2015 Mathematics. Higher. Finalised Marking Instructions

Transcription

1 05 Mathematics Higher Finalised Marking Instructions Scottish Qualifications Authority 05 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. 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 should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ 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 when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 General Comments These marking instructions are for use with the 05 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 indicates the rationale for which each mark is awarded. In general, markers should use the Illustrative Scheme and only use the Generic Scheme where a candidate has used a method not covered in the Illustrative Scheme. All markers should apply the following general marking principles throughout their marking: Marks must be assigned in accordance with these marking instructions. In principle, marks are awarded for what is correct, rather than deducted for what is wrong. One mark is available for each. There are no half marks. Working subsequent to an error must be followed through, with possible full marks for the subsequent 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 through mark cannot be awarded. As indicated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Throughout this paper, unless specifically mentioned in the marking instructions, a correct answer with no working receives no credit. 5 In general, as a consequence of an error perceived to be trivial, casual or insignificant, e.g. candidates lose the opportunity of gaining a mark. But note the second example in comment 7. 6 Where a transcription error (paper to script or within script) occurs, the candidate should be penalised, e.g. This is a transcription error and so the mark is not awarded. Eased as no longer a solution of a quadratic equation. Exceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt. x x x ~~ x x 0 ( x )( x ) 0 x or Page

3 7 Vertical/horizontal marking Where a question results in two pairs of solutions, this technique should be applied, but only if indicated in the detailed marking instructions for the question. Example: Point of intersection of line with curve 5 6 Illustrative Scheme: 5 x y 5 6 x y 7 Markers should choose whichever method benefits the candidate, but not a combination of both. 8 In final answers, numerical values should be simplified as far as possible, unless specifically mentioned in the detailed marking instructions. Examples: should be simplified to or should be simplified to should be simplified to 50 must be simplified to 8 should be simplified to The square root of perfect squares up to and including 00 must be known. 9 Commonly Observed Responses (COR) are shown in the marking instructions to help mark common and/or non-routine solutions. CORs may also be used as a guide when marking similar non-routine candidate responses. 0 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, e.g. angles in degrees rounded to nearest degree; Omission of units; Bad form (bad form only becomes bad form if subsequent working is correct), e.g. ( x x x )(x ) written as ( x x x ) x x x 6x x x x x x 5x 8x 7x gains full credit; Repeated error within a question, but not between questions. In any Show that... question, where the candidate has to arrive at a required result, the last mark of that part is not available as a follow through from a previous error unless specifically stated otherwise in the detailed marking instructions. All working should be carefully checked, even where a fundamental misunderstanding is apparent early in the candidate s response. Marks may still be available later in the question so reference must be made continually to the marking instructions. Page

4 All working must be checked: the appearance of the correct answer does not necessarily indicate that the candidate has gained all the available marks. If you are in serious doubt whether a mark should or should not be awarded, consult your Team Leader (TL). 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 should be marked. 5 Where a candidate has made multiple attempts using the same strategy, mark all attempts and award the lowest mark. Where a candidate has tried different strategies, apply the above ruling to attempts within each strategy and then award the highest resultant mark. For example: Strategy attempt is worth marks Strategy attempt is worth marks From the attempts using strategy, the resultant mark would be. Strategy attempt is worth mark Strategy attempt is worth 5 marks From the attempts using strategy, the resultant mark would be. In this case, award marks. 6 In cases of difficulty, covered neither in detail nor in principle in these instructions, markers should contact their TL in the first instance. Page

5 Paper Section A Question Answer C B D A 5 C 6 B 7 C 8 D 9 B 0 D C C B D 5 A 6 B 7 D 8 C 9 B 0 A Summary A B 6 C 6 D 5 Page 5

6 Paper - Section B Question Generic Scheme Illustrative Scheme Max Mark (a). know to use x interpret result and state conclusion Method 6() 9() 0 ( x ) is a factor. Method remainder 0 ( x ) is a factor. x Method x x x x 6 9 x x 0 ( x ) is a factor. state quadratic factor factorise completely x 5x stated or implied by x x Page 6

7 . Communication at must be consistent with working at that stage i.e. a candidate s working must arrive legitimately at 0 before is awarded.. Accept any of the following for : f() 0 so ( x )is a factor since remainder is 0, it is a factor the 0 from the table linked to the word factor by e.g. so, hence,,,. Do not accept any of the following for : double underlining the zero or boxing the zero without comment x is a factor, ( x ) is a factor, x is a root, ( x ) is a root, ( x ) is a root the word factor only, with no link x x x in any order.. At the expression may be written as 5. An incorrect quadratic correctly factorised may gain. 6. Where the quadratic factor obtained is irreducible, candidates must clearly demonstrate that b ac 0 to gain must appear at or for to be awarded. 8. For candidates who do not arrive at 0 at the stage are not available. 9. Do not penalise candidates who attempt to solve a cubic equation. However, within this working there may be evidence of the correct factorisation of the cubic. 0. Evidence for & may appear in part (b). Page 7

8 (b)(i). 5 know to and differentiate 5 x x 6 find gradient 6 7 state equation of tangent 7 yx. 7 is only available if an attempt has been made to find the gradient from differentiation.. At mark 7 accept y ( x ), yx or any other rearrangement of the equation. 5 6 using y mx c x y m c c 7 y x (b)(ii). 8 set ycurve yline 8 x 6x x x 9 arrange equation in standard cubic form 9 x 6x 9x 0 0 identify x coordinate of B and calculate y coordinate 0 (,9). 9 is only available if 0 appears in at least one arrangement of the equation.. Solutions at 0 must be consistent with working at and Candidates who obtain three distinct factors at can gain 8 and 9 but 0 is unavailable. 6. For 0 accept x, y Do not penalise the appearance of (,). Page 8

9 . arrange in differentiable form f ( x) x x start differentiation complete differentiation and set f( x) 0 evaluate f at stationary point x x 8 or 8 0 x, f ( x) 5 consider end-points 5 7 f() 5, f() (or 5) 6 state max and min values 6 max5, min 6. Candidates must attempt to differentiate a term with a ve or fractional power by for to be awarded.. is not available for simply stating f '( x) 0. A clear link between the candidates derivative and f '( x) 0 is required.. For candidates who integrate, but clearly believe they are finding the derivative 5 6 are available (see CORs K, L, and M). In other instances where candidates have integrated then only and 5 are available. A numerical approach can only gain and 6 are not available to candidates who consider stationary points only. 5. Treat maximum (,5) and minimum (,) as bad form. 6. Vertical marking is not applicable to 5 and The appearance of (,) following any nd derivative or nature table gains. 8. If at the stage a value of x is obtained outwith the given interval then is unavailable, but 6 may still be gained (see CORs F and G). 9. Candidates who consider the end values but do not evaluate the stationary value cannot gain 6. Candidate B Candidate C Candidate D f ( x) x x f x x '( ) 8 for stationary points f '( x) 0 With no further working f ( x) x x ^ for stationary ^ points f '( x) 0 With no further working f ( x) x x f x x '( ) 8 for stationary ^ dy points 0 dx With no further working f ( x) x x 8x for stationary points f '( x) 0 ^ With no further working Page 9

10 Candidate E Solely a numerical attempt f() 5, f(), 7 f(), f() 9 Award only 5 For any similar attempt which includes the evaluation of f for a value outwith the range award 0. Candidate F f ( x) x x f '( x) 8 x... f '( x) x f( ) f() 5, f() min, max 55 8 Candidate G f ( x) x x f '( x) 8 x... f '( x)... 0 x f( ) f() 5, f() min, max Candidate H Candidate I Candidate J f ( x) x x f x x '( ) 8... f '( x)... 0 x 8 f ( x ) 8 6 f() 5, f() min, max f ( x) x '( ) 0 0 f x x x 0 f ( x ) undefined 5 ^ 6 ^ f '( x) x 0 x f() 5, f() 5 6 ^ Candidate K Candidate L Candidate M f ( x) x x x f '( x) x x f() f() 5, f() 5 6 f ( x) x x x x x x 0 x f() f() 5, f() 5 6 x x x x 0 x f() f() 5, f() 5 6 min, max 5 min, max 5 min, max 5 Page 0

11 . collect log terms use laws of logs use laws of logs Method log x7 log x stated or implied by x 7 x log solve for x x 7 x x Method log x 7 log log x stated or implied by log x7 log log ( x) x log 7 log 8( x ). For accept x 7 log log 8. x x Candidate B x x x x log 7 log log log 7 x log x 7 x x 7 8x 8 x 7 x log (x7) log ( x) log (x7) log ( x) log (x7)( x) (x 7)( x) x x5 0 (x 5)( x ) 0 5 x or x discard as x x 5 is not available for candidates who do not discard x Page

12 . interpret the values of a, b and c and substitute know to use discriminant 0 Method simplify or factorise quadratic inequation state range of values of k Method simplify or factorise quadratic expression k 9k... 0 Method 9 k k k or k Method 9 6k 0 k, evidence and range of values of k graph or other evidence leading to k. The 0 must appear at least once at the or stage for to be awarded.. If an x appears in the candidate s discriminant only may be awarded.. The use of any expression masquerading as the discriminant can gain only at most.. Award to candidates who write BOTH 9 6k 0 AND 9 6k For candidates who at simplify or factorise an equation can only be awarded if evidence of solving an inequation (for example a graph) appears. 6. At stage, quoting b ac 0 is not sufficient. 7. At stage, in Method, solutions for k need not be simplified. Page

13 5(a) know to use Theorem of Pythagoras or distance formula process to obtain result Be wary of fudged solutions! D t 5 t 0 Beware D t 0t 5 t 0t 00 5t 0t5 D y y x x ( 5) 0 0 ( 0) D t t D t t See note in the general instructions 5(b) write in differentiable form start differentiation 5 complete differentiation A(t 5, 0) B(0, t 0) 0 t5 t5 AB t0 0 t0 D (t 5) ( t 0) D t t t t D 5t 0t 5 NB: This is an exception to note in the general instructions 5t 0t t t t 0 6 substitute t = 5 and interpret result 6 0 D(5) 0 increasing 50. is only available for differentiating an expression of the form (trinomial) proper fraction.. Do not penalise the use of x instead of t.. 6 is only available to candidates who substitute into a derivative. Page

14 Candidate B D t t t ( ) ( ) D t t t... (0t 0) 0 D(5) 0decreasing 50 Candidate C see note. 5 6 D( t) 5t 0t 5 D( t) 5t 0t (0t 0) 5 D(5) 0decreasing 50 Candidate D 5 6 D( t) 5t 0t 5 D( t) 5t 0t (0t 0) D (5) increasing Candidate E D( t) 5t 0t 5 D( t) 5t 0t 5 0t 0 D(5) 0 increasing Calculating Distance t 5 D 50 t D 5 so distance is increasing Award 0 marks as answer is not from differentiation D( t) 5t 0t 5 D( t) 5t 0t 5 0t 0 D( t) 5t 5t 0t D(5) 0 0decreasing 50 Candidate F Alternative Method D t t dd 0t 0 dt dd t dt dd dt D is increasing Alternative Response 0 0 D is increasing D t t D 5( 8 5) 5[( t ) 9] Graph together with a statement indicating that when t 5, D is increasing and therefore D is increasing. Min TP at x 5 6 Page

15 Paper Question Generic Scheme Illustrative Scheme Max Mark (a) calculate gradient of AB m AB use property of perpendicular lines malt substitute into general equation of a line y x demonstrate result... xy. is only available as a consequence of trying to find and use a perpendicular gradient.. is only available if there is/are appropriate intermediate lines of working between and.. The ONLY acceptable variations for the final equation for the line in are x y, y x, y x. Candidate B 5 For mab 5 7 y x malt y x y x is not available y x - not acceptable y x - not acceptable xy Page 5

16 (b) 5 calculate midpoint of AC 5 M AC (,5) 6 calculate gradient of median 6 mbm 7 determine equation of median 7 yx. 6 and 7 are not available to candidates who do not use a midpoint is only available as a consequence of using a non-perpendicular gradient and a midpoint. 6. Candidates who find either the median through A or the median through C or a side of the triangle gain mark out of. 7. At 7 accept y ( 5) ( x ( )), y5 ( x ), y x 0 or any other rearrangement of the equation. Median through A Median through C M (6, ) M (,) BC 8 m AM 8 8 y ( x 6) or y 7 ( x 5) Award / (c) 8 calculate x or y coordinate AB m CM 8 y ( x ) or y ( x ) 8 8 Award / 8 x or y 9 calculate remaining coordinate of the point of 9 y or x intersection 8. If the candidate s median is either a vertical or horizontal line then award out of if both coordinates are correct, otherwise award 0. For candidates who find the altitude through B in part (b) y 5 ( x ) 7 (b) x 5 7 y yx -error xy (c) y x Leading to x7 and y 8 9 Page 6

17 (a) interpret notation state a correct expression f (( x)( x) ) stated or implied by 0 ( x)( x) stated or implied by. is not available for g f x g(0 x) but may be awarded for x x f ( g( x)) ' g( f ( x))' (a) ( 0 x)( (0 x)) (b) (c) 75 8x x or x 8x 75 ( x 8 ( x 9) x ( x 9) 6 ( x 9) 6 0 x 9 6 or Candidate B f ( g( x)) 0(( x) ( x)) Candidate C f ( g( x)) 0(( x)( x) ) 0 (0 ). ^ ^ (b) write f ( g( x)) in quadratic form Method identify common factor 5 complete the square Method expand completed square form and equate coefficients 5 x x or x x 5 Method ( x x stated or implied by 5 ( x ) 6 5 Method px pqx pq r and p, 5 process for form q and r and write in required 5 q and r 6 Note if p 5 is not available Page 7

18 . Accept 6 ( x ) or ( x ) 6 at 5. ( x x5) Candidate B 5 xx x x5 ( x x5) px pqx pq r and p ( x ) 6 5 q r 6 eased Candidate D Candidate E 5 Candidate C x x 5 ( x )... ( x ) Candidate F 5 5 xx x x5 ( x ) 6 5 eased Eased, unitary coefficient of (lower level skill) (c) x 6 identify critical condition 5 xx x x5 ( x ) 6 so 5 x x ( x ) 6 6 x x 5 ( x )... ( x ) 6 ( x ) 6 0 or f (( g( x)) 0 7 identify critical values 7 5 and. Any communication indicating that the denominator cannot be zero gains 6.. Accept x 5 and x or x 5 and x at If x 5 and x appear without working award /. 5 Candidate B 5 ( x ) 6 x 5 (a) 6 7 ^ f ( g( x)) f ( g( x)) 0 x, x 5 x x determine the value of the required term 9 or or 75. Do not penalise the inclusion of incorrect units.. Accept rounded and unsimplified answers following evidence of correct substitution. Page 8

19 (b) Method (Considering both limits) know how to calculate limit know how to calculate limit or or Method L L L L calculate limit 5 calculate limit 6 interpret limits and state conclusion Method (Frog first then numerical for toad) know how to calculate limit calculate limit determine the value of the highest term less than 50 5 determine the value of the lowest term greater than 50 6 interpret information and state conclusion Method (Numerical method for toad only) continues numerical strategy exact value determine the value of the highest term less than 50 5 determine the value of the lowest term greater than 50 6 interpret information and state conclusion Method (Limit method for toad only) & know how to calculate limit toad will escape Method or L L toad will escape Method numerical strategy toad will escape & Method or L L & 5 calculate limit 6 interpret limit and state conclusion & toad will escape 5 Page 9

20 . 6 is unavailable for candidates who do not consider the toad in their conclusion.. For candidates who only consider the frog numerically award /5 for the strategy. Error with frogs limit Frog Only LF LF frog will escape. Using Method - Toad Only Using Method - Toad Only Toad Conclusions Limit = 5 This is greater than the height of the well and so the toad will escape award 6. However Limit =5 and so the toad escapes - 6 ^. missing ^ missing ^ rounding error so the toad escapes so the toad escapes. Using Method - Toad Only 97.. rounding error so the toad escapes. Iterations f f f f f f f f f t t t t t t t t t t t t Page 0

21 (a) know to equate f ( x) and g( x ) x x x x 5 solve for x x. and are not available to candidates who: (i) equate zeros, (ii) give answer only without working, (iii) arrive at x with erroneous working. Candidate B y x x x x y x x 5 x x 5 subtract to get 0 x x x In this case the candidate has equated zeros Candidate C f ( x) x x g( x) x x 5 f '( x) x g '( x) x.. x x x Page

22 (b) know to integrate interpret limits 5 use upper lower ( x x ) ( x x ) dx 6 integrate 6 7 x x accept 8 unsimplified integral 7 substitute limits 8 evaluate area between f ( x) and h( x ) 9 state total area If limits x0 and x appear ex nihilo award.. If a candidate differentiates at 6 then 6, 7 and 8 are not available. However, 9 is still available. 5. Candidates who substitute at 7, without attempting to integrate at 6, cannot gain 6, 7 or 8. However, 9 is still available. 6. Evidence for 8 may be implied by is a strategy mark and should be awarded for correctly multiplying their solution at 8, or for any other valid strategy applied to previous working. 8. For 5 both a term containing a variable and the constant term must be dealt with correctly. 9. In cases where 5 is not awarded, 6 may be gained for integrating an expression of equivalent difficulty i.e. a polynomial of at least degree two. 6 is unavailable for the integration of a linear expression must be as a consequence of substituting into a term where the power of x is not equalto or 0. Page

23 Valid Strategy Candidates who use the strategy: Total Area = Area A + Area B A B Then mark as follows: Mark Area A for to 8 then mark Area B for to 8 and award the higher of the two. 9 is available for correctly adding two equal areas. Candidate B Invalid Strategy For example, candidates who integrate each of the four functions separately within an invalid strategy Gain if limits correct on 5 is unavailable Gain 6 for calculating either f ( x) and h( x) or g( x) and k( x) f ( x) or g( x) and h( x) or k( x) Gain 7 for correctly substituting at least twice Gain 8 for evaluating at least two integrals correctly 9 is unavailable Candidate C ( x x 8 x x ) dx ~~~~~~~~~ ( x x) dx 8 5 x x 8 6 Candidate E... cannot be negative so 8 however, so Area 8 Candidate D 0 0 ( x x x x ) dx 9 8 ( x x 6) dx 8 8 Candidate F 0 x x 6x 9 ( x x 8 x x ) dx ~~~~ 7 ( 8 x ) 0 x x 7 8 x dx Page

24 5(a) state centre of C state radius of C calculate distance between centres of C and C (, 5) 5 0 calculate radius of C 5. For to be awarded radius of C must be greater than the radius of C.. Beware of candidates who arrive at the correct solution by finding the point of contact by an invalid strategy.. is for Distance cc rc but only if the answer obtained is greater than r c. Page

25 5(b) 5 find ratio in which centre of C divides line joining centres of C and C 6 determine centre of C 7 calculate radius of C 8 state equation of C 5 : 6 (6,7) 7 r 0 (answer must be consistent with distance between centres) 8 ( x 6) ( y 7) 00. For 5 accept ratios :, :, :, : (or the appearance of ) is for r c +r c. 6. Where candidates arrive at an incorrect centre or radius from working then 8 is available. However 8 is not available if either centre or radius appear ex nihilo (see note 5). 7. Do not accept 0 for For candidates finding the centre by stepping out the following is the minimum evidence for 5 and 6 : 5 6 Correct follow through using the ratio : 6 Correct answer using the ratio : Candidate B using the mid-point of centres: C (, 5) C (9,) r 0 5 centre C (,) 6 : radius of C C ( x ) ( y ) note 6 C (0, ) 7 x ( y) 00 8 Candidate C touches C internally only Candidate D - touches C internally only 5 5 centre C (,) centre C (,) 6 7 radius of C radius of C 5 8 ( x) ( y) 5 Candidate E centre C collinear with C,C 5 e.g. centre C (,7) 6 radius of C 5 (touch C internally only) 7 8 ( x ) ( y 7) 05 Page 5 6 radius of C radius of C ( x) ( y) 5

26 6(a) Expands p.q p.r Evaluate p.q Completes evaluation For p.( q r) pq pr with no other working is not available. 6(b) correct expression -qp r or equivalent 6(c) 5 correct substitution 6 start evaluation 5 -q.q q.p q.r r cos0 o 9 7 find expression for r. Award 5 for -q +q.p q.r 9 -q.q q.p q.r = 9 9 o 9 9 r cos50 9 r cos r cos0 Candidate B 9 -q.q q.p q.r = 9 9 o 9 9 r cos 0 9 r Page 6

27 7(a) integrate a term complete integration with constant sin x OR x c sin x c x 7(b) substitute for cos x substitute for and complete x x cos sin... or...(sin x cos x)... sin x cos x cos x sin x. Any valid substitution for cosx is acceptable for.. Candidates who show that cos x sin x cos x may gain both marks.. Candidates who quote the formula for cosx in terms of A but do not use in the context of the question cannot gain. cos x (cos x ) (cos x ) ( sin x) cos xsin x Candidate B cos x sin x (cos x ) ( cos x) cos x 7(c) 5 6 interpret link state result sin x x c sin x cos x dx (cos x ) dx sin x x c 5 6 Page 7

28 8. use compound angle formula compare coefficients process for k process for a k sin5t cos a k cos 5t sin a k cos a = 6, k sin a = 5 stated explicitly k = 9 a rad or 6 5 equates expression for h to 00 6 write in standard format and attempt to solve 7 solve equation for 5t 8 process solutions for t 5 9sin 5t sin 5t t sin t 508 and 8 t = 006 and Treat k sin5t cos a cos 5t sina as bad form only if the equations at the stage both contain k.. 9sin5t cos a 9cos5t sina 9 sin5t cos a cos 5tsina is acceptable for and or.. Accept kcosa 6 and ksina 5 for.. is not available for k cos5t 6 and k sin5t 5, however, is still available. 5. is only available for a single value of k, k is only available for a single value of a. 7. The angle at must be consistent with the equations at even when this leads to an angle outwith the required range. 8. Candidates who identify and use any form of the wave equation may gain, and, however, is only available if the value of a is interpreted for the form k sin( 5 t a). 9. Candidates who work consistently in degrees cannot gain Do not penalise additional solutions at 8.. On this occasion accept any answers which round to 0 and 6 ( significant figures required). Page 8

29 Response : Missing information in working. Candidate B 9cosa 6 cosa 6 9sin a 5 ^ ^ sina tan a tan a ^ 6 6 a 0 979rad or 6 a 0 979rad or 6 Does not satisfy equations at Candidate C k sin5t cos a k cos5t sina kcosa 6, ksina 5 k 9 or 9 5 tan a 6 a 0 979rad or 6 or a rad or 06 Response : Correct expansion of k sin( x + a)º and possible errors for and Candidate D Candidate E Candidate F kcosa 6 kcosa5 kcosa 6 ksina5 ksina 6 ksina5 6 tan a 6 tan a tan a 6 a 76rad or 67 a 76rad or 67 a 5 888rad or 7 Response : Labelling incorrect, sin (A B) = sin A cos B cos A sin B from formula list. Candidate G Candidate H Candidate I ksin A cos B kcos A sin B kcosa 6 ksina5 5 tan a 6 a 0 979rad or 6 Candidate J ksin A cos B kcos A sin B kcos5t 6 ksin5t 5 5 tan5 t 6 5 t 0 979rad or 6 Candidate K ksin A cos B kcos A sin B k cos B 6 k sin B 5 5 tan B 6 B 0 979rad or 6 9sin(5t 095) sin(5t 095) t 095 sin sin(5t 095) t 095 sin [END OF MARKING INSTRUCTIONS] Page 9