AS Mathematics. MPC2 Pure Core 2 Mark scheme June Version: 1.0 Final

Transcription

1 AS Mathematics MPC Pure Core Mark scheme 660 June 07 Version:.0 Final

2 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 07 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.

3 MARK SCHEME AS MATHEMATICS MPC JUNE 07 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 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) 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

4 MARK SCHEME AS MATHEMATICS MPC JUNE 07 Q Solution Mark Total Comment (a) Perimeter of sector = 888 M r + r + r used for the perimeter PI by eg 8 6 OE 0.75 A A correct value for. eg 6/8 NMS scores / (b) (Area of sector) = r = (8 ) M r seen, or used, for the sector area OE eg rl with L r.. = (cm ) A ; NMS scores / Total Condone absent or incorrect units in this question Q Solution Mark Total Comment (a) sin C sin0 M OE eg next line in soln sin dm sin C OE eg or eg C ( to nearest degree) A Must see a value for C as 8.9 or AWRT 8.90 to 8.97 inclusive before seeing 9 (b) (Method using angle B) Angle B = (.08...) B or.0 or.0 or AWRT.05 Value may be seen on the diagram M OE (Area=) 66sin B =.5 (cm ) (to sf) A CAO.5 only NMS scores 0/ Alt (Method not using angle B) 6 6 AC (6) ACcos0 AC 9 (B) 9 OE or. or. Value may be seen on the diagram (Area=) 9 6sin C (M) OE =.5 (cm ) (to sf) (A) () CAO.5 only ; NMS scores 0/ Total 6 Condone absent or incorrect units in this question (a) Verification using 9: Dep on which formula is used, M or MdM can be scored but then A0. (a)(b) If different labels are used for the angles, look for later evidence before applying any penalty; eg cand states a/sina=b/sinb; 6/sin0=6/sinB; then correct rearrangement and calculation to B=8.95 = 9; In (b) cand states Area=/ absinc, finds C= and has /x6x6xsin=.5. No penalty, (a)/ (b)/ unless contradiction eg check diagram does not have values 9 and placed incorrectly at B and C. (b) NB sin(08.95). 5 scores 0/ of

5 MARK SCHEME AS MATHEMATICS MPC JUNE 07 Q Solution Mark Total Comment (a) x 0.5x B x 7 Seen or used. eg log 7 =.5x log. x M kx x kxx kxx 7 OE or 0.5(x ) x = or p log kxlog (x )log (b) x 0.5x ) (OE Accept form p = ) A OE (= 8 B Seen or used; or 0.5x = x B p Expression for p need not be 0.5(x ) x simplified. eg NMS / p log8 or log OE must be exact and from correct work. NMS scores 0/ Total 5 (a) Consult TL if cand has changed both numerator and denominator into form eg 9 f(x) then applied index/log law. Q Solution Mark Total Comment (a) u 08 ; u 7 B; B (b) a c's u c' s u valuefrom(a) a { S =} or / r M or r c' s u valuefrom(a) AF { S =} A Correct exact value for S. NMS scores marks unless FIW. (c) k u n = S u n M OE eg u n = S S k Alt k n n k.5 AF nk Condone < replaced by either = or. OE If incorrect, ft on c s + ve value for S from (b); ineq/eq can be in an unsimplified form but only unknown is k. (Smallest value of ) k is A SC mark for i) NMS k= ii) using uk un r n k k n (M) nk u n = S S k to get k= k (A) OE Condone < replaced by either = or 86.5 (Smallest value of) k is (A) () Total 8 (a)(b) 6 6 Eg u 08 ; B u 7 B; (b) (nothing yet) = M A0F = 86 A0 r / (c) u n = S S n is M0, we need to see n replaced by k before M can be awarded. 5 of

6 MARK SCHEME AS MATHEMATICS MPC JUNE 07 (c) u k= with sufficient evidence eg can score / r / Q5 Solution Mark Total Comment (a) For st pt, x x 0 M x x 0 ( x x ) (Since 0 x ) x x A x= as the only value of x. [Give BOD if x x 0 appears after x x 0 in working] (b) (c) d y x dx d When x, 0 dx a minimum point y so curve has.5.5 x x x xdx = ( c).5 (y = ) 5 M A.5 x x (+ c) A Differentiating one term correctly. ACF A AG Must be using hence. Be convinced. eg shows that the value of the second derivative is at x=, states >0 (or states d y 0) so min. dx M dy Attempt to integrate with at least one d x of the two terms integrated correctly..5 x x OE ; condone unsimplified 5 When x =, y =.5 5 c dm Subst. x = or c s positive x value from part (a), and y = into y = F(x)+ c in an attempt to find the constant of integration.5 6 y x x 5 5 A ACF of the equation with signs and coefficients simplified Total 9 6 of

7 MARK SCHEME AS MATHEMATICS MPC JUNE 07 7 of

8 MARK SCHEME AS MATHEMATICS MPC JUNE 07 Q6 Solution Mark Total Comment (a)(i) h = 0.5 B h = 0.5 OE stated or used. (PI by x-values 0, 0.5, 0.5, 0.75, provided no contradiction) x f(x) = I M h/{f(0)+f()+[f(0.5)+f(0.5)+f(0.75)]} h OE summing of areas of the trapezia.. {f(0)+f()+[f(0.5)+f(0.5)+f(0.75)]} ( M0 if using an incorrect f(x). ) h with{ }= 8 = = = (I ) (=.7 ) =. (to dp) 9 A OE Accept sf rounded or truncated or better evidence for surds. Can be implied by later correct work provided > term or a single term which rounds to. A CAO Must be. SC 5 strips used: Max B0MA0;. A (a)(ii) Increase the number of ordinates E OE eg increase the number of strips (a)(iii) (Area=) k 0 x d x M PI by eg the next line = 8c s answer to (a)(i) dm Do not award if c s (a)(i) is 8 =.56 A CAO Must be.56 SC mark for final answer.56 coming from (b)(i) (Translation) 0 B,,0 B for 0. If not B award B for c, where c 0 OE. 0 (b)(ii) (Stretch) scale factor in y-direction B,,0 OE If not B award B for either correct sf or correct direction of stretch (c) log log 7 x M OE eg x log 7 x =.69. =.7 (to sf) A.7 Condone >sf ie.69(8.) rounded or truncated If logs not used explicitly then 0/. Total (a)(i) For guidance, separate trapezia, 0.5( ) (7 ) (..) +.59(6 ) (a)(i); (c) If relevant brackets are missing, look at later work for further evidence of recovery. (a)(ii) Eg Use more decimal places E0 (b)(i) Must be given as a vector. 8 of

9 MARK SCHEME AS MATHEMATICS MPC JUNE 07 Q7 Solution Mark Total Comment (a) (Area) = 7x 6 Area expressed as a correct definite (dx) B integral. PI by fully correct integration and x correct use of correct limits 7x 6 (dx) = 7 x M Correct integration of two of the terms 6x x A Correct integration of all terms, can be x left unsimplified. 8 7 M F() F(), where F(x) is not the (Area) = ( ) 6 integrand A 5 AG Be convinced = (b) Gradient of the line y 8x is B OE. PI by later work eg grad tang=/ Condone any errors in the rearrangement of constant term d y B 7 x dx At Q, dy 7 x M c s expression = negative reciprocal dx of c s numerical gradient of given line OE 7 x, x 8 A Correct exact x -value y A Correct exact y -value ACF with signs simplified Normal at Q: y x A 6 eg y 8x 0 Total 9 of

10 MARK SCHEME AS MATHEMATICS MPC JUNE 07 Q8 Solution Mark Total Comment (a) 8, B B (b)(i) sin tan sin sin cos cos cos cos 0 cos cos( cos ) cos cos cos cos cos 0 (b)(ii) cos cos M 8 Condone 8., 8. CAO Ignore values outside the given interval. If more than values in given interval deduct mark for each extra (to min of 0) sin tan used cos sin by cos to either correctly express tan sin in terms of cos or to obtain cos cos( cos dm Replacing ) A AG Be convinced. (= 0) B Since cos, cos so is the only value for cos. (c) (cosx 0) cosx M E Correct factorisation or cos, Valid explanation that would eliminate one of the c s values, with only one value or an indication of which value is rejected. cosx. Ft on c s value in (b)(ii) provided cos. PI eg by finding solns for cos and clear attempt to divide values by x 8,, 08, 67 A x equal to or rounding to OE to the four integer values 8,, 08, 67 seen (x =), 78, 0, 68 B,,0 If not B award B if either correct or AWRT three of these values. If more than four values in given interval, deduct mark for each extra, to a min of B0. Ignore values outside 0 x 80. NMS Max /. Total Condone missing degree symbols NB Prem approx for / in (a) may lead to solns 9 and (B0 B0); In (c), if the 9 is used for the same values for x should be obtained and we will award a possible max of MA0B. (b)(i) Condone three non-zero terms written in a different order eg cos cos 0. (b)(ii) If using a letter for cos, the letter should be defined eg let y cos, y y (=0) scores B (b)(ii) Explanation must be based on cos x, then to justify the elimination of one invalid ft value from cand s 0 of

11 MARK SCHEME AS MATHEMATICS MPC JUNE 07 two values. Condone between - and. Examples Math error, impossible, can t be negative E0 Q9 Solution Mark Total Comment log c c log k M log c log c c log c k c log log c k M B A Either log A log B log or B log B log B used with correct A and B; if cand is using their expansion for c in place of c, ignore any errors in the expansion in awarding this M mark log stated or used at any stage. This also includes the step p log fc, k p fc, k. c c k c 6c c 8 c k (*) see below B,,0 c 6c c 8 c seen or used at any stage; B if of the terms are correct. May have to check correct collecting of like terms at a later stage in soln. [See below for altn for these two B marks] 6c c 8 k A OE Correct equation with no logs and no c term. 6( c c ) k k k ( c ) = 6 A 7 ACF for the expression in k. Total 7 Altn: for the two B marks using the difference of two cubes ie X Y X Y X XY Y c c c c{ c c c c } B (PI by next line) = c c c c c OE B log f c, k log, crossing out both log to get f c, k we will condone log f c, k f c, k log f c, k log,, to get f c, k will result in FIW A0 A0 log (*) (*) of