H240/01 Mark Scheme June 2022 3 2. Subject-specific Marking Instructions for A Level Mathematics A 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 2022 4 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. 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. H240/01 Mark Scheme June 2022 5 • 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 2022 6 Question Answer Mark s AO Guidance 1 (a) 5 21 0.5 0.5 0 2 2 2 3 2 2        + + + +             B1 1.1a State the 4 correct non-zero y- values and no others Exact values (including unsimplified) or decimal equivs (0, 1.12, 1.73, 2.29, 2.83) – 3sf or better B0 if other ordinates seen unless clearly not intended to be used M1* 1.1a Attempt to find area between x = 1 and x = 3, using k{y0 + yn + 2(y1 + ... + yn-1)} Big brackets need to be seen or implied y-values must be correctly placed Must be using attempts for at least 4 y- values (but no need to see y = 0 explicitly) Condone using other than 4 intervals as long as values equally spaced between x =1 and x = 3 M1d* 1.1a Use k = 0.5 × 0.5 soi Dep on previous M1 Or using k = 0.5h, h consistent with their different number of intervals = 3.28 A1 1.1 Obtain 3.28, or better Allow answers to > 3sf, as long as they round to 3.28 [4] 1 (b) Under-estimate, as the tops of the trapezia are below the curve B1 3.2b Under-estimate, with any valid explanation Condone just ‘trapezia under curve’ Or curve is concave / decreasing gradient (not decreasing function) Accept explanation on diagrams Allow comparing to true value (3.36) B0 if any additional incorrect or contradictory statements [1] H240/01 Mark Scheme June 2022 7 Question Answer Mark s AO Guidance 1 (c) Use more trapezia, of a lesser width, between the same limits B1 3.2b Convincing reason Condone just ‘more trapezia’ or ‘narrower trapezia’ Could refer to strips or intervals [1] 2 (a) eg 1 > – 2, but 12 < (– 2)2 as 1 < 4 B1 2.1 Any correct counterexample, and contradiction identified Initial inequality soi and then contradiction eg –3 > –4 but 9 < 16 (or 9 ≯ 16) [1] 2 (b) (i) eg sin150o = 0.5 as well B1 2.3 Any correct statement Identifies that sinx = 0.5 could give values of x other than 30o Either specific example or general statement eg ‘many to one’ function [1] 2 (b) (ii) sinxo = 0.5 ⇐ xo = 30o B1 2.5 Any correct relationship If attempting to write general solution then must be fully correct eg x = 30o + 360no, x = 150o + 360no Condone ← instead of ⇐ [1] 2 (c) (4n) + (4n + 4) + (4n + 8) + (4n + 12), where n is an integer B1* 2.1 Four consecutive multiples of 4 written correctly in terms of n, or any other variable Allow BOD if n not explicitly stated to be an integer Sufficient to just list the 4 terms, rather than as a sum Not necessarily starting on 4n Could also define k as a multiple of 4 and then have k, k + 4 etc = 16n + 24 = 8(2n + 3) M1 dep* 2.1 Correctly sum terms, and correctly take out common factor of 8 Or sum and then consider each term separately Could be a different factor if using k H240/01 Mark Scheme June 2022 8 Question Answer Mark s AO Guidance 2n +3 is an integer, so 8(2n + 3) is a multiple of 8 A1 2.4 Conclude appropriately Allow BOD if 2n + 3 not explicitly stated to be an integer If using k… expect 8(0.5k + 3) then justify 0.5k as an integer, or 4(k + 6) then justify k + 6 is a multiple of 2 [3] 3 (a) DR 2x2 – 8x + 6 = 0 x2 – 4x + 3 = 0 M1 1.1 Equate, and rearrange to three term quadratic Attempt to gather like terms, but not necessarily on same side of equation Condone no ‘= 0’ (x – 1)(x – 3) = 0 M1 1.1 Attempt to solve quadratic 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 x = 1, x = 3 A1 1.1 Obtain both correct x values Or one correct (x, y) coordinate following a correct factorisation oe (1, 0) and (3, 4) A1 1.1 Obtain both correct pairs of coordinates Allow as eg x = 1, y = 0 as long as pairings are clear H240/01 Mark Scheme June 2022 9 Question Answer Mark s AO Guidance [4] SC If no method shown for solving quadratic then allow M1 for obtaining 3 term quadratic A1 for x = 1, x = 3 A1 for (1, 0) and (3, 4) SC If no method at all shown then allow B1 for both (1, 0) and (3, 4) 3 (b) M1 1.1 Attempt graph of y = 2x – 2, with positive gradient and negative intercept No need for line to actually intersect with negative y-axis as long as it goes beneath positive x-axis A1 1.1 Graph of y = 2x – 2 passing through both points of intersection of the two quadratic graphs Must pass through both points [2] y x H240/01 Mark Scheme June 2022 10 Question Answer Mark s AO Guidance 3 (c) R B1FT 2.2a Correct region labelled with R, or otherwise clearly identified FT any straight line that splits the overlap area into two finite regions, with the lower region identified Allow for straight line with negative gradient as well, but not x = k [1] 4 (a) 2(x + 1.5)2 + 2.5 B1 1.1a p = 2 Could be implied by 2(x + q)2 + r B1 1.1a q = 1.5 Could be implied by p(x + 1.5)2 + r B1FT 1.1a r = 2.5 FT on their p and q ie 7 – pq2 [3] 4 (b) (– 1.5, 2.5) B1FT 1.1 Correct x-coordinate FT on their (a) Could come from differentiation B1FT 1.1 Correct y-coordinate FT on their (a) No FT on incorrect x-value from differentiation [2] 4 (c) minimum value of the function = 2.5 B1FT 3.1a FT on their minimum value Allow BOD if different answers in (a) and (b) 2.5 must be stated as, or clearly intended to be, the minimum value Just (... , 2.5) is insufficient H240/01 Mark Scheme June 2022 11 Question Answer Mark s AO Guidance tanθ = –1.5 θ = – 56.3o M1 3.1a Attempt to solve tanθ = their (– 1.5) To obtain numerical value for θ Allow an angle in radians (expect –0.983 rad) Allow BOD if different answers in (a) and (b) θ = 124o A1 1.1 Obtain 124o, or better A0 if additional solutions [3] Condone approaches other than ‘hence’ eg B1 – attempt to solve tanθ = – 1.5, from correct derivative (expect 4tanθsec2θ + 6sec2θ = 0) B1 – obtain θ = 124o B1 – obtain min value of 2.5 (no FT) 5 (a) (i) 4 units in the negative x-direction M1 1.1 Indicate horizontal translation (in either direction) in some way with magnitude of 4 (‘units’ not required) B1 for 4 0    Condone informal language as long as intent is clear eg ‘left’ (or even ‘right’, as either direction allowed) M0 if ambiguous eg ‘in’ or ‘on’ the x- axis A1 2.5 or 4 in negative x-direction Correct language needed B2 for 4 0 −       Must now be correct language so A0 for eg ‘along’ the x-axis or ‘left’ Allow ‘parallel to the x-axis’ or ‘horizontal’ [2] H240/01 Mark Scheme June 2022 12 Question Answer Mark s AO Guidance 5 (a) (ii) in the y-direction with sf 16 B1 3.1a Identify direction - correct language needed Allow ‘x-axis invariant’, ‘parallel to the y-axis’ or ‘vertical’ Condone ‘positive’ y-direction (as given function > 0) B1 1.1 or 24 ‘scale factor’ or ‘factor’ needed (condone ‘stretch’ factor) Not dep on previous B1, but must have indicated vertical stretch in some way, including informal language such as ‘upwards’ Cannot be ambiguous language, such as ‘in’, ‘on’, ‘across’ the y-axis [2] 5 (b) DR log2(8x(1– x)) = 1 M1 1.1a Correctly combine two correct log terms Or log2(8x) = log2 2 1 x − Or 3 + log2(x(1– x)) = 1 Or log2(4x(1– x)) = 0 OR use indices base 2 on both sides (ie 2 1 log (1 ) 8 2 x x − − = ) and use rules of indices to split eg ( ) 2 log 1 8 2 2 x x − − =  H240/01 Mark Scheme June 2022 13 Question Answer Mark s AO Guidance 8x(1– x) = 2 M1 1.1a Correct method to remove logs Correctly used on equation of form log2f(x) = log2g(x) or log2f(x) = k OR correct method to deal with log term – expect 8x = 2 1 x − eg 8x2 – 8x + 2 = 0 or 8x(1– x) = 2 or 8x = 2 1 x − A1 1.1 Any correct equation not involving logarithms Could still contain brackets and / or fractions x = 0.5 A1 1.1 Obtain x = 0.5 A0 if additional solutions [4] DR so no credit for answer only 6 (a) 35 + 5×34×(2x) = 243 + 810x B1 1.1 Obtain 243 + 810x Condone 35 + 810x Allow terms not written as a sum eg written separately, or linked with a comma 10×33×(2x)2 or 10×32×(2x)3 M1 1.1a Attempt at least one further term – product of correct binomial coeff, power of 3 and attempted power of 2x, with powers totalling 5 Binomial coeff must be numerical; 5C2 is not yet enough Allow BOD if brackets missing when index is applied to 2x, even if never recovered eg 540x2 or 180x3 + 1080x2 A1 1.1 Obtain correct third term Coefficient simplified Terms separate, listed or summed H240/01 Mark Scheme June 2022 14 Question Answer Mark s AO Guidance + 720x3 A1 1.1 Obtain correct fourth term Coefficient simplified Could be separate term, part of a list or part of a sum If expanding brackets then mark as above, but all 5 sets of brackets must be considered (allow irrelevant terms to be discarded) Alternative method Expanding ( ) 5 2 3 3 1 x   +   243 + 810x or 243( 10 3 1 x + ) B1 First two terms correct Allow with 243 still outside the bracket 2 40 9 243( ) x or 3 80 27 243( ) x M1 Attempt one further term Condone just 3 not 35 being used, but must be the correct binomial coeff and an attempt at the correct power of 2 3 x , but allow BOD if no brackets A1 Either 3rd or 4th term correct Allow with 243 still outside the bracket 243 + 810x + 1080x2 + 720x3 A1 Fully correct expansion With the 243 now multiplied into the expansion [4] 6 (b) x = y + 2y2 B1 3.1a Identify correct substitution Could be stated, or implied by use in their binomial expansion 1080(y + 2y2)2 + 720( y + 2y2)3 M1 1.1a Attempt to use binomial from (a) with their 2 term substitution Must substitute into at least the x2 and x3 terms from their (a) Allow M1 if using 2y + 4y2 as their substitution 4320y3 + 720y3 M1 1.1a Attempt expansion to obtain the two relevant terms in y3 M0 if any other y3 terms Expect 4(their 1080) and (their 720) Allow M1 if using 2y + 4y2 as their substitution - expect 16(their 1080) and 8(their 720) H240/01 Mark Scheme June 2022 15 Question Answer Mark s AO Guidance coeff of y3 is 5040 A1 1.1 Allow 5040y3 Ignore any other non-cubic terms Alternative method 1 Attempting binomial expansion of (3 + (2y + 4y2))5 or ((3 + 2y) + 4y2)5 eg ((3 + 2y) + (4y2)) B1 Group into two expressions, and attempt to use them eg (3 + 2y)5 + 5(3 + 2y)4(4y2) M1 Use their groups to obtain the appropriate two elements of their binomial expansion (ie those that would give y3 terms) eg (… + 720y3 + …) + 5(… 4.33.2y…)(4y2) = 720y3 + 4320y3 M1 Expand to attempt the two y3 terms, and no others coeff of y3 is 5040 A1 Obtain 5040 Alternative method 2 Attempting to expand all 5 brackets eg (3 + 2y + 4y2)5 = (81 + 216y + 648y2 + 960y3 … )(3 + 2y + 4y2) M1 Attempt to use all 5 brackets An attempt to use all 5 is sufficient (216y × 4y2) + (648y2 × 2y) + (960y3 × 3) M1 Attempt all products that would give a y-cubed term Condone additional terms, even those that would give another y3 term Irrelevant terms (ie powers greater than 3) may never be seen H240/01 Mark Scheme June 2022 16 Question Answer Mark s AO Guidance 864y3 + 1296y3 + 2880y3 A1 Obtain correct terms or coefficients, with no more than one incorrect They must have attempted all of the expected y3 terms, and no more, with no more than one coefficient error If (3 + 2y + 4y2)4 × (3 + 2y + 4y2) then expect 2880 + 1296 + 864, If (3 + 2y + 4y2)3 × (3 + 2y + 4y2)2 then expect 1368 + 1728 + 1512 + 432 If they have not yet combined like terms then this A mark can only be implied by a later correct answer or relevant correct combination of terms coeff of y3 is 5040 A1 Obtain 5040 [4] 7 6x2 + 6y + 6x d d y x – 6y d d y x = 0 M1 1.1a Attempt implicit differentiation Either of the two d d y x terms correct, allowing sign errors Condone 6x2dx + 6ydx + 6xdy – 6ydy B1 1.1a Use product rule correctly on middle term Both terms correct Must now be 6y + 6x d d y x , or implied in a correct expression for d d y x H240/01 Mark Scheme June 2022 17 Question Answer Mark s AO Guidance A1 1.1 Obtain correct derivative on LHS Condone missing or incorrect RHS Must now have d d y x and not just dy or dx in terms 6x2 + 6y + 6x – 6y = 0 M1 3.1a Use d d y x = 1 in their equation Must now be equation, but RHS could be incorrect (eg ‘= 2’) x2 + x = 0 x = 0, x = –1 B1 1.1a Solve correct quadratic in x to obtain two correct roots (possibly BC) Quadratic must come from correct implicit differentiation B0 if x ‘cancelled’ in quadratic to give x = –1 as only root, but M1A1 still available x = 0 gives 3y2 = –2, but y2 has to be ≥ 0, so no solutions B1 2.3 Explicitly reject x = 0, with reasoning x = 0 must come from x2 + x = 0 eg negative numbers cannot be square rooted or y2 ≠ – 2 3 2 as y is real (just y2 ≠ – 2 3 is insufficient) Must be sensible reason and not just ‘math error’ or ‘not possible’ Could say that there are only imaginary (or not real) roots – condone ‘complex’ roots x = –1 gives 3y2 + 6y + 4 = 0 b2 – 4ac = 36 – 48 = –12 M1 2.1 Attempt to determine the number of real roots of their 3 term quadratic in y From substituting their x value into the equation of the curve Consider discriminant, or use quadratic formula, or attempt minimum value of function –12 < 0 hence no (real) roots A1 2.4 Obtain correct discriminant from correct quadratic and conclude appropriately x = –1 must come from x2 + x = 0 If using quadratic formula then it must be fully correct and attention drawn to why there are no real roots [8] H240/01 Mark Scheme June 2022 18 Question Answer Mark s AO Guidance 8 (a) 20 (minutes) B1 3.3 Obtain t = 20 Allow [19.9, 20.1] from setting up and using exponential model 97.2 (grams) B1 3.4 Obtain m = 97.2 Allow [97.1, 97.3] from setting up and using exponential model [2] 8 (b) (i) 160e –0.055t = 80 B1 3.4 Equate given model to 80 soi, so could be e–0.055t = 0.5 H240/01 Mark Scheme June 2022 19 Question Answer Mark s AO Guidance e –0.055t = 0.5 –0.055t = ln0.5 M1 3.4 Attempt correct process to find value of t, as far as dealing with exponential term Rearrange to e –0.055t = k, and hence obtain –0.055t = lnk If introducing logs straight away then need to get as far as ln160 – 0.055t = ln(their 80) t = 12.6 (minutes) A1 1.1 Obtain t = 12.6, or better If more sig fig given, then allow answers which round to 12.60 (the more accurate answer is 12.602676..) [3] 8 (b) (ii) d d m t = –8.8e–0.055t B1 3.4 Correct derivative soi Allow unsimplified No need to see d d m t = –8.8e–0.055 × 15 M1 3.4 Substitute t = 15 into their derivative Must be of the form ke-0.055t, with k ≠ 160 Possibly still with k unsimplified Substitution sufficient, no need to evaluate for M1 = –3.86, hence rate of decay is 3.86 grams/minute A1 1.1 Units required, and positive answer Must follow correct derivative ie negative coefficient No need to see –3.86 first, but A0 if clear error Accept 3.9 grams/minute Accept g/m for grams/minute [3] H240/01 Mark Scheme June 2022 20 Question Answer Mark s AO Guidance 8 (c) For A, d d m t = –63.9e–0.0511t Rate of decrease at t = 15 is 29.7 g/min hence A decaying at a faster rate B1 3.4 State A, with clear comparison Insufficient to just say that A has a greater initial mass – needs to consider decay factor as well Allow solutions that identify that B is decaying faster, with supporting evidence eg after 10 minutes, B’s mass is 92.3g which is 58% of initial mass whereas A is 60% of initial mass so B decaying faster eg A’s half-life is 13.6 so B is decaying faster eg change initial mass in model B to 1250 then when t = 10 B’s mass would be 721g which is less than 750 hence decaying faster eg compare coefficients of t (for A, coeff is – 0.0511); B’s is of a greater magnitude hence decaying faster For either solution, the conclusion and the supporting evidence must be consistent Numerical supporting evidence must be correct, allowing for slight inaccuracies from using different numbers of sig fig (see appendix) H240/01 Mark Scheme June 2022 21 Question Answer Mark s AO Guidance [1] 9 dx = 2cosθdθ M1 1.1a Attempt to link dx and dθ Allow sign error only 2 2 4 d 4 4sin .2cos d x x   − = −   M1 3.1a Attempt to write integrand in terms of θ Must substitute for both function and dx Can follow M0 but do not allow just dx = dθ 2 4cos .2cos d   = 2 4cos d  =  A1 1.1 Obtain correct integrand in terms of cosθ only Condone no dθ, as long as previously seen = ( ) 2cos2 2 d   +  M1 2.1 Attempt use of double angle formula Using cos2θ = ± 2cos2θ ± 1 Integrand must be of form k cos2θ , which must have come from correct method with coefficient errors only = sin2θ + 2θ A1FT 1.1 Integrate to obtain sin2θ + 2θ FT on acos2θ + b only   ( ) ( ) 1 3 1 6 π 2 2 2 2 3 3 6 6 π sin2 2 sin π+ π sin π+ π   + = − = ( ) ( ) 1 2 1 1 2 3 2 3 3 π 3 π + − + M1 2.1 Attempt use of limits Must be correct limits (either x or θ, as long as consistent with their integral), correct order and subtraction Allow M1 for use of limits in any integration attempt in terms of θ Allow M1 for either expressions that still involve sin, or exact equivs M0 for decimal values, even if then stated to be the same as 1 3 π Condone eg 1 2 3 from sin120o, but M0 if degrees used in linear term = 1 3 π A.G. A1 2.1 Obtain given answer of 1 3 π Must see both surd values, or an explanation as to why 2 2 3 6 sin π =sin π [7] H240/01 Mark Scheme June 2022 22 Question Answer Mark s AO Guidance 10 (a) area OMB = ( ) 1 1 2 2 sin r r  B1 1.1 Correct (possibly unsimplified) area of OMB Could use other than r for the radius Could set their variable equal to OM, giving a radius that is double this eg OM = x so area = x2sinθ ( ) ( ) 2 2 2 1 1 1 2 4 4 2 sin 3 sin r r r    − = OR ( ) ( ) 2 2 1 1 2 4 2 5 sin r r   = OR ( ) ( ) 2 2 2 1 1 1 2 2 4 3 5 sin r r r    = − M1 3.1a Attempt to use ratio on two correct areas Using two of OMB ( 2 1 4 sin r ), MAB ( ) 2 2 1 1 2 4 sin r r   − and OAB ( 2 1 2 r ) oe with their variable Must be two correct areas Must be using the correct ratio for their two areas ie 2:3 if using OMB and MAB, 2:5 if using OMB and OAB or 3:5 if using MAB and OAB Allow ratio to be used the wrong way around eg 2OMB = 3MAB A1 2.1 Correct equation, in two variables (ie θ and their r) Any correct statement linking the two areas Could use other than r for the radius Or 2x2θ – x2sinθ 3 1 2 4 sin sin    − = θ = 1.25sinθ A.G. A1 2.1 Simplify to given answer At least one line of working once ratio used [4] H240/01 Mark Scheme June 2022 23 Question Answer Mark s AO Guidance 10 (b) 0.599 B1 1.1a Obtain correct first iterate 3sf or better – more accurate answer is 0.599281923... Condone truncating if more sig fig given 0.705, 0.810 M1 1.1a Attempt correct iterative process to find at least 2 more values M1 is for the correct process for finding θ3 and θ4, but these may be incorrect M0 if working in degrees root = 1.13 A1 1.1 Obtain 1.13 Possibly following B0 if first iterate is wrong but process then self corrects Must follow M1 ie a clear attempt to use the correct iterative process Must be 3sf Once M1 is awarded, allow A1 for 1.13 even if an incorrect iterate seen, as process will recover [3] 10 (c) B1* 3.1a Draw y = θ on diagram Draw straight line, starting at the origin which intersects the graph Allow point of intersection to be greater than θ = 1 2 π Ignore incorrect labels, such as y = x H240/01 Mark Scheme June 2022 24 Question Answer Mark s AO Guidance B1 dep* 2.1 Draw correct iterative process on diagram Vertically into the curve, then horizontally into the straight line, as far as the root Initial value should be before root Needs point of intersection to be before θ = 1 2 π B1 1.2 State ‘staircase’ convergence Mark independently from other parts of question, including an incorrect diagram, as staircase can be deduced from the iterates in (b) [3] 10 (d) B1* 3.1a Draw graph of y = sin-10.8θ, for θ ≥ 0 Just need correct shape for y = sin-1kθ graph – a one to one function that starts at the origin (ignore any θ < 0) and has increasing gradient for all θ B1 dep* 3.2a Draw y = θ, and show staircase divergence from the root found in (b), on at least one side of the root Straight line from the origin to intersect their graph Diagram is sufficient for B1 – no comment or explanation required [2] H240/01 Mark Scheme June 2022 25 Question Answer Mark s AO Guidance 11 (a) 3 2 e d 3 ln d y y x x x =   M1 3.1a Separate variables and attempt integration of at least one side Allow ke3y , with k ≠ 1, as ‘attempt’ at integration of LHS ‘Attempt’ at RHS may not be use of integration by parts Allow BOD on missing integral sign / missing dy / missing dx as long as intention clear 3 3 1 3 e d e y y y =  B1 1.1 Correct LHS B0 if still part of an expression also involving x 2 3 2 3 ln d ln d x x x x x x x = −   M1 3.1a Attempt integration by parts on RHS – must have correct parts As far as attempt at 3 2 ln d x x x x − , possibly with 3 1 d x x x not yet simplified 3 3 1 3 ln x x x c = − + A1 1.1 Correct RHS (condone no + c) Condone no modulus sign on lnx 3 3 3 1 1 3 3 e e lne e c = − + so c = 3 1 3e − M1 1.1a Attempt use of (e, 1) to find c Used in an equation involving x, y and c, following some integration attempt of both sides As far as finding c, either exact or as a decimal M1 can be implied by sight of 3 1 3e − or –6.695… following a correct equation 3 3 3 3 1 1 1 3 3 3 e ln e y x x x = − − 3 3 3 3 e 3 ln e y x x x = − − A1 1.1 Obtain correct equation, in required form Any equivalent form on the RHS, but must be e3y = ... A0 if decimal approximation for e3 H240/01 Mark Scheme June 2022 26 Question Answer Mark s AO Guidance [6] 11 (b) e3y = 3e6lne2 – e6 – e3 = 6e6 – e6 – e3 = 5e6 – e3 M1* 2.1 Substitute x = e2, into their integral involving lnx, and attempt to simplify lnx may be lnxp if any coefficient has been taken into the ln term As far as correctly simplifying the ln term to remove ln Must be working exactly, so M0 if decimals seen before ln dealt with 3y = ln(e3(5e3 – 1)) = 3 + ln(5e3 – 1) M1 dep* 2.1 Introduce logs correctly, and attempt to rearrange to given form Their equation must have two terms, or possibly more, with the terms having a common factor of ek Attempt must go as far as splitting into the sum of two terms, with lnek simplified to k y = 1 + 1 3 ln(5e3 – 1) A1 2.1 Obtain y = 1 + 1 3 ln(5e3 – 1) No need to state a, b and c explicitly [3] H240/01 Mark Scheme June 2022 27 Question Answer Mark s AO Guidance 12 (a) 2 d 1 d x t t − = , d 2 d y t = d d d d d d y t x t y x = M1 1.1a Attempt correct process to find gradient in terms of t or p Correctly combine attempts at two derivatives Need 2 d d x kt t − = and d 2 d y t = SC B1 for gradient of –2x-2 if it is never seen in terms of t or p 2 d 2 d y t x = − A1 2.1 Obtain correct gradient In terms of t or p ( ) 2 1 2 2 p y p p x − = − − M1 1.1a Attempt equation of tangent Condone still working in terms of t Allow mixture of t and p as long as convincingly recovered Using their gradient from a differentiation attempt, but not dependent on first M1 Substitution into y – y1 = m(x – x1) or equation involving c from y = mx + c 2 2 4 y p x p = − + A.G. A1 2.1 Obtain given answer Must now be in terms of p Expand brackets and simplify to given answer, or find c and substitute back into equation [4] H240/01 Mark Scheme June 2022 28 Question Answer Mark s AO Guidance 12 (b) 2 1 2 ' p m = B1FT 1.1a Correct (unsimplified) gradient of normal, following their derivative Gradient in terms of t or p, but not x Could either FT on their incorrect derivative or deduce the gradient from the equation given in (a) ( ) 2 1 1 2 2 p p y p x − = − 2 3 1 1 2 2 2 p p y x p = + − M1 1.1 Attempt equation of normal Attempt to use their gradient and P Allow mixture of t and p as long as convincingly recovered Substitution into y – y1 = m(x – x1) or equation involving c from y = mx + c M1 3.1a Use y = 0 to attempt x-coordinate of B Using their attempt at normal equation As far as finding an expression for x at B, y = 0 so x = ( ) 3 2 3 1 1 2 2 2 4 p p p p p − = − A1 2.1 Correct x-coordinate for B Any equivalent form at A, y = 0 so x = 2 4 2 2 p p p = B1 2.1 Correct x-coordinate for A Any equivalent form PA = ( ) ( ) 2 2 1 2 p p + PB = ( ) ( ) 2 2 3 4 2 p p + M1 3.1a Attempt length of PA or PB Or M1 for attempting one of (PA)2 or (PB)2 Must correct distance formula Using the given P, and their coordinates for A and/or B, which must involve a function of p A1 2.1 Correct PA and PB Or correct (PA)2 and (PB)2 PA : PB = 4 1 4 1 p p + : 4 2 4 1 p p + = 1 : 2 p p = 1 : 2p2 A.G. A1 2.1 Simplify ratio to obtain given answer Must show clear method, such as same expression in each square root before cancelling Could also consider fraction and then cancel to deduce given ratio Could simplify (PA)2 : (PB)2, and then square root to obtain ratio [8] H240/01 Mark Scheme June 2022 29 APPENDIX Supporting Evidence for Q8(c) Need to get in touch? If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in touch with our customer support centre. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2022 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. Registered company number 3484466. OCR is an exempt charity. OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals. OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these resources. Though we make every effort to check our resources, there may be contradictions between published support and the specification, so it is important that you always use information in the latest specification. We indicate any specification changes within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy between the specification and a resource, please contact us. Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more information using our Expression of Interest form. Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.