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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 ) $!
June 2012 " #uestion % &2
June 2012 " #uestion 10 '1
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About this booklet
*his booklet has been produced to support mathematics teachers
deli+erin, the Eathematics specification/ in particular the ore
athematics 1 1 unit.
*his booklet looks at questions from the 1 June 2012 examination
paper. t sho3s
real student responses to these questions/ and ho3 the examinin,
team follo3 themark schemes to demonstrate ho3 the students are
a3arded the marks.
How to use this booklet
Pearson Education Ltd 20122
*otal number ofmarks thestudent 3as
a3arded.
ntroduction to
the question on3hat is expected
from the student
#uestion
number
4lank
question
from the
exam paper
ark schemefor this
question
Examiner
commentar5
on the o+erallperformance
on thequestion
Student
response
Examiner
commentar5on the
students
response
6aper 7series
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Core Mathematics 1 (6663)
June 2012 - General !aminer "ee#back
This paper proved a good test of candidates knowledge and
candidates understanding of Core 1
material. There were plenty of easily accessible marks available
for candidates who were
competent in topics such as differentiation, integration,
recurrence relations, arithmetic series andbasic transformation of
curves. Therefore, a typical E grade candidate had enough
opportunity to
gain marks across the majority of uestions. !t the other end of
the scale, there was sufficient
material, particularly in "# and "$ 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.
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$uestion 1
GC Core Mathematics 1% June 2012
$uestion 1
This uestion tests a candidates ability to apply the integration
rule of ++=
+
671
d1
cn
xxx
nn
to three different types of term. )t also tests a candidates
understanding of indices, especially whenintegrating .
.
xwith respect tox.
!bout three8uarters of the candidature achieved all - marks in
this uestion.
%hile most candidates were able to integrate both .*x and &
correctly, a significant minority
struggled to integrate ..
xcorrectly, giving incorrect answers such as
x or x.
1 or 1- x
or - x .
)ncorrect simplification of either
*
x to x or 1. x to x.1
4 or not simplifying 1. + x to
1. x and the omission of the constant of integration were also
other common errors.
)t was pleasing to see very few candidates who differentiated
all three terms.
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$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
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$uestion 1
tu#ent Attemt A
91;
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$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.
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$uestion 1
tu#ent Attemt C
91; There are two ways this response could gain the 91 mark
here. This is for either .*x
becoming x or for & integrating to x& .
!+; This response does not achieve either
*.
* x
x or1
.. 1
.
x
x
which can be un8simplified or
simplified. [email protected]; Even though .x appears in this response 8 it
is the incorrect result of
combining the incorrect integration of .*x with the incorrect
integration of ..
x.
!+; The first two terms are integrated incorrectly.
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$uestion 2
GC Core Mathematics 1% June 2012
$uestion 2
This uestion tests a candidates application of the law of
indices for rational eAponents. )n part 'a(,
the euivalence of nm
a asmn a (' is tested. Bart 'b( is more demanding and tests
n
m
a , where
=
-& -xa is a bracketed term. )n this case the candidate is
reuired to reciprocate and suare root
all three elements in the brackets.
This uestion proved discriminating with about a third of the
candidature gaining all - marks.
)n part 'a(, the majority of candidates were able to evaluate
&
. as #. 9any of those who were
unable to achieve #, were able to score one mark by rewriting
&
. as either ( )& . or & . .
Those candidates who chose to cube first to give 3*# were
usually unable to find & .3*# .
Common errors in this part included rewriting &
. as either .& or & . or evaluating .
as *.
Bart 'b( proved more challenging than part 'a(, with the
majority of candidates managing to obtain
at least one of the two marks available by demonstrating the
correct use of either the reciprocal or
suare root on
-
& -x. The most able candidates 'who usually reciprocated
first before suare
rooting( were able to proceed efficiently to the correct answer.
The most common mistake was for
candidates not to suare root or not to reciprocate all three
elements in the brackets. )t was common
for candidates to give any of the following incorrect
answers;
.
.
& x ,.
-.& x , .
3
&. x , .1++ x , -
&
.
xor
&
-x .
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$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
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$uestion 2
tu#ent Attemt A
(a) 91;
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$uestion 2
tu#ent Attemt C
(a) 91;
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$uestion 3
GC Core Mathematics 1% June 2012
$uestion 3
This is an unstructured uestion testing the use and manipulation
of surds. To solve this problem
candidates have to apply two methods D one of rationalising the
denominator and one of simplifying
the surds 1. and #
. These methods can be applied in any order, resulting in a
number ofdifferent solutions.
This uestion proved discriminating, with just under a half of
the candidature gaining all & marks.
!bout a third of the candidature gained only two marks by either
correctly rationalising the
denominator or by correctly simplifying 1. and # .
! significant number of candidates answered this uestion by
multiplying #1..
by (#1.'(#1.'
+
+
to obtain #1. .1
.
1 + and proceeded no further. ! small number of candidates
multiplied the
denominator incorrectly to give either #.+.+1. + or #3.3.1. +
.
Those candidates who realised that .1. = and ..# = usually
obtained the correct answer
of . + , although a few proceeded to the correct answer by
writing #1..
1
.
1 + as
#1.-
1
-
1+ .
! number of candidates started this uestion by firstly
simplifying 1. and # to give ....
,
but many did not take the easy route of cancelling this to.
1
before rationalising. !lthough
some candidates wrote.
1
as . , those candidates who rationalised this usually
obtained
the correct answer.
! small minority of candidates attempted to rationalise the
denominator incorrectly by multiplying
#1.
.
by either (#1.'(#1.'
+
or(#1.'
(#1.'
+
+
.
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$uestion 3
Mark cheme
Question
NumberScheme Mark
3.
( )
( )
( )
1 #
1 # 1 # 1 #
+ = +
%riting this is sufficient for 91. 91
( ){ } 1 #1 #
+=
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$uestion 3
tu#ent Attemt A
91;
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$uestion 3
tu#ent Attemt *
9+; 9ultiplying by(#1.'
(#1.'
+
is an incorrect method for rationalising the denominator.
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$uestion 3
tu#ent Attemt C
9+; ?o attempt to rationalise the denominator.
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$uestion &
GC Core Mathematics 1% June 2012
$uestion &
This uestion tests a candidates ability to apply the
differentiation rule of1
d
d (' = nnx
nxx to four
different types of term, as well as a candidates understanding
of.
.
d
d
x
y. )t also tests a candidates
understanding of indices, especially when differentiating -
*x and later 1
#
x with respect tox.
This uestion was well answered with about two8thirds of the
candidature achieving all * marks.
)n part 'a(, a small minority of candidates struggled to deal
with the fractional power when
differentiating -
*x . :ome candidates incorrectly reduced the power by 1 to give
a term in .
x 4
whilst other candidates struggled to multiply * by -
or incorrectly multiplied * by 1
. ! few
candidates did not simplify 1
.- x to give
1
#x or integrated throughout, or added a constant to
their differentiated eApression.
)n part 'b(, the most common error was for candidates to
differentiate 1
#x to give 1
#
x . !
few candidates did not understand the notation for the second
derivative, with some integrating their
differentiated result in part 'a( to achieve an answer similar
to the eApression given in the uestion.
:lips occurred with the omission ofxfrom fractional power terms,
with some candidates writing
1
# in part 'a( or.
#
in part 'b(.
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$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
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$uestion &
tu#ent Attemt A
(a) 91;
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$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 .
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$uestion &
tu#ent Attemt C
(a) 91;
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$uestion '
GC Core Mathematics 1% June 2012
$uestion '
This uestion tests a candidates understanding and use of
recurrence relations of the form('f1 nn xx =+ . ecause part 'b( has
its answer given, the candidate can see if they are applying
the
recurrence relation formula correctly. Bart 'c( is more
demanding because it links together thetopics of sigma notation,
recurrence relations and ineualities.
This uestion was answered more successfully by candidates than
similar ones in the past. Thenotation did not appear to be such a
mystery with most candidates realising that this uestion tested
the topic of recurrence relations and not arithmetic seuences.
!bout two8thirds of the candidates
gained at least * out of the 3 marks available.
Bart 'a( was generally very well answered and indeed most who
got this part right went on to score
most of the marks in the uestion. Those who were unsuccessful
often tried to work back from the
given ca 1. = to arrive at ca * = .
)n part 'b(, most candidates scored full marks. @ccasionally
problems were caused by the incorrect
use of brackets.
)n part 'c(, the majority of candidates were able to find -a ,
sum the first four terms of the seuence
and write their sum or F . :ome candidates found -a incorrectly
by omitting brackets
resulting in ccca --1- == . )nstead of summing the first four
terms, a number of
candidates solved - a or put each term . ! few candidates summed
by adding either
1a , .a and a , or .a , a and -a or even .a , a , -a and &a
. :ome arithmetic errors were
made in summing the four terms and a number of candidates
miscopied -a as 3 D 3c. ! number
of candidates used the formula for the sum to nterms of an
arithmetic series in order to sum their
four terms. ! significant number of candidates, who achieved
.11-& c , did not know that
dividing by a negative number reverses the sign of the
ineuality. Those who rearranged.11-& c into c11.-& were
more successful in achieving the correct result.
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$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
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$uestion '
tu#ent Attemt A
(a) 1; Correct answer of c('. .
(b) 9+; The candidate does not substitute their ca = ('
correctly into caa = .
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$uestion '
tu#ent Attemt *
(a) 1; Correct answer of * c .
(b) 91;
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$uestion '
tu#ent Attemt C
(a) 1; Correct answer of * c.
(b) 9+; The candidate does not substitute their ca =* correctly
into caa = .!+;
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$uestion 6
GC Core Mathematics 1% June 2012
$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 $&.
@ther candidates who failed to get full credit miAed nF 1&
and nF *+, used dF &, or found *+T
instead of *+S .
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$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
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$uestion 6
tu#ent Attemt A
(a) 91 !1;
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$uestion 6
tu#ent Attemt *
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$uestion 6
(a) 9+;
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$uestion 6
tu#ent Attemt C
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$uestion 6
(a) 91;
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$uestion +
GC Core Mathematics 1% June 2012
$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(.
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$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
! $y x= !1'-(
(b) ( )
11
1 1
1
*f ' (
'( ' (
x xx x c
++
= + + or euivalent 91 !1
{ }
1*
f '-( 1 1'( '-( 1- c= + + = d91{ }- - 1 1 3c c + + = =
:o, { }
1
1
*f ' ( 3
'( ' (
x xx x= + + !1 cso
?E; f ' ( 1 3-
xx x x
= + +
'-(
# 9arks
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$uestion +
tu#ent Attemt A
(a) 91 !1;
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$uestion +
tu#ent Attemt *
(a) 91;
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$uestion +
tu#ent Attemt C
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$uestion +
(a) 91 !1;
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$uestion ,
GC Core Mathematics 1% June 2012
$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 $(.' . x . !
successful strategy for
some candidates involved negating the uadratic to get &-. +
xx which was usually manipulated
correctly to 1(.' . +x . %hilst a good number negated this
correctly to 1(.' . x some
candidates wrote down incorrect results such as 1(.' . + x ,
1(.' . +x or 1(.' . + x .
%hilst a number of candidates stopped after multiplying out .('
pxq , those who attempted touse the method of euating coefficients
were less successful.
)n part 'b(, most candidates wrote down acb -. for the
discriminant and the majority achieved
the correct answer of -, although some incorrectly evaluated -D
-'1('&( as *. The most
common error was for candidates to substitute the incorrect
values of a F -, b F & and c F 1 into
acb -. . Those candidates who applied the uadratic formula
gained no credit unless they could
identify the discriminant part of the formula.
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$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 '+, &(.
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$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
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47/70
$uestion ,
tu#ent Attemt A
Pearson Education Ltd 2012&6
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48/70
$uestion ,
(a) The candidate uses an alternative method and attempts to
complete the suare on
&-. + xx .
91;
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49/70
$uestion ,
tu#ent Attemt *
Pearson Education Ltd 2012&,
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$uestion ,
(a) 9+; )ncorrect method of completing the suare.
!+ !+;
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51/70
$uestion ,
tu#ent Attemt C
Pearson Education Ltd 2012'0
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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;
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53/70
$uestion
GC Core Mathematics 1% June 2012
$uestion
This structured coordinate geometry uestion tests finding a
point on a line4 finding a line which is
perpendicular to a given line4 finding the intersection of two
lines4 and using surds to find the
distance between two points. Bart 'e( reuires synthesising
information, drawing a diagram andusing higher level problem
solving skills in order to find the reuired area.
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$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.
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$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 $.&
p= 1
'1(
(b){ } 1
1 - ' ( or
- -
xy x y m L
+ = = = 91 !1
:o ' ( m L = 1ft
; - ' (L y x = 91
; # +L x y+ =
or
; 1 # +L x y+ = !1
'&(
(c){ }1 L L= -'# ( x x + = or
1 #
-x x + = 91
.&, 1x y= = !1 !1 cso
'(
(d) ( ) ( ) G.&G G1G -CD = + =91>
( ) ( )
G.&G G1G -CD= + !1 ft
1.& 1.& 1 1.& & or &= + = + = (*) !1
cso'(
(e) !rea triangle triangleABC ABE= +1 1
& #+ & #+
= +
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$uestion
Pearson Education Ltd 2012 ''
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57/70
$uestion
tu#ent Attemt A
Pearson Education Ltd 2012'6
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58/70
$uestion
(a) 1; Correct value ofp.
(b) 91 !1 1 91 !1;
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59/70
$uestion
tu#ent Attemt *
Pearson Education Ltd 2012',
-
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60/70
$uestion
(a) 1; Correct value ofp.
(b) 91 !1; -y F xrearranged and correct gradient of 1L
found.
+; )ncorrect method for finding a perpendicular gradient.
91;
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61/70
$uestion
tu#ent Attemt C
Pearson Education Ltd 201260
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62/70
$uestion
(a) 1; Correct value ofp.
(b) 91 !1 1 91 !1;
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63/70
$uestion 10
GC Core Mathematics 1% June 2012
$uestion 10
)n this uestion candidates are given the graph ofyF f 'x( Fx'$ 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 .
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$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+(.
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$uestion 10
Mark cheme
Question
NumberScheme Mark
1%. (a) 7Coordinates ofAare6 '-.&, +(. 1
'1(
(b)(i)
OoriMontal translation 91
8 and their ft 1.& on postitivex8aAis !1 ft
9aAimum at 3 marked on the y8aAis 1
'(
(b)(ii)
Correct shape, minimum at '+, +( and
a maAimum within the first uadrant. 91
1.& onx8aAis !1 ft
9aAimum at '1, 3( 1
'(
(c) 7k = 6 13 1'1(
# 9arks
Pearson Education Ltd 20126&
1.&
3
O8
y
x
1.&
3
O
y
x
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$uestion 10
tu#ent Attemt A
(a) 1;
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$uestion 10
tu#ent Attemt *
(a) +; '*, +( is incorrect.
(b)(i) 91;
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68/70
$uestion 10
tu#ent Attemt C
Pearson Education Ltd 2012 6+
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8/13/2019 GCE Maths Examiner Feedback C1
69/70
$uestion 10
(a) +; '3, +( is incorrect.
(b)(i) 91;
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f 5ou ha+e an5 sub?ect specific questions about the content of
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*his 4ooklet has been 3ritten to pro+ide teachers 3ith
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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
