fp2 june 2009 ms (new spec)

Mark Scheme (Final)

Summer 2009

GCE

GCE Further Pure Mathematics FP2 (6668/01)

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH

General Marking Guidance

• All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last.

• Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.

• Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie.

• There is no ceiling on achievement. All marks on the mark scheme should be used appropriately.

• All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

• Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited.

• When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, the team leader must be consulted.

• Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

6668/01 Further Pure Mathematics FP2 27th June 2009 L Cope June 2009 Advanced Level in GCE Mathematics Version 3: FINAL MARK SCHEME

1

June 2009 6668 Further Pure Mathematics 2

Mark Scheme Question Number Scheme Marks

1. (a) 1 1 1

( 2) 2 2( 2)r r r r= −

+ + 1 1

2 2( 2)r r−

+ B1 aef

(1)

(b) 1 1

4 2 2( 2) 2

n n

r rr r r r= =

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠

∑ ∑

2 2 2 2 ......

1 3 2 4

2 2 2 2..............1 1 2n n n n

⎛ ⎞ ⎛ ⎞= − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− + +⎝ ⎠ ⎝ ⎠

List the first two terms and the last two terms M1

Includes the first two underlined terms and includes the final two

underlined terms.M1

2 2 2 2;1 2 1 2n n

= + − −+ +

2 2 2 21 2 1 2n n+ − −

+ +A1

2 23

1 2n n= − −

+ +

3( 1)( 2) 2( 2) 2( 1)

( 1)( 2)n n n n

n n+ + − + − +

=+ +

23 9 6 2 4 2 2

( 1)( 2)n n n n

n n+ + − − − −

=+ +

Attempt to combine to an at least 3 term fraction to a single fraction

and an attempt to take out the brackets from their numerator.

M1

23 5

( 1)( 2)n n

n n+

=+ +

(3 5)

( 1)( 2)n n

n n+

=+ +

Correct Result A1 cso AG

[5] 6 marks


2

Question Number Scheme Marks

2. (a) 3 4 2 4 2 iz = − , π θ π− < „

( ) ( )2 24 2 4 2 32 32 64 8r = + − = + = =

( )4 2144 2

tan πθ −= − = −

A valid attempt to find the modulus and argument of 4 2 4 2 i.− M1

( ) ( )( )3

4 48 cos isinz π π= − + −

( )1

4 43So, 8 cos isin3 3

zπ π⎛ ⎞− −⎛ ⎞ ⎛ ⎞= +⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ Taking the cube root of the modulus

and dividing the argument by 3. M1

( ) ( )( )12 122 cos isinz π π⇒ = − + − ( ) ( )( )12 122 cos isinπ π− + − A1 Also, ( ) ( )( )3 7 7

4 48 cos isinz π π= +

or ( ) ( )( )3 9 94 48 cos isinz π π= − + −

Adding or subtracting 2π to the argument for 3z in order to find

other roots.M1

Any one of the final two roots A1 ( )7 7

12 122 cos isinz π π⇒ = +

( ) ( )( )3 34 4and 2 cos isinz π π− −= + Both of the final two roots. A1

[6] 6 marks

Special Case 1: Award SC: M1M1A1M1A0A0 for ALL three of ( )12 122 cos isin ,π π+ ( )3 3

4 42 cos isinπ π+ and

( ) ( )( )7 712 122 cos isinπ π− −+ .

Special Case 2: If r is incorrect and candidate states the brackets ( ) correctly then give the first accuracy mark ONLY where this is applicable.

O arg z α 4 2

4 2

y

x

( 4 2 , 4 2)−


3


3. dsin cos sin 2 sindyx y x x xx− =

d cos sin 2 sind sin siny y x x xx x x− =

An attempt to divide every term in the differential equation by sin .x

Can be implied. See appendix.M1

d cos sin 2

d siny y x xx x− =

( )cos d

sinex xx

±∫ or ( )their P( ) de

x x∫ dM1 Integrating factor

cos d ln sinsine ex x xx

− −∫= = ln sine x− or ln cosece x A1 aef

1sin x

= 1sin x

or 1(sin )x − or cosec x A1 aef

2

1 d cos sin 2sin d sin sin

y y x xx x x x

⎛ ⎞ − =⎜ ⎟⎝ ⎠

d 1sin 2d sin sin

y xx x x⎛ ⎞

= ×⎜ ⎟⎝ ⎠

( )d their I.F. sin 2 their I.F.d

y xx

× = × M1

d 2cosd sin

y xx x⎛ ⎞

=⎜ ⎟⎝ ⎠

d 2cosd sin

y xx x⎛ ⎞

=⎜ ⎟⎝ ⎠

or ( )2cos dsin

y x xx= ∫ A1

2cos dsin

y x xx= ∫

2sinsin

y x Kx= + A credible attempt to integrate the RHS

with/without K+ dddM1

22sin siny x K x= + 22sin siny x K x= + A1 cao [8] 8 marks


4


4. ( )2

2

0

1 3cos d2

A aπ

θ θ= +∫ Applies ( )2

2

0

1 dθ2

rπ

∫ with correct limits.

Ignore dθ .

B1

2 2 2( 3cos ) 6 cos 9cosa a aθ θ θ+ = + +

2 1 cos2cos2

θθ ± ±= M1

2 1 cos26 cos 92

a a θθ +⎛ ⎞= + + ⎜ ⎟⎝ ⎠

Correct underlined expression. A1

2

2

0

1 9 96 cos cos2 d2 2 2

A a aπ

θ θ θ⎛ ⎞= + + +⎜ ⎟⎝ ⎠∫

Integrated expression with at least 3 out of

4 terms of the form sin sin 2A B C Dθ θ θ θ± ± ± ± .

Ignore the 12 . Ignore limits.

M1*

22

0

1 9 96 sin sin 22 2 4

a aπ

θ θ θ θ⎛ ⎞ ⎡ ⎤= + + +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

2 6 sina aθ θ+ + correct ft integration. Ignore the 1

2 . Ignore limits. A1 ft

( ) ( )21 2 0 9 0 0

2aπ π⎡ ⎤= + + + −⎣ ⎦

2 9

2a ππ= + 2 9

2a ππ + A1

Hence, 2 9 1072 2

a ππ π+ = Integrated expression equal to 1072 .π dM1*

2 9 1072 2

a + =

2 49a = As 0,a > 7a = 7a = A1 [8] 8 marks


5


5. 2 2sec (sec )y x x= =

(a) 1 2d 2(sec ) (sec tan ) 2sec tandy x x x x xx= =

Either 12(sec ) (sec tan )x x x or 22sec tanx x

B1 aef

Apply product rule: 2

2 2

2sec tand d4sec tan secd d

u x v xu vx x xx x

⎧ ⎫= =⎪ ⎪⎨ ⎬

= =⎪ ⎪⎩ ⎭

Two terms added with one of either 2 2sec tanA x x or 4secB x

in the correct form.M1

2

2 2 42

d 4sec tan 2secd

y x x xx

= +

Correct differentiation A1 2 2 44sec (sec 1) 2secx x x= − +

Hence, 2

4 22

d 6sec 4secd

y x xx

= − Applies 2 2tan sec 1x x= − leading to the correct result. A1 AG

[4]

(b) ( )4

22 2,yπ = = ( )

4

2d 2 2 (1) 4dyx π

⎛ ⎞ = =⎜ ⎟⎝ ⎠

Both 4

2yπ = and 4

d 4dyx π

⎛ ⎞ =⎜ ⎟⎝ ⎠

B1

( ) ( )4

2 4 2

2

d 6 2 4 2 24 8 16d

yx π

⎛ ⎞= − = − =⎜ ⎟

⎝ ⎠

Attempts to substitute 4x π= into both

terms in the expression for 2

2

d .d

yx

M1

3

33

d 24sec (sec tan ) 8sec (sec tan )d

y x x x x x xx

= − Two terms differentiated with either

424sec tanx x or 28sec tanx x− being correct

M1

4 224sec tan 8sec tanx x x x= −

( ) ( )4

2 4 2

2d 24 2 (1) 8 2 (1) 96 16 80d

yx π

⎛ ⎞= − = − =⎜ ⎟

⎝ ⎠

4

3

3

d 80d

yx π

⎛ ⎞=⎜ ⎟

⎝ ⎠ B1

Applies a Taylor expansion with at

least 3 out of 4 terms ft correctly. M1 ( ) ( ) ( )2 316 80

4 2 4 6 4sec 2 4 ...x x x xπ π π≈ + − + − + − + Correct Taylor series expansion. A1

[6] ( ) ( ) ( ){ }2 340

4 4 3 4sec 2 4 8 ...x x x xπ π π≈ + − + − + − +

10 marks


6


6. , ii

zw zz

= = −+

Complete method of rearranging to

make z the subject. M1 (a) ( i) i iw z z wz w z w z wz+ = ⇒ + = ⇒ = − ii (1 )

(1 )ww z w z

w⇒ = − ⇒ =

− i

(1 )wz

w=

− A1 aef

i3 3

1wz

w= ⇒ =

− Putting in terms of their 3z w = dM1

2 2

2 2

i 3 1 3 1 9 1

i 9 i 1

w w w w w w

u v u v

⎧ ⎫= − ⇒ = − ⇒ = −⎪ ⎪⎨ ⎬⇒ + = + −⎪ ⎪⎩ ⎭

Applies iw u v= + , and uses Pythagoras correctly to get an

equation in terms of u and v without any i 's.

ddM1

2 2 2 29 ( 1)u v u v⎡ ⎤⇒ + = − +⎣ ⎦

Correct equation. A1 2 2 2 2

2 2

9 18 9 90 8 18 8 9u v u u v

u u v

⎧ ⎫⇒ + = − + +⎪ ⎪⎨ ⎬⇒ = − + +⎪ ⎪⎩ ⎭

2 29 9

4 80 u u v⇒ = − + + Simplifies down to 2 2 0.u v u vα β δ+ ± ± ± = dddM1

( )2 29 81 9

8 64 8 0u v⇒ − − + + =

( )2 29 9

8 64u v⇒ − + =

One of centre or radius correct. A1 {Circle} centre ( )9

8 , 0 , radius 38

Both centre and radius correct. A1 [8]

Circle indicated on the Argand diagram in the correct position in

follow through quadrants. Ignore plotted coordinates.

B1ft

(b)

Region outside a circle indicated only.

B1

[2] 10 marks

v

u O


7


7. 2 2 , 1y x a a= − >

Correct Shape. Ignore cusps. B1

(a)

Correct coordinates. B1

[2](b) 2 2 2 , 1x a a x a− = − >

{ },x a> 2 2 2x a a x− = − 2 2 2x a a x− = − M1 aef

2 22 0x x a⇒ + − =

21 1 4(1)( 2 )

2a

x− ± − −

⇒ = Applies the quadratic formula or

completes the square in order to find the roots.

M1

21 1 8

2a

x− ± +

⇒ = Both correct “simplified down” solutions. A1

{ },x a< 2 2 2x a a x− + = − 2 2 2x a a x− + = − or 2 2 2x a x a− = − M1 aef { }2 0 ( 1) 0x x x x⇒ − = ⇒ − =

0x = B1 0 , 1x⇒ = 1x = A1

[6]

(c) 2 2 2 , 1x a a x a− > − >

x is less than their least value B1 ft 21 1 82

ax

− − +< {or}

21 1 82

ax

− + +> x is greater than their maximum value B1 ft

For{ },x a< Lowest Highestx< < M1

{or} 0 1x< < 0 1x< < A1

[4] 12 marks

y

x a− a

2a

O


8


8. 2

2

d d d5 6 2e , 0, 2 at 0.d d d

tx x xx x tt t t

−+ + = = = =

(a) 2AE, 5 6 0 ( 3)( 2) 0

3, 2.m m m m

m+ + = ⇒ + + =

⇒ = − −

1 2e em t m tA B+ , where 1 2 .m m≠ M1

So, 3 2CF e et tx A B− −= + 3 2e et tA B− −+ A1

2

2

d de e ed d

t t tx xx k k kt t

− − −⎧ ⎫= ⇒ = − ⇒ =⎨ ⎬

⎩ ⎭

Substitutes e tk − into the differential

equation given in the question.M1

e 5( e ) 6 e 2e 2 e 2e

1

t t t t t tk k k kk

− − − − − −⇒ + − + = ⇒ =⇒ =

Finds 1.k = A1

{ }PISo, e tx −=

So, 3 2e e et t tx A B− − −= + + CF PItheir theirx x+ M1*

3 2d 3 e 2 e ed

t t tx A Bt

− − −= = − − − Finds ddxt

by differentiating

CF PItheir and theirx xdM1*

0, 0 0 1

d0, 2 2 3 2 1d

t x A Bxt A Bt

= = ⇒ = + +

= = ⇒ = − − −

Applies 0, 0t x= = to x

and d0, 2dxtt

= = to ddxt

to form

simultaneous equations.

ddM1*

2 2 2

3 2 3A BA B+ = −⎧ ⎫

⎨ ⎬− − =⎩ ⎭

1, 0A B⇒ = − = So, 3e et tx − −= − + 3e et tx − −= − + A1 cao [8]


9


8. 3e et tx − −= − +

(b) 3d 3e e 0d

t txt

− −= − = Differentiates their x to give d

dxt

and puts ddxt

equal to 0. M1

A credible attempt to solve. dM1

23 e 0t− = 1

2 ln 3t⇒ = 12 ln 3t = or ln 3t = or awrt 0.55 A1

So, 3 1

3 1 2 22 2ln 3 ln 3 ln 3 ln 3e e e ex

− −− −= − + = − + 3 1

2 23 3x − −= − + Substitutes their t back into x and an

attempt to eliminate out the ln’s. ddM1 1 1 2 2 3

93 3 3 3 3= − + = = uses exact values to give 2 3

9A1 AG

2

32

d 9e ed

t txt

− −= − +

At 12 ln 3,t =

3 12 2

2ln 3 ln 3

2

d 9e ed

xt

− −= − +

Finds2

2

dd

xt

and substitutes their t into2

2

dd

xt

dddM1

3 1

2 29 1 3 19(3) 3

3 3 3 3 3− −= − + = − + = − +

As 2

2

d 9 1 2 0d 3 3 3 3

xt

⎧ ⎫= − + = − <⎨ ⎬

⎩ ⎭

then x is maximum.

9 1 03 3 3

− + < and maximum

conclusion.A1

[7] 15 marks


10

June 2009 6668 Further Pure Mathematics FP2

Appendix List of Abbreviations • dM1 denotes a method mark which is dependent upon the award of the previous method mark. • ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. • dM1∗ denotes a method mark which is dependent upon the award of M1∗ . • ft or denotes “follow through” • cao denotes “correct answer only” • aef denotes “any equivalent form” • cso denotes “correct solution only” • AG denotes “answer given” (in the question paper.) • awrt denotes “anything that rounds to” • aliter denotes “alternative methods” • SC denotes “special case” Extra Solutions If a candidate makes more than one attempt at any question:

• If all but one attempt is crossed out, mark the attempt which is NOT crossed out. • If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single

attempt.


11

Question 1 Question Number Scheme

1. (a)

B1 1 12 2( 2)r r

−+

or 1 12 2

( 2)r r−

+ aef

(b) M1 List the first two terms and the last two ( ) terms. Note that A and B must be found. Note that you need

to be convinced that they are attempting to substitute 1r = and 2r = into their series expansion. M1 Includes (usually adds) the first two underlined terms and includes (usually subtracts) the final two

underlined terms from 2 2 2 2 2 2 2 2......1 3 2 4 1 1 2n n n n

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − + − + + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

to give for

example 2 2 2 2;1 2 1 2n n+ − −

+ +. Do not award this mark if there are too few terms or any extra non-

allowable terms. A1

For 2 2 2 21 2 1 2n n+ − −

+ + or 2 23

1 2n n− −

+ +.

M1 Attempt to combine to an at least 3 term fraction to a single fraction and an attempt to take out the brackets from their numerator.

A1 Correct result of (3 5) .

( 1)( 2)n n

n n+

+ + Note that the answer is given in the question.

Note that an alternative mark scheme has been produced for those candidates who get their partial

fractions the wrong way round in part (a) and then apply them to part (b).


12


Aliter

(b) Way 2 1 1

1 1 1( 2) 2 2( 2)

n n

r rr r r r

= =

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠

∑ ∑

1 1 1 1 ......

2 6 4 8

1 1 1 1..............2( 1) 2( 1) 2 2( 2)n n n n

⎛ ⎞ ⎛ ⎞= − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− + +⎝ ⎠ ⎝ ⎠



underlined terms.M1

1 1 1 12 4 2( 1) 2( 2)n n

= + − −+ +

1 1 1 12 4 2( 1) 2( 2)n n+ − −

+ + A1

3 1 1

4 2( 1) 2( 2)n n= − −

+ +

3( 1)( 2) 2( 2) 2( 1)

4( 1)( 2)n n n n

n n+ + − + − +

=+ +

23 9 6 2 4 2 2

4( 1)( 2)n n n n

n n+ + − − − −

=+ +

Attempt to combine to an at least 3 term fraction to a single fraction and an attempt to take

out the brackets from their numerator.

M1

23 5

4( 1)( 2)n n

n n+

=+ +

2

1

1 3 54 4( 2) 4( 1)( 2)

n

r

n nr r n n

=

⎛ ⎞+= ⎜ ⎟+ + +⎝ ⎠

∑

(3 5)

( 1)( 2)n n

n n+

=+ +

Correct Result A1 cso AG

[5]


13

Mark Scheme for Alternative “Incorrect” response. Question Number Scheme Marks

1. (a) 1 1 1

( 2) 2( 2) 2r r r r= −

+ + Incorrect answer B0

(1)

(b) 1 1

4 2 2( 2) ( 2)

n n

r rr r r r= =

⎛ ⎞= −⎜ ⎟+ +⎝ ⎠

∑ ∑

2 2 2 2 ......

3 1 4 2

2 2 2 2..............1 1 2n n n n

⎛ ⎞ ⎛ ⎞= − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ − +⎝ ⎠ ⎝ ⎠



underlined terms.M1

2 2 2 21 2 1 2n n

= + − −+ +

Cannot award this accuracy mark. A0

2 2 3

1 2n n= + −

+ +

2( 2) 2( 1) 3( 1)( 2)

( 1)( 2)n n n n

n n+ + + − + +

=+ +

22 4 2 2 3 9 6

( 1)( 2)n n n n

n n+ + + − − −

=+ +

Attempt to combine to an at least 3 term fraction to a single fraction

and an attempt to take out the brackets from their numerator.

M1

23 5

( 1)( 2)n n

n n− −

=+ +

(3 5)

( 1)( 2)n n

n n− +

=+ +

Cannot award this accuracy mark. A0


14


2. (a) M1 A valid attempt to find the modulus and argument of 4 2 4 2 i.−

Note there must be an attempt to find both the modulus and argument. M1 Taking the cube root of the modulus and dividing the argument by 3. A1 For ( ) ( )( )12 122 cos isin .π π− + − Allow

i122e .π−

M1 Way 1: Adding or subtracting 2π to the argument for 3z in order to find other roots or Way 2: Adding or subtracting 2

3π to the argument for z in order to find the other roots.

Note that for Way 2 the candidate needs to divide their argument for 3z by 3 and then add and subtract multiples of 2

3π from this new argument.

A1 Any one of ( )7 712 122 cos isinπ π+ or ( ) ( )( ) 7 i 3 i

12 43 34 42 cos isin or 2e or 2e .

π ππ π −− −+

A1 Both ( )7 712 122 cos isinπ π+ and ( ) ( )( )3 3

4 42 cos isin .π π− −+

Do not allow any/both of the roots for z expressed in an exponential form for the final mark. Do not award the final A1 mark if there are any extra root(s) within the range .π θ π− < „

Note that the first accuracy is dependent upon the first two method marks.

Note that the final two accuracy marks are dependent on the third method mark. Note that any/both of the first two method marks can be implied. There is an alternative method that appears in the appendix. In Way 2, candidates pick up some of the

earlier marks later on in their solution. You need to award the marks on ePEN, however, in the same order as they appear for Way 1 on the mark scheme.

Special Case 1: For a candidate who finds ALL three of ( )12 122 cos isin ,π π+ ( )3 3

4 42 cos isinπ π+ and

( ) ( )( )7 712 122 cos isinπ π− −+ then award SC: M1M1A1M1A0A0.

Special Case 2: If r is incorrect and candidate states the brackets ( ) correctly then give the first

accuracy mark ONLY where this is applicable. Note that ( )12 122 cos isinπ π− or ( )( )12 122 cos isinπ π− − are also fine for the first accuracy mark.

Note also that ( )334 42 cos isin ππ − or ( )( )3 3

4 42 cos isinπ π− − are also fine for the second accuracy mark.

Withhold the final accuracy mark, however, if all three roots are not in the form ( )cos isinr θ θ+ . Note that there needs to be a “+” between cosθ and isinθ and the angle θ must be consistent (ie. the same) for cos and sin inside the brackets for the award of the final accuracy mark.


15


2. (a) 3 4 2 4 2 iz = − , π θ π− < „

( ) ( )2 24 2 4 2 32 32 64 8r = + − = + = =

( )4 2144 2

tan πθ −= − = −

A valid attempt to find the modulus and argument of 4 2 4 2 i.− M1

( ) ( )( )3

4 48 cos isinz π π= − + −

Decide to award M1 here!! M1 Decide to award A1 here!! A1

( ) ( )( )34 48 cos 2 isin 2z k kπ ππ π= − + + − +

Adding or subtracting 2( )k π to the argument for 3z in order to find

other roots.M1

( )1

4 432 2

So, 8 cos isin3 3

k kz

π ππ π⎛ ⎞− + − +⎛ ⎞ ⎛ ⎞= +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Taking the cube root of the modulus and dividing the argument by 3.

Award above

( ) ( )( )2 2

12 3 12 32 cos isink kz π π π π⇒ = − + + − + ( ) ( )( )12 120, 2 cos isink z π π= = − + − ( ) ( )( )12 122 cos isinπ π− + − Award

Above

Any one of the final two roots A1 ( )7 712 121, 2 cos isink z π π= = +

( ) ( )( )3 34 41, 2 cos isink z π π− −= − = + Both of the final two roots. A1

[6] 6 marks

Special Case 1: Award SC: M1M1A1M1A0A0 for ALL three of ( )12 122 cos isin ,π π+ ( )3 34 42 cos isinπ π+ and

( ) ( )( )7 712 122 cos isinπ π− −+ .

Special Case 2: If r is incorrect and candidate states the brackets ( ) correctly then give the first accuracy mark ONLY where this is applicable.

O arg z α

4 2

y

x

( 4 2 , 4 2)−


16


3. (a) M1 An attempt to divide every term in the differential equation by sin .x Can be implied. You can be

generous with the result when the candidate divides cosy x− by sin .x

Hence, the resulting DE must be in the form d sin 2 sinP( )d siny x xx yx x± = , where P( )x is a function of x.

For example, P( )x can be tan .x

You can imply the first M1 mark if the candidate writes down an integrating factor of the form ( )cos d

sinex xx

±∫ .

M1 For applying an integrating factor ( )cos d

sinex xx

±∫ or ( )their P( ) de

x x∫ . Ignore d .x A1 ln sine x− or ln cosece x or any equivalent.

A1 1

sin x or 1(sin )x − or cosec x

M1 ( )d their I.F. sin 2 their I.Fd

y xx

× = × or ( ) ( )their I.F. sin 2 their I.F .y x dx× = ×∫

A1 d 2cosd sin

y xx x⎛ ⎞

=⎜ ⎟⎝ ⎠

or ( )2cos dsin

y x xx= ∫ , ignoring the omission of d .x

dddM1 A credible attempt to integrate the RHS with/without .K+ Note that this mark is dependent upon the candidate receiving the first three method marks.

A1 22sin siny x K x= + , cao. If the candidate writes down the integrating factor of either ln sine x− or ln cosece x (with no working), then you

can imply the first two method marks.


17


4. (a)

B1 For ( )

22

0

1 3cos d2

aπ

θ θ+∫ , with correct limits and ignoring dθ .

Note that 2 29cosa θ+ is acceptable for 2.r

M1 Replacing 2cos θ with an expression of the form 2 1 cos2cos .2

θθ ± ±=

A1 Achieving 2 1 cos26 cos 9 .

2a a θθ +⎛ ⎞+ + ⎜ ⎟

⎝ ⎠

The underlined expression can appear over a few lines of a candidate’s working. M1* Integrate to get an expression with at least 3 out of 4 terms in the form sin sin 2 .A B C Dθ θ θ θ± ± ± ±

Ignore the 12 and ignore the limits.

A1ft 2 6 sina aθ θ+ + correct ft integration. Ignore the 12 and ignore the limits.

A1 For 2 9 .2

a ππ +

dM1* Candidates need to put their integrated expression equal to 1072 .π

Note that is method mark is dependent upon the M1* mark being awarded. A1 7.a = Do not allow 7a = ± without reference to a being equal to 7.


18

Question 5 Note that the marks in question 5 are now: (a) B1 M1 A1 A1 (b) B1 M1 M1 B1 M1 A1 Question Number Scheme

5. (a) B1 Correct differentiation of 2sec .y x=

Examples that are acceptable are 12(sec ) (sec tan )x x x or 22sec tanx x or 22(cos ) ( sin )x x−− − or

22sin (cos )x x − or 2

2sincos

xx

, etc.

M1 Two terms added with one of either 2 2sec tanA x x or 4secB x in the correct form. Special Case: M1 can also be awarded in this part if candidate differentiates 2sec tany x x′ = to give the “correct” answer of 3 22sec 2sec tan .y x x x′′ = +

A1 Correct differentiation of give 2 2 44sec tan 2sec .x x x+ A1 Applies 2 2tan sec 1x x= − leading to the correct result. Note that the answer is given in the question.

(b)

B1 Both 4

2yπ = and 4

d 4dyx π

⎛ ⎞ =⎜ ⎟⎝ ⎠

.

M1 Attempts to substitute 4x π= (or even 4x π= − ) into both terms in the expression for

2

2

dd

yx

given in the

question or even their expression for 2

2

d .d

yx

( ) ( )4 26 2 4 2− or ( ) ( )4 2

4 46sec 4secπ π− are sufficient for the award of M1 here.

Note that 4f ( )π′′ by itself is not sufficient. M1 Two terms with either 424sec tanx x or 28sec tanx x− being correct.

B1 4

3

3

d 80d

yx π

⎛ ⎞=⎜ ⎟

⎝ ⎠

M1 Applies a Taylor expansion ( ) ( ) ( ) ( ) ( ) ( ) ( )2 34 4

4 4 4 4 4

f ff f ...

2! 3!x x x

π ππ π π π π

′′ ′′′′+ − + − + − + with at least 3

out of 4 terms followed through correctly with their ( ) ( ) ( ) ( )4 4 4 4f , f , f , f .π π π π′ ′′ ′′′

A1 Correct Taylor series expansion of ( ) ( ) ( )2 316 804 2 4 6 42 4 ...x x xπ π π+ − + − + − +


19


6. (a) M1 Complete method of rearranging to make z the subject.

A1 i

(1 )wz

w=

− or -i

( 1)wz

w=

−

dM1 Putting in terms of their 3.z w = Note that this mark is dependent upon the first M1 mark being awarded.

ddM1 Applies iw u v= + , and uses Pythagoras correctly (on grouped real and imaginary parts) to get an equation in terms of u and v without any i 's. Effectively they need “to square and add”. Note that 2 2 2 23 ( 1)u v u v⎡ ⎤+ = − +⎣ ⎦ would be fine for M1. (Allow working in x and y where candidate

applies i .w x y= + ) Note that this mark is dependent upon the first two M1 marks being awarded.

A1 Correct equation. Aef, for example 2 2 2 29 (1 ) .u v u v⎡ ⎤+ = − +⎣ ⎦

dddM1 Simplifies down to 2 2 0.u v u vα β δ+ ± ± ± = Note this needs to be an EQUATION, with the coefficients of 2u and 2v either both 1 or both 1 ,− but 2u and 2v need to be on the same side. Also note that equations of the form 2 2 ,u v u vα β δ+ = ± ± ± 2 2u v u vα β δ+ ± = ± ± are acceptable for M1. Note that this mark is dependent upon the first three M1 marks being awarded.

A1 One of centre ( )98 , 0 or radius 3

8 correct.

A1 Both the centre ( )98 , 0 and radius 3

8 correct.

(b) B1ft Circle indicated on the Argand diagram in the correct position in follow through quadrants.

This means that candidates must have stated either (however, incorrect) the equation of the circle or the centre and radius of the circle in either parts (a) or (b). Also DO NOT allow the B1 mark if a candidate draws a circle centre (0, 0). Ignore plotted coordinates.

B1 Region outside a circle indicated only. If a candidate draws a circle and shades in the region outside of it then this is OK for B1. The candidate for this mark DOES NOT have to state their equation of the circle or their centre and radius of the circle in either parts (a) or (b).


20

Question Number Example

6. (b) EG 1

Comment: In (b), this response would score B0 B1.

EG 2

Comment: In (a), this response would score Way 2: B0 B1, as the circle should not be touching the v-axis.


21

Question 7 Note that the marks in question 7 are now: (a) B1 B1 (b) M1 M1 A1 M1 B1 A1 (c) B1 B1 M1 A1 Question Number Scheme

7. (a) B1 Correct Shape. Ignore cusps. B1 Note you can award this mark if they state all three of 2, anda a a− on the curve in the correct place.

Note that the coordinates must be stated as ( ) ( ) 2, 0 , , 0 and (0, )a a a− if they are not referred (marked on) the graph. Accept, however, the coordinates interchanged for x and y if they are marked on the curve in the correct place.

(b) M1 Taking out the modulus and writing down 2 2 2x a a x− = − is sufficient for this mark. M1 Applies the quadratic formula with correct “a, b and c” for their equation.

Or completes the square in order to find the roots. If they decide to complete the square it must be done on 2x x+ . So ( )2 21 1

2 4 2 0x a+ − + = and an attempt to solve for x is sufficient for M1.

A1 Both correct “simplified down” solutions of 21 1 8

2a− ± + or

21 812 2

a+− ± or 21

41 22

a− ± + , etc.

M1 Either 2 2 2x a a x− + = − or 2 2 2 .x a x a− = − B1 Answer of 0x = seen anywhere in part (b). Note that this mark is independent of method. A1 1x = . Note that this mark is only dependent on the third method mark being awarded.

If the candidate gives all four solutions of 21 1 8

, 0 , 12

ax

− ± += and includes any extra additional

solutions then deduct the final accuracy mark. Note that the candidate cannot get full marks in this part for additional extra solutions.

If a candidate rejects a solution then do not mark the rejected solution. Note that an alternative method of squaring both sides is detailed in this appendix.

(c) B1ft x is less than their least value. Note that this is a follow through mark. B1ft x is greater than their maximum value. Note that this is a follow through mark. M1 When the “negative” modulus is taken { },x a< Lowest Solution Highest Solutionx< < A1 0 1x< <

Note that the greatest value of x is the largest one of the two solutions found when the positive modulus

(for x a> ) is taken and the least value of x is smallest one of the two solutions found when the positive modulus (for x a> ) is taken.

SC 1: For x is less than or equal their least value and x is greater than or equal their maximum value then award B1B0.

SC 2: For 0 1x≤ ≤ , then award M1 A0.


22


Aliter (b) 2 2 2 , 1x a a x a− = − >

Way 2 2 2 2 2 2( ) ( )x a a x− = − 2 2 2 2 2( ) ( )x a a x− = − M1

4 2 2 4 4 2 22 2x a x a a a x x− + = − + 4 2 2 4 4 2 22 2x a x a a a x x− + = − + 4 2 2 4 4 2 22 2 0x a x a a a x x− + − + − = Rearranges equation onto one side = 0 M1 4 2 2 2 22 2 0x a x x a x− − + = 3 2 2( 2 2 ) 0x x a x x a− − + = Hence, 0x = 0x = appearing anywhere in part (b) B1 So, 3 2 2( 2 2 ) 0x a x x a− − + = 3 2 2( (2 1) 2 ) 0x a x a− + + = 2 2( 1)( 2 ) 0x x x a− + − = Factorises ( 1)x − from their cubic

expression. M1

Hence, 1x = 1x = A1 21 1 4(1)( 2 )

2a

x− ± − −

=

21 1 8

2a

x− ± +

= Both correct

“simplified down” solutions. A1

[6]

For Way 2, the marks are M1 M1 B1 M1 A1 A1. They should be awarded in this order on ePEN.


23


8. (a)

M1 Writes down a solution in the form, 1 2e em t m tA B+ , where 1 2 .m m≠ Note that they must put a constant in front of each exponential term.

A1 3 2e et tA B− −+ . Aef, which usually means 2 3e e .t tA B− −+ M1 Substitutes e tk − into the differential equation given in the question. A1 Finds 1.k =

M1* CF PItheir theirx x+ . They must add together CFtheir x and PItheir x which they have found. Allow this mark generously they have found 1k = and then write 2 3e e 1.t tA B− −+ +

dM1* Finds d

dxt

by differentiating CF PItheir and theirx x .

Note that this method mark is dependent upon the award of M1*. ddM1*

Applies 0, 0t x= = to x and d0, 2dxtt

= = to ddxt

to form simultaneous equations.

Note that this method mark is dependent upon the award of the previous 2 method marks. A1 For the correct answer of 3e et tx − −= − + .

(b)

M1 Differentiates their x to give ddxt

and puts ddxt

equal to 0.

dM1 A credible attempt to solve. Multiplying through by 3e t to give 23 e 0t− = or trying to form a quadratic in e t− and factorising this quadratic or using the formula on this quadratic is enough for the method mark here.

A1 12 ln 3t = or ln 3t = or awrt 0.55

ddM1 Substitutes their t back into x and an attempt to eliminate out the ln’s.

A1 Uses exact values to give 2 3 .9

dddM1 Finds2

2

dd

xt

and substitutes their t into2

2

dd

xt

A1 9 1

3 3 3− + or 3 1

3 3− + or 2 2 3or

33− − or awrt 1.2− and 0< and maximum conclusion.

In part (b), each method mark is dependent upon all previous method marks being awarded.


24

Alternative Way with constants A and B the other way round. Question Number Scheme Marks

8. 2

2

d d d5 6 2e , 0, 2 at 0.d d d

tx x xx x tt t t

−+ + = = = =

(a) 2AE, 5 6 0 ( 3)( 2) 0

2, 3.m m m m

m+ + = ⇒ + + =

⇒ = − −

1 2e em t m tA B+ , where 1 2 .m m≠ M1

So, 2 3CF e et tx A B− −= + 2 3e et tA B− −+ A1

2

2

d de e ed d

t t tx xx k k kt t

− − −⎧ ⎫= ⇒ = − ⇒ =⎨ ⎬

⎩ ⎭

Substitutes e tk − into the differential

equation given in the question.M1

e 5( e ) 6 e 2e 2 e 2e

1

t t t t t tk k k kk

− − − − − −⇒ + − + = ⇒ =⇒ =

Finds 1.k = A1

{ }PISo, e tx −=

So, 2 3e e et t tx A B− − −= + + CF PItheir theirx x+ M1*

2 3d 2 e 3 e ed

t t tx A Bt

− − −= = − − − Finds ddxt

by differentiating

CF PItheir and theirx xdM1*

0, 0 0 1

d0, 2 2 2 3 1d

t x A Bxt A Bt

= = ⇒ = + +

= = ⇒ = − − −

Applies 0, 0t x= = to x

and d0, 2dxtt

= = to ddxt

to form

simultaneous equations.

ddM1*

2 2 2

2 3 3A BA B+ = −⎧ ⎫

⎨ ⎬− − =⎩ ⎭

0, 1A B⇒ = = − So, 3e et tx − −= − + 3e et tx − −= − + A1 [8]

fp2 june 2009 ms (new spec)

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