paper reference(s) 6666/01 edexcel gce edexcel gce core mathematics c4

Bronze 1 This publication may only be reproduced in accordance with Edexcel Limited copyright policy. ©2009–2013 Edexcel Limited.

Paper Reference(s)

6666/01

Edexcel GCE Core Mathematics C4

Bronze Level B1

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 9 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

73 69 60 54 49 45

Bronze 2: 2/12 2

1.

Figure 1

Figure 1 shows part of the curve with equation y = 25.0e x . The finite region R, shown shaded

in Figure 1, is bounded by the curve, the x-axis, the y-axis and the line x = 2.

(a) Copy and complete the table with the values of y corresponding to x = 0.8 and x = 1.6.

x 0 0.4 0.8 1.2 1.6 2

y e0 e

0.08 e

0.72 e

2

(1)

(b) Use the trapezium rule with all the values in the table to find an approximate value for the

area of R, giving your answer to 4 significant figures.

(3)

June 2008


2.

Figure 1

Figure 1 shows the finite region R bounded by the x-axis, the y-axis and the curve with

equation y = 3 cos

3

x, 0 x

2

3.

The table shows corresponding values of x and y for y = 3 cos

3

x.

x 0 8

3

4

3

8

9

2

3

y 3 2.77164 2.12132 0

(a) Copy and complete the table above giving the missing value of y to 5 decimal places.

(1)

(b) Using the trapezium rule, with all the values of y from the completed table, find an

approximation for the area of R, giving your answer to 3 decimal places.

(4)

(c) Use integration to find the exact area of R.

(3)

June 2009

Bronze 2: 2/12 4

3.

Figure 1

Figure 1 shows the finite region R bounded by the x-axis, the y-axis, the line x = 2

and the

curve with equation

y = 1

sec2

x

, 0 ≤ x ≤ 2

The table shows corresponding values of x and y for y = 1

sec2

x

.

x 0 6

3

2

y 1 1.035276 1.414214

(a) Complete the table above giving the missing value of y to 6 decimal places.

(1)

(b) Using the trapezium rule, with all of the values of y from the completed table, find an

approximation for the area of R, giving your answer to 4 decimal places.

(3)

Region R is rotated through 2π radians about the x-axis.

(c) Use calculus to find the exact volume of the solid formed.

(4)

June 2013


4. The curve C has the equation ye–2x

= 2x + y2.

(a) Find x

y

d

d in terms of x and y.

(5)

The point P on C has coordinates (0, 1).

(b) Find the equation of the normal to C at P, giving your answer in the form ax + by + c = 0,

where a, b and c are integers.

(4)

June 2009

5. A curve is described by the equation

x3 4y

2 = 12xy.

(a) Find the coordinates of the two points on the curve where x = –8.

(3)

(b) Find the gradient of the curve at each of these points.

(6)

January 2008

Bronze 2: 2/12 6

6.

Figure 2

Figure 2 shows a sketch of the curve C with parametric equations

x = 4 sin

6

t , y = 3 cos 2t, 0 t < 2.

(a) Find an expression for x

y

d

d in terms of t.

(3)

(b) Find the coordinates of all the points on C where x

y

d

d = 0.

(5)

January 2012


7. The curve C has parametric equations

x = ln t, y = t2 −2, t > 0.

Find

(a) an equation of the normal to C at the point where t = 3,

(6)

(b) a cartesian equation of C.

(3)

Figure 1

The finite area R, shown in Figure 1, is bounded by C, the x-axis, the line x = ln 2 and the line

x = ln 4. The area R is rotated through 360° about the x-axis.

(c) Use calculus to find the exact volume of the solid generated.

(6)

January 2011

Bronze 2: 2/12 8

8. I =

5

2

d)1(4

1x

x.

(a) Given that y = )1(4

1

x, copy and complete the table below with values of y

corresponding to x = 3 and x = 5 . Give your values to 4 decimal places.

x 2 3 4 5

y 0.2 0.1745

(2)

(b) Use the trapezium rule, with all of the values of y in the completed table, to obtain an

estimate of I, giving your answer to 3 decimal places.

(4)

(c) Using the substitution x = (u − 4)2 + 1, or otherwise, and integrating, find the exact value

of I.

(8)

January 2011

TOTAL FOR PAPER: 75 MARKS

END


Question

Number Scheme Marks

1. (a)

B1 (1)

(b) 0 0.08 0.32 0.72 1.28 21Area 0.4 ; e 2 e e e e e

2

B1; M1

0.2 24.61203164... 4.922406... 4.922 (4sf) A1 (3)

(4 marks)

2. (a) 1.14805 awrt 1.14805 B1 (1)

(b) 1 3

... 2 8

A

B1

... 3 2 2.77164 2.12132 1.14805 0 0 can be implied M1

3

3 2 2.77164 2.12132 1.1480516

ft their (a) A1ft

3

15.08202 ... 16

8.884 cao A1 (4)

(c)

3sin3

3cos d13

3

x

xx

M1 A1

9sin3

x

3

2

0

9sin 9 0 93

xA

cao A1 (3)

(8 marks)

x 0 0.4 0.8 1.2 1.6 2

y 0e 0.08e 0.32e 0.72e 1.28e 2e

or y 1 1.08329

…

1.37713

…

2.05443

… 3.59664…

7.38906

…

Bronze 2: 2/12 10

Question

Number Scheme Marks

3. (a) 1.154701 B1 cao

(1)

(b) 1

Area ; 1 2 1.035276 + their 1.154701 1.4142142 6

B1; M1

6.794168 1.778709023... 1.778712

(4 dp) 1.7787 or awrt 1.7787 A1

(3)

(c)

22

0

sec d2

xV x

For

2

sec2

x

.

Ignore limits and dx .

Can be implied.

B1

2

0

2tan2

x

tan2

x

M1

2 tan2

x

or

equivalent

A1

2 2

A1 cao

cso

(4) [8]

4. (a) 2 2d de 2 e 2 2

d d

x xy yy y

x x

A1 correct RHS M1 A1

2 2 2d de e 2 e

d d

x x xyy y

x x

B1

2 2de 2 2 2 e

d

x xyy y

x

M1

2

2

d 2 2 e

d e 2

x

x

y y

x y

A1 (5)

(b) At P, 0

0

d 2 2e4

d e 2

y

x

M1

Using 1mm 1

4m M1

1

1 04

y x M1

4 4 0x y or any integer multiple A1 (4)

(9 marks)


Question

Number Scheme Marks

5. (a) 3 24 12x y xy ( eqn )

2

2

8 512 4 12( 8)

512 4 96

x y y

y y

Substitutes 8x (at least once) into *

to obtain a three term quadratic in y .

Condone the loss of 0.

M1

2

2

4 96 512 0

24 128 0

y y

y y

( 16)( 8) 0y y

24 576 4(128)

2y

An attempt to solve the quadratic in y

by either factorising or by the formula

or by completing the square.

dM1

16 or 8.y y Both 16 and 8.y y

or 8, 8 and 8, 16 . A1

[3]

(b) 2d d d

3 8 ; 12 12d d d

y y yx y y x

x x x

Differentiates implicitly to include

either d d

d dor 12

y y

x xky x . Ignore d

d...

y

x

M1

Correct LHS equation; A1;

Correct application of product rule (B1)

2d 3 12

d 12 8

y x y

x x y

not necessarily required.

d 3(64) 12(8) 96@ 8, 8 , 3,

d 12( 8) 8(8) 32

d 3(64) 12(16) 0@ 8, 16 , 0.

d 12( 8) 8(16) 32

y

x

y

x

Substitutes 8x and at least one of

their y-values to attempt to find any one

of d

d.

y

x

dM1

One gradient found. A1

Both gradients of -3 and 0 correctly

found. A1 cso

[6]

9 marks

Bronze 2: 2/12 12

Question

Number Scheme Marks

6. 4sin , 3cos2 , 0 26

x t y t t

(a) d

4cosd 6

xt

t

,

d6sin 2

d

yt

t B1 B1

So,

d 6sin 2

d4cos

6

y t

xt

B1 oe

(3)

(b) d

0 6sin 2 0d

yt

x M1 oe

@ 0, 4sin 2 , 3cos 0 3 (2, 3)

6

2 4 3@ , 4sin , 3cos 3 (2 3 , 3)

2 3 2

7@ , 4sin 2 , 3cos 2 3 ( 2, 3)

6

3 5 4( 3 )@ , 4sin , 3cos 3 3 ( 2 3 , 3)

2 3 2

t x y

t x y

t x y

t x y

M1

A1 A1 A1

(5)

(8 marks)


Question

Number Scheme Marks

7.

(a) d 1

d

x

t t ,

d2

d

yt

t

2d2

d

yt

x M1 A1

Using 1mm , at 3t

1

18m M1 A1

1

7 ln318

y x M1 A1 (6)

(b) ln exx t t B1

2e 2xy M1 A1 (3)

(c) 2

2e 2 dxV x M1

2

2 4 2e 2 d e 4e 4 dx x xx x M1

4 2e 4e

44 2

x x

x M1 A1

ln 4

4 2

ln 2

e 4e4 64 32 4ln 4 4 8 4ln 2

4 2

x x

x

M1

36 4ln 2 A1

(6)

[15]

Bronze 2: 2/12 14

Question

Number Scheme Marks

8.

(a) 3 0.1847x y awrt B1

5 0.1667x y awrt or 16

B1

(2)

(b) 1

0.2 0.1667 2 0.1847 0.17452

I B1 M1 A1ft

0.543 0.542 or 0.543 A1 (4)

(c) d

2 4d

xu

u B1

1 1

d 2 4 d4 1

x u uux

M1

8

2 duu

A1

2 8lnu u M1 A1

2 5x u , 5 6x u B1

6

52 8ln 12 8ln 6 10 8ln 5u u M1

5

2 8ln6

A1

(8)

[14]


Question 1

A significant majority of candidates were able to score full marks on this question. In part (a), very few candidates failed to

find either one or both of the y-coordinates required. In part (b), some candidates incorrectly used the formula b a

nh

, with 6n instead of 5n to give the width of each trapezium as 13

. Many candidates, however, were able to look at

the given table and deduce the value of h. A few candidates wrote down 0e as 0 instead of 1. Nearly all answers were given

to 4 significant figures as requested in the question.

Question 2

Most candidates could gain the mark in part (a) although 2.99937, which arises from the incorrect angle mode, was seen

occasionally. The main error seen in part (b) was finding the width of the trapezium incorrectly, 3

10

being commonly seen

instead of 3

8

. This resulted from confusing the number of values of the ordinate, 5, with the number of strips, 4. Nearly all

candidates gave the answer to the specified accuracy. In part (c), the great majority of candidates recognised that they needed

to find 3cos d3

xx

and most could integrate correctly. However sin x , 9sin x , 3sin3

x

, 9sin3

x

,

sin3

x

and 3sin

3

x

were all seen from time to time. Candidates did not seem concerned if their answers to part

(b) and part (c) were quite different, possibly not connecting the parts of the question. Despite these difficulties, full marks

were common and, generally, the work on these topics was sound.

Question 3 This was a straightforward question with about 55% of candidates gaining all 8 marks with an overwhelmingly majority of

candidates realising that they needed to use radians in this question.

Although most candidates correctly computed 1.154701 in part (a), a significant number wrote 1.154700, suggesting that

truncation rather than rounding was applied by some at this stage. Also an answer of 1.000004 was occasionally seen (a

consequence of having the calculator in degrees mode).

In applying the trapezium rule in part (b), a small minority of candidates multiplied 1

2 by

8

instead of

1

2 by .

6

Whilst

the table of values clearly shows an interval width of 6

, the application of a formula

b ah

n

with 4n instead of

3n sometimes caused this error. Other errors included the occasional bracketing mistake, use of the x-value of 0 rather

than the ordinate of 1, and the occasional calculation error following a correctly written expression.

In part (c), the majority of candidates were able to apply volume formula 2dy x , although a number of candidates used

incorrect formulae such as 22 dy x or

2dy x or even d .y x Few candidates incorrectly wrote

2

sec2

x

as

either

2

sec4

x

or

22sec

4

x

. A minority of candidates integrated 2sec

2

x

incorrectly to give expressions such as

1tan

2 2

x

or tan .2

x

The majority of candidates, however, were able to apply the limits correctly and recognised the

need to give an exact final answer.

Question 4

Bronze 2: 2/12 16

Work on this topic has shown a marked improvement and the median mark scored by candidates on this question was 8 out

of 9. The only errors frequently seen were in differentiating 2e xy

implicitly with respect to x. A few candidates failed to

read the question correctly and found the equation of the tangent instead of the normal or failed to give their answer to part

(b) in the form requested.

Question 5

This question was generally well done with many candidates scoring at least seven or eight of the nine marks available.

In part (a), the majority of candidates were able to use algebra to gain all three marks available with ease. It was

disappointing, however, to see a significant minority of candidates at A2 level who were unable to correctly substitute

8y into the given equation or solve the resulting quadratic to find the correct values for y.

In part (b), implicit differentiation was well handled, with most candidates appreciating the need to apply the product rule to

the 12xy term although errors in sign occurred particularly with those candidates who had initially rearranged the given

equation so that all terms were on the LHS. A few candidates made errors in rearranging their correctly differentiated

equation to make d

d

y

x the subject. Also some candidates lost either one or two marks when manipulating their correctly

substituted d

d

y

x expressions to find the gradients.

Question 6

This question was generally well answered with about 40% of the candidature gaining all 8 marks.

Whilst a large number of fully correct solutions were seen in part (a), there were a significant number of candidates who

struggled to differentiate sine and cosine functions, with expressions such as d

4cosd 6

xt

t

or

d 3sin 2

d 2

yt

t being encountered frequently. Most candidates were able to apply the chain rule to find an expression for

d

d

y

x, although the application of

d d

d d

y x

t t was occasionally seen. A small proportion of candidates used the compound

angle formula to rewrite x as 2 3 sin 2cos ,t t or likewise the double angle formulae to rewrite y as either

23(2cos 1)t or 2 23(cos sin )t t or

23(1 2sin )t before going on to find theird

d

y

x.

Candidates found part (b) more challenging and a variable range of marks was awarded in this part. Although a few

candidates could not proceed further from setting their d

d

y

x to zero, most candidates appreciated that the numerator from

their d

d

y

x expression needed to be equated to zero, so resulting in the first method mark. A number of candidates who solved

sin 2t = 0, found only one value of t (usually 0t ) and then one point (usually (2, 3) ). A surprisingly large number of

candidates found all four correct values of t, but did not realise that they needed to use these values in order to find four sets

of coordinates for ( , ).x y Candidates who found more than one value for t often relied on the symmetry of the diagram to

find all four points rather than a full solution, and this was permitted. A surprising number of stronger candidates did not

relate the diagram to the question in part (b). These candidates stopped at finding only two or three sets of coordinates when

it was clear from the diagram that there was a total of four points where d

0d

y

x . Few candidates also set the denominator

equal to zero and used the resulting values of t to find erroneous coordinates.

Question 7


Although there were many correct solutions to part (a), a surprising number of candidates made mistakes in establishing d

d

y

x

from their d

d

y

t and their

d

d

x

t. Both 2 and

32t were seen and, in many cases, the method used was not clearly shown and this

resulted in the loss of both the method and the accuracy marks. If d

d

y

x was correctly found, the majority were able to

complete this part correctly. A significant number, however, failed to read the question and gave the equation of the tangent

rather than that of the normal. Part (b) was well done and nearly all could eliminate the parameter. Quite a number of

candidates thought that (ex)2 was 2

ex and this often caused a major loss of marks in part (c).

In part (c), the majority of candidates knew the volume formula but the attempts at integration were of a very variable quality.

Many used the lead given in part (b) but the resulting squaring out of the brackets was often incorrect. Examples of errors

seen are 22

2 4e ex x and 2

2 4e 2 e 4x x or 4e 4x . There were also attempts at direct integration, for

example

32

22

e 2e 2 d

3

x

x x

. Those who used parameters often made similar mistakes and sometimes the d

d

x

t

was omitted. The choice of limits also gave some difficulty; those who integrated using the variable x using the t limits and

vice versa. Despite these frequent mistakes, there were many completely correct solutions.

Question 8

Parts (a) and (b) were usually fully correct and few lost marks through failing to work to the accuracy specified in the

question. The trapezium rule is well known and the only error commonly seen was obtaining an incorrect width of an

individual trapezium.

Part (c) proved more demanding. Many got off to a bad start. The substitution was deliberately given in the form

2

4 1x u so that the essential d

2 4d

xu

u could be found easily. Many, however, rearranged the

substitution and obtained 12

d 11

d 2

xx

u

and the resulting algebraic manipulations often proved beyond candidates.

Those who did complete the substitution often failed to complete the definite integral. An unexpected difficulty was that a

number of candidates failed, in the context, to simplify 4 4u to u.

Many did not see that 2 8u

u

reduced to

82

u and embarked upon complicated solutions using integration by parts

which, although theoretically possible, were rarely completed. The choice of limits also gave some difficulties. The

convention is used that a surd is taken as the positive square root. If this were not the case the expression given at the head of

the question would be ambiguous. Some however produce limits resulting from negative square roots, 3 and 2, as well as the

correct 5 and 6, and did not know which to choose. As in question 6, there was also some confusion in choosing the limits,

some choosing the x limits when the u limits were appropriate and vice versa. Despite these difficulties, nearly 34% of the

candidates gained full marks for this question.

Statistics for C4 Practice Paper Bronze Level B1

Mean score for students achieving grade:

Qu Max score

Modal score

Mean %

ALL A* A B C D E U

1 4 86 3.43 3.77 3.53 3.27 2.96 2.62 1.98

2 8 78 6.24 7.53 6.69 5.57 4.33 3.17 1.95

3 8 8 81 6.44 7.83 7.51 6.98 6.13 5.01 3.68 1.94

4 9 74 6.68 8.20 7.21 6.01 4.46 2.87 1.25

5 9 83 7.50 8.55 7.84 6.81 5.98 4.85 2.86

6 8 69 5.53 7.62 6.51 5.21 4.09 3.28 2.54 1.20

7 15 75 11.27 14.75 12.91 10.99 8.77 6.67 4.26 2.17

8 14 75 10.54 13.82 11.96 9.77 7.64 6.04 5.14 3.45

Bronze 2: 2/12 18

75 77 57.63 66.94 58.22 48.29 38.73 29.13 16.80 Paper Reference(s)

6666/01


Bronze Level B2

Time: 1 hour 30 minutes Materials required for examination Items included with question papers







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

70 63 57 53 48 42


1. Given

f(x) = (2 + 3x)–3

, |x| < 3

2,

find the binomial expansion of f(x), in ascending powers of x, up to and including the term

in x3.

Give each coefficient as a simplified fraction.

(5)

January 2013

Bronze 3: 3/12 20

2.

Figure 1

Figure 1 shows a sketch of the curve with equation y = x ln x, x 1. The finite region R,

shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 4.

The table shows corresponding values of x and y for y = x ln x.

x 1 1.5 2 2.5 3 3.5 4

y 0 0.608 3.296 4.385 5.545

(a) Copy and complete the table with the values of y corresponding to x = 2 and x = 2.5,

giving your answers to 3 decimal places.

(2)

(b) Use the trapezium rule, with all the values of y in the completed table, to obtain an

estimate for the area of R, giving your answer to 2 decimal places.

(4)

(c) (i) Use integration by parts to find

xxx dln .

(ii) Hence find the exact area of R, giving your answer in the form 4

1(a ln 2 + b), where

a and b are integers.

(7)

January 2010


3. f(x) = )3)(1)(12(

24

xxx

x =

)12( x

A +

)1( x

B +

)3( x

C .

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

(4)

(b) (i) Hence find

xx d)(f .

(3)

(ii) Find

2

0

d)(f xx in the form ln k, where k is a constant.

(3)

June 2009

4. )12)(12(

)14(2 2

xx

x A +

)12( x

B +

)12( x

C.

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

(4)

(b) Hence show that the exact value of

2

1

2

)12)(12(

)14(2

xx

x dx is 2 + ln k, giving the value of

the constant k.

(6)

June 2007

5. (a) Expand )34(

1

x, where x <

34 , in ascending powers of x up to and including the term

in x2. Simplify each term.

(5)

(b) Hence, or otherwise, find the first 3 terms in the expansion of )34(

8

x

x

as a series in

ascending powers of x.

(4)

June 2008

Bronze 3: 3/12 22

6.

Figure 3

Figure 3 shows a sketch of part of the curve with equation y = 1 – 2 cos x, where x is

measured in radians. The curve crosses the x-axis at the point A and at the point B.

(a) Find, in terms of , the x coordinate of the point A and the x coordinate of the point B.

(3)

The finite region S enclosed by the curve and the x-axis is shown shaded in Figure 3. The

region S is rotated through 2 radians about the x-axis.

(b) Find, by integration, the exact value of the volume of the solid generated.

(6)

January 2013


7.

Figure 1

Figure 1 shows part of the curve with equation y = (tan x). The finite region R, which is

bounded by the curve, the x-axis and the line x = 4

, is shown shaded in Figure 1.

(a) Given that y = (tan x), copy and complete the table with the values of y corresponding to

x = 16

,

8

and

16

3, giving your answers to 5 decimal places.

x 0 16

8

16

3

4

y 0 1

(3)

(b) Use the trapezium rule with all the values of y in the completed table to obtain an

estimate for the area of the shaded region R, giving your answer to 4 decimal places.

(4)

The region R is rotated through 2 radians around the x-axis to generate a solid of revolution.

(c) Use integration to find an exact value for the volume of the solid generated.

(4)

June 2007

Bronze 3: 3/12 24

8. (a) Find

yy d)34( 2

1

.

(2)

(b) Given that y =1.5 at x = – 2, solve the differential equation

x

y

d

d =

2

)34(

x

y ,

giving your answer in the form y = f(x).

(6)

June 2011


END


Question

Number

Scheme

Marks

1. 3 3

33 3 1 3(2 3 ) 2 1 1

2 8 2

x xx

3(2) or

1

8 B1

2 31 ( 3)( 4) ( 3)( 4)( 5)

1 ( 3)( ) ( ) ( ) ...8 2! 3!

k x k x k x

see notes M1 A1

2 31 3 ( 3)( 4) 3 ( 3)( 4)( 5) 3

1 ( 3) ...8 2 2! 2 3! 2

x x x

2 31 9 27 135

1 ; ...8 2 2 4

x x x

See notes below!

2 31 9 27 135; ...

8 16 16 32x x x A1; A1

[5]

5

Q2 (a) 1.386, 2.291 awrt 1.386, 2.291 B1 B1 (2)

(b) 1

0.5 ... 2

A B1

... 0 2 0.608 1.386 2.291 3.296 4.385 5.545 M1

0.25 0 2 0.608 1.386 2.291 3.296 4.385 5.545 ft their (a) A1ft

0.25 29.477 ... 7.37 cao A1 (4)

(c)(i) 2 2 1

ln d ln d2 2

x xx x x x x

x

M1 A1

2

ln d2 2

x xx x

2 2

ln2 4

x xx C M1 A1

(ii) 4

2 2

1

1ln 8ln 4 4

2 4 4

x xx

M1

15

8ln 44

15

8 2ln 24

ln 4 2ln 2 seen or implied M1

1

64ln 2 154

64, 15a b A1 (7)

[13]

Bronze 3: 3/12 26

Question

Number Scheme Marks

3. (a)

4 2f

2 1 1 3 2 1 1 3

x A B Cx

x x x x x x

4 2 1 3 2 1 3 2 1 1x A x x B x x C x x M1

A method for evaluating one constant M1

12

x , 512 2

5 4A A any one correct constant A1

1x , 6 1 2 3B B

3x , 10 5 2 1C C all three constants correct A1 (4)

(b) (i) 4 3 1

d2 1 1 3

xx x x

4

ln 2 1 3ln 1 ln 32

x x x C A1 two ln terms correct M1 A1ft

All three ln terms correct and “+C” ; ft constants A1ft (3)

(ii) 2

02ln 2 1 3ln 1 ln 3x x x

2ln5 3ln3 ln5 2ln1 3ln1 ln3 M1

3ln 5 4ln 3

3

4

5ln

3

M1

125

ln81

A1 (3)

(10 marks)


Question

Number Scheme Marks

4. (a) A method of long division gives,

22(4 1) 4

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

x

x x x x

2A B1

4

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

B C

x x x x

4 (2 1) (2 1)B x C x

or their remainder, (2 1) (2 1)Dx E B x C x

Forming any one of these two

identities. Can be implied. M1

Let 12

x , 4 2 2B B

See note below

Let 12

x , 4 2 2C C either one of 2B or 2C A1

both B and C correct A1

[4]

4. (b) 22(4 1) 2 2

d 2 d(2 1)(2 1) (2 1) (2 1)

xx x

x x x x

2 22 2

2 ln(2 1) ln(2 1) ( )x x x c

Either ln(2 1)p x or ln(2 1)q x

or either ln2 1p x or ln2 1q x

M1

A Ax B1

2 22 2ln(2 1) ln(2 1)x x

or ln(2 1) ln(2 1)x x

See note below.

A1

cso & aef

2

22

1

1

2(4 1)d 2 ln(2 1) ln(2 1)

(2 1)(2 1)

xx x x x

x x

4 ln5 ln3 2 ln3 ln1

Substitutes limits of 2 and 1

and subtracts the correct way

round. (Invisible brackets

okay.)

depM1

2 ln3 ln3 ln5

3(3)

2 ln5

Use of correct product

(or power) and/or quotient laws

for logarithms to obtain a single

logarithmic term for their

numerical expression.

M1

9

2 ln5

92 ln

5

A1

Or 5

92 ln

and k stated as 9

5 . [6]

10 marks

Bronze 3: 3/12 28

Question

Number Scheme Marks

5. (a)

1 12 2

1122

1 3 1 3(4 3 ) 4 1 1

4 2 4(4 3 )

x xx

x

B1

3122 21 1

2 2

( )( )1 ( )(** ); (** ) ...

2!x x

with ** 1

M1; A1 ft

31

22 23 312 4 4

( )( )11 ( )( ) ( ) ...

2 2!

x x

23 2712 8 128

1 ; ...x x A1 A1 (5)

(b) 21 3 27

( 8) ...2 16 256

x x x

M1

2312 16

23 272 32

.....

4 .....

x x

x x

M1

233

4 2 ; ...32

x x A1; A1 (4)

(9 marks)


Question

Number Scheme Marks

6. (a) 0 1 2cos 0y x 1 2cos 0x , seen or implied. M1

5

,3 3

x

At least one correct value of x. (See notes). A1

Both 5

and3 3

A1 cso

[3]

(b) V

5

3 2

3

(1 2cos ) dx x

For 2(1 2cos )x .

Ignore limits and dx

B1

2(1 2cos ) dx x2(1 4cos 4cos )dx x x

1 cos21 4cos 4 d

2

xx x

2cos 2 2cos 1x x

See notes. M1

3 4cos 2cos2 dx x x

2sin 23 4sin

2

xx x

Attempts 2y to give any two of

,A Ax cos sinB x B x or

cos2 sin2x x .

M1

Correct integration. A1

10 23 35 5

3 3 3 3

2sin 2sin3 4sin 3 4sin

2 2V

Applying limits

the correct way

round. Ignore

.

M1

3 35 2 3 2 3

2 2

18.3060... 0.5435... 17.7625 55.80

24 3 3 or 4 3 3 Two term exact answer. A1

[6] 9

Bronze 3: 3/12 30

Question

Number Scheme Marks

7. (a)

x 0 16

8 3

16

4

y 0 0.44599592

7…

0.64359425

2…

0.817421946

… 1

0.446 or awrt

0.44600 B1

awrt 0.64359 B1

awrt 0.81742 B1

[3]

(b)

1

Area ; 0 2 0.44600 0.64359 0.81742 12 16

Outside brackets 12 16

or 32

For structure of

trapezium rule

............. ;

Correct expression

inside brackets which

all must be multiplied

by2h .

B1

M1

A1

4.81402... 0.472615308... 0.472632

(4dp) for seeing 0.4726 A1 cao

[4]

(c)

Volume

4 4

2

0 0

tan d tan dx x x x

2

tan x dx or

tan x dx

Can be implied.

Ignore limits and

M1

4

0lnsec x

or

4

0lncos x

tan lnsecx x

or tan lncosx x

A1

4lnsec lnsec0

or

4

lncos lncos0

dM1

1

2

1 11

ln ln ln 2 ln1

or

1

2ln ln 1

ln 2

or

2

2ln

or 12

ln2 or

1

2ln

or

12 2ln

ln 2or

2

2ln

or 12

ln2 or

1

2ln

or 1

2 2ln

A1 aef

must be exact. [4]

11 marks


Question

Number Scheme Marks

8.

(a)

121

2

12

4 34 3 d

4

yy x C

M1 A1 (2)

121

24 3y C

(b) 2

1 1d d

4 3y x

y x

B1

12 24 3 d dy y x x

12

1 14 3

2y C

x M1

Using 2,1.5 12

1 14 1.5 3

2 2C

M1

leading to 1C A1

12

1 14 3 1

2y

x

12

24 3 2y

x M1

21 2 3

24 4

yx

or equivalent A1 (6)

[8]

Bronze 3: 3/12 32

Question 1

This question was generally well answered with about 70% of candidates obtaining all of the 5 marks available.

A minority of candidates were unable to carry out the first step of writing (2 + 3x)–3 as

31 3

18 2

x

, with the

1

8outside

the brackets usually written incorrectly as either 2 or 1 . Many candidates were able to use a correct method for expanding a

binomial expression of the form (1 )nax . A variety of incorrect values of a were seen, with the most common being

either 3 , 2

3 or 1. Some candidates, having correctly expanded

33

1 ,2

x

forgot to multiply their expansion by

1.

8

Errors seen included sign errors, bracketing errors, missing factorials (for example, 2! or 3!) and simplification errors.

Question 2

Nearly all candidates gained both marks in part (a). As is usual, the main error seen in part (b) was finding the width of the

trapezium incorrectly. There were fewer errors in bracketing than had been noted in some recent examinations and nearly all

candidates gave the answer to the specified accuracy. The integration by parts in part (c) was well done and the majority of

candidates had been well prepared for this topic.

Some failed to simplify

2 1d

2

xx

x

to d2

xx

and either gave up or produced

313

2

x

x.

In evaluating the definite integral some either overlooked the requirement to give the answer in the form 1

ln 24

a b or

were unable to use the appropriate rule of logarithms correctly.

Question 3

Part (a) was well done with the majority choosing to substitute values of x into an appropriate identity and obtaining the

values of A, B and C correctly. The only error commonly seen was failing to solve 5

54

A for A correctly. Those who

formed simultaneous equations in three unknowns tended to be less successful. Any incorrect constants obtained in part (a)

were followed through for full marks in part (b)(i). Most candidates obtained logs in part (b)(i). The commonest error was,

predictably, giving 4

d 4ln 2 12 1

x xx

, although this error was seen less frequently than in some previous

examinations. In indefinite integrals, candidates are expected to give a constant of integration but its omission is not

penalised repeatedly throughout the paper. In part (b)(ii) most applied the limits correctly although a minority just ignored the

lower limit 0. The application of log rules in simplifying the answer was less successful. Many otherwise completely correct

solutions gave 3ln 3 as ln 9 and some “simplified” 3ln 5 4ln 3 to 3 5

ln4 3

.


Question 5

This question was also generally well tackled with about 50% of candidates obtaining at least 8 of the 9 marks available. A

substantial minority of candidates were unable to carry out the first step of writing 1

(4 3 )x as

121 3

12 4

x

, with the

12

outside the brackets usually written incorrectly as either 2 or 4. Many candidates were able to use a correct method for

expanding a binomial expression of the form (1 )nax . A variety of incorrect values of a and n, however, were seen by

examiners with the most common being a as 34

, 3 and 3 and n as 12, 1 and 2. Some candidates, having correctly

expanded

123

1 ,4

x

forgot to multiply their expansion by 1

2. As expected, sign errors, bracketing errors and

simplification errors were also seen in this part. A significant minority of candidates expanded as far as 3,x and were not

penalised on this occasion.

In part (b), most candidates realised that they needed to multiply ( 8)x by their expansion from part (a) although a small

minority attempted to divide ( 8)x by their expansion. A surprising number of candidates attempted to expand (x+8) to

obtain a power series. Other candidates omitted the brackets around 8x although they progressed as if “invisible”

brackets were there. The mark scheme allowed candidates to score 2 marks out of 4 even if their answer in (a) was incorrect

and many candidates were able to achieve this.

Bronze 3: 3/12 34

Question 6

This question was answered well across all abilities.

In Q6(a), most candidates solved 1 2cos 0x to obtain .3

x

A number of candidates, however, struggled to find

the second value of 5

.3

x

A variety of incorrect second values were seen, the most common being 2 4

,3 3

or

7.

3

In Q6(b), the majority of candidates were able to apply volume formula of 2dy x , although a number of candidates

used incorrect formulae such as 22 dy x or

2dy x or even d .y x Some candidates incorrectly expanded

2(1 2cos )x as either 21 4cos 2x or

21 4cos 2cosx x or 21 4cos 4cosx x . Others

attempted to integrate 2(1 2cos )x directly to give incorrect expressions such as

3(1 2cos ).

3(2sin )

x

x

When the integral

included a term in 2cos x , a few candidates integrated this incorrectly to give expressions such as

3cos

3

x or

3cos

3sin

x

x.

The majority, however, realised the need for using 2cos2 2cos 1x x . Whilst this double angle formula was

generally correctly quoted, this did not always lead to a correct expression for integration as a result of sign or coefficient

errors. The integration of an expanded trigonometric expression was generally well done, as was the substitution of the limits

found in Q6(a). Candidates are advised to show some evidence of how they have substituted their limits, because this allows

some credit to be given if errors occur later in the calculation. Although this question specified an exact answer, decimal

answers were occasionally given.

Question 8

The majority of candidates realised that the answer to part (a) was of the form 124 3k y and although the value 2k

was common, most did obtain12

k . In part (b), the majority of candidates knew that they needed to separate the variables,

although this was not always done correctly. Those who separated correctly usually were able to integrate 2

1

x correctly,

although 2ln x was seen from time to time. A significant number of candidates did not use a constant of integration and


could gain no further marks in the question. It is disappointing to report than many otherwise correct solutions were spoilt by

elementary algebraic errors. Many candidates obtained a correct expression, for example, 12

24 3 2y

x but were

unable to make y the subject of the formula correctly. For examples, 2

44 3 4y

x and, even,

24 3 2y

x

were often seen.



Qu Max score

Modal score

Mean %

ALL A* A B C D E U

1 5 5 88 4.42 4.90 4.71 4.49 4.22 3.72 3.33 2.46

2 13 77 10.06 11.88 9.64 8.27 6.96 4.98 3.39

3 10 77 7.66 9.01 7.97 7.00 5.97 4.76 3.04

4 10 73 7.31 8.92 7.56 6.44 5.17 3.96 2.28

5 9 76 6.80 8.12 7.16 6.26 5.13 3.70 1.89

6 9 9 67 6.02 8.47 7.04 5.84 4.45 3.41 2.76 1.38

7 11 75 8.23 9.84 8.60 7.43 6.09 4.88 3.11

8 8 52 4.19 7.32 5.79 4.04 2.50 1.48 0.91 0.52

75 73 54.69 65.31 55.30 46.57 37.93 29.28 18.07

Bronze 3: 3/12 36

Paper Reference(s)

6666/01


Bronze Level B3


















A* A B C D E

69 61 54 49 43 37

P41860A 37 Turn over

1. A curve C has the equation y2 – 3y = x

3 + 8.

(a) Find x

y

d


(4)

(b) Hence find the gradient of C at the point where y = 3.

(3)

January 2009

2. (a) Use integration by parts to find

xx x de .

(3)

(b) Hence find

xx x de2 .

(3)

June 2008

3. (a) Expand

2)52(

1

x, x <

5

2,

in ascending powers of x, up to and including the term in x2, giving each term as a

simplified fraction.

(5)

Given that the binomial expansion of 2)52(

2

x

kx

, x <

5

2, is

2

1 + x

4

7 + Ax

2 + . . .,

(b) find the value of the constant k,

(2)

(c) find the value of the constant A.

(2)

January 2012

Bronze 3: 3/12 38

4. A curve has equation 3x2 – y

2 + xy = 4. The points P and Q lie on the curve. The gradient of

the tangent to the curve is 38 at P and at Q.

(a) Use implicit differentiation to show that y – 2x = 0 at P and at Q.

(6)

(b) Find the coordinates of P and Q.

(3)

June 2008


5.

Figure 3

Figure 3 shows a sketch of the curve with equation y = )cos1(

2sin2

x

x

, 0 x

2

.

The finite region R, shown shaded in Figure 3, is bounded by the curve and the x-axis.

The table below shows corresponding values of x and y for y = )cos1(

2sin2

x

x

.

x 0 8

4

8

3

2

y 0 1.17157 1.02280 0

(a) Complete the table above giving the missing value of y to 5 decimal places.

(1)

(b) Use the trapezium rule, with all the values of y in the completed table, to obtain an

estimate for the area of R, giving your answer to 4 decimal places.

(3)

(c) Using the substitution u = 1 + cos x, or otherwise, show that

xx

xd

)cos1(

2sin2

= 4 ln (1 + cos x) – 4 cos x + k,

where k is a constant.

(5)

(d) Hence calculate the error of the estimate in part (b), giving your answer to 2 significant

figures.

(3)

January 2012

Bronze 3: 3/12 40

6. Relative to a fixed origin O, the point A has position vector 21i – 17j + 6k and the point B has

position vector 25i – 14j + 18k.

The line l has vector equation

6

10 1

a

b c

r

where a, b and c are constants and λ is a parameter.

Given that the point A lies on the line l,

(a) find the value of a.

(3)

Given also that the vector AB is perpendicular to l,

(b) find the values of b and c,

(5)

(c) find the distance AB.

(2)

The image of the point B after reflection in the line l is the point B´.

(d) Find the position vector of the point B´.

(2)

June 2013 (R)


7.

Figure 3

Figure 3 shows a sketch of part of the curve with equation y = 2

1

x ln 2x.

The finite region R, shown shaded in Figure 3, is bounded by the curve, the x-axis and the

lines x = 1 and x = 4.

(a) Use the trapezium rule, with 3 strips of equal width, to find an estimate for the area of R,

giving your answer to 2 decimal places.

(4)

(b) Find xxx d2ln2

1

.

(4)

(c) Hence find the exact area of R, giving your answer in the form a ln 2 + b, where a and b

are exact constants.

(3)

June 2012

Bronze 4: 4/12 42

8. A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains

constant at 3 °C and t minutes after the bottle is placed in the refrigerator the temperature of

the water in the bottle is °C.

The rate of change of the temperature of the water in the bottle is modelled by the differential

equation

td

d =

125

)3( .

(a) By solving the differential equation, show that

= Ae–0.008t

+ 3,

where A is a constant.

(4)

Given that the temperature of the water in the bottle when it was put in the refrigerator

was 16 °C,

(b) find the time taken for the temperature of the water in the bottle to fall to 10 °C, giving

your answer to the nearest minute.

(5)

January 2013


END


Question Number

Scheme Marks

1 (a) C: 2 33 8y y x

2d d d

2 3 3d d d

y y yy x

x x x

Differentiates implicitly to include either

d

d

yky

x or

d3

d

y

x . (Ignore

d

d

y

x

.)

M1

Correct equation. A1

2d2 3 3

d

yy x

x

A correct (condoning sign error) attempt to

combine or factorise their ‘d d

2 3d d

y yy

x x ’.

Can be implied.

M1

2d 3

d 2 3

y x

x y

23

2 3

x

y A1 oe

(4)

(b) 33 9 3(3) 8y x Substitutes 3y into C. M1

3 8 2x x Only 2x A1

d 3(4) d

( 2,3) 4d 6 3 d

y y

x x

d

4d

y

x from correct working.

Also can be ft using their ‘x’ value and 3y

in the correct part (a) of 2d 3

d 2 3

y x

x y

A1

(3) [7]

Question

Number Scheme Marks

2. (a)

dd

dd

1

e e

ux

x xvx

u x

v

e d e e .1 dx x xx x x x M1 A1

e e dx xx x

e ex xx c A1 (3)

(b)

2 dd

dd

2

e e

ux

x xvx

u x x

v

2 2e d e e .2 dx x xx x x x x M1 A1

2e 2 e dx xx x x

2e 2 e ex x xx x c A1 (3)

(6 marks)

Bronze 4: 4/12 44

Question

Number Scheme Marks

3. (a) 2 2

22

2

1 5 1 5(2 5 ) 2 1 1

(2 5 ) 2 4 2

x xx

x

2(2) or

1

4 B1

21 ( 2)( 3)

1 ( 2)(** ) (** ) ...4 2!

x x M1 A1ft

21 5 ( 2)( 3) 5

1 ( 2) ...4 2 2! 2

x x

21 75

1 5 ; ...4 4

x x

21 5 75

; ...4 4 16

x x A1; A1

[5]

(b) 2

2

2 1 5 75(2 ) ...

(2 5 ) 4 4 16

k xkx x x

x Can be implied by later work

even in part (c). M1

x terms: 2(5 ) 7

4 4 4

x k x x

giving, 10 7 3 k k 3 k A1

[2]

(c) 2x terms: 2 2150 5

16 4

x k x M1

So, 75 5( 3) 75 15 45

8 4 8 4 8

A

45

8 or

55

8 or 5.625 A1

[2]

(9 marks)

4. (a) 2 23 4x y xy ( eqn )

d d d

6 2 0d d d

y y yx y y x

x x x

M1 B1 A1

d 8 6 8

d 3 2 3

y x y

x x y

M1

giving 18 3 8 16x y x y

giving 13 26y x M1

Hence, 2 2 0y x y x A1 cso (6)

(b) At P & Q, 2y x . Substituting into eqn

gives 2 23 (2 ) (2 ) 4x x x x M1

Simplifying gives, 2 4 2x x A1

2 4y x y , hence coordinates are (2,4) and ( 2, 4) A1 (3)

(9 marks)


Question

Number Scheme Marks

5. (a) 0.73508 B1 cao

(1)

(b) 1

Area ; 0 2 their 0.73508 1.17157 1.02280 02 8

B1 M1

5.8589... 1.150392325... 1.150416

(4 dp) awrt 1.1504 A1 (3)

(c) d

1 cos sind

u

u x xx

B1

2sin 2 2(2sin cos )d d

(1 cos ) (1 cos )

x x x

x xx x

sin2 2sin cosx x x B1

4( 1) (1 ).( 1)d 4 d

u uu u

u u

M1

14 1 d 4 ln

u u u cu

dM1

4ln 1 cos 4 1 cosx x c 4ln 1 cos 4cosx x k

AG

A1 cso

(5)

(d) 4ln 1 cos 4cos 4ln 1 cos0 4cos02 2

Applying limits 2

x

and 0x either way

round.

M1

4ln1 0 4ln2 4

4 4ln 2 1.227411278... 4(1 ln2) or

(4 4ln2) or awrt 1.2 A1

Error 4 4ln2 1.1504...

0.0770112776... 0.077 (2sf )

awrt 0.077

or awrt 6.3(%) A1 cso (3)

(12 marks)

Bronze 4: 4/12 46

Question

Number Scheme Marks

6.

6

:

10 1

a

l b c

r ,

21

17

6

OA

,

25

14

18

OB

(a) A is on l, so

21 6

17

6 10 1

a

b c

: 10 6 4 k 4 B1

: 6 21 6(4) 21a a i

Substitutes their

value of into 6 21a

M1

3a 3a A1

cao

(3)

(b) 25 21

14 17

18 6

AB

21 25

17 14

6 18

BA

Finds the

difference between

OA and OB .

Ignore labelling.

M1

4

3

12

AB

4

3

12

BA

0AB l AB d

4 6

3 24 3 12 0; 4

12 1

c c c

M1;

A1 ft

: 17 ( 4)(4) 17 ; 1b c b b j

ddM1;

A1 cso

cao (5)

(c) 2 2 2 2 2 24 3 12 or ( 4) ( 3) ( 12)AB AB M1

So, 13AB A1

cao

(2)

(d) 21 4 17

17 3 ; 20

6 12 6

OB OA BA

M1;A1

cao

(2) [12]


Question

Number Scheme Marks

7. (a)

x 1 2 3 4

y ln2 2 ln 4 3 ln 6 2ln8

0.6931 1.9605 3.1034 4.1589

M1

1

Area 1 ...2

B1

... 0.6931 2 1.9605 3.1034 4.1589 M1

1

14.97989 ... 7.492

7.49 cao A1 (4)

(b) 3 31

2 2 22 2 1

ln 2 d ln 2 d3 3

x x x x x x xx

M1 A1

3 12 2

2 2ln 2 d

3 3x x x x

3 32 2

2 4ln 2

3 9x x x C M1 A1 (4)

(c) 3 3 3 32 2 2 2

4

1

2 4 2 4 2 4ln 2 4 ln8 4 ln 2

3 9 3 9 3 9x x x

M1

16ln 2 ... ... Using or implying ln 2 ln 2n n M1

46 28

ln 23 9

A1 (3)

[11]

Bronze 4: 4/12 48

Question

Number Scheme

Mark

s

8. (a)

d (3 )

d 125t

1 1d d

3 125t

or

125d d

3t

B1

1ln 3

125t c or

1

ln 3125

t c

See notes. M1

A1

1ln 3

125t c

Correct completion to0.008e 3tA .

1 1

125 1253 e or e et c t

c

A1 0.008e 3tA *

[4]

(b) 0 , 16t 0.008(0)16 e 3A ;

13A See notes.

M1;

A1

0.00810 13e 3t

Substitutes 10 into

an equation

of the form 0.008e 3,tA

or equivalent.

M1

0.008 7

e13

t 7

0.008 ln13

t

Correct algebra to

0.008 lnt k , where k is

a positive value.

M1

7ln

1377.3799... 77 nearest minute

0.008t

awrt 77 A1

[5]

9


Question 1

A significant majority of candidates were able to score full marks on this question. In part (a), many candidates were able to

differentiate implicitly and examiners noticed fewer candidates differentiating 8 incorrectly with respect to x to give 8. In

part (b), many candidates were able to substitute 3y into C leading to the correct x-coordinate of 2. Several

candidates either rearranged their C equation incorrectly to give 2x or had difficulty finding the cube root of 8 .

Some weaker candidates did not substitute 3y into C, but substituted 3y into the d

d

y

xexpression to give a gradient

of 2.x

Question 2

In part (a), many candidates were able to use integration by parts in the right direction to produce a correct solution.

Common errors included integrating ex incorrectly to give ln x or applying the by parts formula in the wrong direction by

assigning u as ex to be differentiated and

dd

vx as x to be integrated.

Many candidates were able to make a good start to part (b), by assigning u as 2x and

dd

vx

as ex and again correctly

applying the integration by parts formula. At this point, when faced with integrating 2 exx , some candidates did not make

the connection with their answer to part (a) and made little progress, whilst others independently applied the by parts formula

again. A significant proportion of candidates made a bracketing error and usually gave an incorrect answer of 2e ( 2 2) .x x x c

In part (b), a few candidates proceeded by assigning u as x and dd

vx

as exx and then used their answer to part (a) to obtain v.

These candidates were usually produced a correct solution.

Question 3

This question was also generally well answered with about 50% of candidates obtaining all of the 9 marks available.

In part (a), a minority of candidates were unable to carry out the first step of writing 2

1

(2 5 )x as

21 5

14 2

x

, with

the 1

4outside the brackets usually written incorrectly as either 1 or

1

2. Many candidates were able to use a correct method

for expanding a binomial expression of the form (1 )nax . A variety of incorrect values of a were seen, with the most

common being either 5

2, 5 or 5. Some candidates, having correctly expanded

25

1 ,2

x

forgot to multiply their

expansion by 1

.4

As expected, sign errors, bracketing errors, and simplification errors were also seen in this part.

In parts (b) and (c), most candidates realised that they needed to multiply (2 )k x by their binomial expansion from part

(a) and equate their x and 2x coefficients in order to find both k and A. A small minority, however, attempted to divide

(2 )k x by their part (a) expansion. Other candidates omitted the brackets around 2 ,k x although they progressed as

if these “invisible” brackets were really there.

In part (b), a significant minority of candidates used an incorrect method of multiplying (2 )k x by the first term (usually

41

) of their binomial expansion, and equating the result to 1

2 in order to find k. In part (c), these candidates also multiplied

(2 )k x by the third term (usually 2

16

75x ) of their binomial expansion and equated this to

2Ax in order to find A.

Bronze 4: 4/12 50

A few candidates in parts (b) and (c) applied an alternative method of multiplying out 2 21 7

(2 5 ) ...2 4

x x A x

and equating the result to (2 )k x , in order to correctly find both k and A.

Question 4

This question was generally well done with a majority of candidates scoring at least 6 of the 9 marks available.

In part (a), implicit differentiation was well handled with most candidates appreciating the need to apply the product rule to

the xy term. A few candidates failed to differentiate the constant term and some wrote d

d" ..."

y

x before starting to

differentiate the equation. After differentiating implicitly, the majority of candidates rearranged the resulting equation to

make d

d

y

xthe subject before substituting

d

d

y

xas

83

rather than substituting 83

for d

d

y

xin their differentiated equation. Many

candidates were able to prove the result of 2 0y x . A surprising number of candidates when faced with manipulating

the equation 6 8

2 3

x y

y x

, separated the fraction to incorrectly form two equations 6 8x y & 2 3y x and

then proceeded to solve these equations simultaneously.

Some candidates, who were unsuccessful in completing part (a), gave up on the whole question even though it was still

possible for them to obtain full marks in part (b). Other candidates, however, did not realise that they were expected to

substitute 2y x into the equation of the curve and made no creditable progress with this part. Those candidates who used

the substitution 2y x made fewer errors than those who used the substitution2.

yx The most common errors in this

part were for candidates to rewrite 2y as either

24x or 22 ;x or to solve the equation

2 4x to give only 2x or

even 4.x On finding 2x , some candidates went onto substitute these values back into the equation of the curve,

forming a quadratic equation and usually finding “extra” unwanted points rather than simply doubling their two values of x to

find the corresponding two values for y. Most candidates who progressed this far were able to link their values of x and y

together, usually as coordinates.

Question 5

In part (a), virtually all candidates were able to find the y-value corresponding to8

x

.

In part (b), most candidates were able to apply the trapezium rule correctly to find the correct estimate with the most common

errors being candidates writing h as either 10

,

4

or

16

; or candidates rounding incorrectly to give 1.1503. Few

bracketing errors in part (b) were encountered in this session.

Part (c) provided a diverse range of solutions. Most candidates followed the advice given in the question to use the

substitution of 1 cosu x , so obtaining d

sind

ux

x (or occasionally sin x ), as well as using the double angle

formula for sine to process the numerator of the integral. Whilst some students found the conversion of the given integral to

an expression in u beyond them, many more were able to reach an integral of the form ( 1)

d .u

k uu

Whilst most

candidates reaching this stage then correctly divided through by u and integrated term by term to reach an expression of the

form k(ln u – u), a few resorted to integration by parts and were generally less successful. A significant proportion of

candidates lost the final accuracy mark as a result of not showing how their constant of integration could be combined with

the –4 from their integration to give the stated k in the question; some found a value for k (usually 4) or some simply failed to

state the final result.

In part (d), those candidates who were unable to complete part (c) often realised that they were still able to attempt part (d).

The use of limits for either x or for u was generally successfully completed to obtain the value 1.227 or 4 – 4 ln 2, but the

final step of finding the error was not so successfully tackled.

Question 6 This was a well answered question on vectors with about 56% of candidates gaining at least 10 of the 12 marks available and

about 23% of candidates gaining all 12 marks. Part (d) required the more able candidates to think for themselves.


In part (a), most candidates set the line l equal to the point A and equated the k components to find the value of λ. This was

followed by equating the i components to find the correct value of a, although some candidates, however, incorrectly found a

= 3 from a + 6(4) = 21. A small minority of candidates wrote down equations for the i and k components and solved these

simultaneously to find a.

In part (b), almost all candidates found AB and many applied

1

6

cAB . = 0 in order to find the value of c and

proceeded to find the value of b. Some candidates incorrectly applied

10

63

cb .

1

6

c = 0 and thus made little

progress. Some candidates incorrectly found c = 4 or 3 from a correct 24 + 3c – 12 = 0 whilst others incorrectly found b = –

33, 33 or 1 from a correct b = (–4)(4) = –17.

In part (c), most candidates were able to find the correct distance AB and few errors were seen.

In part (d), the majority of candidates were not able to use the information given earlier in the question and many of them left

this part blank. The most common error of those who attempted this part was to write down B as –25i+ 14j – 18k. 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. Some candidates, by deducing that A was the midpoint of

B and B were able to write down 2

25x = 21,

2

14y = –17 and

2

18z = 6 in order to find the position vector of B.

Question 7

Part (a) was accessible to most candidates. This was the first time in recent years that candidates had to produce their own

table. This, in general, they did well although the number of decimal places recorded often seemed too few to be working

towards a final accuracy of 2 decimal places. To obtain an answer of this accuracy, you should tabulate figures to at least 3

decimal places. Of course, the examiners have no means of knowing what figures the candidate has in their calculator and as

long as there was some tabulation, or the exact expressions ln 2, 2 ln 4, 3 ln 6 and 2ln8 were given, and the

working showed that the correct formula was known, then, if 7.49 was given as the answer, the candidate was given the

benefit of the doubt. A few candidates confused the number of strips with the number of ordinates but these were fewer than

in some recent examinations.

In part (b), most knew that they had to use integration by parts and most attempted this in the “right direction” attempting to

integrate 12x , which was usually correct, and differentiate ln 2x , for which the incorrect

1

2x was often seen. Many who

reached the intermediate stage correctly had difficulty with 32

2 1. d

3x x

x

, failing to divide 32x by x. Fully correct solutions

to part (b) were not common.

If they had an answer to part (b) in the correct form, then most candidates showed that they could complete the question by

using the limits correctly and then using the power rule for logs to obtain an answer in the form specified. There were many

errors of detail in candidates’ solutions to this question but more than 50% of candidates gained eight or more of the available

eleven marks..

Question 8

Some candidates did not attempt to separate the variables in Q8(a). They were also not able to deal with the context of the

question in Q8(b).

In Q8(a), those candidates who were able to separate the variables, were usually able to integrate both sides correctly,

although a number made a sign error by integrating 1

3 to obtain ln 3 . A significant number of candidates

omitted the constant of integration “ c ”and so were not able to gain the final mark. A significant number of candidates did

not show sufficient steps in order to progress from 1

ln 3125

t c to the result0.008e 3tA .

Common errors included candidates removing their logarithms incorrectly to give an equation of the form 1

1253 et

A

or candidates stating the constant A as 1.

Bronze 4: 4/12 52

Q8(b) was often better answered with some candidates scoring no marks in q8(a) and full marks in Q8(b). Those candidates

who used 0.008e 3tA were more successful in this part. They were usually able to write down the condition

16 when 0t in order to find 13.A Some candidates misinterpreted the context of the question to write down

the condition 6 when 0t , yielding the result of 3.A Other incorrect values of A seen by examiners included

1, 16 or 1. Many candidates who found A correctly were usually able to substitute 10 into 0.00816e 3t

and manipulate the result correctly in order to find the correct time.



Qu Max score

Modal score

Mean %

ALL A* A B C D E U

1 7 86 6.02 6.76 6.35 5.50 4.60 3.52 2.07

2 6 76 4.53 5.40 4.87 4.24 3.38 2.32 1.11

3 9 78 7.00 8.87 8.08 6.89 5.87 4.48 4.04 1.83

4 9 71 6.43 8.15 6.86 5.52 4.01 2.59 1.18

5 12 70 8.34 11.49 10.12 7.73 5.92 4.44 3.52 1.60

6 12 69 8.24 11.17 9.31 7.40 4.92 3.69 2.70 0.96

7 11 67 7.32 10.54 9.14 7.50 5.85 4.39 3.04 1.60

8 9 0 54 4.85 8.15 6.01 3.95 2.55 1.60 0.99 0.34

75 70 52.73 62.97 51.55 40.37 30.59 22.72 10.69


Paper Reference(s)

6666/01


Bronze Level B4


















A* A B C D E

67 59 52 47 41 35

Bronze 4: 4/12 54

1.

Figure 1

The curve shown in Figure 1 has equation ex(sin x), 0 x . The finite region R bounded

by the curve and the x-axis is shown shaded in Figure 1.

(a) Copy and complete the table below with the values of y corresponding to x = 4

and x =

2

, giving your answers to 5 decimal places.

x 0 4

2

4

3

y 0 8.87207 0

(2)

(b) Use the trapezium rule, with all the values in the completed table, to obtain an estimate

for the area of the region R. Give your answer to 4 decimal places.

(4)

January 2008

2. The curve C has equation

3x–1

+ xy –y2 +5 = 0

Show that d

d

y

x at the point (1, 3) on the curve C can be written in the form 31

ln( e )

, where λ

and μ are integers to be found.

(7)

June 2013 (R)

Bronze 1: 1/12 55

3. The curve C has equation

cos 2x + cos 3y = 1, –4

x

4

, 0 y

6

.

(a) Find x

y

d


(3)

The point P lies on C where x = 6

.

(b) Find the value of y at P.

(3)

(c) Find the equation of the tangent to C at P, giving your answer in the form ax + by + c = 0,

where a, b and c are integers.

(3)

January 2010

4. Relative to a fixed origin O, the point A has position vector i − 3j + 2k and the point B has

position vector −2i + 2j − k. The points A and B lie on a straight line l.

(a) Find AB .

(2)

(b) Find a vector equation of l.

(2)

The point C has position vector 2i + pj − 4k with respect to O, where p is a constant.

Given that AC is perpendicular to l, find

(c) the value of p,

(4)

(d) the distance AC.

(2)

January 2011

Bronze 4: 4/12 56

5.

Figure 1

Figure 1 shows part of the curve with equation 134 e 3t

x t

. The finite region R shown

shaded in Figure 1 is bounded by the curve, the x-axis, the t-axis and the line t = 8.

(a) Complete the table with the value of x corresponding to t = 6, giving your answer to

3 decimal places.

t 0 2 4 6 8

x 3 7.107 7.218 5.223

(1)

(b) Use the trapezium rule with all the values of x in the completed table to obtain an

estimate for the area of the region R, giving your answer to 2 decimal places.

(3)

(c) Use calculus to find the exact value for the area of R.

(6)

(d) Find the difference between the values obtained in part (b) and part (c), giving your

answer to 2 decimal places.

(1)

June 2013 (R)

Bronze 1: 1/12 57

6. (a) Find

xx d)5( .

(2)

Figure 3

Figure 3 shows a sketch of the curve with equation

y = (x – 1)√(5 – x), 1 x 5

(b) (i) Using integration by parts, or otherwise, find

xxx d)5()1( .

(4)

(ii) Hence find

5

1

.d)5( )1( xxx .

(2)

June 2009

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

l1 : r = (9i + 13j – 3k) + (i + 4j – 2k)

l2 : r = (2i – j + k) + (2i + j + k)

where and are scalar parameters.

(a) Given that l1 and l2 meet, find the position vector of their point of intersection.

(5)

(b) Find the acute angle between l1 and l2, giving your answer in degrees to 1 decimal place.

(3)

Given that the point A has position vector 4i + 16j – 3k and that the point P lies on l1 such that

AP is perpendicular to l1,

(c) find the exact coordinates of P.

(5)

January 2013

Bronze 4: 4/12 58

8. Relative to a fixed origin O, the point A has position vector (10i + 2j + 3k), and the point B

has position vector (8i + 3j + 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)

The point C has position vector (3i + 12j + 3k) .

The point P lies on l. Given that the vector CP is perpendicular to l,

(c) find the position vector of the point P.

(6)

June 2012


END

Bronze 1: 1/12 59

Question

Number Scheme Marks

1. (a)

x 0 4

2 3

4

y 0 1.84432133

2…

4.81047738

1… 8.87207 0

awrt 1.84432 B1

awrt 4.81048 or

4.81047 B1

0 can be

implied

[2]

For structure of

trapezium rule

............. ; M1

Correct expression

inside brackets which

all must be multiplied

by their “outside

constant”.

A1

31.05374... 12.19477518... 12.1948

8

(4dp)

12.1948 A1 cao

[4]

6

marks

2. 1 23 5 0x xy y

1d d d

3 ln3 2 0d d d

xy y y

y x yx x x

(ignore)

1 13 3 ln 3x x B1 oe

Differentiates implicitly to

include either

d dor

d d

y yx ky

x x .

M1*

d

d

yxy y x

x B1

d d... 2 0

d d

y yy x y

x x A1

1. 1, 3

(1 1) d d3 ln3 3 (1) 2(3) 0

d d

y y

x x

Substitutes 1, 3x y into

their differentiated equation or

expression.

dM1*

2. d d dln3 3 6 0 3 ln3 5

d d d

y y y

x x x

d 3 ln3

d 5

y

x

dM1*

3 3d 1 1

ln e ln3 ln 3ed 5 5

y

x

Uses 33 ln e to achieve

3d 1ln 3e

d 5

y

x

A1

cso

[7]

Bronze 4: 4/12 60

Question Number

Scheme Marks

Q3 (a) d

2sin 2 3sin 3 0d

yx y

x M1 A1

d 2sin 2

d 3sin 3

y x

x y Accept

2sin 2

3sin3

x

y,

2sin 2

3sin3

x

y

A1 (3)

(b) At 6

x

, 2

cos cos3 16

y

M1

1

cos32

y A1

3

3 9y y

awrt 0.349

A1 (3)

(c) At ,6 9

,

6 3

9 3

2sin 2 2sind 2

d 3sin 3 3sin 3

y

x

M1

2

9 3 6y x

M1

Leading to 6 9 2 0x y A1 (3)

[9]

Bronze 1: 1/12 61

Question Number

Scheme Marks

4.

(a) 2 2 3 2 3 5 3AB i j k i j k i j k M1 A1 (2)

(b) 3 2 3 5 3 r i j k i j k M1 A1ft (2)

or 2 2 3 5 3 r i j k i j k

(c) 2 4 3 2AC p i j k i j k

3 6p i j k or CA B1

1 3

. 3 . 5 0

6 3

AC AB p

M1

3 5 15 18 0p

Leading to 6p M1 A1 (4)

(d) 2 2 22 2 1 6 3 4 2 46AC M1

46AC accept awrt 6.8 A1

(2)

[10]

Bronze 4: 4/12 62

Question

Number Scheme Marks

5. (a) 6.248046798... = 6.248 (3dp) 6.248 or awrt 6.248 B1

(1)

(b)

1Area 2 ; 3 2 7.107 7.218 + their 6.248 5.223

2

B1; M1

49.369 49.37 (2 dp) 49.37 or awrt 49.37 A1

(3)

(c)

1 1 1

3 3 3(4 e 3) d 12 e 12e dt t t

t t t t

3t

1 1

3 3e e d , 0, 0t t

At B t A B

M1

A1

3 3t B1

1 1

3 312 e 36e 3t t

t t

1 1

3 312 e 36et t

t

A1

81 1

3 3

0

12 e 36e 3t t

t t

1 1 1 1(8) (8) (0) (0)

3 3 3 312(8)e 36e 3(8) 12(0)e 36e 3(0)

Substitutes

limits of 8 and

0 into an

integrated

function of the

form of either 1 1

3 3e et t

t

or 1 1

3 3e et t

t Bt

and subtracts

the correct way

round.

dM1

8 8

3 396e 36e 24 0 36 0

8

360 132e

8

360 132e

A1

(6)

(d)

Difference 8

360 132e 49.37 1.458184439... 1.46 (2 dp)

1.46 or awrt

1.46 B1

(1) [11]

Bronze 1: 1/12 63

Qn Scheme Marks

6. (a)

321

2

32

55 d 5 d

xx x x x C

M1 A1 (2)

32

25

3x C

(b)(i) 3 32 2

2 21 5 d 1 5 5 d

3 3x x x x x x x

M1 A1ft

= …

52

52

52

3

xC

M1

3 52 2

2 41 5 5

3 15x x x C A1 (4)

(ii) 3 52 2

5

1

2 41 5 5

3 15x x x

52

40 0 0 4

15

128 8

8 8.5315 15

awrt 8.53 M1 A1 (2)

(8 marks)

Bronze 4: 4/12 64

Question

Number Scheme Marks

7. (a)

9 2 2

13 4 1

3 2 1

i: (1)

j: (2)

k : (3)

Any two equations.

(Allow one slip). M1

Eg: : 16 6 2 (2) (3) or

4 : 23 9 7 (2) (1) An attempt to eliminate

one of the parameters. M1

Leading to 3 or 2 Either 3 or 2 A1

1

9 1 6

: 13 3 4 1

3 2 3

l

r or

2

2 2 6

: 1 2 1 1

1 1 3

l

r

M1 A1

[5]

(b)

1

4

2

1d , 2

2

1

1

d

1 2

4 1

2 1

Realisation that the dot

product is required

between A 1d and

2B d .

M1

2 2 2 2 2 2

2 4 2cos

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

Correct equation. A1

4

cos 69.1238974... 69.1 (1 dp)21. 6

awrt 69.1 A1

[3]

(c)

4

16

3

OA

,

9 1 9

13 4 13 4

3 2 3 2

OP

9 4 5

13 4 16 4 3

3 2 3 2

AP

M1 A1

5 1

0 4 3 4 5 16 12 4 0

2 2

AP

1d M1

leading to 1

21 7 03

1

3 A1

Position vector

1 289

3 39 11 1 43

13 4 14 or3 3 3

3 22 11

33 3

OP

M1 A1

[6] 14

Bronze 1: 1/12 65

Question Number

Scheme Marks

8. (a)

8 10 2

3 2 1

4 3 1

AB

M1 A1 (2)

(b)

10 2

2 1

3 1

t

r

8 2

3 1

4 1

t

r M1 A1ft (2)

(c)

10 2 3 7 2

2 12 10

3 3

t t

CP t t

t t

M1 A1

7 2 2

10 . 1 14 4 10 0

1

t

t t t t

t

M1

Leading to 4t A1

Position vector of P is

10 8 2

2 4 6

3 4 7

M1 A1 (6)

[10]

Bronze 4: 4/12 66

Question 1

A significant majority of candidates were able to score full marks on this question. In part (a), some candidates struggled to

find either one or both of the y-ordinates required. A few of these candidates did not change their calculator to radian mode.

In part (b), some candidates incorrectly stated the width of each of the trapezia as either 14

or5

. Nearly all answers were

given to 4 decimal places as requested in the question.

Question 2 This question was generally well answered with about 62% of candidates gaining at least 6 of the 7 marks available and

about 27% of candidates gaining all 7 marks.

Most candidates were able to differentiate implicitly but a significant minority struggled to differentiate 3x – 1 and a few

candidates did not apply the product rule correctly on xy. At this point a minority of candidates substituted x = 1 and y = 3 in

their differentiated equation, but the majority of candidates proceeded to find an expression for x

y

d

d in terms of x and y, to

give any equation of the form x

y

d

d=

xy

y x

2

3ln3 1

before substituting in these values. Although the majority of candidates

were able to write x

y

d

d as

5

3ln3, most of them did not show sufficient working (i.e. writing 3 as ln e3) in order to prove

that this could be manipulated to give 51 ln (3e3).

Question 3

As has been noted in earlier reports, the quality of work in the topic of implicit differentiation has improved in recent years

and many candidates successfully differentiated the equation and rearranged it to find d

d

y

x. Some, however, forgot to

differentiate the constant. A not infrequent, error was candidates writing d d

2sin 2 3sin3d d

y yx y

x x and then

incorporating the superfluous d

d

y

x on the left hand side of the equation into their answer. Errors like

d 1

cos3 sin 3d 3

y yy

were also seen. Part (b) was very well done. A few candidates gave the answer 20 , not

recognising that the question required radians. Nearly all knew how to tackle part (c) although a few, as in Q2, spoilt

otherwise completely correct solutions by not giving the answer in the form specified by the question.

Question 4

This question was well done and full marks were common. Part (a) was almost always correct and such errors as were seen

were errors of arithmetic. Even at this level 2 3 1 is seen from time to time. For (b), the majority knew the form of a

straight line. A few got the vectors the wrong way round or produced an answer of the form r a b . The commonest

error was to give an answer of the form

1 3

3 5

2 3

, not recognising that, as the question asks for an equation, this

expression must be preceded by ...r .

Part (c) was more demanding and many were unable to choose vectors in the appropriate directions. Almost all knew that they

had to form an equation by equating a scalar product to zero. However, often one of the position vectors was used; finding the

scalar product of OC with AB being a common choice. Some formed the scalar product of AB with the vector equation

of AC obtaining equations involving parameters as well as p. Almost all knew the appropriate method for part (d).

Question 5 This question was generally well answered although part (c) was found challenging by a number of candidates. About 56% of

candidates gained at least 10 of the 12 marks available and about 23% of candidates gained all 12 marks.

Bronze 1: 1/12 67

In part (a), most candidates correctly computed 6.248 to 3 decimal places, but a few gave their answer to only 2 decimal

places.

In applying the trapezium rule in part (b), a small minority of candidates multiplied 21 by

58

instead of 21 by 2. Whilst the

table of values clearly shows an interval width of 2, the application of a formula h = n

ab with n = 5 instead of n = 4

sometimes caused this error. Other errors included the occasional bracketing mistake, use of the t-value of 0 rather than the

ordinate of 3, and the occasional calculation error following a correctly written expression.

In part (c), a significant minority were only able to integrate 3 to give 3t but struggled to apply the correct strategy to

integrate 3

1

e4

t with respect to t. On the other hand, some candidates applied integration by parts correctly to give

3

1

e12

t – 3

1

e36

but then omitted the integration of 3 to give 3t. A common error was t

3

1

e

being integrated to give

t3

1

31 e-

. There were a number of able students who, having integrated correctly, could not simplify their answer to give the

exact value of 60 – 3

8

e132

.

Most candidates who answered parts (b) and (c) correctly were able to find the correct difference of 1.46 in part (d), although

a few found the percentage error of the estimate. Some candidates who had not answered part (c) correctly applied numerical

integration in their calculator to give 50.828... and so were able to find the correct answer to part (d).

Question 6

Throughout this question sign errors were particularly common. In part (a), nearly all recognised that 2

3

)5( x formed part

of the answer, and this gained the method mark, but 2

3

)5(2

3x ,

2

3

)5(2

3x and

2

3

)5(3

2x , instead of the correct

2

3

)5(3

2x , were all frequently seen. Candidates who made these errors could still gain 3 out of the 4 marks in part (b)(i)

if they proceeded correctly. Most candidates integrated by parts the “right way round” and were able to complete the

question. Further sign errors were, however, common.

Question 7

In this question the majority of candidates were able to score full marks in Q7(a) and Q7(b).

Q7(a) was generally well answered, with most candidates gaining full marks. Having successfully found at least one of either

or , some candidates proceeded no further and did not attempt to find the point of intersection. Some candidates used

the third equation to prove that the two lines intersected, not realising that this was not required. Mistakes included minor

errors in constructing the simultaneous equations or algebraic errors when solving their simultaneous equations.

In Q7(b), a large majority of candidates were able to find the correct acute angle by taking the dot product between the

direction vectors of 1l and

2l . The majority worked in degrees with only a few answers given in radians. Some candidates

applied the dot product formula between multiples of the direction vectors (using their and from part (a)), which

usually led to an obtuse angle; however, most realised that they needed to subtract this angle from 180 in order to find the

correct acute angle. A number of candidates used incorrect vectors such as either 9 13 3 i j k , 2 i j k or

6 3 i j k in their dot product equation.

The majority of candidates found Q7(c) challenging. To make progress with this question it is necessary to use the scalar

product formula to obtain an equation in a single parameter, not the three unknowns x, y and z. So candidates who initially

found the vector AP in terms of and who applied

1

4 0

2

AP

were those who were more successful in gaining

some or all of the 6 marks available. It was noticeable, however, that some of these candidates found incorrect values of

such as 3 or even 1

3 following from a correct 21 7 0 . It was also common to see a number of candidates who

Bronze 4: 4/12 68

incorrectly tried to apply 0OP OA or

1

4 0

2

OP

in order to find their . Candidates who found by a

correct method were usually able to substitute it into the equation for 1l in order to find the exact coordinates of P.

Question 8

In part (a), both marks were gained by nearly all candidates. Only a minority added the components or subtracted them the

wrong way round. If BA was given instead of AB , one mark was lost here but all other marks in the question could be

gained. Again, in part (b), most candidates gained both marks but a substantial number of candidates failed to realise that the

equation of a line must actually be an equation and, for example, omitted the “ r ” from

8 2

3 1

4 1

r .

Part (c) was very demanding and often proved to be an all or nothing affair. Correct methods were not often seen but, when

they were, full marks were usually gained. Most knew that they should equate a scalar product to zero but the wrong vectors

were often used. It would have helped candidates if they had drawn a diagram showing the points O, A, B, C and P. The scalar

products OP . AB or OP . OC were often used instead of CP . AB . To make progress with this question it is

necessary to use the correct scalar product to obtain an equation in a single parameter, not the three unknowns x, y and z. The

most commonly seen successful method was to use the candidate’s answer to (b) to obtain a vector expression for CP in

terms of, say, . If the equation of the line is that given in the previous paragraph, this gives. CP =

1

9

25

. The

scalar product of this vector with the answer to part (a) gives an equation for . The value of the parameter given by this

equation can then be substituted into the answer to part (b) to complete the question. There are alternative approaches to this

question using differentiation or using the cosine rule to find the ratio, for example, :AB AP but these were very rarely

seen.



Qu Max score

Modal score

Mean %

ALL A* A B C D E U

1 6 84 5.04 5.60 5.06 4.37 3.94 3.20 2.06

2 7 75 5.25 6.62 5.79 5.00 3.89 3.19 2.21 0.71

3 9 75 6.71 7.91 6.70 5.71 4.08 2.79 1.74

4 10 70 6.97 9.76 8.14 6.33 4.56 3.16 2.77 1.64

5 11 69 7.57 10.25 8.39 6.69 5.18 4.39 3.52 2.15

6 8 65 5.16 6.79 5.42 4.02 2.86 1.83 0.98

7 14 8 62 8.70 13.31 10.54 7.96 5.85 4.48 3.27 1.47

8 10 50 4.98 9.12 5.93 4.35 3.51 2.95 2.32 1.53

75 67 50.38 59.09 47.51 37.09 29.05 21.91 12.28

paper reference(s) 6666/01 edexcel gce edexcel gce core mathematics c4

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