General Certificate of Education
Advanced Level Examination
January 2010
Mathematics
MPC3
Unit Pure Core 3
Friday 15 January 2010
1.30 pm to 3.00 pm
For this paper you must have:
*
an 8-page answer book
*
the blue AQA booklet of formulae and statistical tables
*
an insert for use in Question 2 (enclosed).
You may use a graphics calculator.
Time allowed
*
1 hour 30 minutes
Instructions
*
Use black ink or black ball-point pen. Pencil should only be used for drawing.
*
Write the information required on the front of your answer book. The Examining Body for
this paper is AQA. The Paper Reference is MPC3.
*
Answer all questions.
*
Show all necessary working; otherwise marks for method may be lost.
*
Fill in the boxes at the top of the insert.
Information
*
The marks for questions are shown in brackets.
*
The maximum mark for this paper is 75.
Advice
*
Unless stated otherwise, you may quote formulae, without proof, from the booklet.
P20901/Jan10/MPC3 6/6/6/
MPC3
2
Answer all questions.
1 A curve has equation y ¼ e4x ðx 2 þ 2x 2Þ .
(a) Show that
dy
¼ 2e4x ð5 3x 2x 2 Þ .
dx
(3 marks)
(b) Find the exact values of the coordinates of the stationary points of the curve.
(5 marks)
2 [Figure 1, printed on the insert, is provided for use in this question.]
(a)
(i) Sketch the graph of y ¼ sin1 x , where y is in radians. State the coordinates of
the end points of the graph.
(3 marks)
(ii) By drawing a suitable straight line on your sketch, show that the equation
1
sin1 x ¼ 4 x þ 1
has only one solution.
(2 marks)
1
(b) The root of the equation sin1 x ¼ 4 x þ 1 is a . Show that 0:5 < a < 1 .
(2 marks)
1
1
(c) The equation sin1 x ¼ 4 x þ 1 can be rewritten as x ¼ sin 4 x þ 1 .
1
(i) Use the iteration xnþ1 ¼ sin 4 xn þ 1 with x1 ¼ 0:5 to find the values of
(2 marks)
x2 and x3 , giving your answers to three decimal places.
(ii)
1
The sketch on Figure 1 shows parts of the graphs of y ¼ sin 4 x þ 1
and
y ¼ x , and the position of x1 .
On Figure 1, draw a cobweb or staircase diagram to show how convergence takes
(2 marks)
place, indicating the positions of x2 and x3 on the x-axis.
P20901/Jan10/MPC3
3
3
(a) Solve the equation
cosec x ¼ 3
giving all values of x in radians to two decimal places, in the interval 0 4 x 4 2p .
(2 marks)
(b) By using a suitable trigonometric identity, solve the equation
cot 2 x ¼ 11 cosec x
giving all values of x in radians to two decimal places, in the interval 0 4 x 4 2p .
(6 marks)
4
(a) Sketch the graph of y ¼ j 8 2x j .
(2 marks)
(b) Solve the equation j 8 2x j ¼ 4 .
(2 marks)
(c) Solve the inequality j 8 2x j > 4 .
(2 marks)
ð 12
5
(a) Use the mid-ordinate rule with four strips to find an estimate for
giving your answer to three significant figures.
0
lnðx 2 þ 5Þ dx ,
(4 marks)
(b) A curve has equation y ¼ lnðx 2 þ 5Þ .
(i) Show that this equation can be rewritten as x 2 ¼ e y 5 .
(1 mark)
(ii) The region bounded by the curve, the lines y ¼ 5 and y ¼ 10 and the y-axis is
rotated through 360° about the y-axis. Find the exact value of the volume of the
solid generated.
(4 marks)
(c) The graph with equation y ¼ lnðx 2 þ 5Þ is stretched with scale factor 4 parallel to the
0
to give the graph with equation y ¼ f ðxÞ .
x-axis, and then translated through
3
Write down an expression for f ðxÞ .
(3 marks)
Turn over for the next question
P20901/Jan10/MPC3
s
Turn over
4
6 The functions f and g are defined with their respective domains by
f ðxÞ ¼ e2x 3 , for all real values of x
gðxÞ ¼
1
,
3x þ 4
for real values of x, x 6¼ 4
3
(a) Find the range of f .
(2 marks)
(b) The inverse of f is f 1 .
(c)
(i) Find f 1 ðxÞ .
(3 marks)
(ii) Solve the equation f 1 ðxÞ ¼ 0 .
(2 marks)
(i) Find an expression for gf ðxÞ .
(1 mark)
(ii) Solve the equation gf ðxÞ ¼ 1 , giving your answer in an exact form.
(3 marks)
7 It is given that y ¼ tan 4x .
sin 4x
dy
, use the quotient rule to show that
¼ pð1 þ tan2 4xÞ ,
cos 4x
dx
where p is a number to be determined.
(3 marks)
(a) By writing tan 4x as
(b) Show that
d2 y
¼ qyð1 þ y 2 Þ , where q is a number to be determined.
dx 2
ð
8
(a) Using integration by parts, find
x sinð2x 1Þ dx .
(b) Use the substitution u ¼ 2x 1 to find
ð
P20901/Jan10/MPC3
(5 marks)
x2
dx , giving your answer in terms of x.
2x 1
(6 marks)
END OF QUESTIONS
Copyright Ó 2010 AQA and its licensors. All rights reserved.
(5 marks)
MPC3 - AQA GCE Mark Scheme 2010 January series
Key to mark scheme and abbreviations used in marking
M
m or dM
A
B
E
or ft or F
CAO
CSO
AWFW
AWRT
ACF
AG
SC
OE
A2,1
–x EE
NMS
PI
SCA
mark is for method
mark is dependent on one or more M marks and is for method
mark is dependent on M or m marks and is for accuracy
mark is independent of M or m marks and is for method and accuracy
mark is for explanation
follow through from previous
incorrect result
correct answer only
correct solution only
anything which falls within
anything which rounds to
any correct form
answer given
special case
or equivalent
2 or 1 (or 0) accuracy marks
deduct x marks for each error
no method shown
possibly implied
substantially correct approach
MC
MR
RA
FW
ISW
FIW
BOD
WR
FB
NOS
G
c
sf
dp
mis-copy
mis-read
required accuracy
further work
ignore subsequent work
from incorrect work
given benefit of doubt
work replaced by candidate
formulae book
not on scheme
graph
candidate
significant figure(s)
decimal place(s)
No Method Shown
Where the question specifically requires a particular method to be used, we must usually see evidence of use of this
method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate,
particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be
provided on the mark scheme.
Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can
be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates
showing no working is that incorrect answers, however close, earn no marks.
Where a question asks the candidate to state or write down a result, no method need be shown for full marks.
Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct
answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark
scheme, when it gains no marks.
Otherwise we require evidence of a correct method for any marks to be awarded.
3
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3
Q
1(a)
y′ = e
−4 x
Solution
( 2 x + 2 ) − 4e−4 x ( x 2 + 2 x − 2 )
Marks
M1
Total
Comments
y ′ = Ae ( ax + b ) ± Be –4 x ( x 2 + 2 x – 2 )
where A and B are non-zero constants
All correct
–4 x
A1
= e −4 x ( 2 x + 2 − 4 x 2 − 8 x + 8 )
or -4 x 2 e-4 x - 6 xe-4 x +10e-4 x
= 2e −4 x ( 5 − 3 x − 2 x 2 )
A1
3
AG; all correct with no errors,
2nd line (OE) must be seen
Condone incorrect order on final line
or
y = x 2 e – 4 x + 2 xe – 4 x – 2e – 4 x
y ′ = − 4 x 2 e – 4 x + 2 xe – 4x + 2 x. – 4e – 4 x
+ 2e –4 x + 8e –4x
2 – 4x
= – 4x e
– 6 xe
– 4x
= 2e −4 x ( 5 − 3x − 2 x 2 )
(b)
+ 10e
–4x
+ De –4 x + Ee – 4x
All correct
– 4x
− ( 2 x + 5 )( x − 1) (= 0)
x=
Ax 2 e – 4 x + Bxe – 4x + Cxe
(M1)
(A1)
−5
,1
2
(A1)
AG; all correct with no errors,
3rd line (OE) must be seen
M1
OE
A1
Both correct and no errors
Attempt at factorisation
(±2 x ± 5)(± x ± 1)
or formula with at most one error
SC x = 1 only scores M1A0
x =1, y = e
−4
5
⎛ 3⎞
x = − , y = e10 ⎜ − ⎟
2
⎝ 4⎠
m1
For y = aeb attempted
A1F
Either correct, follow through only from
incorrect sign for x
A1
5
CSO 2 solutions only
Note: withhold final mark for extra
solutions
Note: approximate values only for y can
score m1 only
Total
8
4
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3 (cont)
Q
2(a)(i)
Solution
Marks
Total
Comments
A
B1
correct shape passing through origin
and stopping at A and B
B
⎛ π⎞
A ⎜ 1, ⎟
⎝ 2⎠
π⎞
⎛
B ⎜ −1, − ⎟
2⎠
⎝
(ii) line intersecting their curve (positive
gradient, positive y intercept)
Correct statement
B1
B1
3
SC A(1, 90) and B ( –1, – 90 ) scores B1
M1
A1
2
one solution only, stated or indicated on
sketch - must be in the first quadrant
(ie curve intersects line once)
Must have scored B1 for graph in (a)(i)
(b)
LHS ( 0.5 ) = 0.5
RHS ( 0.5 ) = 1.1⎞
⎟
LHS (1) = 1.6
RHS (1) = 1.3 ⎟⎠
At 0.5 LHS < RHS , At 1 LHS > RHS
∴ 0.5 < α < 1
or
1
f ( x ) = sin −1 ( x ) − x −1
4
f ( 0.5 ) = − 0.6 ⎫⎪
⎬ AWRT
f (1) = 0.3
⎪⎭
Change of sign ⇒ 0.5 < α < 1
M1
A1
2
CSO
f (x) must be defined
(M1)
Allow f ( 0.5 ) < 0 f (1) > 0
(A1)
or
f (x) must be defined
⎛1
⎞
f ( x ) = sin ⎜ x + 1⎟ – x
4
⎝
⎠
f ( 0.5 ) = 0.4 ⎫⎪
⎬ Attempt
f (1) = – 0.1 ⎪⎭
Change of sign ⇒ 0.5 < α < 1
(M1)
(A1)
or
f ( x ) = 4 sin – 1 x – x – 4
f ( 0.5 ) = – 2.4 ⎫⎪
⎬ attempt
f (1) = 1.3 ⎪⎭
Change of sign ⇒ 0.5 < α < 1
f (x) must be defined
(M1)
(A1)
5
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3 (cont)
Q
2(c)(i)
Solution
x2 = 0.902
x3 = 0.941
Marks
M1
Total
A1
2
Comments
Sight of AWRT 0.902 or AWRT 0.941
These values only
(ii)
Staircase, (vertical line) from x1 to curve,
horizontal to line, vertical to curve
M1
O
x1
x2 x3
A1
Total
2
11
6
x2 , x3 approx correct position on x-axis
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3 (cont)
Q
3(a)
Solution
1
sin x = ,
3
or sight of ± 0.34, ± 0.11π or ± 19.47
(or better)
x = 0.34, 2.8 ( 0 )
(b)
AWRT
Marks
Total
Comments
M1
A1
2
Penalise if incorrect answers in range;
ignore answers outside range
Correct use of cot 2 x = cos ec 2 x – 1
cosec 2 x −1 =11 − cosec x
M1
cosec 2 x + cosec x − 12 ( = 0 )
A1
( cosec x + 4 )( cosec x − 3)( = 0 )
m1
Attempt at Factors
Gives cosec x or – 12 when expanded
Formula one error condoned
A1
Either Line
cosec x = − 4, 3⎫
⎪
1 1 ⎬
sin x = − ,
4 3 ⎪⎭
1
sin x = −
4
⇒ x = 3.39, 6.03
0.34, 2.8 ( 0 )
AWRT
B1F
AWRT
B1
Alternative
cos 2 x
1
= 11 –
2
sin x
sin x
2
2
cos x = 11 sin x – sin x
6
3 correct or their two answers from (a)
and 3.39, 6.03
4 correct and no extras in range
ignore answers outside range
SC 19.47, 160.53, 194.48, 345.52
B1
Correct use of trig ratios and multiplying
by sin 2 x
(M1)
1 – sin 2 x = 11sin 2 x – sin x
0 = 12 sin 2 x – sin x – 1
(A1)
0 = ( 4sin x + 1)( 3sin x – 1)
(m1)
1 1
sin x = – ,
4 3
(A1)
Attempt at factors as above
(B1F)
(B1)
Total
As above
8
7
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3 (cont)
Q
Solution
4(a)
Marks
A1
x
x =2
x= 6
B1
B1
x >6
x< 2
B1
B1
Total
5(a)
x
1.5
4.5
7.5
10.5
y
1.98100
3.22883
4.11496
4.74710
(ii)
2
One correct answer
Second correct answer and no extras
Condone answers shown on the graph, if
clearly indicated
One correct answer
Second correct answer and no extras and
no further incorrect statement eg
6 < x < 2 or 2 < x > 6
SC x ≥ 6 , x ≤ 2 scores B1
6
x values correct PI
M1
3+ y values correct to 2sf or better or
exact values
A1
1.981, 3.228 / 9, 4.114 / 5, 4.747 for y
A1
( or better )
4
(Note: 42.2 with evidence of mid-ordinate
rule with four strips scores 4/4)
y = ln ( x 2 + 5 )
OE
e y = x2 + 5
x2 = e y – 5
B1
( π ) ∫ ( e y − 5 ) ( dy )
M1
= ( π ) ⎡⎣e y − 5 y ⎤⎦
(10 )
1
AG Must see middle line, and no errors
Condone omission of brackets around
f (y) throughout
A1
( 5)
= ( π ) ⎡⎣( e10 − 50 ) − ( e5 − 25 ) ⎤⎦
(c)
2
B1
∫ = 3× ∑y
= 42.2
(b)(i)
2
Modulus graph V shape in 1st quad going
into 2nd quad, touching x-axis. Must cross
y-axis
Condone not ruled
4 and 8 labelled
M1
4
(c)
Comments
y
8
(b)
Total
m1
F (10 ) – F ( 5 )
V = π ⎡⎣e10 − e5 − 25⎤⎦
A1
CSO
including correct notation – must see dy
ISW if evaluated
⎡⎛ x ⎞ 2
⎤
( y =) ln ⎢⎜ ⎟ + 5⎥ + 3
⎥⎦
⎣⎢⎝ 4 ⎠
M1
x2
x
seen, condone ln
+ ..........
4
4
B1
A1
…+3
CSO mark final answer (no ISW)
Total
8
4
3
12
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3 (cont)
Q
6(a)
(b)(i)
Solution
f ( x) > − 3
Marks
M1
Total
A1
2
Comments
‘ > − 3 ’, ‘ x > − 3 ’ or ‘ f ( x ) ≥ − 3 ’
Allow y > – 3
2x
y =e − 3
y + 3 = e2 x
ln ( y + 3) = 2 x
M1
swap x and y
M1
( f ( x ) ) = 12 ln ( x + 3)
−1
A1
Alternative
x → × 2 → e → -3
÷ 2 ← ln ← + 3 ← x
(M1)
(M1)
ln( x + 3)
y=
2
(ii)
(c)(i)
(ii)
3
(A1)
for putting their p ( x ) =1 from
x + 3 =1
M1
x=− 2
A1
2
B1
1
1
⎫
3 e − 3 + 4 ⎪⎪
⎬ either
1
⎪
(= ) 2 x
⎪
3e − 5
⎭
(g f ( x ) = )
(
2x
)
attempt to isolate: ln ( y ± A ) = Bx or
reverse
OE with no further incorrect working
Condone y = …..
OE
k ln ( p ( x ) ) in their part (b)(i)
CSO
SC: B2 x = -2 with no working, if full
marks gained in part (b)(i)
substituting f into g
ISW
1
=1
3e − 5
1 = 3e 2 x − 5
2x
OE
M1
Correct removal of their fraction
m1
Correct use of logs leading to kx = ln
e2 x = 2
2 x = ln 2
1
x = ln 2
2
OE
A1
Total
3
11
9
CSO No ISW except for numerical
evaluation
a
b
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3 (cont)
Q
7(a)
Solution
cos 4 x . 4cos 4 x − sin 4 x . − 4sin 4 x
cos 2 4 x
⎛ dy
⎜
⎝ dx
⎞
=⎟
⎠
4cos 2 4 x + 4sin 2 4 x
=
or better
cos 2 4 x
= 4 (1 + tan 2 4 x )
or
⎛ dy
⎜
⎝ dx
=
(b)
Total
M1
A1
CSO
⎞ cos 4 x . 4 cos 4 x − sin 4 x . − 4 sin 4 x
=⎟
cos 2 4 x
⎠
4 cos 4 x cos 4 x 4 sin 4 x sin 4 x
+
cos 4 x cos 4 x
cos 4 x cos 4 x
A1
Comments
± A cos 2 4 x ± B sin 2 4 x
cos 2 4 x
Both terms correct
3
All correct
± A cos 2 4 x ± B sin 2 4 x
cos 2 4 x
(M1)
(A1)
or better
CSO
= 4 (1 + tan 2 4 x )
Marks
d2 y
= 4 × 2 tan 4 x × ……
dx 2
4sec 2 4x
= 32 tan 4 x sec 2 4 x
(A1)
All correct
M1
A tan 4 x × f(4x)
m1
f(4x) = B sec2 4 x
ft 8 × their p from part (a)
Previous two method marks must have
been earned
A1F
= 32 tan 4 x (1+ tan 4 x )
m1
= 32 y (1+ y 2 )
A1
2
5
CSO
Alternative Solutions
sin 2 4 x
cos 2 4 x
y ′ = 4 + 4 tan 2 4 x = 4 + 4
y ′′ = 4 ×
⎡ ⎡ cos 2 4 x 2 sin 4 x 4 cos 4 x + s in 2 4 x 2 cos 4 x 4 sin 4 x ⎤ ⎤
⎦⎥
⎢⎣
cos 4 4 x
⎢⎣
⎥⎦
4 × 8 sin 4 x cos 4 x ⎡⎣cos 4 x + sin 4 x ⎤⎦
cos 4 4 x
= 32 tan 4 x sec 2 4 x
2
=
= 32 y (1 + y 2 )
or
dy
= 4 sec 2 4 x
dx
d2 y
= 4 × 2 sec 4 x.4 sec 4 x tan 4 x
dx 2
= 32 sec 2 4 x tan 4 x
(m1)
A cos3 4 x ± B sin 3 4 x
where A and B are
cos 4 4 x
constants or trig functions.
Where A is msin4x and B is ncos4x
(A1F)
ft 8 × their p from part (a)
(m1)
k tan 4 x sec 2 4 x
(A1)
CSO
(M1)
(m1)
A sec 4 x × f(4x)
f(4x) = B sec 4 x tan 4 x
ft 8 × their p from part (a)
Previous two method marks must have
been earned
(M1)
2
(A1F)
= 32 (1 + tan 4 x ) tan 4 x
(m1)
= 32 y (1 + y 2 )
(A1)
2
10
CSO
MPC3 - AQA GCE Mark Scheme 2010 January series
MPC3 (cont)
Q
7(b) or
Solution
Marks
dy
= 4(1 + tan 2 4 x)
dx
dy
u = tan 4 x
= 4 + 4u 2
dx
2
d y
du
= (8)u
2
dx
dx
du
= 4 + 4 tan 2 4 x = 4 + 4u 2
dx
d2 y
= 8u (4 + 4u 2 )
2
dx
= 32u (1 + u 2 )
(m1)
(A1)
(m1)
= 32 y (1 + y )
(A1)
Total
∫ x sin ( 2 x −1) dx
dv
= sin ( 2 x − 1)
dx
du
1
= 1 v = − cos ( 2 x − 1)
dx
2
x
( ∫ = ) − cos ( 2 x −1)
2
1
− ∫ − cos ( 2 x −1) (dx)
2
1
x
= − cos ( 2 x − 1) + ∫ cos ( 2 x − 1) (dx)
2
2
x
1
= − cos ( 2 x − 1) + sin ( 2 x −1) + c
2
4
u=x
(b)
u = 2x −1
'du = 2dx '
8
( u +1) du
x2
dx = ∫
2x −1
4u 2
2
⎛ 1 ⎞ u + 2u +1
=⎜ ⎟ ∫
du
u
⎝8⎠
1
⎛1⎞
= ⎜ ⎟ ∫ u + 2 + du
u
⎝8⎠
d
( x ) attempted
dx
M1
∫ sin f ( x ) ,
A1
All correct – condone omission of brackets
m1
correct substitution of their terms into parts
A1
All correct – condone omission of brackets
A1
5
M1
m1
A1
2
∫
Comments
(M1)
2
8(a)
Total
CSO condone missing + c and dx
Condone missing brackets around 2x – 1 if
recovered in final line ISW
OE
All in terms of u
All correct
PI from later working
A1
2
⎤
⎛ 1 ⎞ ⎡u
= ⎜ ⎟ ⎢ + 2u + ln u ⎥
⎝8⎠⎣ 2
⎦
B1
2
⎤
1 ⎡ ( 2 x − 1)
= ⎢
+ 2 ( 2 x − 1) + ln ( 2 x − 1) ⎥ + c
8 ⎢⎣
2
⎥⎦
A1
Total
TOTAL
2
⎤
⎛ 1 ⎞ ⎡ (u + 2)
+ ln u ⎥
or ⎜ ⎟ ⎢
⎝ 8 ⎠ ⎢⎣ 2
⎥⎦
6
11
75
11
2
⎤
1 ⎡ ( 2 x + 1)
+ ln ( 2 x − 1) ⎥ + c
or = ⎢
8 ⎢⎣
2
⎥⎦
CSO condone missing + c only
ISW
klm
Scaled mark component grade boundaries - January 2010 exams
A2 units (legacy)
Component
Code
Component Title
Maximum
Scaled Mark
A
GCE MATHEMATICS UNIT PC3
GCE MATHEMATICS UNIT PC4
GCE MATHEMATICS UNIT S1A WRITTEN
GCE MATHEMATICS UNIT S1A CWK
GCE MATHEMATICS UNIT S1B
GCE MATHEMATICS UNIT S2B
GCE MATHEMATICS UNIT XMCAS
GCE MATHEMATICS UNIT XMCA2
75
75
75
25
75
75
125
125
57
65
59
20
62
60
101
94
49
58
51
17
55
52
89
82
41
51
44
14
48
44
77
70
34
44
37
12
41
36
65
59
27
37
30
10
34
28
54
48
MED4
MED5
MED6
GCE MEDIA STUDIES UNIT 4
GCE MEDIA STUDIES UNIT 5
GCE MEDIA STUDIES UNIT 6
60
60
60
42
50
42
37
41
37
3
33
32
32
3
29
23
27
25
15
22
HEB2
GCE MODERN HEBREW UNIT 2
100
70
61
52
44
36
MUS4
MUS6
GCE MUSIC UNIT 4
GCE MUSIC UNIT 6
120
40
78
36
70
33
62
30
55
27
48
25
PAN2
GCE PANJABI UNIT 2
100
77
66
55
45
35
PLY4
PLY5
PLY6
GCE PHILOSOPHY UNIT 4
GCE PHILOSOPHY UNIT 5
GCE PHILOSOPHY UNIT 6
50
50
60
28
31
42
25
27
37
22
24
32
19
21
27
16
18
22
GCE PHYSICS A UNIT 4 OTQ
GCE PHYSICS A UNIT 4 WRITTEN
GCE PHYSICS A UNITS 5-9 PRACTICAL
GCE PHYSICS A UNITS 5-9 CWK
GCE PHYSICS A UNIT 5 WRITTEN
30
45
30
30
60
22
30
22
26
45
19
27
20
23
40
16
24
18
20
35
13
21
16
17
30
10
19
14
14
26
MPC3
MPC4
MS/SS1A/W
MS/SS1A/C
MS1B
MS2B
XMCAS
XMCA2
PA04/1
PA04/2
PHAP
PHAC
PHA5/W
Scaled Mark Grade Boundaries
B
C
D
E