1. No category

FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C1
Paper K
MARKING GUIDE
This guide is intended to be as helpful as possible to teachers by providing
concise solutions and indicating how marks could be awarded. There are
obviously alternative methods that would also gain full marks.
Method marks (M) are awarded for knowing and using a method.
Accuracy marks (A) can only be awarded when a correct method has been used.
(B) marks are independent of method marks.
Written by Shaun Armstrong
 Solomon Press
These sheets may be copied for use solely by the purchaser’s institute.
C1 Paper K – Marking Guide
1.
2.
(22)y + 3 = 23
2y + 6 = 3
y = − 32
=
∫
(3x2 +
= x3 −
3.
EH
AD
=
1+ 5
3− 5
1
2
x−2) dx
EF
AB
=
∴
1+ 5
3− 5
M1 A2
EH
5
×
=
(a)
1+ 5
3− 5
3+ 5
3+ 5
5 (2 +
3
5 3 5+5
= + +
=2+
9−5
5
M2 A1
5)=5+ 2 5
M1 A1
B2
1
x
B2
O
x
3 solutions
2
x − 4x +
1
x
B1
2
= 0 ⇒ x − 4x = −
1
x
and the graphs of y = x2 − 4x and y = −
(a)
1
x
intersect at 3 points
M1
A1
x + k = ± k2 − 4
M1
x = −k ± k − 4
k = 3 ∴ x = −3 ±
= −3 ±
6.
32 − 4
M1
5
A1
AP: a = 77, l = −70
S50 = 50
[77 + (−70)] = 25 × 7 = 175
2
(b)
AP: a = 2, d =
=
(6)
B1
M1 A1
1
2
n
[4 + 12 (n − 1)]
2
1
n[8 + (n − 1)] = 14
4
B2
M1
n(n + 7)
[k=
1
4
]
x − 3y + 7 = 0 ⇒ x = 3y − 7
sub. into x2 + 2xy − y2 = 7
(3y − 7)2 + 2y(3y − 7) − y2 = 7
y2 − 4y + 3 = 0
(y − 1)(y − 3) = 0
y = 1, 3
∴ x = −4, y = 1 or x = 2, y = 3
C1K MARKS page 2
(6)
A1
(a)
Sn =
7.
B1
(x + k)2 − k2 + 4 = 0
(x + k)2 = k2 − 4
2
(b)
(6)
y
y=−
5.
(4)
M1
y = x2 − 4x
(b)
(3)
B1
x−1 + c
1
2
∴ EH =
4.
M1
M1
A1
 Solomon Press
A1
(7)
M1
M1
A1
M1
A1
M1 A1
(7)
8.
(a)
(b)
9.
(a)
(b)
dy
= 1 − 4x−3
dx
B1
d2 y
= 12x−4
dx 2
M1 A1
y=
∫
(1 − 4x−3) dx
y = x + 2x−2 + c
x = −1, y = 0 ∴ 0 = −1 + 2 + c
c = −1
y = x + 2x−2 − 1
when x = 2, y = 2 + 12 − 1 = 32
M1
y=x− 6 x +9
M1 A1
dy
−1
3
= 1 − 3x 2 = 1 −
x
dx
M1 A1
x=4 ∴y=1
grad of tangent = 1 −
grad of normal =
−1
− 12
3
2
M1 A2
M1 A1
= − 12
M1
=2
M1 A1
∴ y − 1 = 2(x − 4)
y = 2x − 7
(c)
at intersect:
(9)
M1
A1
x − 6 x + 9 = 2x − 7
x + 6 x − 16 = 0
M1
( x + 8)( x − 2) = 0
M1
x = −8, 2
A1
x = 2 ⇒ x = 4 (at P)
x = −8 ⇒ no real solutions ∴ normal does not intersect again
10.
A1
(a)
y − 4 = 3(x + 6)
y = 3x + 22
M1
A1
(b)
at B, x = 0 ∴ y = 2 ⇒ B (0, 2)
at C, x − 7(3x + 22) + 14 = 0
x = −7
∴ C (−7, 1)
B1
M1
A1
A1
(c)
grad AB =
2−4
0 − (−6)
grad AC =
1− 4
−7 − (−6)
(d)
= − 13
(13)
M1 A1
=3
grad AB × grad AC = − 13 × 3 = −1
M1
∴ AB perp to AC ∴ ∠BAC = 90°
A1
AB =
(0 + 6)2 + (2 − 4)2 =
36 + 4 =
AC =
(−7 + 6)2 + (1 − 4)2 =
1+ 9 =
area =
1
2
40 = 2 10
M1 A1
10
× 2 10 × 10 = 10
 Solomon Press
M1 A1
(14)
Total
(75)
C1K MARKS page 3
Performance Record – C1 Paper K
Question no.
1
2
3
4
5
6
7
8
9
10
Topic(s)
indices
integr.
surds
curve
sketch
compl.
square
AP
simul.
eqn
diff.,
integr.
diff.,
normal
straight
lines
Marks
3
4
6
6
6
7
7
9
13
14
Student
C1K MARKS page 4
 Solomon Press
Total
75