Mark Scheme (Results) January 2008
GCE
GCE Mathematics (6665/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
January 2008 6665 Core Mathematics C3 Mark Scheme Question Number
Scheme
Marks
1.
x2 −1
2 x4 2 x4
−1 2x 2 2 − 3x + x + 1 − 2x2 − x2 + x + 1 − x2 +1 x
a = 2 stated or implied c = −1 stated or implied
M1 A1 A1
a = 2, b = 0, c = −1, d = 1, e = 0 d = 1 and b = 0, e = 0 stated or implied
A1
2 x2 − 1 +
x x −1 2
[4] 2.
(a) dy = 2 e 2 x tan x + e 2 x sec 2 x dx
M1 A1+A1
dy = 0 ⇒ 2 e 2 x tan x + e 2 x sec 2 x = 0 dx
M1
2 tan x + 1 + tan 2 x = 0 2 ( tan x + 1) = 0 tan x = −1 ¿
(b)
⎛ dy ⎞ ⎜ ⎟ =1 ⎝ dx ⎠ 0
Equation of tangent at ( 0, 0 ) is y = x
A1 cso
A1
(6)
M1 A1
(2) [8]
Question Number
3.
Scheme
(a)
Marks
f ( 2 ) = 0.38 … f ( 3) = −0.39 …
Change of sign (and continuity) ⇒ root in (b)
(c)
M1
( 2, 3)
cso
¿
x1 = ln 4.5 + 1 ≈ 2.504 08 x2 ≈ 2.50498 x3 ≈ 2.50518
A1
(2)
M1 A1 A1
(3)
Selecting [ 2.5045, 2.5055] , or appropriate tighter range, and evaluating at both ends. f ( 2.5045 ) ≈ 6 × 10−4
M1
f ( 2.5055 ) ≈ −2 × 10−4
Change of sign (and continuity) ⇒ root ∈ ( 2.5045, 2.5055
)
⇒ root = 2.505 to 3 dp ¿ Note: The root, correct to 5 dp, is 2.50524
cso
A1
(2) [7]
Question Number
4.
Scheme
Marks
(a)
( −5, 4 )
y
( 5, 4 )
x
O
Shape ( 5, 4 )
B1 B1
( −5, 4 )
B1
(b) For the purpose of marking this paper, the graph is identical to (a) Shape ( 5, 4 )
B1 B1
( −5, 4 )
B1
(3)
(3)
(c)
y
O
( 4, 8)
x
( −6, − 8) General shape – unchanged B1 Translation to left B1 ( 4, 8 ) B1
( −6, − 8 ) In all parts of this question ignore any drawing outside the domains shown in the diagrams above.
B1
(4) [10]
Question Number
5.
Scheme
Marks
(a) 1000 (b)
(c)
B1
1000 e −5730 c = 500 1 e−5730 c = 2 1 −5730c = ln 2 c = 0.000121
cao
R = 1000 e−22920 c = 62.5
Accept 62-63
(1)
M1
A1 M1 A1
(4)
M1 A1
(2)
(d) R 1000
Shape 1000 O
t
B1 B1
(2) [9]
Question Number
6.
Scheme
Marks
(a) cos ( 2 x + x ) = cos 2 x cos x − sin 2 x sin x
M1
= ( 2 cos 2 x − 1) cos x − ( 2sin x cos x ) sin x
M1
= ( 2 cos 2 x − 1) cos x − 2(1 − cos 2 x) cos x any correct expression
A1
= 4 cos x − 3cos x
A1
3
2 cos x 1 + sin x cos x + (1 + sin x ) (b)(i) + = 1 + sin x cos x (1 + sin x ) cos x
=
(c)
2 (1 + sin x ) (1 + sin x ) cos x
=
2 = 2sec x cos x
sec x = 2 or cos x =
x=
π 5π 3
,
2
M1
cos 2 x + 1 + 2sin x + sin 2 x (1 + sin x ) cos x
=
3
(4)
A1
M1 ¿
cso A1
1 2
(4)
M1 accept awrt 1.05, 5.24
A1, A1
(3) [11]
7.
(a)
dy = 6 cos 2 x − 8sin 2 x dx ⎛ dy ⎞ ⎜ ⎟ =6 ⎝ dx ⎠ 0 1 y−4= − x 6
M1 A1 B1 or equivalent
(b) R = √ ( 32 + 42 ) = 5 4 tan α = , α ≈ 0.927 3
M1 A1
(5)
M1 A1 awrt 0.927
M1 A1
(4)
(c) sin ( 2 x + their α ) = 0
M1
x = −2.03, − 0.46, 1.11, 2.68 First A1 any correct solution; second A1 a second correct solution; third A1 all four correct and to the specified accuracy or better. Ignore the y-coordinate.
A1 A1 A1 (4) [13]
Question Number
8.
Scheme
(a) x = 1 − 2 y
⎛ 1− x ⎞ f :xa⎜ ⎟ ⎝ 2 ⎠ −1
(b) gf ( x ) = =
⎛ 1− x ⎞ y=⎜ ⎟ ⎝ 2 ⎠
⇒
3
1
1
3
or
3
1− x 2
3
3 −4 1 − 2 x3 3 − 4 (1 − 2 x3 )
8 x3 − 1 = 0 1 x= 2
(2)
M1 A1 M1 cso
A1
(4)
Ignore domain Attempting solution of numerator = 0 M1 Correct answer and no additional answers A1
3 2 3 2 d y (1 − 2 x ) × 24 x + ( 8 x − 1) × 6 x (d) = 2 dx (1 − 2 x3 )
=
M1 A1 Ignore domain
1 − 2 x3 8 x3 − 1 = ¿ 1 − 2 x3 8 x3 − 1 gf : x a 1 − 2 x3 (c)
Marks
18 x 2
(1 − 2 x )
3 2
Solving their numerator = 0 and substituting to find y.
x = 0, y = −1
(2)
M1 A1
A1 M1 A1
(5) [13]