Paper 1 Further Pure Mathematics For Examination from 2016 SPECIMEN MARK SCHEME 3 hours MAXIMUM MARK: 120
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1 Cabridge International Exainations Cabridge Pre-U Certificate FURTHER MATHEMATICS (PRINCIPAL) 9795/ Paper Further Pure Matheatics For Exaination fro SPECIMEN MARK SCHEME hours MAXIMUM MARK: The sllabus is approved for use in England, Wales and Northern Ireland as a Cabridge International Level Pre-U Certificate. This docuent consists of printed pages. UCLES [Turn over
2 Mark Schee Notes Marks are of the following three tpes: M A B Method ark, awarded for a valid ethod applied to the proble. Method arks are not lost for nuerical errors, algebraic slips or errors in units. However it is not usuall sufficient for a candidate just to indicate an intention of using soe ethod or just to quote a forula; the forula or idea ust be applied to the specific proble in hand, e.g. b substituting the relevant quantities into the forula. Correct application of a forula without the forula being quoted obviousl earns the M ark and in soe cases an M ark can be iplied fro a correct answer. Accurac ark, awarded for a correct answer or interediate step correctl obtained. Accurac arks cannot be given unless the associated ethod ark is earned (or iplied). Mark for a correct result or stateent independent of ethod arks. The following abbreviations a be used in a ark schee: AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) CAO Correct Answer Onl (ephasising that no follow through fro a previous error is allowed) aef An equivalent for art Answers rounding to cwo Correct working onl (ephasising that there ust be no incorrect working in the solution) ft Follow through fro previous error is allowed o.e. Or equivalent UCLES 9795//SM/
3 n ( r r ) r r r n n r r n Splitting suation and use of given results M r n n )(n ) n( n ) n st for Σr ; nd for Σr & Σ n B B ( n n ) legitiatel A ( A k ( cosθ ) dθ including squaring attept; ignore liits and k M ( sin θ ) dθ for use of the double-angle forula OR integration of cosθ as k sin θ or k cos θ B π / θ cos θ ft (constants onl) in the integration; MUST be separate ters A π A d (i) (sinh x) (sinh x) d. cosh x OR sinh x cosh x M A A (ii) d. B separating variables in (i) s answer M x d A But x sinh t so t sinh (t ) C condone issing C A ALT. Set t sinhθ, t dt coshθ dθ M Full substitution t coshθ dt dθ A t.dθ θ sinh (t ) A sinh θ ALT. Set t tanθ, t dt sec θ dθ M Full substitution t sec θ dt dθ A t sec θ.dθ tan θ ln secθ tanθ ln t t A UCLES 9795//SM/ [Turn over
4 (i) x x.x x ( ) Creating a quadratic in x M For real x, ( ) [ Considering the discriinant M Y Creating a quadratic inequalit M For real x, ( )( ) Y Factorising/solving a -ter quadratic M Y Y CAO A (ii) substituted back (x x ) x [ ] M A substituted back (x x 9) x [ ] M A 5 (i) (a) (b) cosα sinα cos β sin β sinα cosα sin β cos β B B (ii) cosφ sinφ sinφ cosθ cosφ cosθ ultiplication of reflection atrices. Correct order. M M cos φ cosθ sinφ sinφ cosθ cosφ cosφ sinφ cosθ cosφ cosθ sinφ cos( φ θ ) sin( φ θ ) sin( φ θ ) cos( φ θ ) Use of the addition forulae; correctl done M A giving a Rotation (about O) through (φ θ ) M A (i) Possible orders are,,,, & B B Lagrange s Theore, the order of an eleent divides the order of the group (since the order of an eleent the order of the subgroup generated b that eleent) B (ii) E.g. xx. x xx. x b M x. x. x.. b M x. x. ( ) [b ] x. (x ) x. e b (iii) M s for first, correct uses of different conditions; the A for the rd condition used to clinch the result. Proving G not abelian: [e.g. b xx but x e] G not cclic OR establishing a contradiction A B B UCLES 9795//SM/
5 5 7 (i) cosθ i (c is) Use of de Moivre s Theore M c c.is c.i s c.i s i s Binoial expansion attepted M cosθ c c s s and c s cs Equating Re & I parts M sin θ c s cs tanθ cos θ c c s s t t Dividing throughout b c to get t t legitiatel M A (ii) t 5 tanθ B 9 ( ) 9 π tan 9 tan 9 9 Noting that this is tan(tan ) so that tan 5 5 π tan 9 M A A 8 (i) Substituting x, f() and f () into (*) f () 5 M A x (ii) { x f ( x) xf ( x) } {( x ) f ( x) f ( x) } f ( x) e Product Rule used twice; at least one bracket correct Substituting x, f () and f () 5 into this f () ft their f () M A M A (iii) f(x) f() f ()(x ) f ()(x ) f ()(x ) Use of the Talor series (x ) 5 (x ) (x ) st two ters CAO; nd two ters ft (i) & (ii) s answers M A A (iv) Substituting x. f(.). to d.p. CAO M A 9 (i) d x is a Bernouilli (differential) equation u du d B Then d x d du becoes 9x u 9x AG M A UCLES 9795//SM/ [Turn over
6 (ii) Method IF is e x ue x 9xe x x xe e x Use of parts M A M M (x )e x C A General solution is u x Ce x ft B x Ce ft B x Using x, to find C C 7 or x x 7e M A Method Auxiliar equation u C Ae x is the copleentar function M A For particular integral tr u P ax b, u P a Substituting u P ax b and u P a into the d.e. and coparing ters M M a ax b 9x a, b i.e. u P x A General solution is u x Ae x ft particular integral copleentar function provided particular integral has no arbitrar constants and copleentar function has one B x Ae ft B x Using x, to find A A 7 or x x 7e M A (i) λ Substituting λ into plane equation; i.e. λ OR an point on line (since given ) λ λ k λ M k λ 8 λ 8λ A UCLES 9795//SM/
7 7 UCLES 9795//SM/ [Turn over (ii) Working with vector. Substituting into the plane equation: k B M Solving a linear equation in : 9 9 M Q (,, ) A Shortest distance is or PQ 9 8 A (iii) Finding points in the plane: e.g. A(,, ), B(,, 8), C(,, ) M Then vectors in (// to) plane: e.g. 5, 5 5 9, BC AC AB OR B B for an two vectors in the plane M Vector product of an two of these to get noral to plane: 9 (an non-zero ultiple) M A d 9 (an position vector) 9 e.g. 9 x 9 z 9 CAO (aef) M A ALTERNATE SOLUTION ax b cz d contains λ λ λ and... so a aλ bλ b c cλ d and a b c d M B Then a b c d and a b c (λ ters) i.e. equating ters M Eliinating (e.g.) c fro st two equations 9a b M Choosing a, b 9 c and d 9 i.e. x 9 z 9 CAO M A
8 8 or M A (i) w ( ) ( ) 8 arg(w) tan 5 tan ( ) π (ii) (a) z (, π ), (, π ), (, or π ) w ; arg(w) 5, π π These ethod arks can be earned for just the first root 9 9 z ( ), (, π ), (, π ) (b) z, z, z the roots of z.z.z w z z z w ( ) i( ) ALT. Multipling the roots together in an for A arks for the nd & rd roots: r e^(iθ) fors equall acceptable M A M M A A M A (c) Three points in approx. correct places All equall spaced around a circle, centre O, radius (Explained that equilateral) ( π ) l sin or b the Cosine Rule M M A M A (d) k exp{ i. π} or exp{ i. π} or exp{ i. π} B UCLES 9795//SM/
9 9 (i) I n x n x x Correct splitting and use of parts M ( x ) x n. / / n ( x ) ( n ) x A n 5 x n x x Method to get nd integral of correct for n ( ) M n n.5 { I } n I n [i.e. reverting to I s in nd integral ft] M I n n.5 (n ) I n (n ) I n Collecting up I n s M (n ) I n 5 n (n ) I n AG A (ii) (a) Spiral (with r increasing) B Fro O to just short of θ π B (b) r θ dr dθ θ and dr 8 r θ θ M A dθ θ ( ) I L θ Now ( x ) M A / I B 5 I 5 9 or and ( ) 7 Use of given reduction forula M so that L or or awrt.7 ft onl fro suitable k I A UCLES 9795//SM/ [Turn over
10 Base-line case: for n 5, 579 R contains a string of (5 ) 7s B 579 R 58779, 579 R , etc. or for of st & last digits B Assue that, for soe k [ 5, 579 R k Induction hpothesis (k ) 7 s M Then, for n k, 579 R k 579(R k ) M Give the M ark for the ke observation that R k R k or k R k, even if not subsequentl used (k ) 7 s (k ) 7 s which contains a string of (k ) 7s, as required. Proof follows b induction (usual round-up). A A UCLES 9795//SM/
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