General Certificate of Education Advanced Level Examination January 2012
Mathematics
MFP3
Unit Further Pure 3 Monday 23 January 2012
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 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.
P46349/Jan12/MFP3 6/6/6/
MFP3
2
The function yðxÞ satisfies the differential equation
1
dy ¼ f ðx, yÞ dx where
yx f ðx, yÞ ¼ 2 y þx yð1Þ ¼ 2
and (a)
Use the Euler formula yrþ1 ¼ yr þ h f ðxr , yr Þ with h ¼ 0:1 , to obtain an approximation to yð1:1Þ .
(b)
(3 marks)
Use the formula yrþ1 ¼ yr1 þ 2h f ðxr , yr Þ with your answer to part (a), to obtain an approximation to yð1:2Þ , giving your answer to three decimal places. (3 marks)
2
Find pffiffiffiffiffiffiffiffiffiffiffi 4þx2 x!0 x þ x2 lim
3
(3 marks)
Solve the differential equation d2 y dy þ 10y ¼ 26e x þ 2 2 dx dx given that y ¼ 5 and y ¼ f ðxÞ .
(02)
dy ¼ 11 when x ¼ 0 . Give your answer in the form dx (10 marks)
P46349/Jan12/MFP3
3 4 (a)
By using an integrating factor, find the general solution of the differential equation dy 2 þ y ¼ ln x dx x
(b)
Hence, given that y ! 0 as x ! 0 , find the value of y when x ¼ 1 .
ð1 5 (a)
(b)
Explain why
1 2
xð1 2xÞ dx is an improper integral. x 2 þ 3e4x
ð1
6 (a) (b)
(3 marks)
(1 mark)
By using the substitution u ¼ x 2 e4x þ 3 , find ð
(c)
(7 marks)
Hence evaluate
1 2
xð1 2xÞ dx x 2 þ 3e4x
(3 marks)
xð1 2xÞ dx , showing the limiting process used. x 2 þ 3e4x
Given that y ¼ ln cos 2x , find
d4 y . dx 4
(4 marks)
(6 marks)
Use Maclaurin’s theorem to show that the first two non-zero terms in the expansion, 4 (3 marks) in ascending powers of x, of ln cos 2x are 2x 2 x 4 . 3
(c)
Turn over
s
(03)
Hence find the first two non-zero terms in the expansion, in ascending powers of x, (2 marks) of ln sec2 2x .
P46349/Jan12/MFP3
4
It is given that, for x 6¼ 0 , y satisfies the differential equation
7
d2 y dy x 2 þ 2ð3x þ 1Þ þ 3yð3x þ 2Þ ¼ 18x dx dx Show that the substitution u ¼ xy transforms this differential equation into
(a)
d2 u du þ 9u ¼ 18x þ 6 dx 2 dx
(4 marks)
Hence find the general solution of the differential equation
(b)
x
d2 y dy þ 2ð3x þ 1Þ þ 3yð3x þ 2Þ ¼ 18x 2 dx dx
giving your answer in the form y ¼ f ðxÞ .
(8 marks)
The diagram shows a sketch of the curve C with polar equation
8
r ¼ 3 þ 2 cos y ,
0 4 y 4 2p
Initial line O
(a)
Find the area of the region bounded by the curve C.
(b)
A circle, whose cartesian equation is ðx 4Þ2 þ y 2 ¼ 16 , intersects the curve C at the points A and B. (i)
(6 marks)
Find, in surd form, the length of AB .
(6 marks)
(ii) Find the perimeter of the segment AOB of the circle, where O is the pole.
(3 marks)
Copyright ª 2012 AQA and its licensors. All rights reserved.
(04)
P46349/Jan12/MFP3
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.
Q
Solution 2 − 1 1(a) y(1.1) = y(1) + 0.1 4 + 1 = 2 + 0.02 = 2.02
Marks
Total
M1A1 A1
(b) y(1.2) = y(1) + 2(0.1){f[1.1, y(1.1)]}
3
M1
2.02 − 1.1 = 2 + 2(0.1) 2 2.02 + 1.1 = 2.035518… = 2.036 to 3dp
A1F A1 Total
ft on c’s answer to (a) 3 6
M1
x 2 1 4 + x − 2 4 + O( x ) 4 + O( x) = = 2 2 x + x x + x 1 + x
m1
Division by x stage before taking the limit
4 + x − 2 1 lim = 2 x →0 x+x 4
A1
1 x x 2 4 + x = 21 + = 2 1 + + O( x 2 ) 4 24
Total 3
CAO Must be 2.036 Attempt to use binomial theorem OE The notation O(xn) can be replaced by a term of the form kxn
1
2
Comments
3
CSO NMS 0/3
3
m + 2m + 10 = 0 m = −1 ± 3i
M1 A1
PI
Complementary function is (y =) e − x ( A cos 3 x + B sin 3 x)
A1F
OE Ft on incorrect complex value of m
Particular integral: try y = kex k + 2k + 10k = 26 ⇒ k = 2
M1 A1
(GS y =) e − x ( A cos 3 x + B sin 3 x) + 2ex
B1F
c’s CF+ c’s non-zero PI but must have 2 arb consts
x = 0, y = 5 ⇒ 5 = A + 2 so A = 3
B1F
ft c’s k ie A = 5 − k, k ≠ 0
dy = dx e − x (−3 A sin 3 x + 3B cos3 x − A cos3 x − B sin 3 x) + 2e x
M1
Attempt to differentiate c’s GS (ie CF + PI)
11 = 3B − A + 2 ( B = 4) y = e − x (3 cos 3x + 4 sin 3x) + 2ex
A1 A1
2
Total
10 10
CSO
Q
Solution 4(a) IF is exp (
Marks
2
∫ x dx)
= e2lnx = x2
[ ]
d yx 2 = x 2 ln x dx
⇒ yx2 =
=
∫
(ln x)
d x3 dx 3
x3 x2 ln x − dx 3 3
∫
x3 x3 ln x − +A 3 9 x x { y = ln x − + Ax − 2 } 3 9
Total
Comments
M1
and with integration attempted
A1 A1
PI
M1
LHS; PI
M1
Attempt integration by parts in correct direction to integrate xp lnx
A1
RHS
yx2 =
A1
7
(b) Now, as x → 0, xk lnx → 0
E1
Must be stated explicitly for a value of k>0
As x → 0, y → 0 ⇒ A = 0
B1
Const of int = 0 must be convincing
yx2 =
x3 x3 ln x − 3 9
When x = 1, y = −
1 9
B1F Total
3 10
ft on one slip but must have made a realistic attempt to find A
Q
Solution 5(a) The interval of integration is infinite
Marks E1
(b) = u x 2 e −4 x + 3 ⇒ du = (2xe–4x – 4x2e–4x) dx
M1
1 2 x(1 − 2 x)e −4 x ∫ 2 × x 2e−4 x + 3 dx 1 1 = × du 2 u 1 1 = ln u + c = ln ( x 2 e −4 x + 3) {+ c} 2 2
A1
Total 1
Comments OE du/dx or ‘better’
x(1 − 2 x) ∫ x 2 + 3e4 x dx =
∫
(c) I =
∫
∞ 1 2
a
= lim
a→∞
=
3
OE Condone missing c. Accept later substitution back if explicit
x(1 − 2 x) dx x 2 + 3e 4 x
= lim ∫ 1 a →∞
A1
2
x(1 − 2 x) dx x 2 + 3e 4 x
(
)
(
)
M1
1 e −2 {ln a 2 e −4 a + 3 – ln( + 3)} 2 4
M1
Uses part (b) and F(a) – F(1/2)
E1
Stated explicitly (could be in general form)
1 1 e −2 ln{ lim a 2 e −4 a + 3 } – ln( + 3) a→∞ 2 2 4
(
Now lim a 2 e −4 a a→∞
I=
)
=0
1 1 e −2 ln 3 – ln( + 3) 2 2 4
A1 Total
4 8
CSO ACF
Q
Solution 1 (−2 sin 2 x) 6(a) y = ln cos 2x ⇒ y′(x) = cos 2 x
Marks M1 A1
Total
Comments Chain rule
y′′ (x) = − 4sec22x
m1
λ sec22x OE
y′′′ (x) = − 8sec 2x (2sec 2x tan 2x)
M1
K sec22x tan2x OE
{y′′′ (x) = −16tan 2x (sec2 2x)} y′′′′(x) = −16[2sec2 2x(sec2 2x) + tan 2x(2sec 2x (2sec 2x tan 2x))]
M1 A1
(b) y(0) = 0, y′(0) = 0, y′′(0) = −4, y′′′(0) = 0, y′′′′(0) = −32 x2 x4 ln cos 2x ≈ 0 + 0 + (−4) + 0 + (−32) 2 4! 4 ≈ − 2x 2 − x 4 3
B1F
(c) ln (sec2 2x) = − 2ln (cos 2x) 8 ≈ 4x 2 + x 4 3
M1
Product rule OE ACF ft c’s derivatives
M1 A1
A1 Total
6
3
CSO throughout parts (a) and (b) AG PI
2 11
Q
Solution 7(a) u = xy dy du = y+x dx dx 2 dy d2 y d u dy = + ( ) + x dx dx dx 2 dx 2
Marks
Total
Comments
M1 A1
Product rule OE OE
A1
OE
d2 y dy + 2(3 x + 1) + 3 y (3 x + 2) = 18 x 2 dx dx dy d2 y dy + y ) + 9 xy = 18 x ( x 2 + 2 ) + 6( x dx dx dx x
(b)
d 2u du +6 + 9u = 18 x 2 dx dx
A1
d 2u du +6 + 9u = 18 x 2 dx dx CF: Aux eqn m 2 + 6m + 9 = 0 (m + 3) 2 = 0 so m = −3 CF: (u =) e−3x (Ax + B)
M1
PI
A1 A1F
PI
M1
PI. Must be more than just stated
PI: Try (u =) px + q 0 + 6p + 9(px + q) = 18x 9p = 18, 6p + 9q = 0 12 p=2; q=− 9 4 u = e −3 x ( Ax + B ) + 2 x − 3 xy = e −3 x ( Ax + B) + 2 x −
4
CSO AG Be convinced
m1 A1
Both
B1F
c’s CF + c’s PI but must have 2 constants, also must be in the form u = f(x)
4 3
1 4 y = {e −3 x ( Ax + B ) + 2 x − } x 3
A1 Total
8 12
Q
Solution
Marks
Comments π 1 2 2 d θ or r ∫0 r dθ 2∫
1 (3 + 2cos θ ) 2 dθ 2∫
M1
Use of
1 (9 + 12cos θ + 4cos 2 θ ) dθ ∫ 20
B1 B1
Correct expn of [3 + 2cosθ ]2 Correct limits
M1
Attempt to write cos 2 θ in terms of cos 2θ
A1F
Correct integration ft wrong coefficients
8(a) Area = 2π
=
Total
2π
= ∫ (4.5 + 6cos θ + (1 + cos 2θ )) dθ 0
2π
1 = 4.5θ + 6sin θ + θ + sin 2θ 2 0 = 11π
(b)(i)
A1
x 2 + y 2 − 8 x + 16 = 16
M1
r 2 − 8r cos θ + 16 = 16 ⇒ r = 8 cos θ
A1
At intersection, 8 cos θ = 3 + 2 cos θ 3 ⇒ cos θ = 6 π Points 4, and 3 π AB = 2 × 4sin 3 = 4 3
5π 4, 3
Length of arc AOB of circle = 4 × Perimeter of segment AOB =
2π 3
2π 3
8π +4 3 3 Total
CSO Use of any two of x = r cosθ, y = r sinθ , x 2 + y 2 = r 2
M1
Equating rs or equating cosθ s with a further step to solve eqn. (OE eg 4r = 12 + r ⇒ 4r − r = 12)
A1
OE
M1
Valid method to find AB, ft c’s r and θ values
A1
(ii) Let M=centre of circle then ∠ AMB =
6
6
OE surd π 3
B1
Accept equiv eg ∠ AMO =
M1
Use of arc = 4 × ( ∠ AMB in rads)
A1
3
15 Alternative to (b)(i): Writing r= 3 + 2cos θ in cartesian form (M1A1) Finding cartesian coordinates of points A and B ie (2, ± 2 2) (M1A1) Finding length AB (M1A1) TOTAL 75
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
