Mark Scheme (Results) January 0 GCE Further Pure FP (6667) Paper 0 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers.
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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.
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 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) 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 4. All A marks are correct answer only (cao.), unless shown, for example, as 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.
General Principals 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), leading to x =. Completing the square Solving + bx + c = 0 x b : x ± ± q ± c, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n. Integration Power of at least one term increased by. ( n x x ) + 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.
January 0 6667 Further Pure Mathematics FP Mark Scheme Question (a) (b) (c) arg z = arctan() arctan() or arctan() or arctan( ) π = or -45 or awrt -0.785 (oe e.g 7 π ) 4 4 Correct answer only / () z z ( i)( 4 i) i 4i 4i z z = + = + At least correct terms (Unsimplified) = 7 + i cao + ( + ) ( + 4i) ( 4i). i = = i i. + i (+ 4i). ( + i) = 7 = + i or z ( i) i Special case + 7i. 4i = = z +4i 4 i. 4i Multiply top and bottom by ( + i) + i i = ( + ) ( ) Allow A0A0 Correct answers only in (b) and (c) scores no marks Total 7 () ()
Question 4 f ( x) = x + x (a) f(0.5) = -0.475 (- 7 6 ) f() = Sign change (positive, negative) (and f ( x) is continuous) therefore (a root) α is between x = 0.5 and x =.0 (b) Either any one of f(0.5) = awrt -0.4 or f() = f(0.5) = awrt -0.4 and f() =, sign change and conclusion f(0.75) = 0.0664065( 7 ) Attempt f(0.75) 56 9 f(0.65) = -0.40975( ) f(0.75) = awrt 0.07 and f(0.65) = awrt -0. 4096 0.65 α 0.75 (c) f ( x) = 4x + x = 0.75 x 0.65 α 0.75 or 0.65 < α < 0.75 or [ 0.65, 0.75] or ( 0.65, 0.75 ). or equivalent in words. In (b) there is no credit for linear interpolation and a correct answer with no working scores no marks. Correct derivative (May be implied later by e.g. 4(0.75) + ) f (0.75) 0.0664065 = 0.75 = 0.75 Attempt Newton-Raphson f '(0.75).6875 4 6 Correct first application a correct B () () 499 x = 0.759(06976...) = 688 numerical expression e.g. or awrt 0.75 (may be implied) 7 0.75 56 4 6 x 499 0.000578978 = 0.7449 688.56468 Awrt 0.74 ( α ) = 0.74 cao A final answer of 0.74 with evidence of NR applied twice with no incorrect work should score 5/5 (5) Total 0
Question (a) Focus ( 4,0 ) B Directrix x + 4 = 0 (b) dy y = 4x = x dx y dy = 6 x y = 6 d x or d y dy dt =. = 8. dx dt dx 8t x + 4 = 0 or x = - 4 x + 4 = 0 or x = - 4 dy = k x dx dy ky c dx = dy their dt dx their dt () dy dy dy = x or y = 6or = 8. Correct differentiation dx dx dx 8 t At P, gradient of normal = -t y t = t x t 8 ( 4 ) y tx t t Correct normal gradient with no errors seen. Applies y 8t their mn ( x 4t ) ( their ) = or y = m x + c using N x = 4t and y = 8t in an attempt to find c. Their m N must be different from their m T and must be a function of t. + = 8 + 4 * cso **given answer** Special case if the correct gradient is quoted could score M0A0A0 (5) Total 8
Question 4(a) (b) 0 4 = 0 4 T has coordinates (,), (,) and (4,) 4 4 or,, NOT just Reflection in the line y = x Attempt to multiply the right way round with at least 4 correct elements Correct coordinates or vectors Reflection y = x Allow in the axis about the line y = x etc. Provided both features are mentioned ignore any reference to the origin unless there is a clear contradiction. B B () (c) (d) 4 0 correct elements QR = = 4 0 Correct matrix 4 0 4 Note that RQ = = 4 4 0 scores M0A0 in (c) but allow all the marks in (d) and (e) ( QR ) det = 0 = 4 Answer only scores / det QR scores M0 - x 0 x 0-4 () () () (e) Area of T = = Correct area for T B Attempt at " " "4" ± Area of T = 4 = 6 6 or follow through their det(qr) x Their triangle area provided area > 0 ft () Total
Question 5(a) ( z ) = i B Attempt to expand ( z ( + i))( z ( i)) or any valid method to establish the quadratic factor e.g. z = ± i z = ± i z 6z + 9 = ( z ( + i))( z ( i)) = z 6z + 0 6 ± 4 z = ± = b = 6, c = 0 Sum of roots 6, product of roots 0 z 6z + 0 5(b) Attempt at linear factor with their cd in ( z + az + c)( z + d) = ± 0 Or ( z 6z + 0)( z + a) 0a = 0 Or attempts f() z = ( z 6z + 0)( z ) = 0 Showing that f() = 0 is equivalent to scoring both M s so it is possible to gain all 4 marks quite easily e.g. z = i B, shows f() = 0 M, z =. Answers only can score 4/4 (4) Im.5 Argand Diagram 0.5, B B 0 0 0.5.5, 0.5.5 Re -0.5 -, - -.5 First B for plotting (, ) and (, -) correctly with an indication of scale or labelled with coordinates (allow points/lines/crosses/vectors etc.) Allow i/-i for /- marked on imaginary axis. Second B for plotting (, 0) correctly relative to the conjugate pair with an indication of scale or labelled with coordinates or just () Total 6
Question 6(a) n =, LHS = =, RHS = = 4 Shows both LHS = and RHS = B Assume true for n = k When n = k + k + r = k ( k + ) + ( k + ) 4 Adds (k + ) to the given result r= (b) (c) (c) Way [ 4] Attempt to factorise out = 4 k + k + k + Correct expression with 4 k + d 4 k + factorised out. = 4 k k Fully complete proof with no errors and Must see 4 things: true for n =, assumption true for n = k, said true for n = k + and therefore true for all n comment. All the previous marks must have been scored. cso See extra notes for alternative approaches (5) ( r ) = r Attempt two sums r n is M0 = + Correct expression 4 n n n n Completion to printed answer with no = ( n + n + n 8) 4 * errors seen. Attempt S 50 S 0 or S 50 S 9 and = 50 50 9 ( r ) = 004 759 substitutes into a correct expression at = 0 4 4 least once. = 6555 606 Correct numerical expression (unsimplified) = 589 46 cao r r 50 9 ( r ) = r () = 5 0 4 4 r= 50 r= 50 r= 50 r= 0 r= 0 r = 0 for (S 50 S 0 or S 50 S 9 for cubes) (x0 or x) correct numerical expression () () Total = 589 46
Question 7(a) u =, u = 7 B, B (b) At n =, u = = and so result true for n = k Assume true for n = k; u = k B () and so u ( u ) k k + = k + = ( ) + k + k + u k + = + = Correct completion to Substitutes u k into u k+ (must see this line) Correct expression k + u k + = Must see 4 things: true for n =, assumption true for n = k, said true for n = k + and therefore true for all n Fully complete proof with no errors and comment. All the previous marks in (b) must have been scored. cso Ignore any subsequent attempts e.g. k+ uk+ = uk + + = ( ) + etc. (5) Total 7
Question 8(a) (b) (b) Way det( A ) = 0 = Correct attempt at the determinant det( A ) 0 (so A is non singular) det(a) = - and some reference to zero det ( A ) scores M0 BA = A BA = I B = A - Recognising that A - is required At least correct terms in 0 B = * * 0 Bft their det(a) * * Fully correct answer Correct answer only score 4/4 (4) Ignore poor matrix algebra notation if the intention is clear Total 6 A = 6 Correct matrix B a b 0 a + 6b = 0 c + 6d = = or c d 6 a + b = c + d = a =, b =, c =, d = 0 equations in a and b or equations in c and d Solves for a and b or c and d All 4 values correct () (b) Way A = 6 Correct matrix B ( ) "" " " = "" "" "" "6" " 6" "" Attempt inverse of A 0 0 or Attempts A A = 4 6 4 6 A( A ) or ( A ) A B = 0 Fully correct answer (b) Way 4 BA = I Recognising that BA = I B a b 0 0 b = d = 0 equations in a and b or = or c d 0 a + b = 0 c + d = equations in c and d Solves for a and b a =, b =, c =, d = 0 or c and d All 4 values correct
Question 9 (a) dy y = 9x = 9x dx dy = k x dx dy xy = 9 x + y = 0 dx dy dy dt or =. =. dx dt dx p Correct use of product rule. The sum of two terms, one of which is correct. their dy dt dx their dt dy 9 dy 0 dy. = x or x + y = or = Correct differentiation. dx dx dx p y = ( x p) p p = or p Applies y (their m) ( x p) ( their ) y = m x + c using x = p and y = in an attempt to find c. p Their m must be a function of p and come from their dy/dx. x p y 6 p + = * Cso **given answer** Special case if the correct gradient is quoted could score M0A0 (4) (b) x q y 6q + = Allow this to score here or in (c) B (c) 6 p p y = 6q q y Attempt to obtain an equation in one variable x or y 6 ( q p) y( q p ) = 6( q p) y = q p Attempt to isolate x or y must reach 6 pq ( q p) x( q p ) = 6 pq( q p) x = x or y = f(p, q) or f(p) or f(q) q p () y = x = 6 p + q 6 pq p + q One correct simplified coordinate Both coordinates correct and simplified (4) Total 9 Extra Notes
6(a) To show equivalence between 4 k k + + k + and 4 k + k + 4 k ( k + ) + ( k + ) = k + k + k + k + 4 4 4 Attempt to expand one correct expression up to a quartic 4 4 4 4 ( k + ) ( k + ) = k + k + k + k + Attempt to expand both correct expressions up to a quartic One expansion completely correct (dependent on both M s) Both expansions correct and conclusion Or ( k + ) ( k + ) k ( k + ) = k + 4 4 To show ( k ) ( k ) k ( k ) 4 4 + + + Attempt to subtract ( k + ) ( k + ) k ( k + ) = k + k + k + 4 4 Obtains a cubic expression Correct expression ( k + ) ( k + ) k ( k + ) = ( k + ) 4 4 Correct completion and comment 8(b) Way Attempting inverse of A needs to be recognisable as an attempt at an inverse Their Det E.g ( A ) = ( A changed A ) ( A )
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