Prelim Examination 2011 / 2012 (Assessing Units 1 & 2) MATHEMATICS. Advanced Higher Grade. Time allowed - 2 hours

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Francine Carmella Golden
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1 Prlim Eamination / (Assssing Units & ) MATHEMATICS Advancd Highr Grad Tim allowd - hours Rad Carfull. Calculators ma b usd in this papr.. Candidats should answr all qustions. Full crdit will onl b givn whr th solution contains appropriat working

2 Answr all th qustions.. Find th cofficint of u v in th pansion of. u v. A curv C is dfind in trms of th paramtr t b th quations t 9t, t. (a) Find th quations of th tangnts to th curv C at th point (,). d (b) Find a formula for. Eprss our answr in th form t a bt ( ), ct dt whr a, b, c and d ar constants.. (a) Eprss i in polar form. n nπ nπ n (b) Us d Moivr s thorm to show that ( i) cos i sin. (c) Givn that z i is a root of th quation z ( i) z p qi, whr p and q ar ral numbrs, find th valus of p and q.. Us Gaussian limination to rduc th sstm of quations - α z - z z to uppr triangular form. Eplain what happns whn α. Find th solution corrsponding to α. [Turn ovr

3 . (a) Factoris. (b) Hnc valuat. a Eprss our answr in th form ln c, b whr a, b and c ar constants. 8. a, a, a ar th first thr trms of a gomtric squnc. Find: (a) th valu of th first trm and th common ratio of th squnc (b) th valu of th sith trm of th squnc (c) th sum to infinit of th gomtric sris (a ) (a ) (a ). Us implicit diffrntiation to find a formula for, in trms of and, givn that ln. Us our formula to find th gradint of th tangnt to th curv with quation ln at th point (, ). 8. (a) Show that sc θ tan θ. (b) Us th substitution scθ to show that 9 ( π). [Turn ovr

4 9. Prov b induction that, for all natural numbrs n, n n 8 is divisibl b.. Shown is part of th graph of f ( ),. Dtrmin algbraicall th rang of th function ( ),. f [END OF QUESTION PAPER]

5 Marking Schm Advancd Highr Grad / Prlim (Assssing Units & ) ans: 8, marks applis binomial thorm corrctl r r ( u ) r r v r simplifis corrctl r r u ( ) r r r v corrct valu for r r substituts corrctl ( ) 8, (a) ans: 9, 9 marks finds corrctl solvs corrctl for t whn solvs corrctl for t whn corrct gradint corrct quation corrct quation t t t 9t t,, t t, or or 9 9 or 9 (b) ans: 9t t ( t t) marks attmpts to substitut into corrct formula Substituts corrctl into quotint formula substituts corrctl into corrct formula Multiplis corrctl b dt/ attmpt to substitut into t t t t ( ) ( ) ( ) ( ) [ ( t) ] ( t t ) t( t t) ( t t ) 9t t( t t) Quotint formula Multipl b dt/

6 (a) π π ans: cos i sin π π or cos i sin marks corrct modulus corrct argumnt (b) ans: Proof mark corrct proof π ( i) n n n π π cos i sin nπ nπ cos i sin nπ nπ cos i sin n (c) ans: p, q substituts corrctl substituts corrctl simplifis corrctl solvs corrctl solvs corrctl marks π π cos i sin... π π... ( i ) cos isin... i i i p qi ( ) ( )( ) p q Crdit givn for corrct working with z lft in gomtric form. ( - ).. z z - p - q -

7 ans: 8 α α (or quivalnt) marks corrct augmntd matri corrct modifid sstm corrct modifid sstm ans: α sstm is inconsistnt mark ans:,, z mark corrct solution α α (or quivalnt) α (or quivalnt) 8 α α (or quivalnt) inconsistnt (accpt no solution or no answr ),, z

8 (a) ans: ( )( )( ) mark ( )( )( ) (or quivalnt) (b) ans: ln 8 8 marks starts corrctl; i.. attmpts to algbraic long division rwrits corrctl continus corrctl; i.. attmpts to us partial fractions continus corrctl rwrits corrctl intgrats corrctl substituts corrctl A B C A, B, C ( )( ) ( )( ) ( )( ) (or quivalnt) ln ln ln ( ln ln ln ) ( ln ln ln ) ln 8

9 (a) ans: a, r marks starts corrctl corrct first trm continus corrctl corrct common ratio a a a r a a (b) ans: marks knows how to find sith trm (c) ans: knows how to find sum to infinit marks

10 ans: marks rarrangs corrctl diffrntiats corrctl rarrangs corrctl ans: marks substituts corrctl... ln ln

11 8 (a) ans: Proof mark cos θ corrct proof sc θ cos θ cos θ sin θ tan θ cos θ 8 (b) ans: Proof marks substituts corrctl changs limits corrctl simplifis corrctl intgrats corrctl substituts corrctl simplifis corrctl 9 ans: Proof marks 9 tan θ scθ tanθdθ scθ π... π dθ tanθ θ ( sc θ ) [ ] π π π π tan tan π π π π ( π ) π starts corrctl; i.. provs rsult is tru for (.g.) n stats corrct assumption for n k stats corrct rsult for n k (E.g.) 8 which is divisibl b Tru for n k k k k 8 is divisibl b ; i.. 8 t (for t Z ) 8 k k is divisibl b ; i.. k k 8 s (for s Z ) continus proof corrctl

12 complts proof corrctl 8 k k ( ) k k k k k k 8( t) 8 8 k k k ( 8t) ( 8t ) ( ) k ( t ) s k 8 Sinc rsult is tru for n and, sinc (rsult is tru for n k) (rsult is tru for n k ), th rsult is tru for all natural numbrs n ans: & marks knows to diffrntiat diffrntiats corrctl knows to find stationar points; i.. sts f ( ) & attmpts to solv solvs corrctl for solvs corrctl for solvs corrctl for f ( )... ( ) ( )( ) ( )..., & (accpt & ) TOTAL MARKS

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark.

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark. . (a) Eithr y = or ( 0, ) (b) Whn =, y = ( 0 + ) = 0 = 0 ( + ) = 0 ( )( ) = 0 Eithr = (for possibly abov) or = A 3. Not If th candidat blivs that = 0 solvs to = 0 or givs an tra solution of = 0, thn withhold