Mark Scheme (Results) January International Advanced Level in Core Mathematics C12 (WMA01/01)

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1 Mark (Results) January 06 International Advanced Level in Core Mathematics C (WMA0/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere 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 50 years, and by working across 0 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: January 06 Publications Code IA05 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 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 5.. 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) should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. 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. 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

5 subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be 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. If you are using the annotation facility on epen, indicate this action by MR in the body of the script. 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.. Ignore wrong working or incorrect statements following a correct answer. 8. for each question are scored by clicking in the marking grids that appear below each student response on epen. The maximum mark allocation for each question/part question(item) is set out in the marking grid and you should allocate a score of 0 or for each mark, or trait, as shown: 0 am aa bm ba bb bm ba 9. Be careful when scoring a response that is either all correct or all incorrect. It is very easy to click down the 0 column when it was meant to be and all correct.

6 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. 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 correct formula (with values for a, b and c).. Completing the square Solving x bx c 0 : b ( x ) q c, 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 ) 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 mistakes 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. Answers without working The rubric says that these may not gain full credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice.

7 January 06 Core Mathematics C Mark (a) u 6, u ( ) 6 0 or u (6) 6 0 M A ui ( ) ( 0) M i [] + ( 6) Aft = 6 A (a) [] 5 marks M: Attempt to use the given formula correctly at least once. This may be implied by a correct value for u or a value for u which follows through from their u or implied by correct answer for u A: u correct and no incorrect work seen M: Uses sum of the numerical terms from part (a) (may be implied by correct answer for their terms). Attempting to sum an AP here is M0. Aft: obtains u correctly (may be attempted in part (a)) and adds to sum of the first three terms from part (a) A: 6 cao ( 6 implies both A marks) Special Cases: Some candidates attempt u uu u5in part allow M only Some candidates mis-copy one of their terms from part (a) into part allow M only

8 (i) Way : Way : Way : 9 log a 9 a 9 M log or 9 9 a a log a oe or see answer A [] (ii) Way : Way : 0( 8 ) ( 8 )( 8 ) (5 0)( 8 ) M B * 8 0 = 0 so 5 0* 8 Acso [] 5 marks (i) Way : Way : Way : M: Subtracts their powers of M: Cancels fraction to M: Correct use of logs to obtain a and adds their powers of correct expression for a A: cao (answer only is marks) Do not allow work with inexact decimals for this mark e.g log a.5scores MA0 (ii) Way : Way : M: Multiply numerator and denominator by 8 or equivalent. The statement 0( 8 ) is sufficient but do not ( 8 )( 8 ) 0( 8 ) allow unless missing 8 ( 8 ) brackets are implied by subsequent work. B: Correctly obtains ± in the denominator (Must follow M i.e. treat as A). May be 0( 8 ) implied by e.g A: Correct result with no errors seen and 8 used before their final answer. Note that for Way, correct work leading to 5 8 0followed by 5 0 with no intermediate step would lose the final mark M: Attempts to expand (5 0)( 8 ) to obtain at least (not necessarily correct) terms B: All terms correct (Must follow M i.e. treat as A) A: Obtains 0 with no errors and 8 seen or implied by e.g and conclusion that states the given answer i.e. not just 0 = 0

9 . 6x dx = x 6x x x ( c) M A A 6x x x ( c) (8) ( ) n n M: Attempt to integrate original f(x) at least one power increased x x A: Two of the three terms correct un-simplified or simplified (Constant not required) A: All three terms correct un-simplified or simplified (Constant not required) M: Substitutes limits and into their changed function and subtracts the right way round A: cao ( + c is A0) The question requires the use of calculus so a correct answer only scores no marks) M A [5] 5 marks Question. a + d = OR a + d = AND 6 ( a 5 d ) M A 6 ( a 5 d ) A Eliminates one variable to find a or d from equations in a and d dm Obtains a = or d = A Obtains a = and d = A [6] 6 marks MA: Writes down a correct (possibly un-simplified) equation for th term or for sum of the first 6 terms. Allow the individual terms to be added for the sum e.g. a ad ad ad ad a5d Acao: A correct equation for th term and a correct equation for the sum (allow either to be un-simplified) dm: Eliminates one variable from two equations in a and d to find either a or d (see note below) A: One variable correct (This implies previous M mark) A: Both variables correct Note that if both equations are correct and there is no working and the values of a and d are both incorrect, this scores dm0. Also if either or both equations is/are incorrect and values of a and d are obtained with no working this also scores dm0.

10 number 5(a) x 0 Sketch of a positive sine curve- passing through O with at least one complete cycle from O. Condone different amplitudes above and below the x-axis. Correct shape with one and a half cycles as shown (from O to ) and crossing the x axis at and y B B [] Uses h May be implied by use of e.g.... (0 ) M B (a) awrt 0.9 A [] 5 marks as above B: Correct shape with positive gradient through O B: Need not see endpoints labelled. Ignore any part of the curve to the left of the origin but if the curve extends beyond x then then x must be labelled on the diagram. Labels for and may be on the diagram or in the text but not just in a table of values and must be in radians not degrees. (Allow awrt.5 and.) The amplitudes must not be significantly different above and below the x-axis. B: Need ½ of or to see or ½ of 0.6. M: requires first bracket to contain first plus last values and second bracket to include no additional values from the two in the table. If values used in brackets are x values instead of y values this scores M0. A: for awrt 0.9 Separate trapezia may be used: B for and M for ½h(a + b) used times Special Case: Bracketing mistake: i.e. (0 ) scores B M A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). Need to see trapezium rule used so answer only (with no working) is 0/

11 6. f(x) = x + x x 8 (a) Attempts f ( ) M {f(-)=} 0 so (x + ) is a factor of f(x). A x x x8 ( x)( x... M [] or 8 ( )( 6) x x x x x x x x x x x x 8 ( )( )( ) oe A (x =) B [] (a) x or by completion of square ( x ) so x x x 0 x or M A [] marks M: As on scheme must use the factor theorem A: for seeing 0 and conclusion which may be in a preamble and may be minimal e.g. QED, proven, true, tick etc. There must be no obvious errors but need to see at least 80for A but allow 8 0 provided there are no obvious errors. invisible brackets e.g. M: Uses (x + ) as a factor and obtains correct first term of quadratic factor by division or any other method e.g. comparing coefficients or finding roots and factorising A: Correct quadratic and writes ( x )( x x 6) or ( x )( x )( x ) oe Note that this work may be done in part (a) and the result re-stated here. B: States M: Method for finding their roots. Allow the usual rules applied to their quadratic. This mark is for finding the roots and not for just finding factors. You may need to check their roots if no working is shown e.g. if they give decimal answers (.65, -.65 ) 8 A: need both roots. Correct answer implies M mark. Allow x If they give extra roots e.g. x,, 8, lose the final A mark (BMA0)

12 (a) (a) kx kx kx kx M 8 kx, 8 k x, 56 k x... B, A, A Sets "56 k " 5and obtains 5 k M A 56 A So k = [] marks M: The method mark is awarded for an attempt at the Binomial expansion to get the third and/or fourth term. The correct binomial coefficient needs to be combined with the correct power of x. Ignore bracket 8 errors and omission of or incorrect powers of k. Accept any notation for 8 C or 8 C, e.g. or 8 or 8 or 56 from Pascal s triangle. This mark may be given if no working is shown, but either or both of 8k x and 56k xis found. 8 B: This is for + 8kx and not for just A: is cao and is for 8k x or for kx 8( kx ) A: is cao and is for 56k x or for 56( kx ) Any extra terms in higher powers of x should be ignored. Allow terms separated by commas or given as a list for all the marks. n M: Sets their coefficient of x 5 and obtains k.. where n is or 5 A: k or equivalent e.g. (May be implied by their final answer) 56 A: k = cao (± is A0) Note can be marked independently of part (a) so part (a) might be incorrect or not attempted but they have 56k 5 etc. in []

13 8(a) (a) sinx cosx tan x B tan 0 Bft tan " " M One of awrt 8 or awrt or awrt 6 or awrt 5 A Follow through any of their final s for ± 90n within range Aft All of 86.6, 6.6, 66.6, 56.6 A [5] 6 marks B: ( tan x ) or exact equivalent so accept recurring decimal ( ) but not rounded answer Bft: Correct equation as shown or follow through their value for tan x from part (a). Must be tan 0 but 0may be implied later by an attempt to subtract 0 and then divide by. If the processing is unclear or incorrect and 0is never seen, score B0 here. M: Finds arctan of their. Could be implied by their value e.g..9.. or just tan " " A: For one of either awrt 8 or awrt or awrt 6 or awrt 5 Aft: Follow through any of their final answers to which an integer multiple of 90 has been added or subtracted to give another solution in range but not for adding a multiple of 90 to just α. A: For all correct answers to the required accuracy as stated in the scheme. Ignore extra answers outside range but lose last A mark for extra answers inside range. []

14 9.(a) 0000 (.0) 600* or % 600 and * B N Uses 0000 (.0) or (r =).0 B N 0000 (.0) M [] [] N So (.0) A ( N )log 0 (.0) log0 or ( N )log 0 (.0) log0 or ( N ) log.0 or ( N ) log.0 M (d) log log (.0) 0 N * Acso 0 [] (N = ) B [] marks (a) B: A reason must be provided for this mark as the answer is printed. Allow both 0000 % and % as both give the correct answer when entered this way on a calculator. But not 0000 % B: For.0 oe e.g. allow 5 50 M: Correct inequality or equality may use r or their r or.0 and may use N or n. N n A: (.0) cao. Allow (.0) M: Correct use of logs power rule on their previous line which must have come from using the n th term of a GP. Condone missing brackets for this mark e.g. N log 0 (.0) log0. (May follow use of = instead of > or use of r instead of.0 or use of N instead of N - ). These cases can get M0A0M. Allow the base to be absent or just ln for this mark. If the inequality sign is reversed at this point, still allow the M. A*: Answer is exactly as printed (including the bases) and all inequality work should be correct and all previous marks scored and no missing brackets earlier. Allow this mark to score from a correct previous line provided the power rule is used. So fully correct work leading to log0 ( N ) log 0 (.0) log0 N scores the final MA but log 0 (.0) N log0 (.0) N scores M0A0 (no explicit use of power rule) log (.0) (d) 0 B: Only need N = may follow trial and error or uses logs to a different base. Do not allow N or N or N =.0

15 0.(a) 5 5 y x x dy 5 0 x x M A dx 8 [] 5 Put 0 n x x0 so x k n, k 0 M 8 x dm x 8 (Ignore x = 0 if given as a second solution) A 5 So y (8) 5 (8) 000 i.e. y = 9.5 dma 8 [5] d y dx = 5 5 x Bft 9 Substitutes their non-zero x (positive or negative) into their second derivative. M Obtains maximum after correctly substituting 8 into correct second derivative to give correct 5 negative quantity o.e. or decimal e.g (see note below) and considers negative 6 sign deducing maximum. A d y Note that a correct second derivative followed by x = 8 8 therefore dx 9 maximum scores BMA0 here. [] 0 marks (a) n n M: Attempt to differentiate power reduced by one x x (but not just 000 0) A: Two correct terms and no extra terms. Terms may be un-simplified. n M: Puts derivative = 0 and attempts to solve to obtain an equation of the form x k where n is real and k is non-zero dm: Correct processing to obtain a value for x. (Dependent on the first method mark). This mark can only be awarded for processing an equation of the form ax correct powers of x. E.g. 0 ax bx x a bx x k or bx 0 i.e. their derivative must have the ax bx 0 ax bx px qx x k Do not allow incorrect squaring e.g. ax bx 0 px qx 0etc. A: cao dm: Substitutes their positive value for x into y = and not into d y... (Dependent on the first dx method mark) A: cao If x = 8 appears from no working following a correct derivative score MM0A0 then allow full recovery. Bft: Correct follow through second derivative M: Substitutes their non-zero x (positive or negative) into their second derivative. d y Note: Solving 0 is M0 dx Acso: Completely correct work ( o.e.). Note that o.e. could be or 5 5 or or -0..but it has to be correct for the final mark.

16 (a) 6 0 0cos YXZ M 0 6 cosyxz or or A BOC.6(08..) (N.B. 9.8 degrees is A0) A Uses s 5 with their from part (a) M [] awrt 8. A Perimeter r 8, 8 their arc length M awrt 6. A area of sector (5) Bft area of triangle 0sin ( 59.9 or 59.9) Bft [] Area of shaded region = 0sin = 9. (cm ) (5) = M A (a) M: Uses cosine rule must be a correct statement A: Correct value or correct numerical expression for cos YXZ A: accept awrt.6 and must be seen in part (a) (answer in degrees is A0 (9.865 )) M: Uses s 5 with their in radians, or correct formula for degrees if working in degrees ( marks) A: Accept awrt 8. (may be implied by their perimeter) M: Adds their arc length to 8 or ( ) A: Accept awrt 6. do not need units (ignore any given) Bft: This formula used with their θ in radians or correct formula for degrees Bft: Correct formula for area used may use half base times height (may be implied by a correct answer (59.9 )) M: Subtracts their sector area from their triangle area this way round. A: awrt 9. do not need units (ignore any given) Alternative approach to finding angle YXZ and area of triangle: Let foot of perpendicular from X to YZ be W and XW = h and YW = x so WZ = 6 x: , 6, M: Correct work leading to values of and YXZ sin sin.6 A:Correct expression for YXZ, A: awrt.6 80 h x h x x h x h The B for the triangle area in can then score for 6 " 99 ". Note this is 99 8 []

17 (a) and can be marked together (a) 6 f( ) x 9 x x M x f( x) 6x x 9 MAA f( x) 6x x M A (a) When x =, y = 5 B f() M 8 5 Equation of tangent is y 5 ( x) M A [] 0 marks M: expands numerator into a three (or four) term quadratic in x (allow x for x ) M: Divides at least one term in numerator by x correctly following an attempt at expansion. May just be 6 x. A: Two correct terms A: All terms correct M: Evidence of differentiation k k x x k x n n x of an expression of the form Ax or k Bx so and not just C 0. Differentiating top and bottom separately is M0. x x Note this is a hence and so attempts at e.g. use of the quotient rule scores M0. A: cao and cso (May be un-simplified) Note: An incorrect constant in part (a) (e.g. instead of 9) will fortuitously give the same derivative so scores MA0 if otherwise correct. B: 5 only M: Substitute x = into their derived function M: Uses their 5 and their gradient which has come from calculus (not the normal gradient) and x = to give correct ft equation of line. If using y = mx + c must at least obtain a value for c A: any correct form e.g. 5 y x5, 5xy0 0 y 5 5 BUT NOT JUST, this scores MA0 x Note: An incorrect constant in part (a) (e.g. instead of 9) will fortuitously give the correct answer in and will lose the final A mark if otherwise correct. or [] []

18 (a) (a) kx (8k 6) x 9k 0 or kx 8kx 6x 9k 0 B Uses b ac with a k, b 8k 6 and c 9k M k 0k6 0 or 6 k 0k o.e. Reached with no errors A k 0k9 0 * A* [] Attempts to solve k 0k9 0 to give k = M Critical values, k, A k (or) k M Acao [] 8 marks B: Multiplies by k and collects terms to one side in any order. Allow the x terms not to be combined and the = 0 may be implied by use of a correct discriminant. M: Attempts b ac with a k, b 8k 6 and c 9k or uses quadratic formula with b ac seen to solve their equation or uses b ac or e.g. b ac. There must be no x s. A: Obtains a correct three term quadratic inequality that is not the printed answer with no errors seen. A: Correct answer with no errors 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 the given three term quadratic A: Obtains k, accept awrt - 0. M: Chooses outside region (k < Their Lower Limit k Their Upper Limit ) for a term quadratic inequality. Do not award simply for diagram or table. A: k (or) k must be exact here but allow 0. for. Allow other notation such as,, k and k and k score MA0 ISW if possible e.g. k, k followed by k can score MA k, k followed by k (or) k can score MA Allow to be solved in terms of x for the first marks but the final A mark needs the regions in terms of k. Fully correct answer with no working scores full marks. Answers that are otherwise correct but use, lose final mark.

19 (i) x loga x loga loga so loga Or log a x loga loga log a a so log ax loga a Or loga x loga loga loga9 so loga ax loga9 M A x a M x 9a or A a 9 (ii) x x 0 and attempt to solve to give x = or log5 y (implied by correct M answers) x (or log 5 y) and A (i) y 5 or 5 dm y = 5 and 65 A [] 8 marks M: Uses sum or difference of logs correctly e.g. log x log log x or log log log 9 or log log x log etc. x or writes as log a a A: Uses two rules correctly to obtain correct log equation M: Removes logs correctly to obtain an equation connecting x and a A: Correct simplified answer Note that some candidates interpret loga as loga. This can score a maximum of out of if they have log x log logx loga x x Note that loga x loga loga so a etc. scores MA0M0A0 loga loga x loga x Note that loga x loga loga so a etc. scores no marks log a [] (ii) M: Recognise and attempt to solve quadratic A: Obtain both and (Both correct implies MA) dm: Uses powers correctly to find a value for y (Dependent on first method mark) A: Both values correct

20 5 (a) Mid-point of AB = (, - ) M A ( r ) ( "") ( " ") or ( r ) ( 8 "") ( 8 " ") or ( d ) ( 8 ) ( 8 ) M r 5 A "5" ( x"") ( y" ") M 5 ( ) ( ) x y A 8 " " gradient from "(, )" to (, 8), "" M ZM has gradient m M [6] Either : y 8 " " x or: y " " x c 8 and 8 " " c c "8 " ddm x + y 96 = 0 A [] (0marks) (a) M: Uses midpoint formula, or implied by y coordinate of - or x coordinate of A: cao M: Finds radius or radius, diameter or diameter using any valid method probably distance from centre to one of the points. Need not state r = so ignore lhs you are just looking for correct use of Pythagoras with or without the square root so ignore how they reference it for this mark. A: for any equivalent r 5 or r 5 (.8...) etc. Their numeric answer must be identified here as either r or r (may be implied by their equation). If they halve it or double it, this is M A0. M: Attempt to use a true equation for circle with their centre and radius or the letter r, allow sign slips in brackets but do not allow use or r instead of r in the equation. So must be using r ( x...) ( y...) A: correct answer only (Allow 5 5 instead of 5 but not 5 5 ) M: States or uses gradient equation correctly with their centre and (, 8). Must be using their centre and (, 8). If no method is shown and gradient incorrect for their values score M0. M: Finds negative reciprocal. Follow through their gradient ddm: Correct straight line method with (, 8) and perpendicular gradient. Dependent on both previous method marks having been scored. A: cao accept multiples of this equation (Note integer coefficients not required) A common error here is to use the diameter to find the gradient. This usually scores M0MddM0A0 i.e. just one mark for the perpendicular gradient rule. Alternative uses implicit differentiation: e.g. dy 5 ( x) ( y) ( x) ( y) 0 M(correct implicit differentiation) oe dx dy x M(Substitution) dx y 8 Then follow the scheme.

21 6(a) x x x M x 9x0 x = ½ or x = dm A y = 5/ or y = dm A [5] Curve meets x-axis at x and at x = (No need to see y = 0) M A Way Way Way NOTE that the subscripted A s refer to areas on the diagram given at the end of the scheme. All the method marks are for their x = /,, and x x d x x x x M A Use limits and ½ [( () () ) ( ( ).( ) ( ))] A M Use limits and [( () () ()) ( ().() ())] A M Area of trapezium = 5 a b h x x x x 6 ( ) ( ) ( )... or ( )d ( ) ( ) ( ) (may be implied by correct final answer) A Uses correct combination of correct areas. Area of region = Area of trapezium A A Dependent on all previous method marks ddddm 9.5 or A [8] Alternative method using line curve and subtracting area below x- axis 9 x 9 9 x 9 x xdx x x or x xdx x x MA Use limits ½ and on this subtracted integration A A A5 A M x x d x x x x M Use limits and on their integrated curve to obtain A6 MA Uses correct combination of correct areas. Area of region = A A A5 A6 A6 Dependent on all previous method marks ddddm A [8] Alternative method using line curve for areas A and A and adding smaller trapezium 9 x 9 9 x 9 x xdx x x or x xdx x x MA 9 9 Use limits and ½ [( () () ) ( () () ] A M 9 9 Use limits and [( () () ) ( () () ] A M Area of trapezium = 5 9 ( ab) h ( ) ( )... or ( x)d x x x ( ) ( )... M = A Uses correct combination of correct areas. Area of region = A A A 5 Dependent on all previous method marks ddddm A [8] M []

22 Way (a) Way Way Alternative method: Finds area of larger trapezium and subtracts A + A which is found by integrating quadratic between ½ and and adding area below x-axis x x d x x x x M A Use limits and ½ [( () () ) ( ( ).( ) ( ))] A A A6 M AND Use limits and [( () () ) ( ().() ())] A6 Area of trapezium = 5 a b h x x x x 6 ( ) ( ) ( )... or ( )d ( ) ( )....5 ( ) 6 (may be implied by correct final answer) A Uses correct combination of correct areas. Area of region =.5 A A A A Dependent on all previous method marks 6 6 ddddm A M [8] 5 marks M: Puts equations equal or finds x in terms of y and substitutes or substitutes for x dm: Solves three term quadratic in x to obtain x = or in y to obtain y = (Dependent on first M) A: Both answers correct dm: Obtains at least one value for y or x (Dependent on first M) A: Both correct 9 Note: Allow candidates to obtain x x0and solve as xx0 x, The coordinates do not need to be paired M: Attempts to solve 0 x x according to the usual rules A: cao Attempts by T&I can score both marks for x = and x =. If one solution is obtained by this, score MA0 For do not allow mixed methods. For their strategy, they must be finding the appropriate areas but apply the method for the scheme that gives the most credit for the candidate. n n x x at least once M: Attempt at integration of the given quadratic expression A: Correct integration of the given quadratic expression M: Finds area of A M: Finds area of A M: Finds area of appropriate trapezium A: Correct area of trapezium.5 ( ) ddddm: correct final combination A: any correct form of this exact answer 6 n n M: Attempt at integration of ±(the given quadratic expression the given line) x x at least once A: Correct integration as shown in the mark scheme. Allow correct answer even if terms not collected nor simplified. If there are sign errors when subtracting before valid attempt at integration, score MA0 M: Uses the limits ½ and on their subtracted integration M: Attempts to integrate curve M: Uses the limits and on the integrated curve C A: Obtains A6 ddddm: correct final combination A: any correct form of this exact answer Note: A common error with this method is to use the limits ½ and on their subtracted integration and then stop (this should give an area of ). This will usually score /8 in 8

23 Way Way n n M: Attempt at integration of ±(the given quadratic expression the given line) x x at least once A: Correct integration as shown in the mark scheme. Allow correct answer even if terms not collected nor simplified. If there are sign errors when subtracting before valid attempt at integration, score MA0 M: Uses the limits ½ and on their subtracted integration M: Uses the limits and on their subtracted integration M: Finds area of appropriate trapezium A: Correct area of trapezium ddddm: correct final combination A: any correct form of this exact answer n n x x at least once M: Attempt at integration of the given quadratic expression A: Correct integration of the given quadratic expression M: Finds area of A A A6by using the limits ½ and and finds area of A 6 by using the limits and M: Finds area of appropriate trapezium A: Correct area of trapezium.5 ( ) ddddm: correct final combination A: any correct form of this exact answer 6 Diagram for Question 6, 5 A A, 9 8 A5 A 6, 5 A A,

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