response does not seem to be covered by either the principles or detailed Marking Instructions, and you are uncertain how to ae it, you must seek guidance from your Team Leader. You can do this by ing/phoning your team leader. Alternatively, you can refer the iue directly to your Team Leader by checking the Referral bo on the marking screen.. Marking should always be positive, i.e. marks should be awarded for what is correct and not deducted for errors or omiions.. Award one mark for each. Each error should be underlined in red at the point where it first occurs, and not at any subsequent stage of the working. The total mark for each question should be entered in red in the outer right hand margin, opposite the end of the working concerned. Only the mark, as a whole number, should be written; do not use fractions. The marks should correspond to those on the question paper and these marking instructions. Where a candidate has scored zero for any question, or part of a question, 0 should be written in the right hand margin against their answer. Working subsequent to an error must be followed through; with poible full marks for the subsequent working, provided that the level of diffulty involved is approimately similar. Where, subsequent to an error, the working is eased, a deduction of mark(s) should be made. There is no such thing as a transcription error, a trivial error, a casual error or an insignifant error. In general, as a consequence of one of these errors, candidates lose the opportunity of gaining the appropriate mark or pd mark. As indated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Throughout this paper, unle specifally mentioned in the marking scheme, a correct answer with no working receives no credit. Page

5 Normally, do not penalise: Detailed Marking Instructions : Higher Mathemats 00 Vs Working subsequent to a correct answer; Omiion of units; Legitimate variations in numeral answers; Bad form; Correct working in the wrong part of a question; unle specifally mentioned in the marking scheme. 0 No piece of working should be ignored without careful checking even where a fundamental misunderstanding is apparent early in the answer. Reference should always be made to the marking scheme. Answers whh are widely off-beam are unlikely to include anything of relevance but in the vast majority of cases candidates still have the opportunity of gaining the odd mark or two; provided it satisfies the criteria for the marks. If in doubt between two marks, give an intermediate mark, but without fractions. When in doubt between consecutive numbers, give the higher mark. No marks at this stage should be deducted for carele or badly arranged work. It is of great importance that the utmost care should be eercised in adding up the marks. Using the Electron Marks Capture (EMC) screen to tally marks for you is not recommended. A manual check of the total, using the grid iued with this marking scheme, can be confirmed by the EMC system. In cases of diffulty, covered neither in detail nor in principle in these instructions, attention may be directed to the aement of partular answers by making a referral to the Principal Aeor (P.A.) Please see the general Instructions for P.A. referrals. If a candidate presents multiple solutions for a question and it is not clear whh is intended as their final attempt, mark each one and award the lowest mark. Page

6 Marking Scripts No comments, words or acronyms should be written on scripts. Please use the following and nothing else. Tk when a piece of working is correct and gains a mark. You are not epected to tk every line of working but you must check through the whole of a response. Underline each error and place a cro at the end of the line where it occurs. A cro-tk should be used to indate correct working where a mark is awarded as a result of follow through from an error. A double cro-tk should be used to indate correct working whh is inadequate to score any marks e.g. incorrect method whh is mathematally correct or eased working. N.B. Bullets may be used along with the signs opposite to indate whh marks are being awarded. Margins Eample dy = d = 0 = y = Eample = = A tilde should be used to indate a minor transgreion whh is not being penalised, e.g. bad form. Eample sin = 0 = sin 0 = Use a roof to show that something is miing such as a crucial step in the working or part of a solution. These are eential for later stages of the SQA procedures. marks go in this column as whole numbers only Page

7 Paper Section A Detailed Marking Instructions : Higher Mathemats 00 Vs Question Answer A C D A B D C B C 0 B D A B C C A B B C 0 A Summary A B C D Page

8 Paper Section B Triangle ABC has vertes A(, 0), B(,) and C(, 0),as shown in the diagram opposite. Medians AP and CR intersect at the point T(,). B y R P T Q C (a) Find the equation of median BQ. O A (b) Verify that T lies on BQ. (a) know and find midpoint of AC pd calculate gradient of BQ state equation (, 0) or equivalent y = or y = ( ( )) 0 ( ). Candidates who do not use a midpoint lose. There is no need to simplify m for BQ and Do not award for + y = 0, although then + y = 0.. If m cannot be simplified, due to an error, then BQ. is available for using. Accept y = It must, however, be simplified before can be awarded. would be awarded for + y = 0 is still available. and y = m + c where m = c =.. Candidates who find the equations of AP or CR can only gain mark. AP : y 0 = ( ) or y = ( ) CR : y 0 = ( ) or y = ( ) (b) substitute in for T and complete + = + = e.g. Substitution : () () 0 Gradients : m BT 0 = = = m 0 Vectors : BT =, TQ = and BT = TQ BQ. is available as follow through with an appropriate communation statement, e.g. 'T does not lie onbq'.. Statements such as PA, RC and BQ are all medians and therefore all share the same point T do not gain. Since only mark is available here, do not penalise the omiion of any reference to a common point or parallel. 0 Gradient approach : (b) m = = = m leading to : in (c), without further working, gains and BT BQ but loses. but (b) m = and m = leading to m = m so T lies on BQ leading to : in (c), BQ TQ BQ TQ without further working, loses and but gains.. Page

9 Detailed Marking Instructions : Higher Mathemats 00 Vs Triangle ABC has vertes A(, 0), B(,) and C(, 0),as shown in the diagram opposite. Medians AP and CR intersect at the point T(,). B y R P T Q C O (c) Find the ratio in whh T divides BQ. A (c) valid method for finding the ratio complete to simplified ratio Method : Vector approach 0 e.g. BT = and TQ = : Method : Stepping out approach For : without working, only is awarded. Be aware that the working may appear in (b). Some candidates obtain : from erroneous working thus losing. e.g. : Method : Distance Formula approach = = BT TQ e.g. d and d : 0 B T Q or B T Q 0 0. Any evidence of appropriate steps, e.g. 0 and or and but not and, can be awarded to e.g., B T Q is not suffient on its own and so loses. : with no further simplifation may be awarded but not but gains. In this question working for (c) may appear in (b), where the working appears for... leading Response Response Response 0 0 (b) BT = TQ = (b) QT = BQ = = QT (b) QT = TB = BT = TQ 0 (c) : (c) : (c) = so : marks out of marks out of marks out of Response 0 (b) BT = TQ = so BT = TQ (c) : but : would have gained Page

10 (a) (i) Show that ( ) is a factor of f ( ) = + +. (ii) Hence factorise f( ) fully. (b) Solve = (a) Method : Using synthet division = know to use complete evaluation 0 state conclusion pd find quadrat factor pd factorise completely ( ) is a factor see note + stated, or implied by + ( )( )( ) stated eplitly Method : Using substitution and inspection = + + = know to use 0 ( ) is a factor see note + ( )( ) + ( )( )( ) stated eplitly. Communation at For. At must be consistent with working at. i.e. candidate s working must arrive legitimately at zero before is awarded. If the remainder is not 0 then an appropriate statement would be '( ) is not a factor'., minimum acceptable statement is factor. Unacceptable statements : = is a factor, ( + ) the epreion may be written as ( ) ( ) +. is a factor, = is a root, ( ) is a root etc. (b) state solutions = = and or or These may appear in the working at (a). + leading to, then (, 0 ) and (, 0 ) + leading to (, 0 ) and (, 0 ). From (a) ( )( )( ) However, ( )( )( ). From (a) ( )( + ) only leading to ( )( + )( ) leading to = = gains, only does not gain = = does not gain =, = and = gains.. as equation solved is not a cub, but as follow through from a cub equation. Page 0

11 Detailed Marking Instructions : Higher Mathemats 00 Vs (c) The line with equation y = is a tangent to the curve with equation y = + + at the point G. Find the coordinates of G. (d) This tangent meets the curve again at the point H. Write down the coordinates of H. (c) Method : Equating curve and line set y = CURVE 0 G G LINE epre in standard form compare with (a) or factorise identify pd evaluate y Method : Differentiation 0 y know to and differentiate curve set derivative to gradient of line pd solve quadrat equation proce to identify complete to y CURVE G = y LINE Method : Equating curve and line + + = = stated eplitly ( )( )( ) y = Method : Differentiation + + = = 0 y = and at = evaluate y = 0 see note CURVE andy LINE from both curve and line In method :.. is only available if = 0 appears at either the, 0 and or are only available via the working from. If ( )( )( + ) does not appear at. At a quadrat used from (a) may gain 0. If G and H are interchanged then stage. and. stage, it can be implied by, 0 is lost but. Candidates who obtain three distinct factors at and.. A repeated factor at In both methods: or and stage is required for and can gain and but a quadrat from are still available. 0. may gain and for evaluating all y values, but lose 0 to be awarded without justifation. only.. All marks in (c) are available as a result of differentiating + + and solving this equal to (from method ). Only marks and (from method ) are available to those candidates who choose to differentiate + + and solve this equal to 0.. Candidates may choose a combination of making equations equal and differentiation. 0 (d) pd state solution ( ), may appear in (c). Method from (c) would not yield a value for H and so is not available. Page

12 (a) Diagram shows a right angled triangle, where the line OA has equation y = 0. (i) Show that tan a =. (ii) Find the value of sin a. a O (b) A second right angled triangle is added y Diagram as shown in Diagram. The line OB has equation y = 0. y Find the values of sinb andcos b. O a b A A Diagram B (a) write in slope/intercept form connect gradient and tana pd calculate hypotenuse state value of sine ratio y = or y = stated eplitly m = and tan a = or m = tan a and tana = stated, or implied by or may not appear until (c). is only available if sina.. Only numeral answers are acceptable for and. Response a marks out of y = 0 () () = 0 tana = sina = (b) determine tanb know to complete triangle pd determine hypotenuse state values of sine and cosine ratios tan b stated, or implied by = right angled triangle with and correctly shown stated, or implied by = = sin b and cos b may not appear until (c).. is only available if sinb b and cos b. sin b = and cos = without working is awarded marks only.. Only numeral answers are acceptable for and. Page

13 (c) (i) Find the value of sin( a b). Detailed Marking Instructions : Higher Mathemats 00 Vs (ii) State the value of sin( b a). (c) 0 know to use addition formula substitute into epansion pd evaluate sine of compound angle = use sin( ) sin a b sin cos cosasinb 0. sin(a B) = sinacosb cosasinb, or just sinacosb cosasinb, with no further working does not gain. Candidates should not be penalised further at to.. Candidates who use sin( a b) = sina sinb lose 0,, and 0 and. for values of sine and cosine outside the range but can gain a non-zero answer whh is obtained from the result sin( ) = sin.. Treat sin cos cos sin as bad form only if sin and cos subsequently disappear. 0. It is acceptable to work through the whole epansion again for sin( b a) = 0 = =., as follow through, only for Response Response sin( a b) = sina sinb = 0 sina = cosa = sinb = cosb = Marks lost in (a) or (b) = 0 sin( a b) = sinacosb cosasinb 0 0 marks out of sin( b a) = marks out of Eased - not dealing with fraction containing a surd. Response Response From (a) and (b) sina = = (i) sin( a b) = sinasinb cosacos b (c) (i) (ii) cosa cosb sinb = = sin( a b) = sinacosb cosasinb = 0 = sin( b a) = sinbcosa cosbsina = = marks out of (ii) sin( b a) = Here the working was not neceary; the answer would gain = = marks out of 0, provided it is non zero. Page

14 Paper The diagram shows a cuboid OPQR,STUV relative to the coordinate aes. P is the point (,0,0), z V Q is(,, 0) and U is (,, ). U(,, ) y N M is the midpoint of OR. S T R Q(,, 0) N is the point on UQ such that UN = UQ. M O P(, 0, 0) (a) State the coordinates of M and N. (b) Epre the vectors VM and VN in component form. Treat as bad form, coordinates written as components and ve versa, throughout this question. (a) interpret midpoint for M interpret ratio for N (0,, 0) (,, ) (b) intepret diagram pd proce vectors 0 VM = VN = 0 Using evidence from (a) or may have been taken directly from diagram.. V is the point (0,, ), whh may or may not appear in the working to (b). Incorrect Response Response Response V stated (a) M(, 0, 0) N(,, ) (b) V(0,, ) (a) M(0,, 0) N(,,) (b) 0 marks out of VM = Consistent with (a) VN = 0 From diagram marks out of 0 VM = VN = 0 mark out of (b) VM = 0 0 VN = 0 mark out of V(,, ) 0 marks out of used in both but not stated Page

15 Detailed Marking Instructions : Higher Mathemats 00 Vs The diagram shows a cuboid OPQR,STUV relative to the coordinate aes. P is the point (,0,0), z V Q is(,, 0) and U is (,, ). U(,, ) y N M is the midpoint of OR. S T R Q(,, 0) N is the point on UQ such that UN = UQ. M O P(, 0, 0) (c) Calculate the size of angle MVN. Treat as bad form, coordinates written as components and ve versa, throughout this question. (c) Method : Vector Approach know to use scalar product pd find scalar product pd find magnitude of a vector pd find magnitude of a vector pd evaluate angle Method : Cosine Rule Approach know to use cosine rule pd find magnitude of a side pd find magnitude of a side pd find magnitude of a side pd evaluate angle Method : Vector Approach VM.VN cosmvn ˆ = VM VN stated, or implied by VM.VN = VM = 0 VN = or rads or grads Method : Cosine Rule Approach ˆ VM + VN MN cosmvn = stated, or implied by VM VN VM 0 = = VN = MN or rads or grads. is not available to candidates who choose to evaluate an incorrect angle.. For candidates who do not attempt in the question.. should be awarded for any answer that rounds to signifant figures.), then is only available if the formula quoted relates to the labelling or rads or grads (i.e. correct to two Response Response ˆ OM.ON VM.VN cosmon = Wrong angle ˆ cosmvn = Going OM ON VM VN directly to OM.ON = VM.VN = 0 0 from Eased because only one would lose OM = VM = non-zero component. and. ON = VN = or 0 rads or grads marks out of 0 or equivalent marks out of Page

16 (a) cos sin can be epreed in the form kcos( + a), where k > 0 and 0 a < 0. Calculate the values of k and a. (a) use addition formula compare coeffients pd proce k pd proce a k a k a k a a k a = k a = k a = cos cos sin sin or (cos cos sin cos ) stated eplitly cos and sin or sin stated eplitly no justifation required, but do not accept accept any answer whh rounds to. Do not penalise the omiion of the degree symbol.. Treat kcos cosa sin sina as bad form only if the equations at the. cos cosa sin sin a or (cos cosa sin sin a ) is acceptable for. is not available for kcos = and ksin = or ksin =, however, stage both contain k. and. is still available... is lost to candidates who give a in radians only. may be gained only as a consequence of using evidence at stage.. Candidates may use any form of the wave equation for value of a is interpreted for the form kcos( + a)., and, however is only available if the Response A Response B Response k(cos cosa sin sin a ) kcoscosa ksinsina kcos( a) sina = k = = kcoscosa + ksinsina = coscosa + sinsina cosa = tana = cosa = sina = tana = a = then a = See note a = cos( + ) or a = See note marks out of marks out of marks out of Response A Response B Response kcos cosa ksina = kcosa = sin sina k = tana = a = kcos cosa sin sina ksina = kcosa = k = tana = a = kcos cosa ksin sina kcosa = ksina = k = tana = a = See note marks out of marks out of marks out of Page

17 Detailed Marking Instructions : Higher Mathemats 00 Vs (b) (i) Hence state the maimum and minimum values of cos sin. (ii) Determine the values of, in the interval 0 < 0, at whh these maimum and minimum values occur. (b) state maimum and minimum, find corresponding to ma. value pd find corresponding to min. value maimum at and no others minimum at and no others or and and no others maimum at and minimum at. is available for and only if has been penalised at. Accept answers whh round to and for and 0. Candidates who continue to work in radian measure should not be penalised further.. Etra solutions, correct or incorrect, should be penalised at. and.. or but not both. are not available to candidates who work with cos( + ) = 0 or cos( + ) =.. Candidates who use cos( ) from a correct (a) lose but is still available. Response Response From (a) a = From (a) cos( + ) ma/ min = ± ma at min at ma = min = ma at min at 0 marks out of marks out of already penalised at (a) Response A cos( + ) ma at min at 0 mark out of Insuffient evidence for Response B cos( + ) mark out of ma at min at 0 N.B. Candidates who use differentiation in (b) can gain only, as a direct result of their response in (a). This question is in degrees and so calculus is not appropriate for and. Page

18 (a) (i) Show that the line with equationy = is a tangent to the circle with equation y y = 0. (ii) Find the coordinates of the point of contact, P. (a) substitute ( ) ( ) = pd epre in standard form start proof complete proof Method : Factorising ( )( ) = 0 see note equal roots so line is a tangent Method : Discriminant = b ac 0 so line is a tangent = stated eplitly pd coordinates of P, y = = For method :. is only available if = 0 appears at either or stage.. Alternative wording for could be e.g. repeated roots, repeated factor, only one solution, only one point of contact along with line is a tangent. For both methods :. Candidates must work with a quadrat equation at the and stages.. Simply stating the tangency condition without supporting working cannot gain. For candidates who obtain two distinct roots, b ac so line is not a tangent', but ' 0. is still available for 'not equal roots so not a tangent' or is not available. Page

19 Detailed Marking Instructions : Higher Mathemats 00 Vs (b) Relative to a suitable set of coordinate aes, the diagram below shows the circle from (a) and a second smaller circle with centre C. The line y = is a common tangent at the point P. C P The radius of the larger circle is three times the radius of the smaller circle. Find the equation of the smaller circle. + y + + y = 0 (b) Method : via centre and radius 0 state centre of larger circle find radius of larger circle pd find radius of smaller circle strategy for finding centre interpret centre of smaller circle state equation Method : via centre and radius 0 (, ) see note see note stated, or implied by see note e.g " Stepping out" (, ) ( ) ( y ) or y y 0 + = + + = Method : via ratios 0 state centre of larger circle strategy for finding centre state centre of smaller circle strategy for finding radius pd find radius of smaller circle state equation Method : via ratios 0 (, ) see note e.g. "Stepping out" (, ) + see note 0 + y = + y y + = ( ) ( ) or 0 For method :. Acceptable alternatives for. Acceptable alternatives for. (, ) without working gains are or decimal equivalent whh rounds to i.e. to two signifant figures. are 0 but loses. or or decimal equivalent whh rounds to. For method :. (, ) without working gains 0. Acceptable alternatives for but loses. 0 are or decimal equivalent whh rounds to. In both methods:. If m = is used in a stepping out method the centre of the larger circle need not be stated eplitly for.. For the smaller circle, candidates who gue values for either the centre or radius cannot be awarded.. At e.g., are unacceptable, but any decimal whh rounds to is acceptable.. is not available to candidates who divide the coordinates of the centre of the larger circle by. Page

20 Solve cos cos = 0 for 0 < π. know to use double angle formula epre as quadrat in cos start to solve pd reduce to equations in cos only pd complete solutions to include only one where cos = k with k > Method : Using factorisation (cos )... cos cos + ( cos )(cos ) Method : Using quadrat formula (cos )... cos cos 0 = cos = In both methods : ± ( ) ( ) ( ) = = cos and cos, and no solution or = cos and no solution = cos and, = 0 must appear at either of these lines to gain.. is not available for simply stating that. Substituting cosa = cos A or cosa = cos A with no further working. cosa cos a. In the event of cos sin or sin being substituted for cos, equation reduces to a quadrat in cos.. Candidates may epre the quadrat equation obtained at the. 0 gain + = etc. For candidates who do not solve a trig. equation at. = etc. should be treated as bad form throughout. cannot be given until the stage in the form c c + = 0,, cos must appear eplitly to and are only available as a consequence of solving a quadrat equation subsequent to a substitution.. Any attempt to solve acos + bcos = c loses, and.. Accept answers given as decimals whh round to and.. There must be an indation after cos = that there are no solutions to this equation. Acceptable evidence : e.g. cos =, NA, out of range, invalid and cos = no, cos = Unacceptable evidence : e.g. cos =, cos =???, Maths Error.. is not available to candidates who work throughout in degrees and do not convert their answer into radian measure. π π 0. Do not accept e.g.,,,, π Ignore correct solution outside the interval 0 < π. Page 0

21 Detailed Marking Instructions : Higher Mathemats 00 Vs Response cos... cos cos = 0 ( ) ( ) ( ) ± cos = cos = and cos =, and no solution is only available to candidates where one, but not both, of their equations has no solution for cos. marks out of Response cos... cos cos = 0 (cos + )(cos ) = 0 = = π π, cos and cos = cos with no further working cannot gain quadrat in cos subsequently appears then is not available. ; however if a is awarded but marks out of is lost hereas it is not clear whether the candidate has used cos or cos as their substitution. Response A cos cos = 0 cos cos = 0 ± + 0 cos = cos = 0 and cos = =, and undefined Response B cos = cos cos cos = 0 cos cos = 0 ± + 0 cos = cos = 0 and cos = =, and undefined marks out of marks out of Page

22 (0 ) The parabolas with equations y = 0 and y = are shown in the diagram below. y A rectangle PQRS is placed between the two parabolas as shown, so that: Q and R lie on the upper parabola. RQ and SP are parallel to the -ais. T, the turning point of the lower parabola, lies on SP. (a) (i) If TP = units, find an epreion for the length of PQ. (ii) Hence show that the area, A, of rectangle PQRS is given by A ( ) = (b) Find the maimum area of this rectangle. y R S O T Q P = 0 y = ( ) 0 (a) know to and find OT obtain an epreion for PQ complete area evaluation or (0,) stated, or implied by 0 ( ) =. The evidence for and may appear on a sketch.. No marks are available to candidates who work backwards from the area formula.. is only available if has been awarded. (b) know to and start to differentiate pd complete differentiation set derivative to zero pd obtain A ( ) =... stated, or implied by = 0 ( ) or decimal equivalent ignore inclusion of justify nature of stationary point interpret result and evaluate area A ( ) + 0 ( Note : accept in lieu of A ( ) in the nature table. ) Ma and or decimal equivalent N.B. To conclude a maimum the evidence must come from.. At accept any answer whh rounds to.. Throughout this question treat the use of f ( ) or dy as bad form. d. At. At the nature can be determined using the second derivative. accept any answer whh rounds to or. Page

23 Detailed Marking Instructions : Higher Mathemats 00 Vs Response A ( ) = A ( ) = 0 = A ( ) = 0 on its own would not be suffient for. Response At stationary points, A'( ) = 0 = Response A ( ) = = Bad form Response A Response B Response C miing A'( ) + 0 A'( ) A'( ) or slope etc. signs or values are neceary Response Maimum at = y = = Area = = Treat this as an error subsequent to a correct answer. Page

24 (a) A curve has equation y = ( ). Show that the equation of the tangent to this curve at the point where = is. y = y (b) Diagram shows part of the curve and the tangent. y = The curve cuts the -ais at the point A. y = ( ) O A Find the coordinates of point A. Diagram (a) y = CURVE know to and start to differentiate pd complete chain rule derivative pd gradient via differentiation pd obtain at state equation and complete ( )... = ( ) and complete to = y y. is only available as a consequence of differentiating equation of the curve.. Candidates must arrive at the equation of the tangent via the point (, ) and not the origin.. For accept. Response Response Candidates who equate derivatives: Candidates who intersect curve and line: ( ) = leading to = and y = from curve Also obtaining y = from line and so line is a tangent marks out of ( ) = = + = 0 ( ) marks out of Factorising or using discriminant Equal roots or b ac = 0 so line is a tangent (b) ( ) obtain coordinates of A, 0. Accept =, y = 0 where 0 y = may appear from ( ) = 0.. For (, 0 ) without working is awarded, but from erroneous working is lost (see response below). Response = 0 = 0 = 0 = Here cannot be awarded due to the erroneous working. Page

25 Detailed Marking Instructions : Higher Mathemats 00 Vs (c) Calculate the shaded area shown in diagram. y y = y = ( ) O A Diagram (c) Method : Area of triangle area under curve Method : Area of triangle area under curve 0 strategy for finding shaded area A ( ) know to integrate pd start integration pd complete integration limits and pd substitute limits = Shaded area Area of large ( ) d ( ) 0... and ( ) 0 Area under curve pd evaluate area and complete strategy = or or Method : Area between line and curve 0 strategy for finding shaded area A ( ) know to integrate pd start integration pd complete integration limits and pd ' upper lower' and substitute limits pd evaluate area and complete strategy Method : Area between line and curve Area of small + area between line and curve...( ) ( ) and d + = or or ( ) ( ) ( ).. At may not be obvious until the final line of working and may be implied by final answer or a diagram. the value of must lie between 0 and eclusively, however, A <. A. Full marks are available to candidates who integrate with respect to y. and are only available if You may find the following helpful in marking this question: Area between curve and line from. to = Area of = or LARGER Area of = Area under curve from. to = SMALLER Page

26 (a) Given that log = P, show that log = P. (a) convert from log to eponential form know to and convert back to log form pd proce and complete = P log = log P log = P log and complete. No marks are available to candidates who simply substitute in values and verify the result. e.g. log = and log = log = P and log = P Response Response Response log log = P = = P P P = = P log P = P P = = P = log log log = P = P = P log = P = = mark out of marks out of marks out of p P = or P P Without this step would be lost but is still available as follow through. Response P = log = kp log kp = = log P kp log P = = = log = P kp = k = k = log = P as = marks out of Response Beware that some candidates give a circular argument. This is only worth. log = P then log = P P P = = P P = = = = = = log log log log log Plog log Plog log P log P mark out of Response log log log = = = log P mark out of Using change of base result. Page

27 Detailed Marking Instructions : Higher Mathemats 00 Vs (b) Solve log + log =. (b) use appropriate strategy pd start solving proce pd complete proce via log to epo form log log log + log = log + log = = = = ( = ) = ( = ) or or Q + Q = Q + Q = Q = Q = log = log = = ( = ) = ( = ). At any letter ecept may be used in lieu of Q.. Candidates who use a trial and improvement technique by substituting values for gain no marks.. The answer with no working gains no marks. Response Response Response Q + Q = or log + log = log + log = Q = log = log + log = Q = log = log = log = = = = = = = marks out of = marks out of = Response log = = = The marks allocated are dependent on what substitution is used for Q. = marks out of Without justifation, are not available. and Page