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Marking Scheme of Term Exam of F5 Mathematics Compulsory Part 2014 – 15 Section A1 Solution Marks Remarks 1.
15
4
96
610
96
235
yx
yxyx
yxyx
1M
1M
1A
For ppp baab )(
For qp
q
p
aaa
2.
nnppq
nqqnppnqnpqp
nqp
qp
3524
35243254
3254
1M
1M
1A
For moving terms
3. (a)
)76)(76(4936 22
nmnmnm
(b)
)476)(76()67(4)76)(76(
2428)76)(76(24284936 22
nmnmmnnmnmmnnmnm
mnnm
1A
1M
1A
For using the result of (a)
4. Let x be the number of $100-notes and y be the number of $500-notes
255
120002400130001005005000500100
xy
yyxyx
∴The actual number of $100-notes is 25.
1A
1A
1M
1A
pp – 1 for any undefined symbol
5.
(a)
1337
37132493104
24)31(3)52(2
4231
352
x
xxx
xx
xx
(b) 1, 2
1M + 1A
1A
1A
6. 0425 2 xpx
20400
0)25)(4(42
2
pp
p
1M + 1A
1A + 1A
1M for using Δ
7. (a) Let x kg be Ron’s original weight.
60
7.56)1.01)(05.01(
xx
∴ Ron’s weight was 60 kg.
(b) weight loss
kg3.6
7.5663
1M
1A
1M
1A
1M for 1.05 or 0.9
u – 1 for missing unit
1M for 63
u – 1 for missing unit
8. (a) 5
23
2
xxf
Put x = 2, 52
)2(322
f
81 f
Put x = 4, 52
)4(324
f
112 f
(b) 52
)2(322
xxf
53)( xxf
1M
1A
1A
1M
1A
9. (a) A’ = (3, 9) B’ = (5, –2)
(b) Let P be (x, y)
061224442510811896
)2()5()9()3(2222
2222
yxyyxxyyxx
yxyx
2A
1M + 1A
1A
1M for either side correct
Section A2
Solution Marks Remarks
10
11 (a) : = 3 : 5
and ACB : BAC = 3 : 5 (arc prop. to s)
Let ACB = 3x. Then BAC = 5x.
In ABC,
3x + 5x + 60 = 180 ( sum of )
8x = 120
x = 15
BAC = 5(15)
= 75
(b) BOC= 2BAC ( at centre twice at ce)
= 2(75)
= 150
In BOC,
OB = OC (radii of the same circle)
OBC = BCO (base s, isos. )
OCB =
2150180
( sum of )
= 15
1M
1M + 1A
1A
1A
1M
1A
12 (a) Since the graph passes through the origin, the y-intercept is 0,
i.e., c = 0
The axis of symmetry is x = 5. 1025 k
(b) Since the graphs passes through the point (2, –16) and (10, 0), we have
)2()2(16)10()10(0
2
2
baba
)2.........(28)1(..........100
baba
(1) – (2) gives 1 a and 10b
(c) The x-coordinate of the vertex = 5
The y-coordinate of the vertex
)5(10)5( 2
25
The coordinates of the vertex are (5, –25).
1A
1A
1M
1A + 1A
1M
1A
For either one correct
13
14 (a) Slope of BC
41
)4(4)2(0
BCAD
Slope of AD 4
411
The equations of AD:
)0(46 xy
064 yx
Slope of AC 23
0460
ACBE
Slope of BE 32
231
The equation of BE:
8263
)]4([32)2(
xy
xy
0232 yx
1M
1A
1A
(b)
)2..(..........0232)1....(..........064
yxyx
From (1),
)3....(..........46 xy
Substituting (3) into (2),
1614021218202)46(32
xxx
xx
78
x
Substituting 78
x into (3),
7846y
710
The coordinates of P are .7
10 ,78
1M
1A
(c) Slope of CQ
Slope of CP 21
478
07
10
Slope of AB 2)4(0)2(6
Slope of CQ Slope of AB 1221
CQ is perpendicular to AB.
(d) Orthocentre
1M
1A
1A
15 (a) )4)(32( ii
232128 iii i1011
(b) )5)(2()4)(32( iibiia
)5)(2()1011( ibiia
bibiaia 1021011 2 ibaba )1010()211(
)5)(2()4)(32( iibiia
ii 4)4(6
ibaba )1010()211(
i1024
By comparing the real and imaginary parts,
10101024211
baba
)2.(....................1)1....(..........24211
baba
:)2(2)1(
2613
22422211
ababa
2a
Substituting 2a into (2),
12 b 1b
1A
1M
1A
1A
For either one correct
16. 0922 xx Sum of roots
12
2
Product of roots
9
(a)
22
2)( 2
)9(
)9(2)2( 2
922
(b) Sum of roots
922
Product of roots
1
The required equation is
019222
xx
09229 2 xx
1A
1M
1A
1A
1A
For both sum and product
17
18 (a)(i) 854550180PTQ
PTQTQTPTQTPPQ cos))((2222
85cos)24)(7(2247 22
4.24PQ (cor. to 3 sig. fig.)
The distance between P and Q is 24.4 m
(ii) PQT
PTPTQ
PQ
sinsin
PQT
sin
785sin
40728.24
6.16PQT
The compass bearing of P from Q is N 4.28 W.
1M
1A
1M
1A
1A
(b) Let x m be the perpendicular distance from T to PQ.
x 4073.242185sin247
21
86.6x which is greater than 6.
Calvin’s claim is not correct.
1M
1A
19 (a) TQR = 100 ( in alt. seg.)
PTQ + 50 = TQR (ext. of )
PTQ + 50 = 100
PTQ = 50
PTQ = TPQ = 50
PQ = QT (sides opp. equal s)
PQT is an isosceles triangle. Marking Scheme
Case 1 Any correct proof with correct reasons. 3
Case 2 Any correct proof without correct reasons. 2
Case 3 Any relevant correct argument with correct reasons. 1
19 (b) (i) ATR = TPR + PRT (ext. of )
100 = 50 + PRT
PRT = 50
PRT = TPR = 50
RT = PT (sides opp. equal s) Marking Scheme
Case 1 Any correct proof with correct reasons. 2
Case 2 Any correct proof without correct reasons. 1
(ii) PTQ = RPT = 50 (proved)
TPQ = PRT = 50 (proved)
PQT = 180 PTQ TPQ
= 180 RPT PRT
= RTP
PRT TPQ (A.A.A.)
PQRT
=
TPPR
(corr. sides, ~s)
PQ6 =
6QRPQ
36 = PQ(PQ + 3)
PQ 2 + 3PQ 36 = 0
PQ =
2)36)(1(433 2 cm
= 4.68 cm (correct to 2 d.p.)
1M
1A
19 (c) Join OR and OQ.
QTR = 180 ATR PTQ (adj. s on st. line)
= 180 100 50
= 30
QOP = 2QTR ( at centre twice at ce)
= 2(30)
= 60
OR = OQ (radii of the same circle)
RQO = QRO (base s, isos. )
RQO + QRO + 60= 180 ( sum of )
2RQO = 120
RQO = 60
ROQ is an equilateral triangle. The length of QR equals the radius of the circle.
Marking Scheme
Case 1 Any correct proof with correct reasons. 3
Case 2 Any correct proof without correct reasons. 2
Case 3 Any relevant correct argument with correct reasons. 1