physicsandmathstutor.com 4727 Mark Scheme June 2010

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1 477 Mark Scheme June 00 Direction of l = k[7, 0, 0] Direction of l = k[,, ] EITHER n = [7, 0, 0] [,, ] For both directions [ x, y, z]. [7,0, 0] = 0 7x 0z = 0 OR [ x, y, z]. [,, ] = 0 x y z = 0 n = k[0,, 7] METHOD Vector ( a b) from l to l =± [4,, 0] OR ± [ 4,,] OR ± [,, 9] OR ± [,,0] For a correct vector For finding vector product of directions of l and l OR for using scalar products and obtaining equations For correct n ( a b ).n * For finding ( a b ).n d = = n 0 For n in denominator OR for using n d =.94 7 METHOD Planes containing l and l perp. to n * For finding planes and p p seen are r. [0,, 7] = p = 70, r. [0,, 7] = p = 4 For p = 70k and p = 4k 70 4 d = = =.94 For n in denominator OR for using n 0 0 METHOD r = [7 λ, 0,0 0 λ] OR [7 7 λ, 0, 0 λ] r = [4 μ, μ, μ] OR [ μ, μ, μ] For correct points on l and l using different parameters 7λ 0α μ = 4 4 α μ = * For setting up linear equations from 0λ 7α μ = r αn = r and solving for α α = n = 0 For n seen multiplying α d = 0 =.94 7

2 477 Mark Scheme June 00 (i) ar = r a rar= r a Pre-multiply ar = r a by r r (ii) METHOD = e rar= a Use For n =, rar k Assume r ar k = a OR For n = 0, = a EITHER Assumption OR OR k k k k k k r ar = r r ar r = rar = a k k k k k k 0 0 r ar r = e and obtain answer AG = a For stating true for n = OR for n = 0 r ar = rar = a For attempt to prove true for k r ar = r rar r = r ar = a For obtaining correct form Hence true for all n 4 For statement of induction conclusion METHOD r ar = r. rar. r = ra r = a, similarly for rar = a 4 4 r ar = r r ar r = rar = a, similarly for r ar = a For attempt to prove for n =, For proving true for n =,, 4, r ar = eae= a For showing true for n = For n >, r METHOD n nmod n n n n r ar = r rar r = r, hence true for all n For using n mod and correct conclusion n n n n n n OR r ar = r. r a. r = r ar = r n n ar Starting from n, for attempt to prove true for n For proving true for n n n = r ar = For continuation from rar a = n downwards = = For final use of rar a SR can be done in reverse METHOD 4 ar = r a ar = r ar = r 0 a etc. n n For attempt to derive ar = r a n n ar = r a For correct equation SR may be stated without proof n n n r ar = r a n For pre-multiplication by r = ea = a For obtaining a ( r = e may be implied)

3 477 Mark Scheme June 00 (i) (ii) w = cos 4π isin 4π cos isin * cos isin cos 8 isin 8 Allow cis k π and For correct value w = π π For correct value w = π π For w* seen or implied = π π 4 For correct value * k i e π throughout SR For exponential form with i missing, award B0 first time, allow others For w in approximately correct position 4 For AB BC CD For BC, CD equally inclined to Im axis For E at the origin (iii) z = 0 OR Allow points joined by arcs, or not joined Labels not essential 4 z z z z z = 0 For correct equation AEF (in any variable) Allow factorised forms using w, exp or trig 9 4 (i) y = xz y d z = z x dx dx For correct differentiation of substitution dz xz x xz = xcos z d z x cos z dx dx For substituting into DE For DE in variables separable form For attempt at integration sec z dz = dx x to ln form on LHS ln ( sec z tan z) = ln kx OR ln tan ( z π 4 ) = ln kx For correct integration (k not required here) y y sec tan = kx For correct solution x x (ln x) c AEF including RHS = e y OR tan π = kx x 4 (ii) (4, π) sec π tan π= 4k 4 4 For substituting (4, π ) OR tan ( π π 8 4 ) = 4k into their solution (with k) y y sec tan = 4 ( ) x For correct solution AEF x x Allow decimal equivalent 0.0 x y OR tan π = ( tan π ) x or ln ( ) x Allow e x for x x 8

4 477 Mark Scheme June 00 (i) iθ iθ iθ C is = e e e 4 8 = = e e iθ iθ (ii) θ i ( e ) C i S = iθ iθ ( e )( e ) θ i 4 e 4 cosθ isinθ = = 4 e e 4 4cosθ iθ iθ ( ) 4 cosθ C =, 4cos θ 4 sinθ S = 4cos θ 4 8 For using cos n isin n e n θ θ θ = at least once for n For correct series For using sum of infinite GP For correct expression AG SR For omission of st stage award up to M0 A0 OEW For multiplying top and bottom by complex conjugate For reverting to cosθ and sinθ and equating Re OR Im parts For correct expression for C AG For correct expression for S i (i) Aux. equation m m = ± 4i m 7( = 0) CF ( y ) e x ( Acos 4x Bsin 4x) For attempting to solve correct auxiliary equation For correct roots A cos (4 x ε ) ) sin = For correct CF (allow (trig terms required, not e ) f.t. from their m with arbitrary constants PI ( y = ) px q p 7( px q) = 7x For stating and substituting PI of correct form p = For correct value of p and q = For correct value of q x GS y = e ( Acos4x Bsin4x) x For GS. f.t. from their CFPI with 7 arbitrary constants in CF and none in PI. Requires y =. (ii) x x 0 e 0OR very small For correct statement. Allow graph y = x approximately For correct equation Allow, and in words Allow relevant f.t. from linear part of GS 9 ± 4ix 4

5 477 Mark Scheme June 00 7 (i) (,,) and (,,) ± [4,,0] in Π For finding a vector in Π n = [,,] [4,,0] = k[,4,] For finding vector product of direction vectors of l and a line in Π For correct n r. [, 4, ] = 4 For correct equation. Allow multiples (ii) METHOD Perpendicular to Π through ( 7,,0) meets Π For using perpendicular from point on l to Π Award mark for kn used where ( 7 k) 4( 4 k) ( k) = For substituting parametric line coords k = d = 4 = 9. METHOD 4 into Π Π is x 4y z = For attempt to use formula for perpendicular distance ( 7) 4( ) (0) d = = 9. 4 METHOD m = [,,] [ 7,,0] = ( ± )[8,,] OR = [,, ] [ 7,, 0] = ( ± )[,, ] m. [,4,] 4 d = = = 9. 4 METHOD 4 For substituting a point on l into plane equation For finding a vector from l to Π For finding m.n As Method, using parametric form of Π [ 7,, 0] k[, 4, ] = [,, ] s[,, ] t[4,, 0] For using perpendicular from point on l to Π Award mark for kn used k s 4t = 8 4k s t = k = ( 4) For setting up and solving equations s =, t = k s = d = 4 = 9. METHOD d = = For attempt to find distance from O to Π 4 OR from O to parallel plane containing l [ 7,,0]. [,4,] 9 d = = 4 ( 9) For finding d d d d = d = = 9. (iii) ( 7,,0) k (,4,) State or imply coordinates of a point on the reflected line Use k = 4 State or imply distance from (ii) Allow k =± 4 OR ± 4 f.t. from (ii) b = [,, ] For stating correct direction a = [,,8] 4 For correct point seen in equation r = a tb r = [,,8] t[,,] AEF in this form

6 477 Mark Scheme June 00 8 (i) {A, D} OR {A, E} OR {A, F} For stating any one subgroup (ii) A is the identity is not a factor of OR elements can be only of order,,, For identifying A as the identity For reference to factors of (iii) For finding BE and EB AND using ω = 0 0 ω For correct BE (D or matrix) BE = = D 0, EB = = F ω 0 For correct EB (F or matrix) 0 Dor 0 ω, 0 F or M ω 0 closure property satisfied 4 For justifying closure (iv) B ω 0 For correct method of finding either inverse = = C 0 ω For correct B = C Allow ω 0 0 ω E 0 ω = = E For correct E = E Allow 0 ω ω 0 ω 0 (v) METHOD M is not commutative e.g. from BE EB in part (iii) For justification of M being not commutative N is commutative (as mod 9 is commutative) For statement that N is commutative M and N not isomorphic # For correct conclusion METHOD Elements of M have orders,,,,, Elements of N have orders,,,,, Different orders OR self-inverse elements M and N not isomorphic METHOD M has no generator since there is no element of order * For all orders of one group correct For sufficient orders of the other group correct # For correct conclusion SR Award up to if the selfinverse elements are sufficiently well identified for the groups to be nonisomorphic For all orders of M shown correctly N has OR as a generator For stating that N has generator OR M and N not isomorphic # For correct conclusion METHOD 4 M A B C D E F A A B C D E F B B C A F D E C C A B E F D D D E F A B C E E F D C A B F F D E B C A N * For stating correctly all squared elements of one group For stating correctly sufficient squared elements of the other group M and N not isomorphic # For correct conclusion # In all Methods, the last is dependent on at least one preceding

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

GCE. Mathematics. Mark Scheme for June Advanced GCE 4727 Further Pure Mathematics 3. Oxford Cambridge and RSA Examinations GCE Mathematics Advanced GCE 477 Further Pure Mathematics Mark Scheme for June 010 Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range