GCE Examinations
Advanced Subsidiary / Advanced Level
Pure Mathematics
Module P1
Paper A
MARKING GUIDE
This guide is intended to be as helpful as possible to teachers by providing
concise solutions and indicating how marks should 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 & Chris Huffer
 Solomon Press
These sheets may be copied for use solely by the purchaser’s institute.
P1 Paper A – Marking Guide
1.
x=
1
2
(1 + y); (1 + y)2 + 4y + y2 = 9
M1
2
y + 3y – 4 = 0
(y – 1)(y + 4) = 0
y = 1 or – 4
sub in giving (1, 1) and ( – 32 , – 4)
2.
3.
(a)
b2 – 4ac > 0
(m – 1)2 – 4(m + 2) > 0
m2 – 6m – 7 > 0
B1
M1
A1
(b)
(m + 1)(m – 7) > 0
suitable method to get m < – 1 or m > 7
M1
M1 A1
(a)
(5p – 1) – (2p – 2) = (2p – 2) – (3p – 5)
3p + 1 = – p + 3
p = 12
M1
M1
A1
(b)
A.P., a = –
7
2
, d=
S10 = 5[ – 7 + (9 ×
4.
(a)
(b)
5.
A1
M1
A1
M1 A1
5
2
5
2
, n = 10
M1 A1
155
2
M1 A1
)] =
2(1 – cos2x) – cos x = 1
2 – 2 cos2x – cos x = 1
2 cos2x + cos x – 1 = 0
M1
(2cos x – 1)(cos x + 1) = 0
cos x = 12 or cos x = – 1
M1
A1
x=
π
3
, 53π or x = π
x=
π
3
, π or
(6)
(6)
(7)
A1
M2 A1
5π
3
A1
(a)
f(2) = 23 – 5(22) + 7(2) – 2
= 0; ∴ x = 2 is a solution
M1
A1
(b)
algebraic division by (x – 2), giving x2 – 3x + 1
(x – 2)(x2 – 3x + 1) = 0
M1 A1
M1
already know x = 2; other solutions from x = 3 ± 5
(8)
M2
2
x = 0.38, 2.62 (2dp)
A2
 Solomon Press
P1A MARKS page 2
(9)
6.
(a)
dy
−1
= 2x 2 − 1
dx
M1 A1
−1
for max, 2 x 2 − 1 = 0
x = 2; x = 4
sub in to get y = 4; A is (4, 4)
(b)
1
1
M1
A1
A1
1
at B, 4 x 2 − x = 0 ; x 2 (4 − x 2 ) = 0
1
2
x = 0 (at origin) or x = 4
x = 16
(c)
=
7.
(a)
∫
area =
[
8
3
M1
A1
1
M1
4 x 2 − x dx
3
x 2 − 12 x 2
]
16
M1 A1
0
= [( 83 × 64) – ( 12 × 256)] – [0]
M1
=
A1
128
3
grad = 4 − 2 =
– 1
−1 − 5
y–2=
– 1
3
(13)
M1
3
(x – 5)
M1
x + 3y – 11 = 0
A1
( 52−1 , 2+2 4 ) = (2, 3)
M1 A1
(c)
grad of perp. bisector = 3
y – 3 = 3(x – 2)
y = 3x – 3
M1
M1
A1
(d)
D must lie on y = 3x – 3 with coords (x, 3x – 3)
AD 2 = (x – 5)2 + (3x – 3 – 2)2 = 10x2 – 40x + 50
CD 2 = (x – 3)2 + (3x – 3 – 4)2 = 10x2 – 48x + 58
AD 2 = CD 2 leading to x = 1 and y = 0; D is (1, 0)
M1
M1
A1
M1 A1
area = 3 × 12 r 2θ =
32
θ = 2
r
M1
(b)
8.
16
0
M1
(a)
3
2
r 2θ = 48
M1 A1
(b)
P = 3(2r + rθ ) = 6r + 3rθ
sub in giving P = 6r + 96r – 1
M1
M1 A1
(c)
dP
= 6 − 96r − 2
dr
M1 A1
for S.P. = 0
r2 = 16; r = 4 cm
sub in, P = 48 cm
M1
A1
A1
(d)
(13)
d 2P
= 192r −3
dr 2
when r = 4, +ve, ∴minimum
M1
A1
(13)
Total
(75)
 Solomon Press
P1A MARKS page 3
Performance Record – P1 Paper A from Solomon Press
Question no.
Topic(s)
Marks
1
2
simul.
eqns.
quad.
roots,
proof
6
6
3
A.P.
7
4
5
trig.
equation
factor
thm.,
alg. div.,
quad.
form
8
9
Student
 Solomon Press
P1A MARKS page 4
6
7
S.P.,
area by
integr.
coord.
geom.
13
13
8
Total
formulae,
max/min
13
75