© 2013 Mathematics Higher Finalised Marking Instructions Scottish Qualifications Authority 2013 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial 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 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. General Comments These marking instructions are for use with the 2013 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: 1 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. 2 Award one mark for each . There are no half marks. 3 The mark awarded for each part of a question should be entered 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. Only the mark, as a whole number, should be written. 2 2/3 Marks in this column whole numbers only Do not record marks on scripts in this manner. 4 Where a candidate has not been awarded any marks for an attempt at a question, or part of a question, 0 should be written in the right hand margin against their answer. It should not be left blank. If absolutely no attempt at a question, or part of a question, has been made, ie a completely empty space, then NR should be written in the outer margin. 5 Every page of a candidate’s script should be checked for working. Unless blank, every page which is devoid of a marking symbol should have a tick placed in the bottom right hand margin. 6 Where the solution to part of a question is fragmented and continues later in the script, the marks should be recorded at the end of the solution. This should be indicated with a down arrow ( ), in the margin, at the earlier stages. 7 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. 8 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. Page 2 9 Marking Symbols No comments or words should be written on scripts. Please use the following symbols and those indicated on the welcome letter and from comment 6 on the previous page. A tick should be used where a piece of working is correct and gains a mark. Markers must check through the whole of a response, ticking the work only where a mark is awarded. At the point where an error occurs, the error should be underlined and a cross used to indicate where a mark has not been awarded. If no mark is lost the error should only be underlined, i.e. a cross is only used where a mark is not awarded. x A cross-tick should be used to indicate “correct” working where a mark is awarded as a result of follow through from an error. A double cross-tick should be used to indicate correct working which is irrelevant or insufficient to score any marks. This should also be used for working which has been eased. A tilde should be used to indicate a minor error which 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 missing, such as part of a solution or a crucial step in the working. These will help markers to maintain consistency in their marking and essential for the later stages of SQA procedures. The examples below illustrate the use of the marking symbols . Example 1 y x 6x 3 Example 2 2 A(4, 4,0), B(2, 2,6), C(2, 2,0) dy 3 x 2 12 dx x 3 x 2 12 0 x2 y 16 1 x 3 2 4 Example 3 5 6 AB b a 6 6 x 6 AC 6 0 (repeated error) 2 1 Example 4 3 sin x 5 cos x k sin x cos a cos x sin a 1 k cos a 3, k sin a 5 2 must be a factor. Since the remainder is 0, Page 3 10 In general, as a consequence of an error perceived to be trivial, casual or insignificant, e.g. 6 6 12, candidates lose the opportunity of gaining a mark. But note example 4 in comment 9 and the second example in comment 11. 11 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. x 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. 12 Cross 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 Illustrative Scheme: 5 x 2, x 4 6 Cross marked: 5 x 2, y 5 y 5, y 7 6 x 4, y 7 Markers should choose whichever method benefits the candidate, but not a combination of both. 13 In final answers, numerical values should be simplified as far as possible. 43 15 Examples: should be simplified to 45 or 1 41 should be simplified to 43 1 12 15 0 3 64 4 should be simplified to 50 must be simplified to 8 5 3 should be simplified to 4 15 The square root of perfect squares up to and including 100 must be known. 14 Regularly occurring responses (ROR) are shown in the marking instructions to help mark common and/or non-routine solutions. RORs may also be used as a guide in marking similar non-routine candidate responses. 15 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; Repeated error within a question, but not between questions or papers. Page 4 16 In any ‘Show that . . .’ question, where the candidate has to arrive at a formula, the last mark of that part is not available as a follow through from a previous error. 17 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. All working must be checked: the appearance of the correct answer does not necessarily indicate that the candidate has gained all the available marks. 18 In the exceptional circumstance where you are in doubt whether a mark should or should not be awarded, consult your Team Leader (TL). 19 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. 20 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 1 attempt 1 is worth 3 marks Strategy 1 attempt 2 is worth 4 marks From the attempts using strategy 1, the resultant mark would be 3. Strategy 2 attempt 1 is worth 1 mark Strategy 2 attempt 2 is worth 5 marks From the attempts using strategy 2, the resultant mark would be 1. In this case, award 3 marks. 21 It is of great importance that the utmost care should be exercised in totalling the marks. A tried and tested procedure is as follows: Step 1 Manually calculate the total from the candidate’s script. Step 2 Check this total using the grid issued with these marking instructions. Step 3 In SCORIS, enter the marks and obtain a total, which should now be compared to the manual total. This procedure enables markers to identify and rectify any errors in data entry before submitting each candidate’s marks. 22 The candidate’s script for Paper 2 should be placed inside the script for Paper 1, and the candidate’s total score (i.e. Paper 1 Section B + Paper 2) written in the space provided on the front cover of the script for Paper 1. 23 In cases of difficulty, covered neither in detail nor in principle in these instructions, markers should contact their TL in the first instance. A referral to the Principal Assessor (PA) should only be made in consultation with the TL. Further details of PA Referrals can be found in The General Marking Instructions. Page 5 Paper 1 Section A Summary Question 1 Answer A 2 B 3 B 4 A 5 D 6 C 7 B 8 C 9 A 10 D 11 B 12 C 13 A 14 B 15 C 16 C 17 C 18 D 19 B 20 D A 4 B 6 C 6 D 4 Page 6 Paper 1- Section B Question Generic Scheme 21 Max Mark Express 2x2 + 12x + 1 in the form a(x + b)2 + c. ss identify common factor 2 3 ss pd complete the square process for c 1 Illustrative Scheme 2 2 3 1 ss expands completed square form 2 ss equates coefficients 3 pd process for b and c and write in required form Notes: 1. Correct answer without working gains full credit. Regularly Occurring Responses: Candidate A 2(x2 + 6x + 9 – 9 + ) 2 2(x + 3)2 .... 2(x + 3)2 – 17 Method 2 1 2 3 3 2(x + 3)2 – 17 Candidate B 2x2 + 12x +1= (x + 6)2 – 36 + 1 1 2(x2 + 6x + ) Method 1 2(x + 6x… stated or implied by 1 = (x + 6)2 – 35 2 1 2 3 3 2(x + 3)2 – 8 Candidate C Candidate D 1 1 2 2 3 3 Candidate E 1 2 3 awarded as all working relates to completed square form 3 Candidate F ( ) 1 Page 7 3 is lost as no reference is made to completed square form Question 22 Generic Scheme Illustrative Scheme Max Mark A circle C1 has equation x2 + y2 + 2x + 4y – 27 = 0. a 1 2 ic pd 22 b Write down the centre and calculate the radius of C1. 1 state centre find radius (–1, –2) 32 2 Notes: 1. Do not penalise candidates who use -1 and -2 for g and f when calculating the radius. However, candidates who use -1 and 2 or 1 and -2 lose 2 2. √ need not be simplified. 2 The point P(3, 2) lies on the circle C1. Find the equation of the tangent at P. 3 ss find m radius 3 1 4 ic state m tangent 4 –1 5 ic state equation of tangent 5 y – 2 = –1 (x – 3) Notes: 3. 5 is only available as a result of using a perpendicular gradient. Regularly Occurring Responses: Candidate A =1 3 3 Candidate B =1 3 so 4 4 equation of tangent is 5 5 Page 8 Question 22 Generic Scheme Illustrative Scheme Max Mark A second circle C2 has centre (10, –1). c The radius of C2 is half of the radius of C1. x2 + y2 – 20x + 2y + 93 = 0. Show that the equation of C2 is 6 pd find radius 6 √ 7 ic state equation of circle 7 (x – 10)2 + (y + 1)2 = (√ ) 8 pd expand and complete 8 x2 - 20 + 100 + y2 + 2y + 1 = 8 and complete stated or implied by 7 3 Accept Centre 6 Centre 7 √ 8 √ √ √ √ √ √ radius of C2 Regularly Occurring Responses: Candidate C C2 centre is (10,-1) g = -10, f = 1 2g = -20, 2f = 2 Candidate D Candidate E , centre (10, -1) radius = √ , centre (10, -1) radius = √ √ 6 7 8 Candidate F , centre (10, -1) radius = √ 7 8 =√ = √ =√ so radius of C2 = of radius of C1 √ = 4√ 6 7 8 Candidate G 6 7 , centre (10, -1) radius = √ =√ =√ which is half of √ 6 =√ 6 7 8 Page 9 …. 8 Question 22 d Generic Scheme Illustrative Scheme Show that the tangent found in part (b) is also a tangent to circle C2. Method 1 Substituting for y 9 ss substitute y = 5 – x (or x = 5 – y) 10 pd express in standard quadratic form 10 11 ic start proof 12 ic complete proof 11 12 equal roots tangent 9 Max Mark 11 12 b2 – 4ac = 0 tangent or Substituting for 9 10 11 12 equal roots tangent 4 11 12 b2 – 4ac = 0 tangent Method 2 9 ss uses perpendicular gradients 9 given line = 10 pd find equation of radius 10 11 ic starts proof 11 12 ic completes proof 12 , leading to =1 and complete Notes: Method 1 4. = 0 must appear at 9 or 10 stage to gain 10. 5. Candidates who arrive at a quadratic equation which does not have equal roots cannot gain 12 as follow through. ( See General Comments Note 16). 6. Where candidates do not arrive at a quadratic equation in Method 1, marks 10, 11 and 12 are not available. 7. Acceptable communication for 12, ‘only one answer so implies tangent’, ‘discriminant is 0 so tangent’, twice so tangent’, or equivalent relating to tangency. Page 10 Question 23 a 1 ss 2 ic 3 pd 4 pd Notes: Generic Scheme Illustrative Scheme Max Mark The expression 3 sin xo – cos xo can be written in the form k sin(x – a)o, where k >0 and 0 < a < 360. Calculate the values of k and a. use compound angle formula 1 k sin xº cos aº – k cos xº sin aº stated explicitly compare coefficients 2 k cos aº =√ and k sin aº = 1 stated explicitly process for k 3 2 ( do not accept √ ) process for a 4 30 4 1. Treat k sin xº cos aº - cos xº sin aº as bad form only if the equations at the 2 stage both contain k. 2. 2sin xº cos aº - 2cos xº sin aº or 2( sin xº cos aº - cos xº sin aº ) is acceptable for 1 and 3. 3. Accept k cos aº = √ and - k sin aº = - 1 for 2. 4. 2 is not available for k cos xº = √ and k sin xº = 1 , however, 4 is still available. 5. 3 is only available for a single value of k, k > 0. 6. 4 is only available for a single value of a expressed in degrees. 7. Candidates who identify and use any form of the wave equation may gain 1 , 2 and 3 , however, 4 is only available if the value of a is interpreted for the form k sin ( x – a)º. 8. Do not penalise omission of degree sign at 1 or 2 . Regularly Occurring Responses: Response 1: Missing information in working. Candidate A Candidate B 1 1 2cos a = √ 2sin a = 1 tan a = 3 √ a = 30 Candidate C cos a = √ sin a = 1 2 4 tan a = 2 3 √ 4 a = 30 Not consistent with evidence at 2 k sin xº cos aº - k cos xº sin aº 1 k cos a = √ , k sin a = 2 k = 2 or 2 3 tan a = 4 √ a = 30 or 210 However candidates who then write o √ sin xº cos xº = 2 sin(x-30) would 3 4 gain and Response 2: Labelling incorrect, sin (A – B) = sin A cos B – cos A sin B from formula list. Candidate D k sin A cos B – k cos A sin B 1 √ a = 30 2 2 4 1 k cos B= √ k sin B = 1 tan x = tan a = Candidate F k sin A cos B – k cos A sin B k cos x = √ k sin x = 1 k cos a = √ k sin a = 1 Candidate E k sin A cos B – k cos A sin B 1 2 √ 4 x = 30 tan B = √ B = 30 Page 11 4 Question 23 Generic Scheme Illustrative Scheme Max Mark b Determine the maximum value of 4 + 5 cos xo – 5 3 sin xo, where 0 < x < 360. ic interpret expression 5 4 – 5 × 2 sin (x – 30)º 6 pd state maximum 6 14 Notes: 9. A solution using calculus gains no marks unless angles are converted to radian measure before differentiating. 10. ‘Maximum = 14’ with no working gains no marks. 11. 5 is awarded for demonstrating a clear link between the expression in (b) and the wave in part (a) 12. Candidates who start afresh, and use any form of the wave function to arrive at 4 ± 10cos (…) or 4 ± 10 sin (…) correctly, can gain both 5 and 6 . 13. 6 is only available if, at the 5 stage, the candidate’s answer in (a) is multiplied by an integer k, 14. Candidates who equate the given expression to 0 and attempt to solve gain 0 marks. Regularly Occurring Responses: 5 Candidate J Candidate K 5 5 Max = 14 Max 2 + 4 6 Max = 6 Page 12 6 2 . Question Generic Scheme Illustrative Scheme Max Mark Show that the points A( –7, –8, 1), T(3, 2, 5) and B(18, 17, 11) are collinear. 24 a i 24 1 a ii ss 2 ic 3 ic Find the ratio in which T divides AB. use vector approach 4 ic Notes: 1 ⃗⃗⃗⃗⃗ compare two vectors 2 complete proof state ratio 4 ⃗⃗⃗⃗⃗ or ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ or equivalent ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ are parallel and since there is a common point A, B and T are collinear 2:3 stated explicitly (see Note 4) 3 ( ) or ⃗⃗⃗⃗⃗ ( ) 4 1. Any appropriate combination of vectors is acceptable. 2. 3 can only be awarded if a candidate has stated, common point, parallel (common direction) and collinear. 3. Treat ( ) written as as bad form. 4. Accept or 3 5. requires evidence of vectors being parallel, simply stating parallel is insufficient. Regularly Occurring Responses: Candidate A ⃗⃗⃗⃗⃗ Candidate B ( ) ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ( ) 2 ⃗⃗⃗⃗⃗ ( ) Candidate C ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ( ) 1 ⃗⃗⃗⃗⃗ 2 ⃗⃗⃗⃗⃗ TB and AT are parallel, T is a common point so A, T and B are ( ) ⃗⃗⃗⃗⃗ ( ⃗⃗⃗⃗⃗ 2 TB and AT are parallel, T is a common point so A,T and B are collinear. 3 collinear. 3 AT:TB = 2:3 4 AT:TB = 3:2 4 Candidate D ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ( ⃗⃗⃗⃗⃗ ) ( 1 ) ⃗⃗⃗⃗⃗ 2 TB and AT are parallel. T is a common point so A, T and B are collinear. A(-7,-8,1) 3 15 10 T(3,2,5) 10 B(18,17,11) 15 4 ) 1 6 10:15 = 10:15 = 4:6 = 2:3 Page 13 4 Question 24 b Generic Scheme Illustrative Scheme Max Mark The point C lies on the x-axis. If TB and TC are perpendicular, find the coordinates of C. 7 ss know to use scalar product equal to 0 7 Method 1 (c, 0, 0) c 3 T C 2 5 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 8 9 pd pd start to solve complete 8 9 15(c – 3) + 15 × (–2) + 6 × (–5) … c=7 5 6 Method 1 ic pd 5 interpret C use vector approach 6 5 6 ic pd interpret C use vector approach 5 (c, 0, 0) 6 c 3 T C 2 5 7 ss know to use Pythagoras and calculate 7 |⃗⃗⃗⃗⃗ | √ 8 pd | T C | or | T B | calculate the other two lengths 8 |⃗⃗⃗⃗⃗ | √ |⃗⃗⃗⃗⃗ | √ 9 pd 5 Method 2 Method 2 9 complete and c=7 Notes: 6. In Method 1, = 0 must appear at 7 or 8 for 9 to be available. 7. In Method 1, candidates who use ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ can gain a maximum of 4 marks. 8. C must appear in coordinate form at 5 or 9 for 5 to be awarded. 9. If C has more than one non-zero coordinate 9 is not available. 10. 8 is only available for expressions with an unknown. Regularly Occurring Responses: Candidate E Candidate F 5 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ( ) 6 7 =7 (7, 0, 0) Gains full marks Candidate G ⃗⃗⃗⃗⃗ 6 ( ) ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 8 8 9 It is not clear at 6 what is meant by ‘c’ so 8 cannot be awarded as follow through. However ⃗⃗⃗⃗⃗ Page 14 ( ) 6 Paper 2 Question 1 Generic Scheme Illustrative Scheme Max Mark The first three terms of a sequence are 4, 7 and 16. The sequence is generated by the recurrence relation un + 1 = mun + c, with u1 = 4. Find the values of m and c. 1 ic interpret recurrence relation 1 7 = 4m + c 2 ic interpret recurrence relation 2 16 = 7m + c 4 ss know to use simultaneous equation 7m + c = 16 4m + c = 7 leading to 4 pd find m and c 4 m = 3, c = –5 3 3 Notes: 1. Treat equations like Regularly Occurring Responses: Candidate A or as bad form. Candidate B Candidate C No working = 3 and or Only one equation Partial verification = 3 and 1 mark out of 4 Candidate D 2 marks out of 4 Candidate E = 3 and by verification = 3 and and 3 marks out of 4 4 marks out of 4 Page 15 2 marks out of 4 Question 2 a Generic Scheme Illustrative Scheme Max Mark The diagram shows rectangle PQRS with P(7, 2) and Q(5, 6). y Find the equation of QR. Q(5, 6) R P(7, 2) O x S 1 ss know to find gradient 1 mPQ = –2 2 ic use perpendicular gradient 2 mQR = 3 ic state equation of line 3 y–6= 3 1 2 1 (x – 5) 2 Notes: 1. 3 is only available as a consequence of using a perpendicular gradient and the point Q. 2. 2 appearing ex nihilo leading to the correct equation for QR gains 0 marks. b y The line from P with the equation x + 3y = 13 intersects QR at T. Q(5, 6) T R P(7, 2) O x S Find the coordinates of T. 4 ss prepare to solve 4 x + 3y = 13 and x – 2y = –7 5 pd solve for one variable 5 x = 1 or y = 4 6 pd solve for second variable 6 3 y = 4 or x = 1 Notes: 3. Subsequent to making an error in rearranging the equation of QR, 4 can still be awarded but 5 is lost. 4. Stepping out from P to Q and then the reverse from Q is not a valid strategy for obtaining T. 5. 4, 5 and 6 are not available to candidates who: (i) equate zeroes , (ii) give answers only without working. Regularly Occurring Responses: Candidate A leading to 4 5 6 Page 16 Question 2 c Generic Scheme Illustrative Scheme Max Mark Given that T is the midpoint of QR, find the coordinates of R and S. 7 ss valid method eg vectors or stepping out or mid-point formula 7 eg ⃗⃗⃗⃗⃗ ( 3 ss know how to find R R (–3, 2) 9 ss know how to find S using ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ( ) 9 S (–1, –2) 8 ) 8 Notes: 6. Any strategy that relies upon the rectangle being composed of two congruent squares can only be given credit if this fact has been justified. Candidates who have already been penalised in 2(b) for making this assumption can gain full credit in (c). 7. If R appear without working then 7, 8 and 9 are not available. and S Regularly Occurring Responses: Response 1: Examples of evidence for stepping out. 4, 2 (5,6) 4 4 (1,4) (-3,2) (5,6) 4, 2 (1,4) (-3,2) (5,6) (1,4) 2 (-3,2) (5,6) (1,4) 2 4 2 (1,4) Candidate B (-3,2) 5 1 4 4 -3 (5,6) 7 (1,4) (-3,2) 8 =1 4 7 8 Page 17 6 -2 4 -3 -2 2 Similar evidence is required for finding S. Response 2: Examples of insufficient evidence for stepping out. (5,6) 1 (-3,2) 4 4 5 -4 -4 4 -4 2 (5,6) 2 -4 (1,4) (-3,2) 2 2 2 -4 -4 Question 3 a Generic Scheme Illustrative Scheme Max Mark Given that (x – 1) is a factor of x3 + 3x2 + x – 5, factorise this cubic fully. 1 ss know to use x = 1 in synthetic division 1 1 1 3 3 1 -5 1 4 5 4 5 0 2 pd complete evaluation 2 3 ic state quadratic factor 3 x2 + 4x + 5 4 ic valid reason for irreducible quadratic 4 (x – 1) (x2 + 4x + 5) with valid reason 1 4 Notes: 1. Accept any of the following for 4 a) b) c) 2. Do not accept any of the following for 4 a) b) c) cannot factorise further. gain marks 1, 2 and 3. 3. Candidates who use algebraic long division to arrive at 4. Candidates who complete the square and make a relative comment regarding no real roots gain 4. , with a valid reason, as bad form for 4. 5. Treat Regularly Occurring Responses: Candidate A Candidate B 3 4 1 3 With a valid reason. Candidate C 3 = 16 – 20 < 0 so does not factorise. 4 Page 18 2 4 Question 3 b Generic Scheme Illustrative Scheme Max Mark Show that the curve with equation y = x4 + 4x3 + 2x2 – 20x + 3 has only one stationary point. Find the x-coordinate and determine the nature of this point. 5 ss start to differentiate 5 two non-zero terms correct 6 pd complete derivative and equate to 0 6 4x3 + 12x2 + 4x – 20 = 0 7 ic factorise 7 4 (x – 1) (x2 + 4x + 5) 8 pd process for x 8 x=1 9 ic justify nature and state conclusion 9 nature table and minimum 5 Notes: 6. = 0 must appear at 6 or 7 for mark 6 to be gained. 7. 9 can be gained using the second derivative to determine the nature. 8. Candidates who incorrectly obtain more than one linear factor in (a) and use this result in (b) must solve to get more than one solution in order to gain 8. Mark 9 is not available. 9. If the equation solved at 8 is not a cubic then 8 and 9 are not available. Regularly Occurring Responses: Candidate D Candidate E ….. 1 ……. from (a) leading to 5 6 Min 7 8 Minimum acceptable response. 9 Page 19 9 Question 4 Generic Scheme Illustrative Scheme Max Mark The line with equation y = 2x + 3 is a tangent to the curve with equation y = x3 + 3x2 + 2x + 3 at A(0, 3), as shown in the diagram. 𝑦 The line meets the curve again at B. Show that B is the point (–3, –3) and find the area enclosed by A(0, 3) the line and the curve. O 𝑥 B 1 ss know how to show that B is the point of intersection of the line and curve. 1 and or solving simultaneous equations 2 6 ∫ ss 3 ic know to integrate and interpret limits. use “ upper – lower” 4 pd integrate 4 5 1 4 pd substitute limits 5 0 – 6 pd evaluate area 6 27 2 3 ∫ x4 + x3 1 4 3 3 3 4 units2 4 Notes: 1. Where a candidate differentiates one or more terms at 4 then 5 and 6 are not available. 2. Candidates who substitute without integrating at 3 do not gain 4, 5 and 6. 3. Candidates must show evidence that they have considered the upper limit 0 at 5. 4. Where candidates show no evidence for both 4 and 5, but arrive at the correct area, then 4, 5 and 6 are not available. 5. The omission of at 3 should not be penalised. Regularly Occurring Responses: Candidate A 3 ∫ 6 Page 20 Candidate B Candidate C ∫ 3 ∫ 4 so 6 6 so Area 6 6 cannot be negative However …. Reference to ‘Area’ must be made. Candidate D 2 ∫ [ 4 ] =[ ] [ ] 5 = See Candidate C Candidate E 6 ∫ 2 [1 3 3 4 ] 4 [ ] [ 5 ] 6 units 2 Candidate F 2 ∫ [ 4 ] [ ] ] 5 [ 6 = 90 units2 Candidate G ∫ 2 ∫ [1 4 ] 4 [ ] [ 3 5 ] units2 6 Page 21 3 Question 5 Generic Scheme Illustrative Scheme Max Mark Solve the equation log5 (3 – 2x) + log5 (2 + x) = 1, where x is a real number. 1 ss use correct law of logs 1 log5 [(3 – 2x) (2 + x)] = 1 stated or implied by 2 2 ic know to and convert to exponential form 2 (3 – 2x) (2 + x) = 51 3 pd express as an equation in standard quadratic form 3 2x2 + x – 1 = 0 4 ic solve quadratic 4 x = 1 , x = –1 4 2 Notes: 1. For 2 accept …….. = log55. 2. Where candidates discard an acceptable solution either by crossing out or by explicit statement, then 4 is not available. Regularly Occurring Responses: Candidate A Candidate B 4 3 4 Candidate C Candidate D incorrect working leading to 1 4 Here the discard of the original question. 3 Candidate E [ 4 Candidate F ] 1 1 2 2 3 4 not available 3 4 4 is not awarded since 2 is valid in the context of is not a valid solution. Page 22 Question 6 Generic Scheme Given that a 5 sin 3x dx 0 Illustrative Scheme Max Mark 10 , 0 < a < π, calculate the value of a. 3 1 ss integrate correctly 1 2 pd process limits 2 3 pd evaluate and form a correct equation 4 pd start to solve equation 4 5 pd solve for a 5 [ ] 5 3 a= 3 Notes: 1. Candidates who include solutions outwith the range cannot gain 5. 2. The inclusion of + c at 1 should be treated as bad form. 3. 5 is only available for a valid numerical answer. 4. Where the candidate differentiates 1 and 2 are not available. See Candidate A. 5. Where candidate integrate incorrectly 2, 3, 4 and 5 are still available. 6. The value of must be given in radians. Regularly Occurring Responses: Candidate A Candidate B [ ] no solutions 1 [ ] 2 2 3 3 4 4 5 5 Ignore other solutions in given interval Candidate D Candidate C 1 1 2 2 3 3 no solutions 1 4 4 5 5 Ignore other solutions in given interval Page 23 Question 7 Generic Scheme Illustrative Scheme Max Mark A manufacturer is asked to design an open-ended shelter, as shown, subject to the following conditions. Condition 1 The frame of a shelter is to be made of rods of two different lengths: 𝑦 x metres for top and bottom edges; y metres for each sloping edge. 𝑥 Condition 2 The frame is to be covered by a rectangular sheet of material. The total area of the sheet is 24 m2. a Show that the total length, L metres, of the rods used in a shelter is given by L = 3x + 48 . x 1 ss identify expression for L in x and y 1 2 ic identify expression for y in terms of x 2 3 pd complete proof 3 Notes: 1. The substitution for y at 3 must be clearly shown. Page 24 3x + 4y 3 3x + 4 and complete Question 7 b Generic Scheme Illustrative Scheme Max Mark These rods cost £8·25 per metre. To minimise production costs, the total length of rods used for a frame should be as small as possible. i Find the value of x for which L is a minimum. ii Calculate the minimum cost of a frame. 4 pd prepare to differentiate 4 …48x–1 5 pd differentiate 5 3 – 48x–2 6 pd equate derivative to 0 6 3 – 48x–2 = 0 7 pd process for x 7 x=4 8 ic verify nature 8 nature table or 2nd derivative 9 ic evaluate 9 10 pd evaluate cost 10 7 = 24 cost 24 £825 = £198 Notes: at 7. However candidates who process 2. Do not penalise the non-appearance of do not gain 9. 3. is not awarded 9. Regularly Occurring Responses: Candidate A 4 to obtain 5 Candidate B Candidate C Min Min Minimum acceptable response Do not penalise the inclusion of Page 25 Question 8 Generic Scheme Illustrative Scheme for 0 < x < 2π sin 2x = 2 cos2x Solve algebraically the equation Max Mark 1 ss use correct double angle formulae 1 Method 1 2 sin x cos x 2 ss form correct equation 2 2 sin x cos x – 2cos2x = 0 3 ss take out common factor 3 2 cos x (sin x – cos x) = 0 4 ic proceed to solve 4 cos x = 0 and sin x = cos x 5 6 5 pd find solutions 5 6 pd find remaining solutions 6 ss use double angle formula 1 2 ss form correct equation 2 3 ss express as a single trig function 3 4 ic proceed to solve 4 1 6 Method 2 ( √ ( ) ) √ 5 5 pd find solutions 5 6 pd find solutions 6 6 Notes: 1. In Method 1, = 0 must appear at stage 2 or 3 for 2 to be available. 2. Accept the use of the wave function to solve sin x – cos x= 0 at stage 4 in Method 1. 3. Accept as evidence for 2. 4. For candidates who obtain all four solutions in degrees 6 can be gained but 5 is not available. Regularly Occurring Responses: Candidate A Candidate B Correct working leading to Correct working leading to 5 6 Page 26 5 6 Question 9 a Generic Scheme Illustrative Scheme Max Mark The concentration of the pesticide, Xpesto, in soil can be modelled by the equation Pt = P0e–kt where: P0 is the initial concentration; Pt is the concentration at time t; t is the time, in days, after the application of the pesticide. Once in the soil, the half-life of a pesticide is the time taken for its concentration to be reduced to one half of its initial value. If the half-life of Xpesto is 25 days, find the value of k to 2 significant figures. 1 ic 1 interpret half-life 1 P0 = P0e–25k 2 2 3 stated or implied by 2 pd 2 process equation e-25k = 1 2 4 ss write in logarithmic form 3 loge 1 = –25k 2 4 pd 4 process for k k 0·028 Notes: 1. Do not penalise candidates who substitute a numerical value for Regularly Occurring Responses: Candidate A 1 2 ( ) 3 4 Page 27 in part (a). Question 9 b Generic Scheme Illustrative Scheme Max Mark Eighty days after the initial application, what is the percentage decrease in concentration of Xpesto? 5 ic interpret equation 5 Pt = P0e–80 0028 6 pd process 6 Pt 0·1065P0 7 ic state percentage decrease 7 89% 3 Notes: 2. For candidates who use a value of k which does not round to , 5 is not available unless already penalised in part(a). 3. For a value of ex-nihilo then 5, 6 and 7 are not available. 4. 6 is only available for candidates who express Pt as a multiple of P0. 5. Beware of candidates using proportion. This is not a valid strategy. Regularly Occurring Responses: Candidate B Candidate C leading to % 5 5 6 6 7 …% 7 Candidate D Candidate E 5 Pt = P0e–80 0028 5 Let P0 be 100 and Pt = 100 × 0 1065 6 6 P0 7 Candidate F Pt = 10 65 Pt = 100e–80 0028 5 Pt = 10 65 89 35% Candidate G 7 Pt = P0e–80 0028 5 Percentage decrease is 100 - 10 65 = 89 35% 7 6 Candidate H Pt = P0e–80 0028 Pt = 1 × e–80 0028 Pt = 10 65 89 35% decrease 5 Pt = …. e–80 0028 6 Pt = 0.1065 P0 7 89 35% decrease [END OF MARKING INSTRUCTIONS] Page 28 6 7

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