1. No category

FOR OCR
GCE Examinations
Advanced / Advanced Subsidiary
Core Mathematics C3
Paper B
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.
C3 Paper B – Marking Guide
1.
2.
(2x − 3)2 > (x + 2)2
3x2 − 16x + 5 > 0
(3x − 1)(x − 5) > 0
x < 13 or x > 5
1
3
M1
A1
M1
A2
5
3(cosec2 x − 1) − 4 cosec x + cosec2 x = 0
4 cosec2 x − 4 cosec x − 3 = 0
(2 cosec x + 1)(2 cosec x − 3) = 0
cosec x = − 12 or 32
M1
M1
A1
2
3
sin x = −2 (no solutions) or
M1
x = 0.73, π − 0.7297
x = 0.73, 2.41 (2dp)
3.
(i)
dx
= 2y −
dy
A2
3
y
2 y2 − 3
y
=
dy
dx
=1÷
=
dy
dx
(ii)
1
2
y=
, x=
∴ y−
1
2
=
− 15
y
A1
2 y2 − 3
(x −
1
4
B1
)
M1
20y − 10 = −4x + 1
4x + 20y − 11 = 0
4.
(i)
x
xe2x
I ≈
1
3
A1
0
0.25
0.5
0.75
1
0
0.4122 1.3591 3.3613 7.3891
× 0.25 × [0 + 7.3891 + 4(0.4122 + 3.3613) + 2(1.3591)]
= 2.10 (3sf)
(ii)
= [ − 12 e
(i)
(ii)
5
=
∫1
]1
=
2
3
(4 − 2) =
= π∫
5
1
1
3x + 1
1
2
(1 − e−1)
M1 A1
1
dx = [ 23 (3x + 1) 2 ] 15
=
4
3
M1 A1
dx
π(ln 16 − ln 4) =
M1 A1
1
3
π ln 4 =
2
3
π ln 2
[k=
 Solomon Press
C3B MARKS page 2
(7)
M1 A1
= π[ 13 ln3x + 1] 15
1
3
M1
M1
M1 A1
2
1
3x + 1
(6)
A1
1 − 2x 1
= − 12 (e−1 − 1) =
5.
(6)
M1 A1
, grad = − 15
1
4
(5)
2
3
]
M1 A1
(8)
6.
(a)
1
3
V=
(b)
πr2h =
2
πh × h =
3
r
h
, r=
h
3
M1
πh3
1
9
V = 8 × 120 = 960 =
∴
(i)
1
3
=
A1
dV
dV
= 120,
= 13 πh2
dt
dh
dV
dV
dh
dh
=
,
120 = 13 πh2
,
×
dt
dh
dt
dt
dh
= 3.18 cm s−1 (2dp)
when h = 6,
dt
(i)
(ii)
7.
1
3
let radius = r, ∴ tan 30° =
1
9
πh3 ∴ h =
3
B1
dh
360
= 2
πh
dt
M1 A1
9 × 960
π
= 14.011
dh
= 0.58 cm s−1 (2dp)
dt
LHS ≡ 2 sin x cos x −
sin x
cos x
M1 A1
2
sin x
≡ sin x(2cos x − 1) ≡
× cos 2x ≡ tan x cos 2x ≡ RHS
cos x
(i)
cos x
tan x cos 2x = 2 cos 2x
cos 2x (tan x − 2) = 0
cos 2x = 0 or tan x = 2
2x = 90, 270 or x = 63.4
x = 45°, 63.4° (3sf), 135°
A2
t = 0, m = 480
∴ t = 10, m = 0.998 × 480 = 479.04
∴ 479.04 = 400 + 80e−10k
e−10k = 79.04
80
475 = 400 + 80e−kt,
e−kt =
9.
(i)
(ii)
(iii)
A1
M1 A1
75
80
M1
(iv)
(v)
A1
dm
= −80ke−kt
dt
dm
= −80ke−100k = −0.0856
t = 100,
dt
∴ decreasing at rate of 0.0856 g yr−1 (3sf)
M1 A1
M1
A1
f(x) < 3
(11)
B1
= f(2) = 3 − e
4
2x
M1 A1
2x
y=3−e ,
−1
(9)
B1
M1
75
= 53.5 (3sf)
t = − 1k ln 80
(iii)
M1 A1
M1
A1
k = − 101 ln 79.04
= 0.00121 (3sf)
80
(ii)
(9)
M1
cos x
8.
M1
A1
2
≡ 2sin x cos x − sin x
(ii)
M1 A1
e = 3 − y,
∴ f (x) =
1
2
e.g.
−1
2x = ln (3 − y),
ln (3 − x), x ∈
x=
1
2
ln (3 − y)
, x<3
y = f (x) is the reflection of y = f(x) in the line y = x so they
intersect on the line y = x, hence f −1(x) = f(x) ⇒ f −1(x) = x
x1 = 0.4581, x2 = 0.4664, x3 = 0.4648, x4 = 0.4651, x5 = 0.4651
∴ α = 0.465 (3sf)
M1
A2
B2
M1 A1
A1
(11)
Total
(72)
 Solomon Press
C3B MARKS page 3
Performance Record – C3 Paper B
Question no.
Topic(s)
Marks
1
2
3
4
5
6
7
8
9
Total
functions trigonometry differentiation Simpson’s integration connected trigonometry exponentials functions,
rule,
rates
and
numerical
integration
logarithms, methods
differentiation
5
6
6
7
8
Student
 Solomon Press
C3B MARKS page 4
9
9
11
11
72