FSMQ. Additional FSMQ. Mark Scheme for June Free Standing Mathematics Qualification. 6993: Additional Mathematics

Transcription

1 FSMQ Additional FSMQ Free Standing Mathematics Qualification 699: Additional Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

2 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 00 Annesley NOTTINGHAM NG 0DL Telephone: Facsimile: publications@ocr.org.uk

3 699 Mark Scheme June 0. Subject-specific Marking Instructions. M (method) marks are not lost for purely numerical errors. A (accuracy) marks depend on preceding M (method) marks. Therefore M0 cannot be awarded. B (independent) marks are independent of M (method) marks and are awarded for a correct final answer or a correct intermediate stage.. Subject to, two situations may be indicated on the mark scheme conditioning the award of A marks or B marks: i. Correct answer correctly obtained (no symbol) ii. Follows correctly from a previous answer whether correct or not (FT on mark scheme and on the annotations tool).. Always mark the greatest number of significant figures seen, even if this is then rounded or truncated in the answer.. Where there is clear evidence of a misread, a penalty of mark is generally appropriate. This may be achieved by awarding M marks but not an A mark, or awarding one mark less than the maximum.. For methods not provided for in the mark scheme give as far as possible equivalent marks for equivalent work. If in doubt, consult your team leader. 6. Where a follow through (FT) mark is indicated on the mark scheme for a particular part question, you must ensure that you refer back to the answer of the previous part question if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.. Abbreviations The following abbreviations are commonly found in Mathematics mark schemes. Where you see oe in the mark scheme it means or equivalent. Where you see cao in the mark scheme it means correct answer only. Where you see soi in the mark scheme it means seen or implied. Where you see www in the mark scheme it means without wrong working. Where you see rot in the mark scheme it means rounded or truncated.

4 699 Mark Scheme June 0 Where you see seen in the mark scheme it means that you should award the mark if that number/expression is seen anywhere in the answer space, even if it is not in the method leading to the final answer. Where you see figs 7, for example, this means any answer with only these digits. You should ignore leading or trailing zeros and any decimal point e.g. 7000, 7, 70, would be acceptable but 070 or 7 would not..

5 699 Mark Scheme June 0 Section A For (, ) use x + y = 9 Substitute or use Pythagoras soi or 9 so inside dy x x dx At x gradient y 9 "their" 9( x) y 9x8 oe i eg cosp.8.7 oe Conclusion (dependent on awarded) Differentiate All three terms correct 9 isw (dep on first ) Find line using correct point and their 9 Cosine formula correctly used to find any angle Anything that rounds to 7., 8. or 8. As usual only award A marks if the M mark has been awarded. Alternative method: Sub of x = or y = in x + y = 0 to find y or x y = or x = 6 At least one power decreased by. their 9 means: the value of the derivative Only terms P 7. For identifying correct angle Anything that rounds to 7.

6 699 Mark Scheme June 0 ii Area 7 8 sin( their angle P) Use of formula Can be at any vertex Correct substitution from their (i) 6.8 Anything that rounds to 6.8 sinx cosx tanx 0. x.8, 0.8 x 0.9, 00.9 Also x 90.9, 80.9 a - 9 Mis, whichis, b 9 Gradient of AC is Gradient of BM is oe c i Isosceles Use of tan Any nd value rd and th values (ignore extra solutions) One gradient Second gradient Accept complete alternative methods allow tanx = 0. for Alternative method Use of Pythagoras to get sin x or cosx 9 9 and the last three marks are still available, ignore extra solutions Their m m Eg One is the negative reciprocal of the other Allow right-angled isosceles Accept wrong spelling Do not accept right-angled triangle

7 699 Mark Scheme June 0 c ii AB 7 0 Using Pythagoras on AB and BC Attempt by vectors AB and BC BC x x 7 two sides equal in length Boundaries x, x 7 Or fully labelled diagram with correct sides shown Or use of correct formula (allow one error in substitution) or correct shaped graph seen soi Alternative If answer to (c)(i) was right-angled, then accept proof that it is (requires all three lengths.) Alternative: If (c)(i) was equilateral or scalene then (only) for attempt at all three sides. NB If nothing is written in (i) then no credit in this part. x 7 7 a i Attempt to find f() by substitution of = 0, So Yes 7 a ii f( ) = = so no. 7 b i f(x) = (x )(x +x ) = (x )(x + )(x ) B Accept x, x 7 for, Condone < or > Remainder theorem or attempt to divide (justification is sight of x x ) Or: attempt to factorise, justification is sight of (x ) Correct working only Sight of or correct evidence, conclusion required Attempt to factorise or use long division (justifications as in (a)(i)) Sight of correct quadratic soi Answer Alternative: Use of Remainder theorem Sight of nd factor All correct 7 b ii x =,, FT their brackets Must be three roots

8 699 Mark Scheme June 0 8 i for one line for correct shading for other line for correct shading for correct shading to give x 0, y 0 If there is work here that is not crossed out, then mark it and ignore anything on Page 8. Helpful hint: Lines go through (0, ) and (,0) (0, 0) and (6, 0) Intersection at (., 7.) NB If intercepts are within small square of the correct points then give the marks for the lines If B0 for a line allow for shading if negative gradient and lines intersect 8 ii 6x + y is minimum at (0, ) (can be implied by correct answer) So is 9 dy dx x x ( y) x x x c (,) satisfies 8 8 c c 0 0 y x x x 0 sinθ sin θ 0.6 cos θ cos θ cosθ oe Integrate Ignore c (dep on st mark) Substitute cao Use of Pythagoras cos eg 0.8 or isw At least one term with power increased by. (NB do not accept multiplying throughout by x) (ie must be y =.) Sight of a triangle with sides,, acceptable for Then A for cos NB M0 if calculator used to find in order to find cos 6

9 699 Mark Scheme June 0 Section B i P(0) = (0.9) 6 Correct p plus correct power = 0.7(0989 ) Not sf ii P() = 6 (0.9) (0.0) = 0.(8 ) Correct p and q and powers add to 6 Coefficient soi Correct powers for correct p and q soi Coefficient may be missing iii P( st box contains or more eggs) = (their(i) + their (ii)) = ( ) = = 0.08 Accept anything rounding to 0.0 Alternative P(accepted) Ans(ii) Ans(i) soi (Accept 0.7) P( nd box has any cracked eggs) = their (i) = 0.69 P( consignment is rejected) = their (ii) = = Accept anything rounding to 0.6 In either method, accept answers which lie between 0.09 and 0.09 (dep) Add to this Ans(i) (Accept 0.906) P(consignment is rejected) = =

10 699 Mark Scheme June 0 a s ut at with u 0 and a Constant acceleration formulae or integrate twice ignore c s t b t () v t Integrate t t s Integrate Ignore c c i t t t t t 6 Equate their functions c ii 6 6 Substitute their non-zero (c)(i) in their (a) s 6 or s or (b) soi Displacement = 6 (m) d One clearly straight line through origin with positive gradient Ignore labels Other clearly a curve through the origin of correct shape with first part below the line as per diagram 8

11 699 Mark Scheme June 0 i AO x x x Correct application of Pythagoras on the base or AC x x 8x h AO AE h x x h Algebra must be convincing ii V base area height x h Formula seen including x 0h h iii dv 0 6h Differentiation dh cao 0 when 0 6h 0 dep Set (numerator) = 0 h h.89 Any of these answers is acceptable iv d V h dh Or alternatives: Complete method to investigate value of 0 so maximum derivative Or: complete method to investigate the value of V either side and at the turning point v At this point sineao Use of a correct ratio with their h (and/or h x) = Angle EAO =. Accept. which comes from h =.88 NB Answer is given Care: the answer is given SC h =.89 with either dv or missing dh Numerical value must be.89 Accept h 9

12 699 Mark Scheme June 0 a i Max value = Not from any use of 0. from graph a ii Height = 0. (m) or 0 cm b c 6 x x x x y x 6 x x x Area = 6.d 0 x x x x x x x x x Area of cross section = 0.m = 00cm 0 B A 7 each error Dep on B convincing algebra (means sight of an extra correct step www) Integrate (ignore c) A if one error, if two errors (Dep on st ) Deal with limits correctly (Putting x = 0 does not need to be seen) Units An error is signs, powers, coefficients, failure to include the at end Alternative method: Integrate original function is OK, but in dealing with limits x = 0 must then be seen. Omission of is one error. Multiply by x or, ie x integrating gives A0 0

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