GCE Examinations
Advanced Subsidiary / Advanced Level
Further Pure Mathematics
Module FP2
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 Rosemary Smith & Shaun Armstrong
 Solomon Press
These sheets may be copied for use solely by the purchaser’s institute.
FP2 Paper A – Marking Guide
1.
dy
d2 y
= 1 + 4x + 15x2,
= 4 + 30x
dx
dx 2
dy
d2 y
when x = 0,
= 1,
=4
dx
dx 2
y = x + 2x2 + 5x3,
M1 A1
A1
3
(1 + 12 ) 2
= 2 2 = 1
ρ=
4
4
2
2.
M1 A1
u = x, u′ = 1; v′ = sech2x, v = tanh x
∫
ln 2
0
2
x sech x dx = [x tanh x]
ln 2
0
−
M1
∫
ln 2
0
tanh x dx
A1
= [x tanh x − ln (cosh x)] ln0 2
e
e −e
− ln 
ln 2
− ln 2
e +e

= (ln 2 ×
= ln 2 ×
=
3.
(a)
(b)
3
5
M1 A1
− ln 2
ln 2
ln 2
+e
2
− ln 2

 ) − (0 − ln 1)

2 − 12
 2+ 12 
−
ln


2+ 12
 2 
ln 2 − ln 54
=
2
1 − 4x 2
4x
2
1− 4x
4
f ′′(x) =
1 − 4x 2
A1
(8)
M1
M1 A1
+ 2 arcsin 2x +
+ 2 arcsin 2x −
1
2
4x
1 − 4 x2
(−8x)(1 − 4x2)
− 12
= 2 arcsin 2x
M1 A1
A1
A1
f ′′(x)[f(x) − xf ′(x)] =
=
4
2
(2x arcsin 2x +
1 − 4x
4
× 1 − 4x 2 = 4
2
1 − 4x
1 − 4x 2 − 2x arcsin 2x)
 Solomon Press
FP2A MARKS page 2
M1 A1
M1
let y = arcsin 2x ∴ sin y = 2x
dy
∴ cos y
=2
dx
dy
2
2
=
=
2
dx
1 − 4x 2
1 − sin y
f ′(x) = 2x ×
(5)
M1
A1
(9)
4.
(a)
t(3 − t2) = 0
1
3
at A, y = 0 ∴
M1
t = 0 (at O) or ± √3; t > 0 ∴ a = √3
(b)
dx
dy
= 2t,
= 1 − t2
dt
dt
A=
∫
3
0
= 2π ∫
1 3
3 t )
4t 2 + (1 − t 2 )2 dt
3
1 3
3 t )
1 + 2t 2 + t 4 dt
(t −
0
3
= 2π ∫
0
= 2π ∫
0
(t −
3
= 2π[(
(a)
(b)
6.
3
2
M1
2π(t −
1 4
6t
3
2
+
1 3
3 t )(1
2 3
3t
t+
= 2π[ 12 t2 +
5.
A1
−
3
2
1
2
=
1
2
+ t2) dt
1 5
3 t
M1
dt
A1
1 6 3
18 t ] 0
A1
) − 0] = 3π
M1 A1
(ex + e−x)2 − 1
M1
−
2 cosh2x − 1 = 2 ×
=
−
M1 A1
1
4
(e2x + 2 + e−2x) − 1
−2x
2x
(e + e ) + 1 − 1 =
M1
1
2
2x
−2x
(e + e ) = cosh 2x
A1
2(2 cosh2x − 1) = 13 cosh x − 12
4 cosh2x − 13 cosh x + 10 = 0
(4 cosh x − 5)(cosh x − 2) = 0
cosh x = 54 or 2
M1
A1
M1
A1
x = ln ( 54 +
M1 A1
25
16
− 1 ) or ln (2 + √3)
x = ln 2 or ln (2 + √3)
A1
(a)
x2 − 10x + 41 ≡ (x − 5)2 − 25 + 41
≡ (x − 5)2 + 16 ∴ a = −5, b = 16
M1
A1
(b)
∫
9
5
x
2
x − 10 x + 41
u = x − 5,
dx =
∫
9
( x − 5) 2 + 16
∫
4
u+5
0
=
∫
0
u 2 + 16
u
+
2
u + 16
(10)
dx
du
=1
dx
=
4
x
5
(10)
M1
du
A1
5
du
2
u + 16
4
M1
= [ u 2 + 16 + 5 arsinh ( u4 )] 0
M1 A2
= (√32 + 5 arsinh 1) − (4 + 0)
= 4√2 − 4 + 5 ln (1 + √2)
= 4(√2 − 1) + 5 ln (1 + √2) ∴ p = 4, q = 5, r = 1 + √2
M1
A1
(10)
 Solomon Press
FP2A MARKS page 3
7.
(a)
u = xn, u′ = nxn−1; v′ = cos x, v = sin x
π
2
In = [x sin x] 0 − n ∫
n
π
2
0
x
n−1
M1
sin x dx
A1
u = xn−1, u′ =(n − 1)xn−2; v′ = sin x, v = −cos x
π
2
n
In = [x sin x] 0 − n{ [−x
In = (
In = (
(b)
8.
(a)
I0 =
∫
π n
2 )
π n
2 )
n−1
π
2
cos x] 0 +
∫
π
2
0
(n − 1)x
n−2
cos x dx }
M1
A1
− 0 − n(0 − 0) − n(n − 1)In−2
π
2
− n(n − 1)In−2
A1
π
cos x dx = [sin x] 02 = 1
0
I2 = (
π 2
2 )
− 2 × 1 × I0 =
I4 = (
π 4
2 )
− 4 × 3 × ( 14 π2 − 2) =
1
4
M1 A1
π2 − 2
M1 A1
1
16
π4 − 3π2 + 24
2
2
dy
= − c2
y = cx ,
x
dx
2
dy
= − 2c 2 = − 12
at P,
p
c
p
dx
M1 A1
M1
A1
eqn. is y − cp = − 12 (x − cp)
M1
p
2
giving x + yp = 2cp
A1
(b)
X is (cq, 0)
∴ cq + 0 = 2cp giving q = 2p
B1
M1 A1
(c)
P(cp, cp ), Q(2cp, 2cp )
M1
3
2
x-coord of M =
1
2
(cp + 2cp) =
cp
A1
y-coord of M =
1
2
( cp + 2cp ) = 43cp
A1
x×y=
3
2
2
cp × 43cp ∴ eqn is xy = 98c
 Solomon Press
FP2A MARKS page 4
(11)
M1 A1
(12)
Total
(75)
Performance Record – FP2 Paper A
Question no.
Topic(s)
Marks
1
rad. of
curv.
5
2
integr.
hyp. fns
8
3
diff. inv.
trig.
9
4
surface
area
10
5
eqn. in
hyp. fns.
10
6
integr.
std.
forms
10
7
8
reduction
formula
hyperbola,
tangent,
loci
11
12
Total
Student
 Solomon Press
FP2A MARKS page 5
75