FOR OCR
GCE Examinations
Advanced / Advanced Subsidiary
Core Mathematics C2
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 using a valid 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.
C2 Paper K – Marking Guide
1.
4x + 3
=2
x −1
4x + 3
= 52 = 25
x −1
log5
M1
M1
4x + 3 = 25(x − 1)
21x = 28
x = 43
2.
3.
A1
6
(1 − x)6 = 1 + 6(−x) +   (−x)2 + ... = 1 − 6x + 15x2
2
6
(1 + x)(1 − x) = (1 + x)(1 − 6x + 15x2 + ...)
coeff. of x2 = 15 − 6 = 9
M1 A1
(i)
AP: a = 77, l = −70
S50 = 50
[77 + (−70)] = 25 × 7 = 175
2
B1
M1 A1
(ii)
AP: a = 2, d =
Sn =
=
4.
M1
(i)
M1 A2
1
2
(5)
B2
n
[4 + 12 (n − 1)]
2
1
n[8 + (n − 1)] = 14
4
cos2 P = 1 − ( 23 )2 =
M1
n(n + 7)
[k=
1
4
]
5
9
acute ∴ cos ∠QPR =
(ii)
(4)
A1
(7)
M1
5
9
=
1
3
5
QR2 = 72 + ( 3 5 )2 − (2 × 7 × 3 5 ×
A1
1
3
5)
M1
2
QR = 49 + 45 − 70 = 24
QR = 24 = 4 × 6 = 2 6
(iii)
sin Q
3 5
=
2
3
M1
2 6
sin Q =
5.
M1 A1
5
6
∠PQR = 65.9° (1dp)
A1
(i)
= 4x2 + x−2 + c
M1 A2
(ii)
(1, 1) ∴ 1 = 4 + 1 + c
c = −4
y = 4x2 + x−2 − 4
M1
A1
y = (2x −
C2K MARKS page 2
1 2
)
x
[ a = 2, b = −1 ]
M1 A1
 Solomon Press
(7)
(7)
6.
(i)
−27 + 63 − 3p − 6 = 0
p = 10
M1
A1
(ii)
remainder = f(2) = 8 + 28 + 20 − 6 = 50
M1 A1
(iii)
x = −3 is a solution ∴ (x + 3) is a factor
B1
x2 + 4x
x3 + 7x2
x3 + 3x2
4x2
4x2
x+3
− 2
+ 10x − 6
+
+
−
−
10x
12x
2x − 6
2x − 6
M1 A1
∴ (x + 3)(x2 + 4x − 2) = 0
x = −3 or x2 + 4x − 2 = 0
other solutions: x = −4 ± 16 + 8 = −4.45, 0.45
2
7.
=
log3 42
log3 4
=
2 log3 4
log3 4
=2
M2 A1
=
log3 22
2
=
2 log3 2
2
= log3 2
M1 A1
log3 16
log3 4
(i)
r=
(ii)
ar = log3 4
a=
(iii)
S6 =
log3 4
2
(26 − 1) log3 2
2 −1
= 63 log3 2
lg 2
lg 3
∴ S6 = 63 ×
A1
= 39.7 (3sf)
A1
(10)
(i)
2x = 1.2490, π + 1.2490 = 1.2490, 4.3906
x = 0.62, 2.20 (2dp)
B1 M1
M1 A1
(ii)
2 sin y cos y = sin y
sin y (2 cos y − 1) = 0
sin y = 0 or cos y = 12
M1
M1
A1
π
3
y = 0, π or
(i)
(ii)
π
3
, 2π −
π
3
B2 M1
5π
3
A1
x = 4 ∴ y = 12 − 8 + 2 = 6
dy
−1
= 3 − 2x 2
dx
grad = 3 − 1 = 2
∴ y − 6 = 2(x − 4)
y − 6 = 2x − 8
y = 2x − 2
B1
y = 0,
9.
M1
lg 2
lg 3
y=
, π,
4
area =
∫0
=
∫0
4
(11)
M1 A1
M1
M1
A1
[(3x − 4 x + 2) − (2x − 2)] dx
M1
(x − 4 x + 4) dx
= [ 12 x2 −
= (8 −
(9)
M1 A1
y
let y = log3 2 ∴ 3 = 2
8.
M1 A1
64
3
8
3
3
x 2 + 4x] 04
+ 16) − (0) =
M1 A2
8
3
 Solomon Press
M1 A1
(12)
Total
(72)
C2K MARKS page 3
Performance Record – C2 Paper K
Question no.
1
2
3
4
5
6
7
8
9
Topic(s)
logs
binomial
AP
cosine
rule,
sine rule
integr.
remain.
theorem,
alg. div.
GP,
logs
trig. eqn
area by
integr.
Marks
4
5
7
7
7
9
10
11
12
Student
C2K MARKS page 4
 Solomon Press
Total
72