PMT. Mark Scheme (Results) Summer Pearson Edexcel GCE in Further Pure Mathematics FP3 (6669/01)

Transcription

1 Mark Scheme (Results) Summer 4 Pearson Edecel GCE in Further Pure Mathematics FP3 (6669/)

2 Edecel and BTEC Qualifications Edecel 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 5 years, and by working across 7 countries, in 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: Summer 4 Publications Code UA38879 All the material in this publication is copyright Pearson Education Ltd 4

3 General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly 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. Eaminers 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. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers 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 eemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75. The Edecel 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

5 4. All A marks are correct answer only (cao.), unless shown, for eample, 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. 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. 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 Further Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving 3 term quadratic:. Factorisation ( b c) ( p)( q), where pq c, leading to = ( a b c) ( m p)( n q), where pq c and mn a, leading to =. Formula Attempt to use the correct formula (with values for a, b and c). 3. Completing the square b Solving b c : q c, q, leading to = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration Power of at least one term increased by. ( n n )

7 Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent eaminers 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. Eact answers Eaminers reports have emphasised that where, for eample, an eact 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 Scheme Marks.(a) P(,, 3) and r.( i jk ) 3 An appreciation that l is parallel to i j k i j kis the direction of the line (may be implied). " r" ij3kt( ijk ) A correct vector equation in any form. (Allow any multiple of the direction vector.) or r(i j3 k) ijk or r( ijk) (ij3 k) ijk or r( ijk) 5i5j5k (b) t t 3 3t (c) Way (c) Way t t 3t 3 t 4t3 t 3 t... t l meets at r 3ijk PQ 3i jk ( i j3k ) i-j-k Substitutes a parametric form of their line from part (a) into the equation of the plane. This statement is sufficient. Correct equation (allow unsimplified) Solves to find a value for t Dependent on the first M Correct position vector (allow as coordinates (3, -, )) Attempts the vector PQ or QP and correct Pythagoras Allow awrt.45 or t "" PQ "" i-j-k i-j-k Attempts the vector PQ using their value for t and their normal vector and correct Pythagoras. 6 Allow awrt.45 or 6 6 d () (4) () () Total 8

9 Question Scheme Marks.(a) MM T I 5 M not orthogonal. (b) 4 5 T T Attempts MM or MMor scalar product of at least one pair of columns or attempts magnitude of at least one column or finds detm or attempts M - or scalar product or magnitude or detm ± (must see ±) or M - T M and conclusion. Note that not all of T MM or M - is necessary and there may be errors but there must be some correct work (at least one correct relevant element). NB detm = -5. See etra notes for M - This statement is sufficient. (allow other brackets provided the determinant is implied later) ( ) 4 5 Attempts characteristic equation (= may be implied by their value(s) for ) Allow one slip e.g. usually the omission of the -5 ( )((4 )( ) 5) with no errors : Attempts to find the other eigenvalues from their characteristic equation by solving a 3 term quadratic. 45 5, : 5, (c) A correct statement for the 4 y y or 3 y eigenvalue. (May be 5 z z 5 z implied by correct equations) i jk where is a constant Any vector of this form. (d) t t 4 t 7t 5 t t Cartesian equation y z 7 : Attempt to multiply the parametric form or direction of the line by M. Condone use of for the component but the line must pass through the origin. : Correct image vector with or without t : Correct method to convert to cartesian form of a straight line passing through the origin. : Correct equations (any multiple) cso () (5) () (4) Total 3

10 Question Scheme Marks 3.(a) 3 ( ) d arsinh f( ) ( ) arsinh arsinh (b) e sinh e e e (e 3 e ) e e e e (e e )d 3e e e e 6 3 : Attempt to complete the square. Allow ( ) k, k : Correct epression Allow ln f ( ) f ( ) (>) Any equivalent eact form. Allow 3 ln but no other terms e.g. arsinh() Substitutes the correct eponential of sinh Correct epression with powers of e combined. p p e d qe at least once and some correct use of the limits and and subtracts the right way round. Any eact equivalent (allow e ) but all like terms collected but isw following a correct answer. (b) Integration by parts way I e sinh e cosh d e sinh 4e cosh 4I 3 4 e I sinh 4e cosh : Parts twice in the correct direction : A correct epression for I or any constant multiple of I 4 e sinh d e sinh e cosh oe : Correct use of limits having integrated by parts twice : Correct epression (oe) (b) Integration by parts way e cosh e cosh d e cosh e sinh I 4 I 3I e cosh e sinh : Parts twice in the correct direction : A correct epression for I or any constant multiple of I e sinh d e coshe sinh 3 : Correct use of limits having integrated by parts twice : Correct epression (oe) (4) (4) Total 8

11 Question Scheme Marks 4.(a) e e e e e tanh or or e e e e e Use of the correct eponential form of tanh (b) e e (e e ) (e +e ) tanh e e (e e ) Attempts tanh with their tanh, obtains a common denominator and a b epands the numerator correctly three terms from at least once d e.e (e e ) 4 sec h * (e e ) Correct completion with no errors * Allow candidates to process both sides and meet in the middle Note that it is possible to start from sech and obtain tanh by reversing the above work (3) Ignore any imaginary solutions in (b) e e e e Substitutes the correct eponential forms for sinh and cosh e 7e 6 e 6e 7 Obtains a quadratic in e (e )(e 7) e... Attempt to solve their 3TQ in e as far as e =.. ln 7 or awrt.95 (4) Total 7

12 Alternatives for (b) 4sinh 3cosh 3 4sinh 3 3cosh 7cosh 8cosh5 : Attempt to square correctly and obtains a quadratic in sinh 7 cosh 8cosh 5 7 cosh 5 cosh 5 5 cosh e e 7 7 7e 5e 7 Uses the correct form of cosh in terms of eponentials to obtain a 3TQ in e 7e 5e 7 7e e 7 e... Attempt to solve their 3TQ as far as e =.. ln 7 or awrt.95 No other values 4sinh 3cosh 3 4sinh 3 3cosh 7sinh 4sinh : Attempt to square correctly and obtains a quadratic in cosh 7sinh 4sinh sinh 7sinh sinh e e 7 7 7e 48e 7 Uses the correct form of sinh in terms of eponentials to obtain a 3TQ in e 7e 48e 7 7e e 7 e... Attempt to solve their 3TQ as far as e =.. ln 7 or awrt.95 No other values 4sinh 3cosh 3 4 tanh 3 3sech 5tanh 4 tanh : Attempt to square correctly and obtains a quadratic in tanh 5 tanh 4 tanh tanh 5 tanh tanh e e 5 5 e 49 e e Uses the correct form of tanh in terms of eponentials to obtain a TQ in e e 49 e... Attempt to solve their TQ as far as e =.. ln 7 or awrt.95 No other values

13 Question Scheme Marks : Correct form for the derivative of artanh dy ( ) ( ). using. d ( ) 5. : Correct quotient or M ( ) NB product rule on ( ) 3 ( ) ( ) : Completely correct epression ` **ag** Correct solution with no errors seen Alternative y artanh tanh y d y ( ) ( ) sech y d ( ) : Divides by the correct form of sech y or their simplified sech y in dy ( ) ( ). terms of d ( ) : Correct quotient or product rule : Completely correct epression (4) Total 4 M Then as above Alternative y artanh tanh y tanh y 3 d ( ) tanh y sech y y d ( ) 3 dy ( ) ( ) d ( ) : Divides by the correct form of sech y and tanhy in terms of : Correct quotient or product rule : Completely correct epression M Then as above

14 Question Scheme Marks 6(a) Gradient In this question condone the use of a and/or b for α and β sin sin m 3cos 3cos Correct attempt at chord gradient do not allow slips unless a correct method is clear ysin m( 3cos ) or ysin m( 3cos ) A correct straight line method using their chord gradient and the or y m c and attempts to find c point P or the point Q using P or Q sin sin ysin ( 3cos ) 3cos 3cos sin sin ysin ( 3cos ) 3cos 3cos 4cos sin ysin ( 3cos ) 6sin sin sin sin 3cos sin sin y sin 3cos 3cos 3cos 3cos cos cos cos y sin 3sin sin A correct equation for the chord in any form. 3ysin ( ) cos ( ) 6(coscos ( ) sinsin ( )) or 3ysin ( ) cos ( ) 6(cos cos ( ) sinsin ( )) y cos ( ) sin ( ) cos ( ) **ag** cso 3 This is cso there must no errors in applying the factor formulae and sufficient working must be shown to justify the printed answer but allow cos cos ( ) sin sin ( ) cos (b) 3cos 3cos sin sin, or (3cos ( )cos ( ),sin ( )cos ( )) or (3cos ( )cos ( ),sin ( )cos ( )) Correct coordinates of mid-point in any form Coordinates must be in this order but condone outer brackets missing B (4) ()

15 Question Scheme Marks (c) Centre of chord is (3cos ( )cos ( ),sin ( )cos ( )) Attempt factor formulae on both coordinates of mid-point at any stage in (c) y May be implied by their below sin ( )cos ( ) sin ( ) y 3cos ( ) cos ( ) 3cos ( ) Or 3cos ( ) cos ( ) 3cos ( ) y sin ( )cos ( ) sin ( ) : Obtains an epression for k or k or k or k Dependent on the previous (factor formulae must have been used) : Correct epression in any form cos ( ) m Must be seen or used in (c) 3sin ( ) d B sin ( ) So cos ( ) 3m y 4 k 3 3m 9m cso (5) Total

16 Question Scheme Marks 7.(a) dy dy y r ( r ) p( r ) d d Or Or dy dy y p qy d d Attempts to differentiate eplicitly or implicitly to give one of the given forms Substitutes their derivative into dy or dy d r y d r r * r r cso * This is cso and so there must be no errors e.g. d y could give the correct d y answer but loses the but allow to show equivalence of lhs and rhs (3) (b) dy : Use of y d d using their answer to part (a) S y r (must be y and not y ) d r not required here : Correct epression including (may be implied by later work but must appear before any integration) r r Substitutes for y in terms of. d r Dependent on first M. d Substitutes the limits r and r or and r into an epression of the r form krand subtracts. The use r or r r of the limit can be taken on trust dd if omitted. Dependent on both previous method marks. If they reach r correctly then double, then some justification is needed e.g. some mention of symmetry 4 r * cso (5) r r Note that S y dfollowed by correct work could r r score full marks as could the correct use of S y d y (c) arc length Ignore any working B () Total 9

17 Question Scheme Marks 8.(a) OA, OB,OC AB, BC =, AC = 3 : Attempt vector product for two ABAC i jk sides of the triangle. If the method is Or e.g. unclear, at least components must be BABC i jk correct. : Correct vector (b) Attempts theirab AC Area ABC = Dependent on the first M 6 Accept equivalents or awrt. Note that triangles OAB and OBC have the same area but score /4 It must be triangle ABC bc( ijk) ( jk) ijk ( ).( ) ( ) 6 i j -i-j+k 6 Attempt bc. If the method is unclear, at least components must be correct. : Attempt scalar product of a with their bc to obtain a number not a vector. : Obtains = with no errors (allow omission of 6 for all 3 marks) Just a.( -i-j+k) would lose the Alternative Writes this statement (allow other a.b c brackets provided the determinant is implied later) : Clear attempt at determinant : Obtains = with no errors (allow omission of 6 for all 3 marks) (c) Volume of tetrahedron (OABC) = a b coe or c b aoe b c is perpendicular to a or a is parallel to CB All vectors/points lie in the same plane OABC is a parallelogram a, b and c are linearly dependent Do not isw if there are contradictory or wrong statements award B d (4) (3) B () Total 8

18 Question Scheme Marks ( ) d ( ) n( ) d : Integration by parts in the correct direction : Correct epression (If the parts formula is not quoted and the epression is wrong, score MA) 9(a) n n n (b) ( ) n n n ( ) d n n n ( ) n ( ) ( ) d I ni ni ( ) n n n n I n n n n ( ) n In ( ) I Use of or equivalent. Dependent on the previous method mark. Correctly replaces n n ( ) dand ( ) d by I n and I n+. Dependent on both previous method marks. Correct completion to the printed answer with no errors. Correct application of the given reduction formula using n = only I k (must be and not just for arctan) I d arctan I = arctan C I arctan C Cao (constant not needed) ( ) d dd cso (5) (3) Total 8

19 Etra Notes 5 8. (a) M T MM (b) Parts once then eponentials e e e e I e sinh e coshd e e d integrates by parts and writes cosh as eponentials Correct epression e e 3 3 e e e 3e e 3e e 3e e 4 p p e d qe at least once and correct use of the limits and 3 e e 6 3 Any eact equivalent (allow e ) but all like terms collected but isw following a correct answer.

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