GCE Mathematics Unit 477: Further Pure Mathematics 3 Advanced GCE Mark Scheme for June 04 Oxford Cambridge and RSA Examinations
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477 Mark Scheme June 04 (i) 3 7 requires evidence of method for cross product or at least correct 5 7 7 7 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 =
477 Mark Scheme June 04 (ii) 5 4 7 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 4 4 3 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 ±
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 modulus-argument 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 co-ordinates 8i ag complete argument including start line Alternative: 6 3 4 5 6 ( 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
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
477 Mark Scheme June 04 4 (i) B or more Ignore element () 3 7 9 3 7 9 B 4 or more inverse () 7 3 9 7 3 9 B all 7 correct 4 (ii) ( has order ) 9,,9 have order 4 [3] Correctly identifies order of all elements 3 9 3 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
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 non-zero 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
477 Mark Scheme June 04 6 (i) 4 4 l 3 so 3 0 l dot product of correct vectors = 0 5 5 (,, 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
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 3 8 0, 0.899 [4] and on the 8
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 non-singularity 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) 0 0 0 0 0 0 0 0 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
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