Version 1.0. General Certificate of Education (A-level) June 2012 MPC2. Mathematics. (Specification 6360) Pure Core 2. Mark Scheme

Transcription

1 Version.0 General Certificate of Education (A-level) June 0 Mathematics MPC (Specification 660) Pure Core Mark Scheme

2 Mark schemes are prepared by the Principal Eaminer 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 eaminers participate in and is the scheme which was used by them in this eamination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every eaminer understands and applies it in the same correct way. As preparation for standardisation each eaminer 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, eaminers encounter unusual answers which have not been raised they are required to refer these to the Principal Eaminer. It must be stressed that a mark scheme is a working document, in many cases further developed and epanded 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 eamination paper. Further copies of this Mark Scheme are available from: aqa.org.uk Copyright 0 AQA and its licensors. All rights reserved. Copyright 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 eception: 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. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 67) and a registered charity (registered charity number 07). Registered address: AQA, Devas Street, Manchester 5 6EX.

3 Key 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 accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for eplanation 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 EE deduct 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) 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.

4 General Certificate of Education MPC June 0 Q Solution Marks Total Comments (a) (common difference) = 9 B 9 (b) (00th term) = + (00 ) d + (00 ) d or better seen (or used with d = 9 or with d = c s answer (a)) = 9 A 9 NMS mark as B or B0 80 (c) (Sum of series) = 5 80 (80 )(9) } {or Substitution of n = 80, l = 5, a = (or c s value of a used in (b)), n a l n PI or a ( n ) d PI d = 9 (or c s answer to (a)) into = A NMS mark as B or B0 (a) (Area) = 6.5sin 6.5 Total 5 5 = 57.5 (cm ) A 6.5 sin ( ) stated or used OE eg 5 Condone AWRT NMS: 57.5 or AWRT scores B (b) cos B OE eact fraction (c) {AC =} cos () = = = 56.5 m AC = =.5 (cm) A (Alternative) {AC =} (6 sin ) + (.5 6 cos ) () = (m) AC = =.5 (cm) (A) () Total 6 RHS of cosine rule Correct order of evaluation. Do not award if evaluation leads to or would lead to RHS value being outside interval 0 to 95.5 OE with no sight of premature approimation clearly used

5 (a) = = B,,0 B for or (B fully correct unsimplified epression. seen eg or B for either OE seen or OE seen or B for OE seen) (b) 5 =.5 (+ c) BF Ft on correct integration of all non terms (at least two) in c s epression. in (a) 5 Integration of a k as (ie power correct) {= (+ c)} AF Correct integration of c s term(s) ACF (c) 5 ( ) F() F() attempted following =.5 integration. If F() incorrect, ft c s answer.5 to (b) provided integration attempted 9.95 OE eg 89/0 {= =. 0.5} =.95 A 5 0 Since Hence NMS scores 0/ Total 7

6 (a) u B CAO Must be u 8 BF If not correct, ft on c s 6 u (b) r BF Only ft on r = (c s u ) (c s u ) if r <. Answers may be in equivalent fraction form or eact decimal form. If other notation used award the mark if correct or ft value confirmed in (c) (c) ( S u ) r = 6 AF a Use of, ft on c s u and c s r in (a) r and (b) if not recovered, provided r < If not 6, ft on c s u and c s r in (a) and (b) provided r <. (d) n u n = S u n OE eg RHS S ( u u u ) n ( 0.5 ) u ( or u n = ) B Either result, or better eg u n =5.75 n 0.5 n n u = 0.5 A NMS scores 0/ n (Alternative) u ( u n = ) () r n ( u (= 0.875) ) 6 (B) ( u n = ) 6 (A) () (NMS scores 0/) n SC For c s scoring 0/ in (d); Award B to candidates who used S S for n u and obtained the answer 6 OE n Total 8

7 5(a) {Arc =} r r seen or used for the arc length π = 8 π (m) A π (b)(i) π B π OE epression which simplifies to π (ii) {Area of sector =} π r = 8 r seen or used for the sector area If not eact accept sf or better = 08 π (=9.(9..)) A PI by final correct answer π TP 8 OE Correct method (PI) to find either TP or tan {or tan } 8 TP TQ (=TP) or OT or PQ or PQ. If not π {or PQ = 8sin }{or PQ = π 8sin } / then ft c s value for in (b)(i). If c finds two of TP/TQ, OT and PQ/ PQ and gets π 8 8 or cos or sin OT OT one correct, one wrong, mark correct one ie A (A0 possible if no correct length) TP=8 =.769 eact or. to. incl} Correct TP or TQ or PQ or PQ or OT {or PQ=8 = eact or. to. incl} A either eact value or in range indicated {or OT = 6}; PI by value 56 to 56. inclusive for the area of the kite. PQ = 9 or 5.5 to 5.6 incl} { Area of kite PTQO = 8 TP π {or Area = (8 ) sin sin TP } {or area kite = 8 cos π PQ } {or area kite = 8sin π OT } {=8 }{= 6 };{ + 8 } Alternative OE valid method to find area of kite, down to a correct epression with no more than unknown length; ft on c s value of. For method using > one unknown length this M is dependent on previous M for length PI by value or a numerical epression which simplifies to ; or a value 56 to 56. inclusive for the area of the kite. Can also be implied by award of the final A Area triangle PTQ = OE Alternative: Award this method mark TP sin and if both area of triangle PTQ (= ) () Area triangle POQ = and area of triangle POQ (=8 ) are 8 sin(/) found with or without finding area of kite Area of shaded region = 56.(8...) 08 π = If not, condone value from.7 to.89 = (m ) to sf A 6.0 inclusive Alternative Area of shaded region = (08 8 )=.89 = (m ) to sf (A) (6) Total 9

8 6(a)(i) (When = ) d y = = 0 B AG Must see intermediate evaluations (ii) d y = 6 8 When =, d y {so } B d A y = + 8/8 = A, seen in (a)(ii) or earlier. PI by ± 8 term in answer Correct powers of correctly obtained from differentiating the first two terms 6 8 ACF (iii) Since Ft on c s value of y () in (a)(ii) but must d y see reference to sign of y () either > 0, P is a minimum point. EF eplicitly or as inequality, as well as the correct ft conclusion (b) dy ( c) Attempt to integrate with at least two d of the three terms integrated correctly (y =) (+ c) A For OE even unsimplified When =, y = = c Substituting. =, y = into y = F() + c in attempt to find constant of integration, where F() follows attempted dy integration of epression for y A ACF Total 0

9 7(a) tan B sin cos sin tan used on sin cos sin tan cos cos or forms and solves a correct quadratic in sin or cos and then uses to find tan tan or tan 0 tan = A or tan tan 0 or tan or tan tan A Both (b) tan, tan, tan ( =) 5º, ( =) 60º, ( =) 0º A,,0 Uses part (a), at least as far as attempting to solve tan = k, where k is any one of c s values for tan If not A for all three correct, award A for two values correct Special Case If tan in part (a) and scored in (a) and in (b) then apply ft in part (b) ie AF for = 5º, 0º, 50º. (AF if two of these ft values) Special Case: If M0 then award B for any two correct values provided no incorrect etras in given interval. If > answers in the given interval, deduct mark for each etra in the given interval from any A marks awarded in (b). Ignore any answers outside 0 80 Total 7

10 8(a) B Correct shape, curve in st two quadrants only, crossing positive y-ais once and asymptotic to negative -ais. (0, ) O B Coordinates (0, ). Accept y-intercept indicated as on diagram or stated as intercept = B0 if graph clearly drawn crossing aes at more than one point (b)(i) y y OE; 7 OE Eliminates either or y correctly 7 (y ) (y+) (= 0) ; (7 )(7 ) (= 0) A Since y (=7 ) > 0, [ y (=7 ) ] (there is eactly one point of intersection) y-coordinate is B E Correct factors or y = 7 = 9 or better 9 or better or Clear indication that c s negative solution(s) has/have been considered and rejected (ii) 7 so log 7 = log [or log 7 ] = 0.7( ) = 0.7 to SF A OE ft on 7 k, where k is positive, to either log 7 = log k or log 7 k Condone > three significant figures. If use of logarithms not eplicitly seen then score 0/ Total 8

11 9(a) h = 0.5 B PI f() = log0 ( ) I h/{ } {.} = f (0) + f () + [f (0.5) + f (0.5) + f (0.75) ] {.} = OE summing of areas of the trapezia log log log log log 6 6 A OE Accept sf evidence = ( ) = ( ) = (I ) 0.5 [0.957 ] = 0.7 (to SF) A CAO Must be 0.7 (b) 0 B (c)(i) (ii) Condone missing bases for M mark. log0 (0 ) log0 0 log0 Accept log replaced by log in line AG. Bases must be included or statement = log 0 A log 0 0 = given. Condone missing bases in (c)(ii) & (c)(iii) y log0 = log0 (0 ) PI Either y = log0 ( 0 ) (to compare y log ) A or both y log and y log ( 0 0 (Stretch) parallel to -ais, sf 0 ) 0 OE B,,0 Writing in correct form so that stretch details can be stated directly B for correct direction and scale factor ACF (B for correct eact scale factor ACF) (or B for -direction, scale factor /0 ) (or Bfor -direction, scale factor 0 ) Apply ISW if dec follows eact values. (OE scale factor must be in eact form) (iii) log0 (0 ) = log0 ( ) PI by 0 = + or correct (0 = +, 9 = A OE stated or used; accept, and since > 0) (y-coordinate of P) y log A PI by log OE for the gradient of OP 9 Or y log 9 Gradient of OP = A log ; Accept a=000, b=79 log0 = log Total 5 TOTAL 75