General Certificate of Education Advanced Level Examination January 2013
Mathematics
MFP2
Unit Further Pure 2 Wednesday 23 January 2013
9.00 am to 10.30 am
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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
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Time allowed * 1 hour 30 minutes
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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.
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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.
P59577/Jan13/MFP2 6/6/6/
MFP2
2 1 (a)
Show that 12 cosh x 4 sinh x ¼ 4e x þ 8ex
(b)
(2 marks)
Solve the equation 12 cosh x 4 sinh x ¼ 33 giving your answers in the form k ln 2 .
(5 marks)
Two loci, L1 and L2 , in an Argand diagram are given by pffiffiffi L1 : jz þ 6 5ij ¼ 4 2
2
L2 : argðz þ iÞ ¼
3p 4
The point P represents the complex number 2 þ i . (a)
Verify that the point P is a point of intersection of L1 and L2 .
(2 marks)
(b)
Sketch L1 and L2 on one Argand diagram.
(6 marks)
(c)
The point Q is also a point of intersection of L1 and L2 . Find the complex number that is represented by Q. (2 marks)
3 (a)
(b)
Show that
1 1 A ¼ , stating the value of the constant A. 5r 2 5r þ 3 ð5r 2Þð5r þ 3Þ (2 marks)
Hence use the method of differences to show that n X
1 n ¼ ð5r 2Þð5r þ 3Þ 3ð5n þ 3Þ r¼1
(c)
Find the value of 1 X
1 ð5r 2Þð5r þ 3Þ r¼1
(02)
(4 marks)
(1 mark)
P59577/Jan13/MFP2
3
The roots of the equation
4
z 3 5z 2 þ kz 4 ¼ 0 are a, b and g . (a) (i)
Write down the value of a þ b þ g and the value of abg .
(ii) Hence find the value of a 2 bg þ ab 2 g þ abg 2 .
(2 marks) (2 marks)
The value of a 2 b 2 þ b 2 g 2 þ g 2 a 2 is 4 .
(b) (i)
Explain why a, b and g cannot all be real.
(ii) By considering ðab þ bg þ gaÞ2 , find the possible values of k.
5 (a)
e y ey Using the definition tanh y ¼ y , show that, for jxj < 1 , e þ ey 1þx 1 1 tanh x ¼ ln 2 1x d 1 ðtanh1 xÞ ¼ . dx 1 x2
(b)
Hence, or otherwise, show that
(c)
Use integration by parts to show that ð1 2
4 tanh
1
0
(1 mark) (4 marks)
(3 marks)
(3 marks)
m 3 x dx ¼ ln n 2
where m and n are positive integers.
(5 marks)
A curve is defined parametrically by
6
x ¼ t3 þ 5 ,
y ¼ 6t 2 1
The arc length between the points where t ¼ 0 and t ¼ 3 on the curve is s. (a)
Show that s ¼
ð3 3t
pffiffiffiffiffiffiffiffiffiffiffiffiffi t 2 þ A dt , stating the value of the constant A.
(4 marks)
0
(b)
(4 marks) Turn over
s
(03)
Hence show that s ¼ 61 .
P59577/Jan13/MFP2
4
The polynomial pðnÞ is given by pðnÞ ¼ ðn 1Þ3 þ n3 þ ðn þ 1Þ3 .
7 (a) (i)
Show that pðk þ 1Þ pðkÞ , where k is a positive integer, is a multiple of 9. (3 marks)
(ii) Prove by induction that pðnÞ is a multiple of 9 for all integers n 5 1 . (b)
(4 marks)
Using the result from part (a)(ii), show that nðn2 þ 2Þ is a multiple of 3 for any positive integer n. (2 marks)
pffiffiffi Express 4 þ 4 3 i in the form reiy , where r > 0 and p < y 4 p . (3 marks) pffiffiffi (b) (i) Solve the equation z 3 ¼ 4 þ 4 3 i , giving your answers in the form reiy , where r > 0 and p < y 4 p . (4 marks) pffiffiffi (ii) The roots of the equation z 3 ¼ 4 þ 4 3 i are represented by the points P, Q and R on an Argand diagram. pffiffiffi Find the area of the triangle PQR, giving your answer in the form k 3 , where k is an integer. (3 marks) pffiffiffi (c) By considering the roots of the equation z 3 ¼ 4 þ 4 3 i , show that
8 (a)
cos
2p 4p 8p þ cos þ cos ¼0 9 9 9
(4 marks)
Copyright Ó 2013 AQA and its licensors. All rights reserved.
(04)
P59577/Jan13/MFP2
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.
MFP2 - AQA GCE Mark Scheme 2013 January series
MFP2 Q 1(a)
Solution 1 x cosh x (e e x ) 2 1 or sinh x (e x e x ) 2 12cosh x 4sinh x =
Marks
Total
Comments or 12cosh x 6(e x e x ) or 4sinh x 2(e x e x )
M1
6(e x e x ) 2(e x e x ) 12cosh x 4sinh x 4e x 8e x (b)
A1 cso
2
AG
4e x 8e x 33
4e 33e 8 2x
x
( 0)
M1
(e x 8)(4e x 1) ( 0) e x 8,
attempt to multiply by ex to form quadratic in ex
e x
1 4
x 3ln 2 x 2ln 2 Total
m1
factorisation attempt (see below) or correct use of formula
A1
correct roots
A1 A1
5 7
MFP2 - AQA GCE Mark Scheme 2013 January series
MFP2 (cont) Q 2(a)
Solution
Marks
4 4i 16 16 32 4 2
B1
3 4
B1
arg(–2+ 2i) = tan 1 (1) =
Total
Comments
verification that 2 i 6 5i 4 2 2
verification that arg (z+ i) =
3 4
Im
Re
(b) Circle
M1
freehand circle sketched
Centre at – 6 + 5i
A1
clear from diagram or centre stated
Cutting Re axis but not cutting Im axis
A1
“Straight” line
M1
freehand line
Half line from 0 – i
A1
not horizontal or vertical but end point at 0 – i must be clear from diagram/stated
gradient –1 (approx)
A1
6
making 45 to negative Re axis and positive Im axis
(c) Calculation based on fact that L2 passes through centre of L1
Q represents
A1
–10 + 9i
Total
4 idea of vector from centre 4
M1 2
10
must write as a complex number
MFP2 - AQA GCE Mark Scheme 2013 January series
MFP2 (cont) Q 3(a)
Solution
Marks
5r 3 5r 2 1 1 5r 2 5r 3 5r 2 5r 3
M1
(b)
5
5r 2 5r 3
Attempt to use method of differences
1 1 k 3 5n 3 5n 3 3 k 3(5n 3)
S
1 15
Total
Comments
condone omission of brackets for M1 2
A=5
M1
at least 2 terms of correct form seen
A1
correct cancellation leaving correct two fractions
m1
1 5n 3 3 n Sn 5 3(5n 3) 3(5n 3) (c)
A1cso
Total
A1cso
4
B1
1 7
attempt to write with common denominator 1 AG k used correctly throughout 5
MFP2 - AQA GCE Mark Scheme 2013 January series
MFP2 (cont) Q 4(a)(i)
(ii)
Solution 5 4
Marks
Total
B1 B1
2
2 2 2 ( )
M1
= 5 4 20 (b)(i)
A1
Comments
2
FT their results from (a)(i)
1
argument must be sound
If , , are all real then
2 2 2 2 2 2 0 Hence , , cannot all be real
E1
k
B1
k
M1
correct identity for
(ii)
PI
2
2 2( 2 2 2 ) 2
4 2(20)
A1 A1 cso
k 6
Total
4
9
2
substituting their result from (a)(ii) must see k=…
MFP2 - AQA GCE Mark Scheme 2013 January series
MFP2 (cont) Q
Solution e e y x tanh y y e e y xe y xe y e y e y
Marks
Total
Comments
y
5(a)
M1
or
xe 2 y x e 2 y 1
( x 1)e y e y (1 x) ( x 1) e2 y (1 x) e2 y
(b)
A1
1 x 1 1 x y ln 1 x 2 1 x
A1cso
1 1 y ln(1 x) ln(1 x) 2 2 dy 1 1 d x 2(1 x) 2(1 x) 1 x 1 x 2 1 2 2(1 x)(1 x) 2(1 x ) 1 x 2
3
AG
3
AG
M1 A1 A1cso
Alternative 1 d y 1 (1 x) d 1 x M1 d x 2 (1 x) d x 1 x
d y 1 (1 x) (1 x) (1 x) A1 d x 2 (1 x) (1 x)2 dy 1 A1 cso d x 1 x2 (c)
4 tanh
1
4x dx 1 x2 4 x tanh 1 x 2ln(1 x 2 )
x dx 4 x tanh 1 x
A1
12 ln 3
B1
must simplify logarithm to ln3
ln 3 2ln 34
A1
any correct form
tanh Value of integral =
1 1 2
M1
3 ln 4 2 3
Total
A1cso
5 11
all working must be correct
MFP2 - AQA GCE Mark Scheme 2013 January series
MFP2 (cont) Q 6(a)
Solution
dx 3t 2 dt
Marks
dy 12t dt
Comments
B1
both correct
dx dy 4 2 9t 144t dt dt
M1
dx dy ‘their’ dt dt
9t 4 144t 2 dt
A1
OE
2
s
2
2
3
s 3t t 2 16 dt
A1cso
0
(b)
Total
3
4
3
A1
(t 2 16) 2 3
A = 16
where k is a constant; ft their A
M1
k (t 2 A) 2
3
m1
25 2 16 2 = 61
A1 cso Total
2
F(3) – F(0) 4 8
AG
MFP2 - AQA GCE Mark Scheme 2013 January series
MPC1 (cont) Q
Solution
Marks
Total
Comments
7(a)(i) p(k 1) p(k ) k 3 (k 1)3 (k 2)3
(k 1)3 k 3 (k 1)3
M1
(k 2)3 (k 1)3 k 3 6k 2 12k 8 (k 2 3k 2 3k 1) 9k 2 9k 9 9(k 2 k 1) which is a multiple of 9 (since k2 + k + 1 is an integer )
A1
A1cso
(ii) p(1) = 1 + 8 = 9 p(1) is a multiple of 9
p(k+1) = p(k) + 9(k 2 k 1) or p(k+1) = p(k) + 9N Assume p(k) is a multiple of 9 so p(k) = 9M, where M is integer p( k 1) = 9M + 9N = 9(M + N) p(k 1) is a multiple of 9
multiplied out & correct unsimplified
3
correct algebra plus statement
B1
result true for n = 1
M1
p(k+1) = … and result from part (i) considered and mention of divisible by 9 must have word such as “assume” for A1
A1
Result true for n = 1 therefore true for n = 2, n = 3 etc by induction. ( or p(n) is a multiple of 9 for all integers n 1 )
E1
convincingly shown 4
(b) p(n) (n 1)3 n3 (n 1)3
must earn previous 3 marks before E1 is scored
need to see this OE as evidence
= 3n 6n 3
or 3n(n 2 2)
B1
p(n) = 3n(n 2 2) & p(n) is a multiple of 9. Therefore n(n 2 2) is a multiple of 3 (for any positive integer n.) Total
both of these required E1
2 9
plus concluding statement
MFP2 - AQA GCE Mark Scheme 2013 January series
MFP2 (cont) Q 8(a)
Solution
Marks
r 8
Total
Comments
B1
tan 1
4 3 or seen 4 3
or
2 3
A1
(b)(i) modulus of each root = 2
3
4 4 3i 8 e
4 2 8 , , 9 9 9
A2
1 2 (ii) Area = 3 PO OR sin 2 3 1 2 3 2 2 sin 2 3 =3 3
4
A1 if 3 “correct” values not all in requested interval
()4 2 8 2 cos cos cos 9 9 9
e
cos
, 2e
i
2 9
, 2e
i
8 9
A1
correct values of lengths in formula 3
E1 M1
must be stated explicitly in form r (cos i sin )
A1
isolating real terms ; correct and with “2”
4 4 cos explicitly stated to 9 9 earn final A1 mark
or cos
A1cso Total TOTAL
4 9
Correct expression for area of triangle PQR
4 4 cos isin seen earlier 9 9
2 4 8 cos cos 0 9 9 9
i
M1
A1cso
(c) Sum of roots (of cubic) = 0 Sum of 3 roots including Im terms
4 9
2 3
use of De Moivre – dividing argument by 3
2e
i
i
B1 M1
marked as angle to Im axis with 6 “vector” in second quadrant on Arg diag
M1
4 14 75
AG
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