Mark Scheme (Results)
Summer 2008
GCE
GCE Mathematics (6666/01)
Edexcel Limited. Registered in England and Wales No. 4496750
Registered Office: One90 High Holborn, London WC1V 7BH
1
June 2008
6666 Core Mathematics C4
Mark Scheme
Question
1. (a)
Scheme
x
y
e
or y
1
Marks
0
0.4
0.8
1.2
1.6
2
0
0.08
0.32
0.72
1.28
e2
e
1.08329
…
e
e
1.37713…
e
2.05443…
3.59664…
7.38906…
Either e0.32 and e1.28 or
awrt 1.38 and 3.60 B1
(or a mixture of e’s and
decimals)
[1]
(b)
Way 1
Outside brackets
B1;
1
× 0.4 or 0.2
2
For structure of
trapezium
M1
rule [ ............. ] ;
1
Area ≈ × 0.4 ; ×⎡⎣ e0 + 2 ( e0.08 + e0.32 + e0.72 + e1.28 ) + e2 ⎤⎦
2
= 0.2 × 24.61203164... = 4.922406... = 4.922 (4sf)
Aliter
(b)
Way 2
Area ≈ 0.4 × ⎡ e
⎣
0
+ e0.08
2
+
e0.08 + e0.32
2
+
e0.32 + e0.72
2
+
e0.72 + e1.28
2
4.922
+
e1.28 + e2
2
A1 cao
[3]
⎤ 0.4 and a divisor of 2 on B1
⎦ all terms inside brackets.
One of first and last
ordinates, two of the
middle ordinates inside M1
brackets ignoring the 2.
which is equivalent to:
1
Area ≈ × 0.4 ; ×⎡⎣ e0 + 2 ( e0.08 + e0.32 + e0.72 + e1.28 ) + e 2 ⎤⎦
2
= 0.2 × 24.61203164... = 4.922406... = 4.922 (4sf)
4.922
A1 cao
[3]
4 marks
1
Note an expression like Area ≈ × 0.4 + e0 + 2 ( e0.08 + e0.32 + e0.72 + e1.28 ) + e 2 would score B1M1A0
2
Allow one term missing (slip!) in the
(
)
brackets for
The M1 mark for structure is for the material found in the curly brackets ie
⎡⎣ first y ordinate + 2 ( intermediate
2 ft y ordinate ) + final y ordinate ⎤⎦
Question
Number
2. (a)
Scheme
Marks
⎧⎪u = x ⇒ ddux = 1⎫⎪
⎨ dv
x
x⎬
⎪⎩ dx = e ⇒ v = e ⎪⎭
∫ xe
x
Use of ‘integration by parts’
formula in the correct direction. M1
(See note.)
Correct expression. (Ignore dx) A1
dx = x e x − ∫ e x .1 dx
= x e x − ∫ e x dx
= x ex − ex ( + c )
Correct integration with/without + c A1
[3]
(b)
⎧⎪u = x 2 ⇒
⎨ dv
x
⎩⎪ dx = e ⇒
= 2 x ⎫⎪
⎬
v = e x ⎭⎪
du
dx
Use of ‘integration by parts’ formula
M1
in the correct direction.
Correct expression. (Ignore dx) A1
2 x
2 x
x
∫ x e dx = x e − ∫ e .2 x dx
= x 2 e x − 2 ∫ x e x dx
Correct expression including + c.
(seen at any stage! in part (b)) A1 ISW
You can ignore subsequent working.
= x2e x − 2 ( x e x − e x ) + c
⎧⎪= x e − 2 x e + 2e + c ⎫⎪
⎨ x 2
⎬
⎩⎪= e ( x − 2 x + 2 ) + c ⎭⎪
2 x
x
[3]
x
Ignore subsequent working
6 marks
Note integration by parts in the correct direction means that u
and ddvx must be assigned/used as u = x and ddvx = e x in part (a)
for example
+ c is not required in part
(a).
i
i di
t (b)
3
Question
Number
3. (a)
Scheme
From question,
Marks
dA
= 0.032 seen
B1
dt
or implied from working.
dA
= 0.032
dt
2π x by itself seen
B1
or implied from working
dA ⎫
⎧
2
= ⎬ 2π x
⎨A = π x ⇒
dx ⎭
⎩
dx dA dA
=
÷
=
dt
dt
dx
When x = 2cm ,
Hence,
(b)
( 0.032 )
⎧ 0.016 ⎫
; ⎨=
⎬
2π x ⎩ π x ⎭
1
0.032 ÷ Candidate's
dx 0.016
=
dt
2π
dx
= 0.002546479... (cm s-1)
dt
awrt 0.00255 A1 cso
[4]
V = π x 2 (5 x) = 5 π x3
V = π x 2 (5 x) or 5π x3
B1
dV
= 15 π x 2
dx
B1
or ft from candidate’s V
in one variable
dV
= 15π x 2
dx
⎛ 0.016 ⎞
dV dV d x
=
×
= 15π x 2 . ⎜
⎟ ; {= 0.24 x}
dt
dx dt
⎝ πx ⎠
When x = 2cm ,
dA
; M1;
dx
Candidate’s
dV
= 0.24(2) = 0.48 (cm3 s −1 )
dt
dV dx
×
; M1
dx dt
0.48 or awrt 0.48 A1 cso
[4]
8 marks
4
Question
Number
Marks
Scheme
3 x 2 − y 2 + xy = 4
4. (a)
( eqn ∗ )
Differentiates implicitly to include either
± ky ddyx or x ddxy . (Ignore
⎧ dy ⎫
dy ⎛
dy ⎞
= ⎬ 6x − 2 y
+ ⎜y + x ⎟= 0
⎨
dx ⎜⎝
dx ⎟⎠
⎩ dx ⎭
⎧ dy
−6 x − y ⎫
=
⎨
⎬ or
x − 2y ⎭
⎩ dx
Correct application
( 3x
⎧ dy
6x + y ⎫
=
⎨
⎬
2y − x ⎭
⎩ dx
2
( )
(
dy
dx
)
= )
M1
of product rule B1
⎛
dy ⎞
− y2 ) → ⎜ 6x − 2 y
⎟ and
dx ⎠
⎝
( 4 → 0)
A1
not necessarily required.
dy 8
−6 x − y
8
= ⇒
=
dx 3
x − 2y
3
Substituting
dy 8
= into their
M1 ∗
dx 3
equation.
giving −18 x − 3 y = 8 x − 16 y
giving
Attempt to combine either terms in x
or terms in y together to give either dM1 ∗
ax or by.
13 y = 26 x
simplifying to give y − 2 x = 0 AG A1 cso
Hence, y = 2 x ⇒ y − 2 x = 0
[6]
(b)
At P & Q, y = 2 x . Substituting into eqn ∗
gives 3 x 2 − (2 x) 2 + x(2 x) = 4
Simplifying gives, x 2 = 4 ⇒ x = ± 2
Attempt replacing y by 2x
in at least one of the y terms in eqn ∗
M1
Either x = 2 or x = −2
A1
Both (2, 4) and (−2, − 4)
A1
y = 2x ⇒ y = ± 4
Hence coordinates are (2, 4) and (−2, − 4)
[3]
9 marks
5
Question
Number
Scheme
Marks
** represents a constant (which must be consistent for first accuracy mark)
1
3x ⎞
−1 ⎛
−1
= (4 − 3 x) 2 = ( 4 ) 2 ⎜1 − ⎟
4 ⎠
(4 − 3 x)
⎝
5. (a)
− 12
1 ⎛ 3x ⎞
= ⎜1 − ⎟
2⎝
4 ⎠
− 12
(4)
− 12
or
1
2
outside brackets B1
−1
Expands (1 + ** x) 2 to give a
simplified or an un-simplified M1;
1 + (− 12 )(** x) ;
(− 1 )( − 23 )
⎡
⎤
= 12 ⎢ 1 + (− 12 )(** x); + 2
(** x) 2 + ... ⎥
2!
⎣
⎦
with ** ≠ 1
=
(− 1 )(− 23 ) 3 x 2
⎤
1⎡
1 + (− 12 )(− 34x ) + 2
(− 4 ) + ... ⎥
⎢
2⎣
2!
⎦
A correct simplified or an unsimplified [ .......... ] expansion
with candidate’s followed A1
through (** x )
Award SC M1 if you see
(− 12 )(** x) +
1
2
= 12 ⎡⎣ 1 + 83 x ; +
27
128
x 2 + ... ⎤⎦
3
27 2
⎧ 1
⎫
x; +
x + ...⎬
⎨= +
256
⎩ 2 16
⎭
(b)
1
2
x+
+ 4 + 32 x +
= 4 + 2 x; +
3
16
27
32
27
128
x 2 + ... ⎤⎦
27 2
⎡⎣ ........; 128
x ⎤⎦
A1 isw
A1 isw
Ignore subsequent working
[5]
Writing ( x + 8) multiplied by
candidate’s part (a) M1
expansion.
3
27 2
⎛1
⎞
( x + 8) ⎜ +
x +
x + ... ⎟
2
16
256
⎝
⎠
=
3
⎣⎡ 1 + 8 x ; ... ⎦⎤
SC: K ⎡⎣ 1 + 83 x +
1
2
(− 12 )(− 23 )
(** x) 2
2!
Multiply out brackets to find
a constant term, two x terms
and two x 2 terms.
x 2 + .....
x 2 + .....
M1
Anything that cancels to
33 2 A1; A1
4 + 2 x;
x
32
33 2
x + ...
32
[4]
9 marks
6
Question
Number
6. (a)
Scheme
Marks
Lines meet where:
⎛ −9 ⎞
⎛2⎞
⎛3⎞
⎛3⎞
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ 0 ⎟ + λ ⎜ 1 ⎟ = ⎜ 1 ⎟ + µ ⎜ −1⎟
⎜ 10 ⎟
⎜ −1⎟
⎜ 17 ⎟
⎜5⎟
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠
Any two of
i : − 9 + 2λ = 3 + 3µ
j:
λ =1 − µ
k : 10 − λ = 17 + 5µ
(1) – 2(2) gives:
(2) gives:
⎛ −9 ⎞
⎛2⎞
⎜ ⎟
⎜ ⎟
r = ⎜ 0 ⎟ + 3⎜ 1 ⎟
⎜ 10 ⎟
⎜ −1⎟
⎝ ⎠
⎝ ⎠
−9 = 1 + 5µ
(1)
(2)
(3)
Need any two of these correct
equations seen anywhere in part M1
(a).
Attempts to solve simultaneous
equations to find one of dM1
either λ or µ
⇒ µ = −2
Both λ = 3 & µ = − 2
λ =1−− 2 = 3
Substitutes their value of either
λ or µ into the line l1 or l2
respectively. This mark can be ddM1
implied by any two correct
components of ( −3, 3, 7 ) .
⎛3⎞
⎛3⎞
⎜ ⎟
⎜ ⎟
or r = ⎜ 1 ⎟ − 2 ⎜ −1⎟
⎜ 17 ⎟
⎜5⎟
⎝ ⎠
⎝ ⎠
⎛ −3 ⎞
⎜ ⎟
⎜ 3 ⎟ or −3i + 3 j + 7k
⎜7⎟
⎝ ⎠
⎛ −3 ⎞
⎜ ⎟
Intersect at r = ⎜ 3 ⎟ or r = −3i + 3 j + 7k
⎜7⎟
⎝ ⎠
Either check k:
λ = 3 : LHS = 10 − λ
Either check that λ = 3 , µ = − 2
in a third equation or check
that λ = 3 ,
B1
µ = − 2 give the same
coordinates on the other line.
Conclusion not needed.
= 10 − 3 = 7
(As LHS = RHS then the lines intersect.)
d1 = 2i + j − k ,
A1
or ( −3, 3, 7 )
µ = − 2 : RHS = 17 + 5µ = 17 − 10 = 7
(b)
A1
[6]
d 2 = 3i − j + 5k
⎛2⎞ ⎛3⎞
⎜ ⎟ ⎜ ⎟
As d1 • d 2 = ⎜ 1 ⎟ • ⎜ −1⎟ = (2 × 3) + (1 ×− 1) + (−1 × 5) = 0
⎜ −1⎟ ⎜ 5 ⎟
⎝ ⎠ ⎝ ⎠
Then l1 is perpendicular to l2.
7
Dot product calculation between the
two direction vectors:
(2 × 3) + (1 ×− 1) + (−1 × 5) M1
or 6 − 1 − 5
Result ‘=0’ and
A1
appropriate conclusion
[2]
Question
Number
6. (c)
Scheme
Equating i ;
− 9 + 2λ = 5
Marks
⇒ λ=7
⎛ −9 ⎞
⎛ 2 ⎞ ⎛ 5⎞
⎜ ⎟
⎜ ⎟ ⎜ ⎟
r = ⎜ 0 ⎟ + 7⎜ 1 ⎟ = ⎜ 7⎟
⎜ 10 ⎟
⎜ −1⎟ ⎜ 3 ⎟
⎝ ⎠
⎝ ⎠ ⎝ ⎠
uuur
( = OA. Hence the point A lies on l1.)
Substitutes candidate’s λ = 7 into
the line l1 and finds 5 i + 7 j + 3k .
B1
The conclusion on this occasion is
not needed.
[1]
(d)
uuur
Let OX = − 3i + 3 j + 7k be point of intersection
Finding the difference between
uuur
their OX (can be implied) and
uuur
OA .
⎛ ⎛ −3 ⎞ ⎛ 5 ⎞ ⎞ M1
uuur
⎜⎜ ⎟ ⎜ ⎟⎟
AX = ± ⎜ ⎜ 3 ⎟ − ⎜ 7 ⎟ ⎟
⎜⎜ 7 ⎟ ⎜ 3⎟⎟
⎝⎝ ⎠ ⎝ ⎠⎠
⎛ −3 ⎞ ⎛ 5 ⎞ ⎛ −8 ⎞
uuur uuur
uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
AX = OX − OA = ⎜ 3 ⎟ − ⎜ 7 ⎟ = ⎜ −4 ⎟
⎜ 7 ⎟ ⎜ 3⎟ ⎜ 4 ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
±
uuur uuur uuur
uuur
uuur
OB = OA + AB = OA + 2 AX
⎛ 5⎞
uuur ⎜ ⎟
OB = ⎜ 7 ⎟ +
⎜ 3⎟
⎝ ⎠
⎛ −8 ⎞
⎜ ⎟
2 ⎜ −4 ⎟
⎜ 4⎟
⎝ ⎠
⎛5⎞
⎜ ⎟
⎜7⎟ +
⎜ 3⎟
⎝ ⎠
⎛
⎞
uuur ⎟
⎜
2 ⎜ their AX ⎟
⎜
⎟
⎝
⎠
⎛ −11⎞
⎜
⎟
⎜ −1 ⎟ or −11i − j + 11k
⎜ 11 ⎟
⎝
⎠
⎛ −11⎞
uuur ⎜
uuur
⎟
Hence, OB = ⎜ −1 ⎟ or OB = −11i − j + 11k
⎜ 11 ⎟
⎝
⎠
dM1
A1
or ( −11, − 1, 11)
[3]
12 marks
8
Question
Number
7. (a)
Scheme
Marks
2
2
A
B
≡
≡
+
2
4− y
(2 − y )(2 + y )
(2 − y ) (2 + y )
Forming this identity.
NB: A & B are not assigned in M1
this question
2 ≡ A(2 + y ) + B(2 − y )
Let y = −2,
2 = B ( 4) ⇒ B =
1
2
Let y = 2,
2 = A( 4) ⇒ A =
1
2
giving
1
2
(2 − y )
+
Either one of A =
1
2
1
2
(2 + y )
(2 − y )
(If no working seen, but candidate writes down
correct partial fraction then award all three marks. If
no working is seen but one of A or B is incorrect then
M0A0A0.)
9
+
1
2
or B =
1
2
(2 + y )
1
2
A1
, aef A1 cao
[3]
Question
Number
7. (b)
Marks
Scheme
∫
2
dy =
4 − y2
∫
∫
1
2
(2 − y )
Separates variables as shown.
Can be implied. Ignore the B1
integral signs, and the ‘2’.
1
dx
cot x
+
1
2
(2 + y )
dy =
∫
tan x dx
ln(sec x) or − ln(cos x)
Either ± a ln(λ − y ) or ± b ln(λ + y )
∴ − 12 ln(2 − y ) + 12 ln(2 + y ) = ln(sec x) + ( c )
B1
M1;
their ∫ cot1 x dx = LHS correct with ft
for their A and B and no error A1
with the “2” with or without + c
y = 0, x =
π
3
{0 = ln 2 + c
⇒ − ln 2 +
1
2
1
2
ln 2 = ln
( ( )) + c
1
cos π3
Use of y = 0 and x = π3 in an
integrated equation containing c M1*
;
}
⇒ c = − ln 2
− 12 ln(2 − y ) + 12 ln(2 + y ) = ln(sec x) − ln 2
Using either the quotient (or
product) or power laws for M1
logarithms CORRECTLY.
1 ⎛ 2+ y⎞
⎛ sec x ⎞
ln ⎜
⎟ = ln ⎜
⎟
2 ⎝2− y⎠
⎝ 2 ⎠
⎛2+ y⎞
⎛ sec x ⎞
ln ⎜
⎟ = 2ln ⎜
⎟
⎝ 2 ⎠
⎝2− y⎠
⎛2+ y⎞
⎛ sec x ⎞
ln ⎜
⎟ = ln ⎜
⎟
⎝ 2 ⎠
⎝2− y⎠
Using the log laws correctly to
obtain a single log term on both dM1*
sides of the equation.
2
2 + y sec 2 x
=
2− y
4
Hence,
sec 2 x =
8 + 4y
2− y
sec 2 x =
8 + 4y
2− y
A1 aef
[8]
11 marks
10
Question
Number
8. (a)
Scheme
At P (4, 2 3) either 4 = 8cos t
or 2 3 = 4sin 2t
⇒ only solution is t = π3 where 0 „ t „
Marks
4 = 8cos t or 2 3 = 4sin 2t
t = π3 or awrt 1.05 (radians) only
π
2
stated in the range 0 „ t „
π
M1
A1
2
[2]
x = 8cos t ,
(b)
dx
= − 8sin t ,
dt
y = 4sin 2t
Attempt to differentiate both x and y
wrt t to give ± p sin t and M1
± q cos 2t respectively
dy
= 8cos 2t
dt
Correct
At P,
( )
−1
Uses m(N) = −
1
3
A1
1
. dM1*
their m(T)
Uses y − 2 3 = ( their mN )( x − 4 )
or finds c using x = 4 and
dM1*
y = 2 3 and uses
y = (their m N ) x + " c " .
N: y − 2 3 = − 3 ( x − 4 )
or
dy
dt
You may need to check
candidate’s substitutions for
M1*
Note the next two method
marks are dependent on M1*
⎧
⎫
8 ( − 12 )
1
⎪
⎪
=
=
=
awrt
0.58
⎨
⎬
3
3
⎪ ( −8 ) 2
⎪
⎩
⎭
N: y = − 3 x + 6 3
and
Divides in correct way round
and attempts to substitute their
value of t (in degrees or M1*
radians) into their ddyx
expression.
dy 8cos ( 23π )
=
dx −8sin ( π3 )
Hence m(N) = − 3 or
dx
dt
y = − 3x + 6 3
AG
A1 cso
AG
2 3 = − 3 ( 4) + c ⇒ c = 2 3 + 4 3 = 6 3
so N: ⎡⎣ y = − 3x + 6 3 ⎤⎦
[6]
11
Scheme
Question
3
A = ∫ y dx = ∫ 4sin 2t. ( −8sin t ) dt
π
0
2
π
π
3
A = ∫ −32sin 2t.sin t dt =
π
2
∫
dx
dt M1
dt
correct expression
A1
(ignore limits and dt )
attempt at A =
π
4
8. (c)
Marks
3
∫ −32 ( 2sin t cos t ) .sin t dt
π
2
y
Seeing sin 2t = 2sin t cos t
M1
anywhere in PART (c).
π
A=
3
∫ −64.sin
2
t cos t dt
Correct proof. Appreciation
of how the negative sign
affects the limits. A1 AG
Note that the answer is
given in the question.
π
2
π
A=
2
∫ 64.sin
2
t cos t dt
π
3
[4]
(d)
{Using substitution u = sin t ⇒ ddut = cos t }
{change limits:
when t = π3 , u = 23 & when t = π2 , u = 1 }
π
⎡ sin 3 t ⎤ 2
A = 64 ⎢
⎥
⎣ 3 ⎦ π3
or
⎡ u3 ⎤
A = 64 ⎢ ⎥
⎣3⎦
k sin 3 t or ku 3 with u = sin t M1
Correct integration
A1
ignoring limits.
1
3
2
Substitutes limits of either
( t = π2 and t = π3 ) or
( u = 1 and u = )
⎡ 1 ⎛ 1 3 3 3 ⎞⎤
A = 64 ⎢ − ⎜⎜ . . .
⎟⎟ ⎥
⎣⎢ 3 ⎝ 3 2 2 2 ⎠ ⎥⎦
3
2
subtracts the correct way
round.
64
−8 3
3
64
⎛1 1 ⎞
A = 64 ⎜ −
3⎟ =
−8 3
3
⎝3 8 ⎠
(Note that a =
64
3
and dM1
A1 aef
isw
[4]
Aef in the form a + b 3 ,
with awrt 21.3 and anything
that cancels to a = 643 and
b = − 8.
, b = − 8)
16
marks
12