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Mark Scheme (Final)

Summer 2009

GCE

GCE Further Pure Mathematics FP2 (6668/01)

Edexcel Limited. Registered in England and Wales No. 449 6750Registered Office: One90 High Holborn, London WC1V 7BH

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6668/01 Further Pure Mathematics FP2 27thJune2009 L CopeJune 2009 Advanced Level in GCE Mathematics Version 3: FINAL MARK SCHEME

1

June 20096668 Further Pure Mathematics 2

Mark Scheme

Question

NumberScheme Marks

1. (a)1 1 1

( 2) 2 2( 2)r r r r =

+ +

1 1

2 2( 2)r r

+B1aef

(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 termsM1

Includes the first two underlined

terms and includes the final two

underlined terms.

M12 2 2 2

;1 2 1 2n n

= + + +

2 2 2 2

1 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

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2

Question

NumberScheme Marks

2. (a) 3 4 2 4 2 iz = , <

( ) ( )2 2

4 2 4 2 32 32 64 8r= + = + = =

( )4 21 44 2tan = = A valid attempt to find the modulus

and argument of 4 2 4 2 i. M1

( ) ( )( )3 4 48 cos isinz = +

( )1

4 43So, 8 cos isin3 3

z

= +

Taking the cube root of the modulus

and dividing the argument by 3.M1

( ) ( )( )12 122 cos isinz = + ( ) ( )( )12 122 cos isin + A1

Also, ( ) ( )( )3 7 74 48 cos isinz = +

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

Adding or subtracting 2 to the

argument for 3z in order to find

other roots.

M1

Any one of the final two roots A1( )7 712 122 cos isinz = +

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

[6]

6 marks

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

+ and

( ) ( )( )7 712 122 cos isin + .

Special Case 2: If ris incorrect and candidate states the brackets ( ) correctly then give the first accuracy mark

ONLY where this is applicable.

O argz 4 2

4 2

y

x

( 4 2 , 4 2)

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3

Question

NumberScheme Marks

3. dsin cos sin 2 sindy y x x x =

d cos sin 2 sin

d sin sin

y y x x x

x x =

An attempt to divide every term in the differential

equation by sin .x

Can be implied. See appendix.

M1

d cossin2

d sin

y y x

x x =

( )cos

dsin

e

xx

x

or

( )their P ( ) d

e

x x

dM1

Integrating factor

cosd

lnsinsine e

xx

xx

= = lnsine x or lncosece x A1 aef

1

sin=

1

sinxor 1(sin )

or cosec A1 aef

2

1 d cos sin 2

sin d sin sin

y y x x

x x x x

=

d 1sin2

d sin sin

yx

x x

=

( )d

their I.F. sin 2 their I.F.

d

y x

x

= M1

d2cos

d sin

y

x x

=

d2cos

d sin

y

x x

=

or ( )2cos d

sin

yx x

x= A1

2cos dsin

yx

x=

2sinsin

yK

x= +

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

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4

Question

NumberScheme Marks

4. ( )

2

2

0

1 3cos d2

A a

= + Applies ( )

2

2

0

1

d2 r

with correct limits.

Ignore d .

B1

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

2 1 cos 2cos2

= M12 1 cos26 cos 9

2a a

+ = + +

Correct underlined expression. A1

2

2

0

1 9 96 cos cos2 d2 2 2

A a a

= + + +

Integrated expression with at least 3 out of

4 terms of the form

sin sin 2A B C D .

Ignore the 12 . Ignore limits.

M1*2

2

0

1 9 96 sin sin 2

2 2 4a a

= + + +

2 6 sina a + + correct ft integration.

Ignore the 12 . Ignore limits.

A1 ft

( ) ( )21 2 0 9 0 02

a = + + +

2 9

2a

= + 2

9

2a

+ A1

Hence, 29 107

2 2a

+ = Integrated expression equal to 1072 . dM1*

2 9 107

2 2a + =

2 49a =

As 0,a> 7a= 7a= A1[8]

8 marks

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5

QuestionNumber

Scheme Marks

5.2 2sec (sec )y x x= =

(a)1 2d 2(sec ) (sec tan ) 2sec tan

d

yx x x x

x= =

Either 12(sec ) (sec tan )x x

or 22sec tanx x B1 aef

Apply product rule:2

2 2

2sec tan

d d4sec tan sec

d d

u x v x

u vx x x

x x

= =

= =

Two terms added with one of

either 2 2sec tanA x x or 4secB x

in the correct form.

M122 2 4

2

d

4sec tan 2secd

y

x xx = +

Correct differentiation A1

2 2 44sec (sec 1) 2secx x x= +

Hence,2

4 2

2

d6sec 4sec

d

yx

x=

Applies 2 2tan sec 1x x=

leading to the correct result. A1 AG

[4]

(b) ( )4

2

2 2,y = = ( )4

2d2 2 (1) 4

d

y

x

= =

Both

4

2y = and4

d4

d

y

x

=

B1

( ) ( )4

2 4 2

2

d6 2 4 2 24 8 16

d

y

x

= = =

Attempts to substitute4

x = into both

terms in the expression for2

2

d.

d

y

x

M1

33

3

d24sec (sec tan ) 8sec (sec tan )

d

yx x x x x

x=

Two terms differentiated with either424sec tanx x or

28sec tanx x being correctM1

4 224sec tan 8sec tanx x x x=

( ) ( )4

2 4 2

2

d24 2 (1) 8 2 (1) 96 16 80

d

y

x

= = =

4

3

3

d80

d

y

x

=

B1

Applies a Taylor expansion with at

least 3 out of 4 terms ft correctly.M1

( ) ( ) ( )2 3

16 804 2 4 6 4

sec 2 4 ...x x x x + + + + Correct Taylor series expansion. A1

[6]

( ) ( ) ( ){ }2 3404 4 3 4sec 2 4 8 ...x x x x + + + +

10 marks

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6

Question

NumberScheme Marks

6. , ii

zw z

z= =

+

Complete method of rearranging to

make zthe 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

1

wz

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 uand vwithout

any i's.

ddM12 2 2 29 ( 1)u v u v + = +

Correct equation. A12 2 2 2

2 2

9 18 9 9

0 8 18 8 9

u v u u v

u u v

+ = + +

= + +

2 29 94 8

0 u u v = + + Simplifies down to

2 20.u v u v + =

dddM1

( )2 29 81 9

8 64 80u v + + =

( )2 29 9

8 64u v + =

One of centre or radius correct. A1{Circle} centre ( )98 , 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

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7

Question

NumberScheme Marks

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 )

2

ax

=

Applies the quadratic formula or

completes the square in order to find

the roots.

M1

2

1 1 82

ax + = Both correctsimplified 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= B10 , 1x =

1x= A1

[6]

(c)2 2 2 , 1x a a x a > >

xis less than their least value B1 ft21 1 8

2

ax

+< {or}

21 1 8

2

ax

+ +>

xis greater than their maximum value B1 ft

For{ },x a< Lowest Highestx< < M1{or} 0 1x< <

0 1x< < A1

[4]

12 marks

y

xa a

2a

O

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9

Question

NumberScheme Marks

8.3e et tx = +

(b)3d

3e e 0d

t tx

t

= =

Differentiates theirxto gived

dt

and putsd

dtequal to 0.

M1

A credible attempt to solve. dM123 e 0t = 12ln3t = 12 ln3t= or ln 3t= or awrt 0.55 A1

So,

3 13 1 2 22 2

ln 3 ln 3 ln 3 ln 3

e e e ex

= + = +

3 12 23 3x

= +

Substitutes their tback intoxand an

attempt to eliminate out the lns. ddM1

1 1 2 2 3

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

2 3

9A1 AG

23

2

d9e e

d

t tx

t

= +

At 12ln3,t=

3 12 2

2ln3 ln 3

2

d9e e

d

x

t

= +

Finds2

2

d

dt

and substitutes their tinto2

2ddt

dddM1

3 12 2

9 1 3 19(3) 3

3 3 3 3 3

= + = + = +

As2

2

d 9 1 20

d 3 3 3 3

x

t

= + =

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10

June 20096668 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.

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12

Question

NumberScheme Marks

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

= + +

+ +

+ +

List the first two termsand the last two terms

M1

Includes the first two underlinedterms and includes the final two

underlined terms.

M11 1 1 1

2 4 2( 1) 2( 2)n n= + + + 1 1 1 1

2 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 atleast 3 term fraction to a singlefraction and an attempt to take

out the brackets from theirnumerator.

M1

23 5

4( 1)( 2)

n n

n n

+=

+ +

2

1

1 3 54 4

( 2) 4( 1)( 2)

n

r

n n

r r n n=

+=

+ + +

(3 5)( 1)( 2)

n nn n+= + +

Correct Result A1 csoAG

[5]

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13

Mark Scheme for Alternative Incorrect response.

QuestionNumber

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

= + +

+ + + +

List the first two termsand the last two terms

M1

Includes the first two underlinedterms and includes the final two

underlined terms.M1

2 2 2 2

1 2 1 2n n= +

+ +

Cannot award this accuracy mark. A0

2 23

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 3term fraction to a single fraction

and an attempt to take out thebrackets 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

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14

Question 2

QuestionNumber

Scheme

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 . + Allowi

122e .

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

Way 2: Adding or subtracting 23 to the argument forzin order to find the other roots.

Note that for Way 2 the candidate needs to divide their argument for 3zby 3 and then add and subtract

multiples of 23 from this new argument.

A1 Any one of ( )7 7

12 122 cos isin

+ or ( ) ( )( )7 i 3 i12 43 3

4 42 cos isin or 2e or 2e .

+ A1 Both ( )7 712 122 cos isin

+ and ( ) ( )( )3 34 42 cos isin . + Do not allow any/both of the roots forzexpressed 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 . <

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 sameorder 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 , + ( )3 34 42 cos isin

+ and

( ) ( )( )7 712 122 cos isin + then award SC: M1M1A1M1A0A0.

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

accuracy mark ONLY where this is applicable.

Note that ( )12 122 cos isin or ( )( )12 122 cos isin are also fine for the first accuracy mark.

Note also that ( )33

4 42 cos isin or ( )( )

3 34 4

2 cos isin 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 + .Note that there needs to be a + between cos and isin and the angle must be consistent (ie. the

same) for cos and sin inside the brackets for the award of the final accuracy mark.

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15

QuestionNumber

Scheme Marks

2. (a) 3 4 2 4 2 iz = , <

( ) ( )2 2

4 2 4 2 32 32 64 8r= + = + = =

( )4 21 44 2tan = = A valid attempt to find the modulus

and argument of 4 2 4 2 i. M1

( ) ( )( )3 4 48 cos isinz = + Decide to award M1 here!! M1Decide to award A1 here!! A1

( ) ( )( )3 4 48 cos 2 isin 2z k k = + + + Adding or subtracting 2 ( )k to the

argument for 3z in order to findother roots.

M1

( )1

4 432 2

So, 8 cos isin3 3

k kz

+ + = +

Taking the cube root of the modulus

and dividing the argument by 3.Award

above

( ) ( )( )2 212 3 12 32 cos isink kz = + + +

( ) ( )( )12 120, 2 cos isink z = = + ( ) ( )( )12 122 cos isin + Award

Above

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

( ) ( )( )3 34 41, 2 cos isink z = = + Both of the final two roots. A1[6]

6 marks

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

+ and

( ) ( )( )7 712 122 cos isin + .

Special Case 2: If ris incorrect and candidate states the brackets ( ) correctly then give the first accuracy mark

ONLY where this is applicable.

O argz

4 2

y

x

(4 2 , 4 2)

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16

Question 3

QuestionNumber

Scheme

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

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

Hence, the resulting DE must be in the formd sin 2 sin

P( )d sin

y x xx y

x = , where P( )x is a function ofx.

For example, P( ) can be tan .x

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

cosd

sinex

xx

.

M1 For applying an integrating factor( )

cosd

sinex

xx

or

( )their P ( ) de

x x . Ignore d .x

A1 lnsine x or lncosece x or any equivalent.

A11

sinor 1(sin )x or cosecx

M1 ( )d

their I.F. sin 2 their I.Fd

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

A1d

2cosd sin

y

x x

=

or ( )2cos d

sin

yx x

x= , ignoring the omission of d .x

dddM1A credible attempt to integrate the RHS with/without .K+

Note that this mark is dependent upon the candidate receiving the first three method marks.

A122sin siny x K x= + , cao.

If the candidate writes down the integrating factor of either lnsine x or lncosece x (with no working), then you

can imply the first two method marks.

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17

Question 4

Question

NumberScheme

4. (a)

B1For ( )

2

2

0

13cos d

2a

+ , with correct limits and ignoring d .

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

M1 Replacing2cos with an expression of the form 2

1 cos2cos .

2

=

A1Achieving 2

1 cos26 cos 9 .

2a a

+ + +

The underlined expression can appear over a few lines of a candidates working.

M1* Integrate to get an expression with at least 3 out of 4 terms in the form sin sin2 .A B C D Ignore the 12 and ignore the limits.

A1ft2 6 sina a + + correct ft integration. Ignore the 12 and ignore the limits.

A1 For2 9 .

2a

+

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

Note that is method mark is dependent upon the M1* mark being awarded.

A1 7.a= Do not allow 7a= without reference to abeing equal to 7.

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18

Question 5

Note that the marks in question 5 are now: (a) B1 M1 A1 A1 (b) B1 M1 M1B1M1 A1

Question

Number Scheme

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

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

22sin (cos )x x or2

2sin

cos

x, etc.

M1 Two terms added with one of either 2 2sec tanx x or 4secB x in the correct form.

Special Case: M1 can also be awarded in this part if candidate differentiates 2 sec tany x x= to give the

correct answer of 3 22sec 2sec tan .y x x x= +

A1 Correct differentiation of give2 2 44sec tan 2sec .x x x+

A1 Applies2 2tan sec 1x x= leading to the correct result. Note that the answer is given in the question.

(b)

B1 Both4

2y = and4

d4

d

y

x

=

.

M1Attempts to substitute 4x

= (or even4

x = ) into both terms in the expression for2

2

d

d

y

xgiven in the

question or even their expression for2

2

d.

d

y

( ) ( )4 2

6 2 4 2 or ( ) ( )4 24 46sec 4sec are sufficient for the award of M1 here.

Note that4

f ( ) by itself is not sufficient.

M1 Two terms with either 424sec tanx x or 28sec tanx being correct.

B14

3

3

d80

d

y

x

=

M1Applies a Taylor expansion ( ) ( ) ( ) ( )

( )( )

( )2 34 44 4 4 4 4

f ff f ...

2! 3!x x x

+ + + + with at least 3

out of 4 termsfollowed through correctlywith their ( ) ( ) ( ) ( )4 4 4 4f , f , f , f .

A1 Correct Taylor series expansion of ( ) ( ) ( )2 3

16 804 2 4 6 42 4 ...x x x + + + +

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19

Question 6

QuestionNumber

Scheme

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

A1i

(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 uand vwithout 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 inxandywhere 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 example2 2 2 29 (1 ) .u v u v + = +

dddM1 Simplifies down to 2 2 0.u v u v + = 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 v + = 2 2u v u v + = 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 38 correct.

A1 Both the centre ( )98 , 0 and radius 38 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 thecentre 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 circleor their centre and radius of the circle in either parts (a) or (b).

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20

QuestionNumber

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 betouching the v-axis.

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21

Question 7

Note that the marks in question 7 are now: (a) B1 B1 (b) M1 M1 A1 M1B1A1 (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 (markedon) the graph.Accept, however, the coordinates interchanged forxandyif they are marked on the curve in the correctplace.

(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,band 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 2 x+ . So ( )2 21 1

2 42 0x a+ + = and an

attempt to solve forxis sufficient for M1.

A1 Both correct simplified down solutions of21 1 8

2

a + or

21 81

2 2

a+ or 21

4

12

2a + , 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 of21 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 partfor 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 xis less than their least value. Note that this is a follow through mark.B1ft xis 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 ofxis the largest one of the two solutions found when the positive modulus

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

modulus (for a> ) is taken.

SC 1: Forxis less than or equal their least value andxis greater than or equal their maximum value thenaward B1B0.

SC 2: For 0 1x , then award M1 A0.

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22

QuestionNumber

Scheme Marks

Aliter

(b)2 2 2 , 1x a a x a = >

Way 22 2 2 2 2( ) ( )a a x = 2 2 2 2 2( ) ( )a a x = M1

4 2 2 4 4 2 22 2a x a a a x x + = +

4 2 2 4 4 2 22 2a 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 2

2 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 )

2

ax

=

21 1 8

2

ax

+=

Both correctsimplified down solutions.

A1

[6]

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

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23

Question 8

QuestionNumber

Scheme

8. (a)

M1Writes down a solution in the form, 1 2e em t m t A B+ , where 1 2 .m m Note that they must put a constant in

front of each exponential term.

A13 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 their x x+ . They must add together CFtheirx and PItheir 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

dtby differentiating CF PItheir and their x x .

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

Applies 0, 0t x= = tox andd

0, 2d

xt

t= = to

d

dtto form simultaneous equations.

Note that this method mark is dependent upon the award of the previous 2 method marks.

A1 For the correct answer of3e et tx = + .

(b)

M1 Differentiates theirxto gived

dtand puts

d

dtequal to 0.

dM1 A credible attempt to solve. Multiplying through by 3e tto 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 markhere.

A1 12ln3t= or ln 3t= or awrt 0.55

ddM1 Substitutes their tback intoxand an attempt to eliminate out the lns.

A1 Uses exact values to give2 3

.9

dddM1 Finds2

2

d

d

x

tand substitutes their tinto

2

2

d

d

x

t

A19 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.

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Alternative Way with constantsAandBthe other way round.

QuestionNumber

Scheme Marks

8.2

2d 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 t A B+ , where 1 2 .m m M1So, 2 3CF e e

t tx A B = + 2 3e et tA B + A1

2

2

d d

e e ed d

t t tx x

x k k kt t

= = =

Substitutes e tk into the differential

equation given in the question.M1e 5( e ) 6 e 2e 2 e 2e

1

t t t t t t k k k k

k

+ + = =

=

Finds 1.k= A1

{ }PISo, e tx =

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

2 3d 2 e 3 e ed

t t tx A Bt

= = Finds

d

d

x

tby differentiating

CF PItheir and their x x

dM1*

0, 0 0 1

d0, 2 2 2 3 1

d

t x A B

xt A B

t

= = = + +

= = =

Applies 0, 0t x= = tox

andd

0, 2d

xt

t= = to

d

d

x

tto form

simultaneous equations.

ddM1*

2 2 2

2 3 3

A B

A B

+ = =

0, 1A B = =

So, 3e et tx = + 3e et tx = + A1

[8]

fp2 june 09 ms.pdf

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