2014 Mathematics. Advanced Higher. Finalised Marking Instructions

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1 0 Mathematics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 0 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 Assessment team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of eamination 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 Assessment team may be able to direct you to the secondary sources. These ing Instructions have been prepared by Eamination Teams for use by SQA Appointed ers when marking Eternal Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 Part One: General ing Principles for Mathematics Advanced 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 specific ing Instructions for each question. (a) (b) s for each candidate response must always be assigned in line with these general marking principles and the specific ing Instructions for the relevant question. ing should always be positive ie, marks should be awarded for what is correct and not deducted for errors or omissions. GENERAL MARKING ADVICE: Mathematics Advanced Higher The marking schemes are written to assist in determining the minimal acceptable answer rather than listing every possible correct and incorrect answer. The following notes are offered to support ers in making judgements on candidates evidence and apply to marking both end of unit assessments and course assessments. General ing Principles These principles describe the approach taken when marking Advanced Higher Mathematics papers. For more detailed guidance please refer to the detailed ing Instructions. The main principle is to give credit for the skills demonstrated and the criteria met. Failure to have a correct method may not preclude a candidate gaining credit for their solution. The answer to one part of a question, even if incorrect, can be accepted as a basis for subsequent dependent parts of the question. 3 The following are not penalised: working subsequent to a correct answer (unless it provides firm evidence that the requirements of the question have not been met) legitimate variation in numerical values/algebraic epressions. Full credit will only be given where the solution contains appropriate working. Where the correct answer might be obtained by inspection or mentally, credit may be given. 5 Sometimes the method to be used in a particular question is eplicitly stated; no credit will be given where a candidate obtains the correct answer by an alternative method. 6 Where the method to be used in a particular question is not eplicitly stated in the question paper, full credit is available for an alternative valid method. (Some likely alternatives are included but these should not be assumed to be the only acceptable ones.) In the detailed ing Instructions which follow, marks are shown alongside the line for which they are awarded. When marking, no comments at all should be made on the script. The total mark for each question should appear in one of the right-hand margins. The following codes should be used where applicable: correct; X wrong; working underlined or circled wrong; tickcross mark(s) awarded for follow-through from previous answer; ^ ^ - mark(s) lost through omission of essential working or incomplete answer; wavy or broken underline bad form, but not penalised. Page

3 Part Two: ing Instructions for each Question Question Epected Answer/s Ma. (a) f.. 3 Knows to use quotient (or product) rule., Correct derivative, using either rule, unsimplified. 3 3 f( ) f '( ) ( )( )... f '( ) ( ) ( ) Simplifies to answer. By polynomial division (or inspection) correctly simplifies f(). Correctly completes first step in integration. Applies chain rule and simplifies to answer.. (b) 6 3 tan y 3 6 = + 9 dy d sec y. 6 dy 6 6 d sec y sec tan 3 3 Correct form of denominator. d d Multiplies by Processes to remove brackets correctly. Correctly processes from tan - to tan. Correctly differentiates implicitly on both sides Isolates dy on LHS and d epresses in terms of only. Notes:. Evidence of method: Statement of the rule and evidence of progress in applying it. Application showing the difference of two terms, both involving differentiation and a denominator.. Accept use of product use with equivalent criteria for. Page 3

4 Question Epected Answer/s Ma. r 0r 0r r 0 0 r r 5 Unsimplified form of general term, correct (either form). r 0r 0 0 r r 0r r 0 r r 0r 0 0 r 0r r r 0r 0 r r r r r 0r 0 0 r r r 0r 0r 0r r r 0 0 r r 3r0 03r 0r 0r 0 0 r r 3 r 0 r 0 03 r r 0 3 Correct simplification of coefficients.,3 Correct simplification of indices of.,3 For term in ie 3 : r = 7 r = Obtains appropriate value for r from simplified epression.,5 3 = Correct evaluation of above epression. 5 Notes:. No simplification required, but must be stated eplicitly, as required in the question.. Negative indices may be written in denominator with positive indices. 3 Coefficients must be collected to a single epression. eg separate powers of both and or multiple powers of are not permitted. Where an incorrect, simplified epression leads to a non-integer value for r, is not available. 5 Final answer obtained from epansion, with no general term, only and 5 available [ma out of 5]. Page

5 Question Epected Answer/s Ma 3. 3 R R R 5R R 3 6 Sets up augmented matri R R R R 3 Correctly obtains zeroes in first elements of second and third rows., Completes elimination to upper triangular form. 3 ( + λ)z = 3 z 3 Obtains simplified epression for z.,5 z, y, Correct statement based on epression at,6 Correct solution based on matri at 3. Notes: 3. Row operations commentary not required for full credit. Methods not using augmented matri may be acceptable. 3. Not necessary to have unitary values for second elements for. 3.3 Accept lower triangular form. 3. If lower triangular form used, will need to have simplified epression for z Accept z = 3.6 Also accept: When λ = there are no solutions; λ < and λ > ; λ < or λ >. 3.7 Do NOT accept: < λ >. Page 5

6 Question Epected Answer/s Ma. d t dt t 3 Correct differentiation of either y or evidence of knowing to differentiate both equations w.r.t. t. dy t dt t Correct completion of both differentiations. dy t t d t t dy d t t dy d + t + t 3 Processes to answer.,3 Notes:. d dy For eample and. dt t dt t. Although t 0 applies, do not penalise omission..3 Epressed as a single fraction.. Failure to employ chain rule renders unavailable, but out of 3 is possible for t t. Page 6

7 Question Epected Answer/s Ma 5. i j k 3 3 Setting up cross product correctly., = 3i + 5j + 7k Correctly evaluates cross product. 5 3 u. v w Correctly evaluates dot product with u and vector from answer. u lies in the same plane as the one containing both v and w. u is parallel to the plane containing v and w. u is perpendicular to the normal of v and w. All points lie in the same plane. u is perpendicular to v w. Volume of parallelepiped is zero. u, v and w are coplanar/linearly dependent. Any one correct statement.,3,6 u v w Setting up combined product correctly Correctly processes determinant. 5 3 Correctly evaluates = 0 determinant to reach 0. Notes: 5. Alternative layouts and methods possible for full credit. 5. Do NOT accept: Vectors are perpendicular [must specify which vectors]. 5.3 only available where statement is consistent with Rows and 3 are not interchangeable. 5.5 This line of this version may be omitted with and 3 awarded for final answer of Where incorrect answer at 3 is some k 0, accept: Volume of parallelepiped = k or negation of any one of the other statements. Page 7

8 Question Epected Answer/s Ma 6. y = n( 3 cos ) y = n( 3 ) + n(cos ) 3 Correctly takes logs and correctly separates RHS terms. dy 3 sin d cos dy 3 tan d a = 3, b =. 3 LHS and either term of derivative on RHS correct. Rest of derivative correct and values of a and b. y = n( 3 cos ) Correctly takes logs and evidence of: d f ' n f. d f 3 dy 3 cos sin cos 3 d cos dy 3 tan d a = 3, b =. 3 Completes differentiation correctly (unsimplified). Simplifies to obtain correct form and values of a and b. 3 e 3 cos sin cos y dy d Evidence of implicit differentiation. Completes differentiation correctly. 3 dy 3 cos sin cos 3 d cos dy 3 tan d a = 3, b =. 3 Divides by correctly and simplifies to obtain correct form and values of a and b. Notes: Page 8

9 Question Epected Answer/s Ma 7. For n = a RHS = 0 a 0 A Substituting n =. LHS A ARHS. Assume true for n = k, A k k k a = 0 Consider n = k +, A k+ = A k A [ A A k ] k k a a 0 0 See note 5. Inductive hypothesis (must include Assume true for n = k or equivalent phrase) and k epansion of. A.,5 k k k.. aa 0 a a a 0 k k k.. k k k a 0 k k a 0 Hence, if true for n = k, then true for n = k +, but since true for n =, then by induction true for all positive integers n. 3 * Correct multiplication of two matrices and accurate manipulation of indices and brackets. 3 Line * and statement of result in terms of (k + ) and valid statement of conclusion.,6 Page 9

10 Question 7 Notes: for Acceptable phrases include: If true for... ; Suppose true for ; Assume true for. However, not acceptable: Consider n=k and True for n=k. 7.3 No access to 3 or without correct matri multiplication. This includes any evidence that k. = k 7. Correct substitution necessary for. Accept starting with A and proceeding via a whether subsequently corrected or not, loses both 3 (if occurring before line *) and. 7. Minimum acceptable form for : Then true for n = k+, but since true for n =, then true for all n or equivalent. 7.5 This epression (or equivalent) must appear somewhere for the award of. However, it may appear in later working. 7.6 Need meaningful prior working for award of. Page 0

11 Question Epected Answer/s Ma 8. m m + = 0 (m ) = 0 6 Correct auiliary equation. m Correct solution of auiliary equation. C.F./G.S. y Ae Be 3 Statement of general solution/complementary function. 3,,5,6 y = when = 0 gives = A.+0, so A = dy Ae Be Be d dy 3 when = 0 gives d 3 Ae Be B.0. e B, so B So P.S. is: y = e e 5 6 Correct evaluation of A., Correct differentiation of G.S. Substitution to obtain B and particular solution. Notes: 8. Or equivalent. 8. Accept calculation of A after differentiation. 8.3 Incorrectly using y Ae Be, but correctly carrying out (simplified) differentiation, leading to inconsistent equations: A + B = and A + B = 6, gains (for two equations). ie ma 3 (out of 6). 8. Incorrect factorisation of auiliary equation with real, distinct roots not awarded. Follow through marks available for 3, and 6. Since working is eased, 5 not available. ie ma out of 6. When comple roots result, 5 IS available since working is certainly not eased. ie ma 5 out of 6. Incorrect, equal roots ma 5 out of 6 as working eased, but not significantly. 8.5 Starting at y Ae Be with no prior working loses and. 8.6 However, if A.E. appears (i.e. m + m + = 0) and jump straight to 3 y Ae Be then award 3. Page

12 Question Epected Answer/s Ma 9. cos...!! Correct statement of series for cos., 3 3 cos3...!! f ( ) cos3 f (0) f '( ) 3sin3 f '(0) 0 f ''( ) 9cos3 f ''(0) 9 f '''( ) 7sin3 f '''(0) 0 f ''''( ) 8cos3 f ''''(0) 8 3 e...! 3! Substitution and evaluation of coefficients. 3 Correct differentiation and evaluation if doing from first principles e cos Correctly stating series with correct substitution. 3 Knows to multiply the two previously obtained series together. Correctly multiplies out brackets Simplifies to lowest terms. 3, Page

13 Question Epected Answer/s Ma 9. f ( ) e cos3 f (0) '( ) f e cos3 3e sin3 f '(0) f ''( ) e ( 5cos3 sin3 ) f ''(0) 5 f '''( ) e ( 6cos3 9sin3 ) f '''(0) 6 3 ( 5) ( 6) e cos3...! 3! 5 Either all three derivatives correct first derivative and first two evaluations (above ) all evaluations [not eased if at least one each of e and sin/cos3] last two derivatives and last two evaluations correct. Remainder correct e cos Correct substitution of coefficients obtained at into formula and simplifies to lowest terms. Notes: 9. Award for substitution of 3 into series for cos. 9. Must have at least 3 terms for if no further working. 9.3 Candidates may differentiate from first principles for any or all of the three required series for full credit. 9. For 5 and 6 ignore additional terms in or higher. Page 3

14 Question Epected Answer/s Ma 0. y V 3 y d d Correctly identifies circle equation.,5,7,8, Correct form of integral and applies correct limits., 3 Substitutes correct epression for y.,8, Integrates function 8, 0, correctly. 3 9 units 5 Correctly evaluates epression.,6,8,9, Notes: 0. awarded for correct circle equation and if incorrectly manipulated thereafter, 3 not awarded. 0. If awarded, may award 5 for an approimate answer (8 3 or more accurate: ). 0.3 Need to have a positive final value for volume to qualify for d essential. 0.5 May translate semi-circle one unit left without penalty, if done correctly. 0.6 Correct evaluation of any epression to a positive final answer earns Accept any version of the equation of the circle. 0.8 N.B. several wrong methods still lead to 9π. Take care to ensure that the method used is valid. 0.9 Evaluations of epressions involving logs are most likely to go wrong (especially when missing absolute value signs) at some point and will be penalised. eg log = log loses not available if working eased significantly, eg when integrating only a linear function to a quadratic function Halving value at any point or at end leading to units loses 5. Page

15 Question Epected Answer/s Ma. (a) Correct shape and behaviour approaching y asymptote. Asymptote parallel to original.,3 5 c 3 5 marked on y-ais at asymptote and c on -ais. Symmetry. O c 5. (b) y = 3 5 Correct statement of equation of asymptote. 5. (c) From the diagram, the two curves/graphs intersect y f ( ) intersects y y f ( ) intersects y f f So = f (f ()) 6 Valid reason from observation of graph or algebraic. 6,7 Note: a. Possible to award without y = on diagram. a. Where asymptotes meet or are clearly not parallel, do not award. a.3 Statement of equation of asymptote not necessary for. a. Accept asymptote of inverse passing through ( 5,0) for 3 only if labelled as y = + 5. b.5 Award mark for any form of straight line equation. c.6 Two lines intersect not sufficient for 6 unless lines referred to are specified. c.7 Where sketch indicates neither curve crossing y = or not crossing each other, do not award 6, even with an assertion of the form = f(f()) has no solutions as this contradicts the statement in the question. Page 5

16 Question Epected Answer/s Ma. = tanθ d sec d d = sec θdθ = and = 0 become and θ = sec d 3 tan sec sec d Correctly differentiating substitution epression. Processing substitution to obtain both limits for θ., Correctly replacing all terms. Replaces + tan θ with sec θ. cos d 0 0 [ sin] 5 6 Simplifies to integrable form. Integrates and evaluates correctly.,3, = tanθ so tan d d so d d = and = 0 become and θ = 0. ( ) d d = 0 3 = 0 0 ( ) d tan 3 Correctly differentiating rearranged epression. Processing substitution to obtain both limits for θ., Correctly replacing all terms and epressing in terms of only. Notes:. Full credit available to candidates progressing clearly with limits and substituting to arrive at function of and using limits then.. Approimations are not acceptable since the eact value was specified..3 Alternative eact values such as should be awarded 6.. Where candidate keeps limits as = 0 & =, may earn later by replacing sin with without right-angled triangle justification, without penalty., with or Page 6

17 Question Epected Answer/s Ma 3. df 0 e cos sin e sin cos d e sin df For S.P.s, e 0, so sin 0 d 0 Evidence of differentiation and one term correct. A second correct term. 3 Third term correct. Then Hence sin andsosin = 3, Leading to values of F = 9, Sets to 0 and solves to obtain value for sin. Correctly obtains two solutions for. Then obtains the two related values for F. 0 s0so0 7 Identifies correct upper and lower limits for. 0, F 6;, F 5 Greatest efficiency 5 km/litre at 00 km/h. Least efficiency 5 km/litre at 0 km/h Correctly obtains values for F at both endpoints. 6 Correctly worded statement for greatest or least efficiency, with units. 3,0 Correctly worded statement for other etreme.,0 Page 7

18 Notes for question 3: 3. Alternatively, award for evidence of inputting both 0 and π into F to establish endpoints. 3. For only one value for and the correctly obtained value for F, award 6 but not To qualify for 9, need to compare at least 3 positive values. 3. To qualify for 0, need to compare at least positive values. 3.5 Award 5, where answers given in degrees. 3.6 Where candidate has used degrees, negative answers for F are likely. In which case, no follow through mark available for Where a candidate has only solution for, they will not be awarded 5 and 0 is not available. Ma 8 3 out of 0. Where only solution is, loses 5 and 9 not available. 3.8 Where a candidate has two solutions for, but in degrees, they will not be awarded 6. Also, 9 and 0 are not available. Maimum mark: 7 out of Where a candidate has only one solution for, but in degrees, they will not be awarded 5 or 6. Also, 9 and 0 are not available. Maimum mark: 6 out of Appearance of units for both efficiency and speed in either (or both) of greatest and least statements necessary to achieve 9. Appropriate speed to be stated for award of 9 and 0. Page 8

19 Question Epected Answer/s Ma. (a) 3 r r r... r Correct statement of sum. 3r 3r 3r 3 r Valid rearrangement of epression., 5 3r 3r... 3r 3 Makes correct substitution for r in series at.,3,5 3r 9r.. 3r, r 3 Correct statement of range. 5 Page 9

20 Question Epected Answer/s Ma. (b) A B 3r 5r 3r r A r B 3r ; B A 3 3 3r 5r 3r r 3 3r r 6 5 Correct form of partial fractions. 7 6 Either coefficient correct. 7 7 Second coefficient correct. 7 3r 9r 5r 9r r r... 3r and r, so r Recognising form and manipulating correctly. Simplifying to obtain first three terms. Correct statement of range of convergence. 6 f ( ) (3r 5r ) f (0) 5 f '( ) (3r 5r ) (6r 5) f '(0) f ''( ) 6(3r 5r ) (3r 5r ) (6r 5) 9 f ''(0) 5r 9r f( ) Either first two lines correct both derivatives correct evaluation of all three epressions. Completes to obtain series. Note:. First three terms required. Ignore any subsequent terms.. Other arrangements possible for full credit. However, some arrangements will have different ranges of convergence and/or need to eclude r = 0 to avoid division by 0..3 Evaluation of coefficients must not be significantly eased, so requires evaluation of two terms in third line. 0 not available where Maclaurin used to obtain series in (b).. May award when series obtained using Maclaurin s or binomial theorems, but not, 3 or..5 Range must be consistent with fractions in 8. 0 not available if final range given is r..6 Where denominator factorised incorrectly, follow-through marks for 6 and 7 still available. Page 0

21 Question Epected Answer/s Ma 5. (a) e cos d e cos e cos d d d e cos e sin d e cos e sin e cos d Evidence of application of integration by parts. 3 Completes first application. Completes nd application. 3 e cos d e sin e cos + c e cos d e sin cos + c e cos d e sin e cos d d e sin e sin d e sin ( e cos e cos d) e sin e cos e cos d Recognises form of remaining integral and completes manipulation correctly Evidence of application of integration by parts. 3 Completes first application. Completes nd application. 3 e cos d e sin e cos + c e cos d e sin cos + c Recognises form of remaining integral and completes manipulation correctly. Page

22 Question Epected Answer/s Ma 5. (b) I e cos n e ( nsin n ) d n e cos n e nsin n d 5 Completes first application on I n. e cos n e nsin n n e cos n d n sin cos n I ne n e n +c e In nsin n cos n +c n I n e sin n e sin n d n n Completes second application on I n. 3 Correctly identifies and starts to manipulate I n terms. Simplifies epression Completes first application on I n. e sin n e cos n e cos n d n n n I sin cos n e n e n +c n n n In e sin n e cos n +c n n n 6 7 Completes second application on I n. 3 Correctly identifies and starts to manipulate I n terms. n sin cos In e n e n +c n n n e In nsin n cos n n +c 8 Simplifies epression. Page

23 Question Epected Answer/s Ma 5. (c) e I8 8 8sin 8 cos8 0 9 Correct substitution of value of n into epression obtained in part (b) or equivalent epression. 5,6 65 e 0 Processes to statement of answer.,6 Notes: 5. Repeating errors such as sin d cos should not be penalised twice. Award follow-through marks where appropriate. 5. Accept approimations to 3s.f. or better, ie (or better: ) 5.3 Not necessarily including simplification of + and signs. 5. It may be that some candidates try both the given methods. In this case, mark positively, awarding marks in either portion, wherever the criteria for the marks are met. 5.5 Where epression from part (a) has been altered by inserting n in one or more places, 9 is available for correct evaluation (subject to not being significantly eased see note 5.6). 5.6 Where epression being evaluated is eased, then a correct evaluation will earn 0, but not 9. Consider anything containing all of: e, sin8 and cos8 as being of equivalent difficulty. Page 3

24 Question Epected Answer/s Ma 6. (a) r cos isin cos isin 3 Polar form. cos k isin k 3, 3 3 z cos isin, cos isin, 3 3 cos isin, cos isin 3 3π z cos isin, cos isin 3 Demonstrates understanding of method for th roots. Obtains all four correct values.,3 6. (b) z i, Any two solutions. 5 Remaining two. 6. (c) i 6 Diagram showing all solutions to (a) and (b). i Page

25 Question Epected Answer/s Ma 6. (d) z 8 z z Then the solutions to z 0 and the solutions to z 8 = 0. z 0are also 7 Factorises into two correct factors. 8 Statement. 5 (e) Observe that z 6 z z z z 8 z z z z 9 Factorises to obtain either form. Si solutions are those above ecept z = ± 0 Statement of solutions eplicitly or as here. Notes: 6. Accept θ in degrees or radians. Polar form only. 6. Accept angles epressed in radians: or 0 and degrees or Accept in Cartesian form: i. Do not accept decimal approimations. 6. Argand diagram must include all candidate s solutions (at least 5 correct) as well as either one ais correctly scaled or all solutions labelled. Argand diagram need not include a circle or regular polygon. 6.5 Accept verification that all 8 answers from (a) and (b) inde 8 =. For between four and seven solutions, all correctly verified: award 7, but not 8. [END OF MARKING INSTRUCTIONS] Page 5

2013 Mathematics. Advanced Higher. Finalised Marking Instructions

2013 Mathematics. Advanced Higher. Finalised Marking Instructions 0 Mathematics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial