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SULIT 3472/1
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MODUL KECEMERLANGAN PERCUBAAN SPM 2016
TINGKATAN 5
ADDITIONAL MATHEMATICS
Kertas 1
2 Jam
JANGAN BUKA KERTAS SOALAN INI
SEHINGGA DIBERITAHU
1. Tulis nombor kad pengenalan dan angka
giliran anda dalam ruang yang
disediakan.
2. Kertas soalan ini adalah dalam dwibahasa.
3. Soalan dalam bahasa Inggeris medahului
soalan yang sepadan dalam bahasa
Melayu.
4. Calon dibenarkan menjawab keseluruhan
atau sebahagian soalan sama ada dalam
bahasa Inggeris atau bahasa Melayu.
5. Calon dikehendaki membaca maklumat di
halaman belakang kertas soalan ini.
Kertas soalan ini mengandungi 20 halaman bercetak.
Examiner’s code
Question
Full Marks
obatined Marks Acquired
1 3
2 2
3 4
4 3
5 3
6 3
7 3
8 3
9 4
10 3
11 4
12 3
13 3
14 2
15 3
16 3
17 3
18 4
19 3
20 3
21 4
22 4
23 3
24 3
25 4
Total 80
3472/1
Tingkatan 5
Additional Mathematics
Kertas 1
Ogos/Sep
2 Jam
NAMA :_________________________________
KELAS :________________________________
ANGKA GILIRAN:______________________
NO KP:_________________________________
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ALGEBRA
1 a
acbbx
2
42 8
a
bb
c
ca
log
loglog
2 am x a
n = a
m + n 9 Tn = a + (n -1)d
3 am
an = a
m – n 10 Sn = 2 ( 1)
2
na n d
4 ( am
)n = a
m n 11 Tn = 1nar
5 nmmn aaa logloglog 12 ( 1) (1 )
, 11 1
n n
n
a r a rS r
r r
6 nmn
maaa logloglog 13 , 1
1
aS r
r
7 log a mn = n log a m
CALCULUS
KALKULUS
1 dx
duv
dx
dvu
dx
dyuvy ,
4 Area under a curve
=b
a
y dx or (atau)
= b
a
x dy
2 2
,v
dx
dvu
dx
duv
dx
dy
v
uy
3 dx
du
du
dy
dx
dy
5 Volume generated
= 2
b
a
y dx or (atau)
= 2
b
a
x dy
Luas di bawah lengkung
Isipadu kisaran
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STATISTICS
STATISTIK
1 N
xx
8 !
( )!
nr
nP
n r
2 f
fxx
9 !
( )! !
nr
nC
n r r
3 2
22
xN
x
N
xx
10 ( ) ( ) ( ) ( )P A B P A P B P A B
4 2
22
xf
fx
f
xxf
11 P(X = r) = , 1n r n rrC p q p q
5 Cf
FN
Lmm
2
1
12 Mean/Min , 𝜇 = np
6 1
0
100Q
IQ
13 npq
7 i
ii
W
IWI
14
XZ
GEOMETRY
GEOMETRI
1 Distance/Jarak = 2 22 1 2 1( ) ( )x x y y
4 Area of a triangle/Luas segi tiga
= 21
)()( 312312133221 yxyxyxyxyxyx
5. 22
~yxr
2 Midpoint/Titik Tengah
(𝑥 , 𝑦) =
2,
2
2121 yyxx
6. 22
~~
~
^
yx
jyix
r
3 A point dividing a segment of a line
(x , y ) = n+m
my+ny,
n+m
mx+nx 2121
Titik yang membahagi suatu tembereng garis
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TRIGONOMETRY
TRIGONOMETRI
1 Arc length, s =r
Panjang lengkok,𝑠 = 𝑗𝜃
8 sin(𝐴 ± 𝐵) = 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠 𝐵 ± cos𝐴 sin𝐵
sin(𝐴 ± 𝐵) = 𝑠𝑖𝑛𝐴𝑘𝑜𝑠 𝐵 ± kos𝐴 sin𝐵
2 Area of a sector, 21
2A r
Luas sektor , 𝐿 = 1
2 𝑗2𝜃
9 cos(𝐴 ± 𝐵) = cos𝐴 𝑐𝑜𝑠 𝐵 ∓ sin𝐴 sin𝐵
kos(𝐴 ± 𝐵) = kos𝐴 𝑘𝑜𝑠 𝐵 ∓ sin𝐴 sin𝐵
3 . 𝑠𝑖𝑛2𝐴 + 𝑐𝑜𝑠2𝐴 = 1
𝑠𝑖𝑛2𝐴 + 𝑘𝑜𝑠2𝐴 = 1 10 tan tan
tan( )1 tan tan
A BA B
A B
4 2 2sec 1 tanA A
𝑠𝑒𝑘2𝐴 = 1 + 𝑡𝑎𝑛2𝐴
11 2
2tantan 2
1 tan
AA
A
5 2 2cos 1 cotec A A
𝑘𝑜𝑠𝑒𝑘2𝐴 = 1 + 𝑘𝑜𝑡2𝐴 12
sin sin sin
a b c
A B C
6 sin 2A = 2 sin A cos A sin2𝐴 = 2 sin𝐴 𝑘𝑜𝑠 𝐴
13 2 2 2 2 cosa b c bc A
𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 𝑘𝑜𝑠 𝐴
7 cos 2A = 2 2cos sinA A
= 22cos 1A
= 21 2sin A
𝑘𝑜𝑠 2𝐴 = 𝑘𝑜𝑠2𝐴 − 𝑠𝑖𝑛2 𝐴
= 2 𝑘𝑜𝑠2𝐴 − 1 = 1 − 2𝑠𝑖𝑛2 𝐴
14 Area of triangle/Luas segi tiga
= 1
sin2
ab C
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1
Answer all questions
Jawab semua soalan The relation from set A ={ -2, -1, 2, 4, 5 } to set B = {1, 4, 16, 25 } is as shown in the ordered pairs below: {( -2, 4 ), ( -1, 1 ), ( 2, 4 ), ( 4, 16 ), ( 5, 25 )} Hubungan dari set A.={ -2, -1, 2, 4, 5} ke set B = {1, 4,16, 25 } adalah ditunjukkan dalam set pasangan bertertib berikut: {( -2, 4 ), (-1, 1), ( 2, 4 ), ( 4, 16 ), ( 5, 25 ) }
State Nyatakan (a) the objects of 4, objek-objek bagi 4, (b) the codomain of the relation, kodomain hubungan itu, (c) whether this relation is a function. [3 marks]
sama ada hubungan ini ialah satu fungsi. [3 markah]
Answer/Jawapan: (a) (b) (c)
2
Given bx→x:f 2 ,where b is a constant and 𝑓2 (3) = 18 , find the value of b.
Diberi bx→x:f 2 ,dengan keadaan b ialah pemalar dan 𝑓2 (3) = 18 , cari nilai b.
[2 marks] [2 markah]
Answer /Jawapan:
3
1
2
2
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3 It is given the functions ℎ(𝑥) = 2 + 5𝑥 and 𝑘(𝑥) = 𝑝𝑥 − 10, where 𝑝 is a constant. Diberi fungsi ℎ(𝑥) = 2 + 5𝑥 dan 𝑘(𝑥) = 𝑝𝑥 − 10, dengan keadaan 𝑝 ialah pemalar. Find Cari
(a) ℎ−1 (7),
(b) the value of 𝑝 such that 𝑘−1(2) = ℎ(2) nilai p dengan keadaan 𝑘−1(2) = ℎ(2) . . [4 marks] [4 markah] Answer/Jawapan:
(a)
(b)
4
It is given that x-axis is the tangent to the function 𝑓(𝑥) = 𝑚𝑥2 + (3𝑚 − 5)𝑥 + 𝑚 at its minimum point. Find the possible values of m.
Diberi paksi-x adalah tangen kepada fungsi 𝑓(𝑥) = 𝑚𝑥2 + (3𝑚 − 5)𝑥 + 𝑚 pada titik minimumnya. Cari nilai-nilai yang mungkin bagi m. [3 marks] [3 markah]
Answer/Jawapan:
3
4
4
3
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5 Diagram 5 shows the graph of a quadratic function 62)3()( 2 kxxf , where k is
a constant.
Rajah 5 menunjukkan graf fungsi kuadratik 62)3()( 2 kxxf , dengan keadaan k
ialah pemalar.
(a) State the equation of the axis of symmetry of the curve. Nyatakan persamaan paksi simetri bagi lengkung itu. (b) Given that the minimum value of the function is 5, find the value of k. Diberi nilai minimum bagi fungsi itu ialah 5, cari nilai k.
[3 marks] Answer/Jawapan: [3 markah]
(a)
(b)
6
It is given that a right cylinder has a height 6 cm and a radius of r cm. Find the range of values of r for the total surface area of the cylinder to exceed
224𝜋 cm2. Diberi satu silinder tegak dengan ketinggian 6 cm dan jejari r cm.
Cari julat nilai r dengan keadaan jumlah luas permukaan silinder itu melebihi 224𝜋 cm2. [3 marks] [3 markah] Answer/Jawapan:
3
5
3
6
Diagram 5
Rajah 5
x
f(x)
O
5 .
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7 Solve the equation
2𝑥−3
8−𝑥 = 32
2 𝑥
[3 marks]
Selesaikan persamaan 2𝑥−3
8−𝑥 = 32
2 𝑥 [3 markah]
Answer /Jawapan:
8
Express 2 𝑙𝑜𝑔103√𝑥 − 𝑙𝑜𝑔10 (4
3𝑥2) + 𝑙𝑜𝑔1004𝑥3 in the form 𝑎 𝑙𝑜𝑔10 𝑥 + 𝑙𝑜𝑔10𝑏,
where a and b are constants. [3 marks]
Ungkapkan 2 𝑙𝑜𝑔103√𝑥 − 𝑙𝑜𝑔10 (4
3𝑥2) + 𝑙𝑜𝑔1004𝑥3 dalam bentuk 𝑎 𝑙𝑜𝑔10 𝑥 + 𝑙𝑜𝑔10𝑏,
dengan keadaan a and b ialah pemalar . [3 markah]
Answer /Jawapan:
3
7
3
8
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9
Firdaus and Aishah decided to save some money each month, starting January. Firdaus saved 80 cents each month. Aishah saved in such a way that she had 1 cent at the end of January A total of 2 cents at the end of February, A total of 4 cents at the end of March, and so on, so that the total amount she had saved at the end of each month twice the total amount from the previous month.. Firdaus dan Aishah membuat keputusan untuk menyimpan duit mulai bulan Januari. Firdaus menyimpan 80 sen setiap bulan. Aishah pula menyimpan duit dengan keadaan bahawa pada akhir bulan Januari, dia menyimpan 1 sen, jumlah 2 sen pada akhir bulan Februari dan 4 sen pada akhir bulan Mac dan seterusnya supaya jumlah duit disimpan pada setiap akhir bulan adalah dua kali ganda jumlah simpanan pada bulan sebelumnya. Find/Cari (a) how much, in cents, did Aishah save during June ? berapakah, dalam sen,yang Aisiah simpan semasa bulan Jun ? (b) who will save more at the end of the year ? By how much? siapa akan menyimpan lebih banyak pada akhir tahun? Melebihi berapa banyak? [4 marks] [4 markah] Answer/Jawapan: (a) (b)
10
Diagram 10 shows a pendulum released from a position 𝑂𝑃. Rajah 10 menunjukan sebuah bandul dilepaskan dari posisi 𝑂𝑃.
The pendulum will swing freely through angles in the sequence of 70°, 56°, 44.8°, 35.84°,... Calculate the total angles that the pendulum covers before it stops completely. Bandul tersebut akan mengayun dengan bebas melalui sudut dalam jujukan 70°, 56°, 44.8°, 35.84°,... Hitungkan jumlah sudut yang diliputi oleh bandul tersebut sebelum ia berhenti sepenuhnya. [3 marks] [3 markah] Answer/Jawapan :
4
9
3
10
70°
56°
44.8°
O
P
Diagram 10 Rajah 10
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11 Diagram 11 shows a straight line graph of 𝑙𝑜𝑔10𝑦 against 𝑙𝑜𝑔10𝑥 . Variables x and y
are related by the equation 𝑦 = ℎ𝑥3 where ℎ is a constant. Rajah 11 menunjukkan satu graf garis lurus yang diplotkan 𝑙𝑜𝑔10𝑦 melawan 𝑙𝑜𝑔10𝑥 .
Pembolehubah x dan y dihubungkan dengan persamaan 𝑦 = ℎ𝑥3 dengan keadaan ℎ ialah pemalar.
(a) Convert 𝑦 = ℎ𝑥3 in linear form.
Tukarkan 𝑦 = ℎ𝑥3 dalam bentuk linear. (b) Find the value of Cari nilai (i) 𝑙𝑜𝑔10ℎ
(ii) 𝑘 [4 marks] [4 markah]
Answer/Jawapan:
(a) (b) (i) (ii)
12
Solve the equation 2 sin(2𝑥 − 10°) = 1.2516 for 0° ≤ 𝑥 ≤ 360° Selesaikan persamaan 2 sin(2𝑥 − 10°) = 1.2516 𝑢𝑛𝑡𝑢𝑘 0° ≤ 𝑥 ≤ 360° . [3 marks] [3 markah] Answer/Jawapan:
4
11
𝑙𝑜𝑔10𝑦
𝑙𝑜𝑔10𝑥
(k, 6)
2
Diagram 11 Rajah 11
3
12
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13 Given that 𝜃 is an acute angle such that cos 𝜃 =1
2. Without using calculator,
find the value of
Diberi bahawa 𝜃 adalah satu sudut tirus dengan keadaan kos 𝜃 =1
2. Tanpa
menggunakan kalkulator, cari nilai bagi
(a) 𝑐𝑜𝑠𝜃 sin(90° − θ), 𝑘𝑜𝑠𝜃 sin(90° − θ),
(b) 2 tan𝜃 + tan(90° − 𝜃). [3 marks] [3 markah] Answer/Jawapan: (a) (b)
14
Diagram 14 shows a sector POQ of a circle with centre O. Rajah 14 menunjukkan sektor POQ bagi sebuah bulatan berpusat O. Given that the area of the sector POQ is 36 cm2 and arc length of PQ is 12 cm.
Find the value of 𝜃, in radian. Diberi bahawa luas sektor POQ ialah 36 cm2 dan panjang lengkok PQ ialah 12 cm.
Cari nilai 𝜃, dalam radian. [2 marks] [2 markah] Answer/Jawapan:
3
13
2
14
𝜃
P
Q O
Diagram 14 Rajah 14
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15 Diagram 15 shows the position of two houses, 𝑃 and 𝑄.
Rajah 15 menunjukkan kedudukan dua buah rumah , 𝑃 dan 𝑄.
The coordinates of house P and house Q are (-1,4) and (3,6) respectively. There is a rose garden situated along the straight road which connects the two houses. The distance of the rose garden R to house P and to house Q are in the ratio of 2:1. Find the coordinates of the rose garden R. [3 marks]
Koordinat bagi rumah P dan rumah Q masing-masing ialah (-1,4) dan (3,6). Sepanjang jalan lurus yang menyambungi rumah P dan rumah Q, terdapat sebuah taman bunga ros R. Jarak taman bunga ros R dari rumah P dan rumah Q adalah dalam nisbah 2 : 1. Cari koordinat untuk bunga ros R. [3 markah] Answer/Jawapan:
3
15
House Q Rumah Q
0 x
y
Rose Garden R taman bunga ros R
Diagram 15 Rajah 15
House P Rumah P
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16
17
Diagram 16 shows 𝑂𝑃 on a cartesian plane.
Rajah 16 menunjukkan 𝑂𝑃 pada satah Cartesan .
(a) Given that 𝑂𝑄 = 2 𝑂𝑃 , express 𝑂𝑄 in the form of (𝑥𝑦) .
Diberi bahawa 𝑂𝑄 = 2 𝑂𝑃 , ungkapkan 𝑂𝑄 dalam bentuk (𝑥𝑦).
(b) Given that point P1 is the reflection of point P in the y-axis.
Find 𝑄𝑃1 in the form of 𝑥𝑖 + 𝑦𝑗. [3 marks]
Diberi bahawa titik P1 ialah pantulan titik P pada paksi-y.
Cari 𝑄𝑃1 dalam sebutan 𝑥𝑖 + 𝑦𝑗. [3 markah]
Answer/Jawapan : (a) (b)
Diagram 17 shows a triangle POR with 𝑂𝑅 = 3𝑏 and 𝑂𝑃 = 2𝑎.
Rajah 17 menunjukkan segi tiga POR dengan 𝑂𝑅 = 3𝑏 and 𝑂𝑃 = 2𝑎.
Given that 2𝑃𝑄 = 𝑄𝑅 , express 𝑄𝑅 in terms of 𝑎 𝑎𝑛𝑑 𝑏 . [ 3 marks]
Diberi bahawa 2𝑃𝑄 = 𝑄𝑅 , ungkapkan 𝑄𝑅 𝑑𝑎𝑙𝑎𝑚 𝑠𝑒𝑏𝑢𝑡𝑎𝑛 𝑎 𝑑𝑎𝑛 𝑏 . [3 markah]
Answer/Jawapan :
P(3,4)
O x
y
Diagram 16 Rajah 16
3
16
R
Q
P O
Diagram 17 Rajah 17
3
17
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18
Given that ∫ ℎ(𝑥)𝑑𝑥 = 45
1, find the value of
Diberi ∫ ℎ(𝑥)𝑑𝑥 = 45
1, cari nilai bagi
(a) ∫ 2ℎ(𝑥)𝑑𝑥5
1
(b) 𝑘 if ∫ ℎ(𝑥)𝑑𝑥 + ∫ [ℎ(𝑥) + 𝑘𝑥]𝑑𝑥 = 115
2
2
1
𝑘 jika ∫ ℎ(𝑥)𝑑𝑥 + ∫ [ℎ(𝑥) + 𝑘𝑥}𝑑𝑥 = 115
2
2
1 [ 4 marks]
[4 markah] Answer/Jawapan:
(a) (b)
19 Given that 𝑦 = 2𝑥(𝑥 − 6), find Diberi 𝑦 = 2𝑥(𝑥 − 6), cari
(a) 𝑑𝑦
𝑑𝑥 ,
(b) the value of x when y is minimum, nilai x apabila y adalah minimum, (c) the minimum value of y. nilai minimum bagi y. [3 marks] [3 markah] Answer/Jawapan:
(a) (b) (c)
3
19
4
18
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20 The surface area of a sphere increases at the rate of 15𝜋 cm2s-1. Luas permukaan sebuah sfera bertambah pada kadar 15𝜋 cm2s-1. Find the rate of change of the radius, in cm s-1 at the instant when its radius is 10 cm. Cari kadar perubahan jejari sfera itu dalam cm s-1 pada ketika jejarinya ialah 10 cm. [surface area of sphere/luas permukaan sfera = 4𝜋r2 ] [3 marks] [3 markah] Answer/Jawapan:
21 Diagram 21 shows seven cards of different letters Rajah 21 menunjukkan tujuh kad dengan huruf yang berlainan.
If four letters are to be chosen, find Jika empat huruf hendak dipilih, cari (a) the number of possible arrangements, in a row, of the cards. bilangan cara susunan yang mungkin dalam satu baris, bagi kad-kad tersebut. (b) the number of these arrangements in which it ends with a vowel. bilangan cara susunan itu dengan keadaan ia berakhir dengan huruf vokal. [ 4 marks]
[4 markah] Answer/Jawapan:
(a)
(b)
3
20
B A U X I T E
Diagram 21 Rajah 21
4
21
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22 The number of errors, x, on each of 200 pages of an e-book was monitored. The results were summarised and shown below: Bilangan kesilapan, x, dari setiap 200 muka surat daripada sebuah e-buku telah dipantau. Keputusan tersebut telah diringkaskan dan ditunjukkan di bawah:
∑𝑥 = 920 ∑𝑥2 = 5032
(a) Calculate the mean and the standard deviation of the number of errors per page. Hitungkan nilai min dan sisihan piawai untuk bilangan kesilapan untuk setiap muka surat. (b) A further 50 pages of the e-book was monitored and it was found that the mean was 4.4 errors and the standard deviation was 2.2 errors. 50 muka surat dari e-buku dipantau lagi dan didapati min ialah 4.4 kesilapan dan sisihan piawai ialah 2.2 kesilapan. Find the mean and standard deviation of the number of errors per page for the 250 pages. Cari min dan sishan piawai untuk bilangan kesilapan untuk setiap muka surat untuk 250 muka surat. [4 marks] [4 markah] Answer/Jawapan:
(a)
(b)
4
22
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23 In a class of 20 students, 4 of the 9 boys and 3 of the 11 girls are in the athletics team.
A person from the class is chosen to be in the long jump event on Sports Day. Dalam satu kelas yang mempunyai 20 murid-murid, 4 dari 9 murid lelaki dan 3 dari 11 murid perempuan berada dalam pasukan olahraga.Seorang murid akan dipilih untuk acara lompat jauh pada hari sukan. Find the probability that the person chosen is Cari kebarangkalian murid yang dipilih itu adalah
(a) a female member of the athletics team,
seorang murid perempuan dari pasukan olahraga,
(b) a female or a member in the athletics team. [3 marks]
seorang murid perempuan atau murid di pasukan olahraga [3 markah] Answer/Jawapan: (a) (b)
24
The discrete random variable X has a binomial probability distribution with n= 4, where n is the number of trials. Diagram 24 shows the probablity of X. Pemboleh ubah rawak diskret X mempunyai satu taburan kebarangkalian binomial dengan n =4, di mana n ialah bilangan percubaan. Rajah 24 menunjukkan taburan kebarangkalian X.
Find/Cari
(a) the value of 𝑘, nilai 𝑘, (b) P ( 𝑋 ≥ 3) [3 marks] [3 markah] Answer/Jawapan: (a) (b)
3
24
3
23
0 2 3 4 1 x
k
4k
6k
P(X=x)
Diagram 24 Rajah 24
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25 X is a continuous random variable representing the time taken, in minutes, for a group of students to solve a calculus problem.
It is given that 𝑋 ~ 𝑁( 6.7, 1.44). X ialah pembolehubah rawak selanjar yang mewakili masa yang diambil, dalam minit, bagi satu kumpulan murid untuk menyelesaikan satu masalah kalkulus. Diberi 𝑋 ~ 𝑁( 6.7, 1.44). Find Cari (a) the z-score when X = 8.2 skor-z apabila X = 8.2 (b) the value of k when P( 𝑧 ≤ 𝑘)= 0.8729
nilai k apabila P( 𝑧 ≤ 𝑘)= 0.8729
[4 marks] [4 markah]
Answer/Jawapan : (a) (b)
END OF QUESTION PAPER
KERTAS SOALAN TAMAT
4
25
SULIT 3472/1
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3472/1 SULIT
19
THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1 )
KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)
z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
Minus / Tolak
0.0
0.1
0.2
0.3
0.4
0.5000
0.4602
0.4207
0.3821
0.3446
0.4960
0.4562
0.4168
0.3783
0.3409
0.4920
0.4522
0.4129
0.3745
0.3372
0.4880
0.4483
0.4090
0.3707
0.3336
0.4840
0.4443
0.4052
0.3669
0.3300
0.4801
0.4404
0.4013
0.3632
0.3264
0.4761
0.4364
0.3974
0.3594
0.3228
0.4721
0.4325
0.3936
0.3557
0.3192
0.4681
0.4286
0.3897
0.3520
0.3156
0.4641
0.4247
0.3859
0.3483
0.3121
4
4
4
4
4
8
8
8
7
7
12
12
12
11
11
16
16
15
15
14
20
20
19
19
18
24
24
23
22
22
28
28
27
26
25
32
32
31
30
29
36
36
35
34
32
0.5
0.6
0.7
0.8
0.9
0.3085
0.2743
0.2420
0.2119
0.1841
0.3050
0.2709
0.2389
0.2090
0.1814
0.3015
0.2676
0.2358
0.2061
0.1788
0.2981
0.2643
0.2327
0.2033
0.1762
0.2946
0.2611
0.2296
0.2005
0.1736
0.2912
0.2578
0.2266
0.1977
0.1711
0.2877
0.2546
0.2236
0.1949
0.1685
0.2843
0.2514
0.2206
0.1922
0.1660
0.2810
0.2483
0.2177
0.1894
0.1635
0.2776
0.2451
0.2148
0.1867
0.1611
3
3
3
3
3
7
7
6
5
5
10
10
9
8
8
14
13
12
11
10
17
16
15
14
13
20
19
18
16
15
24
23
21
19
18
27
26
24
22
20
31
29
27
25
23
1.0
1.1
1.2
1.3
1.4
0.1587
0.1357
0.1151
0.0968
0.0808
0.1562
0.1335
0.1131
0.0951
0.0793
0.1539
0.1314
0.1112
0.0934
0.0778
0.1515
0.1292
0.1093
0.0918
0.0764
0.1492
0.1271
0.1075
0.0901
0.0749
0.1469
0.1251
0.1056
0.0885
0.0735
0.1446
0.1230
0.1038
0.0869
0.0721
0.1423
0.1210
0.1020
0.0853
0.0708
0.1401
0.1190
0.1003
0.0838
0.0694
0.1379
0.1170
0.0985
0.0823
0.0681
2
2
2
2
1
5
4
4
3
3
7
6
6
5
4
9
8
7
6
6
12
10
9
8
7
14
12
11
10
8
16
14
13
11
10
19
16
15
13
11
21
18
17
14
13
1.5
1.6
1.7
1.8
1.9
0.0668
0.0548
0.0446
0.0359
0.0287
0.0655
0.0537
0.0436
0.0351
0.0281
0.0643
0.0526
0.0427
0.0344
0.0274
0.0630
0.0516
0.0418
0.0336
0.0268
0.0618
0.0505
0.0409
0.0329
0.0262
0.0606
0.0495
0.0401
0.0322
0.0256
0.0594
0.0485
0.0392
0.0314
0.0250
0.0582
0..0475
0.0384
0.0307
0.0244
0.0571
0.0465
0.0375
0.0301
0.0239
0.0559
0.0455
0.0367
0.0294
0.0233
1
1
1
1
1
2
2
2
1
1
4
3
3
2
2
5
4
4
3
2
6
5
4
4
3
7
6
5
4
4
8
7
6
5
4
10
8
7
6
5
11
9
8
6
5
2.0
2.1
2.2
2.3
0.0228
0.0179
0.0139
0.0107
0.0222
0.0174
0.0136
0.0104
0.0217
0.0170
0.0132
0.0102
0.0212
0.0166
0.0129
0.00990
0.0207
0.0162
0.0125
0.00964
0.0202
0.0158
0.0122
0.00939
0.0197
0.0154
0.0119
0.00914
0.0192
0.0150
0.0116
0.00889
0.0188
0.0146
0.0113
0.00866
0.0183
0.0143
0.0110
0.00842
0
0
0
0
3
2
1
1
1
1
5
5
1
1
1
1
8
7
2
2
1
1
10
9
2
2
2
1
13
12
3
2
2
2
15
14
3
3
2
2
18
16
4
3
3
2
20
18
4
4
3
2
23
21
2.4 0.00820 0.00798 0.00776 0.00755 0.00734
0.00714
0.00695
0.00676
0.00657
0.00639
2
2
4
4
6
6
8
7
11
9
13
11
15
13
17
15
19
17
2.5
2.6
2.7
2.8
2.9
0.00621
0.00466
0.00347
0.00256
0.00187
0.00604
0.00453
0.00336
0.00248
0.00181
0.00587
0.00440
0.00326
0.00240
0.00175
0.00570
0.00427
0.00317
0.00233
0.00169
0.00554
0.00415
0.00307
0.00226
0.00164
0.00539
0.00402
0.00298
0.00219
0.00159
0.00523
0.00391
0.00289
0.00212
0.00154
0.00508
0.00379
0.00280
0.00205
0.00149
0.00494
0.00368
0.00272
0.00199
0.00144
0.00480
0.00357
0.00264
0.00193
0.00139
2
1
1
1
0
3
2
2
1
1
5
3
3
2
1
6
5
4
3
2
8
6
5
4
2
9
7
6
4
3
11
8
7
5
3
12
9
8
6
4
14
10
9
6
4
3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4
For negative z use relation:
Bagi z negatif guna hubungan:
Q(z)
z
f (z)
O k
Example / Contoh:
If X ~ N(0, 1), then
Jika X ~ N(0, 1), maka
P(X > k) = Q(k)
P(X > 2.1) = Q(2.1) = 0.0179
SULIT 3472/1
3472/1 SULIT
20
INFORMATION FOR CANDIDATES
MAKLUMAT UNTUK CALON
1. This question paper consists of 25 questions.
Kertas soalan ini mengandungi 25 soalan.
2. Answer ALL questions.
Jawab semua soalan.
3. Write your answers in the spaces provided in the question paper.
Tulis jawapan anda dalam ruang yang disediakan dalam kertas soalan.
4. Show your working. It may help you to get marks.
Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu anda
untuk mendapatkan markah.
5. If you wish to change your answer, cross out the answer that you have done. Then write
down the new answer.
Sekiranya anda hendak menukar jawapan, batalkan jawapan yang telah dibuat. Kemudian
tulis jawapan yang baharu.
6. The diagrams in the questions provided are not drawn to scale unless stated.
Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.
7. The marks allocated for each question are shown in brackets.
Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan.
8. A list of formulae is provided on page 2 to 4.
Satu senarai rumus disediakan di halaman 2 hingga 4.
9. The Upper Tail Probability Q(z) For The Normal Distribution N(0,1) Table is provided on
page 19.
Jadual Kebarangkalian Hujung Atas Q(z) Bagi Taburan Normal N(0,1) disediakan di
halaman 19.
10. You may use a scientific calculator.
Anda dibenarkan menggunakan kalkulator saintifik.
11. Hand in this question paper to the invigilator at the end of the examination.
Serahkan kertas soalan ini kepada pengawas peperiksaan di akhir peperiksaan.
SULIT 3472/1
[See overleaf
3472/1 SULIT
21
MODUL KECEMERLANGAN PERCUBAAN SPM 2016
ADDITIONAL MATHEMATICS
Tingkatan 5
Kertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. Kertas soalan ini adalah dalam dwibahasa.
2. Soalan dalam Bahasa Inggeris mendahului soalan yang sepadan dalam Bahasa
Melayu.
3. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini.
Kertas soalan ini mengandungi 20 halaman bercetak.
The following formulae may be helpful in answering the questions. The symbols given are the ones
commonly used.
Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi adalah
yang biasa digunakan.
3472/2 ADDITIONAL
MATHEMATICS
Paper 2
August /September 2016 2 ½ hours
SULIT 3472/1
3472/1 SULIT
22
ALGEBRA
1 a
acbbx
2
42 8
a
bb
c
ca
log
loglog
2 amx a
n = a
m + n 9 ( 1)nT a n d
3 am a
n = a
m – n 10. 2 ( 1)
2n
nS a n d
4 ( am
)n = a
m n 11 1n
nT a r
5 nmmn aaa logloglog 12 1 1
, 11 1
n n
n
a r a rS r
r r
6 nmn
maaa logloglog 13 , 1
1
aS r
r
7 log a mn = n log a m
CALCULUS
KALKULUS
1 y = uv , dx
duv
dx
dvu
dx
dy
4 Area under a curve
Luas di bawah lengkung
= ( )
b b
a a
y dx or atau x dy
2 2
,v
dx
dvu
dx
duv
dx
dy
v
uy
5 Volume of revolution
Isipadu kisaran
b
a
dxy 2 or (atau) b
a
dyx 2
3 dx
du
du
dy
dx
dy
STATISTICS
STATISTIK
1 N
xx
7
i
ii
W
IWI
2 f
fxx
8
!
( )!
n
r
nP
n r
SULIT 3472/1
[See overleaf
3472/1 SULIT
23
3 2
22
xN
x
N
xx
9
!
( )! !
n
r
nC
n r r
4 2
22
xf
fx
f
xxf
10 ( ) ( ) ( ) ( )P A B P A P B P A B
5 Cf
FN
Lmm
2
1
11 ( ) , 1n r n r
rP X r C p q p q
12 / min,Mean np
6 1
0
100Q
IQ
13 npq
14 x
Z
GEOMETRY
GEOMETRI
1 Distance/jarak
= 212
2
12 yyxx
4 Area of a triangle/ Luas segitiga =
3123121332212
1yxyxyxyxyxyx
2 Mid point / Titik tengah
2,
2, 2121 yyxxyx
5 2 2
~r x y
3 A point dividing a segment of a line
Titik yang membahagi suatu
tembereng garis
nm
myny
nm
mxnxyx 2121 ,,
6 ^
~ ~
2 2~
x i y j
rx y
TRIGONOMETRY
TRIGONOMETRI
1 Arc length, s = r
Panjang lengkok, s= j
8 sin sin cos cos sinA B A B A B
sin sin s sinA B A kos B ko A B
2 Area of a sector, 21
2A r
Luas sektor, L =21
2j
9 cos cos cos sin sinA B A B A B
s os os sin sinko A B k A k B A B
SULIT 3472/1
3472/1 SULIT
24
3 2 2cos 1sin A A
2 2os 1sin A k A
10 tan tan
tan1 tan
A BA B
A tanB
4 2 2sec 1 tanA A
2 2se 1 tank A A
11 2
2 tantan 2
1 tan
AA
A
5 2 2sec 1 cotco A A
2 2se 1 otko k A k A
12 sin sin sin
a b c
A B C
6 sin 2A = 2sin A cos A
sin 2A = 2sin A kos A
13 2 2 2 2 cosa b c bc A
2 2 2 2a b c bc kos A
7 cos 2A = cos2 A – sin
2 A
= 2cos2A – 1
= 1 – 2sin2 A
kos 2A = kos2 A – sin
2 A
= 2kos2A – 1
= 1 – 2sin2 A
14 Area of triangle/ Luas segitiga
= 1
sin2
ab C
SULIT 3472/1
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3472/1 SULIT
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Section A Bahagian A
[40 marks] [40makah]
Answer all questions. Jawab semua soalan.
1. Solve the simultaneous equations 2𝑥 + 𝑦 − 5 = 0 and 𝑦2 − 2𝑥 = 5𝑥𝑦.
Give your answers correct to three decimal places. [5 marks]
Selesaikan persamaan serentak 2𝑥 + 𝑦 − 5 = 0 dan 𝑦2 − 2𝑥 = 5𝑥𝑦.
Beri jawapan anda betul kepada tiga tempat perpuluhan. [5 markah]
2. (a) The atmospheric pressure, 𝑥, measured in mm of mercury for the altitude,
ℎ(𝑥), measured in metres above sea level is given by
ℎ(𝑥) = (30𝑇 + 8000) log10
(760
𝑥)
where 𝑇 is the temperature in degree Celsius. Find the atmospheric pressure on
Mount Everest which has an altitude of approximately 8900 metres if the
temperature is 1C. Give your answer correct to one decimal place.[3 marks ]
Tekanan atmosfera , 𝑥, diukur dalam mm merkuri bagi ketinggian,ℎ(𝑥), diukur dalam meter di atas paras laut diberikan oleh
ℎ(𝑥) = (30𝑇 + 8000) log10
(760
𝑥)
di mana T ialah suhu dalam darjah Celsius. Cari tekanan atmosfera di Gunung
Everest pada suatu ketinggian lebih kurang 8900 meter jika suhu adalah 1C. Beri jawapan anda betul kepada satu tempat perpuluhan. [3 markah]
(b) Alfred deposited RM 4000 in a mutual trust fund. The total amount of his savings
after n years is given by 4000(1 + 𝑟)𝑛, where r is the interest rate per year. Calculate the minimum number of years it will take for his savings to exceed RM 7000 if the interest rate is 2.5% per year. [3 marks]
Alfred menyimpan RM 4000 dalam suatu tabung amanah bersama. Amaun
keseluruhan bagi simpanannya selepas n tahun adalah 4000(1 + 𝑟)𝑛, dimana r ialah kadar bunga setahun. Kirakan bilangan tahun minimum supaya simpanannya melebihi RM 7000 jika kadar bunga ialah 2.5% setahun.
[3 markah]
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26
3. Solution by scale drawing is not accepted.
Penyelesaian secara lukisan berskala tidak diterima.
Diagram 3 shows three points, P, Q and S in a Cartesian plane. The straight line PQ is perpendicular to the straight line QS which intersect the y-axis at S. The equation of the line QS is y = 2x – 5. Rajah 3 menunjukkan tiga titik P, Q dan S pada satu satah Cartesan. Garis lurus PQ berserenjang dengan garis lurus QS yang bersilang dengan paksi-y pada titik S.
Persamaan garis lurus QS ialah y = 2x – 5.
(a) Find the equation of the straight line PQ. [2 marks] Cari persamaan garis lurus PQ. [2 markah]
(b) The straight line PQ is extended to a point R such that PQ : QR = 2 : 3. Find the
coordinates of R. [3 marks] Garislurus PQ dipanjangkan ke titik R dengan keadaan PQ:QR = 2 : 3. Cari koordinat titik R. [3 markah]
(c) Point M moves such that its distance from P is equal to its distance from Q. Find the
equation of the locus M. [2 marks] Titik M bergerak dengan keadaan jaraknya dari P adalah sama dengn jaraknya dari Q. Cari persamaan lokus M. [2 markah]
S
x O
y = 2x – 5
Q
P(3, 6)
y
Diagram 3 Rajah 3
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4. (a) Sketch the graph of 𝑦 = −4 sin3
2𝑥 for 0 ≤ 𝑥 ≤ 2𝜋 . [4 marks]
Lakar graf bagi 𝑦 = −4sin3
2𝑥 untuk 0 ≤ 𝑥 ≤ 2𝜋 . [4 markah]
(b) Hence, using tha same axes, sketch a suitable graph to find the number of
solutions for the equation 4 sin3
2𝑥 −
𝜋
𝑥= 0 for 0 ≤ 𝑥 ≤ 2𝜋 .
State the number of solutions. [3 marks] Seterusnya, dengan menggunakan paksi yang sama, lakar satu graf yang sesuai
untuk mencari bilangan penyelesaian bagi persamaan
4 sin3
2𝑥 −
𝜋
𝑥= 0 untuk 0 ≤ 𝑥 ≤ 2𝜋 .
Nyatakan bilangan penyelesaian itu. [3 markah] 5. A teacher bought a house at 𝑥 . The value of the house appreciates by 7% annually.
It is given that the value of the house after 2 years is RM110000.
Seorang guru membeli sebuah rumah dengan harga 𝑅𝑀𝑥 . Nilai rumah meningkat dengan kadar 7 % setiap tahun. Diberi bahawa nilai rumah selepas 2 tahun adalah RM110000 . Find Cari
(a) the value of 𝑥, [2 marks] nilai 𝑥 [2 markah]
(b) the estimated value of the house 8 years after purchase, [2 marks] nilai anggaran rumah 8 tahun selepas pembeliannya, [2 markah]
(c) the minimum number of years for the value of the house to be double its original value. [3 marks]
bilangan tahun minimum untuk nilai rumah menjadi dua kali ganda nilai asalnya. [3 markah]
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28
6. (a) Mean and median income of overtime per month for 100 workers at a
factory are RM200 and RM150 respectively. Upon review, it was found that one
worker whose overtime income was RM230, but was mistakenly recorded as
RM320.
Min dan median pendapatan lebih masa sebulan bagi 100 orang pekerja di sebuah
kilang masing-masing ialah RM200 dan RM150. Setelah disemak, didapati bahawa
seorang pekerja yang pendapatan lebih masanya RM230 telah salah dicatat
sebagai RM320.
Find
Cari
(i) the new mean overtime income, [3 marks]
min baru pendapatan lebih masa, [3 markah]
(ii) the new median income for overtime. [1 mark]
median baru pendapatan kerja lebih masa. [1 markah]
(b) Table 6 shows the performance of two students who participated in an archery
competition which consists of 10 rounds.
Jadual berikut menunjukkan prestasi dua orang pelajar yang mengambil bahagian
dalam suatu pertandingan memanah yang terdiri daripada 10 pusingan.
Student
Pelajar
Scores
Skor
A 7 , 8 , 6 , 8 , 6 , 5 , 9 , 10 , 7 , 4
B 9 , 5 , 7 , 8 , 7 , 6 , 8 , 6 , 7 , 7
Table 6
Jadual 6
Between student A and B, who has a more consistent performance?
Comment on your answer. [4 marks]
Antara pelajar A dan pelajar B, siapakah yang mempunyai prestasi yang lebih konsisten?
Berikan komen tentang jawapan anda. [4 markah]
SULIT 3472/1
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3472/1 SULIT
29
Section B Bahagian B
[40 marks] [40markah]
Answer any four questions from this section. Jawab mana-mana empat soalan daripada bahagian ini .
7. Use the graph paper to answer this question.
Gunakan kertas graf untuk menjawab soalan ini.
Table 7 shows the values of two variables, 𝑥 and 𝑦, obtained from an experiment.
Variable 𝑥 and 𝑦 are related by the equation = 𝑚𝑥𝑛 , where 𝑚 and 𝑛 are constants.
Jadua 7 menunjukkan nilai-nilai bagi dua pemboleh ubah,𝑥 dan 𝑦, yang diperoleh
daripada suatu eksperimen. Pemboleh ubah 𝑥 dan 𝑦,dihubungkan oleh persamaan
𝑦 = 𝑚𝑥𝑛, dengan keadaan 𝑚 dan 𝑛 ialah pemalar.
𝑥 1.26 1.6 2 3 4 5
𝑦 0.73 1.0 1.4 2.6 4.0 5.6
Table 7
Jadual 7
(a) Based on Table 7, construct a table for the values of log10
𝑦 and log10
𝑥 .
[2 marks]
Berdasarkan Jadual 7, bina satu jadual bagi nilai-nilai log10 𝑦 dan log10 𝑥.
[2 markah]
(b) Plot log10
𝑦 against log10
𝑥 , using a scale of 2 cm to 0.1 unit on both axes.
Hence, draw the line of best fit. [3 marks]
Plot log10 𝑦 melawan log10 𝑥 , dengan menggunakan skala 2 cm kepada 0.1 unit pada
kedua- dua paksi.
Seterusnya, lukis garis lurus penyuaian terbaik. [3 markah]
(c) Using the graph in 7(b), find the value of 𝑚 and 𝑛. [5 marks]
Menggunakan graf di 7(b), cari nilai 𝑚 dan 𝑛. [5 markah]
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8. Diagram 8 shows the curve 𝑦 = (𝑥 + 3)2 intersects the straight line 𝑦 + 3𝑥 = 1
at point A and touches the x-axis.
Rajah 8 menunjukkan lengkung 𝑦 = (𝑥 + 3)2 bersilang dengan garis lurus 𝑦 + 3𝑥 =
1 pada titik A dan menyentuh paksi-x.
Find
Cari
(a) the coordinates of A, [3 marks]
koordinat A, [3 markah]
(b) the area of the shaded region, [4 marks]
luas rantau berlorek [4 markah]
(c) the volume generated, in terms of , when the region bounded by the curve, the
x-axis and the straight line 𝑥 = −1 is revolved through 360 about the x-axis.
[3 marks]
Isipadu janaan, dalam sebutan , apabila rantau yang dibatasi oleh lengkung, paksi x
dan garis lurus 𝑥 = −1 dikisarkan melalui 360 pada paksi x.
[3 markah]
𝑦 + 3𝑥 = 1
O x
A
y
𝑦 = (𝑥 + 3)2
Diagram 8
Rajah 8
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9. In diagram 9, AOBDE, is a semicircle with centre O and has a radius of 5cm. ABC is
a right angle triangle.
Dalam Rajah 9, AOBDE ialah sebuah semi bulatan berpusat O dan mempunyai jejari
5 cm. ABC ialah sebuah segitiga tepat.
It is given that 𝐴𝐷
𝐷𝐶= 3.85 and DC = 2.31 cm.
Diberi bahawa 𝐴𝐷
𝐷𝐶= 3.85 dan DC = 2.31 cm.
[Use / Guna = 3.142 ]
Calculate
Hitung
(a) the value of , in radians, [2 marks]
nilai , dalam radian, [2 markah]
(b) the perimeter, in cm, of the segment ADE, [3 marks]
perimeter, dalam cm, tembereng ADE, [3 markah]
(c) the area, in cm2, of the shaded region BCDF. [5 marks]
luas, dalam cm2, rantau berlorek BCDF. [5 markah]
O
F
D E C
B A
Diagram 9
Rajah 9
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10. Diagram 10 shows a triangle OYX. Point C lies on XY and point A lies on OY.
It is given that 𝑂𝑋 = 3𝑥 , 𝑂𝑌 = 4𝑦, 𝑂𝐴 = 3𝐴𝑌, 𝑋𝐶 ∶ 𝐶𝑌 = 2 ∶ 3 and 𝑂𝐷 = 𝑂𝑋 .
Rajah 10 menunjukkan segitiga OYX. Titik C terletak pada XY dan titik A terletak pada
OY.
Diberi bahawa, 𝑂𝑋 = 3𝑥, 𝑂𝑌 = 4𝑦, 𝑂𝐴 = 3𝐴𝑌, 𝑋𝐶 ∶ 𝐶𝑌 = 2 ∶ 3 𝑑𝑎𝑛 𝑂𝐷 = 𝑂𝑋 .
(a) Express 𝑂𝐶 in terms of 𝑥 and 𝑦. [3 marks]
Ungkapkan 𝑂𝐶 dalam sebutan 𝑥 dan 𝑦. [3 markah]
(b) Given that B is midpoint of AX, express 𝑂𝐵 in terms of 𝑥 and 𝑦.
[3 marks]
Diberi bahawa B ialah titik tengah AX, ungkapkan 𝑂𝐵 dalam sebutan 𝑥 dan 𝑦.
[3 markah]
(c) If points A, C and D are collinear, find the value of . [4 marks]
Jika titik A, C dan D adalah segaris, cari nilai . [4 markah]
B
C
D X O
A
Y
Diagram 10 Rajah10
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11. (a) It is found that 80% of the students in a driving school passed their driving
test. If a sample of 9 students from the school are selected at random, find
the probability that
Didapati bahawa 80% daripada pelajar-pelajar di sebuah sekolah memandu
lulus ujian memandu mereka. Jika 9 orang pelajar dari sekolah itu dipilih secara
rawak, cari kebarangkalian bahawa
(i) all the students in the sample passed the driving test,
kesemua pelajar dalam sampel itu lulus ujian memandu,
(ii) less than 3 students failed the driving test,
kurang daripada 3 orang pelajar gagal ujian memandu,
[4 marks] [4 markah]
(b) The masses of durians produced at a plantation follows a normal distribution with
a mean 1200 g and a standard deviation of 80 g. It is given that the durians with
masses of 1350 g or more will be exported while those with masses between
1000 g and 1350 g will be marketed locally.
Jisim durian yang dihasilkan di sebuah ladang adalah mengikut taburan normal
dengan min 1200 g dan sisihan piawai 80 g. Diberikan bahawa durian yang
mempunyai jisim 1350 g atau lebih akan diekspot manakala durian yang mempunyai
jisim antara 1000 g dan 1350 g akan dipasarkan dalam negara.
(i) If a durian is chosen at random, find the probability that the durian will be
marketed locally.
Jika sebiji durian dipilih secara rawak, cari kebarangkalian bahawa durian itu
akan dipasarkan dalam negara.
(ii) Find the number of durians that are not exported if there are 5000 durians
produced in a certain season.
Cari bilangan durian yang tidak akan diekspot jika 5000 biji durian yang
dihasilkan pada musim yang tertentu.
[6 marks] [6 markah]
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Section C Bahagian C
[20 Marks] [20Markah]
Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini.
12. A particle moves along a straight line with an initial velocity of 24 ms-1. Its acceleration, a ms-2, is given by 𝑎 = 5 − 2𝑡, where t is the time, in seconds, after
passing through a fixed point O.
Suatu zarah bergerak di sepanjang suatu garis lurus dengan halaju awal 24 ms-1
. Pecutannya, a ms
-2, diberi oleh 𝑎 = 5 − 2𝑡 dengan keadaan t ialah masa, dalam saat,
selepas melalui titik tetap O.
Find Cari
(a) the time, in seconds, when its acceleration is zero, [1 marks]
masa, dalam saat, ketika pecutannya sifar, [1 markah]
(b) the maximum velocity, in ms-1, of the particle, [3 marks]
halaju maksimum, dalam ms-1
, bagi zarah itu, [3 markah]
(c) the time, in seconds, when the particle stops instantaneously, [2 marks]
masa dalam saat, apabila zarah berhenti seketika, [2 markah]
(d) the total distance, in m, travelled by the particle in the first 9 seconds.
[4 marks] Jumlah jarak, dalam m, yang dilalui oleh zarah itu dalam 9 saat pertama. [4 markah]
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13. Table 13 shows the price and price indices of five components A, B, C, D and E used in the production of a kind of electrical item.
Jadual 13 menunjukkan harga dan indeks harga bagi lima jenis komponen A, B, C, D dan E yang digunakan dalam penghasilan satu jenis item elektrik.
Component Komponen
Price (RM) Harga (RM)
Price Index in the year 2005 based on the year 2000
Indeks harga pada tahun 2005
berasaskan tahun 2000
Year Tahun
2000
Year Tahun
2005
A x 2.70 135
B 1.80 2.25 125
C 1.50 1.35 y
D 3.00 3.30 110
E 0.50 z 128
Table 13 Jadual 13
(a) Calculate the value of 𝑥, 𝑦 and 𝑧 . [4 marks]
Hitungkan nilai 𝑥, 𝑦 dan z. [4 markah]
(b) If the electrical item uses 10% of component A, 30% of component B, 15% of
componenet C, 40% of component D and 5% of component E. Calculate the composite index for the production cost of the electrical item in the year 2005 based on the year 2000. [2 marks] Jika item elektrik itu menggunkan 10% komponen A, 30% komponen B, 15% komponen C, 40% komponen D dan 5% komponen E. Hitung indeks gubahan bagi kos pengeluaran item elektrik itu pada tahun 2005 berasaskan tahun 2000. [2 markah]
(c) The total cost for these components in the year 2000 is RM2400. Find the
corresponding total cost in the year 2005. [2 marks]
Jumlah kos bagi semua komponen pada tahun 2000 ialah RM2400. Hitung jumlah kos
yang sepadan bagi semua komponen itu pada tahun 2005. [2 markah]
(d) The cost of all the components increase by 20% from the year 2005 to the year 2006. Find the composite index in the year 2006 based on the year 2000 . [2 marks]
Kos bagi semua komponen itu meningkat sebanyak 20% dari tahun 2005 ke tahun 2006. Cari indeks gubahan bagi tahun 2006 berasaskan tahun 2000. [2 markah]
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14. Solution by scale drawing will not be accepted. Penyelesaian secara lukisan berskala tidak akan diterima.
In the diagram 14, BCD is a straight line and ACB is an acute angle. The area of triangle ADE is 32 cm2.
Dalam rajah 14, BCD ialah satu garis lurus dan ACB ialah sudut tirus. Luas segitiga ADE ialah 32 cm
2.
Given AB = 8 cm, AE = 12 cm, AC = 5 cm, CD = 7.5 cm and ∠𝐴𝐵𝐶 = 350. Diberi AB = 8 cm, AE = 12 cm, AC = 5 cm, CD = 7.5 cm dan ∠𝐴𝐵𝐶 = 350.
(a) Calculate Kirakan
(i) ACB, [2 marks/ markah]
(ii) the length, in cm, of AD, [2 marks]
panjang, dalam cm, bagi AD, [2 markah]
(iii) DAE, [2 marks/ markah]
(iv) the area, in cm2, of the quadrilateral ABDE. [3 marks] luas, dalam 𝑐𝑚2, bagi sisi empat ABDE. [3 markah]
(b) Point E’ lies on AE such that E’D = ED. Sketch triangle AE’D. [1mark]
Titik E’ terletak pada AE dengan keadaan E’D =ED.Lakarkan segi tiga AE’D.[1markah]
35
12 cm
8 cm
7.5 cm 5 cm A
B
C
D
E
Diagram 14 Rajah 14
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15. Use graph paper to answer this question. Gunakan kertas graf untuk menjawab soalan ini.
A shop produces two types of sport shoes, M and N. A pair of sport shoes M requires 25 minutes to be produced by machine P and 35 minutes to be produced by machine Q. A pair of sport shoes N requires 15 minutes to be produced by machine P and 45 minutes to be produced by machine Q. Machines P and Q can only operate for 1200 minutes and 2520 minutes a day respectively. The factory produces x pairs of sport shoes M and y pairs of sport shoes N in a day. The number of pairs of sport shoes M produced is not more than twice the number of pairs of sport shoes N. Sebuah kedai menghasilkan dua jenis kasut sukan, M dan N. Sepasang kasut sukan M memerlukan 25 minit untuk dihasilkan oleh mesin P dan 35 minit untuk dihasilkan oleh mesin Q. Sepasang kasut sukan N pula memerlukan 15 minit untuk dihasilkan oleh mesin P dan 45 minit untuk dihasilkan oleh mesin Q. Mesin P dan Q masing-masing hanya dapat beroperasi selama 1200 minit dan 2520 minit.Kilang tersebut menghasilkan x pasang kasut sukan M dan y pasang kasut sukan N sehari. Bilangan kasut M yang dihasilkan tidak melebihi dua kali ganda kasut N.
(a) Write down three inequalities, other than 𝑥 ≥ 0, 𝑦 ≥ 0, which satisfy all the
above constraints. Tuliskan tiga ketaksamaan, selain x ≥ 0, y ≥ 0, yang memenuhi semua
kekangan diatas. [3 marks] [3 markah]
(b) Using a scale of 2 cm to 10 pairs on both axes, construct and shade the
region R which satisfy all the above constraints. Menggunakan skala 2 cm kepada 10 pasang pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua kekangan di atas.
[3 marks] [3 markah]
(c) Using the graph constructed in 15(b), find Dengan menggunakan graf yang dibina di 15(b), cari
(i) the range of the number of pairs of sport shoes N produced if exactly 30
pairs of sport shoes M are produced in a day julat bilangan pasang kasut sukan N yang perlu dihasilkan jika bilangan kasut sukan M yang dihasilkan dalam satu hari ialah 30 pasang.
(ii) the maximum total profit per day if the profit from a pair of sport shoes M
and a pair of sport shoes N are RM30 and RM 35 respectively. Jumlah keuntungan maksimum sehari jika Keuntungan daripada jualan sepasang kasut sukan M dan sepasang kasut sukan N masing-masing ialah RM30 dan RM35.
[4 marks] [4 markah]
END OF QUESTION PAPER KERTAS SOALAN TAMAT
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THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1 )
KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)
z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
Minus / Tolak
0.0
0.1
0.2
0.3
0.4
0.5000
0.4602
0.4207
0.3821
0.3446
0.4960
0.4562
0.4168
0.3783
0.3409
0.4920
0.4522
0.4129
0.3745
0.3372
0.4880
0.4483
0.4090
0.3707
0.3336
0.4840
0.4443
0.4052
0.3669
0.3300
0.4801
0.4404
0.4013
0.3632
0.3264
0.4761
0.4364
0.3974
0.3594
0.3228
0.4721
0.4325
0.3936
0.3557
0.3192
0.4681
0.4286
0.3897
0.3520
0.3156
0.4641
0.4247
0.3859
0.3483
0.3121
4
4
4
4
4
8
8
8
7
7
12
12
12
11
11
16
16
15
15
14
20
20
19
19
18
24
24
23
22
22
28
28
27
26
25
32
32
31
30
29
36
36
35
34
32
0.5
0.6
0.7
0.8
0.9
0.3085
0.2743
0.2420
0.2119
0.1841
0.3050
0.2709
0.2389
0.2090
0.1814
0.3015
0.2676
0.2358
0.2061
0.1788
0.2981
0.2643
0.2327
0.2033
0.1762
0.2946
0.2611
0.2296
0.2005
0.1736
0.2912
0.2578
0.2266
0.1977
0.1711
0.2877
0.2546
0.2236
0.1949
0.1685
0.2843
0.2514
0.2206
0.1922
0.1660
0.2810
0.2483
0.2177
0.1894
0.1635
0.2776
0.2451
0.2148
0.1867
0.1611
3
3
3
3
3
7
7
6
5
5
10
10
9
8
8
14
13
12
11
10
17
16
15
14
13
20
19
18
16
15
24
23
21
19
18
27
26
24
22
20
31
29
27
25
23
1.0
1.1
1.2
1.3
1.4
0.1587
0.1357
0.1151
0.0968
0.0808
0.1562
0.1335
0.1131
0.0951
0.0793
0.1539
0.1314
0.1112
0.0934
0.0778
0.1515
0.1292
0.1093
0.0918
0.0764
0.1492
0.1271
0.1075
0.0901
0.0749
0.1469
0.1251
0.1056
0.0885
0.0735
0.1446
0.1230
0.1038
0.0869
0.0721
0.1423
0.1210
0.1020
0.0853
0.0708
0.1401
0.1190
0.1003
0.0838
0.0694
0.1379
0.1170
0.0985
0.0823
0.0681
2
2
2
2
1
5
4
4
3
3
7
6
6
5
4
9
8
7
6
6
12
10
9
8
7
14
12
11
10
8
16
14
13
11
10
19
16
15
13
11
21
18
17
14
13
1.5
1.6
1.7
1.8
1.9
0.0668
0.0548
0.0446
0.0359
0.0287
0.0655
0.0537
0.0436
0.0351
0.0281
0.0643
0.0526
0.0427
0.0344
0.0274
0.0630
0.0516
0.0418
0.0336
0.0268
0.0618
0.0505
0.0409
0.0329
0.0262
0.0606
0.0495
0.0401
0.0322
0.0256
0.0594
0.0485
0.0392
0.0314
0.0250
0.0582
0..0475
0.0384
0.0307
0.0244
0.0571
0.0465
0.0375
0.0301
0.0239
0.0559
0.0455
0.0367
0.0294
0.0233
1
1
1
1
1
2
2
2
1
1
4
3
3
2
2
5
4
4
3
2
6
5
4
4
3
7
6
5
4
4
8
7
6
5
4
10
8
7
6
5
11
9
8
6
5
2.0
2.1
2.2
2.3
0.0228
0.0179
0.0139
0.0107
0.0222
0.0174
0.0136
0.0104
0.0217
0.0170
0.0132
0.0102
0.0212
0.0166
0.0129
0.00990
0.0207
0.0162
0.0125
0.00964
0.0202
0.0158
0.0122
0.00939
0.0197
0.0154
0.0119
0.00914
0.0192
0.0150
0.0116
0.00889
0.0188
0.0146
0.0113
0.00866
0.0183
0.0143
0.0110
0.00842
0
0
0
0
3
2
1
1
1
1
5
5
1
1
1
1
8
7
2
2
1
1
10
9
2
2
2
1
13
12
3
2
2
2
15
14
3
3
2
2
18
16
4
3
3
2
20
18
4
4
3
2
23
21
2.4 0.00820 0.00798 0.00776 0.00755 0.00734
0.00714
0.00695
0.00676
0.00657
0.00639
2
2
4
4
6
6
8
7
11
9
13
11
15
13
17
15
19
17
2.5
2.6
2.7
2.8
2.9
0.00621
0.00466
0.00347
0.00256
0.00187
0.00604
0.00453
0.00336
0.00248
0.00181
0.00587
0.00440
0.00326
0.00240
0.00175
0.00570
0.00427
0.00317
0.00233
0.00169
0.00554
0.00415
0.00307
0.00226
0.00164
0.00539
0.00402
0.00298
0.00219
0.00159
0.00523
0.00391
0.00289
0.00212
0.00154
0.00508
0.00379
0.00280
0.00205
0.00149
0.00494
0.00368
0.00272
0.00199
0.00144
0.00480
0.00357
0.00264
0.00193
0.00139
2
1
1
1
0
3
2
2
1
1
5
3
3
2
1
6
5
4
3
2
8
6
5
4
2
9
7
6
4
3
11
8
7
5
3
12
9
8
6
4
14
10
9
6
4
3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4
INFORMATION FOR CANDIDATES
MAKLUMAT UNTUK CALON
For negative z use relation:
Bagi z negative guna hubungan:
Q(z)
z
f (z)
O k
Example / Contoh:
If X ~ N(0, 1), then
JikaX ~ N(0, 1), maka
P(X>k) = Q(k)
P(X>2.1) = Q(2.1) = 0.0179
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1. This question paper consists of three sections, Section A, Section B and Section C.
Kertas soalan ini mengandungi tiga bahagian : Bahagian A, Bahagian B danBahagian C.
2. Answer all questions in Section A, any four questions from Section B and any two
questions from Section C. Jawab semua soalan dalam Bahagian A, mana-mana empat soalan daripada Bahagian B dan mana-mana dua soalan daripada Bahagian C.
3. Show your working. It may help you to get marks.
Tunjukkan langkah-langkah penting dalam kerja mengira anda.Ini boleh membantu anda
untuk mendapat markah.
4. The diagrams provided in the questions are not drawn to scale unless stated.
Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.
5. The marks allocated for each question and sub-part of the question are shown in
brackets. Markah yang diperuntukkan bagi setiap soalan dan ceraian soalan ditunjukkan dalam kurungan.
6. A list of formulae is provided on page 2 to 4.
Satu senarai rumus ada disediakan di halaman 2 hingga 4.
7. You may use a scientific calculator.
Anda dibenarkan menggunakan kalkulator saintifik.
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MODUL KECEMERLANGAN PERCUBAAN SPM
TAHUN 2016
ADDITIONAL MATHEMATICS
Tingkatan 5
KERTAS 1
PERATURAN PEMARKAHAN
UNTUK KEGUNAAN PEMERIKSA SAHAJA
3472/1(PP)
Tingkatan
Lima
Additional
Mathematics
Kertas 1
Peraturan
Pemarkahan
Ogos/Sep
2016
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ADDITIONAL MATHEMATICS TRIAL SPM PARER 1 2016
Question Solutions and marking scheme Sub
Marks
Full
Marks
1(a)
(b)
(c)
-2 and 2
{1, 4, 16, 25 } yes
1
1
1
3
2
2
B1 : bbx )2(2 =18
2
2
3(a)
(b)
1
B1 : 2+ 5x =7 or ℎ−1(𝑥) =𝑥−2
5
1
B1 : 12𝑝 − 10 = 2 or h(2) =12
2
2
4
4 1 or 5
B2 : (m-5)(m-1)=0
B1: 0))((4)53( 2 mmm
3 3
5(a)
(b)
3x
11
2
B1 : 2k-6 = 5
1
2
3
6
r > 8 cm only
B2: (r-8)(r+14)>0
B1: 𝑟2 + 6𝑟 − 112 >0
3
3
7
5
8
B2: xxx 533 or equivalent
B1: xxx or 5)3(3 22 or 2𝑥−3+𝑥 or 25+(−3𝑥)
3
3
8
9
2𝑙𝑜𝑔10𝑥 + 𝑙𝑜𝑔10 (
27
2)
B2 : 𝑙𝑜𝑔10(9𝑥(3𝑥2)(2𝑥
32)
4
B1 : 𝑙𝑜𝑔10 4𝑥3
𝑙𝑜𝑔10 100 or 𝑙𝑜𝑔10(
9𝑥(3𝑥2)
4) or equivalent
3
3
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Question Solutions and marking scheme Sub
Marks
Full
Marks
9(a)
(b)
16
B1 : 1(25) − 1(24)
Aishah, 1088 cents or RM10.88 (both)
B1 : 211 = 2048 𝑐𝑒𝑛𝑡𝑠 𝑜𝑟 80 + (11)80 = 960 𝑐𝑒𝑛𝑡𝑠
2
2
4
10
350°
B2 : 70°
1−0.8
B1 : r = 0.8
3
3
11(a)
(b)(i)
(ii)
𝑙𝑜𝑔10𝑦 = 𝑙𝑜𝑔10ℎ + 3𝑙𝑜𝑔10𝑥 2 4
3
B1 : 6−2
𝑘−0= 3
1
1
2
4
12
24.37°, 75.63°, 204.37°, 255.63°
B2 : 2x-10° = 38.74°, 141.26°, 398.74°, 501.26°
B1 : 38.74°
3
3
13(a)
(b)
1
4
7√3
3 𝑜𝑟
7
√3
B1 : tan 𝜃 = √3 or tan ( 90° − 𝜃) =1
√3
1
2
3
14 2 radian
B1 : 1
2( 12) 𝑟 = 36
2 2
15
( 5
3 ,
16
3 )
B2 : (2(3)+1(−1)
3 ,
2(6)+1(4)
3 )
B1 : 2(3)+1(−1)
3 𝑜𝑟
2(6)+1(4)
3
3
3
16(a)
(b)
(68)
-9i -4j
B1 : 𝑄𝑃1 = 𝑄𝑂 + 𝑂𝑃1
or -6i-8j+(-3i +4j)
1
2
3
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Question Solutions and marking scheme Sub
Marks
Full
Marks
17 𝑄𝑅 = −
4
3𝑎 + 2𝑏
B2 : 𝑃𝑅 = 3𝑏 − 2𝑎
B1 : 𝑃𝑅 = 𝑃𝑂 + 𝑂𝑅
3 3
18(a)
(b)
8
2
3 or 0.6667
B2 : 25
2𝑘 −
4
2𝑘 or 4 +[
𝑘𝑥2
2] = 11
B1 : ∫ 𝑔(𝑥)𝑑𝑥 = 4 𝑜𝑟 𝑘𝑥2
2
5
1
1
3
4
19(a)
(b)
(c)
4x-12
3
-18
1
1
1
3
20
3
16
B2 : 15𝜋 = 8𝜋 (10) ×𝑑𝑟
𝑑𝑡
B1 : 8𝜋𝑟 or 𝑑𝐴
𝑑𝑡= 15𝜋
3
3
21 (a)
(b)
840
B1 : 7P4 or 7x6x5x4
480
B1 : 6P3 x
4P1 or 6x5x4x4
2
2
4
22(a)
(b)
��= 4.6
𝜎 = 2
�� = 4.56 and 𝜎 = 2.04
B1 : �� = 4.56 or 𝜎 = 2.04 or ∑𝑥 = 220 ∑ 𝑥2 = 1210
1
1
2
4
23(a)
(b)
3
20
15
20 or
3
4 or 0.75
B1 : P(A ⋃𝐹)= 7
20+
11
20−
3
20
1
2
3
5
2
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Question Solutions and marking scheme Sub
Marks
Full
Marks
24(a)
(b)
1
16
B1 : 6k +2(4k) +2k =1 5
16
2
1
3
25(a)
(b)
1.25
B1 : 8.2−6.7
1.2
1.14
B1 : P( z≥ 𝑘 ) = 0.1271
2
2
4
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3472/2 ADDITIONAL
MATHEMATICS
Paper 2
August /September 2016 2 ½ hours
MODUL KECEMERLANGAN PERCUBAAN SPM 2016
ADDITIONAL MATHEMATICS
PERATURAN PERMARKAHAN Tingkatan 5
Kertas 2
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No Solution Sub Mark
Total Mark
1
𝑦 = 5 − 2𝑥 or 𝑥 =5−𝑦
2
1
5
(5 − 2𝑥)2 − 2𝑥 = 5𝑥(5 − 2𝑥) or 𝑦2 − 2(5−𝑦
2) = 5𝑦 (
5−𝑦
2) 1
𝑥 =−(−47)±√(−47)2−4(14)(25)
2(14) or
𝑦 =−(−23)±√(−23)2−4(7)(−10)
2(7)
1
𝑥 = 2.694 @ 𝑦 = −0.388 or 𝑦 = −0.389 @ 𝑥 = 2.695 1
𝑥 = 0.663 @ 𝑦 = 3.674 1
2(a) 8900 = [30(1) + 8000] log10 (760
𝑥) 1
6
760
𝑥= 12.82 or 12.83 (direct) 1
𝑥 = 59.3 or 59.2 1
2(b) 4000(1 + 0.025)𝑛 > 7000 1
𝑛 log10 1.025 > log10 (7
4) 1
𝑛 = 23 1
3(a) 𝑚𝑃𝑄 = −1
2 or 𝑦 − 6 = −
1
2(𝑥 − 3) 1
7
𝑦 = −1
2𝑥 +
15
2 or equivalent 1
3(b) 𝑄(5,5) 1
5 =3(3)+2ℎ
5 or 5 =
3(6)+2𝑘
5
1
𝑅(8,7
2) 1
3(c) √(𝑥 − 3)2 + (𝑦 − 6)2 = √(𝑥 − 5)2 + (𝑦 − 5)2 1
4𝑥 − 2𝑦 − 5 = 0 1
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No Solution Sub Mark
4(a)
4(b)
Shape of graph of 𝑦 = sin 𝑥 , 0 ≤ x ≤ 2π [must begin from
origin, O(0, 0)]
Shape of graph of 𝑦 = sin3
2𝑥 , 0 ≤ x ≤2π [exactly 1 and
1
2
cycles]
Amplitude of graph = 4
Reflection of graph 𝑦 = −4sin3
2𝑥 at x – axis, 0 ≤ x ≤ 2π [all
correct]
𝑦 = −𝜋
𝑥 [Equation of curve]
Draw the curve passing through (𝜋
3, −3) or (, −1) or (2, −
1
2)
Number of solutions = 4
1
1
1
1
1
1
1
7
5(a) 𝑥(1.07)2 = 110000
1
7
𝑥 = 96078.26 1
5(b) (96078.26)(1.07)8 1
165080.34 1
5(c)
(96078.26)(1.07)𝑛−1 ≥ 2(96078.26)
1
(𝑛 − 1) log10(1.07) ≥ log102 1
11 𝑦𝑒𝑎𝑟𝑠 after purchased 1
O 4𝜋
3
𝜋
3
- 4
4
2 x
y
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No Solution Sub Mark
Total Mark
6(a)(i)
∑𝑥
100= 200
1
8
∑𝑥 = 20000 + 230 − 320 1
�� = 199.1 1
6(a)(ii) 150 1
6(b) ��𝐴 = 7 or ��𝐵 = 7 1
𝜎𝐴 = 1.732 or 𝜎𝐵 = 1.095 1
Student B because smaller value of standard deviation and the scores are not scatted widely from the mean Pelajar B kerana nilai sisihan piawai lebih kecil dan skor tidak bertabur jauh daripada min.
2
7(a) log10 𝑥 0.10 0.20 0.30 0.48 0.60 0.70
log10 𝑦 0.14 0.00 0.15 0.41 0.60 0.75
1 1
7(b) plot log10 𝑦 against log10 𝑥 and correct axes and uniform scales
1
10
6 points plotted correctly 1
Line of best fit (Attachment, pg. 10) 1
7(c) log10 𝑦 = 𝑛 log10 𝑥 + log10 𝑚 can be implied 1
y-intercept = log10 𝑚 = −0.3 1
m = 0.5012 1
Gradient = n = 1.5 1+1
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No Solution Sub Mark
Total Mark
8(a) (𝑥 + 3)2 = 1 − 3𝑥 1
10
(𝑥 + 1)(𝑥 + 8) = 0 1
𝐴(−1,4) 1
8(b) ∫ (𝑥 + 3)2𝑑𝑥 −
1
2
0
−1(4 + 1)(1) 1
[(𝑥 + 3)3
3]−1
0
−5
2 1
[(0 + 3)3
3−
(−1 + 3)3
3] −
5
2 1
23
6 or 3
5
6 1
8(c) 𝜋 [
(𝑥 + 3)5
5]−3
−1
1
𝜋 [(−1 + 3)5
5−
(−3 + 3)5
5] 1
32
5𝜋 or 6
2
5𝜋 1
9(a) 𝜃 = 𝑐𝑜𝑠−1 (10
11.2035) = 26.8⁰ 𝑂𝑅 𝑐𝑜𝑠𝐷𝑂𝐴 =
52+52−8.89352
2(5)(5) 1
10
𝜃 = 0.4678 OR 𝜃 = 0.4748 1
9(b) 𝐴𝑂𝐷 = 180 − 2(26.8) = 126.4 = 2.206 𝑟𝑎𝑑 OR 2.1924 rad
1
5(2.206) = 11.03 or 8.8935 given OR 5(2.1924) 1
Perimeter = 5(2.206)+8.8935 = 19.92 cm OR 5(2.1924)+8.8935 = 19.86 cm
1
9(c) BC=√11.20352 − 102 = 5.052 or
1
2(10)(5.052) 1
1
2(5)2(0.9356) or
1
2(5)2 sin 126.4
OR 1
2(5)2(0.9496) or or
1
2(5)2 sin125.6
1
1
2(5)2(0.9356) and
1
2(5)2 sin126.4
OR 1
2(5)2(0.9496)𝑎𝑛𝑑
1
2(15)2 sin125.6
1
Area BCDF = 1
2(10)(5.052) −
1
2(5)2 sin126.4 −
1
2(5)2(0.9356)
OR 1
2(10)(5.052) −
1
2(5)2 sin125.6 −
1
2(5)2(0.9496)
1
3.50 cm2 OR 3.2
1
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No Solution Sub Mark
Total Mark
10(a)
𝑂𝐶 = 𝑂𝑌 + 𝑌𝐶 𝑎𝑡𝑎𝑢 setara 1
10
𝑂𝐶 = 4𝑦 +3
5(−4𝑦 + 3𝑥) 1
𝑂𝑆 =9
5𝑥 +
8
5𝑦 1
10(b) 𝑂𝐵 = 𝑂𝐴 + 𝐴𝐵 𝑎𝑡𝑎𝑢 setara 1
𝑂𝐵 =3
4(4𝑦) +
1
2(−3𝑦 + 3𝑥) 1
𝑂𝐵 =3
2𝑦 +
3
2𝑥 1
10(c) 𝐴𝐶 =9
5𝑥 −
7
5𝑦 or 𝐴𝐷 = 3𝑥 − 3𝑦 1
9
5𝑥 −
7
5𝑦 = 𝑚(3𝑥 − 3𝑦) 1
3𝑚 = 9
5 or 3𝑚 =
7
5 1
= 9
7 1
11(a)(i) P(X = 9) = 9C9(0.8)9(0.2)0 1
10
0.1342 1
11(a)(ii) 9C7(0.8)7(0.2)2 + 9C8(0.8)8(0.2)1 + 9C9(0.8)9(0.2)0 1
0.7382 1
11(b)(i) 𝑃 (
1000 − 1200
80< 𝑧 <
1350 − 1200
80) = 𝑃(−2.5 < 𝑧
< 1.875) 1
1 − 𝑃(𝑧 > 2.5) − 𝑃(𝑧 > 1.875) = 1 − 0.00621 − 0.0303 1
0.9635 1
11(b)(ii) 𝑃 (𝑧 <
1350 − 1200
80) = 𝑃(𝑧 < 1.875) 1
0.9696 1
0.9696 5000 = 4848 1
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No Solution Sub Mark
Total Mark
12(a) 5 − 2𝑡 = 0, 𝑡 =5
2 1
10
12(b) 𝑣 = 5𝑡 − 𝑡2 + 𝑐 1
𝑡 = 0, 𝑣 = 24, 𝑐 = 24 1
𝑣𝑚𝑎𝑥 = 5(5
2) − (
5
2)2
+ 24 =121
4 𝑜𝑟 30
1
4 1
12(c) 5𝑡 − 𝑡2 + 24 = 0 1
(𝑡 + 3)(𝑡 − 8) = 0, 𝑡 = 8 1
12(d) ∫ −𝑡2 + 5𝑡 + 24
8
0 𝑑𝑥 or |∫ −𝑡2 + 5𝑡 + 24
9
8 𝑑𝑥| 1
[−𝑡3
3+
5𝑡2
2+ 24𝑡]
0
8
or |[−𝑡3
3+
5𝑡2
2+ 24𝑡]
8
9
| 1
[−𝑡3
3+
5𝑡2
2+ 24𝑡]
0
8
+ |[−𝑡3
3+
5𝑡2
2+ 24𝑡]
8
9
| 1
1123
6 or 187
1
6 1
13(a) 2.7
𝑥× 100 = 135 or
1.35
1.50× 100 = 𝑦 or
𝑧
0.50× 100 = 128 1
10
𝑥 = 𝑅𝑀2 , 𝑦 = 90 , 𝑧 = 𝑅𝑀0.64 1+1+1
13(b) 135(10) + 125(30) + 90(15) + 110(40) + 128(5)
100 1
114.9 1
13(c) 114.5 =
𝑃2005
2400× 100 1
RM 2757.60 1
13(d) 114.5 × 120% or equivalent 1
137.88 1
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No Solution Sub Mark
Total Mark
14(a) (i)
sin∠𝐴𝐶𝐵
8=
sin35
5 1
10
∠𝐴𝐶𝐵 = 66.60𝑜 1
(ii) ∠𝐴𝐶𝐷 = 180𝑜 − 66.60𝑜 = 113.4𝑜
𝐴𝐷2 = 52 + 7.52 − 2(5)(7.5)𝑐𝑜𝑠∠113.4 1
𝐴𝐷 = 10.54𝑐𝑚 1
(iii) 1
2(12)(10.54) sin 𝐷𝐴𝐸 = 32 1
𝐷𝐴𝐸 = 30.40° 1
(iv) ∠𝐵𝐴𝐶 = 180𝑜 − 66.60𝑜 − 35 = 78.4𝑜
sin 78.4
𝐵𝐶=
sin35
5 𝒐𝒓 𝐵𝐶 = 8.539 𝒐𝒓 𝐵𝐷 = 8.539 +
7.5 = 16.039 1
Area of ABDE =1
2(8)(16.039)𝑠𝑖𝑛35𝑜 + 32 1
Area = 68.798 cm2 1
14(b)
1
D
E
A
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No Solution Sub Mark
Total Mark
15(a) I 25𝑥 + 15𝑦 ≤ 1200 or 5𝑥 + 3𝑦 ≤ 240 1
10
II 35𝑥 + 45𝑦 ≤ 2520 or 7𝑥 + 9𝑦 ≥ 504 1
III 𝑥 ≤ 2𝑦 1
15(b) Graph (Attachment, pg.11)
At least one straight line is drawn correctly from the
inequalities involving 𝑥and 𝑦. 1
All the straight lines are drawn correctly. 1
Region is shaded correctly. 1
15(c) (i) 15 ≤ 𝑦 ≤ 30 1
(ii) maximum point = (27, 35) 1
maximum profit = 30(27) + 35(35) 1
RM2035 1
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54
No.7(b) log10 𝑦
-
0.2
-0.1
0
0.1
0.2
0.3
0.4
-0.3
0.5
0.6
0.7
x
x
x
x
x
x
0.1
4.5
0.2
4.5
0.3
4.5
0.4
4.5
0.5
4.5
0.6
4.5
0.7
4.5
log10 𝑥
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55
No.14(b)
x
x
(Sport shoes
M)
x x
x
x
5𝑥 + 3𝑦 = 240
𝑦 =1
2𝑥
7𝑥 + 9𝑦= 504
10 20 30 40 50 60
10
20
30
40
50
60
70
𝑦
0 70
7
80
90
100
(Sport shoes
N)
80
R
(27, 35)