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Paper Reference(s)

6666/01

Edexcel GCECore Mathematics C4

Gold Level (Hardest) G4

Time: 1 hour 0 mi!utes

Materials re"uired #or exami!atio! $tems i!cluded %ith "uestio! &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.

$!structio!s to Ca!didates

Write the name of the examining body (Edexcel) your centre number candidate number the

unit 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$

uested rade *ou!daries #or this &a&er:

'+ ' , C - E

6. . 4. 6 0 .6

Gold 4 his publication may only be reproduced in accordance ,ith Edexcel -imited copyright policy$3455064578 Edexcel -imited$

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1 (a) Find4 dxx e x $

()

(b) 9ence find the exact +alue of7

45

dxx e x $(.)

June 2013

. :se the substitution u ; 4xto find the exact +alue of

+

7

5

4)74(

4x

x

dx$

!"

u!e .002

3iure .

Figure 4 sho,s a right circular cylindrical metal rod ,hich is expanding as it is heated$ %ftert seconds the radius of the rod isx cm and the length of the rod is 1x cm$

he cross

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u!e .00

Gold 4 his publication may only be reproduced in accordance ,ith Edexcel -imited copyright policy$3455064578 Edexcel -imited$

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(i) Find

xx

d4

ln $

#"

(ii) Find the exact +alue of

4

"

4 dsin

xx $

$"

a!uar5 .00

(a) Find +

xx

xd

#? x @ 5$

(.)

(b) Gi+en thaty ; / atx ;7 sol+e the differential e.uation

xyd

d ;x

yx 87

)#?( +

gi+ing your ans,er in the formy4; g(x)$

(6)

a!uar5 .010

f() ; " cos4 6 8sin4

(a) ho, that f() ;47 A

40 cos 4$

()

(b) 9ence using calculus find the exact +alue of 4

5

df

)( $

(2)

u!e .010

Gold " =74>74 "

8/13/2019 C4 Gold 4

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Relati+e to a fixed origin O the pointA has position +ector (/i A 786 47) the pointB has position +ector (75i A

7"6 "7) and the point C has position +ector (?i A ? A #7)$

he line lpasses through the pointsA andB$

(a) Find a +ector e.uation for the line l$

()

(b) Find CB $

(.)

(c) Find the siBe of the acute angle bet,een the line segment CB and the line l gi+ing your ans,er in degrees to

7 decimal place$

()

(d) Find the shortest distance from the point C to the line l$

()

he pointX lies on l$ Gi+en that the +ector CX is perpendicular to l

(e) find the area of the triangle CXB gi+ing your ans,er to 8 significant figures$

()

u!e .008

Gold " =74>74 1

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-i.uid is pouring into a large +ertical circular cylinder at a constant rate of 7#55 cm 8s67and is lea&ing out of a

hole in the base at a rate proportional to the s.uare root of the height of the li.uid already in the cylinder$ he

area of the circular cross section of the cylinder is "555 cm 4$

(a) ho, that at time t seconds the height h cm of li.uid in the cylinder satisfies the differential e.uation

t

h

d

d; 5$" kh

,here k is a positi+e constant$

()

When h = 41 ,ater is lea&ing out of the hole at "55 cm8s67$

(b) ho, that k = 5$54$

(1)

(c) eparate the +ariables of the differential e.uation

t

h

d

d

; 5$" 5$54h

to sho, that the time ta&en to fill the cylinder from empty to a height of 755 cm is gi+en by

755

5

d45

15h

h$

(.)

:sing the substitution h; (45 x)4 or other,ise

(d) find the exact +alue of

755

5

d45

15

hh $(6)

(e) 9ence find the time ta&en to fill the cylinder from empty to a height of 755 cm gi+ing your ans,er inminutes and seconds to the nearest second$

(1)

a!uar5 .00

T9T'L 39 ;';E: 2 M'

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uestio!

um*ercheme Mar7s

1(a)

4e dxx x 7st%pplication=4 d 4

d

de e

d

x x

uu x x

x

vv

x

= = = =

4nd%pplication=

d7

d

de e

d

x x

uu x

x

vv

x

= = = =

4e 4 e dx xx x x= { }4e e dx xx x x 5> M7

{ }4e 4 e dx xx x x %7 oe

( )4

e 4 e e d

x x x

x x x=

Either { }4e e e dx x xAx Bx C x

or for{ } { }( )e d e e dx x xK x x K x x

M7

{ }4e 4( e e )x x xx x c= +4e e ex x xAx Bx C M7

!orrect ans,er ,ith>,ithout c+ %7()

(*){ }

( ) ( )

74

5

4 7 7 7 4 5 5 5

e 4( e e )

7 e 4(7e e ) 5 e 4(5e e )

x x xx x

=

%pplies limits of 7 and 5 to anexpression of the form

4e e e x x xAx Bx C 5 5A B and

5C and subtracts the correct ,ay

round$

M7

e 4= e 4 cso %7 oe (.)

?2@

Gold " =74>74 0

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oner

Scheme Marks

1

2

0

2d

(2 1)

x

xx

+ , with substitution 2xu=

d2 .ln2

d

xu

x

= d 1

d 2 .ln2xx

u =

dd

2 .ln2xux = or dd .ln2

u

x u=

or ( ) d1 d ln2u

u x =

B1

2 2

2 1 1d d

ln2(2 1) ( 1)

x

xx u

u

= + + 21

d( 1)

k uu+

where kis constant

M1

1 1

ln2 ( 1)c

u

= + +

2 1

2 1

( 1) ( 1)

( 1) 1.( 1)

u a u

u u

+ +

+ +

M1

A1

chane limits! when x" 0 # x" 1 then u" 1 # u" 2

1 2

2

10

2 1 1d

ln2 ( 1)(2 1)

x

xx

u

= ++

1 1 1

ln2 $ 2

=

%orrect use o& limitsu" 1 and u" 2

de'M1

1

ln2= 1ln2 or

1 1ln ln*

or1 1

2ln2 $n2 A1 ae&

Exact value only! ?6@Alternati+el candidate can re+ert back to x-

1 1

2

00

2 1 1d

ln2(2 1) (2 1)

x

x xx

= + +

1 1 1

ln2 $ 2

=

%orrect use o& limits

x" 0 and x" 1de'M1

1

ln2= 1ln2 or

1 1ln ln*

or1 1

2ln 2 $ ln 2 A1 ae&

Exact value only!

6 marks

Gold "= 74>74 /

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(a) From .uestiond

5$584d

A

t= C7

4 d 4d

AA x x

x

= =

C7

( )d d d 7 5$57#

5$584 D

d d d 4

x A A

t t x x x

= = =

M7

When 4cmx= d 5$57#

d 4

x

t =

9enced

5$5541"#"0?$$$d

x

t= (cm s74 ?

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Gold "= 74>74 75

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Gold " =74>74 77

uestion

Numbercheme Mar&s

4 (i) ( ) ( )4 4ln d 7$ln dx xx x=

( )74 7

4

4

dln

d

d7

d

xxx

uu

x

vv x

x

= = = = =

( ) ( ) 74 4ln d ln $ dx x

xx x x x=

:se of 'integration by

parts* formula in the

correct direction$

M7

!orrect expression$ %7

( )4ln 7 dxx x= %n attempt to multiplyxbya candidate*s ax or 7bx or 7x $ dM7

( )4ln xx x c= + !orrect integration ,ith A

c%7 aef

?4

(ii)4

"

4sin dx x

( )4 4 74NC= cos4 7 4sin or sin 7 cos4x x x x = = !onsideration of double

angle formula forcos4xM7

( )4 4

" "

7 cos 4 7d 7 cos4 d

4 4

xx x x

= =

4

"

74

7sin4

4x x

=

ntegrating to gi+esin4ax b x D 5a b

dM7

!orrect result of anything

e.ui+alent to7 74 "

sin4x x %7

( ) ( )( )4sinsin( )74 4 4 " 4

7 74 4 " 4

( 5) ( )

=

=

ubstitutes limits of 4

and

"

and subtracts the correct

,ay round$

ddM7

( )7 7 74 " 4 / " = + = + ( )

7 7 7 44 " 4 / " / /

or or + + + %7 ae#cso

!andidate must collect

their term and constantterm together for %7

?

No flu&ed ans,ers hence

cso$

8 mar7s

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QuestionNumber

Scheme Marks

6 (a) ( ) 4 4

f "cos 8sin =

7 7 7 7

" cos 4 8 cos 44 4 4 4

= +

M7 M7

7 0

cos44 4

= + cso %7 ()

(b)7 7

cos 4 d sin 4 sin 4 d4 4

= M7 %7

7 7sin 4 cos 4

4 " = + %7

( ) 47 0 0

f d sin 4 cos 4" " /

= + + M7 %7

4

4

50 05 5 5

7# / /$$$

= + + + M7

4 0

7# "

= %7 (2)

?10@

Gold " =74>74 78

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(a)

75 / 4

7" 78 7

" 4 4

AB OB OA

= = =

uuur uuur uuur

or

4

7

4

BA

=

uuur

M7

/ 4

78 7

4 4

= + r or

75 4

7" 7

" 4

= + r accept e.ui+alents M7 %7ft (8)

(b)

75 ? 7

7" ? 1

" # 75

CB OB OC

= = =

uuur uuur uuur

or

7

1

75

BC

=

uuur

( )( ) ( ) ( )44 47 1 75 74# 8 7" 77$4CB = + + = = a,rt 77$4 M7 %7 (4)

(c) $ cosCB AB CB AB =

( ) ( )4 1 45 74# ?cos + + = M7 %7

8

cos 8#$07"

= a,rt 8#$0 %7 (8)

(d) sin74#

d

= M7 %7ft

( )8 1 #$0d = a,rt #$0 %7 (8)(e) 4 4 4 74# "1 /7BX BC d= = = M7

( )7 7 40 1? 8 1 85$44 4 4

CBX BX d = = = ! a,rt 85$7 or 85$4 M7 %7 (8)

(14 mar7s)

Gold "= 74>74 7"

l

X

B

C

d 74#

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n

rcheme Mar&s

d d7#55 or 7#55

d d

V Vc h k h

t t= = Either of these statements M7

( )d

"555 "555d

VV h

h= = d "555

d

V

h= or

d 7

d "555

h

V= M7

dd

dd

d d d

d d d

Vt

Vh

h h V

t V t= =

Eitherd 7#55 7#55

5$"d "555 "555 "555

h c h c hk h

t

= = =

!on+incing proof ofd

d

h

t%7 'G

or

d 7#55 7#55

5$"d "555 "555 "555

h k h k h

k ht

= = = ?@

When 41h= ,ater leaks out such thatd

"55d

V

t=

"55 "55 41 "55 (1) /5c h c c c= = = =

From abo+eD/5

5$54"555 "555

ck= = = as re.uired Proof that 5$54k= C7 'G

?1@

d d5$"

d 5$"

h hk h dt

t k h= =

Separates the variables,ith

d

5$"

h

k h and dt on eitherside ,ith integral signs not

necessary$

M7 oe

755

5

7 5$54time re.uired d

5$545$" 5$54h

h

=

755

5

15

time re.uired d45 hh= !orrect proof %7 'G?.@

Gold " =74>74 71

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n

rcheme Mar&s

755

5

15d

45h

h ,ith substitution 4(45 )h x=

d4(45 )( 7)

d

hx

x= or

d4(45 )

d

hx

x= !orrect

d

d

h

x C7 aef

4(45 ) 45 45h x h x x h= = = 45

dx

xx

or45

d45 (45 )

xx

x

,here is a constant

15 15d $ 4(45 ) d

45h x x

xh=

M7

45755 d

xx

x

=

45

755 7 dxx

=

( ) ( )755 45lnx x c= +ln D 5x x M7

755 4555lnx x %7

change limits= ,hen 5 then 45h x= = and ,hen 755 then 75h x= =

[ ]755

75

455

15d 755 4555ln

45h x x

h=

or ( ) ( )

755 755

55

15d 755 45 4555ln 45

45h h h

h

=

!orrect use of limits ie$

puttingthem in the correct ,ay

round

( ) ( )7555 4555ln75 4555 4555ln 45= Either75x= and 45x=

or 755h= and 5h= ddM7

4555ln45 4555ln75 7555= !ombining logs to gi+e$$$4555ln4 7555

%7 ae#4555ln4 7555= or ( )744555ln 7555

?6@

ime re.uired 4555ln4 7555 8/#$4?"8#77$$$ sec= =

; 8/# seconds (nearest second)

; # minutes and 4# seconds (nearest second) # minutes 4# seconds C7

?1@

1 mar7s

Gold "= 74>74 7#

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Question 1

his .uestion ,as generally ,ell ans,ered ,ith about 08 of candidates gaining at least 1 of the 0 mar&sa+ailable and about "" of candidates gaining all 0 mar&s$ %lmost all candidates attempted this .uestion ,ith

about 78 of them unable to gain any mar&s$

% significant minority of candidates performed integration by parts the ,rong ,ay round in part (a) to gi+e

8 87 7e e d8 8

x xx x x and proceeded by attempting to integrate

87e $

8

xx ome candidates failed to realise that

integration by parts ,as re.uired and ,rote do,n ans,ers such as87 e $

8

xx c+ Fe, candidates integrated ex to

gi+e47

4ex or applied the product rule of differentiation to gi+e

4e 4 e $x xx x+ he maHority of candidates ho,e+er

,ere able to apply the first stage of integration by parts to gi+e 4e 4 e d $x xx x x Many candidates realised that

they needed to apply integration by parts for a second time in order to find 4 e dxx x or in some cases e dxx x $

hose that failed to realise that a second application of integrating by parts ,as re.uired either integrated to gi+e

the final ans,er as a t,o term expression or Hust remo+ed the integral sign$ % significant number of candidates

did not organise their solution effecti+ely and made a brac&eting error ,hich often led to a sign error leading to

the final incorrect ans,er of4e 4 e 4e $x x xx x c +

n part (b) candidates ,ith an incorrect sign in the final term of their integrated expression often proceeded to

use the limits correctly to obtain an incorrect ans,er of 8e 4$ + Errors in part (b) included not substituting thelimit of 5 correctly into their integrated expressionD incorrectly dealing ,ith double negati+esD e+aluating 54e as

7 or failing to e+aluate 5e $ Most candidates ,ho scored full mar&s in part (a) achie+ed the correct ans,er of

e 4 in part (b)$

Gold " =74>74 70

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Question 3

%t the outset a significant minority of candidates struggled to extract some or all of the information from the .uestion$

hese candidates ,ere unable to ,rite do,n the rate at ,hich this cross

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%bout 75 of candidates ,ere able to produce a fully correct solution to this .uestion$

Question 4

t ,as clear to examiners that a significant proportion of candidates found part (i) unfamiliar and thereby struggled to

ans,er this part$ Wea&er candidates confused the integral of lnx ,ith the differential of ln $x t ,as therefore commonfor these candidates to ,rite do,n the integral of lnx as 7 x or the integral of ( )4ln

xas either 4x or

" $x % significant

proportion of those candidates ,ho proceeded ,ith the expected by parts strategy differentiated ( )4ln x

incorrectly to gi+e

either4x or

74x and usually lost half the mar&s a+ailable in this part$ ome candidates decided from the outset to re,rite

( )4ln x

as J ln ln 4 Jx and proceeded to integrate each term and ,ere usually more successful ,ith integrating lnx thanln4$ t is pleasing to report that a fe, determined candidates ,ere able to produce correct solutions by using a method of

integration by substitution$ hey proceeded by either using the substitution as 4xu= or ( )4ln $

xu=

% significant minority of candidates omitted the constant of integration in their ans,er to part (i) and ,ere penalised by

losing the final accuracy mar& in this part$

n part (ii) the maHority of candidates realised that they needed to consider the identity4cos 4 7 4sinx x and so gained

the first method mar&$ ome candidates mis.uoted this formula or incorrectly rearranged it$ % maHority of candidates ,ere

then able to integrate ( )74 7 cos 4x substitute the limits correctly and arri+e at the correct exact ans,er$

here ,ere ho,e+er a fe, candidates ,ho used the method of integration by parts in this part but these candidates ,ere

usually not successful in their attempts$

>uestio!

Part (a) of this .uestion pro+ed a,&,ard for many$ he integral can be carried out simply by decomposition

using techni.ues a+ailable in module !7$ t ,as not unusual to see integration by parts attempted$ his method

,ill ,or& if it is &no,n ho, to integrate lnx but this re.uires a further integration by parts and complicates the

.uestion unnecessarily$ n part (b) most could separate the +ariables correctly but the integration of 78

7

y again a

!7 topic ,as fre.uently incorrect$

Wea&ness in algebra sometimes caused those ,ho could other,ise complete the .uestion to lose the last mar& as

they could not proceed from48 # "ln 4y x x= + to ( )

84 # "ln 4y x x= + $ ncorrect ans,ers such as

4 8 847# #"ln /y x x= + ,ere common in other,ise correct solutions$

Gold " =74>74 7?

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Question 6

!andidates tended either to get part (a) fully correct or ma&e no progress at all$ If those ,ho ,ere successful

most replaced the 4cos and 4sin directly ,ith the appropriate double angle formula$ 9o,e+er many good

ans,ers ,ere seen ,hich ,or&ed successfully +ia 40cos 8 or 4" 0sin $

Part (b) pro+ed demanding and there ,ere candidates ,ho did not understand the notation ( )f

$ ome Hustintegrated ( )f and others thought that ( )f meant that the argument 4 in cos4 should be replaced by

and integrated7 0

cos4 4

+ $ % fe, candidates started by ,riting ( ) ( )f d f d = treating as a

constant$ %nother error seen se+eral times ,as ( ) 47 0f d cos 4 d

4 4

= +

$

Many candidates correctly identified that integration by parts ,as necessary and most of these ,ere able to

demonstrate a complete method of sol+ing the problem$ 9o,e+er there ,ere many errors of detail the correct

manipulation of the negati+e signs that occur in both integrating by parts and in integrating trigonometric

functions pro+ing particularly difficult$ Inly about 71 of candidates completed the .uestion correctly$

Question 7

his pro+ed the most demanding .uestion on the paper$ Nearly all candidates could ma&e some progress ,ith the

first three parts but although there ,ere many often lengthy attempts success ,ith part (d) and (e) ,as

uncommon$ Part (a) ,as .uite ,ell ans,ered most finding AB or BA and ,riting do,n OAAAB or ane.ui+alent$ %n e.uation does ho,e+er need an e.uals sign and a subHect and many lost the final % mar& in this

part by omitting the K =r L from say ( )/ 78 4 4 4= + + + r i 7 i 7 $ n part (b) those ,ho realised that amagnitude or length ,as re.uired ,ere usually successful$ n part (c) nearly all candidates &ne, ho, to

e+aluate a scalar product and obtain an e.uation in cos and so gain the method mar&s but the +ectors chosen,ere not al,ays the right ones and a fe, candidates ga+e the obtuse angle$ Fe, made any real progress ,ith

parts (d) and (e)$ %s has been stated in pre+ious reports a clear diagram helps a candidate to appraise the

situation and choose a suitable method$ n this case gi+en the earlier parts of the .uestion +ector methodsalthough possible are not really appropriate to these parts ,hich are best sol+ed using elementary trigonometry

and Pythagoras* theorem$ hose ,ho did attempt +ector methods ,ere often +ery unclear ,hich +ectors ,ere

perpendicular to each other and e+en the minority ,ho ,ere successful often ,asted +aluable time ,hichsometimes led to poor attempts at .uestion /$ t ,as particularly surprising to see .uite a large number of

solutions attempting to find a +ector CXsay perpendicular to l ,hich ne+er used the coordinates or the position

+ector of C$

Gold "= 74>74 45

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