GCE Mathematics Advanced GCE Unit 477: Further Pure Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations
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477 Mark Scheme June 0 (i) 4 0 0 4 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
477 Mark Scheme June 0 (i) 7 7 7 7 7 7 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
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]
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
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 []
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
477 Mark Scheme June 0 7 (i) 6 6 6 ( 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
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,,,, 0 0 0 0 0 0 cos, cos,cos,cos 0 0 0 0 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 0 + 80 + cos π = = 0 8 A Dep on full marks in (ii) [] M 8
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