GCSE Edexcel GCE Core Mathematics C4 (6666)
Summer 2005
Core Mathematics C4 (6666)
Edexcel GCE
Mark Scheme (Results)
Final Version June 2005 6666 Core C4 Mark Scheme Question Number
Scheme
Marks
1
1.
( 4 − 9x)
1 2
⎛ 9x ⎞2 = 2 ⎜1 − ⎟ 4 ⎠ ⎝ ⎛ 12 ⎛ 9 x ⎞ 12 ( − 12 ) ⎛ 9 x ⎞ 2 = 2 ⎜1 + ⎜ − ⎟ + ⎜− ⎟ + ⎜ 1⎝ 4 ⎠ 1.2 ⎝ 4 ⎠ ⎝ 81 2 729 3 ⎛ 9 ⎞ = 2 ⎜1 − x − x − x + ... ⎟ 128 1024 ⎝ 8 ⎠ 9 81 729 3 = 2 − x, − x 2 , − x + ... 4 64 512
B1 1 2
( − 12 ) ( − 23 ) ⎛
3 ⎞ 9x ⎞ ⎜ − ⎟ + ... ⎟⎟ ⎝ 4 ⎠ ⎠
1.2.3
M1
A1, A1, A1 [5]
Note The M1 is gained for
1 2
( − 12 ) 1.2
( ... )
2
or
1 2
( − 12 ) ( − 32 ) 1.2.3
( ... )
3
Special Case
81 2 729 3 ⎛ 9 ⎞ x − x + ... ⎟ and goes no further If the candidate reaches = 2 ⎜1 − x − 128 1024 ⎝ 8 ⎠ allow A1 A0 A0
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
Question Number
Scheme
⎛ dy ⎞ dy =0 2x + ⎜ 2x + 2 y ⎟ − 6 y dx ⎝ dx ⎠ dy =0 ⇒ x+ y =0 dx
2.
Marks
M1 (A1) A1 or equivalent
Eliminating either variable and solving for at least one value of x or y. y 2 − 2 y 2 − 3 y 2 + 16 = 0 or the same equation in x y = ±2 or x = ± 2 ( 2, − 2 ) , ( −2, 2 )
M1 M1 A1 A1 [7]
dy x + y = Note: dx 3 y − x Alternative
3 y 2 − 2 xy − ( x 2 + 16 ) = 0 y=
2 x ± √ (16 x 2 + 192 )
6 dy 1 1 8x = ± . dx 3 3 √ (16 x 2 + 192 )
dy =0 ⇒ dx
8x =±1 2 √ (16 x + 192 )
64 x 2 = 16 x 2 + 192 x=± 2 ( 2, − 2 ) , ( −2, 2 )
M1 A1± A1
M1
M1 A1 A1 [7]
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
Question Number
3.
Scheme
(a)
Marks
5x + 3 A B = + ( 2 x − 3)( x + 2 ) 2 x − 3 x + 2 5 x + 3 = A ( x + 2 ) + B ( 2 x − 3)
Substituting x = −2 or x = 32 and obtaining A or B; or equating coefficients and solving a pair of simultaneous equations to obtain A or B. A = 3, B = 1
M1 A1, A1 (3)
If the cover-up rule is used, give M1 A1 for the first of A or B found, A1 for the second. (b)
∫
5x + 3 3 dx = ln ( 2 x − 3) + ln ( x + 2 ) 2 ( 2 x − 3)( x + 2 ) 6 ⎡ ... ⎤ = 3 ln 9 + ln 2 ⎣ ⎦2 2 = ln 54
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1 A1ft
M1 A1 cao
A1
(5) [8]
Question Number
4.
Scheme
∫(
1
1 − x2 )
1 2
dx =
∫(
1
1 − sin 2 θ )
3 2
cos θ dθ
Marks
Use of x = sin θ and
dx = cos θ dθ
∫
1 dθ cos 2 θ = ∫ sec2 θ dθ = tan θ
=
Using the limits 0 and
π 6
M1 A1 M1 A1
to evaluate integral
[ tan θ ] 0 π
6
=
M1
M1
1 ⎛ √3⎞ ⎜= ⎟ √3 ⎝ 3 ⎠
cao
A1 [7]
Alternative for final M1 A1
Returning to the variable x and using the limits 0 and
1 2
to evaluate integral
M1
1
⎡ ⎤2 1 ⎛ √3⎞ x ⎢ ⎥ = ⎜= ⎟ 2 ⎢⎣ √ (1 − x ) ⎥⎦ 0 √ 3 ⎝ 3 ⎠
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
cao
A1
Question Number
5.
Scheme
∫ xe
(a)
2x
1 2
dx = x e − 1 2
2x
Marks
∫
1 e 2 x dx Attempting parts in the right direction 2
1 4
= x e2 x − e2 x
M1 A1 A1
1
1 1 2 ⎡ 1 2x 1 2x ⎤ ⎢⎣ 2 x e − 4 e ⎥⎦ = 4 + 4 e 0
M1 A1 (5)
x = 0.4 ⇒ y ≈ 0.890 22 x = 0.8 ⇒ y ≈ 3.962 43
(b)
Both are required to 5 d.p
B1
.
(c)
I ≈ × 0.2 × [ ... 1 2
]
≈ ... × ⎡⎣ 0+7.389 06+2 ( 0.29836+.890 22+1.992 07+3.962 43) ⎤⎦ ft their answers to (b) ≈ 0.1× 21.675 22 ≈ 2.168 cao
Note
1 1 2 + e ≈ 2.097 … 4 4
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
(1)
B1 M1 A1ft
A1
(4) [10]
Question Number
6.
Scheme
(a)
Marks
dx dy = −2 cosec 2 t , = 4sin t cos t dt dt d y −2sin t cos t = ( = −2sin 3 t cos t ) dx cosec 2 t
both
M1 A1 (4)
At t = π4 , x = 2, y = 1
(b)
Substitutes t = π4 into an attempt at
both x and y
dy ⎛ 1⎞ to obtain gradient ⎜ − ⎟ dx ⎝ 2⎠
Equation of tangent is y − 1 = −
M1 A1
B1 M1
1 ( x − 2) 2
M1 A1
Accept x + 2 y = 4 or any correct equivalent (c)
Uses 1 + cot 2 t = cosec2 t , or equivalent, to eliminate t
(4)
M1
2
2 ⎛ x⎞ 1+ ⎜ ⎟ = y ⎝2⎠
y=
correctly eliminates t
A1
cao
A1
8 4 + x2
The domain is x …0
B1
An alternative in (c) 1
1
x x ⎛ y ⎞2 ⎛ y ⎞2 sin t = ⎜ ⎟ ; cos t = sin t = ⎜ ⎟ 2 2⎝ 2⎠ ⎝2⎠ y x2 y sin 2 t + cos 2 t = 1 ⇒ + × =1 2 4 2 8 Leading to y = 4 + x2
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1 A1 A1
(4) [12]
Question Number
7.
Scheme
(a) k component
Marks
2 + 4λ = −2 ⇒ λ = −1
Note µ = 2 Substituting their λ (or µ ) into equation of line and obtaining B B: ( 2, 2, − 2 )
Accept vector forms
M1 A1 M1 A1 (4)
⎛1⎞ ⎜ ⎟ ⎜ −1⎟ = √ 18; ⎜4⎟ ⎝ ⎠
(b)
⎛1⎞ ⎜ ⎟ ⎜ −1⎟ = √ 2 ⎜0⎟ ⎝ ⎠
⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −1⎟ ⋅ ⎜ −1⎟ = 1 + 1 + 0 ( = 2 ) ⎜4⎟ ⎜0⎟ ⎝ ⎠ ⎝ ⎠ 2 1 cos θ = = √ 18 √ 2 3
(c)
(d)
uuur uuur 2 uuur AB = −i + j − 4k ⇒ AB = 18 or AB = √ 18 uuur uuur 2 uuur BC = 3i − 3 j ⇒ BC = 18 or BC = √ 18 uuur uuur Hence AB = BC ¿
uuur OD = 6i − 2 j + 2k
both
B1
B1
cao
M1 A1 (4)
ignore direction of vector
M1
ignore direction of vector
M1
Allow first B1 for any two correct Accept column form or coordinates
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
A1
(3)
B1 B1 (2) [13]
Question Number
8.
Scheme
(a)
Marks
dV is the rate of increase of volume (with respect to time) dt − kV : k is constant of proportionality and the negative shows decrease (or loss) dV = 20 − kV ¿ giving These Bs are to be awarded independently dt
B1
B1 (2)
∫
(b)
1 dV = ∫ 1dt 20 − kV
separating variables
1 − ln ( 20 − kV ) = t ( +C ) k Using V = 0, t = 0 to evaluate the constant of integration 1 c = − ln 20 k 1 ⎛ 20 ⎞ t = ln ⎜ ⎟ k ⎝ 20 − kV ⎠ Obtaining answer in the form V = A + B e − kt 20 20 − kt 20 V= − e Accept (1 − e− kt ) k k k dV = 20 e − kt dt
(c)
Can be implied
dV = 10, t = 5 ⇒ 10 = 20 e − kt dt 75 At t = 10, V = ln 2
1 ⇒ k = ln 2 ≈ 0.139 5
M1 M1 A1 M1
M1 A1
(6)
M1 M1 A1
awrt 108
M1 A1 (5) [13]
Alternative to (b) Using printed answer and differentiating
dV = − kB e − kt dt
M1
Substituting into differential equation − kB e − kt = 20 − kA − kB e − kt 20 A= k Using V = 0, t = 0 in printed answer to obtain A + B = 0 20 B=− k
6666 Core C4 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1 M1 A1 M1 A1
(6)