2011 Mathematics. Higher. Finalised Marking Instructions

Transcription

1 Mathemats Higher Finalised Marking Instructions Scottish Qualifations Authorit The information in this publation ma be reproduced to support SQA qualifations onl on a noncommercial basis. If it is to be used for an other purposes written permiion must be obtained from SQA s NQ Deliver: Eam Operations team. Where the publation includes materials from sources other than SQA (secondar copright), this material should onl be reproduced for the purposes of eamination or aement. If it needs to be reproduced for an other purpose it is the centre s responsibilit to obtain the necear copright clearance. SQA s NQ Deliver: Eam Operations team ma be able to direct ou to the secondar sources. These Marking Instructions have been prepared b Eamination Teams for use b SQA Appointed Markers when marking Eternal Course Aements. This publation must not be reproduced for commercial or trade purposes.

2 Detailed Marking Instructions : Higher Mathemats Final General Comments These marking instructions are for use with the Higher Mathemats Eamination. For each question the marking instructions are in two sections, namel and Gener Scheme. The covers methods whh ou will commonl see throughout our marking. The Gener Scheme indates the rationale for whh each mark is awarded. In general ou should use the and onl use the Gener Scheme where a candidate has used a method not covered in the. All markers should appl the following general marking principles throughout their marking: Marks must be aigned in accordance with these marking instructions. In principle, marks are awarded for what is correct, rather than deducted for what is wrong. Award one mark for each. There are no half marks. Working subsequent to an error must be followed through, with poible full marks for the subsequent working, provided that the level of diffult involved is approimatel similar. Where, subsequent to an error, the working is eased, a deduction of mark(s) should be made. Marking Smbols No comments, words or acronms should be written on scripts. Please use the following and nothing else. A tk should be used where a piece of working is correct and gains a mark. You are not epected to tk ever line of working but ou must check through the whole of a response. Where a mark is lost, the error should be underlined in red at the point where it first occurs, and not at an subsequent stage of the working. 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 irrelevant or insuffient to score an marks. This should also be used for working whh has been eased. A tilde should be used to indate a minor transgreion whh is not being penalised, e.g. 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 miing, such as a crucial step in the working or part of a solution. These will help ou to maintain consistenc in our marking and will aist the eaminers in the later stages of SQA procedures. Higher Mathemats Page

3 Detailed Marking Instructions : Higher Mathemats Final Regularl Occurring Responses (ROR) are shown on the marking scheme to help mark common solutions that are non-routine. RORs ma also be used as a guide in marking other non-routine candidate responses. The mark for each part of a question should be entered in red in the outer right hand margin, opposite the end of the working concerned. The marks should correspond to those on the question paper and these marking instructions. Onl the mark, as a single number, should be written. / Marks in this column - single numbers onl Do not record marks on scripts in this manner. Where a candidate has scored zero for an question, or part of a question, should be written in the right hand margin beside their answer. Ever page of a candidate s script should be checked for working. Unle blank, ever page whh is devoid of a marking smbol, should have a tk placed in the bottom right hand margin. Where a solution is spread over several pages the marks should be recorded at the end of the solution. This should be indated with a down arrow ( ), in the margin, at the earlier stages. The eamples below illustrate the use of the marking smbols. Eample Eample = d = d = = = A(,,), B(,,), C(,,) AB = b + a = AC = Eample Eample Find intersection of + = and = (repeated error) sin cos ksincosa coina kcosa =, ksina = = + = = = Strateg mark awarded. (despite two errors) The subsequent pd mark is lost (Note ) Higher Mathemats Page

4 Detailed Marking Instructions : Higher Mathemats Final Where a transcription error (paper to script or within script) occurs, a mark is lost. e.g. This is a transcription error and so mark is lost. Eased as not solution of a quadrat equation. + + = + + = = Eceptionall this error is not treated as a transcription error as the candidate deals with the intended quadrat equation. The candidate has been given the benefit of the doubt. + + = + + = ( )( ) = = or In general, as a consequence of an error perceived to be trivial, casual or insignifant, e.g. =, candidates lose the opportunit of gaining the appropriate or pd mark. A proceing error made at a strateg mark stage is penalised at the net pd or mark available within that part of the question. The strateg mark ma still be awarded. As indated on the front of the question paper, full credit should onl be given where the solution contains appropriate working. Throughout this paper, unle specifall mentioned in the marking scheme, a correct answer with no working receives no credit. Unle specifall mentioned in the marking scheme, do not penalise: Working subsequent to a correct answer; Correct working in the wrong part of a question; Legitimate variations in numeral answers, e.g. angles in degrees rounded to nearest degree; Omiion of units; Bad form. No piece of working should be ignored without careful checking even where a fundamental misunderstanding is apparent earl in the answer. Reference should alwas be made to the marking scheme. Answers whh are widel off-beam are unlikel to include anthing of relevance, but candidates ma still have the opportunit of gaining the odd mark or two, provided it satisfies the criteria for the marks. In the eceptional circumstance where ou are in doubt whether a mark should or should not be awarded, err on the generous side and award the mark. Higher Mathemats Page

5 Detailed Marking Instructions : Higher Mathemats Final Scored out or erased working whh has not been replaced should be marked where still legible. However, if the scored out or erased working has been replaced, onl the work whh has not been scored out should be marked. A valid approach, within Mathematal problem solving, is to tr different strategies. Where this occurs, all working should be marked. The mark awarded to the candidate is from the highest scoring strateg. This is distinctl different from the candidate who gives two or more solutions to a question/part of a question, deliberatel leaving all solutions, hoping to gain some benefit. All such contradtor responses should be marked and the lowest mark given. It is of great importance that the utmost care should be eercised in adding up the marks. The recommended procedure is as follows: Step Manuall calculate the total from the candidate s script. Step Check this total using the grid iued with these marking instructions. Step Input the scores and obtain confirmation of our total from the EMC screen. (This should highlight an discrepancies hitherto undiscovered.) Place the candidate s script for Paper inside the script for Paper and write the candidate s total score (i.e. Paper Section B + Paper ) in the space provided on the front cover of the script for Paper. In cases of diffult, covered neither in detail nor in principle in these instructions, contact our Team Leader (TL) in the first instance. A referral to the Principal Aeor (PA) should onl be made in consultation with our TL. Please see the General Marking Instructions for PA Referrals. Higher Mathemats Page

6 Paper Section A Detailed Marking Instructions : Higher Mathemats Final Question Answer C B D D A C D A B D D C C B B A A C C D Summar A B C D Higher Mathemats Page

7 Paper Section B A quadrilateral has vertes A(, ), B(, ), C(, ) and D(, ) as shown in the diagram. Detailed Marking Instructions : Higher Mathemats Final C (a) Find the equation of diagonal BD. (b) The equation of diagonal AC is + =. Find the coordinates of E, the point of intersection of the diagonals. D A O E B (a) Gener Scheme pd find gradient of BD state equation of BD or equivalent ( ) ( ) ( ) = or =. There is no need to simplif m for BD. If m cannot be simplified, due to an error, then BD. Candidates who determine the equation of AC lose. Candidates lose and Regularl occurring responses ; however, it must be simplified before is still available. but ma still gain. for the equation of an side of the quadrilateral. can be awarded. Response Response Using = m + c = + c = + c or = + c c = marks out of m m AC BD = = ( ) = ( ) mark out of Candidate has aumed diagonals are perpendular - without evidence. (b) start solution of simultaneous equations pd solve for one variable pd solve for second variable = + = e.g. and or = + or + ( ) = = or = = or =. Candidates who find the equation of AC in (a), correctl or incorrectl, lose. An other incorrect answer from (a) ma still gain Regularl occurring responses, and, and as follow through. in (b). Response = and + = Subsequent to gaining an error was made in simplifing the equation in (a), but strateg mark still awarded in (b). = = Error going from (a) to (b) is penalised at first pd (or ) mark. marks out of Higher Mathemats Page

8 Detailed Marking Instructions : Higher Mathemats Final (c) (i) Find the equation of the perpendular bisector of AB. (ii) Show that this line paes through E. (c) Gener Scheme know and find midpoint of AB (,) pd find gradient of AB interpret perpendular gradient state equation of perp. bisector justifation of point on line or equivalent or equivalent stated, or implied b = ( ) but not = ( ) = = + = when, or + =. Candidates who do not simplif the gradient in (a) and (c) should onl be penalised once.. is onl available as a consequence of using a midpoint and perpendular gradient.. Candidates who use = ( ) at stage, lose and. Candidates who show that the point of intersection of BD or AC and the perpendular bisector is E gain Regularl occurring responses Response m m PERP BISECTOR ''ME'' = = = So perpendular bisector goes through E Response From (i) equation of perpendular bisector is = +, using (, ). Then in (ii) using m = and E(,) leads to = +. Same equation so E lies on line.. There must be reference to the midpoint being a common point to gain this mark.. Response From (b) E(, ) =, = =, so line does not pa through E. Comment must be consistent with E from (b). Higher Mathemats Page

9 Detailed Marking Instructions : Higher Mathemats Final A function f is defined on the set of real numbers b f( ) = ( )( + ). (a) Find where the graph of = f( ) cuts: (i) the -ais; (ii) the -ais. (b) Find the coordinates of the stationar points on the curve with equation = f( ) and determine their nature. (a) Gener Scheme interpret intercept interpret intercept. Candidates who obtain etra -ais intercepts lose. Candidates who obtain etra -ais intercepts lose (, ) (minimum response "(i) ").. Candidates who interchange intercepts can gain at most one mark.. (, ) (minimum response "(ii) ") (b) write in differentiable form know to and start to differentiate pd complete derivative and equate to pd factorise derivative pd proce for pd evaluate -coordinates justif nature of stationar points interpret and state conclusions + + = and = and = and = and =... or and f ( ) or ( )( ) + = f ( ) + + ma min Accept a valid epreion in lieu of f ( )... is onl available if " = " appears at or before,. At.. and stage. are the onl marks available to candidates who solve the nature can be determined using the second derivative. f ( ) + + ma min ma min =. is onl available if the nature table is consistent with the candidate s derivative. is awarded for correct interpretation of the candidate s nature table in words. Regularl occurring responses Response A Response B Response C d d miing slope signs or values are necear This question ma be marked vertall. The dotted rectangle shows what is required for. Higher Mathemats Page

10 Detailed Marking Instructions : Higher Mathemats Final (c) On separate diagrams sketch the graphs: (i) = f( ) ; (ii) = f( ). Gener Scheme (c) (i) = f( ) (ii) = f( ) O ( ), (, ) (,), O ( ). curve showing points from (a) and (b) without annotation cub curve showing all intercepts and stationar points annotated curve from (i) reflected in ais sketch sketch reflected sketch is for an curve consistent with all points found in (a) and (b). Ignore an etra crital points.. In (c)(ii), the minimum requirement is the curve from (c)(i) reflected in -ais showing at least one -intercept unchanged and at least one stationar point correctl annotated. Regularl occurring responses Follow through from candidate s work in (a) and (b). Response Response Response O (, ) (, ) (,) (, ) O (, ) (,) O O O (, ) marks out of mark out of O (, ) marks out of Response No marks available for a quadrat. Not a cub due to apparent change of concavit. Higher Mathemats Page

11 Detailed Marking Instructions : Higher Mathemats Final (a) Solve cos cos + = for <. (a) Gener Scheme know to use double angle formula epre as a quadrat in cos start to solve Method : Using factorisation cos stated, or implied b + cos cos ( cos )(cos ) = must appear at either of these lines to gain.. pd reduce to equations in cos onl proce solutions in given domain is not available for simpl stating that Method : Using quadrat formula + = cos cos cos ( ) ± ( ) In both methods : = = cos and cos, and or = = cos and = = cos and or cosa = cos A with no further working. stated eplitl. In the event of cos sin or sin being substituted for cos, cannot be awarded until the equation reduces to a quadrat in cos.. Substituting cosa = cos A or = etc. should be treated as bad form throughout. cosa cos a. Candidates ma epre the quadrat equation obtained at the. eplitl to gain and stage in the form + etc. For candidates who do not solve a trigonometr quadrat equation at. are onl available as a consequence of solving a quadrat equation.. An attempt to solve. a + b = closes, and. c c + or, cosmust appear is not available to candidates who work in radian measure and do not convert their answers into degree measure. Regularl occurring responses Response Response A Response B (Reading cos as cos ) (See note above) Candidates who include lose (See note above) cos cos + = cos cos + = cos cos + = (cos )(cos ) = cos = or cos = cos cos = cos ( cos ) = cos cos + = cos cos = no solution = cos = or cos = cos ( cos ) = = = cos = or cos = marks out of = = mark out of Candidates who include lose marks out of Higher Mathemats Page

12 Detailed Marking Instructions : Higher Mathemats Final (b) Hence solve cos cos + = for <. (b) Gener Scheme interpret relationship with (a) interpret periodit. Do not penalise the inclusion of in (b). = and and,,,, and. Ignore etra answers, correct or incorrect, outside the given interval, but penalise incorrect answers within the interval.. Do not penalise candidates who use radians in (b) if the have alread been penalised in (a).. Candidates who go back to first principles for (b) can onl gain valid solutions as stated in the. Regularl occurring responses Response A From (a) =,, (b) cos cos + = (cos cos + ) = =,,,,, and Response B From(a) =,, (b) for a correct method leading to =,,,,, mark out of mark out of Response A From (a) =,, (b) =,, Response B From (a) =,, (b) =,,,,, marks out of mark out of Response Response From (a) =,, From (a) =,, (b) cos(. ) cos + = (b) period =,,,,, so =,,,,,,, marks out of marks out of Response From (a) =,, (b) repeats ever =,,, +, + =,,,, marks out of Response (Wrong angles from (a)) e.g. =,, (b) =,, =,,,,, marks out of Higher Mathemats Page

13 Paper Detailed Marking Instructions : Higher Mathemats Final D,OABC is a square based pramid as shown in the diagram below. O is the origin, D is the point (,, ) and OA = units. M is the mid-point of OA. (a) State the coordinates of B. (b) Epre DB and DM C B O M A Throughout this question, coordinates written as components and ve versa are treated as bad form. z D (,, ) Gener Scheme (a) state coordinates of B (,,) (b) pd state components of DB state coordinates of M (,, ) stated, or impliedb pd state components of DM Regularl occurring responses Response A (Transcription error for D) Response B (Transcription error for D) DB = = DM = = DB = and DM = with no working. marks out of marks out of Response A DB = d + b = DM = d + m = Response B DB = and DM = with no working. marks out of marks out of Higher Mathemats Page

14 Detailed Marking Instructions : Higher Mathemats Final (c) Find the size of angle BDM. (c) Gener Scheme know to use scalar product pd find scalar product pd find magnitude of a vector pd find magnitude of a vector pd evaluate angle BDM ˆ DB.DM cos BDM = DB DM stated, or implied b DB.DM = DB = DM = or rads. is not available to candidates who evaluate the wrong angle.. If candidates do not attempt. question., then should be awarded to an answer whh rounds to or rads is onl available if the formula quoted relates to the labelling in the. In the event that both magnitudes are equal or there is onl one non-zero component, Regularl occurring responses. is not available. Response A Response B Response C Response ˆ OB.OM OB.OD cos BOM = cos BOD ˆ = ˆ BD.BM BD.DM cos DBM = cos BDM ˆ = OB OM OB OD BD BM BD DM OB.OM = OB.OD = BD.BM = BD.DM = OB = OB = BD = BD = OM = OD = BM = DM = marks out of marks out of marks out of marks out of Response A Response B Response (Cosine rule) (Scalar Product is ) In (b) DB =, DM = In (b) DB =, DM = ˆ DB.DM DB.DM cos BDM = cos BDM ˆ = DB DM DB DM DB.DM = DB.DM = DB = so perpendular DM = ˆ DB + DM BM cos BDM = DB DM DB = DM = BM = marks out of marks out of marks out of Higher Mathemats Page

15 Detailed Marking Instructions : Higher Mathemats Final Functions f, g and h are defined on the set of real numbers b f( ) = g ( ) = + h ( ) = (a) Find g( f( )). (b) Show that g( f( )) + h ( ) = +. (a) Gener Scheme interpret notation complete proce g ( ) stated, or implied b ( ) +. without working gains onl mark.. f( g ( )) loses but will gain. f( ) g ( ) loses both and for. ( + ). (b) substitute and complete or or or Regularl occurring responses ( ) ( ) + + = + stated eplitl ( ) + + = + stated eplitl + ( ) = + stated eplitl + CAVE : Watch out for erroneous working leading to the required cub. = + stated eplitl Response Response Response + ( + ) = + + ( ) = + From (a) ( + ) In (b) ( ) + + = + + = + As the form of the answer was given in the question, this mark is not available. Response A Note : From (a) g( f( )) = In (b) ( ) = h + is not available to candidates who leave their answer as +. Response B From (a) g( f( )) = In (b) + = + Higher Mathemats Page

16 (c) (i) Show that ( ) is a factor of Detailed Marking Instructions : Higher Mathemats Final +. (ii) Factorise + full. (d) Hence solve g( f( )) + h ( ) =. (c) Gener Scheme = know to use pd complete evaluation state conclusion find quadrat factor pd factorise completel Method : Using snthet division "remainder is zero so ( ) is a factor", accept "( ) is a factor" + + stated, or implied b ( )( )( ) + + If onl the word 'factor' appears, it must be linked to the in the table. The link could be as little as 'so', ' ', ' ', ' ' or 'hence'. The word 'factor' onl, with no link, does not gain stated eplitl.. Method : Using substitution and inspection = know to use + = ( ) is a factor ( )( + + ) stated, or implied b ( )( )( ) + + is onl available as a consequence of the evidence for and. stated eplitl. Communation at must be consistent with working at. i.e. candidate s working must arrive legitimatel at zero before is awarded. If the remainder is not then an appropriate statement would be '( ) is not a factor'.. Unacceptable statements : = is a factor, ( + ). cannot be awarded for solving + = in (c). is a factor, = is a root, ( ) is a root etc. (d) interpret and solve equation in (d), and These must appear eplitl here at (d). + + leading to, and then (, ), (, ) and (, ). From (c) ( )( )( ) However, ( )( )( ) = = = gains + + leading to (, ), (, ) and (, ) onl does not gain... From (c) ( + )( + ) onl, leading to =, = does not gain as equation solved is not a cub. Higher Mathemats Page

17 Detailed Marking Instructions : Higher Mathemats Final (a) A sequence is defined b u with. n+ = u n u = Write down the values of (b) A second sequence is given b,,,,.... u and u. It is generated b the recurrence relation v with. n+ = pv n + q v = Find the values of p and q. (a) Gener Scheme pd find terms of sequence u u = and = Accept and (b) interpret sequence solve for one variable pd state second variable. Candidates ma use p + q = as one of their equations at e.g. p q and p q + = + = p or q = = q or p = =.. Treat equations like p + q = or p() + q = as bad form.. Candidates should not be penalised for using u = pu + q. n+ n Regularl occurring responses Response A (No working) Response B (Onl one equation) Response C (B verifation) p = and q = p + q = p = and q = (e nihilo) or v = v p = and q = v = = n+ n and v = =. mark out of mark out of marks out of Higher Mathemats Page

18 Detailed Marking Instructions : Higher Mathemats Final (c) Either the sequence in (a) or the sequence in (b) has a limit. (i) Calculate this limit. (ii) Wh does the other sequence not have a limit? (c) Gener Scheme know how to find valid limit pd calculate a valid limit onl l = l or l = l = ( ) state reason b. Just stating that l = al + b or l = is not suffient for. a < p < outside interval. An calculations based on formulae masquerading as a limit rule cannot gain. For candidates who use ' b = ', is onl available to those who simplif. Accept > or p > for. This ma be epreed in words.. Candidates who use a without reference to p or cannot gain. and to.. Using b l =, a a = and b =, without substituting, and stating l =, gains but not. Regularl occurring responses Response Response A Response B l = From (b) p = and q = From (b) p = and q = l = See note marks out of In (c) and l = l + l = l l = l = and In (c) l = l + l = l l = l = Both limits eist as < < and < < Impoible marks out of marks out of Response A Response B l = and l = l = and l = l = l = l = l = First has limit because < < Second sequence has no limit as < < not true mark out of marks out of Response A Response B l = l and l = l l = l and l = l l = l = l = l = st has limit because < < Second sequence has no limit because is not between and marks out of marks out of Higher Mathemats Page

19 Detailed Marking Instructions : Higher Mathemats Final The diagram shows the curve with equation and the line with equation = +. = + = + = + The curve and the line intersect at the points (, ), (, ) and (, ). Calculate the total shaded area. O Gener Scheme know to integrate know to deal with areas on each side of ais... or attempt integration Evidence of attempting to interpret the diagram to left of ais separatel from diagram to the right. interpret limits of one area e.g. with no other use "upper lower" ( ) ( ) ( ) ( + ) pd integrate + substitute in limits ( ( ) ( ) ( ) ) ( () () () ) + pd evaluate the area on one side Evidence for ma be implied b, but must be consistent with. interpret integrand with limits of the other area ( ) ( ) d ( ) ( ) d pd evaluate the area on the other side state total area or or or or. The evidence for. The evidence for ma not appear until stage. ma appear in a diagram e.g.. Where a candidate differentiates at and are not available.. Candidates who substitute at still available., then,,, without integrating at lose A B, and. However, and are Higher Mathemats Page

20 Detailed Marking Instructions : Higher Mathemats Final Regularl occurring responses General comment to markers In this question ou should scan the entire response before starting to mark. Where errors occur in the integration/evaluation, use to to mark the better solution and and to mark the poorer solution. A tabular approach to allocating marks is partularl useful in questions like this, where a candidate s response is spread over several pages, or contains working whh appears randoml set out. Response indates the approach to take here. Response Response upper lower... = d + d =.. + d = Total area = + + = The appearance of this integral is Response + d N.B. If due to an error the evaluation is negative it must be dealt with correctl. The responses below illustrate what is required under this circumstance. If both integrals lead to negative values onl Response A Response B Response C = = + d Area = + = suffient to lose and. = = + d Area = + = = = = Area = + d + = or can't be negative Candidates who evaluate an integral and obtain a negative answer must deal with this correctl. The minimum evidence would be e.g. = A = or Area = is lost. Higher Mathemats Page

21 Detailed Marking Instructions : Higher Mathemats Final Variables and are related b n the equation = k. log The graph of log against log is a straight line through the points (, ) and (, ), as shown in the diagram. (, ) (, ) O Find the values of k and n. log Method Gener Scheme introduce logarithms to = use laws of logarithms interpret intercept solve for k interpret gradient k n Method log n = log = nlog + log k = or = n log log log Accept without working k = or = k k stated eplitl stated eplitl Accept without working Method state linear equation introduce logarithms use laws of logarithms use laws of logarithms interpret result Method log = log + log = log + + log or + log log = log + log = log + log = = = log log = = Method interpret point on log. graph convert from log. to ep. form interpret point and convert know to substitute points interpret result. Omiion of base is treated as bad form at the. Gradient ( m ) = is not suffient for.. Throughout this question accept in lieu of. Method log and log = = = and = log and log = = = and = ( ) n n = k and = k (from = k. ) n = and stage.. Markers should not pk and choose within methods. Use the method whh gives the candidate the highest mark. Higher Mathemats Page

22 Regularl occurring responses Response A With no working k = n = Response B With no working k = n = Detailed Marking Instructions : Higher Mathemats Final marks out of marks out of Response (Method ) log k = k = n = marks out of Response (Variation of Method and Response A) log = log + = + k =, n = marks out of log = log + Response (Variation of Method and Response A) = + log = log + log log = = = k = n = marks out of Higher Mathemats Page

23 Detailed Marking Instructions : Higher Mathemats Final (a) The epreion sin cos can be written in the form Rsin( + a) where R > and a < π. Calculate the values of R and a. (a) Gener Scheme use compound angle formula compare coeffients pd proce R pd proce a sin cos + R a Rcoina R a = R a = cos and sin stated eplitl stated eplitl (Accept ) with or without working (Accept ) must be consistent with. Treat as bad form the use of k instead of R.. Treat Rsincosa + coina as bad form onl if the equations at the. sincosa + coin a or (sincosa + coin a) is acceptable for. is not available forrcos and Rsin = =, however, stage both contain R. is still available. and.. is onl available for a single value of a.. Candidates who work in degrees and don t convert to radian measure lose. Do not accept π π or.. Candidates ma use an form of the wave equation for value of a is interpreted for the form Rsin( + a)., and, however, is onl available if the Regularl occurring responses For and Response A Response B Response C Rcosa = Rsina = Rcosa = Rsina = Rcosa = Rsina = tana = a = tana = a = tana = a = Response Response Rsin( a) = Rsincosa Rcoina ksincosa + kcoina Rcosa = Rsina = cosa = sina = R = R = a = a = See note Not consistent with working at marks out of marks out of Higher Mathemats Page

24 (b) Hence find the value of t, where t, for whh t Detailed Marking Instructions : Higher Mathemats Final (cos + sin ) d = (b) Gener Scheme pd integrate given epreion substitute limits pd proce limits know to use wave equation write in standard format start to solve equation pd complete and state solution. The inclusion of + c at sin cos t t (sin cos ) (sin cos) t t + sin cos t + + sin( t + ) = t + = t = sin( ) and stage should be treated as bad form.. For those candidates who use a as or..., follow through their working for to. to are available to candidates who chose to write this integrand as new wave function.. Candidates who use degree measure in (a) lose and are available (see also response A and B below.) Regularl occurring responses Response (No integration) Response and if the continue to do so in (b), onl,, t cos + sind = sin( + ) lose,, and then sin( + ) = + =,, = Needs to be in terms of t sin cos sint cost sin( t + ) sin( t + ) = t + =,, t = Response A (Misreading question) sin( + ) d Response B (Misreading question) sin( + ) d = cos( t + ) = = cos( + ) cos( t + ) = = cos( t + ) + cos t + =,, cos( t + ) + t = i.e. no solution cos( t + ) = If a is left in t + =,, degrees no marks t = i.e. no solution are available. marks out of marks out of t Higher Mathemats Page

25 Circle C has equation ( + ) + ( ) =. Detailed Marking Instructions : Higher Mathemats Final A circle C with equation p = is drawn inside C. The circles have no points of contact. What is the range of values of p? Gener Scheme state centre of C state radius of C state centre of C pd find radius of C in terms of p interpret upper bound for p find distance between centres ( d) identif relevant relationship develop relationship b squaring pd find lower bound for p. Treat as bad form the use of c in lieu of p.. The evidence for p < (, ) Do not accept (, ) p < r + d < r < p < p > must involve an inequalit, but ma be in words. p Accept c in lieu of p or or stated eplitl. Treat p as bad form as long as it is clear that the candidate is using p.. Candidates who are onl working with an equation lose both. and, however, is onl available to candidates who solve an inequation involving a negative coeffient of p. Regularl occurring responses ma still be available. Response A Response B Response Marks to gained C = (, ) C = (, ) For marks to r = r = + p p < d = p < + p < + p < p < p < + p < p < p < p < p > p > Response (see note ) Response Response p = p p = p p = p = p > p p Penalise the use of and/or once onl. < p < < p < < p < so or p < and p > < p < Higher Mathemats Page

26 Detailed Marking Instructions : Higher Mathemats Final Regularl occurring responses Response ( ) + ( + ) = p < p p > Higher Mathemats Page