General Certificate of Education Advanced Level Examination June 2010
Mathematics
MFP3
Unit Further Pure 3 Friday 11 June 2010
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.
P27889/Jun10/MFP3 6/6/
MFP3
2
The function yðxÞ satisfies the differential equation
1
dy ¼ f ðx, yÞ dx f ðx, yÞ ¼ x þ 3 þ sin y
where
yð1Þ ¼ 1
and (a)
Use the Euler formula yrþ1 ¼ yr þ hf ðxr , yr Þ with h ¼ 0:1 , to obtain an approximation to yð1:1Þ , giving your answer to four decimal places. (3 marks)
(b)
Use the formula yrþ1 ¼ yr1 þ 2hf 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 (a)
Find the value of the constant k for which k sin 2x is a particular integral of the differential equation d2 y þ y ¼ sin 2x dx 2
(b)
Hence find the general solution of this differential equation.
ð1 3 (a)
Explain why
4xe4x dx is an improper integral.
(3 marks) (4 marks)
(1 mark)
1
ð (b)
Find
4xe4x dx . ð1
(c)
Hence evaluate
(3 marks) 4xe4x dx , showing the limiting process used.
(3 marks)
1
P27889/Jun10/MFP3
3
By using an integrating factor, find the solution of the differential equation
4
3 dy 3 þ y ¼ ðx 4 þ 3Þ2 dx x
1
given that y ¼ 5 when x ¼ 1 .
(9 marks)
Write down the expansion of cos 4x in ascending powers of x up to and including (2 marks) the term in x 4 . Give your answer in its simplest form.
5 (a)
(b) (i)
Given that y ¼ lnð2 e x Þ , find
dy d2 y d3 y , 2 and 3 . dx dx dx
ðYou may leave your expression for
d3 y unsimplified.Þ dx3
(6 marks)
(ii) Hence, by using Maclaurin’s theorem, show that the first three non-zero terms in the
expansion, in ascending powers of x, of lnð2 e x Þ are x x 2 x 3
(2 marks)
x lim x lnð2 e Þ x ! 0 1 cos 4x
(3 marks)
Find
(c)
The polar equation of a curve C1 is
6
r ¼ 2ðcos y sin yÞ, (a) (i)
Find the cartesian equation of C1 .
0 4 y 4 2p (4 marks)
(ii) Deduce that C1 is a circle and find its radius and the cartesian coordinates of its
centre.
(3 marks)
s
Turn over
P27889/Jun10/MFP3
4
The diagram shows the curve C2 with polar equation
(b)
r ¼ 4 þ sin y,
0 4 y 4 2p
O
(i)
Initial line
Find the area of the region that is bounded by C2 .
(6 marks)
(ii) Prove that the curves C1 and C2 do not intersect.
(4 marks)
(iii) Find the area of the region that is outside C1 but inside C2 .
(2 marks)
1
Given that x ¼ t 2 , x > 0 , t > 0 and y is a function of x, show that:
7 (a) (i)
1 dy dy ¼ 2t 2 ; dx dt
(2 marks)
(ii)
d2 y d2 y dy . ¼ 4t þ 2 dx 2 dt 2 dt
(3 marks) 1
(b)
Hence show that the substitution x ¼ t 2 transforms the differential equation x
d2 y dy ð8x 2 þ 1Þ þ 12x 3 y ¼ 12x 5 2 dx dx
into d2 y dy þ 3y ¼ 3t 4 dt 2 dt (c)
(2 marks)
Hence find the general solution of the differential equation x
d2 y dy ð8x 2 þ 1Þ þ 12x 3 y ¼ 12x 5 2 dx dx
giving your answer in the form y ¼ f ðxÞ .
(7 marks)
END OF QUESTIONS Copyright Ó 2010 AQA and its licensors. All rights reserved.
P27889/Jun10/MFP3
MFP3 - AQA GCE Mark Scheme 2010 June 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 2010 June series
Q
Solution 1(a) y(1.1) = y(1) + 0.1 [1 + 3 + sin 1] = 1 + 0.1×4.84147.= 1.4841(47..) = 1.4841 to 4dp (b) y(1.2) = y(1) + 2(0.1){f[1.1, y(1.1)]}
Marks M1A1
Total
Comments
A1
3
Condone > 4dp
3
Ft on cand’s answer to (a) CAO Must be 2.019
M1
…. = 1 + 2(0.1){1.1+3+sin[1.4841(47..)]} = 2.019 to 3dp
A1F A1
Note: If using degrees max mark is 4/6 ie M1A1A0;M1A1FA0 Total 2(a)
− 4k sin 2 x + k sin 2 x = sin 2 x
k =−
1 3
A1
(b) (Aux. eqn m2 + 1= 0) m = ± i CF: A cos x + B sin x
1 3
Total 3(a) The interval of integration is infinite
∫ 4 xe
−4 x
dx = –xe
–4x
− ∫ − e −4 x dx
∞
3
B1F
4
c’s CF +c’s PI but must have 2 constants
7
E1
A1F
lim a I= ∫ 4 xe dx = 4 xe − 4 x dx ∫ a→∞ 1 1 lim 1 ⎡ 5 ⎤ {–ae– 4a − e– 4a}− ⎢− e −4 ⎥ a→∞ 4 ⎣ 4 ⎦
Accept correct PI PI M0 if m is real OE Ft on incorrect complex values for m For the A1F do not accept if left in the form Aeix + Be−ix
1
OE kxe–4x − ∫ ke − 4 x dx for non-zero k
M1 A1
1 = –xe– 4x − e– 4x {+c} 4 (c)
Substituting into the differential equation
B1 M1 A1F
(GS: y =) A cos x + B sin x − sin 2 x
(b)
6 M1 A1
3
Condone absence of +c
−4 x
lim a e −4 a = 0 a→∞ 5 I = e −4 4
M1
F(a) – F(1) with an indication of limit ‘a → ∞’
M1
For statement with limit/ limiting process shown
A1
3 7
Total
4
CSO
MFP3 - AQA GCE Mark Scheme 2010 June series
MFP3 (cont) Q 4
Solution
Marks
IF is exp ( ∫ 3 dx) x 3lnx =e = x3 3 d ⎡ yx 3 ⎤ = x 3 ( x 4 + 3) 2 ⎦ dx ⎣
(
⇒ yx3 = 1 x 4 + 3 10
)
5 2
+A
5 2
⇒ 1 = 1 (4) + A 5 10 (*)
(
)
5 2
−3
Comments
M1
and with integration attempted
A1 A1
PI
M1 A1
LHS. Use of c’s IF. PI
m1 A1
k x4 + 3
(
A1
Total
5 2
Use of boundary conditions in attempt to find constant after intgr. Dep on two M marks, not dep on m 9
9
5
)
Condone missing ‘A’
m1
⇒ A = –3;
1 4 ⇒ yx3 = x +3 10
Total
ACF. The A1 can be awarded at line (*) provided a correct earlier eqn in y, x and ‘A’ is seen immediately before boundary conditions are substituted.
MFP3 - AQA GCE Mark Scheme 2010 June series
MFP3 (cont) Q 5(a)
Solution
(4 x )
cos 4 x ≈ 1 −
2
2 2
≈ 1 − 8x + (b)(i)
(4 x )
4
+
4!
…
32 4 x … 3
1 dy = × (−e x ) x dx 2 − e d 2 y (2 − e x )(− e x ) − (− e x )(− e x ) = dx 2 (2 − e x )2
Total
A1
2
Comments Clear attempt to replace x by 4x in expansion of cos x…condone missing brackets for the M mark
M1 A1
Chain rule
M1 A1
Quotient rule OE ACF
m1
All necessary rules attempted (dep on previous 2 M marks)
− 2e x
=
(2 − e ) d y (2 − e ) (− 2e ) − (− 2e )2(2 − e )(− e ) = dx (2 − e ) x 2
x 2
3
x
x
3
x
x
x 4
(ii) y(0) = 0; y′(0) = –1; y′′(0) = –2; y′′′(0) = –6 Ln(2−ex)≈y(0)+xy ′(0)+
2
(
A1
6
M1
ACF At least three attempted
3
x x y′′(0)+ y′′′(0)… 2 6
…. ≈ − x − x2 − x3 …
(c)
Marks M1
A1
2
CSO AG (The previous 7 marks must have been awarded and no double errors seen)
)
⎡ x ln 2 − e x ⎤ − x 2 − x 3 − x 4 ... ⎢ ⎥≈ 32 4 ⎣ 1 − cos 4 x ⎦ 8x 2 − x 3 lim − x 2 − o( x 3 ) Limit = x → 0 8 x 2 − o( x 4 )
M1
The notation o(xn) can be replaced by a term of the form kxn
lim − 1 − o( x) x → 0 8 − o( x 2 )
…..
=
……
= −
Using the expansions
Division by x2 stage before taking the limit
m1
1 8
A1
Total
6
3
13
CSO
MFP3 - AQA GCE Mark Scheme 2010 June series
MFP3 (cont) Q
Solution 2 2 2 x + y = r , x = r cos θ , y = r sin θ 6(a)(i)
Marks B2,1,0
r 2 = 2r (cos θ − sin θ )
(ii)
Comments B1 for one stated or used
M1
x 2 + y 2 = 2( x − y )
A1
(x − 1)2 + ( y + 1)2 = 2
M1 A1F
Centre (1, −1); radius √2
A1F
(b)(i)
Total
4
ACF
3
1 (4 + sin θ ) 2 dθ ∫ 2
M1
Use of
1 (16 + 8 sin θ + sin 2 θ ) dθ ∫ 2 0
B1 B1
Correct expn of [4+sinθ ]2 Correct limits
M1
Attempt to write sin 2 θ in terms of cos 2θ
= ⎢8θ − 4 cos θ + θ − sin 2θ ⎥
A1F
Correct integration ft wrong coefficients
= 16.5π
A1
Area =
1 2 r dθ . 2∫
2π
=
2π
= ∫ (8 + 4 sin θ + 0.25(1 − cos 2θ )) dθ 0
⎡ ⎣
1 4
1 8
⎤ 2π ⎦ 0
(ii) For the curves to intersect, the eqn 2(cos θ − sin θ ) = 4 + sin θ must have a solution. 2 cos θ − 3 sin θ = 4 R cos(θ + α ) = 4 , where R =
2 2 + 3 2 and cosα =
cos(θ + α ) =
4 13
2 R
6
CSO
M1
Equating rs and simplifying to a suitable form
M1
OE. Forming a relevant eqn from which valid explanation can be stated directly
A1
OE. Correct relevant equation
> 1 . Since must have
− 1 ≤ cos X ≤ 1 there are no solutions of the equation 2(cos θ − sin θ ) = 4 + sin θ
so the two curves do not intersect.
E1
4
Accept other valid explanations.
2 19
Ft on (a)(ii) and (b)(i)
(iii) Required area =
answer (b)(i) − π (radius of C1 ) = 16.5π − 2π = 14.5π Total 2
M1 A1F
7
MFP3 - AQA GCE Mark Scheme 2010 June series
MFP3 (cont) Q
7(a)(i)
(a)(ii)
Solution
Marks
dx dy dy = dt dx dt 1 1 1 t − 2 dy = dy so dy = 2t 2 dy 2 dx dt dx dt
Total
M1 A1
1 1 d 2 y d ⎛⎜ 2 dy ⎞⎟ dt d ⎛⎜ 2 dy ⎞⎟ 2 2 t t = = dt ⎟⎠ dx dt ⎜⎝ dt ⎟⎠ dx 2 dx ⎜⎝
Comments OE
2
Chain rule
CSO A.G.
d dt d ( f (t ) ) = ( f (t ) ) dx dx dt
M1
O.E. eg
d dx d ( g ( x) ) ( g ( x) ) = dt dt dx ⎡ 12 d 2 y − 12 dy ⎤ +t ⎥ ⎢2t 2 dt ⎦ d t ⎣ d2 y d2 y dy = +2 4 t 2 2 dt dx dt 1
d2 y = 2t 2 dx 2
(b)
1 ⎡ d2 y dy ⎤ dy t ⎢4t 2 + 2 ⎥ − (8t + 1)2t 2 dt ⎦ dt ⎣ dt
1 2
3
m1
A1
Product rule O.E. used dep on previous M1 being awarded at some stage 3
M1
CSO
A.G.
Subst. using (a)(i), (a)(ii) into given DE to eliminate all x
5
+ 12 t 2 y = 12 t 2 3
4t 2
3
3
5
dy d2 y − 16t 2 + 12t 2 y = 12t 2 2 dt dt 3
Divide by 4t 2 gives
d2 y dy − 4 + 3 y = 3t 2 dt dt
(c)
A1
d2 y dy − 4 + 3 y = 3t (*) 2 dt dt 2 Auxl. Eqn. m −4m + 3 = 0 (m − 1)(m − 3) = 0 m = 1 and 3
2
CSO
A.G.
Solving
CF
M1 A1 M1
Aet + Be3t
For PI try
y = pt + q
PI Condone x for t here; ft c’s 2 real values for ‘m’ OE
M1
− 4 p + 3 pt + 3q = 3t ⇒ p = 1, q =
GS of (*) is y = Aet + Be3t + t +
4 3
4 3
GS of d2 y dy x 2 − (8 x 2 + 1) + 12 x 3 y = 12 x 5 dx dx 2 2 4 is y = Ae x + Be 3 x + x 2 + 3
A1 CF + PI with 2 arb. constants and both CF and PI functions of t only
B1F
A1
Total TOTAL
7 14 75
8
AQA – Further pure 3 – Jun 2010 – Answers Question 1:
Exam report
dy = f ( x, y ) = x + 3 + Sin( y ) dx a ) yr += yr + hf ( xr , yr ) 1
with y (1) = 1
1)
= x0 1 and= y1 y= (1) 1 = x1 1.1 and = y2 y= (1.1) 1 + 0.1(1 + 3 + sin(1)) y2 = 1.4841 to 4 decimal places b) y= yr −1 + 2hf ( xr , yr ) r +1
y (1.2)= y (1) + 2 × 0.1× (1.1 + 3 + sin(1.4841) )
This was the best answered question on the paper. Numerical solutions of first order differential equations continue to be a good source of marks for all candidates. The most common loss of marks was due to calculators being set in degree mode. It is worth recording that slightly more candidates than in recent series have slipped back into not showing the necessary working. Without such working, wrong answers cannot be awarded credit.
y (1.2) = 2.019 to 3 d . p. Question 2:
a) y = kSin(2 x)
Exam report
dy = 2kCos (2 x) dx
2
d y = −4kSin(2 x) dx 2
d2y +y= Sin(2 x) becomes dx 2 Sin(2 x) −4kSin(2 x) + kSin(2 x) =
The equation
1 3 b) The auxiliary equation associated with the differential equation is −3Sin(2 x) = Sin(2 x)
λ 2 + 1 =0
so k = −
λ =i or λ =−i
The complementary function = is y ACos ( x) + BSin( x) 1 The general solution is y = ACos ( x) + BSin( x) − Sin(2 x) 3
A, B ∈ A, B ∈
Question 3: a ) The integral is improper because
Exam report
the interval of integration is infinite. 1 b) ∫ 4 xe −4 x dx = − xe −4 x − ∫ −e −4 x dx =− xe −4 x − e −4 x + c 4 N
N 1 1 1 c) ∫ 4 xe −4 x dx = − xe −4 x − e −4 x = − Ne −4 N − e −4 N + e −4 + e −4 1 4 4 4 1
lim e −4 N = 0 and lim Ne −4 N = 0 so
N →∞
N →∞
∫
∞
1
4 xe −4 x dx =
This was another very good source of marks for candidates, with many fully correct solutions presented. In part (a) some candidates decided to ignore the given form of the particular integral and worked with pcos2x + qsin2x . Such an approach was not penalised by examiners provided the candidate showed that both p = 0 1 and q = − . The vast majority of candidates 3 showed that they knew the methods to solve the second order differential equation but errors in forming and solving the auxiliary equation were sources of loss of marks. Real solutions from the auxiliary equation were more heavily penalised than other errors.
5 −4 e 4
∫
∞
1
4 xe −4 x dx exists and
Improper integrals and limiting processes continue to cause problems for a significant minority of candidates. Part (a) was generally not answered well with too many candidates making a statement which they then contradicted in part (c). As highlighted in the MFP3 examiners’ reports for January 2008 and January 2010, an integral, with ∞ as a limit, is improper because the interval of integration is infinite. Most candidates correctly applied integration by parts to score the three marks available in part (b). In part (c) examiners expected to see the infinite upper limit replaced by, for example, a, the integration carried out and then consideration of the limiting process as a → ∞ . It was not uncommon to see in solutions the statement “as 1 −4 a −4 a " a → ∞, e → 0 so e ( − a − ) → 0" without 4 seemingly any particular analysis of lack of analysis was penalised.
lim ae −4 a . This a →∞
Question 4:
Exam report
3 dy 3 1 + y = ( x 4 + 3) 2 and y (1) = dx x 5 3
∫x ln( x ) An integrating factor = I e= e= x3 The equation becomes: x3
dx
3
3 dy + 3 x 2= x3 ( x 4 + 3) 2 dx
3 d 3 = x y ) x 3 ( x 4 + 3) 2 ( dx
x3 y=
∫ x (x 3
3
4
+ 3) 2 dx=
3 1 4 x3 ( x 4 + 3) 2 dx ∫ 4
5 1 2 x 3 y = × ( x 4 + 3) 2 + c 4 5 5 1 c 4 2 + = y x + 3 ( ) 3 10 x x3 1 For x = 1, y = this gives 5 1 1 = × 32 + c so c = −3 5 10 5 1 3 4 2 − = y x + 3 ( ) 3 10 x x3
Question 5:
Approximately half the candidates scored full marks for this question. Almost all 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 minority did not recognise the form of the integral of
∫ (x x
3
4
+3
)
3 2
(
dx as k x + 3 4
)
5 2
or did not apply a suitable
substitution, so made very little further progress.
Exam report
a ) From the formulae book x2 x4 + + .... 2 24 (4 x) 2 (4 x) 4 + + ... so Cos (4 x) = 1− 2 24 32 Cos (4 x) = 1 − 8 x 2 + x 4 + ... 3 −e x dy x = b) i= ) y ln(2 − e ) dx 2 − e x −2e x d 2 y −e x (2 − e x ) − e 2 x = 2 x 2 dx (2 − e ) (2 − e x )2 Cos ( x) =− 1
d 3 y −2e x (2 − e x ) 2 − (−2e x ) × 2(−e x )(2 − e x ) = dx3 (2 − e x ) 4 −2e x (2 − e x ) − 4e 2 x −4e x − 2e 2 x = (2 − e x )3 (2 − e x )3 ii ) y (0) = ln ( 2 − e0 ) = 0 −1 −2 y '(0) = = −1 y ''(0) = = −2 2 −1 (2 − 1) 2 −4 − 2 = −6 y (3) (0) = (2 − 1)3 −2 2 −6 3 Conclusion : ln(2 − e x ) =0 − 1x + x + x + ... 2! 3! ln(2 − e x ) =− x − x 2 − x 3 + ... − x 2 − x3 − x 4 + ... c) x ln(2 − e x ) = x(− x − x 2 − x 3 + ...) = 1 − Cos (4 x) = 1 − 1 + 8 x 2 + ... = 8 x 2 + ... Therefore,
x ln(2 − e x ) − x 2 + ... 1 = 2 =− + ... 1 − cos(4 x) 8 x + ... 8 x ln(2 − e x ) 1 = − x → 0 1 − cos(4 x ) 8
Conclusion : lim
Candidates’ differentiation skills have improved significantly over recent series. Almost all candidates could correctly quote the series expansion for cos4x although some failed to give their answer in its simplest form. The methods required to obtain the three derivatives were well understood and, although some algebraic errors were seen, there were many correct expressions presented. Most candidates displayed good knowledge of Maclaurin’s theorem but only those who had made no errors in earlier differentiations could score both marks for showing the printed result in part (b)(ii). There was a pleasing improvement in explicitly reaching the stage of a constant term in both the numerator and denominator before taking the limit as x→0 in the final part of the question.
Question 6:
C= 2(Cosθ − Sinθ ) 1 :r
0 ≤ θ ≤ 2π
Exam report
a )i ) Multiplying by r : = r 2 2rCosθ − 2rSinθ x2 + y 2 = 2 x − 2 y
x2 + = y 2 2( x − y )
ii ) x 2 + y 2 − 2 x + 2 y = 0 ( x − 1) 2 − 1 + ( y + 1) 2 − 1 =0 ( x − 1) 2 + ( y + 1) 2 = 2 This is the circle centre (1, − 1), radius 2 . b) C2 : r = 4 + Sinθ 0 ≤ θ ≤ 2π 1 2π 1 2π 2 (4 + Sin θ ) d θ = (16 + 8Sinθ + Sin 2θ )dθ ∫ ∫ 0 0 2 2 2 π 1 1 1 = Area (16 + 8Sinθ + − Cos (2θ ))dθ ∫ 0 2 2 2
i ) Area =
2π
1 33 33 = θ − 4 Cos ( θ ) − Sin (2 θ ) = π 4 8 2 0 ii ) r = 4 + Sinθ and r = 2(Cosθ − Sinθ ) so 4 + Sinθ = 2Cosθ − 2 Sinθ 3Sinθ − 2Cosθ = −4 R=
32 + 22 = 13 3 2 4 Sinθ − Cosθ = − 13 13 13 4 Sin(α − θ ) = − < −1 13 No solution for θ . 2 33 29 iii ) Area = π − π 2 = π 2 2
( )
This question on polar coordinates proved to be the most demanding question on the paper. In part (a), the better candidates had no problems in finding the cartesian equation for C 1 and rearranging it by completing the squares so as to be able to deduce that the curve was a circle. Many other candidates failed to eliminate θ even after a page or more of working and abandoned part (a) of the question. Many candidates scored heavily on the more familiar part (b)(i), finding the area bounded by the second curve C2. Proving that the two curves did not intersect in part (b)(ii) was a challenge, even for the better candidates, and it was rare to award all four marks for this part of the question. However, some excellent solutions were seen which used a variety of methods including forming and solving quadratic equations in either sin x, cos x, tan x, sec x , use of calculus and use of the R, α form. In the final part of the question, those who used part (a)(ii) to find the area enclosed by C 1 and used it with their answer to part (b)(i) had no problems scoring both marks. Those who tried to use integration to find the area enclosed by C 1 did not analyse the problem fully and failed to score. A common misinterpretation of the word ‘intersect’ is indicated by the not uncommon answer ‘Since C 1 and C 2 do not intersect the required area is just 16.5π , the area bounded by C 2 .’
Question 7: 1 2
a) = x t=
Exam report
t
dy dy dt = × =2 dx dt dx d 2 y d dy = ii ) 2 = dx dx dx i)
dx 1 dt = and = 2 t dt 2 t dx dy t dt d dy dt 2 t × dt dt dx dy dy d2y t × 2 = t 2 + 4 t dt dt dt 2
1 dy = +2 t dt d2y dy b) x 2 − (8 x 2 + 1) + 12 x3 y = 12 x5 becomes dx dx dy d2y dy 2 t 2 + 4t 2 − (8t + 1) 2 t + 12t t y = 12t t (÷ t ) dt dt dt dy d2y dy dy 12t × y 12t 2 (÷4t ) + 4t 2 − 16t − 2 += dt dt dt dt d2y dy − 4 + 3 y = 3t (E) 2 dt dt c) The auxiliary equation associated with the diff.eq. is 2
λ 2 − 4λ + 3 = 0 (λ − 3)(λ − 1) = 0 = λ 3= or λ 1 The complementary function is y = Ae3t + Bet A, B ∈ The particular integral is of the form dy d2y 0 a = = dt dt 2 ( E ) becomes 0 − 4a + 3at + 3b = 3t 4 This = gives a 1= and b 3 4 y = t + + Ae3t + Bet and with t = x2 3 2 2 4 so y x 2 + + Ae3 x + Be x = 3 y= at + b
Grade Mark
Max 75
A* 69
Grade boundaries A 64
B 56
Candidates generally answered part (a)(i) correctly but showing the printed result involving the second derivatives in part (a)(ii) proved to be more difficult, although candidates’ attempts displayed a significant improvement over candidates’ performances on previous papers. Although most candidates realised what was required to answer part (b), careless work resulted in many less than convincing solutions. In the final part of the question, the majority of candidates started out correctly to solve the differential equation for y in terms of t with many obtaining the correct form of the complementary function (although some used x instead of t), but finding the correct particular integral proved to be more problematic. A very common careless error resulted in candidates solving the equation “−4p + 3pt + q = 3t” instead of the correct equation “−4p + 3pt + 3q = 3t”. However, correct answers were quite frequently seen and it was particularly pleasing to see a higher proportion of candidates correctly converting back from t to x.
C 49
D 42
E 35
