Mark Scheme (Results) - Mathsaurus
Mark Scheme (Results)Summer 2018Pearson Edexcel GCE A Level MathematicsPure Mathematics Paper 2 (9MA0/02)
Edexcel and BTEC QualificationsEdexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body.We provide a wide range of qualifications including academic, vocational, occupational andspecific programmes for employers. For further information visit our qualifications websitesat www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us usingthe details on our contact us page at www.edexcel.com/contactus.Pearson: helping people progress, everywherePearson aspires to be the world’s leading learning company. Our aim is to help everyoneprogress in their lives through education. We believe in every kind of learning, for all kindsof people, wherever they are in the world. We’ve been involved in education for over 150years, and by working across 70 countries, in 100 languages, we have built an internationalreputation for our commitment to high standards and raising achievement throughinnovation in education. Find out more about how we can help you and your students at:www.pearson.com/ukSummer 2018Publications Code 9MA0 02 1806 MSAll the material in this publication is copyright Pearson Education Ltd 2018
General Marking Guidance All candidates must receive the same treatment. Examiners must mark thelast candidate in exactly the same way as they mark the first. Mark schemes should be applied positively. Candidates must be rewarded forwhat they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to theirperception of where the grade boundaries may lie. All the marks on the mark scheme are designed to be awarded. Examinersshould always award full marks if deserved, i.e. if the answer matches themark scheme. Examiners should also be prepared to award zero marks if thecandidate’s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principlesby which marks will be awarded and exemplification/indicative content willnot be exhaustive. When examiners are in doubt regarding the application of the mark schemeto a candidate’s response, a senior examiner must be consulted before a markis awarded. Crossed out work should be marked UNLESS the candidate has replaced itwith an alternative response.
PEARSON EDEXCEL GCE MATHEMATICSGeneral Instructions for Marking1. The total number of marks for the paper is 100.2. These mark schemes use the following types of marks: M marks: Method marks are awarded for ‘knowing a method and attempting to apply it’,unless otherwise indicated.A marks: Accuracy marks can only be awarded if the relevant method (M) marks havebeen earned.B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided. 3. AbbreviationsThese are some of the traditional marking abbreviations that will appear in the markschemes. bod – benefit of doubt ft – follow through the symbol cao – correct answer only cso - correct solution only. There must be no errors in this part of the question toobtain this mark isw – ignore subsequent working awrt – answers which round to SC: special case o.e. – or equivalent (and appropriate) d or dep – dependent indep – independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer givenwill be used for correct ft4. All M marks are follow through.A marks are ‘correct answer only’ (cao), unless shown, for example, as A1 ft toindicate that previous wrong working is to be followed through. After a misread,however, the subsequent A marks affected are treated as A ft, but answers that don’tlogically make sense e.g. if an answer given for a probability is 1 or 0, should neverbe awarded A marks.
5. For misreading which does not alter the character of a question or materially simplify it,deduct two from any A or B marks gained, in that part of the question affected.6. Where a candidate has made multiple responses and indicates which response they wishto submit, examiners should mark this response.If there are several attempts at a question which have not been crossed out, examinersshould mark the final answer which is the answer that is the most complete.7. Ignore wrong working or incorrect statements following a correct answer.8. Mark schemes will firstly show the solution judged to be the most common responseexpected from candidates. Where appropriate, alternative answers are provided in thenotes. If examiners are not sure if an answer is acceptable, they will check the markscheme to see if an alternative answer is given for the method used. If no suchalternative answer is provided but the response is deemed to be valid, examiners mustescalate the response for a senior examiner to review.
General Principles for Pure Mathematics Marking(But note that specific mark schemes may sometimes override these general principles)Method mark for solving 3 term quadratic:1. Factorisation( x2 bx c) ( x p)( x q), where pq c , leading to x .(ax2 bx c) (mx p)(nx q), where pq c and mn a , leading to x .2. FormulaAttempt to use the correct formula (with values for a, b and c)3. Completing the square2b Solving x bx c 0 : x q c 0, q 0 , leading to x .2 2Method marks for differentiation and integration:1. DifferentiationPower of at least one term decreased by 1. ( x xnn 1)2. IntegrationPower of at least one term increased by 1. ( x xnn 1)Use of a formulaWhere a method involves using a formula that has been learnt, the advicegiven in recent examiners’ reports is that the formula should be quotedfirst.Normal marking procedure is as follows:Method mark for quoting a correct formula and attempting to use it, even ifthere are small errors in the substitution of values.Where the formula is not quoted, the method mark can be gained byimplication from correct working with values but may be lost if there is anymistake in the working.Exact answersExaminers’ reports have emphasised that where, for example, an exactanswer is asked for, or working with surds is clearly required, marks willnormally be lost if the candidate resorts to using rounded decimals.Answers without workingThe rubric says that these may not gain full credit. Individual mark schemes willgive details of what happens in particular cases. General policy is that if it couldbe done “in your head”, detailed working would not be required. Most candidatesdo show working, but there are occasional awkward cases and if the markscheme does not cover this, please contact your team leader for advice.
QuestionSchemeg( x) 1(a)g(5) Way 1MarksAOsM11.1bA11.1b2x 5, x 5x 32(5) 52("7.5") 5 7.5 gg(5) 5 3"7.5" 3 40 4gg(5) or 4 or 4.4 9 9 (2)(a)Way 2 2x 5 2 5x 3 gg( x) gg(5) 2x 5 x 3 3 2(5) 5 2 5 (5) 3 2(5) 5 (5) 3 3 40 4or4or4.4 9 9 gg(5) M11.1bA11.1b(2)(b){Range:} 2 y 152B11.1b(1)(c)Way 12x 5y yx 3 y 2 x 5 yx 2 x 3 y 5x 33y 5 3x 5 x( y 2) 3 y 5 x or y y 2 x 2 g 1 ( x) 3x 5,x 22 x 152M11.1bM12.1A1ft2.5(3)(c)Way 22 x 6 111111 y 2 y 2 x 3x 3x 3111111 x 3 x 3 or y 3 y 2y 2x 2 y g 1 ( x) 11 3,x 22 x 152M11.1bM12.1A1ft2.5(3)(6 marks)Notes for Question 1(a)M1:Note:A1:Note:Full method of attempting g(5) and substituting the result into g 2x 5 2 5x 3 9x 5 Way 2: Attempts to substitute x 5 into, o.e. Note that gg( x) 14 x 2x 5 x 3 3 404Obtainsor 4 or 4.4 or an exact equivalent99 404Give A0 for 4.4 or 4.444. without reference toor 4 or 4.499
Notes for Question 1 Continued(b)151515 15 Accept any of 2 g , 2 g( x) , 2, 222 2 B1:States 2 y Note:Accept g( x) 2 and g( x) (c)Way 1M1:M1:A1ft:15o.e.2Correct method of cross multiplication followed by an attempt to collect terms in x orterms in a swapped yA complete method (i.e. as above and also factorising and dividing) to find the inverseUses correct notation to correctly define the inverse function g 1 , where the domain ofg 1 stated correctly or correctly followed through (using correct notation) on the values shown intheir range in part (b). Allow g 1 : x . Condone g 1 . Do not accept y .Note:Correct notation is required when stating the domain of g 1 ( x) . Allow 2 x 15 15 or 2, 2 2 1515, 2 g 1 ( x) 22Do not allow A1ft for following through their range in (b) to give a domain for g 1 as x Do not allow any of e.g. 2 g Note:(c)Way 2M1:M1:A1ft:Note:2x 5kin the form y 2 , k 0 and rearranges to isolate y and 2 on one sidex 3x 3of their equation. Note: Allow the equivalent method with x swapped with yA complete method to find the inverseAs in Way 1If a candidate scores no marks in part (c), but states the domain of g 1 correctly, orWrites y states a domain of g 1 which is correctly followed through on the values shown in theirrange in part (b)then give special case (SC) M1 M0 A0
QuestionSchemeMarksAOsM13.1aA11.1bOA 2i 3j 4k , OB 4i 2 j 3k , OC ai 5j 2k , a 02AB BD , AB 4E.g. OD OB BD OB AB(a)or OD OB BD OB AB OB OB OA 2OB OAor OD OB BD OB AB OA AB AB OA 2 AB 2 3 4 4 4 2 2 2 3 3 3 4 4 2 or 3 2 2 4 3 2 2 3 2 5 4 7 4 2 2 5 3 7 6 7 or 6i 7 j 10 k 10 (2)(a 2)2 (5 3)2 ( 2 4)2(b) AC 4 (a 2)2 (5 3)2 ( 2 4)2 (4)2 (a 2) 8 a . or2 a 4a 4 0 a .2(as a 0 ) a 2 2 2 (or a 2 8 )M11.1bdM12.1A11.1b(3)(5 marks)Notes for Question 2(a)M1:A1:Note:Complete applied strategy to find a vector expression for ODSee schemeGive M0 for subtracting the wrong way wrong to give e.g.(4i 2 j 3k ) (2i 3j 4k) (4i 2 j 3k ) (4i 2 j 3k) ( 2i 5j 7k) (2i 3j 4k)Note:Note:Note:Note:(b)M1:Writing e.g. OD OB AB or OD 2OB OA with no other work is M0Finding coordinates, i.e. (6, 7, 10) without reference to the correct position vectors is A0Allow M1A1 for writing down 6i 7 j 10 k with no workingM1 can be implied for at least two correct components in their position vector of DdM1:Complete method of correctly applying Pythagoras’ Theorem on AC 4 and using a correctNote:A1:Note:Note:Note:Finds the difference between OA and OC , then squares and adds each of the 3 componentsNote: Ignore labellingmethod of solving their resulting quadratic equation to find at least one of a .Condone at least one of either awrt 4.8 or awrt 0.83 for the dM markObtains only one exact value, a 2 2 2Writing a 2 2 2 , without evidence of rejecting a 2 2 2 is A04 32for A1, and isw can be applied2Writing a 0.828., without reference to a correct exact value is A0Allow exact alternatives such as 2 8 or
QuestionScheme3Statement: “If m and n are irrational numbers, where m n,then mn is also irrational.”(a)E.g. m 3 , n 12 mn 3 MarksAOsM11.1bA12.4 12 6 statement untrue or 6 is not irrational or 6 is rational(2)(b)(i),(ii) Way 1yV shaped graph {reasonably}symmetrical about the y-axiswith vertical interpret(0, 3) or 3 stated or markedon the positive y-axisSuperimposes thegraph of y x 3 on topOB11.1bM13.1aA12.4of the graph of y x 3xthe graph of y x 3 is either the same or above the graph ofy x 3 {for corresponding values of x}or when x 0, both graphs are equal (or the same)when x 0, the graph of y x 3 is above the graph of y x 3(3)(b)(ii)Way 2Reason 1When x 0, x 3 x 3Reason 2When x 0, x 3 x 3Any one of Reason 1 or Reason 2M13.1aBoth Reason 1 and Reason 2A12.4(5 marks)Notes for Question 3(a)M1:A1:Note:(b)(i)B1:(b)(ii)M1:States or uses any pair of different numbers that will disprove the statement.14E.g. 3 , 12 ; 2 , 8 ; 5 , 5 ; , 2 ; 3e, ;5e Uses correct reasoning to disprove the given statement, with a correct conclusion 4 12Writing 3e untrue is sufficient for M1A1 5e 5See schemeFor constructing a method of comparing x 3 with x 3 . See scheme.A1:Explains fully why x 3 x 3 . See scheme.Note:NoteNote:Do not allow either x 0, x 3 x 3 or x 0, x 3 x 3 as a valid reasonx 0 (or where necessary x 3) need to be considered in their solutions for A1Do not allow an incorrect statement such as x 0, x 3 x 3 for A1
Notes for Question 3 Continued(b)(ii)Note:Note:Allow M1A1 for x 0, x 3 x 3 and for x 0, x 3 x 3Allow M1 for any of x is positive, x 3 x 3 x is negative, x 3 x 3 x 0, x 3 x 3 x 0, x 3 x 3 x 0, x 3 and x 3 are equal x 0, x 3 and x 3 are equal when x 0, both graphs are equal for positive values x 3 and x 3 are the sameCondone for M1 x 0, x 3 x 3(b)(ii)Way 3 x 0, x 3 x 3 For x 0, x 3 x 3 For 3 x 0, as x 3 3 and {0 } x 3 3,M13.1aA12.4then x 3 x 3 For x 3, as x 3 x 3 and x 3 x 3,then x 3 x 3
QuestionScheme164(i) 3 5r 2 131 798 ;r(ii) u1 , u2 , u3 , ., : un 1 r 1(i)Way 1 AOsM13.1aM1M11.1b1.1bA1*2.112, u1 un3161616 r3 5r 2 3 5r 2r r 1 r 1 r 1162(216 1) (2(8) 15(5)) 22 1 Marks 728 131070 131798 *(4)(i)Way 2 3 5r 2r r 1 16 3 5r 2 1.1b1.1b2.132, u3 , . (can be implied by later working)23M11.1b 2 3 2 3 50 50 or 50 3 2 3 2 M12.2aA11.1br 1 (3 16) r 1r 1162(216 1)(2(5) 15(5)) 22 1 48 680 131070 131798 *(i)Way 3(ii)Sum 10 17 26 39 60 97 166 299 560 1077 2106 4159 8260 16457 32846 65619 131 798 *2 u1 , u2 3 100 ur r 1 325 11300650 or108or108.3oror 3 3126 (3)(7 marks)
Notes for Question 4(i)16M1: 3 5r 2 to be foundrUses a correct methodical strategy to enable the given sum,r 1Allow M1 for any of the following: expressing the given sum as either161616 3 5r 2r ,r 1 r 1r 1163 16 5r 2r r 1r 116 attempting to find both 16r 1r r 1r 116 5r and 2 separatelyrr 1r 1M1:A1*:Way 1: Correct method for finding the sum of an AP with a 8, d 5, n 16Way 2: (3 16) and a correct method for finding the sum of an APCorrect method for finding the sum of a GP with a 2, r 2, n 16For all steps fully shown (with correct formulae used) leading to 131798Note:Way 1: Give 2nd M1 for writing16 3 5r asr 1Note:Way 2: Give 2 M1 for writing16Note:Give 3rd M1 for writing r 11616r 1r 116(8 83)2 3 5r as 48 2 (5 80) or 48 680 2r as162(1 216 )or 2(216 1) or (217 2)1 2(i)Way 3M1:M1:M1:A1*:(ii)At least 6 correct terms and 16 terms shownAt least 10 correct terms (may not be 16 terms)At least 15 correct terms (may not be 16 terms)All 16 terms correct and an indication that the sum is 131798M1:For some indication that the next two terms of this sequence areM1:A1:Note:Note:Note:Note:Note:r 1r(3 16) and attempting to find bothndr 3 5r and 2 separately16M1:16 2 3 516r 1 16or3 2,2 3 2 3 2 3 For deducing that the sum can be found by applying 50 50 or 50 , o.e. 3 2 3 2 3251Obtainsor 108 or 108.3 or an exact equivalent3332Allow 1st M1 for u2 (or equivalent) and u3 (or equivalent)322 3 2Allow 1st M1 for the first 3 terms written as , , ,.3 2 33 2Allow 1st M1 for the 2nd and 3rd terms written as , ,. in the correct order2 32Condone written as 0.66 or awrt 0.67 for the 1st M1 mark3 3251Give A0 for 108.3 or 108.333. without reference toor 108 or 108.333
QuestionScheme5The equation 2 x3 x2 1 0 has exactly one real root f ( x) 2x(a)f ( xn ) xn 1 xn f ( x ) n 3 x 2 1 f ( x) 6 x 2 2 x x xn 1n 2 xn3 xn 2 16 xn 2 2 xnxn (6 xn 2 2 xn ) (2 xn 3 xn 2 1)6 xn 2 2 xn xn 1 4 xn 3 xn 2 16 xn 2 2 xn*MarksAOsB11.1bM11.1bA1*2.1(3)(b) x1 1 4(1)3 (1)2 12(1)3 (1) 2 1x2 orx 1 26(1)2 2(1)6(1)2 2(1)32 x2 , x3 43M11.1bA11.1b(2)(c)Accept any reasons why the Newton-Raphson method cannot be usedwith x1 0 which refer or allude to either the stationary point or thetangent. E.g. There is a stationary point at x 0 Tangent to the curve (or y 2 x3 x2 1) would not meet the x-axisB12.3 Tangent to the curve (or y 2 x3 x2 1) is horizontal(1)(6 marks)Notes for Question 5(a)B1:States that f ( x) 6 x2 2 x or states that f ( xn ) 6 xn 2 2 xn (CondoneM1:Substitutes f ( xn ) 2 xn3 xn 2 1 and their f ( xn ) into xn 1 xn A1*:Note:Note:NoteNotedy 6 x2 2 x )dxf ( xn )f ( xn )A correct intermediate step of making a common denominator which leads to the given answerf ( xn )Allow B1 if f ( x) 6 x2 2 x is applied as f ( xn ) (or f ( x)) in the NR formula xn 1 xn f ( xn )Allow M1A1 for4 xn3 xn 2 12 x3 x 2 1x(6 x 2 2 x) (2 x3 x 2 1) xn 1 xn 1 x 6x2 2x6x2 2x6 xn 2 2 xn2 x3 x 2 1for M1"6 x 2 2 x "2 xn3 xn 2 12 x3 x 2 1Condone xn or x (i.e. no xn 1 . ) for M12"6 xn 2 xn ""6 x 2 2 x "Condone x x Note:2 xn3 xn 2 1f ( xn )32Give M0 for xn 1 xn followed by xn 1 2 xn xn 1 6 xn 2 2 xnf ( xn )Note:Correct notation, i.e. xn 1 and xn must be seen in their final answer for A1*
Notes for Question 5 Continued(b)M1:An attempt to use the given or their formula once. Can be implied byNote:Allow one slip in substituting x1 132x2 and x3 433Condone x2 and x3 awrt 0.667 for A143 2Condone , listed in a correct order ignoring subscripts4 3A1:Note:Note:(c)B1:Note:See schemeGive B0 for the following isolated reasons: e.g. You cannot divide by 0 The fraction (or the NR formula) is undefined at x 0 At x 0, f ( x1 ) 0 x1 cannot be 0 6 x2 2 x cannot be 0the denominator is 0 which cannot happenif x1 0, 6 x2 2 x 04(1)3 (1)2 1or 0.75 o.e.6(1)2 2(1)
Question6(a)Scheme(a) f ( x) 3x3 8x2 9 x 10, x (i) {f (2) 24 32 18 10 } f (2) 0(ii) {f ( x) } ( x 2)( 3x2 2 x 5) or (2 x)(3x2 2 x 5)(b)MarksAOsB1M1A1(3)1.1b2.2a1.1bM12.4A12.1 3 y 6 8 y 4 9 y 2 10 0 ( y 2 2)( 3 y 4 2 y 2 5) 0Gives a partial explanation by explaining that 3 y 4 2 y 2 5 0 has no {real} solutions with areason, e.g. b2 4ac (2)2 4( 3)( 5) 56 0 or stating that y 2 2 has 2 {real} solutions or y 2 {only}Complete proof that the given equation has exactly two {real} solutions(2)(c)3tan 8tan 9tan 10 0; 7 10 32{Deduces that} there are 3 solutionsB1(1)2.2a(6 marks)Notes for Question 6(a)(i)B1:(a)(ii)M1:f (2) 0 or 0 stated by itself in part (a)(i)Deduces that ( x 2) or (2 x) is a factor and attempts to find the other quadratic factor by using long division to obtain either 3x2 kx ., k value 0 or 3x2 x , value 0, can be 0 factorising to obtain their quadratic factor in the form ( 3x2 k x c), k value 0,c can be 0, or in the form ( 3x2 x ), value 0, can be 0A1:(b)M1:A1:Note:Note:Note:Note:Note:Note:( x 2)( 3x2 2 x 5), (2 x)(3x2 2 x 5) or ( x 2)(3x2 2 x 5) stated together as a productSee schemeSee scheme. Proof must be correct with no errors, e.g. giving an incorrect discriminant valueCorrect calculation e.g. (2)2 4( 3)( 5), 4 60 or 56 must be given for the first explanationNote that M1 can be allowed for a correct follow through calculation for the discriminant of their “ 3 y 4 2 y 2 5 ”which would lead to a value 0 together with an explanation that 3 y 4 2 y 2 5 0 hasno {real} solutions or for the omiss
Pure Mathematics Paper 2 (9MA0/02) Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites
CSEC English A Specimen Papers and Mark Schemes: Paper 01 92 Mark Scheme 107 Paper 02 108 Mark Scheme 131 Paper 032 146 Mark Scheme 154 CSEC English B Specimen Papers and Mark Schemes: Paper 01 159 Mark Scheme 178 Paper 02 180 Mark Scheme 197 Paper 032 232 Mark Scheme 240 CSEC English A Subject Reports: January 2004 June 2004
Matthew 27 Matthew 28 Mark 1 Mark 2 Mark 3 Mark 4 Mark 5 Mark 6 Mark 7 Mark 8 Mark 9 Mark 10 Mark 11 Mark 12 Mark 13 Mark 14 Mark 15 Mark 16 Catch-up Day CORAMDEOBIBLE.CHURCH/TOGETHER PAGE 1 OF 1 MAY 16 . Proverbs 2—3 Psalms 13—15 Psalms 16—17 Psalm 18 Psalms 19—21 Psalms
Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. All marks on the mark scheme should be used appropriately. All marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme.
H567/03 Mark Scheme June 2018 6 USING THE MARK SCHEME Please study this Mark Scheme carefully. The Mark Scheme is an integral part of the process that begins with the setting of the question paper and ends with the awarding of grades. Question papers and Mark Schemes are d
the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.
Mark scheme for Paper 1 Tiers 3-5, 4-6, 5-7 and 6-8 National curriculum assessments ALL TIERS Ma KEY STAGE 3 2009. 2009 KS3 Mathematics test mark scheme: Paper 1 Introduction 2 Introduction This booklet contains the mark scheme for paper 1 at all tiers. The paper 2 mark scheme is printed in a separate booklet. Questions have been given names so that each one has a unique identifi er .
the mark scheme, the own figure rule can be used with the accuracy mark. Of Own Figure rule Accuracy marks can be awarded where the candidates’ answer does not match the mark scheme, though is accurate based on their valid method. cao Correct Answer Only rule Accuracy marks will only be awarded if the candidates’ answer is correct, and in line with the mark scheme. oe Or Equivalent rule .
Abrasive jet Machining consists of 1. Gas propulsion system 2. Abrasive feeder 3. Machining Chamber 4. AJM Nozzle 5. Abrasives Gas Propulsion System Supplies clean and dry air. Air, Nitrogen and carbon dioxide to propel the abrasive particles. Gas may be supplied either from a compressor or a cylinder. In case of a compressor, air filter cum drier should be used to avoid water or oil .
