Mark Scheme (Results) Summer 2019
Mark Scheme (Results)Summer 2019Pearson Edexcel GCEIn Mathematics (9MA0) Paper 2Pure Mathematics 2
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 2019Publications Code 9MA0 02 1906 MSAll the material in this publication is copyright Pearson Education Ltd 2019
General Marking Guidance All candidates must receive the same treatment. Examiners must mark the firstcandidate in exactly the same way as they mark the last.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.There is no ceiling on achievement. All marks on the mark scheme should beused appropriately.All the marks on the mark scheme are designed to be awarded. Examiners shouldalways award full marks if deserved, i.e. if the answer matches the markscheme. 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 principles bywhich marks will be awarded and exemplification may be limited.When examiners are in doubt regarding the application of the mark scheme to acandidate’s response, the team leader must be consulted.Crossed out work should be marked UNLESS the candidate has replaced it withan alternative response.
PEARSON EDEXCEL GCE MATHEMATICSGeneral Instructions for Marking1. The total number of marks for the paper is 100.2. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for ‘knowing a method and attempting to applyit’, 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 doubtft – follow throughthe symbolwill be used for correct ftcao – correct answer onlycso - correct solution only. There must be no errors in this part of the question toobtain this markisw – ignore subsequent workingawrt – answers which round toSC: special caseo.e. – or equivalent (and appropriate)dep – dependentindep – independentdp decimal placessf significant figures The answer is printed on the paperThe second mark is dependent on gaining the first mark4. 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.5. 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.6. Ignore wrong working or incorrect statements following a correct answer.7. 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.
General Principles for Further 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 square2Solvingbx2 bx c 0 : x q c 0, q 0 , leading to x . 2 Method marks for differentiation and integration:1. DifferentiationPower of at least one term decreased by 1. ( x n x n 1 )2. IntegrationPower of at least one term increased by 1. ( x n x n 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.
Assessment ObjectivesAssessment ObjectiveA01A02A03DefinitionUse and apply standard techniquesReason, interpret and communicate mathematicallySolve problems within mathematics and in other lect routine proceduresCorrectly carry out routine proceduresAccurately recall facts, terminology and definitionsConstruct rigorous mathematical arguments (including proofs)Make deductionsMake inferencesAssess the validity of mathematical argumentsExplain their reasoningUses mathematical language and notation correctlyTranslate problems in mathematical contexts into mathematical processesTranslate problems in non-mathematical contexts into mathematical processesInterpret solutions to problems in their original contextEvaluate (the) accuracy and limitations (of solutions to problems)Translate situations in context into mathematical modelsUse mathematical modelsEvaluate the outcomes of modelling in contextRecognise the limitations of modelsWhere appropriate, explain how to refine (models)
b 2 2 2 4 If 0 marks are scored on application of the mark scheme then allowSpecial Case B1 M0 A0 (total of 1 mark) for any of3 x 112x y 2xy 2x 4 y 2x 2 y 2 4 422x 2 22x 4 y 1 log2x log4 y x log2 y log4 or x log2 2 y log2 ln 2x ln 4 y x ln 2 y ln 4 or x ln 2 2 y ln 2 1 y log x o.e. {base of 4 omitted} 2 2 2 Way 12 x 22 y 2 32 323 y .2131E.g. y x or y (2 x 3)2442x 2 y 2 x 2y (3)Way 2 1 log( 2 x 4 y ) log 2 2 1 log 2 x log 4 y log 2 2 x log 2 y log 4 log1 log(2 2) y .y log(2 2) x log 2log 413 y x 24 B11.1bM12.1A11.1b(3)Way 3 1 log( 2 x 4 y ) log 2 2 1 1 xlog 2 x log 4 y log log 2 y log 4 log y . 2 2 2 2 1 xlog log (2 )2213 y y x log 424 B11.1bM12.1A11.1b(3)Way 4 1 log 2 ( 2 x 4 y ) log 2 2 2 1 3log 2 2 x log 2 4 y log 2 y . x 2 y 2 2 2 131E.g. y x or y (2 x 3)244B11.1bM12.1A11.1b(3)(3 marks)
QuestionScheme1Way 5x42 4y 41x y 34 3413x y y .24131E.g. y x or y (2 x 3)24442 4 1:A1:Note:Note:Note:Notes for Question 1Way 1Writes a correct equation in powers of 2 onlyComplete process of writing a correct equation in powers of 2 only and using correct index laws toobtain y written as a function of x.13y x o.e.24Way 2, Way 3 and Way 4Writes a correct equation involving logarithmsComplete process of writing a correct equation involving logarithms and using correct log laws toobtain y written as a function of x. 1 xlog log (2 )22 ln(2 2) x ln 2 log(2 2) x log 2 or y y or y ln 4log 4log 4131or y x or y (2 x 3) o.e.244Way 5Writes a correct equation in powers of 4 onlyComplete process of writing a correct equation in powers of 4 only and using correct index laws toobtain y written as a function of x.13y x o.e.24Allow equivalent results for A1 where y is written as a function of xYou can ignore subsequent working following on from a correct answer. 1 11Allow B1 for 2 x 4 y 4y x log 4 (4 y ) log 4 x 2 222 2 2 2 2 2 1 2 x followed by M1 A1 for y log 4 x or y log 4 or y log 4 x 4(2 ) 2 2 2 2 2 x 3 or y log 4 2 2 or y log 4 ( 2 (2 x 1 ))
Question2(a)SchemeTime (s)Speed (m s 1 2.42542Uses an allowable method to estimate the area under the curve. E.g.Way 1: an attempt at the trapezium rule (see below) 2 42 Way 2: { s } (25) { 550} 2 Way 3: 42 2 25(a) a 1.6 s 2(25) (0.5)(1.6)(25)2{ 550}Way 4: { d }(2)(5) 5(5) 10(5) 18(5) 28(5)Way 5: { d } 5(5) 10(5) 18(5) 28(5) 42(5) 63(5) 315 103(5) 515 315 515 415 2 2 5 10 18 28 42 Way 7: { d } (25) 437.5 6 11 (5) [2 2(5 10 18 28) 42] or [ "315" "515" ]22 415 {m}Way 6: { d }(b)Alt 1Uses a Way 1, Way 2, Way 3, Way 5, Way 6 or Way 7 method in (a)Overestimate and a relevant explanation e.g. {top of} trapezia lie above the curve Area of trapezia area under curve An appropriate diagram which gives reference to the extra area Curve is convexd2 y 0 dx 2 Acceleration is {continually} increasing The gradient of the curve is {continually} increasing All the rectangles are above the curve (Way 5)(1)(b)Alt 2Uses a Way 4 method in (a)Underestimate and a relevant explanation e.g. All the rectangles are below the curveB1ft2.4(1)(4 marks)Notes for Question 2(a)M1:A low-level problem-solving mark for using an allowable method to estimate the area under thecurve. E.g.Way 1: See scheme. Allow (2 2(5 10 18 28) 42); 0 for 1st M1 u v Way 2: Uses s t which is equivalent to finding the area of a large trapezium 2 Way 3: Complete method using a uniform acceleration equation.Way 4: Sums rectangles lying below the curve. Condone a slip on one of the speeds.Way 5: Sums rectangles lying above the curve. Condone a slip on one of the speeds.Way 6: Average the result of Way 3 and Way 4. Equivalent to Way 1.Way 7: Applies (average speed) (time)
ote:Notes for Question 2 ContinuedcontinuedCorrect trapezium rule method with h 5. Condone a slip on one of the speeds.The ‘2’ and ‘42’ should be in the correct place in the [.].415Units do not have to be statedGive final A0 for giving a final answer with incorrect units.e.g. give final A0 for 415 km or 415ms 1Only the 1st M1 can only be scored for Way 2, Way 3, Way 4, Way 5 and Way 7 methodsFull marks in part (a) can only be scored by using a Way 1 or a Way 6 method.Give M0 M0 A0 for { d } 2(5) 5(5) 10(5) 18(5) 28(5) 42(5) 105(5) 525 (i.e. using too many rectangles)(10 18)(18 28)(28 42) (2 10)(10) (5) (5) (5) 395 mCondone M1 M0 A0 for 222 2 (5 10)(10 18)(18 28) (28 42) (2 5) Give M1 M1 A1 for 5 415 m2222 2 5Give M1 M1 A1 for (2 42) 5(5 10 18 28) 415 m2Bracketing mistake:Unless the final calculated answer implies that the method has been applied correctly5give M1 M0 A0 for (2) 2(5 10 18 28) 42 { 169 }25give M1 M0 A0 for (2 42) 2(5 10 18 28) { 232 }2Give M0 M0 A0 for a Simpson’s Rule MethodAlt 1This mark depends on both an answer to part (a) being obtained and the first M in part (a)See schemeAllow the explanation “curve concaves upwards”Do not allow explanations such as “curve is concave” or “curve concaves downwards”Do not allow explanation “gradient of the curve is positive”Do not allow explanations which refer to “friction” or “air resistance”The diagram opposite is sufficient as anexplanation. It must show the top of atrapezium lying above the curve.Alt 2This mark depends on both an answer to part (a) being obtained and the first M in part (a)See schemeDo not allow explanations which refer to “friction” or “air-resistance”
Question3 (a)SchemeMarksAOsB12.3Allow explanations such as student should have worked in radians they did not convert degrees to radians 40 should be in radians should be in radians2 40 angle (or ) should beor9180 correct formula is r 2 {where is in degrees}360 40 correct formula is r 2 360 (1)(b)Way 1{Area of sector } 1 2 2 (5 ) 2 9 25 {cm2 }9or awrt 8.73 {cm2 }M11.1bA11.1b(2)(b)Way 2{Area of sector } 40 360 (52 ) 25 {cm2 }9or awrt 8.73 {cm2 }M11.1bA11.1b(2)(3 marks)Notes for Question 3(a)B1:(b)M1:Note:A1*:(b)M1:A1:Note:Note:Explains that the formula use is only valid when angle AOB is applied in radians.See scheme for examples of suitable explanations.Way 1Correct application of the sector formula using a correct value for in radians40 Allow exact equivalents for e.g. or in the range [0.68, 0.71]18025Accept or awrt 8.73 Note: Ignore the units9Way 2Correct application of the sector formula in degrees25Accept or awrt 8.73 Note: Ignore the units.950Allow exact equivalents such as 18 25Allow M1 A1 for 500 {cm2 } or awrt 8.73 {cm2 } 1809
QuestionScheme4C1 : x 10cos t , y 4 2 sin t , 0 t 2 ; C2 : x 2 y 2 66Way 1(10cos t )2 (4 2 sin t )2 66100(1 sin 2 t ) 32sin 2 t 66100 68sin 2 t 66 sin 2 t 100cos2 t 32(1 cos2 t ) 661268cos2 t 32 66 cos2 t 12 cos t . sin t .Substitutes their solution back into the relevant original equation(s)to get the value of the x-coordinate and value of thecorresponding y-coordinate.Note: These may not be in the correct quadrantS (5 2 , 4) or x 5 2 , y 4 or S (awrt 7.07, ay 22 x y 22{cos2 t sin 2 t 1 } 1 { 32 x 100 y 3200} 10 4 2 os t )2 (4 2 sin t )2 66M13.1a 1 cos 2t 1 cos 2t 100 32 6622 50 50cos 2t 16 16cos 2t 66 34cos 2t 66 66 cos 2t .Substitutes their solution back into the original equation(s) to get thevalue of the x-coordinate and value of the y-coordinate.Note: These may not be in the correct quadrantM1A12.11.1bdM11.1bM11.1bS (5 2 , 4) or x 5 2 , y 4 or S (awrt 7.07, 4)A13.2ax266 x 266 y 2y2 1 110032100322112 32 y 2 100 y 2 320032 x 2 6600 100 x 2 3200x2 50 x .y 2 16 y .Substitutes their solution back into the relevant original equation(s) toget the value of the corresponding x-coordinate or y-coordinate.Note: These may not be in the correct quadrantS (5 2 , 4) or x 5 2 , y 4 or S (awrt 7.07, 4)Way 3{C2 : x2 y 2 66 } x 66 cos , y 66 sin {C1 C2 } 10cos t 66 cos ,4 2 sin t 66 sin 2 10cos t 4 2 sin t {cos sin 1 } 1 66 66 then continue with applying the mark scheme for Way 122Way 42(6)Note: Give final A0 for writing x 5 2 , y 4followed by S ( 4, 5 2 )(6 marks)Notes for Question 4
A1:M1:M1:M1:Note:A1:dM1:M1:A1:Way 1Begins to solve the problem by applying an appropriate strategy.E.g. Way 1: A complete process of combining equations for C1 and C 2 by substituting theparametric equation into the Cartesian equation to give an equation in one variable (i.e. t) only.Uses the identity sin 2 t cos2 t 1 to achieve an equation in sin 2 t only or cos2 t onlyA correct equation in sin 2 t only or cos2 t onlydependent on both the previous M marksRearranges to make sin t . where 1 sin t 1 or cos t . where 1 cos t 111Condone 3rd M1 for sin 2 t sin t 24See schemeSelects the correct coordinates for SAllow either S (5 2 , 4) or S (awrt 7.07, 4)Way 2Begins to solve the problem by applying an appropriate strategy.E.g. Way 2: A complete process of using cos2 t sin 2 t 1 to convert the parametric equationfor C1 into a Cartesian equation for C1Complete valid attempt to write an equation in terms of x only or y only not involvingtrigonometryA correct equation in x only or y only not involving trigonometrydependent on both the previous M marksRearranges to make x . or y .their x2 or their y 2 must be 0 for this markSee schemetheir x2 and their y 2 must be 0 for this markSelects the correct coordinates for S 10 Allow either S (5 2 , 4) or S (awrt 7.07, 4) or S ( 50 , 4) or S , 4 2 Way 3Begins to solve the problem by applying an appropriate strategy.E.g. Way 3: A complete process of writing C 2 in parametric form, combining the parametricequations of C1 and C 2 and applying cos2 sin 2 1 to give an equation in one variable(i.e. t ) only.then continue with applying the mark scheme for Way 1Way 4Begins to solve the problem by applying an appropriate strategy.E.g. Way 4: A complete process of combining equations for C1 and C 2 by substituting theparametric equation into the Cartesian equation to give an equation in one variable (i.e. t) only.Uses the identities cos 2t 2cos2 t 1 and cos 2t 1 2sin 2 t to achieve an equation in cos 2t onlyAt least one of cos 2t 2cos2 t 1 or cos 2t 1 2sin 2 t must be correct for this mark.A correct equation in cos 2t onlydependent on both the previous M marksRearranges to make cos 2t . where 1 cos 2t 1See schemeSelects the correct coordinates for S 10 Allow either S (5 2 , 4) or S (awrt 7.07, 4) or S ( 50 , 4) or S , 4 2 QuestionSchemeMarksAOs
4C1 : x 10cos t , y 4 2 sin t , 0 t 2 ; C2 : x 2 y 2 66Way 5(10cos t )2 (4 2 sin t )2 66M13.1a(10cos t )2 (4 2 sin t )2 66(sin 2 t cos2 t )M1A12.11.1bdM11.1bM11.1bA13.2a100cos2 t 32sin 2 t 66sin 2 t 66cos2 t 34cos2 t 34sin 2 t tan t .Substitutes their solution back into the relevant original equation(s)to get the value of the x-coordinate and value of thecorresponding y-coordinate.Note: These may not be in the correct quadrantS (5 2 , 4) or x 5 2 , y 4 or S (awrt 7.07, 4)(6)M1:M1:A1:dM1:M1:A1:Way 5Begins to solve the problem by applying an appropriate strategy.E.g. Way 5: A complete process of combining equations for C1 and C 2 by substituting theparametric equation into the Cartesian equation to give an equation in one variable (i.e. t) only.Uses the identity sin 2 t cos2 t 1 to achieve an equation in sin 2 t only and cos2 t onlywith no constant termA correct equation in sin 2 t and cos2 t containing no constant termdependent on both the previous M marksRearranges to make tan t .See schemeSelects the correct coordinates for S 10 Allow either S (5 2 , 4) or S (awrt 7.07, 4) or S ( 50 , 4) or S , 4 2
Question5Scheme States lim x 0 x x is 9x 4 MarksAOsB11.2M11.1bA11.1b9x dx49 2 3 x2 3 42 32 2 3254 16 9 4 3333382 or 12or awrt 12.733 (3)(3 marks)Notes for Question 5B1:States 9x dx with or without the 'dx '43M1:A1:Integrates x to give x 2 ; 0See scheme9Note:Note:33 3 You can imply B1 for x 2 or for 9 2 4 2 4Give B0 for 9x dx 1Note:Note: 3x dx or for1 9x dx without reference to
Summer 2019 Pearson Edexcel GCE In Mathematics (9MA0) Paper 2 Pure Mathematics 2 . Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest a
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.
Aug 15, 2019 · Mark Scheme (Results) Summer 2019 Pearson International Advanced Subsidiary Level In Chemistry (WCH11) Paper 01 Structure, . perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All th
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 .
