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1 Paper Reference(s) 6666/0 Edexcel GCE Core Mahemaics C4 Silver Level S4 Time: hour 30 minues Maerials required for examinaion papers Mahemaical Formulae (Green) Iems included wih quesion Nil Candidaes may use any calculaor allowed by he regulaions of he Join Council for Qualificaions. Calculaors mus no have he faciliy for symbolic algebra manipulaion, differeniaion and inegraion, or have rerievable mahemaical formulas sored in hem. Insrucions o Candidaes Wrie he name of he examining body (Edexcel), your cenre number, candidae number, he uni ile (Core Mahemaics C4), he paper reference (6666), your surname, iniials and signaure. Informaion for Candidaes A bookle Mahemaical Formulae and Saisical Tables is provided. Full marks may be obained for answers o ALL quesions. There are 8 quesions in his quesion paper. The oal mark for his paper is 75. Advice o Candidaes You mus ensure ha your answers o pars of quesions are clearly labelled. You mus show sufficien working o make your mehods clear o he Examiner. Answers wihou working may gain no credi. Suggesed grade boundaries for his paper: A* A B C D E Silver 4 This publicaion may only be reproduced in accordance wih Edexcel Limied copyrigh policy Edexcel Limied.

2 . (a) Find he binomial expansion of ( 8x), x < 8, in ascending powers of x up o and including he erm in x 3, simplifying each erm. (4) 3 (b) Show ha, when x =, he exac value of ( 8x) is () (c) Subsiue x = 00 ino he binomial expansion in par (a) and hence obain an approximaion o 3. Give your answer o 5 decimal places. (3) January 00. f (x) = (9 + 4x 3, x <. ) Find he firs hree non-zero erms of he binomial expansion of f(x) in ascending powers of x. Give each coefficien as a simplified fracion. (6) June 0 Silver 4: 8/

3. Figure A hollow hemispherical bowl is shown in Figure. Waer is flowing ino he bowl. When he deph of he waer is h m, he volume V m 3 is given by V = π h (3 4h), 0 h 0.5.

3 3. Figure A hollow hemispherical bowl is shown in Figure. Waer is flowing ino he bowl. When he deph of he waer is h m, he volume V m 3 is given by V = π h (3 4h), 0 h 0.5. (a) Find, in erms of π, dv dh when h = 0.. (4) π Waer flows ino he bowl a a rae of m 3 s. 800 (b) Find he rae of change of h, in m s, when h = 0.. () June 0 Silver 4: 8/ 3

4 4. Figure Figure shows he curve wih equaion y = x, x 0. 3x + 4 The finie region S, shown shaded in Figure, is bounded by he curve, he x-axis and he line x =. The region S is roaed 360 abou he x-axis. Use inegraion o find he exac value of he volume of he solid generaed, giving your answer in he form k ln a, where k and a are consans. (5) January 0 Silver 4: 8/ 4

5 5. Figure Figure shows a skech of par of he curve C wih parameric equaions x =, y =. The curve crosses he y-axis a he poin A and crosses he x-axis a he poin B. (a) Show ha A has coordinaes (0, 3). (b) Find he x-coordinae of he poin B. (c) Find an equaion of he normal o C a he poin A. () () (5) The region R, as shown shaded in Figure, is bounded by he curve C, he line x = and he x-axis. (d) Use inegraion o find he exac area of R. (6) January 03 Silver 4: 8/ 5

6 6. A curve has parameric equaions x = an, y = sin, 0 < < π. (a) Find an expression for dy in erms of. You need no simplify your answer. (3) (b) Find an equaion of he angen o he curve a he poin where = 4 π. Give your answer in he form y = ax + b, where a and b are consans o be deermined. (5) (c) Find a caresian equaion of he curve in he form y = f(x). (4) June A curve is described by he equaion x + 4xy + y + 7 = 0 (a) Find d y in erms of x and y. (5) A poin Q lies on he curve. The angen o he curve a Q is parallel o he y-axis. Given ha he x-coordinae of Q is negaive, (b) use your answer o par (a) o find he coordinaes of Q. (7) June 03 Silver 4: 8/ 6

7 8. (a) Using he ideniy cos θ = sin θ, find sin θ d θ. () Figure 4 Figure 4 shows par of he curve C wih parameric equaions x = an θ, y = sin θ, 0 θ < π. The finie shaded region S shown in Figure 4 is bounded by C, he line x = and he x-axis. 3 This shaded region is roaed hrough π radians abou he x-axis o form a solid of revoluion. (b) Show ha he volume of he solid of revoluion formed is given by he inegral where k is a consan. π k 6 sin 0 θ d θ, (c) Hence find he exac value for his volume, giving your answer in he form pπ + qπ 3, where p and q are consans. (3) (5) June 009 END TOTAL FOR PAPER: 75 MARKS Silver 4: 8/ 7

8 Quesion Number Scheme Marks Q (a) ( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) 3 3 8x = + 8x + 8x + 8x + M A 3! 3 = 4x 8 x ; 3x A; A (4) (b) ( x) cso 8 8 = = = = M A () 3 3 (c) 4x 8x 3x = 4( 0.0) 8( 0.0) 3( 0.0) = = M cao 3 = M = A (3) [9]. ( ) ( ) f x = M ( kx ) ( )... = n nkx... 3, 3 or + = + + n no a naural number, k ( kx ) ( )( ) ( kx ) 3 + =... + f heir k 9 B M A f x = x + x A f = A (6) [6] 4 ( x) x x Silver 4: 8/ 8

9 3. (a) dv dh h h = π π or equivalen M A h =, ( ) ( ) A 0. d V = π 0. π 0. = 0.04π dh π 5 M A (4) dh dv dv π (b) = = d d dh 800 πh πh or π heir (a) 800 M A h = 0., h π = 5 = d 800 π 3 awr 0.03 A () [6] 4. Volume So Volume x Use of 3x = π ( ) ( ) ln 3 π x 4 = + 3 = ( π ) ln6 ln = π ln =. B V π y ( x ) ± kln M ln ( 3 x + 4 ) A 3 Subsiues limis of and 0 and subracs he correc way dm round. ln 4 3 π or ln 3 π A oe isw [5] (5 marks) Silver 4: 8/ 9

10 Quesion Number Scheme Marks 5. Working paramerically: ln x =, y = or y = e (a) { x = 0 } 0 = = Applies x = 0 o obain a value for. M A When =, y = = 3 Correc value for y. [] (b) { y } (c) = 0 0 = = 0 dy ln = Applies y = 0 o obain a value for. (Mus be seen in par (b)). When = 0, x = (0) = x = A [] d = and eiher dy ln d = or B dy ln e ln d = Aemps heir d y divided by heir d A A, = "", so m( T) = 8ln m( N ) = 8ln Applies = "" and d x. d m( N) = m( T) y 3 = ( x 0) or y = 3 + x or equivalen. M A oe 8ln 8ln cso [5] Area( R) =. d Complee subsiuion for boh y and M x = = 4 and x = = 0 B Eiher ln ( ) or ( ) M* = ± α (ln ) ln ± α (ln )( ) (d) ( ) ln = ln ln 4 5 = ln or ( ) ( ) M M M A ln Depends on he previous mehod mark. Subsiues heir changed limis in and subracs eiher way round. dm* or equivalen. A 5 ln [6] 5 Silver 4: 8/ 0

11 Quesion Number 6. (a) x Scheme y = = an, sin Marks d y d = (an )sec, d cos = Correc d x d and d y d B dy = d cos x ansec (b) When π, 4 = 4 cos = sin x =, y = (need values) π π dy cos 4 When =, m(t) = = π 4 an sec π 4 4 The poin ( ) ± cos d heir x d + cos d heir x, or (, awr 0.7 ) These coordinaes can be implied. π y = sin is no sufficien for ( ( ) 4 d B) M A [3] B, B = = = = = any of he five underlined.()() 4 8.() expressions or awr 0.8.() ( ) ( ) B aef T: y = ( x ) 4 Finding an equaion of a angen wih heir poin and heir angen gradien or finds c by using y = (heir gradien) x + " c". M aef 3 3 T: y = x + or y = x + Correc simplified EXACT equaion of angen A aef cso or 3 = ( ) + c c = = Hence T: y = x + or y = x [5] Silver 4: 8/

12 Quesion Number Scheme Marks 7. x + 4xy + y + 7 = 0 (a) dy dy dy = x + 4y + 4x + y = 0 M A B dy x + 4 y + (4x + y) = 0 dm dy x 4y x y = = 4x+ y x + y A cso oe (5) (b) 4x + y = 0 M y = x x = y A x + 4 x( x) + ( x) + 7 = y + 4 y y + y + 7 = M* 3 3x + 7 = y + = x = 9 y = 36 dm* x = 3 y = 6 A When x = 3, When y = 6, x = (6) ddm* y = ( 3) y = 6 x = 3 A cso (7) [] Silver 4: 8/

13 M A () 8. (a) sin θdθ = ( cos θ) dθ = θ sin θ ( + C) 4 an dθ sec x = θ = θ d d sin sec d dθ M A (b) y x= y = ( ) π π θ π θ θ θ ( sinθcosθ) = π dθ M cos θ 6 sin d = π θ θ k = 6π A x = 0 anθ = 0 θ = 0, π x = anθ = θ = B (5) (c) V π 6 = 6π sin θdθ 0 V sin θ = 6π θ 4 π 6 0 π π = 6π sin ( 0 0) 4 3 π 3 4 = = 8 3 6π π π 3 Use of correc limis M M 4 p =, q = A (3) 3 (0 marks) Silver 4: 8/ 3

14 Quesion This proved a suiable saring quesion and here were many compleely correc soluions. The majoriy of candidaes could complee par (a) successfully. In par (b), hose who realised ha working in common (vulgar) fracions was needed usually gained he mehod mark bu, as noed in he inroducion, he working needed o esablish he prined resul was 8 3 frequenly incomplee. I is insufficien o wrie down =. The examiners 00 5 acceped, for example, = = =. In par (c), mos candidaes realised ha hey had o evaluae heir answer o par (a) wih x = 0.0. However many failed o recognise he implicaion of par (b), ha his evaluaion needed o be muliplied by 5. I was no uncommon for candidaes o confuse pars (b) and (c) wih he expansion and decimal calculaion appearing in (b) and fracion work leading o 3 appearing in (c). Quesion Many candidaes go off o a very bad sar o his quesion by wriing ( x ) ( ) 9 4x ( 3 x) 9+ 4 = 3+ x or + = +. Such errors in algebra are heavily penalised as he resuling binomial expansions are significanly simplified and, in his case, gave answers in incorrec powers 4 of x. Those who obained 3 9 x + showed ha hey undersood he binomial heorem bu here were many errors in signs, ofen due o he failure o use brackes correcly. Some 4 candidaes seemed o lose he hread of he quesion and, having expanded + x 9 correcly, failed o muliply by 6. I was no unusual o see an, ofen correc, erm in x 3 provided. The examiners ignore his bu such addiional work does lose ime. Quesion 3 This quesion was well done and full marks were common. Candidaes were roughly equally divided beween hose who expanded and differeniaed and hose who differeniaed using he produc rule. The laer mehod was he more complicaed and more subjec o error bu many correc soluions were seen using boh mehods. If he differeniaion was correc, nearly all compleed par (a) correcly. Raher oddly, a number of cases were seen where π ( 3 4 ) h h was misread as π ( 3 4 ) h h. Par (b) was generally well done alhough here were a minoriy of sudens who made no aemp a i a all. The large majoriy π dv π dv correcly inerpreed as and realised hey had o divide by 800 d 800 dh. Invering d V dh did cause difficuly for some candidaes. For example, = was seen 0.5πh πh 0.5πh πh Silver 4: 8/ 4

15 from ime o ime and relaively common. π 5 = 5π leading o he answer π, insead of he correc 3 3, was Quesion 4 A leas 90% of he candidaure was able o apply he volume of revoluion formula correcly. Only a few candidaes did no include π in heir volume formula or did no square he expression for y. The inegraion was well aemped and he majoriy of candidaes recognised ha he inegral f( x) could be manipulaed ino he form d x and inegraed o give heir resul in he form f( x) k ln (3x + 4) usually wih k =. A variey of incorrec values of k were seen wih he mos 3 common being eiher 3 or. A significan number of candidaes inegraed incorrecly o give answers such as x ln (3x + 4) or x ln (3x + 4). Those candidaes who applied he subsiuion u = 3x + 4 proceeded o achieve ln u, and 3 changed heir x-limis of 0 and o give correc u-limis of 4 and 6. Oher subsiuions of u = 3x or u = x, were also used, usually successfully. Unproducive aemps were seen by a minoriy of candidaes, such as inegraion by pars or x x x simplifying o give, 3x + 4 3x + 4 or inegraing x and 3x + 4 separaely and hen muliplying or dividing he wo resuls ogeher. The majoriy of candidaes were able o apply he limis correcly and examiners observed he correc answer in a variey of differen forms. Quesion 5 This quesion, and in paricular he final Q5(d), proved challenging for a large number of candidaes, wih abou 8% of he candidaure scoring a leas of he 5 marks available and only abou 7% scoring all 5 marks. Q5(a) and Q5(b) were almos invariably compleed correcly, he main source of error in Q5(b) being ha a very small number of candidaes did no realise ha = 0 follows from =. Many correc soluions o Q5(c) were seen. The principal reason for loss of marks came from dy candidaes being unable o find he derivaive of. Dividing by d d i.e. dividing by proved challenging for a number of candidaes. Some candidaes, having correcly esablished d y as being ( ) ln, hen proceeded incorrecly o equae his Silver 4: 8/ 5

16 o 4 ln. Mos knew how o obain he gradien of he normal, and could wrie down he equaion of a sraigh line. Q5(d) was answered well by small number of candidaes, and, alhough a significan number 0 could wrie he area as (. ) d, many were unable o perform he inegraion of 4 wih respec o. Some wroe as, hus simplifying he problem, whils aemps such + as were no uncommon. Candidaes who were unable o make an aemp a he + inegraion of were unable o access he final 4 marks in his par. Approaches ha faciliaed inegraion included re-wriing ln as e or subsiuing u =, leading o du = ln = u ln, and hereby circumvening a direc inegraion of. d (or x Oher candidaes used a caresian approach, giving he area as ( ) equivalen), bu again a number were unable o carry ou he inegraion. Quesion 7 This quesion discriminaed well beween candidaes of all abiliies, wih abou 80% of candidaes gaining a leas 5 marks of he marks available and abou 45% gaining a leas 8 Silver 4: 8/ 6

17 marks. Only abou 5% of candidaes gained all marks. Par (a) was answered well wih full marks commonly awarded. Par (b) was far more challenging wih only a small minoriy presening a complee and correc soluion. In par (a), many candidaes were able o differeniae correcly, facorise ou d y, and rearrange heir equaion o arrive a a correc expression for he gradien funcion. A minoriy did no apply he produc rule correcly when differeniaing 4 xy, whils a small number lef he consan erm of 7 in heir differeniaed equaion. In par (b), hose small proporion of candidaes who realised hey needed o se he denominaor of heir d y expression equal o zero usually wen on o answer his par correcly. Some, however, did no aemp his par, while a majoriy aemped o solve dy = 0, and many proceeded o obain coordinaes of ( 6, 3) for he poin Q, despie a number of hem iniially skeching a curve wih a verical angen. A smaller proporion solved d y, presumably because he digi is wrien as a verical line; whils ohers eiher subsiued y = 0 or x = 0 ino heir d y expression. Manipulaion and brackeing errors someimes led o candidaes wriing equaions such as y = A or x = A, where A was negaive. Examiners were surprised ha a fair number of candidaes, having obained heir value of x (or y), hen proceeded o subsiue his ino x + 4xy + y + 7 = 0, raher han using he much simpler y = kx or x = ky. Quesion 8 The responses o his quesion were very variable and many los marks hrough errors in manipulaion or noaion, possibly hrough menal iredness. For examples, many made errors in manipulaion and could no proceed correcly from he prined cos θ = sin θ o x θ sin θ = cos θ and he answer sin θ was ofen seen, insead of sin θ. In 4 4 par (b), many never found d x or realised ha he appropriae form for he volume was dθ π y d θ. dθ However he majoriy did find a correc inegral in erms of θ alhough some were unable o use he ideniy sin θ = sinθcosθ o simplify heir inegral. The incorrec value k = 8π was very common, resuling from a failure o square he facor in sin θ = sinθcosθ. Candidaes were expeced o demonsrae he correc change of limis. Minimally a reference π o he resul an =, or an equivalen, was required. Those who had complee soluions 6 3 usually gained he wo mehod marks in par (c) bu earlier errors ofen led o incorrec answers. Silver 4: 8/ 7

18 Saisics for C4 Pracice Paper Silver Level S4 Mean score for sudens achieving grade: Qu Max Modal Mean score score % ALL A* A B C D E U Silver 4: 8/ 8

Mark Scheme (Results) January GCE Mathematics 6666 Core Mathematics 4

Mark Scheme (Results) January GCE Mathematics 6666 Core Mathematics 4 Mark Scheme (Resuls) January 0 GCE Mahemaics 6666 Core Mahemaics 4 Edecel and BTEC Qualificaions Edecel and BTEC qualificaions come from Pearson, he world s leading learning company. We provide a wide