A-LEVEL Mathematics MPC4 Pure Core 4 Mark scheme 660 June 06 Version:.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 06 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 MPC4 JUNE 06 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 explanation 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 x EE deduct x 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
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q Solution Mark Total Comment (a) 9x = A( 4x) + B( + x) Correct equation and attempt to find A or B A = 5 A e.g. Using x = or or simultaneous 4 equation such as 9 = 4A + B and = A + B B = 9 A NMS or cover up rule scores SC for A = 5 or B = 9 or SC for both A = 5 and B = 9 (b)(i) ( + x) = x + kx provided k 0 = x + 4x A Accept +(x) ( 4x) = ( 4 x) B ACF for eg or 0. or 0. ( 4 x) = + ( ) ( 4 ( )( ) x) + ( 4! x) = ( + 4 6 x + 9 x ) 9x ( + x)( 4x) = A Condone poor use of or missing brackets. PI by later work 5 ( x + 4x ) + 9. ( + 4 6 x + 9 x ) PI by correct answer ft on candidate s A and B with their relevant series. = + 7x x A 7 Must have, 7 or 7. Alt. ( 4x) = + ( ) ( 4x) + ( )( ) ( 4x) =! If A and B are correct full expansions are 5 + 5x 0x and + x + 8 x + 4 9 x + 6 7 x A Alt. for combined expansions without using PF s. 9x = (9x )( + (+x)( 4x) x) ( 4x) = (9x )( x + 4x ) ( + 4 6 x + 9 7 x ) Attempt to multiply any two of their three brackets together as far as the term in x then A Condone unsimplified fractions in the binomial expansion(s), but final answer must be fully simplified. A and B could be included in the series e.g. 9 = 9 ( 4x) 6 8x = 9 6 ( 8x 6 ) could still score B A. (b)(ii) < x < only B Strictly < : Accept x < Do NOT accept < x < or x < or inclusion of the equality sign. If < x < is also seen, it must be clear that < x < 4 4 Total is the answer. 4 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q Solution Mark Total Comment cosθ = cos θ used B PI: Correct expression in terms of cosθ used. (cos θ ) 5cosθ + (= 0) Attempt to use identity for cosθ of the form acos θ + b to obtain a quadratic in cosθ. 6cos θ 5cosθ = 0 (cosθ )(6cosθ + ) = 0 m Attempt to factorise their quadratic or correct use of quadratic formula. (cosθ = ) cosθ = 6 θ = 99. 6 0, 60. 4 0 A A Either correct CAO Both correct and no extra values in the interval but ignore any values outside of the interval including 0 0 and 60 0. Total 5 To earn the m mark, candidate s factors must give their 6cos θ and their i.e. the first and last terms of their quadratic. If the quadratic formula is used it must be used correctly for their quadratic. If they get the correct quadratic and NMS, both correct answers for cosθ implies the m mark, or If they get the correct quadratic and NMS, one correct answer for θ implies the m mark. If they get a wrong quadratic, they must show the working to (possibly) score the m mark. Interval is specified to be 0 0 < θ < 60 0 ; hence the reason for ignoring solutions 0 0 and 60 0 that might come fromcosθ =. Degree signs are not required; 99.6 and 60.4 are sufficient for the A marks. Allow SC if both rounded correctly to greater accuracy 99.5940 and 60.4059 if A0 A0. 5 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q Solution Mark Total Comment (a) + x 6x x = Ax + B + C x + x 6x = Ax(x ) + B(x ) + C Correct above equation and attempt to find one of A, B or C or an attempt at long division A = B = C = 9 A A A 4 e.g. using x = in an attempt to find C or forming simultaneous equations and attempt to solve. If long division is used award once x + has been obtained but only award the A marks once the values are clearly identified or it is written in the required form of Ax + B + C Alternative method of division +x 6x x = x(x ) 9x+x+ x = x + 4x+ x x. for x +. For the A marks, A, B and C must be clearly identified or seen in the required form of Ax + B + NMS scores B for one correct value, B for correct values and B4 for all three correct values C. x (b) +x 6x x dx= x + + 9 x dx = px + qx + r ln (x ) = x + x + 9 ln (x ) Aft A x + Bx + C ln (x ) Correct use of F(6) F() = [. 6 +.6 + 9 ln ( )] [. +. + 9 ln (6 )] m Correct substitution of limits for their p, q and r. = 69 + 9 ln A 4 OE The Aft and m can be earned even if left in terms of A, B and C or if invented value(s) for A, B and C are used. Condone missing brackets from the ln (x ) term for the mark but only award the Aft mark if they have clearly recovered; PI by sight of ln9 or ln after using the limits or a correct final answer. Treat a decimal answer (should be 9.55 ) after a correct exact form as ISW but award A0 if an exact answer is not seen. Total 8 6 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q4 Solution Mark Total Comment (a)(i) m = m 0 k t Using m =, m 0 = 4 and t = 8 k 8 = 8 or k = ( 0.5) = 0.97004 B = 4k 8 OE e.g. k = ( ) 8 Must see a correct exact expression for k or k 8 or k=0.9700404(.) to at least 8 d. p. AG be convinced Note that AG so to earn the mark they must show us a correct exact expression for k or k 8. Accept such as k = e (ln0.5 8 ) or e 0.08664 or ( ) 8 or 0.9700404( ) as sufficient evidence but withhold the 8 mark if a clear error has been made e.g. k = 0.5. Candidates who work with logs must reach an expression such as log k = log log 4 8 first. (a)(ii) = m 0 (0.97004) 60 m 0 = 8 A or m 0 = (0.97004) 60 PI by A later Must be 8 no ISW NMS scores SC for 8 only but sight of greater accuracy (8.098 ) implies if 8 not seen. (b) m = m 0 k t 8.06 = 0 k 00 00 k = 0.806 m 0 = m 0 k t OE A OE: e.g. k = e ln (0.806) 00 k t = t log k = log ( ) t = log ( ) log k A linear equation in t from k t = e.g. t = log k (0.5) = 0 A 4 Must be 0 No ISW For guidance, for first A, k = 0.9979 PI by later correct work. The first is for a correct interpretation of the information given so could equally be awarded for an expression involving logs of k such as ln 8.06 = ln0 + 00 ln k then A for a correct expression for lnk such as lnk = ln8.06 ln0 00 or, using base 0, logk = log8.06. 00 Those who use the value of k from (a) could only score M0 A0 A0. NMS scores SC4 for 0 only but sight of greater accuracy (0.006 ) implies A if 0 not seen. Total 7 7 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q5 Solution Mark Total Comment (a)(i) Use of sin B + cos B = ( 5 ) + cos B = Or use of right-angled triangle with opp = and hyp = 5 to get adj = 4 or. cos B = 5 B cos(sin ( )) = is B0 5 5 AG ; must see evidence of working (a)(ii) (sin B = sin B cos B) = 5 5 = 4 5 A Correct identity (PI) and substitution AG so line above must be seen. (b)(i) cos A = exact value sin(a B) = sin A cos B cos A sin B Use of 5 = 5 5 5 = 5 5 or (5 5) 5 0 5 5 OE seen B m A 4 cosa = seen or used (not 0.667 etc.) ft on their value of cosa 5 5 5 term becoming before final answer 5 OE seen and be convinced You must see justification between the use of the identity and the final answer to earn the m A. (b)(ii) cos (A B) = cos A cos B + sin A sin B = 5 + 5 5 A ft on their value of cos A fully correct = + 4 5 5 A Total 0 OE for and 4 5+4 5 but not left as 5 5 8 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q6 Solution Mark Total Comment NO MISREADS ARE ALLOWED IN THIS QUESTION (a) 4 4 AB = [ ] B or BA = [ ] 0 0 AB. [ 5] = (4 ) + ( 5) + ( 0 ) Bft Correctly finding scalar product using their AB and direction vector of l Accept + 60 0 or 5 5= 4 + + 0 + 5 + cosθ Correct use of their a. b = a b cosθ 5 cosθ = 560 5 = 5 A A 5 CAO OE all correct in this form or better. The B mark for AB or BA or any multiple could be PI by its use in the scalar product e.g. [ ] etc. 5 The Bft mark is for the scalar product of their AB with the direction vector of l. The mark is for a clear attempt at the scalar product definition of a.b = a b cos θ in any form using their AB with the direction vector of l. As in the MS, there is no need for minus signs in squared terms. Provided they earn the mark with the correct values included, it is possible to score both A marks for cos θ = without the intermediate form being seen. 5 0 4 (b) Line AB: (r =) [ 6] + μ [ ] 9 0 + λ = 0 + 4 μ 5 5 λ = 6 μ + λ = 9 0μ λ = μ = Verifying that all three equations are satisfied and conclusion e.g. intersect. AA E Set up two correct equations for their l but with correct l and attempt to eliminate λ or μ. Accept these three equations in column vector form Clear checking of λ and μ in unused equation or showing P lies on both lines. Dependent on correct co-ordinates of P. P(, 0, ) A 5 Accept as a column vector Look out for any alternative correct versions for the vector equation of l, e.g. If B(4, 6, ) is used as known point in l this leads to λ = and μ = but also look out for 4 multiples of [ ] being used as the direction vector. 0 9 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q6 Solution Mark Total Comment (c) Using points C and D CD = t [ 5] B Since C and D lie on l AB = CD 560 = 5t t = (±) 4 B or 560 = t 5 4 seen ( OC/D ) = OP + t [ 5] Uses P the mid-point of AB and CD in this method. = [ 0] + [ 5] or [ 0] [ 5] Use of their OP and t [ 5] (4, 0, ) A Accept as a column vector. (0, 0, 5 ) A 5 Accept as a column vector. Candidate could just get t = 4 0 giving point(4,, 0 ) and then find (0,, 5 ) by symmetry. (c) R is any point on l + λ (OR ) = [ 5 5λ] + λ + λ + λ (PR ) = [ 5 5λ] [ 0] (B) [ 5 5λ] but award at OR OP stage + λ + λ If R is at C or D then AB = CD = 6 PR 560=6 ( + λ) + (5 5λ) + ( + λ) 560 = 6(5λ 70λ + 5) (B) or better 9λ 8λ + 5 = 0 (λ )(λ 5) = 0 () Factorising (their 9λ and +5 terms) or attempting to solve their quadratic equation correctly PI by correct values of λ. λ = or 5 (0, 0, 5 ) (A) Accept as a column vector. (4, 0, ) (A) (5) Accept as a column vector. Total 5 0 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q7 Solution Mark Total Comment (a) ( dx e 6t ) = ( 6) B ACF : e 6t dt 4 ( dy dt ) = pet. t + qe t (t) From quotient rule = et. t e t. (t) A ACF : dy dx = (9.. e e )/4 e = e4 m A 5 dy found using product rule : dy = dt dt pet. t + qe t. t For guidance, when t =, dx dt = e and dy dt = 4 e = e t t et t A (= et 9tet et 9t, e t t et t etc. Using dy = dy / dx and clear evidence of dx dt dt an attempt to substitute t = (must be this) done in either order. CAO : Accept this or e4 or 0.5e4 only t et t ) (b) x = 4 e 6t 4 4x = 4 e 6t leading to e 6t = 4 4x e Any correct expression for e 6t or e 6t. e e 6t = 4( x) e t e = x A AG Must see inversion step and be convinced they haven t worked backwards Alternative for the mark is a correct expression for t or e t or e t OE. 4x = 4 e 6t 6t = ln (4 4x) t = ln (4 4x) or 4x = 4 e 6t e 6t = 4 4x e t = 4 4x OE e.g. further correct working needed before printed answer for A. e e t = x (OE) for, then (c) From (b) e t = e x ln(e t e ) = ln ( x ) t = ln e ln ( x) Find t (or t) in terms of x; must have used laws of logs correctly on both sides (possibly lne = at this stage) e y = A From y = et ; must be in this form x [ ln x ] t Alt. From x, t = ( ln (4 4x)) or an expression for t then A for printed answer. 6 Total 9 of
MARK SCHEME A-LEVEL MATHEMATICS MPC4 JUNE 06 Q8 Solution Mark Total Comment (a) θ = tan ( x ) tan θ = x sec θ. dθ dx = B or sec θ = dx dθ sec θ = + tan θ = + ( x ) dθ = = dx (+ 9x 4 ) Use of correct identity to get sec θ in terms of x ; condone missing bracket. 6 4+9x A Correct algebra with k = 6. k 4+9x is given so be convinced (b) If implicit differentiation not used then 0/. Embedded answer of k = 6 is sufficient. The and A marks are available even if the candidate has dy as their derivative. dx 9y(4 + 9x ) dy = cosec y dx 9y siny dy = 4+9x dx B Correct separation including dy and dx at ends of integrals (do not penalise unless the dy or dx is directly under the fraction bar), the integral signs at the front and the integrands. The 9 can be on RHS as 9. LHS : parts with u = (9)y and dv = sin y for p y cos y + q cos y dy (9) ( y cos y + cos y dy) A ycosy + siny A If 9 used on the LHS RHS: 6 tan ( x ) Bft ft on k from (a) i.e. k tan ( x ). ycosy + siny = 6 tan ( x ) + c Using x = 0 and y = π to find c sin π. π. cos π = 0 + c gives c = π Must have an expression of form psiny + qycosy = rtan ( x ) + c and use x=0 and y = π to find c. PI by a correct ft value for c. ycosy + siny = 6 tan ( x ) + π A 7 OE but must be a complete, correct expression Correct separation must be seen on a single line but accept 9y cosecy dy on LHS. With the 9 on the RHS, the indefinite integral line should read 9 sin y y cos y = 54 tan ( x ) + k. Total 0 of