January 015 (IAL) Mark Scheme (Results) January 015 Pearson Edexcel International A Level in Further Pure Mathematics F1 (WFM01/01)
January 015 (IAL) 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 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 150 years, and by working across 70 countries, in 100 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 January 015 Publications Code IA0056 All the material in this publication is copyright Pearson Education Ltd 015
January 015 (IAL) 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.
January 015 (IAL) PEARSON EDEXCEL IAL MATHEMATICS General Instructions for Marking 1. 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. 3. 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
January 015 (IAL) Question Number Scheme 1. 3 f ( x) x x 9x 9x60 Notes 1 i is also a root Seen anywhere B1 x x + 5 : Attempt to expand ( x (1 i))( x(1 i)) or any valid method to establish the quadratic factor Marks : x x 5 : Attempt other quadratic factor f( x) x x5x x 1 : x x 1 x x1 xx3 x... Attempt to solve their other quadratic factor. x = and x = 3 Both values correct (7) Total 7 Alternative using Factor Theorem f (3)... or f ( )... : Attempts f(3) or f(-) 0f(3) 0 or f( =) : Shows or states f(3) = 0 or f(-) = 0 f (3)... and f ( )... f(3) 0 and f( ) 0 NB : Attempts f(3) and f(-) or f(3) and g(-) where g(x) = f(x)/(x - 3) or f(-) and h(3) where h(x) = f(x)/(x + ) : Shows or states f(3) = 0 and f(-) = 0 or shows or states f(3) = 0 and g(-) = 0 where g(x) = f(x)/(x - 3) or shows or states f(-) = 0 and h(3) = 0 where h(x) = f(x)/(x + ) 3 3 g x x x 3x0, h x x 5x 11x 15 x = 3 or x = One of x = 3 or x = clearly stated as a root x = 3 and x = Both x = 3 and x = clearly stated as roots x 1 i B1
January 015 (IAL) Question Number 3 (a) Scheme Notes Marks 1 f( x) x 3x 5 x f()... and f(3)... Attempts both f() and f(3) (b) f () 1.9116.., f (3).03... Sign change (and f( x) is continuous) therefore a root exists between x and x 3 5 f( x) 3x 6x x 3.5.03075015 3 8.9737081 Both values correct : f () 1.9116.. (awrt 1.9), and f (3).03... (awrt 3.0 or e.g. ), sign change (or 5 equivalent) and conclusion : x n n 1 x : 3x 6x 5 3.5 : or equivalent un-simplified x and no other terms (+ c loses this mark) Correct attempt at Newton-Raphson using their values of f (3) andf (3)..77 Cao (Ignore any subsequent applications) Correct derivative followed by correct answer scores full marks in (b) Correct answer with no working scores no marks in (b) NB if the answer is incorrect it must be clear that both f(3)andf (3) are being used in f 3 the Newton-Raphson process. So that just 3 with an incorrect answer and no f '3 other evidence scores M0. () (5) Total 7
Question Number Scheme Notes Marks 3 z z i * i 1 1i z* x iy B1 xiyi xiyi... Substitutes for z and their z* and attempts to expand x x y x y y i i x y xi x y 1 and x 1 x1 x... x3, y Way z z zz z z x yx y x y x y Compares real and imaginary parts (allow sign errors only) Solves real and imaginary parts to obtain at least one value of x or y x 3 cso y cso i * i * i * Attempt to expand i i i i i z* x iy (may be implied) B1 x y x i x y 1 and x 1 x1 x... x3, y PhysicsAndMathsTutor.com Compares real and imaginary parts (allow sign errors only) Solves real and imaginary parts to obtain at least one value of x or y x 3 cso y cso January 015 (IAL), (6) Total 6, Total 6 Way 3 z z zz z z zz * i z z * 1 1i zz * 1, z z * 1 i * i * i * Attempt to expand * * Compares real and imaginary parts (allow sign errors only) z 6z50 or z 6z 5 0 Correct quadratic B1 * * z 6z50 or z 6z 50 z *... or z... Solves to obtain at least one value of z or z * z 3, i x 3 cso y cso, Total 6
January 015 (IAL) Question Number Scheme Notes Marks (a) 1 1 dy 1 dy 1 y 1x y 1x 1x kx dx dx dy dy y 1xy 1 y d x d x dy dy dp 1. 6. their d y 1 dx dp dx 6p dp dx their d p d y 1 1 d d 1 1x or y y 1or y 6. dx dx dx 6p Correct differentiation or equivalent expressions 1 mt mn p Correct perpendicular gradient rule p y 6p their m x 3p or (b) (c) y 6p p x 3p y px p p 3 6 3 * p yx 1 y y y 36 6 6y16 0 N y mxc m p p with their N and(3,6 )in an attempt tofind' c'. Their m N must have come from calculus and should be a function of p which is not their tangent gradient. Achieves printed answer with no errors Substitutes the given value of p into the normal Substitutes to obtain an equation in one variable (x, y or q ) y18 y1 0 y Solves their 3TQ y 18 x 7 : One correct coordinate : Both coordinates correct, (5) Focus is (3, 0) or a = 3 or OS = 3 Must be seen or used in (c) B1 y 0 x 18 1 1 : Correct attempt at area A 18 31 18 318 : Correct expression A = 5 Correct area () Total 1 * (5)
January 015 (IAL) Question Number Scheme Notes Marks 5(a) 3 1, B1, B1 () (b) 9 1 1 :Use of 16 16 : 1 16 () (c) 9 Sum 3 Attempt numerical sum Product 17 x 17 1 x 0 9 x 9 16 0 Attempt numerical product Uses x (sum)x + (prod) with sum, prod numerical (= 0 not reqd.) x Any multiple (including = 0) Alternative: Finds roots explicitly (a) 3 7 x i 8 8 3 7 3 7 3 i i= B1 8 8 8 8 3 7 3 7 1 i i = 8 8 8 8 B1 (b) 3 7 3 7 1 i i 8 8 8 8 16 (c) 9 5 7 9 5 7 i, i 8 8 8 8 9 5 7 9 5 7 f( x) x i x i 8 8 8 8 () Total 8 () : Substitutes their α and β and attempt to square and add both brackets : 1 cso (allow 0.065) 16 () Uses x x With numerical values (May expand first) 9 5 7 9 5 7 9 5 7 9 5 7 f( x) x x i x i i i 8 8 8 8 8 8 8 8 Attempt to expand (may occur in terms of α and β but must be numerical for both M s) 9 x x 0 Collects terms (= 0 not reqd.) x x Any multiple (including = 0) 9 16 0 () Total 8
January 015 (IAL) Question Number 6(i)(a) (b) Scheme Notes Marks A: Stretch scale factor 3 parallel to the x-axis B: Rotation 10 degrees (anticlockwise) about (0, 0) or about O B1: Stretch B1: SF 3 parallel to (or along) x-axis Allow e.g. horizontal stretch SF 3 (Ignore any reference to the origin) B1: Rotation about (0, 0) B1: 10 degrees (anticlockwise) (or equivalent e.g. -150 o or 150 o clockwise). Allow equivalents in radians. (c) 3 1 3 0 C=BA Attempts BA (This statement is sufficient) 1 3 0 1 3 3 1 3 3 (ii) detm = k5. k1 k 5k (ii) b ac5 3 b ac 0 So no real roots so detm 0 Way k5. k1k 5k (ii) Way 3 k5. k1k 5k Correct matrix : Correct attempt at determinant : Correct determinant (allow unsimplified) Attempts discriminant or uses quadratic formula Convincing explanation and conclusion with no previous errors B1B1 B1B1 () () () () Total 10 : Correct attempt at determinant : Correct determinant (allow unsimplified) 5 7 k 8 Attempts to complete the square: detm> 0 k Convincing explanation and conclusion Therefore detm 0 with no previous errors : Correct attempt at determinant : Correct determinant (allow unsimplified) M k 50 k 5 7 k dt e M Attempts coordinates of turning point d det 5 dk 8 Minimum detm is 7 8 detm 0 therefore Convincing explanation and conclusion with no previous errors
January 015 (IAL) Question Number Scheme Notes Marks n 7 rarb nn11n1 r1 (a) rarbr rarb ab B1 (b) n 1 1 r1 6 : Attempt to use one of the standard formulae correctly : 1 nn1n1ab 1 nn 1 6 B1: abn 1 n n 1 n 1 3 a b n 1 6ab 1 n n 11 n 1 6 6 n1 n1 3 ab n1 6abn 9n 11 1 6 r ar b nn 1n 1a b nn 1abn B1 n 3n13 ab n1 6abn 9n 11 : Compares coefficients to obtain at 33a3b9, 3a3b16ab11 least one equation in a and b ab, ab3 : One correct equation : Both equations correct Both values correct. This can be b1, a 3 withheld if b 3, a 1 is not rejected. 0 0 r 9 r ar b rarbf (0) f (8or 9) Use of f (0) f (8or 9) r9 1 1 6 6 05119 877 Correct (possibly un-simplified) numerical expression =330 5 = 978 cao (3) Total 11 (8)
January 015 (IAL) Question Number 8(i) When n = 1 When n = (ii) Scheme Notes Marks 1 1 u1 3 5 u 3 13 Both True for n = 1 and n = k k k 1 k 1 Assume uk 3 and u k 1 3 : Attempts u k+ in terms of u k+1 k1 k1 k k uk 5uk 16uk 5 3 6 3 and u k : Correct expression k1 k1 k k 5. 5.3 6. 6.3 5. 3. 5.3.3 k1 k1 k1 k1 So u k k1 k1. 3.3 3 or 3 k1 1 k1 1 k k If true for k and k + 1 then shown true for k + and as true for n = 1 and n =, true for n Z + f () 7 8 1 30 Attempt u k+ in terms of f(k) and 3 f(k) only Correct expression with no errors Full conclusion with all previous marks scored So true for n = Shows true for n = B1 Assume f( k) 7 k 8k1 30p for some integer p k f( k1) f( k) 7 8 k1 1 k 7 8k1 Attempt f(k+1) f(k) k k 7 7 8 k 7 91 8 8f ( k) 8 k 830 p 30k f ( k1) 930 p 30k If true for k then shown true for k + 1 and as true for n =, true for n (n Z) : Attempt rhs in terms of k f(k) or 7 8k 1 : Correct expression which is a multiple of 30 Obtains f(k+1) as a correct multiple of 30 with no errors Full conclusion with all previous marks scored B1 (6) (6) Total 1
January 015 (IAL)
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