GCE Mathematics. Mark Scheme for June Unit 4727: Further Pure Mathematics 3. Advanced GCE. Oxford Cambridge and RSA Examinations


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1 GCE Mathematics Unit 477: Further Pure Mathematics 3 Advanced GCE Mark Scheme for June 04 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, Cambridge Nationals, Cambridge Technicals, 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 notforprofit 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 04
3 477 Mark Scheme June 04 (i) 3 7 requires evidence of method for cross product or at least correct values calculated (eg) z 0 x y 4,3x 5y 3 x, y or any valid point e.g.(0, 3, ), (3, 0, ) r 0 Alternative: Find one point Find a second point and vector between points multiple of r 0 Alternative: Solve simultaneously Point and direction found r 0 [4] oe vector form to at least expressions for x,y,z parametrically, or two relationship between variables. Must have full equation including r =
4 477 Mark Scheme June 04 (ii) Condone lack of absolute signs for.86 with no workings scores 6 oe surd form. isw Alternative: find parameter for perpendicular meets For complete method with calculation look for λ = 7/6 plane and use to find distance errors [] du dy Correctly finds u y y dy du Or u dx dx dx dx dy x or for complete unsimplified so DE y 4y e dx substitution du x Can be implied by next 4u e dx exp 4d e 4 x I x ft Must have form du ( ) ( ) d f x u g x for this mark and x any further marks Can be implied by subsequent work e x du 4e x u e x * Multiples through by IF of form e kx, dx simplifying RHS 4x 3x ue e ( A ) *dep* Integrates 3 x 4 u A x dep * Rearranges to make u or y the subject No more than numerical error at 3 e e x y 3 e A e 4 x Alternative from 4 th mark to 6 th mark 4x CF: (u= ) Ae x du x PI: u ke, ke dx ke x 4ke x e x, k 3 * dep* [8] Cao PI chosen & differentiated correctly Substitutes and solves this step ignore use of ±
5 477 Mark Scheme June 04 3 (i) 6 i/6 z z e k 0,,,3,4,5 Oe exactly 6 roots accept roots,  given as integers. Diagram B 6 roots in right quadrant, 3 (ii) 4 6 i ( i) e 6 B [4] correct angles and moduli Attempts modulusargument form, getting at least correct 6 4 i 8e for (mod) 6 and arg x 6 as evidenced by labels, circles, or accurate diagram, or by coordinates 8i ag complete argument including start line Alternative: ( i) 6i 5i 0i 5i 6i i 6i 5 0i 5 6i no more than term wrong Sc for only lines & 3correct 8i ag Alternative: ( i) i 6 3 ( i) i 8i ag [3] 3
6 477 Mark Scheme June 04 3 (iii) 6 k i/6 z 8i z ( i)e i 4 k i/6 e e i (/4 k/3) e, k 0,,,3,4,5 or equivalent k Alternative: i (/4 k/3) z 3 i ( k) 6 8e e, k 0,,,3,4,5 [3] i ( / k /3) or equivalent: e.g. e accept unsimplified modulus 4
7 477 Mark Scheme June 04 4 (i) B or more Ignore element () B 4 or more inverse () B all 7 correct 4 (ii) ( has order ) 9,,9 have order 4 [3] Correctly identifies order of all elements so order 4 similarly 7,3,7 order 4 B justifies order for at least element of order 4 Allow one error must show workings towards a 4 for demonstration that these elements are order 4` no element of order 8 so not cyclic www condone no generator in place of no element or order 8 4 (iii) [3] For two sets which both contain and all (4) elements inverses B One subgroup of order 4 {,3, 9, 7} and {, 3, 9, 7} for correspondence of their elements of same order, 9 9,3 3, 7 7 or 3 7, 7 3 [5] 5
8 477 Mark Scheme June 04 5 AE: 5 6 0, 3 B CF: PI: 3 e x x A B e Bft e x y a Bft x x x x ae 5ae 6a e e Differentiate and substitute a a GS: x x 3x ( y ) e Ae B e ft x 0, y 0 A B 0 Use condition on GS x x 3x y e A e 3 B e x 0, y 0 A 3B 0 * Differentiate their GS of form x mx nx y k e Ae B e where k, m, n are nonzero constants and m, n not A, B dep* Use condition and attempt to find A, B x x 3x e e e ft must be of form k e x plus a standard CF form with arbitrary constants Must have arbitrary constants y www Must have y = [0] 6
9 477 Mark Scheme June 04 6 (i) 4 4 l 3 so 3 0 l dot product of correct vectors = (,, 7) on l but 4 7 8so not in substitute point on line into and calculate d hence l not in Full argument includes key Argument can be about a general components point on line [3] 6 (ii) 4 ( r ) B 7 closest point where meets 4( 4 ) ( ) (7 ) 8 parametric form of (x, y, z) substituted into plane 6 (iii) ft 3 5 r 3 [4] 3 5 r 3 Bft oe 3 5 [] must have r = 7
10 477 Mark Scheme June 04 7 (i) i i isin e e B any equivalent form If use z, must define it in isin n e e in 5 i i isin e e 5 i5 i3 i i i3 i5 e 5e 0e 0e 5e e * binomial expansion can be unsimplified 5 5i 5i 3i 3i i i 3i sin ( e e ) 5( e e ) 0( e e ) isin5 5 isin3 0 isin 5 6 sin sin 5 5sin 3 0sin dep* [4] grouping terms AG Award B then sc for candidates who omit this stage from otherwise complete argument must convince on the 6 elimination of i 7 (ii) 5 6sin 0sin sin5 5sin3 * Attempts to eliminate sin5θ and sin3θ 5 6sin 6sin 0 Or 6sin 5 θ =6sin θ sin 0, dep* must have 3 values for sin θ , [4] and on the 8
11 477 Mark Scheme June 04 8 (i) 0 is identity 0 B for, must at least get matrix in form a b a b G x y b a a b b a, or state existence of inverse y x due to nonsingularity a b c d ac bd bc ad b a d c bc ad ac bd and ( ac bd) ( bc ad) a c b d b c a d ( a b )( c d ) 0 [6] Must not attempt to prove commutativity in (i) 8 (ii) c d a b ac bd bc ad d c b a bc ad ac bd a b c d so commutative b a d c 8 (iii) order 4 [] g must be correct allow error in getting g 4 [3] must also consider matrices reversed, but may be seen in (i) 9
12 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge CB EU OCR Customer Contact Centre Education and Learning Telephone: Facsimile: 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; Hills Road, Cambridge, CB EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 04
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