1. No category

FOR OCR
GCE Examinations
Advanced / Advanced Subsidiary
Core Mathematics C2
Paper C
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 C – Marking Guide
1.
tan2 θ =
1
3
1
3
tan θ = ±
θ =
2.
π
6
π
6
,
M1
A1
π
6
− π or π −
π
6
π
6
, − π6
B1 M1
θ = −
5π
6
(i)
= log2 (32 × 5)
= 2 log2 3 + log2 5 = 2p + q
(ii)
= log2
, − ,
5π
6
,
3
5× 2
A1
B1
M1 A1
= log2 3 − log2 5 − log2 2
=p−q−1
3.
(i)
=1+
(ii)
4.
(i)
(ii)
1
4
1
4
nx +
1
32
n(n − 1)
2
1
32
(i)
( 14 x)2 + ...
A1
n(n − 1)
M1
8n = n(n − 1)
n[8 − (n − 1)] = 0
n≠0 ∴n=9
M1
A1
n=
7 − 2x − 3x2 =
M1
A1
x = −2 is a solution ∴ (x + 2) is a factor
B1
+ 1
− 7x + 2
− 7x
− 8x
x + 2
x + 2
M1
A1
∴ ( 13 , 6), (1, 2)
A1
(−
4
x3
f(x) = 2x + c
(−1, 3) ∴ 3 = 2 + c
c=1
−2
f(x) = 2x + 1
=
4
∫1
M1 A1
M1
A1
(2x−2 + 1) dx
= [−2x−1 + x] 14
=
( − 12
C2C MARKS page 2
(8)
) dx
−2
(ii)
M1 A1
∴ (x + 2)(3x2 − 4x + 1) = 0
(x + 2)(3x − 1)(x − 1) = 0
x = −2 (at P), 13 , 1
∫
(6)
2
x
7x − 2x2 − 3x3 = 2
3x3 + 2x2 − 7x + 2 = 0
f(x) =
(6)
B1 M1
n(n − 1)x2 + ...
3x2 − 4x
x + 2 3x3 + 2x2
3x3 + 6x2
2
− 4x
2
− 4x
5.
M1
B1 A1
= 1 + n( 14 x) +
(5)
+ 4) − (−2 + 1) =
M1 A1
4 12
M1 A1
 Solomon Press
(8)
6.
(i)
sin A
8
sin1.7
14
=
4
7
sin A =
(ii)
7.
(a)
(b)
M1
sin 1.7
∠BAC = 0.6025
∠ACB = π − (1.7 + 0.6025) = 0.839 (3sf)
A1
M1 A1
AB2 = 82 + 142 − (2 × 8 × 14 × cos 0.8391)
AB = 10.50
P = 10.50 + (14 − 8) + (8 × 0.8391) = 23.2 cm (3sf)
M1
A1
M1 A1
= 31 × 3x = 3y
= 3−1 × (3x)2 =
(i)
(ii)
1
3
3y −
1
3
M1 A1
M1 A1
y2
y2 = 6
y2 − 9y + 18 = 0
(y − 3)(y − 6) = 0
y = 3, 6
3x = 3, 6
M1
A1
lg 6
lg 3
x = 1,
B1 M1
x = 1, 1.63 (3sf)
8.
(i)
3
∫1
A1
(x2 − 2x + k) dx = [ 13 x3 − x2 + kx] 13
= (9 − 9 + 3k) − ( 13 − 1 + k)
= 2k +
2
3
∴ 2k +
= lim [ −4x
− 32
k →∞
] k2
M2 A1
k →∞
M1
2 2
k2
= lim ( 2 −
4
)=
3
2
A1
r =
6
−48
=
3
a=
−48
− 12
=
(iii)
Sn =
M1
A1
= 96
96
1 − (− 12 )
(ii)
B1
− 18
− 18 = − 12
r=
A1
= 64
96[1 − (− 12 )n ]
1 − (− 12 )
(11)
k2
ar = −48, ar4 = 6
3
M1
M1 A1
k →∞
(i)
M1 A2
= 8 23
4
4
= lim {− 3 − (−
)}
9.
(9)
2
3
k=4
(ii)
(8)
M1 A1
= 64[1 − ( − 12 )n]
S∞ − Sn = 64 − 64[1 − ( − 12 )n]
M1
= 64( − 12 )n = 26 × (−1)n × 2−n = (−1)n × 26 − n
difference is magnitude, ∴ = 2
M1 A1
6−n
 Solomon Press
M1
A1
(11)
Total
(72)
C2C MARKS page 3
Performance Record – C2 Paper C
Question no.
1
2
3
4
5
6
7
8
9
Topic(s)
trig. eqn
logs
binomial
factor
theorem,
alg. div.
area by
integr.
sine rule,
sector
of a
circle
logs
integr.
GP
Marks
5
6
6
8
8
8
9
11
11
Student
C2C MARKS page 4
 Solomon Press
Total
72