Mark Scheme (Results) January 2010
GCE
GCE Core Mathematics C4 (6666/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edexcel’s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on 0844 576 0025, our GCSE team on 0844 576 0027, or visit our website at www.edexcel.com.
If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Expert email service helpful. Ask The Expert can be accessed online at the following link: http://www.edexcel.com/Aboutus/contact-us/
January 2010 Publications Code UA022713 All the material in this publication is copyright Š Edexcel Ltd 2010
January 2010 6666 Core Mathematics C4 Mark Scheme
Question Number
Q1
Scheme
(a) (1 − 8 x ) 2 = 1 + ( 12 ) ( −8 x ) + 1
( 12 )( − 12 ) 2
Marks
( −8 x ) + 2
( 12 )( − 12 ) ( − 23 ) 3!
( −8 x )
3
+…
= 1 − 4 x − 8 x 2 ; − 32 x 3 − …
(b)
(1 − 8 x ) = =
A1; A1
8 ⎞ ⎛ ⎜1 − ⎟ ⎝ 100 ⎠ 92 23 23 = = 100 25 5
(4)
M1
¿
cso
(c) 1 − 4 x − 8 x 2 − 32 x3 = 1 − 4 ( 0.01) − 8 ( 0.01) − 32 ( 0.01) = 1 − 0.04 − 0.0008 − 0.000 032 = 0.959 168 2
23 = 5 × 0.959 168 = 4.795 84
GCE Core Mathematics C4 (6666) January 2010
M1 A1
A1
(2)
3
M1 M1
cao
3
A1
(3) [9]
Question Number Q2
Scheme
(a)
1.386, 2.291
Marks
awrt 1.386, 2.291
1 (b) A ≈ × 0.5 ( ... ) 2 = ... ( 0 + 2 ( 0.608 + 1.386 + 2.291 + 3.296 + 4.385 ) + 5.545 )
B1 B1 B1 M1
= 0.25 ( 0 + 2 ( 0.608 + 1.386 + 2.291 + 3.296 + 4.385 ) + 5.545 ) ft their (a) A1ft = 0.25 × 29.477 ... ≈ 7.37 cao A1
(c)(i)
x2 x2 1 ln x − ⌠ ⎮ × dx 2 ⌡ 2 x 2 x x = ln x − ⌠ ⎮ dx ⌡2 2 2 x x2 = ln x − ( +C ) 2 4
∫ x ln x dx =
(2)
(4)
M1 A1
M1 A1
4
⎡ x2 x2 ⎤ ⎛ 1⎞ (ii) ⎢ ln x − ⎥ = ( 8ln 4 − 4 ) − ⎜ − ⎟ 4 ⎦1 ⎝ 4⎠ ⎣2 15 = 8ln 4 − 4 15 = 8 ( 2 ln 2 ) − 4 1 = ( 64 ln 2 − 15 ) 4
GCE Core Mathematics C4 (6666) January 2010
4
M1
ln 4 = 2 ln 2 seen or implied M1 a = 64, b = −15
A1
(7) [13]
Question Number
Q3
Scheme
(a) −2sin 2 x − 3sin 3 y
(b) At x =
π 6
,
dy =0 dx dy 2sin 2 x =− dx 3sin 3 y ⎛ 2π cos ⎜ ⎝ 6
GCE Core Mathematics C4 (6666) January 2010
Accept
2sin 2 x −2sin 2 x , −3sin 3 y 3sin 3 y
π
3
⇒y=
π 9
2sin 2 ( π6 ) 2sin π3 dy 2 =− = − =− π π dx 3sin 3 ( 9 ) 3sin 3 3 2⎛ π⎞ =− ⎜x− ⎟ 9 3⎝ 6⎠ 6 x + 9 y − 2π = 0
y− Leading to
M1 A1
⎞ ⎟ + cos 3 y = 1 ⎠ 1 cos 3 y = 2
3y = ⎛π π ⎞ (c) At ⎜ , ⎟ , ⎝6 9⎠
Marks
π
5
A1
(3)
M1 A1
awrt 0.349 A1
(3)
M1 M1 A1
(3) [9]
Question Number Q4
Scheme
Marks
(a) A: ( −6, 4, − 1)
(b)
Accept vector forms B1
⎛ 4⎞ ⎛ 3⎞ 2 2 ⎜ ⎟⎜ ⎟ 2 2 2 2 ⎜ −1 ⎟ . ⎜ −4 ⎟ = 12 + 4 + 3 = 4 + ( −1) + 3 3 + ( −4 ) + 1 cos θ ⎜ 3⎟ ⎜ 1⎟ ⎝ ⎠⎝ ⎠ 19 awrt 0.73 cos θ = 26
(c) X: (10, 0, 11) ⎛ 10 ⎞ ⎛ −6 ⎞ (d) AX = ⎜⎜ 0 ⎟⎟ − ⎜⎜ 4 ⎟⎟ ⎜ 11 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 16 ⎞ ⎜ ⎟ = ⎜ −4 ⎟ ⎜ 12 ⎟ ⎝ ⎠
(e)
(3)
Accept vector forms
B1
(1)
Either order
M1
cao
A1
2
Do not penalise if consistent incorrect signs in (d)
A1
Use of correct right angled triangle AX = cos θ d 4 26 d = 19 ≈ 27.9 awrt 27.9
M1
(2)
l1
X 4 26
θ d
(2)
M1
= 416 = 16 × 26 = 4 26 ¿
A
M1 A1
A1
AX = 162 + ( −4 ) + 122
(f)
(1)
Y
l2
M1 A1
(3)
26
[12]
GCE Core Mathematics C4 (6666) January 2010
6
Question Number
Q5
Scheme
(a)
Marks
⌠ 9 x + 6 dx = ⌠ ⎛ 9 + 6 ⎞ dx ⎮ ⎮⎜ ⎟ ⌡ x x⎠ ⌡⎝ = 9 x + 6 ln x ( +C )
⌠ 1 ⌠ 9 x + 6 dx ⎮ 13 d y = ⎮ ⌡ x ⌡y −1 ⌠ 9x + 6 ∫ y 3 d y = ⎮⌡ x dx
(b)
M1 A1
(2)
Integral signs not necessary B1
2
y3
= 9 x + 6 ln x ( +C )
± ky 3 = their ( a )
3 23 y = 9 x + 6 ln x ( +C ) 2
ft their ( a )
2 3
y = 8, x =1
3 23 8 = 9 + 61n 1 + C 2 C = −3 2 2 y 3 = ( 9 x + 6 ln x − 3) 3
y 2 = ( 6 x + 4 ln x − 2 )
3
2
M1 A1ft
M1 A1
( = 8 (3x + 2 ln x − 1) ) 3
A1
(6) [8]
GCE Core Mathematics C4 (6666) January 2010
7
Question Number
Q6
Scheme
dA = 1.5 dt dA A = π r2 ⇒ = 2π r dr When A = 2 2 2 = π r2 ⇒ r =
π
Marks
B1 B1
( = 0.797 884 ... )
M1
d A d A dr = × dt dr dt dr 1.5 = 2π r dt
M1
dr 1.5 = ≈ 0.299 dt 2π π2
awrt 0.299
A1 [5]
GCE Core Mathematics C4 (6666) January 2010
8
Question Number Q7
Scheme
(a)
Marks
y = 0 ⇒ t ( 9 − t 2 ) = t ( 3 − t )( 3 + t ) = 0 t = 0, 3, − 3
Any one correct value B1
At t = 0 , x = 5 ( 0 ) − 4 = −4 2
Method for finding one value of x
M1
At A, x = −4 ; at B, x = 41
Both
A1
dx = 10t dt
Seen or implied
B1
At t = 3 , x = 5 ( 3) − 4 = 41 2
( At t = −3,
)
x = 5 ( −3) − 4 = 41 2
(b)
⌠ ⌠ dx 2 ⎮ y dx = ⎮ y dt = ∫ t ( 9 − t )10t dt d t ⌡ ⌡
(3)
M1 A1
= ∫ ( 90t 2 − 10t 4 ) dt =
90t 3 10t 5 − 3 5
( = 30t
( +C )
3
⎡ 90t 3 10t 5 ⎤ 3 5 ⎢ 3 − 5 ⎥ = 30 × 3 − 2 × 3 ⎣ ⎦0
A = 2∫ y dx = 648
GCE Core Mathematics C4 (6666) January 2010
( units ) 2
3
− 2t 5 ( +C ) )
( = 324 )
A1 M1
A1
(6) [9]
9
Question Number
Q8
Scheme
(a)
Marks
dx = −2sin u du
B1
⌠ 1 1 ⌠ dx = ⎮ × −2sin u du ⎮ 2 2 2 2 ⌡ x 4− x ⌡ ( 2 cos u ) 4 − ( 2 cos u ) −2sin u ⌠ du =⎮ ⌡ 4 cos 2 u 4sin 2 u 1 1 =− ⌠ du ⎮ 4 ⌡ cos 2 u 1 = − tan u ( +C ) 4
x = 2 ⇒ 2 = 2 cos u ⇒ u = x = 1 ⇒ 1 = 2 cos u ⇒ u =
π
M1
Use of 1 − cos 2 u = sin 2 u
M1
1 ±k ⌠ du ⎮ ⌡ cos 2 u
M1
± k tan u
M1
π 4 M1
3
π
1⎛ π π⎞ ⎡ 1 ⎤4 − tan u ⎢⎣ 4 ⎥⎦ π = − 4 ⎝⎜ tan 4 − tan 3 ⎠⎟ 3
=−
(b)
⌠ V =π⎮ ⎮ ⌡1
2
(
1 1− 3 4
)
⎛ 4 ⎜ 1 ⎜ x ( 4 − x2 ) 4 ⎝
⎛ 3 −1 ⎞ ⎜⎜ = ⎟ 4 ⎟⎠ ⎝
A1
(7)
2
⎞ ⎟ dx ⎟ ⎠
M1
2
1 ⌠ dx = 16π ⎮ ⌡1 x 2 4 − x 2 ⎛ 3 −1 ⎞ = 16π ⎜⎜ ⎟⎟ ⎝ 4 ⎠
16π × integral in (a) M1 16π × their answer to part (a)
A1ft
(3) [10]
GCE Core Mathematics C4 (6666) January 2010
10
GCE Core Mathematics C4 (6666) January 2010
11
Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 01623 467467 Fax 01623 450481 Email publications@linneydirect.com Order Code UA022713 January 2010 For more information on Edexcel qualifications, please visit www.edexcel.com/quals Edexcel Limited. Registered in England and Wales no.4496750 Registered Office: One90 High Holborn, London, WC1V 7BH