Mark Scheme (Results) Summer 2010
GCE
Core Mathematics C4 (6666)
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/
Summer 2010 Publications Code UA023705 All the material in this publication is copyright © Edexcel Ltd 2010
June 2010 6666 Core Mathematics C4 Mark Scheme
Question Number
1.
Scheme
(a)
(b)(i)
(ii)
⎛π ⎞ ⎛π ⎞ y ⎜ ⎟ ≈ 1.2247, y ⎜ ⎟ = 1.1180 ⎝6⎠ ⎝4⎠
Marks
accept awrt 4 d.p.
⎛π ⎞ I ≈ ⎜ ⎟ (1.3229 + 2 ×1.2247 + 1) ⎝ 12 ⎠ ≈ 1.249 ⎛π ⎞ I ≈ ⎜ ⎟ (1.3229 + 2 × (1.2973 + 1.2247 + 1.1180 ) + 1) ⎝ 24 ⎠ ≈ 1.257
GCE Core Mathematics C4 (6666) Summer 2010
B1 for
B1 for
π 12 cao
π 24 cao
B1 B1
(2)
B1 M1
A1 B1 M1 A1
(6) [8]
Question Number
Scheme
2.
∫ sin x e
Marks
du = − sin x dx
B1
dx = − ∫ eu du
M1 A1
cos x +1
= − eu = − ecos x+1
ft sign error
A1ft
π
⎡⎣ − ecos x +1 ⎤⎦ 2 = − e1 − ( − e 2 ) 0
or equivalent with u
M1
= e ( e − 1) ¿
cso
A1
GCE Core Mathematics C4 (6666) Summer 2010
(6) [6]
Question Number
3.
Scheme
Marks
d x 2 ) = ln 2.2 x ( dx dy dy = 2 y + 2x ln 2.2 x + 2 y dx dx
B1
M1 A1= A1
Substituting ( 3, 2 ) dy dy = 4+6 dx dx dy = 4 ln 2 − 2 dx
8ln 2 + 4
M1 Accept exact equivalents
M1 A1
(7) [7]
GCE Core Mathematics C4 (6666) Summer 2010
Question Number
4.
Scheme
(a)
dx dy = 2sin t cos t , = 2sec 2 t dt dt
B1 B1
1 ⎛ ⎞ ⎜= 3 ⎟ ⎝ sin t cos t ⎠
or equivalent M1 A1
dy sec 2 t = dx sin t cos t (b)
Marks
At t =
π 3
, x=
3 , y = 2√3 4
(4)
B1
π
sec2 dy 3 = 16 = dx sin π cos π √ 3 3 3 16 ⎛ 3⎞ y − 2√3 = ⎜x− ⎟ 4⎠ √3⎝ 3 y=0 ⇒ x= 8
M1 A1
M1 M1 A1
(6) [10]
GCE Core Mathematics C4 (6666) Summer 2010
Question Number
5.
Scheme
(a)
Marks
A=2 2 x + 5 x − 10 = A ( x − 1)( x + 2 ) + B ( x + 2 ) + C ( x − 1) x →1 −3 = 3B ⇒ B = −1 x → −2 −12 = −3C ⇒ C = 4
B1
2
(b)
2 x 2 + 5 x − 10 −1 ⎛ x⎞ = 2 + (1 − x ) + 2 ⎜ 1 + ⎟ ( x − 1)( x + 2 ) ⎝ 2⎠
(1 − x )
−1
M1 A1 A1
(4)
−1
M1
= 1 + x + x 2 + ...
B1
−1
x x2 ⎛ x⎞ + = − + + ... 1 1 ⎜ ⎟ 2 4 ⎝ 2⎠ 2 x 2 + 5 x − 10 ⎛ 1⎞ = ( 2 + 1 + 2 ) + (1 − 1) x + ⎜1 + ⎟ x 2 + ... ( x − 1)( x + 2 ) ⎝ 2⎠ ft their A − B + 12 C
= 5 + ...
= ... +
3 2 x + ... 2
0 x stated or implied
B1 M1 A1 ft A1 A1
(7) [11]
GCE Core Mathematics C4 (6666) Summer 2010
Question Number
6.
Scheme
(a)
Marks
f (θ ) = 4 cos 2 θ − 3sin 2 θ
⎛1 1 ⎞ ⎛1 1 ⎞ = 4 ⎜ + cos 2θ ⎟ − 3 ⎜ − cos 2θ ⎟ ⎝2 2 ⎠ ⎝2 2 ⎠ 1 7 = + cos 2θ ¿ 2 2 (b)
∫ θ cos 2θ dθ = 2 θ sin 2θ − 2 ∫ sin 2θ dθ 1
1
1 1 = θ sin 2θ + cos 2θ 2 4 1 2 7 7 ∫ θ f (θ ) dθ = 4 θ + 4 θ sin 2θ + 8 cos 2θ
M1 M1 cso
A1
(3)
M1 A1 A1 M1 A1
π
⎡ ⎣
...
⎡π 2 7⎤ 7 ⎤ 2 = ⎢ + 0 − ⎥ − ⎡⎢0 + 0 + ⎤⎥ ⎦0 8⎦ ⎣ 8⎦ ⎣ 16 2 π 7 = − 16 4
M1 A1
(7) [10]
GCE Core Mathematics C4 (6666) Summer 2010
Question Number
7.
Scheme
(a)
j components
3 + 2λ = 9 ⇒ λ = 3
Leading to
C : ( 5, 9, − 1)
( µ = 1) accept vector forms
JJJG JJJG Choosing correct directions or finding AC and BC
(b)
⎛1⎞ ⎛ 5⎞ ⎜ ⎟⎜ ⎟ ⎜ 2 ⎟ . ⎜ 0 ⎟ = 5 + 2 = √ 6 √ 29 cos ∠ACB ⎜1⎟ ⎜ 2⎟ ⎝ ⎠⎝ ⎠ ∠ACB = 57.95°
(c)
Marks
use of scalar product awrt 57.95°
M1 A1 A1
(3)
M1 M1 A1 A1
(4)
A : ( 2, 3, − 4 ) B : ( −5, 9, − 5 ) ⎛ 3⎞ ⎛ 10 ⎞ JJJG ⎜ ⎟ JJJG ⎜ ⎟ AC = ⎜ 6 ⎟ , BC = ⎜ 0 ⎟ ⎜ 3⎟ ⎜4⎟ ⎝ ⎠ ⎝ ⎠ 2 2 2 2 AC = 3 + 6 + 3 ⇒ AC = 3 √ 6
BC 2 = 102 + 42 ⇒ BC = 2 √ 29 1 ABC = AC × BC sin ∠ACB 2 1 = 3 √ 6 × 2 √ 29sin ∠ACB ≈ 33.5 2
M1 A1 A1
15 √ 5 , awrt 34
M1 A1
(5) [12]
Alternative method for (b) and (c) (b) A : ( 2, 3, − 4 ) B : ( −5, 9, − 5 ) C : ( 5, 9, − 1)
AB 2 = 7 2 + 62 + 12 = 86 AC 2 = 32 + 62 + 32 = 54 BC 2 = 102 + 02 + 42 = 116 Finding all three sides 116 + 54 − 86 cos ∠ACB = ( = 0.530 66 ...) 2 √ 116 √ 54 ∠ACB = 57.95° awrt 57.95° If this method is used some of the working may gain credit in part (c) and appropriate marks may be awarded if there is an attempt at part (c).
GCE Core Mathematics C4 (6666) Summer 2010
M1 M1 A1 A1
(4)
Question Number
8.
Scheme
(a)
Marks
dV = 0.48π − 0.6π h dt dV dh = 9π V = 9π h ⇒ dt dt dh 9π = 0.48π − 0.6π h dt dh Leading to 75 = 4 − 5h ¿ dt
(b)
∫
75 dh = ∫ 1dt 4 − 5h −15ln ( 4 − 5h ) = t ( +C )
M1 A1 B1 M1 cso
A1
separating variables
M1
(5)
M1 A1
−15ln ( 4 − 5h ) = t + C
When t = 0 , h = 0.2
−15ln 3 = C t = 15ln 3 − 15ln ( 4 − 5h )
M1
When h = 0.5 ⎛ 3 ⎞ t = 15ln 3 − 15ln1.5 = 15ln ⎜ ⎟ = 15ln 2 ⎝ 1.5 ⎠
awrt 10.4
M1 A1
Alternative for last 3 marks t = ⎡⎣ −15ln ( 4 − 5h ) ⎤⎦
0.5 0.2
= −15ln1.5 + 15ln 3 ⎛ 3 ⎞ = 15ln ⎜ ⎟ = 15ln 2 ⎝ 1.5 ⎠
GCE Core Mathematics C4 (6666) Summer 2010
M1 M1 awrt 10.4
A1
(6)
Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 01623 467467 Fax 01623 450481 Email
[email protected] Order Code UA023705 Summer 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
