General Certificate of Education June 2008 Advanced Level Examination
MATHEMATICS Unit Further Pure 3 Monday 16 June 2008
MFP3
1.30 pm to 3.00 pm
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 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. *
P5520/Jun08/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 lnðx þ yÞ yð2Þ ¼ 3
and Use the improved Euler formula
1
yrþ1 ¼ yr þ 2ðk1 þ k2 Þ where k1 ¼ hf ðxr , yr Þ and k2 ¼ hf ðxr þ h, yr þ k1 Þ and h ¼ 0:1 , to obtain an approximation to yð2:1Þ , giving your answer to four decimal places. (6 marks)
2
(a) Find the values of the constants a, b, c and d for which a þ bx þ c sin x þ d cos x is a particular integral of the differential equation dy 3y ¼ 10 sin x 3x dx (b) Hence find the general solution of this differential equation.
3
(a) Show that x 2 ¼ 1 2y can be written in the form x 2 þ y 2 ¼ ð1 yÞ2 .
(4 marks) (3 marks)
(1 mark)
(b) A curve has cartesian equation x 2 ¼ 1 2y . Find its polar equation in the form r ¼ f ðyÞ , given that r > 0 .
P5520/Jun08/MFP3
(5 marks)
3
4
(a) A differential equation is given by d2 y dy x 2 ¼ 3x 2 dx dx Show that the substitution u¼
dy dx
transforms this differential equation into du 1 u ¼ 3x dx x
(2 marks)
(b) By using an integrating factor, find the general solution of du 1 u ¼ 3x dx x giving your answer in the form u ¼ f ðxÞ .
(6 marks)
(c) Hence find the general solution of the differential equation x
d2 y dy ¼ 3x 2 dx 2 dx
giving your answer in the form y ¼ gðxÞ . ð 5
(a) Find
(2 marks)
x 3 ln x dx . ðe
(b) Explain why
(3 marks)
x 3 ln x dx is an improper integral.
(1 mark)
0
ðe (c) Evaluate
x 3 ln x dx , showing the limiting process used.
(3 marks)
0
6
(a) Find the general solution of the differential equation d2 y dy 2 3y ¼ 10e2x 9 2 dx dx (b) Hence express y in terms of x, given that y ¼ 7 when x ¼ 0 and that as x ! 1 .
(10 marks) dy !0 dx (4 marks)
Turn over for the next question P5520/Jun08/MFP3
s
Turn over
4
7
(a) Write down the expansion of sin 2x in ascending powers of x up to and including the (1 mark) term in x 3 . (b)
(i) Given that y ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffi dy d2 y and when x ¼ 0 . (5 marks) 3 þ e x , find the values of dx dx 2
(ii) Using Maclaurin’s theorem, show that, for small values of x, pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 7 3 þ e x 2 þ 4 x þ 64 x 2
(2 marks)
(c) Find pffiffiffiffiffiffiffiffiffiffiffiffiffix 3þe 2 lim x!0 sin 2x
(3 marks)
8 The polar equation of a curve C is r ¼ 5 þ 2 cos y,
p 4 y 4 p
(a) Verify that the points A and B, with polar coordinates ð7, 0Þ and ð3, pÞ respectively, lie on the curve C. (2 marks) (b) Sketch the curve C.
(2 marks)
(c) Find the area of the region bounded by the curve C.
(6 marks)
p (d) The point P is the point on the curve C for which y ¼ a , where 0 < a 4 . The 2 point Q lies on the curve such that POQ is a straight line, where the point O is the pole. Find, in terms of a , the area of triangle OQB . (4 marks)
END OF QUESTIONS
Copyright Ó 2008 AQA and its licensors. All rights reserved.
P5520/Jun08/MFP3
MFP3 - AQA GCE Mark Scheme 2008 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 2008 June series
MFP3 Q 1
Solution k1 = 0.1 × ln ( 2 + 3) = 0.1609(4379…) (= *)
Marks M1 A1
Total
Comments
PI
k2 = 0.1 × f (2.1, 3 + *...) … = 0.1 × ln(2.1 + 3.16094…)]
M1
…. = 0.1660(31…)
A1
PI
m1
Dep on previous two Ms and numerical values for k’s
1 [ k1 + k2 ] 2 = 3 + 0.5 × 0.3269748….
y(2.1) = y(2) +
= 3.163487… = 3.1635 to 4dp
A1 Total
2(a) PI: yPI = a + bx + c sin x + d cos x ′ = b + c cos x − d sin x yPI b + c cos x − d sin x − 3a − 3bx − 3c sin x
6 6
Must be 3.1635
M1
Substituting into DE
M1
Equating coefficients (at least 2 eqns)
− 3d cos x = 10sin x − 3x b−3a= 0; −3b= −3; c−3d=0; −d−3c=10 1 a = ; b = 1 ; c = −3; d = −1 3 1 yPI = + x − 3sin x − cos x 3 (b) Aux. eqn. m − 3 = 0
4
A1 for any two correct
∫y
Altn.
M1
3x
Ae
−1
dy = ∫ 3 dx OE (M1)
OE
( yCF =) Ae3 x
A1
1 ( yGS =) Ae3 x + + x − 3sin x − cos x 3
B1F
3
(c’s CF + c’s PI ) with 1 arbitrary constant
B1
7 1
AG
Total 3(a) x + y = 1 − 2y + y ⇒ x + y = (1 − y)2 2
(b)
A2,1
2
2
2
2
x2 + y 2 = r 2 y = r sin θ x2 = 1 − 2y so x2 + y2 = (1 − y)2 ⇒ r 2 = (1 − r sinθ ) 2
M1 M1
Or x = r cosθ
A1
OE eg r 2 cos2θ = 1 − 2r sinθ PI by the next line
r = 1 − r sin θ or r = −(1 − r sin θ ) r (1 + sin θ ) = 1 or r (1 − sin θ ) = −1 1 r > 0 so r = 1 + sin θ
m1
Either
A1 Total
5 6
4
CSO
MFP3 - AQA GCE Mark Scheme 2008 June series
MFP3 (cont) Q 4(a)
Solution dy du d 2 y ⇒ = u= dx dx dx 2 du du 1 x − u = 3x 2 ⇒ − u = 3x dx dx x
Marks
A1
2
M1
= e− ln x
(c)
Comments
M1
1 (b) IF is exp ( ∫ − dx) x
AG Substitution into LHS of DE and completion
and with integration attempted
A1
1 or x
= x−1
Total
A1
or multiple of x−1
d ⎡ux −1 ⎤⎦ = 3 dx ⎣
M1
LHS as differential of u × IF . PI
⇒ ux−1 = 3x + A
m1
Must have an arbitrary constant (Dep. on previous M1 only)
u = 3x 2 + Ax
A1
dy = 3x 2 + Ax dx
M1
6 Replaces u by
dy and attempts to dx
integrate y = x3 +
Ax 2 +B 2
A1F Total
5(a)
3 ∫ x ln x dx =
4
4
x x ln x − ∫ 4 4
⎛1⎞ ⎜ ⎟ dx ⎝ x⎠
2
ft on cand’s u but solution must have two arbitrary constants
10
...= kx 4 ln x ± ∫ f ( x ) , with f(x) not
M1
involving the ‘original’ ln x A1
…… =
x4 x4 ln x − + c 4 16
(b) Integrand is not defined at x = 0 (c)
∫ =
e 0
{
e
3
Condone absence of ‘+ c’
E1
1
OE
}
x3 ln x dx = lim ∫ x3 ln x dx a →0
a
⎡ a4 3e 4 a4 ⎤ − lim ⎢ ln a − ⎥ 16 a →0 ⎣ 4 16 ⎦
But lim a 4 ln a = 0 a →0
So
A1
∫
e 0
x 3 ln x dx exists and =
M1
F(e) – F(a)
B1
Accept a general form eg lim x k ln x = 0 x →0
4
3e 16
A1 Total
3 7
5
CSO
MFP3 - AQA GCE Mark Scheme 2008 June series
MFP3 (cont) Q Solution 2 6(a) Aux eqn: m − 2m − 3 = 0 m = −1, 3
Marks M1 A1
CF ( yC =) Ae3 x + Be − x
M1
Try ( yPI =) a e −2 x ( + b) dy = −2ae −2 x dx d2 y = 4ae −2 x dx 2
M1
⇒a=2 b=3
A1 B1
( yGS =) Ae3 x + Be − x + 2e −2 x + 3
B1F
When A = 0, 5 = 0 + B + 3 ⇒ B = 2 y = 2e − x + 2e −2 x + 3 Total
PI
A1
M1
dy = 3 Ae3 x − Be − x − 4e −2 x dx dy As x → ∞ , e − kx → 0 , → 0 so A = 0 dx
Comments
A1
Substitute into DE gives 4ae −2 x + 4ae −2 x − 3ae −2 x − 3b = 10e −2 x − 9
(b) x = 0, y = 7 ⇒ 7 = A + B + 2 + 3
Total
10
(c’s CF+c’s PI) with 2 arbitrary constants Only ft if exponentials in GS and two arbitrary constants remain
B1F
B1 B1F A1
6
4 14
Must be using ‘A’ = 0 CSO
MFP3 - AQA GCE Mark Scheme 2008 June series
MFP3 (cont) Q 7(a)
(b)(i)
Solution
sin 2 x ≈ 2 x −
( 2x) 3!
3
+ .. = 2 x −
4 3 x + .. 3
Total
B1
1
Comments
1 − dy 1 = ( 3 + e x ) 2 (e x ) dx 2
M1 A1
Chain rule
1 3 d2 y 1 x 1 x −2 x −2 = + − + e 3 e 3 e (e 2 x ) ( ) ( ) dx 2 2 4
M1 A1
Product rule OE OE
1 1 1 7 y′(0) = ; y″(0) = − = 4 4 32 32 1 1 1 7 = (ii) y(0) = 2; y′(0) = ; y″(0) = − 4 4 32 32 x2 McC. Thm: y(0) + x y′(0) + y″(0) 2 1 7 2 3 + ex ≈ 2 + x + x 4 64
(c)
Marks
1 7 2 ⎡ ⎤ ⎡ 3 + e x − 2 ⎤ ⎢ 2 + 4 x + 64 x − 2 ⎥ ⎢ ⎥=⎢ ⎥ 4 ⎥ ⎣⎢ sin 2 x ⎦⎥ ⎢ 2 x − x3 3 ⎣ ⎦ ⎡1 7 ⎤ ⎢ 4 + 64 x + ... ⎥ = ⎢ ⎥ ⎢ 2 − 4 x 2 + .. ⎥ 3 ⎣ ⎦ 1 ⎡ 3 + ex − 2 ⎤ 1 lim ⎢ ⎥= 4 = x →0 sin 2 2 8 x ⎢⎣ ⎥⎦
A1
5
CSO
M1 A1
2
CSO; AG
M1
Dividing numerator and denominator by x to get constant term in each
m1
A1F Total
3 11
7
Ft on cand’s answer to (a) provided of the form ax+bx3
MFP3 - AQA GCE Mark Scheme 2008 June series
MFP3 (cont) Q Solution 8(a) θ = 0, r = 5 + 2cos0 = 7 {A lies on C}
θ = π, r = 5 + 2cos π = 3 {B lies on C} (b)
Marks B1
Total
B1
2
Comments
5
B1 3
Closed single loop curve, with (indication of) symmetry
7
B1
2
Critical values, 3,5,7 indicated
5
(c)
Area =
1 (5 + 2cos θ ) 2 dθ ∫ 2
π
=
1 ( 25 + 20cosθ + 4cos2 θ ) dθ 2 −∫π
Use of
B1 B1
OE for correct expansion of (5 + 2cos θ ) 2 For correct limits
M1
Attempt to write cos 2 θ in terms of cos 2θ
π
1 = ∫ ( 25 + 20cos θ + 2(cos 2θ + 1) ) dθ 2 −π 1 π = [ 27θ + 20sin θ + sin 2θ ] − π 2 = 27π (d) Triangle OBQ with OB = 3 and angle BOQ = α
OQ = 5 + 2 cos(−π + α)
Area of triangle OQB = =
1 OB × OQ sin α 2
3 ( 5 − 2cos α ) sin α 2
1 2 r dθ 2∫
M1
A1F A1
6
Correct integration ft wrong non-zero coefficients in a + bcosθ + ccos2θ CSO
B1
PI
M1
OE
m1
Dep. on correct method to find OQ
A1 Total TOTAL
4 14 75
8
CSO
AQA – Further pure 3 – Jun 2008 – Answers Question 1:
Exam report
dy ln( x + y ) f ( x, y ) = and = dx
3 y (2) =
k1 = 0.1× ln(2 + 3) = 0.16094 y0 + k1 =3 + 0.16094 =3.16094
0.1× ln ( 2.1 + 3.16094 ) = 0.16603 k2 = 1 3.1635 3 + ( 0.16094 + 0.16603) = y (2.1) = 2 Question 2:
The majority of candidates were able to correctly use the given improved Euler formula to find the approximate value of y(2.1) to four decimal places. Otherwise, the most common error was to use y r + h instead of 1 y r +k 1 in writing k 2 as 0.1ln (2.1 + 3.1). There was a minority of candidates who failed to gain marks because they gave the wrong answer for y(2.1) and showed no method in their working, instead just giving a table of incorrect values.
a ) y =a + bx + cSinx + dCosx dy = b + cCosx − dSinx dx dy − 3= y 10 Sinx − 3 x becomes dx b + cCosx − dSinx − 3(a + bx + cSinx + dCosx = ) 10 Sinx − 3 x (−d − 3c) Sinx + (c − 3d )Cosx − 3bx + b −= 3a 10 Sinx − 3 x
−d − 3c =10 −3b =−3 and This gives: = = c − 3d 0 b − 3a 0 1 −1, c = −3= d= , b 1= ,a 3 1 The particular integral is y = + x − 3Sinx − Cosx 3 b) The complementary function: The auxiliary equation = : λ − 3 0= λ 3 y = Ae3 x The general solution y =
Exam report
Most candidates differentiated the given expression correctly and substituted the result into the given differential equation. Subsequently, however, there were many cases of an incorrect expansion for the term –3y which lead to the values for two of the four constants being incorrect. In part (b), it was pleasing to find only a small number of candidates trying to use an integrating factor with the original differential equation. The successful candidates either used the auxiliary equation m – 3 = 0 or used the reduced equation and then separated the variables. A common error, after finding m = 3, was to assume that this represented repeated roots, leading to a complementary 3x function of the form e (Ax + B), with two arbitrary constants for this first order differential equation. The error for those using the separation of variables method was to omit the arbitrary constant.
1 + x − 3Sinx − Cosx + Ae3 x 3
Question 3:
Exam report
a ) x = 1 − 2 y by adding y on both sides 2
2
x 2 += y 2 y 2 − 2 y + 1 = ( y − 1) 2 b) x 2 + y 2 = ( y − 1) 2 r (rSinθ − 1) = r= rSinθ − 1 or r = −rSinθ + 1 r (1 − Sinθ ) = 1 −1 or r (1 + Sinθ ) = 2
2
1 1 or r = Sinθ − 1 1 + Sinθ We wnat r > 0, so we keep the expression 1 r= 1 + Sinθ r
Part (a) was generally answered correctly, with most candidates showing sufficient detail in reaching the printed result. Although most candidates started their solution for part (b) by successfully substituting rcosθ for x and rsinθ for y into either the given equation or the alternative form given in part (a), many failed to go on to score full marks because they did not consider and eliminate the negative square root (or the second solution of the quadratic equation).
Question 4:
Exam report
2
d y dy dy a) = x 2 − u 3x 2 = dx dx dx du x −u = 3x 2 dx du 1 − u= 3x dx x b) An integrating factor= is I
e∫
1 ln x
1 − dx x
= e=
1 x
The equation becomes: 1 du 1 3 − u= x dx x 2 d u u = 3x + A = 3 dx x x
Most candidates scored full marks for their solution to part (a). Again, part (b) was answered well by the majority of candidates, but it was not uncommon to find solutions which used the wrong integrating factor, x, or lacked an arbitrary constant. Those candidates who started part (c) by equating their answer for (b) to
dy dx
and integrating normally scored
both marks, although some lost the accuracy mark because their general solution of this second order differential equation did not contain two arbitrary constants.
u 3 x 2 + Ax = dy A x3 + x 2 + B c)= u = 3 x 2 + Ax so y = dx 2 Question 5:
Exam report
1 4 1 1 x ln x − ∫ x 4 × dx 4 4 x 1 4 1 3 x ln x − ∫ x dx = 4 4 1 4 1 4 = x ln x − x + c 4 16
a ) ∫ x3 ln x dx =
e
b) ∫ x 3 ln xdx is an improper integral because 0
the function x3 ln x is not defiend at x = 0. c) ∫
e
a
e
1 1 e4 e4 1 1 x ln xdx = x 4 ln x − x 4 = − − a 4 ln a + a 4 16 a 4 16 4 16 4 3
lim a 4 ln a = 0 and lim a 4 = 0 x →0
∫
e
0
x3 ln xdx exists and
x →0
∫
e
0
x 3 ln xdx =
3e 4 16
Most candidates applied integration by parts accurately to obtain the correct answer for the integral 3 of x ln x. Although many candidates gave a correct explanation in part (b) for why the integral was improper, there were others whose incorrect explanations centred on the limit e or the interval of integration being infinite. For full marks in part (c), candidates were expected to pay particular attention 4 to the value of the limit of, for example, a ln a as a tended to 0.
Question 6: d2y dy a ) 2 − 2 − 3= y 10e −2 x − 9 dx dx The auxiliary equation is λ 2 − 2λ − 3 = 0 (λ − 3)(λ + 1) = 0 λ = 3 or λ = −1
Exam report
Ae3 x + Be − x •The complementary function is yc = ae −2 x + b •The particular integral y = dy d2y = −2ae −2 x = 4ae −2 x dx dx 2 d2y dy 10e −2 x − 9 becomes − 2 − 3y = 2 dx dx 4ae −2 x + 4ae −2 x − 3ae −2 x − 3= b 10e −2 x − 9 5ae −2 x − 3= b 10e −2 x − 9 This= gives a 2= and b 3 The particular integral = y 2e −2 x + 3 The general solution is y = 3 + 2e −2 x + Ae3 x + Be − x b) When x = 0, y = 7 so 7 = 3 + 2 + A + B A+ B = 2
In part (a), many candidates scored high marks for finding the general solution of the second order differential equation, with the most common slip being a wrong expansion of brackets, which led to −9 instead of −3 for the constant part of the particular integral. A more serious but less common error was −2x to look for a particular integral of the form axe + b . Part (b) proved, as expected, to be more of a challenge to candidates, with many unable to deal with the boundary condition expressed as a limit. This was the first time that this had appeared on an examination paper for this unit and, in general, only the better candidates could handle it.
dy = −4e −2 x + 3 Ae3 x − Be − x dx dy When x → ∞, we want → 0, this mean A =0. dx ( Because if A ≠ 0, Ae3 x → ∞) and B = 2 x →∞ y= 3 + 2e −2 x + 2e − x
Question 7: a ) Sin( x) =x −
Exam report 3
3
(2 x) 4 x + ... so Sin(2 x) =2 x − =2 x − x3 + ... 6 6 3 1
b)i ) y = 3 + e x = ( 3 + e x ) 2 1 − dy 1 x = × e × (3 + ex ) 2 dx 2 1 3 1 x 1 x d2y 1 x x −2 x −2 3 3 e e e e e = + + × − + ( ) ( ) 2 2 dx 2 2 0) = 4 2 y (= 1 1 1 − = ( 4) 2 2 4 1 1 − 32 7 y ''(0) = − (4) = 4 4 32
y '(0) =
x 2+ Conclusion : 3 + e=
1 7 x + x 2 + ... 4 64
1 7 x + x 2 + ... 4 64 4 2 Sin(2 x) =2 x − x + ... 3 1 7 1 7 x + x 2 + ... + x + ... 3 + ex − 2 4 64 so = = 4 64 4 4 Sin(2 x) 2 x − x 2 + ... 2 − x + ... 3 3
c) 3 + e x − 2=
lim x →0
3 + ex − 2 1 = 8 Si n ( 2 x )
The examiners expected candidates to evaluate 3! in the expansion for sin 2x, although credit was given retrospectively if the evaluation was left until part (c). Most candidates gave the correct expansion in part (a), and it was pleasing to find a greater proportion of the candidates than last summer applying the chain rule and product rule correctly in part (b)(i). Although Maclaurin’s theorem was well known, a significant number of candidates who had obtained a wrong value for the second derivative in part (b)(i) tried to convince the examiners that it led to the printed result in part (b)(ii). Such candidates would have been better advised to look for their error in part (b)(i). Although many candidates scored the three marks in part (c), there were others who did not show the division of the numerator and denominator by x before finding the limit as x tended to zero.
Question 8:
r = 5 + 2Cosθ
Exam report
−π ≤ θ ≤ π
a ) For θ = 0, r = 5 + 2Cos 0 = 7 For θ == π , r 5 + 2Cosπ = 3 A(7, 0) and B(3, π ) belong to the curve C. b) 1 π 1 π (5 + 2Cosθ ) 2 dθ = ∫ 25 + 4Cos 2θ + 20Cosθ dθ ∫ 2 −π 2 −π 1 π = A 25 + 2Cos 2θ + 2 + 20Cosθ dθ 2 ∫−π 1 π A = ∫ 27 + 2Cos 2θ + 20Cosθ dθ 2 −π c) A =
π
1 27 = A θ + Sin 2θ + 10 Sinθ = 27π 2 2 −π d ) B(3, π ) P(5 + 2Cosα , α ) and Q(5 + 2Cos (−π + α ), −π + α ) Notice cos(−π + α ) =−Cos (α ) The length OB= 3, the length OQ= 5 − 2Cosα
method seen involved the integral
The angle BOQ is α
which no credit was awarded.
1 OB × OQ × Sin( BOQ) 2 1 15 3 Area = × 3 × ( 5 − 2Cosα ) Sinα = Sinα − Sin 2α 2 2 2 3 Area = ( 5Sinα − Sin 2α ) 2
The area of the triangle BOQ is
Grade Mark
Grade boundaries Max 75
A 63
Parts (a) and (b) were generally answered well, but some candidates did not identify the critical value of 5 on their sketches. In part (c), although the usual errors were seen — for example the wrong expansion 2 2 (5 + 2cosθ ) = 25 + 10cosθ + 4cos θ or a sign error in 2 the identity for 4cos θ in terms of cos2θ — most candidates had a thorough understanding of the method required to find the area of the region bounded by the curve. Although only a minority of candidates scored full marks for the final part of this last question, it is pleasing to report that a significant number of other candidates were awarded partial credit for finding an expression for OQ, although many lost at least one mark because they used π −α instead of α − π for θ. The most common wrong
B 55
C 47
D 39
π
1
∫ π α 2 r dθ , for 2
− +
E 31
