Mark Scheme (Results) October Pearson Edexcel IAL in Core Mathematics 12 (WMA01/01)

Transcription

1 Mark Scheme (Results) October 06 Pearson Edexcel IAL in Core Mathematics (WMA0/0)

2 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 at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 0 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: October 06 Publications Code WMA0_0_60_MS All the material in this publication is copyright Pearson Education Ltd 06

3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their 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 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. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 PEARSON EDEXCEL IAL MATHEMATICS General Instructions for Marking. The total number of marks for the paper is. The Edexcel Mathematics 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 have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe 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 given or d The second mark is dependent on gaining the first mark

5 4. All A marks are correct answer only (cao.), unless shown, for example, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft.. 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. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.

6 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quaatic:. Factorisation ( x + bx + c) = ( x + p)( x + q), where pq = c, leading to x = ( ax + bx + c) = ( mx + p)( nx + q), where pq = c and mn = a, leading to x =. Formula Attempt to use the correct formula (with values for a, b and c).. Completing the square b Solving x + bx + c = 0 : x ± ± q ± c = 0, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n+ )

7 Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are small errors in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.

8 . f( x) = x + x 4x + 6x ( ) + + x x 4x 6x x x 4x 6x dx= + + ( + c) M AAA x = x + 8x x + c A [] marks M: Attempt to integrate original f(x) one power increased n n x x + A: Two of the four terms in x correct un simplified or simplified (ignore no constant here). They may be listed. + x x x is acceptable for an un simplified term BUT x isn't + A: Three terms correct (may be) unsimplified. They may be listed separately A: All four terms correct (may be) unsimplified on a single line. A cao: All four terms correct simplified with constant of integration on a single line. You may isw after sight of correct answer.

9 . x log 7 = log4 or x log 49 = log4 or x = log74 log4 x = = awrt MA log 7 M () x + = M 8 So x = or 0. A () M: Uses logs and brings down x correctly M: Makes x the subject correctly. This must follow a method that did involve taking logs A: Accept awrt (N.B. Correct answer with no working implies two previous marks) marks M: Uses powers correctly to undo log. Accept x + = or equivalent such as x + = 0.04 A: Correct answer (Correct answer implies method mark). Accept 0.0

10 (i) (i) (ii) 0 = = ( 6) LHS = ( + 6)( 6) = oe = = * 4 = M A* M A [] A* [] marks M: Shows at least one term on LHS as multiple of with a correct intermediate step Look for 4 = 9 or =, or even 4= or 9 followed by 4 = = or = 4 or = = 6 6 or 6 0 = 80 = 6 = 6 or even 80 = followed by 80 = 6 A*: All three terms must have the intermediate step with followed by Special Case: Score M A0 for = = without the intermediate steps Alternative method: M: Multiplies all terms by to achieve = and simplifies any one of the above terms to, -0, 0 or showing the intermediate step A: All terms simplified showing the intermediate step (see main scheme on how to apply) followed by =, and minimal conclusion eg. hence true (ii) M: Multiply numerator and denominator by 6or A: Multiplies out to a correct (unsimplified) answer. For example allow = A: The denominator must be simplified so or similar such as 4 4 is seen before you see the given answer. There is no need to 'split' into two separate fractions. Alternative method: M: Alternatively multiplies the rhs by ( + 6)( ) A: Correct unsimplified rhs Accept A*: Simpifies rhs to 7 and gives a minimal conclusion e.g. hence true or hence ( + 6) =

11 4. f (x) = 6x 7x 4x+ 0 (i) Attempts f( ± ) Or Use long division as far as remainder M Remainder = 49 A (ii) Attempts f ( ± ) Or Use long division as far as remainder M Remainder = 0 A 6x 7x 4x+ 0 = ( x )(6x + x 0) M A [4] x + x = ax + b cx + d ac = bd = M = ( x )(x+ )(x ) A (6 0) ( )( ) where "6"and " 0" [4] (i) M: Attempts f( ) ± or attempts long division x + x+ 6x 7x 4x marks and achieves a numerical R A: cao Accept f = 49 R or even just 49 for both marks If the candidate has attempted long division they must be stating the remainder = 49 or R = 49 (ii) M: Attempts f ( ± ) Or attempts long division. See above for application of this mark. This time quotient must start 6x A: cao Accept f () = 0 or even just 0 for both marks If the candidate has attempted long division they must be stating the remainder = 0 or R = 0 M: Recognises (x - ) is factor and obtains quaatic factor with two correct terms by any correct method. If division is used look for a minimum of the first two terms 6x ± x... x 6x 7x 4x+ 0 6x 8 x If factorisation is used look for correct first and last terms 6x 7x 4x+ 0 = ( x )(6 x... x± 0) A: Correct quaatic M: Attempt to factorise their quaatic A: cao need all three factors together. Do not penalise candidates who go on to state the roots. Allow 6( x ) x+ x following ( x )(6x + x 0) Note: There may be candidates who just write down the factors from their GC. The question did state hence so we need to be careful here and see some correct work. 6x 7x 4x+ 0 = ( x ) x x+ presumably from the roots is M0A0M0A0 6x 7x 4x+ 0 = 6( x ) x x+ with no working can score MA0MA0

12 . ax 4 ax ax ax = = 4, ax + a x a x a = a M 4 M B, A, A 80 a = = 8 or equivalent A 4 a = A [] 7 marks M: The method mark is awarded for an attempt at Binomial to get the second and/or third and/or fourth term. You need to see the correct binomial coefficient combined with correct power of x. e.g... x Condone bracket errors. Accept any notation for C, C and C, e.g., and or, 0 and 0 from Pascal s triangle. The mark can be applied in the same way if is taken out as a factor. B: For the first term of 4. (writing just is B0 ) A: is cao and is for two correct and simplified terms from ax, + a x and a x Allow two correct from ( ax), + ( ax) and ( ax)... with the brackets. 4 Allow decimals. Allow lists A: is c.a.o and is for all of the terms correct and simplified. 4 Allow + ( ax) and ( ax)... (ignore x 4 terms) 4 Allow decimal equivalents 0.ax a x. a x... Allow listing. M: Puts their coefficient of x equal to their coefficient of x (There should be no x terms) A: This is cao for obtaining a or a correctly (may be unsimplified) A: This is cao for a = Condone a = ± We will condone all marks to be scored in from a solution in where all signs are +ve 40 4 = 4 + ax + a x + a x... 4 [4]

13 6. u = 4, u = 6 and u = r = 4 M, A [] B [] (c) 0 ( ) 0 u = ar = r. M 6 (d) ( ) 6 (e) u 6 ui = or i= i i= 0 0 = ar = 6 = 6( ) 6 u = = = 0.64 A 4 = ui = u + u6 M i= M: Attempt to use formula correctly at least twice. It may be seen for example in u and u4 A: All three correct exact simplified answers. Allow 0.6 B: Accept (c) 0 M: Uses ( ) 0 4 or equivalent such as Allow awrt u = ar = r with their r 6 Acao [] [] M A [] 9 marks A: Accept awrt 0.64 or (d) M: Uses correct sum formula with a = 6 and their r or alternatively for adding their first six terms. FYI Sight of 6, 4, 6, 0.7, 7., 4.7 followed by 98. implies this mark. (You may only see the first 4 terms in part a) 4 A: Obtains = 98 7 (must be exact). For information (e) 660 = Allow M: Uses correct sum to infinity formula with a = 6 and either A: Obtains 08 (must be exact) r = or their r as long as r <

14 y 7. Shape and position correct (0, /9) correct B B O x [] State h = 0., or use of 0. ; B aef { 0.9 ( )}... ; MA For structure of { } 0. {0.476} = awrt.6 A [4] 6 marks B: Curve just in quaant one and two with a gradient that is approaching zero at the lhs and increases as x increases. Curves that just cross the y axis into quaant may be penalised. As a rule of thumb expect it reach at least as far as x = -. B: The point (0, /9) lies on the curve. Accept /9 marked on the y axis. Accept a statement when x = 0, y = /9 Do not accept or 0.. Condone 0,0. B: For using 0. or h = 0. or equivalent such as (-0.) M: Scored for the sight of the correct structure for the outer bracket. You need to see the first y value plus the last y value plus times a bracket containing the sum of the remaining y values with no additional values. If the only mistake is a copying error or is to omit one of the remaining y values then this may be regarded as a slip and the M mark can be allowed (An extra repeated term forfeits the M mark however) ( ) or awrt 8.08implies this mark A: For { ( )} or { ( ) + ( )} oe A: For answer which rounds to.6. Correct answer implies all 4 marks NB: Separate trapezia may be used: B for 0., M for / h(a + b) used 4 or times followed by A (if it is all correct ) and A as before.

15 8. sin D sin. = 6 M sin D = so D = 0.84 M, A B = π ( ) =.0 * A* [4] Uses angle DBC = π. = awrt.94 B Area of sector is r θ = 6 '.94' or Area of triangle ABD = 6 sin. M ( = 4.9) ( = 4.0) Total area is 6 '.94' + 6 sin. dm = 48.9cm A [4] 8 marks M: Uses sine rule the sides and angles must be in the correct positions M: Makes sin D the subject and uses inverse sine (in degrees or radians) A: Accept awrt 0.84 or in degrees accept answers truncating or rounding to 48.0 A*: Answer is printed so should see either π (.+ awrt 0.84) or π. awrt 0.84 before you see.0 If the question was changed to degrees look for accuracy to one decimal places throughout the question π for the final A mark. So. rads = awrt 6.0 and ( 80 awrt 6.0 awrt 48.0) = awrt =.0 80 There are many ways to attempt this question: For example M: Uses cosine rule 6 = + x xcos. (where x =AD) and attempts to solve to find x. For information x their '6.9' M: Uses cosine rule cos B = ( awrt 6.9) A: Achieves cos B = 6 A:.0* B: Uses angles on a straight line formula. Score for π. or allow awrt.94 as evidence. If converted to degrees accept awrt. as evidence M: Uses a correct area formula for the sector or a correct area formula for the triangle. You may follow through on an incorrectly found angle DBC For example π. is acceptable but 80. is not as it is using mixed units. If the angle was found in degrees, the correct formula must be used. For the triangle the correct combinations of sides and angle should be attempted. area of triangle ABD = their 6.9 sin. or 6 their 6.9 sin their ADB dm: Adds together a correct area formula for the sector and a correct area formula for the triangle. You may follow through on an incorrectly found angle DBC or ADB A: Accept awrt 48.9 (do not need units) e.g. You may see the ( ) ( ) ( )

16 n 9. Uses ( a+ ( n ) d) with n = 0 to give 0a + 4d = 9 * n Uses ( a+ ( n ) d) with n = 8 and S=97 M Obtain 8a+ d = 97 or a+ 7d = 0 A [] (c) Solves simultaneous equations to find either a or d M a = 6 and d = A, A [] (d) Uses a+ ( n ) d with n = 0 M = 8 A [] 8 marks Mark the whole question as one. B: Use the correct formula for the sum of an AP with n = 0, S =9 AND proceeds to the given answer. It is acceptable for the 9 to appear just at the answer stage. Could use formula with n = 0, S =9 and l = a+9d It is OK to list but minimum would be a+ a+ d + a+ d... + a+ 9d = 9 8 M: Obtain a correct second equation e.g. 97 = ( a+ (8 ) d) or equivalent. Condone a slip on the 97. Note that if the candidate reads 97 as 97 they will only have access to M marks in this question. This is due to the fact that with this number, the values of a and d would be fractional and this could not occur as they must be integers A: A simplified equation so accept either 8a+ d = 97 or a+ 7d = 0 Sight of one of these scores both marks. (c) M: Solves simultaneous equations to find either a or d. Do not concern yourself with the process as calculators are allowed on this paper so score if they proceed to either a and/or d A: Obtains correct a or d (just one) A: Obtains correct a and d (both) (d) M: Uses correct formula for n th term using their a and d but with n = 0. Look for ' a' + 9' d' A: Correct answer B* []

17 0. Use sin x tan x cos x = to give 8sin x= cos x M Use cos x= sin x i.e. 8sin x= ( sin x) M So 8sin x= + sin x and sin x 8sin x = 0 * A * [] Solves the three term quaatic sin x 8sin x = 0 M So ( sin x ) = (or ) A ( θ ) = 9.47or or 40. dm θ = 99.7, 70., 79.7 or 0. A, A [] M: Use sin x tan x cos x = to give 8sin x= cos xor equivalent M: Use cos x= sin x i.e. 8sin x= ( sin x) May also be seen 8 tan x= cos x 8 tan x= sin x A: Proceeds to given answer with no errors. (This is a given answer so do not tolerate bracketing or notation errors such as cos x written as cos x or sin x appearing as sin ) M: Solving quaatic by usual methods (see notes). If the formula is quoted it must be correct but allow solutions from calculators. A: You only need to see. This is an intermediate answer so condone appearing as awrt 0. Condone errors on the lhs so accept for this mark x/ a/ θ =, sin x=, sin x= dm: Uses inverse sine to obtain an answer for θ. This may appear as answers for x. The only stipulation is that invsin k, k < It is dependent upon seeing a correct method of solving their quaatic Accept answers rounding to dp for θ e.g. awrt 9. or 99. or 40.. It may also be implied by a correct answer for θ e.g. awrt 9.7 or 99.7 or 70. A: Two correct, awrt one dp θ = 99.7, 70., 79.7 or 0. A: All four correct, awrt one dp θ = 99.7, 70., 79.7 or 0. 8 marks

18 . (k ) x kx 6 = 0 or ( k) x + kx + 6 = 0 B Uses b 4ac with a =± k ±, b = ± k and c = ± 6 M And states b 4ac > 0 with a =± ( k ), b = ± k and c = ± 6 Aft Proceeds correctly with no errors to 6k + k > 0 * A* Attempts to solve 6k + k = 0 to give k = M Critical values, k =, A 6k k 0 + > gives k > (or) k < M A [4] 8 marks B: Expresses equation as three term quaatic in x. (k ) x kx 6 = 0 oe. The equals 0 may be implied by subsequent work. Allow ( k) x + kx + 6 = 0 Allow an equation of the form kx x kx 6( = 0) as long as it is followed by a = k... M: Attempts b 4ac with a =± k ±, b = ± k and c = ± 6 or uses quaatic formula to solve equation or uses the discriminant on two sides of an equation or inequation e.g. b = 4ac or b < 4ac A: Uses the discriminant condition, eg b 4ac > 0 or b > 4ac with a =± k ±, b = ± k and c = ± 6 A*: Proceeds to given answer with no errors. AG. Condone missing = 0 on the equation Condone a solution where (k ) x kx 6 = 0 is followed by 44k + 4(k ) > 0 Watch for a = k, b = + k and c = 6 which does give the correct inequality but loses the final A* M: Uses factorisation, formula, or completion of square method to find two values for k, or finds two correct answers with no obvious method for their three term quaatic A: Obtains accept -., 0. (awrt) here but need exact answer for final A. k =, Also condone x =, for this mark. M: Chooses outside region ( k < Their Lower Limit k > Their Upper Limit inequality. Do not award simply for diagram or table. Award if final answer is k (or) k or < k < Condone x appearing instead of k must be exact and must be k. A: k > (or) k < k Must be two separate inequalities and not be k > and k < ) for appropriate term quaatic [4]

19 x 9x 8x f( x) = = 0 xx ( 9x 8) = 0 M 7 9 ± 8+ 4 x = dm 9 ± 40 9± 9 A A x = or x = [4] Differentiates (usual rules), correctly and sets = 0 f ( x) = x 8x 8 = 0 M, A Solves f( x) = 0(or multiple) x = 9 and - dm A Substitutes one of their values for x into f(x) ddm x = 9 y = -7 and x =- y = A [6] (c) a = 9 B [] marks M: Attempts to solve f(x) = 0, by taking out a factor of (/cancelling by ) x and obtaining a quaatic factor. x 9x 8 Allow on x = 0 or just the numerator xx ( 9x 8) = This is implied by sight of x 9x 8 = 0 dm: Uses formula or completion of square method to find at least one value for x, for their three term quaatic. Factorisation is M0. Note that their term quaatic equation may be x x = A: One correct solution need not be fully simplified. So allow x = but not x = A: Two correct solutions need not be simplified or attributed correctly to A or B. Special case: If a candidate takes out a common factor of x and uses a calculator to write down the exact surd answers to the quaatic they have used (a limited) amount of algebra. Decimals would not be awarded for this SC. We will therefore score this SC M M A0 A0 for out of 4. down the answers with no working scores 0 marks M: Differentiates f(x) to a term quaatic You may see confusion over the 7 but score for f( x) A: Differentiates correctly and sets correct derivative = 0 x 8x 8 0 9± 9 xx ( 9x 8) = 0 x= Just writing being a term quaatic = or any multiple thereof. For example it may be common to see dm: Solves quaatic to give two solutions. It is dependent upon the previous M. Allow any appropriate method including the use of a calculator. Condone x x = 0 ( x 9)( x+ ) = 0 9 x 8x 8 = A : Gives both 9 and ddm: Substitute at least one of their values of x (obtained from a solution of f( x) = 0) into f(x) to give y =. A: Gives both 7 and (arising from x values of 9 and ) (Do not require coordinates). Again they do not need to be attributed correctly to C or D (c) B: For a = 9 only (no ft)

20 (c) See ( x ) ( y ) r ± + ± = Or see Attempt (8 ) + ( ( )) or (8 ) + ( ( )) ( x ) ( y ), 0 x + y ± x± 6y+ c= 0 M Substitute ( 8, ) into equation M + + = x + y x+ 6y 40 = 0 Gradient of AP = 7 So gradient of tangent is 7 Equation of tangent is ( y +) = 7(x 8) y = 7x + 4 or m = -7, c =4 Way Way y = x + 6 meets circle when As tangent has gradient AQ has ( x ) + ( x+ 9) = 0 or when y ( ) gradient - and ( y 7) + ( y+ ) = 0 i.e. x + 6x+ = 0 or when y 8y+ 8 = 0 Solve to give x or y = x = A, A [4] B M dm A M y+ x= A Solve y+ x= with y = x + 6 or alternatively solve y+ x= with the equation of the circle to give x or y = M [4] Substitute to give y = (or x = ) dm ( A 4, ) only [] marks M : Scored for centre at (, ) ( x± ) + ( y± ) =... or x + y ± x± 6 y+... = 0 M: Scored for an attempt at finding the radius or the radius (see scheme). It need not be in the equation It can be implied by 0 or or 0 If the form x + y ± x± 6y+ c= 0is used it is for substituting ( 8, ) into the equation A: LHS or RHS correct ( x ) + ( y + ), =... or ( x ± a) + ( y ± b), = 0 x + y x+ 6 y... = 0 A: Correct equation. Accept ( x ) + ( y + ) = 0 or x + y x+ 6y 40 = 0 or x + y x+ 6y = 40 B : Obtain /7. Implied by use of 7 in their tangent M: Uses negative reciprocal dm: Linear equation through point (8, -) with their negative reciprocal gradient A: cao (c) M: Eliminates x or y from two relevant equations, that is whose intersection is Q. A: Correct quaatic in x or in y M: Solves (with usual rules) to give first variable. The first M must have been scored dm: Substitute in either (relevant) equation to give second coordinate, dependent upon both previous M's A: Correct answer accept x = 4, y =. Withhold this if two answers given

21 4. y x x = dy x 6 dx = + and substitutes x = to give gradient = m = 4 M A Normal has gradient = m 4 Equation of normal is ( y+ ) = " "( x ) so x 4y 7 = 0 4 x x x + 6 x 8d x = x M The Line meets the x-axis at 7 The Curve meets the x-axis at 4 Uses correct limits correctly for their integral x x 4 4 i.e x = ( ) 4 M dm A [] Finds area above line, using area of triangle or integration = ("7" ) M Area of R = 8 + = 9 A [6] marks B B M

22 M: Differentiates to give d y =± x ± 6and substitutes x = dx A: Obtains answer 4. M: Uses negative reciprocal of their numerical d y (follow through). M must have been awarded dx dm: Linear equation through point (, ) with their changed gradient. Dependent upon the first M, so you would allow for ( y+ ) = 4( x ) following an answer of 4 A: cao accept kx ( 4y 7) = 0 where k is a positive or negative integer Candidates who work with a gradient of ± from their d y =± x ± 6will score 0 marks in this part of the dx question. M: Integrates a quaatic expression correctly. If they integrate (line -curve) follow through on their new quaatic The terms including the coefficients must be correct for their quaatic B: Obtains 7 for the point where the line meets the x - axis B: Finds that the curve meets the x axis at 4. You may score this for y = 0 x=, 4 ignoring even an incorrect Also allow for a limit in the integral. You may even score this if 4 appears (in the correct place) on the diagram M: Uses the limits 4 and in their integrated function If a candidate writes down ( ) 4 4 ± x + 6x 8dx=± (from a GC) we will allow them to score this mark. M: Finds appropriate area above the line for their attempted integral, so if they integrate just curve look for area of triangle = " their 7 " or ''7 '' ''7 '' x ''dx= x x 7 7 '' if they integrate (line - curve) from 4 to, then the triangle would be = their '' '' " their 7 4" 4 A: correct work leading to 9 A candidate who does the integration on a GC can potentially score M0 B B M M A0

23 00 = πr + πrh + rh M A 00 π r ( h = ) π r+ r or 00 π r ( rh = ) π + dm V = π rh= dv 00π π r = 4+ π 00π π r = 0 or 4+ π πr (00 πr ) πr(00 πr ) V = = ( r+ πr) 4 + π dv Accept awrt 6..9r = 00 0 π π r = leading to r = M * A cso * M A dm [] 00 r = or answers which round to 4.6 dm A π V = 88 B (c) d V 6π r d V =, and sign considered Accept 4 + π = awrt.8r M [6] d V d r r=.. = 7 < 0 and therefore maximum A marks []

24 M: Sets total surface area equal to 00 with at least two correct terms. Note that 00 = πr + πrh or even 00 = πr + πrh + πr does not mean that two terms are correct. A: Completely correct 00 = πr + πrh + rh dm: Makes h or rh the subject of their formula which must have had two terms in h This is dependent upon the previous M M: Gives formula for volume. This may be implied by sight of V = π r their h πr(00 πr ) A*: cso substitutes for r or for rh correctly and proceeds correctly to V = 4+ π Parts b and c can be scored together dv M: Attempts to differentiate V or numerator of V Accept A Br = ± dv You may see ( 4+ π ) = A ± Br if candidates multiply by ( 4+ π ) first A: Accept any equivalent correct answer or correct numerator if only this was considered. Also accept decimals. dm: Setting d V =0 and finding a value for r using correct mathematics (May be implied by answer). d r Note that you may not see dy r. It is acceptable to go straight to r. Allow = 0 dx dm: Using square root to find r. Dependent upon all previous M's. An answer of for r following a correct derivative may imply this mark as some candidates find r to the nearest cm rather than V to the nearest cm. If you don t see incorrect work you may award this mark. 00 A : For any equivalent correct answer. Accept r = π or awrt 4.6 Correct answer implies previous two M marks B : Obtain V= 88 Exact answer only. Do not accept, for example, 87.8 (c) d V M: Score for either a second derivative of = ± Cr and considers the sign. πr(00 πr ) It can be implied by A ± Br ± Cr and a consideration of the sign 4+ π d V Or a second derivative of = ± Cr and substitutes in their value of 'r' from d V 6π r d V Or a completely correct second derivative = accept = awrt.76r 4 + π A: Clear statements and conclusion. For both marks d V () must be correct (see above), not just the numerator. () A statement (which could be implied) that when their r (which does not need to be correct) is d V d V substituted into then is either negative or < 0 () and a minimal conclusion such as hence maximum d V d V For example, accept for both marks =.76r When r = 4. < 0, hence max

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