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Mark Scheme (Results)Summer 2007

GCE

GCE Mathematics (6666/01)

Edexcel Limited. Registered in England and Wales No. 4496750Registered Ofce: One90 Hig Hol!orn" London W#$% 7&H

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QuestionNumber Scheme Marks

Aliter 1. 3f(&) (3 2&)−= +

Way 2

3 $ % 2

3

( 3)( $)(3) ( 3)(3) (* * &)' (3) (* * &)

2( 3)( $)( %)

(3) (* * &) """3

− − −

−

− −+ − + = − − − + +

with * * 1≠

127 or 3(3) − (See note ↓ )

&pands 3(3 2&)−+ togi e an un+simp ified or

simp ified3 $(3) ( 3)(3) (* * &)− −+ − '

, correct un+simp ifiedor simp ified }{""""""""""

e&pansion -ithcandidate.s fo o-ed

thro. ( )* * x

#1

M1

,1

3 $ % 2

3

( 3)( $)(3) ( 3)(3) (2&)' (3) (2&)

2( 3)( $)( %)

(3) (2&) """3

− − −

−

− −+ − + = − − − + +

21 1 127 /1 2$3

3172

( 3)( )(2&)' ( )( )($& )

( 10)( )(/& ) """

+ − + = + − +

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

27 27 /1 72= − + − +

,n!thing that

cance s to1 2&

'27 27

−

Simp ified 2 3/ & /0&

/1 72−

,1'

,1

[5]

5 marks

,ttempts using Mac aurin e&pansions need to be esca ated up to !our team eader"

6f !ou fee the mark scheme does not app ! fair ! to a candidate p ease esca ate the response up to !our teameader"

01 4ore Maths 4$ 25 th June 2007June 2007 A !ance Su"si iar#/A !ance $e!el in GCE Mathematics %ersi&n ' *E +,-A$ %ERS,.-

(

Special Case 5 6f !ou see theconstant 1

27 in a candidate.sfina binomia e&pression then!ou can a-ard #1

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2.1

2

0

2d

(2 1)

x

x x

+∫ -ith substitution 2 x u =

d2 " n2

d x u

x =

d 1d 2 " n 2 x

x u

= d

d 2 " n 2 x u x = or d

d " n2u x u=

or ( ) d1d n2u

u x = #1

2 2

2 1 1d d

*n2(2 1) ( 1)

x

x x u

u = ÷+ + ∫ ∫ 2

1d

( 1)k u

u +∫ -here k is constant

M1

1 1

n2 ( 1) c u −

= + ÷ ÷ +

2 1

2 1

( 1) ( 1)

( 1) 1"( 1)

u a u

u u

− −

− −

+ → ++ → − +

M1

,1

change imits5 -hen x 9 0 : x 9 1 then u 9 1 : u 9 2

1 2

210

2 1 1d

n 2 ( 1)(2 1)

x

x x u

−= ++ ∫

1 1 1n2 3 2

= − − − ÷ ÷ 4orrect use of imits

u 9 1 and u 9 2 depM1

1n2

= 1n2 or

1 1n $ n /− or

1 12 n2 3n2− ,1 aef

Exact al!e only" [6] , ternati e ! candidate can re ert back to x ;

1 1

200

2 1 1d

n2(2 1) (2 1)

x

x x x −= + + ∫

1 1 1

n2 3 2

= − − − ÷ ÷ 4orrect use of imits

x 9 0 and x 9 1 depM1

1n2

= 1n2 or

1 1n$ n/− or

1 12 n 2 3 n 2− ,1 aef

Exact al!e only"# marks


4

6f !ou see this inte$ration app ied an!-here in acandidate.s -orking then !ou can

a-ard M1 ,1

There are other acceptab eans-ers for ,1 eg5

1 12*n/ *n $or

N#5 <se !our ca cu ator to checkeg" 0"2$0$$ ;

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%.(a)dd

d 1

d 2

1

cos2 sin 2

u x

v

x

u x

x v x

= = = = (see note below)

1 12 26nt cos 2 d sin2 sin2 "1 d x x x x x x x = = −∫ ∫

<se of =integration b!parts. formu a in the

correct direction"4orrect e&pression"

M1

,1

( )1 1 12 2 2sin2 cos2 x x x c = − − +

12sin2 cos 2 x x → −

or 1sin cosk kx kx → −-ith 1 0k k ≠ >

dM1

1 12 $sin2 cos 2 x x x c = + + 4orrect e&pression -ith > c ,1

[4]

(b) ( )cos 2 122cos d d x x x x x x +=∫ ∫

Substitutes correctly for 2cos x in the

gi en integraM1

1 1

cos 2 d d2 2

x x x x x = +∫ ∫ 1 1 1 1sin 2 cos 2 ' d

2 2 $ 2 x x x x x = + + ÷ ∫ ( )1 their ans-er to (a) '2

or under ined e&pression ,1'

21 1 1sin2 cos2 ( )

$ / $ x x x x c = + + + 4omp ete ! correct

e&pression -ith -ithout > c ,1

[3]

& marks

Notes:

(b) 1 12 26nt cos 2 d sin2 sin2 "1 d x x x x x x x = = ±∫ ∫ This is acceptab e for M1 M1

dd

dd

1

cos2 sin2

u x

v x

u x

x v x λ

= = = =

6nt cos 2 d sin2 sin2 "1 d x x x x x x x λ λ = = ±∫ ∫ This is a soacceptab e for M1 M1


5

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Aliter

%.(b) Way 2 ( )cos 2 12

2cos d d x x x x x x +=

∫ ∫ Substitutes correctly

for2

cos x in thegi en integra ;M1

dd

d 1 1 1 1d 2 2 $ 2

1

cos2 sin2

u x

v x

u x

x v x x

= = = + = +

; oru x = and d 1 1

d 2 2cos2v x

x = +

( )21 1 1 1$ 2 $ 2sin2 sin2 d x x x x x x = + − +∫

2 21 1 1 1$ 2 / $sin2 cos2 x x x x x c = + + − + ( )

1their ans-er to (a) '2

or under ined e&pression ,1

21 1 1sin2 cos2 ( )



[3]

Aliter (b)

Way %( )2cos 2 d 2cos 1 d x x x x x x = −∫ ∫

Substitutes correctly for cos2 x in

cos 2 d x x x

∫

M1

2 1 12 $2 cos d d sin2 cos2 x x x x x x x x c − = + +∫ ∫

2 1 1 1 1cos d sin2 cos 2 ' d

2 2 $ 2 x x x x x x x x = + + ÷ ∫ ∫

( )1their ans-er to (a) '

2or under ined e&pression

,1'

21 1 1sin2 cos2 ( )



[3]

& marks


6

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'. (a) , method of ong di ision gi esWay 1

22($ 1) $2

(2 1)(2 1) (2 1)(2 1) x

x x x x + ≡ ++ − + −

2 A = #1

$(2 1)(2 1) (2 1) (2 1)

B C x x x x

≡ ++ − + −

$ (2 1) (2 1)B x C x ≡ − + +or their remainder (2 1) (2 1)Dx E B x C x + ≡ − + +

?orming an! one of these t-oidentities" 4an be imp ied" M1

@et 12&= − $ 2 2B B= − = −

See note below

@et 12&= $ 2 2C C = = either one of 2B = − or 2C = ,1

both B and C correct ,1[4]

Aliter

'. (a)22($ 1)

(2 1)(2 1) (2 1) (2 1)

x B C A

x x x x

+ ≡ + ++ − + −Way 2

See below for the award of B1decide to award B1 here!! …

… for 2 A = #1

22($ 1) (2 1)(2 1) (2 1) (2 1) x A x x B x C x + ≡ + − + − + + ?orming this identit!"4an be imp ied" M1

Auate & 2 / $ 2 A A= =

@et 12&= − $ 2 2B B= − = −

See note below

@et 12&= $ 2 2C C = =either one of 2B = − or 2C = ,1

both B and C correct ,1[4]


7

6f a candidate states one of either B or C correct ! then the methodmark M1 can be imp ied"

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Aliter 5. (a)

Way %6fl 1 and l 2 intersect then5

1 1 1 20 1 3 11 0 1

÷ ÷ ÷ ÷+ λ = + µ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷− −

,n! t-o of

5 1 1 2 (1)5 3 (2)5 1 (3)

+ λ = + µλ = + µ

− = − µ

i *k

Erites do-n an! t-o of theseeAuations M1

(1) : (2) !ie*ds 3(3) !ie*ds 7

µ µ

==

either one of the µ .s correct ,1both of the µ .s correct ,1

ither5 These eAuations are then inconsistentBr5 3 7≠Br5 @ines l 1 and l 2 do not intersect

4omp ete method gi ing rise toan! one of these three

e&p anations"#1

[4]

Aliter 5. (a)

Way ' ,n! t-o of

5 1 1 2 (1)5 3 (2)5 1 (3)

+ λ = + µλ = + µ

− = − µ

i *k

Erites do-n an! t-o of theseeAuations M1

(1) : (2) !ie ds 3(3) FDS 3 3

µ == − =

3 µ = ,1FDS of (3) 9 3 ,1

(3) !ie ds 1 3− ≠ 4omp ete method gi ing rise tothis e&p anation" #1

[4]


$0

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5. (b)2 %

1 1 : 2 %1 $

OA OBλ µ ÷ ÷= = = = ÷ ÷ ÷ ÷−

***r ***r

-nly one o either211

OA ÷= ÷ ÷−

***r or

%%$

OB

÷= ÷ ÷

***r

or

(2 1 1) A − or (% % $)B "(can be imp ied)

#1

% 2 3% 1 $$ 1 %

AB OB OA ÷ ÷ ÷= − = − = ÷ ÷ ÷ ÷ ÷ ÷−

***r ***r ***r or

3$%

BA− ÷= − ÷ ÷−

***r ?inding the difference bet-eentheir B# and B, "(can be imp ied)

M1

,pp !ing the dot productformu a bet-een Ga o-ab eH

ectors" See notes below.M1

,# 3 $ %= + +i * k " 1 0= + +( i * k : θ is ang e

1

1

3 $ 0cos

%0 " 2"

AB

ABθ

• + += = ± ÷ ÷ (

(***r

,pp ies dot product formu abet-een 1( and their ,#"± M1

4orrect e&pression" ,1

710cos θ = 7

10 or 0"7 or

7

100 ,1 caobut not

7%0 2 [6]

1) marks

Note: 6f candidate use cases 2 3 $ and % the! cannot gain the fina three marks for this part"Note: 4andidate can on ! gain some a of the fina three marks if the! use case 1"


$$

4andidates can score this mark if there is a comp ete method for finding thedot product bet-een their ectors in the fo o-ing cases5

Case 2 5 1 0= + +( i * k

and 2 2 1= + −( i * k

2 1 0cos2 "

θ + + =

Case % 5 1 0= + +( i * k

and 2 2(2 1 )= + −( i * k

$ 2 0cos2 " 2$

θ + + =

Case 1 5 their ft ( ) ,# 3 $ %± = ± + +i * k

and 1 0= + +( i * k

3 $ 0cos

% 0 " 2θ

+ + =± ÷ ÷

Case ' 5 their ft ( ) ,# 3 $ %± = ± + +i * k

and 2 2= + −( i * k

$ %cos

% 0 " θ

+ − =± ÷ ÷

Case 5 5 their ft B, 2 1 1= + −i * k

and their ft B# % % $= + +i * k

10 % $cos

"θ

+ − = ± ÷ ÷

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#. (c)2

22

sintan

cost

x t t

= = siny t =

Way 1

2

2

sin1 sin

t x

t = −

<ses 2 2cos 1 sint t = − M1

2

21y

x y

= −iminates = t’ to -rite an eAuation

in o ing x and y " M1

2 2 2 2(1 ) x y y x xy y − = − =

2 2 2 (1 ) x y xy x y x = + = +Fearranging and factorising -ith

an attempt to make 2y the subKect" ddM1

2

1 x

y x

= + 1 x

x + ,1

[4] Aliter #. (c) 2 21 cot sect co t + = <ses 2 21 cot sect co t + = M1

Way 2

2

1sin t

= <ses 22

1cos

sinec t

t = M1 imp ied

Dence 2

1 11

x y + = iminates = t’ to -rite an eAuation

in o ing x and y " ddM1

Dence2 1

1 or (1 ) 1

x y

x x = − + +

11 or

(1 ) 1 x

x x − + + ,1

[4]


$4

1

11 x + is an acceptab e response for the fina accurac! ,1 mark"

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Aliter #. (c) 2tan x t = siny t =

Way % 2 21 tan sect t + = <ses 2 21 tan sect t + = M1

2

1cos t

= <ses 22

1sec

cost

t = M1

2

11 sin t

= −

Dence 2

11

1 x

y + = −

iminates = t’ to -rite an eAuationin o ing x and y " ddM1

Dence2 1

1 or (1 ) 1

x y

x x = − + +

11 or

(1 ) 1 x

x x − + + ,1

[4] Aliter #. (c) 2 2 2sin 1 cosy t t = = − <ses 2 2sin 1 cost t = − M1

Way '

2

11

sec t = − <ses 2

2

1cos

sect

t = M1

2

11

(1 tan )t = − + then uses 2 2sec 1 tant t = + ddM1

Dence 2 11 or

(1 ) 1 x

y x x

= − + +1

1 or (1 ) 1

x x x

− + + ,1

[4]


$5

1


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Aliter #. (c) 2tan x t = siny t =

Way 5 2tan tan x t t x = =

Lra-s a right+ang ed triang e andp aces both x and 1 on the

triang e

<ses !thagoras to deduce theh!potenuse

M1

M1

Dence sin1

x y t x

= =+

iminates = t’ to -rite an eAuationin o ing x and y " ddM1

Dence 2

1 x

y x

= + 1 x

x + ,1

[4]

12 marks


1

x (1 ) x +

t

$6

1

1


There are so man! -a!s that a candidate can proceed -ith part (c)" 6f a candidate produces a correct so utionthen p ease a-ard a four marks" 6f the! use a method commensurate -ith the fi e -a!s as detai ed on themark scheme then a-ard the marks appropriate !" 6f !ou are unsure of ho- to app ! the scheme p easeesca ate !our response up to !our team eader"

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&. (a)& 0 1

π /π 3

1π

$π

! 0 0"$$% % 27; 0" $3% $2%2; 0"/17$21 $ ; 1

nter marks into e N in the correct order" 0"$$ or a-rt 0"$$ 00 #1a-rt 0" $3% #1a-rt 0"/17$2 #1

0 can beimp ied

[3]

(b)Way 1 ( ) }{1

,rea ' 0 2 0"$$ 00 0" $3% 0"/17$2 1

2 1

π≈ × × + + + +

Butside brackets12 1

π× or 32π

?or structure of trape ium ru e { }"""""""""""""

'4orrect e&pressioninside brackets -hich amust be mu tip ied b! 2

h"

#1

M1

,1

$"/1$02""" 0"$72 1%30/""" 0"$7232π= × = = ($dp) for seeing 0"$72 ,1 cao

[4]

Aliter (b)

Way 2

}{ 0 0"$$ 00 0"$$ 00 0" $3% 0" $3% 0"/17$2 0"/17$2 11 2 2 2 2 ,rea + + + +π≈ × + + +

-hich is eAui a ent to5

( ) }{1 ,rea ' 0 2 0"$$ 00 0" $3% 0"/17$2 1

2 1π≈ × × + + + +

1π

and a di isor of 2 ona terms inside brackets"

Bne of first and astordinates t-o of the

midd e ordinates insidebrackets ignoring the 2"

4orrect e&pression insidebrackets if 1

2 -as to befactorised out"

#1

M1

,1

2"$0701""" 0"$72 1%30/""" 0"$721π= × = = 0"$72 ,1 cao

[4]

Question

Number

Scheme Marks


$7

6n (a) for 1 x π = -riting 0"$$% % ; then 0"$% 00 gains #1 for a-rt 0"$$ 00 e en though 0"$% 00 is incorrect"

6n b ou can fo o- thou h a candidate.s a ues from art a to a-ard M1 ft ,1 ft

{ }12 20 0 2(0"$$ 00 0" $3% 0"/17$2) 1 0"37/1 Area π = × × + + + + = gains #0M1,1,0

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&. (c) Oo ume ( ) ( ) ( )$ $

2

0 0

tan d tan d x x x x

π π

π π = =∫ ∫ ( )

2tan & d&∫ or tan& d&∫

4an be imp ied"6gnore imits and ( )π

M1

( ) [ ] $0nsec x π

π = or ( ) [ ] $0ncos x π

π = −tan nsec x x

→or tan ncos x x → − ,1

( ) ( ) ( )$nsec nsec 0π π = − or

( ) ( ) ( )$ncos ncos 0π π = − −

The correct use of imitson a function other than

tan &' ie$ x π = ‘ minus. 0 x = "

n(sec 0) 0= ma! beimp ied" 6gnore ( )π

dM1

( ) ( )12

1 11n n n 2 n1π π = − = −

or

( ) ( )12

n n 1π = − −

n 2π = or 22

nπ or 12 n2π or ( )1

2*nπ − or

( )12 2nπ

n 2π or 2

2nπ

or12 n2π or ( )1

2*nπ −

or ( )12 2nπ

,1 aef

m!st be exact. [4]

11 marks

eware: 6n part (c) the factor of π is not needed for the first three marks"

eware: 6n part (b) a candidate can a so add up indi idua trape ia in this -a!5

( ) ( ) ( ) ( )1 1 1 12 1 2 1 2 1 2 1 ,rea " 0 0"$$ 00 " 0"$$ 00 0" $3% " 0" $3% 0"/17$2 " 0"/17$2 1π π π π≈ + + + + + + +


$)

6f a candidate gi es the correct e&act ans-er and then -rites 1"0//77 ; then such a candidate canbe a-arded ,1 (aef)" The subseAuent -orking -ou d then be ignored" (is-)

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3. (a)ddP

kP t

= and 00 (1)t P P = =

dd

P k t

P =∫ ∫

Separates the ariab es

-ithdP P ∫ and dk t ∫ on

either side -ith integrasigns not necessar!"

M1

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

4orrect eAuation-ith -ithout > c"

,1

Ehen 0 00 nt P P P c = = =( )0or kt P Ae P A= =

<se of boundar! condition(1) to attempt to find theconstant of integration"

M1

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

Dence 0kt P P e= 0

kt P P e= ,1

[4]

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

0 02 t P P e = Substitutes 02P P = into an

e&pression in o ing P M1

2"% 2"%2 n n2t t e e= = or 2"% n2t =;or 2 n n2kt kt e e= = or n2kt =

iminates 0P and takes

n of both sidesM1

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

0"2772%//72""" 2$ 0 3 "2%277 """ minutest = × × =

3 mint = or

hr 3 minst = (to nearest minute)

a-rt 3t = or

hr 3 mins ,1[3]


$9

0kt P P e= -ritten do-n -ithout the first M1 mark gi en scores a four marks in part (a)"

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''

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d

cosdP

P t t

λ λ = and 00 (1)t P P = =

Aliter 3. +c,

Way 2d cos dP t t

P λ

λ =∫ ∫ Separates the ariab es

-ithdP

P λ ∫ and cos dt t λ ∫ on either side -ith integrasigns not necessar!"

M1

( )1 1n sin 'P t c λ λ λ = +Must see 1 n P λ and

1 sin t λ λ '4orrect eAuation-ith -ithout > c"

,1

Ehen 10 00 nt P P P c λ = = =

( )sin 0or t

P Ae P Aλ = =

<se of boundar! condition(1) to attempt to find the

constant of integration"M1

1 1 10 0n sin n n sin nP t P P t P λ λ λ λ λ = + = +

0 0sin n nn sin "t P P P t e e e eλ λ + = =

Dencesin

0t P P e λ = sin

0t P P e λ = ,1

[4]


'(

0kt P P e= -ritten do-n -ithout the first M1 mark gi en scores a four marks in part (a)"

sin0

t P P e λ = -ritten do-n -ithout the first M1 mark gi en scores a four marks in part (c)"

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