General Certificate of Education January 2007 Advanced Level Examination
MATHEMATICS Unit Further Pure 3 Friday 26 January 2007
MFP3
1.30 pm to 3.00 pm
For this paper you must have: * an 8-page answer book * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
Time allowed: 1 hour 30 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MFP3. Answer all questions. Show all necessary working; otherwise marks for method may be lost. * *
* *
Information The maximum mark for this paper is 75. The marks for questions are shown in brackets. * *
Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. *
P89991/Jan07/MFP3 6/6/
MFP3
2
Answer all questions.
1 The function yðxÞ satisfies the differential equation dy ¼ f ðx, yÞ dx where
f ðx, yÞ ¼ lnð1 þ x 2 þ yÞ yð1Þ ¼ 0:6
and (a) Use the Euler formula
yrþ1 ¼ yr þ h f ðxr , yr Þ with h ¼ 0:05 , to obtain an approximation to yð1:05Þ , giving your answer to four decimal places. (3 marks) (b) Use the improved Euler formula 1
yrþ1 ¼ yr þ 2ðk1 þ k2 Þ where k1 ¼ h f ðxr , yr Þ and k2 ¼ h f ðxr þ h, yr þ k1 Þ and h ¼ 0:05 , to obtain an approximation to yð1:05Þ , giving your answer to four decimal places. (6 marks)
2 A curve has polar equation rð1 sin yÞ ¼ 4 . Find its cartesian equation in the form y ¼ f ðxÞ .
3
(6 marks)
(a) Show that x 2 is an integrating factor for the first-order differential equation 1 dy 2 þ y ¼ 3ðx 3 þ 1Þ2 dx x
(b) Solve this differential equation, given that y ¼ 1 when x ¼ 2 .
P89991/Jan07/MFP3
(3 marks) (6 marks)
3
ðe
4
ln x pffiffiffi dx is an improper integral. x 0 ð 1 (b) Use integration by parts to find x 2 ln x dx . (a) Explain why
(1 mark)
(3 marks)
ðe (c) Show that
ln x pffiffiffi dx exists and find its value. x 0
(4 marks)
5 Find the general solution of the differential equation d2 y dy 4 þ 3y ¼ 6 þ 5 sin x 2 dx dx
(12 marks)
1
6 The function f is defined by f ðxÞ ¼ ð1 þ 2xÞ2 . (a)
(i) Find f 0 0 0 ðxÞ .
(4 marks)
(ii) Using Maclaurin’s theorem, show that, for small values of x , 1
1
f ðxÞ 1 þ x 2 x 2 þ 2 x 3
(4 marks)
(b) Use the expansion of e x together with the result in part (a)(ii) to show that, for small values of x , 1
e x ð1 þ 2xÞ2 1 þ 2x þ x 2 þ kx 3 where k is a rational number to be found.
(3 marks)
(c) Write down the first four terms in the expansion, in ascending powers of x, of e 2x . (1 mark) (d) Find 1
lim x!0
e x ð1 þ 2xÞ2
e2x 1 cos x
(4 marks)
Turn over for the next question
P89991/Jan07/MFP3
s
Turn over
4
7 A curve C has polar equation r ¼ 6 þ 4 cos y,
p4y4p
The diagram shows a sketch of the curve C, the pole O and the initial line.
Initial line O
(a) Calculate the area of the region bounded by the curve C . (b) The point P is the point on the curve C for which y ¼
(6 marks)
2p . 3
The point Q is the point on C for which y ¼ p . Show that QP is parallel to the line y ¼
p . 2
(4 marks)
(c) The line PQ intersects the curve C again at a point R . The line RO intersects C again at a point S . (i) Find, in surd form, the length of PS . (ii) Show that the angle OPS is a right angle.
END OF QUESTIONS
Copyright Ó 2007 AQA and its licensors. All rights reserved.
P89991/Jan07/MFP3
(4 marks) (1 mark)
MFP3 - AQA GCE Mark Scheme 2007 January series
Key to mark scheme and abbreviations used in marking 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
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
MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp
mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph 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. 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.
3
MFP3 - AQA GCE Mark Scheme 2007 January series
MFP3 Q 1(a)
(b)
Solution y (1.05) = 0.6 + 0.05 × [ln (1 + 1 + 0.6) ] = 0.6477 (7557..) = 0.6478 to 4dp
k1 = 0.05 × ln (1 + 1 + 0.6 ) = 0.0477 ( 75…)
k2 = 0.05 × f (1.05, 0.6477…)
… = 0.0505 ( 85…)
M1 A1F
Condone >4 dp
PI ft candidate’s evaluation in (a)
PI
1 [ k1 + k2 ] 2 = 0.6 + 0.5 × 0.09836…
m1
Dep on previous two Ms and numerical values for k’s
= 0.6492 to 4dp
A1F
r − r sin θ = 4 r−y=4 r = y+4
Total
x 2 + y 2 = ( y + 4) 2
M1 B1 A1 M1
x 2 + y 2 = y 2 + 8 y + 16
A1F
x − 16 8
6 9
Must be 4 dp… ft one slip
r sinθ = y stated or used r 2 = x 2 + y 2 used ft one slip
2
A1 Total
3(a) IF is exp ⎛ 2 dx ⎞ ⎜∫ ⎟ ⎝ x ⎠ = e 2ln x = x2 (b)
3
Comments
A1F
y (1.05 ) = y (1) +
y=
Total
M1
… = 0.05 × ln (1 + 1.052 + 0.6477…) ⎤⎦
2
Marks M1A1 A1
6 6
M1 A1 A1
1 d ⎡⎣ yx 2 ⎤⎦ = 3x 2 ( x3 + 1) 2 dx 3 2 ⇒ yx 2 = ( x3 + 1) 2 + A 3
And with integration attempted 3
PI
M1A1
3 2
2 ⇒ 4 = (9) + A 3
3
m1
k ( x 3 + 1) 2
A1
Condone missing ‘A’
m1
Use of boundary conditions to find constant
⇒ A = −14 3 ⎧2 ⎫ ⇒ y = x −2 ⎨ ( x 3 + 1) 2 − 14 ⎬ ⎩3 ⎭
CSO AG be convinced
A1 Total
6 9
4
Any correct form
MFP3 - AQA GCE Mark Scheme 2007 January series
MFP3 (cont) Q Solution 4(a) Integrand is not defined at x = 0 (b)
∫x
−
1 2
1
Marks E1
Total 1
1
⎛1⎞ ln x dx = 2 x 2 ln x − ∫ 2 x 2 ⎜ ⎟ dx ⎝ x⎠
1
...= kx 2 ln x ± ∫ f ( x ) , with f ( x ) not
M1
involving the ‘original’ ln x
A1 1
1
…… = 2 x 2 ln x − 4 x 2 ( +c )
(c)
∫
e
ln x
0
x
dx =
A1
e ln x lim dx ∫ a→0 a x
So
e ln
∫0
x x
3
Condone absence of ‘+ c’
M1
1 ⎡ 12 ⎤ 2 ⎢ 2a ln a − 4a ⎥ a→0 ⎣ ⎦ 1 lim But a 2 ln a = 0 a→0 1
= −2e 2 –
Comments
OE
lim
M1
F(b) – F(a)
B1
Accept a general form e.g. lim x k ln x = 0 x→0
1
dx exists and = −2e 2
A1 Total
4 8
M1
PI
m = 3 and 1 CF is A e3x + B ex PI Try y = a + b sin x + c c os x y ′(x) = b cos x − c sin x
A1 A1F M1 A1
PI
y ′′(x) = −b sin x − c cos x Substitute into DE gives a=2 4c + 2b = 5 and 2c – 4b = 0
A1F M1 B1 A1
ft can be consistent sign error(s)
b = 0.5, c=1
A1F A1F
ft a slip ft a slip
GS: y = A e3x + B ex + 2 + 0.5sinx + cosx
B1F
5 Auxl. eqn m 2 − 4m + 3 = 0
Total
Condone ‘a’ missing here
12
12
5
y = candidate’s CF and candidate’s PI (must have exactly two arbitrary constants)
MFP3 - AQA GCE Mark Scheme 2007 January series
MFP3 (cont) Q 6(a)(i)
Solution
1 f ′ ( x ) = (1 + 2 x) 2
1 − 2
f ′′ ( x ) = − (1 + 2 x )
−
f ′′′ ( x ) = 3 (1 + 2 x ) (ii)
(2) = (1 + 2 x)
Marks 1 − 2
3 2
Total
M1A1 A1F
5 − 2
A1
1 2
f ( x ) = (1 + 2 x ) ⇒ f ( 0 ) = 1;
B1
f ′(0) = 1; f ′′(0) = –1; f ′′′(0) = 3
M1
ft a slip 4
All three attempted ft on k (1 + 2 x )′′′
A1F 2
f (x) = f(0) + xf ′(0) +
3
x x f ′′(0)+ f ′′′(0) 2 6
x 2 x3 + 2 2 1 e x (1 + 2 x ) 2 ≈
… ≈1+ x − (b)
Comments
⎛ x 2 x3 ⎞ ⎛ x 2 x3 ⎞ 1 1 + x + + + x − + ⎟ ⎜ ⎟⎜ 2 6 ⎠⎝ 2 2⎠ ⎝ 2 ≈1+x (1+1) + x (–0.5 + 1 + 0.5) ⎛1 1 1 1⎞ + x3 ⎜ − + + ⎟ ⎝2 2 2 6⎠ 2 ≈ 1 + 2 x + x 2 + x3 3 (2 x) 2 (2 x)3 + +… (c) e2x = 1 + 2 x + 2 6 4 = 1 + 2x + 2x2 + x 3 + … 3 1 2 4 (d) 1 − cos x = x + {o( x )} 2
A1
4
M1
CSO AG
Attempt to expand needed
A1
A1
3
B1
1
CSO
B1
1 2
e x (1 + 2 x) − e 2 x = 1 − cos x
2 3 ⎡ 4 ⎤ x − ⎢1 + 2 x + 2 x 2 + x3 ⎥ 3 3 ⎦ ⎣ 1 2 4 x + {o( x )} 2 2 3 lim lim − x + {o( x )} = ….. = x→0 x→0 1 2 4 x + {o( x )} 2 lim −1 + o( x) =–2 x→0 1 2 + o( x ) 2 Total
1 + 2 x + x2 +
M1
Series used
A1F
A1F
4 16
6
ft a slip but must see the intermediate stage
MFP3 - AQA GCE Mark Scheme 2007 January series
MFP3 (cont) Q 7(a)
Solution
Area =
Marks
1 (6 + 4cos θ ) 2 dθ 2∫
π ⎞ 1⎛ = ⎜ ∫ 36 + 48cosθ + 16cos 2 θ ⎟ dθ 2 ⎝ -π ⎠
⎛π ⎞ = ⎜ ∫ 18 + 24cos θ + 4(cos 2θ + 1) ⎟ dθ ⎝ -π ⎠
= [ 22θ + 24 sin θ + 2 sin 2θ ]
Total
Comments
1 2 r dθ 2∫
M1
use of
B1 B1
for correct expansion of [6 + 4cosθ )]2 for limits
M1
Attempt to write cos 2 θ in terms of cos 2θ
A1F
correct integration ft wrong coefficients
π -π
= 44π
A1
(b) At P, r = 4; At Q, r = 2; 2π = −2 P {x = } r cos θ = 4 cos 3 Q {x = } r cos θ = 2 cos π = −2
Since P and Q have same ‘x’, PQ is vertical so QP is parallel to the vertical π line θ = 2
6
CSO
B1
PI
M1 A1
Attempt to use r cosθ Both
E1
4
(c)(i) OP = 4; OS = 8;
B1
π 3
B1
or S (4, 4√3) and P (–2, 2√3)
M1
Cosine rule used in triangle POS OE PS 2 = (4 + 2) 2 + (4 3 − 2 3) 2
Angle POS =
PS 2 = 42 + 82 − 2 × 4 × 8 × cos
PS = 48 (ii)
π oe 3
{= 4 3}
Since 82 = 42 +
(
)
2
48 ,
A1
4
E1
1
OS 2 = OP 2 + PS 2 ⇒ OPS is a right angle. (Converse of Pythagoras Theorem)
Total TOTAL
15 75
7
Accept valid equivalents e.g. PR = 2PQ = 2(2√3) = PS. π ∠SRP = ∠RSP = ∠RPO = 6 ⇒ OPS is a right angle
AQA January Examinations 2007 Scaled Mark Component Grade Boundaries (GCE Specifications) Generally, scaled mark boundaries are the same as raw mark boundaries as there is no scaling of marks. However, they may be different if a unit of assessment consists of more than one component.
Component Code LAT1 LAT2 LAT3 LAT4 LAT5 LAT6 LAW1 LAW2 LAW3 LAW4 LAW5 MD01 MD02 MFP1 MFP2 MFP3 MFP4 MM1A/W MM1A/C MM1B MM2A/W MM2A/C
Maximum Scaled Mark 35 50 50 35 70 50
A 24 29 29 21 45 32
LAW UNIT 1 LAW UNIT 2 LAW UNIT 3 LAW UNIT 4 LAW UNIT 5
65 65 65 85 85
48 45 44 60 58
43 40 39 55 54
38 36 34 50 50
33 32 29 45 47
29 28 25 40 44
MATHEMATICS UNIT MD01 MATHEMATICS UNIT MD02 MATHEMATICS UNIT MFP1 MATHEMATICS UNIT MFP2 MATHEMATICS UNIT MFP3 MATHEMATICS UNIT MFP4 MECHANICS 1A - WRITTEN MECHANICS 1A - COURSEWORK MATHEMATICS UNIT MM1B MECHANICS 2A - WRITTEN MECHANICS 2A - COURSEWORK
75 75 75 75 75 75 75 25 75 75 25
61 60 60 61 61 61 60 20 60 61 20
53 52 53 53 53 53 51 18 51 54 18
45 44 46 45 45 45 44 15 43 46 15
38 37 39 37 38 38 36 13 35 40 13
31 30 32 29 31 31 29 10 27 34 10
Component Title LATIN UNIT 1 LATIN UNIT 2 LATIN UNIT 3 LATIN UNIT 4 LATIN UNIT 5 LATIN UNIT 6
Page 9 of 13
Scaled Mark Grade Boundaries B C D 21 18 15 26 23 20 26 23 20 18 16 14 39 33 28 29 26 23
E 12 17 18 12 23 20
