General Certificate of Education Advanced Level Examination January 2012
Mathematics
MFP2
Unit Further Pure 2 Friday 20 January 2012
1.30 pm to 3.00 pm
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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 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.
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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.
P46339/Jan12/MFP2 6/6/6/
MFP2
2
Show, by means of a sketch, that the curves with equations
1 (a)
y ¼ sinh x y ¼ sech x
and
have exactly one point of intersection.
(4 marks)
Find the x-coordinate of this point of intersection, giving your answer in the form a ln b . (4 marks)
(b)
p . 6 (1 mark)
Draw on an Argand diagram the locus L of points satisfying the equation arg z ¼
2 (a)
(b) (i)
A circle C, of radius 6, has its centre lying on L and touches the line ReðzÞ ¼ 0 . (2 marks) Draw C on your Argand diagram from part (a).
(ii) Find the equation of C, giving your answer in the form j z z0 j ¼ k .
(3 marks)
(iii) The complex number z1 lies on C and is such that arg z1 has its least possible value.
Find arg z1 , giving your answer in the form pp, where 1 < p 4 1 .
(2 marks)
A curve has cartesian equation
3
1
y ¼ 2 lnðtanh xÞ (a)
Show that dy 1 ¼ dx sinh 2x
(b)
(02)
(4 marks)
The points A and B on the curve have x-coordinates ln 2 and ln 4 respectively. Find the arc length AB , giving your answer in the form p ln q, where p and q are rational numbers. (8 marks)
P46339/Jan12/MFP2
3
The sequence u1 , u2 , u3 , … is defined by
4
u1 ¼
3 4
unþ1 ¼
3 4 un
Prove by induction that, for all n 5 1 , 3nþ1 3 un ¼ nþ1 3 1
(6 marks)
Find the smallest positive integer values of p and q for which
5
p p p cos þ i sin 8 8 ¼i p p q cos i sin 12 12
6 (a)
(b)
(7 marks)
Express 7 þ 4x 2x 2 in the form a bðx cÞ2 , where a, b and c are integers. (2 marks) By means of a suitable substitution, or otherwise, find the exact value of ð5
dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 7 þ 4x 2x 2 2
Turn over
s
(03)
(6 marks)
P46339/Jan12/MFP2
4
The numbers a , b and g satisfy the equations
7
a 2 þ b 2 þ g 2 ¼ 10 12i ab þ bg þ ga ¼ 5 þ 6i (a)
Show that a þ b þ g ¼ 0 .
(b)
The numbers a , b and g are also the roots of the equation
(2 marks)
z 3 þ pz2 þ qz þ r ¼ 0 Write down the value of p and the value of q.
(2 marks)
It is also given that a ¼ 3i .
(c) (i)
Find the value of r.
(3 marks)
(ii) Show that b and g are the roots of the equation
z 2 þ 3iz 4 þ 6i ¼ 0
(2 marks)
(iii) Given that b is real, find the values of b and g .
(3 marks)
8 (a)
(b)
Write down the five roots of the equation z5 ¼ 1 , giving your answers in the (1 mark) form eiy , where p < y 4 p . Hence find the four linear factors of z4 þ z3 þ z2 þ z þ 1
(c)
Deduce that 2
z þzþ1þz
(d)
(3 marks)
1
þz
2
¼
2p z 2 cos þ z 1 5
4p z 2 cos þ z 1 5
2p ¼ Use the substitution z þ z 1 ¼ w to show that cos 5
pffiffiffi 51 . 4
(4 marks)
(6 marks)
Copyright ª 2012 AQA and its licensors. All rights reserved.
(04)
P46339/Jan12/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 January 2012
Q
Solution y
1(a)
Marks
Total
Comments
x
Sketch y = sinh x
B1
gradient > 0 at (0, 0); no asymptotes
Sketch y = sech x : Symmetry about x = 0 with max point Asymptote y = 0 Point (0, 1) marked or implied
B1 B1 B1
must not cross x-axis
1 cosh x sinh 2 x = 2 Use of ln 1 = x ln 2 + 5 2 or 1 2x e − e −2 x ) = 2 OE ( 2 e 4 x − 4e 2 x − 1 = 0 Correct use of formula Result
(b)
sinh x =
(
4
M1 M1 m1
)
A1
Total
use of double angle formula dependent on previous M2 4
(M1)
incorrect sinh x, cosh x M0 (no marks)
(M1) (m1) (A1)
ie multiply by e 2 x and rewrite (4) 8
MFP2 January 2012
Q
Solution 2(a)
Marks
Total
Comments
B1
1
condone a short line, ie it stops at or inside circle
B1 B1
2
not touching Re axis
Im
Re z1
Half-line with gradient < 1
(b)(i) Circle centre on L, x-coord 6 indicated touching Re z = 0 not at (0, 0) (ii) y-coord of centre is 2 3 or
6 3
B1
z0= 6 + 2 3 i, k=6
B1F, B1
(iii) Point z1 shown 1 argπz1 = − 6
ft error in coords of centre 3
B1 B1 Total
3(a)
OE; PI
dy 1 = dx 2 tanh x
PI 2 8
B1 × sech 2 x
1 2sinh x cosh x 1 = sinh 2x
=
B1 for expressing in terms of sinh x and cosh x
M1 A1
4
M1
use of formula; accept √ inserted at any stage
m1
relevant use of cosh 2 − sinh 2 = 1
A1
OE
2
(b)
1 dy 1+ = 1+ sinh 2 2 x dx cosh 2 2 x sinh 2 2 x cosh 2 x = sinh 2 x 1 Integral is ln sinh 2 x 2 255 15 sinh(2ln 4) = sinh(2ln 2) = 32 8 1 17 s = ln 2 4 Total =
AG; PI by previous line
M1A1
M1 for ln sinh
B1B1
PI
A1F
8 12
ft error in
1 2
MFP2 January 2012
Q
Solution 4 Assume result true for n = k 3 Then uk +1 = 3k +1 − 3 4 − k +1 3 −1
=
Marks
3(3k +1 − 1) 4(3 − 1) − (3k +1 − 3)
A1
4 × 3k +1 − 3k +1 = 3k + 2 3k + 2 − 3 uk +1 = k + 2 3 −1 32 − 3 3 n =1 = = u1 32 − 1 4 Induction proof set out properly
A1
Fraction = e
B1 E1
pπi 8
πi
− qπi 12
M1
(3 p + 2 q )
Alternative 1 Numerator = cos p8π + isin − qπ 12
pπ 8
(
m1
OE
A1F A1
ft errors of sign or arithmetic slips CAO
7
(B1) − qπ 12
Denominator = cos + isin Fraction = qπ qπ cos p8π + isin p8π cos 12 + isin 12
)(
qπ pπ needs more than just cos 12 − sin 12
(B1)
)
= cos 24π ( 3p + 2q ) + isin 24π ( 3p + 2q )
(M1) (A1)
( 3p + 2q ) = 0
3p + 2q = 12 = p 2,= q 3 Alternative 2 LHS cos p8π + isin
RHS i cos
allow for attempt to subtract powers
A1
12πi
= i if cos
must have earned previous 5 marks
B1
i = e 24 3p + 2q = 12 = p 2,= q 3
π 24
6 6
B1
pπi qπi + 8 12
= e 24
clearly shown
A1
Total
Denominator = e
Comments
M1
k +1
5 Numerator = e
Total
qπ 12 qπ 12
+ sin
or sin 24π ( 3p + 2q ) = 1
(A1F) (A1) pπ 8 qπ 12
p π π qπ = − 8 2 12 3p + 2q = 12 = p 2,= q 3
(7)
CAO
(7)
CAO (correct answers, insufficient working 3/7 only)
(B1) (B1)
qπ = cos p8π sin = or sin p8π cos 12
Introduction of
(m1)
(M1)
π 2
(m1) (A1) (A1F) (A1) Total
7
MFP2 January 2012
Q 6(a)
Solution 7 + 4 x − 2 x 2 =9 − 2( x − 1) 2
(b) Put = u
Marks M1A1
2( x − 1)
A1F
1 du ∫ 2 9 − u2
1 u sin −1 3 2 Change limits or replace u π 4 2
or
Alternative – if integration is attempted without substitution: sin −1 1
2 (x – 1) 2 3 Substitution of limits π 4 2
(6)
CAO
(A1F)
(A1F) (m1) (A1)
7(a) Use of ( ∑ α ) =∑ α + 2 ∑ αβ p= 0, q= 5 + 6i
use 3iβγ = –r
−27i + 15i − 18 + r = 0 or βγ = 5 + 6i + α 2
= r 18 + 12i
(ii) Cubic is ( z − 3i)( z 2 + 3iz − 4 + 6i) or use of βγ and β + γ (iii)
CAO
(A1)
2
(c)(i) Substitute 3i for z or
6
(M1)
Total
(b)
provided sin −1
m1 A1
1 2
outside integrand u for sin −1 p
A1
π 2 8
2
ft error in (a); must have u 2 only, ie
A1F
=
=
Comments
u k ( x − 1) any k allow=
M1
du = 2 dx I=
Total 2
f (−2) = 0 or equate imaginary parts
8 M1 A1
2
B1,B1
2
AG
allow for 3iβγ = r
M1 A1 A1F
3
any form one error
M1A1
2
clearly shown
3
correct answers no working and no check B1 only
M1
β= −2, γ = 2 − 3i
A1,A1F
Total
12
MFP2 January 2012
Q
Solution 8(a)
(b)
2πi 5
4πi 5
1, e , e , e
−2πi 5
,e
−4πi 5
z5 − 1 = z4 + z3 + z 2 + z + 1 z −1 2πi 4πi −2πi −4πi 5 5 5 5 = − − − − z e z e z e z e
Marks B1
Total 1
B1 M1A1
(c) Correct grouping of linear factors 2πi −2πi 2π e 5 +e 5 = 2cos 5 2π 4π 2 2 z − 2cos z + 1 z − 2cos z + 1 5 5 2 ÷ z to give answer
M1
(d) Substitute into LHS to give w2 + w − 1 2π 4π RHS w − 2cos w − 2cos 5 5 2 Solve w + w − 1 = 0 −1 ± 5 w= 2 2π 5 −1 cos = 5 4 with reasons for choice Total TOTAL
B1
Comments accept e
B0 if assumed 3
A1
6πi
clearly shown
4
B1 M1 A1 A1 E1
6 14 75
8πi
accept if e 5 , e 5 used here
A1 A1
0
AG
Scaled mark unit grade boundaries - January 2012 exams A-level Max. Scaled Mark
Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E
Code
Title
LAW02 LAW03
LAW UNIT 2 LAW UNIT 3
94 80
69
MD01 MFP1 MM1A MM1B MPC1
MATHEMATICS UNIT MD01 MATHEMATICS UNIT MFP1 MATHEMATICS UNIT MM1A MATHEMATICS UNIT MM1B MATHEMATICS UNIT MPC1
75 75 100 75 75
-
MS1A MATHEMATICS UNIT MS1A MS/SS1A/W MATHEMATICS UNIT S1A - WRITTEN MS/SS1A/C MATHEMATICS UNIT S1A - COURSEWORK
-
73 63
66 57
59 51
52 45
46 40
62 56 50 44 67 60 53 46 no candidates were entered for this unit 59 52 46 40 61 55 49 43
39 39 34 37
100 75 25
-
74 54 20
65
56
47
38 28 10
MS1B MD02 MFP2 MM2B MPC2 MS2B MFP3 MPC3 MFP4 MPC4
MATHEMATICS UNIT MS1B MATHEMATICS UNIT MD02 MATHEMATICS UNIT MFP2 MATHEMATICS UNIT MM2B MATHEMATICS UNIT MPC2 MATHEMATICS UNIT MS2B MATHEMATICS UNIT MFP3 MATHEMATICS UNIT MPC3 MATHEMATICS UNIT MFP4 MATHEMATICS UNIT MPC4
75 75 75 75 75 75 75 75 75 75
69 59 69 69 67 64 60 63
56 64 52 63 66 63 60 57 54 57
49 57 45 55 59 55 52 50 48 51
42 50 38 47 52 47 44 43 42 45
36 44 31 39 46 40 37 37 37 39
30 38 25 32 40 33 30 31 32 33
MEST1 MEST2 MEST3 MEST4
MEDIA STUDIES UNIT 1 MEDIA STUDIES UNIT 2 MEDIA STUDIES UNIT 3 MEDIA STUDIES UNIT 4
80 80 80 80
67 74
55 63 57 68
47 54 47 56
40 45 37 45
33 36 27 34
26 28 18 23
PHIL1
PHILOSOPHY UNIT 1
90
-
55
49
43
37
32
