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 01 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 candidates 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 01

3 699 Mark Scheme June 01 Annotations Annotation Meaning Tick Cross Benefit of doubt Follow through Ignore subsequent working Method mark awarded 0 Method mark awarded 1 Accuracy mark awarded zero Accuracy mark awarded 1 Independent mark zero Independent mark 1 Special case Omission mark Misread 1

4 699 Mark Scheme June 01 Subject-specific Marking Instructions 1 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 independent marks: i. Correct answer correctly obtained (no symbol) ii. Follows correctly from a previous answer whether correct or not ( 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 1 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. 5 Where a follow through () 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; 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 eg 7000, 7, 70, would be acceptable but 070 or 7 would not.

5 699 Mark Scheme June 01 Section A 1 (i) Attempt to rearrange to to y =... Gradient = 1.5 Mark final answer (Ans = 1.5x is A0) Find two points on line and then the gradient by y x (The two points must lie on the line) Ans 1 (ii) Gradient = soi Use (,1) and their normal gradient in a standard form for a line There must be an attempt to find a normal gradient for this method mark to be earned y = x oe terms only x y = k Sub (,1) Ans

6 699 Mark Scheme June 01 7 < x + 1< 1 Subtract either side by 1 (or 1 if done the other way round) 8 < x < 11 Divide throughout by < x < Or < x and x<, 1, 0,1,, Mark final answer 11 8 SC1 for x < or x > but not both Can be done by trial: One missing or one extra B Mark final answer John's age, x soi Ages in 5 yrs x + 5, x + 5 soi B all correct Condone use of their own letters x + 5 = (x + 5) x + 5 = x + 15 x = 10 No ISW Forming simultaneous equations: Correct equations implies, Eg j = p and j + 5 = (p + 5) Give only if there is an attempt to eliminate one variable. SC Answer only with no equation formed or incorrect algebra

7 699 Mark Scheme June 01 5 cos θ = 1 sin θ = 1 = 9 9 Any use of calculators to approximate gets A0 cosθ = 1 tan θ = 5 ISW Find third side of triangle = Ans 5

8 699 Mark Scheme June 01 5 (i) Diffn all powers reduced by 1 allow one error y = x x 6x+ (beware dividing by x) dy = x x 6 dx All terms correct x x+ = 0 x x = 0 ( )( 1) Set = 0 and solve (dependent on first M mark) x =, 1 Both x y = 7, 6.5 (, 7), ( 1,6.5) Both pairs (Allow this mark if the coordinated are explicitly stated in 5(ii). ) 5 5 (ii) Correct shape (Cubic the right way up and two turning points, does not need three intercepts on x-axis) Through (0, ), (, 7) and ( 1, 6.5). Dep on 1st B mark Allow y-intercept in range [,] At x = allow y in range [ 6, 8] At x = 1 allow y in range[6, 8] 6

9 699 Mark Scheme June 01 6 (i) P( from ) = pq 1 term with correct fractions and powers 1 5 = = 7 or oe ISW 6 (ii) P(at least 1) = 1 P(0) = 1 q Correct coefficient of attached to their probability term soi At least sf Correct answer only B 5 = = or At least sf P(1) + P() + P() Add three terms = Three terms correct soi = sin 0 sin B Sin rule: = 6 Sin rule or complete method via the perpendicular to find angle 6 sin 0 sin B = ( = ) Correctly applied in this case soi o B = 7.6 ( ) or 105. One value o and 105(. ) or 7.6 FT 180 their B, unless B = 90 7

10 699 Mark Scheme June x Increase in power of 1 in at least one term (beware multiplying ( 8x x ) dx= x by x) 0 0 Both terms 8 = 8 = 6 Substitute limits of 0 and their upper limit following integration 56 1 = or 85 or 85. or better (NB Limits the wrong way round A0) Total area = Dep on previous M, ratio of their ans / 160 Proportion = = = 5.% Dep on previous (NOT 5) NB The answer is given. 6 5% of 160 dep on all other M marks = 8.8 The two answers related, eg both corrected to sf = 85 Dep on previous 9 (i) x + 8x+ 19 = ( x+ ) = + + ( x ) 9 (ii) When x = their a Value is their b Attempt to complete square or expand rhs to give quadratic expression For www For NB For completion of square at least ( x ± ) seen FT FT 9 (iii) 1 their b 1 FT 8

11 699 Mark Scheme June (i) Angle CAB = tan = 67. or bearing = Mark final answer (Don't allow 67) Distance = = 600 Can be awarded if seen in part (ii) (Rounds to 600 to sf) 10 (ii) 1000 AD = or anything that rounds to 17 cos55 Don t accept premature approximation CD = 1000tan 55 or anything that rounds to 18 DB = 00 CD = 97 AD + DB AB = anything that rounds to 115(m) cao www Sine rule on right-hand triangle x y 600 = = sin.6 sin1. sin15 (all values seen, angles rounding to 1. and.6) x = y = both values Answer Correct answer www B Accept a combination 9

12 699 Mark Scheme June 01 Section B 11 (a) Radius 10 Centre (, 0) 11 (b) (i) ( ) 11 (b) (ii) 11 (b) (iii) Substitute y = x+ 6 into x + y = 100 Substitute ( x ) ( x ) = 100 5x + 0x+ 0 = 100 x + x 1 = 0 term quadratic ( x+ 6)( x ) = 0 Solve a term quadratic Either both x or both y or one pair A is ( 6, 6 ) and B is (,10) Both pairs Midpoint AB =, (,) 1 ( ) ( ) AB = = = 0 = 8 5 or (c) distance = radius half their (iii) = = 0 = 5 or.7 Their centre to their midpoint = ( ) + = 0 10

13 699 Mark Scheme June 01 1 (a) v= 8t t Diffn v Final answer v= 0 when 8t = t t = Set = 0 or t=, dep on first or when t =, v = 8 = = 0 SC for confirming = 0 when v divided by 1 (b) (i) v= 8t t Diffn their v a= 8 t When t = 0, a= 8 Set t = 0 dep on 1 st 1 (b) (ii) 18 Substitute t= t = gives s= 6 = 1... oe 1 (c) Inverted parabola implies symmetry about t = through (0,0) and (,0) (sight of correct vertex not necessary) Straight line through (, 0) and (0, their 8) (, their 8) 11

14 699 Mark Scheme June 01 1 (i) x 0, y 0 Condone < for and > for Ignore extras (including for instance, x + y 50 ) y x y x 1 (ii) 1 (iii) The line is 0x + 50y = x = 0 y = 0 y = x y = x Shading Ignore any shading that may relate to 1(iii) accept inequality soi (accept written in (ii) ) Draw line (Correct line seen on graph is B) (0, ) Total = www Point soi Total (Total of seen with nothing else is B) 1

15 699 Mark Scheme June 01 1 (i) dy = 1x 10x dx Diffn and sub x = 1 When x= 1, m= www 1 gradient of normal = FT their numerical m soi 1 Equation is y 0 = ( x 1) Find eqn of line dep on and their normal gradient y+ x= oe eg y = x 5 1 (ii) 1 ( ) x = x x + Substitute their equation 1 x= 8x 10x + 8x 10x + x+ 1 = 0 At least 1 correct intermediate step seen, beware answer given 1 (iii) For line: y = 1 = = oe Substitute into correct line For curve: y = = + 1 = Substitute into curve 1

16 699 Mark Scheme June 01 1 (iv) 1 f ( x) = 0 when x= 1 and f(x) = (x 1)(x 1)(ax + b) oe f ( x) = ( x 1)(x 1)(x+ 1) or f(x) = (x 1) (x 1 ) (cx + d) 1 x = 5 y = C is, www 8 Long division by (x 1) and (x 1) or by one given factor plus attempt to factorise resulting quadratic or by (x x+1) Giving third root Answer NB The working for this part may appear elsewhere but can only be credited if their final answer is seen here. Evidence of long division on cubic is correct first line of long division plus kx in the quotient 1

17 OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge C EU OCR Customer Contact Centre Education and Learning Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, C EU Registered Company Number: 866 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 01