EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK
SCHEME. 42. Question number. Scheme. Marks. 1. a = 7, d = 2. B1. S20 = 2. 1 ´
20 ´ (2 ...
EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK SCHEME
Question number
Scheme
Marks
a = 7, d = 2
1.
S20 =
1 2
B1
´ 20 ´ (2 ´ 7 + 19 ´ 2) = 520
M1 A1 (3 marks)
2
ó (5 x + 3 Ö x) dx = 5 x + 2 x 32 + C ô 2 õ
2.
M1 A1 A1 B1 (4 marks)
3.
(a)
Ö80 = 4Ö5
(b)
(4 - Ö5) = 16 - 8Ö5 + 5 = 21 - 8Ö5
B1
2
(1)
M1 A1 A1 (3) (4 marks)
Gradient of AB =
4.
Gradient of l = y-4=
4 - (-6) æ 5ö ç= - ÷ 3-7 è 2ø
M1 A1
2 5 2 (x - 3) 5
M1 2x - 5y + 14 = 0
M1 A1
(5)
(5 marks) 5.
(a)
y
O (b)
x
y
O
Position, Shape
B1
(0, 2), (2, 0)
B1 B1
Position, Shape
B1
æ1 ö æ3 ö (0, 1), ç , 2 ÷, ç , 0 ÷ è2 ø è2 ø
B2 (1, 0)
(3)
(3)
x (6 marks)
42
EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK SCHEME
Question number 6.
Scheme
(a)
Marks
5 - 2x = 2x2 - 3x - 16
2x2 - x - 21 = 0
(2x - 7)(x + 3) = 0
x = -3, x =
7 2
M1 A1
y = 11, y = -2 (b)
M1 A1
M1 A1ft
Using critical values x = -3,
x=
7 2
M1
x < -3,
x>
7 2
M1 A1ft
(6)
(3)
(9 marks) 7.
(a)
a + (n - 1)d = 250 + (10 ´ 50) = £750
M1 A1
(b)
1 1 n [2a + (n - 1) d ] = ´ 20 ´ (500 + 19 ´ 50), 2 2
(c)
B:
1 ´ 20 ´ (2A + 19 ´ 60) 2
= £14500
[= 10(2A + 1140)], = “14500”
Solve for A: A = 155
(2)
M1 A1, A1 (3) B1, M1 M1 A1
(4)
(9 marks) 8.
(a) (b)
(x + 5)2 - 25 + 36
a = 5, b2 - 4ac = 100 - 144,
< 0, therefore no real
(2)
k =25
M1 A1
(2)
Shape, position
B1 B1
(-5, 0) (0, 25)
B1 B1ft
4k = 100
y
O
B1, M1 A1 (3) M1 A1
roots
(c) Equal roots if b2 - 4ac = 0 (d)
b = 11
(4)
x (11 marks)
43
EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK SCHEME
Question number 9.
(a)
Scheme
Marks
f(x)= x3 - 4x2 + 6x + C
M1 A1
5 = 27 - 36 + 18 + C
C=-4
y = 8 - 16 + 12 - 4 = 0
M1 A1
(4)
M1 A1
(2)
(b)
x = 2:
(c)
f ¢(3) = 27 - 24 + 6 = 9, Parallel therefore equal gradient
B1, M1
3x2 - 8x + 6 = 9
M1
3x2 - 8x - 3 = 0
(3x + 1)(x - 3) = 0
Q: x = -
1 3
M1 A1
(5)
(11 marks) 10.
(a)
dy = 3x2 - 5 - 2x-2 dx
M1 A2(1,0) 2 dy =3´1-5dx 1
At both A and B, (b)
Gradient of normal = y - (-2) =
(c)
(= -4)
1 4
(5)
M1 A1ft
1 (x - 1) 4
4y = x - 9
M1 A1 y= -
Normal at A meets y-axis where x = 0: Similarly for normal at B:
M1 A1
4y = x + 9
Length of PQ =
y=
9 9 9 + = 4 4 2
9 4
9 4
(4)
B1 M1 A1 A1
(4) (13 marks)
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