General Certificate of Education January 2006 Advanced Level Examination

MATHEMATICS Unit Further Pure 3 Friday 27 January 2006

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. All necessary working should be shown; 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, formulae may be quoted, without proof, from the booklet. *

P80786/Jan06/MFP3 6/6/6/

MFP3

2

Answer all questions.

1

(a) Find the roots of the equation m2 ‡ 2m ‡ 2 ˆ 0 in the form a ‡ ib. (b)

(i) Find the general solution of the differential equation d2 y dy ‡ 2 ‡ 2y ˆ 4x 2 dx dx (ii) Hence express y in terms of x, given that y ˆ 1 and

2

(a) Find

…a 0

(6 marks) dy ˆ 2 when x ˆ 0. dx

xe 2x dx , where a > 0.

a!1

(c) Hence find

…1 0

xe 2x dx .

(a) Show that y ˆ x3

(4 marks)

(5 marks)

(b) Write down the value of lim ak e 2a , where k is a positive constant.

3

(2 marks)

(1 mark) (2 marks)

x is a particular integral of the differential equation dy 2xy ‡ 2 ˆ 5x 2 dx x 1

1

(3 marks)

(b) By differentiating …x 2

1†y ˆ c implicitly, where y is a function of x and c is a c is a solution of the differential equation constant, show that y ˆ 2 x 1 dy 2xy ‡ 2 ˆ0 dx x 1

(3 marks)

(c) Hence find the general solution of dy 2xy ‡ 2 ˆ 5x 2 dx x 1

P80786/Jan06/MFP3

1

(2 marks)

3

4

(a) Use the series expansion ln…1 ‡ x† ˆ x

1 x 2 ‡ 1 x3 2 3

1 x4 ‡ . . . 4

to write down the first four terms in the expansion, in ascending powers of x, of ln…1 x†. (1 mark) (b) The function f is defined by f …x† ˆ esin x Use Maclaurin's theorem to show that when f …x† is expanded in ascending powers of x: (i) the first three terms are 1 ‡ x ‡ 12 x 2

(6 marks)

(ii) the coefficient of x3 is zero.

(3 marks)

(c) Find sin x lim e

x!0

1 ‡ ln…1 x 2 sin x

x†

(4 marks)

Turn over for the next question

P80786/Jan06/MFP3

s

Turn over

4

5

(a) The function y…x† satisfies the differential equation dy ˆ f …x, y† dx where

f …x, y† ˆ x ln x ‡

and

y x

y …1† ˆ 1

(i) Use the Euler formula yr‡1 ˆ yr ‡ h f …xr , yr † with h ˆ 0:1, to obtain an approximation to y …1.1†.

(3 marks)

(ii) Use the formula yr‡1 ˆ yr 1 ‡ 2h f …xr , yr † with your answer to part (a)(i) to obtain an approximation to y …1.2†, giving your answer to three decimal places. (4 marks) (b)

(i) Show that

1 is an integrating factor for the first-order differential equation x dy dx

1 y ˆ x ln x x

(ii) Solve this differential equation, given that y ˆ 1 when x ˆ 1.

(3 marks) (6 marks)

(iii) Calculate the value of y when x ˆ 1:2, giving your answer to three decimal places. (1 mark)

P80786/Jan06/MFP3

5

6

(a) A circle C1 has cartesian equation x 2 ‡ …y of C1 is r ˆ 12 sin y.

6†2 ˆ 36. Show that the polar equation (4 marks)

(b) A curve C2 with polar equation r ˆ 2 sin y ‡ 5, 0 4 y 4 2p is shown in the diagram.

O

c

Initial line

Calculate the area bounded by C2 .

(6 marks)

(c) The circle C1 intersects the curve C2 at the points P and Q. Find, in surd form, the area of the quadrilateral OPMQ, where M is the centre of the circle and O is the pole. (6 marks)

END OF QUESTIONS

P80786/Jan06/MFP3

MFP3 – AQA GCE Mark Scheme, 2006 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.

2

AQA GCE Mark Scheme, 2006 January series – MFP3

MFP3 Q

Solution 1(a)

(m + 1)

2

= −1 m = −1 ± i

(b)(i) CF is e–x(A cos x + B sin x) {or e–xA cos(x + B) (–1+i)x (–1–i)x + Be } but not Ae

Marks M1

Total

A1

2

Comments Completing sq or formula

M1 A1

If m is real give M0 On wrong a’s and b’s but roots must be complex. OE

p = 2, q = –2 GS y = e–x(Acosx + Bsinx) +2x–2

M1 A1 A1 B1

(ii) x=0, y=1 ⇒ A = 3 y′(x) = – e–x (Acosx + Bsinx) + + e–x (– Asinx + Bcosx) + 2 y′(0) = 2 ⇒ 2 = –A+B+2 ⇒ B = 3

B1 M1 A1 A1

{P.Int.} try y = px + q

2 p + 2( px + q ) = 4 x

y = 3e–x(cosx + sinx) +2x–2 −2 x ∫ xe dx = –

=– a

1 – 2x 1 – 2x xe − e {+c} 2 4

∫ xe

−2 x

dx =–

0

= (b)

1 –2a 1 –2a 1 ae − e – (0– ) 2 4 4

1 1 –2a 1 –2a – ae − e 4 2 4

lim a k e −2 a = 0 a→∞ ∞

(c)

1 –2x 1 xe − ∫ - e − 2 x dx 2 2

∫ xe

−2 x

dx

Slips 4 12

Total 2(a)

6

On one slip Their CF + their PI with two arbitrary constants. Provided an M1gained in (b)(i) Product rule used

M1 A1

Reasonable attempt at parts

A1

Condone absence of +c

M1

F(a) – F(0)

A1

5

B1

1

=

0

lim

1 1 –2a 1 –2a – ae − e } a→∞ 4 2 4 1 1 = –0–0= 4 4 =

M1

{

A1 Total

If this line oe is missing then 0/2

2 8

3

On candidate’s “1/4” in part (a). B1 must have been earned

MFP3 – AQA GCE Mark Scheme, 2006 January series

MFP3 Q 3(a)

Solution 3

Marks B1

2

y = x − x ⇒ y′(x) = 3 x − 1

(

2x x3 − x dy 2 xy + 2 = 3x 2 − 1 + dx x − 1 x2 −1 = 3x 2 − 1 + (b)

[

(

)

M1

)

2x 2 x 2 − 1 = 5x 2 − 1 2 x −1

A1

]

d dy ( x 2 − 1) y = 2 xy + ( x 2 − 1) dx dx

(

)

Total

Substitution into LHS of DE

3

SC Differentiated but not implicitly give max of 1/3 for complete solution

dy =0 dx

c is a soln. of x −1 dy 2 xy + 2 =0 dx x − 1

⇒y=

Completion. If using general cubic all unknown constants must be found

M1A1

Differentiating x 2 −1 y = c wrt x leads to 2 xy + ( x 2 − 1)

Comments Accept general cubic.

2

A1

3

Be generous

c is a soln with one arb. x −1 dy 2 xy constant of + 2 =0 dx x − 1 c ⇒y= 2 is a CF of the DE x −1

(c) ⇒ y =

2

GS is CF + PI

M1

c y= 2 + x3 − x x −1

A1

Total

4

2

8

Must be using ‘hence’; CF and PI functions of x only CSO Must have explicitly considered the link between one arbitrary constant and the GS of a first order differential equation.

AQA GCE Mark Scheme, 2006 January series – MFP3

MFP3 Q 4(a)

Solution

ln (1 − x ) = − x −

(b)(i) f (x) = e

sin x

Marks

Total

B1

1

1 2 1 3 1 4 x − x − x ... 2 3 4

⇒ f(0) = 1

Comments

B1

f ′(x) = cos x e sin x ⇒ f ′(0) = 1

M1A1

f′′ (x) = – sin x e sin x + cos2x e sin x f′′ (0) = 1

Product rule used M1A1

x2 f′′(0) 2 1 so 1st three terms are 1 + x + x 2 2 Maclaurin f (x)= f(0)+xf ′(0)+

(ii) f′′′(x) = cos x(cos2x–sinx) e sin x +

A1

CSO AG

3

CSO AG SC for (b): Use of series expansions….max of 4/9

M1A1

+{2cosx(–sinx)–cos x} e sin x f′′′(0) = 0 so the coefficient of x3 in the series is zero

A1

(c) sinx ≈ x.

e sin x − 1 + ln (1 − x ) = x 2 sin x

6

1 − x 3 + o( x 4 ) 3 x3

B1

Ignore higher power terms in sinx expansion

M1 A1

Series from (a) & (b) used Numerator kx3 (+…)

1 + o( x ) = 3 1 + o( x 2 ) −

Condone if this step is missing

e sin x − 1 + ln (1 − x ) 1 =− 2 x→0 3 x sin x

lim

A1

Total

4 14

5

On candidate’s x3 coefficient in (a) provided lower powers cancel

MFP3 – AQA GCE Mark Scheme, 2006 January series

MFP3 Q

Solution

Marks

5(a)(i) y(1.1) = y(1) + 0.1[1ln1+1/1]

Total

Comments

M1A1

= 1+0.1 = 1.1

A1

(ii) y(1.2) = y(1) + 2(0.1)[f(1.1, y(1.1)]

3

M1A1

…. = 1+2(0.1)[1.1ln1.1+(1.1)/1.1]

A1

On answer to (a)(i)

…. = 1+0.2×1.104841198…. …. = 1.22096824 = 1.221 to 3dp (b)(i)

− IF is e ∫

A1

1 dx x

– lnx

= e

(ii)

CAO

Condone e

M1

∫

1 dx x for

M mark

A1

=e

ln x −1

4

= x− 1 =

1 x

A1

3

AG (be convinced) (b)(i) Solutions using the printed answer must be convincing before any marks are awarded

d ⎛ y⎞ ⎜ ⎟ = ln x dx ⎝ x ⎠

M1A1

y ⎛1⎞ = ∫ ln x dx = x ln x − ∫ x⎜ ⎟ dx x ⎝ x⎠

M1

Integration by parts for xk lnx

y = x ln x − x + c x

A1

Condone missing c.

y(1) = 1 ⇒ 1 = ln1 – 1 + c

m1

Dependent on at least one of the two previous M marks

⇒ c = 2 ⇒ y = x 2 ln x − x 2 + 2 x

A1

6

(iii) y(1.2) =1.222543…= 1.223 to 3dp

B1

1

Total

6

17

OE eg

y = x ln x − x + 2 x

AQA GCE Mark Scheme, 2006 January series – MFP3

MFP3 Q 6(a)

Solution 2

Marks

Total

Comments

2

x + y − 12 y + 36 = 36 r 2 − 12r sin θ + 36 = 36 ⇒ r = 12 sin θ

A1

1 (2 sin θ + 5) 2 dθ . ∫ 2

(b) Area =

Use of y = r sin θ (x = r cosθ PI) Use of x 2 + y 2 = r 2

M1 M1 m1 4

CSO AG

1 2 r dθ . 2∫

M1

Use of

B1 B1

Correct expn. of (2sinθ +5)2 Correct limits

M1

Attempt to write sin 2 θ in terms of cos 2θ.

A1

Correct integration ft wrong coeffs

2π

1 .. = ∫ (4 sin 2 θ + 20 sin θ + 25 )dθ 2 0 2π

1 = ∫ (2(1 − cos 2θ ) + 20 sin θ + 25 )d 2 0 θ

2π 1 [27θ − sin 2θ − 20 cosθ ] 0 2 = 27π. =

(c) At intersection 12 sin θ = 2 sin θ + 5

A1

6

M1

OE eg r = 6(r − 5)

⇒ sin θ =

A1

OE eg r = 6

⎛ 5π ⎞ Points ⎜ 6, ⎟ and ⎜ 6, ⎟ ⎝ 6⎠ ⎝ 6 ⎠ OPMQ is a rhombus of side 6

A1

OE

5 10 ⎛ π⎞

Area = 6 × 6 × sin

Or two equilateral triangles of side 6

2π oe 3

M1 A1

= 18 3

Any valid complete method to find the area (or half area) of quadrilateral.

A1

6

Total Total

16 75

Extra notes:

The SC for Q4 2

sin x

e

CSO

3

⎛ 1⎛ x3 ⎞ 1 ⎛ x3 ⎞ x3 ⎞ = 1 + ⎜⎜ x − ...⎟⎟ + ⎜⎜ x − ...⎟⎟ + ⎜⎜ x − ... ⎟⎟ ... 3! ⎠ 2! ⎝ 3! ⎠ 3! ⎝ 3! ⎠ ⎝

M1 for 1st 3 terms ignoring any higher powers than those shown.

7

Accept unsimplified surd

MFP3 – AQA GCE Mark Scheme, 2006 January series

A1 for all 4 terms (could be treated separately ie last term often only comes into (b)(ii)

= 1+ x − = 1+ x +

1 x3 1 2 + x − .... + x 3 − .... 6 2 6

1 2 x 2

(

) (

)

A1 (be convinced…..ignore any powers of x above power 2)

Coefficient of x3: −

x3 1 3 + x 6 6

=0

A1 (be convinced…..ignore any powers of x above power 3)

Quite often the 2nd A mark is awarded before the 1st A1

8

klm

AQA January 2006 Examinations Scaled Mark Component Grade Boundaries (GCE Specifications) Comp. Code HS2O HS2Q HS2R HS2U

Component Title GCE HISTORY UNIT 2 ALTERNATIVE O GCE HISTORY UNIT 2 ALTERNATIVE Q GCE HISTORY UNIT 2 ALTERNATIVE R GCE HISTORY UNIT 2 ALTERNATIVE U

Maximum Scaled Mark 50 50 50 50

A 34 35 37 36

Scaled Mark Grade Boundaries B C D 29 24 20 31 27 23 32 28 24 32 28 24

E 16 20 20 20

ICT1 ICT2 ICT3 ICT4 ICT5 ICT6

GCE INFORMATION AND COMM TECH UNIT 1 GCE INFORMATION AND COMM TECH UNIT 2 GCE INFORMATION AND COMM TECH UNIT 3 GCE INFORMATION AND COMM TECH UNIT 4 GCE INFORMATION AND COMM TECH UNIT 5 GCE INFORMATION AND COMM TECH UNIT 6

60 60 60 90 90 90

39 45 42 56 54 59

34 40 36 50 48 50

29 36 30 44 42 42

24 32 24 38 36 34

20 28 18 32 30 26

LAW1 LAW2 LAW3 LAW4 LAW5

GCE LAW UNIT 1 GCE LAW UNIT 2 GCE LAW UNIT 3 GCE LAW UNIT 4 GCE LAW UNIT 5

65 65 65 85 85

49 49 45 59 56

45 45 40 54 52

41 41 35 49 48

37 37 31 44 44

34 33 27 40 40

GCE MATHEMATICS UNIT D01 GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP1 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M1A - WRITTEN GCE MATHEMATICS UNIT M1A - COURSEWORK GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT M2A - WRITTEN GCE MATHEMATICS UNIT M2A - COURSEWORK

75 75 75 75 75 75 75 25 75 75 25

59 61 62 58 60 59 58 20 59 58 20

52 53 54 51 52 51 50 18 52 50 18

45 45 46 44 45 44 43 15 45 43 15

38 38 39 38 38 37 36 13 38 35 13

32 31 32 32 31 30 30 10 31 29 10

MD01 MD02 MFP1 MFP2 MFP3 MFP4 MM1A/W MM1A/C MM1B MM2A/W MM2A/C

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