2013 Mathematics. Advanced Higher. Finalised Marking Instructions

Transcription

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. 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 i.e, 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. 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 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. Write down the binomial epansion of and simplify your answer. 0 0 C C C 0 C C Correct binomial coefficients. Correct powers of and Simplifies indices. Completes simplification of coefficients.. Accept negative indices Award ncr or ( ) form.. Including signs. + - or - - : do not award. Epanding wrong epression: ( ), only are available..5 Epanding ( ), only are available.. Differentiate f() = e cos sin. f () = e cos ( sin).sin + e cos.sincos Uses product rule. First term correct. Second term correct. Simplified alternatives. = e cos sin + e cos.sincos = e cos (sin sin ) = e cos sin(cos sin ). Evidence of method: Statement of the rule and evidence of progress in applying it. OR Application showing the sum of two terms, both involving differentiation.. Signs switched: available for e cos sin - e cos.sincos or equivalent. Page

4 . p Matrices A and B are defined by A 6 and B (a) Find A. (b) Find the value of p for which A is singular. (c) Find the values of p and if B = A. a A p p 6 p 8 p p p Correct answer., 6 p 5p = 0 p Improved alternative. b A is singular when deta = 0 Property stated or implied. (6 p)( p) + 50p = 0 6 p + p + 50p = 0 p + 6p + 6 = 0 (p + ) = 0 p = Correct value of p. 5, OR A is singular when A is singular, [i.e. when deta = 0] Eplicitly states property. c A ' p 6 p + p p [not essential, but preferred] = 0 = Correct value of p A transpose (A T ) correct. Does not have to be eplicitly stated. 6 p 6, p 5 Values of p and correct.,. For (a) and (c), statement of answers only: award full marks. For (b), p = only, award only ( out of ). Misinterpretation of A T as inverse leading to p = 0 and = OR to p = and = OR to p = and = OR any other set of inconsistent equations: do not award or 5 i.e. 0 out of.. Accept unsimplified answers.. Usually implied by net line..5 For any equation based on answer to (a), correctly obtaining all possible solutions, including comple, both available. No solutions, not possible etc. not available, even if true. Page

5 . The velocity, υ, of a particle P at time t is given by υ = e t + e t. (a) Find the acceleration of P at time t. (b) Find the distance covered by P between t = 0 and t = ln. a dv a dt Evidence of knowing to differentiate. = e t + e t Correct completion. b ln t t s v dt e e dt ln 0 0 e t t e ln ln e e ln ln ln e e Correctly set up integral. Integrating correctly. 8 or or equivalent 5 Evaluates correctly.,. Accept rounded answers without working between and 5 to s.f. or better.. Accept 6, but not 6, or.. Eceptionally, accept statement of formula as sufficient evidence for.. Evaluation of any incorrect function may be awarded 5 if evaluation of at least one e lnq involved.. Candidates may integrate v to obtain an epression for s and evaluate from there. Do not penalise the omission of + c. Page 5

6 5. Use the Euclidean algorithm to obtain the greatest common divisor of 0 and 8, epressing it in the form 0a + 8b, where a and b are integers. 0 = = = = 7 so gcd is 7 7 = 7 9 = 7 (8 7) = = 9(0 8) 8 = (a = 9, b = ) Starting correctly. Obtains GCD. Accept (8, 0) = 7 Equates GCD from and evidence of correct back substitution., Correct form of final answer = 7 9 not sufficient for. 5. a, b do not need to be stated eplicitly. 5. Accept a properly laid out grid approach. 5. Where candidate incorrectly starts 0= and correctly completes, available. Leads to a = - 9 and b =7 (or to a = 9 and b = - 7). 5.5 For stating a = 9, b = and nothing else, the question has not been answered, so 0/. 5.6 Where 7 = 9 0 8, or arithmetically correct equivalent following from a wrong GCD with an inappropriate method or without supporting working, is available, but is not. Page 6

7 6. Question Epected Answer/s Ma sec Integrate with respectto. tan f Evidence knows correct f form of integral.... Coefficient correct..ln. Use of ln or log e... tan Completes, including use of ln tan c OR u = + tan OR u = tan Correct substitution. du sec d Differentiates accurately. du sec OR d Correct substitution of du and f(u) into integral. ln u c OR ln u c = ln tan c Integrates correctly and substitutes back.,, 6. Do not penalise omission of + c. 6. symbols necessary for 6. Accept log for full marks. 6. Accept answer without working for full marks. 6.5 Award marks out of. Page 7

8 7. Given that z i, write down z and epress z in polar form. z z i cos π i sin Correct statement of conjugate. One of r, θ correct. Second correct and accurate substitution. z cos isin cos isin Processes to answer.,5 OR z i i i z i r cos θ i sin π r, θ, z π π cos i sin Obtains in Cartesian form. One of r, θ correct. Second correct and accurate substitution. 7. Accept cis π for &. 7. Accept angles epressed in degrees, i.e. 60, Where a candidate has applied de Moivre s theorem to k(cos θ isin θ), do not penalise. 7. Correct polar form only. Answer in form k(cos θ isin θ) loses unless correct form appears also. 7.5 Accept answers from π to π as being in polar form. For answers outside this range, do not award. 7.6 Since it is possible to use the conjugate of to find award for cis or cis and for cis or cis, but only for cis or cis. Page 8

9 8. Use integration by parts to obtain cos d. 5 Evidence of integration by. sin sin d parts. Correct choice of u, v. Accuracy of both epressions. sin cos cos d Correct second application. 9 9 sin cos sin c 5 Final integration and 9 7 simplification. 8. Selection of parts wrong way round, leading to - and no further, gains &. 8. & 5 available for follow through marks with a valid epression of equivalent difficulty. 8. Do not penalise omission of + c in this case. 8. For a follow through mark to be awarded here, fractions and three (or more) trig functions are required. Page 9

10 9. Prove by induction that, for all positive integers n, n r r r r n n For n = L.H.S n r r r r. R.H.S n n = + + = 8 = = 8 true for n 6 Evaluation of both sides independently to 8. 8 Assume true for n = k, k r r r r k k Consider n = k +, k r r k r k r r r r r k k k k k k k k k k k k k k k k k k k k k k k k 8k k k k 6k k 8 Inductive hypothesis (must include Assume true or equivalent phrase)., Addition of (k + )th term. 5 5 Use of inductive hypothesis and first step in factorisation process.,6 Processing and simplifying to arrive at second factor. k k k k 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. 6 Statement of result in terms of (k + ) and valid statement of conclusion.,7 9. ers to take etra care to ensure that no steps are omitted in the algebra required for, 5 and Alternative approach manipulating final form to Aim: [or target] k + 7k + 8k + 0k + 8 full marks are available, provided a valid conclusion is stated and working towards achieving aim is clear. 9. Acceptable phrases include: If true for... ; Suppose true for ; Assume true for ; Assume for n = k. However, not acceptable would include: Consider n=k ; Assume n = k ; and True for n=k. 9. Correct statement of RHS in following lines where k(k + ) replaces, award. 9.5 (k + )th term may appear later in working, but still achieves award of. 9.6 Aim: [or target] k + 7k + 8k + 0k + 8 award. 9.7 Acceptable form for 6 : If true for n = k, then true for n = k+, but since true for n =, then true for all positive integers, n or equivalent. Final line may be omitted if final line of algebra (k + )(k + ) appears as aim/target. 9.8 RHS = 8, LHS = 8 or true for n = are insufficient, on their own, for. Page 0

11 0. Describe the loci in the comple plane given by: (a) z i (b) z z 5 a Circle... Observation that locus will be a circle. 8...centre (0, ) [or i], radius Identification of centre OR z + i = + iy + i = + i(y + ) (in either form) and radius. 8 y i Correct epression for modulus in Cartesian form. 5 y Circle centre (0, ), radius Statement that locus is a circle, centre (in either form) and radius. OR - o Sketch of a circle Identification of centre and radius.,6 0a. (0, - i) not acceptable. 0a. For diagrammatic approach, radius must be stated or clearly just touch -ais, with centre identified as (0, - ). 0a.5 also acceptable for. 0a.6 Where there is no point given and no point marked on the y-ais, do not award. 0a.7 Where point on y-ais is identified as i, may be awarded. 0a.8 Correct statement of centre and radius alone not sufficient for. Must eplicitly state locus a circle or sketch. Page

12 0. (continued) b Set of points equidistant from (, 0) and ( 5, 0) Straight line Observation that equidistant from specified points. Identifies form of locus. = 5 Statement of equation. OR z = z 5 ( ) y = ( 5) y = = = Collects real and imaginary parts and equates moduli.,8 Accurately processes to reach equation. which is a straight line 5 Eplicitly states form of locus. OR = o 5 Sketch of aes with any straight line drawn. Vertical line to left of y-ais. Eplicitly states equation OR identifies the point (-, 0) as being on the line. 0b. Statement line = - only, award full () marks. Statement = - only loses and 5, i.e. out of. 0b. and no further, award, Equates moduli, only. 0b.8 For this statement with squared omitted from both sides, may be awarded. Page

13 . A curve has equation 6 Find the values of at the point ( -, ). Differentiates and first product. Differentiates y + = 0 correctly. Evaluates. OR Evaluates. Differentiating (Δ): ( ) 5 Differentiates first three terms of (Δ) correctly, including a product. Differentiates final product of (Δ) correctly. + ( - ) 6 Evaluates. OR Differentiating ( ): ( ) ( ) 5 Evidence of valid application of quotient (or product) rule. Differentiates correctly. ( ) ( ) 6 Evaluates.,. Where a carried error is incorporated, there must be at least one instance of evaluating to earn 6. Where the candidate has erroneously prefied the question with = and subsequently ignores = 0, leading to and = award 5 out of 6, losing.. Where the candidate has simplified subsequent working with an error in first three marks, any three terms including a product could earn. A second product would make 5 also available.. Evaluation includes any simplification of differentiated epression in both parts. Therefore and 6 awarded for correct evaluation of unsimplified derivative in each case. Page

14 . Let n be a natural number. For each of the following statements, decide whether it is true or false. If true, give a proof, if false, give a countereample. A If n is a multiple of 9 then so is n. B If n is a multiple of 9 then so is n. A Suppose n = 9m for some natural number [positive integer], m. Generalisation, using different letter., 6 Then n = 8m = 9(9m ) Correct multiplication and 9 etracted as a factor. Hence n is a multiple of 9, so A is true. Conclusion of proof and state A true. B False. Accept any valid countereample: n =, 6,, 5, etc Valid countereample and conclusion. 5. Final mark ( ) not available unless evidence of a proof attempted.. No credit given for numerical eamples without generalisation.. Do not penalise failure to specify that m is a natural number.. Any number of numerical eamples on their own secures no marks..5 Countereamples must have n as a natural number (positive integer)..6 Starting i.e. using the same letter on both sides, leads ineorably to 0/..7 A word-based proof is very unlikely to be awarded full marks for A as the criteria for will be very difficult to achieve with words alone. A clear, logical answer of this type can access and. Page

15 . Part of the straight line graph of a function f() is shown. y (0, c) O (, 0) (a) Sketch the graph of f (), showing points of intersection with the aes. (b) State the value of k for which f() + k is an odd function. (c) Find the value of h for which h even function. f is an a y (0, ) Straight line with negative gradient crossing the positive sections of the - and y-aes. O (c, 0) Both intersections correctly annotated. b y f c is odd. k = c Correctly stated. c y Sketch of with point of reflection marked. O (, 0) y f is even h = 5 Eplicit statement of answer.. Answer h = only, no other working or diagram, award full marks [ out of ].. Where a candidate has clearly used their diagram from part (a) as the basis for (b) and (c), leading to k = - and h = c (with working/further diagram) award 5 and not ( out of the three marks for (a) and (b)). Statement of above answers only, zero out of.. An accurate diagram of y = f( + ), on its own, gains no marks. Page 5

16 . Solve the differential equation d y dy 6 9 y e d d dy when = 0 d, given that y = and m 6m + 9 = 0 (m ) = 0 m = C.F. y = Ae + Be Correct auiliary equation (or equivalent). Correct solution of auiliary equation and statement of complimentary function. P.I. Try y = C e dy Ce C d e Correct form of particular integral.,7 Correct first derivative of P.I., d y Ce 6Ce 6Ce 9C e d Ce 6Ce 6Ce 9C e - 6 Ce C e 9C e e Ce e C Correct differentiation of first derivative. For correctly substituting epressions for both derivatives. For correctly solving to obtain C. 5 G.S. y = Ae + Be + e 8 Correct collation of above answers to obtain full General Solution. 6 dy Ae Be Be e 6 d When = 0, y = A = dy d B B e 9 0 Derivative of G.S. Use of i.c.s to find first constant correctly. Second constant. P.S. y = e e + e States solution. 6 Page 6

17 Question. Accept (C + D + E)e or equivalent for.. Accept correctly differentiated version of (C + D + E)e or equivalent for.. Must be of equivalent difficulty if wrong P.I. used to obtain. i.e. contains at least one use of prod/quot rule.. Must be of equivalent difficulty if wrong st derivative used to obtain 5 i.e. contains at least one use of prod/quot rule..5 And other constants, e.g. D and E = 0, where using alternative P.I.s as note...6 Not needed if subsequent lines incorporate information, especially 9 or..7 Incorrectly using for P.I. leading to A = and B = -, still available. i.e. ma 7 (out of ). To be awarded 6, a correct differentiation of the first derivative is required as well as correct substitution..8 Incorrectly using for P.I. leading to A = and B = -, 9 0 still available. i.e. ma 5 (out of )..9 Incorrectly using and using for P.I., 9 still available. i.e. ma (out of )..0 Incorrectly using and using for P.I., 9 still available. i.e. ma (out of ).. Do not penalise omission of = 0 in first two lines. Page 7

18 5. (a) Find an equation of the plane π, through the points A(0,, ), B(, 0, ) and C(0, 0, 5). (b) π is the plane through A with normal in the direction j + k. Find an equation of the plane π. (c) Determine the acute angle between planes π and π. a AB 0 0 AC OR - BC 0 Any two correct vectors. i AB AC j k 0 or equivalent Evidence of appropriate method. 0 = i j + k Obtains vector product (any form). y + z = 0 + : y + z = 5 OR r = ( ) + ( )+ ( ) or equivalent Obtains constant and states equation of plane. b ( ) ( ) + = : y + z = 5 6 Evidence of appropriate method. Processes to obtain equation of second plane. c Normal vectors: 0 n andn, n 9, n 7 Obtains two correct lengths. cos (angle between normals) = n. n 0 8 Evidence knows how to n n use formula. Angle = 5º acute angle between planes is 5º ( ). 9 Processes to statement of answer. 5 P.T.O. for alternative method for question 5(c) and marking notes for all parts of question 5. Page 8

19 5. c OR = i j + k, so i j + k = and j + k = 7 States vector and obtains moduli. = n. n.cos θ =.cos θ 8 Evidence knows how to use formula. cos θ = so θ = (or 5 ) 9 Processes to statement of answer. 5. i.e. non-parallel. 5. Although unconventional, accept vectors written horizontally. 5. Do not award where co-ordinates of A/B/C used. 5. Award mark for obtaining value of constant = (or follow-through). 5.5 Where candidate uses 90 - θ, only 7 and 8 available. So θ = 5 so acute angle = 90-5 =5 scores ma of /. Page 9

20 6. In an environment without enough resources to support a population greater than 000, the population P(t) at time t is governed by Verhurst s law Show that dp P000 P. dt P ln 000 t C 000 P for some constant C. Hence show that 000K P( t) for some constant K. -000t K e Given that P(0) = 00, determine at what time t, P(t) = 900. dp P000 P dt dp So P 000 P dt Separates variables. 5 P A B Appropriate form of partial fractions. 000 P P 000 P A, B dp dt 000 P 000 P Obtains correct values of both A and B. P 000t c lnp ln 000 Integrates correctly, including +c. 6 P ln 000 P 000t c P 000 P Ke 000t where K c e 5 Accurately converts to eponential form. Page 0

21 6. (continued) P = 000Ke 000t PKe 000t, P + PKe 000t =000Ke 000t, 6 Multiplies out fractions and collects P terms. 000t 000Ke P 000t Ke ( ) 7 Factorises and divides to obtain required form. Since P(0) = 00, 000K 00 K K or Equates and process to obtain value of K. Require e -000t 9 Inserts value of K and equates e -000t = 50 e 000t = 6 000t = ln6 t ln6 000 [or (sf)] 0 Solves to obtain value for t. 6. Eplanation of new constant not required. Do not penalise bad form when changing constants. E.g., for RHS = e 000t + c, award Both steps required as final result given in question. 6. Accept statement of value of K (consistent with previous working). 6. Accept approimation (sf or better), ie or. 6.5 Do not penalise omission of integration symbols for. 6.6 Candidates putting lnp + lose. Page

22 7. Write down the sums to infinity of the geometric series and + + Valid for <. Assuming that it is permitted to integrate an infinite series term by term, show that, for <, ln Show how this series can be used to evaluate ln. Hence determine the value of ln correct to decimal places Correct statement of sum. + + Correct statement of sum. Integrating the first of these gives: 5... ln 5 c Correct integration of both sides. Putting = 0 gives c = 0. Correct evaluation of c. Similarly,... ln Adding together gives: 5... ln 5 ln as required. OR = ln Correct integration of both sides. Evidence of appropriate method. Appropriate intermediate step. Adds series. Integrates LHS + c 5 Integrates ln( + ) 6 Integrates ln( ) Page

23 7. (continued) Putting = 0 gives c = 0. 7 Correct evaluation of c., ln as required. OR = ( ) or equivalent ( ) ( ) as required. Now choose such that, ie + =, so So ln = 0 69 to d.p Evidence of appropriate use of Maclaurin. 5,7 All five derivatives correct OR first two derivatives and first three evaluations correct. 5 All si evaluations correct OR final three derivatives correct and final three evaluations correct. 5 Correctly substitutes obtained values into Maclaurin. Simplification en route to required result. 8 States appropriate equation. Correctly solves equation. Obtains accurate approimation.,6 7. Do not penalise omission of +c for or 5 (or 6 in alternative method). for transferable: for the eplicit evaluation of constant of integration on either (or both) of the integrals, award (or 7 in alternative method). 7. To illustrate that series used (and not calculator), all terms must appear for this mark to be awarded. For approimations, four or more decimal places in all four terms are required. 7. Any assumption that the two constants of integration are equal and can therefore be cancelled out loses. 7. For an unsupported statement that award 8 and For attempts using Maclaurin s Theorem, only terms up to and including are required for,,and Attempts utilising Maclaurin s Theorem do not need to continue to the term in, as the wording of the question implies continuation of the established pattern. Hence errors in calculating the terms in and may be considered working subsequent to a correct answer. 7.7 Differentiation of either ln( ) or both of ln( + ) and ln( ). Also accept differentiation of either ln epression and subsequent substitution of (- ) for to obtain the other (for,, and 5 ), 6 for combining and 7 for simplifying. Substitution of ( ) for is not accepted as it leads to an epansion which cannot be approimated in the same way. 7.8 Simplification to penultimate line or equivalent required for award of 7. [END OF MARKING INSTRUCTIONS] Page