Mark Scheme (Results) Summer 05 Pearson Edexcel GCE in Core Mathematics 4 (6666/0)
Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding bo. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. 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 50 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: www.pearson.com/uk Summer 05 Publications Code UA040 All the material in this publication is copyright Pearson Education Ltd 05
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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.
EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. 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) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark dm denotes a method mark which is dependent upon the award of the previous method mark. 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, 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. 5
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 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 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 )
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
Question Number. (a) x June 05 6666/0 Core Mathematics 4 Mark Scheme Scheme 5x 5x (4 5 ) 4 4 4 (kx) ( )( ) (kx)...! (4) or B Marks see notes M Aft 5x 4 ( )( ) 5x! 4... 5 5 x x... 8 8 See notes below! 5 5 x; x... 4 64 isw A; A (b) x 0 (4 5(0.)) 4.5 9 [5] or k or.5 o.e. B [] (c) or.5 or 5 4 0 5 64 0 So, 54 56 or 54 56 yields,. (a) B M 8 or 56 8 8...... See notes M 8 8 Question s or 6 56 or 54 84 or 56 8 etc. A oe [] 8 (4) or outside brackets or as candidate s constant term in their binomial expansion. Expands... kx to give any terms out of terms simplified or un-simplified, Eg: (kx) or (kx) ( )( ) (kx) or... ( )( ) (kx)!! where k is a numerical value and where k. A A correct simplified or un-simplified (kx) ( )( ) (kx) expansion with consistent (kx).! (kx), k, must be consistent (on the RHS, not necessarily on the LHS) in a candidate s expansion. 9
. (a) ctd. Award BMA0 for 5x ( )( ) 5x! 4... because (kx) is not consistent. Incorrect bracketing: 4 ( )( ) 5x! 4... is BMA0 unless recovered. A 5 4 x (simplified fractions) or allow.5x x 4 A 5 Accept only 64 x 0.9065 x SC If a candidate would otherwise score nd A0, rd A0 then allow Special Case nd A for either 5 SC: ;... 8 x 5 or SC:... 8 x... 5 8 x 5 8 x... or SC: 5 8 x 5 8 x... (where can be or omitted), where each term in the... is a simplified fraction or a decimal, OR SC: for 0 8 x 50 8 x... (i.e. for not simplifying their correct coefficients.) Candidates who write 4 ( )( ) 5x! 4... 4 and not 5 4 and achieve 5 4 x 5 64 x... will get BMAA0A Ignore extra terms beyond the term in x. You can ignore subsequent working following a correct answer. (b) B or.5 or k or.5 o.e. (Ignore how k is found.) (c) M Substitutes x or 0. into their binomial expansion found in part (a) which must contain both 0 an x term and an x term (or even an x term) and equates this to either or their k from (b), where k is a numerical value. M can be implied by or.5 or = awrt. M can be implied by 54 their k 56, with their k found in part (b). A M cannot be implied by k their 54 56, with their k found in part (b). 8 6 54 56 or any equivalent fraction, eg: or. Also allow or any equivalent fraction. 8 56 84 8 Also allow A for p 8, q 8 or p 8, q 8 or p 56, q 8 or p 56, q 8, where You can recover work for part (c) in part (b). You cannot recover part (b) work in part (c). Candidates are allowed to restart and gain all marks in part (c) from an incorrect part (b). Award M A for the correct answer from no working. 0
. (a) Alternative methods for part (a) Alternative method : Candidates can apply an alternative form of the binomial expansion. (4 5x) (4) (4) (5x) ( )( ) (4) (5x)! B 4 or M A A A Any two of three (un-simplified) terms correct. All three (un-simplified) terms correct. 5 4 x (simplified fractions) or allow.5x or 4 x 5 Accept only 64 x or 0.9065 x The terms in C need to be evaluated. So C0(4) C(4) (5 x); C(4) (5 x) without further working is B0M0A0. Alternative Method : Maclaurin Expansion f(x) (4 5x) f (x) 5 (4 5x) Correct f (x) 4 f(x) (4 5x) (5) B a(4 5x) ; a M (4 5x) (5) A oe f(0), f(0) 5 4 and f (0) 5 So, f(x) 5 5 x; 4 64 x... A; A
Question Number. (a) (b) (a) x xy 4y 64 0 x y x Scheme 8y 0 x y x 8y or y x x 8y o.e. Marks MA dm M A cso [5] 0 x y 0 M y x x y Aft x x x 4 x 64 0 y yy 4y 64 0 dm 6 5 9 9 5 x x x 64 0 x 64 0 y y 4y 64 0 y 64 0 9 9 4 4 x 576 5 x 4 5 or 4 y 56 5 5 y 6 5 or 6 5 A cso 4 When x, y 4 5 5 and 4 5 When 6 y, x 6 5 5 and 6 5 4 5, 6 5 4 and 5, 6 5 or x 4 5, y 6 5 Alternative method for part (a) x y x 8y 0 (x y) x 8y 0 dm. (a) General Writing down x y ( x 8 y) 0 x y x 8y or y x x 8y Writing down and x 4 5, y 6 5 Question s x y x 8y or y x from no working is full marks x 8y x y x 8y or y x x 8y from no working is MA0BMA0 Few candidates will write x y x 8y 0 leading to This should get full marks. cso o.e. x y x 8y, o.e. ddm A MA M [6] A cso [5]
d d. (a) M Differentiates implicitly to include either y y x or 4y ky. (Ignore ). A Both x x and... 4y 64 0 8y 0 If an extra term appears then award A0. M xy x x y x y or x 8y y or x y or x y x y x 8y will get st A (implied) as the " 0" can be implied by the rearrangement of their equation. dm dependent on the FIRST method mark being awarded. An attempt to factorise out all the terms in d y y i.e.... ( x 8 y)... or... (x 8 y). (Allow combining in variable). d x A x y y x or x 8y x 8y or equivalent. cso If the candidate s solution is not completely correct, then do not give this mark. You cannot recover work for part (a) in part (b).. (b) M Sets their numerator of their equal to zero (or the denominator of their equal to zero) o.e. Aft dm A st M can also be gained by setting equal to zero in their x y x 8y 0 If their numerator involves one variable only then only the st M mark is possible in part (b). If their numerator is a constant then no marks are available in part (b) If their numerator is in the form ax by 0orax by 0 then the first marks are possible in part (b). x y 0 is not sufficient for M. x 8y Either Sets x y to zero and obtains either y x or x y the follow through result of making either y or x the subject from setting their numerator of their equal to zero dependent on the first method mark being awarded. Substitutes either their y x or their x y into the original equation to give an equation in one variable only. 4 4 6 6 Obtains either x or or y or, (or equivalent) by correct solution only. 5 5 5 5 i.e. You can allow for example x 48 or 4.8, etc. 0 x 576 5 (not simplified) or y 56 5 (not simplified) is not sufficient for A.
. (b) ddm dependent on both previous method marks being awarded in this part. ctd Method Either: substitutes their x into their y x or substitutes their y into their substitutes the other of their y x or their x y and achieves either: x y, or into the original equation, exactly two sets of two coordinates or exactly two distinct values for x and exactly two distinct values for y. Method Either: substitutes their first x-value, x into x xy 4y 64 0 to obtain one y-value, y and substitutes their second x-value, x into x xy 4y 64 0 to obtain y-value y or substitutes their first y-value, y into x xy 4y 64 0 to obtain one x-value x and substitutes their second y-value, y into x xy 4y 64 0 to obtain one x-value x. Three or more sets of coordinates given (without identification of two sets of coordinates) is ddm0. A 4 Both 5, 6 4 5 and 5, 6 5, only by cso. that decimal equivalents are fine. Also allow x 4 5, y 6 5 4 and x 5, y 6 5 all seen in their working to part (b). Allow x 4 5, y 6 5 for rd A. 6 followed by eg. 5, 4 5 and 6 5, 4 x 4 5, y 6 5 5 (eg. coordinates stated the wrong way round) is rd A0. It is possible for a candidate who does not achieve full marks in part (a), (but has a correct numerator for ) to gain all 6 marks in part (b). Decimal equivalents to fractions are fine in part (b). i.e. 4.8,. and 4.8,.. 4 5, 6 5 and 4 5, 6 5 from no working is M0A0M0A0M0A0. Candidates could potentially lose the final marks for setting both their numerator and denominator to zero. No credit in this part can be gained by only setting the denominator to zero. 4
Question Number. (a) (b) (c) Scheme y 0 4x x e x 0 x(4 e x ) 0 e x 4 x A 4ln x e x xe x e x xe x 4e x c Attempts to solve e x 4 giving x... in terms of ln where 0 4ln cao (Ignore x 0 ) xe x e xe x e x x x x M A, 0, 0 M, with or without xe 4e o.e. with or without c 4x x 4x x or 4x 0 4ln (4x x e x ) x xe x 4e x 0 4ln or ln6 or their limits Marks [] A (M on epen) A o.e. B [] (4ln) (4ln )e (4ln) 4e (ln ) (ln ) 6 4 (4ln) (0) (0)e (0) 4e (0) See notes M (ln ) (ln ) (ln) (ln), see notes A Question s. (a) M Attempts to solve e x 4 giving x... in terms of ln where 0 A 4ln cao stated in part (a) only (Ignore x 0 ) (b) NOT E Part (b) appears as MMA on epen, but is now marked as MAA. M Integration by parts is applied in the form xe e x, where 0, 0. (must be in this form) with or without A xe e or equivalent, with or without. Can be un-simplified. A isw SC xe x 4e x or equivalent with or without + c. Can be un-simplified. You can also allow e x (x ) or e x (x 4) for the final A. You can ignore subsequent working following on from a correct solution. SPECIAL CASE: A candidate who uses u x, dv e x, writes down the correct by parts formula, but makes only one error when applying it can be awarded Special Case M. (Applying their v counts for one consistent error.) [] 8 5
. (c) B 4x x or 4x oe M Complete method of applying limits of their x A and 0 to all terms of an expression of the form Ax Bx e x Ce x (where A 0, B 0 and C 0 ) and subtracting the correct way round. Evidence of a proper consideration of the limit of 0 is needed for M. So subtracting 0 is M0. A ln6 or ln4 or equivalent is fine as an upper limit. A correct three term exact quadratic expression in ln. For example allow for A (ln) (ln) 8(ln ) 8(4ln ) (4ln) (ln) (4ln) (4ln)e (4ln) that the constant term of needs to be combined from 4e 4e o.e. Also allow ln(ln ) or ln ln ln for A. Do not apply ignore subsequent working for incorrect simplification. Eg: (ln ) (ln) 64(ln) (ln) or (ln4) (ln) Bracketing error: ln (ln), unless recovered is final A0. Notation: Allow (ln ) (ln) for the final A. 5.978 without seeing (ln) (ln) is A0. 5.978 following from a correct x xe 4e is MA0. 5.978 from no working is M0A0. 6
Question Number Scheme 5 0 8 l : r, l : r 5 4. Let acute angle between l and l. p 5 : You can mark parts (a) and (b) together. l l i : 5 8 Finds and substitutes their into l 4. (a) Marks M So, (b) : 5 4 (c) j 5 4( ) 4 k : p 5 p (4) 5() p 5 or k : p p (4) p 5 0 d, d 4 5 cos K cos 0 4 5 0() ()(4) ()(5) (0) () (). () (4) (5) 5i j k or 5 or (5,, ) Equates j components, substitutes their and solves to give... Equates k components, substitutes their and their and solves to give p... or equates k components to give their " p the k value of A found in part (a), substitutes their and solves to give p... p 5 Realisation that the dot product is required between A and Bd. d An attempt to apply the dot product formula between Ad and Bd. 9 0. 50.806....8 ( dp) anything that rounds to.8 A A M M A M dm (A on epen) [] [] [] (d) See notes M Writes down a correct trigonometric equation involving d sin d 0 the shortest distance, d. Eg: sin, oe. dm their AB d 0 sin.8... d 7.45654075... 7.46 (sf ) anything that rounds to 7.46 A [] 7
4. (b) Alternative method for part (b) j: 9 5 p 9 7 k : p 5 p 9 7() p 5 Eliminates to write down an equation in p and Substitutes their and solves to give p... p 5 M M A 4. (d) Alternative Methods for part (d) Let X be the foot of the perpendicular from B onto l 0 d, Method 9 leading to 0 78 0 5 (Allow a sign slip in copying d ) Applies and solves the resulting equation to find a value for. Substitutes their value of into their. : This mark is dependent upon the previous M mark. M dm d BX 7 6 5 5 7.45654075... Method So d 0 56 0 9 d 5 awrt 7.46 Finds in terms of, finds d d and sets this result equal to 0 and finds a value for. A M Substitutes their value of into their : This mark is dependent upon the previous M mark. 78 d BX 7.45654075... awrt 7.46 A 5 dm 8
Question 4 s 4. (a) M Finds and substitutes their into l A 5 Point of intersection of 5i j k. Allow or (5,, ). You cannot recover the answer for part (a) in part (c) or part (d). (b) M Equates j components, substitutes their and solves to give... M Equates k components, substitutes their and their and solves to give p... or equates k components to give their " p the k value of A found in part (b). A p 5 (c) NOTE Part (c) appears as MAA on epen, but now is marked as MMA. M Realisation that the dot product is required between Ad and Bd. Allow one slip in candidates copying down their direction vectors, d and d. dm dependent on the FIRST method mark being awarded. An attempt to apply the dot product formula between Ad and Bd. A anything that rounds to.8. This can also be achieved by 80 48.796... awrt.8 0.555... c is A0. 0 6 60 76 MA for cos (0) (4) (). () ( 4) (5) 60. 50 Alternative Method: Vector Cross Product Only apply this scheme if it is clear that a candidate is applying a vector cross product method. 0 i j k Realisation that the vector d d 4 0 7i 9j k cross product is required M 5 4 5 between Ad and Bd. sin sin 9 0. 50 (7) (9) () An attempt to apply the (0) () (). () (4) (5) vector cross product formula.806....8 ( dp) anything that rounds to.8 A (d) M Full method for finding B and for finding the magnitude of or the magnitude of dm dependent on the first method mark being awarded. Writes down correct trigonometric equation involving the shortest distance, d. d d Eg: sin or cos(90 ), o.e., where their AB is a value. their AB their AB and "their " or stated as A anything that rounds to 7.46 dm (A on epen) 9
Question Number Scheme Marks 5. : You can mark parts (a) and (b) together. (a) x 4t, 5 y 4t 8 t dt 4, 5 4 t Both dt dt 4ordt 4 and 5 4 t dt B 5 4 t 5 5 So, t Candidate s d y divided by a candidate s d x M dt dt o.e. 4 8 8t d y When, 7 7 t or 0.8475 cao A [] Way : Cartesian Method 0 (x ) When t, x 7 Way : Cartesian Method (x )(x ) (x x 5) (x ) x 6x (x ) When t, x 7 0 (x ), simplified or un-simplifed. B (x ), 0, 0 M 7 or 0.8475 cao A [] Correct expression for, simplified or un-simplified. B f(x)(x ) f (x), (x ) M where f(x) their "x ax b", g(x) x 7 or 0.8475 cao A [] (b) t x 4 y 4 x 4 8 5 x 4 y x 8 0 x (x )(x ) 8(x ) 0 y x (x 5)(x ) 0 or y or y x y x x 5 x, { a and b 5} or y(x ) (x )(x ) 8(x ) 0 (x 5)(x ) x 0 x x y Eliminates t to achieve an equation in only x and y See notes Correct algebra leading to x 5 or a and b 5 x M dm A cso [] 6 0
Question Number Scheme 5. (b) Alternative Method of Equating Coefficients y x ax b y(x ) x ax b x y(x ) (4t ) (4t ) 5 6t t 0 x ax b (4t ) a(4t ) b (4t ) a(4t ) b 6t t 0 t : 4 4a a constant: 9 a b 0 b 5 5. (b) Alternative Method of Equating Coefficients t x 4 y 4 x 4 8 5 x 4 y x 8 0 x y x 5 0 (x ) y(x ) (x 5)(x ) 0 x ax b (x 5)(x ) 0 y x x 5 x or equating coefficients to give a and b 5 x y Correct method of obtaining an equation in only t, a and b Equates their coefficients in t and finds both a... and b... a and b 5 Eliminates t to achieve an equation in only x and y Correct algebra leading to x 5 or a and b 5 x Marks M dm A [] M dm A cso []
Question 5 s 5. (a) B 4 dt dt 4 5 t or dt 8t 5 or t dt 4 5(t) (), etc. can be simplified or un-simplified. dt You can imply the B mark by later working. M Candidate s d y divided by a candidate s d x or d y multiplied by a candidate s dt dt dt dt M can be also be obtained by substituting t into both their and their and then dt dt dividing their values the correct way round. A 7 or 0.8475 cao (b) M Eliminates t to achieve an equation in only x and y. dm dependent on the first method mark being awarded. Either: (ignoring sign slips or constant slips, noting that k can be ) 0 Combining all three parts of their x 8 x to form a single fraction with a common denominator of k(x ). Accept three separate fractions with the same denominator. 0 x Combining both parts of their x 5 x, (where x 5 is their 4 4 8), to form a single fraction with a common denominator of k(x ). Accept two separate fractions with the same denominator. 0 Multiplies both sides of their y x 8 x or their y x 5 0 x by k(x ). that all terms in their equation must be multiplied by k(x ). Condone invisible brackets for dm. A Correct algebra with no incorrect working leading to y x x 5 or a and b 5 x Some examples for the award of dm in (b): 0 (x )(x ) 8 0 dm0 for y x 8 y. Should be... 8(x )... x x dm0 for y x dm0 for y x 5 dm0 for y x 5 y x 5 0 x 0 x 0 x 0 x y x x 5 x (x )(x ) 0 y. The 8 part has been omitted. x y x(x ) 5 0 x. Should be... 5(x )... y(x ) x(x ) 5(x ) 0(x ). Should be just 0. with no intermediate working is dma.
Question Number Scheme 6. (a) A ( x)(x ), x sin 0 d cos ( x)(x ) or ( x x ) ( sin ) ( sin ) cos d sin sin cos d 4 4sin cos d 4 4( cos cos 4 cos d, k 4 d or 4cos cos d cos or cos used correctly d in their working. Can be implied. Substitutes for both x and, where d. Ignore d Applies cos sin see notes 4 cos d or 4cos d : d is required here. Marks B M M A 0 sin or sin or sin 6 and sin or sin or sin See notes B cos (b) k cos d k k 4 sin So 4 cos d sin 6 6 d Applies cos cos to their integral Integrates to give sin, 0, 0 or k( sin ) M [5] M (A on epen) sin sin 6 6 4 4 or 8 6 A cao cso [] 8
Question 6 s 6. (a) B cos d. Also allow cos d. This mark can be implied by later working. You can give B for cos used correctly in their working. M Substitutes x sin and their from their rearranged d into ( x)(x ). Condone bracketing errors here. d. For example d. Condone substituting cos for the st M after a correct d x cos or cos d d M Applies either sin cos sin or ( sin ) cos 4 4sin 4 cos cos 4cos to their expression where is a numerical value. A Correctly proves that ( x)(x ) is equal to 4 cos d or 4cos d All three previous marks must have been awarded before A can be awarded. Their final answer must include d. You can ignore limits for the final A mark. B Evidence of a correct equation in sin or sin for both x-values leading to both values. Eg: 0 sin or sin or sin which then leads to 6, and sin or sin or sin which then leads to Allow B for x sin 6 Allow B for sin x x or sin followed by x 0, 6 ; x, (b) NOTE Part (b) appears as MAA on epen, but is now marked as MMA. M Writes down a correct equation involving cos and cos Eg: cos cos or cos cos or cos cos and applies it to their integral. : Allow M for a correctly stated formula (via an incorrect rearrangement) being applied to their integral. M Integrates to give an expression of the form sin or k( sin ), 0, 0 (can be simplified or un-simplified). A A correct solution in part (b) leading to a two term exact answer. 4 8 Eg: or 6 or 8 6 5.05485 from no working is M0M0A0. Candidates can work in terms of k (note that k is not given in (a)) for the MM marks in part (b). If they incorrectly obtain 4 cos d in part (a) (or guess k 4) then the final A is available for a correct solution in part (b) only. 6 4
Question Number 7. (a) (b) Scheme A B PP ( ) P ( P) A( P) BP Can be implied. M A, B Either one. A giving ( P ) P dp P ( P )cos t dt dp cost dt PP ( ) See notes. cao, aef A can be implied by later working B oe ln ( P ) ln P sin t c t 0, P ln ln 0 c c ln or ln( ) ln(p ) ln P, 0, 0 Marks M ln ( P ) ln P sin t A See notes M ln ( P ) ln P sin t ln ( P ) ln sin t P Starting from an equation of the form ln(p ) ln P Ksint c, ( P ) e sin t,,, K, 0, applies a fully correct method to P eliminate their logarithms. Must have a constant of integration that need not be evaluated (see note) sin sin ( ) e t t P P P 6 Pe A complete method of rearranging to make P the subject. gives P Pe sint 6 P( e sint ) 6 Must have a constant of integration 6 that need not be evaluated (see note) P * sin ( e t ) Correct proof. (c) population = 4000 P 4 States P 4 or applies P 4 M (4 ) sin t ln ln 4 t 0.478700467... Obtains sint ln k or sint ln k, 0, k 0where and k are numerical values and can be anything that rounds to 0.47 Do not apply isw here M dm [] A * cso M A [7] [] 5
Question Number Scheme 7. (b) ln ( P ) ln P sin t c Marks Method for Q7(b) As before for BMA (P ) ln P sin t c Starting from an equation of the form ln(p ) ln P Ksint c, (P ) e sint c (P ) or Ae sint,,, K, 0, applies a fully correct rd M P P (P ) APe sint P APe sin t P( Ae sint ) P t 0, P ( Ae sin (0) ) ( A) A P P e sint ( Ae sint ) method to eliminate their logarithms. Must have a constant of integration that need not be evaluated (see note) A complete method of rearranging to make P the subject. Condone sign slips or constant errors. Must have a constant of integration that need not be evaluated (see note) See notes (Allocate this mark as the nd M mark on epen). 4 th dm nd M 6 ( e sin t ) * Correct proof. A * cso Question 7 s 7. (a) M A B Forming a correct identity. For example, A(P ) BP from PP ( ) P ( P) A and B are not referred to in question. A Either one of A or B. A ( P ) P or any equivalent form. This answer cannot be recovered from part (b). MAA can also be given for a candidate who finds both A andb and A B P ( P ) is seen in their working. Candidates can use cover-up rule to write down, so as to gain all three marks. ( P ) P Equating coefficients from A(P ) BP gives A B, A A, B 6
7. (b) B Separates variables as shown on the Mark Scheme. dpand dt should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs. Eg: P P dp cost dt or cost dt o.e. are also fine for B. st M ln(p ) ln P, 0, 0. Also allow ln( M (P )) ln NP; M, N can be. Condone ln(p ) ln P or ln(p(p )) or ln(p P) or ln(p P) st A Correct result of ln(p ) ln P sin t or ln(p ) ln P sint o.e. with or without c nd M Some evidence of using both t 0 and P in an integrated equation containing a constant of integration. Eg: c or A, etc. rd M Starting from an equation of the form ln(p ) ln P K sint c,,,, K, 0, applies a fully correct method to eliminate their logarithms. 4 th M dependent on the third method mark being awarded. A complete method of rearranging to make P the subject. Condone sign slips or constant errors. For the rd M and 4 th M marks, a candidate needs to have included a constant of integration, in their working. eg. c, A, ln A or an evaluated constant of integration. 6 nd A Correct proof of P. : This answer is given in the question. ( e sin t ) (P ) ln P (P ) sin t c followed by P (P ) ln P (P ) sint c e sint c P e sint e c is rd M0, 4 th M0, nd A0. (P ) P e sin t e c is final MM0A0 4 th M for making P the subject there are three type of manipulations here which are considered acceptable for making P the subject. ( P ) () M for e sint ( P ) Pe sint P 6 Pe sint P( e sint ) 6 P 6 P ( e sint ) ( P ) () M for e sint 6 P P e sint e sint 6 P P 6 ( e sint ) () M for ln(p ) ln P sint ln P(P ) e sint P P e sint (P ) e sin t leading to P.. (c) M States P 4 or applies P 4 M Obtains sint ln k or sint ln k, where and k are numerical values and can be A anything that rounds to 0.47. (Do not apply isw here) Do not apply ignore subsequent working for A. (Eg: 0.47 followed by 47 years is A0.) Use of P 4000 : Without the mention of P 4, sin t ln.9985 or sin t ln.9985 or sint.9... will usually imply M0MA0 Use of Degrees: t awrt 7. will usually imply MMA0 7
Question Scheme Number 8. (a) y x x ln Either T: y 9 ln( x ) or T: y ( ln)x 9 8ln, where 9 ( ln)() c {Cuts x-axis y 0 } x ln or lne xln or y ln B See notes Sets y 0 in their tangent equation 9 9ln(x ) or 0 ( ln)x 9 8ln, and progresses to x... So, x ln ln or ln o.e. ln (b) V x (b) or x or 9 x x ln or 9x ln9 V x x ln0 0 V x with or without, which can be implied x x x Eg: either or (ln) (ln ) M M Marks A cso B o.e. or 9 x 9x (ln9) or (ln9)9x, x x 9 x or 9 x or e xln ln ln 9 ln ex ln A o.e. Dependent on the previous 4 ln ln 40 method mark. Substitutes ln x and x 0 and subtracts dm the correct way round. Vcone (9) their ( a ). See notes. Bft V cone (9) ln 7 ln Vol(S) 40 ln 7 or 6 6 or etc., isw A o.e. ln ln ln ln 9 ln Eg: p, q ln [6] 0 Alternative Method : Use of a substitution V x B o.e. u x du x ln uln V u ln V x 0 u 9 ln then apply the main scheme. u uln du 9 ln ln u ln du x u (ln) or (ln)u, where u x 40 ln x u, where u x A (ln) Substitutes limits of 9 and in u (or and 0 in x) and subtracts the correct way round. M M dm [4] 8
Question 8 s 8. (a) B x ln or lne xln or y ln. Can be implied by later working. M Substitutes either x or y 9 into their d y which is a function of x or y to find mt and either applies y 9 their m T x, where mt is a numerical value. or applies y their m T x their c, where mt is a numerical value and c is found by solving 9 their m T () c The first M mark can be implied from later working. M Sets y 0 in their tangent equation, where mt is a numerical value, (seen or implied) and progresses to x... A An exact value of ln or ln ln9 or ln ln by a correct solution only. Allow A for ln or (ln ) (ln9 ) or or, where is an integer, ln ln ln and ignore subsequent working. Using a changed gradient (i.e. applying or their their ) is M0 M0 in part (a). Candidates who invent a value for m T (which bears no resemblance to their gradient function) cannot gain the st M and nd M mark in part (a). A decimal answer of.08976077 (without a correct exact answer) is A0. 8. (b) B A correct expression for the volume with or without Eg: Allow B for x or x or 9 x or e x ln or e x ln or e x ln9 with or without M Either x x x or (ln) or 9 x 9x or (ln9)9x (ln ) (ln9) e xln ln ex (ln) or (ln)ex ln or e x ln9 ex ln9 (ln9) or (ln9)ex ln9, etc where x x (ln) or 9x 9x (ln) are allowed for M x x x or 9x 9x x are both M0 M can be given for 9 x 9x (ln9) or (ln9)9x A Correct integration of x. Eg: x x ln or x x ln9 or 9 9 x or e xln ln 9 ln ex ln dm dependent on the previous method mark being awarded. Attempts to apply x and x 0 to integrated expression and subtracts the correct way round. Evidence of a proper consideration of the limit of 0 is needed for M. So subtracting 0 is M0. 9
dm dependent on the previous method mark being awarded. Attempts to apply x and x 0 to integrated expression and subtracts the correct way round. Evidence of a proper consideration of the limit of 0 is needed for M. So subtracting 0 is M0. Bft V cone. (9) theiranswer to part (a) Sight of 7 implies the B mark. ln Alternatively they can apply the volume formula to the line segment. They need to achieve the result highlighted by **** on either page 9 or page 0 in order to obtain the Bft mark. 6 6 p A or or, etc., where their answer is in the form ln ln 9 ln q The in the volume formula is only needed for the st B mark and the final A mark. A decimal answer of 7.7488 (without a correct exact answer) is A0. A candidate who applies x will either get B0 M0 A0 M0 B0 A0 or B0 M0 A0 M0 B A0 x unless recovered is B0. Be careful! A correct answer may follow from incorrect working V x 0 (9) ln x ln would score B0 M0 A0 dm0 M A0. 0 7 ln 4 ln ln 7 ln ln 8. (b) nd Bft mark for finding the Volume of a Cone V cone 9x ln 8ln 9 ln 9x ln 8ln 9 7 ln or their part (a)answer ln **** 9 ln 8ln 9 8ln 8ln 9 ln 7ln 7ln 79 8ln 9 8ln 9 7ln 7ln 7 ln Award Bft here where their lower limit is or their ln part (a) answer. 0
nd Bft mark for finding the Volume of a Cone 8. (b) Alternative method : cone 9 ln8ln 9 d ln ln V x x 8x ln 4x ln 6xln 4ln 4(ln ) 8 7x ln 6x ln 8x ln 4xln 4 x(ln ) 8x ln **** 6ln 648ln 4ln 648ln 648(ln ) 6 7 ln 6 ln 8 ln ln ln ln 4 ln 4 (ln) 8 ln ln ln Award Bft here where their lower limit is ln or their part (a) answer. 6 4 7 8 ln 6 4 ln ln ln ln ln ln 4 6ln 4ln 6 8 4 ln 4 ln ln ln ln 4 (ln) 8 ln ln 7 6ln 4ln 6 648ln 648ln 6 ln 8 6ln 4ln 6 4ln 4 648ln 4 ln 8 648ln 4ln 6 ln 7 6 ln 4ln 6 6 ln 4ln 6 ln 7 ln
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