A-LEVEL Mathematics. Further Pure 4 MFP4 Mark scheme June Version/Stage: 1.0 Final

Transcription

1 A-LEVEL Mathematics Further Pure 4 MFP4 Mark scheme 6360 June 205 Version/Stage:.0 Final

2 Mark schemes are prepared b the Lead Assessment Writer and considered, together with the relevant questions, b a panel of subject teachers. This mark scheme includes an amendments made at the standardisation events which all associates participate in and is the scheme which was used b them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that ever associate understands and applies it in the same correct wa. As preparation for standardisation each associate analses a number of students scripts: alternative answers not alread covered b the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised the are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in man cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one ear s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this Mark Scheme are available from aqa.org.uk Copright 205 AQA and its licensors. All rights reserved. AQA retains the copright on all its publications. However, registered schools/colleges for AQA are permitted to cop material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocop an material that is acknowledged to a third part even for internal use within the centre.

3 Ke to mark scheme abbreviations 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 accurac B mark is independent of M or m marks and is for method and accurac E mark is for explanation or ft or F follow through from previous incorrect result CAO correct answer onl CSO correct solution onl AWFW anthing which falls within AWRT anthing which rounds to ACF an correct form AG answer given SC special case OE or equivalent A2, 2 or (or 0) accurac marks x EE deduct x marks for each error NMS no method shown PI possibl implied SCA substantiall correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specificall requires a particular method to be used, we must usuall see evidence of use of this method for an marks to be awarded. Where the answer can be reasonabl obtained without showing working and it is ver unlikel that the correct answer can be obtained b using an incorrect method, we must award full marks. However, the obvious penalt 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 reasonabl allow the solution of the question directl, the correct answer without working earns full marks, unless it is given to less than the degree of accurac accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for an marks to be awarded. 3 of 4

4 Q Solution Mark Total Comment (a) 3 a a 3 a Correct expansion of triple scalar product or ( u v ) or correct vector product 2a CAO (b)(i) 2a a 4 (ii) c 4 d c 3 d 2 BF Sets their expression equal to 0 and solves the resulting linear equation correctl Forming a sstem of equations and solving to correctl find either c or d Both c and d correct u 3v+ 2w 3 Correct linear combination stated NMS u 3v+ 2w scores 3 marks Total 6 4 of 4

5 Q2 Solution Mark Total Comment (a + 2b 6c) x (a b + 3c) axa axb +3axc +2bxa 2bxb +6bxc 6cxa +6cxb 8cxc Expansion of brackets at least six terms correct with or axb+3axc+2bxa+6bxc-6cxa+6cxb Expansion full correct unsimplified and use of axa bxb cxc 0 (seen or implied) Use of (axb bxa or cxa axc) and cxb bxc 9cxa + 3bxa or 3j + 3( 2i) + 2(3j) 6(2i) 8i +9j, 5 each term Note candidates who do not use vector product smbols 2 eg a ab + 3 ac +... or attempt to use components of vectors score M0 Total 5 5 of 4

6 Q3 Solution Mark Total Comment (a) c replaced b c + c 2 gives a+ b c b c bc b+ a c a c ca c+ a+ b a+ b ab Combining columns or rows sensibl, working towards first factor b c bc ( a+ b c) a c ca a + b ab First factor correctl extracted r 2 replaced b r 2 r r 3 replaced b r 3 r b c bc 0 a b ca + bc 0 a + c ab + bc b c bc 0 a b c( a b) 0 a+ c b( a+ c) Combining rows or columns sensibl, working towards second factor b c bc ( a b)( a+ c)( a+ b c) 0 c 0 b b c bc 0 c b+ c 0 b Three factors correctl extracted and remaining determinant correct Correct expansion to obtain final factor- dependent on previous and Hence full factorisation ( a+ b c)( a b)( a+ c)( b+ c) 6 Full correct - CSO (b) Comparing gives c 2 and b 3 Attempting to substitute c 2 and b 3 into their answer from part (a) Hence 5( a+ )( a 3)( a+ 2) ( 0) F Correct factors PI b correct values, provided FT is cubic equation in a with three linear factors a 2,, 3 3 CSO must have 6 marks in part (a) Total 9 6 of 4

7 Q3 Solution Mark Total Comment 3 ALTERNATIVE to (a) r 2 replaced b r 2 r r 3 replaced b r 3 r a b c bc b a a b c( b a) c a a+ c bc ( + a) a b c bc ( a b)( a+ c) c b () () () Combining columns or rows sensibl, working towards first factor Combining rows or columns sensibl, working towards second factor First factor correctl extracted r 3 replaced b r 3 r 2 a b c bc c 0 0 b+ c a b c bc ( b+ c) c 0 0 a b c bc c a+ b c 0 0 () () Third factor correctl extracted Correct expansion to obtain final factor- dependent on previous and Hence full factorisation ( a+ b c)( a b)( a+ c)( b+ c) () (6) Full correct CSO ALTERNATIVE to (b) a 6 3 a 2 2a 2 a+ 3 3a a 2 2a 6 6 a 3 2 a a a a a 2 2 a () Correctl expanding determinant 5( a+ )( a 3)( a+ 2) ( 0) () CAO a 2,, 3 () (3) CSO 7 of 4

8 Q4 Solution Mark Total Comment (a) λ 2 4 λ ( 0) ( λ)(4 λ) + 2 ( 0) ( λ 2)( λ 3) ( 0) λ 2, 3, each eigenvalue When λ 2, x or x+ 0 OE Correct equation used to find eigenvector for either λ 2 or λ 3 or an multiple A correct eigenvalue found for λ 2 When λ 3, 2 x or 2x+ 0 2 or an multiple 6 A correct eigenvalue found for λ 3 (b) Using vectors above, required matrix columns must be multiples of 2 and 4 b Comparing with a 2 Attempt to compare their eigenvectors with given matrix in correct order PI b correct value of a or b gives a 8 and b 2 3 each value Total 9 8 of 4

9 Q5 Solution Mark Total Comment (a) from original or or Method row reduction to stage as above Method 2 elimination of one variable to obtain 35+ 7z 7 etc see above Two equations that are multiples of each other () () Having row of 0s or stating one row is multiple of another stating or showing one row is multiple of another or reducing both to same equ n (b) Let λ Then z 5λ x 2 2λ 5 Setting one variable equal to a parameter and obtaining expressions for both other variables each variable Other possibilities, eg x 85 2 x α ; + β 2 z 0 5 z 4 0 The equations represent three planes which meet in a line/form a sheaf. E Must earn at least two marks in part (a) Total 6 9 of 4

10 Q6 Solution Mark Total Comment (a) does not match Substituting and comparing with direction vector of line direction ratios of line or B or showing vector product is not zero (b) Direction vectors for plane are 3 2, B Correct identification of one direction vector for plane, an multiples of these c Both vectors correct (ma have multiples, in an order) attempted Correct for their vector product Watch signs if terms in vector product are in a different order Use of their normal vector and correct point eg (4,,-2) to find value for c Plane is r.5 7 OE 5 CSO 0 of 4

11 Q6 contd Solution Mark Total Comment (c) Equation of line perpendicular to plane and containing D is 8 r 2 + t 5 6 Meets plane when (8 + t) +5(-2 +5t) +(6 +t ) 7 Equation of line through D using their normal from (b) Correct use of their line and their plane to obtain linear equation in t t 9 Correct t value obtained Hence for reflected point, t 2 9 BF Doubling their t value Reflected point coordinates correct Total of 4

12 Q7 Solution Mark Total Comment (a) a b 5 c d 3 gives a+ b 5 c+ d 3 a b c d gives a b c d B B both equations correct both equations correct Hence a 3, b 2, c 2, d Solving to get at least two correct values 3 2 Matrix is 2 4 (b)(i) p q r s 2.2 p q Hence r s.2 2 Multiplication of matrices in correct order to form matrix equation - accept TS A Rearranging - correct order on RHS accept T AS p q r s p q r s BF 4 CAO Correct inverse of their matrix from (a) seen anwhere ALTERNATIVE p q r s 2.2 3p 2q 2 p q r 2s 2r s.2 3p 2q 3.4 and 2p - q 2 Gives p 0.6 and q -0.8 () () Multiplication of matrices in correct order to form matrix equation LHS full correct 3r -2s.2 and 2r s Gives r 0.8 and s 0.6 () Solving to find all correct values p q r s () (4) CAO (ii) (Anticlockwise) rotation Matrix must be correct in part (b)(i) through (about O) 2 Correct angle Total 0 2 of 4

13 Q8 Solution Mark Total Comment 2 k (a)(i) k Correct expansion of 3 b 3 determinant k ( 5 + 3k ) Correct unsimplified, brackets removed k 5 3 Correct conclusion (ii) k 2 k k k k 4 k A2 one row or column correct at least six terms correct A2 all correct Transpose of their matrix dependent on previous M - 7 k k 4 k 4 3k Full correct (b) When k, determinant of M 2 B or det M /2 Hence volume scale factor 2 Image volume 6 2 Correct use of ± " their" volume scale factor to find image volume 0.5 (cm 3 ) 3 CAO must be positive (c) 2 5 x x+ 2+ 5z z + z x+ z - Substituting k 5 and multipling - at least two components correct all correct x' ' + z' ( x z) (3+ 4 z) + ( x+ z) 0 Therefore each point lies in the plane x + z 0 3 AG be convinced Must see either first three lines or concluding statement when top line is missing 3 of 4

14 Q8cont d Solution Mark Total Comment (d) 2 k x x z z x kz x 3 4z + x+ z z 2 + kz z 0 x+ 2z 0 Use of Mv v with at least two equations correct Full correct with terms combined Hence k 4 Equation of line is x z OE Total 9 TOTAL 75 Using their equations to obtain Cartesian equations of line CSO 4 of 4