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Paper Reference(s)

6666/01

Edexcel GCECore Mathematics C4

Silver Level S4

Time: 1 hour 30 minutes

Materials reuired !or examination "tems included #ith uestion

$a$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.

"nstructions to Candidates

Write the name of the examining body (Edexcel) your centre number candidate numberthe unit title (!ore Mathematics !") the paper reference (####) your surname initialsand signature$

"n!ormation !or Candidates

% 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 Candidates

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 'oundaries !or this $a$er:

%( % ) C * E

6+ +, +0 44 3- 31

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

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1. (a) Find the binomial expansion of

(7 6 /x) x /

7

in ascending po,ers ofxup to and including the term inx8 simplifying each term$4

(b) ho, that ,henx9755

7 the exact +alue of (7 6 /x) is

1

48$

2

(c) ubstitute x9755

7 into the binomial expansion in part (a) and hence obtain an

approximation to 48$ Gi+e your ans,er to 1 decimal places$

3

anuar 2010

2. f (x) 9)":(

74x+

|x|;4

8$

Find the first three non

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3.

5i&ure 1

% hollo, hemispherical bo,l is sho,n in Figure 7$ Water is flo,ing into the bo,l$

When the depth of the ,ater is h m the +olume V m8is gi+en by

V974

7h4(8 6 "h) 5 h5$41$

(a) Find in terms of h

V

d

d,hen h 9 5$7$

4

Water flo,s into the bo,l at a rate of/55

m8s67$

(b) Find the rate of change of h in m s67 ,hen h 9 5$7$2

une 2011

il+er "> /?74 8

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4.

5i&ure 1

Figure 7 sho,s the cur+e ,ith e.uation

y9

+"84

4x

x x5$

he finite region S sho,n shaded in Figure 7 is bounded by the cur+e the x

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+.

5i&ure 2

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

x9 7 64

7t y9 4t6 7$

he cur+e crosses they

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6. % cur+e has parametric e.uations

x9 tan4t y9 sin t 5 ; t;4

$

(a) Find an expression forx

y

d

din terms oft. 2ou need not simplify your ans,er$

(3)

(b) Find an e.uation of the tangent to the cur+e at the point ,here t9"

.

Gi+e your ans,er in the formy9 ax + b ,here a andb are constants to be determined$(5)

(c) Find a cartesian e.uation of the cur+e in the formy49 f(x)$()

June !""#

-. % cur+e is described by the e.uation

x4B "xy By4B 40 9 5

(a) Findd

d

y

xin terms ofx andy.

+

% point Q lies on the cur+e$

he tangent to the cur+e at Q is parallel to they /?74 #

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,. (a) Asing the identity cos 49 7 6 4 sin4 find

$sin

4 d

2

5i&ure 4

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

x 9 tan y 9 4 sin 4 5 ;4

$

he finite shaded region S sho,n in Figure " is bounded by C the linex 98

7

and thex /?74 78

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?uestion 1

his pro+ed a suitable starting .uestion and there ,ere many completely correct solutions$he maority of candidates could complete part (a) successfully$ On part (b) those ,horealised that ,or&ing in common (+ulgar) fractions ,as needed usually gained the methodmar& but as noted in the introduction the ,or&ing needed to establish the printed result ,as

fre.uently incomplete$ Ot is insufficient to ,rite do,n / 487 755 1 = $ he examiners

accepted for example/ :4 48 48

7755 755 41 1

= = = $ On part (c) most candidates realised

that they had to e+aluate their ans,er to part (a) ,ith 5$57x= $ Co,e+er many failed torecognise the implication of part (b) that this e+aluation needed to be multiplied by 1$ Ot ,asnot uncommon for candidates to confuse parts (b) and (c) ,ith the expansion and decimalcalculation appearing in (b) and fraction ,or& leading to 48 appearing in (c)$

?uestion 2

Many candidates got off to a +ery bad start to this .uestion by ,riting ( )4: " 8 4x x+ = + or

( ) ( )74 74: " 8 4x x

+ = + $ uch errors in algebra are hea+ily penalised as the resulting binomial

expansions are significantly simplified and in this case ga+e ans,ers in incorrect po,ers

ofx$ hose ,ho obtained

7

447 "

78 :

x

+

sho,ed that they understood the binomial theorem

but there ,ere many errors in signs often due to the failure to use brac&ets correctly$ ome

candidates seemed to lose the thread of the .uestion and ha+ing expanded

7

4

4"7 :x

+

correctly failed to multiply by7

8$ Ot ,as not unusual to see an often correct term in #x

pro+ided$ he examiners ignore this but such additional ,or& does lose time$

?uestion 3

his .uestion ,as ,ell done and full mar&s ,ere common$ !andidates ,ere roughly e.uallydi+ided bet,een those ,ho expanded and differentiated and those ,ho differentiated usingthe product rule$ he latter method ,as the more complicated and more subect to error butmany correct solutions ,ere seen using both methods$ Of the differentiation ,as correctnearly all completed part (a) correctly$ Rather oddly a number of cases ,ere seen ,here

( )47

8 "74

h h ,as misread as ( )47

8 "4

h h $ Part (b) ,as generally ,ell done although

there ,ere a minority of students ,ho made no attempt at it at all$ he large maority

correctly interpreted/55

as

d

d

V

tand realised they had to di+ide

/55

byd

d

V

h$ On+erting

d

d

V

h

did cause difficulty for some candidates$ For example4 4

7 7 7

5$1 5$1h h h h

=

,as seen

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from time to time and41

741

= leading to the ans,er

4

84

instead of the correct

7

84 ,as

relati+ely common$

?uestion 4

%t least :5 of the candidature ,as able to apply the +olume of re+olution formula correctly$Qnly a fe, candidates did not include in their +olume formula or did not s.uare theexpression fory$he integration ,as ,ell attempted and the maority of candidates recognised that the integral

could be manipulated into the formf ( )

df ( )

xx

x

and integrated to gi+e their result in the form

kln (8x4B ") usually ,ith78

k= $ % +ariety of incorrect +alues of k,ere seen ,ith the most

common being either 8 or 7$ % significant number of candidates integrated incorrectly to gi+e

ans,ers such asx4ln (8x4B ") or 4xln (8x4B ")$

hose candidates ,ho applied the substitution 48 "u x= + proceeded to achie+e7

ln8

u and

changed theirx

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ha+ing correctly establishedd

d

y

xas being ( 4)4 ln 4t then proceeded incorrectly to e.uate this

to " ln 4t $ Most &ne, ho, to obtain the gradient of the normal and could ,rite do,n thee.uation of a straight line$

J1(d) ,as ans,ered ,ell by small number of candidates and although a significant number

could ,rite the area as ( )5

"

74 7 $ d

4t t

many ,ere unable to perform the integration of

4t ,ith respect to t$ ome ,rote 4t as 4t thus simplifying the problem ,hilst attempts such

as74

7

t

t

+

+ ,ere not uncommon$ !andidates ,ho ,ere unable to ma&e an attempt at the

integration of 4t ,ere unable to access the final " mar&s in this part$ %pproaches thatfacilitated integration included re

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Question 7

his .uestion discriminated ,ell bet,een candidates of all abilities ,ith about /5 ofcandidates gaining at least 1 mar&s of the 74 mar&s a+ailable and about "1 gaining at least /mar&s$ Qnly about 71 of candidates gained all 74 mar&s$ Part (a) ,as ans,ered ,ell ,ithfull mar&s commonly a,arded$ Part (b) ,as far more challenging ,ith only a small minority

presenting a complete and correct solution$

On part (a) many candidates ,ere able to differentiate correctly factorise outd

d

y

x and

rearrange their e.uation to arri+e at a correct expression for the gradient function$ % minoritydid not apply the product rule correctly ,hen differentiating " xy ,hilst a small number leftthe constant term of 40 in their differentiated e.uation$

On part (b) those small proportion of candidates ,ho realised they needed to set the

denominator of theirdd

y

x expression e.ual to =ero usually ,ent on to ans,er this part

correctly$ ome ho,e+er did not attempt this part ,hile a maority attempted to sol+ed

5d

y

x= and many proceeded to obtain coordinates of ( # 8) for the point Q despite a

number of them initially s&etching a cur+e ,ith a +ertical tangent$ % smaller proportion

sol+edd

7d

y

x= presumably because the digit 7 is ,ritten as a +ertical line ,hilst others either

substituted 5y= or 5x= into their

d

d

y

x expression$ Manipulation and brac&eting errorssometimes led to candidates ,riting e.uations such as 4y A= or 4x A= ,here A ,as

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negati+e$ Examiners ,ere surprised that a fair number of candidates ha+ing obtained their+alue of x (ory) then proceeded to substitute this into 4 4" 40 5x xy y+ + + = rather thanusing the much simpler y kx= or $x ky=

Question 8

he responses to this .uestion ,ere +ery +ariable and many lost mar&s through errors inmanipulation or notation possibly through mental tiredness$ For examples many made errorsin manipulation and could not proceed correctly from the printed 4cos 4 7 4sin = to

4 7 7sin cos 44 4

= and the ans,er7

sin44 "

x ,as often seen instead of

7sin4

4 "

$ On

part (b) many ne+er foundd

d

x

or realised that the appropriate form for the +olume ,as

4 d dd

xy

$

Co,e+er the maority did find a correct integral in terms of although some ,ere unable touse the identity sin4 4sin cos = to simplify their integral$ he incorrect +alue /k = ,as +ery common resulting from a failure to s.uare the factor 4 in sin 4 4sin cos = $!andidates ,ere expected to demonstrate the correct change of limits$ Minimally a reference

to the result7

tan# 8

= or an e.ui+alent ,as re.uired$ hose ,ho had complete solutions

usually gained the t,o method mar&s in part (c) but earlier errors often led to incorrectans,ers$

il+er "> /?74 7/

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Statistics !or C4 9ractice 9a$er Silver Level S4

Mean score for students achieving grade:

QuMaxscore

Modalscore

Mean%

ALL A* A B C D E U

# 78 7.0" 7.#$ 7.01 $."" %.78 %.20 ".$8

! $ $# 4.14 %.%4 4.%% 4.1% ".7" ".2" 2.%# 1.$$

" $ $$ ".#4 %.78 %.11 4.12 2.#" 1.#0 1.17 0.%8

# % $1 ".07 4.7$ ".7$ 2.72 1.#" 1.1% 0.7" 0.22

$ 1% 11 $0 8.#8 1".00 10.01 8.%0 7.11 %.8# 4.77 2.8$

12 %7 $.87 #.42 $.87 %.1$ ".%$ 2.1% 0.#8

7 12 8 %4 $.%1 10.%7 7.## $.48 %.2 4.08 2.#" 1.47

& 10 "# ".#" $.21 "."$ 1.## 1.0$ 0.%4 0.1#

7$ $' ##(#7 $$() #"(! "#("& !($ !)()& (#