SKOR A SPM (Addmaths) Penerbitan Pelangi Sdn Bhd

2KERTAS MODEL

SIJIL PELAJARAN MALAYSIA 3472/1

Additional Mathematics

Kertas 1

2 jam Dua jam

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. Jawab semua soalan.
4. Anda dibenarkan menggunakan kalkulator saintifik.


1. Given that f(x) = 4 + 3 sin 2x for 0 < x < 2π. State [1 mark / markah] For
[1 mark / markah] Examiner’s
Diberi f(x) = 4 + 3 sin 2x untuk 0 < x < 2π. Nyatakan [1 mark / markah]
Use
(a) the amplitude of f, 1(a)

amplitud bagi f, 1

(b) the period of f, 1(b)

tempoh bagi f,

(c) the maximum and minimum values of f.

nilai maksimum dan nilai minimum bagi f.

Answer / Jawapan : (b) (c) 1
(a)

1(c)

2. (a) Show that cot sec x x = sin x. 1
x+ tan 2(a)
sek x
Tunjukkan bahawa kot x+ tan x = sin x. [2 marks / markah] 2
[2 marks / markah] 2(b)
(b) Hence, solve sec 2x =– 1 for 0 < x < 360°.
cot 2x + tan 2x 2 2
sek 2x 1
Seterusnya, selesaikan kot 2x + tan 2x =– 2 bagi 0 < x < 360°.

Answer / Jawapan : (b)
(a)

Total

7

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KERTAS MODEL SPM 2  3472/1

3. Table 3 shows the amount of donation collected for 10 classes. For
Examiner’s
Jadual 3 menunjukkan jumlah kutipan sumbangan bagi 10 kelas.
Use
Class / Kelas A BCDE FGH I J
3(a)
Amount of donation 1

collected 210 340 157 38 141 105 183 302 245 146 3(b)
2
Jumlah kutipan sumbangan

(RM)

Table 3 / Jadual 3 [1 mark / markah]
Find [2 marks / markah]

Cari

(a) the median of the donation collected,

median bagi jumlah kutipan sumbangan,

(b) the interquatile range of the donation collected.
julat antara kuartil bagi jumlah kutipan sumbangan.

Answer / Jawapan : (b)
(a)

4. Find the value of lim x2 – 4 .
x + 2
x → –2 4
2
Cari nilai bagi had x2 – 4 . [2 marks / markah]
x → –2 x+ 2

Answer / Jawapan :

5. gf(x) Diagram 5 shows the composite function gf(x) that maps

p to –5. Given f : x → 8 1, x . 1 and g : x → x – 6,
3x – 3
find 5(a)
1
q p –5 Rajah 5 menunjukkan fungsi gubahan gf(x) yang memetakan p
5(b)
f(x) g(x) kepada –5. Diberi f : x → 8 1 , x . 1 dan g : x → x – 6, cari 1
3x – 3
(a) the value of p, Total
Diagram 5 / Rajah 5 nilai p, [1 mark / markah] 7

(b) the value of q. [1 mark / markah]
nilai q.

Answer / Jawapan : (b)
(a)

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3472/1 KERTAS MODEL SPM 2

6. Given the function g(x) = (x – 3)2, determine whether g–1(x) is a function. Give a reason. For
Diberi fungsi g(x) = (x – 3)2, tentukan sama ada g–1(x) ialah satu fungsi. Beri satu sebab. Examiner’s
[2 marks / markah]
Use
Answer / Jawapan :
6

2

7. The sum of the first n terms of a progression is 2n logp (q2n). [2 marks / markah] 7(a)
Hasil tambah n sebutan yang pertama bagi satu janjang ialah 2n logp (q2n). 2

(a) Find the nth terms of the progression. 7(b)
2
Cari sebutan ke-n bagi janjang tersebut.

(b) Determine whether the progression is an arithmetic progression. Give a reason.

Tentukan sama ada janjang tersebut ialah satu janjang aritmetik. Beri satu sebab.

[2 marks / markah]

Answer / Jawapan : (b)
(a)

..
8. A repeating decimal 0.523 (0.5232323 …) can be expressed as a sum of a constant p and one

Sinaftiunnitoemgbeoor pmerepturiluchparnobgerreuslasniogn0..52. 3. (0.5232323 …) boleh diungkapkan sebagai satu hasil tambah pemalar 8(a)
1
p dan satu janjang geometri ketakterhinggaan.
8(b)
(a) State the value of p. 3
Nyatakan nilai p. [1 mark / markah]
.. [3 marks / markah]
(b) UEnxgpkraepsksa0n.05.2532. 3. assebaagfraaicstaitounpeincathhaen lowest terms.

dalam sebutan terendah.

Answer / Jawapan : (b)
(a)

9. Given y = 5(25x) – 126(5x), find the values of x when y = –25. [3 marks / markah] 9
3
Diberi y = 5(25x) – 126(5x), cari nilai-nilai x apabila y = –25.
Total
Answer / Jawapan : 13

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KERTAS MODEL SPM 2  3472/1

10. Express log5 x . log2 25 + 3 log2 y as a single logarithm in base 2. [3 marks / markah] 10
3
Ungkapkan log5 x . log2 25 + 3 log2 y sebagai satu logaritma tunggal dalam asas 2.

Answer / Jawapan :

11. Given that the roots of the quadratic equation 2x2 – 21x + p = 0 where p is a constant are in

the ratio 1 : 2. Find the value of p.

Diberi bahawa punca-punca bagi persamaan kuadratik 2x2 – 21x + p = 0 dengan keadaan p ialah pemalar,

adalah dalam nisbah 1 : 2. Cari nilai p. [3 marks / markah]

Answer / Jawapan : 11
3

12. y Diagram 12 shows the graph of the quadratic function
f(x) = –(x + p)2 + q, where p and q are constants. The point
(4,3)
(4, 3) is the maximum point of the graph. Find
O x 12(a)
Rajah 12 menunjukkan graf bagi fungsi kuadratik f(x) = –(x + p)2 + q, 2
dengan keadaan p dan q ialah pemalar. Titik (4, 3) ialah titik maksimum
graf itu. Cari 12(b)
2
Diagram 12 / Rajah 12 (a) the value of p and of q, [2 marks / markah]

Answer / Jawapan : nilai p dan nilai q,
(a)
(b) the y-intercept. [2 marks / markah]

pintasan-y.

(b)

13. Find the range of k if the straight line y = 4kx + 3 does not intersect the curve 13
y = –9x2 + (5k – 1)x – 1. 3

Cari julat nilai k jika garis lurus y = 4kx + 3 tidak menyilang lengkung y = –9x2 + (5k – 1)x – 1. Total
13
[3 marks / markah]
Answer / Jawapan :

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3472/1 KERTAS MODEL SPM 2

14. A vector ∼q has a magnitude of 30 units and has the same direction as 6∼i – 8∼j. Find vector ∼q For
in terms of ∼i and ∼j. Examiner’s

Satu vektor ∼q mempunyai magnitud 30 unit dan sama arah dengan 6∼i – 8∼j. Cari vektor ∼q. da[2lammsaerbkust /a mn ∼ai drkaanh∼j]. Use
Answer / Jawapan :
14

2

15. B Diagram 15 shows a triangle OAB. The straigh→t line CE intersects
→
with the straight line AB at point D. Given OC : OB = 2 : 3,
CD →→ → → → →→
AD : AB = 1 : 2, OE = p OA and CE = qCD. If OA = ∼x and
→
A E OB = ∼y, find the value of p and of q.
O Rajah 15 menunjukkan segi tiga OAB. Garis lurus CE bersilang dengan garis
lurus AB di titik D. Diberi bahawa O→C : O→B = 2 : 3, A→D : A→B = 1 : 2, O→E = pO→A dan
Diagram 15 / Rajah 15 C→E = qC→D. Jika O→A = ∼x dan O→B = ∼y, cari nilai p dan nilai q. [4 marks / markah]

Answer / Jawapan : 15
4

16. A P B Diagram 16 shows an equilateral triangle ABC. Sector PAQ, BPR
and CQR are three identical sectors. Given AB = 12 cm, find
Rajah 16 menunjukkan sebuah segi tiga sama ABC. Sektor PAQ, BPR dan 16(a)
CQR adalah tiga sektor yang sama. Diberi AB = 12 cm, cari 1

Q R [Use / Guna π = 3.142] 16(b)
3
C (a) the value of /PBR in terms of π, [1 mark / markah]
[3 marks / markah]
Diagram 16 / Rajah 16 nilai /PBR dalam sebutan π,
Answer / Jawapan :
(b) the area, in cm2, of the shaded region.
(a)
luas, dalam cm2, bagi kawasan berlorek.

(b)

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10

SPM Additional Mathematics

KERTAS MODEL SPM 2  3472/1

17. y2 Diagram 17 shows the straight line graph obtained by plotting For
Examiner’s
y2 against x . The variables x and y are related by the equation
Use
y2 = ax + b, where a and b are constants. Given the gradient
3 17(a)
(8,13) of the straight line is 2 . 2

Rajah 17 menunjukkan graf garis lurus yang diperoleh dengan memplot 17(b)
2
0x y2 melawan x . Pemboleh ubah x dan y dihubungkan oleh persamaan

Diagram 17 / Rajah 17 y2 = ax + b, dengan keadaan a dan b ialah pemalar. Diberi kecerunan
3
garis lurus tersebut ialah 2 .

(a) Express the equation y2 = ax + b in linear form used to obtain the straight line graph

shown in Diagram 17.
Ungkapkan persamaan y2 = ax + b dalam bentuk linear yang digunakan untuk memperoleh graf garis

lurus seperti ditunjukkan dalam Rajah 17. [2 marks / markah]

(b) Find the value of x when y = 4. [2 marks / markah]

Cari nilai x apabila y = 4.

Answer / Jawapan :
(a)

(b)

18. Vijay’s house y Diagram 18 shows the locations of the house of
Rumah Vijay Vijay, Chang and Mat. Given their houses are situated

Chang’s house along the same straight road which is perpendicular
Rumah Chang to the junction at the origin from Chang’s house. The

(2, 4) Mat’s house distances between Vijay’s house and Mat’s house
Rumah Mat from Chang’s house are 20 km respectively and

the location of Chang’s house is (2, 4). Calculate the

Ox location of Vijay’s house and Mat’s house. 18
4
Diagram 18 / Rajah 18 Rajah 18 menunjukkan kedudukan bagi rumah Vijay, Chang dan
Mat. Diberi rumah mereka berada pada satu jalan lurus yang

sama di mana ia berserenjang dengan simpang di asalan dari rumah Chang. Jarak antara rumah Vijay dan
rumah Mat dari rumah Chang masing-masing ialah 20 km dan kedudukan rumah Chang adalah pada (2,

4). Hitung kedudukan rumah Vijay dan Mat.

[4 marks / markah]

Answer / Jawapan :

Total
8

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3472/1 KERTAS MODEL SPM 2

∫ 19.   m 1 dx = 2 , find the value of m. For
1 – 3 Examiner’s

Given that (2x 1)—23 Use

∫ Diberi m 1 dx = 2 , cari nilai m. [3 marks / markah] 19
1 (2x – 1)—23 3

Answer / Jawapan :

3

20. Given that y = (x – 3)(2x + 1)3. Find the small change in y in terms of h, when the value of

x changes from 3 to 3 + h.
Diberi y = (x – 3)(2x + 1)3. Cari perubahan kecil dalam y, dalam sebutan h, apabila nilai x berubah dari 3 ke
3 + h. [4 marks / markah] 20
4
Answer / Jawapan :

21. 6 2 4 7 5 8 Diagram 21 shows a 6-digit number. Find the number of different
ways to form a 6-digit odd number which is greater than 400 000 if
Diagram 21 / Rajah 21 there is no repetition of digit.

Rajah 21 menunjukkan satu nombor 6 digit. Cari bilangan cara yang berlainan 21
untuk membentuk satu nombor 6 digit ganjil yang melebihi 400 000 jika tiada 3

digit berulang. [3 marks / markah]

Answer / Jawapan :

22. The events A and B are independent. Given P(A) = 0.4 and P(A9 ∩ B9) = 0.42, find P(B). 22
2
Peristiwa A dan B adalah tidak bersandar. Diberi P(A) = 0.4 dan P(A9 ∩ B9) = 0.42, cari P(B).

[2 marks / markah]

Answer / Jawapan :

23. The probability of Rosland wining a toy at a game station is 0.25. 23
Find the probability that he wins his second toy on his 6th game. 3

Kebarangkalian Rosland memenangi satu anak patung pada suatu stesen permainan ialah 0.25. Total
Cari kebarangkalian bahawa dia memenangi anak patung yang kedua pada permainan yang ke-6. 15

[3 marks / markah]
Answer / Jawapan :

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KERTAS MODEL SPM 2  3472/1

24. f(z) Diagram 24 shows a standard normal distribution graph of For
the mass of flour, X, in boxes produced by a factory. Examiner’s

Given that the mean and variance of the mass of flour Use

are 400 g and 60 g respectively. If the percentage of the 24
4
probability represented by the area of the shaded region is

O 1.291 k z 9.338%, find, to the nearest integer, the mass of the flour,
X, when the z-score is k.
Rajah 24 menunjukkan satu graf taburan normal piawai bagi jisim
Diagram 24 / Rajah 24 tepung, X, dalam kotak yang dihasilkan oleh sebuah kilang. Diberi min
dan varians jisim tepung masing-masing ialah 400 g dan 60 g. Jika
peratus kebarangkalian yang diwakili oleh luas kawasan berlorek ialah 9.338%, cari, kepada integer yang
terdekat, jisim tepung, X, apabila skor-z ialah k.

[4 marks / markah]
Answer / Jawapan :

25. d (cm) Diagram 25 shows the displacement-time graph of
a particle X which moves in a straight line such
60 that, t seconds after passing a fixed point O, its
50 displacement from O is d cm. On the axes provided
40 in the answer space, draw the velocity-time graph
of X.

30 Rajah 25 menunjukkan graf sesaran-masa bagi satu zarah
X yang bergerak dalam satu garis lurus dengan keadaan, t
20 saat selepas melalui satu titik tetap O, sesarannya dari O ialah
15 d cm. Pada paksi yang diberi dalam ruang jawapan, lukis graf 25
10 halaju-masa bagi X. 3

0 5 10 15 t (s) [3 marks / markah]
Diagram 25 / Rajah 25

Answer / Jawapan :

v (cm s–1)

t (s) Total
0 5 10 15 7

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2KERTAS MODEL

SIJIL PELAJARAN MALAYSIA 3472/2

Additional Mathematics

Kertas 2

2 jam 30 minit 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. Tunjukkan langkah-langkah penting dalam kerja mengira anda.
4. Anda dibenarkan menggunakan kalkulator saintifik.

Section A
Bahagian A
[40 marks]
[40 markah]
Answer all questions in this section.
Jawab semua soalan di bahagian ini.

1. Diagram 1 shows a curve with equation y = 30 – 28y that intersects the straight line 7x – 8y = 12. Find the
coordinates of the points A and B. x

Rajah 1 menunjukkan satu lengkung dengan persamaan y = 30 – 28y yang bersilang dengan garis lurus 7x – 8y = 12. Cari
x
koordinat-koordinat bagi titik A dan titik B [6 marks / markah]

y 7x – 8y = 12
B y = 30 –
28y
x

O x
A

Diagram 1 / Rajah 1

2. The length and the width of a rectangle is x cm and y cm respectively. The length of the rectangle is increasing
at a rate of 2 cm s–1 and the width of the rectangle is decreasing at a rate such that the area of the rectangle
is always 64 cm2.

Panjang dan lebar satu segi empat masing-masing ialah x cm dan y cm. Panjang segi empat tersebut bertambah pada kadar
2 cm s–1 dan lebarnya menyusut pada satu kadar dengan keadaan keluasan segi empat tersebut adalah sentiasa 64 cm2.

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KERTAS MODEL SPM 2  3472/2

(a) Show that the width of the rectangle is decreasing at time t s. [2 marks / markah]

Tunjukkan bahawa lebar segi empat tersebut menyusut pada masa t s.

(b) Find, in cm, the length and the width of the rectangle when the perimeter of the rectangle is minimum.
Cari, dalam cm, panjang dan lebar segi empat tersebut apabila perimeternya adalah minimum. [6 marks / markah]

3. Diagram 3 shows a circle with centre O. The points A, B, C and D lie on the circumference of the circle
such that AB is parallel to DC.

Rajah 3 menunjukkan satu bulatan berpusat O. Titik-titik A, B, C dan D terletak di atas lilitan bulatan dengan keadaan AB selari

dengan DC.

BC

O D
A

Diagram 3 / Rajah 3

Given the total length of minor arc AB and minor arc DC is 13.2 cm and the total area of sector AOB and

sector COD is 52.8 cm2. Find

Diberi jumlah panjang lengkok minor AB dan panjang lengkok minor DC ialah 13.2 cm dan jumlah keluasan sektor AOB dan
sektor COD ialah 52.8 cm2. Cari

(a) the radius, in cm, of the circle, [4 marks / markah]

jejari, dalam cm, bagi bulatan tersebut,

(b) the angle, in degree and minute, of AOD. [3 marks / markah]

sudut, dalam darjah dan minit, bagi AOD.

4. Given that a sin θ – b cos θ = a2 + b2 sin 3θ – tan–11 b 24.
kos θ = a2 + b2 sin 3θ – a
Diberi bahawa a sin θ – b tan–11 b 24.
a

(a) Express 8 cos2 θ – 6 sin θ cos θ in the form p + q sin (2θ + a) where p, q and a (0° , a , 90°) are

constants.

Ungkapkan 8 kos2 θ – 6 sin θ kos θ dalam bentuk p + q sin (2θ + a), dengan keadaan p, q dan a (0° , a , 90°) ialah

pemalar. [3 marks / markah]

(b) Find the value of a and of b if a < 8 cos2 θ – 6 sin θ cos θ < b. [3 marks / markah]

Cari nilai a dan nilai b jika a < 8 kos­2 θ – 6 sin θ kos θ < b.

5. Solution by scale drawing is not accepted.

Penyelesaian secara lukisan berskala tidak diterima.

The equation of two parallel lines l1 and l2 are y – x – 4 = 0 and y – x + 6 = 0 respectively.

Persamaan dua garis lurus l1 dan l2 yang selari masing-masing ialah y – x – 4 = 0 dan y – x + 6 = 0.

(a) Find the equation of the line that passes through the origin and perpendicular to l1 and l2.
Cari persamaan garis lurus yang melalui asalan dan berserenjang dengan l1 dan l2. [2 marks / markah]

(b) Calculate the shortest distance between the lines l1 and l2. [3 marks / markah]

Hitung jarak terdekat antara garis l1 dan garis l2.

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3472/2 KERTAS MODEL SPM 2
6. Diagram 6 shows the trajectory of a shot-put thrown by Kamal.
Rajah 6 menunjukkan trajektori bagi suatu lontaran peluru oleh Kamal.

H

Ground / Tanah
d

Diagram 6 / Rajah 6

The height, H cm, of the shot-put above the ground is expressed by the formula H = 1.6 + 0.6d – 0.0025d2
where d is the horizontal distance of the shot from Kamal, measured in cm.

Given that the shot travelled in a vertical plane until it hits the ground.

Ketinggian, H cm, lontaran peluru tersebut di atas tanah boleh diungkapkan oleh formula H = 1.6 + 0.6d – 0.0025d2 dengan
keadaan d ialah jarak mengufuk peluru tersebut dari Kamal, diukur dalam cm.
Diberi laluan lontaran peluru tersebut adalah pada satah mencancang sehingga ia jatuh ke tanah.

(a) Find, in cm, the horizontal distance travelled by the shot when it hits the ground.
Cari, dalam cm, jarak mengufuk, laluan lontaran peluru ketika ia jatuh ke tanah. [3 marks / markah]

(b) (i) Express H = 1.6 + 0.6d – 0.0025d2 in the form a – p(d – q)2 where a, p and q are constants.

Ungkapkan H = 1.6 + 0.6d – 0.0025d2 dalam bentuk a – p(d – q)2 dengan keadaan a, p dan q ialah pemalar.

[3 marks / markah]

(ii) Find, in cm, the maximum height of the shot above the ground. [1 mark / markah]
Cari, dalam cm, ketinggian maksimum peluru di atas tanah.

(iii) Find, in cm, the horizontal distance of the shot from Kamal when it is at its maximum height.
Cari, dalam cm, jarak mengufuk peluru tersebut dari Kamal ketika ia berada pada ketinggian maksimum.

[1 mark / markah]

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KERTAS MODEL SPM 2  3472/2

Section B
Bahagian B
[40 marks]
[40 markah]
Answer any four questions from this section.
Jawab mana-mana empat soalan daripada bahagian ini.

7. Solution by scale drawing is not accepted.

Penyelesaian secara lukisan berskala tidak diterima.

Diagram 7 shows the position vector of two boats, A and B, relative to a fixed point O are –2∼i + 8∼j and q∼i – 2∼j
respectively.

Rajah 7 menunjukkan vektor kedudukan bagi dua perahu, A dan B, relatif kepada satu titik tetap O, masing-masing ialah

–2∼i + 8∼j dan q∼i – 2∼j.

y

Boat A
Perahu A

Ox

Boat B
Perahu B

Diagram 7 / Rajah 7

Boat A moves with a velocity of (4∼i + p∼j) m min–1 and boat B moves with a velocity of (3∼i – ∼j) m min–1
at the same time.
Perahu A bergerak dengan kelajuan (4∼i + p∼j) m min–1 dan perahu B bergerak dengan kelajuan (3∼i – ∼j) m min–1 pada masa
yang sama.

(a) Given boat A moves with a speed of 5 m min–1, find the value of p where p , 0.
Diberi perahu A bergerak dengan satu kelajuan 5 m min–1, cari nilai p dengan keadaan p , 0. [2 marks / markah]

(b) Find the direction of boat A that moves with a velocity of (4∼i + p∼j) m min–1, correct to the nearest
degree.
Cari arah perahu A yan g bergerak dengan h alaju (4∼i + p∼j) m min–1, betul kepada darjah yang terdekat.

[3 marks / markah]

(c) Find the time taken, in minutes, when boat A meets boat B. Hence, find the position vector where the

two boats meet.
Cari masa yang diambil, dalam minit, apabila perahu A bertemu dengan perabu B. Seterusnya, cari kedudukan vektor di
mana kedua-dua perahu tersebut bertemu. [5 marks / markah]

8. The gradient of a curve is given by dy = k – 2 + 3–k = 0, where k  . Find
dx (x – 2)2
dy 3–k
Kecerunan bagi satu lengkung diberi oleh dx =k–2+ (x – 2)2 = 0, dengan keadaan k  . Cari

(a) the set of values of k given that the curve has two distinct turning points, [6 marks / markah]

set nilai-nilai k diberi bahawa lengkung tersebut mempunyai dua titik pusingan yang berbeza,

© Penerbitan Pelangi Sdn. Bhd. KM2–12 SPM Additional Mathematics

3472/2 KERTAS MODEL SPM 2
(b) the equation of the curve if k = 4 and it passes through the point (3, 2).
[4 marks / markah]
persamaan lengkung tersebut jika k = 4 dan ia melalui titik (3, 2).

9. Given Tn is the nth term of an arithmetic progression.
Diberi Tn ialah sebutan ke-n satu janjang aritmetik.

(a) Show that Un = 23[4 – Tn] is the nth term of a geometric progression. [5 marks / markah]

Tunjukkan bahawa Un = 23[4 – Tn] ialah sebutan ke-n satu janjang geometri. [3 marks / markah]
[2 marks / markah]
(b) If Tn = 1 (7n – 2),
3
1
Jika Tn = 3 (7n – 2),

(i) find the first three terms of the geometric progression,

cari tiga sebutan pertama bagi janjang geometri tersebut,

(ii) find the sum to infinity of the geometric progression.

cari hasil tambah hingga ketakterhinggaan bagi janjang geometri tersebut.

10. (a) In a family of six members, the ratio of the probability that two female members chosen to the probabilty
of four female members chosen at random is 4 : 1. Find

Dalam satu keluarga dengan enam ahli keluarga, nisbah kebarangkalian bahawa dua orang perempuan dipilih kepada
kebarangkalian bahawa empat orang perempuan dipilih secara rawak adalah 4 : 1. Cari

(i) the probability that a female member is chosen at random, [3 marks / markah]

kebarangkalian bahawa seorang perempuan dipilih secara rawak,

(ii) the mean and standard deviation of the distribution. [2 marks / markah]

min dan sisihan piawai bagi taburan tersebut.

(b) The random variable X is normally distributed with mean µ and standard deviation σ. Given that
P(X , 2.746) = 13% and P(X , 7.254) = 87%.

Pemboleh ubah rawak X tertabur secara normal dengan min µ dan sisihan piawai σ. Diberi bahawa P(X , 2.746) = 13%
dan P(X , 7.254) = 87%.

(i) Find the mean of variable X.
Cari min bagi pemboleh ubah X. [3 marks / markah]

(ii) Find the variance of variable X. [2 marks / markah]

Cari varians bagi pemboleh ubah X.

11. Use a graph paper to answer this question.

Gunakan kertas graf untuk menjawab soalan ini.

Table 11 shows the values of two variables, x and y, obtained from an experiment. Variables x and y are
related by the equation py2 = qx3 + 1, where p and q are constants.

Jadual 11 menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada suatu eksperimen. Pemboleh ubah
x dan y dihubungkan oleh persamaan py2 = qx3 + 1, dengan keadaan p dan q adalah pemalar.

x 1.0 1.5 2.0 2.5 3.0 3.5

y 2.24 2.72 3.46 4.43 5.57 6.85

Table 11 / Jadual 11 [2 marks / markah]
(a) Based on Table 11, construct a table for values of x3 and y2.

Berdasarkan Jadual 11, bina satu jadual bagi nilai-nilai x3 dan y2.

© Penerbitan Pelangi Sdn. Bhd. KM2–13 SPM Additional Mathematics

KERTAS MODEL SPM 2  3472/2

(b) Plot y2 against x3, using a scale of 2 cm to 5 units on both axes. Hence, draw the line of best fit.

Plot y2 melawan x3, dengan menggunakan skala 2 cm kepada 5 unit pada kedua-dua paksi. Seterusnya, lukis garis lurus

penyuaian terbaik. [3 marks / markah]

(c) Using the graph in 11(b), find the value of

Menggunakan graf di 11(b), cari nilai

(i) p, [2 marks / markah]
(ii) q. [3 marks / markah]

Section C
Bahagian C
[20 marks]
[20 markah]
Answer any two questions from this section.
Jawab mana-mana dua soalan daripada bahagian ini.

12. Use a graph paper to answer this question.

Gunakan kertas graf untuk menjawab soalan ini.

Miss Wee plan to buy x hair dryers and y irons as door gifts for her schools annual dinner. The price of a
hair dryer and an iron is RM60 and RM70 respectively. The purchased of the items is based on the following
constraints:

Cik Wee merancang untuk membeli x pengering rambut dan y seterika sebagai buah tangan untuk jamuan tahunan sekolahnya.
Harga bagi seunit pengering rambut dan seunit seterika masing-masing ialah RM60 dan RM70. Pembelian barangan tersebut
adalah berdasarkan kepada kekangan berikut:

I The minimum allocation is RM5 600.

Jumlah peruntukan minimum ialah RM5 600.

II The number of iron is not more than 4 times the number of hair dryer.

Bilangan seterika adalah tidak melebihi 4 kali bilangan pengering rambut.

III The total number of door gift is at most 160 if the number of hair dryer and iron its at the ratio 1 : 2.

Jumlah bilangan buah tangan adalah tidak melebihi 160 jika nisbah pengering rambut kepada seterika adalah 1 : 2.

(a) Write three inequalities, other than x > 0 and y > 0, which satisfy all the above constraints.
Tulis tiga ketaksamaan, selain daripada x > 0 dan y > 0, yang memenuhi semua kekangan di atas.
[3 marks / markah]


(b) Using a scale of 2 cm to 10 hair dryers on the horizontal axis and 2 cm to 20 irons on the vertical axis,

construct and shade the region R which satisfy all the above constraints.

Menggunakan skala 2 cm kepada 10 pengering rambut pada paksi mengufuk dan 2 cm kepada 20 seterika pada paksi

mencancang, bina dan lorek rantau R yang memenuhi semua kekangan di atas. [3 marks / markah]

(c) Using the graph constructed in 12(b), find

Menggunakan graf yang dibina di 12(b), cari

(i) the minimum number of irons can buy by Miss Wee if the number of hair dryer is 60,

bilangan minimum seterika yang dibeli oleh Cik Wee jika bilangan pengering rambut ialah 60,

[1 mark / markah]

(ii) the maximum total number of hair dryer and iron that can be purchased by Miss Wee.
jumlah maksimum bilangan pengering rambut dan seterika yang boleh dibeli oleh Cik Wee. [3 marks / markah]

© Penerbitan Pelangi Sdn. Bhd. KM2–14 SPM Additional Mathematics

3472/2 KERTAS MODEL SPM 2

13. Table 13 shows the price indices for the year 2016 and 2018 based on the year 2010 of Bushra’s monthly
expenses on room rental, petrol and food.

Jadual 13 menunjukkan indeks harga bagi tahun 2016 and 2018 berasaskan tahun 2010 untuk perbelanjaan bulanan Bushra
bagi sewa bilik, petrol dan makanan.

Monthly Expenses Price index in 2016 based on 2010 Price index in 2018 based on 2010
Indeks harga pada tahun 2016 Indeks harga pada tahun 2018
Perbelanjaan Bulanan berasaskan tahun 2010 berasaskan tahun 2010

(RM) 124 135

Room rental

Sewa bilik

Petrol 115 p

Petrol

Food 108 116

Makanan

Table 13 / Jadual 13

(a) The monthly expenses on petrol in the year 2010 and 2018 is RM150 and RM210 respectively. Find

Perbelanjaan bulanan untuk petrol pada tahun 2010 dan 2018 masing-masing ialah RM150 dan RM210. Cari

(i) the value of p, [1 mark / markah]

nilai p.

(ii) the monthly expenses on petrol in the year 2016. [2 marks / markah]

perbelanjaan bulanan untuk petrol pada tahun 2016.

(b) The composite index for the monthly expenses in the year 2016 based on the year 2010 is 116.8 and
the weightage of the monthly expenses on room rental, petrol and food is in the ratio n : 3 : 1. Find

Indeks gubahan untuk perbelanjaan bulanan pada tahun 2016 berasaskan tahun 2010 ialah 116.8 dan pemberat bagi
perbelanjaan bulanan untuk sewa bilik, petrol dan makanan adalah dalam nisbah n : 3 : 1. Cari

(i) the value of n, to the nearest integer, [3 marks / markah]

nilai n, kepada integer terdekat,

(ii) the corresponding monthly expenses in the year 2016 if the monthly expenses in the year 2010 is

RM1 500.

perbelanjaan bulanan sepadan pada tahun 2016 jika perbelanjaan bulanan pada tahun 2010 ialah

RM1 500. [2 marks / markah]

(c) Find the price index of the monthly expenses on food in year 2018 based on year 2016.

Cari indeks harga bagi perbelanjaan bulanan untuk makanan pada tahun 2018 berasaskan tahun 2016.

[2 marks / markah]

14. A particle moves along a straight line and passes through a fixed point O. Its velocity, v m s–1, is given

by v = pt2 + 7 t + 1 , where p is a constant and t is the time, in second, after passing through O. Given the
5 2
acceleration of the particle is –1 m s–2 when t = 4 s.

Suatu zarah bergerak di sepanjang suatu garis lurus dan melalui satu titik tetap O. Halajunya, v m s–1 diberi oleh

v = pt2 + 7 t + 1 , dengan keadaan p ialah pemalar dan t ialah masa, dalam saat, selepas melalui O. Diberi pecutan zarah itu
5 2
ialah –1 m s–2 apabila t = 4 s.

[Assume motion to the right is positive]

[Anggapkan gerakan ke arah kanan sebagai positif]

© Penerbitan Pelangi Sdn. Bhd. KM2–15 SPM Additional Mathematics

KERTAS MODEL SPM 2  3472/2
[3 marks / markah]
(a) Find the value of p. [2 marks / markah]
[5 marks / markah]
Cari nilai p.

(b) Find the time t, in seconds, when the particle stop instantaneously.

oCari masa t, dalam saat, apabila zarah itu berhenti seketika.

(c) Find the total distance, in m, travelled by the particle in the first 10 seconds.

Cari jumlah jarak, dalam m, yang dilalui oleh zarah itu dalam 10 saat yang pertama.

15. Solution by scale drawing is not accepted.

Penyelesaian secara lukisan berskala tidak diterima.

Diagram 15 shows the locations of town A and town B.

Rajah 15 menunjukkan kedudukan bandar A dan bandar B.

Town B
Bandar B

12 km

Town A
Bandar A

Diagram 15 / Rajah 15

Town B is due north of town A and the distance between town A and town B is 12 km. The bearing of town
C from town B is 140°. The distance between town A and town C is 8 km.

Bandar B adalah ke utara bandar A dan jarak di antara bandar A dan bandar B ialah 12 km. Bearing bandar C dari bandar B
ialah 140°. Jarak di antara bandar A dan bandar C ialah 8 km.

(a) Copy Diagram 15, sketch and mark the possible position for town C and town C9 where town C is

nearer to town B.

Salin Rajah 15, lakar dan tandakan kedudukan yang mungkin untuk bandar C dan bandar C9 dengan keadaan bandar C

lebih dekat dengan bandar B. [2 marks / markah]

(b) Find /AC9B and /ACB. [3 marks / markah]

Cari /AC9B dan /ACB.​

(c) Calculate the length of BC9, in km. [2 marks / markah]

Hitung panjang BC9, dalam km.

(d) Find the difference of area ABC9 and the area ACC9. [3 marks / markah]

Cari beza antara luas ABC9 dan luas ACC9.

© Penerbitan Pelangi Sdn. Bhd. END OF QUESTION PAPER SPM Additional Mathematics
KERTAS PEPERIKSAAN TAMAT

KM2–16

ANSWERS

KERTAS MODEL SPM 1 No. Solution and Mark Scheme Sub Total
• B = Be given mark(s) Marks Marks
Paper 1
12
1. 3 cos 2a = 7 cos a 1. B1: 6 cos2 a – 7 cos a – 3 = 0 1

3 cos 2a – 7 cos a = 0

3(2 cos2 a – 1) – 7 cos a = 0 cos a =– 1
3
6 cos2 a – 7 cos a – 3 = 0

( 3 cos a + 1)(2 cos a – 3) = 0

3 cos a + 1 = 0   or   2 cos a – 3 = 0

cos a = – 1 cos a ≠ 3
3 2

1 2 2. sin2q – 1 π = 1    ,   0<q<π 2. B1: π or 5 π 12
3 2 6 6 1

2q – 1 π = π , 5 π 13
3 6 6 1
1
2q = π + 1 π , 5 π + 1 π π 7
6 3 6 3 4 12
q = , π
π 7
2q = 2 , 6 π

\ q = π , 7 π y
4 12

5π x
6

π
6
O

3. gf(x) + 3 = 0 3. B1: 6x2 – 8x + 3 – k = 0
2(3x2 – 4x) – k + 3 = 0 B2: (–8)2 – 4(6)(3 – k) = 0
6x2 – 8x + 3 – k = 0 k= 1
3
a = 6 , b = –8 , c = 3 – k

For equal roots

b2 – 4ac = 0

(–8)2 – 4(6)(3 – k) = 0

64 – 24(3 – k) = 0

64 = 24(3 – k)

64 = 3 – k
24

8 = 3 – k
3
k = 3 – 8

3
k = 1

3

© Penerbitan Pelangi Sdn. Bhd. A – 1 SPM Additional Mathematics

ANSWERS

4. 4. (a) 34 12

M (b) –2~p – ~q 1
5. B1: (15.4 × 20) or (20.2 × 30) 14
qB 1
–2p O ~ B2: Mean = 18.28 kg 1
B3: 20 [3.52 + 15.42] or 1
~ 30 [2.62 + 20.22]
Standard deviation = 3.81 kg 13
–q p 1
~ ~ 1

A

(a) O→M = 32+ 52 = 34
(b) O→M = –2~p – ~q

5. Group A: Group B:

∑x = 15.4 ∑y = 20.2
20 30

∑x = 308 ∑y = 606

Total mean = 308 + 606 = 18.28 kg
50
∑x2
σ2A = 20 – (–xA)2 = 3.52

∑x2 – (15.4)2 = 3.52
20
∑x2 = 20[3.52 + (15.4)­2]

= 4 988.2

σ2B = ∑y2 – (–xB)2 = 2.62
30

∑y2 – (20.2)2 = 2.62
30
∑y2 = 30[2.62 + (20.2)­2]

= 12 444

σ2 = 4 988.2 + 12 444 – (18.28)2
total 50

= 14.4856

\ σtotal = 14.4856 = 3.81 kg


6. x2 – 16 > (x + 4)(x2 – 5x + 5) 6. B1: (–x2 + 6x – 9)
B2: (x + 4)(x – 3)2 < 0
(x + 4)(x – 4) > (x + 4)(x2 – 5x + 5) x < –4
(x + 4)(x – 4) – (x + 4)(x2 – 5x + 5) > 0
(x + 4)[(x – 4) – (x2 – 5x + 5)] > 0
(x + 4)[x – 4 – x2 + 5x – 5] > 0
(x + 4)(–x2 + 6x – 9) > 0
(x + 4)(x2 – 6x + 9) < 0
(x + 4)(x – 3)2 < 0

\ (x – 3)2 > 0   ,   x + 4 < 0
x < –4


© Penerbitan Pelangi Sdn. Bhd. A – 2 SPM Additional Mathematics

ANSWERS

11.1 + 13.7 + 12 + 16.5 + 17.2 + 18.9 7. (a) Mean = 18.78 14
Median = 18.05 1
7. (a) Mean = + 19.8 + 23.6 + 27.3 + 27.7
10

=18170.8

= 18.78



Arrange the data in ascending order:

11.1, 12.0, 13.7, 16.5, 1174.22, 148.39, 19.8, 23.6, 27.3, 27.7

Median = 17.2 + 18.9 = 18.05
2

(b) The distribution is skewed to the right. Median is (b) The distribution is skewed to the 1
more suitable as a measure of central tendency for right. 1
the distribution. 1
Median is more appropriate as a
measure of central tendency. 1

8. π3 radians = 60° 1 8. 1 r2 π sin 2
B1: 2 3 – 60° = 1.45 3

Area of shaded region = 1.45

1 r2 (θ – sin θ) = 1.45 r = 4 cm
2

1  1 r2 π
2 3 – sin 60° = 1.45

1 r2 (0.1812) = 1.45
2
2 × 1.45
r = 0.1812

r = 4.00 cm

9. log3 m = p, log15 m = q 9. q 1
(a) B1: log5 3 = p – q 1

(a) log5 m =p pq 1
log5 3 p–q

log5 m =p log5 3

log5 m = q
log5 15

p log5 3 = q
log5 (3 × 5)

p log5 3 = q
log5 3 + 1

p log5 3 = q(log5 3 + 1)

p log5 3 – q log5 3 = q

log5 3(p – q) = q q

log5 3 = p – q

q pq
p–q = p–q
3 4log5 m = p

(b) log5 75 = log5 (25 × 3) q
(b) 2 + p – q
= log5 52 + log5 3
q
=2+ p–q

© Penerbitan Pelangi Sdn. Bhd. A – 3 SPM Additional Mathematics

ANSWERS

10. (a) d 2y = –36 10. (a) The stationary point is a minimum 1 4
dx2 x3 d 2y 3
because dx2 . 0 1
d 2y –36 1
at (–3, 17), dx2 = (–3)3 1

=43 . 0

\ The stationary point is a minimum because

d 2y . 0.
dx2

d 2y = –36x –3 (b) B1: 18 +c =0
(b) dx2 x2

∫ d y = –36x –3 dx
dx c = –2

dy = –36x –2 + c
dx –2

m = dy = 18 + c. dy = 18 –2
dx x2 dx x2

at (–3, 17), 18 + c = 0
x2

18 + c = 0
(–3)2

c = –2

\ The equation of gradient is dy = 18 – 2.
dx x2

11. P(2h + 7, 3h – 8) , Q(3h + 2, 4h – 1) 11. B1: (h – 5) or (h + 7) 1
Distance = (x1–x2)2 + (y1–y2)2 B2: 2h2 + 4h – 96 = 0 1
h = 6, –8 1
PQ = 170
[3h + 2 – (2h + 7)]2 + [4h – 1 – (3h – 8)]2 = 170

(h – 5)2 + (h + 7)2 = 170

h2 – 10h + 25 + h2 + 14h + 49 = 170

2h2 + 4h – 96 = 0

2(h2 + 2h – 48) = 0

2 ≠ 0   ,   h2 + 2h – 48 = 0 or h + 8 = 0
(h – 6)(h + 8) = 0 h = –8
h – 6 = 0
h = 6

12. (a) To find the first three terms in terms of n, substitute 12. (a) B1: 4n2, 4(n + 1)2, 4(n + 2)2, … 1 4
n with (n + 1) and (n + 2) for the subsequent term. 1
yn = 8n + 4
xn = 4n2 + 1)2
xxnn = 4(n + 2)2
+1 = 4(n

+ 2

\ 4n2, 4(n + 1)2, 4(n + 2)2, …

yn = 4(n + 1)2 – 4n2 4n2
= 4(n2 + 2n + 1) –
= 4n2 + 8n + 4 – 4n2
= 8n + 4

© Penerbitan Pelangi Sdn. Bhd. A – 4 SPM Additional Mathematics

(b) From (a), yn = 8n + 4 (b) B1: 12, 20, 28, … ANSWERS
When n = 1, y1 = 8(1) + 4 = 12 Sn = xn + yn – 4 1
When n = 2, y2 = 8(2) + 4 = 20 1
When n = 3, y3 = 8(3) + 4 = 28
\ 12, 20, 28, … 13
1
\ a = 12, d = 20 – 12 = 28 – 20 = 8 1
12
Hence, y is an arithmetic progression. 1
n 14
\ Sn = 2 [2a + (n – 1)d] 1
1
=n2 [2(12) + 8(n – 1)] 1

=2n [24 + 8n – 8]

=2n [8n + 16]

= 4n2 + 8n + 4 – 4
=xn + yn – 4


13. 81x – 1 = 1 . 27y + 2 13. B1: 34(x – 1) or 33(y + 2)
9
34(x – 1) = 3–2 . 33(y + 2)
B2: 4x – 8 = 3y
\ 4(x – 1) = –2 + 3(y + 2)
4x – 4 = –2 + 3y + 6 y= 4x – 8
4x – 8 = 3y 3

y = 4x – 8
3

14. 3, 1, 7, 1 14. B1: 1 × 3!
6
11    
1 × 3! = 3 × 2 = 6 ways

15. (a) x = 3 15. (a) x = 3
(b) B1: p = 24
(b) Given (3, –25) is the minimum point:
p
x – 8 = 0

x = 8pp = 3 B2: –p2 + q = –25
= 24 16

Minimum value = –25 p = 24 and q = 11
–1p62 + q = –25
–12642 + q = –25
–36 + q = –25

q = –25 + 36

q = 11

16. The relation is not a function because all the values of x 16. Not a function 1 2
has two values of y except x = 0. Because all the values of x has two 1
values of y except x = 0.

© Penerbitan Pelangi Sdn. Bhd. A – 5 SPM Additional Mathematics

ANSWERS

17. Given Sn = 3 – 2–4n, 17. (a) B1: Sn – 1 = 3 – 2–4(n – 1) 14
B2: Un = 15[2–4n] 1
(a) Sn – 1 = 3 – 2–4(n – 1) 15[2–4(n + 1)] 1
Un = Sn – Sn – 1


= 3 – 2–4n – [3 – 2–4(n – 1)]
= –2–4n + 2–4(n – 1)
= +2–4n . 24 – 2–4n

= 2–4n[24 – 1]

= 2–4n[15]

\ Un + = 15[2–4(n + 1)]
1

(b) Un + 1 = 15 . 2–4n – 4 (b) UUn +n 1 = 1 1
= 15 . 2–4n . 2–4 16
13
1
= Un . 1 ⇒ UUn +n 1 = 1
16 16
The sequence is a geometric
\ The sequence is a geometric progression. progression.

18. (a) y = 3 + 2x = 3x –1 + 2x 18. (a) B1: dy = –3 +2
x dx x2

Gradient at P = –1

ddyx = –3x–2 + 2 = –1 x=1

–3 = – 3
x2

x2 = 1
x = ±1
x = 1, x ≠ –1

\ x-coordinate of P is 1.

(b) dy = dy . dx    (b) dy = –0.05 1
dt dx dt dt 14

= –1 . (0.05)

= –0.05

19. (1, 2.4) , (4.5, 0.3) y – y 19. (a) –0.6
2 – 1
(a) gradient = m =
x2 x1

=04.3.5––21.4

= –0.6

(b) y= h = hk–x (b) B1: k = e0.6 1
kx B2: c = ln h = 3 1
ln y = ln (hk–x) k = 1.82 and h = 20.09 1
ln y = ln h + ln k–x
ln y = ln h – x ln k

Compare with Y = mX + c
\ m = –ln k = –0.6
\ k = e0.6 = 1.82
c = ln h


Substitute point (1, 2.4) and m = –0.6 into
Y = mX + c
2.4 = –0.6(1) + c
2.4 = –0.6 + c
c = 2.4 + 0.6

= 3
\ c = ln h = 3
h = e3 = 20.09

© Penerbitan Pelangi Sdn. Bhd. A – 6 SPM Additional Mathematics

ANSWERS

20. The probability Ismail wins the 1st set = p 81 20. B1: 1p × 4 2 1 4
100 5 1 3
Given the probability Ismail wins the match = 1
Let w be the probability Ismail wins the set. 1

P(w1  w2) + P(w1  w9  w3) + P(w19  w  w3) B2: 1
2 2 1
1p 1 3 2 3(1 3 4 4 1
= 81 × 5 × 5 or – p) × 5 × 5
100

1p × 4 2 + 1p × 1 × 3 2 + 3(1 – p) × 3 × 4 4 = 81
5 5 5 5 5 100

4 p + 3 p + (1 – p)1 12 2 = 81 B3: 4 p + 3 p + (1 – p)1 12 2 = 81
5 25 25 100 5 25 25 100

20 p + 3 p + 12 – 12 p = 81 3
25 25 25 25 100 4

\ 11 p = 81 – 12
25 100 25

\ 11 p = 33
25 100

\ p = 33 × 25
100 11

= 3
4

\ The probability Ismail won the first set is 3 .
4

21. y= k2 +x 21. B1: k2 [–(x – 3)–2]
(x – 3)

y = k2(x – 3)–1 + x –k2
(x – 3)2
dy B2: + 1 = 0
dx
For stationary points, =0

\ dy = k2 [–(x – 3)–2] + 1 = 0 x=3±k
dx –k2
– 3)2
(x = –1

k2 = (x – 3)2

\ ±k = (x – 3)

\ x – 3 = ±k

x = 3 ± k.

22. 32x – 1­ = 4x 22. B1: (2x – 1) lg 3 = x lg 4 13
1
lg 32x – 1 = lg 4x 1
(2x – 1) lg 3 = x lg 4 14
1
2x – 1 = lg 4 2x – 1
x lg 3 x
2x – B2: = 1.2619
x
1 = 1.2619

2x – 1 = 1.2619x x = 1.35

0.7381x = 1

x = 1.35

23. Let X be the marble produced by the factory. 23. B1: P(X < m) = 0.2

X ~ N(46, 25) B2: P(Z > 0.842) = 0.2
or
P(X . m) = 80% Z = –0.842
1 – P(X < m) = 0.8
P(X < m) = 0.2
m
1 2 PZ < – 46 = 0.2
5

© Penerbitan Pelangi Sdn. Bhd. A – 7 SPM Additional Mathematics

ANSWERS

f(z) f(z) m – 46 1
B3: 5 = –0.842 1

m = 42 g 1
1
–z 0 z 0 zz
1
From normal distribution table, P(Z > 0.842) = 0.2 1
m – 46

\ 5 = –0.842
m – 46 = –4.21
m = 46 – 4.21

= 41.79
 42 g

24. (a) x = 0, P(x = 0) = k|0 – 4| = 4k 24. (a) B1: 4k or 3k or 2k or k is seen 4
x = 1, P(x = 1) = k|1 – 4| = 3k 4
k = 1
x = 2, P(x = 2) = k|2 – 4| = 2k 12

x = 3, P(x = 3) = k|3 – 4| = k

x = 4, P(x = 4) = 2k

Total probability = 1

P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = 1

4k + 3k + 2k + k + 2k = 1

12k = 1

k = 1
12

1 2 1 2 1 2(b) 4× 1 1 × 1 × 1 1 2 1 2(b) × 1 × 1
Mean = 0 12 + 3 12 +2 2 12 + B1: 0 4 12 + 1 3 12 +

31 1 2 1+ 4 2 × 1 2 212 × 1 2 + 31 1 2 + 412 × 1 2
12 12 12 12 12

=0 + 3 +4+ 3 + 8
12
1.5
= 1.5

25. Ping pong balls M N Total 25. B1: 8C1 × 7C2 or 5C2 × 10C1 1
White 3 47 15C3 15C3 1
Orange 2 68 1
Total 5 10 15 1 2 1 2B2: 6C1 × 3C2 + 2C1 × 3C1 × 4C1
15C3 15C3

Let X = the event that an orange ping pong ball selected, B3: 24 + 20 – 6
Y = the event that 2 ping pong balls from brand M 65 91 65

selected

Siti selects three ping pong balls at random
\ P(X  Y) = P(X) + P(Y) – P(X  Y)

P(X) = P(an orange ball × 2 white balls)
=8C1 × 7C2
=6254 15C3

© Penerbitan Pelangi Sdn. Bhd. A – 8 SPM Additional Mathematics

P(Y) = P(2 balls brand M × 1 ball brand N) 226 or 0.497 ANSWERS
455 1

=5C2 × 10C1 = 20
15C3 91

P(X  Y) = P(1 orange ball brand N × 2 white ball
brand M) +
P(1 orange ball brand M × 1 white ball
brand M × 1 white ball brand N)

1 2 1 2 =6C1 × 3C2 + 2C1 × 3C1 × 4C1
15C3 15C3

= 18 + 24 – 6
455 455 65

P(X  Y) = P(X) + P(Y) – P(X  Y)

=2654 + 2901 – 6 = 226 or 0.497
65 455

Paper 2 No. Solution and Mark Scheme Mark Sub Total
Scheme Marks Marks

• P = Knowledge, K = Method, N = Value

1. (a) sin x + sin 2x 1. (a) sin 2x = 2 sin x cos x or K1 2 8
1 + cos x + cos 2x cos 2x = 2 cos2 x – 1

= 1 sin x + 2 sin x cos x 1 sin x (1 + 2 cos x) N1
+ cos x + 2 cos2 x – cos x (1 + 2 cos x)

= sin x (1 + 2 cos x)
cos x (1 + 2 cos x)

= tan x

(b) (i) y (b) (i) Shape of tangent graph, N1 3
points (0, 0), (π, 0) and
6 asymptotes
4
2 x = π and x = 3π N1
2 2
0 x
–2 π π 3π
–4 2 2 N1
–6
All correct

(ii) sin x + sin 2x + x =4 (ii) y=4– x P1 3
1 + cos x + cos 2x 2π 2π
x
y = 4 – 2π Sketch straight line with K1
negative gradient and
x0π correct y-intercept
y 4 3.5

\ From graph, the number of solutions 2 N1
 = 2.

© Penerbitan Pelangi Sdn. Bhd. A – 9 SPM Additional Mathematics

ANSWERS

2. Let the number of pens bought by Ramesh = x 2. 6 000 P1 6 6
x P1
The cost of each pen = 6 000 cents. K1
x K1
6 000 K1
If 10 more pens could have been bought, the cost (x + 10) N1

of each pen = 6 000 cents. P1 4 7
(x + 10) K1
K1
The discount of each pen is 20 cents. 6 000 – 6 000 = 20 N1
x (x + 10)
6 000 – 6 000 = 20 P1 3
x (x + 10) K1
N1
3 46 20x2 + 200x – 60 000 = 0
000 – 6 000 = 20 × [x(x + 10)]
x (x + 10)

6 000(x + 10) – 6 000x = 20x(x + 10) (x – 50)(x + 60) = 0
50
6 000x + 60 000 – 6 000x = 20x2 + 200x

\ 20x2 + 200x – 60 000 = 0

20(x2 + 10x – 3 000) = 0

20 ≠ 0 , x2 + 10x – 3 000 = 0

(x – 50)(x + 60) = 0

x – 50 = 0   or x + 60 = 0

x = 50 x ≠ –60

\ The number of pens bought by Ramesh is 50.

3. (a) Given the angle of each sector forms a 3. (a) a = 3 , d = 2
progression.
n [2(3) + 2(n – 1)] = 360
3°, 5°, 7°, … 2

It is an arithmetic progression.
a=3,d=5–3=2

Total angle in a circle = 360° (n + 20)(n – 18) = 0
n n = 18
Sn = 2 [2a + (n – 1)d] = 360°

n [2(3) + 2(n – 1)] = 360
2
n[6 + 2n – 2] = 720

n[2n + 4] = 720

2n2 + 4n – 720 = 0

2(n2 + 2n – 360) = 0
2 ≠ 0 , n2 + 2n – 360 = 0
(n + 20)(n – 18) = 0
n + 20 = 0 or n – 18 = 0
n ≠ –20 n = 18

(b) Area of the sector = 3.3π cm2 (b) 1 (6)2θ = 3.3p
2
1 r2θ = 3.3p
2

1 (6)2θ = 3.3p θ = 3.3 × 180° × 2
2 36
3.3p × 2
θ = 36

θ = 3.3 × 180° × 2 33°
36

θ = 33°

The angle of the sector is 33°.

© Penerbitan Pelangi Sdn. Bhd. A – 10 SPM Additional Mathematics

4. 4. ANSWERS
Time of 6
arrival (t)
Mid- Number Waiting Cumulative
point of days time frequency
(min)

0906 – 0910 0908 5 0 5

0911 – 0915 0913 10 5 15

0916 – 0920 0918 15 10 30

0921 – 0925 0923 26 15 56

0926 – 0930 0928 21 20 77

0931 – 0935 0933 15 25 92

0936 – 0940 0938 8 30 100

(a) Average waiting time 5(0) + 10(5) + 15(10) +
26(15) + 21(20) + 15(25) +
5(0) + 10(5) + 15(10) + 26(15)
8(30)
= + 21(20) + 15(25) + 8(30) (a) 5 + 10 + 15 + 26 + 21 + 15 + 8 P1 3
5 + 10 + 15 + 26 + 21 + 15 + 8 K1
1 625 N1
= 1 625 100
100
16.25 minutes
= 16.25 minutes

(b) Percentage of days that the express bus is (b) 100 – 77 P1 3
late K1
(100 – 77) × 100% N1
=(10010–077) × 100% 100

= 23% 23%

5. (a) log3 x + 2 log9 y = 3 5. (a) log3 x + 2 log3 y =3 K1 2 6
log3 x + log9 y2 = 3 2 N1

log3 x + log3 y2 = 3 x = 27
log3 32 y

\ log3 x + 2 log3 y =3
2
\ log3 (xy) = 3


xy = 33 = 27

x = 27
y

© Penerbitan Pelangi Sdn. Bhd. A – 11 SPM Additional Mathematics

ANSWERS

(b) log12 (x + y) = 1..................... a (b) 27 + y2 = 12 K1 4
y K1
From (a), log3 x + 2 log9 y = 3 ⇒ x = 27 N1
y N1
27
Substitute x = y into equation a 7
K1 2
 2 log12 27 (y – 9)(y – 3) = 0 N1
y + y =1 x = 3, y = 9
x = 9, y = 3 K1 4
27 + y2 = 121
y SPM Additional Mathematics

\ y2 + 27 = 12y

y2 – 12y + 27 = 0

(y – 9)(y – 3) = 0

y – 9 = 0   or   y – 3 = 0

y = 9 y=3

When y = 9, x = 27 =3
9

When y = 3, x = 27 =9
3

\ x = 3, y = 9 or x = 9, y = 3

6. Q 6.

N

y M
~

O ~x P 3~x R

(a) Given 5O →N = 8O →M 3 4(a) 8 ~x + 1 (–~x + ~y)
5 2
O →N 8 O →M
= 5

=85[ O →P + P →M] 4 [~x + ~y]
5
3 4 P →Q
=58 ~x + 1
2

3 4
=58 ~x + 12(P →O + O →Q)

3 4= 8 ~x + 21 (–~x + ~y)
5

3 4= 8 1 ~x + 1 y
5 2 2 ~

= 4 [~x + ~y]
5

(b) Q→R = Q→O + O→R (b) Q→N = 1 (4~x – ~y)
5
= 4–~~xy + 4y~x
= – ~

Q→N = Q→O + O→N
4 4
= –~y + 5 ~x + 5 y
~
=54~x – 1
5 ~y

= 1 (4~x – ~y)
5

© Penerbitan Pelangi Sdn. Bhd. A – 12

ANSWERS

N→R = N→O + O→R N→R = 45(4~x – ~y) K1
N1
= – 4  (~x + ~y) + 4~x
5

= – 4 ~x – 4 ~y + 4~x Q→N 1 N→R
5 5 4
=
16 4
= 5 ~x – 5 ~y

= 45(4~x – ~y) Points Q, N and R are N1
collinear. Hence, N lies on the
Q→N 1 (4~x – ~y) line QR.
N→R 5
\ =
4 (4~x ~y)
5 –

=41

\ Q→N = 1 N→R
4

Points Q, N and R are collinear.
Hence, N lies on the line QR.

(c) From (b) Q→N = 1 (c) Q→N : N→R = 1 : 4 N1 1
N→R 4

\ Q→N : N→R = 1 : 4

7. (a) x 0.1 1.5 2.5 3.5 4.5 6.5 7. (a) xy 0.39 3.60 4.50 N1 1 10
y 3.9 2.4 1.8 1.5 1.3 1.0
xy 0.39 3.60 4.50 5.25 5.85 6.50
xy 5.25 5.85 6.50

(b) (b) Label correct axes and use K1 3
uniform scale
xy
8.3 All points plotted perfectly N1
Draw the line of best fit N1
8

7

6

5

4
3.8

3

2

1
y

0 0.5 1 1.5 2 2.25 2.5 3 3.5 4

© Penerbitan Pelangi Sdn. Bhd. A – 13 SPM Additional Mathematics

ANSWERS

(c) y = x p q (c) (i) xy = –qy + p P1 4
y(x q) = p + P1
+ p = c = 8.3 K1
xy + qy = p N1
xy = –qy + p
Y = mX + c 8.3 – 1.3 = –q P1 2
0 – 3.5 N1

(i) p = c = 8.3 q = 2 10

m = –q = 8.3 – 1.3 = –2
0 – 3.5

\ q = 2

(ii) y= x p (ii) 1 = 1
+q p 8.3
x+q
1 = p p = 0.1205
y

1 = 1 x + q
y p p

\ gradient of straight line:

1 = 1
p 8.3

\ p = 0.1205

8. House B (–1,10) House A (h,k) 8.

House M (1,6)

House C (x,y)

 2(a) x, y x y
M= –1 + 10 + = (1,6) (a) –1 + = 1 or 10 + = 6 K1 2
2 2 2 2 N1

–1 + x =1   ,    10 + y =6 P1 6
2 2 K1
–1 + x = 2 10 + y = 12
C = (3, 2)

 \ x = 3 y = 2

\ House C = (3, 2)

(b) When the road AM perpendicular to the road (b) k–6 × 2 – 10 = –1
BC h–1 3 – (–1)

mAM × mBC = –1

k–6 × 32––(1–01) = –1 h = 2k – 11
h–1

k–6 × –8 = –1
h–1 4

k – 6 = 1
h – 1 2

2k – 12 = h – 1

h = 2k – 11.................. a

© Penerbitan Pelangi Sdn. Bhd. A – 14 SPM Additional Mathematics

Gradient AB =– 1 k – 10 = – 1 P1 ANSWERS
3 h – (–1) 3 K1
k – 10 N1 2
h – (–1) = – 1 N1 5 10
3
3(k – 10) = –(h + 1) P1
N1
3k – 30 = –h – 1 3k = –h + 29
P1
3k = –h + 29.......................... b P1

Substitute h = 2k – 11 into equation b h=5
3k = –(2k – 11) + 29 k=8
3k = –2k + 11 + 29
5k = 40

k = 40 =8
5

Substitute k = 8 into equation a
h = 2(8) – 11
  = 5

\ h = 5, k = 8

(c) Area bounded by the triangle ABC   1 5  –1  3  5
  =21 2 8  10  2  8
5  –1  3  5 (c)
8  10  2  8

=12 (50 – 2 + 24) – (–8 + 30 + 10) 20 unit2

=21 |40|

= 20 unit2

9. y 9. (a) Volume A = 1 πx2(5 – y) unit3
3
x2 = 4y

5

A Volume B = 2πy2 unit3
y
A (x,y)
B

Ox x

y = mx + 5

(a) Given the volume generated = 22 π unit3
3

V­ olume A = volume of a cone

=13 πr2h

=31πx2(5 – y) unit3

∫ Volume B = π y x2 dy
0

∫=π y 4y dy
0
3 4= 4π y2 y
2 0

= 2πy2 unit3

\ Volume A + Volume B = 22 π
3

© Penerbitan Pelangi Sdn. Bhd. A – 15 SPM Additional Mathematics

ANSWERS

1 πx2(5 – y) + 2πy2 = 22 π 2y2 + 20y – 22 = 0 K1
3 3 (y + 11)(y – 1) = 0 K1
A = (2, 1) N1
Substitute x2 = 4y into the equation

3 4 1 y) 2y2 22
π 3 (4y)(5 – + = 3 π

3 4
3× 43 y(5 – y) + 2y2 = 22 × 3
3
4y(5 – y) + 6y2 = 22
20y – 4y2 + 6y2 – 22 = 0
2y2 + 20y – 22 = 0
2(y2 + 10y – 11) = 0
2 ≠ 0  ,    y2 + 10y – 11 = 0
(y + 11)(y – 1) = 0
y + 11 = 0  or y – 1 = 0
y ≠ –11 y = 1

\ When y = 1, x2 = 4(1), x = ±2,
\ x = 2, x ≠ –2
\ point A = (2, 1)

(b) (i) (0, 5), (2, 1) (b) (i) m = 1 – 5
2 – 0
m = 1 – 5 = –2 K1 2
2 – 0 N1
–2 P1 3

(ii) Area of the shaded region ∫ 3 4 2 1 x2 K1
(ii) 0 (–2x + 5) – 4 dx N1
∫ 3 4 = 2 1 dx P1 2 10
0 (–2x + 5) – 4 x2 N1
K1 3
=–22x2 x3 K1
3 1 24 1 3 2 N1
+ 5x – 4   0 3 1 24–2x2 1 x3 2
2 4 3 0
1 + 5x –  
12
= –(2)2 + 5(2) – (2)3

= 5 1 unit2 513 unit2
3

1 2 10. (a) (i) p 1 X B 1 1 2 1 210. 1 2 4 6
= 5   ,  ~ 8, 5 (a) (i) 8C2 5 5

1 2 1 2 1 2 4 6
P(X = 2) = 8C2 5 5
0.2936
= 0.2936

1 2 X 1 1 2 4 n
(ii) ~ B  n, 5 (ii) lg 5 , lg 0.1

P(X > 1) . 0.9

1 – P(X = 0) . 0.9 n . 10.32

P(X = 0) , 0.1

1 2 1 2 nC0 1 0 4 n , 0.1 n = 11
5 5

1 2lg 4 n , lg 0.1
5


n lg 0.8 , lg 0.1

n(–0.0969) , –1

n . 0.01969

n . 10.32

\ n = 11

© Penerbitan Pelangi Sdn. Bhd. A – 16 SPM Additional Mathematics

(b) X ~ N(60, 52) (b) 60 – a – 60 or 60 + a – 60 P1 ANSWERS
P[(60 – a) , x , (60 + a)] = 0.95 5 5 5
K1
P31 60 – a – 60 2 1 60 + a – 60 24 1 2 –a a K1
5 , Z , 5 = 0.95 P 5 ,Z, 5 = 0.95 K1
N1
1 2 –a a P1Z a 2
0.95 \P 5 ,Z, 5 = 0.95 . 5 = 0.025

1 2 a 1 – 0.95 Z = 1.96
0.025 P Z. 5 = 2

–z z P1Z . a 2 = 0.025 9.8
5

From normal distribution table, a = 1.96
5
\ a = 9.8

11. B 11. 10

M N (a) AB = 25 m N1 2
Aθ MN = 6 m N1
7m 24 m

O

(a) Given AO = 7 m = AN (sector OAN),
BO = 24 m = BM (sector OBM).

\ AB2= AO2 + OB2
= 72 + 242

AB = 625 = 25 m

\ AB = AN + BM – MN = 25
7 + 24 – 25 = MN
\ MN = 6 m



(b) tan /OAB = 24  2(b) tan–1 24 K1 2
7 7 N1
 2 24
/OAB = tan–1 7

= 73.7°

= 73.7° × 3.142 OAB = 1.2865 rad.
180°

= 1.2865 rad.

(c) /ABO = 3.142 – 1.2865 – π = 0.2847 rad. (c) 24(0.2847) or 7(1.2865) P1 3
2 24(0.2847) + 7(1.2865) + 6 K1
21.8383 m N1
Length of fencing of the shaded region

= perimeter of shaded region
=SOM + SON + MN
=rBM (0.2847) + rAN (1.2865) + 6
= 24(0.2847) + 7(1.2865) + 6

= 21.8383 m

© Penerbitan Pelangi Sdn. Bhd. A – 17 SPM Additional Mathematics

ANSWERS

(d) Area of ∆AOB = 1 × 7 × 24 = 84 m2 (d) 81.994 m2 or 31.519 m2 K1 3
2 81.994 + 31.519 – 84 K1
1 29.513 m2 N1 10
Area of sector OBM = 2 r2θ 3
N1 3
=12 × 242 × 0.2847 N1
N1
= 81.994 m2 K1
K1
Area of sector OAN = 1 × 72 × 1.2865 N1
2

= 31.519 m2

\ Area of shaded region
= Area of sector OBM + area of sector
OAN – area of ∆AOB

= 81.994 + 31.519 – 84

= 29.513 m2

12. Types of Unleaded Leaded 12.
petrol petrol petrol
Diesel (a) 5x + 3y > 150
Station 5 000 4 000 4x + 5y > 200
Station P, x 3 000 5 000 1 000 x + 2y > 60
2 000 (b) 1 straight line drawn correclty
Station Q, y 3 straight lines drawn correclty
Correct shaded region
Total 150 000 200 000 60 000

(a) 5 000x + 3 000y > 150 000
5x + 3y > 150


4 000x + 5 000y > 200 000
4x + 5y > 200


1 000x + 2 000y > 60 000
x + 2y > 60


(b) 5x + 3y > 150 x 0 30
y 50 0

4x + 5y > 200 x 0 50
y 40 0

x + 2y > 60 x 0 60
y 30 0
y

70

60

50 (0, 50) R
40

30

20

10 k = 2x + y x + 2y = 60

x

0 10 20 30 40 50 60 70

5x + 3y = 150 4x + 5y = 200

© Penerbitan Pelangi Sdn. Bhd. A – 18 SPM Additional Mathematics

(c) Given the operating cost for station P and (c) k = 1 000x + 500y K1 ANSWERS
station Q are 1 000x and 500y respectively. K1 4
draw the correct straight line N1
If the total operating cost for the 2 stations is k = 1 000x + 500y N1 10
k, then k = 1 000x + 500y. (0, 50) 2
RM25 000 N1 3
In order to draw the straight line, let k be N1
10 000. K1
N1
1 000x + 500y = 10 000 N1
⇒ 2x + y = 20

x 0 10
y 20 0

From the graph, the minimum point is (0, 50).

Hence, the minimum operating cost for the

two stations = RM1 000(0) + RM500 (50)

= RM25 000

13. H 10 m G 13.

125 m (a) FH = 125 m
E F 5m FC = 89 m

116° 8m
17.51 m

89 m
8m

DC

5m

A 10 m B

(a) FH2 = GH2 + GF2  ,  FC2 = GF2 + GC2

= 102 + 52 = 52 + 82

= 125 = 89

\ FH = 125 m \ FC = 89 m

(b) Area of plane CFH = 47.4 m2 (b) sin θ = 47.4 × 2
125 . 89

\ 1  ab sin θ = 47.4
2
12
× 125 × 89 sin θ = 47.4 θ = 64°
sin 116°
θ = 47.4 × 2
125 . 89

\ θ = 64°

\ Obtuse angle of /HFC = 180° – 64°

= 116°

(c) Use cosine rule, (c) 125 2 + 89 2 – 2(125 )(89 ) K1 5
a2 = b2 + c2 – 2bc cos A cos 116°
CH2 = 125 2 + 89 2 – 2(125 )(89 ) cos 116°
= 306.4745 CH = 17.51 m N1

\ CH = 306.4745

= 17.51 m

© Penerbitan Pelangi Sdn. Bhd. A – 19 SPM Additional Mathematics

ANSWERS

Use sine rule, sin 116° = sin /FHC K1
17.51 89
sin 116° = sin /FHC N1
17.51 89 N1

\ sin /FHC = 89 sin 116° /FHC = 28°589
17.51 /FCH = 35°29
\ /FHC = 28°589

/FCH = 180° – 116° – 28°589 = 35°29

14. (a) (i) Total sales of drinks in 2012 = RM108.5 14. (a) (i) a(25) + 1.20(40) + 0.8(35) K1 2 10
N1 2
= 108.5 K1 2
a(25) + 1.20(40) + 0.8(35) = 108.5 a = RM1.30 N1 2
25a + 76 = 108.5 K1
25a = 108.5 – 76 N1 2
K1
25a = 32.5 N1

\ a = RM1.30 P1

N1

(ii) b = Price index for tea in the year 2015 (ii) 1.60 × 100
based on the year 2012 1.30

=RRMM11..6300 × 100 b = 123.08

= 123.08

(b) (i) Given composite index for the cost of 123.08(25) + c(40)
making the drinks in year 2015 based
on year 2012 = 139 (b) (i) + 137.5(35) = 139
100

123.08(25) + c(40) + 137.5(35) = 139
100
7 889.5 + 40c = 13 900 c = 150.26
40c = 6 010.50
c = 150.26



(ii) I15/12 = P15 × 100 = 139 (ii) 139 × 75
P12 100

P15 × 100 = 139
75
139 × 75
\ P15 = 100 RM104.25

= 104.25

\ The cost of making the drinks in
year 2015 is RM104.25.

(c) –I 17/15 = PP1175 × 100 = 120......................... a (c) P = 100 × P17 or
15 120

–I 15/12 = PP1152 × 100 = 139......................... b P15 =139 × P12
from a, P15 = 100
100 × P17
120

from b, P15 = 139 × P12 166.80
100

\ 100 × P17 = 139 × P12
120 100

PP1127 × 100 = 139 × 120
\ –I 17/12 = 100

166.80

© Penerbitan Pelangi Sdn. Bhd. A – 20 SPM Additional Mathematics

15. (a) Given v = t2 – 5t + 4 ∫(a) (t2 – 5t + 4) dt ANSWERS
P1 3 10
∫ Displacement, s = v dt K1
N1
∫=(t2 – 5t + 4) dt t3 – 5t2 + 4t + c
t3 5t2 3 2 N1 4
=3 – 2 N1
+ 4t + c
K1
when t = 0, s = 0, c = 0 s = 1 t3 – 5 t2 + 4t N1
3 2
t3 5
\s= 3 – 2 t2 + 4t

(b) The robotic car stops instantaneously at P (b) t = 1 min, t = 4 min
and Q,
11 8
v = t2 – 5t + 4 = 0 6 3
(t – 4)(t – 1) = 0 s = m or s = – m
t – 4 = 0   or   t – 1 = 0
t = 4 min   , t = 1 min 11 + 8
6 3
at P, t = 1 min, s =13 (1)3 – 5 (1)2 + 4(1)
2
PQ = 4.5 m
=161 m

at Q, t = 4 min, s = 1 (4)3 – 5 (4)2 + 4(4)
3 2
8
= – 3 m

Robotic car

QRO P

s= – 8 s=0 s = 11
3 t=0 6
t=4 t=1

\ The distance PQ = 11 + 8 = 4.5 m
6 3

(c) at R, acceleration = 0 (c) t = 2.5 min N1 3
RQ = 2.25 m or N1
\a= dv =0 RO = 0.4167 m
dt N1
R is closer to O
a= d (t2 – 5t + 4) = 0
dt
2t – 5 = 0

t = 2.5 min
when t = 2.5 min,

s = 1 (2.5)3 – 5 (2.5)2 + 4(2.5)
3 2

= –0.4167 m

\ RQ = 8 – 0.4167 = 2.25 m
3

RO = 0.4167 m
\ RO , RQ
Hence, R is closer to O.

© Penerbitan Pelangi Sdn. Bhd. A – 21 SPM Additional Mathematics