H640/02 Mark Scheme June 2022 3 2. Subject-specific Marking Instructions for AS Level Mathematics B (MEI) 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. H640/02 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. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. 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. H640/02 Mark Scheme June 2022 5 f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) 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 2 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 A the rubric specifies 3 s.f. as standard, so this statement reads “3 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. E marks are lost unless, by chance, the given results are established by equivalent working. 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” and “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. H640/02 Mark Scheme June 2022 6 Question Answer Marks AO Guidance 1 𝑅2 = 12 + √3 2 M1 1.1 may be implied by correct answer tan𝛼= √3 1 or sin𝛼= √3 2 or cos𝛼= 1 2 soi M1 1.1 may see eg 𝛼= tan−1 (√3 1 ) may be implied by correct answer R = 2 or 𝛼= 𝜋 3 or 𝛼= 60 ֠ seen A1 1.1 2cos(θ ‒ 𝜋 3 ) or 2cos(θ ‒ 60 ֠ ) isw A1 1.1 [4] Alternatively cos𝜃+ √3sin𝜃= 𝑅cos𝜃cos𝛼+ 𝑅sin𝜃sin𝛼 so 1 = 𝑅cos𝛼 and √3 = 𝑅sin𝛼 M1 for equating coefficients 1 cos𝛼 = √3 sin𝛼 M1 for eliminating R 𝛼= 𝜋 3 or 𝛼= 60 ֠ seen A1 2cos(θ ‒ 𝜋 3 ) or 2cos(θ ‒ 60 ֠ ) isw A1 2 50 1−0.5 soi M1 1.1 100 A1 1.1 [2] H640/02 Mark Scheme June 2022 7 Question Answer Marks AO Guidance 3 (a) M1 B1 A1 1.1 1.1 1.1 decreasing concave up curve in 1st and 2nd quadrants which does not cut the x-axis; mark intent decreasing curve with intercept (0,3); may be in one quadrant only smooth curve from (‒0.5, a) through (2.5, b), where 4.5 ≤ a ≤ 5 and 0 < b < 0.5 [3] 3 (b) log (3 × 0.4𝑥) = log (0.8) oe M1 3.1a taking logarithms in any base 𝑥log0.4 = log0.8 −log3 oe M1 1.1 3rd law of logs used correctly 1.44 cao A1 1.1 if M0M0 allow SC1 for 1.44 unsupported Alternatively 0.4𝑥= 0.8 3 M1 𝑥= log0.4 ( 0.8 3 ) M1 may see 𝑥log0.4 = log ( 0.8 3 ) oe x = 1.44 cao A1 [3] H640/02 Mark Scheme June 2022 8 Question Answer Marks AO Guidance 4 (a) 0.22 J S 0.31 0.23 M1 A1 1.1 1.1 Venn diagram with 2 overlapping regions and 0.22 correctly placed; condone incorrect or no labelling all probabilities or percentages correctly placed and correctly labelled; ignore values in intersection allow if no box drawn if labels are eg A and B¸ A and B need to be defined [2] 4 (b) 0.31 + 0.23 + p(J∩S) = 1 ‒ 0.22 oe M1 1.1a may be implied by correct answer or by 24% 0.24 oe isw A1 1.1 do not allow 24% [2] 5 2𝑛−1 correctly evaluated for any odd positive integer B1 1.1 n ≥ 3 B0 if only rounded number in standard form seen 2𝑛−1 correctly evaluated for any odd positive integer for which Tom’s conjecture is false B1 2.1 eg 29 −1 = 511, eg 215 −1 = 32767 eg 221 −1 = 2097151 eg 511 is divisible by 7 with 9 seen [so not prime] B1 2.2a NB 32767 and 2097151 both divisible by 7; 2047 divisible by 23 correct value of n may be embedded in formula NB B0 if answer spoiled by eg so 511 is prime [3] H640/02 Mark Scheme June 2022 9 Question Answer Marks AO Guidance 6 M1 A1 3.1a 1.1 area identified which is symmetrical about the mean 𝑥= 𝜇± 𝜎 at points of inflection [2] 7 0.4 × 20 + 1.3 × 10 + 3.6 × 5 + 2 × 10 + 0.8 × 15 M1 1.1 allow one incorrect frequency density and/or one incorrect class width NB 8 + 13 + 18 + 20 +12 soi with four of five correct implies M1 may be implied by 71 71 A1 1.1 if M0 allow SC1 for 8, 13, 18, 20, 12 and no others seen [2] 8 (a) population since all distances of at least 120 km are used oe B1 1.2 [1] 8 (b) 161-163 km B1 1.1 [1] 8 (c) we need to see two elements: Ali’s complaint is justified oe and correct numerical reasoning with reference to upper tail B1 2.2a eg 10 riders rode more than 160 km or 14 rode 156 km or more or the reserve should have ridden (approximately) 161 km so (Ali’s complaint is) justified oe [1] H640/02 Mark Scheme June 2022 10 Question Answer Marks AO Guidance 9 (a) mean 112.4 isw or 112 isw B1 1.1 variance 8.8 or √8.82 cao isw B1 1.1 B0 for 8.757 explicitly rounded to 8.8 [2] 9 (b) N(their 112.4, their 8.8) M1 3.3 allow M1 for 8.8² or √8.8 N(a, b) A1 1.1 a = 112.4 or 112 and b = 8.8 or 2.972 [2] 9 (c) P(mark < 104.5) or P(mark < 105) found from their distribution in part (b) M1 3.4 may see N(−∞, 104.5,112.4, √8.8) NB 0.00387 or 0.0063(06) implies M1 NB 0.00573 or 0.00914 implies M1 NB 0.00379(69..) or 0.00619(81…) may imply M1 FT use of variance = 8.757 NB 0.200(199…) and 0.184(665…) may imply M1 FT use of sd = 8.8 if probability is correctly found to be 0 eg from use of N(112.4, 8.8 205) allow M1 only – no further marks available 205 × their non-zero 0.00387 M1 3.1a or compare 1 205 (≈0.00488) with their non-zero 0.00387 0.79 to 0.794 or 1.17 to 1.175 so consistent oe A1 3.5a or probabilities similar so consistent oe [3] H640/02 Mark Scheme June 2022 11 Question Answer Marks AO Guidance Alternatively InvNorm( 1 205 , 112.4, √8.8) or InvNorm( 1 205 , 112, √8.8) used to find their mark M1 FT their distribution compares their mark with 105 104.7 or 104.3 is close to 105 so good fit M1 A1 9 (d) P(mark between 114.5 and 115.5) found M1 3.4 NB awrt 0.0915 or awrt 0.0807 implies M1 18.75 to 18.77 so allow 18 or 19 or 16.5 to 16.534 so allow 16 or 17 A1 3.5a unsupported answers score M0 [2] H640/02 Mark Scheme June 2022 12 Question Answer Marks AO Guidance 10 (a) (𝑥−2) = 5cosθ and (𝑦−1) = 5sinθ M1 3.1a allow sign errors (𝑥−2)2 + (𝑦−1)2 = (5cos𝜃)2 + (5sin𝜃)2 oe M1 1.1 or ( 𝒙−𝟐 𝟓) 𝟐 + ( 𝒚−𝟏 𝟓) 𝟐 = cos2𝜃+ sin2𝜃 oe (𝑥−2)2 + (𝑦−1)2 = 52 oe isw or (𝑥−2) 5² 2 + (𝑦−1) 5² 2 = 1 oe isw A1 1.1 may see eg 𝑥2 −4𝑥+ 𝑦2 −2𝑦= 20 if M0M0 allow SC1 for 𝑦= 1 + 5sin (cos−1 ( 𝑥−2 5 )) or 𝑥= 2 + 5cos (sin−1 ( 𝑦−1 5 )) [3] Alternatively 𝑥2 = (2 + 5cos𝜃)2and 𝑦2 = (1 + 5sin𝜃)2 𝑥2 + 𝑦2 = 5 + 20cos𝜃+ 10sin𝜃+ 25sin2𝜃+ 25cos2𝜃 𝑥2 + 𝑦2 = 20 + 4𝑥+ 2𝑦 oe isw M1 M1 A1 if only seen in expanded form, allow one coefficient error; allow sign errors must have terms in cos𝜃 and sin𝜃 Alternatively radius = 5 and centre is (2, 1) (𝑥−2)2 + (𝑦−1)2 = 52 M1 M1 A1 allow sign error in coordinates of centre FT their centre all correct H640/02 Mark Scheme June 2022 13 Question Answer Marks AO Guidance 10 (b) gradient of radius is −4 3 B1 3.1a gradient of tangent is 3 4 M1 2.1 FT 1 ÷ 𝑡ℎ𝑒𝑖𝑟− 4 3 (𝑦−−3) = 3 4 (𝑥−5) oe M1 2.4 allow one sign error; FT their 3 4 may see −3 = 3 4 × 5 + 𝑐 3𝑥−4𝑦−27 = 0 𝑜𝑟−3𝑥+ 4𝑦+ 27 = 0 A1 1.1 [4] Alternatively d𝑦 d𝑥= 5cos𝜃 −5sin𝜃 oe B1 or d𝑦 d𝑥= 2−𝑥 𝑦−1 oe eg 2(𝑥−2) + 2(𝑦−1) d𝑦 d𝑥= 0 substitution of cos𝜃= 3 5 and sin𝜃= − 4 5 oe or (5,‒3) in their d𝑦 d𝑥 M1 d𝑦 d𝑥= 3 5 ⁄ −(−4 5 ⁄ ) 𝐨𝐫 2−5 −3−1 oe; allow one sign error; (𝑦−−3) = 3 4 (𝑥−5) oe M1 allow one sign error; FT their 3 4 may see −3 = 3 4 × 5 + 𝑐 3𝑥−4𝑦−27 = 0 𝑜𝑟−3𝑥+ 4𝑦+ 27 = 0 A1 [4] H640/02 Mark Scheme June 2022 14 Question Answer Marks AO Guidance 11 (a) Nina’s, because hers is the largest sample size oe B1 2.2a allow eg Nina’s, because with a larger sample size the probabilities get closer to the true probabilities oe [1] 11 (b) 11p + kp =1 M1 3.1a p = 1 11+𝑘 A1 1.1 [2] 11 (c) 𝑡ℎ𝑒𝑖𝑟 1 11+𝑘× 𝑘 or 𝑡ℎ𝑒𝑖𝑟 1 11+𝑘× 120 M1 2.1 multiply by k or by 120; may be embedded 120 × 𝑡ℎ𝑒𝑖𝑟 𝑘 11+𝑘 M1 1.2 multiplying by both k and 120 120𝑘 11+𝑘 oe A1 1.1 [3] 11 (d) 32 = their 120𝑘 11+𝑘 oe M1 1.1 k = 4 A1 1.1 [2] H640/02 Mark Scheme June 2022 15 Question Answer Marks AO Guidance Alternatively 11𝑝= 1 − 32 120 may be implied by 𝑝= 1 15 (from (𝑃(𝑋≠12)) k = 4 M1 A1 or 𝑘𝑓 120 = 32 120 (from 11f = 120 ‒ 32 = 88 so f = 8 and so 𝑘𝑝= ⋯) k = 4 11 (e) 𝑌~B (30, 𝑡ℎ𝑒𝑖𝑟 4 11+4) or 𝑌~B (30, 32 120) used to find P(Y = 8) M1 3.1a Y is the number of 12s obtained in 30 rolls; 0.16 – 0.163 BC A1 1.1 allow B2 for 0.1628 – 0.163 unsupported [2] 12 (a) H0 : μ = 1.5 H1 : μ < 1.5 B1 1.1 both hypotheses in terms of μ μ is the population mean weight of flour in a bag B1 2.5 allow μ is the population mean weight of a bag of flour [2] H640/02 Mark Scheme June 2022 16 Question Answer Marks AO Guidance 12 (b) N(𝑎, 𝑏) M1 3.3 a and b are numerical values a = 1.5 or b = 0.242 32 or 0.0018 A1 2.2a N(1.5, 0.242 32 ) isw or N(1.5,0.0018) isw A1 3.1a allow 0.0424² for variance A0 if answer spoiled by wrong variable quoted eg 𝜇~N(1.5, 0.242 32 ) or 𝑋~N(1.5, 0.242 32 ); allow only 𝑋̅ oe if variable included [3] 12 (c) 0.0786 > 0.05 or ‒1.4142 > ‒1.645 M1 3.4 or 1.44 > 1.43(02586 … ) NB 1.43(02586…) is from InvNorm(0.05, 1.5,0.0424) do not reject H0 A1 1.1 allow accept H0 or not significant or reject H1 there is insufficient evidence at the 5% level to suggest that the mean weight of the flour in the bags is less than 1.5 kg A1 2.2b do not allow eg conclude / prove / indicate or other assertive statement instead of suggest if calculated values are used full marks may be awarded for awrt 0.07865 or 0.0786, or ‒ 1.415 ≤ z ≤ ‒ 1.414 ; otherwise award a maximum of M1A1 for 0.07…or ‒1.4… other calculated values score M0 [3] H640/02 Mark Scheme June 2022 17 Question Answer Marks AO Guidance 13 (a) it can’t be fully justified because eg different samples may lead to different conclusions oe eg the proportion could be 0.35 and 61/140 may have arisen by chance oe eg the sample may not be representative oe eg the researcher used a sample not a population oe B1 2.4 do not allow eg the sample is too small eg the sample is too small to be representative [1] H640/02 Mark Scheme June 2022 18 Question Answer Marks AO Guidance 13 (b) H0 : p = 0.35 H1 : p > 0.35 B1 1.1 allow equivalent in words; do not allow percentages p is the probability that a baby (selected at random) is born without wisdom teeth B1 2.5 or p is the proportion of babies that are born without wisdom teeth B1B1 if other symbol instead of p used if correctly defined P(X ≥ k) found using B(140, 0.35), where k = 60, 61 or 62 M1* 3.3 or critical region is X ≥ k found from calculation of probability; allow k = 58, 59 or 60 NB P(X ≥ 60) = 0.03272 – 0.033 or P(X ≥ 62) = 0.01438 – 0.015 NB 0.967…, 0.978… and 0.985…imply M1 P(X ≥ 61) = 0.02197 – 0.022 A1 1.1 or critical region is X ≥ 59 from 0.0475…. or 0.048 their 0.022 correctly compared with 0.05 M1dep* 3.4 or 61 correctly compared with their 59; allow their 0.978 correctly compared with 0.95 do not accept H0 or reject H0 or accept H1 or significant A1FT 1.1 A0 if their 0.022 > 0.05 or 61 < their 59 sufficient evidence at the 5% level to suggest that the probability that a baby is born without wisdom teeth is more than 0.35 A1 1.1 dependent on award of all other marks apart from second B1 do not allow eg conclude / prove / indicate or other assertive statement instead of suggest [7] H640/02 Mark Scheme June 2022 19 Question Answer Marks AO Guidance 14 (a) 0.2 × {0.96154 + 0.86207 + 0.73529 + 0.60976 + 0.5} soi M1 2.1 allow M1A1 for calculation of exact values using formula in parts (a) and (b) 0.73373… ≈ 0.7337 AG A1 2.4 need to see 0.73373… as well as 0.7337 for A1 [2] 14 (b) 0.2 × {1 + 0.96154 + 0.86207 + 0.73529 + 0.60976} M1 1.1 or (3.66866 - 0.5 + 1) × 0.2 0.8337 correct to 4 dp A1 1.1 [2] 14 (c) 0.1 B1 1.1 FT their 0.8337(32) ‒ 0.7337(32), dependent on award of M1 in part (b) [1] 14 (d) 0.79162 ‒ 0.77912 M1 3.1a if M0 allow SC1 for correct interval identified eg 0.77912 to 0.79162 0.0125 A1 2.4 [2] 14 (e) increase n oe use rectangles of smaller width oe B1 2.2a do not allow eg reduce interval eg just ‘smaller’ rectangles – need to specify width reduction [1] 15 (a) 51.635 or 51.64 or 51.6 B1 3.4 [1] H640/02 Mark Scheme June 2022 20 Question Answer Marks AO Guidance 15 (b) 1995 estimate (probably) reliable since it is interpolation B1 2.2b allow eg the first estimate.. 2025 estimate (probably) not reliable since it is extrapolation B1 2.2b allow eg the second estimate… [2] 15 (c) No, because trends in life expectancy at birth may vary considerably between nations B1 2.4 LDS advantage [1] 15 (d) series 2 (the top one) is Italy – life expectancy (generally) higher in Europe (than Africa) B1 2.4 LDS advantage the values are decreasing (from 1990) in South Africa (– unusual since most show an upward trend) or little (or no) overall increase in South Africa (since 1970) or South Africa has lower life expectancy (than most developed countries) B1 2.4 LDS advantage [2] H640/02 Mark Scheme June 2022 21 Question Answer Marks AO Guidance 15 (e) B1 1.1 Point at (700, 47.56) ringed LDS advantage [1] 15 (f) the diagram supports this statement for values of GDP per capita from k to n where 0 < k ≤ 20 000 and 40 000 ≤ n ≤ 60 000 since there appears to be positive correlation oe B1 2.3 must give specific range of values ; must say supports statement oe for values of GDP per capita ≥ K where 40 000 ≤ K ≤ 60 000 there appears to be no association between GDP per capita and life expectancy at birth so the diagram does not support Sundip’s statement for these values B1 2.2b the range may be implied by reference to a specific range identified for the first mark; must say does not support statement oe [2] 40.00 50.00 60.00 70.00 80.00 90.00 0 50000 100000 150000 Life expectancy at birth in 2010 GDP per capita in US $ Scatter diagram of life expectancy at birth in 2010 against GDP per capita in US $ H640/02 Mark Scheme June 2022 22 Question Answer Marks AO Guidance 16 (a) 𝑑𝑦 𝑑𝑥= 24𝑥3 + 24𝑥2 −42𝑥+ 12 M1 3.1a allow one sign or coefficient error; must be four terms their 𝑑𝑦 𝑑𝑥= 0 M1 1.1 at least two terms correct f(k) evaluated, where k is a factor of ±12 or ± 𝑎 12, where a = 1,2,3,4 or 6 M1 2.1 may be implied by x = ‒2 seen unsupported or (𝑥+ 2) identified as factor (𝑥+ 2)(4𝑥2 −4𝑥+ 1) or (2𝑥−1)(2𝑥2 + 3𝑥−2) M1 3.1a by inspection or long division; allow one sign error or one coefficient error in trinomial may be implied by x = ½ seen unsupported or (2𝑥−1) oe identified as factor 𝑥= −2 and 𝑥= 1 2 and no others A1 1.1 may see 𝑥= 1 2 (repeated) A0 for 𝑥= −2 (repeated) ( 1 2 , − 31 8 ) and (−2 , −82) and no others A1 1.1 𝑑²𝑦 𝑑𝑥² = 72𝑥2 + 48𝑥−42 M1* 1.1 allow one sign or one coefficient error, FT their 𝑑𝑦 𝑑𝑥 ; allow M1 for 12𝑥2 + 8𝑥−7 𝑑²𝑦 𝑑𝑥² = 150 when x = ‒2 so minimum value or eg x ‒2.1 (‒2) ‒1.9 𝑑𝑦 𝑑𝑥 ‒16.224 (0) 13.824 A1 1.1 NB test indecisive at x = ½ A0 for just eg 𝑑²𝑦 𝑑𝑥² > 0 so minimum award M1A1 for consideration of gradient either side of ‒2, values must be correct to at least 2sf for A1 H640/02 Mark Scheme June 2022 23 Question Answer Marks AO Guidance eg dependent on at least two terms correct in derivative; must see values x 0 (½) 1 𝑑𝑦 𝑑𝑥 12 (0) 18 M1 3.1a or eg or eg x 0 (½) 1 y ‒6 (−3.875) ‒1 x 0 (½) 1 𝑑²𝑦 𝑑𝑥² ‒42 (0) 78 inflection at ( 1 2 , − 31 8 ) A1 3.2a values in table must be correct their 72𝑥2 + 48𝑥−42 = 0 M1dep* 1.1 x = − 7 6 isw A1 1.1 ignore calculation of associated y-value allow any correct decimals to 3 sf or more [12] 16 (b) M1 B1 A1 1.1 1.1 1.1 curve with a minimum in 3rd quadrant and stationary point of inflection in 4th quadrant and no other stationary points (0, ‒6) identified as y-intercept (intercept must be below the x-axis and above ‒20) correct curve with intercepts at (‒a,0) and (b,0), where ‒3 < a < ‒2.6 and 0.8 < b < 1.2; minimum at (‒2, y) where ‒90 < y < ‒80 and inflection for 0 < x < 1 and y is between the x-axis and the y-intercept [3] Need to get in touch? 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