H240/01 Mark Scheme June 2023 5 5. Subject Specific Marking Instructions a. Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^) is sufficient, but not required. For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. Award NR (No Response) - if there is nothing written at all in the answer space and no attempt elsewhere in the script - OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’) - OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question. Note: Award 0 marks only for an attempt that earns no credit (including copying out the question). If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt unless it is ambiguous for which part it is intended. b. An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. If you are in any doubt whatsoever you should contact your Team Leader. H240/01 Mark Scheme June 2023 6 c. The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. e. The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. H240/01 Mark Scheme June 2023 7 Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f. We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value. • When a value is not given in the paper accept any answer that agrees with the correct value to 3 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. NB for Specification B (MEI) the rubric is not specific about the level of accuracy required, so this statement reads “2 s.f”. Follow through should be used so that only one mark in any question is lost for each distinct accuracy error. Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. g. Rules for replaced work and multiple attempts: • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others. • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete. • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately. h. For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i. If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the command words “Show” or “Determine”. Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. j. If in any case the scheme operates with considerable unfairness consult your Team Leader. H240/01 Mark Scheme June 2023 8 Question Answer Marks AO Guidance 1 (a) BC2 = 62 + 152 – 2×6×15×cos30o M1 1.1a Attempt use of cosine rule Allow either omission of 2, or + not –, but no other errors Allow other fully complete methods, such as basic trigonometry, possibly combined with Pythagoras BC = 10.3 cm A1 1.1 Obtain 10.3cm, or better If > 3sf then allow 10.25, or answers that round to 10.25 Condone no units [2] 1 (b) sin30 sin 4 6 D = M1 1.1 Attempt use of sine rule Correct equation seen, with fractions either way up Could also be implied by method eg sin-1(0.75) is M1, but just 0.75 is M0 Allow other fully complete methods D = 48.6o A1 1.1 Obtain D = 48.6o, or better D = 48.590377… Allow D = 0.848 radians or D = 131o A1FT 3.1a Obtain D = 131o, or better FT their first angle as long as <150o D = 131.409622… A0 if additional angles given as well Allow D = 2.29 radians (could be FT on incorrect acute angle in radians, as long as D < 2.618) [3] H240/01 Mark Scheme June 2023 9 Question Answer Mark s AO Guidance 2 (a) (i) ( )( ) 3 2 3 2 3 2 3 2 x x x x + + − − + M1 1.1 Attempt to rewrite fractions using correct common denominator Common denominator could just appear as 9 – 4x Must include correct attempt at numerators as well 6 9 4x − A1 2.1 Obtain correct simplified fraction No need to state values for a, b and c explicitly www – if middle terms shown for expansion of denominator, then these must be correct ISW any further attempt to ‘simplify’ SC B1 for answer only, with no method shown [2] 2 (a) (ii) 6 2 9 4x = − 6 = 18 – 8x 8x = 12 M1 1.1a Attempt to solve equation – as far as clearing the fraction and combining constant terms M1 for using their fraction, as long as of correct form Correct method to clear fraction, so M0 for eg 6 = 18 – 4x, but allow sign error when combining constant terms x = 3 2 A1 1.1 Obtain x = 3 2 aef, but fractions must be simplified [2] H240/01 Mark Scheme June 2023 10 Question Answer Mark s AO Guidance 2 (b) DR ( )( ) 2 8 2 1 y y − + M1 3.1a Attempt to solve disguised quadratic in 2y If factorising then expansion should give x2 and one other term correct Quadratic formula should be correct – allow one slip when substituting as long as general formula already seen as correct Completing the square needs to go as far as x – p = ±√q 2y = 8, 2y = – 1 A1 2.1 Obtain two correct roots (could still be in terms of eg u if substitution used) SC If no method shown then award B1 in place of M1A1 for both correct roots (final two marks can still be awarded) y = log28 = 3 M1 1.1 Attempt to solve 2y = k, where k > 0 May just see y = 3, with no explicit use of log2 Allow BOD if attempt at solving 2y = – 1 still present If k ≠ 8 then solution method must be seen, even if k is a power of 2 y = 3 only; 2y = – 1 has no solutions as 2y > 0 for all y A1 2.3 Obtain y = 3, having rejected 2y = – 1 with some reasoning Must have some reason, eg ‘2y is always positive’, ‘2y cannot be negative’, ‘cannot take log of a negative number’, ‘not defined’, ‘not real’, ‘no solutions’ A0 for ‘math error’, ‘does not work’, ‘not possible’ [4] SC If no method at all shown then allow B1 for y = 3, with no other solutions H240/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 3 (a) f(x + h) – f(x) = ((x + h)2 + 2(x + h)) – (x2 + 2x) M1 2.1 Attempt expression for f(x + h) – f(x) Allow sign error from no bracket around final term, ie (x + h)2 + 2(x + h) – x2 + 2x is M1, but no other errors allowed If considering x2 and 2x separately then expressions for both must be seen = x2 + 2xh + h2 + 2x + 2h – x2 – 2x = 2xh + h2 + 2h M1 2.1 Expand and simplify f(x + h) – f(x) Expand and gather like terms (either separately, or single expression) Condone sign errors only, so M0 if collecting like terms after an incorrect attempt to divide by h Allow BOD if 2x… + 2x becomes 0 rather than 4x 2 f( ) f( ) 2 2 x h x xh h h h h + − + + = 2 2 x h = + + M1 2.1 Attempt f( ) f( ) x h x h + − Divide all terms by h Allow BOD if previous error results in a term with a denominator of h 0 f '( ) lim(2 2) 2 2 h x x h x → = + + = + A1 2.5 Complete proof by considering limit as h → 0 www, including correct signs throughout Must divide by h before h → 0 Must see ‘lim’, ‘h → 0’, and f '( ) x at some point in their solution and not just when quoting the generic formula, but allow BOD for f ( ) f ( ) f '( ) x h x h x + − = followed by =…, =…, =… on subsequent lines A0 if ‘lim’ still in final answer Condone d d y x in place of f '( )x [4] H240/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 3 (b) y = x2 + 2x + c B1 2.2a State or imply correct equation, including + c ‘y =’ could be implied by use of 5 c may be implied by later work 5 = 1 – 2 + c c = 6 M1 1.1 Attempt c using (– 1, 5) Allow M1 if equation incorrect, as long as from attempt at integrating 2x + 2 ie of form y = kx2 + 2x + c c may be implied by method eg y = x2 + 2x, followed by 5 = 1 – 2 and then an attempt to ‘balance’ the sides Must use x and y the correct way around As far as attempting a value for c y = x2 + 2x + 6 A1 1.1 Obtain correct equation, including y = ... Equation must be stated, and not just implied by c = 6 seen Allow f(x) = … A0 for ‘equation’ = x2 + 2x + 6 [3] Just stating y = x2 + 2x + 6 or y = (x + 1)2 + 5 gets full marks (may come from observing that (-1, 5) is the minimum point) H240/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 4 (a) AB = 2 2 2 4 20 2 5 + = = B1 1.1 Correct length aef Condone 4.47 or better Allow isw eg 20 4 5 = Allow BOD on signs eg AB = – 2i – 4j seen [1] 4 (b) (p – 3)2 + (p – 5)2 = 20 M1 1.1a Attempt correct equation for length BC Using their attempt at length of AB Condone error on RHS eg having 20 not 20 p2 – 8p + 7 = 0 p = 7 A1 1.1 BC Solve correct quadratic to obtain at least p = 7 If second value of p stated then it must be correct C is 7i + 7j A1 1.1 Correct position vector for C; it could be given as column vector, but not coordinate No need to discard p = 1 [3] If M0, question is ‘determine’ so some evidence needed for full marks – either justifying lengths are equal, or use of components of 2 and 4 7i + 7j with some explanation B3 7i + 7j with no explanation B2 (7, 7) with some explanation B2 (7, 7) with no explanation B1 H240/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 4 (c) OM is 4i + 4j OR BM is i – j B1 1.1 Correct midpoint soi Could instead find vector BM Allow M seen as coordinate, as it is part of their method and not a requested answer Condone M = 4i + 4j, but penalise clear error eg AM = 4i + 4j is B0 Could be soi on a diagram D is 6i + 2j B1 1.1 Correct position vector (not coordinate) for D Do not penalise D given as coordinate if already penalised in part (b) [2] Answer only is B0B1 4 (d) Kite B1* 2.2a Mark independently of reason B1dep * 2.2a Evidence is required to support statements made All relevant evidence quoted must be correct eg two pairs of adjacent sides of same length AD = CD = 26 (or compare components of vectors) ; condone not stating AB = BC as given in question Sides must be defined as adjacent, so B0 for just ‘two pairs of equal sides’, but allow BOD if clarified on an explicit diagram seen in part (d) eg diagonals are perpendicular AC has gradient of 1, BD has gradient of –1 If using a geometrical argument, then identify that ABC is isosceles, M is mid-point of AC hence perpendicular bisector eg BD being a line of symmetry AM = MC, with perpendicular argument as above B0 for reasoning using angles (ie a pair of facing equal angles) unless justified. [2] H240/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 5 (a) (i) a = 2 B1 1.1 Either stated or embedded in equation eg |2x – b| seen ignore any other values seen B0 for a = – 2, unless subsequently corrected b = 6 B1 1.1 Either stated or embedded in equation eg |ax – 6| seen ignore any other values seen c = 1 B1 1.1 Either stated or embedded in equation eg |ax – b| + 1 seen ignore any other values seen [3] 5 (a) (ii) Because f is a many to one function eg f(0) = f(6) B1 1.2 Any correct reason Condone no explicit example Could also say ‘because f is not one to one’ B1 BOD for ‘it is not one to one’ If referring to ‘one to many’ or ‘many to one’ it must be clear whether this is f or f -1 (just ‘it’ or ‘the function’ is not enough) Allow implication of function eg ‘as it is a many to one function there is no inverse function’ May also refer to the ‘horizontal line test’, but need to state outcome eg horizontal line would cross graph of y = f(x) twice [1] H240/01 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 5 (b) (i) y = px – q px = y + q x = 1 p (y + q) M1 3.1a Complete attempt to find inverse function of f(x) = px – q Correct order of operations, allow sign error only Could use coordinate geometry and reflection in y = x Allow M1 BOD if more than one function is being considered ( ) 1 1 g q p p x x − = + A1 1.1 Obtain correct inverse, in terms of x Could be single term ie ( ) 1 g x q p x + − = A1 for just 1 q p p x + , ie ( ) 1 g x − can be omitted If LHS seen, it must be ( ) 1 g x − or y (allow BOD for 1 g−, or using f not g) BOD if modulus sign included A0 if additional equations given x ≥ 0 B1 1.2 Correct domain B0 for x > 0 Independent of the first two marks If in words then must be correct, so B1 for ‘any non-negative x’ but B0 for ‘any positive x’ g -1(x) ≥ 0 is B0 Condone incorrect set notation as long as intention is clear [3] 5 (b) (ii) 0 < p ≤ 1 B1 3.1a Correct set of values, any notation No need for 0 < p as specified in question, so B1 for p ≤ 1 B0 for p < 1 B0 for any additional incorrect values B0 if just single example and not set of values Condone incorrect set notation as long as intention is clear H240/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance [1] Question Answer Marks AO Guidance 6 (a) ( ) 2 3 d 2 3 e d x x y x x + = + M1 1.1a Attempt to differentiate using the chain rule Obtain derivative of form 2 3 f( )ex x x + Could also split into two terms and use product rule to obtain derivative of form 2 2 3 3 f( )e e e e x x x x x k + (k ≠ 0) M0 if attempt to split results in sum not product ( ) 2 3 2 3 ex x x + + A1 1.1 Obtain correct derivative Brackets must be seen, or implied by later work aef eg ( )( ) ( )( ) 2 2 3 3 2 e e + e 3e x x x x x from splitting into two terms first Could be in terms of u, as long as u clearly defined ( ) 2 3 2 3 e 0 x x x + + = 2x + 3 = 0 x = – 3 2 A1 1.1 Equate correct derivative to 0 soi and solve to obtain x = – 3 2 ISW any y-coordinates if given A0 if any additional solutions for x Must see differentiation, so x = – 3 2 with no supporting method gets no credit (as question is ‘determine’) H240/01 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 2 3 e 0 x x +  for all x or 2 3 e 0 x x +  or x2 + 3x = ln0, but this is not possible B1FT 2.4 Indicate no solutions from the exponential term FT their derivative as long as of form 2 3 f( )ex x x + or f( )eu x Allow BOD for explanations such as ex > 0 for all x Must have some reason, eg ‘ 2 3 ex x + is always positive’, ‘ 2 3 ex x + cannot be negative’, ‘cannot take ln of a negative number’, ‘not defined’, ‘not real’, ‘no solutions’ A0 for ‘math error’ or ‘doesn’t work’ [4] 6 (a) Alternative method lny = x2 + 3x 1 d 2 3 d y x y x = + M1 Take ln and attempt implicit differentiation Must deal correctly with lny ( ) d 2 3 d y y x x = + A1 Obtain correct derivative May still have 1 y on LHS 2x + 3 = 0 x = – 3 2 A1 Equate correct derivative to 0 soi and solve to obtain x = – 3 2 2 3 e 0 x x +  B1 Indicate no solutions from the exponential term See main MS for guidance Could also explain why no solutions from 1 y H240/01 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 6 (b) ( ) 2 2 2 2 3 3 2 d 2e 2 3 e d x x x x y x x + + = + + M1 3.1a Attempt to differentiate again using the product rule correctly Obtain derivative of form ( ) 2 2 3 ex x ax bx c + + + aef ( ) ( ) 2 2 2 3 2 d 2+ 2 3 e d x x y x x + = + A1 1.1 Obtain correct derivative aef eg (depending on method) ( ) ( ) 2 2 2 3 3 3 2e 2 2 3 e 3 2 3 e x x x x x x x x x + + + + + + + or ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 3 3 3 2e 4 e e 2 e 3e 6 e e 9e e x x x x x x x x x x x x + + + + Could be in terms of u, as long as u clearly defined convex means 2 2 d 0 d y x  B1 1.2 State, or clearly imply, correct condition at any point in proof Must be general statement, and not just > 0 from testing the stationary point H240/01 Mark Scheme June 2023 20 Question Answer Marks AO Guidance (2x+3)2 ≥ 0 hence 2 + (2x+3)2 > 0 M1 3.1a Explain why correct quadratic is always positive Could note minimum value of 2 as completed square form Could use expanded quadratic, which should be 4x2 +12x + 11 If showing no real roots then must also say that it is a positive quadratic Condone > / ≥ muddles for M1 only Could show that there are no points of inflection and 2 2 d 0 d y x  for at least one point g( ) e 0 x  for all x; quadratic > 0 for all x hence curve is always convex A1 2.4 Full and convincing proof to show that curve is convex for all x www [5] 6 (b) Alt method for first 2 marks ( ) d 2 3 d y y x x = + ( ) 2 2 d d 2 2 3 d d y y y x x x = + + M1 Attempt second derivative, using implicit differentiation and the product rule If still 1 d 2 3 d y x y x = + then must be a correct attempt to differentiate the LHS A1 Obtain correct derivative aef Then B1 M1 A1 as above Will need to use ( ) d 2 3 d y y x x = + to make further progress H240/01 Mark Scheme June 2023 21 Question Answer Marks AO Guidance Question Answer Marks AO Guidance 7 (a) cos(A – B) = cosAcos(–B) – sinAsin(–B) M1 2.1 Replace B with –B in given identity cos(–B) = cosB, sin(–B) = – sinB, cos(A – B) = cosAcosB – sinA(– sinB) cos(A – B) = cosAcosB + sinAsinB A.G. A1 2.4 State cos(–B) = cosB and sin(–B) = – sinB, and conclude with correct identity Condone – sinAsin(–B) becoming sinAsinB with no intermediate step cos(–B) = cosB, sin(–B) = – sinB must be stated, but no justification needed [2] H240/01 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 7 (b) ( )( ) 3 3 1 1 2 2 2 2 cos sin cos sin     − + B1 2.1 Use correct identities, with exact trig values, to obtain a correct expression Allow BOD for ambiguous positioning of + and – signs in a product, but penalise explicit errors if a single identity is seen in isolation If expansion done before exact trig values used, then the expression must still be correct at the point that the B1 is awarded 2 2 3 1 4 4 cos sin   − M1 2.1 Expand brackets May be recognised as difference of two squares so no need to see 3 3 4 4 cos sin cos sin     − To obtain answer of form 2 2 cos sin a b   − (a > 0, b > 0), with possibly cos sin cos sin c c     − also present ( ) 2 2 3 1 4 4 cos 1 cos   − − 2 1 4 cos − A.G. A1 2.1 Use Pythagorean identity and simplify to given answer www eg if middle terms shown for expansion, then these must be correct [3] 7 (c) (i) max value is 3 4 B1 1.1 Correct max value when θ is 180o B1 1.1 Correct angle B0 if any extra angles given Must be ‘positive’ so B0 for 0o Must be in degrees [2] Marks are independent 7 (c) (ii) min value is 1 4 − B1 1.1 Correct min value H240/01 Mark Scheme June 2023 23 Question Answer Marks AO Guidance when θ is 90o B1 1.1 Correct angle B0 if any extra angles given Must be in degrees SC If angles in both parts are correct, but in radians, then penalise only once (mark as B0 in (i) and B1 in (ii)) [2] Marks are independent Question Answer Marks AO Guidance 8 (a) ( ) ( )( ) 3 2 3 3 3 4 2 4 1 1 x x + = + B1 1.1 Correct first two terms Allow unsimplified Expect 9 8 1 x + H240/01 Mark Scheme June 2023 24 Question Answer Marks AO Guidance ( )( ) ( ) 3 1 2 2 2 3 4 2 x + M1 1.1 Attempt third term Condone lack of brackets when attempting to square ie 2 3 4 x Coefficient must be ( )( ) 3 1 2 2 2 or equiv A1 1.1 Obtain correct third term Allow unsimplified 2 3 4 x is A0 unless recovered by later work Expect 2 27 128 x ( ) ( ) 3 3 2 2 2 3 27 4 16 4 3 8 1 8 9 x x x x + = + = + + B1FT 1.1a Multiply their 3 term expansion by 8 Bracket expanded and coefficients simplified If B1M1A1 awarded, but attempt to simplify then goes wrong, B1FT is not also awarded ISW once correct expansion seen [4] 8 (b) 4 3 x  or 4 4 3 3 x −  B1 1.1 Could also be 4 3 x  or 4 4 3 3 x − , as n > 0 Must be condition for x, not kx [1] 8 (c) ( )( ) 2 2 2 27 16 8 9 1 2 x x ax a x + + + + coeff of x2 is 8a2 + 18a + 27 16 M1 3.1a Expand (1 + ax)2 and attempt at least one coeff of x2 Allow ax as middle term, and/or ax2 as third term Attempt at x2 term could be part of a fuller expansion H240/01 Mark Scheme June 2023 25 Question Answer Marks AO Guidance M1 1.1 Attempt all three coeff of x2, and no others If part of fuller expansion then M1 awarded when only three relevant terms used 8a2 + 18a + 27 16 = 107 16 2 8 18 5 0 a a + −= A1 3.1a Equate to 107 16 to obtain correct quadratic aef, including unsimplified A0 if a mix of terms and coefficients, but can be recovered ( )( ) 2 5 4 1 0 a a + − = a = 5 2 − and a = 1 4 A1 1.1 Solve quadratic, possibly BC, to obtain a = 5 2 − and a = 1 4 [4] H240/01 Mark Scheme June 2023 26 Question Answer Marks AO Guidance 9 (a) The rate at which the number of bees changes relative to the rate at which the number of flowers changes or Rate of increase of bees per increase in wildflower plants B1 3.3 Must mention ‘rate of change’ of bees, or equiv, and how this relates to wildflower plants B0 for ‘change’ not ‘rate of change’ B1 for rate of change of bees with respect to plants B1 for rate of change of bees as the plants change B1 BOD for rate of change of bees compared to the number of plants State or imply that it is bees compared to flowers, so B0 if other way around Must relate to bees and plants, and not just B and F [1] See appendix for further examples 9 (b) 0.1 d 5e d t F t = B1 1.1 Correct derivative No need to see d d F t notation Could be unsimplified d 2 3sin3 d B t t = − B1 1.1 Correct derivative No need to see d d B t notation NB watch out for 2 – 3sint 0.1 d d d 2 3sin3 d d d 5e t B B t t F t F − =  = M1 3.4 Correct method to combine their two derivatives – algebraic or numerical B0B0M1 is possible Must have e0.1t in denominator, or e– 0.1t in the numerator, but allow muddles with the placing of the 5 0.4 d 2 3sin12 0.484 d 5e B F − = = A1 3.4 Substitute t = 4 to obtain 0.484, or better [4] H240/01 Mark Scheme June 2023 27 Question Answer Marks AO Guidance 9 (b) Alternative method 10ln 50 F t   =     20 20ln cos 30ln 50 50 F F B       = + +             B1 Correct expression for B as a function of F d 20 30 sin 30ln d 50 B F F F F     = −         M1 Attempt differentiation May see ln terms split first (possibly even including use of cos(A – B) A1 Obtain correct derivative aef 0.4 0.4 0.4 d 20 30 50e sin 30ln d 50e 50e 50 B F     = −         = 0.484 A1 Substitute F = 0.4 50e to obtain 0.484, or better Could use t = 4 if derivative now in terms of t 9 (c) The data comes from the summer, so taking it beyond 12 weeks is unlikely to be reliable B1 3.5b Summer will be over so pattern may not continue Summer is not greater than 12 weeks Fewer bees and/or flowers in autumn/winter Any reason referring to a change in season having an effect on bees and/or flowers B0 for just considering long-term behaviour eg flowers will not continue to increase exponentially Reasons must reference seasons / different time of year (could be implied by ‘weather getting colder’) [1] H240/01 Mark Scheme June 2023 28 Question Answer Marks AO Guidance 10 (a) Both f(0) and f(1) are positive so no sign change will be seen B1 2.3 Identify both y-values being positive and state ‘no sign change’ or equiv Could also evaluate f(0) as 1 and f(1) as 2.9 (or better), and refer to no sign change – these are both positive so no need to include > 0 Could also refer to the asymptote / discontinuity within this range (x = 0 to x = 1) Also allow ‘graph is not continuous in this interval’ B0 for no reference to interval Could say that the two points chosen are not on the same part of the curve [1] 10 (b) 2 e 2 4 1 x x = − − 2 e 8 2 x x = − + 2 8 2 ex x = − M1 1.1 Attempt rearrangement, as far as kx2 = … Allow sign error(s) only 2 16 4 2ex x = − 4 4 2ex x = − ( ) 1 4 2e 4 x x = − A.G. A1 1.1 Obtain given answer convincingly If 1 1 e 4 8 x x = − then an additional line of working needed before given answer (eg show common denominator of 16) [2] H240/01 Mark Scheme June 2023 29 Question Answer Marks AO Guidance 10 (c) x2 = 0.285074813… B1 1.1 Correct first iterate (at least 4sf) State 0.2851 or better 0.28943, 0.28817, 0.28853, 0.28843, 0.28846, 0.28845... M1 1.1 Correct iterative process (at least 3 more values) Allow M1 for 3sf – expect 0.289, 0.288 and then 0.288 or 0.289 depending whether truncating or rounding α = 0.2885 A1 1.1 Correct root, given to 4sf, following 2 iterates that agree to 4sf ie at least 7 iterations needed, given to at least 4sf A0 for eg x8 = 0.2885 (implies 8th iterate and not root) Process self corrects so B0M1A1 possible; or B1M1A1 if error in term other than x2 [3] 10 (d) 2 16 F ( ) 2 8 x x x −  = − M1 1.1a Attempt differentiation using the chain rule Obtain derivative of form 2 2 8 kx x − Condone subscripts still present in derivative F (0.3) 3.75  = − M1 1.1 Attempt F (0.3)  – not dependent on previous M1, but must follow some attempt at differentiation M1 can be implied by correct –3.75 (from correct derivative), but explicit substitution must be seen if F ( )x  is incorrect Must come from differentiating F(x) and not a different function H240/01 Mark Scheme June 2023 30 Question Answer Marks AO Guidance For convergence F ( ) 1   , but – 3.75 < – 1, so iteration will not find root A1 2.5 Correct reasoning, following correct F (0.3)  Allow F ( ) 1   −, hence will not converge Condone F ( )x  not F ( )   [3] No credit for just testing the given iterative formula Question Answer Marks AO Guidance 11 (a) log10S = log10(abt) log10S = log10a + log10bt M1 3.3 Attempt to show reduction to linear form Introduce logs on both sides, and correctly split to the sum of two terms log10S = tlog10b + log10a A1 3.3 Obtain correct equation Condone no base; any bases seen must be 10 A0 for log10bt unless previously seen as tlog10b which is of the form Y = mX + c A1 3.3 Link to equation of straight line Could instead refer to linear relationship [3] If M0 then allow SC B1 for statement such as S against t is an exponential function so logS against t will give a straight line 11 (a) Alternative method log10S = mt + c S = 10 mt + c M1 Attempt equation of straight line, and attempt expression for S Must be using log10S against t Must use base of 10 S = 10 mt × 10c A1 Correctly split into two terms which is of the form S = abt A1 Link to exponential model H240/01 Mark Scheme June 2023 31 Question Answer Marks AO Guidance 11 (b) m = log10b = 0.06 so b = 100.06 = 1.15 B1 2.1 Link gradient of line of best fit to linear form and confirm b ≈1.15 Allow m in range [0.055, 0.065] Or log101.15 = 0.06 and compare to gradient c = log10a = 2.08 so a = 102.08 = 120 B1 2.1 Link intercept of line of best fit to linear form and confirm a ≈ 120 Allow c in range [2.075, 2.085] Or log10120 = 2.08 and compare to intercept [2] Plotted points are linear so may not see line of best fit drawn If substituting into formula (either given model or linear reduction) then B1 for finding and verifying any point that would be on the line of best fit B1 for finding and verifying a second point 11 (c) S = 120 × 1.157 M1 3.4 Substitute t = 7 into given model soi predicted sales are 319 items A1 3.4 Conclude with integer value Accept 320 items [2] H240/01 Mark Scheme June 2023 32 Question Answer Marks AO Guidance 11 (d) (i) GP with a = 138 and r = 1.15 B1 3.1b State or imply sum of GP with a as 120 or 138, and r as 1.15 Could be implied by attempt to use GP sum formula (but not just nth term) – allow slip as long as clearly sum being considered 138(1 1.15 ) 70000 1 1.15 t − = − M1* 3.1b Attempt sum of GP, with a = 120 or 138 and r = 1.15, related to 70000 Must be correct sum formula May have n not t throughout Allow r = 1.15t + 1 with a = 120 but this is B1 M1 only, as not a valid method) 1.15t = 77.087 M1 dep* 1.1 Attempt to rearrange equation as far as 1.15t = Must now have a = 138 (or equiv) Allow sign errors only Allow T&I as not DR t = 31.088… hence 32 months A1 3.2a Obtain 32 (‘months not required’) www If 32 given as answer only then allow full marks; if any method shown then mark using the main scheme [4] Allow BOD with any inequality signs H240/01 Mark Scheme June 2023 33 Question Answer Marks AO Guidance 11 (d) (ii) Unlikely to be reliable as sales may not continue in same pattern as market could become saturated B1 3.2b State or imply that the model is unlikely to be valid, with a sensible reason why – could refer to reason why pattern may not continue or extrapolation not being reliable Decrease in demand Increase in competition No values beyond t = 6 so pattern unknown Reason why sales are likely to level off / plateau or unlikely to continue to increase (‘other factors’ not enough) Item sales may vary according to season [1] Question Answer Marks AO Guidance 12 (a) du = exdx B1 1.1 Correct statement linking du and dx or 1 d d 2 x u u = + ( ) 2 7 2 8 1 d 2 u u u u + − +  M1 1.1 Use ex = u + 2 to attempt integrand in terms of u Must see clear evidence of substitution, including how exdx is dealt with M0 for going straight from 7ex – 8 to 7u + 6 with no justification Must include du 2 2 7 14 8 7 6 d d ( 2) ( 2) u u u u u u u u + − + = = + +   A1 2.1 Correct integrand Including both integral sign and du throughout, as AG [3] H240/01 Mark Scheme June 2023 34 Question Answer Marks AO Guidance 12 (b) 2 2 7 6 2 ( 2) A B C u u u u u u + + + = + + 2 ( 2) ( 2) 7 6 Au u B u Cu u + + + + = + M1 3.1a Attempt correct partial fractions May have 2 2 Au B C u u + + + but M0 for just 2 B u with no A u Correct method to combine correct fractions, and at least one constant attempted If considering 2 2 7 8 ( 2) u u u − + + then must use partial fractions on the second term to get credit 2 2 3 2 2 u u u + − + A1 2.1 Correct partial fractions May have 2 2 3 2 2 u u u + − + Possibly implied by their A, B, and C values ie A = 2, B = 3, C = – 2 2ln|u| – 2ln|u + 2| – 3u-1 M1 1.1 Attempt integration of 2 B u and at least one of A u or 2 C u + , and no others Allow errors in coefficients only Allow M1 if only two fractions, as long as of required form If using 2 Au B u + then it must be a correct integration attempt (ie split into two fractions first) A1FT 2.1 FT on their two or three fractions as long as ku-2 and one or two fractions each with a linear denominator Condone brackets not modulus Condone no brackets as long as implied by later working, eg when limits are used (2ln4 – 2ln6 – 3 4 ) – (2ln2 – 2ln4 – 3 2 ) M1 1.1a Attempt use of correct limits – correct order and subtraction; u or x but commensurate with their integral Allow substitution into any function that is clearly attempt at integration H240/01 Mark Scheme June 2023 35 Question Answer Marks AO Guidance 2 3 3 4 4 ln 2 4 6 2      − +          M1 3.1a Attempt to rearrange correct numerical integral to required form Must be correct numerical expression from correct working Terms may have been combined before use of limits, but must still be correct expression to gain M1 Correct attempt to combine ln terms ie deal with coefficients and correct product / quotient for the sum / differences Allow one slip 3 16 ln 4 9 + A1 2.1 Obtain 3 16 ln 4 9 + Condone 3 4 2ln 4 3 + Fractions must be simplified ISW an incorrect attempt to write this answer in a different form, but A0 if further work done eg multiplying by a constant to clear the fractions [7] H240/01 Mark Scheme June 2023 36 APPENDIX Exemplar responses for Q9(a) Response Mark Comment The rate of increase in the number of bees regarding the increase in the number of plants B1 BOD ‘regarding’ Rate of bees with respect to flowers B0 no ‘change’ Rate of change in number of bees in terms of the number of flowers B1 BOD ‘in terms of’ The rate of growth in the number of bees when the number of plants increases over time B1 The rate of change of flowers according to the number of bees B0 wrong way around The rate of increase of bees over the rate of increase in wildflowers B0 not ‘over’ – suggests fraction The rate of change of number of bees compared to number of plants B1 BOD ‘number’ Rate at which the number of bees increases as the number of plants increase B1 Includes ‘rate’ and ‘increases’ The change in number of bees in respect to the number of flowers B0 no ‘rate’ The rate of increase in number of bees in accordance to the number of plants B1 BOD ‘in accordance’ How the number of bees vary with the number of flowers B0 not rate of change The rate at which the bees to flowers ratio is changing B1 BOD ‘ratio’ Rate of growth of bees depending on the number of flowers B1 Rate of change of bees per wildflower plant B1 Rate of change of the number of bees in terms of flowers B1 The rate of change between the bees and the plants B0 no dependency implied The rate of change in number of bees against the change in plants B1 The rate of change of bees compared to flowers B1 Rate of change of the number of bees as the number of flowers vary B1 The rate at which the number of bees increase with the number of plants B1 BOD ‘with’ Need to get in touch? 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