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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Pre-U Certificate
MARK SCHEME for the May/June 2013 series
1347 MATHEMATICS (STATISTICS WITH PURE
MATHEMATICS)
1347/01
Paper 1 (Pure Mathematics), maximum raw mark 65
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, Pre-U,
GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level
components.
Pa
age
e2
1
2
Ma
ark
k Sch
S hem
me
e
Prre-U
U – May
M y/J
Jun
ne 20
013
3
(
(i)
1 – 14x + 84x
8 x2
M1
M1
A1
A
(
(ii)
Suubsttitu
ute x = 1//35
M1
M1
Geet 117
1 7/17
75
A1
A
(
(i)
[22]
Syllab
bus
s
13
347
7
1 an
nd corrrect met
m tho
od ffor anoth
her terrm
A l co
All
orreect, ign
norre extr
e ra term
t ms
S bs rea
Sub
r son
nab
ble x in
ntoo their an
nsw
wer
[22]
M1
M1
A1
A
O exttracct frac
Or
f ctio
onaal eqquiivallen
nt
Deccreeasiing cu
D
urvee, not
n bellow
w x--ax
xis
C rrecct iincllud
Cor
ding
g ap
ppaaren
ntly
y asym
a mpttotic to
t axis
a s too
r ht, alloow
righ
w x [ 0 only
o y
[22]
(
(ii)
ln y = ln
n 2 + 0.7x
0 x
x=
3
4
1
(ln
n y − ln
n 2)
ln
l 0.7
M1
M1
A1
A
[22]
F nal ans
Fin
a sweer, aeff, allow
w log
l g0.7 (y/22)
3338 350
3 0
B1
B
[11]
(
(ii)
1 3353
3 40
00
B1
B √
[11]
( × 4,
(i)
4 f.
f.t. from
f m (i)
(
( ) 3338 450
(iii)
4 0
B1
B √
[11]
( + 100
(i)
1 0, f.t. from
f m (i)
(
(
(i)
B1
B
2 seeen aany
27
ywh
herre
M1
M1
R arraangge and
Rea
a d taake ln
k=
1
ln
n 1..5 or
o 00.20
03
2
A1
A
(
(ii)
2
122 ×  
3
=5
5
L w of
Law
o loogss ussed
d co
orreectlly onc
o ce
(
(i)
A = 27
2
[33]
A swer, aeef or
An
o a.r.
a .t. 00.20
03
2
1
3
M1
M1
A1
A
O sub
Or
bstiitutte into
i o fo
orm
mulaa
[22]
A swer, aef, allo
An
a ow 5.333 or
o bett
b ter
M1
M1
U e b2 – 4a
Use
ac I 0
k2 – 12k
1 kI0
A1
A
T is in
Thi
neqquaality
y, ae
a sim
s mplif
ifieed f
2) I 0
k(kk – 12
M1
M1
Metho
Me
od ffor so
oluttion
n (nnot “I
I=0, I 122”)
0 I k I 12
2
A2
A2
(
(i)
Pa
ape
er
0
01
[55]
O e erroor, A1
On
A on
nly
© Ca
amb
brid
dge
e In
nterrnattion
nal Ex
xam
mina
atio
ons
s 20
013
3
Page 3
6
42 + 22
(i)
(ii)
Mark Scheme
Pre-U – May/June 2013
[ Allow
M1 A1
AC = 2 5 – 3
B1√
Their (i) – 3 seen [if circle equation used, need to
reject C′(–6 ÷ 5 , 3 – 6 ÷ 5 )]
M1
Use ∆ and make AD subject of formula
M1
A1√
A1
Multiply by conjugate of q r – p
Correct on their (i) if form q r – p used
CAO, aef provided of form a + b c with a, b, c
integer
AD =
)
44
2 5 −3
=
(
44 2 5 + 3
11
)
=12 + 8 5
∫
4
1
4
3/ 2
2
x1/ 2 + dx =  2 x + 2 ln x 
x
 3
1
[=7.439]
Area of trapezium = 8.25
Difference 0.811 (3 SF)
[5]
M1
B1
B1
M1
Attempt to integrate curve, limits 1, 4
One term correct
Fully correct indefinite integral
Method for trapezium, e.g. integration
[y-coords 3 and 2.5]
A1
A1
8.25 seen or implied
0.811 or better, final answer +ve
[6]
(i)
(ii)
20 or exact equivalent
=2 5
(
8
Paper
01
Use Pythagoras
1
AD × 2 5 − 3 = 22
2
7
Syllabus
1347
 43

= 12 − 2 ln 4


Velocity 20 (+ve x)
B1
Initial position
B1
[2]
One fact about position
(20t – 250)2 + (15t – 500)2
625t2 – 25 000t + 312 500 AG
M1
A1
[2]
Use Pythagoras; correctly simplify to AG, at
least one intermediate line
(iii) 625[t2 – 40t + 500]
= 625[(t – 20)2 + 100]
Minimum distance
625 × 100
One fact about velocity
M1
Take out factor and halve t term
A1
Fully correct, allow 625 omitted
M1
Use their b
= 250
A1√
Time t = 20
A1√
625 × their b
[5]
their a
© Cambridge International Examinations 2013
Page 4
9
(i)
Mark Scheme
Pre-U – May/June 2013
y′ = 2x – 3
M1
Differentiate correctly
=3
A1
Obtain m = 3
x = 3 – 3y
M1
Use
A1
(ii)
y = (3 – 3y)2 – 3(3 – 3y)
(ii)
10y = 9y2 or 3x2 – 8x – 3 = 0
A1
This equation, ae simplified form
 1 10 
− ,

 3 9
A1
Get
x
3
 5 10 
(a) (5, 0),  , 
3 9 
 1 10 
(b) (3, 0),  − , − 
9
 3
[4]
10
1
or − (with or without others)
9
3
Both coordinates, no others
M1 A1√
Coords translated ±2, x or y: M1
M1 A1√
Coords reflected, either axis: M1
 2 20 
(c) (6, 0),  − , 
 3 9 
M1 A1√
[6]
× 2 or ÷ 2, any combination: M1
All √ on their (ii)
d2P
18
= 6v 2 + 3 − 2
2
dv
v
M1
A1
[2]
Differentiate
Fully correct
= 0 at 6v4 + 3v2 – 18 = 0
3(2v2 – 3) (v2 + 2) = 0
M1
M1
Polynomial and equate to 0
Method for solving quadratic in v2
A1
3
seen or implied
2
v = 1.22 (474…)
A1
v = 1.22 or better and nothing else
P = 22.0 (454)
A1
[5]
P = 22.0 or better
d2P
36
= 12v + 3 K 0 ∴ minimum
2
dv
v
B1
[1]
Correctly show minimum, needn’t justify “K 0”,
can use numerical gradients or other complete
argument
v2 =
(iii)
Answer, ae simplified f
Subs their x or y into quadratic
A1
10 (i)
[4]
Paper
01
−1
and method for finding q
m
M1
or x2 – 3x = 1 –
(iii)
Syllabus
1347
3
2
© Cambridge International Examinations 2013