PMT. A-LEVEL Mathematics. MFP3 Further Pure 3 Mark scheme June Version: 1.0 Final

Transcription

1 A-LEVEL Mathematics MFP Further Pure Mark scheme 0 June 0 Version:.0 Final

2 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this eamination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and epanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular eamination paper. Further copies of this mark scheme are available from aqa.org.uk Copyright 0 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important eception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

3 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for eplanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A, or (or 0) accuracy marks EE deduct marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp decimal place(s) Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. of

4 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 Q Solution Mark Total Comment (a) a 5( a b) ( 0) B a 5( a b) OE PI by two correct equations or correct values for a and b. 5a 0, a 5b 0 M Equating coefficients to form two equations, at least one correct. PI by net line. 4 a, b 5 A Correct values for both a and b. (b) Au eqn m 5 0 M PI Or solving y 5y 0 as far as y Ae 5 A OE ( y CF ) ( y GS ) e A BF Total Ae.5 OE. ( ygs ) c s CF 0. 8, must have eactly one arbitrary constant; ft c s non zero values for a and b from part (a). Q Solution Mark Total Comment (a) 5 () () B Correct epansion. ISW in higher powers. sin =...! 5! = 5 (b) 4 B Correct simplified epansion seen or used... ISW in higher powers. p sin = ( p )(...) M Series epansions for sin and attempted and used in the given function = ( p) ( p) ( p ) 0; ( p) q 5 5 p ; m c s coefficient of equated to 0 and c s coeff of 5 equated to q (PI) and an attempt to solve as far as reaching a value for p and a value for q. 8 q 5 A 4 ACF of both eact values. Total 5 4 of

5 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 Q Solution Mark Total Comment (a) DO NOT ALLOW ANY MISREADS IN THIS QUESTION (a) k 0. () ln() 0.ln (=0.09 ) M k 0. (0 ) ln(0 ) OE seen or used k 0. f(0., k) M = 0.0. ln k 0.0. ln0. c' s k OE seen or used = 0.ln(.9 ) = 0.09(99 A AWFW 0.09 to 0.09 inclusive. PI by final answer.08 or.08. y 0. = 0.09(..) 0.09(9..) m c' s k c' s k but dep on previous two Ms scored. PI by.08 or.08. =.08 (to dp) A 5 CAO Must be.08 (b)(i) dy d y dy d d d y Product rule used = ln y ( ) B ln y ln y ( ) ln( y) ( ) y M A seen or used. d y ACF for in terms of and y only. d (ii) y 0, y 0 ln, B y 0 and 0 ln y seen or used. ln y 0 ln d y BF Seen or used, ft on c s, an eact d value at some stage which may be left unsimplified. ( 5ln 4 values must be eact but may be unsimplified y ) ln BF (c' s y0) c' s y 0 (iii) y =.080(479..) =.080 to dp B Must be.080 and dep on values in (b)(ii) being correctly found Total 5 of

6 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 Q4 Solution Mark Total Comment (a) 0 r ; cos r r cos 0 r 0 M r cos used at any stage. r 0 A r 0 or a more suitable correct form to allow elimination of r; or a correct Cartesian equation in ACF 9r ( 0) 9( 5 y ) ( 0) M r y 5( 9y ) 9y 80 used to form a Cartesian equation. 5( 4) 9y 80 m Completing the square 5 9y 40 ( 4) y 0 = 5[( 4) 4 ] 9y OE; this m cannot be awarded retrospectively c, d 0 A 5 CSO; c, d 0 stated (b) Translation 4 0 BF ft is only applied if m scored in (a). Must be translat and vector form. B0F if more than one transformation Total (a) y 0 scores the first marks of

7 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 Q5 Solution Mark Total Comment (a) B Condone A, B (b) du d u d du. M Replacing by dy d and dy d by u I.F. e (d) M e (d) OE PI = = ln ln e AF Aln B ln A e OE ft c s non-zero A and B values. ( ) A correct IF in form d d du d u u u (+c) =0, u=4 c so u dy d d y d y d m A Dep on both previous M s d u candidate' s IF d LHS as A u d m m ln (+d) A ACF y OE PI Dep on previous MMm, replacing u to d form y g( ) seen or used (c could d still be present) q in form p or other valid method to integrate (+) /(+) or c to integrate, where c is a general const of integration or a non-zero value other than =0, y= d y ln A Total ACF eg ( ) y ln( ) ( ) 7 of

8 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 8 of

9 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 Q Solution Mark Total Comment (a) lim ln p lim ln lim lim = a M Changing to k ( ) p p a 0 k ( ) p a 0 a lim M ln p = a k ln a Changing to a k ln a ( ) k a 0 p = 0 A (b) ln d 7 dv u ln, d 7, du, v d M d u, v d k with k or ln d = ln d 7 = ln (+c) A ACF ln d = 7 lim p p ln 7 d lim = p ln p p p 0 lim ln p lim = p p p p 0 M lim F p F OE provided p consistent use of same letter and F follows from an attempt at integration by parts ln d = A 4 Total 7 together with either the split into two relevant limit epressions (eg see previous line in solutions) with indication of 0 evaluations or reference to (a) with general k or k= or an eplicit statement lim ln p for k 0 = 0 OE with k p p general k or k= 9 of

10 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 Q7 Solution Mark Total Comment Au eqn m 4 0 M PI by correct values of m seen/used. ( y CF ) Asin Bcos A Asin Bcos OE Try ( y PI ) ae 4 M bsin c cos M ( y' ' PI ) ae 4 (4b 4c)cos (4c 4b)sin A Correct ( y '' PI ) ae 4 (4b 4c)cos (4c 4b)sin 4 + 4( ae bsin c cos) = = 0e 4 8sin 4cos m Substitution into y 4y, seen or used, dep on nd and rd M and at least one set of differentiations being in the form ke 4 ( p q)cos ( r s)sin for non-zero constants k, p, q, r and s. 0a 0 a c 8 c ; sin terms cancel 4b 4 b ; cos terms cancel ( y PI ) 4 0.5e B 4 0.5e term in PI sin cos B sin cos term in PI with correct sin and cos terms in m line Asin Bcos + ( ygs ) 4 0.5e sin cos y.5, 0 ; B B e y, ; A ; e y =( )cos +sin A e sin 4 e AF Correct ft value of either A or B, the coeffs of sin and cos respectively, ft only on wrong non-zero coefficients in the GS; m must have been scored. A 0 ACF. Must be a correct eqn. and ALL previous 9 marks must be scored Total 0 0 of

11 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 Q8 Solution Mark Total Comment (a) M (OA=) tan tan PI by as final answer OA = A Or as final answer (b)(i) (ii) Polar eqn of AN: r cos OAcos M Or ON OAcos OE r cos AF Ft from (a). OE ON (At B): ( tan ) cos m ( tan )cos c' s OA. OE (cos sin ) A 4 AG Printed result convincingly obtained. sin sin M sin OE eg sin 4 (Since 0 ( B ) A ( ) ) so A Condone B NMS scores 0/ (c) (Area of triangle OAB =) OA tan B sin A B M Valid method as far as a correct epression in terms of known lengths/angles OE eg OA cos tan tan OA eg tan sin 4 OAsin sin A AG = of

12 MARK SCHEME A-LEVEL MATHEMATICS MFP JUNE 0 (d) ( Area = ) tan ( ) B = tan tan sec tan (d) = ln sec (d ) M Use of r (d ) (d) BF Correct epn of [+tan ] and limits M and c s, in terms of such that 0 used with k r (d) tan sec with k r (d) tan A Correct integration of sec tan following use of correct identity Shaded area= B tan (d ) M Condone difference taken in wrong order for this M mark = ln ln... A Any correct numerical epression for the shaded area, can be unsimplified but must be eact = ln ln A 7 Total 7 TOTAL 75 Simplified to the difference of two correct eact terms condoning one of them to be unsimplified; OE factorised form of