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Paper Reference(s)

6666/01

Edexcel GCE Core Mathematics C4

Gold Level (Harder) G2

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 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. Answers

without working may gain no credit.

Suggested grade boundaries for this paper:

A* A B C D E

65 58 47 42 36 28

Gold 3: 11/12 2

1. f(x) = 2)13(

1

xx =

x

A +

)13( x

B +

2)13( x

C.

(a) Find the values of the constants A, B and C.

(4)

(b) (i) Hence find

xx d)(f .

(ii) Find

2

1

d)(f xx , leaving your answer in the form a + ln b, where a and b are constants.

(6)

June 2012

2. The current, I amps, in an electric circuit at time t seconds is given by

I = 16 – 16(0.5)t, t 0.

Use differentiation to find the value of t

I

d

d when t = 3 .

Give your answer in the form ln a, where a is a constant.

(5)

January 2011

3. (a) Use the binomial expansion to show that

21 11

1 2

xx x

x

, |x| < 1

(6)

(b) Substitute 1

26x into

21 11

1 2

xx x

x

to obtain an approximation to √3.

Give your answer in the form a

b where a and b are integers.

(3)

June 2013


4. Given that y = 2 at x = 4

, solve the differential equation

x

y

d

d =

xy 2cos

3.

(5)

June 2012

5.

Figure 2

A container is made in the shape of a hollow inverted right circular cone. The height of the

container is 24 cm and the radius is 16 cm, as shown in Figure 2. Water is flowing into the

container. When the height of water is h cm, the surface of the water has radius r cm and the

volume of water is V cm3.

(a) Show that V = 27

4 3h.

(2)

[The volume V of a right circular cone with vertical height h and base radius r is given by the

formula V = 3

1 r

2h .]

Water flows into the container at a rate of 8 cm3 s

–1.

(b) Find, in terms of π, the rate of change of h when h = 12.

(5)

January 2009

Gold 3: 11/12 4

6. (a) Find

xx dtan 2 .

(2)

(b) Use integration by parts to find

xxx

dln1

3.

(4)

(c) Use the substitution u = 1 + ex to show that

x

x

e1

e3

dx = 2

1e

2x – e

x + ln (1 + e

x) + k,

where k is a constant.

(7)

January 2009


7. (a) Express 24

2

y in partial fractions.

(3)

(b) Hence obtain the solution of

2 cot x x

y

d

d = (4 – y

2)

for which y = 0 at x = 3

, giving your answer in the form sec

2 x = g( y).

(8)

June 2008

Gold 3: 11/12 6

8.

Figure 3

Figure 3 shows part of the curve C with parametric equations

x = tan , y = sin , 0 < 2

.

The point P lies on C and has coordinates

3

2

1,3 .

(a) Find the value of at the point P.

(2)

The line l is a normal to C at P. The normal cuts the x-axis at the point Q.

(b) Show that Q has coordinates (k3, 0), giving the value of the constant k.

(6)

The finite shaded region S shown in Figure 3 is bounded by the curve C, the line x = 3 and the

x-axis. This shaded region is rotated through 2 radians about the x-axis to form a solid of

revolution.

(c) Find the volume of the solid of revolution, giving your answer in the form p 3 + q 2,

where p and q are constants.

(7)

June 2011

TOTAL FOR PAPER: 75 MARKS

END


Question Number

Scheme Marks

1. (a) 2

1 3 1 3 1A x Bx x Cx B1

0x 1 A M1

1

3x 1

31 3C C any two constants

correct A1

Coefficients of 2x

0 9 3 3A B B all three constants

correct A1 (4)

(b) (i)

2

1 3 3d

3 1 3 1x

x x x

13 3

ln ln 3 1 3 13 1 3

x x x C

M1 A1ft A1ft

1

ln ln 3 13 1

x x Cx

(ii) 2

2

11

1f d ln ln 3 1

3 1x x x x

x

1 1

ln 2 ln 5 ln1 ln 25 2

M1

2 2

ln ...5

M1

3 4

ln10 5

A1 (6)

[10]

2. d

16ln 0.5 0.5d

tI

t M1 A1

At 3t 3d16ln 0.5 0.5

d

I

t M1

2ln 0.5 ln 4 M1 A1

[5]

Gold 3: 11/12 8

Question

Number Scheme Marks

3. (a)

1 1

2 21

(1 ) (1 )1

xx x

x

1 1

2 2(1 ) (1 )x x

B1

31 1 1

2 22 2 2 2( )( ) ( )( )1 1

1 ... 1 ( ) ( ) ...2 2! 2 2!

x x x x

M1 A1

A1

2 21 1 1 3

1 ... 1 ...2 8 2 8

x x x x

2 2 21 3 1 1 11 ...

2 8 2 4 8x x x x x M1

211

2x x

Answer is given

in the question. A1 *

(6)

(b)

2126

126

1 1 1 11

1 26 2 26

M1

ie:

3 3 1405

5 1352

B1

so,

70253

4056

7025

4056 A1 cao

(3) [9]

4. 2

3d d

cosy y x

x Can be implied. Ignore integral signs B1

= 23sec dx x

213tan

2y x C M1 A1

2,4

y x

212 3tan

2 4C

M1

Leading to

1C

213tan 1

2y x or equivalent A1 (5)

[5]


Question Number

Scheme Marks

5 (a) Similar triangles 16 2

24 3

r hr

h

Uses similar triangles, ratios or

trigonometry to find either one of

these two expressions oe.

M1

2 3

21 1 2 4

3 3 3 27

h hV r h h

AG

Substitutes 23hr into the formula for

the volume of water V. A1

(2)

(b) From the question, d

8d

V

t

d8

d

V

t B1

2 2d 12 4

d 27 9

V h h

h

2 2d 12 4or

d 27 9

V h h

h

B1

2 2

d d d 9 188

d d 4

h V V

t t dh h h

Candidate’s d d

d d

V V

t h ; M1;

2128

27

h

or 2 2

9 188 or

4 h h

oe

A1

When d 18 1

12,d 144 8

hh

t

18

144 or

1

8 A1 oe

isw

Note the answer must be a one term exact

value.

Note, also you can ignore subsequent

working after 18

144 .

(5)

[7]

Gold 3: 11/12 10

Question Number

Scheme Marks

6 (a) 2tan dx x

2 2 2 2: sec 1 tan gives tan sec 1NB A A A A The correct underlined identity. M1 oe

2sec 1 dx x

tan x x c Correct integration

with/without + c A1

(2)

(b) 3

1ln dx x

x

2

2

d 1d

13dd 2 2

ln ux x

v xx x

u x

x v

2 2

1 1 1ln . d

2 2x x

x x x

Use of ‘integration by

parts’ formula in the correct direction.

Correct direction means that ln .u x

M1

Correct expression. A1

2 3

1 1 1ln d

2 2x x

x x

An attempt to multiply through

n

k

x, n , 2n by 1

x and an

attempt to ...

… “integrate”(process the

result); M1

2 2

1 1 1ln

2 2 2x c

x x

correct solution with/without +

c A1 oe

(4)


7. (a) 2

2 2

4 (2 )(2 ) (2 ) (2 )

A B

y y y y y

so 2 (2 ) (2 )A y B y M1

Let 2,y 12

2 4B B , Let 2,y 12

2 4A A M1

giving 1 12 2

(2 ) (2 )y y

A1 cao (3)

(b) 2

2 1d d

4 coty x

y x

B1

1 12 2 d tan d

(2 ) (2 )y x x

y y

1 12 2ln(2 ) ln(2 ) ln(sec )y y x c B1 M1 A1 ft

30,y x

3

1 1 12 2 cosln 2 ln 2 ln c

M1

0 ln 2 ln 2c c

1 12 2ln(2 ) ln(2 ) ln(sec ) ln 2y y x

1 2 sec

ln ln2 2 2

y x

y

M1

2 sec

ln 2ln2 2

y x

y

22 sec

ln ln2 2

y x

y

M1

22 sec

2 4

y x

y

Hence, 2 8 4sec

2

yx

y

A1 (8)

(11 marks)

Gold 3: 11/12 12

Question

Number Scheme Marks

8. (a) tan 3 or 3

sin2

M1

3

awrt 1.05 A1 (2)

(b) 2dsec

d

x

,

dcos

d

y

3

2

d coscos

d sec

y

x

M1 A1

At P, 3 1cos

3 8m

Can be implied A1

Using 1mm , 8m M1

For normal 12

3 8 3y x M1

At Q, 0y 12

3 8 3x

leading to 1716

3x 1716

k 1.0625 A1 (6)

(c) 2 2 2 2d

d d sin sec dd

xy x y

M1 A1

2tan d A1

2sec 1 d M1

tan C A1

3 32

300d tan 3 0 0V y x

M1

213

3 13

1,p q A1 (7)

[15]


Question 1

The majority of candidates gained full marks in part (a) and it was better done than similar questions in previous years. Most

obtained the identity 2

1 3 1 3 1A x Bx x Cx and found A and C by making an appropriate substitution. Finding

B proved more difficult and the error 0B was not uncommon. In part (b)(i), most could gain the method mark, recognising

that the integral of 1

x was ln x , and those with incorrect non-zero values of A, B and C were able to benefit from follow through

marks in this part. There were, however, many errors when finding 3

d3 1

xx

and

2

3d

3 1x

x

; –3 ln (3x – 1) and 3 ln

(x – 1)2 respectively, being common errors. The majority could start part (b)(ii) and, if they had correct values of A, B and C, then

full marks for the question were common.

Question 2

Those who knew, and often quoted, a formula of the form d

lnd

x xa a ax

usually found this question straightforward.

Those who did not, tried a number of methods and these were frequently incorrect. Errors seen included

1

16 0.5 , 16 0.5 ln and 8 lnt tt t t

. Nearly all candidates substituted 3t into their d

d

I

t but a significant

number of candidates failed to give their answer in the form ln a , as required by the question, leaving their answer in the

form lnn a .

Question 3

This question discriminated well across all abilities, with about 52% of candidates gaining at least 6 of the 9 marks available and

about 20% of candidates gaining all 9 marks.

In part (a), the most popular method was to rewrite 1

1

x

x

as

1 1

2 2(1 ) (1 )x x

and achieve the result

211

2x x by multiplying out the binomial expansion of

1

2(1 )x with the binomial expansion of

1

2(1 ) .x

Some

candidates, however, were not able to formulate a strategy for expressing 1

1

x

x

in a form so that relevant binomial

expansions could be applied. The most common mistake was to express 1

1

x

x

as

1

2

1

2

(1 )

(1 )

x

x

and then try to divide the

two expressions, once expanded, but without success.

Although many candidates had written

1 1

2 2(1 ) (1 )x x

, a significant number did not attempt to multiply the two resulting

expansions together, with several attempting to divide their expansions and some deciding to add their expansions after observing

1

2x both expansions. Many candidates were able to use a correct method for expanding a binomial expression of the form

(1 )nx with some making sign errors when simplifying. The majority of candidates, however, who recognised the need to

multiply the two expressions did so successfully, showing sufficient working and ignoring higher powers of x, to produce the

given result.

Gold 3: 11/12 14

Examples of alternative methods seen from a few candidates in part (a) included:

122 12

2

1 (1 )(1 ) (1 )(1 ) (1 )

1 (1 )(1 ) (1 )

x x x xx x

x x x x

, etc.

or

1

2 21 (1 )(1 )

(1 )(1 )1 (1 )(1 )

x x xx x

x x x

, etc.

In part (b), the main obstacle to success was the lack of realisation that a substitution of 1

26x must be made into both sides

of 21 1

11 2

xx x

x

. A significant number substituted into the RHS alone, assuming the LHS was 3 and

claimed that 1405

31352

. Even when candidates did substitute 1

26x correctly into both sides, many neglected to equate

both sides, and so had little chance of figuring out how to proceed to estimate 3 . Those that equated both sides usually

achieved the correct estimate of 7025

.4036

Few candidates ignored the instruction given in part (b) and equated 1

1

x

x

to 3 , deduced the value of

1

2x and

substituted this into the RHS in order to find an estimate for 3 . Historically there have been a number of past examination

questions that have asked candidates to "use a suitable value of x" and presumably these candidates had decided to do just that.

Question 4

This question was not well done and over one quarter of the candidates gained no marks on this question. In all of the other

questions on the paper fewer than 8% of the candidates gained no marks. Many did not recognise that the question required a

separation of the variables and they could make no progress.

The main error in separation was obtaining, through faulty algebra, 1

dyy

on one side of the equation. A quite unexpected

error was to see those who had, correctly, obtained dy y integrate this to ln y . The only explanation of this appears to be that

many questions on this topic do result in logarithms and candidates were following an expected pattern rather than actually solving

the question set. Those who could deal with the y side of the equation often had trouble with 2

1d

cosx

x

, not rewriting this as

2sec dx x and proceeding to tan x . Many long and fruitless attempts were made using a double angle formula. Those who

did integrate sometimes left it there, not realising that it was necessary to evaluate a constant of integration. About one third of

candidates did obtain full marks.

Question 5

A considerable number of candidates did not attempt part (a), but of those who did, the most common method was to use similar

triangles to obtain 23hr and substitute r into

213

V r h to give 34

27.V h Some candidates used trigonometry to


find the semi-vertical angle of the cone and obtained 23hr

from this. A few candidates correctly used similar shapes to

compare volumes by writing down the equation

3

213

.(16) 24 24

V h

Part (b) discriminated well between many candidates who were able to gain full marks with ease and some candidates who were

able to gain just the first one or two marks. Some incorrectly differentiated 21

3V r h to give

2d 1

d 3

Vr

h . Most of the

successful candidates used the chain rule to find d

d

h

t by applying

d d.

d

V V

t dh The final answer

1

8 was sometimes

carelessly written as 1

.8 Occasionally, some candidates solved the differential equation

d8

d

V

t and equated their solution

to

34

27

h and then found

d

d

t

h or differentiated implicitly to find

d

d

h

t.

Question 6

In part (a), a surprisingly large number of candidates did not know how to integrate 2tan .x Examiners were confronted with

some strange attempts involving either double angle formulae or logarithmic answers such as 2ln(sec )x or

4ln(sec ).x

Those candidates who realised that the needed the identity 2 2sec 1 tanx x sometimes wrote it down incorrectly.

Part (b) was probably the best attempted of the three parts in the question. This was a tricky integration by parts question owing

to the term of 3

1 ,x

meaning that candidates had to be especially careful when using negative powers. Many candidates applied the

integration by parts formula correctly and then went on to integrate an expression of the form 3

k

xto gain 3 out of the 4 marks

available. A significant number of candidates failed to gain the final accuracy mark owing 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 dd

lnvx

x implied 1x

v .

In part (c), most candidates correctly differentiated the substitution to gain the first mark. A significant proportion of candidates

found the substitution to obtain an integral in terms of u more demanding. Some candidates did not realise that 2e x

and 3e x

are

2(e )xand

3(e )x respectively and hence

2 1u , rather than 2( 1)u was a frequently encountered error seen in the

numerator of the substituted expression. Fewer than half of the candidates simplified their substituted expression to arrive at the

correct result of

2( 1)d .

uu

u

Some candidates could not proceed further at this point but the majority of the candidates who

achieved this result were able to multiply out the numerator, divide by u, integrate and substitute back for u. At this point some

candidates struggled to achieve the expression required. The most common misconception was that the constant of integration

was a fixed constant to be determined, and so many candidates concluded that 32.k Many candidates did not realise that

32

when added to c combined to make another arbitrary constant k.

Question 7

In part (a), many candidates realised that they needed to factorise the denominator to give two linear factors, and usually

proceeded to give a fully correct solution. A few candidates, however, thought that 24 y was an example of a repeated linear

factor and tried to split up their fraction up accordingly. Some candidates struggled with factorising 24 y giving answers such

Gold 3: 11/12 16

as (4 )(4 )y y or ( 2)( 2).y y The majority of candidates were able to write down the correct identity to find their

constants, although a noticeable number of candidates, when solving 4 2A found 2.A

A significant minority of candidates who completed part (a) correctly made no attempt at part (b). About half of the candidates in

part (b) were able to separate out the variables correctly. Many of these candidates spotted the link with part (a). It was pleasing

that candidates who progressed this far were able to correctly integrate tan x and correctly find the two ln terms by integrating

their partial fraction. Common errors at this point were integrating tan x to give 2sec x and the sign error involved when

integrating 2K

y. A significant number of candidates at this point did not attempt to find a constant of integration. Other

candidates substituted 3

x and 0y into an integrated equation which did not contain a constant of integration. A majority

of candidates who found the constant of integration struggled to simplify their equation down to an equation with a single ln term

on each side. The most common error of these candidates was to believe that ln ln lnA B C implies .A B C

Of all the 8 questions, this was the most demanding in terms of a need for accuracy. Fewer than 10% of candidates were able to

score all 11 marks in this question, although statistics show that about half of the candidates were able to score at least 5 marks.

Question 8

The majority of candidates were able to complete part (a), although some candidates gave the answer in degrees rather than

radians. Most could start part (b) correctly and, apart from a few errors in sign, obtain 2

d cos

d sec

y

x

, although this was often

simplified to cos rather than the correct 3cos . The majority of candidates were able to demonstrate the correct method of

finding the equation of the normal and to complete the part by substituting 0y and solving for x. A small number of

candidates eliminated , successfully differentiated the cartesian equation and completed the question.

Part (c) proved challenging for many candidates and a substantial number of candidates thought that the volume was given by

2sin , often ignoring d or dx . Among those who recognised that the appropriate integral was2 2sin sec d , many

were unable to rewrite this in a form which could be integrated. ln sec and 3sin were among the erroneous attempts seen.

Those who realised that 2 2 2 2sin sec tan sec 1 usually completed the question correctly, although a few used

x limits rather than limits. There are a number of possible alternative approaches to this question and there were some

successful attempts using integration by parts. A number of candidates attempted to use the cartesian form of the equation but few

of these were able to establish a method of integrating

2

21

x

x.

Statistics for C4 Practice Paper G2

Mean score for students achieving grade:

Qu Max score

Modal score

Mean %

ALL A* A B C D E U

1 10 75 7.47 9.68 8.85 7.82 6.62 5.31 3.97 2.22

2 5 57 2.83 4.71 3.30 2.12 1.36 0.94 0.90 0.67

3 9 9 59 5.28 8.16 6.52 5.28 4.26 3.4 2.61 1.64

4 5 57 2.86 4.81 4.10 3.00 1.70 0.81 0.30 0.08

5 7 47 3.26 5.18 2.52 1.20 0.65 0.20 0.11


6 13 55 7.18 10.02 5.96 3.82 2.50 1.38 0.40

7 11 46 5.05 7.44 4.70 3.12 2.10 1.42 0.83

8 15 52 7.86 14.16 10.39 7.32 4.95 3.34 2.04 1.10

75 56 41.79 55.80 38.72 27.03 19.05 12.82 7.05 Paper Reference(s)

6666/01


Gold Level (Harder) G3








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.









A* A B C D E

65 58 47 42 36 28

1. f(x) = (3 + 2x)–3

, x < 23 .

Gold 4: 12/12 18

Find the binomial expansion of f(x), in ascending powers of x, as far as the term in x3.

Give each coefficient as a simplified fraction.

(5)

June 2007

2. Using the substitution u = cos x +1, or otherwise, show that

2

0

1 dsine

xxxcos = e(e – 1).

(6)

June 2010

3. Express )13)(2(

10209 2

xx

xx in partial fractions.

(4)

January 2013


4. With respect to a fixed origin O the lines l1 and l2 are given by the equations

l1 : r =

17

2

11

+

4

1

2

l2 : r =

p

11

5

+ μ

2

2

q

where λ and μ are parameters and p and q are constants. Given that l1 and l2 are perpendicular,

(a) show that q = –3.

(2)

Given further that l1 and l2 intersect, find

(b) the value of p,

(6)

(c) the coordinates of the point of intersection.

(2)

The point A lies on l1 and has position vector

13

3

9

. The point C lies on l2.

Given that a circle, with centre C, cuts the line l1 at the points A and B,

(d) find the position vector of B.

(3)

January 2009

Gold 4: 12/12 20

5.

Figure 2

Figure 2 shows a sketch of the curve with parametric equations

x = 2 cos 2t, y = 6 sin t, 0 t 2

.

(a) Find the gradient of the curve at the point where t = 3

.

(4)

(b) Find a cartesian equation of the curve in the form

y = f(x), –k x k,

stating the value of the constant k.

(4)

(c) Write down the range of f(x).

(2)

January 2009


6. The points A and B have position vectors 2i + 6j – k and 3i + 4j + k respectively.

The line 1l passes through the points A and B.

(a) Find the vector AB .

(2)

(b) Find a vector equation for the line 1l .

(2)

A second line 2l passes through the origin and is parallel to the vector i + k. The line 1l meets the

line 2l at the point C.

(c) Find the acute angle between 1l and 2l .

(3)

(d) Find the position vector of the point C.

(4)

January 2008

7. The line 1l has equation r =

1

2

1

4

3

2

, where λ is a scalar parameter.

The line 2l has equation r =

2

0

5

3

9

0

, where is a scalar parameter.

Given that 1l and 2l meet at the point C, find

(a) the coordinates of C.

(3)

The point A is the point on 1l where λ = 0 and the point B is the point on 2l where μ = –1.

(b) Find the size of the angle ACB. Give your answer in degrees to 2 decimal places.

(4)

(c) Hence, or otherwise, find the area of the triangle ABC.

(5)

June 2010

Gold 4: 12/12 22

8. A population growth is modelled by the differential equation

t

P

d

d = kP,

where P is the population, t is the time measured in days and k is a positive constant.

Given that the initial population is P0,

(a) solve the differential equation, giving P in terms of P0, k and t.

(4)

Given also that k = 2.5,

(b) find the time taken, to the nearest minute, for the population to reach 2P0.

(3)

In an improved model the differential equation is given as

t

P

d

d = P cos t,

where P is the population, t is the time measured in days and is a positive constant.

Given, again, that the initial population is P0 and that time is measured in days,

(c) solve the second differential equation, giving P in terms of P0, and t.

(4)

Given also that = 2.5,

(d) find the time taken, to the nearest minute, for the population to reach 2P0 for the first time,

using the improved model.

(3)

June 2007


END


Question Number

Scheme Marks

** represents a constant

1. (a) 3 3

33 2 1 2f( ) (3 2 ) 3 1 1

3 27 3

x xx x

Takes 3 outside the bracket to give any

of 3(3) or 127

.

See note below.

B1

2 3127

( 3)( 4) ( 3)( 4)( 5)1 ( 3)(* * x); (* * x) (* * x) ...

2! 3!

with * * 1

Expands 3(1 * * )x

to give a simplified or an un-simplified

1 ( 3)(* * x) ;

A correct simplified or an un-simplified

.......... expansion

with candidate’s

followed thro’ * * x

M1;

A1

2 31 2x 2x 2x27 3 3 3

( 3)( 4) ( 3)( 4)( 5)1 ( 3)( ) ( ) ( ) ...

2! 3!

2

3127

8x 801 2x x ...

3 27

2 31 2x 8x 80x; ...

27 27 81 729

Anything that

cancels to 1 2x;

27 27

Simplified 2 38x 80x

81 729

A1;

A1

[5]

5 marks

Gold 4: 12/12 24

2. d

sind

ux

x B1

cos 1sin e d e dx ux x u M1 A1

eu ft sign error A1ft

cos 1e x

cos 1 1 22

0e e ex

or equivalent with u M1

e e 1 cso A1 (6)

[6]

Question

Number Scheme Marks

3.

29 20 10

( 2)(3 1) ( 2) (3 1)

x x B CA

x x x x

3A their constant term 3 B1

29 20 10 ( 2)(3 1) (3 1) ( 2)x x A x x B x C x Forming a correct identity. B1

Either 2 : 9 3 , : 20 5 3

constant: 10 2 2

x A x A B C

A B C

or

2 36 40 10 7 14 7 2x B B B

1 20 7 7 71 10 1

3 3 3 3 3x C C C

Attempts to find the value of

either one of their B or their C

from their identity.

M1

Correct values for

their B and their C, which are

found using a correct identity.

A1

[4]


Question Number

Scheme Marks

4 (a) 1 2 4 d i j k , 2 2 2q d i j k

As 1 2

2

1 2 ( 2 ) (1 2) ( 4 2)

4 2

q

q

d d

Apply dot product calculation

between two direction

vectors, ie.

( 2 ) (1 2) ( 4 2)q

M1

1 2 0 2 2 8 0

2 6 3 AG

q

q q

d d

Sets 1 2 0 d d

and solves to find 3q A1 cso

(2) (b) Lines meet where:

11 2 5

2 1 11 2

17 4 2

q

p

First two of

: 11 2 5 (1)

: 2 11 2 (2)

: 17 4 2 (3)

q

p

i

j

k

Need to see equations

(1) and (2).

Condone one slip.

(Note that 3q .)

M1

(1) + 2(2) gives: 15 17 2 Attempts to solve (1) and (2)

to find one of either or dM1

(2) gives: 2 11 4 5 Any one of 5 or 2 A1

Both 5 and 2 A1

(3) 17 4(5) 2( 2)p

Attempt to substitute their

and into their k component

to give an equation in p

alone.

ddM1

17 20 4 1p p 1p A1 cso

(6)

(c)

11 2 5 3

2 5 1 or 11 2 2

17 4 1 2

r r Substitutes their value of

or into the correct line l1 or

l2 .

M1

Intersect at

1

7 or 1, 7, 3

3

r

1

7

3

or 1, 7, 3 A1

(2)

Gold 4: 12/12 26

5. (a) d

4sin 2d

xt

t ,

d6cos

d

yt

t B1, B1

d 6cos 3

d 4sin 2 4sin

y t

x t t

M1

At 3

t

, 3

2

3 3

4 2m

accept equivalents, awrt 0.87 A1 (4)

(b) Use of 2cos 2 1 2sint t M1

cos 22

xt , sin

6

yt

2

1 22 6

x y

M1

Leading to 18 9 3 2y x x cao A1

2 2x 2k B1 (4)

(c) 0 f 6x either 0 f x or f 6x B1

Fully correct. Accept 0 6y , 0, 6 B1 (2)

(10 marks)


Question

Number Scheme Marks

6. (a)

2 3

6 & 4

1 1

OA OB

3 2 1

4 6 2

1 1 2

AB OB OA

Finding the difference

between OB and OA . M1

Correct answer. A1

[2] An expression of the form

vector vector

(b)

1

2 1

: 6 2

1 2

l

r or

3 1

4 2

1 2

r

1

2 1

: 6 2

1 2

l

r or

3 1

4 2

1 2

r

M1

r = theirOA AB or

r = theirOB AB or

r = theirOA BA or

r = theirOB BA

(r is needed.)

A1

aef

[2]

(c) 2

0 1 1

: 0 0 0

0 1 1

l

r r

1 2 2AB d i j k , 2 0 d i j k & is angle

2

2 2 2 2 2 22

1 1

2 0

2 1cos

. (1) ( 2) (2) . (1) (0) (1)

AB

AB

d

d

Considers dot product

between 2d and their .AB M1

2 2 2 2 2 2

1 0 2cos

(1) ( 2) (2) . (1) (0) (1)

Correct followed through

expression or equation. A1

4

3cos 45 or or awrt 0.79.

3. 2

4

45 or or awrt 0.79 A1 cao

[3]

This means that cos does not

necessarily have to be the subject

of the equation. It could be of the

form 3 2 cos 3.

Gold 4: 12/12 28

Question

Number Scheme Marks

6. (d) If l1 and l2 intersect then:

2 1 1

6 2 0

1 2 1

: 2 (1)

: 6 2 0 (2)

: 1 2 (3)

i

j

k

Either seeing equation (2) written

down correctly with or without

any other equation or seeing

equations (1) and (3) written

down correctly.

M1

(2) yields 3

Any two yields 3, 5

Attempt to solve either equation

(2) or simultaneously solve any

two of the three equations to find

…

dM1

either one of or correct. A1

1

2 1 5 1 5

: 6 3 2 0 5 0 0

1 2 5 1 5

l or

r r

5

0

5

or 5 5i k

Fully correct solution & no

incorrect values of or seen

earlier.

A1 cso

[4]


7. (a) j components 3 2 9 3 1 M1 A1

Leading to : 5, 9, 1C accept vector forms A1 (3)

(b) Choosing correct directions or finding AC and BC M1

1 5

2 . 0 5 2 6 29cos

1 2

ACB

use of scalar product M1 A1

57.95ACB awrt 57.95 A1 (4)

(c) : 2, 3, 4 : 5, 9, 5A B

3 10

6 , 0

3 4

AC BC

2 2 2 23 6 3 3 6AC AC M1 A1

2 2 210 4 2 29BC BC A1

1

sin2

ABC AC BC ACB

1

3 6 2 29sin 33.52

ACB 15 5 , awrt 34 M1 A1 (5)

[12]

Question Scheme Marks

Gold 4: 12/12 30

Number

8. (a) d

d

PkP

t and 00, (1)t P P

d

dP

k tP

Separates the variables

with dP

P and dk t on

either side with integral signs not necessary.

M1

ln ;P kt c Must see lnP and kt ;

Correct equation with/without + c.

A1

When 0 00, lnt P P P c

0or ktP Ae P A

Use of boundary condition (1) to attempt

to find the constant of integration.

M1

0ln lnP kt P 0 0ln lnln .kt P PP kte e e e

Hence, 0

ktP P e 0

ktP P e A1

[4]

(b) 02 & 2.5P P k 2.5

0 02 tP P e Substitutes 02P P into

an expression involving P

M1

2.5 2.52 ln ln2t te e or 2.5 ln2t

…or 2 ln ln2kt kte e or ln2kt

Eliminates 0P and takes

ln of both sides M1

12.5

ln2 0.277258872... dayst

0.277258872... 24 60 399.252776... minutest

399mint or 6 hr 39 minst (to nearest

minute)

awrt 399t or

6 hr 39 mins A1

[3]

Question Number

Scheme Marks

8. (c) d

cosd

PP t

t and 00, (1)t P P


d

cos dP

t tP

Separates the variables

with dP

P and

cos dt t on either

side with integral signs not necessary.

M1

ln sin ;P t c

Must see lnP and sin t ;

Correct equation with/without + c.

A1

When 0 00, lnt P P P c

sin

0or tP Ae P A

Use of boundary condition (1) to attempt

to find the constant of integration.

M1

0ln sin lnP t P 0 0sin ln lnln sin .t P PP te e e e

Hence, sin

0

tP P e sin

0

tP P e A1

[4]

(d) 02 & 2.5P P sin2.5

0 02 tP P e

sin2.5 2 sin2.5 ln2te t

…or … 2 sin ln2te t

Eliminates 0P and

makes sin t or sin2.5t

the subject by taking ln’s

M1

112.5

sin ln2t Then rearranges to make t the subject.

dM1

(must use sin-1)

0.306338477...t

0.306338477... 24 60 441.1274082... minutest

441mint or 7 hr 21 minst (to nearest

minute)

awrt 441t or

7 hr 21 mins A1

[3]

14 marks

Gold 4: 12/12 32

Question 2

This question was generally well done and, helped by the printed answer, many produced fully correct answers. The commonest

error was to omit the negative sign when differentiating cos 1x . The order of the limits gave some difficulty. Instead of the

correct 1

2e du u , an incorrect version

2

1e du u was produced and the resulting expressions manipulated to the printed

result and working like 2 1 2 1e e e e e e 1 was not uncommon.

Some candidates got into serious difficulties when, through incorrect algebraic manipulation, they obtained 2e sin du x u

instead of e du u . This led to expressions such as 2e 2 du u u u and the efforts to integrate this, either by parts twice

or a further substitution, often ran to several supplementary sheets. The time lost here inevitably led to difficulties in finishing the

paper. Candidates need to have some idea of the amount of work and time appropriate to a 6 mark question and, if they find

themselves exceeding this, realise that they have probably made a mistake and that they would be well advised to go on to another

question.

Question 3

This was correctly answered by about 40% of the candidates.

A majority incorrectly expressed

29 20 10

( 2)(3 1)

x x

x x

as

2 1

( 2) (3 1)x x

, having failed to realise that the algebraic

fraction given in the question is improper, thereby losing 3 of the 4 marks available.

For those achieving the correct partial fractions, a process of long division was typically used to find the value of the constant

term, and the resulting remainder, usually 5 4,x became the LHS of the subsequent identity. A minority of them, however,

applied 29 20 10 ( 2)(3 1) (3 1) ( 2)x x A x x B x C x in order to obtain the correct partial

fractions.


Question 4

The majority of candidates identified the need for some form of dot product calculation in part (a). Taking the dot product 1 2.l l ,

was common among candidates who did not correctly proceed, while others did not make any attempt at a calculation, being

unable to identify the vectors required. A number of candidates attempted to equate 1l and

2l at this stage. The majority of

candidates, however, were able to show that 3.q

In part (b), the majority of candidates correctly equated the ,i j and k components of 1l and

2l , and although some candidates

made algebraic errors in solving the resulting simultaneous equations, most correctly found and . In almost all such cases

the value of p and the point of intersection in part (c) was then correctly determined.

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. Those candidates who decided to draw a diagram usually increased their chance of success. Most

candidates who were successful at this part applied a vector approach as detailed in the mark scheme. The easiest vector

approach, adopted by a few candidates, is to realise that 1 at A, 5 at the point of intersection and so 9 at B. So

substitution of 9 into l1 yields the correct position vector 7 11 19 . i j k A few candidates, by deducing that the

intersection point is the midpoint of A and B were able to write down 9

1,2

x

37

2

y and

133,

2

z in order

to find the position vector of B.

Question 5

Nearly all candidates knew the method for solving part (a), although there were many errors in differentiating trig functions. In

particular d

2cos 2d

tt

was often incorrect. It was clear from both this question and question 2 that, for many, the calculus of

trig functions was an area of weakness. Nearly all candidates were able to obtain an exact answer in surd form. In part (b), the

majority of candidates were able to eliminate t but, in manipulating trigonometric identities, many errors, particularly with signs,

were seen. The answer was given in a variety of forms and all exact equivalent answers to that printed in the mark scheme were

accepted. The value of k was often omitted and it is possible that some simply overlooked this. Domain and range remains an

unpopular topic and many did not attempt part (c). In this case, inspection of the printed figure gives the lower limit and was

intended to give candidates a lead to identifying the upper limit.

Question 6

In part (a), a majority of candidates were able to subtract the given position vectors correctly in order to find .AB Common

errors in this part included some candidates subtracting the position vector the wrong way round and a few candidates who could

not deal with the double negative when finding the k component of .AB

In part (b), a significant majority of candidates were able to state a vector equation of l1. A significant number of these

candidates, however, wrote 'Line = ' and omitted the ‘r’ on the left hand side of the vector equation, thereby losing one mark.

Many candidates were able to apply the dot product correctly in part (c) to find the correct angle. Common errors here included

applying a dot product formula between OA and ;OB or applying the dot product between either OA or OB and the

direction vector of l1. Interestingly, a surprising number of candidates either simplified 2 2 2(1) ( 2) (2) to 5 or when

finding the dot product multiplied -2 by 0 to give -2.

Part (d) proved more discriminating. The majority of candidates realised that they needed to put the line 1l equal to line 2l . A

significant number of these candidates, however, were unable to write l2 as ( ) i k or used the same parameter (usually )

as they had used for l1. Such candidates then found difficulty in making further progress with this part.

Gold 4: 12/12 34

Question 7

Part (a) was fully correct in the great majority of cases but the solutions were often unnecessarily long and nearly two pages of

working were not unusual. The simplest method is to equate the j components. This gives one equation in , leading to 3 ,

which can be substituted into the equation of 1l to give the coordinates of C. In practice, the majority of candidates found both

and and many proved that the lines were coincident at C. However the question gave the information that the lines meet at

C and candidates had not been asked to prove this. This appeared to be another case where candidates answered the question that

they had expected to be set, rather than the one that actually had been.

The great majority of candidates demonstrated, in part (b), that they knew how to find the angle between two vectors using a

scalar product. However the use of the position vectors of A and B, instead of vectors in the directions of the lines was common.

Candidates could have used either the vectors

1

2

1

and

5

0

2

, given in the question, or AC and BC . The latter was much

the commoner choice but many made errors in signs. Comparatively few chose to use the cosine rule. In part (c), many continued

with the position vectors they had used incorrectly in part (b) and so found the area of the triangle OAB rather than triangle ABC.

The easiest method of completing part (c) was usually to use the formula Area = 12

sinab C and most chose this. Attempts to

use Area = 12base height were usually fallacious and often assumed that the triangle was isosceles. A few complicated

attempts were seen which used vectors to find the coordinates of the foot of a perpendicular from a vertex to the opposite side. In

principle, this is possible but, in this case, the calculations proved too difficult to carry out correctly under examination conditions.




Qu Max score

Modal score

Mean %

ALL A* A B C D E U

1 5 78 3.88 4.59 4.06 3.64 3.00 2.30 1.38

2 6 64 3.81 5.84 5.13 4.00 2.69 1.71 0.94 0.36

3 4 1 57 2.26 3.53 2.49 2.09 1.73 1.58 1.50 1.15

4 13 61 7.94 10.15 7.19 4.59 3.25 1.74 0.58

5 10 54 5.38 7.41 5.34 3.97 2.72 1.64 0.63

6 11 57 6.30 8.66 5.80 4.15 3.11 1.68 1.27

7 12 54 6.42 10.86 8.23 6.15 4.39 3.01 2.02 1.02

8 14 36 5.09 8.99 3.90 1.81 0.80 0.35 0.09

75 55 41.08 55.65 38.53 26.97 19.18 12.17 6.48

Gold 4: 12/12 36

Paper Reference(s)

6666/01


Gold Level (Hardest) G4








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.









A* A B C D E

62 52 42 36 30 26

1. (a) Find 2 dxx e x .

(5)

(b) Hence find the exact value of 1 2

0dxx e x .

(2)

Gold 2 :10/12 37

June 2013

2. Use the substitution u = 2x to find the exact value of

1

0

2)12(

2x

x

dx.

(6)

June 2007

3.

Figure 2

Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated. After

t seconds the radius of the rod is x cm and the length of the rod is 5x cm.

The cross-sectional area of the rod is increasing at the constant rate of 0.032 cm2

s–1

.

(a) Find t

x

d

d when the radius of the rod is 2 cm, giving your answer to 3 significant figures.

(4)

(b) Find the rate of increase of the volume of the rod when x = 2.

(4)

June 2008

Gold 4: 12/12 38

4. (i) Find

x

xd

2ln .

(4)

(ii) Find the exact value of

2

4

2 dsin

xx .

(5)

January 2008

5. (a) Find

x

x

xd

69, x > 0.

(2)

(b) Given that y = 8 at x =1, solve the differential equation

x

y

d

d =

x

yx 3

1

)69(

giving your answer in the form y 2

= g(x).

(6)

January 2010

6. f(θ) = 4 cos2

θ – 3sin2

θ

(a) Show that f(θ) = 2

1 +

2

7 cos 2θ.

(3)

(b) Hence, using calculus, find the exact value of

2

0

df

)( .

(7)

June 2010

Gold 2 :10/12 39

7. Relative to a fixed origin O, the point A has position vector (8i + 13j – 2k), the point B has

position vector (10i + 14j – 4k), and the point C has position vector (9i + 9j + 6k).

The line l passes through the points A and B.

(a) Find a vector equation for the line l.

(3)

(b) Find CB .

(2)

(c) Find the size of the acute angle between the line segment CB and the line l, giving your

answer in degrees to 1 decimal place.

(3)

(d) Find the shortest distance from the point C to the line l.

(3)

The point X lies on l. Given that the vector CX is perpendicular to l,

(e) find the area of the triangle CXB, giving your answer to 3 significant figures.

(3)

June 2009

Gold 4: 12/12 40

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm3s

–1 and is

leaking out of a hole in the base, at a rate proportional to the square root of the height of the

liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm2.

(a) Show that at time t seconds, the height h cm of liquid in the cylinder satisfies the differential

equation

t

h

d

d= 0.4 kh,

where k is a positive constant.

(3)

When h = 25, water is leaking out of the hole at 400 cm3s

–1.

(b) Show that k = 0.02.

(1)

(c) Separate the variables of the differential equation

t

h

d

d= 0.4 0.02h

to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by

100

0

d20

50h

h.

(2)

Using the substitution h = (20 x)2, or otherwise,

(d) find the exact value of

100

0

d20

50h

h.

(6)

(e) Hence find the time taken to fill the cylinder from empty to a height of 100 cm, giving your

answer in minutes and seconds to the nearest second.

(1)

January 2008


END

Gold 2 :10/12 41

Question

Number Scheme Marks

1. (a)

2e dxx x , 1st Application:

2 d2

d

de e

d

x x

uu x x

x

vv

x

, 2nd

Application:

d1

d

de e

d

x x

uu x

x

vv

x

2e 2 e dx xx x x

2e e dx xx x x , 0 M1

2e 2 e dx xx x x A1 oe

2e 2 e e dx x xx x x

Either 2e e e dx x xAx Bx C x

or for

e d e e dx x xK x x K x x

M1

2e 2( e e )x x xx x c 2e e ex x xAx Bx C M1

Correct answer, with/without c A1

(5)

(b)

12

0

2 1 1 1 2 0 0 0

e 2( e e )

1 e 2(1e e ) 0 e 2(0e e )

x x xx x

Applies limits of 1 and 0 to an

expression of the form 2e e e ,x x xAx Bx C 0 , 0A B

and 0C and subtracts the correct

way round.

M1

e 2 e 2 cso A1 oe

(2) [7]

Gold 4: 12/12 42

Question Number

Scheme Marks

2.

1

2

0

2d

(2 1)

x

xx

, with substitution 2xu

d

2 .ln2d

xu

x

d 1

d 2 .ln2x

x

u

dd

2 .ln2xux or d

d.ln2u

xu

or d1d

ln2uu x

B1

2 2

2 1 1d d

ln2(2 1) ( 1)

x

xx u

u

2

1d

( 1)k u

u

where k is constant

M1

1 1

ln2 ( 1)c

u

2 1

2 1

( 1) ( 1)

( 1) 1.( 1)

u a u

u u

M1

A1

change limits: when x = 0 & x = 1 then u = 1 & u

= 2

1 2

2

10

2 1 1d

ln2 ( 1)(2 1)

x

xx

u

1 1 1

ln2 3 2

Correct use of limits u = 1 and u = 2

depM1

1

6ln2

16ln2

or 1 1ln4 ln8

or 1 12ln2 3n2

A1 aef

Exact value only! [6] Alternatively candidate can revert back to x …

1 1

2

00

2 1 1d

ln2(2 1) (2 1)

x

x xx

1 1 1

ln2 3 2

Correct use of limits x = 0 and x = 1

depM1

1

6ln2 1

6ln2or 1 1

ln4 ln8 or 1 1

2ln2 3ln2 A1 aef

Exact value only!

6 marks

Gold 2 :10/12 43

3. (a) From question, d

0.032d

A

t B1

2 d

2d

AA x x

x

B1

d d d 1 0.016

0.032 ;d d d 2

x A A

t t x x x

M1

When 2cmx , d 0.016

d 2

x

t

Hence, d

0.002546479...d

x

t (cm s

-1) A1 cso (4)

(b) 2 3(5 ) 5V x x x B1

2d15

d

Vx

x B1 ft

2d d d 0.01615 . ; 0.24

d d d

V V xx x

t x t x

M1

When 2cmx , 3 1d0.24(2) 0.48 (cm s )

d

V

t

A1 (4)

(8 marks)

Gold 4: 12/12 44

Question

Number Scheme Marks

4. (i) 2 2ln d 1.ln dx xx x

12 1

2

2

dln

d

d1

d

xxx

uu

x

vv x

x

12 2

ln d ln . dx xx

x x x x

Use of ‘integration by

parts’ formula in the

correct direction.

M1

Correct expression. A1

2ln 1 dxx x An attempt to multiply x by

a candidate’s ax

or 1bx

or 1x

. dM1

2ln xx x c Correct integration with + c A1 aef

[4]

(ii) 2

4

2sin dx x

2 2 12

NB: cos2 1 2sin or sin 1 cos2x x x x

Consideration of double

angle formula for cos2x M1

2 2

4 4

1 cos2 1d 1 cos2 d

2 2

xx x x

2

4

12

1sin 2

2x x

Integrating to give

sin 2ax b x ; , 0a b dM1

Correct result of anything

equivalent to 1 12 4

sin 2x x A1

2sinsin( )1

2 2 2 4 2

1 12 2 4 2

( 0) ( )

Substitutes limits of 2 and

4 and subtracts the correct

way round.

ddM1

1 1 1

2 4 2 8 4 1 1 1 2

2 4 2 8 4 8 8or or A1 aef ,

cso

Candidate must collect

their term and constant

term together for A1

[5]

No fluked answers, hence

cso.

9 marks

Gold 2 :10/12 45

Question Number

Scheme Marks

Q5 (a) 9 6 6

d 9 dx

x xx x

M1

9 6lnx x C A1 (2)

(b) 13

1 9 6d d

xy x

xy

Integral signs not necessary B1

13

9 6d d

xy y x

x

23

23

9 6lny

x x C 23 their aky M1

23

39 6ln

2y x x C ft their a A1ft

8y , 1x

C 1n 61982

33

2

M1

3C A1

23

29 6ln 3

3y x x

32 6 4ln 2y x x 3

8 3 2ln 1x x A1 (6)

[8]

Gold 4: 12/12 46

Question Number

Scheme Marks

6. (a) 2 2f 4cos 3sin

1 1 1 1

4 cos 2 3 cos 22 2 2 2

M1 M1

1 7cos 2

2 2

cso

A1 (3)

(b) 1 1

cos2 d sin 2 sin 2 d2 2

M1 A1

1 1

sin 2 cos 22 4 A1

21 7 7f d sin 2 cos 2

4 4 8 M1 A1

2

2

0

7 70 0 0

16 8 8 ...

M1

2 7

16 4

A1 (7)

[10]

Gold 2 :10/12 47

7. (a)

10 8 2

14 13 1

4 2 2

AB OB OA

or

2

1

2

BA

M1

8 2

13 1

2 2

r or

10 2

14 1

4 2

r accept equivalents M1 A1ft (3)

(b)

10 9 1

14 9 5

4 6 10

CB OB OC

or

1

5

10

BC

22 21 5 10 126 3 14 11.2CB awrt 11.2 M1 A1 (2)

(c) . cosCB AB CB AB

2 5 20 126 9cos M1 A1

3

cos 36.714

awrt 36.7 A1 (3)

(d) sin126

d

M1 A1ft

3 5 6.7d awrt 6.7 A1 (3)

(e) 2 2 2 126 45 81BX BC d M1

1 1 27 5

9 3 5 30.22 2 2

CBX BX d

awrt 30.1 or 30.2 M1 A1 (3)

(14 marks)

l

X

B

C

d 126

Gold 4: 12/12 48

Question

Number Scheme Marks

8. (a) d d

1600 or 1600d d

V Vc h k h

t t , Either of these statements M1

d

4000 4000d

VV h

h

d4000

d

V

h or

d 1

d 4000

h

V M1

dd

dd

d d d

d d d

Vt

Vh

h h V

t V t

Either,

d 1600 16000.4

d 4000 4000 4000

h c h c hk h

t

Convincing proof of d

d

h

t A1 AG

or

d 1600 16000.4

d 4000 4000 4000

h k h k hk h

t

[3]

(b) When 25h water leaks out such that d

400d

V

t

400 400 25 400 (5) 80c h c c c

From above;

800.02

4000 4000

ck as required Proof that 0.02k B1 AG

[1]

(c) d d

0.4d 0.4

h hk h dt

t k h

Separates the variables with

d

0.4

h

k h and dt on either

side with integral signs not

necessary.

M1 oe

100

0

1 0.02time required d

0.020.4 0.02h

h

100

0

50time required d

20h

h

Correct proof A1 AG

[2]

Gold 2 :10/12 49

Question

Number Scheme Marks

8. (d) 100

0

50d

20h

h with substitution 2(20 )h x

d

2(20 )( 1)d

hx

x or

d2(20 )

d

hx

x Correct

d

d

h

x B1 aef

2(20 ) 20 20h x h x x h

20d

xx

x

or

20d

20 (20 )

xx

x

where is a constant

50 50d . 2(20 ) d

20h x x

xh

M1

20

100 dx

xx

20100 1 dx

x

100 20lnx x c

ln ; , 0x x M1

100 2000lnx x A1

change limits: when 0 then 20h x

and when 100 then 10h x

10010

200

50d 100 2000ln

20h x x

h

or 100 100

00

50d 100 20 2000ln 20

20h h h

h Correct use of limits, ie.

putting

them in the correct way

round

1000 2000ln10 2000 2000ln20 Either 10x and 20x

or 100h and 0h ddM1

2000ln20 2000ln10 1000 Combining logs to give...

2000ln 2 1000 A1 aef 2000ln2 1000 or 1

22000ln 1000

[6] (e) Time required 2000ln2 1000 386.2943611... sec

= 386 seconds (nearest second)

= 6 minutes and 26 seconds (nearest second) 6 minutes, 26 seconds B1

[1]

13 marks

Gold 4: 12/12 50

Question 1

This question was generally well answered with about 73% of candidates gaining at least 5 of the

7 marks available and about 44% of candidates gaining all 7 marks. Almost all candidates

attempted this question with about 13% of them unable to gain any marks.

A significant minority of candidates performed integration by parts the wrong way round in part

(a) to give 3 31 1e e d

3 3

x xx x x and proceeded by attempting to integrate 31e .

3

xx Some

candidates failed to realise that integration by parts was required and wrote down answers such as

31e .

3

xx c Few candidates integrated ex to give 21

2ex

or applied the product rule of

differentiation to give 2e 2 e .x xx x The majority of candidates, however, were able to apply the

first stage of integration by parts to give 2e 2 e d .x xx x x Many candidates realised that they

needed to apply integration by parts for a second time in order to find 2 e dxx x , or in some cases

e dxx x . Those that failed to realise that a second application of integrating by parts was required

either integrated to give the final answer as a two term expression or just removed the integral

sign. A significant number of candidates did not organise their solution effectively, and made a

bracketing error which often led to a sign error leading to the final incorrect answer of 2e 2 e 2e .x x xx x c

In part (b), candidates with an incorrect sign in the final term of their integrated expression often

proceeded to use the limits correctly to obtain an incorrect answer of 3e 2. Errors in part (b)

included not substituting the limit of 0 correctly into their integrated expression; incorrectly

dealing with double negatives; evaluating 02e as 1 or failing to evaluate 0e . Most candidates

who scored full marks in part (a), achieved the correct answer of e 2 in part (b).

Gold 2 :10/12 51

Question 3

At the outset, a significant minority of candidates struggled to extract some or all of the information from the question. These

candidates were unable to write down the rate at which this cross-sectional area was increasing, dd

0.032;At or the cross-

sectional area of the cylinder 2A x and its derivative

dd

2Ax

x ; or the volume of the cylinder 35V x and its

derivative 2d

d15 .V

xx

In part (a), some candidates wrote down the volume V of the cylinder as their cross-sectional area A. Another popular error at this

stage was for candidates to find the curved surface area or the total surface area of a cylinder and write down either210A x

or212A x respectively. At this stage many of these candidates were able to set up a correct equation to find

ddxt

and usually

divided 0.032 by their ddAx

and substituted 2x into their expression to gain 2 out of the 4 marks available. Another error

frequently seen in part (a) was for candidates to incorrectly calculate 0.032

4 as 0.0251. Finally, rounding the answer to 3

significant figures proved to be a problem for a surprising number of candidates, with a value of 0.003 being seen quite often;

resulting in loss of the final accuracy mark in part (a) and this sometimes as a consequence led to an inaccurate final answer in

part (b).

Part (b) was tackled more successfully by candidates than part (a) – maybe because the chain rule equation d d d

d d d

V V x

t x t

is rather more straight-forward to use than the one in part (a). Some candidates struggled by introducing an extra variable r in

addition to x and obtained a volume expression such as 2(5 ).V r x Many of these candidates did not realise that r x

and were then unable to correctly differentiate their expression for V. Other candidates incorrectly wrote down the volume as 22 (5 ).V x x Another common error was for candidates to state a correct V, correctly find

ddVx

, then substitute 2x to

arrive at a final answer of approximately 188.5.

About 10% of candidates were able to produce a fully correct solution to this question.

Question 4

Gold 4: 12/12 52

It was clear to examiners that a significant proportion of candidates found part (i) unfamiliar and thereby struggled to answer this

part. Weaker candidates confused the integral of ln x with the differential of ln .x It was therefore common for these

candidates to write down the integral of ln x as 1 ,x

or the integral of 2ln x as either 2

x or 4 .

x A significant proportion of

those candidates, who proceeded with the expected by parts strategy, differentiated 2ln x

incorrectly to give either 2x

or 12x

and usually lost half the marks available in this part. Some candidates decided from the outset to rewrite 2ln x

as

' ln ln 2'x , and proceeded to integrate each term and were usually more successful with integrating ln x than ln 2. It is

pleasing to report that a few determined candidates were able to produce correct solutions by using a method of integration by

substitution. They proceeded by either using the substitution as 2xu or 2

ln .xu

A significant minority of candidates omitted the constant of integration in their answer to part (i) and were penalised by losing the

final accuracy mark in this part.

In part (ii), the majority of candidates realised that they needed to consider the identity 2cos2 1 2sinx x and so gained

the first method mark. Some candidates misquoted this formula or incorrectly rearranged it. A majority of candidates were then

able to integrate 12

1 cos2x , substitute the limits correctly and arrive at the correct exact answer.

There were, however, a few candidates who used the method of integration by parts in this part, but these candidates were usually

not successful in their attempts.

Question 5

Part (a) of this question proved awkward for many. The integral can be carried out simply by decomposition, using techniques

available in module C1. It was not unusual to see integration by parts attempted. This method will work if it is known how to

integrate ln x , but this requires a further integration by parts and complicates the question unnecessarily. In part (b), most could

separate the variables correctly but the integration of 13

1

y, again a C1 topic, was frequently incorrect.

Weakness in algebra sometimes caused those who could otherwise complete the question to lose the last mark as they could not

proceed from 23 6 4ln 2y x x to

32 6 4ln 2y x x . Incorrect answers, such as

2 3 3216 64ln 8y x x , were common in otherwise correct solutions.

Gold 2 :10/12 53

Question 6

Candidates tended either to get part (a) fully correct or make no progress at all. Of those who were successful, most replaced the 2cos and

2sin directly with the appropriate double angle formula. However many good answers were seen which worked

successfully via 27cos 3 or

24 7sin .

Part (b) proved demanding and there were candidates who did not understand the notation f . Some just integrated f

and others thought that f meant that the argument 2 in cos 2 should be replaced by and integrated

1 7cos

2 2 . A few candidates started by writing f d f d , treating as a constant. Another

error seen several times was 21 7f d cos 2 d

2 2

.

Many candidates correctly identified that integration by parts was necessary and most of these were able to demonstrate a

complete method of solving the problem. However there were many errors of detail, the correct manipulation of the negative signs

that occur in both integrating by parts and in integrating trigonometric functions proving particularly difficult. Only about 15% of

candidates completed the question correctly.

Question 7

This proved the most demanding question on the paper. Nearly all candidates could make some progress with the first three parts

but, although there were many, often lengthy attempts, success with part (d) and (e) was uncommon. Part (a) was quite well

answered, most finding AB or BA and writing down OA+λAB, or an equivalent. An equation does, however need an equals sign

and a subject and many lost the final A mark in this part by omitting the “ r ” from, say,

8 13 2 2 2 r i j k i j k . In part (b), those who realised that a magnitude or length was required were usually

successful. In part (c), nearly all candidates knew how to evaluate a scalar product and obtain an equation in cos , and so gain

the method marks, but the vectors chosen were not always the right ones and a few candidates gave the obtuse angle. Few made

any real progress with parts (d) and (e). As has been stated in previous reports, a clear diagram helps a candidate to appraise the

situation and choose a suitable method. In this case, given the earlier parts of the question, vector methods, although possible, are

not really appropriate to these parts, which are best solved using elementary trigonometry and Pythagoras’ theorem. Those who

did attempt vector methods were often very unclear which vectors were perpendicular to each other and, even the minority who

were successful, often wasted valuable time which sometimes led to poor attempts at question 8. It was particularly surprising to

see quite a large number of solutions attempting to find a vector, CX say, perpendicular to l, which never used the coordinates or

the position vector of C.

Gold 4: 12/12 54

Question 8

This proved by far the most difficult question on the paper and discriminated well for those candidates who were above the grade

A threshold for this paper. Only a few candidates were able to score above 8 or 9 marks on this question.

Many ‘fudged’ answers were seen in part (a). A more rigorous approach using the chain rule of d d d

d d d

h h V

t V t was required,

with candidates being expected to state d

d

V

t and

d

d

V

h (or its reciprocal). The constant of proportionality also proved to be a

difficulty in this and the following part.

Few convincing proofs were seen in part (b) with a significant number of candidates not understanding how to represent 3 -1400 cm s algebraically.

Only a minority of candidates were able to correctly separate the variables in part (c). Far too often, expressions including

d0.02 d

0.4

hh t were seen by examiners. There were a significant number of candidates who having written

dd

0.4

ht

k h

could not progress to the given answer by multiplying the integral on the left hand side by 5050

.

Despite struggling with the previous three parts, a majority of candidates were able to attempt part (d), although only a few

candidates were able to produce the correct final exact answer. A majority of candidates who attempted this part managed to

correctly obtain d

2 40d

hx

x and then use this and the given substitution to write down an integral in x. At this point a

significant number of candidates were unable to manipulate the expression 10x

kx

into an expression of the form

201k

x

. The converted limits 10x and 20,x caused an added problem for those candidates who progressed

further, with a significant number of candidates incorrectly applying 10x as their lower limit and 20x as their upper

limit.

A time of 6 minutes 26 seconds was rarity in part (e).



Qu Max score

Modal score

Mean %

ALL A* A B C D E U

1 7 7 69 4.83 6.81 6.31 5.42 4.11 2.74 1.62 0.74

2 6 35 2.09 3.51 1.66 0.88 0.46 0.20 0.09

3 8 36 2.89 4.90 2.22 1.12 0.56 0.27 0.12

4 9 50 4.50 6.53 4.29 2.57 2.15 0.50 0.18

5 8 54 4.28 6.10 3.27 2.02 1.15 0.45 0.29

6 10 44 4.38 9.04 6.25 3.83 2.06 1.03 0.44 0.17

7 14 41 5.75 8.24 5.36 3.73 2.46 1.55 0.89

8 13 27 3.50 5.79 2.58 1.05 0.69 0.31 0.15

75 43 32.22 47.63 28.63 17.54 11.24 5.34 2.63

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