GCE Examinations
Advanced Subsidiary / Advanced Level
Pure Mathematics
Module P5
Paper D
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.
P5 Paper D – Marking Guide
1.
2.
(a)
dy
−( x 2 + 1)cosech x coth x − 2 x cosech x
=
dx
( x 2 + 1) 2
(b)
x = 0.5,
ρ=
dy
= −4.55 (2dp)
dx
ds
= 2(s + a) ∴
dψ
∫
1
s+a
ds =
A1
∫
2 dψ
ln | s + a | = 2ψ + c
s + a = e 2ψ + c = e 2ψ × e c
∴ s = Ae2ψ − a
[where A = ec]
3.
(a)
(b)
sinh 3x ≡ sinh (2x + x)
≡ sinh 2x cosh x + cosh 2x sinh x
≡ 2 sinh x cosh2x + (1 + 2 sinh2x)sinh x
≡ 2 sinh x(1 + sinh2x) + sinh x + 2 sinh3x
≡ 2 sinh x + 2 sinh3x + sinh x + 2 sinh3x
≡ 4 sinh3x + 3 sinh x
4 sinh3x + 3 sinh x = 7 sinh2x
sinh x(4 sinh2x − 7 sinh x + 3) = 0
sinh x(4 sinh x − 3)(sinh x − 1) = 0
sinh x = 0 or 34 or 1
x = 0 or ln
(
3
4
+ 1 + 169
) or ln (1 +
(5)
M1 A1
A1
M1
A1
(5)
M1
A1
M1
M1
A1
M1
M1
A1
1+1
)
x = 0 or ln 2 or ln (1 + √2)
P5D MARKS page 2
M2 A2
M1
A2
 Solomon Press
(11)
4.
(a)
1
∫
9 − 4x
1
2
=
(b)
1− 2x
∫
1
2
dx =
9 − 4 x2
∫
y
1
1
y
∫
dy =
1
9
4
arcsin ( 23x ) +
x
1− 2x
0
9 − 4 x2
1
2
ln | y | =
5.
(a)
1
2
M1
arcsin ( 23x ) + c
1
2
2x
9 − 4 x2
dx
9 − 4x 2 + c
1
2
x
1
2
9 − 4x 2 − ( 12 arcsin 0 +
9 − 4x 2 −
3
2
dy
dy
= 4a ∴
= 2a
y
dx
dx
dy
a
at P,
= 22ap
= 1p
dx
grad of PQ =
grad of PS =
M1 A1
A1
(12)
M1
ap 2 − aq 2
=
2( p − q )
= ( p + q )( p − q ) = p 2+ q
2p
M1 A1
A1
p 2 −1
2p
(c)
)
A1
2 ap − 2 aq
ap 2 − a
3
2
A1
2
2 ap − 0
A1
M1
eqn. is y − 2ap = 1p (x − ap2)
(b)
M1 A1
9 − 4x 2 ] 0
2y
giving yp = x + ap
M1
M1 A1
1
2
arcsin ( 23x ) +
arcsin ( 23x ) +
∫
M1 A1
dx
y
[ln | y |] 1 = [ 12 arcsin ( 23x ) +
ln | y | − ln 1 =
dx
− x2
arcsin ( 23x ) −
1
2
=
(c)
1
2
∫
dx =
2
∴ p 2+ q = 2
p −1
2
p − 1 = p(p + q)
p2 − 1 = p2 + pq
∴ pq = −1
A1
tangent at P: yp = x + ap2 (i)
tangent at Q: yq = x + aq2 (ii)
(i) × q: ypq = xq + ap2q
(ii) × p: ypp = xp + apq2
subtracting 0 = x(p − q) + apq(q − p)
0 = x − apq
pq = −1 ∴ x = −a ∴ meet on directrix
M1
M1
A1
A1
M1
 Solomon Press
(13)
P5D MARKS page 3
6.
(a)
u = secn−2x, u′ = (n − 2)secn−3x sec x tan x; v′ = sec2x, v = tan x
In = [sec
π
4
n−2
x tan x] 0 −
π
4
∫
(n − 2)sec
0
π
4
In = (√2)n−2 − 0 − (n − 2) ∫
In = (√2)n−2 − (n − 2) ∫
π
4
0
n−2
2
x tan x dx
A1
secn−2x(sec2x − 1) dx
secnx dx + (n − 2) ∫
0
π
4
0
M1 A1
secn−2x dx
In = (√2)n−2 − (n − 2)In + (n − 2)In−2
(n − 1)In = (√2)n−2 + (n − 2)In−2
(b)
I1 =
∫
π
4
M1
A1
π
sec x dx = [ln | sec x + tan x |] 04
0
M1
= ln (√2 + 1) − ln (1 + 0) = ln (√2 + 1)
2I3 = (√2)1 + I1 = √2 + ln (√2 + 1)
I3 = 12 √2 + 12 ln (√2 + 1)
7.
(a)
x2 = a2sinh2u, x = a sinh u,
∫
∫
∫
a 2 + x 2 dx =
=
1
2
=
1
2
=
1
2
=
(b)
∫
3
(c)
3
0
s = 2[ 2t
A1
∫
cosh 2u + 1 du
M1
a2[ 12 sinh 2u + u] + c
A1
a
2
a sinh u cosh u +
2
x
a
=
1
2
a×
=
1
2
ax 1 + ax 2 +
=
1
2
x a2 + x2 +
1+ t2 +
1
2
1
2
M1
a cosh u du
2
× 1 + ax 2 +
2
1
2
1
2
1
2
2
au+c
2
a arsinh(
M1
x
a
)+c
M1
a2arsinh( ax ) + c
1
2
a2arsinh( ax ) + c
A1
M1
M1 A1
1 + t 2 dt
s = 2[( 32 √10 +
P5D MARKS page 4
2
4 + 4t 2 dt
0
s=2∫
a 2 + a 2 sinh 2 u (a cosh u) du
2
(13)
M1 A1
dx
dy
= 2; y = t2,
= 2t
dt
dt
x = 2t,
s=
M1 A1
M1 A1
A1
dx
= a cosh u
du
2
M1 A1
A1
arsinh t] 30
M1
arsinh 3) − (0 + 0)] = 3√10 + arsinh 3
 Solomon Press
M1 A1
(16)
Total
(75)
Performance Record – P5 Paper D
Question no.
1
Topic(s)
diff. hyp.
fns
Marks
5
2
intrinsic
coords
5
3
eqn. in
hyp. fns.
4
integr.
std.
forms
11
12
5
6
7
parabola,
tangent
reduction
formula
integr.
using
hyp. sub.,
arc length
13
13
16
Total
75
Student
 Solomon Press
P5D MARKS page 5