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]