General Certificate of Education June 2006 Advanced Level Examination

MATHEMATICS Unit Further Pure 3 Monday 19 June 2006

MFP3 9.00 am to 10.30 am

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. *

P85407/Jun06/MFP3 6/6/6/

MFP3

2

Answer all questions.

1 It is given that y satisfies the differential equation d2 y dx 2

5

dy ‡ 4y ˆ 8x dx

10

10 cos 2x

(a) Show that y ˆ 2x ‡ sin 2x is a particular integral of the given differential equation. (3 marks) (b) Find the general solution of the differential equation. (c) Hence express y in terms of x , given that y ˆ 2 and

(4 marks) dy ˆ 0 when x ˆ 0. dx

(4 marks)

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

f …x, y† ˆ

and

x2 ‡ y2 xy

y…1† ˆ 2

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

(3 marks)

(b) Use the improved Euler formula 1

yr‡1 ˆ yr ‡ 2 …k1 ‡ k2 † where k1 ˆ hf …xr , yr † and k2 ˆ hf …xr ‡ h, yr ‡ k1 † and h ˆ 0:1, to obtain an approximation to y…1:1†, giving your answer to four decimal places. (6 marks)

P85407/Jun06/MFP3

3

3

(a) Show that sin x is an integrating factor for the differential equation dy ‡ …cot x†y ˆ 2 cos x dx

(3 marks)

(b) Solve this differential equation, given that y ˆ 2 when x ˆ

p . 2

(6 marks)

4 The diagram shows the curve C with polar equation r ˆ 6…1

cos y†,

0 4 y < 2p

O

(a) Find the area of the region bounded by the curve C.

Initial line

(6 marks)

(b) The circle with cartesian equation x 2 ‡ y 2 ˆ 9 intersects the curve C at the points A and B.

5

(i) Find the polar coordinates of A and B.

(4 marks)

(ii) Find, in surd form, the length of AB.

(2 marks)

  3a ‡ 2 3 (a) Show that lim ˆ . a ! 1 2a ‡ 3 2 … 1

3 (b) Evaluate 3x ‡ 2 1 rational number.

(2 marks)

 2 dx , giving your answer in the form ln k , where k is a 2x ‡ 3 (5 marks)

P85407/Jun06/MFP3

s

Turn over

4

6

(a) Show that the substitution uˆ

dy ‡ 2y dx

transforms the differential equation d2 y dy ‡ 4 ‡ 4y ˆ e 2x 2 dx dx into du ‡ 2u ˆ e 2x dx

(4 marks)

(b) By using an integrating factor, or otherwise, find the general solution of du ‡ 2u ˆ e 2x dx giving your answer in the form u ˆ f …x† .

(5 marks)

(c) Hence find the general solution of the differential equation d2 y dy ‡ 4 ‡ 4y ˆ e 2x 2 dx dx giving your answer in the form y ˆ g…x† .

P85407/Jun06/MFP3

(5 marks)

5

7

(a)

(i) Write down the first three terms of the binomial expansion of …1 ‡ y† 1 , in ascending powers of y. (1 mark) (ii) By using the expansion cos x ˆ 1

x2 x4 ‡ 2! 4!

:::

and your answer to part (a)(i), or otherwise, show that the first three non-zero terms in the expansion, in ascending powers of x, of sec x are 1‡

x 2 5x 4 ‡ 2 24

(5 marks)

(b) By using Maclaurin's theorem, or otherwise, show that the first two non-zero terms in the expansion, in ascending powers of x, of tan x are x‡

x3 3

 x tan 2x . (c) Hence find lim 1 x ! 0 sec x

(3 marks)



END OF QUESTIONS

P85407/Jun06/MFP3

(4 marks)

MFP3 – AQA GCE Mark Scheme, 2006 June series

Key To Mark Scheme And Abbreviations Used In Marking M

mark is for method

m or dM A B E

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

or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA

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 June series – MFP3

MFP3 Q 1(a)

Solution y = 2 x + sin 2 x ⇒ y ' = 2 + 2cos 2 x ⇒ y '' = −4sin 2 x

Marks

A1

(b) Auxiliary equation m 2 − 5m + 4 = 0 m = 4 and 1 CF: A e4x + B ex GS: y = A e4x + B ex + 2 x + sin 2 x (c) x = 0, y = 2 ⇒ 2 = A + B x = 0, y ' = 0 ⇒ 0 = 4 A + B + 4

M1 A1 M1 B1 B1 B1

Solving the simultaneous equations gives A = −2 and B = 4 y = –2e4x + 4 ex + 2 x + sin 2 x

M1 A1 Total

(b)

⎡1 + 2 y1 = 2 + 0.1 × ⎢ ⎣ 1× 2 = 2 + 0.1 × 2.5 = k1 = 0.1 × 2.5 = 0.25 2

2

⎤ ⎥ ⎦ 2 .25

Comments

Need to attempt both y′ and y″

M1 A1

– 4sin 2x – 5(2 + 2cos2x)+4(2x + sin2x) = 8x – 10 – 10cos2x

2(a)

Total

3

CSO AG Substitute. and confirm correct

4

Their CF + 2x + sin2x Only ft if exponentials in GS Only ft if exponentials in GS and differentiated four terms at least

4 11

M1 A1

k2 = 0.1 × f (1.1, 2.25)

A1 M1 A1 M1

… = 0.1 × 2.53434… = 0.2534(34...)

A1

1 m1 [0.25 + 0.253434...] 2 = 2.2517 to 4dp A1 Total cot xdx 3(a) M1 IF is e ∫ ln sinx = e A1 = sin x A1 (b) d ( y sin x ) = 2sin x cos x M1 A1 dx

3 PI ft from (a) PI

y(1.1) = y(1) +

6 9

If answer not to 4dp withhold this mark

3

AG

y sin x = ∫ sin 2 x dx

M1

Method to integrate 2sinxcosx

1 y sin x = − cos 2 x + c 2

A1

OE

m1

Depending on at least one M

y = 2 when x =

π

π

2

⇒

1 = − cos π + c 2 2 3 1 c = ⇒ y sin x = ( 3 − cos 2 x ) 2 2 2sin

A1 Total

6 9

3

OE eg y sin x = sin2x + 1

MFP3 – AQA GCE Mark Scheme, 2006 June series

MFP3 (cont) Q 4(a)

Solution

Area = …=

1 2

Marks

1 36(1 − cos θ ) 2 dθ ∫ 2 2π

∫ 36(1 − 2cosθ + cos

θ ) dθ

2

0

Total

Comments

1 2 r dθ 2∫

M1

use of

B1 B1

for correct explanation of [6(1–cosθ )]2 for correct limits

M1

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

A1

Correct integration; only ft if integrating a + bcosθ + ccos2θ with non-zero a, b, c. CSO

2π

= 9 ∫ 2 − 4cos θ + (cos 2θ + 1) dθ 0

2π

9 ⎡ ⎤ = ⎢ 27θ − 36sin θ + sin 2θ ⎥ 2 ⎣ ⎦0

(b)(i)

= 54 π

A1

x2 + y 2 = 9 ⇒ r 2 = 9

B1

1 2 ⎛ π ⎞ ⎛ 5π ⎞ Pts of intersection ⎜ 3, ⎟ ; ⎜ 3, ⎟ ⎝ 3⎠ ⎝ 3 ⎠

A & B: 3 = 6 − 6cos θ ⇒ cos θ =

3 =3 3 2

A1 A1 M1

4

OE (accept ‘different’ values of θ not in the given interval)

A1

2

OE exact surd form

12

Total 5(a)

(b)

2⎞ ⎛ ⎜ 3+ a ⎟ 3+0 3 = = ⇒ lim ⎜ ⎟ a →∞ 2+0 2 ⎜2+ 3 ⎟ a⎠ ⎝ ∞

3

2

∫ (3x + 2) − 2 x + 3

PI

M1

(ii) Length AB = 2 × r sin θ

………… = 2 × 3 ×

6

M1 A1

2

dx

1

= [ ln(3x + 2) − ln(2 x + 3)]1

aln(3x + 2) + bln(2x + 3)

M1 A1

∞

∞

⎡ ⎛ 3x + 2 ⎞ ⎤ = ⎢ ln ⎜ ⎟⎥ ⎣ ⎝ 2 x + 3 ⎠⎦ 1

m1

⎧ ⎛ 3a + 2 ⎞ ⎫ = ln ⎨ lim ⎜ ⎟ ⎬ − ln1 a →∞ ⎝ 2a + 3 ⎠ ⎩ ⎭ 3 3 = ln − ln1 = ln 2 2

M1 A1 Total

4

5 7

CSO

AQA GCE Mark Scheme, 2006 June series – MFP3

MFP3 (cont) Q 6(a)

Solution dy du d 2 y dy + 2y ⇒ u= = 2 +2 dx d x dx dx du dy dy LHS of DE⇒ − 2 + 4 + 4y dx dx dx du LHS: + 2(u − 2y) + 4y dx

⇒ (b)

(c)

du + 2u = e −2 x dx

Marks M1 A1

Comments

2 terms correct

M1 A1

2dx IF is e ∫ = e2x

Total

Substitution into LHS of DE as far as no derivatives of y 4

CSO AG

B1

d ⎡ue 2 x ⎤ = 1 ⎦ dx ⎣

M1 A1

⇒ ue2x = x + A

A1

⇒ u = xe– 2x + Ae– 2x Alternative : Those using CF+PI Auxiliary equation, m + 2 = 0 ⇒ uCF = Ae −2 x

A1

For uPI try uPI = kxe −2 x ⇒

M1

ke −2 x − 2kxe −2 x + 2kxe −2 x {= e −2 x }

A1

⇒ k = 1 ⇒ uPI = xe −2 x

A1

⇒ uGS = Ae −2 x + xe −2 x

A1

5

B1

dy + 2 y = xe– 2x + Ae– 2x dx 2dx IF is e ∫ = e2x

⇒

LHS

Use (b) to reach a 1st order DE in y and x

M1 B1

d ⇒ ⎡⎣ ye 2 x ⎤⎦ = x + A dx x2 ⇒ ye2x = + Ax + B 2 ⎛ x2 ⎞ ⇒ y = e– 2x ⎜ + Ax + B ⎟ ⎝ 2 ⎠

A1

A1 A1

Total

5 14

5

MFP3 – AQA GCE Mark Scheme, 2006 June series

MFP3 (cont) Q Solution − 1 7(a)(i) (1 + y ) = 1 − y + y 2 ...... (ii) 1 sec x ≈ 2 x x4 1 − + .... 2 24

Marks B1

Total 1

Comments

B1

−1

⎡ x2 x4 ⎤ = ⎢1 − + ....⎥ = 2 24 ⎦ ⎣ ⎧⎪ ⎛ x 2 x 4 ⎞ ⎛ x 2 x 4 ⎞ 2 ⎫⎪ ⎨1 − ⎜ − + ⎟ + ⎜ − + ⎟ ⎬ ⎪⎩ ⎝ 2 24 ⎠ ⎝ 2 24 ⎠ ⎪⎭ ⎧ x2 x4 x4 ⎫ = ⎨1 + − + + ...⎬ 2 24 4 ⎩ ⎭

M1 M1

x2 5x4 ;+ 24 2 Alternative: Those using Maclaurin f(x) = sec x f(0) = 1; f′(x) = secx tanx; {f′(0) = 0} f′′(x) = secx tan2x + sec3x; f ′′(0) = 1 f′′′(x) = secx tan3x + 5tanx sec3x; f(iv)(x) = secx tan4x +18tan2x sec3x … +5sec5x ⇒ f(iv)(0) = 5 sec x ≈ printed result = 1+

A1;A1

(b) f(x) = tan x; f(0) = 0; f′(x) = sec2x; {f′(0) = 1} f′′(x) = 2secx(secx tan x); f ′′(0) = 0 f′′′(x) = 4secx tanx(secx tanx) + 2sec4x f′′′(0) = 2 2 1 tanx = 0 + 1x + 0x2 + x3 …= x + x3 3! 3 Alternative: Those using otherwise sin x ⎛ x 3 ⎞⎛ x2 ⎞ ≈ ⎜ x − ... ⎟⎜ 1 + ... ⎟ .. = cos x ⎝ 6 ⎠⎝ 2 ⎠ x3 x3 1 = x + − ... = x + x3 .... 2 6 3 (c)

(

3 ⎛ x tan 2 x ⎞ x 2 x + o( x ) = ⎜ ⎟ x2 ⎝ sec x − 1 ⎠ + o( x 4 ) 2 2 + o( x 2 ) = 1 + o( x 2 ) 2 ⎛ x tan 2 x ⎞ lim ⎜ ⎟= 4 x →0 ⎝ sec x − 1 ⎠

5

AG be convinced

(B1) (M1) (m1)

Product rule oe Chain rule with product rule OE

(A2)

CSO AG

B1 M1 A1

Chain rule with product rule oe 3

CSO AG

(M1) (A1) (A1)

)

1 tan2x = 2 x + (2 x)3 3 Condone o(xk) missing

B1 M1 M1

A1

Total TOTAL

6

4 13 75

ft on 2k after B0 for tan2x = kx+…

AQA June Examinations 2006 Scaled Mark Unit Grade Boundaries (GCE Specifications) Unit Code

Unit Title

MD01 MD02 MFP1 MFP2 MFP3 MFP4 MM03 MM04 MM05 MM1A MM1B MM2A MM2B MPC1 MPC2 MPC3 MPC4 MS03 MS04 MS1A MS1B MS2A MS2B

Scaled Mark Grade Boundaries B C D

Maximum Scaled Mark

A

GCE MATHEMATICS UNIT D01

75

61

54

47

40

33

GCE MATHEMATICS UNIT D02 GCE MATHEMATICS UNIT FP1 GCE MATHEMATICS UNIT FP2 GCE MATHEMATICS UNIT FP3 GCE MATHEMATICS UNIT FP4 GCE MATHEMATICS UNIT M03 GCE MATHEMATICS UNIT M04 GCE MATHEMATICS UNIT M05 GCE MATHEMATICS UNIT M1A GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT M2A GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC1 GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S03 GCE MATHEMATICS UNIT S04 GCE MATHEMATICS UNIT S1A GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT S2A GCE MATHEMATICS UNIT S2B

75 75 75 75 75 75 75 75 100 75 100 75 75 75 75 75 75 75 100 75 100 75

64 60 58 62 59 59 59 57 79 61 81 62 61 60 60 61 61 61 80 60 78 60

56 52 51 54 51 51 51 49 69 53 71 54 53 53 53 54 53 53 70 52 68 52

48 44 44 46 44 44 43 41 59 45 61 46 45 46 46 47 45 45 60 44 58 44

41 36 37 39 37 37 36 33 49 37 51 38 38 39 39 40 38 38 50 37 49 37

34 29 30 32 30 30 29 26 40 30 41 31 31 33 32 33 31 31 41 30 40 30

E