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Paper Reference(s)

6666/01

Edexcel GCECore Mathematics C4

Silver Level S2

Time: 1 hour 30 miutes

Materials re!uired "or examiatio #tems icluded $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.

#structios to Cadidates

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

the unit title (!ore Mathematics !") the paper reference (####) your surname initials

and signature$

#"ormatio "or Cadidates

% 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 Cadidates

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$

Su''ested 'rade (oudaries "or this %a%er:

&) & * C + E

66 ,- ,1 4, 3. 33

Silver 2 his publication may only be reproduced in accordance ,ith Edexcel -imited copyright policy$3455/64578 Edexcel -imited$

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1)74()7(

94

4

+ xx

x:

)7( x

A; 4)7( x

B;

)74( +x

C$

Find the +alues of the constantsAB and C$

4

ue 200.

2 (a)

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4

5i'ure 2

Figure 4 sho,s a s&etch of the cur+e ,ith e.uationy :x8ln (x4; 4) x 5$

he finite regionR sho,n shaded in Figure 4 is bounded by the cur+e the x?axis and the

linex : 4$

he table belo, sho,s corresponding +alues ofx andy fory :x8ln (x4; 4)$

x 5"

4

4

4

"

48 4

y 5 5$84"5 8$9475

(a) !omplete the table abo+e gi+ing the missing +alues ofy to " decimal places$

2

(b)

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,))(( 47

7514 4

+

+

xx

xxA;

7x

B;

4+x

C$

(a) Find the +alues of the constantsAB and C$

4

(b) Aence or other,ise expand ))(( 47

7514 4

+

+

xx

xx

in ascending po,ers ofx as far as the term

inx4$ Gi+e each coefficient as a simplified fraction$

ue 2010-

6 he areaA of a circle is increasing at a constant rate of 7$1 cm 4 s67$ Find to 8 significant

figures the rate at ,hich the radius r of the circle is increasing ,hen the area of the circle

is 4 cm4$

,January 2010

Relati+e to a fixed origin O the pointA has position +ector (4i6 ; 17)

the pointB has position +ector (1i ; 4 ; 757)

and the pointD has position +ector (6i ; ; "7)$

he line lpasses through the pointsA andB$

(a) Find the +ector AB $

2

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

2

(c) ho, that the si@e of the angleBAD is 759B to the nearest degree$

4

he pointsAB andD together ,ith a point C are the +ertices of the parallelogram ABCD

,here AB : DC $

(d) Find the position +ector of C$

2

(e) Find the area of the parallelogramABCD gi+ing your ans,er to 8 significant figures$

3

(f) Find the shortest distance from the point D to the line l gi+ing your ans,er to 8

significant figures$

2

January 2012

il+er 4= #>74 "

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- Cn an experiment testing solid roc&et fuel some fuel is burned and the ,aste products are

collected$ hroughout the experiment the sum of the masses of the unburned fuel and ,aste

products remains constant$

-etxbe the mass of ,aste products in &g at time tminutes after the start of the experiment$

Ct is &no,n that at time tminutes the rate of increase of the mass of ,aste products in &g per

minute is ktimes the mass of unburned fuel remaining ,here kis a positi+e constant$

he differential e.uation connectingxand tmay be ,ritten in the form

d( )

d

xk M x

t= ,hereMis a constant$

(a) Explain in the context of the problem ,hatd

d

x

tandMrepresent$

2

Gi+en that initially the mass of ,aste products is @ero

(b) sol+e the differential e.uation expressingxin terms of kMand t$

6

Gi+en also thatx:7

4M ,hen t: ln "

(c) find the +alue ofx,hen t: ln 9 expressingxin terms ofM in its simplest form$

4

ue 2013 8

T9T&L 598 &E8: , M&8;S

E74 1

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Duestion

Numbercheme Mar&s

1 ( ) ( ) ( ) ( )44

9 7 4 7 4 7 7x A x x B x C x= + + + + 7

7x 9 8 8B B= = M7

7

4x

49 8

7" 4

C C = =

%ny t,o ofAB C %7

4x terms 9 4 "A C A= + = %ll three correct %7 4

=4>

3 ( )d 4 ln 4$4d

x x

x= 7

d d

ln 4$4 4 4 4d d

x y yy y xx x

+ = + M7 %7: %7

ubstituting ( )8 4

d d

/ln 4 " " #d d

y y

x x+ = + M7

d

"ln 4 4d

y

x= %ccept exact e.ui+alents M7 %7

=>

il+er 4= #>74 #

2 (a) { }7 7

sin 8 d cos 8 cos 8 d8 8

= x x x x x x x M7 %7{ }

7 7cos8 sin8

8 9= + +x x x c %7

=3>

(b) { }4 47 4

cos8 d sin8 sin8 d8 8

= x x x x x x x x M7 %7{ }4

7 4 7 7sin 8 cos 8 sin 8

8 8 8 9

= + + x x x x x c %7 is$

{ }47 4 4sin8 cos8 sin8

8 9 40

= + +

x x x x x cIgnore

subsequent

working

=3>

6 mar7s

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il+er 4= #>74 0

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Duestion

Numbercheme Mar&s

4 (a) 5$5888 7$819# a,rt 5$5888 7$819# 7 7 2

(b) ( ) [ ]7 4%rea $$$4 "

R 7

( )$$$ 5 4 5$5888 5$84"5 7$819# 8$9475 + + + + M7 7$85 %ccept 7$8 %7 3

(c)4 d4 4

d

uu x x

x= + = 7

( ) ( )4

8 4

5%rea ln 4 dR x x x

= + 7

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

4ln 4 d ln 4 d 4 ln dx x x x x x x u u u

+ = + = M7Aence ( ) ( )

"

4

7

4%rea 4 ln dR u u u=

cso

%7 4

(d) ( )4 4 7

4 ln d 4 ln 4 d4 4

u uu u u u u u u

u

=

M7 %7

4

4 ln 4 d

4 4

u uu u u

=

( )

4 4

4 ln 44 "

u uu u u C

= +

M7 %7

( )

"4 4

4

7%rea 4 ln 4

4 4 "

u uR u u u

=

: ( ) ( )( )74 / / ln " " / 4 " ln 4 7 " + + M7

( )74 4ln 4 7= + 74ln 4 + %7 6 =1,>

il+er 4= #>74 /

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, (a) 4A = 7 ( ) ( ) ( ) ( )44 1 75 7 4 4 7x x A x x B x C x+ = + + + +

7x 8 8 7B B = = M7 %7 4x 74 8 "C C = = %7 4

(b)( ) ( )

( )74

74 1 754 7 4 7

7 4 4

x x xx

x x

+ = + + + +

M7

( ) 7 47 7 $$$x x x = + + + 7

7 4

7 7 $$$4 4 "

x x x

+ = + +

7

( ) ( )

( ) ( )4

44 1 75 74 7 4 7 7 7 $$$7 4 4

x xx x

x x

+ = + + + + + + + M7

1 $$$= + ft their 74A B C + %7 ft

48$$$ $$$

4x= + + 5x stated or implied %7 %7

=11>

Question

NumberScheme Marks

Q6

d

7$1d

A

t = B1

4 d 4

d

AA r r

r = = B1

When 4A=

( )44

4 5$090 //" $$$r r

= = = M1

d d d

d d d

A A r

t r t=

d

7$1 4d

rr

t= M1

4

d 7$15$499

d 4

r

t

= a,rt 5$499 A1

[5]

il+er 4= #>74 9

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?uestio

(b)

4 8

= 7 8

1 1

l

= +

r or

1 8

4 8

75 1

= +

r M7 %7ft

=2>

-et dbe the shortest

distance from Cto l$

(c)

7 4 8 8

7 7 4 or 4

" 1 7 7

AD OD OA DA

= = = =

uuur uuur uuur uuur

M7

4 4 4 4 4 4

8 8

8 4

1 7cos

$ (8) (8) (1) $ ( 8) (4) ( 7)

AB AD

AB AD

= =

+ + + +

uuur uuur

uuur uuur

%pplies dot product

formula bet,een

their ( )orAB BA

and their ( )or $AD DA

M7

4 4 4 4 4 49 # 1cos

(8) (8) (1) $ ( 8) (4) ( 7) + = + + + +

!orrect follo,ed

through expression or

e!uatio$

%7

/cos 759$5491""$$$ 759 (nearest )

"8$ 7"

= = = a,rt 759%7 cso

&G

=4>

(d) ( ) ( )" 8 8 1OC OD DC OD AB= + = + = + + + + +i 6 7 i 6 7

( ) ( )1 4 75 8 4OC OB BC OB AD= + = + = + + + + i 7 i 7 M7

o 4 " 9OC= + +i 7 %7

=2>

(e) ( )74%rea ( "8)( 7")sin759 4 48$79/9"951ABCD = = a,rt 48$4 M7 dM7%7

=3>

(f) sin077"

d= or "8 48$79/9"951$$$=d M7

7"sin07 8$180/5#1#8$$$d = = a,rt 8$1" %7=2>

1, mar7s

il+er 4= #>74 75

7"

B07

d

l

C

759

D

BA

-et

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?uestio

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?uestio 1

he maIority of candidates gained full mar&s on this .uestion$ Most obtained the identity

( ) ( ) ( ) ( )449 7 4 7 4 7 7x A x x B x C x + + + + and found B and C by substituting 7x= and

74

x= $ % significant number of candidates found an incorrect +alue of Cafter ma&ing the

error ( )

48 9

4 " = $ his can arise through the misuse of a calculator$ he +alue of A ,asusually found either by substituting 5x= or e.uating coefficients of 4x $ Relati+ely fe,candidates attempted the .uestion by e.uating all three coefficients to obtain three e.uations

and sol+ing these e.uations simultaneously$ he ,or&ing for this method is rather

complicated and errors ,ere often made$

?uestio 2

his .uestion ,as generally ,ell ans,ered ,ith around 15J of the candidature gaining all #

mar&s$ he maIority of candidates ,ere able to apply the integration by parts formula in the

correct direction$ ome candidates ho,e+er did not assign u andd

d

v

xand then ,rite do,n

theird

d

u

xand v before applying the by parts formula ,hich meant that if errors ,ere made

the method used ,as not al,ays clear$

Cn part (a) sin 8 dx x caused some problems for a minority of candidates ,ho producedresponses such as cos8x or 8cos8x or

7cos8

8x $ %fter correctly applying the by parts

formula a fe, candidates then incorrectly ,rote do,n7

cos 8 d8

x x as 7 cos8# x $Most candidates ,ho could attempt part (a) ,ere able to ma&e a good start to part (b) by

assigning u as 4x andd

d

v

xas cos 8 x and then correctly apply the integration by parts formula$

%t this point ,hen faced ,ith4

sin 8 d8

x x x some candidates did not ma&e the connection,ith their ans,er to part (a) and made little progress$ Kther candidates independently applied

the by parts formula again ,ith a number of them ma&ing a sign error$

Question 3

his .uestion ,as also ,ell ans,ered and the general principles of implicit differentiation

,ere ,ell understood$ y far the commonest source of error ,as in differentiating 4x

examples such as 4x 4 lnx x and 74xx ,ere all regularly seen$ hose ,ho &ne, ho, to

differentiate 4x nearly al,ays completed the .uestion correctly although a fe, had difficulty

in finding ( )d

4d

xyx

correctly$ % minority of candidates attempted the .uestion by ta&ing the

logs of both sides of the printed e.uation or a rearrangement of the e.uation in the form

il+er 4= #>74 74

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44 4x xy y= $ !orrectly done this leads to .uite a neat solution but more fre.uently errors

such as ( )4 4ln 4 ln 4 lnx xy y+ = + ,ere seen$

?uestio 4

Part (a) ,as ,ell done and the only error commonly seen in part (b) ,as using the incorrect

,idth of the trape@ium4

1instead of

4

"$ % fe, candidates made errors often due to a lac&

of clear brac&eting but great maIority completed part (b) correctly and ga+e their ans,er to

the degree of accuracy specified in the .uestion$ Part (c) ,as ,ell done and the maIority ,ere

able to findd

d

u

xand ma&e a complete substitution for the +ariables$ he only common error in

this part ,as simply to ignore the limits and to gi+e no Iustification for the limits becoming 4

and "$ Most recognised that the integral in part (d) re.uired integration by parts and those ,ho

used a method in+ol+ing integrating ( )4u to4

4

4

uu and differentiating lnu usually

reached the half ,ay stage correctly$ he second integration pro+ed more difficult and there

,ere many errors in simplifying the expression

4 74

4

uu

u

before the second integration$

he errors often arose from a failure to use the necessary brac&ets$ here ,ere also many

subse.uent errors in signs and a fe, candidates omitted the 74 from their integration$

hose ,ho at the first stage of integration by parts integrated ( )4u to ( )4

4

4

u ,hich is

of course correct had mar&edly less success ,ith the second integral than those ,ho

integrated to4

44

uu $

% fe, split the integral up into t,o separate integrals ln du u u and ln du u but the secondof these integrals ,as rarely completed correctly$ hose ,ho ignored the hint in the .uestion

and attempted to integrate ,ith respect tox,ere generally unable to deal ,ith1

4d

4

xx

x +

,hich arises after integrating by parts once

Question 5

he first part of .uestion 1 ,as generally ,ell done$ hose ,ho had difficulty generally tried

to sol+e sets of relati+ely complicated simultaneous e.uations or did long di+ision obtaining

an incorrect remainder$ % fe, candidates found B and C correctly but either o+erloo&ed

findingAor did not &no, ho, to find it$ Part (b) pro+ed +ery testing$ Nearly all ,ere able to

ma&e the connection bet,een the parts but there ,ere many errors in expanding both ( ) 7

7x

and ( ) 7

4 x + $ Fe, ,ere able to ,rite ( )

77x as ( )

77 x

and the resulting expansions

,ere incorrect in the maIority of cases both4

7 x x+ and4

7 x x being common$

il+er 4= #>74 78

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method of base perpendicular height$ he most common error in part (e) ,as for candidatesto find the product of lengthsADandAB$

!andidates ,ho ,ere successful in part (f) usually found the shortest distance by multiplying

their AD by sin 07(or e.ui+alent)$ hose candidates ,ho multiplied AB by sin 07did notrecei+e any credit$ % fe, candidates attempted to use +ectors to find DE ,hereEis the

point ,here the perpendicular from Dmeets the line l often spending considerable time forusually little or no re,ard$

Question 8

Cn general this ,as the most poorly ans,ered .uestion on the paper ,ith about 71J of

candidates ,ho failed to score and about 77J of candidates gaining 7 mar& usually in

part (a)$ his .uestion discriminated ,ell bet,een candidates of higher abilities ,ith about

40J of candidates gaining at least / of the 74 mar&s a+ailable and only about 0J of

candidates gaining all 74 mar&s$ Many ,ea&er candidates made little or no progress in part

(b) maybe because of the generalised nature of the differential e.uation$

Cn part (a) a significant number of candidates ,ere not clear or precise in their explanations$% number of them used the ,ord Lmass and it ,as not clear ,hether they ,ere referring to

the mass of the unburned fuel or the mass of the ,aste products$

Cn part (b) those candidates ,ho ,ere able to separate the +ariables ,ere usually able to

integrate both sides correctly although a number of candidates integratedxM

7incorrectly

to gi+e ln (M6 x)$ Many others substituted t: 5x: 5 immediately after integration to find

their constant of integration as 6lnMand most used a +ariety of correct methods to eliminate

logarithms in order to findx:M(7 6 e6kt) (or e.ui+alent)$ % significant number of candidates

ho,e+er correctly rearranged their integrated expression into the form x: M6 Ae6ktbefore

using t : 5 x : 5 to correctly find A$ !ommon errors in this part included omitting the

constant of integration or treating Mas a +ariable$ %lso a number of candidates struggled to

remo+e logarithms correctly and ga+e an e.uation of the formM6x: e6kt; ec,hich ,as then

sometimes manipulated toM6x:Ae6kt$

Cn part (c) some candidates ,ere able to substitute t : ln " x : 47 M into one of their

e.uations in+ol+ing x and t but only a minority ,ere able to find a numeric +alue of k$ Knly

the most able candidates ,ere able to find k : 47 and substitute this into their e.uation

together ,ith t: ln 9 to findx: 84

M$

Statistics "or C4 ractice a%er Silver Level S2

Mean score for students achieving grade:

QuMaxscore

Modalscore

Mean%

ALL A* A B C D E U

4 84 3.34 3.91 3.70 3.45 3.17 2.79 2.36 1.66

! 6 71 4.28 5.83 5.13 4.10 3.12 2.24 1.46 0.51

" 7 74 5.20 6.72 6.02 5.43 4.70 3.95 2.91 1.45

# 15 67 9.99 14.23 12.43 10.19 7.93 5.88 4.41 2.94

$ 11 68 7.49 10.31 8.79 7.52 6.39 5.34 4.19 2.57

5 65 3.23 4.34 3.12 2.26 1.59 0.86 0.56

& 15 60 9.04 13.94 10.95 7.98 6.17 4.54 3.55 1.61

8 12 41 4.93 9.74 5.48 3.21 1.63 1.20 0.95 0.33 &$ " #&'$( $'8# #$'(( "$'"& !&'$" !(') '"

il+er 4= #>74 71