GCE Examinations
Advanced Subsidiary / Advanced Level
Further Pure Mathematics
Module FP2
Paper H
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 H – Marking Guide
1.
2x2 + y2 = 4, 4x + 2y
dy
dy
= 0 or 2x + y
=0
dx
dx
d2 y
dy
+ ( )2 = 0
2
dx
dx
dy
dy
at (1, −√2) 2 − √2
=0 ∴
= √2
dx
dx
d2 y
d2 y
2 − √2 2 + 2 = 0 ∴
= 2√ 2
dx
dx 2
2+y
3
ρ=
2.
(a)
(1 + ( 2) 2 ) 2
2 2
1
2
1
2
ln(xy) = arcosh ( 53 ) = ln
(
(e−x + ex) = f(x) ∴ cosh x is even
5
3
+
25
9
)
− 1 = ln 3
∴ xy = 3
cosh (3x − y) = 1 ∴ 3x − y = 0
3.
∫
A1
A1
M1 A1
M1 A1
A1
B1
∴ 3x = y = 3x , x2 = 1
M1
x > 0 ∴ x = 1, y = 3
A1
1
dx =
13cosh x − 5sinh x
=
∫
∫
∫
13
2
1
dx
(e + e ) − 52 (e x − e − x )
x
−x
1
dx
4e + 9e − x
x
ex
dx
4e2 x + 9
du
u = ex,
= ex
dx
1
= ∫
du
4u 2 + 9
1
= 14 ∫
du
u 2 + 94
=
(8)
M1
A1
M1
M1
A1
M1
=
1
4
× 23 arctan ( 23u ) + c
A1
=
1
6
arctan ( 23 ex) + c
A1
 Solomon Press
FP2H MARKS page 2
(8)
(ex + e−x)
(e−x + e−(−x)) =
1
2
M1 A1
M1 A1
4
2 2
f(x) = cosh x =
f(−x) =
(b)
= 3 3 = 3 6
M1 A1
(8)
4.
(a)
let y = arcsin (2x − 1) ∴ sin y = 2x − 1
dy
=2
∴ cos y
dx
dy
2
2
=
=
2
dx
1 − sin y
1 − (2 x − 1) 2
2
=
(b)
x=
3
4
dy
=
dx
5.
(a)
1 − (4 x − 4 x + 1)
, y = arcsin
1
−
9
16
eqn. is y −
π
6
3
4
=
2
1
2
=
4
3
3
4
(x −
π
6
=
y=
π
6
− √3
4
3
2
4x − 4x
2
=
1
x − x2
A1
B1
4
3
3
16
x=0 ∴y−
M1 A1
π
6
= 1 =
=
M1
M1
)
M1
× (− 34 )
M1
A1
dy
dy
= 4a ∴
= 2a
y
dx
dx
dy
a
at P,
= 22at
= 1t
dx
2y
(9)
M1
A1
eqn. is y − 2at = 1t (x − at2)
giving yt = x + at2
A1
(b)
at Q, y = 0 ∴ x = −at2 ; at R, x = 0 ∴ y = at
∴ M is (− 12 at2, 12 at)
M1 A1
A1
(c)
grad of OM =
1 at − 0
2
− 12 at 2 −0
M1
= − 1t ; grad of OP = 2 at2 −0 = 2t
at −0
OM perp. OP ∴ 2t × − 1t = −1 ∴ t2 = 2
M1 A1
M1 A1
(11)
 Solomon Press
FP2H MARKS page 3
6.
(a)
In − In−2 =
∫
=
∫
=
∫
∴ In =
(b)
cos nθ − cos(n − 2)θ
dθ
sin θ
−2sin(n − 1)θ × sin θ
dθ
sin θ
−2 sin (n − 1)θ dθ
= n2−1 cos (n − 1)θ
M1
2
n −1
A1
cos (n − 1)θ + In−2
4
∫
π
cos θ
dθ = [ln | sin θ | ] 2π
4
sin θ
π
2
π
4
1
2
= ln 1 − ln
= −ln 2
− 12
1
2
=
π
4
π
I5 = [ 24 cos 4θ ] 2π + I3 =
1
2
4
(b)
ae = √3, ae =
1
2
M1 A1
ln 2 =
− (− 12 ) +
1
2
1
2
ln 2 − 1
ln 2 − 1 =
1
2
M1 A1
ln 2
M1 A1
∴ ae × ae = a2 = 4
M1
a>0 ∴a=2
b2 = a2(1 − e2) = a2 − (ae)2 = 4 − 3 = 1
b>0 ∴b=1
A1
M1
A1
x2
4
A=
(c)
M1 A1
ln 2
I3 = [ 22 cos 2θ ] 2π + I1 = −1 − 0 +
(a)
M1 A1
π
In = [ n2−1 cos (n − 1)θ ] 2π + In−2
I1 =
7.
M1
4
3
+ y2 = 1,
∫
2
1
2
x + 2y
2πy 1 +
−2
2
= 2π ∫
−2
= 2π ∫
−2
=
π
2
−2
=
π
2
∫
∫
y
4y
2
2
dx
2
16(1 − x4 ) + x 2 dx
M1
16 − 3x 2 dx
3x2 = 16 sin2θ, x =
π
3
A=
π
2
∫−
=
π
2
∫−
=
8π
3
=
4π
3
∫−
=
4π
3
[θ +
=
4π
3
[
4
3
sin θ,
A1
dx
=
dθ
16 − 16sin 2 θ ×
π
3
π
3
4 cos θ ×
π
3
π
3
π
3
π
3
π
3
M1 A1
16 − 4x 2 + x 2 dx
−2
∫−
M1 A1
x2
dx
16 y 2
16 y 2 + x 2
16 y 2
y
2
dy
dy
=0 ∴
= −4 yx
dx
dx
π
3
+
4
3
4
3
4
3
cos θ
M1 A1
cos θ dθ
M1
cos θ dθ
cos2 θ dθ
A1
1 + cos 2θ dθ
M1
1
2
π
sin 2θ ] −3 π
A1
3
3
4
− (−
π
3
−
3
4
)] =
4π
3
(
2π
3
+
3
2
)=
 Solomon Press
FP2H MARKS page 4
(13)
8
9
π2√3 + 2π
M1 A1
(18)
Total
(75)
Performance Record – FP2 Paper H
Question no.
Topic(s)
Marks
1
rad. of
curv.
8
2
3
4
5
6
7
eqn. in
hyp. fns.
integr.
hyp. fns
diff. inv.
trig.
parabola,
tangent
reduction
formula
8
8
9
11
13
Total
ellipse,
surface
area
18
75
Student
 Solomon Press
FP2H MARKS page 5