Paper 1: Pure Mathematics 1 Mark Scheme Question 15 Continued

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Paper 1: Pure Mathematics 1 Mark SchemeQuestion1(a)SchemeMarksAOsdy(i) 12 x3 24 x 2dxM1A11.1b1.1bd2 y(ii) 36 x 2 48 xdx 2A1ft1.1b(3)(b)Substitutes x 2 into theirShowsdy 12 23 24 22dxdy 0 and states ''hence there is a stationary point''dxM11.1bA12.1(2)(c)d2 ySubstitutes x 2 into their 2 36 22 48 2dxM11.1bd2 y 48 0 and states ''hence the stationary point is a minimum''dx 2A1ft2.2a(2)(7 marks)Notes:(a)(i)M1: Differentiates to a cubic formdyA1: 12 x3 24 x 2dx(a)(ii)A1ft: Achieves a correctd2 ydyfor their 36 x 2 48 x2dxdx(b)M1:A1:Substitutes x 2 into theirdydxdy 0 and states ''hence there is a stationary point'' All aspects of the proofdxmust be correctShows(c)d2 ydx 2Alternatively calculates the gradient of C either side of x 2A1ft: For a correct calculation, a valid reason and a correct conclusion.M1:Substitutes x 2 into theird2 yFollow through on an incorrect 2dxPearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201737

Question2(a)SchemeUses s r 3 r 0.4 OD 7.5 cmMarksAOsM11.2A11.1b(2)(b) Uses angle AOB 0.4 or uses radius is (12 – ‘7.5’) cm11Uses area of sector r 2 (12 7.5)2 ( 0.4)22 27.8cm2M13.1aM11.1bA1ft1.1b(3)(5 marks)Notes:(a)M1:A1:Attempts to use the correct formula s r with s 3 and 0.4OD 7.5 cm (An answer of 7.5cm implies the use of a correct formula and scores bothmarks)(b)M1:AOB 0.4 may be implied by the use of AOB awrt 2.74 or uses radius is(12 – their ‘7.5’)M1: Follow through on their radius (12 – their OD) and their angleA1ft: Allow awrt 27.8 cm2. (Answer 27.75862562). Follow through on their (12 – their ‘7.5’)Note: Do not follow through on a radius that is negative.38Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

Question3(a)SchemeAttempts x 2 y 5 .22Centre (2, 5)MarksAOsM11.1bA11.1b(2)(b)Sets k 22 52 0 k 29M12.2aA1ft1.1b(2)(4 marks)Notes:(a)M1:Attempts to complete the square so allow x 2 y 5 .A1:States the centre is at (2, 5). Also allow written separately x 2, y 522(2, 5) implies both marks(b)Deduces that the right hand side of their x . y . . is 0 or2M1:A1ft: k 29 Also allow k2. 29 Follow through on their rhs of x . y . 2QuestionScheme4Writes t 1d tt0 11 dt and attempts to integratet t ln t c ln 7 2a ln 2a a ln a a ln77with k 222MarksAOsM12.1M11.1bM11.1bA11.1b(4 marks)Notes:M1:Attempts to divide each term by t or alternatively multiply each term by t -1M1:Integrates each term and knowsM1:Substitutes in both limits, subtracts and sets equal to ln7A1:Proceeds to a ln t dt ln t.1The c is not required for this mark77and states k or exact equivalent such as 3.522Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201739

Question5SchemeAttempts to substitute x 1 yinto y 26 x 1 4 7 ( x 1) 2 Attempts to write as a single fraction(2 x 5)( x 1) 6y ( x 1)2 x 2 3x 1y x 1a 3, b 1MarksAOsM12.1M12.1A11.1b(3 marks)Notes:M1:M1:3x 1or equivalent into y 4t 7 t2Award this for an attempt at a single fraction with a correct common denominator.Score for an attempt at substituting t x 1 Their 4 7 term may be simplified first 2 A1:402 x 2 3x 1Correct answer only y x 1a 1 3, b Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

Question6 (a)(i)(ii)Scheme10750 barrelsGives a valid limitation, for example The model shows that the daily volume of oil extractedwould become negative as t increases, which is impossible States when t 10,V 1500 which is impossible64 States that the model will only work for 0 t7MarksAOsB13.4B13.5b(2)(b)(i)Suggests a suitable exponential model, for example V AektUses 0,16000 and 4,9000 in 9000 16000e4k k1 9 ln 4 16 awrt 0.144V 16000e(ii)1 9 ln t4 16 or V 16000e 0.144tUses their exponential model with t 3 V awrt 10400 barrelsM13.3dM13.1bM11.1bA11.1bB1ft3.4(5)(7 marks)Notes:(a)(i)10750 barrelsB1:(a)(ii)B1: See scheme(b)(i)M1: Suggests a suitable exponential model, for example V Aekt , V Ar t or any othersuitable function such as V Aekt b where the candidate chooses a value for b.dM1: Uses both 0,16000 and 4,9000 in their model.With V Aekt candidates need to proceed to 9000 16000e4kWith V Ar t candidates need to proceed to 9000 16000r 4With V Aekt b candidates need to proceed to 9000 M1:A1: 16000 b e4k bwhere b is given16000 .as a positive constant and A b Uses a correct method to find all constants in the model.Gives a suitable equation for the model passing through (or approximately through in thecase of decimal equivalents) both values 0,16000 and 4,9000 . Possible equations forthe model could be for exampleV 16000 0.866V 16000e 0.144t V 15800e 0.146t 200t(b)(ii)B1ft: Follow through on their exponential modelPearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201741

QuestionSchemeAttempts7AC AB BC 2 i 3 j k i 9 j 3 k 3 i 6 j 4 kAttempts to find any one length using 3-d Pythagoras Finds allof ABcos BAC 14, AC 61, BC9114 61 912 14 61angle BAC 105.9 *MarksAOsM13.1aM12.1A1ft1.1bM12.1A1*1.1b(5)(5 marks)Notes:M1:Attempts to find AC by using AC AB BCM1:Attempts to find any one length by use of Pythagoras' TheoremA1ft: Finds all three lengths in the triangle. Follow through on their AC2M1:Attempts to find BAC using cos BAC 2AB AC BC22 AB ACAllow this to be scored for other methods such as cos BAC A1*:42AB. ACAB ACThis is a show that and all aspects must be correct. Angle BAC 105.9 Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

QuestionScheme8 (a)f (3.5) 4.8, f (4) ( )3.1Change of sign and function continuous in interval[3.5, 4] Root *MarksAOsM11.1bA1*2.4(2)(b)Attempts x1 x0 f ( x0 )3.099 x1 4 16.67f ( x0 )M11.1bx1 3.81A11.1by ln(2x – 5)(2)(c) y 30 2 x 2M13.1aA12.42Attempts to sketch both y ln(2x – 5) and y 30 – 2x2States that y ln(2x – 5) meets y 30 – 2x in just one place,therefore y ln(2x – 5) 30 – 2x has just one root f (x) 0 hasjust one root(2)(6 marks)Notes:(a)M1: Attempts f(x) at both x 3.5 and x 4 with at least one correct to 1 significant figureA1*: f (3.5) and f(4) correct to 1 sig figure (rounded or truncated) with a correct reason andconclusion. A reason could be change of sign, or f (3.5) f (4) 0 or similar with f(x)being continuous in this interval. A conclusion could be 'Hence root' or 'Therefore root ininterval'(b)f ( x0 )3.099evidenced by x1 4 16.67f ( x0 )M1:Attempts x x0 1A1:Correct answer only x1 3.81(c)M1:For a valid attempt at showing that there is only one root. This can be achieved byA1: Sketching graphs of y ln(2x – 5) and y 30 – 2x on the same axes2 Showing that f(x) ln(2x – 5) 2x – 30 has no turning points2 Sketching a graph of f(x) ln(2x – 5) 2x – 30Scored for correct conclusion2Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201743

QuestionScheme9(a)tan cot MarksAOssin cos cos sin M12.1sin 2 cos 2 sin cos A11.1bM12.1A1*1.1b11sin 2 2 2cosec2 *(4)1 sin 2 2States tan cot AND no real solutions as 1 sin 2 (b)B112.4(1)(5 marks)Notes:(a)M1:cos Writes tan sin and cot sin cos sin 2 cos 2 sin cos M1: Uses the double angle formula sin 2 2sin cos A1*: Completes proof with no errors. This is a given answer.A1:Achieves a correct intermediate answer ofNote: There are many alternative methods. For exampletan cot tan main scheme.(b)B1:tan 2 1 sec2 111then as the tan tan cos 2 sin cos sin tan cos Scored for sight of sin 2 2 and a reason as to why this equation has no real solutions.Possible reasons could be 1 sin 2 1.and therefore sin 2 2or sin 2 2 2 arcsin 2 which has no answers as 1 sin 2 441Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

Question10SchemeUse ofMarksAOsB12.1M11.1bA11.1bM12.1A1*2.5sin( h) sin ( h) Uses the compound angle identity for sin( A B) with A ,B h sin( h) sin cos h cos sin hAchievessin( h) sin sin cos h cos sin h sin hh sin h cos h 1 cos sin hh sin hcos h 1 1 and 0hhsin( h) sin Hence the limit h 0 cos and the gradient of( h) dythe chord gradient of the curve cos * d Uses h 0,(5 marks)Notes:B1:States or implies that the gradient of the chord is sin( h) sin or similar such asM1:sin( ) sin for a small h or Uses the compound angle identity for sin(A B) with A , B h or hsin cos h cos sin h sin or equivalenthsin hcos h 1M1: Writes their expression in terms ofandhhdyA1*: Uses correct language to explain that cos d A1:ObtainsFor this method they should use all of the given statements h 0,sin( h) sin cos h 1 cos 0 meaning that the limit h 0h( h) and therefore the gradient of the chordsin h 1,h gradient of the curve dyPearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017d cos 45

Question10altSchemeUse ofsin( h) sin ( h) h h h h sin sin sin( h) sin 2 2 2 2 Sets ( h) hand uses the compound angle identity for sin(A B) andMarksAOsB12.1M11.1bA11.1bM12.1A1*2.5hhsin(A – B) with A , B 22Achieves sin( h) sin h h h h h h h h h sin 2 cos 2 cos 2 sin 2 sin 2 cos 2 cos 2 sin 2 h h sin 2 cos h h2 2 h sin h 2 1Uses h 0, h 0 henceand cos cos h2 2 2sin( h) sin Therefore the limit h 0 cos and the gradient of( h) the chord gradient of the curve dy cos d *(5 marks)Additional notes:A1*: Uses correct language to explain thatdy cos . For this method they should use thed h sin h 2 1(adapted) given statement h 0, h 0 hencewith cos cos h2 2 2sin( h) sin meaning that the limit h 0 cos and therefore the gradient of the( h) chord gradient of the curve 46dycos d Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

Question11(a)SchemeSets H 0 1.8 0.4d 0.002d 2 0MarksAOsM13.4dM11.1bA12.2aSolves using an appropriate method, for exampled 0.4 0.4 2 4 0.002 1.8 2 0.002Distance awrt 204 m only(3)(b)States the initial height of the arrow above the ground.B13.4(1) (c) 1.8 0.4d 0.002d 2 0.002 d 2 200d 1.8M11.1b 0.002 (d 100)2 10000 1.8M11.1b 21.8 0.002(d 100)2A11.1b (3)(d)(i)22.1 metresB1ft3.4(ii)100 metresB1ft3.4(2)(9 marks)Notes:(a)M1:M1:Sets H 0 1.8 0.4d 0.002d 2 0Solves using formula, which if stated must be correct, by completing square (look for d 100 2A1:(b)B1:(c)M1:10900 d . ) or even allow answers coming from a graphical calculatorAwrt 204 m onlyStates it is the initial height of the arrow above the ground. Do not allow '' it is the height ofthe archer''Score for taking out a common factor of 0.002 from at least the d 2 and d termsM1:For completing the square for their d 2 200d termA1: 21.8 0.002(d 100)2 or exact equivalent(d)B1ft: For their '21.8 0.3' 22.1mB1ft: For their 100mPearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201747

Question12 (a)SchemeN aT b log10 N log10 a log10 T band c log10 aso m b log10 N log10 a b log10 T MarksAOsM12.1A11.1b(2)(b)Uses the graph to find either a or b a 10intercept or b gradientM13.1bUses the graph to find both a and b a 10intercept and b gradientM11.1bUses T 3 in N aT b with their a and bM13.1bNumber of microbes 800A11.1b(4)(c)N 1000000 log10 N 6M13.4We cannot ‘extrapolate’ the graph and assume that the model stillholdsA13.5b(2)(d)States that 'a' is the number of microbes 1 day after the start of theexperimentB13.2a(1)(9 marks)48Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

Question 12 continuedNotes:(a)M1:Takes logs of both sides and shows the addition lawM1: m b and c log10 aUses the power law, writes log log10 a b log10 T and states10 N(b)M1:Uses the graph to find either a or b a 10intercept or b gradient. This would be implied byb 2.3 or a 101.8 63the sight of M1:Uses the graph to find both a and b a 10intercept and b gradient. This would be impliedb 2.3 and a 101.8 63by the sight of M1:A1:Uses T 3 N aT b with their a and b. This is implied by an attempt at 63 32.3Accept a number of microbes that are approximately 800. Allow 800 150 followingcorrect work.There is an alternative to this using a graphical approach.M1:3 log10 T 0.48Finds the value of log10 T from T 3. Accept as T M1:Then using the line of best fit finds the value of log10 N from their ''0.48''Accept log10 N 2.9M1:Finds the value of N from their value of log10 N log10 N 2.9 N 10'2.9'A1:Accept a number of microbes that are approximately 800. Allow 800 150 followingcorrect work(c)M1For using N 1000000 and stating that log10 N 6A1:Statement to the effect that ''we only have information for values of log N between 1.8 and4.5 so we cannot be certain that the relationship still holds''. ''We cannot extrapolate withany certainty, we could only interpolate''There is an alternative approach that uses the formula.M1:A1: 1000000 log10 63 1.83 .log10 TUse N 1000000 in their N 63 T 2.3 2.3The reason would be similar to the main scheme as we only have log10 T values from 0 to1.2. We cannot ‘extrapolate’ the graph and assume that the model still holds(d)B1:Allow a numerical explanation T 1 N a1b N a giving a is the value of N at T 1Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201749

Question13(a)SchemeMarksAOsdydyAttempts dtdx dxdtM11.1bdy dxA11.1b3 sin 2t 2 3 cos t sin t(2)(b)Substitutes t 2 dyin 3dx3 sin 2t sin tUses gradient of normal 3 1 1 dy 3 dx3 Coordinates of P 1, 2 Correct form of normal y 3 21 x 1 3Completes proof 2 x 2 3 y 1 0 .1b5 1, 6 2M12.45into x 2cos t, y 3 cos 2t ,6M11.1bA11.1bSubstitutes x 2cos t and y 3 cos 2t into 2 x 2 3 y 1 0Uses the identity cos 2t 2cos2 t 1 to produce a quadratic in cost 12cos2 t 4cos t 5 0Finds cos tSubstitutes their cos t 5 7 Q ,3 3 18 (6)(13 marks)50Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

Question 13 continuedNotes:(a)M1:A1:dydysin 2tAttemptsAlternatively candidates may apply the dt and achieves a form ksin tdx dxdtsin t cos tdouble angle identity for cos 2t and achieve a form ksin t3 sin 2tScored for a correct answer, eitheror 2 3 costsin t(b)2 dyin theirwhich must be in terms of t3dxM1:For substituting t M1:Uses the gradient of the normal is the negative reciprocal of the value ofseen in the equation of l.B1:dy. This may bedx3 States or uses (in their tangent or normal) that P 1, 2 3 Uses their numerical value of 1 dy with their 1, to form an equation of the2 dx normal at PA1*: This is a proof and all aspects need to be correct. Correct answer only 2 x 2 3 y 1 0M1:(c)M1:For substituting x 2cos t and y 3 cos 2t into 2 x 2 3 y 1 0 to produce an equationin t. Alternatively candidates could use cos 2t 2cos2 t 1 to set up an equation of the formy Ax 2 B .M1:Uses the identity cos 2t 2cos2 t 1 to produce a quadratic equation in costIn the alternative method it is for combining their y Ax 2 B with 2 x 2 3 y 1 0 to getan equation in just one variableA1:For the correct quadratic equation 12cos2 t 4cos t 5 0Alternatively the equations in x and y are 3x2 2 x 5 0 12 3 y 2 4 y 7 3 0M1:Solves the quadratic equation in cost (or x or y) and rejects the value corresponding to P.M1:Substitutes their cos t 5 5 or their t arccos in x 2cos t and y 3 cos 2t6 6 If a value of x or y has been found it is for finding the other coordinate.A1:57 5 7 Q ,3 . Allow x ,y3 but do not allow decimal equivalents.318 3 18 Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201751

Question14(a)SchemeMarksAOsUses or implies h 0.5B11.1bFor correct form of the trapezium rule M11.1b0.54.393 3 2.2958 2 2.3041 1.9242 1.9089 2A11.1b(3)(b)Any valid statement reason, for example Increase the number of strips Decrease the width of the strips Use more trapeziaB12.4(1)(c)For integration by parts on x 2 ln x dx x3ln x 3 x2dx3 2 x 5 dx x2 5x c M12.1A11.1bB11.1bM12.1M12.1A11.1bAll integration attempted and limits used3Area of S 1x 3 x3 x 2 ln xx3dx ln x x 2 5 x 2 x 5 327 9 x 1Uses correct ln laws, simplifies and writes in required formArea of S 28 ln 27 (a 28, b 27, c 27)27(6)(10 marks)52Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 2017

Question 14 continuedNotes:(a)0.5 . B1:States or uses the strip width h 0.5. This can be implied by the sight ofM1:trapezium ruleFor the correct form of the bracket in the trapezium rule. Must be y values rather than xvalues first y value last y value 2 sum of other y values A1:(b)B1:(c)M1:B1:M1:M1:A1:in the4.393See schemeUses integration by parts the right way around.Look forA1:2x3ln x 3 x2 ln x dx Ax3 ln x Bx 2 dxx2dx3Integrates the 2 x 5 term correctly x2 5xAll integration completed and limits usedaSimplifies using ln law(s) to a form ln cb28Correct answer only ln 2727Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –Issue 1 – April 2017 Pearson Education Limited 201753

Question15(a)SchemeAttempts to differentiate using the quotient rule or otherwisef ( x) e2 x 1 8cos 2 x 4sin 2 x 2e e2 x 1 AOsM12.1A11.1bM12.1A1*1.1b2 x 12Sets f ( x) 0 and divides/ factorises out the eProceeds viaMarks2 x 1termssin 2 x8 to tan 2 x 2*cos 2 x 4 2(4)(b)(i) Solves tan 4 x 2 and attempts to find the 2nd solutionM13.1aA11.1bM13.1aA11.1bx 1.02(ii) Solves tan 2 x 2 and attempts to find the 1st solutionx 0.478(4)(8 marks)Notes:(a)M1:A1:Attempts to differentiate by using the quotient rule with u 4sin 2 x and v ealternatively uses the product rule with u 4sin 2 x and v e1 2xFor achieving a correct f ( x) . For the product rulef ( x) e1 2x 8cos 2 x 4sin 2 x 2e1 2 x 1or2xM1:This is scored for cancelling/ factorising out the exponential term. Look for an equation injust cos 2x and sin 2xA1*: Proceeds to tan 2 x 2 . This is a given answer.(b) (i)M1:Solves tan 4 x 2 attempts to find the 2nd solution. Look for x arctan 24Alternatively finds the 2 solution of tan 2 x 2 and attempts to divide by 2Allow awrt x 1.02 . The correct answer, with

Paper 1: Pure Mathematics 1 Mark Scheme Question Scheme Marks AOs 1(a) (i) 32 d 12 24 d y x x x M1 A1 1.1b 1.1b (ii) 2 2 2 d 36 48 d y xx x A1ft 1.1b (3) (b) Substitutes x 2 M1into their 32 d 12 2 24 2 d y x 1.1b Shows d 0 d y x and states ''hence there is a stationary point'' A1 2.1 (2) (c) Substitutes x 2 into their 2 2 2 d 36 2 48 2 d y x M1 1.1b 2 2 d 48 0 d y x and states ''hence the .

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