AQA MFP2 w MS JUN15

A-LEVEL Mathematics Further Pure – MFP2 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 – MFP2 -JUNE 2015 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 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q1 Solution (a) Mark r + 1= A( r + 2) + B or A( r + 2) B = + r +1 r +1 either A = or B = −1 (b) Total OE with factorials removed M1 correctly obtained A1 1 = − ( r + 2) r ! ( r + 1)! ( r + 2)! A1 1 1 + − +… − 2! 3! 3! 4! M1 1 − ( n + 2)! A1 Total allow if seen in part (b) use of their result from part (a) at least twice 1 − ( n + 1)! ( n + 2)! Sum = Comment must simplify 2! and must have scored at least M1 A1 in part (a) (a) Alternative Method Substituting two values of r to obtain two correct equations in A and B with factorials evaluated correctly B A B ; r=1 ⇒ = + earns M1 then A1, A1 as in main scheme r = ⇒ = A+ 2 1 − earns marks ( r + 1)! ( r + 2)! However, using an incorrect expression resulting from poor algebra such as = A( r + 2)!+ B ( r + 1)! with candidate often fluking A = 1, B = –1 scores M0 ie zero marks which you should denote as FIW These candidates can then score a maximum of M1 in part (b) NMS (b) ISW for incorrect simplification after correct answer seen of 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q2 Solution (a) Mark Total Comment y O x –1 (b) Graph roughly correct through O M1 Correct behaviour as x → ±∞ & grad at O A1 Asymptotes have equations y = & y = −1 B1 sech x = e x − e− x ; x = x −x e +e e + e− x condone infinite gradient at O for M1 both correct ACF or correct squares of these expressions seen B1 x 22 + ( e x − e ) ( sech x + x = ) ( e x + e− x ) must state equations −x 2 x = sech x + (c) e x + + e −2 x = e x + + e −2 x A1 6(1 − x ) = + x x + x − ( = 0) 2 1+ k  x = k ⇒ x = ln   1− k  1 , x = ln x = ln 2 x = , attempt to combine their squared terms with correct single denominator M1 x = − AG valid proof convincingly shown to equal including LHS seen B1 M1 correct use of identity from part (b) A1 obtained from correct quadratic forming quadratic in x A1F A1 Total FT a value of k provided k < both solutions correct and no others any equivalent form involving ln 11 (a) Actual asymptotes need not be shown, but if asymptotes are drawn then curve should not cross them for A1 Gradient should not be infinite at O for A1 (b) Condone trailing equal signs up to final line provided “ sech x + x = ” is seen on previous line for A1 Denominator may be e x + 4e x + + e x + 4e −2 x + e −4 x etc for M1 and A1 Accept sech x + tanh= x 2 (e (e + e− x ) = for A1 x + e− x ) x x −x  sinh x  + ( (e − e ) ) Alternative :  + =  2  cosh x cosh x  ( 12 (e x + e− x ) ) and then A1 for scores B1 M1 x −2 x e + e + 4= , (all like terms combined in any order) + x sech x = x 1 −2 x e + + e 4 of 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q3 (a) Solution Mark dx = 1− dt t B1 dy = dt t   d x 2  d y 2   +   =   dt   dt   (b)   ( 2π ) ( 2ln t )  t −   t (2t ) − (t + 1) t2 ACF squaring and adding their expressions and attempting to multiply out A1 1 dt   OE eg M1 ∫ ( ln t ) 1 + t Comment B1   1− + +  t t t  1  = 1 +   t  1+ + t t 2π Total 1  1  −  t −  ( dt )  t t  t  ∫    1  2π ( 2ln t )  t −  −  2t +   t   t   = 2π (3ln − + 4) = π (6ln − 2) AG be convinced B1 must have 2π , limits and dt M1 integration by parts - clear attempt to integrate + and differentiate 2ln t t A1 correct (may omit limits, π and dt) A1 correct including π (no limits required) A1 Total (b) May have two separate integrals and attempt both using integration by parts for M1 { ( ∫ )} Must see (2π ) 2t ln t − ∫ ( dt ) − 2t −1 ln t − 2t −2 ( dt ) (may omit limits, π and dt) for first A1 and 2π ( 2t ln t − 2t ) − ( 2t −1 ln t + 2t −1 ) for second A1 Condone poor use of brackets if later recovered of 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q4 (a) Solution Mark f ( k + 1)= 24 k +7 + 33k + M1 16 × 24 k +3 convincingly showing k += f ( k + 1) − 16f ( k ) = (81 − 16 × 3) × 33k E1 = 33 × 33k (b) A1 f (1) = 209 therefore f (1) is a multiple of 11 Total Comment must see 16 = OE B1 19 etc f (1)= 209= 11 × 19 or 209 ÷ 11 = therefore true when n=1 f ( k += 1) 16f ( k ) + 33 × 33k M1 = 11M + 11N = 11( M + N ) Therefore f ( k + 1) is a multiple of 11 attempt at f (k + 1) = using their result from part (a) where M and N are integers A1 Assume f ( k ) is a multiple of 11 (*) Since f(1) is multiple of 11 then f(2), f(3),… are multiples of 11 by induction (or is a multiple of 11 for all integers n ≥ ) Total E1 must earn previous marks and have (*) before E1 can be awarded (a) It is possible to score M1 E0 A1 (b) Withhold E1 for conclusion such as “a multiple of 11 for all n ≥ ” or poor notation, etc of 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q5 Solution (a) Im(z) Mark Total Comment Ignore the line OP drawn in full or circles drawn as part of construction for locus L A Re(z) O B –4 (b)(i) +P Straight line Through midpoint of OP, P correct Perpendicular to OP, P correct M1 A1 A1 ( x − 2) + ( y + 4) = x + y M1 2y − x +5 = A(5,0) & B (0, −2.5) 5 5 5 C  , −  ⇒ complex num = − i 2 4 A1 A1 (ii) either α= A1 5 5 − i or k = 4 M1 5 5 + i = 4 A1 z− Total P represents – 4i may have + 0i and – 2.5i allow statement with correct value for centre or radius of circle must have exact surd form (a) Withhold the final A1 (if marks earned) if the straight line does not go beyond the Re(z) axis and negative Im(z) axis The two A1 marks can be awarded independently (b)(i) Alternative 1: grad OP = –2 ⇒ grad L = ( x − 1) OE A1 then A1, A1 as per scheme 0.5 M1 ; y + 2= Alternative 2: substituting z= x (or a) then z= iy ( or ib) into given locus equation Both ( x − 2) + = and + y + 16 = x and 22 + ( y + 4) = y M1; − x + 16 = OE for A1 then A1, A1 as per scheme of 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q6 Solution (a) Mark ( x − 2) (4 − x ) + 4x − x + + x − x2 × 13 (+)  x−2 1−     + x − x2 Total Comment M1 product rule ( condone one error) A1 correct unsimplified dy   =  + 4x − x  dx  k A1cso  −1  x −   ( x − 2) + x − x + 9sin       27 1 + 9sin −1   " their " k  2 = or 3+ π 32 − ( x − ) correct unsimplified k=2 M1 ft “their” k m1 correct sub of limits (simplified at least this far) A1 cso Total last two terms above combined correctly A1 + x − x2 (b) B1 must have earned marks in part(a) to be awarded this mark (a) Second A1 ; may combine all three terms correctly and obtain 10 + x − x + x − x2 of 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q7 (a) (b)(i) Solution Mark αβ + βγ + γα = αβγ = − 27 Total B1 B1 ; αβ = − αβ + αβ + β = 27 May use γ instead of β throughout (b)(i) B1 Clear attempt to eliminate either α or β from “their” equations correct M1 α3 = − 27 or β3 = 27 A1 or β = 3 2 α = − , β= , γ = 3 either α = − (ii) k  − 1=  ∑α = 27  A1  ⇒ k = −27  α = −2i (c)(i) α =−2 − 2i (ii) 27( −2 − 2i) − 2ik + =0 1 +1⇒ z = z y −1 27 12 − +4= ( y − 1) ( y − 1) 27 − 12( y − 1) + 4( y − 1)3 = 27 − 12 y + 12 + 4( y − y + y − 1) = y − 12 y + 35 = ∑α ' β '= 3+ αβ + βγ + γα 3+ ∑α ' = αβγ all roots clearly stated B1 or substituting correct root into equation B1 B1 correctly substituting “their ” α = −2i and “their ” α =−2 − 2i B1 may use any letter instead of y M1 sub their z into cubic equation A1 A1 A1 removing denominators correctly correct and (y–1)3 expanded correctly = (B1) 2(αβ + βγ + γα ) + α + β + γ sum of new roots =3 M1 for either of the other two formulae correct in terms of αβγ , αβ + βγ + γα and α + β +γ (M1) αβγ (A1) =0 ∏= A1 y= Alternative: A1 M1 k= −27 + 25i (d) Comment αβ + βγ + γα + + α + β + γ αβγ −35 = 1+ (A1) y − 12 y + 35 = Total (A1) (5) may use any letter instead of y 17 10 of 11 MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q8 Solution (c) Comment So ω is a root of z = B1 must have conclusion plus verification that ω5 = ω2 , ω3 , ω4 B1 OE powers mod ( must not include 1) or clear statement that sum of roots (of z5 − = ) is zero (ii) (ii) Total ( ω = ) cos 2π + isin 2π = (a)(i) (b)(i) Mark 1+ ω+ ω + ω3= + ω4 − ω5 = 1− ω B1 1  1   ω+  +  ω+  − ω  ω  1 = ω + + + ω+ − ω ω 1+ ω+ ω + ω3 + ω = =0 ω2 2π 2π = cos − isin 5 ω 2π ⇒ ω+ = 2cos ω  Solving quadratic  ω+ ω  correct expansion M1 A1 AG correctly shown to = not allow simply multiplying by ω M1 −1 ± 2π Rejecting negative root since cos >0 2π −1 Hence cos =  =  A1 SC1 if result merely stated M1 must see both values must see this line for final A1 A1 It is possible to score SC1 M1 A1 Total (b)(ii) May replace by ω3 and by ω and/or by ω5 in valid proof ω2 ω 1 Alternative: 1+ ω+ ω + ω3 + ω = ⇒ + +1+ω+ω = M1 ω ω 2 1  1 1 1     ω+  − 2+  ω+  + =0 ⇒  ω+  +  ω+  − =0 A1 ω  ω ω ω    11 of 11 [...]...MARK SCHEME – A-LEVEL MATHEMATICS – MFP2 -JUNE 2015 Q8 Solution (c) Comment So ω is a root of z 5 = 1 B1 1 must have conclusion plus verification that ω5 = 1 ω2 , ω3 , ω4 B1 1 OE powers mod 5 ( must not include 1) 1 or clear statement that sum of roots (of z5 − 1 = 0 ) is zero 5 (ii) (ii) Total ( ω = ) cos 2π... ω  ω  1 1 = ω 2 + 2 + 2 + ω+ − 1 ω ω 1+ ω+ ω 2 + ω3 + ω 4 = =0 ω2 2π 2π 1 = cos − isin 5 5 ω 1 2π ⇒ ω+ = 2cos ω 5 1  Solving quadratic  ω+ ω  correct expansion M1 A1 2 AG correctly shown to = 0 do not allow simply multiplying by ω 2 M1 −1 ± 5 2 2π Rejecting negative root since cos >0 5 2π 5 −1 Hence cos = 5 4  =  A1 SC1 if result merely stated M1 must see both values must see this line for