FOR EDEXCEL
GCE Examinations
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 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.
C2 Paper C – Marking Guide
1.
2.
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
(a)
a[1 − ( 13 ) 4 ]
1 − 13
a = 200 ×
3.
135
1− 13
= 200
27
40
= 202 12
(b)
=
(a)
(−4, 0) ∴ 0 = 4 − 20 + 16k + 128
16k = −112, k = −7
M1
A1
(b)
4 + 5x − 7x2 − 2x3 = 0
x = −4 is a solution ∴ (x + 4) is a factor
B1
(a)
M1 A1
+ 5x
+ 4x
x + 4
x + 4
M1 A1
M1
∴ ( − 12 , 0), (1, 0)
A1
(7)
y
x
B2
(−60, −1), (120, 1)
(ii)
(b)
(5)
+ 1
+ 5x + 4
∴ (x + 4)(1 + x − 2x2) = 0
(x + 4)(1 + 2x)(1 − x) = 0
x = −4 (at A), − 12 , 1
(i)
(4)
A1
O
5.
M1 A1
M1 A1
= 135
−2x2 + x
x + 4 −2x3 − 7x2
2
−2x3 − 8x
x2
x2
4.
M1 A1
B2
x − 30 = −180 − 20.5, 20.5
= −200.5, 20.5
x = −170.5, 50.5 (1dp)
B1 M1
M1 A1
2
(8)
(a)
= 3 − log8 8 3
= 3 − 23 = 73
B1 M1 A1
A1
(b)
(22)x − 3(2 × 2x) = 0
(2x)2 − 6(2x) = 0
2x(2x − 6) = 0
2x = 0 (no solutions) or 6
M1
x=
lg 6
lg 2
C2C MARKS page 2
M1
A1
= 2.58 (3sf)
M1 A1
 Solomon Press
(9)
6.
(a)
f ′(x) = −1 + 2x
f ′′(x) = − 23 x
(b)
− 13
A1
−1 + 2x
for TP,
M1 A1
− 43
− 13
=0
M1
1
3
x =2
x=8
7.
∴ (8, 6)
M1
A1
A1
(c)
1
f ′′(8) = − 24
, f ′′(x) < 0 ∴ maximum
M1 A1
(a)
grad PQ =
(b)
(c)
8− 2
−3− ( −5)
= 3, grad QR =
4 −8
9 − ( −3)
= − 13
M1 A1
grad PQ × grad QR = 3 × ( − 13 ) = −1
M1
∴ PQ perp. to QR, ∴ ∠PQR = 90°
A1
∠PQR = 90° ∴ PR is a diameter
∴ centre = mid-point of PR = ( −52+ 9 ,
2+ 4
2
49 + 1 =
radius = dist. (−5, 2) to (2, 3) =
M1
M1 A1
) = (2, 3)
50
B1
∴ (x − 2)2 + (y − 3)2 = ( 50 )2
x2 − 4x + 4 + y2 − 6y + 9 = 50
x2 + y2 − 4x − 6y = 37
[ k = 37 ]
8.
2π
3
(a)
= 12 × (2π −
(b)
chord = 2 × 12 sin
2π
3
P = (12 ×
3
2
= 24 ×
= 12 3
M1
area of segment = ( 12 × 122 ×
as % of area of circle =
(a)
3
2
88.443
π × 122
2π
3
[k=4]
) − ( 12 × 122 × sin
8
∫2
)
M2
M1 A1
(10)
M1 A1
B1 M1 A1
A1
(1 + 3 x ) dx
3
= [x + 2x 2 ] 82
M1 A1
3
(c)
2π
3
× 100% = 19.6% (1dp)
x
2
4
6
8
1+ 3 x
5.243
7
8.348
9.485
area ≈ 12 × 2 × [5.243 + 9.485 + 2(7 + 8.348)]
=
A1
) = 88.443
= 45.4 (3sf)
(b)
(10)
M1 A1
) + 12 3
−
= 72( 2π
3
9.
A1
M1 A1
= 8π + 12 3 = 4(2π + 3 3 ) cm
(c)
M1
) = 16π cm
π
3
(9)
= [8 + 2( 2 2 ) ] − [2 + 2( 2 2 )]
M1
= (8 + 32 2 ) − (2 + 4 2 )
M1
= 6 + 28 2
A1
= (6 + 28 2) − 45.4 × 100% = 0.43%
M1 A1
(13)
Total
(75)
6 + 28 2
 Solomon Press
C2C MARKS page 3
Performance Record – C2 Paper C
Question no.
1
2
Topic(s)
binomial
GP
Marks
4
5
3
4
trig.
factor
theorem, graph,
alg. div. trig. eqn
7
5
6
7
8
9
logs
SP
circle
sector
of a
circle
trapezium
rule,
area by
integr.
9
9
10
10
13
8
Student
C2C MARKS page 4
 Solomon Press
Total
75