physicsandmathstutor.com Mark Scheme 4727 January 2006

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1 Mark Scheme 77 January

2 77 Mark Scheme January 006 Directions [,, ] and [, 3,] B For identifying both directions (may be implied by working) [,, ]. [, 3, ] θ= cos 3 = cos = 7.0, 7 or.6 rad A For correct answer For using scalar product of their direction vectors For completely correct process for their angle (i) Identities b, 6 B B For correct identities Subgroups {b, d}, {6, } B B For correct subgroups (ii) { abcd,,, } {, 6, 8, } or {8, 6,, } B B For b 6, d B 3 For ac,,8 in either order SR If B0 B0 B0 then A may be awarded for stating the orders of all elements in G and H 7 dy dz 3 (i) 3y = For differentiating substitution dz x + xz = e A For resulting equation in z and x = B For correct IF f.t. for an equation in suitable form ( ) ( ) d x d e 3 x z OR y e = For using IF correctly Integrating factor ( x ) x e e x 3 x ze OR y e = x( + c) A For correct integration (+c not required here) ( ) x 3 e 3 y = x+ c A 6 For correct answer AEF (ii) As x, y 0 B For correct statement i i (i) cos ( e θ θ θ= + e ), iθ iθ sin i ( e e ) iθ iθ iθ iθ cos θsin θ= ( e + e ) ( e e ) 7 B For either expression, seen or implied i θ= z may be used for e θ throughout 6 iθ iθ iθ iθ iθ iθ ( e e ). ( e e 6 e e ) = (( 6i θ 6i θ) ( i θ i θ e e e e ) ( e i θ e i θ) ) A A For expanding terms For the correct expansions SR Allow A A0 for k ( e iθ e iθ i i i i )( e θ e θ 6 e θ e θ) , k 6 = For grouping terms and using multiple angles ( cos 6 cos cos ) = θ θ θ+ AG A 6 For answer obtained correctly 3

3 77 Mark Scheme January 006 (ii) 3 π 0 cos θsin θdθ= 3 π 3 sin 6 sin sin 6 0 = θ θ θ+ θ = π 0 = π A For integrating answer to (i) For all terms correct A 3 For correct answer 9 5 (i) EITHER z = 8cis(k+ ) π, k = 0,,,3 (k+ ) πi 8e, k 0,,,3 B For correct modulus AEF OR z = = B For correct arguments AEF (ii) { } z = + + B For any of ± ± i, i, i, i i z = + i, + i, i, i B For any one value of z correct B For all values of z correct AEFcartesian (may be implied from symmetry or factors) ( z α),( z β),( z γ),( z δ ) B f.t., where α, βγδ,, are answers above (iii) EITHER ( z (+ i))( z ( i)) For combining factors from (ii) in pairs ( z ( + i))( z ( i)) Use of complex conjugate pairs = ( z + z+ 8)( z z+ 8) A For correct answer OR z + 6 = ( z + az+ b)( z + cz+ d) For equating coefficients a + c = 0, b + ac + d = 0, ad + bc = 0, bd = 6 For solving equations Obtain ( z + z+ 8)( z z+ 8) A 3 For correct answer 6 (i) MB = [,, ], OF = [,, ] B For either vector correct (allow multiples) MB OF For finding vector product of their MB and OF = [,, ] OR k[, 6, ] A 3 For correct vector (ii) EITHER Find vector product of any two of ± [,, ], ± [0, 0, ], For finding two relevant vector products ± [,, 0] and any two of ± [, 0, ], ± [,, 0], ± [0,, ] Obtain k [,, 0] A For correct LHS of plane CMG Obtain k [,, ] A For correct LHS of plane OEG For substituting a point into each equation x+ y = and x+ y z = 0 A For both equations correct AEF OR Use ax + by + cz = d with coordinates For use of cartesian equation of plane of C, M, G OR O, E, G substituted Obtain a: b: c = ::0 for CMG A For correct ratio Obtain a: b: c = :: for OEG A For correct ratio For substituting a point into each equation x+ y = and x+ y z = 0 A 5 For both equations correct AEF 9 35

4 77 Mark Scheme January 006 (iii) EITHER Put x, yorz= t in planes OR evaluate k [,, 0] k [,, ] For solving plane equations in terms of a parameter OR for finding vector product of normals to planes from (ii) Obtain r = a+ tbwhere a = [0,,],[,0,] OR [,,0] A Obtain a correct point AEF b = k [,,] A 3 Obtain correct direction AEF m 7 (i) ( x ax) = ( x ax)( x ax) K ( x ax) For considering powers of x ax = x aak ax, associativity, x x = e A A For using associativity and inverse properties m = x a x = x ex when m= n, B For using order of a correctly not m< n = x x A For using property of identity = e order n A 6 For correct conclusion (ii) EITHER ( x ax) z= e For attempt to solve for z AEF axz = xe = x xz = a x A For using pre- or post multiplication z = x a x A For correct answer OR Use ( pq) = q p For applying inverse of a product of OR ( pqr) = r q p elements State ( x ) = x A For stating this property Obtain x a x A 3 For correct answer with no incorrect working SR correct answer with no working scores B only (iii) ax = xa x = a x a Start from commutative property for ax xa = a x A Obtain commutative property for a x 8 (i) m + km+ = 0 For stating and attempting to solve auxiliary eqn m= k± k A For correct solutions, at any stage AEF kt k k (a) x = e Ae + Be For using e t with distinct real roots of A aux eqn For correct answer AEF kt i k t i k t (b) x = e Ae + Be For using e t with complex roots of aux eqn This form may not be seen explicitly but if stated as final answer earns A0 kt x = e A cos k t+ B sin k t A For correct answer OR kt cos x = e C sin k t+α (c) x = e t ( A + B t) A For using e t with equal roots of aux eqn For correct answer. Allow k for 36

5 77 Mark Scheme January 006 (ii)(a) x = B e sin 3t B For using t = 0, x = 0 correctly. f.t. from (b) t x Be For differentiating x & = ( 3cos 3t sin 3t) A For correct expression. f.t. from their x t = 0, x& = 6 B = 3, x = 3e sin 3t A For correct solution AEF SR and AEF OK for x = C e cos 3t+ π ( ) (b) x 0 B For correct statement e 0 and sin( ) is bounded B For both statements 37

physicsandmathstutor.com 4727 Mark Scheme June 2010

physicsandmathstutor.com 4727 Mark Scheme June 2010 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,,