Version 1.0. General Certificate of Education (A-level) January Mathematics MPC4. (Specification 6360) Pure Core 4. Final.

Transcription

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

2 Mark schemes are prepared by the Principal Eaminer and considered, together with the relevant questions, by a panel of subject teachers. his 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. he 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 alrea 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. he 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 AWR 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 MPC: January 0 - Mark scheme Q Solution Marks otal Comments (a) A( ) B( ) Use two values of to find A m and B A B Both (b) (c) 7 6 C D 6 ln( ) ln( ) ln ln5 ln ln ln ln 5 ln 7 (a) Condone poor algebra for if continues correctly. ft ft m 5 otal (b) Complete division for ; obtain a value for C (C) and a remainder a b Complete division leading to values for C and D C D stated or written in epression. SC B C, D not found or wrong; D, C not found or wrong. Use parts (a) and (b) to obtain integrable form ft on C Both correct; ft on A, B and D Condone missing brackets Correct substitution of limits 9 5 p q 7 P Q (c) Form C using candidate s P, Q, C for. Condone missing. C C for ft ISW etra terms eg for first three terms only; ma /5 Candidate s C; must have a value. 6 6 is an integrable form, as a ln is in the formula book, a a a but they must try to integrate to show they know this, or use partial fractions again with 6 for Substitute limits into C mln( ) nln( ), or equivalent, for m; substitution must be completely correct. Condone 9 ln 7 for 5

5 (a) ; equating coefficients A( ) B( ) term A B constant A B A B m Set up simultaneous equations and solve. Both ; cover up rule A (b) B A B and used to find A and B SC NMS A and B both correct / One of A or B correct / C D 7 6 C C D D C D C D stated or written in epression SC B C, D not found or wrong; D, C not found or wrong. Complete method for C and D C =, D = stated or written in epression. SC B C, D not found or wrong; D, C not found or wrong D C D C D Use two values of to set up simultaneous equations C D stated or written in epression. SC B C, D not found or wrong; D, C not found or wrong.

6 (a)(i) tan B Fraction required Allow.. (recurring) (ii) (b),, seen (from Pythagoras) or cot tan tan( ) Remove fractions within fractions = (b) tan( ) m otal 6 Use cos ec cot SC B or Use tan formula Correct manipulation to form a b a b c d integers c d m n or any multiple sin 5 5 cos 5 5 Remove fractions within fractions m Use formulae for sin cos Correct manipulation to form a b a b c d integers c d m n or any multiple and (a)(ii) Special case B for or (b) for substituting candidates values for tan and tan into correct formula. Completely correct or completely correct ft on tan, tan. Special case answer is a or where a is integer or for ma0 9 a

7 (a) 6 6 k Simplified coefficients required (b) B OE replaced by 8 in answer to (a) Condone missing brackets, allow one error. Simplified coefficients required. (c) ( ) 67 6 Use in binomial epansion from part (b) (b) otal 7 OE Condone missing brackets, allow one error. (a)(b) Condone for for Use binomial formula; condone one error and missing brackets.

8 (a) e 8 P Must use t = 60 Nearest thousand required only (b)(i) e 8 t 500 OE ake logs correctly leading to t 8ln 000 epression for t. t 7.6 (minutes) Accept 7.6 (ii) e e e e e 500 e 90e e 00 (e 0 rejected) t 6.8 (minutes) Set up equation; condone one error; allow in t. Condone inequality. 8 e Multiply by and rearrange 500 to AG, be convinced. Solve quadratic equation (retaining positive root). CAO (b)(i) t t t e 000e e 500 t ln000 t 7.6 (minutes) t t t t e 000e ln e ln000 ln e t ln000 t 7.6 (minutes) otal 9 ake logs correctly leading to epression for t. ake logs correctly. t 8 (b)(ii) for solve quadratic equation Let e solve quadratic equation by inspection, 00 seen; 00 0 with 00 and 0 seen; factors complete square 5 05 all correct formula Final answer ; must have t 6.8 for all correct (b)(i) 7.6 as final answer NMS / 7.6 following wrong working AO (FIW) but could still score M mark(s )

9 5(a) y y 8t t t t Substitute and epand t t k 7 (b)(i) 6t dt dt t BB t 6 6 d y Use chain rule 6t t and calculate gradient using t (ii) 8 t y 6 y tangent y 6 6 y Calculate and y using t = Both correct ACF CSO y 6 ISW Substitute into candidate s tangent; calculate y y 8 used to verify 7 5(a) 8 y y y y 7 otal Eliminate t k 7 (b)(i) y y 0 B Product rule attempted; two terms added, one with d y 8 t y 6 y d y y 6 y m Calculate and y using Both correct. Calculate gradient from candidate s epression. t tangent y ACF CSO y 6 ISW

10 Q Solution Marks otal 5(b)(i) 7 y Correct epression for from y candidate s implicit equation. Quotient rule attempted; y and y 7 y y two terms subtracted. y Numerator; first term; y second term y t y t 6 y 6 6 for d 7 y y y y y y B m B Use t to calculate y Evaluate and invert. Use t to calculate ACF CSO Correct epression for from candidate s implicit equation and attempt derivatives

11 6(a) Evaluate f ; not long 6 5 division factor Processing and conclusion. (b) 7 cos cos 5cos 7cos 8 cos cos 9sin sin cos B B Use acf of cos formula Use acf of sin formula All in cosines. (c) 6cos cos 5 0 cos b ac 5 7 b ac 0, no solution to 5 0 only solution is cos m otal 0 Simplification and substitute cos to obtain AG CSO. Factorise f Find discriminant of quadratic factor; or seen in formula Conclusion; CSO Condone is only solution 7 (a) For ; minimum processing seen; ; and no other working is A0 6 minimum conclusion 0 hence factor (b) For mark; cos (eventually) in form sin cos ccos cos to obtain acos b ; 9sin sin in form ccos sin and use (c) k 5 fully correct m candidate s values of a, b, c used in epression for b ac or complete square to obtain b b c a a a b ac correct or and stated to be negative so no solution or solutions are not real (imaginary) Accept imaginary solutions from calculator if stated to be imaginary. Condone 7 is negative, or similar, so no solution. Conclusion is solution, or cos is solution

12 7 sin y B Correct separation and notation; condone missing integral signs B y y sin cos cos d Use parts u sin du dv v k cos with correct substitution into formula cos sin 9 cos sin C y 9 cos sin C C 9 cos sin 0 y 9 9 y cos sin 0 m 9 otal 9 CAO Use y to find C 6 CAO And invert to y 9... CSO, condone first B not given Second finding C; substitute Must calculate a value of C. π y into f y pcos qsin C and evaluate using radians. 6 k where y 9 m for reaching form P cos Q sin R y 9 and inverting to k P cos Q sin R P and Q are or or

13 8 (a)(i) OB OA implied by two AB 0 correct components Allow as,, (ii) 5 AB ft ft on AB cos 6 0 Correct formula for cos with consistent vectors and correct 5 moduli, in form a b c CSO Accept 5., 5.0 (b) p AB BC 5p 0 p p BC 5 p p p 0 p 6 8p 0 6 p p B m SC B 90 following sp = 0 Set up scalar product. p at C. Any letter for p. Clear attempt to find BC in terms of p. BC or CB correct Epand scalar product and solve for p; ( 0 possibly implied) OD OA BC m Correct vector epression to find OD written in components 5 7 D is at,, 6 CAO; condone column vector for last marks OD OC BA 5 otal m 5 7 D is at,, Part (b) NB p can come from wrong working where candidate uses OC his is M0 and scores no further marks, (unless they happen to find and go on to use it correctly). in place of BC.