ΑΣΚΗΣΕΙΣ ΜΑΘΗΜΑΤΙΚΩΝ ΟΞΦΟΡΔΗΣ

Centre Number Candidate Number

Surname

Other Names

Candidate Signature

General Certificate of Education

Advanced Level Examination

January 2012

Mathematics MFP4

Unit Further Pure 4

Friday 27 January 2012 9.00 am to 10.30 am

For this paper you must have:* the blue AQA booklet of formulae and statistical tables.

You may use a graphics calculator.

Time allowed* 1 hour 30 minutes

Instructions* Use black ink or black ball-point pen. Pencil should only be used for

drawing.* Fill in the boxes at the top of this page.* Answer all questions.* Write the question part reference (eg (a), (b)(i) etc) in the left-hand

margin.* You must answer the questions in the spaces provided. Do not write

outside the box around each page.* Show all necessary working; otherwise marks for method may be

lost.* Do all rough work in this book. Cross through any work that you do

not want to be marked.

Information* The marks for questions are shown in brackets.* The maximum mark for this paper is 75.

Advice* Unless stated otherwise, you may quote formulae, without proof,

from the booklet.* You do not necessarily need to use all the space provided.

For Examiner’s Use

Examiner’s Initials

Question Mark

1

2

3

4

5

6

7

8

TOTAL

P45892/Jan12/MFP4 6/6/6/ MFP4(JAN12MFP401)

2

Answer all questions in the spaces provided.

1 The vectors a and b are such that a . b ¼ 21 , ja j ¼ 5ffiffiffi2p

and jb j ¼ 3 .

Determine the exact value of ja� b j . (5 marks)

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(03)

QUESTION

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REFERENCE

4

2 Describe the single transformation represented by each of the matrices:

(a)

0 0 1

0 1 0

1 0 0

264

375 ; (2 marks)

(b)

0:6 0 �0:80 1 0

0:8 0 0:6

264

375 . (3 marks)

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(04)

5

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(05)

QUESTION

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6

3 (a) Find the eigenvalues and corresponding eigenvectors of the matrix M ¼ 4 5

5 4

� �.

(6 marks)

(b) The plane transformation T is given by the matrix M. Write down the coordinates of

the invariant point of T. (1 mark)

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(06)

7

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(07)

QUESTION

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REFERENCE

8

4 Let X ¼ 3 x

�1 7

� �.

(a) Determine XXT . (2 marks)

(b) Show that DetðXXT � XTXÞ4 0 for all real values of x. (4 marks)

(c) Find the value of x for which the matrix ðXXT � XTXÞ is singular. (1 mark)

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(08)

9

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(09)

QUESTION

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10

5 (a) Determine the two values of the integer n for which the system of equations

2xþ nyþ z ¼ 5

3x� yþ nz ¼ 1

�xþ 7yþ z ¼ n

does not have a unique solution. (4 marks)

(b) For the positive value of n found in part (a), determine whether the system is

consistent or inconsistent, and interpret this result geometrically. (6 marks)

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(10)

11

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(11)

QUESTION

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REFERENCE

12

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QUESTION

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REFERENCE

13

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(13)

QUESTION

PART

REFERENCE

14

6 The planes P1 and P2 have equations

r .

2

1

7

2435 ¼ 10 and r .

3

1

�4

24

35 ¼ 7

respectively.

(a) Determine, to the nearest degree, the acute angle between P1 and P2 . (4 marks)

(b) By setting z ¼ t , find cartesian equations for the line of intersection of P1 and P2 in

the form

x� a

l¼ y� b

m¼ z ¼ t (6 marks)

(c) The line L, with equation r ¼20

�17

24

35þ l

1

9

4

2435 , intersects P1 at the point P and P2

at the point Q.

Show that PQ ¼ kffiffiffi2p

, where k is an integer. (6 marks)

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15

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(15)

QUESTION

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16

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QUESTION

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17

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(17)

QUESTION

PART

REFERENCE

18

7 The plane transformation T is a rotation through y radians anticlockwise about O,

and maps points ðx, yÞ onto image points ðX , Y Þ such that

X

Y

� �¼ c �s

s c

� �x

y

� �

where c ¼ cos y and s ¼ sin y .

(a) Write down the inverse of the matrixc �ss c

� �and hence show that

x ¼ cX þ sY and y ¼ �sX þ cY (3 marks)

(b) The curve C has equation x2 � 6xy� 7y2 ¼ 8 .

The image of C under T is the curve C ’ with equation pX 2 þ qXY þ rY2 ¼ 8 .

(i) Use the results of part (a) to show that

q ¼ 6s2 þ 16sc� 6c2

and express p and r similarly in terms of c and s. (4 marks)

(ii) Given that y is an acute angle, find the values of c and s for which q ¼ 0 and hence

in this case express the equation of C ’ in the form

X 2

a2� Y2

b2¼ 1 (8 marks)

(iii) Hence explain why C is a hyperbola. (1 mark)

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19

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20

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21

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QUESTION

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REFERENCE

22

8 For n 6¼ 1, the vectors a, b and c are such that

a ¼1

n

n2

264

375, b ¼

2n

2n2 þ n

�1

264

375 and c ¼

n� 1

n2 � 1

1� n2

264

375

Determine the value of n for which a, b and c are linearly dependent. (9 marks)

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END OF QUESTIONS

Copyright � 2012 AQA and its licensors. All rights reserved.


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QUESTION

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Version 1.0

General Certificate of Education (A-level) January 2012

Mathematics

(Specification 6360)

MFP4

Further Pure 4

Final

Mark Scheme

Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all examiners participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every examiner understands and applies it in the same correct way. As preparation for standardisation each examiner analyses a number of students’ scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, examiners encounter unusual answers which have not been raised they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this Mark Scheme are available from: aqa.org.uk Copyright © 2012 AQA and its licensors. All rights reserved. Copyright AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester M15 6EX.

http://www.aqa.org.uk/

Key to mark scheme abbreviations M mark is for method m or dM mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation

or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A2,1 2 or 1 (or 0) accuracy marks –x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.

MFP4 Q Solution Marks Total Comments

1

Use of ab cosθ = a . b = 21

⇒ 25

7cos =θ

M1 A1

⇒ 25

1sin =θ B1 ft FT exact only

Use of | a × b | = ab sinθ = 3 M1 A1 5 CSO Total 5

2(a) Reflection in x = z M1 A1 2

(b) Rotation about the y-axis M1 A1

Through cos – 1 0.6 (≈ 53.13o) A1 3 Ignore direction Total 5

3(a) Char. Eqn. is λ2 – 8λ – 9 = 0 M1 Attempted Quadratic solved to get two roots dM1 ⇒ λ = 9, –1 A1 Substg. back λ (at least once) : M1

λ = 9 ⇒ –x + y = 0

⇒ λ = 9 has evecs. α

11

A1 any α ≠ 0

λ = –1 ⇒ x + y = 0

⇒ λ = –1 has evecs.β

−11

A1 6 any β ≠ 0

(b) (0, 0) B1 1

Total 7

MFP4 (cont) Q Solution Marks Total Comments

4(a) X XT =

− 713 x

−713

x M1 Attempted multn. with XT correct

=

−−+

50373792

xxx

A1 2

(b)

XTX =

−713

x

− 713 x

=

+−−49737310

2xxx

M1 Good attempt

X XT – XTX =

−++−

2

2

144441

xxxx

M1 Good attempt

Det(X XT – XTX) = 2

2

144441

xxxx−++−

= (x + 1)2 x

x−

−14

41

M1 Good attempt to factorise/expand the determinant

{ }16)1()1( 22 +−+−= xx ≤ 0 for all real x

E1 4 Explained/demonstrated fully

(c) x = –1 B1 1 CSO

Total 7


5(a)

17113

12

−− nn

Expanding the det. of the coefft. mtx. Setting it = 0 Obtaining & solving a quadratic eqn. in n

M1 M1 M1

0 = n2 + 17n – 18 = (n + 18)(n – 1) ⇒ n = 1, –18 A1 4 CSO

(b)

n = 1 gives 2x + y + z = 5 3x – y + z = 1 –x + 7y + z = 1

B1 ft their chosen integer n

Eliminating one variable from a pair of equations, twice M1

e.g. – ⇒ x – 2y = –4 and – ⇒ 4x – 8y = 0

A1 ft A1 ft

Inconsistency clearly demonstrated from fully correct working E1

3 planes have no common intersection (or form a ∆r prism) B1 ft 6

Also ft “3 planes meet in a common line” or “3 planes form a sheaf” if consistency conclusion made

Total 10


6(a) Use of scalar product on

712

and

− 41 3

M1

Sc.Prod. = ± 21 B1 Moduli 54 and 26 correct B1 Accept 7.348… & 5.099… AWRT 56o A1 4 From correct working

(b) 2x + y + 7t = 10 and 3x + y – 4t = 7 noted or used M1

Eliminating (say) y to get x as a fn. of t M1 x = 11t – 3 A1 CAO Substg. back for y M1 y = 16 – 29t A1 CAO

( )tzyx==

−−

=+

2916

113

B1 ft 6

(c)

Attempt at either

+−

+

7419

20

λλ

λ•

712

= 10 or

+−

+

7419

20

λλ

λ•

− 413

= 7

M1

Solving either 40 + 2λ – 1 + 9λ + 49 + 28λ = 10 or 60 + 3λ – 1 + 9λ – 28 – 16λ = 7

M1

λ1 = –2 λ2 = 6 A1A1

P = (18, –19, –1) and Q = (26, 53, 31)

PQ =

256627232728 222 ==++ M1A1 6 NB P, Q not required: d = | λ1 – λ2 | × | i + 9j + 4k | = 8 × 27 = 256 M1 A1

Total 16


7(a)

− cs

sc B1

−

=

YX

cssc

yx

M1

=

+−+

cYsXsYcX

A1 3

(b)(i)

( ) ( )( ) ( )2 26 7 8cX sY cX sY sX cY sX cY+ − + − + − − + =

( ) ( )

( )

2 2 2 2 2 2 2 2

2 2 2 2

2 6

7 2 8

c X csXY s Y c s XY sc Y X

s X csXY c Y

+ + − − + −

− − + = M1 Substn. for x & y in eqn. and multiplying

out

p = c2 + 6sc – 7s2

A1

q = 16cs – 6(c2 – s2) A1 AG

r = s2 – 6sc – 7c2

A1 4

(ii) Factorising:

3s2 + 8sc – 3c2 = (3s – c)(s + 3c) = 0 M1A1 Or by double angles

Deducing a tan value M1

tanθ = 31 (θ acute) A1

cosθ =

103

, sinθ = 101

A1 Both Substg. sensible values back for

p and r M1

2X 2 – 8Y 2 = 8 A1

112 2

2

2

2

=−YX

A1 8 CSO

(iii) Since C′ is a hyperbola, and it is just C

rotated, it follows that C is a hyperbola E1 1

Total 16


8

For considering 22

22

1112121

nnnnnnnn

−−−+−

B1 Or by scalar triple product

= (n – 1)nn

nnnnn

−−−++

1112

121

2

2 M1A1 For 1st factor

2

1 2 1( 1) 2 1

( 1) ( 1)(2 1) 0

nn n n n n

n n n n= − + +

+ + −

3 3 2'R R R= +

M1 Row ops. for 2nd factor

= (n – 1)(n + 1)012

12121

2

−++

nnnnnn

n M1A1

= (n – 1)(n + 1) { }122222 223223 +−−−−−++ nnnnnnnn OR

1 2 1( 1)( 1) 0 1

2 1 0

nn n n

n n= − +

−

2 2 1'R R nR= − =

(n – 1)(n + 1){ }122 22 +−− nnn M1 Full method for remaining factors = (n – 1)(n + 1)(n – 1)2 A1 n = –1 B1 9 CSO

Note: Expanding straightaway scores B1 M1 and then A1 for n4 – 2n3 + 2n – 1 Thereafter, M1 A1 for 1st factor, M1 for 2nd factor attempted and M1 for full method for remaining factors plus A1 and B1 cso at the end, as above.

Total 9 TOTAL 75


Surname

Other Names

Candidate Signature



June 2011

Mathematics MFP4

Unit Further Pure 4

Wednesday 22 June 2011 9.00 am to 10.30 am












from the booklet.



Question Mark

1

2

3

4

5

6

7

8

TOTAL

P39382/Jun11/MFP4 6/6/ MFP4(JUN11MFP401)

2


1 The matrices A and B are given in terms of p by

A ¼1 p 4

�3 2 1

2 �1 1

264

375 and B ¼

p 1 5

9 p �12 0 1

264

375

(a) Find each of detA and detB in terms of p. (3 marks)

(b) Without finding AB, determine all values of p for which AB is singular. (3 marks)

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4

2 The plane transformation T is the composition of a reflection in the line y ¼ x tan afollowed by an anticlockwise rotation about O through an angle b .

Determine the matrix which represents T, and hence describe T as a single

transformation. (6 marks)

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5

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(05)

QUESTION

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REFERENCE

6

3 Given the vectors p ¼1

4

7

2435 , q ¼

7

�24

24

35 and r ¼

2

3

t

2435 , where t is a scalar

parameter, determine the value of t in each of the following cases:

(a) p� q is parallel to r ; (3 marks)

(b) p, q and r are linearly dependent. (3 marks)

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7

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(07)

QUESTION

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REFERENCE

8

4 The system of equations S is given in terms of the real parameters a and b by

2xþ yþ 3z ¼ aþ 1

5x� 2yþ ðaþ 1Þz ¼ 3

axþ 2yþ 4z ¼ b

(a) Find the two values of a for which S does not have a unique solution. (4 marks)

(b) In the case when a ¼ 2 , determine the value of b for which S has infinitely many

solutions. (4 marks)

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9

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QUESTION

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REFERENCE

10

5 (a) (i) Find the eigenvalues and corresponding eigenvectors of A ¼ 1 3

�2 8

� �. (6 marks)

(ii) Hence write down each of the matrices U, D and U�1 such that A ¼ UDU�1 , whereD is a diagonal matrix. (4 marks)

(b) A 2� 2 matrix M has distinct real eigenvalues l and m , with corresponding

eigenvectors v1 and v2 .

(i) By considering the diagonalised form of M, determine the eigenvalues of M3 .

(2 marks)

(ii) Write down the eigenvectors of M3 . (1 mark)

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11

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QUESTION

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REFERENCE

12

6 (a) The transformation U of three-dimensional space is represented by the matrix

1 4 �32 �1 0

1 1 �1

264

375

(i) Write down a vector equation for the line L with cartesian equation

x� 1

2¼ y� 2

3¼ z� 3

6(2 marks)

(ii) Find a vector equation for the image of L under U, and deduce that it is a line

through the origin. (4 marks)

(b) The plane transformation V is represented by the matrix1 4

2 �1

� �.

L1 is the line with equation y ¼ 12xþ k , and L2 is the image of L1 under V.

(i) Find, in the form y ¼ mxþ c , the cartesian equation for L2 . (4 marks)

(ii) Deduce that L2 is parallel to L1 and find, in terms of k, the distance between these

two lines. (3 marks)

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13

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PART

REFERENCE

14

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P39382/Jun11/MFP4

(14)

QUESTION

PART

REFERENCE

15

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(15)

QUESTION

PART

REFERENCE

16

7 Let D ¼nðnþ 1Þ nþ 1 �1

0 1 n

1 �ðnþ 1Þ 1

��

��.

(a) (i) Show that ðn2 þ nþ 1Þ is a factor of D. (2 marks)

(ii) Hence, or otherwise, express D in factorised form. (2 marks)

(b) By expanding D directly, show that

D ¼ ½nðnþ 1Þ�2 þ fðnÞ

where fðnÞ can be expressed as the sum of two squares. (2 marks)

(c) Hence express the number 12 321 as the sum of three squares. (2 marks)

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17

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(17)

QUESTION

PART

REFERENCE

18

8 The diagram shows the plane P and the lines L and L’. The plane P and the line L

have equations

r .

3

�26

24

35 ¼ 37 and r ¼

1

2

�7

24

35þ l

2

1

2

2435

The line L does not lie in P , and intersects it at the point P.

(a) Determine the value of y , the angle between L and P , giving your answer to the

nearest 0.1�. (4 marks)

(b) Find the coordinates of P. (4 marks)

(c) The line L’ lies in P and is such that the angle between L and L’ is y , the angle

between L and P .

(i) Find a vector which is parallel to P and perpendicular to L. (3 marks)

(ii) Hence, or otherwise, find a vector equation for L’ in the form r ¼ aþ mb .(4 marks)

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P39382/Jun11/MFP4

QUESTION

PART

REFERENCE

P

L

Py

L’

(18)

19

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REFERENCE

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Version 1.0

General Certificate of Education (A-level) June 2011

Mathematics


MFP4

Further Pure 4

Final

Mark Scheme

Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all examiners participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the candidates’ responses to questions and that every examiner understands and applies it in the same correct way. As preparation for standardisation each examiner analyses a number of candidates’ scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, examiners encounter unusual answers which have not been raised they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this Mark Scheme are available from: aqa.org.uk Copyright © 2011 AQA and its licensors. All rights reserved. Copyright AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester M15 6EX.



Q Solution Marks Total Comments

1(a) det A = 5p – 1 B1 det B = p2 – 10p – 11 M1A1 3 M1A0 if num error(s) made (b) Use of det(AB) = det A det B B1 PI Finding three values of p M1 Allow correct factors here p = 5

1 , 11, –1 A1F 3 ft numerical errors in (a) Total 6 2

− αα

αα2cos2sin

2sin2cos&

−ββββ

cossinsincos

B1

used or written down

Mult’n of these in the correct order

B1

at least two entries correct

Use of addition formulae

M1

At least once

+−+++

)2cos()2sin()2sin()2cos(βαβαβαβα

A1F

ft only for use of clockwise rot’n and/or mult’n in wrong order

Reflection ... A1F ft as above ... in y = x tan ( )βα 2

1+ A1F 6 ft as above Total 6 3(a) Vector product attempted M1

p × q =

−=

−×

304530

42

7

741

A1

...

−=

232

15 , so t = –2 A1 3

(b) Scalar triple product attempted M1 OE, eg determinant

p × q • r =

•

− t32

232

15 = 15(13 – 2t) A1

... = 0, so t = 6 21 A1 3

ALT: 5p + q = 6r

... ⇒ t = 6 21

B2,0 B1

or any correct linear relationship

Total 6


4(a) 42

125312

aa +− = a2 + 3a – 10

M1 A1

Attempt at det of coeff matrix Correct (accept unsimplified)

Equating to 0 and solving quadratic in a a = 2, –5

m1 A1

4

SC: B1 for verifying a = 2 B1 for verifying a = −5

(b) bzyx

zyxzyx

=++=+−=++

4 2 233 2 533 2

B1

Eliminations leading to two equations in two variables

M1

Further elimination leading to value of b m1

b = 4 A1 4

ALT: Finding two variables in terms of

third M1 eg y = x and z = 1 – x

Substituting into third equation m1

b = 4 A1

8 5(a) (i) Characteristic eqn 01492 =+− λλ M1A1 M1A0 if num error(s) made

λ = 2, 7 A1 Substituting back for at least one eval m1 for λ = 2, –x + 3y = 0 or

for λ = 7, –2x + y = 0

evecs

13

and

21

A1A1 6 or non-zero multiples

(ii) U =

2113

, D =

7002

B1FB1F Columns of U and D are

interchangeable, but must match; ft wrong answers in (i)

U– 1 =

−

−3112

51

B1F B1F

4

1/det U adjoint matrix; ft incorrect U (provided det ≠ 0)

(b) (i) evals of M3 are 33 , µλ B1

since M3 = U D3 U – 1 E1 2

(ii) evecs of M3 are v1 and v2 B1 1 13


6 (a)(i)

+

=

632

321

λr B2,1 2

Any correct vector line equation; B1 if one vector correct, or if both correct but equation not in correct form

(ii) 1 4 3 1 2 42 1 0 2 31 1 1 3 6

λ λλ λλ λ

− + − = − + = − + −

r M1 A1 A1

Attempt at multiplication At least one entry correct All three correct

Clear and valid explanation that this is a line through O

E1

4

(b) (i)

+

− kp

p

2112

41=

−+

kpkp

23

43 B1

M1A1 For LHS

For RHS Answer satisfies y = 2

1 x – 3k A1 4 (ii) Equal gradients, hence parallel E1F ft if previous answer is of the

form y = 21 x + c

Distance = |k – c| cosθ with tanθ = 21 M1 Allow incorrect value of c here

...=

58k A1 3 Allow 3.58k

13 7 (a)(i)

Appropriate row/column operation M1 eg R1′ = R1 + R3 , R3′ = R3 + R1 or C3′ = C3 – nC2

∆ = 1)1(1

100012

+−

++

nn

nn

... = (n2 + n + 1)1)1(1

10001

+− nn A1 2 Factor correctly extracted

(ii) Expanding remaining determinant M1 OE ∆ = (n2 + n + 1)2 A1 2

(b) ∆ = (n2 + n)2 + 2n2 +2n + 1

… = (n2 + n)2 + (n + 1)2 + n2 B1 B1

2

Accept unsimplified

(c) Setting n = 10 M1

1112 = 12321 = 1102 + 112 + 102 A1 2

8


8(a) Use of sin or cosθ =moduli ofproduct

productscalar M1

using

−62

3 and

212

Numerator = 16, denominator = 21 B1B1 Allow numerator 185 sinθ =

2116 ⇒ θ ≈ 49.6o A1 4 Allow AWRT 49.6

(b) 3762

3

7λ22λ1λ2

=

−•

−++

M1

6λ + 3 – 2λ – 4 + 12λ – 42 = 37 ... ⇒ λ = 5

m1 A1

with attempt to solve

giving P = (11, 7, 3) B1F 4 ft wrong value of λ

(c)(i) Use of the vectors

−62

3 and

212

M1

Vector product attempted

Required vector is

−

7610

m1

A1

3

OE

Or a non-zero multiple

(ii)

a =

3711

B1F ft wrong answer in (b)

b =

−

7610

×

−62

3 =

28150

Fully correct equation for L'

M1A1F

A1 4

Or a non-zero multiple; ft wrong answer to (c)(i)

15 TOTAL 75


Surname

Other Names

Candidate Signature



January 2011

Mathematics MFP4

Unit Further Pure 4

Friday 28 January 2011 9.00 am to 10.30 am












from the booklet.



Question Mark

1

2

3

4

5

6

7

8

TOTAL

P38407/Jan11/MFP4 6/6/6/ MFP4(JAN11MFP401)

2


1 Let D ¼1 2 3

x y z

yþ z zþ x xþ y

��

��.

(a) Use a row operation to show that ðxþ yþ zÞ is a factor of D. (2 marks)

(b) Hence, or otherwise, express D as a product of linear factors. (2 marks)

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2 The non-zero vectors a and b have magnitudes a and b respectively.

Let c ¼ ja� bj and d ¼ ja . bj .

By considering the definitions of the vector and scalar products, or otherwise, show

that

c2 þ d 2 ¼ a2b2 (3 marks)

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5

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3 (a) Find the values of t for which the system of equations

txþ 2yþ 3z ¼ a

2xþ 3y� tz ¼ b

3xþ 5yþ ðt þ 1Þz ¼ c

does not have a unique solution. (3 marks)

(b) For the integer value of t found in part (a), find the relationship between a, b and c

such that this system of equations is consistent. (3 marks)

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7

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4 The non-singular matrix X ¼3 1 �32 4 3

�4 2 �1

24

35 .

(a) (i) Show that X2 � X ¼ kI for some integer k. (3 marks)

(ii) Hence show that X�1 ¼ 120ðX� IÞ . (2 marks)

(b) The 3� 3 matrix Y has inverse Y�1 ¼60 0 0

0 0 �100 20 0

24

35 .

Without finding Y , determine the matrix ðXYÞ�1 . (3 marks)

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9

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REFERENCE

10

5 The planes P1 and P2 have vector equations r .

6

2

9

2435 ¼ 5 and r .

10

�1�11

24

35 ¼ 4

respectively.

(a) Write down cartesian equations for P1 and P2 . (1 mark)

(b) Find a vector equation for the line of intersection of P1 and P2 . (5 marks)

(c) The plane P3 has cartesian equation 5xþ 3yþ 11z ¼ 28 .

Use your answer to part (b) to find the coordinates of the point of intersection of

P1 , P2 and P3 . (4 marks)

(d) Determine a vector equation for the plane which passes through the point ð4, 1, 9Þand which is perpendicular to both P1 and P2 . (3 marks)

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11

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13

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REFERENCE

14

6 The plane P has equation r .

12

15

16

24

35 ¼ 11 and the point Q has coordinates ð1, 1, �1Þ.

(a) Show that Q is in P . (1 mark)

(b) (i) Write down cartesian equations for the line l which passes through Q and is

perpendicular to P . (2 marks)

(ii) Deduce the direction cosines of l. (2 marks)

(c) The points M and N are on l, and each is 50 units from P .

Find the coordinates of M and N . (3 marks)

(d) Given that the point Pð5, 1, �4Þ is in P , determine the area of triangle PMN .

(3 marks)

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15

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REFERENCE

16

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QUESTION

PART

REFERENCE

17

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QUESTION

PART

REFERENCE

18

7 Let Y ¼3 �1 1

�1 3 1

1 1 3

24

35 .

(a) Show that 4 is a repeated eigenvalue of Y, and find the other eigenvalue of Y.

(7 marks)

(b) For each eigenvalue of Y, find a full set of eigenvectors. (5 marks)

(c) The matrix Y represents the transformation T.

Describe the geometrical significance of the eigenvectors of Y in relation to T.

(3 marks)

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REFERENCE

(18)

19

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22

8 The plane transformation T is represented by the matrix M ¼ �3 8

�1 3

� �.

(a) The quadrilateral ABCD has image A 0B 0C 0D 0 under T.

Evaluate detM and describe the geometrical significance of both its sign and its

magnitude in relation to ABCD and A 0B 0C 0D 0 . (3 marks)

(b) The line y ¼ px is a line of invariant points of T, and the line y ¼ qx is an invariant

line of T.

Show that p ¼ 12and determine the value of q. (5 marks)

(c) (i) Find the 2� 2 matrix R which represents a reflection in the line y ¼ 12x . (2 marks)

(ii) Given that T is the composition of a shear, with matrix S , followed by a reflection in

the line y ¼ 12x , determine the matrix S and describe the shear as fully as possible.

(5 marks)

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QUESTION

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REFERENCE

Version 1.0

General Certificate of Education (A-level) January 2011

Mathematics


MFP4

Further Pure 4

Mark Scheme

Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all examiners participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the candidates’ responses to questions and that every examiner understands and applies it in the same correct way. As preparation for standardisation each examiner analyses a number of candidates’ scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, examiners encounter unusual answers which have not been raised they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this Mark Scheme are available from: aqa.org.uk Copyright © 2011 AQA and its licensors. All rights reserved. Copyright AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester M15 6EX.

Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011

3




4


1(a) Δ =

1 2 3x y z

x y z y z x z x y+ + + + + +

M1

e.g. R3′ = R3 + R2

=

1 2 3( )

1 1 1x y z x y z+ +

A1

2

(b) Expanding remaining det.

Δ = (x + y + z)(x – 2y + z) M1 A1

2

Total 4 2 c = | a × b | = ab sinθ

d = | a • b | = ab |cosθ | B1 B1

Condone lack of | - |

c2 + d2 = a2b2(cos2θ + sin2θ ) = a2b2 B1 3 Legitimately shown Total 3

3(a) 2 32 33 5 1

tt

t−+

= 8t2 – 7t – 1 = 0 M1

M1

Attempt at det. of coefft. mtx. (or equivalent) Equating to zero and solving a quadratic eqn. in t

t = 1, 18− A1 3

(b)

t = 1 ⇒ 2 3

2 3 3 5 2

x y z ax y z b

x y z c

+ + =+ − =

+ + =

B1

FT any integer value found

E.g. ➀ + ➁ – ➂ ⇒ a + b = c M1 A1 3

Total 6 4(a)

(i) X2 = 23 1 32 24 34 2 19

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

M1 A1

≥ 5 correct for the M All 9 correct for the A

X2 – X = 20I i.e. k = 20 A1 3 Shown legitimately (ii) Multg. X2 – X = 20I by X – 1 M1 Re-arranging X – I = 20 X – 1

to get X – 1 = 120 ( )−X I

A1

2

Legitimately

(b)

X – 1 = 120

2 1 32 3 34 2 2

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥− −⎣ ⎦

B1

Noted or used

(XY) – 1 = Y – 1 X – 1 M1 Incl. attempt at the multn. 6 3 9

2 1 12 3 3

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

A1

3

Total 8


5

MFP4(cont) Q Solution Marks Total Comments

5(a) ∏1 : 6x + 2y + 9z = 5 ∏2 : 10x – y – 11z = 4

B1

1

Both

(b) Method 1

E.g. ∏1 + 2 ∏2 ⇒ 13(2x – z = 1) M1

11 2x z λ+= =

M1 A1

Parametrisation attempt

Using 1st two to find 3rd in terms of λ: x = λ, z = 2λ – 1 ⇒ y = 7 – 12λ M1

r =

0 17 121 2

λ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥+ −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

A1

5

Any correct vector eqn.form

Method 2 (0, 7, –1) or ( )7 1

12 6, 0, or ( )12 , 1, 0 (M1A1) Finding a pt. on the line

6 10 12 1 ( )13 129 11 2

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥× − = ± −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(M1) (A1)

Finding a d.v.

r =

0 17 121 2

λ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥+ −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

(A1)

Any correct vector eqn.form

(c) Substg. x = λ, y = 7 – 12λ, z = 2λ – 1

into 5x + 3y + 11z = 28

M1

Solving a linear eqn. in λ : M1 λ = – 2 A1 CAO ( – 2, 31, – 5) B1 4 FT their λ in their line eqn.

(d) n =

1122

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

B1

FT when used as part of a plane eqn., not a line

d =

4 11 129 2

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥• −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= 10

M1

Attempted

112 102

⎡ ⎤⎢ ⎥• − =⎢ ⎥⎢ ⎥⎣ ⎦

r

A1

3

Any correct vector eqn. form FT

Alt. r =

4 6 101 2 19 9 11

λ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ + −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(M1) (A1) (A1)

Plane eqn. attempt Point + at least 1 d.v. All correct

Total 13


6


6(a) 1 121 15 12 15 16 111 16

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥• = + − =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

shown

B1

1

(b) (i) eqn., or use, of line incorporating

p.v. 111

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

and d.v. 121516

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

M1

1 1 1

12 15 16x y z− − += =

A1

2

(ii) 2 2 212 15 16+ + = 25 B1 FT

12 15 16, ,

25 25 25

B1

2

FT

(c)

Use of r = 111

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

+ λ 121516

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

with λ = ± 2

M1

(25, 31, 31) and (– 23, – 29, – 33) A1 A1 3

(d) Method 1 PQ = 5 B1 Area ΔPMN = 1

2 PQ MN× × = 250 M1 A1 3 (Since Q = midpt.MN) Method 2 20

3035

PM⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

, 283029

PN−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

(M1)

Attempted

9( )20 20

12PM PN

⎡ ⎤⎢ ⎥× = ± −⎢ ⎥⎢ ⎥⎣ ⎦

attempted

within an area formula

(M1)

||gm. or Δ

Area ΔPMN = 250 (A1) CAO Total 11


7


7(a) Attempt at Char.Eqn. λ3 – 9λ2 + 24λ – 16 = 0

M1 A3,2,1

One each following coefft.

Attempt at (at least) one linear factor M1 (λ – 1)(λ – 4)2 = 0 A1 λ = 1, 4, 4 A1 7

(b) λ = 1 ⇒

2 02 0

2 0

x y zx y z

x y z

− + =− + + =

+ + = ⇒

111

α⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

M1 A1

Substg. back and solving

λ = 4 ⇒ x + y – z = 0 B1 Choosing any two independent vectors

satisfying this M1

E.g.s:

101

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

, 011

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

, 112

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A1

5

(c) 1

11

α⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

≡ a line of invariant points

M1 A1

Invariant line LoIPs

e.g.

101

α⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

+011

β⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

≡ an invariant plane

B1

3

Total 15


8


8(a) Det(M) = – 1 B1 Magnitude = 1 ⇒ area invariant B1 FT area s.f. – ve sign ⇒ cyclic order of vertices is

reversed OR “reflection” involved

B1

3

(b) Method 1

Char. Eqn.: λ2 – 1 = 0 ⇒ λ = ± 1 M1 A1 Finding and solving attempt Substg. back: λ = 1 ⇒ y = 1

2 x

and λ = –1 ⇒ y = 14 x

M1 A1

A1

5

Method 2 3 8

1 3x

mx−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

= (8 3)(3 1)

m xm x

−⎡ ⎤⎢ ⎥−⎣ ⎦

(M1)

Attempted

Use of y′ = mx′: 3m – 1 = 8m2 – 3m (M1) Solving a quadratic eqn. in m = 1

4 , 12 (M1A1) From (4m – 1)(2m – 1) = 0

p = 12 and q = 1

4 (A1)

(c) (i) p = 1

2 = tanθ

⇒ cos2θ = 35 and sin2θ = 4

5

M1

For these attempted and used in a reflection matrix

R =

3 45 5

345 5

⎡ ⎤⎢ ⎥⎢ ⎥

−⎢ ⎥⎣ ⎦

A1

2

(ii) Use 3 45 5

345 5−

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

S = 3 81 3

−⎡ ⎤⎢ ⎥−⎣ ⎦

M1

FT their R

S found using inverse matrix M1 Or equivalent method

= 3 45 5

345 5

⎡ ⎤⎢ ⎥⎢ ⎥

−⎢ ⎥⎣ ⎦

3 81 3

−⎡ ⎤⎢ ⎥−⎣ ⎦

= 15

13 369 23

−⎡ ⎤⎢ ⎥−⎣ ⎦

A1

Shear, parallel to y = 12 x B1 CAO

mapping (e.g.) (1, 1) → (4.6, 2.8) B1 5 FT any pt. and its image Total 15 TOTAL 75


Surname

Other Names

Candidate Signature



June 2010

Mathematics MFP4

Unit Further Pure 4

Tuesday 15 June 2010 9.00 am to 10.30 am












from the booklet.



Question Mark

1

2

3

4

5

6

7

8

TOTAL

P28066/Jun10/MFP4 6/6/6/ MFP4(JUN10MFP401)

2


1 The position vectors of the points P, Q and R are, respectively,

p ¼3

4

�1

24

35, q ¼

�1

2

2

24

35 and r ¼

1

4

1

24

35

(a) Show that p, q and r are linearly dependent. (2 marks)

(b) Determine the area of triangle PQR. (4 marks)

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2 Let A ¼ 1 x

2 3

� �, B ¼ 1 �1

2 2

� �and C ¼ 4� 4x 8

8x� 4 4

� �.

(a) Find AB in terms of x. (2 marks)

(b) Show that BTAT ¼ C for some value of x. (5 marks)

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5

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6

3 The plane P1 is perpendicular to the vector 9i� 8jþ 72k and passes through the

point Að2, 10, 1Þ .

(a) Find, in the form r . n ¼ d , a vector equation for P1. (3 marks)

(b) Determine the exact value of the cosine of the acute angle between P1 and the

plane P2 with equation r . ðiþ jþ kÞ ¼ 11 . (4 marks)

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PART

REFERENCE

8

4 The fixed points A and B and the variable point C have position vectors

a ¼3

�4

1

24

35, b ¼

2

1

�3

24

35 and c ¼

2� t

t

5

24

35

respectively, relative to the origin O, where t is a scalar parameter.

(a) Find an equation of the line AB in the form ðr� uÞ � v ¼ 0 . (3 marks)

(b) Determine b� c in terms of t. (4 marks)

(c) (i) Show that a . ðb� cÞ is constant for all values of t, and state the value of this

constant. (2 marks)

(ii) Write down a geometrical conclusion that can be deduced from the answer to

part (c)(i). (1 mark)

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9

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(09)

QUESTION

PART

REFERENCE

10

5 Factorise fully the determinant

x y z

x2 y2 z2

yz zx xy

�� . (8 marks)

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11

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QUESTION

PART

REFERENCE

12

6 The line L and the plane P have vector equations

r ¼7

8

50

24

35þ t

6

2

�9

24

35 and r ¼

�2

0

�25

24

35þ l

5

3

4

24

35þ m

1

6

2

24

35

respectively.

(a) (i) Find direction cosines for L. (2 marks)

(ii) Show that L is perpendicular to P . (3 marks)

(b) For the system of equations

6pþ 5qþ r ¼ 9

2pþ 3qþ 6r ¼ 8

�9pþ 4qþ 2r ¼ 75

form a pair of equations in p and q only, and hence find the unique solution of this

system of equations. (5 marks)

(c) It is given that L meets P at the point P.

(i) Demonstrate how the coordinates of P may be obtained from the system of equations

in part (b). (2 marks)

(ii) Hence determine the coordinates of P. (2 marks)

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13

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15

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QUESTION

PART

REFERENCE

16

7 The transformation T is represented by the matrix M with diagonalised form

M ¼ UDU�1

where U ¼ 4 �1

1 3

� �and D ¼ 27 0

0 1

� �.

(a) (i) State the eigenvalues, and corresponding eigenvectors, of M. (4 marks)

(ii) Find a cartesian equation for the line of invariant points of T. (2 marks)

(b) Write down U�1 , and hence find the matrix M in the form

a b

c d

� �

where a, b, c and d are integers. (5 marks)

(c) By finding the element in the first row, first column position of Mn , prove that

4� 33nþ1 þ 1

is a multiple of 13 for all positive integers n. (5 marks)

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17

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18

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QUESTION

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REFERENCE

19

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QUESTION

PART

REFERENCE

20

8 The matrix12 16

�9 36

� �represents the transformation which is the composition, in

either order, of the two plane transformations

E: an enlargement, centre O and scale factor k ðk > 0Þ

and

S: a shear parallel to the line l which passes through O

Show that k ¼ 24 and find a cartesian equation for l. (7 marks)

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21

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23

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DO NOT WRITE ON THIS PAGEANSWER IN THE SPACES PROVIDED

Version 1.0

General Certificate of Education June 2010

Mathematics MFP4

Further Pure 4

Mark Scheme

klm

Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation meeting attended by all examiners and is the scheme which was used by them in this examination. The standardisation meeting ensures that the mark scheme covers the candidates’ responses to questions and that every examiner understands and applies it in the same correct way. As preparation for the standardisation meeting each examiner analyses a number of candidates’ scripts: alternative answers not already covered by the mark scheme are discussed at the meeting and legislated for. If, after this meeting, examiners encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this Mark Scheme are available to download from the AQA Website: www.aqa.org.uk Copyright © 2010 AQA and its licensors. All rights reserved. COPYRIGHT AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester M15 6EX

MFP4 - AQA GCE Mark Scheme 2010 June series

3

Key to mark scheme and abbreviations used in marking M mark is for method m or dM mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation

or ft or F follow through from previous incorrect result

MC

mis-copy

CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A2,1 2 or 1 (or 0) accuracy marks NOS not on scheme –x EE deduct x marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.


4


1(a) 3 4 11 2 2

1 4 1

−− = 6 + 8 + 4 + 2 – 24 + 4

M1

Good attempt at det M0 for | .. | = 0 and no working

or 3(2 – 8) – 4(–1 – 2) – 1(–4 – 2) etc or 3(2 – 8) + 1(4 + 4) + 1(8 + 2) etc Correctly shown = 0 A1 Or 3p + 4q = 5r (M1)

(A1)

2

(b) For attempt at 2 of (±) PQ , PR , QR

Area ΔPQR = 12 QP QR× e.g.

= 12 4 2 3

2 0 2−−

i j k= 1

2 (4 2 4 )± − +i j k

M1

M1

Formula used with attempt at a vector product of any 2 of the above (ignore missing 1

2 for now)

= 2 2 212 4 2 4+ + M1

Method for finding magnitude of their relevant vector

= 3 A1 4 CSO Total 6

2(a) 2 1 2 18 4

x x+ −⎡ ⎤= ⎢ ⎥⎣ ⎦

AB M1

A1

2

Good attempt (at least one entry in R1 ) All four correct

(b)

BTAT = (AB)T Or 1 21 2

⎡ ⎤⎢ ⎥−⎣ ⎦

1 23x

⎡ ⎤⎢ ⎥⎣ ⎦

= 2 1 82 1 4

xx

+⎡ ⎤⎢ ⎥−⎣ ⎦

M1

A1

Ft their (a) Or CAO

2x + 1 = 4 – 4x Or 2x – 1 = 8x – 4 M1 Ft previous answers x = 1

2 A1 CAO Checking/noting x = 1

2 in other eqn. B1 5 Visibly Total 7

3(a) Clearly identifying n =

98

72

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

d =9 28 10

72 1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− •⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= 10

B1

M1 A1

3

(b)

Use of Sc.prod. of normalsprod. of their moduli

M1 Must be (9i – 8j + 72k), (i + j + k)

or their n from (a) Nr. = 73

Dr. = 73 3 or 15987 B1 B1

Ft their n from (a) only

cosθ =

13

A1

4 CAO Allow unsimplified exact forms

Total 7


5


4(a) (v =) ± (a – b) = ±

15

4

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

M1 A1

M1 A0 if AB± found but not stated/shown this is v

u =

34

1

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

or 213

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

B1

3

(b) b × c =

3 52 1 3 3 16

2 5 3 2

tt

t t t

+⎡ ⎤⎢ ⎥− = −⎢ ⎥

− ⎢ ⎥−⎣ ⎦

i j k

M1

A3,2,1

4

(c) (i) a • b × c =

34

1

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

•3 53 163 2

ttt

+⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦

= 77

M1

A1

2

Or starting again: 3 4 12 1 3

2 5t t

−−

− CAO

(ii) C never lies in plane of O, A, B (or is a fixed distance from it) or Vol. //ppd. OABC always = 77 or Vol. tetrhdrn. OABC always = 77

6 or O is never in plane of A, B, C or OA , OB , OC never co-planar

B1

1 Any suitable geometrical comment

Vectors ; points

Total 10 5

2 2 2

x y z

x y zyz zx xy

Δ =

2 2 2 2 2

( ) ( )

x y x z x

x y x z xyz z x y y x z

− −

= − −− −

M1 M1

By C2′ = C2 – C1 (eg) C3′ = C3 – C1 (eg)

2

1 1

( )( )

x

y x z x x y x z xyz z y

= − − + +− −

A1 A1

First two factors extracted (what’s left has to be correct also)

2

1 0

( )( )

x

y x z x x y x z yyz z z y

= − − + −− −

M1

By C3′ = C3 – C2 (e.g.)

2

1 0

( )( )( ) 11

x

y x z x z y x y xyz z

= − − − +−

A1

3rd factor extracted

( )( )( )( )x y y z z x xy yz zx= − − − + + M1

A1

8

Further R/C ops or expansion of remaining det (almost a dM1) CAO up to equivalents due to re-positioning of the signs

Alternatives using Cyclic Symmetry and the Factor Theorem are fine

Total 8


6

MFP4 (cont)

Q Solution Marks Total Comments 6(a)(i) • = 2 2 26 2 9+ + attempted and

6 2 9, , −⎛ ⎞±⎜ ⎟• • •⎝ ⎠

M1

• = 11 and all correct A1 2 ± (0.545, 0.182, – 0.818) ok

(ii) Either

5 1 63 6 3 24 2 9

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥× = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Explaining that d.v. of L is in dirn. of Π ’s nml. ⇒ L ⊥r Π

M1 A1

B1

Correct vector product only here

Or 6 52 3 09 4

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥• =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

and 6 12 6 09 2

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥• =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

Explaining that d.v. of L is ⊥r to 2 (non-//) vectors in Π ⇒ L ⊥r Π

(M1) (A1)

(B1)

3

Not just stating

(b) E.g. 6×① – ②: 46 = 34p + 27q M1 Eliminating r from any pair of eqns.

2×① – ③: –57 = 21p + 6q

② – 3×③: –217 = 29p – 9q

A1 A1

Any 2 correct eqns (1 mark each)

2×④ + 9×⑤: 605 = – 121p M1 Solving a 2×2 system (any means) in order to get values for p, q, r

p = –5, q = 8, r = –1 A1 5 All 3 CAO

(c)

(i)

7 6 2 58 2 0 3 650 9 25 4 2

ttt

λ μλ μ

λ μ

+ = − + ++ = + +− = − + +

→

9 6 58 2 3 675 9 4 2

tt

t

λ μλ μ

λ μ

= − + += − + += + +

M1

Including re-arrangement attempt

i.e. the above system with p = – t, q = λ and r = μ

A1

2

(ii) Substg. t = 5 into L’s eqn.

Or λ = 8 and μ = – 1 into ∏ ’s eqn. M1 P = (37, 18, 5) A1 2 CAO Total 14


7

MFP4 (cont)

Q Solution Marks Total Comments 7(a)(i) Evals λ = 27 1 B1 Both

Evecs (α)

41⎡ ⎤⎢ ⎥⎣ ⎦

and (β)1

3−⎡ ⎤⎢ ⎥⎣ ⎦

B1 B1 B1

4

Correctly matched up with evals (look out for λ1, v1 notations)

(ii)

y = – 3x from λ = 1

B1

B1

2

Ft 4y = x if evecs mis-matched Must say why they have chosen this one

(b) U – 1 = 1

13

3 11 4

⎡ ⎤⎢ ⎥−⎣ ⎦

B1 B1

Det; mtx

M = U D U – 1

= 113

4 11 3

−⎡ ⎤⎢ ⎥⎣ ⎦

27 00 1

⎡ ⎤⎢ ⎥⎣ ⎦

3 11 4

⎡ ⎤⎢ ⎥−⎣ ⎦

M1 Including attempt to multiply (at least U D …)

= 1

13

4 11 3

−⎡ ⎤⎢ ⎥⎣ ⎦

81 271 4

⎡ ⎤⎢ ⎥−⎣ ⎦

or 113

108 127 3

−⎡ ⎤⎢ ⎥⎣ ⎦

3 11 4

⎡ ⎤⎢ ⎥−⎣ ⎦

A1

Ft incorrect/missing U – 1 for one product; ignore missing 1

13 until the end

=

25 86 3

⎡ ⎤⎢ ⎥⎣ ⎦

A1

5 CAO

(c) Mn = U Dn U – 1

Dn = 27 00 1

n⎡ ⎤⎢ ⎥⎣ ⎦

Mn (1,1) = ( )113 12 27 1n× +

So 3 3 14 3 3 1 4 3 1n n+× × + = × + div. by 13 Since M has all integer elements, each element of Mn is an integer also

M1

B1

A1

E1

E1

5

Including attempt to multiply Legitimately so from their working, from fact that the element is an integer Explaining why it must be an integer

Total 16


8

MFP4 (cont)

Q Solution Marks Total Comments 8 det W = 12.36 + 9.16 = 576 = k2 M1 Attempt at det. = k2

⇒ k = 24 A1 2

124 W =

1 22 3

3 38 2

⎡ ⎤⎢ ⎥−⎣ ⎦

B1

1 22 3

3 38 2

xy

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦

1 22 33 32 8

x yy x

+⎡ ⎤⎢ ⎥−⎣ ⎦

M1 A1

Equating this to

xy⎡ ⎤

=⎢ ⎥⎣ ⎦

M1

y = 34 x A1 CAO

ALT. 1

124 W =

1 22 3

3 38 2

⎡ ⎤⎢ ⎥−⎣ ⎦

(B1)

1 22 3

3 38 2

xmx

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦

( )( )

1 22 33 32 8

m xm x

⎡ ⎤+⎢ ⎥−⎣ ⎦

(M1) (A1)

Setting y′ = mx′ and solving for m (M1) Get (4m – 3)2 = 0 y = 3

4 x (A1) CAO ALT. 2 λ2 – 2λ + 1 = 0

⇒ λ = 1 (twice) (M1) (A1)

This may simply be stated or assumed

λ = 1 ⇒

1 22 33 18 2

00

x yx y

− + =− + =

(M1) (A1)

y = 34 x (A1) 5

Total 7 TOTAL 75

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