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Paper Reference(s)

6666/01

Edexcel GCECore Mathematics C4

Gold Level (Harder) G3

Time: 1 hor 30 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 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$

ested rade *o!daries #or this &a&er:

'+ ' , C - E

6. . 4 4 36

Gold 3 his publication may only be reproduced in accordance ,ith Edexcel -imited copyright policy$3455064578 Edexcel -imited$

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12 f(x) 9 (8 : 4x)68 x; 48

$

Find the binomial expansion of f(x) in ascending po,ers of x as far as the term in x8$

Gi+e each coefficient as a simplified fraction$(5)

!e 00

2 77?74 4

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With respect to a fixed origin O the lines l7and l4are gi+en by the e.uations

l7> r9

70

4

77

:

"

7

4

l4> r9

p

77

1

:

4

4

q

,here and are parameters andp and q are constants$ Gi+en that l7and l4are perpendicular

(a) sho, that q 9 68$

()

Gi+en further that l7and l4intersect find

(b) the +alue ofp

(6)

(c) the coordinates of the point of intersection$

()

he pointA lies on l7and has position +ector

78

8

=

$ he point C lies on l4$

Gi+en that a circle ,ith centre C cuts the line l7at the pointsA andB

(d) find the position +ector ofB$

(3)

a!ar 005

Gold 8> 77?74 8

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ire

Figure 4 sho,s a s&etch of the cur+e ,ith parametric e.uations

x 9 4 cos 4t y 9 # sin t 5 t 4

$

(a) Find the gradient of the cur+e at the point ,here t 98

$

(4)

(b) Find a cartesian e.uation of the cur+e in the form

y 9 f(x) 6k x k

stating the +alue of the constant k$

(4)

(c) Write do,n the range of f(x)$

()

a!ar 005

Gold 8> 77?74 "

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he pointsA andB ha+e position +ectors 4i : #76 8 and 8i : "7 : 8 respecti+ely$

he line 7l passes through the pointsA andB.

(a) Find the +ector AB $

()

(b) Find a +ector e.uation for the line 7l $()

% second line 4l passes through the origin and is parallel to the +ector i : 8$ he line 7l meets the line 4l at

the point C$

(c) Find the acute angle bet,een 7l and 4l $

(3)

(d) Find the position +ector of the point C$

(4)

a!ar 00

he line 7l has e.uation r9

+

7

4

7

"

8

4

,here is a scalar parameter$

he line 4l has e.uation r9

+

45

1

8

=

5

,hereis a scalar parameter$

Gi+en that 7l and 4l meet at the point C find

(a) the coordinates of C$

(3)

he pointA is the point on 7l ,here 9 5 and the pointB is the point on 4l ,here 9 67$

(b) Find the si@e of the angleACB$ Gi+e your ans,er in degrees to 4 decimal places$

(4)

(c) Aence or other,ise find the area of the triangleABC$

(.)

!e 010

Gold 8 >77?74 1

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% population gro,th is modelled by the differential e.uation

t

P

d

d9 kP

,here P is the populationt is the time measured in days andk is a positi+e constant$

Gi+en that the initial population is P5

(a) sol+e the differential e.uation gi+ing P in terms of P5k andt$

(4)

Gi+en also thatk 9 4$1

(b) find the time ta&en to the nearest minute for the population to reach 4P5$

()

Bn an impro+ed model the differential e.uation is gi+en as

t

P

d

d9 P cos t

,here P is the populationt is the time measured in days and is a positi+e constant$

Gi+en again that the initial population is P5and that time is measured in days

(c) sol+e the second differential e.uation gi+ing P in terms of P5 andt$

(4)

Gi+en also that 9 4$1

(d) find the time ta&en to the nearest minute for the population to reach 4P5 for the first time using the

impro+ed model$()

!e 00

T9T'L 9 ;';E: . M' 77?74 #

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oner

Scheme Marks

** represents a constant

) ( )3 3

33 2 1 2f( ) (3 2 ) 3 1 13 27 3

x xx x

= + = + = +

Takes 3 outside thebracket to gie an!

of3(3) or

1

27 "

See note below.

#1

2 31

27

( 3)( $) ( 3)( $)( %)1 ( 3)(* * &)' (* * &) (* * &) """

2 3

= + + + +

with * * 1

&pands3(1 * * )x +

to gie a simpified oran un+simpified

1 ( 3)(* * &)+ '

, correct simpifiedor an un+simpified

}{ """""""""" e&pansion

-ith candidate.sfoo-ed thro. ( )* * x

M1'

,1

2 31 2& 2& 2&

27 3 3 3

( 3)( $) ( 3)( $)( %)1 ( 3)( ) ( ) ( ) """

2 3

= + + + +

231

27

/& /01 2& & """

3 27

= + +

2 31 2& /& /0&' """

27 27 /1 72= + +

,n!thing that

cances to1 2&

'27 27

Simpified2 3

/& /0&

/1 72

,1'

,1

[5]

5 marks

Gold 8 >77?74 0

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2 d

sind

ux

x= C7

cos 7sin e d e dx ux x u+ = M7 %7

eu= ft sign error %7ft cos 7e x+=

( )cos 7 7 445

e e ex

+ = or e.ui+alent ,ith u M7

( )e e 7= cso %7 (6) >6?

o!

ercheme Mar8s

4= 45 75

( 4)(8 7) ( 4) (8 7)

x x B CA

x x x x+ + +

+ + 8A= their constant term 8= C7

4= 45 75 ( 4)(8 7) (8 7) ( 4)x x A x x B x C x+ + + + + Forming a correct identity$ C7

Either4 > = 8 > 45 1 8

constant> 75 4 4

x A x A B C

A B C

= = + + = +

or

4 8# "5 75 0 7" 0 4x B B B= = = =

7 45 0 0 07 75 7

8 8 8 8 8x C C C= + = = =

%ttempts to find the +alue of

either one of theirBor their C

from their identity$

M7

!orrect +alues for

theirBand their C ,hich are

found using a correct identity$

%7

>4?

Gold 8> 77?74 /

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Gold 8 >77?74 =

QuestionNumber

Scheme Marks

4 (a) 7 4 "= + d i 7 8 4 4 4q= + +d i 7 8

%s 7 4

4

7 4 ( 4 ) (7 4) ( " 4)

" 4

q

q

= = + +

d d

%pply dot product calculation

bet,een t,o direction

+ectors ie$( 4 ) (7 4) ( " 4)q + +

M1

7 4 5 4 4 / 5

4 # 8 %G

q

q q

= + = = =

d d ets 7 4 5 =d d

and sol+es to find 8q= A1 cso

(2)(b) -ines meet ,here>

77 4 1

4 7 77 4

70 " 4

q

p

+ = +

First t,o of

> 77 4 1 (7)

> 4 77 4 (4)

> 70 " 4 (8)

q

p

= ++ = + = +

i

7

8

Need to see e.uations

(7) and (4)$

!ondone one slip$

(Note that 8q= $)

M1

(1): 4() gi+es> 71 70 = + 4 = %ttempts to sol+e (7) and (4)to find one of either or

dM1

() gi+es> 4 77 " 1 + = =%ny one of 1 or 4 = = A1

Coth 1 and 4 = = A1

(3) 70 "(1) 4( 4)p = +

%ttempt to substitute their

and

into their 8componentto gi+e an e.uation inp

alone$

ddM1

70 45 " 7p p = + = 7p= A1 cso(6)

(c)

77 4 1 8

4 1 7 or 77 4 4

70 " 7 4

= + =

r rubstitutes their +alue of

or into the correct line l7orl4$

M1

Bntersect at ( )

7

0 or 7 0 88

= r

7

08

or ( )7 0 8 A1

(2)

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a)d

"sin4d

xt

t=

d#cos

d

yt

t= C7 C7

d #cos 8

d "sin 4 "sin

y t

x t t

= = M7

%t 8t

= 84

8 8

" 4m

= = accept e.ui+alents a,rt 5$/0 %7 (")b)

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n

rcheme Mar&s

4 8

# "

7 7

OA OB

= =

uuur uuur

8 4 7

" # 4

7 7 4

AB OB OA

= = =

uuur uuur uuur

Finding the difference

bet,een OB andOA $M7

!orrect ans,er$ %7

>?

%n expression of the form

( ) ( )+ector +ector

7

4 7

> # 4

7 4

l

= +

r or

8 7

" 4

7 4

= +

r

7

4 7

> # 4

7 4

l

= +

r or

8 7

" 4

7 4

= +

r

M7

r 9 ( )theirOA AB or

r 9 ( )theirOB AB orr 9 ( )theirOA BA or

r 9 ( )theirOB BA(r is needed$)

%7

aef

>?

4

5 7 7

> 5 5 5

5 7 7

l

= + =

r r

7 4 4AB= = +d i 7 8 4 5= + +d i 7 8 is angle

( ) ( )4

4 4 4 4 4 44

7 7

4 5

4 7cos

$ (7) ( 4) (4) $ (7) (5) (7)

AB

AB

= =

+ + + +

d

d

uuur

uuur

!onsiders dot product

bet,een 4d and their $ABM7

4 4 4 4 4 4

7 5 4cos

(7) ( 4) (4) $ (7) (5) (7)

+ +=

+ + + +!orrect follo,ed through

expression or e"atio!$%7

"

8cos "1 or or a,rt 5$0=$

8$ 4

= = o"

"1 or or a,rt 5$0== o %7 cao

>3?

Gold 8 >77?74 77

his means that cos does not

necessarily ha+e to be the subectof the e.uation$ Bt could be of the

form 8 4 cos 8$=

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Gold 8> 77?74 74

DuestionNumber

cheme Mar&s

62 (d) Bf l1and l2intersect then>

4 7 7

# 4 5

7 4 7

+ =

> 4 (7)

> # 4 5 (4)

> 7 4 (8)

+ = =

+ =

i

7

8

Eitherseeing e.uation (4) ,ritten

do,n correctly ,ith or ,ithoutany other e.uation or seeing

e.uations (7) and (8) ,ritten

do,n correctly$

M7

(4) yields 8

%ny t,o yields 8 1

== =

%ttempt to sol+e either e.uation

(4) or simultaneously sol+e any

t,o of the three e.uations to find

dM7

either one of or correct$ %7

7

4 7 1 7 1

> # 8 4 5 1 5 5

7 4 1 7 1

l or

= + = = =

r r

1

5

1

or 1 1+i 8

Fully correct solution no

incorrect +alues of or seen

earlier$

%7 cso

>4

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2 (a) 7components 8 4 = 8 + = = ( )7= M7 %7

-eading to ( )> 1 = 7C accept +ector forms %7 (3)

(b) !hoosing correct directions or finding ACand BC M7

7 1

4 $ 5 1 4 # 4=cos

7 4

ACB = + =

use of scalar product M7 %7

10$=1ACB = a,rt 10$=1 %7 (4)

(c) ( ) ( )> 4 8 " > 1 = 1A B

8 75

# 5

8 "

AC BC

= =

uuur uuur

4 4 4 4

8 # 8 8 #AC AC = + + = M7 %7

4 4 475 " 4 4=BC BC = + = %7

7

sin4

ABC AC BC ACB= !

7

8 # 4 4=sin 88$14

ACB = 71 1 a,rt 8" M7 %7 (.)

>1?

Gold 8 >77?74 78

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Gold 8> 77?74 7"

QuestionNumber

Scheme Marks

8. (a)d

d

PkP

t= and 00 (1)t P P= =

d

d

Pk t

P =

Separates the ariabes

-ithdP

P

and dk t

on

either side -ith integrasigns not necessar!"

M1

( )n 'P kt c = +Must see nP and kt '

orrect e4uation-ith5-ithout 6 c"

,1

hen 0 00 nt P P P c = = =

( )0or kt

P Ae P A= =

8se of boundar!condition (1) to attempt

to find the constant of

integration"

M1

0n nP kt P = + 0 0n nn "kt P P P kte e e e+ = =

9ence 0kt

P P e= 0

ktP P e= ,1

>4

(b) 02 : 2"%P P k= = 2"%

0 02 tP P e =

Substitutes 02P P= intoan e&pression inoing

P

M1

2"% 2"%2 n n2t te e= = or 2"% n2t=

;or 2 n n2kt kt

e e= = or n2kt=iminates 0P and takes

n of both sidesM1

1

2"%n2 0"2772%//72""" da!st = =

0"2772%//72""" 2$ 3

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Gold 8 >77?74 71

QuestionNumber

Scheme Marks

8. (c)d

cosd

PP t

t = and 00 (1)t P P= =

dcos d

Pt t

P

=

Separates the ariabes

-ithdP

P andcos dt t

on either

side -ith integra signsnot necessar!"

M1

( )n sin 'P t c= +

Must see nP andsin t '

orrect e4uation-ith5-ithout 6 c"

,1

hen 0 00 nt P P P c = = =

( )sin 0or t

P Ae P A= =

8se of boundar!condition (1) to attempt

to find the constant ofintegration"

M1

0n sin nP t P= + 0 0sin n nn sin "t P PP te e e e + = =

9encesin

0

tP P e

= sin0

tP P e

= ,1>4

(d) 02 : 2"%P P = = sin2"%

0 02 tP P e =

sin2"% 2 sin2"% n2te t= =

;or ; 2 sin n2t

e t = =

iminates 0P and

makes sin t or sin2"%t

the sub=ect b! takingn.s

M1

( )112"% sin n2t = Then rearranges

to make tthe sub=ect"dM1

(must use sin+1)0"30

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Question 2

his .uestion ,as generally ,ell done and helped by the printed ans,er many produced fully correct ans,ers$

he commonest error ,as to omit the negati+e sign ,hen differentiating cos 7x + $ he order of the limits ga+e

some difficulty$ Bnstead of the correct7

4

e du u an incorrect +ersion4

7

e du u ,as produced and the resulting

expressions manipulated to the printed result and ,or&ing li&e ( ) ( )4 7 4 7

e e e e e e 7 = + = ,as notuncommon$

ome candidates got into serious difficulties ,hen through incorrect algebraic manipulation they obtained4

e sin du

x u instead of e du u $ his led to expressions such as ( )4e 4 du u u u and the efforts to integratethis either by parts t,ice or a further substitution often ran to se+eral supplementary sheets$ he time lost here

ine+itably led to difficulties in finishing the paper$ !andidates need to ha+e some idea of the amount of ,or& and

time appropriate to a # mar& .uestion and if they find themsel+es exceeding this realise that they ha+e probablymade a mista&e and that they ,ould be ,ell ad+ised to go on to another .uestion$

@estio! 3

his ,as correctly ans,ered by about "5H of the candidates$

% maority incorrectly expressed

4= 45 75

( 4)(8 7)

x x

x x

+ +

as4 7

( 4) (8 7)x x

+ ha+ing failed to realise that the

algebraic fraction gi+en in the .uestion is improper thereby losing 8 of the " mar&s a+ailable$

For those achie+ing the correct partial fractions a process of long di+ision ,as typically used to find the +alue of

the constant term and the resulting remainder usually 1 "x became the -A of the subse.uent identity$ %minority of them ho,e+er applied

4= 45 75 ( 4)(8 7) (8 7) ( 4)x x A x x B x C x+ + + + + in order to obtain

the correct partial fractions$

Gold 8> 77?74 7#

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Question 4

he maority of candidates identified the need for some form of dot product calculation in part (a)$ a&ing the

dot product 7 4$l l ,as common among candidates ,ho did not correctly proceed ,hile others did not ma&e any

attempt at a calculation being unable to identify the +ectors re.uired$ % number of candidates attempted to

e.uate 7l and 4l at this stage$ he maority of candidates ho,e+er ,ere able to sho, that 8$q=

Bn part (b) the maority of candidates correctly e.uated the i 7 and 8components of 7l and 4l and although

some candidates made algebraic errors in sol+ing the resulting simultaneous e.uations most correctly found and $ Bn almost all such cases the +alue of p and the point of intersection in part (c) ,as then correctlydetermined$

here ,as a failure by many candidates to see the lin& bet,een part (d) and the other three parts of this .uestion,ith the maority of them lea+ing this part blan&$ hose candidates ,ho decided to dra, a diagram usually

increased their chance of success$ Most candidates ,ho ,ere successful at this part applied a +ector approach as

detailed in the mar& scheme$ he easiest +ector approach adopted by a fe, candidates is to realise that 7= atA 1= at the point of intersection and so == atB$ o substitution of == into l7 yields the correct position

+ector 0 77 7= $ + i 7 8 % fe, candidates by deducing that the intersection point is the midpoint of A andB

,ere able to ,rite do,n=

74

x+ = 8

04

y+ = and78

84

z+ = in order to find the position +ector ofB$

Question 5

Nearly all candidates &ne, the method for sol+ing part (a) although there ,ere many errors in differentiating trig

functions$ Bn particular ( )d

4cos4d

tt

,as often incorrect$ Bt ,as clear from both this .uestion and .uestion 4 that

for many the calculus of trig functions ,as an area of ,ea&ness$ Nearly all candidates ,ere able to obtain anexact ans,er in surd form$ Bn part (b) the maority of candidates ,ere able to eliminate tbut in manipulating

trigonometric identities many errors particularly ,ith signs ,ere seen$ he ans,er ,as gi+en in a +ariety of

forms and all exact e.ui+alent ans,ers to that printed in the mar& scheme ,ere accepted$ he +alue of k,asoften omitted and it is possible that some simply o+erloo&ed this$ Iomain and range remains an unpopular topic

and many did not attempt part (c)$ Bn this case inspection of the printed figure gi+es the lo,er limit and ,as

intended to gi+e candidates a lead to identifying the upper limit$

Question 6

Bn part (a) a maority of candidates ,ere able to subtract the gi+en position +ectors correctly in order to find $AB !ommonerrors in this part included some candidates subtracting the position +ector the ,rong ,ay round and a fe, candidates ,ho

could not deal ,ith the double negati+e ,hen finding the 8component of $AB

Bn part (b) a significant maority of candidates ,ere able to state a +ector e.uation of l7$ % significant number of these

candidates ho,e+er ,rote J -ine 9 J and omitted the 'r* on the left hand side of the +ector e.uation thereby losing one

mar&$

Gold 8 >77?74 70

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Many candidates ,ere able to apply the dot product correctly in part (c) to find the correct angle$ !ommon errors here

included applying a dot product formula bet,een OA and KOB or applying the dot product bet,een either OA or OB and

the direction +ector of l7$ Bnterestingly a surprising number of candidates either simplified4 4 4(7) ( 4) (4)+ + to 1 or

,hen finding the dot product multiplied L4 by 5 to gi+e L4$

Part (d) pro+ed more discriminating$ he maority of candidates realised that they needed to put the line 7l e.ual to line 4l $

% significant number of these candidates ho,e+er ,ere unable to ,rite l4as ( ) +i 8 or used the same parameter (usually) as they had used for l7$ uch candidates then found difficulty in ma&ing further progress ,ith this part$

Question 7

Part (a) ,as fully correct in the great maority of cases but the solutions ,ere often unnecessarily long and nearly

t,o pages of ,or&ing ,ere not unusual$ he simplest method is to e.uate the 7components$ his gi+es one

e.uation in leading to 8= ,hich can be substituted into the e.uation of 7l to gi+e the coordinates of C$ Bnpractice the maority of candidates found both and and many pro+ed that the lines ,ere coincident at C$Ao,e+er the .uestion ga+e the information that the lines meet at Cand candidates had not been as&ed to pro+e

this$ his appeared to be another case ,here candidates ans,ered the .uestion that they had expected to be set

rather than the one that actually had been$

he great maority of candidates demonstrated in part (b) that they &ne, ho, to find the angle bet,een t,o

+ectors using a scalar product$ Ao,e+er the use of the position +ectors of A andB instead of +ectors in the

directions of the lines ,as common$ !andidates could ha+e used either the +ectors

7

4

7

and

1

5

4

gi+en in the

.uestion or AC and BC$ he latter ,as much the commoner choice but many made errors in signs$

!omparati+ely fe, chose to use the cosine rule$ Bn part (c) many continued ,ith the position +ectors they hadused incorrectly in part (b) and so found the area of the triangle OAB rather than triangle ABC$ he easiest

method of completing part (c) ,as usually to use the formula %rea 9 74 sinab Cand most chose this$ %ttempts to

use %rea 9 74base height ,ere usually fallacious and often assumed that the triangle ,as isosceles$ % fe,complicated attempts ,ere seen ,hich used +ectors to find the coordinates of the foot of a perpendicular from a

+ertex to the opposite side$ Bn principle this is possible but in this case the calculations pro+ed too difficult to

carry out correctly under examination conditions$

Gold 8> 77?74 7/

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tatistics #or C4 ;ractice ;a&er G3

Mean score for students achieving grade:

uMa!score

Modalscore

Mean"

#$$ #% # & ' ( ) *

1 % 7/ 3"// $"% $"0< 3"