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Mark Scheme (Final)

January 2009

GCE

GCE Core Mathematics C4 (6666/01)

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.

January 2009

6666 Core Mathematics C

Mark Scheme

Question

Number

Scheme Marks

1. (a) C :

2 3 y − 3 y = x + 8

Differentiates implicitly to include either

d

d

y ky x

± or

d 3 d

y

x

±. (Ignore

d

d

y

x

⎛ ⎞

⎜ =⎟

⎝ ⎠

.)

M

d (^) d d 2 2 3 3 d d d

y (^) y y y x x x x

⎧ ⎫

⎨ =^ ⎬ −^ =

⎩ ⎭

Correct equation. A

( )

d 2 2 3 3 d

y y x x

− =

A correct (condoning sign error) attempt to

combine or factorise their ‘

d d 2 3 d d

y y y x x

− ’.

Can be implied.

M

2 d 3

d 2 3

y x

x y

=

−

2 3

x

y −

A1 oe

[4]

(b)

3 y = 3 ⇒ 9 − 3(3) = x + 8 Substitutes y = 3 into C. (^) M

3 x = − 8 ⇒ x = − 2 Only x = − (^2) A

d 3(4) d ( 2,3) 4 d 6 3 d

y y

x x

− ⇒ = ⇒ =

−

d 4 d

y

x

= from correct working.

Also can be ft using their ‘ x ’ value and y = 3 in the

correct part (a) of

2 d 3

d 2 3

y x

x y

=

−

A

[3]

7 marks

1(b) final A1. Note if the candidate inserts their x value and y = 3

into

2 d 3

d 2 3

y x

x y

=

−

, then an answer of

d (^2) their , d

y x x

= may indicate a

correct follow through.

Number

Scheme Marks

2. (a) Area( R ) =^

1 2

2 2

0 0

d 3(1 4 ) d (1 4 )

x x x x

− = +

∫ ∫

Integrating

1 2 3(1 4 ) x

−

• to give

1 2 ± k (1 + 4 ) x.

M

1 2

2

1 (^2 )

3(1 4 )

.

⎡ (^) + x ⎤ = ⎢ ⎥

⎢⎣ ⎥⎦ Correct integration.

Ignore limits.

A

1 2

2 3 2 0

= ⎡^ (1 +4 ) x ⎤ ⎣ ⎦

( ) (^ )

3 3 2 2

= 9 − (1)

Substitutes limits of 2 and 0 into a

changed function and subtracts the

correct way round.

M

9 3 2 2 2

= − = 3 (units) 3 A

[4]

(Answer of 3 with no working scores M0A0M0A0.)

(b) Volume

2 2

0

d (1 4 )

x x

π

⎛ ⎞

= ⎜ ⎟

⎜ ⎟

∫ (^) ⎝ + ⎠

Use of

2

V = π y d x

∫

.

Can be implied. Ignore limits and d. x

B

( )

2

0

d 1 4

x x

= π

+

∫

± k ln 1 + 4 x M

( )

2 9

= π ⎡⎣ 4 ln 1 + 4 x ⎤⎦ 09

4 ln 1^ +^4 x A

( ) ( ) ( )

9 9 4 4

= π^ ⎡^ ln 9^ − ln1⎤

⎣ ⎦

Substitutes limits of 2 and 0

and subtracts the correct way round.

dM

So Volume

9

= 4 πln 9

9

4 π^ ln 9 or^

9

2 π^ ln 3 or^

18

4 π^ ln 3 A1 oe isw

[5]

9 marks

Note the answer must be a one term exact

value. Note, also you can ignore

subsequent working here.

Note that

9 = 4 πln 9 + c (oe.) would be awarded the final A0.

Note that ln1 can be implied as equal to 0.

Number

Scheme Marks

3. (a)

2 2 27 x + 32 x + 16 ≡ A (3 x + 2)(1 − x ) + B (1 − x ) + C (3 x + 2) Forming this identity M

Substitutes either 2 3

x = − or x = 1

into their identity or equates 3

terms or substitutes in values to

write down three simultaneous

equations.

M

2 x = − 3 , ( ) ( )

64 5 20 5 12 − 3 + 16 = 3 B ⇒ 3 = 3 B ⇒ B = 4

x = 1, 27 + 32 + 16 = 25 C ⇒ 75 = 25 C ⇒ C = 3

Both B = 4 and C = (^3) A

( Note the A1 is dependent on

both method marks in this part.)

Equate x

2 :

0

A C A A

A

= − + ⇒ = − + ⇒ = −

⇒ =

0, 16 2 4

x A B C

A A A

= = + +

⇒ = + + ⇒ = ⇒ =

Compares coefficients or

substitutes in a third x -value or

uses simultaneous equations to

show A = 0.

B

[4]

(b) (^2)

f ( ) (3 2) (1 )

x x x

= +

+ −

2 1 4(3 x 2) 3(1 x )

− − = + + −

Moving powers to top on any one

of the two expressions

M

( )

3 2 1 4 2 1 2 x 3(1 x )

− (^) − = ⎡^ + ⎤+ − ⎣ ⎦

( )

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

− (^) − = + + −

3 3 2 2 2

( 2)( 3)

1 1 ( 2)( ); ( ) ...

2!

⎧ (^) x − − x ⎫ = (^) ⎨ + − + + ⎬

⎩ ⎭

Either

3 2

1 ( 2)( )

x ± − or

1 ± ( 1)(− − x )from either first or

second expansions respectively

dM1;

Ignoring 1 and 3, any one

correct (^) {..........} expansion.

A

( 1)( 2) 2

3 1 ( 1)( ); ( ) ...

2!

x x

⎧ − − ⎫

+ ⎨ + − − + − + ⎬

⎩ ⎭

Both (^) {..........} correct. (^) A

{ } { }

27 2 2 = 1 − 3 x + 4 x + ... + 3 1 + x + x +...

39 2 = 4 + 0 x ;+ 4 x

39 2 4 + (0 ) ; x (^) 4 x A1; A

[6]

Number

Scheme Marks

3. (c) (^) 1.08 6.4 16 Actual f (0.2) (6.76)(0.8)

4.341715976...

+ +

= =

= = =

Or

2

Actual f (0.2) (3(0.2) 2) (1 0.2)

3.75 4.341715976...

= = +

+ −

= + = =

Attempt to find the

actual value of f(0.2)

or seeing awrt 4.3 and believing it

is candidate’s actual f(0.2).

Candidates can also attempt to find

the actual value by using

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

A B C

x x x

+ +

+ + −

with their A , B and C.

M

39 2 Estimate f (0.2) 4 4 (0.2)

4 0.39 4.

= = +

= + =

Attempt to find an estimate for

f(0.2) using their answer to (b)

M

4.39 4.341715976...

%age error 100 4.341715976...

−

= ×

their estimate - actual 100 actual

× M

= 1.112095408... = 1.1 % (2sf ) 1.1% A1 cao

[4]

14 marks

Number

Scheme Marks

4. (a) d^ 1 = −^2 i^ +^ j^ −^4 k^ ,^ d^ 2 =^ q i^ +^2 j^ +^2 k

As

1 2

1 2 ( 2 ) (1 2) ( 4 2)

q

q

⎧ ⎛ − ⎞ ⎛ ⎞ ⎫

⎪ ⎜ ⎟ ⎜ ⎟ ⎪

⎨ •^ =^ •^ ⎬=^ −^ ×^ +^ ×^ +^ −^ ×

⎜ ⎟ ⎜ ⎟

⎪ ⎜ ⎟ ⎜ ⎟ ⎪

⎝ − ⎠ ⎝ ⎠

⎩ ⎭

d d

Apply dot product calculation between

two direction vectors, ie.

( 2− × q ) + (1 × 2) + ( 4− × 2)

M

2 6 3 AG

q

q q

• = ⇒ − + − =

− = ⇒ = −

d d (^) Sets d 1 (^) • d 2 = 0

and solves to find q = − 3

A1 cso

[2]

(b) Lines meet where:

q

p

λ μ

⎛ ⎞ ⎛ −^ ⎞ ⎛ − ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

+ = +

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ − ⎠ ⎝ ⎠ ⎝ ⎠

First two of

: 11 2 5 (1)

: 2 11 2 (2)

: 17 4 2 (3)

q

p

λ μ

λ μ

λ μ

− = − +

+ = +

− = +

i

j

k

Need to see equations

(1) and (2).

Condone one slip.

(Note that q = − 3 .)

M

(1) + 2 (2) gives: 15 = 17 + μ ⇒ μ= − 2

Attempts to solve (1) and (2) to find

one of either λ or μ

dM

Any one of λ = 5 or μ= − (^2) A (2) gives: 2 + λ= 11 − 4 ⇒ λ= 5 Both λ= 5 and μ= − (^2) A

(3) ⇒ 17 − 4(5) = p + 2( 2)−

Attempt to substitute their λ and μ

into their k component to give an

equation in p alone.

ddM

⇒ p = 17 − 20 + 4 ⇒ p = 1 p = 1 A1 cso

[6]

(c)

2 5 1 or 11 2 2

⎛ ⎞ ⎛ −^ ⎞ ⎛ −^ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= + = −

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ − ⎠ ⎝ ⎠ ⎝ ⎠

r r

Substitutes their value of λ or μ into

the correct line l 1 or l 2.

M

Intersect at ( )

7 or 1, 7, 3

⎛ ⎞

⎜ ⎟

= −

⎜ ⎟

⎜ ⎟

⎝ − ⎠

r

⎛ ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or ( 1, 7, − 3 ) A

[2]

Number

Scheme Marks

(d) Let OX = i + 7 j − 3 k

uuur be point of intersection

AX OX OA

⎛ ⎞ ⎛ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= − = − =

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ −^ ⎠ ⎝ ⎠ ⎝ − ⎠

uuur uuur uuur

Finding vector AX

uuur by finding the

difference between OX

uuur and OA

uuur

. Can

be ft using candidate’s OX

uuur .

M1 ±

OB = OA + AB = OA + 2 AX

uuur uuur uuur uuur uuur

OB

⎛ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

= +

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ − ⎠

uuur

3 2 their

AX

⎛ ⎞ ⎛^ ⎞

⎜ ⎟ ⎜^ ⎟

+ ⎜ ⎟

⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

uuur

dM

Hence,

OB

⎛ − ⎞

⎜ ⎟

=

⎜ ⎟

⎜ ⎟

⎝ − ⎠

uuur or OB = − 7 i + 11 j − 19 k

uuur

⎛ − ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or − 7 i + 11 j − 19 k

or ( −7, 11, − 19 )

A

[3]

13 marks

Number

Scheme Marks

5. (a) Similar triangles^

r h r h

⇒ = ⇒ =

Uses similar triangles, ratios or

trigonometry to find either one of these

two expressions oe.

M

(^2 ) 1 2 1 2 4

h h V r h h

π

π π

⎛ ⎞

= = ⎜ ⎟ =

⎝ ⎠

AG

Substitutes 2 3

r =^ h into the formula for the

volume of water V.

A

[2]

(b) From the question,

d 8 d

V

t

=

d 8 d

V

t

= B

2 2 d 12 4

d 27 9

V h h

h

π π

= =

2 2 d 12 4 or d 27 9

V h h

h

π π

= B

Candidate’s

d d

d d

V V

t h

÷ ; M1;

2 2

d d d 9 18 8 d d 4

h V V

t t dh π h π h

= ÷ = × =

2 12 8 27

⎛ (^) π h ⎞ ÷ ⎜ ⎟

⎝ ⎠

or (^2 )

8 or 4 π h π h

× oe (^) A

When

d 18 1 12, d 144 8

h h t π π

= = =

144 π

or

8 π A1 oe isw

[5]

7 marks

Note the answer must be a one term exact value.

Note, also you can ignore subsequent working after

144 π

.

Number

Scheme Marks

6. (a)

2 tan x d x ∫

2 2 2 2 ⎡ (^) NB : sec A = 1 + tan A gives tan A = sec A − 1 ⎤ ⎣ ⎦

The correct underlined identity. M1 oe

2 = sec x −1 d x ∫

= tan x − x ( + c )

Correct integration

with/without + c

A

[2]

(b) (^3)

ln x d x ∫ x

2 2

d 1 d

d^31 d (^2 )

ln

u x x

v x x (^) x

u x

x v

− −^ − −

⎧ = ⇒ = ⎫

⎪ ⎪

⎨ ⎬

⎪ =^ ⇒^ =^ = ⎪

⎩ ⎭

Use of ‘integration by parts’ formula

in the correct direction.

M

2 2

ln. d 2 2

x x x x x

= − − −

∫

Correct expression. A

2 3

ln d 2 2

x x x x

= − +

∫

An attempt to multiply through

n

k

x

, n ∈ , n … 2 by

1 x and an

attempt to ...

… “integrate”(process the result); M

2 2 (^ )

ln 2 2 2

x c x x

⎛ ⎞

= − + ⎜ − ⎟ +

⎝ ⎠

correct solution with/without + c A1 oe

[4]

Correct direction means that u =ln x.

Number

Scheme Marks

(c)

3 e d 1 e

x

x

x ∫ +

d d 1 d 1 1 e e , , d d e d 1

x x x

u x x u x u u u

⎧ ⎫

⎨ =^ +^ ⇒^ =^ =^ = ⎬

⎩ − ⎭

Differentiating to find any one of the

three underlined

B

2 2

3

e .e ( 1) .e 1 d. d 1 e e

( 1) 1

or. d ( 1)

x x x

x x

u x u u

u u u u

−

= =

+

−

=

−

∫ ∫

∫

Attempt to substitute for

2 e f ( )

x = u ,

their

d 1

d e

x

x

u

= and 1 e

x u = +

or

3 e f ( )

x = u , their

d 1

d 1

x

u u

=

−

and

1 e

x u = +.

M1*

2 ( 1) d

u u u

−

=

∫

2 ( 1) d

u u u

−

∫

A

2 2 1 d

u u u u

− +

=

∫

u 2 d u u

= − +

∫

An attempt to

multiply out their numerator

to give at least three terms

and divide through each term by u (^) dM1*

( )

2

2 ln 2

u = − u + u + c

Correct integration

with/without +c

A

2 (1 e ) 2(1 e ) ln(1 e ) 2

x x x c

+

= − + + + +

Substitutes 1 e

x u = + back into their

integrated expression with at least

two terms.

dM1*

1 1 2 2 2

e e 2 2e ln(1 e )

x x x x = + + − − + + + c

1 1 2 2 e^2 e^2 2e^ ln(1^ e )

x x x x = + + − − + + + c

1 2 3 2 e^ e^ ln(1^ e ) 2

x x x = − + + − + c

1 2 2 e^ e^ ln(1^ e )

x x x = − + + + k AG

1 2 2 e^ e^ ln(1^ e )

x x x − + + + k

must use a + c and

3 " − 2 "combined. A1 cso

[7]

13 marks

Number

Scheme Marks

7. (a) At A ,

2 x = − 1 + 8 = 7 & y = −( 1) = 1 ⇒ A (7,1) A (7,1)^ B

[1]

(b)

3 2 x = t − 8 , t y = t ,

d (^2) 3 8 d

x t t

= − ,

d 2 d

y t t

=

Their

d d

y t divided by their^

d d

x t M 2

d 2

d 3 8

y t

x t

∴ =

− (^) Correct

d d

y x A

At A , 2

2( 1) 2 2 2

m( ) 3( 1) 8 3 8 5 5

− − −

= = = =

− − − −

T

Substitutes for t to give any of the

four underlined oe:

A

T : y − (^) ( their 1) = mT (^) ( x −( their 7))

or ( )

2 14 9 1 = 5 7 + c ⇒ c = 1 − 5 = − 5

Hence

2 9 5 5

T : y = x −

Finding an equation of a tangent

with their point and their tangent

gradient

or finds c and uses

y = (their gradient) x + " " c.

dM

gives T : 2 x − 5 y − 9 = 0 AG 2 x − 5 y − 9 = (^0) A1 cso

[5]

(c)

3 2 2( t − 8 ) t − 5 t − 9 = 0

Substitution of both

3 x = t − 8 t and 2 y = t into T

M

3 2 2 t − 5 t − 16 t − 9 = 0

{ }

2 ( t + 1) (2 t − 7 t − 9) = 0

( t + 1) ( { t + 1)(2 t − 9) = 0 }

A realisation that

( t^ +^1 )is a factor.^

dM

{ }

9 t = −1 (at A ) t = 2 at B

9 t = (^2) A

Candidate uses their value of t to

find either the x or y coordinate

ddM

One of either x or y correct. A

( ) ( )

( )

9 2 9 729 441 2 2 8 8

9 2 81 2 4

8 36 55.125 or awrt 55.

20.25 or awrt 20.

x

y

= − = − = =

= = =

Both x and y correct. A

Hence (^) ( )

441 81 B (^) 8 , 4 awrt

[6]

12 marks

• Note: 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.

Oe or equivalent.

January 2009

6666 Core Mathematics C

Appendix

Question 1

Question

Number

Scheme Marks

Aliter

1. (a)

Way 2

C :

2 3 y − 3 y = x + 8

Differentiates implicitly to include either

2 d

d

x kx y

±. (Ignore

d

d

x

y

⎛ ⎞

⎜ =⎟

⎝ ⎠

.)

M

d 2 d 2 3 3 d d

x x y x y y

⎧⎪ ⎫⎪

⎨ =^ ⎬ −^ =

⎪⎩ ⎪⎭

Correct equation. A

( )

2 d d

y x

y − = x Applies

( )

d d

d 1

d

y x

x

y

= (^) dM

2 d 3

d 2 3

y x

x y

=

−

2 3

x

y −

A1 oe

[4]

Aliter

1. (a)

Way 3

C :

2 3 y − 3 y = x + 8

gives

3 2 x = y − 3 y − 8

( )

1 2 3 ⇒ x = y − 3 y − 8

Differentiates in the form ( ) ( )

2 1 3 3 f (^ y )^^ f (^ y )

− ′ (^). M ( ) ( )

2 d (^1 ) 3 8 2 3 d 3

x y y y y

− = − − −

Correct differentiation. A

( )

2 2 3

d 2 3

d (^3 3 )

x y

y (^) y y

−

=

− −

( )

2 2 3 d^3 3

d 2 3

y^ y^ y

x y

− −

=

−

Applies

( )

d d

d 1

d x y

y

x

=

dM

( )

2 3 3 d^3

d 2 3

y^ x

x y

=

−

2 d 3

d 2 3

y x

x y

⇒ =

−

( )

2 3 3 3

x

y −

or

2 3

x

y −

A1 oe

[4]

Question

Number

Scheme Marks

Aliter

2. (a)

Way 2

Area( R ) =

1 2

2 2

0 0

d 3(1 4 ) d (1 4 )

x x x x

− = +

∫ ∫

{Using substitution

d (^1 4) d 4

u u = + x ⇒ (^) x = }

{change limits:

When x = 0 , u = 1 & when x = 2 , u = 9 }

So, Area( R ) =

1 2

9 1 4 1

3 u d u

− ∫

Integrating

1

± λ u −^2 to give

1 ± k u^2. M

( )

1 2

2

1 2 0

u

⎡ ⎤

= ⎢^ ⎥

⎢ ⎥

⎣ ⎦

Correct integration.

Ignore limits.

A

1 2

9 3 2 1

= ⎡^ u ⎤ ⎣ ⎦

( ) (^ )

3 3 2 2

= 9 − (1)

Substitutes limits of either

( u = 9 and u = 1 )or

in x , ( x = 2 and x = 0 )into a changed

function and subtracts the correct way

round.

M

9 3 2 = 2 − 2 = 3 (units) 3 A

[4]

Aliter

2. (a)

Way 3

Area( R ) =

1 2

2 2

0 0

d 3(1 4 ) d (1 4 )

x x x x

− = +

∫ ∫

{Using substitution (^2) d 1 (^1 4 2) d 4 2 d d

u u = + x ⇒ u (^) x = ⇒ u u = x }

{change limits:

When x = 0 , u = 1 & when x = 2 , u = 3 }

So, Area( R ) =

3 3 3 1 3 2 2 1 1

u u u d^^ = d u ∫ ∫

3 Integrating ± λto give ± k u. M

1

u

⎡ ⎤

= ⎢ ⎥

⎣ ⎦

Correct integration.

Ignore limits.

A

( ) ( )

3 3 = 2 (3) − 2 (1)

Substitutes limits of either

( u^ =^ 3 and^ u =^1 )or

in x , ( x = 2 and x = 0 )into a changed

function and subtracts the correct way

round.

M

9 3 2 = 2 − 2 = 3 (units) 3 A

[4]

Question

Number

Scheme Marks

Aliter

3. (a)

Way 2

2 2 27 x + 32 x + 16 ≡ A (3 x + 2)(1 − x ) + B (1 − x ) + C (3 x + 2) Forming this identity M

2 terms : 27 3 9 (1)

terms : 32 12 (2)

constants: 16 = 2 4 (3)

x A C

x A B C

A B C

= − +

= − +

+ +

equates 3 terms. M

(2) + (3) gives 48 = 3 A + 16 C (4)

(1) + (4) gives 75 = 25 C ⇒ C = 3

(1) gives 27 = − 3 A + 27 ⇒ 0 = − 3 A ⇒ A = 0

(2) gives 32 = − B + 36 ⇒ B = 36 − 32 = 4 Both B = 4 and C = (^3) A

Decide to award B1 for A = (^0) B

[4]

3. (a) (^) If the candidate assumes A = 0 and writes the identity 27 x^2^ + 32 x + 16 ≡ B (1 − x ) + C (3 x +2)^2

and goes on to find B = 4 and C = 3 then the candidate is awarded M0M1A0B0.

3. (a) (^) If the candidate has the incorrect identity 27 x^2^ + 32 x + 16 ≡ A (3 x + 2) + B (1 − x ) + C (3 x + 2)^2 and

goes on to find B = 4, C = 3 and A = 0 then the candidate is awarded M0M1A0B1.

3. (a) If the candidate has the incorrect identity 2 2 2 27 x + 32 x + 16 ≡ A (3 x + 2) (1 − x ) + B (1 − x ) + C (3 x + 2) and goes on to find B = 4, C = 3 and

A = 0 then the candidate is awarded M0M1A0B1.

Number

Scheme Marks

Aliter

3. (b)

Way 2

2

f ( ) (3 2) (1 )

x x x

= +

+ −

2 1 4(3 x 2) 3(1 x )

− − = + + −

Moving powers to top on any one

of the two expressions

M

2 1 4(2 3 ) x 3(1 x )

− − = + + −

2 3 (^ 2)(^ 3) 4 2

4 (2) ( 2)(2) (3 ); (2) (3 )

2!

x x

⎧ − − − − − ⎫

= ⎨ + − + +⎬

⎩ ⎭

Either

2 3 (2) ( 2)(2) (3 ) x

− − ± − or

1 ± ( 1)(− − x ) from either first or

second expansions respectively

dM1;

Ignoring 1 and 3, any one

correct (^) {..........} expansion.

A

( 1)( 2) 2

3 1 ( 1)( ); ( ) ...

2!

x x

⎧ − − ⎫

+ ⎨ + − − + − + ⎬

⎩ ⎭

Both (^) {..........} correct. (^) A

{ } { }

1 3 27 2 2 = 4 4 − 4 x + 16 x + ... + 3 1 + x + x +...

39 2 = 4 + 0 x ;+ 4 x

39 2 4 + (0 ) ; x (^) 4 x A1; A

[6]

Number

Scheme Marks

Aliter

3. (c)

Way 2

1.08 6.4 16

Actual f (0.2) (6.76)(0.8)

4.341715976...

+ +

= =

= = =

Attempt to find the

actual value of f(0.2)

M

39 2 Estimate f (0.2) 4 4 (0.2)

4 0.39 4.

= = +

= + =

Attempt to find an estimate for f(0.2)

using their answer to (b)

M

%age error 100 100 4.341715976...

⎛ ⎞

= − ⎜ × ⎟

⎝ ⎠

their estimate 100 100 actual

⎛ ⎛ ⎞ ⎞

− ⎜ ⎜ ⎟× ⎟

⎝ ⎝^ ⎠ ⎠

M

= 100 −101.

= − 1.112095408... = 1.1 % (2sf ) 1.1% A1 cao

[4]

3. (c) Note that:

4.39 4.341715976...

%age error 100

−

= ×

= 1.0998638... =1.1 % (2sf )

Should be awarded the final marks of

M0A

3. (c) Also note that:

4.341715976...

%age error 100 100

⎛ ⎞

= − ⎜ × ⎟

⎝ ⎠

= 1.0998638... =1.1 % (2sf )

Should be awarded the final marks of

M0A

…so be wary of 1.0998638…

Question

Number

Scheme Marks

4. (a) −^2 q +^2 −^8 is sufficient for M1.

Aliter

4. (b)

Way 2

Only apply Way 2 if candidate does not find both λ and

μ.

Lines meet where:

q

p

λ μ

⎛ ⎞ ⎛ −^ ⎞ ⎛ − ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

+ = +

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ − ⎠ ⎝ ⎠ ⎝ ⎠

First two of

: 11 2 5 (1)

: 2 11 2 (2)

: 17 4 2 (3)

q

p

λ μ

λ μ

λ μ

− = − +

+ = +

− = +

i

j

k

Need to see equations

(2) and (2).

Condone one slip.

(Note that q = − 3 .)

M

(2) gives λ= 9 + 2 μ

(1) gives 11 − 2(9 + 2 μ) = − 5 − 3 μ

Attempts to solve (1) and (2) to find

one of either λ or μ

dM

11 − 18 − 4 μ= − 5 − 3 μ

gives: 11 − 18 + 5 = μ ⇒ μ= − 2 Any one of^ λ^ =^ 5 or^ μ= −^2 A

(3) gives 17 − 4(9 + 2 μ) = p + 2 μ

Candidate writes down a correct

equation containing p and one of either

λ or μ which has already been found.

A

(3) ⇒ 17 − 4(9 + 2( 2))− = p + 2( 2)−

Attempt to substitute their value for

λ( = 9 + 2 μ)and μ into their

k component to give an equation in

p alone.

ddM

⇒ 17 − 20 = p − 4 ⇒ p = 1 p = 1 A1^ cso

[6]

4. (c)

If no working is shown then any two out of the three

coordinates can imply the first M1 mark.

M

Intersect at ( )

7 or 1, 7, 3

⎛ ⎞

⎜ ⎟

= −

⎜ ⎟

⎜ ⎟

⎝ − ⎠

r

⎛ ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or ( 1, 7, − 3 ) A

[2]

Number

Scheme Marks

Aliter

4. (d) (^) Let OX = i + 7 j − 3 k

uuur be point of intersection

Way 2

AX OX OA

⎛ ⎞ ⎛ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= − = − =

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ −^ ⎠ ⎝ ⎠ ⎝ − ⎠

uuur uuur uuur

Finding the difference between their

OX

uuur (can be implied) and OA

uuur .

AX

⎛ ⎛ ⎞ ⎛ ⎞⎞

⎜ ⎜ ⎟ ⎜ ⎟⎟

= ± −

⎜ ⎜ ⎟ ⎜ ⎟⎟

⎜ ⎜ ⎟ ⎜ ⎟⎟

⎝ ⎝^ − ⎠^ ⎝^ ⎠⎠

uuur M1^ ±

OB = OX + XB = OX + AX

uuur uuur uuur uuur uuur

OB

⎛ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

= +

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ −^ ⎠ ⎝ − ⎠

uuur their OX their AX

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎜ ⎟ +⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

uuur uuur

dM

Hence,

OB

⎛ − ⎞

⎜ ⎟

=

⎜ ⎟

⎜ ⎟

⎝ − ⎠

uuur or OB = − 7 i + 11 j − 19 k

uuur

⎛ − ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or − 7 i + 11 j − 19 k

or ( −7, 11, − 19 )

A

[3]

Aliter At^ A ,^ λ^ =^ 1. At^ X ,^ λ^ =5.

4. (d)

Way 3

Hence at B , λ = 5 + (5 − 1) = 9

λ B = ( their λ X ) + ( their λ X −theirλ A )

λ B = 2 their ( λ X ) − ( theirλ A )

M

OB

⎛ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟

= +

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ − ⎠

uuur Substitutes their value of λ into the

line l 1.

dM

Hence,

OB

⎛ − ⎞

⎜ ⎟

=

⎜ ⎟

⎜ ⎟

⎝ − ⎠

uuur or OB = − 7 i + 11 j − 19 k

uuur

⎛ − ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or − 7 i + 11 j − 19 k

or ( −7, 11, − 19 )

A

[3]

Number

Scheme Marks

Aliter

4. (d)

OA = 9 i + 3 j + 13 k

uuur

and the point of intersection OX = i + 7 j − 3 k

uuur

Way 4

9 Minus 8 1

3 Plus 4 7

13 Minus 16 3

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

→ →

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ − ⎠

Finding the difference

between their OX

uuur (can be

implied) and OA

uuur .

( )

AX

⎛ ⎛ ⎞ ⎛ ⎞⎞

⎜ ⎜ ⎟ ⎜ ⎟⎟

= ± −

⎜ ⎜ ⎟ ⎜ ⎟⎟

⎜ ⎜^ ⎟^ ⎜^ ⎟⎟

⎝ − ⎠ ⎝ ⎠

⎝ ⎠

uuur

M1 ±

1 Minus 8 7

7 Plus 4 11

3 Minus 16 19

⎛ ⎞ ⎛ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

→ →

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ −^ ⎠ ⎝ ⎠ ⎝ − ⎠

their OX their AX

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎜ ⎟ +⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

uuur uuur

dM1

Hence,

OB

⎛ − ⎞

⎜ ⎟

=

⎜ ⎟

⎜ ⎟

⎝ − ⎠

uuur or OB = − 7 i + 11 j − 19 k

uuur

⎛ − ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or − 7 i + 11 j − 19 k

or ( −7, 11, − 19 )

A1

[3]

Aliter

4. (d)

OA = 9 i + 3 j + 13 k

uuur and OB = a i + b j + c k

uuur

and the point of intersection OX = i + 7 j − 3 k

uuur

Way 5

As X is the midpoint of AB , then

( )

1, 7, 3 , ,

⎛ +^^ a^ +^ b^ + c ⎞ − = ⎜ ⎟ ⎝ ⎠

Writing down any two of

these “equations” correctly.

M1

2(1) 9 7

2(7) 3 11

2( 3) 13 19

a

b

c

= − = −

= − =

= − − = −

An attempt to find at least

two of a , b or c.

dM1

Hence,

OB

⎛ − ⎞

⎜ ⎟

=

⎜ ⎟

⎜ ⎟

⎝ − ⎠

uuur or OB = − 7 i + 11 j − 19 k

uuur

⎛ − ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or − 7 i + 11 j − 19 k

or ( −7, 11, − 19 )or

a = − 7, b = 11, c = − 19

A1

[3]

Number

Scheme Marks

Aliter

4. (d) Let OX = i + 7 j − 3 k

uuur be point of intersection

Way 6

AX OX OA

⎛ ⎞ ⎛ ⎞ ⎛ − ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= − = − =

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ −^ ⎠ ⎝ ⎠ ⎝ − ⎠

uuur uuur uuur

and AX = 64 + 16 + 256 = 336 = 4 21

uuur

Finding the difference

between their OX

uuur (can be

implied) and OA

uuur .

AX

⎛ ⎛ ⎞ ⎛ ⎞⎞

⎜ ⎜ ⎟ ⎜ ⎟⎟

= ± −

⎜ ⎜ ⎟ ⎜ ⎟⎟

⎜ ⎜^ ⎟^ ⎜^ ⎟⎟

⎝ − ⎠ ⎝ ⎠

⎝ ⎠

uuur

Note AX = 336

uuur would

imply M1.

M1 ±

BX OX OB

λ λ

λ λ

λ λ

⎛ ⎞ ⎛ −^ ⎞ ⎛ −^ + ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= − = − + = −

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ −^ ⎠ ⎝ −^ ⎠ ⎝ −^ + ⎠

uuur uuur uuur

Hence BX = AX = 336

uuur uuur gives

( ) ( ) ( )

2 2 2

− 10 + 2 λ + 5 − λ + − 20 + 4 λ = 336

Writes distance equation

of

2 BX = 336

uuur where

BX = OX − OB

uuur uuur uuur and

OB

λ

λ

λ

⎛ − ⎞

⎜ ⎟

= +

⎜ ⎟

⎜ − ⎟

⎝ ⎠

uuur

dM1

2 2 2

2

2

2

( 1)( 9) 0

λ λ λ λ λ λ

λ λ

λ λ

λ λ

λ λ

− + + − + + − + =

− + =

− + =

− + =

− − =

At A , λ = 1 and at B λ = 9,so,

11 2(9)

17 4(9)

OB

⎛ − ⎞

⎜ ⎟

= +

⎜ ⎟

⎜ ⎟

⎝ − ⎠

uuur

Hence,

OB

⎛ − ⎞

⎜ ⎟

=

⎜ ⎟

⎜ ⎟

⎝ − ⎠

uuur or OB = − 7 i + 11 j − 19 k

uuur

⎛ − ⎞

⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ − ⎠

or − 7 i + 11 j − 19 k

or ( −7, 11, − 19 )

A1

[3]

Question

Number

Scheme Marks

5. (a) Similar shapes^ ⇒^ either

1 2 3 3 3 1 2 3

(16) 24 24

or (16) 24 24

V h

V h

π

π

⎛ ⎞ ⎛ ⎞

= ⎜ ⎟ =⎜ ⎟

⎝ ⎠ ⎝ ⎠

1 2 3 3 3 1 2 3

(24) 24

or (24) 24

r V h

V h r

π

π

⎛ ⎞ ⎛ ⎞

= ⎜ ⎟ =⎜ ⎟

⎝ ⎠ ⎝ ⎠

Uses similar shapes to find either one of

these two expressions oe.

M1

(^3 3) 4 2048 24 27

h h V

π

π

⎛ ⎞

= × ⎜ ⎟ =

⎝ ⎠

AG

Substitutes their equation to give the

correct formula for the volume

of water V.

A1

[2]

5. (a) Candidates simply writing:

V = × π h or

2 (^1 163)

V π h

⎛ ⎞

= ⎜ ⎟

⎝ ⎠

would be awarded M0A0.

(b) From question,^ ( )

d 8 8 d

V

V t c t

= ⇒ = +

d 8 d

V

t

= or V = 8 t B1

1 1 1 1 3 3 3 3 27 27(8 ) 54 2 3 4 4

V t t t h h π π π π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= ⎜ ⎟ ⇒ = ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 1 1 3 3 3 27(8 ) 54 2 or or 3 4

t t t

π π π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

B1

2 d 3

d

h k t t

− 1 = ±^ ;^ M1; (^3 2) 3

d 2 1 3 d 3

h t

t π

⎛ ⎞ −

= ⎜ ⎟

⎝ ⎠

1 (^3 2) 3

d 2 1 3 d 3

h t

t π

⎛ ⎞ −

= ⎜ ⎟

⎝ ⎠

A1 oe

When

3 12 12, 32 3 2

h t

π

π

⎛ ⎞

= = ⎜ ⎟ × =

⎝ ⎠

So when 1 2 1 3 3 3

3

d 2 1 2 1 12, d 32 1024 8

h h

t π π π π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ =

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

8 π A1 oe

[5]

Question

Number

Scheme Marks

7. (a)

It is acceptable for a candidate to write x = 7, y = 1, to

gain B1.

A (7,1) B1

[1]

Aliter

(c)

Way 2

3 2 x = t − 8 t = t t ( − 8) = t ( y −8)

So,

2 2 2 2 x = t ( y − 8) = y y ( −8)

2 2 2 x − 5 y − 9 = 0 ⇒ 2 x = 5 y + 9 ⇒ 4 x = (5 y +9)

Hence,

2 2 4 y y ( − 8) = (5 y + 9)

Forming an equation in terms of y

only.

M1

2 2 4 ( y y − 16 y + 64) = 25 y + 90 y + 81

3 2 2 4 y − 64 y + 256 y = 25 y + 90 y + 81

3 2 4 y − 89 y + 166 y − 81 = 0

A realisation that

( y^ −^1 )is a factor.^

dM1 ( y − 1)( y − 1)(4 y − 81) = 0

Correct factorisation A1

81 y = 4 = 20.25 (or awrt 20.3) Correct y-coordinate (see below!)

2 81 81 2 x = 4 ( 4 − 8)

Candidate uses their y -coordinate

to find their x -coordinate.

ddM1

Decide to award A1 here for

correct y-coordinate.

A1

441 x = 8 = 55.125 (or awrt 55.1) Correct x -coordinate A1

Hence (^) ( )

441 81 B (^) 8 , 4 [6]

Number

Scheme Marks

Aliter

7. (c)

Way 3

t = y

So (^) ( ) ( )

3 x = y − 8 y

2 x − 5 y − 9 = 0 yields

( ) ( )

3 2 y − 16 y − 5 y − 9 = 0

Forming an equation in terms of y

only.

M1

( ) ( )

3 ⇒ 2 y − 5 y − 16 y − 9 = 0

( y^ +^1 ) (^ { 2 y^ −^7 y −^9 )=^0 }

A realisation that

( y^ +^1 )is a factor.^

dM1

( y^ +^1 ) (^ { y^ +^1 )(^2 y −^9 )=^0 } Correct factorisation.^ A1

81 y = 4 = 20.25 (or awrt 20.3) Correct y-coordinate (see below!)

( ) ( )

3 81 81 x = 4 − (^84)

Candidate uses their y -coordinate

to find their x -coordinate.

ddM1

Decide to award A1 here for

correct y-coordinate.

A1

441 8

x = = 55.125 (or awrt 55.1) Correct x -coordinate A1

Hence (^) ( 441 81 ) 8 4

B , [6]