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Marking Scheme for Additional Mathematics Trial 1 (2009)
Paper 2
Question 1
2 2
2 2
2 2
2
Given the following equationslet 2 24 be (eq 1)and 8 be (eq 2)By substitution,Substitute (eq 2) into (eq 1)
we have, 8 2 24Expanding and solve for 64 16 2 242 18 64 24 02
x y yx y
y y yy
y y y yy yy
2
2
18 40 0Simplifying the quadratics by dividing by 2
9 20 0Using simple factorization,
4 5 04 or 5
Now for each value of we will get thecorresponding value of .For 4 and using (eq 2)
8
y
y y
y yy y
yx
yx
4
4(4, 4)For 5 and using the same method,
8 53
(3,5)Therefore, the possible pair of solutions are(3,5) and (4,4)
x
yxx
Question 2
20
20
20Given that 2 (20 1) 1102
Simplifying the information above we have,10 2 (19) 110
*(actually in mathematics we cannot have twoequations sign on the same line)2 19 11 ............ let th
S a d
S a d
a d
8
is be (eq 1)Second information,
(8 1) 27 2 ............... let this be (eq 2)
T a da d
2 19 11 ............ let this be (eq 1)7 2 ............... let this be (eq 2)
Notice that we can solve for and using the simultaneous equation method.For this, we use elimination method but,of
a da d
a d
15 6
course you can use substitution.But first, (eq 2) 22 14 4 ............ let this be (eq 3)(eq 1 eq 3)5 15
3Substitute 3 into (eq 2)
7(3) 22 2123
(b)15 6( 46 14(3)) ( 46 5(3))2 2
1
a d
dd
daaa
S S
5 ( 46 42) 3( 46 15)2
15 ( 4) 3( 31)230 93
63
Question 3
score
Upper limit Lower limit frequency (f) x fx fx2
10 19 5 14.5 72.5 1051.25 20 29 4 24.5 98 2401 30 39 8 34.5 276 9522 40 49 15 44.5 667.5 29703.75 50 59 43 54.5 2343.5 127720.75 60 69 20 64.5 1290 83205 70 79 5 74.5 372.5 27751.25 100 5120 281355
5120100
51.2
fxx
N
x
x
22
2281355 51.2100
192.1113.8604
fxx
N
1
1
1
1
1
5
1
1 1
1
1 6 1 1
1
1
1
1 7 1
Question 4
3
3
2
2
2
4 4
2 3 42 3 4
1 2 3
2 3
2 2 44
42
2( 1)
14 )( 2)( 1) 2
1 1 2 2
21 2 31 1 2
2 3
dx x dxx
xc
cx
a iix x x xdx dx
x x
dx x x x dxx x x
x x x c
cx x x
Question 5 Please note: For part (a) 1 Mark is given to students who draw a table and plotted at least 6 points. 1 Mark is given when the students label the axis correctly (either in degrees or in radians) 1 Mark is given for the shape of the graph correct. 1 Mark is given for labeling the graph.
4 2
4 2
4 2
4 2
5 3
5 3
( )
Given that the general gradient of an
equation is 15 12 at a certain point .
Thus, 15 12
15 12
15 12
15 125 3
3 4At the point (1, 4)4 3 4
5The equati
bdydx
x x xdy x xdx
dy x x dx
dy x x
x xy c
y x x c
cc
5 3on is 3 4 5y x x For part (b) 1 Mark is given for showing the correct equation of the straight line. 1 Mark is given for sketching the straight line correctly. 1 Mark is given for stating the correct number of solutions in this case the answer is 4 solutions.
1
1
1
1
1
1
1
7
7
2cos( ) 1y x
12xy
●
A B
F
C D E
b
3a
Question 6 From the diagram above, we were given several
clues, , , and that
. We were also told that is parallel
to
(since it is par
3
al
43
DAD b
DE k DB AE
B
CAB a AB
C
AE qBC
lel, this relation is truewhere is a const
(3
an
4343
t)
( )
( )
( )
( )
( )
......... l t
43
et
)
q
AE q
AE
BD
BA AD
BA AD
a
q
AE q
AE q
AE
a
q a b
AE qa qb
ab
DC
AB
his be eq. 1
Now we also know that,
which means; ( )
( )
3
3 ( 1) ....... let this be eq.Comparing eq. 1 and eq
2
3.
3
2
a
DE k DB
DA AE k DB
b AE k
AE ka kb b
AE ka k b
q k
b
and 1 (equating the together)1 3
4 114
From parallel to and
, where is a consta
Next we have to find the value o .
t
f
n .
k q qk k
k
k
EF AB BF mBC
EF pAB p
BF mBC
BE E
m
F
( ) from the above part.
But (refer to diagram above)
m a b
BE BA AE
1(from the first part 4
1 3sub. into 3 we have )4 4
3 3 4 4
3 334 4
9 34 4
9 34 4
9 3( ) ( ) ...... let this be eq. 14 4
But,
EA a b
BE a a b
BE
k
k q
a b
EF a b ma mb
EF m a m b
E
q
F
k
or 3
by comparing this with eq. 19 3( ) 3 and ( ) 04 43 3 ( , not needed)4 2
pAB EF pa
m p m
m p
( )3 2
Given that and 5 7
3 25 7
512
The modulus of is the magnitude of OA
25 144 169
13 units
b
AB OB
AB AO OB
AB
OA
OA
OA
OA
SECTION B Question 7
0
4
01
1 202
4
401
2
4
4Evaluate 1 2
(1 2 )4 1 2 4 1 22
4 (1 2 ) 4 1 3
4 1 3 8
dxx
xx dx
x
1
1
1
1
1
1
1
1 8
1
1
2
2
2
2
3 2
3
33
3
First find the roots of the curve 9When 0, then 9 0
93. Here the roots are ( 3,0) and (3,0)
The area bounded by equation 9 and -axis,
9
93
27 9 27 9 18 18
36
y xy x
xx
y x x A
A x dx
xA x
A
A
2
9
0
922
0
3
unit)
Volume generated about the -axis,
Volume of cone 9
1 (12 )(12) 93 2
81576 81 02
576 40.5535.51682.541 unit
iiy
V y dy
yV y
V
VVV
Question 8
Given that the 100 and 5)
110 100( 110) ( )5
( 2) 1 ( 2)From the normal distribution table1 0.0228 0.9772
xz
a
P X P Z
P Z P Z
6 55
)The probability the a bar of chocolate with massless that 100gm is 0.9772Thus the probability of a box of chocolates which hasless than 5 chocolates < than 100gm1 ( ( 5) ( ( 6))1 ( (0.9772) (0.0
b
P X P XC
6 6 06
4 1 31
228) (0.9772) (0.0228) )1 (6(0.89108)(0.0228) 0.870764 1)1 0.1218997 0.8707640.0073363
)(0.0073363) (0.9926637)
0.028704
C
iiC
Question 9 2
2 3
1
py qxx
y p qx x
From the graph, Please refer to suggested graph. p = 0.5 q = 5.01 Note: 1 mark will be given if the student writes the complete expression of the linear function. 2 marks will be given if the tables for values
2
yx
and 3
1x
are shown.
1 mark will be given for the correct axis shown on the graph. 1 mark will be given for correct plotting of all the points. 1 mark will be given for the best line drawn 1 mark will be given for construction on the line to show the working to get the gradient hence the value for p. 1 mark will be given for the accuracy of the value of p 1 mark will be given for the line to touch the vertical axis 1 mark will be given for the accuracy for the value of q.
x 1 1.5 2 2.5 3 3.5 y 5.5 11.6 20.3 31.5 45.2 61.4
2
yx
5.5 5.16 5.08 5.04 5.02 5.01
3
1x
1 0.29 0.125 0.064 0.037 0.023
1
1
1
1
1
1
1
1 10
1
1
1
1 1
1
1
1 1
1 10
Question 10
1
o
)Consider the triangle where 12, 6
6tan tan12
tan 0.5
tan (0.5)26.5650.4637
)The angle and angle are equalSo, using the property of
180 26.
aAOM
OA OM
PAO MAO
MAO
MAOMAOMAO radians
bPAO APO
AOP
o
2
o
2
2
565 26.565126.872.2146
)1Area of major sector is 2
233.13 or 4.0694 1 12 (4.0694)2292.997
)Length of arc, 12(2.2146)
26.5752
Length of chord 2 sin2
AOPAOP radians
c
r
radians
A
A cmd
ss cm
r
AP
o126.872(12)sin( )2
24 0.89442721.4663
Perimeter of the segment 26.5752 21.4663 48.0415
APAP cm
cmcm
Question 11
1
Given that 1.5 rate of water entering the cone
Rate of water leaving the cone 2
We are asked to find where is the depth of water
at time . *The net rate would be (1.5 2) 0.5
dVdt
dVdt
dh hdt
t cms
Based on similar triangles,
2
2
3
3
2
2
2
1
420420 5
1Volume of cone is 3
13 5
13 25
753
75
25
25 0.5 and when 12
25(288)
0.02762
that is to say the wat
rh
h hr
V r h
hV h
hV
hV
dV hdhdV hdhdh dh dVdt dV dtdh h cmdt hdhdtdh cmsdt
er level is falling atthe rate of 0.0276 every second.cm
4
20 r
h 1
1
1
1
1
1
1
1
1
1 10
1
1
1
1
1
1
1
1
1
1
10
SECTION C Question 12
240 4)
Intial velocity of the ball, 0
40 8
40 /)
To find the range of time the object movesupwards from 0 would be to find whenthe time stops instantaneously; 0
40 840 8 08 40
s t ta
tds v tdtv m sb
tv
v tt
tt
2
5Therefore, the range of time would be 0 5)
The maximum height reached would be when theobject is instantaneously at rest; from ) 5
40 4 when 5200 4(25)200 100100
Maximum distance
st
c
b ts t t tsss m
2
2
2
2
from the ground is 100 96196
)Time taken to travel that distance
40 4 If the object moves 96 m passes O96 40 4
4 40 96 010 24 02 12 0
12It took 12 to reach the ground after passi
mm
d
s t tt t
t tt tt t
t ss
1
ng 40 8 when 1240 96
56
Ov t t svv ms
Question 14
2
1
o o
2 2 2
Given that the area of 18.6 )
1 5.2 7.9 sin 18.62
18.6 2sin5.2 7.937.2sin41.08
sin 0.90555
sin 0.90555
64.897 or 64 53')
Using cosine rule.5.2 7.9 2 5.2 7.9 cos64
PQS
PQS cma
A QPS
QPS
QPS
QPS
QPS
QPSb
QS
o
2 o
2
2
o o o o
.897
27.04 62.41 82.16cos64.89789.45 34.8654.59
54.597.389
)Angle 180 85 58 37Using sine rule7.389sin 58 sin 37
7.389 sin 37sin 58
7.389 0.60180.8480
5.2437 )
Area of
QSQSQS
QSQSc
SQR
SR
SR
SR
SR cmd
Q
2
1 7.389 5.2437 sin852
1 38.7457 0.99619219.29904
QSR
QSR
SR
A
A cm
1
1
1
1
1
1
1
1
1
1 10
1
1
1
1
1
1
1 1
1
1
10
Question 15
9898
96
98
98
98
9696
98
)
100
1101.30 100
110 1.30100
1.43 Index for the year 1996 based on 1998
100
1.30 100 90.911.43
aP IPP
P
P RM
P IP
Question 13
2 1000 ........ inequality 1
2 3 1200 ........ inequal
)8 4 4000
4 6 2400
3 3ity 2
200 ......... inequality 600
3
x y
x y
x y
ax y
x y
x y
c) From the graph, if the 200 units of B are produced, then the maximum of product A would be 300 units. The Profit line 5 8 400x y 1 Mark is given if there is any indication of rough estimation of the profit line (like the above, an example) Drawing the profit line on the graph and translating it up parallel to the vertices that shows the maximum profit gets 1 mark. The maximum profit would be the production of 200 of B and 300 of A
20022002
2000
2002
2002
15 )
110( ) (4) 120(5) 1009
110 4 600 900 10010 4 3004 300 10
75 2.5)
100
1.3 100
130
bIw
Iwx yI
xx y x
x yy x
y xcP IP
x Ix
I
13 b)
5 8x y k
Profit line
1
1
1
1
1
1
1
1
1
1
10
1
1
1
1 1
1
1
1 Area shaded correctly
1 All the straight line graphs are correctly drawn
Indicate the profit point
1
10
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