GCE. Mathematics. Mark Scheme for June Advanced GCE Unit 4727: Further Pure Mathematics 3. physicsandmathstutor.com

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1 GCE Mathematics Advanced GCE Unit 77: Further Pure Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

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3 77 Mark Scheme June 0 (i) [5, 6, 7]. [,, ] sin 5 6 ( 7) ( ) sin 69. ( ,.06) 0 6 [5, 6, 7] [,, ] sin 5 6 ( 7) ( ) 8 sin (ii) METHOD 0 d ( ) 6 METHOD M* M (*dep) A A SR M* M (*dep) A A M A For using scalar product of line and plane vectors For both moduli seen For correct scalar product For correct angle For vector product of line and plane vectors AND finding modulus of result For moduli of line and plane vectors seen For correct modulus 8 For correct angle For use of correct formula For correct distance ( ) (6 ) ( ) 0 M For substituting parametric form into plane d 6 OR distance from (, 6, ) to (5,, 7) 8 ( ) 96 A For correct distance METHOD Plane through (, 6, ) parallel to p is M For finding parallel plane through (, 6, ) 0 6 xyz 6 d 6 6 A For correct distance METHOD e.g. (0, 0, 0) on p M For using any point on p to find vector vector to (, 6, ) = (, 6, 7) and scalar product seen e.g. [, 6, 7]. [,, ] [, 6, 7]. [,, ] A For correct distance d 6 6 METHOD 5 l meets p where ( 5 t) (6 6 t) ( 7 t) 0 For finding t where l meets p t d [5, 6, 7] sin M and linking d with triangle d 0 A For correct distance (i) METHOD EITHER i i e e e e e e i i i i cos isin icot OR in reverse with similar working M M A EITHER For changing LHS terms to e i cos OR in reverse For using cot sin i i cos For either of e e soi sin ()(i) For fully correct proof to AG SR If factors of or i are not clearly seen, award M M A0

4 77 Mark Scheme June 0 (i) METHOD i i i i e e e e EITHER i i e e i i e e (ii) cos isin cos isin OR cos isin cos isin isin isin cos icot cos sin METHOD cos i sin cos cos isin cos isin sin isin cos cos cos isin sin sin icos sin i cos =icot icot sin i cos METHOD t t i cos isin t t cos isin t t i t t M M A M M A M For multiplying top and bottom by complex conjugate in exp or trig form For using both double angle formulae correctly For fully correct proof to AG For using both double angle formulae correctly For appropriate factorisation For fully correct proof to AG For substituting both t formulae correctly it it i ti M For appropriate factorisation icot i i t it t t t t A For fully correct proof to AG METHOD 5 i i i e e e i e i i i i e e e For multiplying top and bottom by e i i e e M i and attempting to divide by e i i i e e OR multiplying top and bottom by e ( cos ) cos cos For using both double angle formulae M isin isin cos isin correctly icot A For fully correct proof to AG M For a circle centre O A For indication of radius = and anticlockwise arrow shown B For locus of w shown as imaginary axis described downwards 6

5 77 Mark Scheme June 0 (i) METHOD x A m ( 0) CF ( y ) e METHOD Separating variables on d y y 0 ln y x x A M A M For correct auxiliary equation (soi) For correct CF For integration to this stage CF ( y ) e A For correct CF (ii) PI ( y ) pcosx qsinx B For stating PI of correct form y psinx qcosx M For substituting y and y into DE ( p q)sin x(p q)cosx 5cosx A For correct equation pq 0 M For equating coeffs and solving p, q p q A A For correct value of p, and of q x GS ( y ) Ae cosx sinx B 7 For GS 5 5 f.t. from their CF+PI with arbitrary constant in CF and none in PI SR Integrating factor method may be used, followed by -stage integration by parts or C+iS method Marks for (i) are awarded only if CF is clearly identified (iii) x e 0, cosx sin x sin x M For considering either term 5 5 cos A For correct range (allow < ) CWO y OR y f.t. as p q y p q from (ii) (i) abc ( ab) c ( ba) c b( ac) M For using commutativity correctly bca ( ) ( bca ) ( cba ) cba A For correct proof (use of associativity may be implied) Minimum working: abc bac bca cba OR abc acb cab cba OR abc bac bca cba (ii) ea},{, eb},{, ec},{, ebc},{, eca},{, eab},{, eabc } B For any 5 subgroups B For the other subgroups and none incorrect (iii) {, eabab,, },{, eacca,, },{, ebcbc,, } B For any subgroups {, e a, bc, abc},{, e b, ca, abc},{, e c, ab, abc } B For more subgroup {, ebccaab,, } B For more subgroup (5 in total) and none incorrect (iv) All elements ( e) have order B* For appropriate reference to order of elements in G OR all are self-inverse OR no element of G has order OR no order subgroup has a generator or is cyclic OR subgroups are of the form{, eabab,, } (the Klein group) all order subgroups are isomorphic B (*dep) 9 For correct conclusion

6 77 Mark Scheme June 0 5 (i) dy k du M ku A u x ku u x u kd k k du k u xu d M For using chain rule For correct d y For substituting for y and d y A For correct equation AG x kx k (ii) k B For correct k (iii) du u x IF d ln e x x x e B For correct IF x x f.t. for IF = x k using k or their numerical value for k d M d u. For u.their IF x.their IF x x u. ( c x x ) A For correct integration both sides y A For correct solution for y cx x 6 (a) Closure ( axb) ( cxd) ( ac) x( b d) B For obtaining correct sum from distinct elements P B For stating result is in P OR is of the correct form SR award this mark if any of the closure result, the identity or the inverse element is stated to be in P OR of the correct form Identity 0x 0 B For stating identity (allow 0) Inverse ax b B For stating inverse (b) (i) Order 9 B* For correct order (ii) x B For correct inverse element (iii) M For considering sums of ax b ( axb) ( axb) ( axb) ax b and obtaining ax b 0x 0 ax b has order a, b (except a b 0 ) A For equating to 0x 0 OR 0 and obtaining order SR For order stated only OR found from incomplete consideration of numerical cases award B Cyclic group of order 9 has element(s) of order 9 M (*dep) For reference to element(s) of order 9 ( Q, (mod)) is not cyclic A For correct conclusion 0 9

7 77 Mark Scheme June 0 7 (i) R Q B For sketch of tetrahedron labelled in some way At least one right angle at O must be indicated or clearly implied (ii) (iii) O P M For using base height OPQ pq, OQR qr, ORP rp A For all areas correct CAO RP RQ RP RQ sinr PQR B For correct justification LHS = pq qr rp PQR B For correct expression = ( ) ( B For PQR in vector form OR ( ) ( OR ( ) ( PQR qri prj pqk M RHS pq qr rp A M A 6 0 For finding vector product of their attempt at PQR For correct expression For using aibjck a b c For completing proof of AG WWW 5

8 77 Mark Scheme June 0 8 (i) For expanding ( c i s) : at least terms and Re( c i s) = cos c 6c s s M* binomial coefficient needed A For correct terms M cos c 6 c ( c ) ( c ) For using s c (*dep) cos 8cos 8cos A For correct expression for cos CAO (ii) cos cos 8c 8c c For multiplying by c 6 6 cos cos 0 cos B to obtain AG WWW (iii) 6 6c c 0c 0 M For factorising sextic c 8c c 0 with ( c),( c) or ( c ) For quartic, b ac 6 0 A For justifying no other roots CWO c only n A For obtaining n AG Note that M A0 A is possible SR For verifying n by substituting c 6 into 6c c 0c 0 B (iv) 6 6c c 0c 0 0 M c 8c c 5 For factorising sextic with c For quartic, b ac 60 0 A For justifying no other roots CWO cos 0 only A For correct condition obtained AG Note that M A0 A is possible SR For verifying cos 0 by substituting c 0 6 into 6c c 0c 0 B SR For verifying and satisfy cos cos B 6

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