9795 FURTHER MATHEMATICS

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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Pre-U Certificate MARK SCHEME for the May/Jue series 9795 FURTHER MATHEMATICS 9795/ Paper (Further Pure Mathematics, maximum raw mark This mark scheme is published as a aid to teachers ad cadidates, to idicate the requiremets of the examiatio. It shows the basis o which Examiers were istructed to award marks. It does ot idicate the details of the discussios that took place at a Examiers meetig before markig bega, which would have cosidered the acceptability of alterative aswers. Mark schemes should be read i cojuctio with the questio paper ad the Pricipal Examier Report for Teachers. Cambridge will ot eter ito discussios about these mark schemes. Cambridge is publishig the mark schemes for the May/Jue series for most IGCSE, Pre-U, GCE Advaced Level ad Advaced Subsidiary Level compoets ad some Ordiary Level compoets.

2 Page Mark Scheme Syllabus Paper Pre-U May/Jue 9795 x 6x + (x + B 6 ( + ( x x ta 6 M ta x st A M A π π π 6 A [] 5 l( + x x x + x x + x +... from the Formula Book 5 5 l( x x x x x x... B + x l x { l( + x l( } x 5 {...} + x + x x tah x from the Formula Book A l( + x valid for < x ad so l( x is valid for x < so LHS valid for < x <, which matches the rage for RHS B M (i dy ( x. ( x ( x +.x ( x + x + ( x + + ( x ( x Use of quotiet rule; correct usimplified or clear explaatio this is < M A E [] ALT: y + x x + dy + ( x ( x + < [] Cambridge Iteratioal Examiatios

3 Page Mark Scheme Syllabus Paper Pre-U May/Jue 9795 (ii Asymptotes y Stated or clear from graph x ± Stated or clear from graph B B Crossig-poits (, ad (, Noted or clearly show o graph B B regios M All correct (icl. o TPs A [6] (i d d attempted i + 5j k (ALT: Use of scalar prods. & attempt to get compoets i terms of the rd M A (ii Sh. Dist. (b a. (b a ± ( i + j 7k (i + 5j k (i + 5j k ( i + j 7 k ft scalar prod. 8 8 cao 8 8 ft M B B B A [] 9 ALT: Solvig + λ µ k 5 to fid closest poits o lie, (, 6, 7 from λ ad (,, from µ givig k ad Sh.D. 8 [5] Cambridge Iteratioal Examiatios

4 Page Mark Scheme Syllabus Paper Pre-U May/Jue (i z z ( cos θ + i. si θ ( cos [ θ ] i. si[ θ ] cosθ i.si θ ( cosθ i. si θ De Moivre s Thm. used for at least z + i si θ Give aswer obtaied from correct uses of de Moivre s Thm. ad correct trig. M A [] (ii 5 z i si 5 θ M z 5 5 z 5z + z + 5 Use of biomial expasio M z z z 5 z 5 z + z 5 z z z Pairig up terms M i si 5θ i si θ + i siθ Use of (i s result ( M si 5 θ 6 si 5θ 6 5 si θ siθ A [5] 6 (i M M r + r siθ x + y + y Squarig ad cacellig: + y y + y + x y ( x A (ii Parabola All correct: Crossig-poits at (±, ad (, M A [] [] (iii π π ( siθ π dθ r dθ Recogitio that this is related to area M π ( x Matchig up with parabola-related regio M x x Igore ve aswer A [] Cambridge Iteratioal Examiatios

5 Page 5 Mark Scheme Syllabus Paper Pre-U May/Jue (i x + y ( x + y xy( x + y or equivalet M A [] (ii (a + β ( ad β ( 9 8 substd. ito (i s result ft + β 9 M A [] (b 9 8 t 7t + 8 (t (t 8, β, M A The 8 + β 9 ( ( + Explicit statemet required A [] 8 (i (a x G x G ad pre-multiplyig by this (or x i the case gives the result (NB Both directios must be dealt with B B [] (b Sice each xg i is distict, ad there are of them, the set xg is just a permutatio of the elemets of G OR metio that it is just a row of the group table ad hece cotais a permutatio of the elemets of G B [] (ii Multiply all elemets together: xg xg xg xg g g g g E (Sice G is abelia x.(g g g g (g g g g E Sice g g g g is a elemet of G, it has a iverse; Pre/post-mult g. by this iverse the gives x e E [] (iii (a Elemets may have a order which divides ito (is a factor of B [] (b Because the chage of the order of mults. i g.g g.g g.g g.g g.(g g g g is oly valid i a abelia group B [] Cambridge Iteratioal Examiatios

6 Page 6 Mark Scheme Syllabus Paper Pre-U May/Jue Reflectio i y x ta 8 π : Allow cos ( π ' s, etc. B Shear // y-axis, mappig (, to (, : B Rotatio through π clockwise about O: B Shear // x-axis, mappig (, to (, : B Multiplyig them together i this order (from right-to-left M A Reflectio i y x M A [8] NB Multiplyig the matrices i the reverse order scores max. B + M ; the B for correct ad M for Reflectio ad A for i x-axis NB Icorrect fial matrices automatically lose the last marks (a y k x cos x d y d y k x si x + k cos x ad k x cos x k si x Attempt at st ad d derivatives usig the Product Rule M Substitutig both of these ito the give DE M k x cos x k si x + k x cos x si x Comparig terms to evaluate k: k M A Aux. Eq. m + solved m ± i M A Comp. F. is y C A cos x + B si x ft Accept y C Ae ix + Be ix here B G. S. is y A cos x + B si x x cos x ft provided y P has o arb. costs. & y C has B Do ot accept fial aswer ivolvig complex umbers [8] Cambridge Iteratioal Examiatios

7 Page 7 Mark Scheme Syllabus Paper Pre-U May/Jue 9795 (b(i x, y & d y dy B x d y Differetiatig + y d y + xy 5x 9 M Use of Product Rule ad implicit differetiatio (at least oce d y + d y dy dy y + y + x + y dy 5 x dy dy FT 78 from ad also from istead of (both 78 M A A A [6] (b(ii Use of y y( + (x.y ( + (x.y ( + 6 (x.y ( + M + (x (x + (x + ft A Substitutig x. ito this series y(.. ft M A (i ( p iq ( p q + + i.pq B Comparig real ad imagiary parts: p q 6 ad pq 6 M Solvig simultaeously : p ± 8, q m i.e. ( 8 i 6 6i ± M A (ii (a Use of z ( + β + γ z + ( β + βγ + γ z ( βγ M [] [] A i, B 6i, C 8 i.e. f(z z ( iz + ( 6iz 8 A A A [] (b Differetiatig to get f (z z 8( iz + ( 6i OR z Az + B ft B 8 8i ± 6( i ( 6i Solvig z usig the quadratic formula M 6 z ( i 6i 6 ± ( i i 6 6i ± Use of (i s result (o the right thig: z ( i ± i(8 i A 5 + i or i M A [5] Cambridge Iteratioal Examiatios

8 Page 8 Mark Scheme Syllabus Paper Pre-U May/Jue 9795 (i y (x (x + e x, y (x (x + e x, y (x (8x + e x, y ( (x (6x + e x B B B B [] d y (ii Cojecture ( x +. e x Oe mark each: coefft. of x, costat B B [] (iii Diffferetiatig their cojectured expressio (must be liear e x M d + + y ( x +. e x + e x FT max / A A ( + x + ( +. (+ e x Show of correct form A Usual iductio roud-up/explaatio of proof, icludig clear demostratio that (+ th formula is i the right form. ( (i (a sech e + e θ ( e + e d (b ( dθ tahθ θ ( e θ e θ ( e θ + e ( e + e ( e + e ( e e ( e e ( e + e tah θ show legitimately sech θ from (a E M A M A [5] [] (ii (a I tah θ.tah θ tah dθ θ ( sech θ dθ M M tah θ I I I (tah M A tah ALT: I I θ ( tah θ dθ tah θ.sech θ dθ M M tah θ (tah M A [] Cambridge Iteratioal Examiatios

9 Page 9 Mark Scheme Syllabus Paper Pre-U May/Jue 9795 (ii (b I dθ l B (c ( I I + ( I I + ( I I ( I I + ( I I + ( I I r ( tah r r r ( r Use of the method of differeces M whe l A r [] I I r ( r r Cacellatio of terms i the summatio M I r ( r l AG A r Igorig method of differeces, but optig for a direct iterative approach scores max / M M A A As, I sice tah < r 5 ( ( ( ( l M r 5 7 r 7 E l ( r + r r Igorig method of differeces, but optig for a direct iterative approach scores max / M M A A A [[7] Cambridge Iteratioal Examiatios

De Moivre s Theorem - ALL

De Moivre s Theorem - ALL. Let x ad y be real umbers, ad be oe of the complex solutios of the equatio =. Evaluate: (a) + + ; (b) ( x + y)( x + y). [6]. (a) Sice is a complex umber which satisfies = 0,.