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
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Mathematical Problem Solving: Integration, Substitution, and Vector Calculus, Study Guides, Projects, Research of Mathematics
A mathematical problem that involves integration, substitution, and vector calculus. The problem includes various equations and expressions, and the goal is to find the values of certain variables or expressions. The document also includes some notes and comments from the solver. This document could be useful for students studying advanced mathematics, particularly calculus and vector calculus.
What you will learn
- How do you find the value of λ and µ when the lines intersect?
- How do you find the equation of the line l1?
- What is the value of the integral ∫1λu2−±du?
- What is the area of the region R?
- What is the distance between points OX and OA?
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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.
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 )
⎡ (^) + x ⎤ = ⎢ ⎥
⎢⎣ ⎥⎦ Correct integration.
Ignore limits.
A
1 2
2 3 2 0
= ⎡^ (1 +4 ) x ⎤ ⎣ ⎦
( ) (^ )
3 3 2 2
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
∫ (^) ⎝ + ⎠
Use of
2
V = π y d x
∫
Can be implied. Ignore limits and d. x
B
2
0
d 1 4
x x
∫
± 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 :
A C A A
A
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
⎧ (^) x − − x ⎫ = (^) ⎨ + − + + ⎬
⎩ ⎭
Either
3 2
x ± − or
1 ± ( 1)(− − x )from either first or
second expansions respectively
dM1;
Ignoring 1 and 3, any one
correct (^) {..........} expansion.
A
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
4. (a) d^ 1 = −^2 i^ +^ j^ −^4 k^ ,^ d^ 2 =^ q i^ +^2 j^ +^2 k
As
1 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
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
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
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
- 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
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
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
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]
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
Lines meet where:
q
p
⎛ ⎞ ⎛ −^ ⎞ ⎛ − ⎞ ⎛ ⎞
First two of
q
p
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
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