2015 Mathematics. Advanced Higher. Finalised Marking Instructions

Size: px

Start display at page:

Download "2015 Mathematics. Advanced Higher. Finalised Marking Instructions"

Madlyn Nichols
5 years ago
Views:

1 015 Mathematics Advanced Higher Finalised ing Instructions Scottish Qualifications Authority 015 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 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 Assessment team may be able to direct you to the secondary sources. These ing Instructions have been prepared by Examination Teams for use by SQA Appointed ers when marking External 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. If a specific candidate response does not seem to be covered by either the principles or detailed ing Instructions, and you are uncertain how to assess it, you must seek guidance from your Principal Assessor. 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. 1 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. 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 expressions. 4 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 explicitly 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 explicitly 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 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 Expected Answer/s Max x 5 x 5 x C0 C1 C x x x 5 x 5 C C4 C5 x x x 4 1 correct unsimplified expansion fully simplified powers of x powers of and binomial coefficients x 10x 40x 80x x x OR powers of correct. 4 Completes and simplifies correctly. 1.1 Accept negative powers of x. 1. Coefficients must be fully processed to simplified fractions and whole numbers. Page

4 a dy 5x x5x 1 x x 5x x 10 1 For using quotient rule and correct denominator. correct differentiation for both parts of numerator simplified form. b x x f ' x e sin x e...sin x cosx 4 evidence of using product rule x e sin x sin x cosx x or e sin x sin 6x 5 6 first term second term.1 Alternative Method Product rule 5 1 y x x 1 dy 5x 1 x.x 5 x 1 1 for evidence of using product rule and one term correct. for second term x 5x x 10. Where a candidate has a wrong, but factorisable expression in the numerator, factorisation is not required for award of this mark.. Where terms are the wrong way round, lose 1 d 5x 1 x... 15x x 10:award..4 0 simplified fraction Page 4

5 0 s0 0 a 19d a19d 5 1 for correct substitution into formula 1. for correct substitution into formula 1. u 7 a 0d a 0d s10 a 9 d 60 a40d 74 1d 4 d a d 1 4 a 1 5 s 10 OR a0d 7 s a 7 a d s a0d 7 4 for correct substitution into formula 1. a d 5 s 10.1 Accept correct equations for 1 and without explicit statement of general formulae. However, simply stating values for a and d, is not sufficient, so do not award 1 or, ie working required.. Candidates can also obtain two simultaneous equations using S 0 and S 1 formulae. One mark each, then follow the above for final marks. Page 5

6 4 4 4 x y x y dy dy 4x 4y dy dy dy 1 1 y x1 eqn.tangent: y= x+ 5 OR or or y x y + x 5 0 dy x x 4y 6 6 4y at A 1, dy x terms and constant. y terms gradient. 4 equation Published form would have 4 at expanded form, not as marked. 4. Rearrangement and explicit statement of dy not required for full marks. d 4. Where candidate asserts that 14 14, 1 not given, but, and 4 all possible, leading to 6y = x + 51 (or equivalent) for 4 5 Singular when det A = 0 p 1 1 p p p p evidence of any valid method for obtaining deta and setting = 0 1. expansion or equivalent method by first row p p 6 0 p p 0 p or p = 4 for polynomial solutions (both) 5.1 = 0 needs to appear for full credit. Page 6

7 6 ln y ln x ln y x ln 1 dy x ln y dy x ln. x 1 evidence of taking logs. for differentiating correctly. for dy in terms of x and y. OR u x y dy u ln du u 1 correct substitution into original equation. for differentiating correctly. dy x ln.x for dy in terms of x and y. OR ln y x ln 1 evidence of taking logs. e ln y e x ln y e x ln for rearranging correctly. dy xln. e x ln for differentiating to obtain in terms of x and y. 6.1 Accept dy x x ln. Page 7

8 starting correctly reach GCD p10 q 4 equates GCD from and evidence of substitution 4 obtains values of p and q. 7.1 GCD does not need to be explicitly stated. 7. p and q must be explicitly stated 8 1 dt dy t 1 1 cosec t dt dy dy dt. dt cosec t t 1 1 obtain dt obtain dy dt 1 obtain dy in terms of t in any correct form. t1.cosec t 8.1 cosec t or equivalent. Page 8

9 9 n n n! n! n!! n!! n n nn nn n n n1!! n!! n n 1 n nn 1n 1 1! 1!!! 4 1 demonstrates understanding of factorial form algebraically 1. using property n! = n (n 1)! 1 correctly expressing with common denominator OR as a single fraction 1,5. n n n 1 n 1n *! n n n n n 6 n 6n 6 n as required OR LHS = 10 1 = 9 RHS = 9. So, true for n =. Assume true for some n = k. ie k k k k k+1 k k k k 4 simplification to answer, line * (or equivalent) essential. 1 Base case and assumptive hypothesis stated. Using identity n n n 1 r r 1 r 1 in context. k k k k + k k 1 k k 1 k k k 1 1 Using property that n! = n(n 1)! 4 Completes proof, including accurate induction statement. Page 9

10 9 (cont) awarded wherever they appear, eg as part of an attempted induction proof. 9. Must include 10-1 to demonstrate application of understanding of how to process factorials. However, if this is satisfactorily demonstrated later, 1 may then be awarded. 9. Although successfully completed in only a few cases, proof by induction may be attempted and marks allocated as above. 9.4 Many attempts at induction are likely to include base case and assumptive hypothesis, but then candidates k k1 attempt to prove that k 1. Award max 4 since this approach does not use,6inductive hypothesis and therefore is not a proof by induction. 9.5 Where candidate starts at this line, all marks may be awarded for being correct so far. However, the lack of working is likely to mean that an incorrect expression here may lose more than one mark. Page 10

11 10 4x 4x 4x x e x e xe x 1 4x 1 4x x e xe 4 8 e 8 1 4x 1 4x 1 4x x e xe e 4 8 4x x e.4. e.e e e e or e knowing and using integration by parts appropriate choice for u and v and correct application Second application,4,6. 4 final integral and substitute limits. 5 exact value 1,5, only available where working required to obtain value has not been eased. eg must have at least three terms and non-zero value resulting from x = Where nd application undoes first and no further progress: max x 4x goes to Where candidate asserts that e = 4 e 80e 18 8,48 or e e 5, lose (wrong) and 5 (eased). Award 4 only if appropriate 4x goes to 8 substitution to exact values appears For wrong signs in either/both by parts operations, award: uv+,uv+ max 5, lose 1 (wrong) and 5 (eased). uv+,uv max 5, lose 1 (wrong) and 5 (eased). 8 e 1 4 uv,uv+, leading to max 5 lose only (wrong) Lose final mark when appropriated to 8 84 when no exact version Where final integration is subtracted, again leading to e 1 lose or ; e lose 5 5 or lose 5 5 or 14 5 lose Page 11

12 M M M M1M for M 1 for M. for M. 1 Reflection in the line y = x 4 correct interpretation MM: 1 Incorrect order gives do not award Incorrect order MM 1 leading to reflection in the line y = x, award Accept, in isolation, correct description of single transformation, eg reflection in line through (0, 0) at 45 to positive direction of the x-axis. Simply stating that x- and y-coordinates swapped not sufficient. 1 Let numbers be n 1, n + 1, n ℵ 1 correct form for any two consecutive odd numbers 1,. n1 n1 4n 4n 1 4n 4n 1 correct expressions squared out. 8n which is divisible by 8 multiple of 8 and communication. 1.1 This line may be omitted and awarded where correct form appears in next line. 1. In most cases, use of two different letters, ie two odd numbers not necessarily consecutive, leads to at most only and being awarded. Page 1

13 1 a z x iy x ixy y x iy 4 x ixy y x y 4 ixy y y 4 y ixy 4 0 y 0, y, ixy 0, x 0 z i 1 b x ixy y i x y 4i x y ixy x y 4 i x y xy xy x y 4, x y 0 4 x y 4 x y, y x 1 writing in form x + iy and either LHS or RHS correct. equating real parts and solving to obtain y equating imaginary parts and x = correct expansion in x and y. 5 equating real and imaginary parts 4. 6 two equations 1. leading to y = x which gives z = 1 i and z = 1 + i. 7 solutions. 1.1 Or obtain one equation and substitute into other. 1. Alternatively for two correct equations equating both real and imaginary parts, without further progress, award. 1. Accept use of a and b (ie a + ib) or other letters, without penalty if used consistently. 1.4 Classify making same mistake in part (b) as in part (a) as a repeated error, so only penalised if eased. Page 1

14 14 g x f x f x g x f x f x f x f x g x since g( x) g( x) function is even. h x f x f x f x f x f x f x hx h x f x f x since h( x) h( x) function is odd. g x h x f x by adding initial equations 4 1 communicating knowledge of an even and an odd function. showing that g(x) is even. showing that h(x) is odd. 1 1 f x g x h x Since g even and h odd, f ( x) is the sum of an even and an odd functions. 4 correct expression and conclusion For writing f x f x is not essential. 14. Award 1 where statements appear at start of answer or as part of individual show thats. Geometric description acceptable for 1, but needs to be watertight, eg function will be odd if unchanged by o 180 / rotation about origin and function is even if unchanged by reflection in y-axis [or line x = 0]. 14. Accept f x 1 [ g x hx ] as bad form without penalty. Page 14

15 15 a 1 u1 i j k direction vector 1 4 u 4i 4 j k direction vector v1 4, v & for vector equations 1,,,6,8. 15 b If they intersect two equations for two parameters z z Since z z, the lines intersect at 1,, two parameter solutions for checking third component in both equations. 6 point of intersection 4,. Page 15

16 15 u 1 u to get normal 4 i j k i j k i j 1 1 or i j k 6i j 1k 6 Point of 6x y 1z. intersection 1 6 So equation of plane is 6x y 1 z 6 OR x y 4 z correct strategy to find normal. correct processing to obtain vector. substituting normal vector into an equation of a plane. May also use either of the given points. finding correct value for constant and correct equation. Note: 15.1 If written in parametric or symmetric form award not 1. Including their statement at the start of 15b. 15. If direction vector and fixed point interchanged in one or both, award 1, but not. 15. Do not penalise use of same parameter at this stage Using same parameter for both equations, leading to, 6, or (, 6, 0) max For L1: i j k, L: 4i 4 j k or equivalent, lose 1 but available (repeated error.) 15.6 Do not penalise vectors written without underlines Acceptable form, without penalty: 4. 1 Page 16

17 16 d y dy 10y e m m10 0 x m 1 i 10 1 correct auxillary equation. solves correctly 1. 1i x 1 x y e Acosx Bsin x OR y = ae Be i x appropriate complementary function. try y Ce d y x dy Ce x 4Ce x 4 5 particular integral for dy and d y 4Ce 4Ce 10Ce e x x x x C y Ae cosx Be sin x e 6 x x x A, A 6 6 dy 1 Ae cosx Ae sin x Be sin x Be cosx e x x x x x A B, B, B for finding C combine CF and PI for general solution. 8 value of A. 9 for differentiating correctly So particular solution is: y e cosx e sin x e x x x 10 value of B and statement of final answer 4. Page 17

18 16. (cont) Note: 16.1 For errors in the solution of the auxiliary equation leading to: A: Complex conjugates, lose 9, but remainder all available, so max 10. B: Two real roots, neither of which is, lose, but -8 7 all available, so max. 10 Lose C: Two real roots, one of which is, lose, but -8 and 10 8 all available, so max 10. Lose 9. 5 D: Ignores RHS completely, ie treats as homogeneous: max 10. Only available ( 8 eased, so not available). 16. Omitting PI from general solution, lose 7 and 8, but 9 and 10 8 both available, so max May award 7 at 9 point if clear that CF and PI have been incorporated to produce GS differentiated May award 10 if GS explicitly stated earlier and values of A and B are clearly identified. Page 18

19 x x x x x x x 6x x x x5 9 1 for knowing to divide and starting division correct division 1. A Bx C x x 1 6x x 5 x x 1 x x 1 x x 1 6x x 5 A x 1 Bx C x for correct form of PFs 5. 4 creating correct equation x 0 5 A C x 50 10A A 5 C 0 x 1 8 A B C 8 10 B B1 5 for any two values,4. 6 for third value x x x x xx 1 x 1 1 x x x k 51n 1n 1 7 for putting into integral and any one term correctly integrated. for any second term. 9 for third term and + k. 8 Note: 17.1 Where candidate has NOT carried out division see COWAs (Commonly Occuring Wrong Answers) below. 17. Do not penalise slightly ambiguous use of +C rather than introducing some new letter. 17. Do not penalise (legitimate) omission of absolute value symbols For incorrect answers, some evidence of provenance of values for A, B and C is required for the award of BOTH 5 and Check here that candidate has included +C in PFs, since omission will usually lead to the correct answer or similar, as C = 0. Omission of +C means COWA D. Page 19

20 17 COWAs 9 If don t A 1 x x A Bx C x x 1 xx 1 1, not available. available. 1 1 x x A x x Bx C x 50 10A A 5 x 0 1 A C C x 1 0 A B C 0 10 B 4 B WRONG x x 1 5 x x x 1 x x 1 x 5ln x x 1 x 1 5ln 1n 1 tan 1 x x x k 4 not available. 5 available 4. 6 available 5. 7 available. available. 9 available. Max B If do then forget to put + Lose 7 8. Max 9. Page 0

21 17 COWAs (cont) C 6x x5 x x x x A B x x 1 x x 1 6x x 5 A x 1 B x x 50 10A A 5 x 0 5 A B B 0 1, available. wrong form of PFs. available. 5 available not available. 6x x5 5 x x x 1 x5ln x k 7 available. available. 9 not available. Max C C without division, leading to: 5ln tan 1 x x C Only available Max 9 D 6 5 x x A Bx x x 1 x x x x A x B x x x0 5 A x 50 10A x 1(say) 8 A B B 1 D with division This will usually lead to the correct answer or similar, but no consideration of +C leading to C = 0. Therefore losing and 6, so Max 7 9 D D without division, leading to a variety of answers. Only available Max 9 Page 1

22 18 a Method 1 Method V = Ah (here of below) V = Ah dh dh dv dv d or Ah dt dv dt dt dt dv dh A*... k h A dh dt dh 1 * dv A 1. Adh k h k h A dt k dh k h h A dt A 1 1 Method Method 4 1 (Ah) in brackets and/or A following line needed for Method, since taking A out as a constant necessary to illustrate understanding of validity of step. One or both of *lines needed for Method 1. dv dv dh V or h dt dh dt A dv k h A dh dh dt dt dt A 1 1 Page

23 18 b dh 0cm / hr when h= 144 dt k 0 = 144 A k 1 A 40k A 40 4 dh Subs in 0 dt and h = 144. Award this mark if substitution appears in part (d). dh k h dt A 1 k 1 1 dh dt OR dh dt h A h 40 k h t c A 144 c c4 k h t4 A k h t1 A k h t1 A 4 separating variables. 5 integrating correctly. 1 h t evaluating constant of integration and completion Page

24 18 c 1 0 t t t 960 hours 960 number days = 40 days knowing to set correct expression to zero Processing to obtain number of days 4 18 d A 400 k 1 A 40 k 10 9 for finding k. 1 h dv 108 dt 10 obtaining h or h Rate to vegetation is 108 cm / hr 11 processing to answer with interpretation dv A dt 1195 cm / hr which comes from taking 4 t. Do not award Using h = 144 in part d leading to 77, do not award 10 or Do not penalise omission of integration symbols Where candidate has used 144 instead of 0 initially, 7 lost, but 8 available if resulting quadratic solved correctly to obtain both t = 0 and t = 190, discarding t = 0 answer and converting to 80 days. [END OF MARKING INSTRUCTIONS] Page 4

2011 Maths. Advanced Higher. Finalised Marking Instructions

2011 Maths. Advanced Higher. Finalised Marking Instructions 0 Maths Advanced Higher Finalised Marking Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis.