EDEXCEL CORE MATHEMATICS C1 (6663) – JUNE 2008
Question
Number
PROVISIONAL MARK SCHEME
Scheme
5
2 x + x3 + c
3
1.
Marks
M1 A1 A1 (3)
(3 marks)
x( x 2 − 9) or ( x ± 0)( x 2 − 9) or ( x − 3)( x 2 + 3 x) or ( x + 3)( x 2 − 3 x)
2.
x( x − 3)( x + 3)
B1
M1 A1 (3)
(3 marks)
3.
(a)
y
10
(7, 3)
x
B1 B1 B1 (3)
y
7
x
B1 B1 (2)
(3.5, 0)
(5 marks)
4.
(a)
f ′( x) = 3 + 3x 2
(b)
3 + 3x 2 = 15
M1 A1 (2)
and start to try and simplify
x2 = k → x = k
x = 2 (ignore x = −2)
(ignore + )
M1
M1
A1 (3)
(8 marks)
EDEXCEL CORE MATHEMATICS C1 (6663) – JUNE 2008
Question
Number
5.
(a)
Scheme
[ x2
Marks
=]a −3
B1 (1)
(b) [ x3 = ] ax2 − 3 or
= a ( a − 3) − 3 ,5 =
(c)
PROVISIONAL MARK SCHEME
a (a − 3) − 3
B1
a 2 − 3a − 3 (*)
A1 cso (2)
a 2 − 3a − 3 = 7
a 2 − 3a − 10 = 0 or a 2 − 3a = 10
M1
(a − 5)(a + 2) = 0
M1
a = 5 or − 2
A1 (3)
(6 marks)
6.
(a)
y
5
B1
–2.5
M1
x
A1 (3)
(b)
2x + 5 =
3
x
M1
2 x 2 + 5 x − 3 [= 0]
or
2x2 + 5x = 3
A1
(2 x − 1)( x + 3) [= 0]
M1
x = −3 or
A1
y=
1
2
3
or 2 × (−3) + 5
−3
or y =
Points are ( −3, −1) and ( 12 , 6)
3
1
2
or 2 × ( 12 ) + 5
M1
(correct pairings)
A1 ft (6)
(10 marks)
EDEXCEL CORE MATHEMATICS C1 (6663) – JUNE 2008
Question
Number
7.
(a)
PROVISIONAL MARK SCHEME
Scheme
Marks
5, 7, 9, 11 or 5 + 2 + 2 + 2 = 11 or 5 + 6 = 11
use a = 5, d = 2, n = 4 and t4 = 5 + 3 × 2 = 11
(b)
tn = a + (n − 1)d with one of a = 5 or d = 2 correct
= 5 + 2(n − 1) or 2n + 3
(c)
B1 (1)
Sn =
or
M1
1 + 2(n + 1)
A1 (2)
n
n
[ 2 × 5 + 2(n − 1)] or use of ( 5 + "their 2n + 3")
2
2
= {n(5 + n – 1)} = n(n + 4)
M1 A1
A1 cso (3)
(*)
(d) 43 = 2n + 3
M1
[n] = 20
(e)
A1 (2)
S 20 = 20 × 24 , = 480 (km)
M1 A1 (2)
(12 marks)
8.
(a)
[No real roots implies b 2 − 4ac < 0 .] b 2 − 4ac = q 2 − 4 × 2q × ( −1)
So q 2 − 4 × 2q × ( −1) < 0 i.e. q 2 + 8q < 0
(b) q(q + 8) = 0 or
M1
A1 cso (2)
(*)
( q ± 4 )2 ± 16 = 0
M1
(q) = 0 or −8
(2 cvs)
−8 < q < 0 or q ∈ (–8, 0) or
q < 0 and q > –8
A1
A1 ft (3)
(5 marks)
9.
(a)
 dy
 dx

=  3kx 2 − 2 x + 1

M1 A1 (2)
7
2
B1
(b) Gradient of line is
When x = − 12 : 3k × ( 14 ) − 2 × ( − 12 ) + 1, =
(c)
7
2
M1
3k 3
= ⇒k =2
4 2
A1 A1 (4)
x = − 12 ⇒ y = k × ( − 18 ) − ( 14 ) − 12 − 5, = −6
M1 A1 (2)
(8 marks)
EDEXCEL CORE MATHEMATICS C1 (6663) – JUNE 2008
Question
Number
10.
(a)
PROVISIONAL MARK SCHEME
Scheme
Marks
( 7 − 1)2 + ( 0 − 3)2
QR =
= 36 + 9 or
M1
A1
45
= 3 5 or a = 3
A1 (3)
(b) Gradient of QR (or l1 ) =
Gradient of l2 is −
(d)
M1 A1
1
or 2
− 12
M1
y − 3 = 2 ( x − 1) or
Equation for l2 is:
(c) P is (0, 1)
3− 0
3
1
or
,= −
1− 7
−6
2
y −3
x −1
= 2 [or y = 2x + 1]
(allow “x = 0, y = 1” but it must be clearly identifiable as P)
(1 − xP )2 + ( 3 − yP )2
PQ =
1 QR × PQ
2
B1 (1)
M1
PQ = 12 + 22 = 5
Area of triangle is
M1 A1 ft (5)
A1
= 12 3 5 × 5, =
15
or 7.5
2
M1 A1 (4)
(13 marks)
11.
(a)
(x
2
(x
2
+3
+3
x
y=
)
2
2
)
= x 4 + 3 x 2 + 3 x 2 + 32
2
=
x4 + 6x2 + 9
= x 2 + 6 + 9 x −2
2
x
x3
9
+ 6 x + x −1 ( +c )
3
−1
20 =
M1
27
9
+ 6×3 − + c
3
3
c = −4
x3
[ y =] + 6 x − 9 x −1 − 4
3
(*)
A1 cso (2)
M1 A1 A1
M1
A1
A1 ft (6)
(8 marks)