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Paper Reference(s)

6663/01
Edexcel GCE
Core Mathematics C1
Gold Level G5
Time: 1 hour 30 minutes
Materials required for examination Items included with question papers
Mathematical Formulae (Green) Nil

Candidates may use any calculator allowed by the regulations of the Joint
Council for Qualifications. Calculators must not have the facility for symbolic
algebra manipulation, differentiation and integration, or have retrievable
mathematical formulas stored in them.

Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number,
the unit title (Core Mathematics C1), the paper reference (6663), your surname, initials
and signature.

Information for Candidates

A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
Full marks may be obtained for answers to ALL questions.
There are 11 questions in this question paper. The total mark for this paper is 75.

Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.
You must show sufficient working to make your methods clear to the Examiner.
Answers without working may gain no credit.

Suggested grade boundaries for this paper:

A* A B C D E
63 55 48 40 32 24

Gold 5 This publication may only be reproduced in accordance with Edexcel Limited copyright policy.
©2007–2016 Edexcel Limited.
 PhysicsAndMathsTutor.com

1. Simplify

(a) (2√5)2,
(1)

2
(b) , giving your answer in the form a + √b, where a and b are integers.
2  5  3 2
(4)
May 2015

2. Factorise completely x − 4x3.

(3)
January 2013

3. Express 82x + 3 in the form 2y, stating y in terms of x.

(2)
January 2013

4. A sequence a1 , a 2 , a 3 , ... is defined by

a1 = 2,
a n  1 = 3a n – c

where c is a constant.

(a) Find an expression for a 2 in terms of c.

(1)

3
Given that a
i 1
i = 0,

(b) find the value of c.

(4)
January 2011

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5. The curve C has equation y = x(5 − x) and the line L has equation 2y = 5x + 4.

(a) Use algebra to show that C and L do not intersect.

(4)
(b) Sketch C and L on the same diagram, showing the coordinates of the points at which C
and L meet the axes.
(4)
January 2012

6. (a) By eliminating y from the equations

y  x  4,

2 x 2  xy  8,
show that
x2  4x  8  0 .
(2)
(b) Hence, or otherwise, solve the simultaneous equations

y  x  4,

2 x 2  xy  8,

giving your answers in the form a ± b3, where a and b are integers.
(5)
May 2007

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7.

Figure 1

2
Figure 1 shows a sketch of the curve with equation y = , x ≠ 0.
x

2
The curve C has equation y = − 5, x ≠ 0, and the line l has equation y = 4x + 2.
x

(a) Sketch and clearly label the graphs of C and l on a single diagram.

On your diagram, show clearly the coordinates of the points where C and l cross the
coordinate axes.
(5)

(b) Write down the equations of the asymptotes of the curve C.

(2)

2
(c) Find the coordinates of the points of intersection of y = − 5 and y = 4x + 2.
x
(5)
January 2013

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8. Given that y = 2x,

(a) express 4x in terms of y.

(1)

(b) Hence, or otherwise, solve

8(4x) – 9(2x) + 1 = 0.
(4)
May 2015

9. The straight line with equation y = 3x – 7 does not cross or touch the curve with equation
y = 2px2 – 6px + 4p, where p is a constant.

(a) Show that 4p2 – 20p + 9 < 0.

(4)
(b) Hence find the set of possible values of p.
(4)
May 2016
___________________________________________________________________________

10. (a) Calculate the sum of all the even numbers from 2 to 100 inclusive,

2 + 4 + 6 + ...... + 100.
(3)
(b) In the arithmetic series

k + 2k + 3k + ...... + 100,

k is a positive integer and k is a factor of 100.

(i) Find, in terms of k, an expression for the number of terms in this series.

(ii) Show that the sum of this series is

5000
50 + .
k
(4)
(c) Find, in terms of k, the 50th term of the arithmetic sequence

(2k + 1), (4k + 4), (6k + 7), … ,

giving your answer in its simplest form.

(2)
May 2011

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11.

Figure 2

Figure 2 shows a sketch of the curve C with equation

1
y=2− , x ≠ 0.
x

The curve crosses the x-axis at the point A.

(a) Find the coordinates of A.

(1)
(b) Show that the equation of the normal to C at A can be written as

2x + 8y −1 = 0.
(6)

The normal to C at A meets C again at the point B, as shown in Figure 2.

(c) Find the coordinates of B.

(4)
January 2012

TOTAL FOR PAPER: 75 MARKS

END

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Question
Scheme Marks
number
1 (a) 20 B1
(1)
(b) 2 2 5 3 2
 M1
2 5 3 2 2 5 3 2
...
 A1
2
Numerator = 2(2 5  3 2)  2 10  6 M1
2 2 10  6
  3  10 A1
2 5 3 2 2
(4)
[5]
2 x(1  4 x 2 ) B1
Accept x(4 x  1) or  x(4 x  1) or  x(1  4 x ) or even
2 2 2

4 x( 14  x 2 ) or equivalent
quadratic (or initial cubic) into two brackets M1
x(1  2 x)(1  2 x) or  x(2 x  1)(2 x  1) or x(2 x  1)(2 x  1) A1
[3]
3 (82 x  3   23 
2x  3
)  23(2 x 3) or 2ax b with a = 6 or b = 9 M1
6x 9
2 or  2 3(2 x  3)
as final answer with no errors
A1
or ( y  )6 x  9 or 3(2x + 3)
[2]
4 (a) B1
(1)
(b) M1
M1
A1ft
A1 oae
(4)
[5]

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Question
Scheme Marks
number
5 (a) 1
x(5  x )  (5 x  4) (o.e.) M1
2
2 x 2  5 x  4( 0) (o.e.) e.g. x 2  2.5 x  2   0  A1

b 2  4ac   5   4  2  4
2
M1

 25  32  0 , so no roots or no intersections or no solutions A1

(4)
 y
(b) Curve:  shape and passing through (0, 0) B1



 shape and passing through (5, 0) B1

Line : +ve gradient and no intersections
x with C. If no C drawn score B0 B1
         Line passing through (0, 2) and

(  0.8, 0) marked on axes B1
 (4)
[8]
6 (a) 2 x  x( x  4)  8
2
M1
x2  4 x  8  0 A1cso
(2)
(b)  4  4 2  4  1  8
or  x  2   4  8  0
2
x M1
2
x = - 2 + (any correct expression) A1

48  16 3  4 3 or 12  4 3  2 3 B1

y  22 3 4  M: Attempt at least one y value M1
x  2  2 3 , y  6  2 3 x  2  2 3 , y  6  2 3 A1
(5)
[7]

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Question
Scheme Marks
number
7 (a) 2
y y is translated up or down. M1
x
2
y   5 is in the correct position. A1
x
Intersection with x-axis at  52 , 0  only
x
Independent mark. B1
y  4 x  2 : attempt at straight line,
with positive gradient with positive y
intercept. B1
Intersection with x-axis at
  12 , 0 and y-axis at 0 , 2 . B1
Check graph in question for possible answers and space below graph for
answers to part (b)
(5)
(b) Asymptotes : x  0
(or y-axis) and y   5. An asymptote stated correctly.
B1
(Lose second B mark for Independent of (a)
extra asymptotes)
y   5. (Lose second B mark
These two lines only. Not ft their graph B1
for extra asymptotes)
(2)
(c) 2 y2 2
Method 1:  5  4x  2 Method 2:  M1
x 4 y 5
4 x2  7 x  2  0  x  y  3 y  18  0  y 
2
dM1
x   2, 14 y   6, 3 A1
When x   2, y   6 ,When When y   6, x   2
M1A1
x , y 3
1
4 When y  3, x  14 .
(5)
[12]
8 (a) 4  y
x 2 B1
Allow y 2 or y  y or “y squared”
"4x  " not required
(1)
(b) 8 y 2  9 y  1  (8 y  1)( y  1)  0  y  ... or
M1
(8(2 x )  1)((2 x )  1)  0  2 x  ...
1
2 x (or y )  , 1 A1
8
x  3 x  0 M1 A1
(4)
[5]

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Question
Scheme Marks
number
9 (a) 2 px 2  6 px  4 p "  "3 x  7
or
2
M1
 y7  y7
y  2p  6p 4p
 3   3 
Examples
2 px 2  6 px  4 p  3 x  7   0  ,  2 px 2  6 px  4 p  3 x  7   0 
2
 y7  y7
2p  6p   4 p  y   0 , 2 py 2  10 p  9  y  8 p   0  dM1
 3   3 
y  2 px 2  6 px  4 p  3 x  7

E.g.

b2  4ac   6 p  3  4 2 p 4 p  7
2
ddM1

b2  4ac  10 p  9  4 2 p 8 p 

2

4p2  20p + 9 < 0 * A1*

(4)
(b)  2p  9 2p 1  0  p ... to obtain p = M1
9 1
p , A1
2 2
1
2
 p  4 12 M1 A1

(4)
[8]

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Question
Scheme Marks
number
10 (a) Series has 50 terms B1
1 1
S  (50)(2  100)  2550 or S  (50)(4  49  2)  2550 M1 A1
2 2
(3)
(b)(i) 100
B1
k
(ii) 1  100  1  100   100  
Sum:  k  100  or   2k    1k  M1 A1
2 k  2  k   k  
5000
 50  (*) A1 cso
k
(4)
(c) 50th term = a  (n  1)d
 (2k  1)  49"(2k  3)" Or 2k + 49(2k) + 1 + 49(3) M1
 100k  148  100k  148 A1
(2)
[9]

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Question
Scheme Marks
number
11 (a) 1 
 , 0 B1
2 
(1)
(b) dy
 x 2 M1A1
dx
2
1 dy  1 
At x  ,   4 (= m) A1
2 dx  2 
1  1
Gradient of normal      M1
m  4
1 1
Equation of normal: y  0    x   M1
4 2
2x + 8y – 1 = 0 (*) A1cso
(6)
(c) 1 1 1
2  x M1
x 4 8
[ = 2 x 2  15x  8  0 ] or [ 8 y 2  17 y  0 ]
( 2 x  1)( x  8)  0 leading to x = … M1
1
x    or  8 A1
2
17
y= (or exact equivalent) A1ft
8
(4)
[11]

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Examiner reports

Question 1
(a) This mark was scored by the majority of candidates. The most common error was to
misinterpret the demand as (2 + √5)2 rather than (2√5)2.

(b) Most candidates rationalised the denominator by the correct expression with a few using
a multiple of 2√5 + 3√2 such as –2√5 – 3√2 or 10(2√5 + 3√2). Some candidates then
evaluated the denominator incorrectly and a significant number of candidates who correctly
obtained (2√10 + 6)/2, simplified incorrectly to √10 + 6, or 2√10 + 3. A small number of
candidates used a slightly more efficient route of multiplying the top and bottom of the
fraction by √2, making the rationalisation a little easier.

Question 2
This question was correctly answered by most of the candidates. The vast majority of
candidates got the first mark for identifying the factor of x or –x (or occasionally 4x), though a
significant number of candidates stopped at this point without taking into account that the
question was worth 3 marks. A minority did not gain this first mark as they wrote erroneous
statements such as x  4 x3  x(4 x 2  1) .

Of the candidates who progressed beyond the initial step, most correctly factorised the
resulting quadratic using a difference of 2 squares correctly in their final factorisation. Some
candidates made errors particularly sign errors. A number of candidates “lost” the 1 and gave
x(–4x2) which demonstrated weak algebraic understanding and some went on to try and solve
for x by setting the equation equal to 0. Some candidates did not distinguish between
factorising and solving.

A small number of candidates gained the first mark, by a correct initial factorisation and then
reversed the negatives in factorising the quadratic to give x(2x + 1)(2x – 1), thus losing the
accuracy marks and gaining just 2 of the 3 marks available.

Question 3
The majority of the candidates answered this question efficiently and correctly and gained the
two marks. Many others did state that 8 = 23 somewhere in their workings, but lacked any
evidence of multiplication of the powers 3 and 2x + 3 to gain the method mark. There were a
1
number of candidates who incorrectly ended up using 8 = 2 3 . Common errors included
dividing by 4, attempting to cube (2x + 3) or expanding 3(2x + 3) wrongly to get 6x + 6 or
6x + 3. The most common error was to add the powers (instead of multiplying them), giving
22x+ 6. A small minority attempted to use logarithms, but this was rare.

Question 4
On the whole, this was a high scoring question, with most candidates understanding the
notation and 45% obtaining full marks. Almost all candidates earned the first mark for 6 – c or
3  2 – c, given as their answer in part (a). A correct expression for the third term was seen
regularly, occasionally followed by incorrect simplification to 18 – 2c or even 18 – c.
Candidates who attempted to use the formula for the sum of an Arithmetic Progression lost the
final three marks. A few candidates simply equated the expression for the third term to zero
and solved to find c, ignoring or not understanding the summation.

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Question 5
Most could start part (a) by attempting to form a suitable equation but slips in simplifying the
equation of the line ( y  52 x  4 was common) often meant that the correct equation was not
obtained. Those who did have a correct quadratic usually used the discriminant (sometimes as
part of the quadratic formula) to complete the question. A sizeable number though simply
tried to factorise and concluded that since the equation did not factorise therefore there were
no roots or C and L do not intersect.

The candidates usually fared better in part (b) and there were many excellent sketches scoring
full marks. Weaker candidates had the parabola the wrong way up and it was not uncommon
to see the line crossing the curve despite the information given in part (a). Very few lost
marks for their line or curve stopping on the axes although some thought that if they drew
their line stopping before it crossed the curve that would satisfy the information in part (a).
Some candidates lost a mark for failing to indicate the coordinates (–0.8, 0) where the line
crossed the x-axis.

Question 6
In part (a) most tried the simple substitution of (x – 4) into the second equation. Some made a
sign error (–4x instead of +4x) and proceeded to use this incorrect equation in part (b). Some
candidates did not realise that part (a) was a first step towards solving the equations and
repeated this work at the start of part (b) (sometimes repairing mistakes made there). The
major loss of marks in part (b) was a failure to find the y values but there were plenty of errors
made in trying to find x too. Those who attempted to complete the square were usually
successful although some made sign errors when rearranging the 2 and some forgot the
+ sign. Of those who used the quadratic formula it was surprising how many incorrect
versions were seen. Even using the correct formula was no guarantee of success as incorrect

cancelling was common: was often simplified to or became

Question 7
In Q7(a) the topic testing transformation of a graph proved to challenging to the candidates as
the graph was given in the specific form rather than the more general form of y = f(x) – 5.
The majority of answers had a correct shaped graph but many varieties of translation, left or
right were quite common. Those that did perform a translation of 5 units down often omitted
to find the x-intercept thus losing a mark. Poor drawing with graphs overlapping or incorrect
curvature also lost marks.

The straight line graph was drawn well and was usually in the correct position, but many
candidates forgot to find the intercepts, particularly the x-intercept which required some
algebraic manipulation.

In Q7(b) candidates were asked to give the equations of the asymptotes. A common error seen
was to confuse the x and y to give the asymptotes as x = –5 and y = 0 instead of x = 0 and
y = –5. A large number of candidates left this section blank and a few stated x ≠ 0 and y ≠ –5
which lost one of the two marks. The asymptote y = –5 was more often given than x = 0 even
though the question asked for the equations of the asymptotes. Those who translated the graph
up, left or right could still obtain the correct asymptotes, as these answers could be obtained
independently and correctly from the equation.

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In Q7(c), many candidates realised that they had to eliminate one variable in order to find the
point of intersection. Most chose to equate the y terms and then demonstrated their
competence in solving the resulting three term quadratic. However many answers contained
algebraic errors and hence incorrect co-ordinates. Candidates would be advised to look for
errors in their working, when they reach an unlikely answer.

Some candidates found manipulating the fractions challenging, but continued after finding
one variable.

Question 8
In part (a) many candidates gave 4x = 2y and then proceeded to use this in (b), thus gaining
no marks for this question. Others probably realised that this was incorrect but were unable
to find an alternative and so left the question blank. In a number of instances there was valid
working but no explicit y2, so no mark in (a) but often the y2 was then used correctly in (b).

In part (b) candidates who had been successful in (a) almost invariably made a good start to
(b). A few did not appear to make the connection between the parts and started afresh in (b),
with varying degrees of success. Those that were able to produce a quadratic usually gave
the correct one and solved it correctly for y. However, several left their answers in terms of y
and then neglected to find the values for x, thus losing the final two marks. Of those who did
go on to find values for x, errors included x = 1 from y = 1 and x = 13 rather than x =  3.
Logs were seen infrequently.

Question 9
This proved to be one of the more challenging questions on the paper although most
candidates were able to score at least half of the available marks.

Most candidates recognised that the first part needed the use of the discriminant of a
quadratic. Unfortunately this was often applied to the given quadratic, with no attempt to
involve the line. When the two equations were connected and terms brought to one side, sign
errors were relatively common. A few solutions had a discriminant involving x and a common
error was to use b = 6p – 3. There were, however, a fair number of efficient and accurate
solutions.

The best solutions for the second part of the question used factorisation, and a sketch was
often drawn to decide on the correct region. Those who used the formula made work for
themselves and often lost accuracy, while only a handful managed to complete the square and
obtain the right answer. Some candidates just gave the critical values without trying to find a
region. Of those who attempted an inequality, many wrote it in a correct form but a few used
x instead of p and some omitted to write ‘and’ when expressing the final answer as two
separate inequalities.

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Question 10
This question was found difficult by many. Part marks 3, 0, 2 were common, although some
did try to use the sum formula correctly in part (b) to obtain the method mark. Relatively few
could establish the number of terms for this part, and proceed to use it correctly.

The majority of candidates knew which formula to use in part (a) and consequently gained the
method mark. The problem was realising there were 50 even numbers, common errors were n
= 100, 99, 98 or even 49. Calculating 25  102 correctly, caused problems for many. Only a
small number of weaker candidates did not use the formula but wrote out all the terms and
attempted to add. They were rarely successful.

Many candidates seemed unclear how to attempt part (b)(i). Often it was not attempted; nk
100
was a common wrong answer. There were a few candidates who got n = , but then failed
k
to use this in part (b)(ii).

In part (b)(ii) many candidates scored only the method mark. Those who chose the ‘1st plus
last’ formula found the easier proof, the other sum formula leading to problems with the
1 100 200
brackets for some students. Some became confused by n = arriving at or 200k or
2 k /2 k
50k. Others attempted to work backwards from the result with little success.

The majority of candidates were successful with part (c) even if they had failed to score many
marks in the previous sections. Many could find d = 2k + 3 and use a correct formula for the
50th term, but several continued after reaching 100k + 148 to rewrite it as 50k + 74 or
25k + 37. Common errors were using a sum formula or making a sign slip when finding d.
This type of question needs to be read carefully

Question 11
This was a substantial question to end the paper and a number of candidates made little
attempt beyond part (a). Part (c) proved quite challenging but there were some clear and
succinct solutions seen.

Some stumbled at the first stage obtaining x = 2 or even 1 instead of 1

2 to the solution of
1
2   0 but most scored the mark for part (a).
x

The key to part (b) was to differentiate to find the gradient of the curve and most attempts did
try this but a number had  x 2 . Some however tried to establish the result without
differentiation and this invariably involved inappropriate use of the printed answer. Those
who did differentiate correctly sometimes struggled to evaluate  12  correctly. A correct
2

“show that” then required clear use of the perpendicular gradient rule and the use of their
answer to part (a) to form the equation of the normal. There were a good number of fully
correct solutions to this part but plenty of cases where multiple slips were made to arrive at
the correct equation.

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Most candidates set up a correct equation at the start of part (c) but simplifying this to a
correct quadratic equation proved too challenging for many. Those who did arrive at
2x2 + 15x – 8 = 0 or 8y2 –17y = 0 were usually able to proceed to find the correct coordinates
1
of B, but there were sometimes slips here in evaluating 2  for example.
8

There were a few candidates who used novel alternative approaches to part (c) such as
substituting xy for 2x – 1 from the equation of the curve into the equation of the normal to
obtain the simple equation 8y + xy = 0 from whence the two intersections y = 0 and x = –8
were obtained.

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Statistics for C1 Practice Paper Gold Level G5

Mean score for students achieving grade:

Max Modal Mean
Qu ALL A* A B C D E U
score score %
1 5 5 82 4.09 4.94 4.79 4.53 4.26 3.98 3.61 2.62
2 3 3 63 1.90 2.96 2.73 2.38 2.16 1.92 1.65 1.35
3 2 0 51 1.01 1.98 1.72 1.41 1.14 0.94 0.86 0.66
4 5 63 3.13 4.88 4.61 4.09 3.49 2.96 2.43 1.65
5 8 54 4.34 7.81 7.18 6.12 5.13 4.18 3.37 1.96
6 7 63 4.41 5.76 5.10 4.70 4.23 3.59 2.32
7 12 12 61 7.32 11.56 11.11 9.90 8.56 7.24 5.71 3.21
8 5 0 50 2.50 4.61 4.07 3.00 2.40 1.90 1.40 0.68
9 8 8 55 4.42 7.41 6.64 5.25 4.25 3.41 2.66 1.52
10 9 55 4.91 8.36 7.44 5.70 4.84 4.15 3.56 2.37
11 11 47 5.12 10.95 10.36 8.63 6.58 4.23 3.19 1.09
75 57.53 43.15 65.46 66.41 56.11 47.51 39.14 32.03 19.43

Gold 5: 15/15 18