Mark Scheme (Results)
January 2009
GCE
GCE Mathematics (6665/01)
Edexcel Limited. Registered in England and Wales No. 4496750
Registered Office: One90 High Holborn, London WC1V 7BH
January 2009
6665 Core Mathematics C3
Mark Scheme
Question
Number
1
Scheme
Marks
)
(
1
d
d
√ ( 5 x − 1) ) =
( 5 x − 1) 2
(
dx
dx
1
−1
= 5 × ( 5 x − 1) 2
2
(a)
M1 A1
dy
5
−1
= 2 x √ ( 5 x − 1) + x 2 ( 5 x − 1) 2
dx
2
At x = 2 ,
(b)
dy
10
10
= 4√9 +
= 12 +
dx
3
√9
46
=
3
M1 A1ft
M1
Accept awrt 15.3
d ⎛ sin 2 x ⎞ 2 x 2 cos 2 x − 2 x sin 2 x
⎜
⎟=
x4
dx ⎝ x 2 ⎠
A1
M1
(6)
A1+A1
A1
(4)
[10]
Alternative to (b)
d
sin 2 x × x −2 ) = 2 cos 2 x × x −2 + sin 2 x × ( −2 ) x −3
(
dx
⎛ 2 cos 2 x 2sin 2 x ⎞
= 2 x −2 cos 2 x − 2 x −3 sin 2 x ⎜ =
−
⎟
x2
x3 ⎠
⎝
6665/01 GCE Mathematics January 2009
2
M1 A1 + A1
A1
(4)
Question
Number
2
Scheme
2x + 2
x +1
2x + 2
x +1
−
=
−
x − 2 x − 3 x − 3 ( x − 3)( x + 1) x − 3
(a)
2
2 x + 2 − ( x + 1)( x + 1)
( x − 3)( x + 1)
=
(b)
Marks
=
( x + 1)(1 − x )
( x − 3)( x + 1)
=
1− x
x−3
M1 A1
M1
Accept −
x −1 x −1
,
x −3 3− x
d ⎛ 1 − x ⎞ ( x − 3)( −1) − (1 − x )1
⎜
⎟=
2
dx ⎝ x − 3 ⎠
( x − 3)
=
− x + 3 −1 + x
( x − 3)
2
=
A1
(4)
M1 A1
2
( x − 3)
2
cso
A1
(3)
[7]
Alternative to (a)
2 ( x + 1)
2x + 2
2
=
=
2
x − 2 x − 3 ( x − 3)( x + 1) x − 3
M1 A1
x + 1 2 − ( x + 1)
2
−
=
x −3 x −3
x−3
1− x
=
x−3
Alternatives to (b)
1− x
2
−1
1
= −1 −
= −1 − 2 ( x − 3 )
f ( x) =
x −3
x−3
−2
f ′ ( x ) = ( −1)( −2 )( x − 3)
=
2
2
( x − 3)
−1
f ( x ) = (1 − x )( x − 3)
−1
−2
f ′ ( x ) = ( −1)( x − 3) + (1 − x )( −1)( x − 3)
− ( x − 3) − (1 − x )
1
1− x
=−
−
=
2
2
x − 3 ( x − 3)
( x − 3)
=
2
2
( x − 3)
6665/01 GCE Mathematics January 2009
M1
A1
M1 A1
cso
A1
3
(3)
M1
A1
A1
2
(4)
(3)
Question
Number
Scheme
Marks
3
(a)
( 3, 6 )
y
( 7, 0 )
O
Shape
( 3, 6 )
B1
( 7, 0 )
B1
Shape
( 3, 5 )
B1
( 7, 2 )
B1
B1
(3)
x
(b)
y
( 3, 5)
( 7, 2 )
6665/01 GCE Mathematics January 2009
4
(3)
[6]
x
O
B1
Question
Number
Scheme
Marks
x = cos ( 2 y + π )
4
dx
= −2sin ( 2 y + π )
dy
dy
1
=−
dx
2sin ( 2 y + π )
At y =
π
4
dy
1
1
=−
=
3π 2
dx
2sin
2
,
y−
π
1
x
4 2
1
π
y = x+
2
4
=
M1 A1
dx
dy
before or after substitution
Follow through their
A1ft
B1
M1
A1
(6)
[6]
6665/01 GCE Mathematics January 2009
5
Question
Number
5
Scheme
Marks
g ( x) ≥ 1
(a)
B1
( ) = 3e
f g ( x ) = f ex
(b)
2
= x 2 + 3e x
( fg : x a x
2
x2
+ lne x
2
+ 3e x
2
)
(c)
fg ( x ) ≥ 3
(d)
2
2
d 2
x + 3e x = 2 x + 6 x e x
dx
(
)
2
2
A1
(2)
B1
(1)
M1 A1
M1
A1
6 x − x2 = 0
x = 0, 6
6665/01 GCE Mathematics January 2009
M1
2
2x + 6x ex = x2 ex + 2x
2
ex ( 6 x − x2 ) = 0
ex ≠ 0 ,
2
(1)
A1 A1
6
(6)
[10]
Question
Number
6
(a)(i)
Scheme
Marks
sin 3θ = sin ( 2θ + θ )
= sin 2θ cos θ + cos 2θ sin θ
= 2sin θ cos θ .cos θ + (1 − 2sin 2 θ ) sin θ
M1 A1
= 2sin θ (1 − sin 2 θ ) + sin θ − 2sin 3 θ
M1
= 3sin θ − 4sin 3 θ
(ii)
(b)
cso
8sin 3 θ − 6sin θ + 1 = 0
−2sin 3θ + 1 = 0
1
sin 3θ =
2
π 5π
3θ = ,
6 6
π 5π
θ= ,
18 18
√3
(4)
M1 A1
M1
A1 A1
sin15° = sin ( 60° − 45° ) = sin 60° cos 45° − cos 60° sin 45°
1 1 1
− ×
2 √2 2 √2
1
1
1
= √ 6 − √ 2 = (√ 6 − √ 2)
4
4
4
=
A1
(5)
M1
×
M1 A1
cso
A1
(4)
[13]
Alternatives to (b)
1 sin15° = sin ( 45° − 30° ) = sin 45° cos 30° − cos 45° sin 30°
1 √3 1 1
×
−
×
√2 2
√2 2
1
1
1
= √ 6 − √ 2 = (√ 6 − √ 2)
4
4
4
M1
=
M1 A1
cso
A1
(4)
2 Using cos 2θ = 1 − 2sin 2 θ , cos 30° = 1 − 2sin 2 15°
2sin 2 15° = 1 − cos 30° = 1 −
sin 2 15° =
2−√3
4
√3
2
M1 A1
2
1
2−√3
⎛1
⎞
⎜ ( √ 6 − √ 2 ) ⎟ = ( 6 + 2 − 2 √ 12 ) =
4
⎝4
⎠ 16
1
Hence
sin15° = ( √ 6 − √ 2 )
4
6665/01 GCE Mathematics January 2009
7
M1
cso
A1
(4)
Question
Number
7
(a)
Scheme
f ′ ( x ) = 3e x + 3 x e x
M1 A1
3e x + 3 x e x = 3e x (1 + x ) = 0
x = −1
f ( −1) = −3e −1 − 1
M1 A1
B1
x1 = 0.2596
x2 = 0.2571
x3 = 0.2578
B1
Choosing ( 0.257 55, 0.257 65 ) or an appropriate tighter interval.
M1
(b)
(c)
Marks
(5)
B1
B1
(3)
f ( 0.257 55 ) = −0.000 379 ...
f ( 0.257 65 ) = 0.000 109 …
Change of sign (and continuity) ⇒ root ∈ ( 0.257 55, 0.257 65 )
( ⇒ x = 0.2576 , is correct to 4 decimal places)
Note: x = 0.257 627 65 … is accurate
6665/01 GCE Mathematics January 2009
8
A1
cso
A1
(3)
[11]
Question
Number
8
(a)
(b)
Scheme
R 2 = 32 + 42
R=5
4
tan α =
3
α = 53 ... °
Maximum value is 5
Marks
M1
A1
M1
awrt 53°
A1
ft their R
B1 ft
At the maximum, cos (θ − α ) = 1 or θ − α = 0
θ = α = 53 ... °
(c)
(d)
(4)
M1
ft their α
A1 ft
(3)
f ( t ) = 10 + 5cos (15t − α ) °
Minimum occurs when cos (15t − α ) ° = −1
M1
The minimum temperature is (10 − 5 ) ° = 5°
A1 ft
15t − α = 180
t = 15.5
6665/01 GCE Mathematics January 2009
awrt 15.5
9
M1
M1 A1
(2)
(3)
[12]