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Paper Reference(s)

6666/01

Edexcel GCECore Mathematics C4

Gold Level (Harder) G2

Time: 1 hor !0 mi"tes

Materials re#ired $or exami"atio" %tems i"clded &ith #estio" 'a'ers

Mathematical Formulae (Green) Nil

Candidates may use any calculator allowed by the regulations of the JointCouncil for Qualifications. Calculators must not have the facility for symbolicalgebra manipulation, differentiation and integration, or have retrievablemathematical formulas stored in them.

%"strctio"s to Ca"didates

Write the name of the examining body (Edexcel) your centre number candidate number theunit title (!ore Mathematics !") the paper reference (####) your surname initials andsignature$

%"$ormatio" $or Ca"didates

% boo&let 'Mathematical Formulae and tatistical ables* is pro+ided$Full mar&s may be obtained for ans,ers to %-- .uestions$here are / .uestions in this .uestion paper$ he total mar& for this paper is 01$

dvice to Ca"didates

2ou must ensure that your ans,ers to parts of .uestions are clearly labelled$2ou must sho, sufficient ,or&ing to ma&e your methods clear to the Examiner$ %ns,ers,ithout ,or&ing may gain no credit$

**ested *rade +o"daries $or this 'a'er:

, - C . E

6 4 42 !6 2

Gold 2 his publication may only be reproduced in accordance ,ith Edexcel -imited copyright policy$3455064578 Edexcel -imited$

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1 f(x) 9 4)78(

7

xx9

x

A:

)78( x

B: 4

)78( xC

$

(a) Find the +alues of the constantsABand C$(4)

(b) (i) ;ence find

xx d)(f $

(ii) Find

4

7

d)(f xx lea+ing your ans,er in the form a : ln b ,here a and b are

constants$(6)

3"e 2012

2 he currentI amps in an electric circuit at time t seconds is gi+en by

I 9 7# 6 7#(5$1)t t 5$

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Gi+e your ans,er in the forma

b,here a and b are integers$

(!)

3"e 201!

4 Gi+en thaty9 4 atx9"

sol+e the differential e.uation

x

y

d

d9

xy 4cos

8$

()

3"e 2012

.

5i*re 2

% container is made in the shape of a hollo, in+erted right circular cone$ he height of thecontainer is 4" cm and the radius is 7# cm as sho,n in Figure 4$ Water is flo,ing into the

container$ When the height of ,ater is h cm the surface of the ,ater has radius r cm and the+olume of ,ater is V cm8$

(a) ho, that V940

" 8h$

(2)

Gold 2 his publication may only be reproduced in accordance ,ith Edexcel -imited copyright policy$3455064578 Edexcel -imited$

7/22/2019 C4 Gold 2

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AThe volume V of a right circular cone with vertical height h and base radius r is given by the

formula V98

7r4h $B

Water flo,s into the container at a rate of / cm8s67$

(b) Find in terms of the rate of change of h ,hen h 9 74$()

3a"ar 200

6 (a) Find

xx dtan 4 $

(2)

(b)

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(a) Express 4"4

y in partial fractions$

(!)

(b) ;ence obtain the solution of

4 cotx xydd 9 (" 6y4)

for ,hichy 9 5 atx98

gi+ing your ans,er in the form sec4x 9 g(y)$

()

June 2008

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5i*re !

Figure 8 sho,s part of the cur+e C ,ith parametric e.uations

x 9 tan y 9 sin 5 =4

$

he pointP lies on C and has coordinates

8

4

78 $

(a) Find the +alue of at the pointP(2)

he line l is a normal to C atP$ he normal cuts thexCaxis at the point !$

(b) ho, that ! has coordinates (k8 5) gi+ing the +alue of the constant k$(6)

he finite shaded region " sho,n in Figure 8 is bounded by the cur+e C the linex 9 8 and thexCaxis$ hisshaded region is rotated through 4radians about thexCaxis to form a solid of re+olution$

(c) Find the +olume of the solid of re+olution gi+ing your ans,er in the form#8 : $4 ,here# and $ areconstants$

()

3"e 2011

T7TL 578 99E8: M8

E;.

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QuestionNumber

Scheme Marks

1 (a) ( ) ( )4

7 8 7 8 7A x Bx x Cx= + + D7

5x ( )7 A= M7 7

8

x 78

7 8C C= = any t,o constantscorrect %7!oefficients of 4x 5 E 8 8A B B= + = all three constantscorrect

%7 (4)

(b) (i)( )

4

7 8 8d

8 7 8 7x

x x x

+

( )( )

( ) ( )78 8

ln ln 8 7 8 78 7 8

x x x C= + +

M7 %7ft %7ft

( ) ( )7

ln ln 8 78 7

x x Cx

= +

(ii) ( ) ( )4

4

77

7f d ln ln 8 7

8 7x x x x

x

=

7 7

ln 4 ln 1 ln7 ln 41 4

=

M7

4 4

ln $$$1

= + M7

8 "

ln75 1

= + %7 (6)

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stio"

m+ercheme Mar?s

(a)7 7

4 47

(7 ) (7 )7

xx x

x

+ = +

7 7

4 4(7 ) (7 )x x

+ D7

87 7 7

4 44 4 4 4( )( ) ( )( )7 77 $$$ 7 ( ) ( ) $$$4 4 4 4

x x x x = + + + + + +

M7 %7%7

4 47 7 7 87 $$$ 7 $$$4 / 4 /x x x x

= + + + + +

4 4 47 8 7 7 77 $$$4 / 4 " /x x x x x= + + + + + M7

4774

x x= + + Answer is givenin the $uestion

%7

(6)

b) ( )

( )

474#

74#

7 7 7 777 4# 4 4#

+ = + + M7

ie?8 8 7"51

1 7814= D7

so0541

8"51#

=0541

"51#%7 cao

(!)

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io

erScheme Marks

(a) imilar triangles7# 4

4" 8

r hr

h

= =

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io

erScheme Marks

(a) 4tan dx x

4 4 4 4? sec 7 tan gi+es tan sec 7%B A A A A = + = he correct underlined identity$ M1 oe

4sec 7 dx x=

( )tanx x c= + !orrect integration,ith@,ithout : c

A1

(2)

(b)8

7ln dx x

x4

4

d 7d

78dd 4 4

ln ux x

v xx x

u x

x v

= =

= = =

4 4

7 7 7ln $ d

4 4x x

x x x=

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(a) 44 4

" (4 )(4 ) (4 ) (4 )

A B

y y y y y +

+ + so4 (4 ) (4 )A y B y + + M7

-et 4y= ( ) 744 "B B= = -et 4y= ( ) 7

44 "A A= = M7

gi+ing7 74 4

(4 ) (4 )y y+

+ %7 cao (8)

(b) 44 7d d

" coty x

y x=

D77 74 4 d tan d

(4 ) (4 )y x x

y y+ =

+ ( )7 74 4ln(4 ) ln(4 ) ln(sec )y y x c + + = + D7 M7 %7 ft

85y x = = ( )( )8

7 7 74 4 cosln 4 ln 4 ln c

+ = + M7

{ }5 ln 4 ln 4c c= + = 7 74 4ln(4 ) ln(4 ) ln(sec ) ln 4y y x + + =

7 4 secln ln4 4 4

y x

y + =

M7

4 secln 4ln

4 4

y x

y

+ = 4

4 secln ln

4 4

y x

y

+ = M7

44 sec

4 "

y x

y

+=

;ence

4 / "sec

4

yx

y

+=

%7 (/)(11 mar?s)

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uestionNumber

cheme Mar&s

(a) tan 8 = or8

sin4

= M7

8= a,rt 7$51 %7 (2)

(b) 4d

secd

x

=

dcos

d

y

=

( )84d cos

cosd sec

y

x

= = M7 %7

%tP 8 7

cos8 /

m = =

!an be implied %7

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>estio" 1

he maority of candidates gained full mar&s in part (a) and it ,as better done than similar .uestions in pre+ious

years$ Most obtained the identity ( ) ( )4

7 8 7 8 7A x Bx x Cx= + + and foundA and Cby ma&ing an appropriatesubstitution$ Finding Bpro+ed more difficult and the error 5B= ,as not uncommon$ Hn part (b)(i) most could

gain the method mar& recognising that the integral of7

x

,as lnx and those ,ith incorrect nonCOero +alues ofA

B and C,ere able to benefit from follo, through mar&s in this part$ here ,ere ho,e+er many errors ,hen

finding8

d8 7

xx

and( )

4

8d

8 7x

x

I 68 ln (8x6 7) and 8 ln (x6 7)4respecti+ely being common errors$ he

maority could start part (b)(ii) and if they had correct +alues ofABand C then full mar&s for the .uestion ,erecommon$

>estio" 2

hose ,ho &ne, and often .uoted a formula of the form ( )d

ln

d

x xa a a

x

= usually found this .uestion

straightfor,ard$ hose ,ho did not tried a number of methods and these ,ere fre.uently incorrect$ Errors seen

included ( ) ( )7

7# 5$1 7# 5$1 ln and / lnt tt t t $ Nearly all candidates substituted 8t= into their

d

d

I

t but a

significant number of candidates failed to gi+e their ans,er in the form ln a as re.uired by the .uestion lea+ingtheir ans,er in the form lnn a $

Question 3

his .uestion discriminated ,ell across all abilities ,ith about 14 of candidates gaining at least # of the mar&s a+ailable and about 45 of candidates gaining all mar&s$

Hn part (a) the most popular method ,as to re,rite7

7

x

x

+

as7 7

4 4(7 ) (7 )x x

+ and achie+e the result

4774

x x+ + by multiplying out the binomial expansion of7

4(7 )x+ ,ith the binomial expansion of7

4(7 ) $x

ome candidates ho,e+er ,ere not able to formulate a strategy for expressing7

7

x

x

+

in a form so that

rele+ant binomial expansions could be applied$ he most common mista&e ,as to express as

7

4

7

4

(7 )

(7 )

x

x

+

and then

try to di+ide the t,o expressions once expanded but ,ithout success$

%lthough many candidates had ,ritten7 7

4 4(7 ) (7 )x x

+ a significant number did not attempt to multiply the

t,o resulting expansions together ,ith se+eral attempting to di+ide their expansions and some deciding to add

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their expansions after obser+ing7

4x both expansions$ Many candidates ,ere able to use a correct method for

expanding a binomial expression of the form (7 )nx+ ,ith some ma&ing sign errors ,hen simplifying$ hemaority of candidates ho,e+er ,ho recognised the need to multiply the t,o expressions did so successfullysho,ing sufficient ,or&ing and ignoring higher po,ers ofx to produce the gi+en result$

Examples of alternati+e methods seen from a fe, candidates in part (a) included?744 74

4

7 (7 )(7 ) (7 )(7 ) (7 )

7 (7 )(7 ) (7 )

x x x xx x

x x x x

+ + = = = + etc$

or

7

4 47 (7 )(7 )

(7 )(7 )7 (7 )(7 )

x x xx x

x x x

+ + += = + +

etc$

Hn part (b) the main obstacle to success ,as the lac& of realisation that a substitution of7

4#x= must be made

into both sides of 47 777 4

x x xx

+ = + + $ % significant number substituted into the R; alone assuming the

-; ,as 8 and claimed that7"51

87814

$ E+en ,hen candidates did substitute7

4#x= correctly into both

sides many neglected to e.uate both sides and so had little chance of figuring out ho, to proceed to estimate

8 $ hose that e.uated both sides usually achie+ed the correct estimate of0541

$"58#

Fe, candidates ignored the instruction gi+en in part (b) and e.uated7

7

x

x

+

to 8 deduced the +alue of

7

4x= and substituted this into the R; in order to find an estimate for 8 $ ;istorically there ha+e been a

number of past examination .uestions that ha+e as&ed candidates to Quse a suitable +alue of xQ and presumablythese candidates had decided to do ust that$

>estio" 4

his .uestion ,as not ,ell done and o+er one .uarter of the candidates gained no mar&s on this .uestion$ Hn all ofthe other .uestions on the paper fe,er than / of the candidates gained no mar&s$ Many did not recognise that

the .uestion re.uired a separation of the +ariables and they could ma&e no progress$

he main error in separation ,as obtaining through faulty algebra7

dyy

on one side of the e.uation$ % .uite

unexpected error ,as to see those ,ho had correctly obtained dy y integrate this to lny $ he only explanationof this appears to be that many .uestions on this topic do result in logarithms and candidates ,ere follo,ing anexpected pattern rather than actually sol+ing the .uestion set$ hose ,ho could deal ,ith the y side of the

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e.uation often had trouble ,ith 47

dcos

xx

not re,riting this as 4sec dx x and proceeding to tanx $ Many longand fruitless attempts ,ere made using a double angle formula$ hose ,ho did integrate sometimes left it therenot realising that it ,as necessary to e+aluate a constant of integration$ %bout one third of candidates did obtainfull mar&s$

Question 5

% considerable number of candidates did not attempt part (a) but of those ,ho did the most common method,as to use similar triangles to obtain 48

hr= and substitute r into 478V r h= to gi+e8"

40$V h= ome

candidates used trigonometry to find the semiC+ertical angle of the cone and obtained48hr= from this$ % fe,

candidates correctly used similar shapes to compare +olumes by ,riting do,n the e.uation8

478

$(7#) 4" 4"

V h

=

Part (b) discriminated ,ell bet,een many candidates ,ho ,ere able to gain full mar&s ,ith ease and some

candidates ,ho ,ere able to gain ust the first one or t,o mar&s$ ome incorrectly differentiated 478V r h= to

gi+e 4d 7

d 8

Vr

h= $ Most of the successful candidates used the chain rule to find

d

d

h

tby applying

d d$

d

V V

t dh he

final ans,er7

/,as sometimes carelessly ,ritten as

7$

/ ccasionally some candidates sol+ed the differential

e.uationd

/d

V

t= and e.uated their solution to

8"

40

hand then found

d

d

t

hor differentiated implicitly to find

d

d

h

t$

Question 6

Hn part (a) a surprisingly large number of candidates did not &no, ho, to integrate 4tan $x Examiners ,ereconfronted ,ith some strange attempts in+ol+ing either double angle formulae or logarithmic ans,ers such as

4ln(sec )x or "ln(sec )$x hose candidates ,ho realised that the needed the identity 4 4sec 7 tanx x= + sometimes ,rote it do,n incorrectly$

Part (b) ,as probably the best attempted of the three parts in the .uestion$ his ,as a tric&y integration by parts

.uestion o,ing to the term of 87 x meaning that candidates had to be especially careful ,hen using negati+epo,ers$ Many candidates applied the integration by parts formula correctly and then ,ent on to integrate an

expression of the form8

k

x to gain 8 out of the " mar&s a+ailable$ % significant number of candidates failed togain the final accuracy mar& o,ing to sign errors or errors ,ith the constants and in

4 4lnx c

x x

+ + $ %

minority of candidates applied the by parts formula in the ',rong direction* and incorrectly stated that dd lnv

xx=

implied 7xv= $

Hn part (c) most candidates correctly differentiated the substitution to gain the first mar&$ % significantproportion of candidates found the substitution to obtain an integral in terms of u more demanding$ ome

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candidates did not realise that 4e x and 8e x are 4(e )x and 8(e )x respecti+ely and hence 4 7u rather than 4( 7)u,as a fre.uently encountered error seen in the numerator of the substituted expression$ Fe,er than half of the

candidates simplified their substituted expression to arri+e at the correct result of4( 7)

d $u

uu

ome candidatescould not proceed further at this point but the maority of the candidates ,ho achie+ed this result ,ere able tomultiply out the numerator di+ide by u integrate and substitute bac& for u$ %t this point some candidatesstruggled to achie+e the expression re.uired$ he most common misconception ,as that the constant ofintegration ,as a fixed constant to be determined and so many candidates concluded that 84 $k= Manycandidates did not realise that 84 ,hen added to ccombined to ma&e another arbitrary constant k$

Question 7

Hn part (a) many candidates realised that they needed to factorise the denominator to gi+e t,o linear factors andusually proceeded to gi+e a fully correct solution$ % fe, candidates ho,e+er thought that 4" y ,as anexample of a repeated linear factor and tried to split up their fraction up accordingly$ ome candidates struggled,ith factorising 4" y gi+ing ans,ers such as (" )(" )y y+ or ( 4)( 4)$y y+ he maority of candidates ,ereable to ,rite do,n the correct identity to find their constants although a noticeable number of candidates ,hen

sol+ing " 4A= found 4$A =% significant minority of candidates ,ho completed part (a) correctly made no attempt at part (b)$ %bout half ofthe candidates in part (b) ,ere able to separate out the +ariables correctly$ Many of these candidates spotted thelin& ,ith part (a)$ Ht ,as pleasing that candidates ,ho progressed this far ,ere able to correctly integrate tanx and correctly find the t,o ln terms by integrating their partial fraction$ !ommon errors at this point ,ereintegrating tanx to gi+e 4sec x and the sign error in+ol+ed ,hen integrating 4

&y $ % significant number of

candidates at this point did not attempt to find a constant of integration$ ther candidates substituted 8x = and

5y= into an integrated e.uation ,hich did not contain a constant of integration$ % maority of candidates ,hofound the constant of integration struggled to simplify their e.uation do,n to an e.uation ,ith a single ln term on

each side$ he most common error of these candidates ,as to belie+e thatln ln ln

A B C+ = implies$A B C+ =

f all the / .uestions this ,as the most demanding in terms of a need for accuracy$ Fe,er than 75 ofcandidates ,ere able to score all 77 mar&s in this .uestion although statistics sho, that about half of thecandidates ,ere able to score at least 1 mar&s$

>estio"

he maority of candidates ,ere able to complete part (a) although some candidates ga+e the ans,er in degrees

rather than radians$ Most could start part (b) correctly and apart from a fe, errors in sign obtain 4d cos

d sec

y

x

=

although this ,as often simplified to cos rather than the correct 8cos $ he maority of candidates ,ere ableto demonstrate the correct method of finding the e.uation of the normal and to complete the part by substituting

5y= and sol+ing forx$ % small number of candidates eliminated successfully differentiated the cartesiane.uation and completed the .uestion$

Part (c) pro+ed challenging for many candidates and a substantial number of candidates thought that the +olume

,as gi+en by 4sin often ignoring d ordx $ %mong those ,ho recognised that the appropriate integral ,as

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4 4sin sec d many ,ere unable to re,rite this in a form ,hich could be integrated$ lnsec and 8sin ,ereamong the erroneous attempts seen$ hose ,ho realised that 4 4 4 4sin sec tan sec 7 = = usually completedthe .uestion correctly although a fe, used x limits rather than limits$ here are a number of possiblealternati+e approaches to this .uestion and there ,ere some successful attempts using integration by parts$ %number of candidates attempted to use the cartesian form of the e.uation but fe, of these ,ere able to establish a

method of integrating4

47

x

x+$

tatistics $or C4 9ractice 9a'er G2

Mean score for students achieving grade:

Qu Maxscore

Modalscore

Mean%

ALL A* A B C D E U

10 75 7.47 9.68 8.85 7.82 6.62 5.31 3.97 2.22

! 5 57 2.83 4.71 3.30 2.12 1.36 0.94 0.90 0.67

3 9 9 59 5.28 8.16 6.52 5.28 4.26 3.4 2.61 1.64

" 5 57 2.86 4.81 4.10 3.00 1.70 0.81 0.30 0.08

# 7 47 3.26 5.18 2.52 1.20 0.65 0.20 0.11

$ 13 55 7.18 10.02 5.96 3.82 2.50 1.38 0.40

11 46 5.05 7.44 4.70 3.12 2.10 1.42 0.83

& 15 52 7.86 14.16 10.39 7.32 4.95 3.34 2.04 1.10

# #$ "'( ##'&) 3&'! !')3 (')# !'&! ')#

Gold 4 ?75@74 70