AQA MFP4 w MS JUN15

A-LEVEL Mathematics Further Pure – MFP4 Mark scheme 6360 June 2015 Version/Stage: 1.0 Final 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 examination 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 expanded 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 examination paper Further copies of this Mark Scheme are available from aqa.org.uk Copyright © 2015 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 exception: 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 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent or (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) decimal place(s) No Method Shown 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 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q1 (a) Solution Mark Total a −4 a a −4 7= −2 +2 −2 −2 −4 2 −2 or (ii) or correct vector product 12a + 48 = a = −4 1  2 =     Correct expansion of triple scalar product  12  ( u × v = )   −10 = 12a + 48 (b)(i) M1 Comment 3   −4    c  −4  + d 7     −2  A1 CAO B1F Sets their expression equal to and solves the resulting linear equation correctly M1 Forming a system of equations and solving to correctly find either c or d A1 Both c and d correct c=3 d =2 = u 3v + w A1 A1 Correct linear combination stated NMS = u 3v + w scores marks Total of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q2 Solution (a + 2b – 6c) x (a – b + 3c) Mark Total Comment = axa – axb +3axc +2bxa – 2bxb +6bxc –6cxa +6cxb –18cxc M1 Expansion of brackets – at least six terms correct with × or ∧ = –axb+3axc+2bxa+6bxc-6cxa+6cxb A1 Expansion fully correct unsimplified and use of axa = bxb = cxc = (seen or implied) m1 Use of (axb = –bxa or cxa = –axc) and cxb = –bxc = –9cxa + 3bxa or 3j + 3(–2i) + 2(3j) – 6(2i) = –18i +9j A1,A1 A1 each term Note candidates who not use vector product symbols eg a − ab + 3ac + or attempt to use components of vectors score M0 Total 5 of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q3 (a) Solution c1 replaced by c1 + c2 gives a+b−c b−c Mark Total Comment −bc b + a − c a − c −ca −c + a + b a + b ab b − c −bc (a + b − c) a − c −ca a + b ab M1 Combining columns or rows sensibly, working towards first factor A1 First factor correctly extracted m1 Combining rows or columns sensibly, working towards second factor A1 Three factors correctly extracted and remaining determinant correct m1 Correct expansion to obtain final factor- dependent on previous M1 and m1 r2 replaced by r2 – r1 r3 replaced by r3 – r1 −bc b−c a − b −ca + bc a + c ab + bc −bc b−c a − b −c ( a − b) a + c b( a + c ) b − c −bc ( a − b)( a + c )( a + b − c ) −c b b − c −bc −c =b + c b Hence full factorisation = (b) (a + b − c)(a − b)(a + c)(b + c) A1 Comparing gives c = and b = M1 Hence 5( a + 1)( a − 3)( a + 2) ( = 0) A1F a =−2, − 1, A1 Total Fully correct - CSO Attempting to substitute c = and b = into their answer from part (a) Correct factors PI by correct values, provided FT is cubic equation in “a” with three linear factors CSO must have marks in part (a) of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q3 Solution Mark Total Comment ALTERNATIVE to (a) r2 replaced by r2 – r1 r3 replaced by r3 – r1 a b−c −bc b − a a − b c(b − a ) − c − a a + c b (c + a ) a b − c −bc (a − b)(a + c) −1 −c −1 b (M1) Combining columns or rows sensibly, working towards first factor (m1) Combining rows or columns sensibly, working towards second factor (A1) First factor correctly extracted (A1) Third factor correctly extracted (m1) Correct expansion to obtain final factor- dependent on previous M1 and m1 r3 replaced by r3 – r2 a b − c −bc −1 −c 0 b+c a b − c −bc (b + c) −1 −c 0 a b − c −bc −1 −c = a + b − c 0 Hence full factorisation = (a + b − c)(a − b)(a + c)(b + c) (A1) (6) Fully correct – CSO ALTERNATIVE to (b) a −6 a − −2a −2 a + 3a = a a − −2a −3 −6 −2 −6 (M1) Correctly expanding determinant 5( a + 1)( a − 3)( a + 2) ( = 0) (A1) CAO a =−2, − 1, (A1) a+3 3a a + 3a a − −2a (3) CSO of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q4 (a) Solution 1− λ Mark Comment −1 ( = 0) 4−λ (1 − λ )(4 − λ ) + ( = 0) M1 (λ − 2)(λ − 3) ( = 0) λ = 2, When Total A1,A1 A1 each eigenvalue λ = 2,  −1 −1  x  0   2   y  = 0      M1 Correct equation used to find eigenvector for either λ = or λ = A1 A correct eigenvalue found for λ=2 A correct eigenvalue found for λ =3 or x + y = OE 1   −1 or any multiple   When λ = 3,  −2 −1  x  0  OE    y  = 0  or x + y =      1   −2  or any multiple   (b) A1 Using vectors above, required matrix columns must be multiples of 1  1   −2  and  −1     4 b  Comparing with    a −2  Attempt to compare their eigenvectors with given matrix in correct order PI by correct value of a or b M1 gives a = −8 A1 and b = A1 Total A1 each value of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q5 Solution Mark Total Comment (a)  −11 −3   35 7     65 13 13  −11 −3  from original  −10 −4    9 −17 −7 11  −11 −3  or  14 14    13 26 26   −11 −3 or  35 −14 56    65 −26 104  Method – row reduction to stage as above M1 A1 Method – elimination of one variable to obtain 35 y + z = etc see above (M1) Two equations that are multiples of each other (A1) stating or showing one row is multiple of another or reducing both to same equ’n Let y = λ Then M1 Setting one variable equal to a parameter and obtaining expressions for both other variables z = − 5λ x= − 2λ (b) Having row of 0s or stating one row is multiple of another A1 A1 The equations represent three planes which meet in a line/form a sheaf Total E1 A1 each variable Other possibilities, eg  x  8 5    x  y = 1  + α  −1 ; =  y          z       z  0  −4   +β       10 −  −      Must earn at least two marks in part (a) of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q6 (a) Solution Mark Total   2     − 1  =   does not match        −2  0   −2  Comment Substituting and comparing with direction vector of line direction ratios of line  2 3 or   ×  −1 =      −2    (b)  −   0  − 10 ≠ 0      −  0 B1 or showing vector product is not zero Direction vectors for plane are 3 2  −1 ,          −2  3  2  −1 ×          −2  2  10     2 2   c =  10  = 14     0    1  Plane is r 5 =   1  OE B1 Correct identification of one direction vector for plane, any multiples of these M1 Both vectors correct (may have multiples, in any order) attempted A1 Correct for their vector product Watch signs if terms in vector product are in a different order M1 Use of their normal vector and correct point eg (4,1,-2) to find value for c A1 CSO 10 of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q6 contd (c) Solution Mark Total Comment Equation of line perpendicular to plane and containing D is   1  r = −2  + t 5       1  Equation of line through D using their normal from (b) M1 Meets plane when (8 + t) +5(-2 +5t) +(6 +t )= t= m1 Correct use of their line and their plane to obtain linear equation in t A1 Correct t value obtained Hence for reflected point, t = 2× B1F 8  1   −2  + 5 =   9  6  1  74  1  −8 9  56  A1 Total Doubling their t value Reflected point coordinates correct 11 11 of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q7 (a) Solution Mark  a b  1    c d  1 =  −3 gives      a+b = c+d = −3 a b   1  1  c d   −1 =  −1 gives      a −b = c−d = −1 Hence a = 3, b = 2, c = −2, d = −1 (b)(i) Total Comment B1 both equations correct B1 both equations correct M1 Solving to get at least two correct values Matrix is  3 2   −2 −1 A1  p q    3.4   r s   −2 −1 = 1.2       M1 Multiplication of matrices in correct order to form matrix equation - accept TS = A m1 Rearranging - correct order on RHS accept T = AS −1 B1F Correct inverse of their matrix from (a) seen anywhere  p q  3.4    Hence  =    r s  1.2   −2 −1 −1  p q  3.4   −1 −2   r s  = 1.2          p q  0.6 −0.8  r s  =  0.8 0.6      A1 CAO ALTERNATIVE  p q    3.4  r s   −2 −1 = 1.2     3 p − 2q p − q  3.4  3r − s 2r − s  = 1.2    2  2  (M1) Multiplication of matrices in correct order to form matrix equation (A1) LHS fully correct (A1) Solving to find all correct values 3p – 2q = 3.4 and 2p - q = Gives p = 0.6 and q = -0.8 3r -2s = 1.2 and 2r – s = Gives r = 0.8 and s = 0.6 (ii)  p q  0.6 −0.8  r s  =  0.8 0.6      (A1) (Anticlockwise) rotation M1 through 53.10 (about O) A1 Total (4) CAO Matrix must be correct in part (b)(i) Correct angle 10 12 of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q8 (a)(i) Solution Mark Total Comment k k −1 3= −1 −1 −1 M1 Correct expansion of by determinant =−3 − − + 3k A1 Correct unsimplified, brackets removed ( =−15 + 3k ) k ≠5 (ii)  −7 k +2  8 − 3k A1 3 −4 k − −3  −4  −7 k + − 3k   −4 k − −4    3  −3  −7 k + − 3k   M = −4 k − −4   3k − 15  −3  Correct conclusion M1 one row or column correct A2 A1 at least six terms correct A2 all correct m1 Transpose of their matrix – dependent on previous M1 -1 (b) When k = 1, determinant of M = –12 Hence volume scale factor = Image volume = ×6 12   x    =     y  −1 −1  z  Fully correct B1 or det M –1 = –1/12 M1 Correct use of ± " their " volume scale factor to find image volume A1  x + y + 5z  3 y + z     − x + y − z  x '− y '+ z ' = ( x + y + 5z ) − (3 y + z ) + ( − x + y − z ) =0 Therefore each point lies in the plane x− y+z= 12 = 0.5 (cm3) (c) A1 CAO – must be positive M1 M1 - Substituting k = and multiplying at least two components correct A1 A1 all correct A1 AG be convinced Must see either first three lines or concluding statement when top line is missing 13 of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q8cont’d (d) Solution Mark  k  x  x     y =  y       −1 −1  z   z   x + y + kz   x  3 y + z  =      y  − x + y − z   z  y + kz = y + 4z = − x + y − z =0 Use of Mv = v with at least two “equations” correct A1 Fully correct with terms combined A1 Equation of line is m1 x y = = z −4 −2 A1 Total TOTAL Comment M1 Hence k = OE Total Using their equations to obtain Cartesian equations of line CSO 19 75 14 of 14 [...]...     −1 1 −1  z   z   x + 2 y + kz   x  3 y + 4 z  =      y  − x + y − z   z  2 y + kz = 0 2 y + 4z = 0 − x + y − 2 z =0 Use of Mv = v with at least two “equations” correct A1 Fully correct with terms combined A1 Equation of line is m1 x y = = z −4 −2 A1 Total TOTAL Comment M1 Hence k = 4 OE Total 5 Using their equations to obtain Cartesian equations of line CSO 19... 0 5 1 12 = 0.5 (cm3) (c) A1 3 CAO – must be positive M1 M1 - Substituting k = 5 and multiplying at least two components correct A1 A1 all correct A1 3 AG be convinced Must see either first three lines or concluding statement when top line is missing 13 of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q8cont’d (d) Solution Mark  1 2 k  x  x   0 3 4   y =  y       −1 1 −1... − 3k   −4 k − 1 −4    3 3  −3  −7 k + 2 8 − 3k  1  M = −4 k − 1 −4   3k − 15  3 −3 3  3 Correct conclusion M1 one row or column correct A2 A1 at least six terms correct A2 all correct m1 Transpose of their matrix – dependent on previous M1 -1 (b) When k = 1, determinant of M = –12 Hence volume scale factor = Image volume = 1 ×6 12  1 2 5  x   0 3 4  =     y  −1 1 −1...MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q6 contd (c) Solution Mark Total Comment Equation of line perpendicular to plane and containing D is  8  1  r = −2  + t 5      6  1  Equation of line through D using their normal from (b) M1 Meets plane when (8 + t) +5(-2 +5t) +(6 +t )= 7 t= 1 9 m1 Correct use of their line and their... of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q7 (a) Solution Mark  a b  1  5   c d  1 =  −3 gives      a+b = 5 c+d = −3 a b   1  1  c d   −1 =  −1 gives      a −b = 1 c−d = −1 Hence a = 3, b = 2, c = −2, d = −1 (b)(i) Total Comment B1 both equations correct B1 both equations correct M1 Solving to get at least two correct values Matrix is  3 2   −2... -2s = 1.2 and 2r – s = 1 Gives r = 0.8 and s = 0.6 (ii)  p q  0.6 −0.8  r s  =  0.8 0.6      (A1) (Anticlockwise) rotation M1 through 53.10 (about O) A1 Total (4) CAO Matrix must be correct in part (b)(i) 2 Correct angle 10 12 of 14 MARK SCHEME – A-LEVEL MATHEMATICS – MFP4 -JUNE 15 Q8 (a)(i) Solution Mark Total Comment 1 2 k 3 4 2 k −1 0 3= 4 1 −1 3 4 −1 1 −1 M1 Correct expansion of 3 by...    M1 Multiplication of matrices in correct order to form matrix equation - accept TS = A m1 Rearranging - correct order on RHS accept T = AS −1 B1F Correct inverse of their matrix from (a) seen anywhere  p q  3.4 2   3 2  Hence  =    r s  1.2 1   −2 −1 4 −1  p q  3.4 2   −1 −2   r s  = 1.2 1   2 3        p q  0.6 −0.8  r s  =  0.8 0.6      A1 4 CAO