Mark Scheme Pure Mathematics Year 1 (AS) Unit Test 1 .
Mark schemeMarksAOsPearsonProgression Stepand ProgressdescriptorAttempt to multiply the numerator and denominator by6 3 4 8 3 k (8 3) . For example,8 3 8 3M11.1b3rdAttempt to multiply out the numerator (at least 3 termscorrect).M11.1aM11.1bA11.1bQ1Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsSchemeRationalise thedenominator of afraction with asimple surddenominator48 3 18 32 4 3Attempt to multiply out the denominator (for example, 3 termscorrect but must be rational or 64 – 3 seen or implied).64 8 3 8 3 3p and q stated or implied (condone if all over 61).44144414or p , q 3 61616161(4 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.1
Mark schemePure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsQ2aSchemeStatement that discriminant is b2 – 4ac, and/or implied bywriting k 8 4 1 8k 1 MarksAOsPearsonProgression Stepand ProgressdescriptorM11.1a4thM11.1bUnderstand anduse thediscriminant;conditions forreal, repeated andno real rootsA11.1b2Attempt to simplify the expression by multiplying out thebrackets. Condone sign errors and one algebraic error (but notmissing k term from squaring brackets and must have k2, k andconstant terms).k 2 8k 8k 64 32k 4 o.e.k 2 16k 60(3)2bKnowledge that two equal roots occur when the discriminantis zero. This can be shown by writing b2 – 4ac 0, or bywriting k 2 16k 60 0M1k 10, k 6A11.1b5th1.1bSolve problemsinvolving thediscriminant incontext andconstruct simpleproofs involvingthe discriminant(2)2cCorrect substitution for k 8: f( x) x 2 16 x 65B12.2a3rdAttempt to complete the square for their expression of f(x).M11.1bSolve quadraticequations by useof formulaA12.3f( x) x 8 12Statement (which can be purely algebraic) that f(x) 0,because, for example, a squared term is always greater than orequal to zero, so one more than a square term must be greaterthan zero or an appeal to a reasonable sketch of y f(x).(3)(8 marks) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.2
Mark schemePure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsNotes2aNot all steps have to be present to award full marks. For example, the second method mark can still be awardedif the answer does not include that step.2bAward full marks for k 6, k 10 seen. Award full marks for valid and complete alternative method (e.g.expanding (x – a)2 comparing coefficients and solving for k).2cAn alternative method is acceptable. For example, students could differentiate to find that the turning point ofthe graph of y f(x) is at (8, 1), and then show that it is a minimum. The minimum can be shown by usingproperties of quadratic curves or by finding the second differential. Students must explain (a sketch will suffice)that this means that the graph lies above the x-axis and reach the appropriate conclusion. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.3
Mark schemePure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsQ3aScheme115 (m) is the height of the cliff (as this is the height of theball when t 0). Accept answer that states 115 (m) is theheight of the cliff plus the height of the person who is ready tothrow the stone or similar sensible comment.MarksAOsPearsonProgression Stepand ProgressdescriptorB13.2a4thUnderstand theconcepts ofdomain and range(1)3bAttempt to factorise the – 4.9 out of the first two (or all)terms.3.1a4thM13.1aSolve simplequadraticequations bycompleting thesquareA13.1aM15 h(t ) 4.9 t 2 2.5t 115 or h(t ) 4.9 t 2 t 1152 h(t ) 4.9 t 1.25 ( 4.9) 1.25 11522or22 5 5 h(t ) 4.9 t ( 4.9) 115 4 4 h(t ) 122.65625 4.9 t 1.25 o.e.2(N.B. 122.65625 3925)32Accept the first term written to 1, 2, 3 or 4 d.p. or the fullanswer as shown.(3)3ci3.1a4thM13.1aForm and solvequadraticequations incontextA13.5aStatement that the stone will reach ground level whenh(t) 0, or 4.9t 2 12.25t 115 0 is seen.M1Valid attempt to solve quadratic equation (could be usingcompleted square form from part b, calculator or formula).Clearly states that t 6.25 s (accept t 6.3 s) is the answer, orcircles that answer and crosses out the other answer, orexplains that t must be positive as you cannot have a negativevalue for time.(3) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.4
Mark scheme3ciiPure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functionshmax awrt 123B1ft3.1a4th3.2aForm and solvequadraticequations incontextft A from part b.B1ft5t or t 1.254ft C from part b.(2)(9 marks)Notes3cAward 4 marks for correct final answer, with some working missing. If not correct B1 for each of A, B and Ccorrect.If the student answered part b by completing the square, award full marks for part c, providing their answer totheir part b was fully correct. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.5
Mark schemePure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsQ4aSchemeAttempt to solve q(x) 0 by completing the square or by usingthe formula.MarksAOsPearsonProgression Stepand ProgressdescriptorM11.1b3rdSolve quadraticequations by useof formulax 2 10 x 20 0 x 5 2 45 0orx 10 100 4(1)( 20)2(1)x 5 3 5 and/or statement that says a 5 and b 5A11.1b(2)4bFigure 1q(0) 20, so y q(x)intersects y-axis at (0, 20)and x-intercepts labelled(accept incorrect values frompart a).B1ft1.1b3rdSketch graphs ofquadraticfunctionsy p(x) intersects y-axisat (0, 3).B11.1by p(x) intersects x-axisat (6, 0).B11.1bGraphs drawn as shown withall axes intercepts labelled.The two graphs should clearlyintersect at two points, one ata negative value of x and oneat a positive value of x. Thesepoints of intersection do notneed to be labelled.B11.1b(4) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.6
Mark scheme4cPure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsStatement indicating that this is the point where p(x) q(x)or x 2 10 x 20 3 M12.2a1x seen.2Their equation factorised, or attempt to solve their equation bycompleting the square.M11.1b 23 11 2 , 4 A11.1b 2, 4 A11.1b2x2 19x – 46 04thSolve morecomplicatedsimultaneousequations whereone is linear andone is quadratic(2x – 23)(x 2) 0(4)4dx – 2 or x B123o.e.2{x : x ¡ , x 2} {x : x ¡ , x 11.5}B12.2a4th2.2aRepresentsolutions toquadraticinequalities usingset notationNB: Must see “or” or (if missing SC1 for just the correctinequalities).(2)(12 marks)Notes4aEquation can be solved by completing the square or by using the quadratic formula. Either method is acceptable.4bAnswers with incorrect coordinates lose accuracy marks as appropriate. However, the graph accuracy marks canbe awarded for correctly labelling their coordinates, even if their coordinates are incorrect.4cIf the student incorrectly writes the initial equation, award 1 method mark for an attempt to solve the incorrectequation. Solving the correct equation by either factorising or completing the square is acceptable. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.7
Mark schemeQ5Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsSchemeFigure 2Asymptote drawn at x 6MarksAOsPearsonProgression Stepand ProgressdescriptorB11.1b5thAsymptote drawn at y 5B11.1b 13 Point 0, labelled. 3 13Condoneclearly on y axis.3B11.1b 26 Point ,0 labelled. 5 B11.1bB11.1bCondoneUnderstand anduse properties ofasymptotes forgraphs of theform y a/x and y a/x226clearly on x axis.5Correctly shaped graph drawnin the correct quadrantsformed by the asymptotes.(5)(5 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.8
Mark schemeQ6aPure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsSchemeFigure 3Evidence of attempt to show1stretch of sf in x direction2MarksAOsPearsonProgression Stepand ProgressdescriptorM11.1b3rdTransform graphsusing stretches(e.g. one correct set ofcoordinates – not (0, –2)).Fully complete graph with allpoints labelled.A11.1b(2) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.9
Mark scheme6bFigure 4Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsEvidence of attempt to showreflection in y axis (e.g. onecorrect set of coordinates – not(0, –2)).M1Fully complete graph with allpoints labelled.A11.1b3rdTransform graphsusing translations1.1b(2)(4 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.10
Mark schemePure Mathematics Year 1 (AS) Unit Test 1: Algebra and FunctionsQ7aSchemeGraph of y 2x 5 drawn.Figure 5MarksAOsPearsonProgression Stepand ProgressdescriptorB11.1b4thGraph of 2y x 6 drawn.B11.1bGraph of y 2 drawn onto thecoordinate grid and thetriangle correctly shaded.B12.2aRepresent linearand quadraticinequalities ongraphs(3)7bAttempt to solve y 2x 5 and 2y x 6 simultaneouslyfor y.M1y 3.4A11.1bBase of triangle 3.5B12.2aM12.2aA11.1bArea of triangle 12 (“3.4” – 2) 3.5Area of triangle is 2.45 (units2).2.2a5thSolve problemsinvolving linearand quadraticinequalities incontext(5)(8 marks)Notes7bIt is possible to find the area of triangle by realising that the two diagonal lines are perpendicular and thereforefinding the length of each line using Pythagoras’ theorem. Award full marks for a correct final answer using thismethod.In this case award the second and third accuracy marks for finding the lengths 2.45 and 9.8 Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.11
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) planeQ1aSchemeUse of the gradient formula to begin attempt to find k.MarksAOsPearsonProgression Stepand ProgressdescriptorM12.2a1st 2 (k 1)3k 1 ( 2)3 (i.e. correct or3k 4 121 (3k 4)2Assumedknowledge.32substitution into gradient formula and equating to ).2k 6 15 9kA1*1.1b21 7kk 3* (must show sufficient, convincing and correct working).(2)1bStudent identifies the coordinates of either A or B. Can be seenor implied, for example, in the subsequent step when studentattempts to find the equation of the line.1.1b2ndM11.1bFind the equationof a straight linegiven the gradientand a point on theline.A11.1bB1A(5, 2) or B(1, 4).Correct substitution of their coordinates into y mx b ory y1 m(x x1) o.e. to find the equation of the line.For example, 3 3 2 5 b or y 2 x 5 or 2 2 3 3 4 1 b or y 4 x 1 2 2 311or 3x 2 y 11 0y x 22(3) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane1cMidpoint of AB is (3, 1) seen or implied.B12.2aSlope of line perpendicular to AB is , seen or implied.B12.2aAttempt to find the equation of the line (i.e. substituting theirmidpoint and gradient into a correct equation). For example,M11.1bA11.1b233rdFind the equationof a perpendicularbisector.2 2 1 3 b or y 1 x 3 3 3 2 x 3 y 3 0 or 3 y 2 x 3 0 . Also accept any multiple of2 x 3 y 3 0 providing a, b and c are still integers.(4)(9 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) planeQ2aSchemem 11 ( 7) 189 6 4 105Correct substitution of (4, 7) or ( 6, 11) and their gradientinto y mx b or y y1 m(x x1) o.e. to find the equationof the line. For example,MarksAOsPearsonProgression Stepand ProgressdescriptorB11.1b2ndM11.1bFind the equationof a straight linegiven two points.A11.1b9 9 7 4 b or y 7 x 4 or5 5 9 9 11 6 b or y 11 x 6 .5 5 5y 9x 1 0 or 5y 9x 1 0 only(3)2b1.1b3rdB11.1bSolve problemsinvolving lengthand area in thecontext of straightline graphs.B11.1by 0, x 11 1 so A ,0 . Award mark for x seen.999 B1x 0, y 11 1 so B 0, . Award mark for y seen.55 5 Area 1 1 1 1 2 5 9 90(3)(6 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) planeQScheme3MarksAOsPearsonProgression Stepand Progressdescriptory mx 2 seen or implied.M11.1b4thSubstitutes their y mx 2 into x2 6 x y 2 8 y 4M13.1aM11.1bUse thediscriminant todetermineconditions for theintersection ofcircles andstraight lines.Uses b2 4ac 0 , 6 12m 4 1 m2 16 0M13.1aRearranges to 20m2 36m 7 0 or any multiple of this.A11.1bAttempts solution using valid method. For example,M12.2aA11.1bx2 6 x mx 2 8(mx 2) 4 o.e.2Rearranges to a 3 term quadratic in x(condone one arithmetic error). 1 m x22 (6 12m) x 16 0 2m m 36 2 4 20 7 2 20 36 9299 2 29or m o.e. (NB decimals A0). 10510(7)(7 marks)Notes2y 2 y 2 y 2 6 y2 8 y 4Elimination of x follows the same scheme. x leading to m m m This leads to (1 m2 ) y 2 (4 6m 8m2 ) y 4 12m 4m2 0 Use of b2 4ac 0 gives 4 6m 8m2 2 4 1 m2 4 12m 4m2 0 which reduces to4m2 20m2 36m 7 0. m cannot equal 0, so this must be discarded as a solution for the final A mark.b2 4ac 0 could be used implicitly within the quadratic equation formula. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) planeQ4aSchemeStudent attempts to complete the square twice for the firstequation (condone sign errors).MarksAOsPearsonProgression Stepand ProgressdescriptorM12.2a4thFind the centreand radius of acircle, given theequation, bycompleting thesquare. x 5 25 y 6 36 3 x 5 2 y 6 2 6422Centre ( 5, 6)A13.2aRadius 8A13.2aStudent attempts to complete the square twice for the secondequation (condone sign errors).M12.2aCentre (3, q)A13.2aRadius 18 q 2A13.2a x 3 2 9 y q 2 q 2 9 x 3 2 y q 2 18 q 2(6)4bUses distance formula for their centres and 80 . Forexample, 5 3 2 6 q 2 80 M12.2a5thSolve coordinategeometryproblemsinvolving circlesin context.2Student simplifies to 3 term quadratic. For example,q 2 12q 20 0M11.1bConcludes that the possible values of q are 2 and 10A11.1b(3)(9 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) planeQ5aSchemeStudent completes the square twice. Condone sign errors.MarksAOsPearsonProgression Stepand ProgressdescriptorM11.1b4th x 4 2 16 y 5 2 25 1 0 x 4 2 y 5 2 40So centre is (4, 5)A11.1band radius is 40A11.1bFind the centreand radius of acircle, given theequation, bycompleting thesquare.(3)5b2.2a5thM11.1bSolve coordinategeometryproblemsinvolving circlesin context.Obtains solutions y 3, y 7 (must give both).A11.1bRejects y 7 giving suitable reason (e.g. 7 5) or ‘itwould be below the centre’ or ‘AQ must slope upwards’ o.e.B12.3M1Substitutes x 10 into equation (in either form).102 8 10 y 2 10 y 1 0 or 10 4 y 5 4022Rearranges to 3 term quadratic in y y 2 10 y 21 0(could be in completed square form y 5 4 )2(4)5cmAQ 3 ( 5) 1 10 43B11.1b5thFind the equationof the tangent to agiven circle at aspecified point.ml2 3 (i.e. 1 over their mAQ )B1ft2.2aSubstitutes their Q into a correct equation of a line. Forexample,M11.1bA11.1b 3 3 10 b or y 3 3 x 10 y 3x 27(4) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane5d3.1auuur 6 AQ o.e. (could just be in coordinate form). 2 M1uuur 2 AP o.e. so student concludes that point P has 6 coordinates (2, 1).M13.1aM12.2aA11.1bSubstitutes their P and their gradient13( mAQ from 5c) into a5thSolve coordinategeometryproblemsinvolving circlesin context.correct equation of a line. For example, 1 1 1 2 b or y 1 x 2 3 3 11y x 33(4)5ePA 4040Uses Pythagoras’ theorem to find EP .9Area of EPA Area 140(could be in two parts). 40 29203B13.1aB12.2aM11.1bA11.1b5thSolve coordinategeometryproblemsinvolving circlesin context.(4)(19 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Mark schemeMarksAOsPearsonProgression Stepand ProgressdescriptorCorrectly shows that either 1 f(3) 0, f( 2) 0 or f 0 2 M13.1a4thDraws the conclusion that (x – 3), (x 2) or (2x 1) musttherefore be a factor.M12.2aEither makes an attempt at long division by setting up thelong division, or makes an attempt to find the remainingfactors by matching coefficients. For example, statingM11.1bA12.2aA11.1bA11.1bQ1Pure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraSchemeDividepolynomials bylinear expressionswith noremainder x 3 ax2 bx c 2 x3 x2 13x 6or x 2 rx2 px q 2x3 x2 13x 6or 2 x 1 ux2 vx w 2 x3 x2 13x 6For the long division, correctly finds the the first twocoefficients.For the matching coefficients method, correctly deduces thata 2 and c 2 or correctly deduces that r 2 and q 3 orcorrectly deduces that u 1 and w –6For the long division, correctly completes all steps in thedivision.For the matching coefficients method, correctly deduces thatb 5 or correctly deduces that p 5 or correctly deducesthat v –1States a fully correct, fully factorised final answer:(x – 3)(2x 1)(x 2)(6 marks) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.1
Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraNotesOther algebraic methods can be used to factorise h(x). For example, if (x – 3) is known to be a factor then2 x3 x2 13x 6 2 x2 ( x 3) 5x( x 3) 2( x 3) by balancing (M1) (2 x2 5x 2)( x 3) by factorising (M1) (2 x 1)( x 2)( x 3) by factorising (A1) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.2
Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraQSchemeMarksAOsPearsonProgression Stepand Progressdescriptor2aStates or implies the expansion of a binomial expression tothe 8th power, up to and including the x3 term.M11.1a5thUnderstand anduse the generalbinomialexpansion forpositive integer n(a b) C0 a C1a b C2 a b C3a b .8888786 285 3or(a b)8 a8 8a7b 28a6b2 56a5b3 .Correctly substitutes 1 and 3x into the formula:M11.1bM1dep1.1bA11.1b(1 3x)8 18 8 17 3x 28 16 3x 56 15 3x .23Makes an attempt to simplify the expression (2 correctcoefficients (other than 1) or both 9x2 and 27x3).(1
If the student incorrectly writes the initial equation, award 1 method mark for an attempt to solve the incorrect equation. Solving the correct equation by either factorising or completing the square is acceptable.
A level Mathematics specimen papers - Paper 2 Pure Mathematics mark scheme 1 9MA0/02: Pure Mathematics Paper 2 Mark scheme Question Scheme Marks AOs 1 1 2) 2 r M1 1.1a A1 225 22. 24 r 1.1b length of minor arc ) dM1 3.1a 36 ab A1 1.1b (4) 1 Alt 1 2(4.8) 2 r M1 1.1
A level Mathematics specimen papers - Paper 1 Pure Mathematics mark scheme 1 9MA0/01: Pure Mathematics Paper 1 Mark scheme Question Scheme Marks AOs 1 (a) 1.09811 ( ) 2 R B1 1.1b M1 1.1b 1 5 4 (3 dp) A1 1.1b (3) (b) Any valid reason, for example Increas
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
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
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.
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 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.
