Mark Scheme (Results) January 2011
GCE
GCE Core Mathematics C4 (6666) Paper 1
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January 2011 Publications Code UA026241 All the material in this publication is copyright © Edexcel Ltd 2011
General Instructions for Marking 1. The total number of marks for the paper is 75. 2. The Edexcel Mathematics mark schemes use the following types of marks: •
M marks: method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.
•
A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.
•
B marks are unconditional accuracy marks (independent of M marks)
•
Marks should not be subdivided.
3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. •
bod – benefit of doubt
•
ft – follow through
•
the symbol
•
cao – correct answer only
•
cso - correct solution only. There must be no errors in this part of the question to obtain this mark
•
isw – ignore subsequent working
•
awrt – answers which round to
•
SC: special case
•
oe – or equivalent (and appropriate)
•
dep – dependent
•
indep – independent
•
dp decimal places
•
sf significant figures
•
¿ The answer is printed on the paper
•
will be used for correct ft
The second mark is dependent on gaining the first mark
January 2011 Core Mathematics C4 6666 Mark Scheme Question Number
∫ x sin 2 x dx = −
1.
=
[
π
...
]02
=
Scheme
Marks
∫
M1 A1 A1
x cos 2 x + 2
...
cos 2 x dx 2 sin 2 x + 4
π
M1 M1 A1
4 [6]
2.
At t = 3
dI = −16 ln ( 0.5 ) 0.5t dt dI = −16 ln ( 0.5 ) 0.53 dt = −2 ln 0.5 = ln 4
M1 A1 M1 M1 A1 [5]
GCE Core Mathematics C4 (6666) January 2011
1
Question Number
Scheme
Marks
3. (a)
5 A B = + ( x − 1)( 3x + 2 ) x − 1 3x + 2 5 = A ( 3 x + 2 ) + B ( x − 1)
5 = 5A ⇒ A = 1 5 5 = − B ⇒ B = −3 3
x →1
x→−
(b)
∫
2 3
5 dx = ( x − 1)( 3x + 2 )
∫
M1 A1 A1
(3)
3 ⎞ ⎛ 1 − ⎜ ⎟ dx ⎝ x − 1 3x + 2 ⎠ = ln ( x − 1) − ln ( 3 x + 2 )
( +C )
ft constants
M1 A1ft A1ft (3)
(c)
∫
∫
⎛1⎞ ⎜ ⎟ dy ⎝ y⎠ ln ( x − 1) − ln ( 3x + 2 ) = ln y
5 dx = ( x − 1)( 3x + 2 )
K ( x − 1) 3x + 2 K 8= 8 64 ( x − 1) y= 3x + 2 y=
Using ( 2, 8 )
M1
( +C )
M1 A1
depends on first two Ms in (c) M1 dep depends on first two Ms in (c)
M1 dep A1
(6) [12]
GCE Core Mathematics C4 (6666) January 2011
2
Question Number 4.
Scheme
Marks
(a)
JJJG AB = −2i + 2 j − k − ( i − 3 j + 2k ) = −3i + 5 j − 3k
M1 A1
(2)
(b)
r = i − 3 j + 2k + λ ( −3i + 5 j − 3k )
M1 A1ft
(2)
or (c)
r = −2i + 2 j − k + λ ( −3i + 5 j − 3k )
JJJG AC = 2i + pj − 4k − ( i − 3 j + 2k )
JJJG or CA
= i + ( p + 3) j − 6k ⎛ 1 ⎞ ⎛ −3 ⎞ JJJG JJJG ⎜ ⎟⎜ ⎟ AC. AB = ⎜ p + 3 ⎟ . ⎜ 5 ⎟ = 0 ⎜ −6 ⎟ ⎜ −3 ⎟ ⎝ ⎠⎝ ⎠ −3 + 5 p + 15 + 18 = 0 Leading to p = −6 (d)
AC 2 = ( 2 − 1) + ( −6 + 3) + ( −4 − 2 ) 2
2
2
M1
M1 A1
( = 46 )
AC = √ 46
B1
(4)
M1
accept awrt 6.8
A1 (2) [10]
GCE Core Mathematics C4 (6666) January 2011
3
Question Number
Scheme
Marks
5. (a)
( 2 − 3x )
−2
⎛ 3 ⎞ = 2 ⎜1 − x ⎟ ⎝ 2 ⎠
−2
−2
B1
−2
⎛ 3 ⎞ ⎛ 3 ⎞ −2. − 3 ⎛ 3 ⎞ −2. − 3. − 4 ⎛ 3 ⎞ ⎜ 1 − x ⎟ = 1 + ( −2 ) ⎜ − x ⎟ + ⎜− x⎟ + ⎜ − x ⎟ + ... 1.2 ⎝ 2 ⎠ 1.2.3 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 27 2 27 3 = 1 + 3x + x + x + ... 4 2 1 3 27 27 −2 ( 2 − 3x ) = + x + x 2 + x3 + ... 4 4 16 8 (b)
2
27 2 27 3 ⎛1 3 ⎞ f ( x ) = ( a + bx ) ⎜ + x + x + x + ... ⎟ 16 8 ⎝4 4 ⎠ 3a b Coefficient of x; + =0 ( 3a + b = 0 ) 4 4 27a 3b 9 Coefficient of x 2 ; + = ( 9a + 4b = 3) 16 4 16 Leading to a = −1, b = 3
(c) Coefficient of x 3 is
27a 27b 27 27 + = × ( −1) + × 3 8 16 8 16 27 = 16
3
M1 A1
M1 A1
(5)
M1
A1 either correct
M1 A1 M1 A1
(5)
M1 A1ft
cao
A1
(3) [13]
GCE Core Mathematics C4 (6666) January 2011
4
Question Number
Scheme
Marks
6. (a)
dx 1 = , dt t
dy = 2t dt dy = 2t 2 dx
M1 A1
Using mm′ = −1 , at t = 3 1 18 1 y − 7 = − ( x − ln 3) 18
m′ = −
(b)
x = ln t ⇒ t = e x y =e −2 V =π
∫ (e
2x
M1 A1
− 2 ) dx 2
∫ (e
M1 A1
(6)
B1 2x
(c)
M1 A1
2x
(3)
M1
− 2 ) dx = 2
=
∫ (e
4x
− 4 e 2 x + 4 ) dx
e4 x 4 e2 x − + 4x 4 2
M1 M1 A1
ln 4
⎡ e4 x 4 e2 x ⎤ π⎢ − + 4 x ⎥ = π ⎡⎣( 64 − 32 + 4 ln 4 ) − ( 4 − 8 + 4 ln 2 ) ⎤⎦ 2 ⎣ 4 ⎦ ln 2
M1
= π ( 36 + 4 ln 2 )
A1 (6) [15]
Alternative to (c) using parameters V =π
∫
∫ (t
2 1⎞ ⎛ 2 ⎜ ( t − 2 ) × ⎟ dt = t⎠ ⎝
2
∫
− 2)
2
dx dt dt
4⎞ ⎛ 3 ⎜ t − 4t + ⎟ dt t⎠ ⎝
t4 − 2t 2 + 4 ln t 4 The limits are t = 2 and t = 4 =
M1 M1 M1 A1
4
⎡t4 ⎤ π ⎢ − 2t 2 + 4 ln t ⎥ = π ⎡⎣( 64 − 32 + 4 ln 4 ) − ( 4 − 8 + 4 ln 2 ) ⎤⎦ ⎣4 ⎦2 = π ( 36 + 4 ln 2 )
M1 A1 (6)
GCE Core Mathematics C4 (6666) January 2011
5
Question Number
Scheme
Marks
7.
x = 3 ⇒ y = 0.1847 x = 5 ⇒ y = 0.1667
(a)
awrt B1 awrt or 16 B1 (2)
(b)
(c)
I≈
1 ⎡ 0.2 + 0.1667 + 2 ( 0.1847 + 0.1745 ) ⎤⎦ 2⎣ ≈ 0.543
dx = 2 (u − 4) du
B1 M1 A1ft
0.542 or 0.543
A1
(4)
B1
∫ 4 + √ ( x −1) ∫ ∫ 1
dx =
1 × 2 ( u − 4 ) du u
8⎞ ⎛ ⎜ 2 − ⎟ du u⎠ ⎝ = 2u − 8ln u x = 2 ⇒ u = 5, x = 5 ⇒ u = 6 =
M1 A1 M1 A1 B1
[ 2u − 8ln u ] 5 = (12 − 8ln 6 ) − (10 − 8ln 5)
M1
⎛5⎞ = 2 + 8ln ⎜ ⎟ ⎝6⎠
A1
6
(8) [14]
GCE Core Mathematics C4 (6666) January 2011
6
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[email protected] Order Code UA026241 January 2011 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
