A-LEVEL Mathematics Further Pure3 – MFP3 Mark scheme 6360 June 2015 Version/Stage: Final Mark Scheme V1 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 – MFP3- 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 MARK SCHEME – A-LEVEL MATHEMATICS – MFP3- JUNE 2015 Q1 (a) Solution Mark Total DO NOT ALLOW ANY MISREADS IN THIS QUESTION M1 + 52 y (2.05) = y (2 ) + 0.05 = + 0.05 × 13.5 = 5.675 y (2.1) = y (2) + × 0.05 f [2.05, y (2.05)] (b) 2.05 + 5.6752 = + × 0.05 × 2.05 = 6.67 to sf Total A1 M1 OE PI Ft on c’s (a) answer A1F A1 Comment CAO Must be 6.67 (b) For the PI if line missing, check to see if evaluation matches 5.1 + Q2 Solution Mark M1 ∫ tan x dx 2 × [answer (a)] to at least 3sf 41 Total Comment I.F e OE eg e − ln cos x OE Only ft sign error in integrating tan x A1 = e ln sec x = sec x A1F dy sec x + sec x(tan x ) y = tan x sec x dx d [ y sec x] = tan x sec x dx ∫ y sec x = tan x sec x (dx) M1 LHS as d [y × candidate' s IF] dx PI A1 m1 ∫ y sec x = t dt PI OE eg y sec x = where u = cos x ∫ u − 1 du , u5 A1 y sec x = tan x (+ c) π π sec = tan + c ; = + c 4 Dep on prev MMm Correct boundary condition applied to obtain an eqn in c m1 with correct exact value for either sec tan y sec x = y= π used tan x + 4 ( cos x + tan x ) A1 Total ACF Condone answer left in a ‘correct’ form different to y= f(x), eg y sec x = tan x + of π or MARK SCHEME – A-LEVEL MATHEMATICS – MFP3- JUNE 2015 Q3 (a)(i) (a)(ii) Solution (2 x) (2 x) (2 x) ln(1 + x ) = x − + − … = 2x − 2x + x3 − 4x … ln[(1 + x )(1 − x )] = ln(1 + x ) + ln(1 − x ) Mark Total B1 Comment ACF Condone correct unsimplified ln(1 + x ) + ln(1 − x ) PI M1 ( ) {or ln − x = −4 x − (b) (− x ) 2 } PI = − x − x … 1 Expansion valid for − < x < 2 A1 x x + x = x 1 + + O( x ) 18 3x − − x x 3x − x + x 18 = ln[(1 + x )(1 − x )] − x − x B1 Correct first two terms in expn of M1 Series expansions used in both numerator and denominator B1 CSO Must be simplified Condone │x│< 9+ x lim 3x − x + x x → ln[(1 + x )(1 − x )] lim − + O( x) = x → − + O( x ) = 24 Dividing numerator and denominator by x2 to get constant term in each, leading to a finite limit Must be at least a total of ‘terms’ divided by x2 m1 A1 Total = 1 NOT → 24 24 of MARK SCHEME – A-LEVEL MATHEMATICS – MFP3- JUNE 2015 Q4 (a) Solution The interval of integration is infinite (b) ∫ (x − 2) e −2 x u = x−2, Mark E1 dv du = e −2 x , = , v = −0.5e −2 x dx dx (x − 2)e −2 x − − e −2 x dx 2 1 = − (x − )e −2 x − e −2 x (+c) ∞ ∫ −2 x lim a→∞ Now ∞ M1 du = , v = k e −2 x with k = ± 0.5 , ± dx − A1 (x − 2)e −2 x − − e −2 x (dx) OE 2 ∫ A1 M1 Evidence of limit ∞ having been replaced lim by a (OE) at any stage and seen or a→∞ taken at any stage with no remaining lim relating to a p e −2 a = , (p>0) E1 General statement or specific statement with p = stated explicitly Each must include the in the exponential −4 e A1 dx = lim a ( x − 2) e a→∞ ∫ −2 x dx −2 a −4 −2 a − (a − )e − e − − e lim a→∞ ∫ (x − 2) e Comment OE dx … = − ∫ ( x − 2) e Total −2 x dx = Total No errors seen in F(a) − F(2) (M1E0A1 is possible) of MARK SCHEME – A-LEVEL MATHEMATICS – MFP3- JUNE 2015 Q5 (a) Solution Mark Aux eqn m + 6m + = Total (m + 3) = M1 ( yCF =) ( Ax + B)e −3 x Try ( y PI =) a sin x + b cos x ( y ' PI =) 3a cos 3x − 3b sin 3x ( y ' ' PI =) − 9a sin 3x − 9b cos 3x A1 a sin x + b cos x or Altn k cos 3x M1 − 9a sin x − 9b cos x + 6(3a cos x − 3b sin x) (b)(i) (b)(ii) + 9(a sin x + b cos x) = 36 sin x m1 − 18b = 36 y PI = −2 cos 3x A1 A1 ( yGS =) ( Ax + B)e −3 x − cos 3x B1F 18a = f ′′(0) + 6f ′(0) + 9f (0) = 36 sin f ′′(0) + 6(0) + 9(0) = ⇒ f ′′(0) = Substitution into DE, dep on previous M and differentiations being in form p cos x + q sin x or Altn − 3k sin 3x and − 9k cos 3x Seen or used Correct y PI seen or used E1 f″′(0) = 108cos0−0−0 = 108 f(iv)(0) = 0−6 f″′(0) −0 = − 648 x3 x2 x (iv) f(x)≈0+x(0)+ (0)+ f″′(0) + f (0)… 4! 3! x3 x4 (108) + (− 648)… f(x)≈ 4! 3! = 18 x − 27 x ( yGS =) c’s CF + c’s PI, must have exactly two arbitrary constants AG Convincingly shown with no errors f″′(0) =108 and f(iv)(0) = −648 seen or used B1 f(x)≈ M1 x3 x (iv) f″′(0) + f (0) used with c’s 4! 3! non-zero values for f″′(0) and f(iv)(0) A1 Altn: Use of answer to part (a) f(x) = (6 x + )e −3 x − cos x = Comment Factorising or using quadratic formula OE on correct aux eqn PI by correct value of ‘m’ seen/used 18 x − 27 x Ignore any extra higher powers of x terms [B1] Correct series for e−3x (at least from x2 terms up to x4 terms inclusive) and cos3x (at least x2 terms and x4 terms) substituted and also product of (px+q) term with e−3x series attempted where p and q are numbers [M1] =(2−2)+(6−6)x+(9−18+9)x2+(27−9)x3+ +(6.75−27−6.75)x4 = 18 x − 27 x [A1] [3] Total If using (a) to answer (b)(i), for guidance, f″(x) = 54 xe 11 −3 x − 18e −3 x + 18 cos 3x of MARK SCHEME – A-LEVEL MATHEMATICS – MFP3- JUNE 2015 Q6 (a) Solution Mark dx dy dy = dt dt dx dy dy dy dy ⇒ 2x = = dx dt dt dx 2 d dy d y dx d dy d y = ; = x x dt dx dt dt dx dx dt d2 y dx dy d2 y + 2x = dt dx dt dx 2e 2t 4x2 d2 y dy d2 y x + = dx dt dx d2 y + x y = x (ln x ) + dx d2 y dy becomes − x + y = (ln x ) + dx dt x Total Comment M1 OE Relevant chain rule eg A1 OE eg dy dt dy = dx dx dt M1 dy − t d y = e dx dt OE.Valid 1st stage to differentiate xy′(x) oe wrt t or to differentiate x −1y′(t) oe wrt x m1 Product rule OE (dep on MM ) to obtain an eqn involving both second derivatives A1 OE eg A1 Or better d y − t − t dy − t d y = e − e + e dt dx 2 dt {Note: e−t could be replaced by } x x5 d2 y dy −2 + y = (2t ) + t t d e dt dy d2 y ⇒ − + y = 4t + 5e −t dt dt ⇒ (b) Auxl eqn m2 −2m + = (m − 1) A1 +1 = (m − 1)2 + k or using quadratic formula on M1 m=1±i CF: ( yC =) e t ( A cos t + B sin t ) A1 B1F P.Int Try ( y P =) a + bt + ct + de −t M1 AG Be convinced correct aux eqn PI by correct values of ‘m’ seen/used Ft on m = p ± qi , p, q≠0 and arb constants in CF Condone x for t here (y ′(t)=) b + 2ct − de −t ; (y ′′(t)=) 2c + de −t Substitute into DE gives 2c + de −t − 2(b + 2ct − de −t ) + M1 + 2(a + bt + ct + de −t ) = 4t + 5e −t d = 1; c = 2b − 4c = and 2c − 2b + 2a = Substitution and comparing coeffs at least once B1 A1 Need both OE PI by c’s b=2×c’s c and c’s a=c’s c provided c’s c≠0 Need both Ft on c’s CF + PI, provided PI is non-zero and CF has two arbitrary constants and RHS is fn of t only b = and a = GS (y=) e t ( A cos t + B sin t ) + + 4t + 2t + e −t [ ( ) ( )] y = x A cos ln x + B sin ln x + + 1 + ln x + (ln x ) + x Total A1 B1F y=f(x) with ACF for f(x) A1 10 17 of MARK SCHEME – A-LEVEL MATHEMATICS – MFP3- JUNE 2015 Q7 Solution Mark Total Comment (2) π (1 + cos 2θ ) (dθ ) ∫( − ) M1 Use of B1 Correct expn of [1+cos2θ ]2 and correct limits M1 cos 2θ = ±1 ± cos 4θ used with k ∫ r dθ A1F Correct integration ft wrong coefficients π (a) Area = r (dθ ) or 2∫ ∫ π r (dθ) π = ∫ 2π (1 + cos 2θ + cos 2θ ) dθ −2 = ∫ (1 + cos 2θ + 0.5 + 0.5 cos 4θ ) dθ 1 π θ + sin 2θ + 0.5θ + sin 4θ 2 −π = π = (b)(i) + sin θ = + cos 2θ + sin θ = + − sin θ (2 sin θ − 1)(sin θ + 1) (=0) sin θ = −1 gives the pole, O A1 Or r (2r − 3) (=0), each PI by correct roots E1 Or r = gives the pt O OE eg finds 2nd pair of coords (0, - π/2) and chooses (3/2, π/6) π 3 π At A, sin θ = 0.5 so A , 2 6 Eqn of line thro’ A parallel to initial line is r sin θ = At B, r = − sin θ = − 2 16r B1F PI Ft on r sin θ = rA sin θ A M1 Solving r sin θ = k and r = + cos 2θ to reach a cubic eqn in r or in sin θ 16r = 32r − 18 A1 Correct cubic eqn in r (2r − 3)(4r − 2r − 3) = A1 A1 OB = rB = + + 48 = r = 1.5 , θ = (or in sin θ eg sin θ = sin θ − ) Or (2 sin θ − 1)(4 sin θ + sin θ − 3) = Since rA = 1.5 and rB > , (b)(iii) CSO Equating rs (or equating sinθ s) followed (or preceded) by cos 2θ = ± (1 ± sin θ ) M1 (b)(ii) A1 ( 13 + 1) AB = ±( rA cosθ A − rB cosθ B ) A2,1,0 A.G Note: A2 requires correct surd for OB and also correct justifications for ignoring the other two roots of the cubic eqn Max of A1 if justification absent OE method to find AB or AB2 OB sin(θ B − θ A ) eg AB = OE single ‘eqn’ sin θ A M1 or AB = rA + OB − 2rAOB cos(θ B − θ A ) or OB = rA + AB − rA AB cosθ A cos θ B = rB = 13 + =(0.758(7 )) AB = 0.425 (to 3sf) A1 Total TOTAL (b)(ii) m1 18 75 2 OE eg solving correct quadratic eg sin θ B = or θ B = 0.709(41 ) 13 + 0.425 Condone >3sf (0.425428….) (2 sin θ − 1)(4 sin θ + sin θ − 3) = sin θ = 0.5 (pt A), eg sin θ < −1 impossible, so sin θ = − + 52 of