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

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1 GCE Mathematics Advanced GCE Unit 477: Further Pure 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 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 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

3 477 Mark Scheme June 0 (i) vectors in plane: two of 4, 6 =, 4 M Differences between two pairs Any multiple 0 4 r = 6 + λ + µ A Aef of parametric equation Must have r =... [] (ii) 0 4 M can be awarded where vector = 8 M Calculate vector product product has method shown or only A or multiple one term wrong r 6. 8 = 0 Or Cartesian form = d with attempt M to compute d x+ 8y z = 9 A Aef of cartesian equation, isw. Alternate method [4] M A MA M A M A EITHER x, y, z in parametric form both parameters in terms of e.g. x, y substitute into parametric form of z OR x, y, z in parametric form equations in x, y, z and one parameter eliminate parameter

4 477 Mark Scheme June 0 (i) B each error From table clearly closed Must be clear they are referring to B tabulated results is identity B,, 7 7 (mod8) B Or appears in every row [] Superfluous fact/s gets (ii) has order and,, 7 all have order B [] (iii) {, }, {, }, {, 7} (and {}) B All correct, no extras Allow {} included or omitted [] (iv) in H 9 (mod0) so not order Shows and states that or that M 7 is not order (or is order 4) no element of order > in G so not isomorphic A Completely correct reasoning [] Or, if zero, then SC for merely stating comparable orders and then saying that orders don t correspond, so not isomorphic" Or table for H with saying not all elements self inverse, so not isomorphic

5 477 Mark Scheme June 0 du dy u y y dy du = u du cos x in DE gives x + u = x A du cos x + u = B Divide Both sides x x ln x I = exp( ) = e x M Correctly integrates Must have form d u f( xu ) g( x) = x A Can be implied by subsequent work du x xu cos x d ( xu ) = cos x xu= sin x ( + A) M Integrate sin x+ A u = x A Or gives GS in implicit form Must include constant at this stage sin x+ A y = x A [8]

6 477 Mark Scheme June 0 4 (i) Sketch Must have axes, A shown across and either scale (or co-ordinates) B with B in rough position, or angle and distance on argand diagram. No inconsistencies π i OA = =, OB = e = and BOA = π M Can be seen on diagram Alt. Attempts AB or second angle hence OAB equilateral A [] 4 (ii) e πi Or e π i. Isw For full marks can use CiS form, or M for evidence they are MA clear polar co-ordinates, in radians. considering BA, or for Not C-iS i [] 4 (iii) π i i π e = e M For mod and arg Hence so must use their e πi ( ) ( π π) = 4 cos isin Aft aef 4 4 = + i B [] Condone use of.. 4

7 477 Mark Scheme June 0 AE: λ + λ + = 0 M λ = ± i A CF: e ( Acosx+ Bsin x) PI: y = ae Aft ae ae + ae = e M 4a = a = A 4 GS: y e ( Acos x Bsin x) 4 B Differentiate & substitute = + + Aft Must have y = dy = e ( + Acos x+ Bsin x 4 ) + e sin + cos M* ( A x B x) dy x= 0, = 0 ( + A 4 ) + B= 0 x= 0, y = 0 + A= 0 A y 4 = B= Aft From their GS 4, 0 ( x) Differentiate their GS of form y = e P + Acos nx + Bsin nx where P is constant or linear term, n not 0 or ( ) Or Ae cos( x+ α) Must be in real form If PI y = axe,then max of M,A,A, B0,M,A0,A0 (since cannot be consistent) M, M, A. Must have a constant in their PI Allow one error *M Use conditions But M0 if leads to solution of y = 0 = A Must have y = and be in real form 4 e cos 6 (i) x= t+, y = t, z = t+ ( t ) ( t ) ( t ) [] B Parameterise = Substitute into plane or B for y and z correctly in terms of x e.g. y = x 7, z = x + Then M for full simultaneous equations method. 0t = 0 t = M Solve Intersect at (, 4, ) A cao Accept vector form []

8 477 Mark Scheme June 0 6 (ii) 0 cos( π θ ) = = Attempt to find angle or its MA 0 complement θ = 0.64 A or 7. [] 6 (iii) If P is point of intersection and Q is required point, Use PQ with right angled triangle or PQ = λ so sinθ = = M* or PQ.n ˆ = ± where n = consider component of PQ in PQ λ 0 direction of normal vector. 0 = Valid method to set up equation in M 0 λ 0 λ alone. λ = ± A points have position vectors 4 ± *M Dep on st M points at (.8,,.4) and (4., 7,.6) A cao (May work from general point on original equation) Alternative: t+ + (t ) ( t+ ) Distance = = + + M* A *M Solve t = 0.4 or.6 A At least one value found (.8,,.4) and (4., 7,.6) A [] Zero if formula used without substitution in of parametric form. 6

9 477 Mark Scheme June 0 7 (i) ( ab) = abab... ab = a b as commutative Some demonstration that they M Must give reason understand commutativity ( ) ( ) = a b = ee = e A Using orders of a and b So ab has order,,, or 6 ( b a ab a ab e so ab not order ) Condone absence of this line ( ab) = a b = eb = b and b not order, Insufficient to merely have M Consider other cases so ab not order simplified all (ab) n ( ab) = a b = aa e = aee = a e, so ab not order (So must be order 6) A AG Complete argument [4] 7 (ii) ac has order 8 B Or abc or generator 8 is LCM of and 9, (so we can use a similar argument or explicit consideration of M to part (i)) other cases So as G has an element of order 8 it must be cyclic. A AG Complete argument 8 (i) ( ) cosθ + isin θ = cosθ + isinθ 4 4 = c + ics 0cs 0ics + cs + is M [] B Or cos θ = re{ cosθ + isin θ } 4 θ = c c s + cs M Take real parts cos 0 ( ) ( ) = c 0c c + c c M = c 0c + 0c + c 0c + c cos θ = 6c 0c + c A AG [] ( ) No more than error, can be unsimplified 7

10 477 Mark Scheme June 0 8 (ii) Multiplying by x gives 6x 0x + x= 0 Hence, so no marks for using quadratic at this stage. letting x = cosα gives cosα = 0 M 7 9 hence α = π, π, π, π, π A 8 (iii) 7 9,,,, cos, cos,cos,cos α = π π π π π cos π = 0 which is not a root A 7 9 so roots x = π π π π A [4] 4 0 ± 80 6x 0x + = 0 x = B Can be gained if seen in (ii) cos decreases between 0 and π so cos π is greatest root 0 so cos π = = 0 8 A Dep on full marks in (ii) [] M 8

11 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: 0 Facsimile: 0 OCR 0

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