MATHEMATICS Unit Further Pure 2

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1 Geeral Certificate of Educatio Jauary 008 Advaced Level Examiatio MATHEMATICS Uit Further Pure MFP Thursday Jauary am to 0.0 am For this paper you must have: * a 8-page aswer book * the blue AQA booklet of formulae ad statistical tables. You may use a graphics calculator. Time allowed: hour 0 miutes Istructios * Use blue or black ik or ball-poit pe. Pecil should oly be used for drawig. * Write the iformatio required o the frot of your aswer book. The Examiig Body for this paper is AQA. The Paper Referece is MFP. * Aswer all questios. * Show all ecessary workig; otherwise marks for method may be lost. Iformatio * The maximum mark for this paper is 75. * The marks for questios are show i brackets. Advice * Uless stated otherwise, you may quote formulae, without proof, from the booklet. P9797/Ja08/MFP 6/6/ MFP

2 Aswer all questios. (a) Express þ i i the form re iy, where r>0 ad p < yp. ( marks) (b) Solve the equatio z 5 ¼ þ i givig your aswers i the form re iy, where r>0 ad p < yp. (5 marks) (a) Show that (b) ðr þ Þ ðr Þ ¼ r þ Hece, usig the method of differeces, show that P r¼ r ¼ ð þ Þð þ Þ 6 ( marks) (6 marks) A circle C ad a half-lie L have equatios p j z ffiffiffi i j¼ ad argðz þ iþ ¼ p 6 respectively. (a) Show that: (i) the circle C passes through the poit where z ¼ i; ( marks) (ii) the half-lie L passes through the cetre of C. ( marks) (b) O oe Argad diagram, sketch C ad L. ( marks) (c) Shade o your sketch the set of poits satisfyig both p j z ffiffiffi i j ad 0 argðz þ iþ p 6 ( marks) P9797/Ja08/MFP

3 The cubic equatio z þ iz þ z ð þ iþ ¼0 has roots a, b ad g. (a) Write dow the value of: (i) a þ b þ g ; ( mark) (ii) ab þ bg þ ga ; ( mark) (iii) abg. ( mark) (b) Fid the value of: (i) a þ b þ g ; ( marks) (ii) a b þ b g þ g a ; ( marks) (iii) a b g. ( marks) (c) Hece write dow a cubic equatio whose roots are a, b ad g. ( marks) 5 Prove by iductio that for all itegers 5 X r¼ ðr þ Þðr!Þ ¼ð þ Þ! (7 marks) Tur over for the ext questio P9797/Ja08/MFP Tur over s

4 6 (a) (i) By applyig De Moivre s theorem to ðcos y þ i si yþ, show that cos y ¼ cos y cos y si y ( marks) (ii) Fid a similar expressio for si y. ( mark) (iii) Deduce that ta y ¼ ta y ta y ta y ( marks) (b) (i) Hece show that ta p is a root of the cubic equatio x x x þ ¼ 0 ( marks) (ii) Fid two other values of y, where 0 < y < p, for which ta y is a root of this cubic equatio. ( marks) (c) Hece show that ta p 5p þ ta ¼ ( marks) 7 (a) Give that y ¼ l tah x, where x>0, show that dy dx ¼ cosech x (6 marks) (b) A curve has equatio y ¼ l tah x, where x>0. The legth of the arc of the curve betwee the poits where x ¼ ad x ¼ is deoted by s. (i) Show that s ¼ ð coth x dx ( marks) (ii) Hece show that s ¼ lð cosh Þ. ( marks) END OF QUESTIONS Copyright Ó 008 AQA ad its licesors. All rights reserved. P9797/Ja08/MFP

5 AQA Further pure Ja 008 Aswers Questio : a)+ i + i e b) Let's write z ( re ) r e i 5 i 5 i5θ 5 iθ 5 5 i5θ + i becomes e 5 so r ad 5θ + k r ad θ + k k,, 0,, r ad θ,,,, th The5 roots of + i are : e z re, i i i i i e, e, e, e Part (a) was doe well by the majority of cadidates. However, resposes to part (b) were less successful. A umber of cadidates gave the roots of z 5 as their aswer to part (b), whilst others left the modulus of the roots as istead of, ad others agai gave solutios outside the rage of θ as specified i the questio. Some cadidates yet agai were either uable to hadle + k or, whe takig the fifth root of 5 + k i 5 wrote e e k + i, Questio : (r+ ) (r ) (8r + r + 6r+ ) (8r r + 6r ) r + ( + ) ( + ) b) r (r ) (r ) (r ) (r ) r + + (+ )( ( + )( + ) 6 r ) Agai part (a) was aswered well, but solutios to part (b) were mixed. Geerally speakig, the best solutios came from cadidates who rewrote part (a) as r ( ( r+ ) ( r ) ) before makig their summatio. Those cadidates who preferred to use part (a) i the form i which it was prited either forgot to sum the s to make or oly partially divided by. A small umber of cadidates used the method of iductio either through cofusig the two methods of summatio or by deliberately choosig a alterative method. Either way, o credit could be give.

6 Questio : ) ) ( ) 6 ai i i i + + b) c) The circle C passes through the poit where z i ii)the cetre of C is the poit where z + i arg( z+ i) arg( + i+ i) arg( + i) Ta. 6 The half-lie L passes through the cetre of C. Lack of clear evidece that cadidates uderstood what they were doig i part (a) caused a loss of marks for this part of the questio. Methods varied. Those cadidates who tured this part of the questio ito a coordiate geometry exercise probably provided the clearest solutios. Those cadidates who evaluated i ad arg( + i) provided less covicig solutios ad i some cases evidet error. For istace it was ot ucommo to see i writte as ( i) Part (b) o the whole was doe well except that i some istaces ot all the results of part (a) were icorporated i the cadidate s Argad diagram. Part (c) was doe well but, if a mistake did occur, it was almost always that the shaded area would be bouded by the real axis rather tha by a lie parallel to the real axis through the poit represeted by the complex umber z i. Questio : z iz z i has roots + + ( + ) 0 αβγ,,. ai )) α + β + γ i ii) αβ + αγ + βγ iii) αβγ + i biα + β + γ α + β + γ αβ + αγ + βγ ) ) ii α β γ + + i 7 ) α β + α γ + β γ ( αβ + αγ + βγ ) ( α βγ + β αγ + γ αβ ) z z i (7 + ) z ( ( αβ αγ βγ ) αβγ ( α β γ ) i 0 ) ( + i)( i) 9 + i+ i αβ + αγ + βγ 7+ i iii) α β γ ( αβγ ) ( + i) α β γ i c) z ( 7) z + (7 + i) z i 0 This questio was probably the most popular questio o this paper ad certaily showed cadidates well prepared to aswer questios o this part of the specificatio. There were may fully correct solutios or correct apart from the odd sig error, the most commo of which was to write dow ( i) as + istead of. If there was a major loss of marks, it was usually i the iability of a cadidate to evaluate Σα β ad i this case the cadidate started by cosiderig ( α ) oly to fid that the evaluatio of Σα posed a serious problem.

7 Questio 5: The propositio P, for all, ( r + )( r!) ( + )! is to be prove by iductio Base case:, (!) (!) r + r + ad ( + )!! the propositio P is true Let's supppose that P is true, ie ( r k k + Let's show that Pk + is true,let's show that ( r + )( r!) ( k+ )( k+ )! k+ k (!) (!)! kk ( + )! + k + k+ ( k+ )! k + )( r!) kk ( + )! r + r r + r + k+ + k+ ( k+ )! k+ k + k+ ( k+ )! k + k+ ( k+ )!( k+ ) ( k+ ) Resposes to this questio varied cosiderably. It was ot, i geeral, that cadidates did ot uderstad the method of iductio but rather that the algebraic maipulatio especially i the hadlig of factorials proved to be a stumblig block. For istace k (k +)! would be writte as (k + k )! ad i a sigificat umber of solutios cadidates, havig maaged to reach (k +)(k + )(k +)!, abadoed their solutios ot realisig that the result was, i fact, (k +)(k + )!. ( k+ )! ( k+ ) Coclusio : If the propositio is true for k, the it is true for k+ ad because it true for, we ca coclude accordig to the iductio pricipal that the propositio is true for all. For all, ( r + )( r!) ( + )! Questio 6: a)) i Cosθ + isiθ Cosθ + isiθ ad also Cosθ + isiθ Cos θ + icos θ Siθ Cosθ Si θ isi θ ( Cos θ Cosθ Si θ) i ( Cos θ Siθ Si θ) + By idetifyig the real ad imagiary parts, we have Cos Cos Cos Si θ θ θ θ ii) Siθ Cos θ Siθ Si θ Si Cos Si Si iii) Taθ Cos Cos Cos Si θ θ θ θ θ θ θ θ Now divide the umerator ad the deomiator by Cos θ Siθ Si θ Cos Cos Ta Ta Ta Ta Ta θ θ θ θ θ θ θ Si θ Ta θ Ta θ Cos θ Resposes to this questio were rather disappoitig. I part (a)(i), although most cadidates correctly quoted cosθ + i si θ (cosθ + i siθ), some immediately wet o to use the multiple agle formulae istead of expadig (cosθ + i siθ). Some of those cadidates who expaded (cosθ + i siθ) did ot seem to realise that the aswers to parts (a)(i) ad (a)(ii) were obtaied by simply equatig real ad imagiary parts. Other cadidates wrote i as +i ad so were uable to reach the correct result of part (a)(ii) ad the prited result i part (a)(iii). Eve those cadidates who worked parts (a)(i) ad (a)(ii) correctly i terms of siθ ad cosθ, havig writte Siθ, did ot realise Cosθ that the divisio of umerator ad deomiator by cosθ would give the prited result, but rather chose to use si θ + cos θ to express umerator ad deomiator i a differet form, with o hope of reachig the prited result.

8 Questio 6:cotiues b) i) For θ, we have Ta Ta Ta. Let x be ta Ta x x the x x x x x x x+ 0 ii) the other values of θ ca be obtaied by solvig Taθ : θ + k θ + k for k 0, θ 5 k, θ 9 k, θ c)the three roots of the equatio x x x+ 0 5 are α Ta, β Ta, γ Ta Although the questio i part (b)(i) started with the word hece few cadidates took up the hit ad replaced θ by i part (a)(iii). If this part was attempted it was ofte doe by solvig the cubic equatio i x to fid its three roots ad the by quotig that Ta ad a correspodig result 5 for Ta ad cosequetly usig these results i part (c), a method ot idicated by the questio. α + β + γ 5 Ta + Ta 5 Ta + Ta Questio 7: x a) y l Tah x > 0 sec x h x Cosh dy dx x x x Tah Cosh Sih dy dx x x x Cosh Sih Sih( ) Sihx dy Cosech x dx Although may cadidates were able to write dow x tah multiplied by sech x, fewer were able to combie these results to obtai cosech x. Eve those cadidates who expressed dy dx etirely i terms of x x cosh ad sih seemed to baulk at the algebra which led to x x sih cosh.

9 Questio 7:cotiues dy b) i) s + dx + Cosech x dx Sih x + Cosh x + Cosech x + Coth x Sih x Sih x Sih x s dx Coth x dx Cosh x ii) s Coth x dx dx l ( Sih x) Sih x s l(sih ) l(sih) Sih s l Sih usig Sihx sih x cosh x, we have sih sihcosh sihcosh s l sih s l cosh Part (b)(i) was doe well ad may cadidates were able to arrive at s l sih l sih i part (b)(ii) but were uable to reach the prited aswer. If the itegral of coth x was performed icorrectly, it was ofte by coth x beig replaced by followed by l tah x or l cosh x as the tah x itegral. Grade boudaries

10 MFP - AQA GCE Mark Scheme 008 Jauary series Key to mark scheme ad abbreviatios used i markig M m or dm A B E mark is for method mark is depedet o oe or more M marks ad is for method mark is depedet o M or m marks ad is for accuracy mark is idepedet of M or m marks ad is for method ad accuracy mark is for explaatio or ft or F follow through from previous icorrect result MC mis-copy CAO correct aswer oly MR mis-read CSO correct solutio oly RA required accuracy AWFW aythig which falls withi FW further work AWRT aythig which rouds to ISW igore subsequet work ACF ay correct form FIW from icorrect work AG aswer give BOD give beefit of doubt SC special case WR work replaced by cadidate OE or equivalet FB formulae book A, or (or 0) accuracy marks NOS ot o scheme x EE deduct x marks for each error G graph NMS o method show c cadidate PI possibly implied sf sigificat figure(s) SCA substatially correct approach dp decimal place(s) No Method Show Where the questio specifically requires a particular method to be used, we must usually see evidece of use of this method for ay marks to be awarded. However, there are situatios i some uits where part marks would be appropriate, particularly whe similar techiques are ivolved. Your Pricipal Examier will alert you to these ad details will be provided o the mark scheme. Where the aswer ca be reasoably obtaied without showig workig ad it is very ulikely that the correct aswer ca be obtaied by usig a icorrect method, we must award full marks. However, the obvious pealty to cadidates showig o workig is that icorrect aswers, however close, ear o marks. Where a questio asks the cadidate to state or write dow a result, o method eed be show for full marks. Where the permitted calculator has fuctios which reasoably allow the solutio of the questio directly, the correct aswer without workig ears full marks, uless it is give to less tha the degree of accuracy accepted i the mark scheme, whe it gais o marks. Otherwise we require evidece of a correct method for ay marks to be awarded.

11 MFP - AQA GCE Mark Scheme 008 Jauary series MFP Q Solutio Marks Total Commets (a) (b) Ay method for fidig r or θ M r, θ AA i 5 z e M eeds some referece to a +k i i k i e 0 + M A for r z 5 icorrect r, θ part (a) AF A for θ AF z i 9 i 7 i 0 0, 0 7 i 5 i 0 0 e, e e, e, e A,,0 F 5 0 Accept r i ay form eg Correct but some aswers outside rage allow A ft icorrect r, θ i part (a) (a) Attempt to expad ( r ) ( r ) (b) r r ( r + ) or ( ) Total 8 + M r expaded A r + A AG MA rows see r r + A Do ot allow M for ( + ) equal to aythig ot r M M for multiplicatio of bracket or takig + out as a factor r A CAO r ( + )( + ) A 6 AG 6 Total 9

12 MFP - AQA GCE Mark Scheme 008 Jauary series MFP (cot) Q Solutio Marks Total Commets (a)(i) z i i + M A i (ii) Cetre of circle is + i B Do ot accept (,) uless attempt to solve usig trig Substitute ito lie M arg ( + i) show 6 A (b) Circle: cetre correct 0, through Half lie: through ( 0, ) B B B through cetre of circle B (c) Shadig iside circle ad below lie BF Bouded by y B Total (a)(i) α i B (ii) αβ B (iii) αβγ + i B (b)(i) α ( α ) αβ used M Allow if sig error or missig ( i) AF 7 AF ft errors i (a) (ii) α β ( αβ ) αβ βγ M Allow if sig error i missig ( αβ ) αβγ α A 9 + i i AF ft errors i (a) 7+ i AF ft errors i (a) α β γ + i i M AF ft sig error i αβγ (iii) z + 7z + 7+ i z i 0 BF (c) BF Total Correct umbers i correct places Correct sigs 5

13 MFP - AQA GCE Mark Scheme 008 Jauary series MFP (cot) Q Solutio Marks Total Commets 5 Assume result true for k k + r + r! The (( k ) )( k ) k( k ) + + +! + +! MA Takig out ( k + )! as factor m ( k! ) ( k k k) A ( k )( k ) + +! A k show ( + )!! B P P ad k k+ 6(a)(i) P true E 7 If all 6 marks eared Total 7 cosθ + isi θ cosθ + isiθ M cos θ + icos θsiθ + i cosθsi θ + isiθ A Real parts: cosθ cos θ cosθsi θ A AG (ii) Imagiary parts: si θ cos θ siθ si θ AF (iii) siθ taθ cosθ M Used cos θsiθ si θ cos θ si θcosθ AF Error i si θ taθ ta θ ta θ ta θ taθ ta θ A AG (b)(i) ta B Used (possibly implied) x x ta is a root of x M Must be hece x x x+ 0 A (ii) 5 9 Other roots are ta, ta BB (c) 5 9 ta + ta + ta M 5 ta + ta A Total Must be hece 6

14 MFP - AQA GCE Mark Scheme 008 Jauary series MFP (cot) Q Solutio Marks Total Commets 7(a) dy dx x tah B x sech B B sih x M cosh x cosh x sih x cosh x sih x cosech x A 6 AG OE ie expressig i sih x ad cosh x M ie use of sih A sih Acosh A Alterative l sih x l cosh x cosh x sih x sih x cosh x cosh x sih x sih x cosh x Use of siha sih A cosh A result (B) (BB) (M) (M) (A) (b)(i) s + cosech x dx M coth x d x A AG (ii) [ l sih ] s x M eeds to be correct l sih l sih A sihcosh l AF must be see sih l ( cosh) A AG Total TOTAL 75 7

Version 1.0: abc. General Certificate of Education. Mathematics MFP2 Further Pure 2. Mark Scheme examination - January series

Version 1.0: abc. General Certificate of Education. Mathematics MFP2 Further Pure 2. Mark Scheme examination - January series Versio.0: 008 abc Geeral Certificate of Educatio Mathematics 660 MFP Further Pure Mark Scheme 008 examiatio - Jauary series Mark schemes are prepared by the Pricipal Examier ad cosidered, together with