Five C4 practice papers produced by Edexcel, with mark schemes and examiners' reports
Paper Reference(s)
6666/01
Edexcel GCECore Mathematics C4Silver Level S1
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Green) Nil
Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Core Mathematics C4), the paper reference (6666), 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 8 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. Answerswithout working may gain no credit.
Find the binomial expansion of f (x) in ascending powers of x, up to and including the term inx3. Give each coefficient as a simplified fraction.
(6)
June 2009
2. (a) Use the binomial theorem to expand
3
1
)38( x , x < 38 ,
in ascending powers of x, up to and including the term in x3, givingeach term as a simplified fraction.
(5)
(b) Use your expansion, with a suitable value of x, to obtain anapproximation to 3(7.7). Give your answer to 7 decimal places.
(2)
January 2008
3. f(x) = )49(
6
x, x <
4
9.
(a) Find the binomial expansion of f(x) in ascending powers of x, up to and including theterm in x3. Give each coefficient in its simplest form.
(6)
Use your answer to part (a) to find the binomial expansion in ascending powers of x, up toand including the term in x3, of
(b) g(x) = )49(
6
x, x <
4
9,
(1)
(c) h(x) = )89(
6
x, x <
8
9.
(2)
June 2012
Silver 1: 5/12 2
4. (a) Find the binomial expansion of
3 (8 9 )x , |x| < 8
9
in ascending powers of x, up to and including the term in x3. Give each coefficient as asimplified fraction.
(6)
(b) Use your expansion to estimate an approximate value for 3 7100 , giving your answer to4 decimal places. State the value of x, which you use in your expansion, and show allyour working.
(3)
June 2013 (R)
5. The curve C has equation
16y3 + 9x2y − 54x = 0.
(a) Find x
y
d
d in terms of x and y.
(5)
(b) Find the coordinates of the points on C where x
y
d
d = 0.
(7)
June 2012
6. Water is being heated in a kettle. At time t seconds, the temperature of the water is θ °C.
The rate of increase of the temperature of the water at any time t is modelled by thedifferential equation
d
dt
= λ(120 – θ), θ ≤ 100
where λ is a positive constant.
Given that θ = 20 when t = 0,
(a) solve this differential equation to show that
θ = 120 – 100e–λt
(8)
When the temperature of the water reaches 100 °C, the kettle switches off.
(b) Given that λ = 0.01, find the time, to the nearest second, when the kettle switches off.(3)
June 2013
Silver 1: 5/12 3
7.
Figure 3
The curve C shown in Figure 3 has parametric equations
x = t 3 – 8t, y = t
2
where t is a parameter. Given that the point A has parameter t = –1,
(a) find the coordinates of A.(1)
The line l is the tangent to C at A.
(b) Show that an equation for l is 2x – 5y – 9 = 0.(5)
The line l also intersects the curve at the point B.
(c) Find the coordinates of B.(6)
January 2009
Silver 1: 5/12 4
8. With respect to a fixed origin O, the line l has equation
13 28 21 1
r , where λ is a scalar parameter.
The point A lies on l and has coordinates (3, – 2, 6).
The point P has position vector (–pi + 2pk) relative to O, where p is a constant.
Given that vector PAuuur
is perpendicular to l,
(a) find the value of p.(4)
Given also that B is a point on l such that <BPA = 45°,
(b) find the coordinates of the two possible positions of B.(5)
June 2013
TOTAL FOR PAPER: 75 MARKS
END
Silver 1: 5/12 5
QuestionNumber
Scheme Marks
1. 12
1f 4
4x x
x
M1
12 ...
4 1 ... ... 1
1 ... 2
or 1
2 1 ... B1
2 33 3 51 12 2 2 2 21
2 ... 1 ...4 2 4 3! 4
x x x M1 A1ft
ft their 4
x
2 31 1 3 5, ...
2 16 256 2048x x x A1, A1 (6)
(6 marks)
Silver 1: 5/12 6
Question Number
Scheme Marks
** represents a constant (which must be consistent for first accuracy mark)
2. (a) 1 13 31 1
3 33 3
8 3 8 1 2 18 8
x xx
Takes 8 outside the
bracket to give any of 13(8)
or 2 .
B1
51 2 1 22 33 3 3 3 31
3
( )( ) ( )( )( )2 1 ( )(** ); (** ) (** ) ...
2! 3!x x x
with ** 1
Expands 13(1 ** )x to give
a simplified or an un-simplified
131 ( )(** )x ;
M1;
A correct simplified or an
un-simplified ..........
expansion withcandidate’s followed
through ** x
A1
51 2 1 22 33 3 3 3 33 3 31
3 8 8 8
( )( ) ( )( )( )2 1 ( )( ) ( ) ( ) ...
2! 3!x x x
2 351 18 64 15362 1 ; ...x x x Either 1
82 1 ........x or
A1;
2 31 1 52 ; ...
4 32 768x x x
anything that
cancels to 1
2 ;4
x
Simplified 2 35132 768x x A1
[5]
(b)13 2 31 1 5
(7.7) 2 (0.1) (0.1) (0.1) ...4 32 768
Attempt to substitute0.1x into a candidate’sbinomial expansion.
M1
2 0.025 0.0003125 0.0000065104166...
1.97468099... awrt 1.9746810 A1 [2
]
7 marks
Silver 1: 5/12 7
QuestionNumber
Scheme Marks
3. (a) 12f ... ... ...x x
M1
1
2
6 9 ...
12
6
9,
6
3, 2 or equivalent B1
3 3 51 1
2 32 2 2 2 212 ... 1 ; ...
2 3!kx kx kx
M1; A1ft
2
2 1 ...9
x or
42
9x A1
2 34 4 402 ...
9 27 729x x x A1 (6)
(b) 2 34 4 40g 2 ...
9 27 729x x x x B1ft (1)
(c) 2 34 4 40h 2 2 2 2 ...
9 27 729x x x x M1 A1 (2)
2 38 16 3202 ...
9 27 729x x x
[9]
Silver 1: 5/12 8
Qn Scheme Mark
4. (a) 1
3 3(8 9 ) 8 9x x Power of 1
3M1
1.
1 11 3 33
9 98 1 2 1
8 8
x x
1
38 or 2 B1
2.
51 2 1 22 33 3 3 3 3( )( ) ( )( )( )1
2 1 ( ) ( ) ( ) ...3 2! 3!
k x k x k x
M1 A1
2 351 2 1 2
3 3 3 3 3( )( ) ( )( )( )1 9 9 92 1 ...
3 8 2! 8 3! 8
x x x 2 33 9 45
2 1 ; ...8 64 512
x x x
2 33 9 452 ; ...
4 32 256x x x A1; A1
(6)
(b) 3 3 37100 10 71 10 (8 9 ) , so 0.1x x Writes down
Substituting either of their values into 3xy to obtain a value of the other variable.
M1
3 3
2, , 2,2 2
both A1 (7)
[12]
Silver 1: 5/12 10
QuestionNumber
Scheme Marks
6. d120 , 100
dt
„
(a)1
d d120
t
or
1d d
120t
B1
ln 120 ; t c or 1ln 120 ; t c
M1 A1;
M1 A1
0 , 20t ln 120 20 (0) c M1
ln100 ln 120 ln100c t then either... or...
ln 120 ln100t ln100 ln 120t
120ln
100t
100
ln120
t
120e
100t
100e
120t
dddM1
100e 120t 120 e 100
120 100e
t
t
A1 * leading to 120 100e t
(8)(b) 0.01 , 100 0.01100 120 100e t M1
0.01 120 100100e 120 100 0.01 ln
100t t
1 120 100ln
0.01 100t
Uses correct order ofoperations by moving
from 0.01100 120 100e t to give ...t and
lnt A B , where 0B
dM1
1 1ln 100ln 5
0.01 5t
160.94379...t = 161 (s) (nearest second) awrt 161 A1
(3)[11]
Silver 1: 5/12 11
Question Number
Scheme Marks
7 (a) At A, 21 8 7 & ( 1) 1 (7,1)x y A (7,1)A B1(1)
(b) 3 28 ,x t t y t ,
2d3 8
d
xt
t ,
d2
d
yt
t
2
d 2
d 3 8
y t
x t
Their d
dyt divided by their d
dxt M1
Correct dd
yx A1
At A, 2
2( 1) 2 2 2m( )
3( 1) 8 3 8 5 5
T Substitutes for t to give any ofthe four underlined oe:
A1
: their 1 their 7Ty m x T
or 92 145 5 51 7 1c c
Hence 925 5: y x T
Finding an equation of a tangentwith their point and their
tangent gradient or finds c and uses
(their gradient) " "y x c .
dM1
gives : 2 5 9 0x y T AG 2 5 9 0x y A1 cso(5)
(c) 3 22( 8 ) 5 9 0t t t Substitution of both 3 8x t t
and 2y t into TM1
3 22 5 16 9 0t t t
2( 1) (2 7 9) 0t t t A realisation that 1t is a factor.
dM1 ( 1) ( 1)(2 9) 0t t t
921 (at ) att A t B 9
2t A1
29 9 729 4412 2 8 8
29 812 4
8 36 55.125 or awrt 55.1
20.25 or awrt 20.3
x
y
Candidate uses their value of tto find either the x or y
coordinate ddM1
One of either x or y correct. A1Both x and y correct. A1
Hence 814418 4,B awrt (6)
[12]
Silver 1: 5/12 12
Silver 1: 5/12 13
QuestionNumber
Scheme Marks
8.
13 2
: 8 2
1 1
l
r , 3, 2, 6A , 0
2
p
OP
p
uuur
(a) 3
2 0
6 2
p
PA
p
uuur 3
0 2
2 6
p
AP
p
uuurFinds the difference
between OAuuur
and OPuuur
.Ignore labelling.
M1
3
2
6 2
p
p
3
2
2 6
p
p
Correct difference. A1
3 2
2 2 6 2 4 6 2 0
6 2 1
p
p p
p
M1
1p A1 cso(4)
(b) 2 2 2 2 2 24 ( 2) 4 or ( 4) 2 ( 4)AP AP M1
So, orPA AP 36 or 6 cao A1 cao
It follows that, "6"AB PA or
"6 2" 2PB PA B1 ft
{Note that "6" 2(the modulus of the direction vector of )AB l }
3 2
2 2 2
6 1
OB
uuur or
13 2
8 3 2
1 1
OB
uuur and
13 2
8 7 2
1 1
OB
uuur
Uses a correct method inorder to find both possible
sets of coordinates of B.M1
7 1
2 and 6
4 8
Both coordinates arecorrect.
A1 cao
(5)[9]
Silver 1: 5/12 14
Question 1
This proved a suitable starting question and the majority of candidates gained 5 or 6 of the available 6 marks. Nearly all could obtain the index as 12 but there
were a minority of candidates who had difficulty in factorising out 4 from the brackets and obtaining the correct multiplying constant of 12
. Candidates’
knowledge of the binomial expansion itself was good and, even if they had an incorrect index, they could gain the method mark here. An unexpected number of
candidates seemed to lose the thread of the question and, having earlier obtained the correct multiplying factor 12
and expanded
12
14
x
correctly,
forgot to multiply their expansion by 2
1.
Question 2
In part (a), a majority of candidates produced correct solutions, but a minority of candidates were unable to carry out the first step of writing 138 3x as
133
2 18
x . Those who did so were able to complete the remainder of this part but some bracketing errors, sign errors and manipulation errors were
seen.
In part (b), many candidates realised that they were required to substitute 0.1x into their binomial expansion. About half of the candidates were able to
offer the correct answer to 7 decimal places, but some candidates made calculation errors even after finding the correct binomial expansion in part (a). A fewcandidates used their calculator to evaluate the cube root of 7.7 and received no credit.
Question 3
A small number of candidates used the binomial expansion with index 1
2 but the great majority used the correct index,
1
2 , and were able to expand an
expression of the form 121 kx
correctly to obtain at least three marks. Although many dealt with the 9 correctly, taking 129 outside a bracket, some
did not combine it correctly with the 6, multiplying their binomial by 18 rather than 2. Full marks were common in part (a). In part (b), most realised that a change
in signs was necessary but many changed the sign of the term in 2x as well as the terms in x and 3x . Part (c) was less well done than part (b) and many
multiplied all three of the terms in x, 2x and 3x by 2 instead of by 2, 4 and 8 respectively. Questions like parts (b) and (c) have rarely been set on these papers
and it was clear that many candidates were not able to think their way into a solution that did not require a practised technique.
Question 4This question was generally well answered with about 76% of candidates gaining at least 6 of the 9 marks available and about 17% of candidates gaining all 9marks. Part (a) was accessible with most candidates scoring all 6 marks and part (b) was discriminating and challenged the more able candidates.
In part (a), most candidates manipulated 3 )98( x to give 3
1
8
912
x , with the 2 outside the brackets sometimes written incorrectly as either
1 or 21 and a few incorrectly used a power of
23
. Many candidates were able to use a correct method for expanding a binomial expression of the form (1 +
ax)n. A variety of incorrect values of a were seen, with the most common being 89 . Some candidates, having correctly expanded
3
1
8
91
x , forgot to
multiply their expansion by 2. Sign errors, bracketing errors, and simplification errors were also seen in this part.
In part (b), the majority of candidates solved the equation 3 )98( x = 3 1700 to give an answer of x = –788. These candidates substituted this
value of x into the answer they had found in part (a), even though the question states that the binomial expansion is only valid for x < 98 . Only a minority of
candidates realised that they needed to simplify 3 1700 to 10 3 1.7 before deducing that they needed to substitute x = 0.1 into their binomial
expansion. Most of these candidates achieved the correct approximation of 19.2201, although a few forgot to multiply by 10 at the end and wrote 1.9220.
Question 5
As has been noticed more than once in recent years, the topic of implicit differentiation is well understood and full marks in part (a) were very common. Mistakes
mainly concerned the differentiation of 29x y , involving a misinterpretation of the product rule.
Part (b) proved a test even for the most able. Most recognised that the numerator of their answer to (a) had to be equated to zero and obtained 3xy or an
equivalent but then many just gave up then immediately. It was disappointing to see a significant minority of those who realised that they should solve the
simultaneous equations 3xy and 3 216 9 54 0y x y x , started by transforming 3xy to 3y x . Those who did start correctly
often had problems with the resulting algebra and had difficulty reaching the correct 4 16x or 4 81
16y . Those who got this far often failed to realise
that these equations have two solutions. Those who had correct values for either x or y could complete quickly by substituting into 3xy but some made
Silver 1: 5/12 15
extra work for themselves by either starting all over again and finding the other variable independently or by substituting into3 216 9 54 0y x y x . The latter was particular unfortunate if x had been found first as this resulted in a cubic in y which is difficult to solve. This
question was a very discriminating and it may be worth noting that the proportion of those who gained full marks on this question was slightly less than theproportion of those gaining the equivalent of a grade A on this paper.
Question 6
This question was generally well answered with about 57% of candidates gaining at least 8 of the 11 marks available and about 37% of candidates gaining all 7marks. A minority of candidates made no creditable attempt in part (a) and then scored full marks in part (b).
In part (a), those candidates who were able to separate the variables, were usually able to integrate both sides correctly, although a number integrated
1
120 incorrectly to give ln 120 . Many candidates substituted 0, 120t immediately after integration, to find their
constant of integration as ln100 and most used a variety of correct methods to eliminate logarithms in order to achieve the printed result. A significant
number of candidates, however, correctly rearranged their integrated expression into the form 120 e ,tA before using
0, 120t to correctly find A. Common errors in this part included omitting the constant of integration, treating as a variable and incorrect
manipulation in order to fudge the printed result. Also, a number of candidates struggled to remove logarithms correctly and gave an equation of the form
120 e et c which was then sometimes manipulated to 120 e .tA
In part (b), most candidates were able to substitute the given values into the printed equation and achieve 161t seconds. Some candidates made careless
errors when manipulating their expressions, whilst a number did not round their answer of 160.94... to the nearest second. Few candidates substituted the
given values into their incorrect answer from part (a).
Question 7
Part (a) was answered correctly by almost all candidates. In part (b), many candidates correctly applied the method of finding a tangent by using parametricdifferentiation to give the answer in the correct form. Few candidates tried to eliminate t to find a Cartesian equation for C, but these candidates were usually notable to find the correct gradient at A.
In part (c), fully correct solutions were much less frequently seen. A significant number of candidates were able to obtain an equation in one variable to score the
first method mark, but were then unsure about how to proceed. Successful candidates mostly formed an equation in t, used the fact that 1t was a factor
and applied the factor theorem in order for them to find t at the point B. They then substituted this t into the parametric equations to find the coordinates of B.Those candidates who initially formed an equation in y only went no further. A common misconception in part (c), was for candidates to believe that the gradient
at the point B would be the same as the gradient at the point A and a significant minority of candidates attempted to solve 2
2 2
3 8 5
t
t
to find t at the
point B.
Question 8
In general, this was the most poorly answered question on the paper with about 25% of candidates failing to score. Some candidates did not seem to have a firmgrasp of what was required and many produced pages of irrelevant working. This question did discriminate well between candidates of average to higherabilities, with about 54% of candidates gaining at least 4 of the 9 marks available and only about 10% of candidates gaining all 9 marks. Part (a) was found to befairly accessible, and part (b) was challenging to all but the most able candidates. Many were unable to think about the question logically or produce a cleardiagram and establish the relationship between the length of AB (and/or PB) and the length of PA.
In part (a), the vector PAuuur
(or APuuur
) was usually found, although sign slips, adding OPuuur
to OAuuur
and mixing up of the i, j and k components of OPuuur
were common errors. Many candidates found the correct value of p by applying the correct scalar product between PAuuur
(or APuuur
) and the direction vector
2 2 i j k and setting the result equal to 0, although some candidates used ,OAuuur
2 2 i j k or 13 8 i j k instead of
2 2 . i j k Other errors included taking the dot product between OAuuur
and OBuuur
or deducing 1p from a correct 4 4 0.p
Those candidates who attempted part (b) usually managed to find the magnitude of PA and many drew a diagram of triangle PAB correctly and deduced
.PA AB From this point, however, many candidates did not know how to proceed further, resulting in a lot of incorrect work which yielded no further
marks. Some candidates, however, were able to form a correct equation in order to find both values of . It was unfortunate that a few, having found the
correct values of 3 and 7 then substituted these into
3 2
2 2
6 1
instead of the equation for the line l. The most popular
method for finding correct values of was for candidates to form and solve a Pythagorean equation in of 6AB or 2 36.AB Other
successful methods for finding included solving 6 2PB or solving a dot product equation between either PAuuur
and PBuuur
or ABuuur
and
PBuuur
.
Few candidates realised that the length AB was twice the length of the direction vector of the line l and applied twice the direction vector 2 2 i j k in
either direction from A in order to find both positions for B.
Materials required for examination Items included with question papersMathematical Formulae (Green) Nil
Candidates may use any calculator allowed by theregulations of the Joint
Council for Qualifications. Calculators must not have thefacility for symbolic
algebra manipulation, differentiation and integration, orhave 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 C4), the paper reference (6666), 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 8 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. Answerswithout working may gain no credit.
Figure 2 shows a sketch of the curve with equation y = x3 ln (x2 + 2), x 0.
The finite region R, shown shaded in Figure 2, is bounded by the curve, the x-axis and theline x = 2.
The table below shows corresponding values of x and y for y = x3 ln (x2 + 2).
x 04
2
2
2
4
23 2
y 0 0.3240 3.9210
(a) Complete the table above giving the missing values of y to 4 decimal places.(2)
(b) Use the trapezium rule, with all the values of y in the completed table, to obtain anestimate for the area of R, giving your answer to 2 decimal places.
(3)
(c) Use the substitution u = x2 + 2 to show that the area of R is
uuu dln)2(2
14
2
.
(4)
(d) Hence, or otherwise, find the exact area of R.(6)
June 2011
Silver 2: 6/12 21
5.))(( 21
1052 2
xx
xx A +
1x
B +
2x
C.
(a) Find the values of the constants A, B and C.(4)
(b) Hence, or otherwise, expand ))(( 21
1052 2
xx
xx in ascending powers of x, as far as the term
in x2. Give each coefficient as a simplified fraction.
(7)
June 20108
6. The area A of a circle is increasing at a constant rate of 1.5 cm2 s–1. Find, to 3 significantfigures, the rate at which the radius r of the circle is increasing when the area of the circleis 2 cm2.
(5)
January 2010
7. Relative to a fixed origin O, the point A has position vector (2i – j + 5k),
the point B has position vector (5i + 2j + 10k),
and the point D has position vector (–i + j + 4k).
The line l passes through the points A and B.
(a) Find the vector AB .(2)
(b) Find a vector equation for the line l.(2)
(c) Show that the size of the angle BAD is 109°, to the nearest degree.(4)
The points A, B and D, together with a point C, are the vertices of the parallelogram ABCD,where AB = DC .
(d) Find the position vector of C.(2)
(e) Find the area of the parallelogram ABCD, giving your answer to 3 significant figures.(3)
(f) Find the shortest distance from the point D to the line l, giving your answer to 3significant figures.
(2)
January 2012
Silver 2: 6/12 22
8. In an experiment testing solid rocket fuel, some fuel is burned and the waste products arecollected. Throughout the experiment the sum of the masses of the unburned fuel and wasteproducts remains constant.
Let x be the mass of waste products, in kg, at time t minutes after the start of the experiment.It is known that at time t minutes, the rate of increase of the mass of waste products, in kg perminute, is k times the mass of unburned fuel remaining, where k is a positive constant.
The differential equation connecting x and t may be written in the form
d( )
d
xk M x
t , where M is a constant.
(a) Explain, in the context of the problem, what d
d
x
t and M represent.
(2)
Given that initially the mass of waste products is zero,
(b) solve the differential equation, expressing x in terms of k, M and t.(6)
Given also that x = 1
2M when t = ln 4,
(c) find the value of x when t = ln 9, expressing x in terms of M, in its simplest form.(4)
M is the total mass of unburned fuel and waste fuel (or the initial mass of unburned fuel)
Any one correctexplanation.
B1
Both explanationsare correct.
B1
(2)
(b)
1d dx k t
M x
or
1d d
( )x t
k M x
B1
ln M x kt c or
1ln M x t c
k
M1 A1
0 , 0t x ln 0 (0)M k c M1
ln ln lnc M M x kt M then either... or...
ln lnkt M x M ln lnkt M M x
lnM x
ktM
ln
Mkt
M x
e kt M x
M ekt M
M x
ddM1
e ktM M x ektM x M
e ktM x M A1 * cso leading to e ktx M M or
(1 e )ktx M oe(6)
(c)1
, ln 42
x M t
ln 41(1 e )
2kM M M1
ln 4 ln 41 11 e ln 4 ln 2
2 2k ke k
So 1
2k A1
1ln9
21 ex M
dM1
2
3x M 2
3x M A1 cso
(4)[12]
Silver 2: 6/12 29
Question 1
The majority of candidates gained full marks on this question. Most obtained the identity
229 1 2 1 2 1 1x A x x B x C x and found B and C by substituting 1x and 12x . A
significant number of candidates found an incorrect value of C after making the error 23 92 4 . This can arise through the misuse
of a calculator. The value of A was usually found either by substituting 0x or equating coefficients of 2x . Relatively few candidates
attempted the question by equating all three coefficients to obtain three equations and solving these equations simultaneously. The workingfor this method is rather complicated and errors were often made.
Question 2
This question was generally well answered with around 50% of the candidature gaining all 6 marks. The majority of candidates were able to
apply the integration by parts formula in the correct direction. Some candidates, however, did not assign u and d
d
v
x and then write down
their d
d
u
x and v before applying the by parts formula, which meant that if errors were made the method used was not always clear.
In part (a), sin 3 dx x caused some problems for a minority of candidates who produced responses such as cos3x or
3cos3x or 1
cos33
x . After correctly applying the by parts formula, a few candidates then incorrectly wrote down
1cos3 d
3x x as
1cos3
6x .
Most candidates who could attempt part (a) were able to make a good start to part (b), by assigning u as 2x and d
d
v
xas cos3 ,x and
then correctly apply the integration by parts formula. At this point, when faced with 2
sin 3 d3
x x x , some candidates did not make the
connection with their answer to part (a) and made little progress. Other candidates independently applied the by parts formula again, with anumber of them making a sign error.
Question 3
This question was also well answered and the general principles of implicit differentiation were well understood. By far the commonest
source of error was in differentiating 2x ; examples such as 2x , 2 lnx x and 12xx were all regularly seen. Those who knew how to
differentiate 2x nearly always completed the question correctly, although a few had difficulty in finding d2
dxy
x correctly. A
minority of candidates attempted the question by taking the logs of both sides of the printed equation or a rearrangement of the equation in
the form 22 2x xy y . Correctly done, this leads to quite a neat solution, but, more frequently, errors, such as
2 2ln 2 ln 2 lnx xy y , were seen.
Question 4
Part (a) was well done and the only error commonly seen in part (b) was using the incorrect width of the trapezium 2
5 instead of
2
4.
A few candidates made errors, often due to a lack of clear bracketing, but great majority completed part (b) correctly and gave their answer to
the degree of accuracy specified in the question. Part (c) was well done and the majority were able to find d
d
u
x and make a complete
substitution for the variables. The only common error in this part was simply to ignore the limits and to give no justification for the limitsbecoming 2 and 4. Most recognised that the integral in part (d) required integration by parts and those who used a method involving
integrating 2u to
2
22
uu and differentiating ln u usually reached the half way stage correctly. The second integration proved
Silver 2: 6/12 30
more difficult and there were many errors in simplifying the expression
2 12
2
uu
u
before the second integration. The errors often
arose from a failure to use the necessary brackets. There were also many subsequent errors in signs and a few candidates omitted the 12 from
their integration.
Those who, at the first stage of integration by parts integrated 2u to 2
2
2
u , which is, of course, correct, had markedly less
success with the second integral than those who integrated to
2
22
uu .
A few split the integral up into two separate integrals, ln du u u and ln du u but the second of these integrals was rarely
completed correctly. Those who ignored the hint in the question and attempted to integrate with respect to x were generally unable to deal
with
5
2d
2
xx
x
, which arises after integrating by parts once
Question 5
The first part of question 5 was generally well done. Those who had difficulty generally tried to solve sets of relatively complicatedsimultaneous equations or did long division obtaining an incorrect remainder. A few candidates found B and C correctly but eitheroverlooked finding A or did not know how to find it. Part (b) proved very testing. Nearly all were able to make the connection between the
parts but there were many errors in expanding both 11x
and 12 x
. Few were able to write 11x
as 11 x
and the resulting expansions were incorrect in the majority of cases, both 21 x x and 21 x x being common.
12 x
was handled better but the constant 1
2 in
11
12 2
x
was frequently incorrect. Most recognised that they should collect
together the terms of the two expansions but a few omitted their value of A when collecting the terms.
Question 6
Connected rates of change is a topic which many find difficult. The examiners reported that the responses to this question were of asomewhat higher standard than had been seen in some recent examinations and the majority of candidates attempted to apply the chain rule
to the data of the question. Among those who obtained a correct relation, d
1.5 2d
rr
t or an equivalent, a common error was to use
2r , instead of using the given 2A to obtain 2
r
. Unexpectedly the use of the incorrect formula for the area of the circle,
22A r , was a relatively common error.
Question 7
This question discriminated well across all abilities, with parts (e) and (f) being the most demanding, and those candidates who drew theirown diagram being the more successful. About 15% of the candidature was able to gain all 15 marks.
Part (a) was well answered with only a few candidates adding OB to OA instead of applying OB – OA . Candidates who failed
to answer part (a) correctly usually struggled to gain few if any marks for the remainder of this question.
In part (b), most candidates were able to write down a correct expression for l, but a number of candidates did not form a correct equation bywriting either r = ... and so lost the final accuracy mark. (After some discussion the examiners also accepted l = ..., which is quite common,though non standard.)
In part (c), most candidates were able to take the correct dot product between either AB and AD or BA and DA to obtain the
correct answer of 109°. The most common error was to obtain an answer of 71° by incorrectly taking the dot product between either ABand DA or BA and AD , and using this answer to obtain an answer of 109° without proper justification. A small minority of
candidates applied the cosine rule correctly to achieve the correct answer. A number of candidates struggled with this part and usually took
the dot product between non-relevant vectors such as OA and OB or AB and BD .
Silver 2: 6/12 31
In part (d), a significant number of candidates were able to obtain the correct position vector of OC = 2i + 4j + 9k by adding either
OD to AB or OB to AD . A few candidates also achieved the correct result by arguing that the midpoints of the two diagonals
of a parallelogram are coincident. Occasionally the incorrect answer of OC = (4i + 2j + k) was given, which is a result of taking the
difference between OD and AB .
Candidates who were successful in part (e) found the area of the parallelogram either by finding the area of triangle ABD using 21 bd sin A
and doubling the result or by applying a method of base perpendicular height. The most common error in part (e) was for candidates to findthe product of lengths AD and AB.
Candidates who were successful in part (f) usually found the shortest distance by multiplying their AD by sin 71 (or equivalent). Those
candidates who multiplied AB by sin 71 did not receive any credit. A few candidates attempted to use vectors to find DE , where
E is the point where the perpendicular from D meets the line l, often spending considerable time for usually little or no reward.
Question 8
In general, this was the most poorly answered question on the paper with about 15% of candidates who failed to score and about 11% ofcandidates gaining 1 mark usually in part (a). This question discriminated well between candidates of higher abilities, with about 27% ofcandidates gaining at least 8 of the 12 marks available and only about 7% of candidates gaining all 12 marks. Many weaker candidates madelittle or no progress in part (b), maybe because of the generalised nature of the differential equation. In part (a), a significant number of candidates were not clear or precise in their explanations. A number of them used the word “mass” and itwas not clear whether they were referring to the mass of the unburned fuel or the mass of the waste products.In part (b), those candidates who were able to separate the variables, were usually able to integrate both sides correctly, although a number of
candidates integrated xM
1 incorrectly to give ln (M – x). Many others substituted t = 0, x = 0 immediately after integration, to find
their constant of integration as –ln M and most used a variety of correct methods to eliminate logarithms in order to find x = M(1 – e–kt) (orequivalent). A significant number of candidates, however, correctly rearranged their integrated expression into the form x = M – Ae–kt beforeusing t = 0, x = 0 to correctly find A. Common errors in this part included omitting the constant of integration or treating M as a variable.Also, a number of candidates struggled to remove logarithms correctly and gave an equation of the form M – x = e–kt + ec which was thensometimes manipulated to M – x = Ae–kt.
In part (c), some candidates were able to substitute t = ln 4, x = 21 M into one of their equations involving x and t, but only a minority were
able to find a numeric value of k. Only the most able candidates were able to find k = 21 and substitute this into their equation together with
Materials required for examination Items included with question papersMathematical Formulae (Green) Nil
Candidates may use any calculator allowed by theregulations of the Joint
Council for Qualifications. Calculators must not have thefacility for symbolic
algebra manipulation, differentiation and integration, orhave 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 C4), the paper reference (6666), 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 8 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. Answerswithout working may gain no credit.
Suggested grade boundaries for this paper:
A* A B C D E
67 59 51 45 38 32
Silver 3: 7/12 34
1. Express in partial fractions
2
5 3
(2 1)( 1)
x
x x
(4)
June 2013 (R)
2. (a) Use integration to find
xx
xdln
13 .
(5)
(b) Hence calculate
xxx
dln12
13
.
(2)
January 2013
3. f(x) = )1()23(
1632272
2
xx
xx
, x <
3
2.
Given that f(x) can be expressed in the form
f(x) = )23( x
A + 2)23( x
B +
)1( x
C
,
(a) find the values of B and C and show that A = 0.(4)
(b) Hence, or otherwise, find the series expansion of f(x), in ascending powers of x, up to andincluding the term in x2. Simplify each term.
(6)
(c) Find the percentage error made in using the series expansion in part (b) to estimate thevalue of f(0.2). Give your answer to 2 significant figures.
(4)
January 2009
Silver 3: 7/12 35
4. A curve C has parametric equations
x = sin2 t, y = 2 tan t , 0 ≤ t < 2
.
(a) Find x
y
d
d in terms of t.
(4)
The tangent to C at the point where t = 3
cuts the x-axis at the point P.
(b) Find the x-coordinate of P.(6)
June 2010
5. Find the gradient of the curve with equation
ln y = 2x ln x, x > 0, y > 0,
at the point on the curve where x = 2. Give your answer as an exact value.(7)
June 2011
6. With respect to a fixed origin O, the lines l1 and l2 are given by the equations
l1: r =
2
3
6
+
3
2
1
, l2: r =
3
15
5
+ μ
1
3
2
,
where μ and are scalar parameters.
(a) Show that l1 and l2 meet and find the position vector of their point of intersection A.(6)
(b) Find, to the nearest 0.1°, the acute angle between l1 and l2.(3)
The point B has position vector
1
1
5
.
(c) Show that B lies on l1.(1)
(d) Find the shortest distance from B to the line l2, giving your answer to 3 significant figures.
(4)
June 2011
Silver 3: 7/12 36
7.
Figure 2
Figure 2 shows a sketch of the curve C with parametric equations
x = 5t 2 − 4, y = t(9 − t 2)
The curve C cuts the x-axis at the points A and B.
(a) Find the x-coordinate at the point A and the x-coordinate at the point B.(3)
The region R, as shown shaded in Figure 2, is enclosed by the loop of the curve.
(b) Use integration to find the area of R.(6)
January 2010
Silver 3: 7/12 37
8. (a) Using the substitution x = 2 cos u, or otherwise, find the exact value of
2
122
d)4(
1x
xx.
(7)
Figure 3
Figure 3 shows a sketch of part of the curve with equation y = 41
)4(
42xx
, 0 < x < 2.
The shaded region S, shown in Figure 3, is bounded by the curve, the x-axis and the lines withequations x = 1 and x = √2. The shaded region S is rotated through 2π radians about the x-axisto form a solid of revolution.
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed.(3)
January 2010
TOTAL FOR PAPER: 75 MARKS
END
Silver 3: 7/12 38
QuestionNumber
Scheme Marks
1.
2 2
5 3
(2 1)( 1) (2 1) ( 1) ( 1)
x A B C
x x x x x
2, 2A C
At least one of “A” or “C” are correct.
B1
Breaks up their partialfraction correctly into
three terms and both " " 2A and
" " 2C .
B1 cso
1.2. 25 3 ( 1) (2 1)( 1) (2 1)x A x B x x C x Writes down a
correct identity andattempts to find thevalue of either one
“A” or “ B” or “C”.
M13. 1 2 2x C C
1 5 1 1 13 2
2 2 4 2 4x A A A
Either2 : 0 2 , constant : 3x A B A B C
: 5 2 3 2x A B C
leading to 1B
Correct value for “B”which is found usinga correct identity and
follows from theirpartial fractiondecomposition.
A1 cso
So,
2 2
5 3 2 1 2
(2 1)( 1) (2 1) ( 1) ( 1)
x
x x x x x
[4]
Silver 3: 7/12 39
2. (a)3
1ln dx x
x , 23
2
d 1ln
d
d 1
d 2 2
uu x
x x
v xx v
x x
2 2
1 1 1ln . d
2 2x x
x x x
In the form 2 2
1 1ln .x
x x x
M1
2
1ln
2x
x
simplified or un-simplified. A1
2
1 1.
2x x
simplified or un-simplified. A1
2 3
1 1 1ln d
2 2x x
x x
2 2
1 1 1ln
2 2 2x c
x x
22
1 1. .x
x x dM1
Correct answer, with/without c A1[5]
(b)2
2 2 2 2 2 21
1 1 1 1 1 1ln ln 2 ln1
2 4 2(2) 4(2) 2(1) 4(1)x
x x
Applies limits of2 and 1 to theirpart (a) answer
and subtracts thecorrect way
round.
M1
1
83 1 3 1
ln 2 or ln 2 or 3 2ln 2 , etc, or awrt 0.116 8 16 16
or equivalent. A1
[2]7
Silver 3: 7/12 40
Question Number
Scheme Marks
3 (a) 2 227 32 16 (3 2)(1 ) (1 ) (3 2)x x A x x B x C x Forming this identity M1
23 ,x 64 5 20 5
3 3 3 312 16 4B B B
1,x 27 32 16 25 75 25 3C C C
Substitutes either 23x or
1x into their identity orequates 3 terms or
substitutes in values towrite down three
simultaneous equations.
M1
Both 4B and 3C A1(Note the A1 is dependenton both method marks in
this part.)
Equate x2: 27 3 9 27 3 27 0 3
0
A C A A
A
0, 16 2 4
16 2 4 12 0 2 0
x A B C
A A A
Compares coefficients orsubstitutes in a third x-
value or uses simultaneousequations to show A = 0.
B1
(4)
(b) 2
4 3f ( )
(3 2) (1 )x
x x
2 14(3 2) 3(1 )x x Moving powers to top on
any one of the twoexpressions
M1
2 1321 1 3(1 )x x
3 3 22 2
( 2)( 3)1 1 ( 2)( ); ( ) ...
2!x x
Either 321 ( 2)( )x or
1 ( 1)( )x from eitherfirst or second expansions
respectively
dM1;
2( 1)( 2)3 1 ( 1)( ); ( ) ...
2!x x
Ignoring 1 and 3, any one
correct .......... expansion.A1
Both .......... correct. A1
2 22741 3 ... 3 1 ...x x x x
23944 0 ;x x 239
44 (0 ) ;x x A1; A1
(6)
Silver 3: 7/12 41
Question Number
Scheme Marks
3 (c)
1.08 6.4 16Actual f (0.2)
(6.76)(0.8)
23.48 29354.341715976...
5.408 676
Or
2
4 3Actual f (0.2)
(3(0.2) 2) (1 0.2)
4 29353.75 4.341715976...
6.76 676
Attempt to find the actual value of f(0.2) or seeing awrt 4.3 and
believing it is candidate’sactual f(0.2).
Candidates can also attemptto find the actual value by
using
2(3 2) (3 2) (1 )
A B C
x x x
with their A, B and C.
M1
2394Estimate f (0.2) 4 (0.2)
4 0.39 4.39
Attempt to find an estimatefor f(0.2) using their answer
to (b)M1
4.39 4.341715976...
%age error 1004.341715976...
their estimate - actual100
actual M1
1.112095408... 1.1%(2sf ) 1.1%A1 cao
(4)
[14]
QuestionNumber
Scheme Marks
4. (a) 2d d
2sin cos , 2secd d
x yt t t
t t B1 B1
2
3
d sec 1
d sin cos sin cos
y t
x t t t t
or equivalent M1 A1 (4)
(b) At 3
t ,
3
4x , 2 3y B1
2secd 163d 3sin cos
3 3
y
x
M1 A1
16 3
2 33 4
y x
M1
3
08
y x M1 A1 (6)
Silver 3: 7/12 42
[10]
Silver 3: 7/12 43
QuestionNumber
Scheme Marks
5. 1 d
d
y
y x … B1
… 1
2ln 2x xx
M1 A1
At 2x , ln 2(2) ln 2y M1
leading to 16y Accept 4ln 2ey A1
At 2,16 1 d
2ln 2 216 d
y
x M1
d16 2 2ln 2
d
y
x A1 (7)
[7]
Silver 3: 7/12 44
QuestionNumber
Scheme Marks
6. (a) i: 6 5 2 j: 3 2 15 3 Any two equations M1 leading to 3 , 4 M1 A1
6 1 3
3 3 2 3
2 3 7
r or
5 2 3
15 4 3 3
3 1 7
r M1 A1
k: LHS 2 3 3 7 , RHS 3 4 1 7 B1 (6)
(As LHS = RHS, lines intersect)
(b)
1 2
2 . 3 2 6 3 14 14cos
3 1
110.92 M1 A1
Acute angle is 69.1 awrt 69.1 A1 (3)
(c)
6 1 5
3 1 2 1
2 3 1
r 1 lies on B l B1 (1)
(d) Let d be shortest distance from B to 2l
5 3 2
1 3 4
1 7 6
AB
uuurM1
2 222 4 6AB uuur
= 56 awrt 7.5 A1
sin56
d
M1
56sin 69.1 6.99d awrt 6.99 A1 (4) [14]
Silver 3: 7/12 45
QuestionNumber
Scheme Marks
Q7 (a) 20 9 3 3 0y t t t t t
0, 3, 3t Any one correct value B1
At 0t , 25 0 4 4x Method for finding one value of x M1
At 3t , 25 3 4 41x
2At 3, 5 3 4 41t x
At A, 4x ; at B, 41x Both A1 (3)
(b) d
10d
xt
t Seen or implied B1
2dd d 9 10 d
d
xy x y t t t t t
t
M1 A1
2 490 10 dt t t
3 590 10
3 5
t tC 3 530 2t t C A1
33 5
3 5
0
90 1030 3 2 3 324
3 5
t t
M1
22 d 648 unitsA y x A1 (6)
[9]
Silver 3: 7/12 46
Question
NumberScheme Marks
Q8 (a) d
2sind
xu
u B1
2 2 2 2
1 1d 2sin d
4 2cos 4 2cosx u u
x x u u
M1
2 2
2sind
4cos 4sin
uu
u u
Use of 2 21 cos sinu u M1
2
1 1d
4 cosu
u
2
1d
cosk u
u
M1
1tan
4u C tank u M1
2 2 2cos4
x u u
1 1 2cos3
x u u M1
4
3
1 1tan tan tan
4 4 4 3u
1 3 11 3
4 4
A1 (7)
(b) 14
22
2
1
4d
4V x
x x
M1
2
2 21
116 d
4x
x x
16 integral in (a) M1
3 1
164
16 their answer to part (a)A1ft (3)
[10]
Silver 3: 7/12 47
Question 1
This question was generally well answered with about 74% of candidates gaining all 4 marks.
The majority of candidates were able to split up 2)1)(12(
35
xx
x in the correct form of
)12( x
A +
)1( x
B2)1( x
C
although some missed the (x + 1) factor to give the incorrect form of )12( x
A + 2)1( x
B. Many candidates were successful in
either substituting values and/or equating coefficients in order to find their constants. A minority who unnecessarily formed threesimultaneous equations by equating coefficients were less successful in finding all three constants correctly.
Question 2
Only about 45% of the candidates were able to gain all 7 marks in this question as it involved a challenging integration by parts, on account
of the term 3
1.
x This meant that candidates had to be especially careful when dealing with negative powers of x.
In Q2(a), the majority of candidates applied the integration by parts formula correctly in the right direction to gain 3 out of the 5 marks
available. Many of them then proceeded to integrate an expression of the form 3x
to give an expression of the form
2x
although a
minority gave an expression of the form 4x
. A significant number of candidates failed to gain the final accuracy mark due to sign errors
or errors with the constants and in 2 2
ln x cx x
. A minority of candidates applied the by parts formula in the ‘wrong
direction’ and incorrectly stated that d
lnd
vx
x implied
1v
x .
In Q2(b), most candidates gained the method mark for substitution of 2x and 1x into their answer in Q2(a) and subtracting the
correct way round. The final mark was largely dependent upon their having obtained the correct answer in Q2(a).
Question 3
Part (a) was tackled well by many candidates. The majority of candidates were able to write down the correct identity. The most popular
strategy at this stage (and the best!) was for candidates to substitute 1x and 23x into their identity to find the values of the
constants B and C. The substitution of 23x caused problems for a few candidates which led them to find an incorrect value for B.
Many candidates demonstrated that constant A was zero by use of a further value of x or by comparing coefficients in their identity. Asignificant minority of candidates manipulated their original identity and then compared coefficients to produce three equations in order tosolve them simultaneously.
In part (b), most candidates were able to rewrite their partial fractions with negative powers and apply the two binomial expansions correctly,
usually leading to the correct answer. A significant minority of candidates found the process of manipulating 24(3 2)x to
2321 x
challenging.
A significant number of candidates were unsure of what to do in part (c). Some candidates found the actual value only. Other candidatesfound the estimated value only. Of those who progressed further, the most common error was to find the difference between these values andthen divide by their estimate rather than the actual value. Some candidates did not follow the instruction to give their final answer correct to2 significant figures and thus lost the final accuracy mark.
Question 4
The majority of candidates knew how to tackle this question and solutions gaining all the method marks were common. However there weremany errors of detail and only about 32% of the candidates gained full marks. In part (a), many candidates had difficult in differentiating
2sin t and 2 tan t . 2 tan t was more often differentiated correctly, possibly because the differential of tan t is given in the
formula book, although 2ln sec t or 2ln sec t were often seen. Many could not differentiate 2sin t correctly. 2cos t , 2cos t
and 2sin t were all common. Nearly all candidates knew they had to divide t
y
d
d by
t
x
d
d, although there was some confusion in
notation, with candidates mixing up their xs and ts. The majority knew how to approach part (b), finding the linear equation of the tangent to
Silver 3: 7/12 48
the curve at 3
, 2 34
, putting 0y and solving for x. Some candidates used 0y prematurely and found the tangent to the
curve at 3
, 04
rather than at 3
, 2 34
.
Question 5
The majority of those who used implicit differentiation were successful on this question. The commonest error was to differentiate
2 lnx x incorrectly and, occasionally, integration by parts was seen. Those who differentiated correctly were usually able to find that
16y when 2x and complete the question. The commonest error at this stage was ln 4ln 2y leading to 8.y
Those who started by making y the subject of the formula rarely made progress beyond the first step of writing 2 lne x xy . This can be
differentiated using the chain rule but the majority made some attempt to transform this expression before differentiating and this was often
done incorrectly, 2 ln 2 lne e ex x x x being a common error. Those who transformed, correctly, to 2xx often differentiated this to
2 12 xx x .
Question 6
In part (a) the majority of candidates were able to set up equations in and and, with a few exceptions were able to solve them
correctly. Substitution into one of the given line equations to obtain the coordinates of A usually followed correctly, although a substantialnumber of candidates were unable, or forgot, to show that the lines did indeed intersect. In part (b), the great majority of candidates realisedthat a scalar product was involved although substantial numbers of candidates selected the position vectors of the fixed points on the linerather than the direction vectors. Among those who did select the correct vectors, the commonest error was to give an obtuse angle rather
than the acute angle which the question asked for. Part (c) was well done although some did not show the consistency of for all three
components.
Part (d) proved very demanding. Those who were able to draw a simple diagram to represent the situation and who remembered that eachpart of a question frequently relies on previous parts were able to find the length of AB, using the results of parts (a) and (c), and the anglefound in part (b) to complete the question using elementary trigonometry, although some used the tangent of the angle rather than the sine.These were however a small minority of candidates and the majority either just left the question blank or thought that AB was the length they
were looking for. Some tried more complicated methods such as finding a general expression for BX and taking the scalar product of this
vector with the direction of 2l and equating to zero. The working for this method gets very complicated 337 but a few correct
solutions of this type were seen.
Question 7
Part (a) was well done. The majority of candidates correctly found the x-coordinates of A and B, by putting 0y , solving for t and then
substituting in 25 4x t . Full marks were common. Part (b) proved difficult. A substantial minority of candidates failed to substitute
for the dx when substituting into dy x or used d
d
t
x rather than
d
d
x
t. A surprising feature of the solutions seen was the number of
candidates who, having obtained the correct 29 10 dt t t t , were unable to remove the brackets correctly to obtain
2 490 10 dt t t . Weaknesses in elementary algebra flawed many otherwise correct solutions. Another source of error was using the
x-coordinates for the limits when the variable in the integral was t. At the end of the question, many failed to realise that
3 2 4
090 10 dt t t gives only half of the required area.
Some candidates made either the whole of the question, or just part (b), more difficult by eliminating parameters and using the cartesianequation. This is a possible method but the indices involved are very complicated and there were very few successful solutions using thismethod.
Question 8
Answers to part (a) were mixed, although most candidates gained some method marks. A surprisingly large number of candidates failed to
deal with 24 4cos u correctly and many did not recognise that 22
1d sec d tan
cosx x x x C
x
in this
context. Nearly all converted the limits correctly. Answers to part (b) were also mixed. Some could not get beyond stating the formula for thevolume of revolution while others gained the first mark, by substituting the equation given in part (b) into this formula, but could not see theconnection with part (a). Candidates could recover here and gain full follow through marks in part (b) after an incorrect attempt at part (a).
Silver 3: 7/12 49
Statistics for C4 Practice Paper Silver Level S3
Mean score for students achieving grade:
QuMax score
Modal score
Mean %
ALL A* A B C D E U
1 4 82 3.29 3.87 3.51 3.24 2.89 2.23 1.97 0.81
2 7 7 71 4.98 6.86 5.97 4.94 3.90 2.81 2.09 0.94
3 14 72 10.04 12.02 9.57 8.16 6.42 5.42 2.69
4 10 65 6.50 9.62 8.38 6.73 4.96 3.47 2.03 0.93
5 7 64 4.47 6.79 5.76 4.56 3.36 2.24 1.42 0.70
6 14 62 8.70 12.92 10.84 8.84 6.78 4.88 3.35 1.75
7 9 60 5.36 6.92 4.68 3.52 2.62 1.58 1.25
8 10 41 4.07 5.71 2.63 1.67 0.85 0.41 0.20
75 63 47.41 59.11 45.19 35.24 25.52 18.27 9.27
Silver 3: 7/12 50
Silver 3: 7/12 51
Paper Reference(s)
6666/01
Edexcel GCECore Mathematics C4Silver Level S4
Time: 1 hour 30 minutes
Materials required for examination Items included with question papersMathematical Formulae (Green) Nil
Candidates may use any calculator allowed by theregulations of the Joint
Council for Qualifications. Calculators must not have thefacility for symbolic
algebra manipulation, differentiation and integration, orhave 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 C4), the paper reference (6666), 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 8 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. Answerswithout working may gain no credit.
Suggested grade boundaries for this paper:
A* A B C D E
65 58 50 44 37 31
Silver 4: 8/12 52
1. (a) Find the binomial expansion of
(1 – 8x), x < 8
1,
in ascending powers of x up to and including the term in x3, simplifying each term.(4)
(b) Show that, when x = 100
1, the exact value of (1 – 8x) is
5
23.
(2)
(c) Substitute x = 100
1 into the binomial expansion in part (a) and hence obtain an
approximation to 23. Give your answer to 5 decimal places.
(3)
January 2010
2. f (x) = )49(
12x
, x < 2
3.
Find the first three non-zero terms of the binomial expansion of f(x) in ascending powers of x.Give each coefficient as a simplified fraction.
(6)
June 2011
Silver 4: 8/12 53
3.
Figure 1
A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl.
When the depth of the water is h m, the volume V m3 is given by
V = 12
1 h2(3 – 4h), 0 h 0.25.
(a) Find, in terms of , h
V
d
d when h = 0.1.
(4)
Water flows into the bowl at a rate of 800
m3 s–1.
(b) Find the rate of change of h, in m s–1, when h = 0.1.(2)
June 2011
Silver 4: 8/12 54
4.
Figure 1
Figure 1 shows the curve with equation
y =
43
22x
x, x 0.
The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and theline x = 2.
The region S is rotated 360° about the x-axis.
Use integration to find the exact value of the volume of the solid generated, giving youranswer in the form k ln a, where k and a are constants.
(5)
January 2012
Silver 4: 8/12 55
5.
Figure 2
Figure 2 shows a sketch of part of the curve C with parametric equations
x = 1 – 2
1t, y = 2t – 1.
The curve crosses the y-axis at the point A and crosses the x-axis at the point B.
(a) Show that A has coordinates (0, 3).(2)
(b) Find the x-coordinate of the point B.(2)
(c) Find an equation of the normal to C at the point A.(5)
The region R, as shown shaded in Figure 2, is bounded by the curve C, the line x = –1 and thex-axis.
(d) Use integration to find the exact area of R.(6)
January 2013
Silver 4: 8/12 56
6. A curve has parametric equations
x = tan2 t, y = sin t, 0 < t < 2
.
(a) Find an expression for x
y
d
d in terms of t. You need not simplify your
answer.(3)
(b) Find an equation of the tangent to the curve at the point where t = 4
.
Give your answer in the form y = ax + b , where a and b are constantsto be determined.
(5)
(c) Find a cartesian equation of the curve in the form y2 = f(x).(4)
June 2007
7. A curve is described by the equation
x2 + 4xy + y2 + 27 = 0
(a) Find d
d
y
x in terms of x and y.
(5)
A point Q lies on the curve.
The tangent to the curve at Q is parallel to the y-axis.
Given that the x-coordinate of Q is negative,
(b) use your answer to part (a) to find the coordinates of Q.(7)
June 2013
Silver 4: 8/12 57
8. (a) Using the identity cos 2θ = 1 – 2 sin2 θ , find
.sin 2 d
(2)
Figure 4
Figure 4 shows part of the curve C with parametric equations
x = tan θ, y = 2 sin 2θ, 0 θ < 2
.
The finite shaded region S shown in Figure 4 is bounded by C, the line x = 3
1
and the x-
axis. This shaded region is rotated through 2 radians about the x-axis to form a solid ofrevolution.
(b) Show that the volume of the solid of revolution formed is given by the integral
k
6
0
2sin
d ,
where k is a constant.(5)
(c) Hence find the exact value for this volume, giving your answer in the form p 2 + q√3,
Substitutes limits of 2 and 0 and subtracts the correct way
round.dM1
So Volume 1
ln 43
1
ln 43
or 2
ln 23
A1 oe isw
[5](5 marks)
Silver 4: 8/12 60
Question Number
Scheme Marks
5. Working parametrically:ln 21
1 , 2 1 or e 12
t tx t y y
(a) 10 0 1 2
2x t t Applies 0x to obtain a value for t. M1
When 2t , 22 1 3y Correct value for y.A1 [2]
(b) 0 0 2 1 0ty t Applies 0y to obtain a value for t.(Must be seen in part (b)).
M1
When 0t , 1
1 (0) 12
x 1x A1 [2]
(c)
d 1
d 2
x
t and either
d2 ln 2
dty
t or
ln 2de ln 2
dty
t
B1
d 2 ln 21d2
ty
x
Attempts their d
d
y
t divided by their
d.
d
x
t
M1
At A, "2",t so 1
( ) 8ln 2 ( )8ln 2
m m T NApplies "2"t and
1( )
( )m
m
NT
M1
13 ( 0)
8ln 2y x or
13
8ln 2y x or equivalent. M1 A1 oe
cso[5]
(d) 1Area( ) 2 1 . d
2tR t Complete substitution for both y and dx M1
1 4 and 1 0x t x t B1
1 2
2 ln 2
t
t
Either 2
2ln 2
tt
or (2 )2 1
(ln 2)
tt t
or 2 1 (ln 2)(2 )t t t
M1*
22 1
ln 2
tt t A1
0
4
1 2 1 1 164
2 ln 2 2 ln 2 ln 2
t
t
Depends on the previous methodmark.
Substitutes their changed limits in t andsubtracts either way round.
dM1*
152
2ln 2 15
22ln 2
or equivalent. A1
[6]15
Silver 4: 8/12 61
QuestionNumber
Scheme Marks
6. (a) 2tan ,x t siny t
2d2(tan )sec
d
xt t
t ,
dcos
d
yt
t Correct
d
d
x
t and
d
dt
y B1
2
d cos
d 2 tan sec
y t
x t t
4cos
2sin
t
t
dd
cos
their xt
t
dd
cos
their xt
t
M1
A1
[3]
(b) When 4 ,t 12
1 ,x y (need values) The point 12
1, or 1, awrt 0.71
These coordinates can be implied. B1, B1
( 4siny is not sufficient forB1)
When ,4
t m(T) = 4
24 4
cosd
d 2 tan sec
y
x
11 22
1 1 12 2 2
211
1 2
2.(1)(2) 84 22.(1)2.(1) any of the five underlined
expressions or awrt 0.18B1 aef
T: 1 12 4 2
1y x
Finding an equation of a tangentwith their point and their tangent
gradient or finds c by using(their gradient) " "y x c .
M1 aef
T: 31
4 2 4 2y x or 2 3 2
8 8y x Correct simplified EXACT equation of tangent
A1 aef cso
or 31 1 1 12 4 2 2 4 2 4 2
1 c c
Hence T: 31
4 2 4 2y x or 2 3 2
8 8y x
[5]
Silver 4: 8/12 62
Question Number
Scheme Marks
7. 2 24 27 0x xy y
(a)d d d
2 4 4 2 0d d d
y y yx y x y
x x x
M1 A1 B1
d
2 4 (4 2 ) 0d
yx y x y
x dM1
d 2 4 2
d 4 2 2
y x y x y
x x y x y
A1 cso oe
(5)(b) 4 2 0x y M1
2y x 1
2x y A1
2 24 ( 2 ) ( 2 ) 27 0x x x x 2
21 14 27 0
2 2y y y y
M1*
23 27 0x 2327 0
4y
2 9x 2 36y dM1*3x 6y A1
When 3x ,2( 3)y When 6y ,
1(6)
2x ddM1*
6y 3x A1 cso (7)
[12]
Silver 4: 8/12 63
8. (a) 2 1 1 1sin d 1 cos 2 d sin 2
2 2 4C M1 A1 (2)
2d
tan secd
xx
(b) 22 2 2dd d 2sin 2 sec d
d
xy x y
M1 A1
2
2
2 2sin cosd
cos
M1
216 sin d 16k A1
0 tan 0 0x , 1 1
tan3 3 6
x
B1 (5)
(c) 6 2
0
16 sin dV
6
0
1 sin 216
2 4V
M1
116 sin 0 0
12 4 3
Use of correct limits M1
23 4
16 2 312 8 3
4
3p , 2q A1 (3)
(10 marks)
Silver 4: 8/12 64
Question 1
This proved a suitable starting question and there were many completely correct solutions. The majority of candidates could complete part(a) successfully. In part (b), those who realised that working in common (vulgar) fractions was needed usually gained the method mark but,as noted in the introduction, the working needed to establish the printed result was frequently incomplete. It is insufficient to write down
8 231
100 5 . The examiners accepted, for example,
8 92 23 231
100 100 25 5 . In part (c), most
candidates realised that they had to evaluate their answer to part (a) with 0.01x . However many failed to recognise the implication of
part (b), that this evaluation needed to be multiplied by 5. It was not uncommon for candidates to confuse parts (b) and (c) with the
expansion and decimal calculation appearing in (b) and fraction work leading to 23 appearing in (c).
Question 2
Many candidates got off to a very bad start to this question by writing 29 4 3 2x x or 12 129 4 3 2x x
.
Such errors in algebra are heavily penalised as the resulting binomial expansions are significantly simplified and, in this case, gave answers
in incorrect powers of x. Those who obtained
1
221 4
13 9
x
showed that they understood the binomial theorem but there were many
errors in signs, often due to the failure to use brackets correctly. Some candidates seemed to lose the thread of the question and, having
expanded
1
224
19
x
correctly, failed to multiply by
1
3. It was not unusual to see an, often correct, term in 6x provided. The
examiners ignore this but such additional work does lose time.
Question 3
This question was well done and full marks were common. Candidates were roughly equally divided between those who expanded anddifferentiated and those who differentiated using the product rule. The latter method was the more complicated and more subject to error butmany correct solutions were seen using both methods. If the differentiation was correct, nearly all completed part (a) correctly. Rather oddly,
a number of cases were seen where 213 4
12h h was misread as 21
3 42
h h . Part (b) was generally well done although
there were a minority of students who made no attempt at it at all. The large majority correctly interpreted 800
as
d
d
V
t and realised they
had to divide 800
by
d
d
V
h. Inverting
d
d
V
h did cause difficulty for some candidates. For example,
2 2
1 1 1
0.5 0.5h h h h
was seen from time to time and
25
125
leading to the answer
2
32
, instead of the correct
1
32, was relatively common.
Question 4
At least 90% of the candidature was able to apply the volume of revolution formula correctly. Only a few candidates did not include intheir volume formula or did not square the expression for y. The integration was well attempted and the majority of candidates recognised that the integral could be manipulated into the form
f ( )d
f ( )
xx
x
and integrated to give their result in the form k ln (3x2 + 4) usually with 1
3k . A variety of incorrect values of k were
seen with the most common being either 3 or 1. A significant number of candidates integrated incorrectly to give answers such as x2 ln (3x2 +4) or 2x ln (3x2 + 4).
Those candidates who applied the substitution 23 4u x proceeded to achieve1
ln3
u , and changed their x-limits of 0 and 2 to give
correct u-limits of 4 and 16. Other substitutions of 23u x or 2u x , were also used, usually successfully.
Silver 4: 8/12 65
Unproductive attempts were seen by a minority of candidates, such as integration by parts or simplifying 2
2
3 4
x
x to give
2
2 2,
3 4
x x
x or integrating 2x and 3x2 + 4 separately and then multiplying or dividing the two results together.
The majority of candidates were able to apply the limits correctly and examiners observed the correct answer in a variety of different forms.
Question 5
This question, and in particular the final Q5(d), proved challenging for a large number of candidates, with about 18% of the candidaturescoring at least 12 of the 15 marks available and only about 7% scoring all 15 marks.
Q5(a) and Q5(b) were almost invariably completed correctly, the main source of error in Q5(b) being that a very small number of candidates
did not realise that 0t follows from 2 1.t
Many correct solutions to Q5(c) were seen. The principal reason for loss of marks came from candidates being unable to find the derivative
of 2 1.t Dividing d
d
y
t by
d
d
x
t
1i.e. dividing by
2
proved challenging for a number of candidates. Some candidates,
having correctly established d
d
y
xas being ( 2)2 ln 2t , then proceeded incorrectly to equate this to 4 ln 2t . Most knew how to
obtain the gradient of the normal, and could write down the equation of a straight line.
Q5(d) was answered well by small number of candidates, and, although a significant number could write the area as
0
4
12 1 . d
2t t , many were unable to perform the integration of 2t with respect to t. Some wrote 2t as 2t , thus
simplifying the problem, whilst attempts such as
12
1
t
t
were not uncommon. Candidates who were unable to make an attempt at the
integration of 2t were unable to access the final 4 marks in this part. Approaches that facilitated integration included re-writing 2t as
ln 2et or substituting 2tu , leading to d
2 ln 2 ln 2d
tuu
t , and thereby circumventing a direct integration of 2 .t
Other candidates used a cartesian approach, giving the area as 1
2 2
1
2 1 dx x
(or equivalent), but again a number were unable to
carry out the integration.
Silver 4: 8/12 66
Question 7
This question discriminated well between candidates of all abilities, with about 80% of candidates gaining at least 5 marks of the 12 marksavailable and about 45% gaining at least 8 marks. Only about 15% of candidates gained all 12 marks. Part (a) was answered well with fullmarks commonly awarded. Part (b) was far more challenging with only a small minority presenting a complete and correct solution.
In part (a), many candidates were able to differentiate correctly, factorise out d
d
y
x, and rearrange their equation to arrive at a correct
expression for the gradient function. A minority did not apply the product rule correctly when differentiating 4 ,xy whilst a small number
left the constant term of 27 in their differentiated equation.
In part (b), those small proportion of candidates who realised they needed to set the denominator of their d
d
y
x expression equal to zero
usually went on to answer this part correctly. Some, however, did not attempt this part, while a majority attempted to solve d
0d
y
x , and
many proceeded to obtain coordinates of ( 6, 3) for the point Q, despite a number of them initially sketching a curve with a vertical
tangent. A smaller proportion solved d
1d
y
x , presumably because the digit 1 is written as a vertical line; whilst others either substituted
0y or 0x into their d
d
y
x expression. Manipulation and bracketing errors sometimes led to candidates writing equations such as
2y A or 2x A , where A was negative. Examiners were surprised that a fair number of candidates, having obtained their value of
x (or y), then proceeded to substitute this into 2 24 27 0x xy y , rather than using the much simpler y kx or
.x ky
Question 8
The responses to this question were very variable and many lost marks through errors in manipulation or notation, possibly through mental
tiredness. For examples, many made errors in manipulation and could not proceed correctly from the printed 2cos 2 1 2sin to
Silver 4: 8/12 67
2 1 1sin cos 2
2 2 and the answer
1sin 2
2 4
x was often seen, instead of 1
sin 22 4
. In part (b), many never
found d
d
x
or realised that the appropriate form for the volume was
2 dd
d
xy
.
However the majority did find a correct integral in terms of although some were unable to use the identity sin 2 2sin cos to simplify their integral. The incorrect value 8k was very common, resulting from a failure to square the factor 2 in
sin 2 2sin cos . Candidates were expected to demonstrate the correct change of limits. Minimally a reference to the result
1tan
6 3
, or an equivalent, was required. Those who had complete solutions usually gained the two method marks in part (c) but
Materials required for examination Items included with question papersMathematical Formulae (Green) Nil
Candidates may use any calculator allowed by theregulations of the Joint
Council for Qualifications. Calculators must not have thefacility for symbolic
algebra manipulation, differentiation and integration, orhave 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 C4), the paper reference (6666), 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 7 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. Answerswithout working may gain no credit.
Suggested grade boundaries for this paper:
A* A B C D E
68 59 50 46 39 31
Silver 4: 8/12 70
1. The curve C has the equation 2x + 3y2 + 3x2 y = 4x2.
The point P on the curve has coordinates (–1, 1).
(a) Find the gradient of the curve at P.(5)
(b) Hence find the equation of the normal to C at P, giving your answer in the formax + by + c = 0, where a, b and c are integers.
(3)
January 2012
Silver 4: 8/12 71
2.
Figure 1
Figure 1 shows part of the curve with equation y = √(0.75 + cos2 x). The finite region R,shown shaded in Figure 1, is bounded by the curve, the y-axis, the x-axis and the line with
equation x = 3
.
(a) Copy and complete the table with values of y corresponding to x = 6
and x =
4
.
x 012
6
4
3
y 1.3229 1.2973 1
(2)
(b) Use the trapezium rule
(i) with the values of y at x = 0, x = 6
and x =
3
to find an estimate of the area of R.
Give your answer to 3 decimal places.
(ii) with the values of y at x = 0, x =12
, x =
6
, x =
4
and x =
3
to find a further
estimate of the area of R. Give your answer to 3 decimal places.
(6)
June 2010
Silver 4: 8/12 72
3.
Figure 2
Figure 2 shows a sketch of the curve C with parametric equations
x = 3 sin 2t, y = 4 cos2 t, 0 t .
(a) Show that x
y
d
d = k3 tan 2t, where k is a constant to be determined.
(5)
(b) Find an equation of the tangent to C at the point where t = 3
.
Give your answer in the form y = ax + b, where a and b are constants.(4)
(c) Find a cartesian equation of C.(3)
June 2012
Silver 4: 8/12 73
4. The line l1 has vector equation
r =
1
4
6
+ λ
3
1
4
and the line l2 has vector equation
r =
1
4
6
+
1
4
3
where λ and μ are parameters.
The lines l1 and l2 intersect at the point A and the acute angle between l1 and l2 is θ.
(a) Write down the coordinates of A.(1)
(b) Find the value of cos θ.(3)
The point X lies on l1 where λ = 4.
(c) Find the coordinates of X.(1)
(d) Find the vector AX .(2)
(e) Hence, or otherwise, show that AX = 4√26.(2)
The point Y lies on l2. Given that the vector YX is perpendicular to l1,
(f) find the length of AY, giving your answer to 3 significant figures.(3)
January 2010
Silver 4: 8/12 74
5. With respect to a fixed origin O, the lines l1 and l2 are given by the equations
l1 : r = (–9i + 10k) + λ(2i + j – k)
l2 : r = (3i + j + 17k) + μ(3i – j + 5k)
where λ and μ are scalar parameters.
(a) Show that l1 and l2 meet and find the position vector of their point of intersection.(6)
(b) Show that l1 and l2 are perpendicular to each other.(2)
The point A has position vector 5i + 7j + 3k.
(c) Show that A lies on l1.(1)
The point B is the image of A after reflection in the line l2.
(d) Find the position vector of B.(3)
June 2008
6. (a) Find
xxx d2cos .
(4)
(b) Hence, using the identity cos 2x = 2 cos2 x – 1, deduce
xxx dcos2 .
(3)
June 2007
Silver 4: 8/12 75
7.
Figure 3
Figure 3 shows the curve C with parametric equations
x = 8 cos t, y = 4 sin 2t, 0 t 2
.
The point P lies on C and has coordinates (4, 2√3).
(a) Find the value of t at the point P.(2)
The line l is a normal to C at P.
(b) Show that an equation for l is y = –x√3 + 6√3.(6)
The finite region R is enclosed by the curve C, the x-axis and the line x = 4, as shown shadedin Figure 3.
(c) Show that the area of R is given by the integral
2
3
2 dcossin64
ttt .
(4)
(d) Use this integral to find the area of R, giving your answer in the form a + b√3, where aand b are constants to be determined.
(4)
June 2008
TOTAL FOR PAPER: 75 MARKS
END
Silver 4: 8/12 76
QuestionNumber
Scheme Marks
1. (a)2d d d
2 6 6 3 8d d d
y y yy x y x x
x x x
M1 A1 B1
2
d 8 2 6
d 6 3
y x xy
x y x
not necessarily required.
At 1, 1 ,P 2
d 8( 1) 2 6( 1)(1) 4m
d 6(1) 3( 1) 9
y
xT
dM1 A1 cso
[5]
(b) So, m(N) = 49
1 9
4
M1
N: 91 1
4y x M1
N: 9 4 13 0x y A1 [3]
(8 marks)
QuestionNumber
Scheme Marks
2. (a) 1.2247, 1.11806 4
y y
accept awrt 4 d.p. B1 B1 (2)
(b)(i) 1.3229 2 1.2247 112
I
B1 for 12
B1 M1
1.249 cao A1
(ii) 1.3229 2 1.2973 1.2247 1.1180 124
I
B1 for24
B1 M1
1.257 cao A1 (6) [8]
Silver 4: 8/12 77
QuestionNumber
Scheme Marks
3. (a) d
2 3 cos 2d
xt
t B1
d
8cos sind
yt t
t M1 A1
d 8cos sin
d 2 3 cos 2
y t t
x t
M1
4sin 2
2 3 cos 2
t
t
d 2
3 tan 2d 3
yt
x
2
3k
A1 (5)
(b) When 3
t
3, 1
2x y can be implied B1
2 23 tan 2
3 3m
M1
3
1 22
y x M1
2 2y x A1 (4)
(c) 3 sin 2 3 2sin cosx t t t M1
2 2 2 2 212sin cos 12 1 cos cosx t t t t
2 12 14 4
y yx
or equivalent M1 A1 (3)
[12]
Silver 4: 8/12 78
A X
Y
1l
2ld
4 26
Question
NumberScheme Marks
Q4 (a) A: 6, 4, 1 Accept vector formsB1 (1)
(b) 2 22 2 2 2
4 3
1 . 4 12 4 3 4 1 3 3 4 1 cos
3 1
M1 A1
19
cos26
awrt 0.73A1 (3)
(c) X: 10, 0,11 Accept vector formsB1 (1)
(d)
10 6
0 4
11 1
AX
Either order M1
16
4
12
caoA1 (2)
(e) 22 216 4 12AX M1
416 16 26 4 26 Do not penalise if consistentA1 (2)
incorrect signs in (d)
(f)
Use of correct right angled triangle M1
cosAX
d M1
1926
4 2627.9d awrt 27.9
A1 (3)
[12]
Silver 4: 8/12 79
5. (a) Lines meet where:
9 2 3 3
0 1 1 1
10 1 17 5
Any two of
: 9 2 3 3 (1)
: 1 (2)
: 10 17 5 (3)
i
j
kM1
(1) – 2(2) gives: 9 1 5 2 M1 (2) gives: 1 2 3 A1
9 2 3 3
0 3 1 or 1 2 1
10 1 17 5
r r M1
Intersect at
3
3 or 3 3 7
7
r r i j k A1
Either check k:
3 : LHS 10 10 3 7
2 : RHS 17 5 17 10 7
B1 (6)
(b) 1 2 d i j k , 2 3 5 d i j k
As 1 2
2 3
1 1 (2 3) (1 1) ( 1 5) 0
1 5
d d
Then l1 is perpendicular to l2.
M1 A1 (2)
(c) Equating i ; 9 2 5 7 9 2 5
0 7 1 7
10 1 3
r ( .OAuuur
Hence the point A lies on l1.) B1 (1)
(d) Let 3 3 7OX i j kuuur
be point of intersection 3 5 8
3 7 4
7 3 4
AX OX OA
uuur uuur uuurM1 ft
2OB OA AB OA AX uuur uuur uuur uuur uuur
5 8
7 2 4
3 4
OB
uuurM1 ft
Hence,
11
1
11
OB
uuuror 11 11OB i j k
uuurA1
(12 marks)
Silver 4: 8/12 80
QuestionNumber
Scheme Marks
6. (a)dd
d 1d 2
1
cos2 sin2
ux
vx
u x
x v x
(see note below)
1 12 2Int cos2 d sin2 sin2 .1 dx x x x x x x
Use of ‘integration byparts’ formula in the
correct direction.Correct expression.
M1
A1
1 1 12 2 2sin2 cos2x x x c
12sin2 cos2x x
or 1sin coskkx kx with 1 , 0k k
dM1
1 12 4sin2 cos2x x x c Correct expression with
+c A1
[4]
(b) cos2 122cos d dxx x x x x
Substitutes correctly for 2cos x in the
given integralM1
1 1
cos2 d d2 2
x x x x x
1 1 1 1
sin2 cos2 ; d2 2 4 2
x x x x x
1their answer to (a) ;
2or underlined
expression
A1;
21 1 1sin2 cos2 ( )
4 8 4x x x x c Completely correct
expression with/without+c
A1
[3]
7 marks
Silver 4: 8/12 81
7. (a) At (4,2 3)P either 4 8cos or 2 3 4sin 2t t M1
only solution is 3t where 20 t „ „ A1
(b) 8cosx t , 4sin 2y td
8sind
xt
t ,
d8cos2
d
yt
t M1 A1
At P,
23
3
8cosd
d 8sin
y
x
M1
12
32
8 1awrt 0.58
38
Hence m(N) = 3 or 13
1M1
N: 2 3 3 4y x M1
N: 3 6 3y x (*) A1 cso (6)
(c) 3
2
4
0
d 4sin 2 . 8sin dA y x t t t
M1 A1
3 3
2 2
32sin 2 .sin d 32 2sin cos .sin dA t t t t t t t
M1
3
2
264.sin cos dA t t t
2
3
264.sin cos dA t t t
(*) A1 (4)
(d)2
3
3sin64
3
tA
or
32
13
643
uA
M1 A1
1 1 3 3 364 . . .
3 3 2 2 2A
M1
1 1 6464 3 8 3
3 8 3A
A1 (4)
(16 marks)
Silver 4: 8/12 82
Question 1
Most candidates attempted this question and many achieved full marks.
In part (a), most candidates were able to differentiate implicitly to gain the first three marks. A minority of candidates struggled to apply the
product rule correctly on 23x y . At this point a minority of candidates substituted 1x and 1y in their differentiated equation,
but the majority of candidates proceeded to find an expression for d
d
y
x in terms of x and ,y before substituting in these values. Although
the majority of candidates were able to find the correct answer of 4
,9
common errors in this part included sign errors either in
rearranging or when substituting 1x and 1y into their d
d
y
x expression. A small number of candidates tried to rearrange the
equation given in the question in order to make y the subject. This resulted in very few, if any, marks being awarded. In part (b), a small minority of candidates either found the equation of the tangent and gained no marks or did not give their equation of thenormal in the form ax + by + c = 0, where a, b and c are integers, and lost the final accuracy mark.
Question 2
This question was a good starting question and over 60% of the candidates gained full marks. A few candidates used a wrong angle modewhen calculating the values in part (a). In part (b), the majority knew the structure of the trapezium rule. The most common errors were to
miscalculate the interval width using, for example, 9
and
15
in place of
12
and
24
. Some were unable to adapt to the situation in
which they did not need all the information given in the question to solve part of it and either used the same interval width for (b)(i) and (b)(ii) or answered b(ii) only. A few answered b(ii) only and proceeded to attempt to find an exact answer using analytic calculus, which in thiscase is impossible. These candidates were apparently answering the question that they expected to be set rather than the one which hadactually been set. In Mathematics, as in all other subjects, carefully reading and answering the question as set are necessary examinationskills.
Question 3
Nearly all candidates gained some marks in part (a) realising that they had to divide d
d
y
t by
d
d
x
t. Most could differentiate 3 sin 2t
correctly, although occasionally dividing by 2, instead of multiplying, was seen. Differentiating 24cos t proved more difficult. Many had
to use a double angle formula and this lead to many errors in signs and constants. 2 2d4cos sin
dt k t
t , where k might be
4, 8 or 2 was also frequently seen. Many who correctly obtained d 2
tan 2d 3
yt
x were unable to transform this correctly
to the form specified in the question and, in this context, surd manipulation was a weak area. Nearly all knew how to solve part (b), a C1topic, and, if they had a correct answer to part (a), gained full marks here.
Part (c) proved demanding and only about 15% of the candidates were able to complete the question correctly. Many realised that they had to
use a double angle formula and gained the first mark, either by writing 2 3 sin cosx t t or cos 2 2y t . Although other
approaches are possible, the most commonly successful method was to use 2 2sin cos 1 , where is either t or 2t, to eliminate
the parameter. There are many alternative forms of the answer to this question.
Some otherwise correct answers, for example, 3 12
yx y
, lost the final mark as the answer only gave half of the curve.
Question 4
The majority of candidates made good attempts at parts (a) to (e) of this question. Many, however, wasted a good deal of time in part (a),
proving correctly that 0 before obtaining the correct answer. When a question starts “Write down ….”, then candidates should
realise that no working is needed to obtain the answer. The majority of candidates knew how to use the scalar product to find the cosine ofthe angle and chose the correct directions for the lines. Parts (c) and (d) were well done. In part (e) , as in Q1(b), the working needed to
establish the printed result was often incomplete. In showing that the printed result is correct, it is insufficient to proceed from 416 to
4 23 without stating 2416 16 26 or 4 26 . Drawing a sketch, which many candidates seem reluctant to do, shows that
part (f) can be solved by simple trigonometry, using the results of parts (b) and (e). Many made no attempt at this part and the majority of
Silver 4: 8/12 83
those who did opted for a method using a zero scalar product. Even correctly carried out, this is very complicated 10419 and it was
impressive to see some fully correct solutions. Much valuable time, however, had been wasted.
Question 5
In part (a), most candidates were able to set up and solve the three equations correctly. Some candidates either did not realise that theyneeded to perform a check for consistency or performed this check incorrectly. A surprising number of candidates did not follow theinstruction in the question to find the position vector of the point of intersection. A few candidates were unable to successfully negotiate the
absence of the j term in ( 9 10 ) i k for l1 and so formed incorrect simultaneous equations.
In part (b), a majority of candidates realised that they needed to apply the dot product formula on the direction vectors of l1 and l2. Many ofthese candidates performed a correct dot product calculation but not all of them wrote a conclusion.
In part (c), a majority of candidates were able to prove that A lies on l1, either by substituting 7 into l1 or by checking that substituting
(5, 7, 3) into l1 gave 7 for all three components.
There was a failure by many candidates to see the link between part (d) and the other three parts of this question with the majority of them
leaving this part blank. The most common error of those who attempted this part was to write down B as 5 7 3 . i j k Those
candidates who decided to draw a diagram usually increased their chance of success. Most candidates who were successful at this partapplied a vector approach as detailed in the mark scheme. The easiest vector approach, adopted by a few candidates, is to realise that
7 at A, 3 at the point of intersection and so 1 at B. So substitution of 1 into l1 yields the correct position
vector 11 11 . i j k A few candidates, by deducing that the intersection point is the midpoint of A and B were able to write down
53,
2
x 7
32
y and 3
7,2
z in order to find the position vector of B.
Question 6
Question 8
In part (a), many candidates were able to give 3 .t Some candidates gave their answer only in degrees instead of radians. Other
candidates substituted the y-value of P into 4sin 2y t and found two values for t, namely 6 3, .t The majority of these
candidates did not go on to reject 6 .t In part (b), many candidates were able to apply the correct formula for finding ddyx
in terms of t,
although some candidates erroneously believed that differentiation of 4sin 2t gave either 8cos 2 , 8cost t or 2cos 2 .t
Some candidates who had differentiated incorrectly, substituted their value of t into ddyx
and tried to “fudge” their answer for ddyx
as 13
,
Silver 4: 8/12 84
after realising from the given answer that the gradient of the normal was 3. The majority of candidates understood the relationship
between the gradient of the tangent and its normal and many were able to produce a fully correct solution to this part.
A few candidates, however, did not realise that parametric differentiation was required in part (b) and some of these candidates tried toconvert the parametric equations into a Cartesian equation. Although some candidates then went on to attempt to differentiate their Cartesianequation, this method was rarely executed correctly.
Few convincing proofs were seen in part (c) with a significant number of candidates who were not aware with the procedure of reversing the
limits and the sign of the integral. Therefore, some candidates conveniently differentiated 8cos t to give “positive” 8sin ,t even
though they had previously differentiated 8cos t correctly in part (a). After completing this part, some candidates had a ‘crisis of
confidence’ with their differentiation rules and then went on to amend their correct solution to part (a) to produce an incorrect solution.
Other candidates differentiated 8cos t correctly but wrote their limits the wrong way round giving
2
3
4sin 2 . 8sin dA t t t
and after stating sin 2 2cos sint t t (as many candidates were able to do in this part) wrote
2
3
264sin cos d .A t t t
These candidates then wrote down the given answer by arguing that all areas should be positive.
Part (d) was unstructured in the sense that the question did not tell the candidates how to integrate the given expression. Only a minority of
candidates spotted that a substitution was required, although some candidates were able to integrate 264sin cost t by inspection. Many
candidates replaced 2sin t with 12 1 cos 2t but then multiplied this out with cos t to give 1 1
2 2cos cos cos 2t t t . Very
few candidates correctly applied the sum-product formula on this expression, but most candidates usually gave up at this point or went on to
produce some incorrect integration. Other candidates replaced 2sin t with 21 cos t , but did not make much progress with this. A
significant number of candidates used integration by parts with a surprising number of them persevering with this technique more than oncebefore deciding they could make no progress. It is possible, however, to use a ‘loop’ method but this was very rarely seen. It was clear toexaminers that a significant number of stronger candidates spent much time trying to unsuccessfully answer this part with a few candidatesproducing at least two pages of working.
