General Certificate of Education Advanced Subsidiary Examination January 2013
Mathematics
MFP1
Unit Further Pure 1 Friday 18 January 2013
1.30 pm to 3.00 pm
d
For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
e
s
Time allowed * 1 hour 30 minutes
n
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 each question in the space provided for that question. If you require extra space, use an AQA supplementary answer book; do not use the space provided for a different question. * 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.
e
d
n
o
C
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.
P56803/Jan13/MFP1 6/6/
MFP1
2
A curve passes through the point ð1, 3Þ and satisfies the differential equation
1
dy x ¼ dx 1 þ x 3 Starting at the point ð1, 3Þ, use a step-by-step method with a step length of 0.1 to estimate the value of y at x ¼ 1:2 . Give your answer to four decimal places. (5 marks)
Solve the equation w 2 þ 6w þ 34 ¼ 0 , giving your answers in the form p þ qi , where p and q are integers. (3 marks)
2 (a)
It is given that z ¼ ið1 þ iÞð2 þ iÞ .
(b) (i)
Express z in the form a þ bi , where a and b are integers.
(ii) Find integers m and n such that z þ mz* ¼ ni .
3 (a)
(3 marks) (3 marks)
Find the general solution of the equation pffiffiffi 3 p sin 2x þ ¼ 4 2 giving your answer in terms of p .
(b)
(6 marks)
Use your general solution to find the exact value of the greatest solution of this equation which is less than 6p . (2 marks)
ð1 4
(02)
Show that the improper integral
1 pffiffiffi dx has a finite value and find that value. 25 x x (4 marks)
P56803/Jan13/MFP1
3
The roots of the quadratic equation
5
x 2 þ 2x 5 ¼ 0 are a and b . (a)
Write down the value of a þ b and the value of ab.
(2 marks)
(b)
Calculate the value of a 2 þ b 2 .
(2 marks)
(c)
Find a quadratic equation which has roots a 3 b þ 1 and ab 3 þ 1 .
(5 marks)
The matrix X is defined by
6 (a)
(i)
m Given that X ¼ 3 2
1 2 . 3 0
2 , find the value of m. 6
(1 mark)
(ii) Show that X3 7X ¼ nI , where n is an integer and I is the 2 2 identity matrix.
(4 marks)
1 0 It is given that A ¼ . 0 1
(b) (i)
Describe the geometrical transformation represented by A.
(1 mark)
(ii) The matrix B represents an anticlockwise rotation through 45 about the origin.
1 1 Show that B ¼ k , where k is a surd. 1 1
(2 marks)
(iii) Find the image of the point Pð1, 2Þ under an anticlockwise rotation through 45
about the origin, followed by the transformation represented by A.
Turn over
s
(03)
(4 marks)
P56803/Jan13/MFP1
4
The variables y and x are related by an equation of the form
7
y ¼ ax n where a and n are constants. Let Y ¼ log10 y and X ¼ log10 x . (a)
Show that there is a linear relationship between Y and X .
(b)
The graph of Y against X is shown in the diagram.
(3 marks)
Y 6 5 4 3 2 1 O
1
2
3
4
5
6
7
Find the value of n and the value of a.
8 (a)
X (4 marks)
Show that n X
2rð2r 2 3r 1Þ ¼ nðn þ pÞðn þ qÞ2
r ¼1
where p and q are integers to be found. (b)
(6 marks)
Hence find the value of 20 X
2rð2r 2 3r 1Þ
(2 marks)
r ¼11
(04)
P56803/Jan13/MFP1
5
An ellipse is shown below.
9
y P O
A
B
x
The ellipse intersects the x-axis at the points A and B. The equation of the ellipse is ðx 4Þ2 þ y2 ¼ 1 4 (a)
Find the x-coordinates of A and B.
(b)
The line y ¼ mx ðm > 0Þ is a tangent to the ellipse, with point of contact P. (i)
(2 marks)
Show that the x-coordinate of P satisfies the equation ð1 þ 4m2 Þx 2 8x þ 12 ¼ 0
(3 marks)
(ii) Hence find the exact value of m.
(4 marks)
(iii) Find the coordinates of P.
(4 marks)
Copyright ª 2013 AQA and its licensors. All rights reserved.
(05)
P56803/Jan13/MFP1
Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp
mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) 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.
MFP1 - AQA GCE Mark Scheme 2013 January series
Q 1
Solution y n h f ( xn )
Marks
/
h y′ (1) = 0.1 y (1) (=0.05)
M1
y (1.1) ≈ 3 + 0.05 = 3.05
A1
y n 1
y(1.2)≈y(1.1)+ 0.1 y / (1.1) =3.05+ 0.1 y / (1.1)
Comments OE Attempt to find h y′ (1). PI by eg 3.05 for y(1.1)
m1
Attempt to find y(1+0.1)+0.1×y′(1+0.1) must see evidence of calculation if correct ft [0.047..+c’s y(1.1)] value not obtained
≈ 3.05 + 0.047(19…..)
A1F
OE; ft on [0.047..+c’s y(1.1)] value; PI
≈ 3.0972 (to 4 d.p.)
A1
≈ 3.05 + 0.1
1.1 1 1.13
1100 3.05 0.1 2331
Total 2(a)
Total
(w=)
6 36 4(34) 6 100 2 2
6 10i 2 = −3 ± 5i
5
5 Correct substitution into quadratic formula OE
M1
B1
=
A1
(b)(i) z = i(1+i)(2+i) = i(2+3i+i2) = 2i +3i2 + i3
Must be 4 dp.
100 = 10i or 3
100 /2 = 5i
−3 ± 5i (p= −3, q = ±5) NMS mark as 3/3 or 0/3
M1
Attempt to expand all brackets.
= 2i +3(−1)+i(−1)
B1
i2 = −1 used at least once
= −3 + i
A1
(ii) z* = −3 − i −3 + i + m(−3 − i) = ni −3−3m=0; 1−m=n
m = −1, n = 2
−3 + i (a = −3, b = 1)
B1F
OE Ft c’s a − bi
M1
Equating both real parts and the imag. parts, PI by next line Both correct
A1
Total
3
3 9
MFP1 - AQA GCE Mark Scheme 2013 January series
Q 3(a)
Solution
sin
2 3 3 2
sin
2x
x
2 n
4
3
24
2x
;
1 2 n 2 3 4
GS: x = n
(b)
;
;
x
4
2 n
2 3
1 2 2 n 2 3 4
x = n
5 24
n = 5 (gives greatest soln 0
A1 4
Total
Dep on correct full GS.
8 M1
(+c)
Both in ACF, but must now be exact and in terms of for A2. A1 if decimal approx used. Applying a correct value for n which gives greatest soln.0 so) m
(iii)
Comments OE Sub y=0 in eqn of ellipse and either eliminate fraction or take sq root, condoning missing ±, ie ( x 4) ()1 2 Both 2 and 6 NMS Mark as B2 or B0
12
4
4 2 x 8 x 12 0 ; 3
Valid method to solve a correct
4 x 2 24 x 36 0 x 2 6x 9 0
x
( 8) 0 ; 8 3
Subst value for m in LHS of eqn (b)(i); ft on c’s value of m.
M1
2
1 2 (1 4 ) x 8 x 12 (=0) 12
quadratic equation; as far as either correct subst into quadratic formula with b2 − 4ac evaluated to 0 or correct factorisation or correct value
m1
(x − 3)2 (= 0)
of x after
better seen.; OE, correct use of −b/2a
x=3
Coordinates of P are 3 ,
4 2 x 8 x 12 0 or 3
A1
3 12
A1 4
Total TOTAL
13 75
Must see earlier justification Correct coordinates with the y-coord in any correct exact form 3 eg . 2 3 NMS SC 1 for 3 , 12
Scaled mark unit grade boundaries - January 2013 exams A-level Maximum Scaled Mark
A*
LAW UNIT 3
80
66
MD01
MATHEMATICS UNIT MD01
75
-
63
57
52
47
42
MD02
MATHEMATICS UNIT MD02
75
68
62
55
49
43
37
MFP1
MATHEMATICS UNIT MFP1
75
-
69
61
54
47
40
MFP2
MATHEMATICS UNIT MFP2
75
67
60
53
47
41
35
MFP3
MATHEMATICS UNIT MFP3
75
68
62
55
48
41
34
MFP4
MATHEMATICS UNIT MFP4
75
68
61
53
45
37
30
MM1B
MATHEMATICS UNIT MM1B
75
-
58
52
46
40
35
MM2B
MATHEMATICS UNIT MM2B
75
66
59
52
46
40
34
MPC1
MATHEMATICS UNIT MPC1
75
-
64
58
52
46
40
MPC2
MATHEMATICS UNIT MPC2
75
-
62
55
48
41
35
MPC3
MATHEMATICS UNIT MPC3
75
69
63
56
49
42
36
MPC4
MATHEMATICS UNIT MPC4
75
58
53
48
43
38
34
MS1A
MATHEMATICS UNIT MS1A
100
-
78
69
60
52
44
MS/SS1A/W
MATHEMATICS UNIT S1A - WRITTEN
75
58
34
MS/SS1A/C
MATHEMATICS UNIT S1A - COURSEWORK
25
20
10
MS1B
MATHEMATICS UNIT MS1B
75
-
60
54
48
42
36
MS2B
MATHEMATICS UNIT MS2B
75
70
66
58
50
42
35
MEST1
MEDIA STUDIES UNIT 1
80
-
54
47
40
33
26
MEST2
MEDIA STUDIES UNIT 2
80
-
63
54
45
36
28
MEST3
MEDIA STUDIES UNIT 3
80
68
58
48
38
28
18
MEST4
MEDIA STUDIES UNIT 4
80
74
68
56
45
34
23
Code LAW03
Title
Scaled Mark Grade Boundaries and A* Conversion Points A B C D E 60
54
48
43
38
