Mark Scheme (Results) Summer 2008
GCE
GCE Mathematics (6665/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
June 2008 6665 Core Mathematics C3 Mark Scheme Question Number
1.
Scheme
e2 x+1 = 2 2 x + 1 = ln 2 1 x = ( ln 2 − 1) 2
(a)
dy = 8e 2 x +1 dx
(b) x=
1 ( ln 2 − 1) ⇒ 2
Marks
M1
A1
(2)
B1 dy = 16 dx
1 ⎛ ⎞ y − 8 = 16 ⎜ x − ( ln 2 − 1) ⎟ 2 ⎝ ⎠ y = 16 x + 16 − 8ln 2
B1
M1 A1
(4) [6]
Question Number
2.
Scheme
R 2 = 52 + 122 R = 13 12 tan α = 5 α ≈ 1.176
(a)
(b)
cos ( x − α ) =
Marks
M1 A1 M1 cao A1
6 13
M1
6 = 1.091 … 13 x = 1.091 … + 1.176 … ≈ 2.267 …
x − α = arccos
A1 awrt 2.3
A1
x − α = −1.091 … accept … = 5.19 … for M M1 x = −1.091 … + 1.176 … ≈ 0.0849 … awrt 0.084 or 0.085 A1
(c)(i)
(4)
Rmax = 13
ft their R
B1 ft
awrt 1.2, ft their α
M1 A1ft
(ii) At the maximum, cos ( x − α ) = 1 or x − α = 0 x = α = 1.176 …
(5)
(3) [12]
Question Number
3.
Scheme
Marks
y
(a)
shape Vertices correctly placed O
(2)
shape B1 Vertex and intersections with axes correctly placed B1
(2)
x
y
(b)
x
O
(c)
(d)
B1 B1
P : ( −1, 2 )
B1
Q : ( 0, 1)
B1
R : (1, 0 )
B1
1 x 2 2 Leading to x = 3 1 x < −1 ; 2 + x +1 = x 2 Leading to x = −6
x > −1 ;
2 − x −1 =
(3)
M1 A1 A1 M1 A1
(5) [12]
Question Number
4.
Scheme
Marks
x 2 − 2 x − 3 = ( x − 3)( x + 1)
(a)
⎛ 2 ( x − 1) x +1 − ⎜⎜ or ⎝ ( x − 3)( x + 1) ( x − 3)( x + 1) x−3 1 ¿ = = ( x − 3)( x + 1) x + 1
f ( x) =
2 ( x − 1) − ( x + 1) ( x − 3)( x + 1)
B1
⎛ 1⎞ ⎜ 0, ⎟ ⎝ 4⎠
(b)
(c)
Let y = f ( x )
M1 A1 cso
1 1 Accept 0 < y < , 0 < f ( x ) < etc. 4 4
1 x +1 1 x= y +1 yx + x = 1 1− x y= x 1 − x f −1( x ) = x
A1
(4)
B1 B1
(2)
y=
⎛ 1⎞ Domain of f −1 is ⎜ 0, ⎟ ⎝ 4⎠ (d)
⎞ ⎟⎟ ⎠
1 2x − 3 +1 1 1 = 2 2x − 2 8 x2 = 5 x = ±√5 fg ( x ) =
or
1 −1 x
M1 A1
ft their part (b) B1 ft
(3)
2
M1 both
A1 A1
(3) [12]
Question Number
5.
Scheme
Marks
sin 2 θ + cos 2 θ = 1 sin 2 θ cos 2 θ 1 + = 2 2 sin θ sin θ sin 2 θ 1 + cot 2 θ = cosec2 θ ¿
(a) ÷ sin 2 θ
M1 cso
A1
(2)
Alternative for (a) 1 + cot 2 θ = 1 +
cos 2 θ sin 2 θ + cos 2 θ 1 = = 2 2 sin θ sin θ sin 2 θ = cosec 2 θ ¿
2 ( cosec 2 θ − 1) − 9 cosec θ = 3
(b)
2 cosec θ − 9 cosec θ − 5 = 0 ( 2 cosec θ + 1)( cosec θ − 5 ) = 0 2
or or
cosec θ = 5 or
5sin θ + 9sin θ − 2 = 0 ( 5sin θ − 1)( sin θ + 2 ) = 0
θ = 11.5°, 168.5°
cso
A1
M1
2
sin θ =
M1
1 5
M1 M1 A1 A1 A1
(6) [8]
Question Number
6.
Scheme
(a)(i)
Marks
d 3x e ( sin x + 2 cos x ) ) = 3e3 x ( sin x + 2 cos x ) + e3 x ( cos x − 2sin x ) ( dx
M1 A1 A1 (3)
( = e ( sin x + 7 cos x ) ) 3x
d 3 5 x3 2 (ii) ( x ln ( 5x + 2 ) ) = 3x ln ( 5x + 2 ) + 5x + 2 dx
M1 A1 A1 (3)
2 d y ( x + 1) ( 6 x + 6 ) − 2 ( x + 1) ( 3 x + 6 x − 7 ) = 4 dx ( x + 1) 2
(b)
M1
( x + 1) ( 6 x 2 + 12 x + 6 − 6 x 2 − 12 x + 14 ) = 4 ( x + 1)
=
(c)
20
( x + 1)
3
M1 cso
¿
A1
d2y 60 15 =− =− 4 2 dx 4 ( x + 1)
M1
( x + 1)
M1
4
= 16
x = 1, − 3
A1 A1
both
A1
Note: The simplification in part (b) can be carried out as follows 2 ( x + 1) ( 6 x + 6 ) − 2 ( x + 1) ( 3x 2 + 6 x − 7 )
(6x =
( x + 1)
3
4
+ 18 x 2 + 18 x + 6 ) − ( 6 x 3 + 18 x 2 − 2 x − 14 )
( x + 1) 20 x + 20 20 ( x + 1) 20 = = = 4 4 3 ( x + 1) ( x + 1) ( x + 1) 4
M1 A1
(5)
(3) [14]
Question Number
7.
Scheme
(a)
Marks
f (1.4 ) = −0.568 … < 0 f (1.45 ) = 0.245 … > 0
M1
Change of sign (and continuity) ⇒ α ∈ (1.4, 1.45 ) (b)
3x3 = 2 x + 6 2x +2 x3 = 3 2 2 x2 = + 3 x ⎛ 2 2⎞ x= ⎜ + ⎟ ¿ ⎝ x 3⎠
√
(c)
A1
(2)
M1 A1 cso
x1 = 1.4371 x2 = 1.4347 x3 = 1.4355
(d) Choosing the interval (1.4345, 1.4355 ) or appropriate tighter interval.
A1
(3)
B1 B1 B1
(3)
M1
f (1.4345 ) = −0.01 ... f (1.4355 ) = 0.003 …
M1
Change of sign (and continuity) ⇒ α ∈ (1.4345, 1.4355 ) ⇒ α = 1.435 , correct to 3 decimal places
Note: α = 1.435 304 553 …
¿ cso
A1
(3) [11]