General Certificate of Education Advanced Level Examination January 2010 Mathematics MFP3 Unit Further Pure 3 Tuesday 19 January 2010 9.00 am to 10.30 am For this paper you must have: * a 12-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 black ink or black 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 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. P21941/Jan10/MFP3 6/6/ MFP3
Transcript
General Certificate of Education
Advanced Level Examination
January 2010
Mathematics MFP3
Unit Further Pure 3
Tuesday 19 January 2010 9.00 am to 10.30 am
For this paper you must have:* a 12-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 black ink or black 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 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.
P21941/Jan10/MFP3 6/6/ MFP3
2
P21941/Jan10/MFP3
Answer all questions.
1 The function yðxÞ satisfies the differential equation
dy
dx¼ f ðx, yÞ
f ðx, yÞ ¼ x lnð2xþ yÞwhere
yð3Þ ¼ 2and
(a) Use the Euler formula
yrþ1 ¼ yr þ h f ðxr, yrÞ
with h ¼ 0:1 , to obtain an approximation to yð3:1Þ , giving your answer to four
decimal places. (3 marks)
(b) Use the improved Euler formula
yrþ1 ¼ yr þ12ðk1 þ k2Þ
where k1 ¼ h f ðxr, yrÞ and k2 ¼ h f ðxr þ h, yr þ k1Þ and h ¼ 0:1 , to obtain an
approximation to yð3:1Þ , giving your answer to four decimal places. (5 marks)
2 (a) Given that y ¼ lnð4þ 3xÞ , find dy
dxand
d2y
dx2. (3 marks)
(b) Hence, by using Maclaurin’s theorem, find the first three terms in the expansion, in
ascending powers of x, of lnð4þ 3xÞ . (2 marks)
(c) Write down the first three terms in the expansion, in ascending powers of x,
of lnð4� 3xÞ . (1 mark)
(d) Show that, for small values of x,
ln
�4þ 3x
4� 3x
�� 3
2x (2 marks)
3
P21941/Jan10/MFP3
3 (a) A differential equation is given by
xd2y
dx2þ 2
dy
dx¼ 3x
Show that the substitution
u ¼ dy
dx
transforms this differential equation into
du
dxþ 2
xu ¼ 3 (2 marks)
(b) Find the general solution of
du
dxþ 2
xu ¼ 3
giving your answer in the form u ¼ f ðxÞ . (5 marks)
(c) Hence find the general solution of the differential equation
xd2y
dx2þ 2
dy
dx¼ 3x
giving your answer in the form y ¼ gðxÞ . (2 marks)
4 (a) Write down the expansion of sin 3x in ascending powers of x up to and including the
term in x3 . (1 mark)
(b) Find
limx! 0
3x cos 2x� sin 3x
5x3
� �(4 marks)
Turn over
s
4
P21941/Jan10/MFP3
5 It is given that y satisfies the differential equation
d2y
dx2þ 3
dy
dxþ 2y ¼ 2e�2x
(a) Find the value of the constant p for which y ¼ pxe�2x is a particular integral of the
given differential equation. (4 marks)
(b) Solve the differential equation, expressing y in terms of x, given that y ¼ 2
anddy
dx¼ 0 when x ¼ 0 . (8 marks)
6 (a) Explain why
ð11
ln x2
x3dx is an improper integral. (1 mark)
(b) (i) Show that the substitution y ¼ 1
xtransforms
ðln x2
x3dx into
ð2y ln y dy .
(2 marks)
(ii) Evaluate
ð102y ln y dy , showing the limiting process used. (5 marks)
(iii) Hence write down the value of
ð11
ln x2
x3dx . (1 mark)
7 Find the general solution of the differential equation
d2y
dx2þ 4y ¼ 8x2 þ 9 sin x (8 marks)
5
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8 The diagram shows a sketch of a curve C and a line L, which is parallel to the initial line
and touches the curve at the points P and Q.
The polar equation of the curve C is
r ¼ 4ð1� sin yÞ, 04y < 2p
and the polar equation of the line L is
r sin y ¼ 1
(a) Show that the polar coordinates of P are 2,p6
� �and find the polar coordinates of Q.
(5 marks)
(b) Find the area of the shaded region R bounded by the line L and the curve C. Give your
answer in the form mffiffiffi3
pþ np , where m and n are integers. (11 marks)
END OF QUESTIONS
C
Initial line
LQ P
R
O
MFP3 - AQA GCE Mark Scheme 2010 January series
3
Key to mark scheme and abbreviations used in marking M mark is for method m or dM mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation
or ft or F follow through from previous incorrect result
MC
mis-copy
CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A2,1 2 or 1 (or 0) accuracy marks NOS not on scheme –x EE deduct x marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach dp 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.
ft one slip; accept terms in π and 3 left unsimplified
Shaded area = Area of triangle OPQ −
( )π
22π6
12 [4 1 sin ]2
θ× −∫ dθ
M1
OE
Shaded area =
3 − 16 π 332 8
⎛ ⎞− +⎜ ⎟⎜ ⎟
⎝ ⎠ = 15 3 − 8 π
A1
11
CSO Accept m = 15, n = −8
Total 16 TOTAL 75
AQA – Further pure 3 – Jan 2010 – Answers Question 1: Exam report
( )
( )( )( )
( )( )
( )( )
( )
0 0
1 1 0
1
0 1
2
ln 2 3 2
) 3.1 3.1 ( , )
2 0.1 3ln 2 3 2
) ( , ) 0.1 3ln 2 3 2 0.6238
2.
(3.1) 2.6238 4 .
(3.1) 2.
6238
0.1 3.1ln 2 3.1 2.6238 0.6750
1(3.1) 2 0.6238 0.67502
6494
r r
r r
dy x x y x and ydxa x and y y y hf x y
b k hf x y
y
y to d p
k
y
y t
k
= + = =
= = = +
= + × +
= = × + =
+ =
= ×
=
+
+
=
=
= +
4 . .o d p
Numerical solutions of first order differential equations continue to be a good source of marks for all candidates and this was the best answered question on the paper. Very few candidates mixed up the x and y values in applying the given formulae. Almost all candidates gave their final answers to the required degree of accuracy.
Question 2: Exam report
( )
( )
( ) ( )
2
2 2
2
2
3 94 3 (4 3 )
3 9ln(4 3 ) ln(
) ln 4 3
) (0) ln 43 9'(0) ''(0)4 16
:
3 9) ln 4 3 ln(4) ... ( ( ))4 32
4 3) ln ln 4 3 ln 4 34 3
ln
4) ...4
(4)
32
a y x
and
b y
y and y
Conclusion
c x x x substituting x wit
dy
h x
xd x x
d yd
x
x x dx x
x x x
= = −+ +
+ = + −
= +
=
= = −
− = − − + −
+ = + − − −
+
= + 2 23 9 3 9... ln(4) ...4 32 4 32
4 3 3 3ln ...4 3 4 4
32
x x x x
x xx
xx
− + − − − + + = + + ≈−
Most candidates were able to find 2
2
dy d yand
dx dxcorrectly,
although some less able candidates failed to apply the chain rule. Although a few candidates in part (b) attempted to use the printed expansion of ln(1 + x) from the formulae booklet instead of applying the prescribed method, the majority of candidates answered the question as instructed and showed good knowledge of Maclaurin’s theorem. Many candidates failed to appreciate that in part (c) they had to replace x with −x, and instead multiplied both their x and x2 terms by −1. A significant minority of candidates in part (d) did not realise that they had to write the expression as the difference of their two expansions.
Question 3: Exam report
( )
2
2
22ln 2
2 2
2 2 2 3
2
2
) 2 3
2 3
) An integrating factor is
The equation becomes 2
2
3
3 integrating
) int
3
dx xx
d y dy dya x x with udx dx dx
dubecomes x u xdx
b I e e xdu
Au x
x xu xdx
d x u x x u x Adx
dy Ac u x so by e
du u
grati
x
ng b
dx x
dx x
+ =
+ = =
+ =
∫=
=
= =
+ =
= = +
= =
+
+
21 ,2
:oth
Ay x B A Bx
sides
= − + ∈
This question was answered well with the majority of candidates able to use the given substitution correctly. Most candidates were able to show that they knew how to find and use an integrating factor to solve a first order differential equation. However, a significant number of candidates failed to write their general solutions with the required number of arbitrary constants.
Question 4: Exam report 3
3
2
3
2 33
3 33
3
0
)sin( ) ...6
(3 )sin(3 ) (3 ) ...6
) In the same manner (2 ) 1 2 ...
3 (2 ) (3 ),5
1 93 (1 2 ) (3 ) ...5 21 9 36 .
9sin(3 ) 3 ...2
3l
.. ...5 2 10
imx
xa x x so
xx x
bCos x x
xCos x Sin xHencex
x x x xx
x x x
x xx
→
= − +
= − +
= − +−
= − − − + = − + + = − +
=
− +
3
(2 ) (3 ) 35 10
xCos x Sin xx− = −
This question on series expansions and the limiting process was generally answered very well, but a significant minority of candidates in part (b) did not explicitly reach the stage of a constant term in both the numerator and denominator before taking the limit as 0x → .
Question 5: Exam report
( )
( ) ( )
( ) ( )
2 2
22
2
2
2
2
2 2 2 2
2
2 2
2 22
3 2 2
)
2
4 4
Substituting in the equation:4 4 3 2 2 2
2)The auxiliary equa
2
2 2 2
2t
x
x
x
x
x
x
x
x
x
x x x x
x
d y dy y edx dxa y pxe
p px e
p px e
p px e p px e pxe e
p
dy pe pxedxd y pe p px edx
e e sob
p
−
−
−
−
− − − −
− −
− −
− −
+ + =
=
= −
= − +
−
= −
= −
+ + − +
−
−
=
−
− ==
2
2 2
2
ion associated to the equation is :3 2 0
( 1)( 2) 01 2
The complementary function is The general solution is 2
0, 2 2 0
0, 0
x x
x x
x
or
y xe Ae Bewhen x y this gives A B
dywhen x this giv
y e e
d
A B
x
λ λλ λλ λ
− −
− − −
= +
+ + =+ + == − = −
= − + += = = + +
= =
2 2
0 2 2
Solving these two equations simultaneously:4 6
The solution wanted is 2 6 4 x x x
es A B
B and Ay xe e e− − −= −
= − − −
= −
+ −
=
In part (a), it was pleasing to see a higher proportion of candidates than in previous papers using the given form of the particular integral rather than introducing the extra term qe−2x. The majority of candidates used the product rule correctly to find the value of the constant p. Most candidates showed that they knew the methods required to solve the second order differential equation, although a minority of candidates made the error of applying the boundary conditions to the complementary function before adding on the particular integral.
Question 6: Exam report 2
31
2
2
3 3
32
1 2
interval of integration is infinit
ln) is an improper integral
because the .1 1 1) ) ,
ln 1 2 ln( )
1 12 ln
)
2 ln(
2 ln( ) ln( )
)
e
a
xa dxx
dxb i y x andx y dy y
x dx x dxx x
becomes y dyy y
ii y y dy y
y y dy
y
∞
= = = −
= × ×
× × ×−
=
=
∫
∫ ∫
∫
∫
∫11 2
12 2
2 2
2 2
0 0
1
0
2
31
1
0
1
1ln( ) 02
1 1ln( )2 2
1lim ln( ) 0 lim 02
Conclusion: 2 ln( )
1) When , 0,
When 1, 1ln, 2 ln(
12 ln( )2
a a
a
a a
y dyy
a a y
a a a
a a and a
y y dy exists and
iii x tends to y tends tox
x yxHence dx y
y y dy
yx
→ →
∞
− ×
= − −
= − +
= =
∞ =
= =
=
= −
∫
∫
∫∫
0 1
1 0) 2 ln( ) 1
2dy y y dy= − =∫ ∫
In part (a), a significant number of candidates failed to give the correct reason for why the integral was improper, with most giving the reason that the integrand was undefined for the upper limit rather than that the interval of integration was infinite. In part (b)(i), a large number of candidates failed to use the substitution correctly. Part (b)(ii) was generally answered well with most candidates integrating by parts correctly to score 3 of the 5 marks available, but some candidates did not gain the final two marks as examiners did not see the lower limit replaced by, for example, a and the consideration of the limiting process as 0a → . Part (b)(iii) was answered incorrectly by the majority of candidates, who usually either stated the same value or the reciprocal value of their answer to part (b)(ii).
Question 7: Exam report
22
2
2
2
2
2
4 8 9 ( )
Complementary function:
The auxiliary equation is 4 02 2
(2 ) (2 )
Particular integral:
sin( ) ( )
2 ( ) ( )
2 ( )
cf
d y y x Sin xdx
i or iy ACos x BSin x
y ax bx c d x eCos xdy ax b dCos x eSin xdxd y a dSin x eCdx
ax bx c a d d Sin x e e Cos x x Sin xThis gives a b c d and e
+ = +
− − + + + + + = +
+ + + + − + + − + = += = = − = =
22 1 3 ( ) (2 ) (2 ) solution is y x Sin x ACos x BSin x= − + + +
This unstructured question was answered well by many of the candidates. However, a significant minority of candidates did not even write down the correct form of the auxiliary equation, or they solved it incorrectly to give real rather than imaginary values. When finding the particular integral, a large number of candidates considered more terms than they needed to, and they did not always go on to show that the relevant coefficients of the extra terms were zero. The majority of candidates knew that they had to find a complementary function and a particular integral and almost all scored the final mark for combining the two to give the general solution.
Question 8: Exam report
2
2
2
)Solving the eqautions simultaneously4(1 ) 1
: 4(1 ) 14 4
5
1 04 4 1 0
(2 1) 012
For = , 4(1 ) 26 65 5For = , 4(1 ) 26
(2,6
6
6
)
6
ar Sin and rSin
gives Sin SinSin SinSin SinSin
Sin so
r Sin
or
P
r Sin
θ θθ θ
θ θ
θ θ
θ
θ
π πθ
π
π πθ θ
πθ
π
= − =− =
− + − =
− + =
− =
= =
= =
=
=
−
− =
( )2 22 2
6 6
)The area shaded =Area of the triangle POQ 21 5 22 2 2 32 6 6 3
1 16 1 8 1 22
1 18 1 2 22
5(2, )
2
6b Areabounded by line OP
AreaOPQ Sin Sin
Area bounded by line OP is Sin d Sin Sin d
Cos Si
Q
π π
π π
π π π
θ θ θ θ θ
θ
π
− ×
= × × × − = =
× − = + −
= + − −
∫ ∫
22
66
3 18 2 22 4
3 3 18 2 4 7 34 12 4 3 6
The area shaded is 3 2(4 7 3) 15 3 8
n d Sin Cos
Sin Cos
ππ
ππ
θ θ θ θ θ
π π ππ π
π π
= − +
= − − + = −
− − = −
∫
Part (a) was generally answered well with most candidates forming an equation in either sinθ or r and solving it correctly to give the coordinates of P and Q. Candidates who tried to verify the coordinates of P generally failed to even verify them in both polar equations. Part (b) was the most demanding question on the paper. The majority of candidates scored at least 4 of the 11 marks by applying the
formula 212
A r dβ
αθ= ∫ to find an
area of a region partly bounded by the curve C and in doing so they correctly expanded (1−sinθ)2, wrote sin2θ in terms of cos2θ and integrated correctly. The errors occurred because many candidates did not use the correct limits. A significant number of candidates also failed to find, or even consider the need for, the area of triangle OPQ and so correct answers for the area of the shaded region R were generally only presented by the most able candidates.
Grade boundaries Grade A B C D E Mark Max 75 63 55 47 39 32
