GCE Maths Examiner Feedback C1

8/13/2019 GCE Maths Examiner Feedback C1

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Core Mathematics 1Examiner Feedback

Senior Examiners feedback on real studentresponses to questions on the June 2012examination papers.

Supporting Edexcel GCE in Mathematics


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Core Mathematics 1 (6663)

Contents

About this booklet 2

June 2012 - eneral examiner feedback !

June 2012 " #uestion 1 $

June 2012 " #uestion 2 %

June 2012 " #uestion ! 1$

June 2012 " #uestion $ 1%

June 2012 " #uestion & 2$

June 2012 " #uestion ' 2%

June 2012 " #uestion ( !(

June 2012 " #uestion ) $!

June 2012 " #uestion % &2

June 2012 " #uestion 10 '1

Pearson Education Ltd 2012 1


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About this booklet

*his booklet has been produced to support mathematics teachers deli+erin, the Eathematics specification/ in particular the ore athematics 1 1 unit.

*his booklet looks at questions from the 1 June 2012 examination paper. t sho3s

real student responses to these questions/ and ho3 the examinin, team follo3 themark schemes to demonstrate ho3 the students are a3arded the marks.

How to use this booklet

Pearson Education Ltd 20122

*otal number ofmarks thestudent 3as

a3arded.

ntroduction to

the question on3hat is expected

from the student

#uestion

number

4lank

question

from the

exam paper

ark schemefor this

question

Examiner

commentar5

on the o+erallperformance

on thequestion

Student

response

Examiner

commentar5on the

students

response

6aper 7series


4/70

Core Mathematics 1 (6663)

June 2012 - General !aminer "ee#back

This paper proved a good test of candidates knowledge and candidates understanding of Core 1

material. There were plenty of easily accessible marks available for candidates who were

competent in topics such as differentiation, integration, recurrence relations, arithmetic series andbasic transformation of curves. Therefore, a typical E grade candidate had enough opportunity to

gain marks across the majority of uestions. !t the other end of the scale, there was sufficient

material, particularly in "# and "$ to stretch and challenge the most able candidates.

%hile standards of algebraic manipulation were generally good, some weakness in this area was

seen in "&'c(, "#'a(, "$'b( and "$'c(. )n "*, it was disappointing to see a number of good

candidates who made arithmetic errors in calculations such as 1+ 1-'&( in part 'a( and

.

*+/ '1+( &$'&( 0 or even + 2 1& in part 'b(. !lso a significant minority of candidates in "3'b(

incorrectly simplified('

*

1

1

x to 1x .

)n summary, "1, "'a(, "-, "&, "*'a(, "*'b(, "$'a(, "$'b( and "1+ were a good source of marks

for the average candidate, mainly testing standard ideas and techniues4 and "'b(, ", "*'c(,

"*'d(, "3, "$'c( and "$'d( were discriminating at the higher grades. "$'e( was the most

demanding uestion with only about 5 of the candidature able to correctly find the area of the

uadrilateralACBE.



5/70

$uestion 1

GC Core Mathematics 1% June 2012

$uestion 1

This uestion tests a candidates ability to apply the integration rule of ++=

+

671

d1

cn

xxx

nn

to three different types of term. )t also tests a candidates understanding of indices, especially whenintegrating .

.

xwith respect tox.

!bout three8uarters of the candidature achieved all - marks in this uestion.

%hile most candidates were able to integrate both .*x and & correctly, a significant minority

struggled to integrate ..

xcorrectly, giving incorrect answers such as

x or x.

1 or 1- x

or - x .

)ncorrect simplification of either

*

x to x or 1. x to x.1

4 or not simplifying 1. + x to

1. x and the omission of the constant of integration were also other common errors.

)t was pleasing to see very few candidates who differentiated all three terms.

Pearson Education Ltd 2012&


6/70

$uestion 1

9ark :cheme

Question

NumberScheme Mark

1.( )

1

* * & d &

1

x xx x x c

x

+ + = + + + 91 !1 1 4 &x x x c= + + !14 !1

- 9arks

Pearson Education Ltd 2012 '


7/70

$uestion 1

tu#ent Attemt A

91;


8/70

$uestion 1

tu#ent Attemt *

91; There are three ways this response could earn the 91 mark. This is for either .*x becoming x 4 or for .

.

xbecoming 1 x or for & integrating to x& .

!1; is missing.

Pearson Education Ltd 2012 +

2/4


9/70

$uestion 1

tu#ent Attemt C

91; There are two ways this response could gain the 91 mark here. This is for either .*x

becoming x or for & integrating to x& .

!+; This response does not achieve either

*.

* x

x or1

.. 1

.

x

x

which can be un8simplified or

simplified. ?@TE; Even though .x appears in this response 8 it is the incorrect result of

combining the incorrect integration of .*x with the incorrect integration of ..

x.

!+; The first two terms are integrated incorrectly.

!1;


10/70

$uestion 2


$uestion 2

This uestion tests a candidates application of the law of indices for rational eAponents. )n part 'a(,

the euivalence of nm

a asmn a (' is tested. Bart 'b( is more demanding and tests n

m

a , where

=

-& -xa is a bracketed term. )n this case the candidate is reuired to reciprocate and suare root

all three elements in the brackets.

This uestion proved discriminating with about a third of the candidature gaining all - marks.

)n part 'a(, the majority of candidates were able to evaluate &

. as #. 9any of those who were

unable to achieve #, were able to score one mark by rewriting &

. as either ( )& . or & . .

Those candidates who chose to cube first to give 3*# were usually unable to find & .3*# .

Common errors in this part included rewriting &

. as either .& or & . or evaluating .

as *.

Bart 'b( proved more challenging than part 'a(, with the majority of candidates managing to obtain

at least one of the two marks available by demonstrating the correct use of either the reciprocal or

suare root on

-

& -x. The most able candidates 'who usually reciprocated first before suare

rooting( were able to proceed efficiently to the correct answer. The most common mistake was for

candidates not to suare root or not to reciprocate all three elements in the brackets. )t was common

for candidates to give any of the following incorrect answers;

.

.

& x ,.

-.& x , .

3

&. x , .1++ x , -

&

.

xor

&

-x .



11/70

$uestion 2

Mark cheme

Question

NumberScheme Mark

2. (a)( ) ( ) ( )

& &&& or or or 3*# =

91

#= !1

'(

(b) 1 1 1-

1--

& - & 1or or

- & &

-

x x

xx

=

91

or

& &x

x

= !1

'(

- 9arks



12/70

$uestion 2

tu#ent Attemt A

(a) 91;


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14/70

$uestion 2

tu#ent Attemt C

(a) 91;


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$uestion 3


$uestion 3

This is an unstructured uestion testing the use and manipulation of surds. To solve this problem

candidates have to apply two methods D one of rationalising the denominator and one of simplifying

the surds 1. and #

. These methods can be applied in any order, resulting in a number ofdifferent solutions.

This uestion proved discriminating, with just under a half of the candidature gaining all & marks.

!bout a third of the candidature gained only two marks by either correctly rationalising the

denominator or by correctly simplifying 1. and # .

! significant number of candidates answered this uestion by multiplying #1..

by (#1.'(#1.'

+

+

to obtain #1. .1

.

1 + and proceeded no further. ! small number of candidates multiplied the

denominator incorrectly to give either #.+.+1. + or #3.3.1. + .

Those candidates who realised that .1. = and ..# = usually obtained the correct answer

of . + , although a few proceeded to the correct answer by writing #1..

1

.

1 + as

#1.-

1

-

1+ .

! number of candidates started this uestion by firstly simplifying 1. and # to give ....

,

but many did not take the easy route of cancelling this to.

1

before rationalising. !lthough

some candidates wrote.

1

as . , those candidates who rationalised this usually obtained

the correct answer.

! small minority of candidates attempted to rationalise the denominator incorrectly by multiplying

#1.

.

by either (#1.'(#1.'

+

or(#1.'

(#1.'

+

+

.



16/70

$uestion 3

Mark cheme

Question

NumberScheme Mark

3.

( )

( )

( )

1 #

1 # 1 # 1 #

+ = +

%riting this is sufficient for 91. 91

( ){ } 1 #1 #

+=


17/70

$uestion 3

tu#ent Attemt A

91;


18/70

$uestion 3

tu#ent Attemt *

9+; 9ultiplying by(#1.'

(#1.'

+

is an incorrect method for rationalising the denominator.

!+;


19/70

$uestion 3

tu#ent Attemt C

9+; ?o attempt to rationalise the denominator.

!+;


20/70

$uestion &


$uestion &

This uestion tests a candidates ability to apply the differentiation rule of1

d

d (' = nnx

nxx to four

different types of term, as well as a candidates understanding of.

.

d

d

x

y. )t also tests a candidates

understanding of indices, especially when differentiating -

*x and later 1

#

x with respect tox.

This uestion was well answered with about two8thirds of the candidature achieving all * marks.

)n part 'a(, a small minority of candidates struggled to deal with the fractional power when

differentiating -

*x . :ome candidates incorrectly reduced the power by 1 to give a term in .

x 4

whilst other candidates struggled to multiply * by -

or incorrectly multiplied * by 1

. ! few

candidates did not simplify 1

.- x to give

1

#x or integrated throughout, or added a constant to

their differentiated eApression.

)n part 'b(, the most common error was for candidates to differentiate 1

#x to give 1

#

x . !

few candidates did not understand the notation for the second derivative, with some integrating their

differentiated result in part 'a( to achieve an answer similar to the eApression given in the uestion.

:lips occurred with the omission ofxfrom fractional power terms, with some candidates writing

1

# in part 'a( or.

#

in part 'b(.



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$uestion &

Mark cheme

Question

NumberScheme Mark

4. (a) -

& * y x x x= + 1

d -&'( * d

yx x

x = +

91

1

1& # x x= + !1 !1 !1

'-(

(b)

d #+

d

yx x

x

=

91 !1

'(

* 9arks



22/70

$uestion &

tu#ent Attemt A

(a) 91;


23/70

$uestion &

tu#ent Attemt *

(a) 91; There are two ways this response could gain the 91 mark here. This is for either

&x becoming .1&x or for x. differentiating to .

!1;


24/70

$uestion &

tu#ent Attemt C

(a) 91;


25/70

$uestion '


$uestion '

This uestion tests a candidates understanding and use of recurrence relations of the form('f1 nn xx =+ . ecause part 'b( has its answer given, the candidate can see if they are applying the

recurrence relation formula correctly. Bart 'c( is more demanding because it links together thetopics of sigma notation, recurrence relations and ineualities.

This uestion was answered more successfully by candidates than similar ones in the past. Thenotation did not appear to be such a mystery with most candidates realising that this uestion tested

the topic of recurrence relations and not arithmetic seuences. !bout two8thirds of the candidates

gained at least * out of the 3 marks available.

Bart 'a( was generally very well answered and indeed most who got this part right went on to score

most of the marks in the uestion. Those who were unsuccessful often tried to work back from the

given ca 1. = to arrive at ca * = .

)n part 'b(, most candidates scored full marks. @ccasionally problems were caused by the incorrect

use of brackets.

)n part 'c(, the majority of candidates were able to find -a , sum the first four terms of the seuence

and write their sum or F . :ome candidates found -a incorrectly by omitting brackets

resulting in ccca --1- == . )nstead of summing the first four terms, a number of

candidates solved - a or put each term . ! few candidates summed by adding either

1a , .a and a , or .a , a and -a or even .a , a , -a and &a . :ome arithmetic errors were

made in summing the four terms and a number of candidates miscopied -a as 3 D 3c. ! number

of candidates used the formula for the sum to nterms of an arithmetic series in order to sum their

four terms. ! significant number of candidates, who achieved .11-& c , did not know that

dividing by a negative number reverses the sign of the ineuality. Those who rearranged.11-& c into c11.-& were more successful in achieving the correct result.



26/70

$uestion '

Mark cheme

Question

NumberScheme Mark

5. (a) 1 1, , 1n na a a c n+= = , cis a constant

{ } or '( or *a c c c=

1'1(

(b) { } ( ) G* Ga c c= 911 c= (*) !1 cso

'(

(c) ( ) { }- G1 G - 3a c c c= = 91-

1

'* ( '1 ( '- 3 (i

i

a c c c

=

= + + + 91

G-& 11 G c or G-& 11 G c = 91

c or c !1 cso'-(

3 9arks

Pearson Education Ltd 2012 2'


27/70

$uestion '

tu#ent Attemt A

(a) 1; Correct answer of c('. .

(b) 9+; The candidate does not substitute their ca = (' correctly into caa = .

!+;


28/70

$uestion '

tu#ent Attemt *

(a) 1; Correct answer of * c .

(b) 91;


29/70

$uestion '

tu#ent Attemt C

(a) 1; Correct answer of * c.

(b) 9+; The candidate does not substitute their ca =* correctly into caa = .!+;


30/70

$uestion 6


$uestion 6

This uestion tests a candidates application of arithmetic series to real8life problems. Bart 'a( and

part 'b( test standard formulae. Bart 'c( is more demanding, reuiring the candidate to unify the

units and reduce the sum formula to an unfamiliar form. )n part 'd(, the solution of' 1( & *,m m + = although trivial, reuires a certain mathematical maturity that some candidatesmay lack.

This uestion was both well answered and discriminating with about three8uarters of the

candidature gaining at least 3 of the 1+ marks available and one8uarter achieving full marks.

Bart 'a( was well answered with the majority using a 1-dand a small minority using &n & inorder to find the 1&thterm. ! few candidates listed each term and a number identified the 1&thterm

correctly. ! small number of candidates found the total amount saved over the 1& weeks.

)n part 'b(, many candidates used ((1'.'. dnan

+ to find the total amount paid over *+ weeks,

although a few applied lF a 'nD 1(d and substituted the result into ' (n a l+ . )t was not

uncommon, however, for arithmetic errors such as mulitplying 1& by + or even adding + to $&.

@ther candidates who failed to get full credit miAed nF 1& and nF *+, used dF &, or found *+T

instead of *+S .



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32/70

$uestion 6

Mark cheme

Question

NumberScheme Mark

6. (a) oys :euence; 1+,1&, +, &, ...

{ }1&

1+, & 1- 1+ 1-'&(4 #+a d T a d = = = + = + = or +.1 1-'+.+&(4 I+.#++ = 91 !1'(

(b){ } [ ]*+

*+'1+( &$'&(

S = + 91 !1

+'1&( $-&+ or I$-.&+= = !1'(

(c){ } ( )1+, 1+ '1+( ' 1('1+( or 1+' 1(

m

m ma d S m m

= = = + +

91 !1

( )* or *++ '1+( ' 1('1+(

mm= + d91

*++ '1+(' 1(

mm= + or 1*++ 1+ ' 1(m m= +

1*+ ' 1(m m= +& * ' 1(m m = + (*) !1 cso

'-(

(d) { } &m= 1'1(

1+ 9arks



33/70

$uestion 6

tu#ent Attemt A

(a) 91 !1;


34/70

$uestion 6

tu#ent Attemt *



35/70

$uestion 6

(a) 9+;


36/70

$uestion 6

tu#ent Attemt C

Pearson Education Ltd 2012 3'


37/70

$uestion 6

(a) 91;


38/70

$uestion +


$uestion +

)n this uestion, candidates are given the derived function f ' (x and a point '-, 1(P lying onf ' ( .y x= Bart 'a( is not mathematically demanding but reuires the candidate to have a clear

understanding that it is the gradient function that they have been given. Bart 'b( is more familiar,reuiring candidates to integrate f ' (x to give an euation for f ' (x containing cand then recognisethat the given point provides the information reuired to calculate c.

This uestion discriminated well with just over half of the candidature gaining at least * of the #

marks available. ! significant number of candidates started by integrating f K 'x( to give f 'x(.

)n most cases they realised that this was valid work for part 'b( and so relabelled their initial work

as 'b(.

)n part 'a(, the majority of candidates evaluated f K'-( and used a correct line formula in order to

find the euation of the tangent. ! few made arithmetic errors when evaluating (-' -*

.1

+ ,

some incorrectly manipulatedy 1 F 'xD -( intoyF x3 and some found the euation of the

normal. :ome candidates rewrote x*

as x* and used this throughout the uestion and

others deduced a tangent gradient of.

1by looking at the coefficient ofxin the f K 'x( eApression.

! significant minority of candidates started part 'a( by differentiating f K 'x( and substituting

xF - into the resulting eApression. @ther candidates incorrectly used '-, 1( by setting their

eApression for f K '-( eual to 1, which resulted in a meaningless euation.

)n part 'b(, most candidates integrated f K'x( to find f 'x( but unfortunately a significant minority

failed to find the value of cusing '-, 1(. @f those who used '-, 1( to find c, a number made

arithmetic errors. @ther candidates found f '-( in terms of cand euated their result to + instead of

1. ! small number of candidates failed to integrate correctly and some candidates either

incorrectly simplified

x

to .x or ( )1

1

*x to .1

x . ! very small number of candidates found the

value of ccorrectly but failed to include this evaluated cin an eApression for f 'x(.

Pearson Education Ltd 2012 3+


39/70

$uestion +

Mark cheme

Question

NumberScheme Mark

7. (a)'-, 1(P lies on Cwhere

1 *f ' ( , +

x x x

x = + >

1 *f '-( '-( 4

- = + = 91 !1

! 1 ' -(y x = d91

! $y x= !1'-(

(b) ( )

11

1 1

1

*f ' (

'( ' (

x xx x c

++

= + + or euivalent 91 !1

{ }

1*

f '-( 1 1'( '-( 1- c= + + = d91{ }- - 1 1 3c c + + = =

:o, { }

1

1

*f ' ( 3

'( ' (

x xx x= + + !1 cso

?E; f ' ( 1 3-

xx x x

= + +

'-(

# 9arks

Pearson Education Ltd 20123,


40/70

$uestion +

tu#ent Attemt A

(a) 91 !1;


41/70

$uestion +

tu#ent Attemt *

(a) 91;


42/70

$uestion +

tu#ent Attemt C

Pearson Education Ltd 2012 &1

3/8


43/70

$uestion +

(a) 91 !1;


44/70

$uestion ,


$uestion ,

This is the first time in the current specification that a negative uadratic has been being tested.

Bart 'a( tests the method of completing the suare and is made more demanding because it is

applied to a negative uadratic. Bart 'b( tests finding the discrimininant and is made more difficultby the fact that the uadratic is given in the form .axcbxy ++= . )n part 'c(, candidates can use

the information gleaned from earlier parts in order to sketch .&- xxy = .

This uestion was poorly answered with about 1+5 of the candidature gaining all # marks.

)n part 'a(, the .x term and the order of the terms in .&- xx created problems for the

majority of candidates. The most popular 'and most successful( method was to complete the

suare. ! large number of candidates struggled to deal with the .x term with bracketing errors

leading to incorrect answers such as $(.' . + x or $(.' . x . ! successful strategy for

some candidates involved negating the uadratic to get &-. + xx which was usually manipulated

correctly to 1(.' . +x . %hilst a good number negated this correctly to 1(.' . x some

candidates wrote down incorrect results such as 1(.' . + x , 1(.' . +x or 1(.' . + x .

%hilst a number of candidates stopped after multiplying out .(' pxq , those who attempted touse the method of euating coefficients were less successful.

)n part 'b(, most candidates wrote down acb -. for the discriminant and the majority achieved

the correct answer of -, although some incorrectly evaluated -D -'1('&( as *. The most

common error was for candidates to substitute the incorrect values of a F -, b F & and c F 1 into

acb -. . Those candidates who applied the uadratic formula gained no credit unless they could

identify the discriminant part of the formula.

Pearson Education Ltd 2012 &3


45/70

$uestion ,

)n part 'c(, a majority of candidates were able to draw the correct shape of the graph and a number

correctly identified they8intercept as &. @nly a small minority were able to correctly position the

maAimum turning point in the fourth uadrant and some labelled it as ', 1(. ! number of correct

sketches followed from either candidates using differentiation or from candidates plotting points

from a table of values. ! number of candidates after correctly identifying the discriminant as -

'including some who stated =that this meant no roots>( could not relate this information to theirsketch with some drawing graphs crossing thex8aAis at either one or two points. Common errors

included drawing graphs of cubics or positive uadratics, drawing negative uadratics with a

maAimum at '&, +( or with a maAimum at '+, &(.

Pearson Education Ltd 2012&&


46/70

$uestion ,

Mark cheme

Question

NumberScheme Mark

". (a) - & ' ( ,x x q x p = p, q are integers.

{ }

- & - & ' ( - & ' ( 1x x x x x x = + = + = + 911 ' (x= !1 !1

'(

(b) { } { } G - G - -' 1(' &( 1* +b ac = = 91-= !1

'(

(c)

Correct shape 919aAimum #ithinthe -thuadrant !1

Curve cuts through 8& or '+, &( marked on they8aAis 1'(

# 9arks

Pearson Education Ltd 2012 &'

O

8 &

y

x


47/70

$uestion ,

tu#ent Attemt A

Pearson Education Ltd 2012&6


48/70

$uestion ,

(a) The candidate uses an alternative method and attempts to complete the suare on

&-. + xx .

91;


49/70

$uestion ,

tu#ent Attemt *

Pearson Education Ltd 2012&,


50/70

$uestion ,

(a) 9+; )ncorrect method of completing the suare.

!+ !+;


51/70

$uestion ,

tu#ent Attemt C

Pearson Education Ltd 2012'0


52/70

$uestion ,

(a) The candidate uses an alternative method and attempts to complete the suare on

&-. + xx .

91; 'xD ( 1 implies a correct method of completing the suare on &-. + xx .

')gnore = F +>(.

!+; )ncorrect value ofpF or result is not written in the form DkD 'xD (. )f the

candidate had instead statedpF they would have gained the !1 mark by implied

working.

!+; )ncorrect value of q.

(b) 91 !1;


53/70

$uestion


$uestion

This structured coordinate geometry uestion tests finding a point on a line4 finding a line which is

perpendicular to a given line4 finding the intersection of two lines4 and using surds to find the

distance between two points. Bart 'e( reuires synthesising information, drawing a diagram andusing higher level problem solving skills in order to find the reuired area.



54/70

$uestion

This uestion proved discriminating across all abilities with about a uarter of the candidature

gaining at least 1 out of the 1& marks available. ! significant number of candidates gave up on

this uestion before they reached part 'e(.

Bart 'a( was well answered by the majority of candidates. !fter the substitution yF -, most were

able to obtain .1$

=p , although some simplified this to #.&.

!gain, part 'b( was well answered with many candidates rearranging xy .- =+ into the form

yF mx cin order to find the gradient of 1L . @ccasionally, the use of two points on 1L was seen

as an alternative approach to finding the gradient of 1L , whilst some preferred to differentiate their

1L after rearranging. 9ost candidates were able to use the perpendicular gradient rule to write

down the gradient of .L and use this gradient to find an euation of .L . 9ethods of approach

were roughly eually divided between those using (' 11 xxmyy = oryF mx c. The majority

of candidates were able to simplify their euation into a correct form of +=++ cbyax , although

some rearranged (.'.- = xy incorrectly to give +. =+ xy . Common errors in this part

included candidates incorrectly finding the gradient of 1L by finding the gradient betweenAand C

or stating the gradient as from looking at the coefficient of xin xy .- =+ .

)n part 'c(, a large number of those with a correct euation of .L found the correct coordinates of

D, with a few, fortunately, using their correct un8simplified version of .L rather than their

incorrect rearrangement. The majority of candidates without a correct part 'b( were able to

demonstrate that they could solve the euations for 1L and .L simultaneously and received some

credit for this. There were a number of candidates who euated their euations for 1L and .L to

give #..- +=+ yxxy . :ome manipulated this into +11- = yx and then gave up4

whilst others continued to setxF + to find a value foryand similarly setyF + to find a value forx.

)n part 'd(, it was pleasing to see many candidates able to make a good attempt at finding thedistance between the points CandD. :ome drew diagrams and others uoted a correct formula.

Jelatively few candidates got miAed up when determining the differences in the x-values and the

differences in they8values, although a few used incorrect formulae such as ( ) 1

1 (' yyxx +++

or ( ) 1

1 (' yyxx . :ome candidates lost the final mark in this part by being unable to

correctly manipulate fractions and surds, whilst others did not provide sufficient working to arrive

at the answer given on the paper.

Bart 'e( was the most challenging uestion on the paper with the majority of candidates not

attempting it and many of those that did only able to offer incomplete solutions. ! significant

number of candidates did not draw a clear diagram, which is essential in understanding the nature ofthis problem. Those that were successful usually summed up the area of two relevant triangles

'usually triangleABCand triangleABE ( or found half the product ofABand CE, although a

significant number of candidates used the incorrect method of finding the product ofABand CE.

! few candidates used other more elaborate methods to find the correct area of -&. :ome

candidates attempted to find lengths of various lines without any apparent purpose and gave no

indication of finding an area. ! small number thought uadrilateralACBEwas a trapeMium.

Pearson Education Ltd 2012 '3


55/70

$uestion

Mark cheme

Question

NumberScheme Mark

$. (a)1 1

1 ; - 4 ' , -( lies on .

-L y x y x A p L+ = =

{ } 1 1$$ or or $.&

p= 1

'1(

(b){ } 1

1 - ' ( or

- -

xy x y m L

+ = = = 91 !1

:o ' ( m L = 1ft

; - ' (L y x = 91

; # +L x y+ =

or

; 1 # +L x y+ = !1

'&(

(c){ }1 L L= -'# ( x x + = or

1 #

-x x + = 91

.&, 1x y= = !1 !1 cso

'(

(d) ( ) ( ) G.&G G1G -CD = + =91>

( ) ( )

G.&G G1G -CD= + !1 ft

1.& 1.& 1 1.& & or &= + = + = (*) !1 cso'(

(e) !rea triangle triangleABC ABE= +1 1

& #+ & #+

= +


56/70

$uestion

Pearson Education Ltd 2012 ''


57/70

$uestion

tu#ent Attemt A



58/70

$uestion

(a) 1; Correct value ofp.

(b) 91 !1 1 91 !1;


59/70

$uestion

tu#ent Attemt *

Pearson Education Ltd 2012',


60/70

$uestion


(b) 91 !1; -y F xrearranged and correct gradient of 1L found.

+; )ncorrect method for finding a perpendicular gradient.

91;


61/70

$uestion

tu#ent Attemt C



62/70

$uestion


(b) 91 !1 1 91 !1;


63/70

$uestion 10


$uestion 10

)n this uestion candidates are given the graph ofyF f 'x( Fx'$ D x( .

Bart 'a( is straightforward and reuires candidates to write down the coordinates where yF f 'x(

crosses thex8aAis.

Bart 'b( is familiar to candidates and tests the effect of simple transformations onyF f 'x( .

Bart 'c( is discriminating and reuires candidates to find the value of k where ', 3( onyF f 'x( is

mapped onto ', 1+( onyF f 'x( k .



64/70

$uestion 10

This uestion was both well answered and discriminating with about two8thirds of the candidature

gaining at least * of the # marks available and about one8third achieving full marks.

)n part 'a(, most candidates solved f 'x( F + to find the correctx-coordinate of.

$forA. :ome

candidates, however, appliedxF + on $ D xand arrived at an incorrect value of $. @ther common

incorrect values forxwere * or 3.

)n part 'b(, most candidates were able to give the correct shape for each of the transformed curves.

)n part 'i(, most translated the graph ofyF f 'x( in the correct direction. Nery few candidates

translatedyF f 'x( to the right, and even fewer translatedyF f 'x( in a vertical direction. :ome

labelled they-intercept correctly as '+, 3( but erroneously drew their maAimum point slightly to the

right of they8aAis in the first uadrant. 9ost realised that the transformed curve would cut the

positivex8aAis at =theirxin part 'a( D >. @ther candidates, who gave no answer to part 'a(,

labelled thisx8intercept asA or some left it unlabelled. @ccasionally the point ', +( was

incorrectly labelled as ', +( although it appeared on the negativex8aAis.

)n part 'ii(, most graphs had their minimum at the origin and their maAimum within the first

uadrant. 9any realised that the transformed curve would cut thex8aAis at . @ther

candidates, who gave no answer to part 'a(, labelled this intercept as A

, whilst some left it

unlabelled. :ome misunderstood the given function notation and stretchedyF f 'x( in thex8

direction with scale factor resulting in a maAimum of '$, 3( and an x8intercept at '1.&, +(. Nery

occasionally a stretch of they8direction4 or a two way stretch4 or even a reflection of yF f 'x( was

seen. )n a few cases there was an attempt by some candidates to make the graph pass through both

'+, +( and '+, 3(.

)n part 'c(, a significant number of candidates wrote down kF 13 whilst some left this part

unanswered. ! number of candidates wrote down the euationyF f 'x( kFx'$ D x( k and

substituted in the point ', 1+( to find the correct value of k. Common incorrect answers included

kF 13 'from 3 1+( or kF 3 'following kF 1+(.



65/70

$uestion 10

Mark cheme

Question

NumberScheme Mark

1%. (a) 7Coordinates ofAare6 '-.&, +(. 1

'1(

(b)(i)

OoriMontal translation 91

8 and their ft 1.& on postitivex8aAis !1 ft

9aAimum at 3 marked on the y8aAis 1

'(

(b)(ii)

Correct shape, minimum at '+, +( and

a maAimum within the first uadrant. 91

1.& onx8aAis !1 ft

9aAimum at '1, 3( 1

'(

(c) 7k = 6 13 1'1(

# 9arks


1.&

3

O8

y

x

1.&

3

O

y

x


66/70

$uestion 10

tu#ent Attemt A

(a) 1;


67/70

$uestion 10

tu#ent Attemt *

(a) +; '*, +( is incorrect.

(b)(i) 91;


68/70

$uestion 10

tu#ent Attemt C

Pearson Education Ltd 2012 6+


69/70

$uestion 10

(a) +; '3, +( is incorrect.

(b)(i) 91;


70/70

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f 5ou ha+e an5 sub?ect specific questions about the content of this bookletthat require the help of a sub?ect specialist/ 5ou ma5 find our Ask *heExpert email ser+ice helpful. Ask *he Expert can be accessed online at the

follo3in, link@ [email protected]

*his 4ooklet has been 3ritten to pro+ide teachers 3ith additional supportfor E athematics. t pro+ides real student responses to past

examination questions 3ith tips and ,uidance from our senior examinin,team.

ther publications in this series@

ore athematics 2 2 Examiner Beedback

ore athematics ! ! Examiner Beedback

ore athematics $ $ Examiner Beedback

echanics 1 1 Examiner Beedback

Statistics 1 S1 Examiner Beedback

Acknowledgements

Edexcel 3ould like to thank 8ee ope 6rincipal Examiner for contributin,his time and expertise to the de+elopment of this publication.

;o+ember 2012

All the material in this publication is cop5ri,ht

Documents

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