ΑΣΚΗΣΕΙΣ ΜΑΘΗΜΑΤΙΚΩΝ ΟΞΦΟΡΔΗΣ
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Transcript of 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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(05)
QUESTION
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REFERENCE
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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(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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(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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(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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(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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QUESTION
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REFERENCE
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(21)
QUESTION
PART
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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QUESTION
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END OF QUESTIONS
Copyright � 2012 AQA and its licensors. All rights reserved.
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P45892/Jan12/MFP4
(24)
QUESTION
PART
REFERENCE
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.
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
MFP4 (cont) Q Solution Marks Total Comments
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
MFP4 (cont) Q Solution Marks Total Comments
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
MFP4 (cont) Q Solution Marks Total Comments
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
MFP4 (cont) Q Solution Marks Total Comments
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
Centre Number Candidate Number
Surname
Other Names
Candidate Signature
General Certificate of Education
Advanced Level Examination
June 2011
Mathematics MFP4
Unit Further Pure 4
Wednesday 22 June 2011 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.
For Examiner’s Use
Examiner’s Initials
Question Mark
1
2
3
4
5
6
7
8
TOTAL
P39382/Jun11/MFP4 6/6/ MFP4(JUN11MFP401)
2
Answer all questions in the spaces provided.
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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3
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(03)
QUESTION
PART
REFERENCE
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
PART
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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QUESTION
PART
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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REFERENCE
(08)
9
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(09)
QUESTION
PART
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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P39382/Jun11/MFP4
QUESTION
PART
REFERENCE
(10)
11
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(11)
QUESTION
PART
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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P39382/Jun11/MFP4
QUESTION
PART
REFERENCE
(12)
13
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PART
REFERENCE
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(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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QUESTION
PART
REFERENCE
(16)
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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QUESTION
PART
REFERENCE
P
L
Py
L’
(18)
19
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QUESTION
PART
REFERENCE
20
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END OF QUESTIONS
Copyright � 2011 AQA and its licensors. All rights reserved.
Do not writeoutside the
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P39382/Jun11/MFP4
(20)
QUESTION
PART
REFERENCE
Version 1.0
General Certificate of Education (A-level) June 2011
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 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.
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.
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
Q Solution Marks Total Comments
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
Q Solution Marks Total Comments
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
Q Solution Marks Total Comments
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
Centre Number Candidate Number
Surname
Other Names
Candidate Signature
General Certificate of Education
Advanced Level Examination
January 2011
Mathematics MFP4
Unit Further Pure 4
Friday 28 January 2011 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.
For Examiner’s Use
Examiner’s Initials
Question Mark
1
2
3
4
5
6
7
8
TOTAL
P38407/Jan11/MFP4 6/6/6/ MFP4(JAN11MFP401)
2
Answer all questions in the spaces provided.
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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REFERENCE
(02)
3
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QUESTION
PART
REFERENCE
4
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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QUESTION
PART
REFERENCE
6
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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QUESTION
PART
REFERENCE
8
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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(08)
9
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(09)
QUESTION
PART
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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QUESTION
PART
REFERENCE
13
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QUESTION
PART
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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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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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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END OF QUESTIONS
Copyright � 2011 AQA and its licensors. All rights reserved.
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Version 1.0
General Certificate of Education (A-level) January 2011
Mathematics
(Specification 6360)
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
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.
Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011
4
MFP4 Q Solution Marks Total Comments
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
Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011
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
Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011
6
MFP4(cont) Q Solution Marks Total Comments
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
Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011
7
MFP4(cont) Q Solution Marks Total Comments
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
Mark Scheme – General Certificate of Education (A-level) Mathematics – Further Pure 4 – January 2011
8
MFP4(cont) Q Solution Marks Total Comments
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
Centre Number Candidate Number
Surname
Other Names
Candidate Signature
General Certificate of Education
Advanced Level Examination
June 2010
Mathematics MFP4
Unit Further Pure 4
Tuesday 15 June 2010 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.
For Examiner’s Use
Examiner’s Initials
Question Mark
1
2
3
4
5
6
7
8
TOTAL
P28066/Jun10/MFP4 6/6/6/ MFP4(JUN10MFP401)
2
Answer all questions in the spaces provided.
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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(03)
QUESTION
PART
REFERENCE
4
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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QUESTION
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REFERENCE
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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7
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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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QUESTION
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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
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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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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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QUESTION
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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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24
There are no questions printed on this page
Copyright � 2010 AQA and its licensors. All rights reserved.
P28066/Jun10/MFP4
(24)
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.
MFP4 - AQA GCE Mark Scheme 2010 June series
4
MFP4 Q Solution Marks Total Comments
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
MFP4 - AQA GCE Mark Scheme 2010 June series
5
MFP4 (cont) Q Solution Marks Total Comments
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
MFP4 - AQA GCE Mark Scheme 2010 June series
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
MFP4 - AQA GCE Mark Scheme 2010 June series
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
MFP4 - AQA GCE Mark Scheme 2010 June series
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