9795 FURTHER MATHEMATICS

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Pre-U Certificate www.xtremepapers.

1 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Pre-U Certificate MARK SCHEME for the Ma/June question paper for the guidance of teachers 9795 FURTHER MATHEMATICS 9795/ Paper (Further Pure Mathematics), maximum raw mark This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptabilit of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the Ma/June question papers for most IGCSE, Pre-U, GCE Advanced Level and Advanced Subsidiar Level sllabuses and some Ordinar Level sllabuses.

2 Page Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June 9795 n ( r r + ) r r + r n n r r n Splitting summation and use of given results r n n + )(n + ) n( n + ) + n st for Σr ; nd for Σr & Σ n B B 6 ( n n + ) legitimatel A [] ( A k ( + cosθ ) dθ including squaring attempt; ignore limits and k ( + sin θ ) dθ for use of the double-angle formula 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 terms A π + A [] d (i) (sinh x) (sinh x) d. cosh x OR sinh x cosh x + A A [] (ii) d +. B Sep g. Vars. in (i) s answer x + d A But x sinh t so + t sinh (t ) + C condone missing + C A ALT. Set t sinhθ, t dt coshθ dθ Full substn. t coshθ dt dθ A + t.dθ θ sinh (t ) A + sinh θ ALT. Set t tanθ, t dt sec θ dθ Full substn. t sec θ dt dθ A + t sec θ.dθ + tan θ ln secθ + tanθ ln t + + t A [] Universit of Cambridge International Examinations

3 Page Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June 9795 (i) x + x +.x x + ( ) Creating a quadratic in x For real x, ( ) Considering the discriminant Creating a quadratic inequalit For real x, (6 + )( ) Factorising/solving a -term quadratic 6 cso NB lack of inequalit earlier with unjustified correct answer loses onl the rd M mark A [5] (ii) substd. back (x x + ) x [ ] A substd. back 6 (x + 6x + 9) x [ 6 6 ] A Allow alternative approach via calculus: d x x + Solving quadratic to find values of x ( x + ) Then A A each pair of correct (x, ) coordinates [] 5 (i) cosα sinα (a) sinα cosα cos β sin β (b) sin β cos β B B [] (ii) cosφ sinφ sinφ cosθ cosφ cosθ Multn. of reflection matrices. Correct order. cos φ cosθ + sinφ cosφ sinφ cosθ sinφ cosθ cosφ cosφ cosθ + sinφ cos( φ θ ) sin( φ θ ) sin( φ θ ) cos( φ θ ) giving a Rotation (about O) Use of the addition formulae; correctl done through (φ θ ) acw [or (θ φ ) cw] A A Those who get the initial matrices in the wrong order, can get 5/6, losing onl that M mark [6] Universit of Cambridge International Examinations

4 Page Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June (i) Possible orders are,,,, 6 & B B Lagrange s Theorem, the order of an element divides the order of the group (since the order of an element the order of the subgroup generated b that element) B [] (ii) E.g. xx. x xx. x b x. x. x.. b x. x. ( ) [b ] x. (x ) x. e b M s for first, correct uses of different conditions; the A for the rd condition used to clinch the result. SPECIAL CASE Allow / for those who correctl argue the converse A [] (iii) Proving G not abelian: [e.g. b xx but x e] G not cclic OR establishing a contradiction B B [] 7 (i) cosθ + i (c + is) Use of de Moivre s Theorem c + c.is + 6c.i s + c.i s + i s Binomial expansion attempted cosθ c 6c s + s and c s cs Equating Re & Im parts sin θ c s cs tanθ cos θ c 6c s + s t t Dividing throughout b c to get 6t + t legitimatel A [5] (ii) t tanθ B 5 9 ( ) + 9 π + tan 9 9 tan 9 Noting that this is tan( tan ) so that tan 5 5 π + tan 9 A A [] Universit of Cambridge International Examinations

5 Page 5 Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June (i) Subst g. x, f() and f () into (*) f () 5 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 A Subst g. x, f () and f () 5 into this f () ft their f () A [] (iii) f(x) f() + f ()(x ) + f ()(x ) + 6 f ()(x ) + Use of the Talor series + (x ) + 5 (x ) (x ) + st two terms cao; nd two terms ft (i) & (ii) s answers A A [] (iv) Subst g. x. f(.). to d.p. cso (i.e. exactl this answer) A [] 9 (i) d + x is a Bernouilli (differential) equation u du d B Then d + x d du becomes 9x u 9x AG A [] (ii) Method IF is e x ue x 9xe x x xe e x Use of parts A (x + )e x + C A Gen. Soln. is u x + + Ce x ft B x + + Ce ft B x Using x, to find C C 7 or x x + + 7e A [9] Universit of Cambridge International Examinations

6 Page 6 Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June 9795 Method Aux. Eqn. m u C Ae x is the Comp. Fn. A For Part. Intgl. tr u P ax + b, u P a Subst g. u P ax + b and u P a into the d.e. and comparing terms a ax b 9x a, b i.e. u P x + A Gen. Soln. is u x + Ae x ft PI + CF provided PI has no arbitrar constants and CF has one B x + + Ae ft B x Using x, to find A A 7 or x x + + 7e A [9] (i) + λ Subst g. + λ into plane equation; i.e. + 6λ OR an point on line (since given ) + λ + λ 6 k + 6λ k + 6λ + 8 λ λ 6 A [] (ii) + m Working with vector 6m. m + m Subst g. into the plane equation: + 6m 6 k m Solving a linear equation in m: + m + 6m + 9m 9 6 m Q (6, 6, ) B A Sh. Dist. is m 6 or PQ Alternate methods that find onl Sh. Dist. but not Q can score A onl A [5] Universit of Cambridge International Examinations

7 Page 7 Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June 9795 (iii) Finding points in the plane: e.g. A(,, ), B(,, 8), C(,, ) Then vectors in (// to) plane: e.g. AB, 6 OR B B for an two vectors in the plane 9 AC 5, 5 6 BC 5 Vector product of an two of these to get normal to plane: (an non-zero multiple) 9 A d 9 (an position vector) 9 e.g. 9 x 9 + z 9 cao (or an correct equivalent form) ALTERNATE SOLUTION A [6] ax + b + cz d contains + λ + λ + 6λ and... so a + aλ + bλ b + c + 6cλ d and a + b c d B Then a b + c d and a + b + 6c (λ terms) i.e. equating terms Eliminating (e.g.) c from st two eqns. 9a + b Choosing a, b 9 c and d 9 i.e. x 9 + z 9 cao A [6] Universit of Cambridge International Examinations

8 Page 8 Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June 9795 or A (i) w ( ) + ( + ) arg(w) tan 5 tan ( + ) π (ii) (a) z (, π ), (, π ), (, or π ) π A [] w ; arg(w) 5, π These method marks can be earned for just the first root 9 9 z ( ), (, π ), (, π ) 6 6 A marks for the nd & rd roots: 6 r e^(iθ) forms equall acceptable A A [] (b) z, z, z the roots of z.z +.z w z z z w ( ) + i( + ) ALT. Multipling the roots together in an form (c) Three points in approx. correct places A [] All equall spaced around a circle, centre O, radius (Explained that equilateral) ( π ) l sin 6 or b the Cosine Rule (d) k exp{ i. π} or exp{ i. π} or exp{ i. π} B A A [5] [] Universit of Cambridge International Examinations

9 Page 9 Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June 9795 (i) I n x n x 6 + x Correct splitting and use of parts ( 6 + x ) x n. / / n ( 6 + x ) ( n ) x A n 5 x n 6 + x 6 + x Method to get nd integral of correct form n ( ) n n.5 { 6 I } n + I n [i.e. reverting to I s in nd integral ft] I n n.5 6(n ) I n (n ) I n Collecting up I n s (n + ) I n 5 n 6(n ) I n AG A [6] (ii) (a) Spiral (with r increasing) B From O to just short of θ π B [] (b) r θ dr dθ θ and dr 8 6 r + θ + θ A 6 dθ θ ( ) I L 6 + θ Now ( + x ) A / 6 I 6 B 5 I or and ( 6 ) 7 Use of given reduction formula so that L or 6 or awrt.7 ft onl from suitable k I 6 A NB The last marks can be earned b integrating in a variet was [7] Universit of Cambridge International Examinations

10 Page Mark Scheme: Teachers version Sllabus Paper Pre-U Ma/June 9795 Base-line case: for n 5, 579 R contains a string of (5 ) 7 s B 579 R , 579 R , etc. or form of st & last digits B Assume that, for some k 5, 579 R k Induction hpothesis (k ) 7 s Then, for n k +, 579 R k + 579(R k + ) Give the M mark 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 + ) 7 s, as required. Proof follows b induction (usual round-up). A E [6] Universit of Cambridge International Examinations

Paper 1 Further Pure Mathematics For Examination from 2016 SPECIMEN MARK SCHEME 3 hours MAXIMUM MARK: 120

Paper 1 Further Pure Mathematics For Examination from 2016 SPECIMEN MARK SCHEME 3 hours MAXIMUM MARK: 120 Cabridge International Exainations Cabridge Pre-U Certificate www.xtreepapers.co FURTHER MATHEMATICS (PRINCIPAL) 9795/ Paper Further Pure Matheatics For Exaination fro SPECIMEN MARK SCHEME hours MAXIMUM