1. No category

FOR OCR
GCE Examinations
Advanced / Advanced Subsidiary
Core Mathematics C2
Paper L
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 L – Marking Guide
1.
(i)
y
x
2
y = tan
B2
y = sin 2x
O
(ii)
2.
x
4 solutions
the graphs intersect at 4 points
area of segment = ( 12 × r2 ×
1 2
rπ
6
=
shaded area =
3.
4.
B2
1
6
π
3
) − ( 12 × r2 × sin
1 2
r
4
−
2
r π − 2( 16 r π −
1 2
rπ
3
=
−
=
1 2
r
2
3 −
+
1 2
rπ
6
π
3
)
3
2
1 2
rπ
6
B1
B1
2
r
1 2
r
2
3)
M1
3
1 2
r (3
6
3 − π)
A1
(i)
u 2 = k2 − 1
u3 = (k2 − 1)2 − 1 = k4 − 2k2
B1
M1 A1
(ii)
k4 − 2k2 + k2 − 1 = 11
k4 − k2 − 12 = 0
(k2 + 3)(k2 − 4) = 0
k2 = −3 (no solutions) or 4
k=±2
M1
M1
A1
A1
(i)
x
0
0.5
1
1.5
2
1
1
0.8
0.5
0.3077
0.2
x2 + 1
area ≈
1
2
× 0.5 × [1 + 0.2 + 2(0.8 + 0.5 + 0.3077)]
= 1.10 (3sf)
(ii)
5.
B1 M2
A1
1
4
=
(i)
(ii)
B1 M1
M1
A1
loga 27 − loga 8 = 3
=3
loga 27
8
M1
, a=
3 27
8
=
3
2
3lg 2 + lg 6
lg 6 − lg 2
C2L MARKS page 2
(7)
M1 A1
(x + 3) lg 2 = (x − 1) lg 6
x(lg 6 − lg 2) = 3 lg 2 + lg 6
x=
(7)
M1 A1
area = 8 × 1.10385 = 70.6464
volume = 2 × 70.6464 = 141 cm3 (3sf)
27
8
(6)
A1
2
a3 =
(6)
M1
M1
= 3.52
M1 A1
 Solomon Press
(7)
6.
(i)
= [2x + x−1] 42
= (8 +
(ii)
y=
y=
∫
1
4
M1 A1
) − (4 +
1
2
3 34
)=
M1 A1
(2x3 + 1) dx
x4 + x + c
1
2
M1 A1
x = 0, y = 3 ∴ c = 3
y = 12 x4 + x + 3
when x = 2, y = 8 + 2 + 3 = 13
7.
(i)
1 − 8x3
x2
=0 ⇒
M1 A1
1 − 8x3 = 0
M1
x3 =
M1
x=
(ii)
B1
1
8
1
2
A1
−2
f(x) = x − 8x
−2
∫ f(x) dx = ∫ (x − 8x) dx
= −x−1 − 4x2 + c
(iii)
8.
(i)
(iii)
M1 A2
= −[−x−1 − 4x2] 21
M1
= −{( − 12 − 16) − (−2 − 1)} = 13 12
M1 A1
2
S6 =
6
2
[3000 + (5 × −x)] = 8100
3000 − 5x = 2700,
(ii)
M1 A1
= 1500 − (7 × 60) = 1500 − 420 = £1080
Sn =
n
2
9.
(i)
(ii)
M1 A1
[3000 − 60(n − 1)]
M1
[ k = 30 ]
M1 A1
the value of sales in a month would become negative
which is not possible
B1
f(2) = 16 − 20 + 2 + 2 = 0 ∴ (x − 2) is a factor
M1 A1
2x2 − x
3
2
x − 2 2x − 5x
3
2x − 4x2
2
− x
2
− x
(10)
1
x + 2
−
+
+ x
+ 2x
− x + 2
− x + 2
M1 A1
f(x) = (x − 2)(2x2 − x − 1)
f(x) = (x − 2)(2x + 1)(x − 1)
M1 A1
(iii)
x = − 12 , 1, 2
B1
(iv)
sin θ = 2 (no solutions), − 12 or 1
θ =π+
θ =
π
2
,
π
6
7π
6
, 2π −
,
π
6
or
(9)
M1 A1
x = 60
= n[1500 − 30(n − 1)]
= 30n[50 − (n − 1)] = 30n(51 − n)
(iv)
(9)
π
2
M1 B1
11π
6
 Solomon Press
A2
(11)
Total
(72)
C2L MARKS page 3
Performance Record – C2 Paper L
Question no.
1
2
Topic(s)
trig.
graphs
sector
of a
circle
Marks
6
6
3
4
sequence trapezium
rule
7
5
6
7
8
9
logs
integr.
area by
integr.
AP
factor
theorem,
alg. div.,
trig. eqn
7
9
9
10
11
7
Student
C2L MARKS page 4
 Solomon Press
Total
72