PRAKTIS HEBAT ADDMATHS TG4

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Practice Makes Perfect!
H E BA T Additional Praktis

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Matematik
Form SPM Mathematics H BA T
4
4 KSSM Tambahan Praktis HEBAT! SPM KSSM E
4
4





Praktis Hebat! SPM is a series of topical practices aiming to SPM
ORM
F
TITLES IN FORM enhance students’ understanding of the latest SPM format
THIS SERIES 4 5 as well as the various assessment requirements. With the
Bahasa Melayu Form
incorporation of i-THINK and HOTS questions, these materials
4
4
English
aim to equip students with the necessary skills to master the
Mathematics/Matematik
subject and thus ace the assessment. KSSM
Additional Mathematics/Matematik Tambahan KSSM
Science/Sains 4
Physics/Fizik FORMAT PEPERIKSAAN MATEMATIK TAMBAHAN 4
FORMAT PEPERIKSAAN
Chemistry/Kimia SPM MULAI 2021
SPM MULAI 2021
Biology/Biologi KERTAS 1 KERTAS 2
PERKARA
Sejarah (3472/1) (3472/2)
Pendidikan Islam Additional Mathematics/Matematik Tambahan
Jenis item • Subjektif Respons Terhad
Ekonomi • Subjektif Berstruktur
Perniagaan
Bilangan Bahagian A Bahagian A
Prinsip Perakaunan
soalan 12 soalan (64 markah) 7 soalan (50 markah) (Jawab semua soalan)
(Jawab semua soalan) Bahagian B
© Penerbitan Pelangi Sdn. Bhd. 2020 Bahagian B 4 soalan (30 markah) (Jawab 3 soalan)
All rights reserved. No part of this book may 3 soalan (16 markah) Additional Mathematics
be reproduced, stored in a retrieval system, Bahagian C
or transmitted in any form or by any means, (Jawab 2 soalan) 4 soalan (20 markah) (Jawab 2 soalan)
electronic, photocopying, mechanical, recording
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Cover Praktis Hebat SPM AddMath Tkt 4.indd 1 08/07/2020 1:55 PM

Pract ice Functions 1
spm 1 Fungsi


Pract ice Quadratic Functions 8
spm 2 Fungsi Kuadratik



Pract ice Systems of Equations 18
spm 3 Sistem Persamaan



Pract ice Indices, Surds and Logarithms 21
spm 4 Indeks, Surd dan Logaritma



Pract ice Progressions 28
spm 5 Janjang



Pract ice Linear Law 38
spm 6 Hukum Linear



Pract ice Coordinate Geometry 50
spm 7 Geometri Koordinat


Pract ice Vectors 61
spm 8 Vektor



Pract ice Solution of Triangles 72
spm 9 Penyelesaian Segi Tiga



Pract ice Index Numbers 79
10 Nombor Indeks
spm


Assessment Paper 92

Answers A1 – A16

FORMAT INSTRUMEN SPM# MULAI 2021
FORMAT INSTRUMEN SPM MULAI 2021
MATEMATIK TAMBAHAN (3472)
MATEMATIK TAMBAHAN ((3472))



BIL. PERKARA KERTAS 1 (3472 / 1) KERTAS 2 (3472 / 2)

1 Jenis Ujian Bertulis
Instrumen

2 Jenis Item • Subjektif Respons Terhad
• Subjektif Berstruktur
3 Bilangan Bahagian A: Bahagian A:
Soalan 12 soalan (64 markah) 7 soalan (50 markah)
(Jawab semua soalan) (Jawab semua soalan)
Bahagian B:
Bahagian B: 4 soalan (30 markah)
3 soalan (16 markah) (Jawab 3 soalan)
(Jawab 2 soalan) Bahagian C:
4 soalan (20 markah)
(Jawab 2 soalan)

Jumlah
4 80 markah 100 markah
Markah
5 Konstruk • Mengingat dan Memahami • Mengingat dan Memahami
• Mengaplikasi • Mengaplikasi
• Menganalisis • Menganalisis
• Menilai • Menilai
• Mencipta • Mencipta

6 Tempoh
Ujian 2 jam 2 jam 30 minit

7 Cakupan Standard kandungan dan standard pembelajaran dalam Dokumen
Konteks Standard Kurikulum dan Pentaksiran (DSKP) KSSM

Aras Rendah : Sederhana : Tinggi
8
Kesukaran 5 : 3 : 2

Kaedah
9 Analitik
Penskoran
Alat
10 Kalkulator saintifik yang tidak boleh diprogram.
Tambahan

Additional Mathematics Form 4 Practice 2 Quadratic Functions
Pract ice


spm 2 Quadratic Functions
Fungsi Kuadratik




FORMULAE

2
–b ± b – 4ac 3. If f(x) = a(x – h) + k, then
2
1. x = 2
2a Jika f(x) = a(x – h) + k, maka
2
2. x – (sum of roots)x + (product of roots) = 0 vertex/ verteks: (h, k)
x – (hasil tambah punca)x + (hasil darab punca) = 0 axis of symmetry/ paksi simetri: x = h.
2
5. It is given that 4 and –1 are the roots of a
PAPER 1
quadratic equation. Form a quadratic equation
Quadratic Equations and which has the roots. PL 3
Inequalities Diberi bahawa 4 dan –1 merupakan punca-
Persamaan Ketaksamaan dan punca bagi suatu persamaan kuadratik.
2.1 Kuadratik TEXTBOOK
pp. 36 – 44 Bentukkan satu persamaan kuadratik yang
mempunyai punca-punca tersebut.
1. By using the method of completing the square,
solve the quadratic equation x + 3x – 9 = 0. 6. Given a quadratic equation has the roots –5
2
1
Give your answers correct to three decimal and —. PL 3
6
places. PL 3 Diberi suatu persamaan kuadratik mempunyai
Dengan menggunakan kaedah penyempurnaan 1
kuasa dua, selesaikan persamaan kuadratik punca-punca –5 dan —.
6
x + 3x – 9 = 0. Berikan jawapan anda betul (a) Find the sum of roots and product of roots
2
kepada tiga tempat perpuluhan. of the quadratic equation.
Cari hasil tambah punca dan hasil darab
2. By using the quadratic formula, solve the punca bagi persamaan kuadratik tersebut.
quadratic equation 13 – 4x = 2x(x – 3). Give
your answers correct to four significant figures. (b) By using the sum of roots and product
Dengan menggunakan rumus kuadratik, selesai of roots obtained in (a), form a quadratic
persamaan kuadratik 13 – 4x = 2x(x – 3). equation which has the roots.
Berikan jawapan anda betul kepada empat Dengan menggunakan hasil tambah punca
dan hasil darab punca yang diperoleh di
angka bererti. PL 3
(a), bentukkan satu persamaan kuadratik
3. Solve the quadratic equation 2x – 8x + 5 = 0. yang mempunyai punca-punca tersebut.
2
Give your answers correct to four significant 7. Find the roots for –2x – 5x + 13 = 0. Give
2
figures. PL 3 your answer correct to three decimal places.
Selesaikan persamaan kuadratik 2x – 8x + 5 = 0. Cari punca bagi –2x – 5x + 13 = 0. Berikan
2
2
Berikan jawapan anda betul kepada empat jawapan anda betul kepada tiga tempat
angka bererti. perpuluhan. PL 3
4. Solve the quadratic equation –2x + 7x – 4 = 0. 8. Given one of the roots for the quadratic
2
1
Give your answers correct to three decimal equation 10x + x + h = 0 is —, find the value
2
places. PL 3 of the constant h. PL 4 2
Selesai persamaan kuadratik –2x + 7x – 4 = 0. Diberi salah satu punca bagi persamaan
2
1
Berikan jawapan anda betul kepada tiga kuadratik 10x + x + h = 0 ialah —, cari nilai
2
tempat perpuluhan. pemalar h. 2
© Penerbitan Pelangi Sdn. Bhd. 8

Additional Mathematics Form 4 Practice 2 Quadratic Functions
9. Form a quadratic equation which has the roots 14. It is given that (p + 1) and 5 are the roots of
1
2
— and –4. PL 3 the quadratic equation x + (q – 2)x – 10 = 0,
2 such that p and q are constants. Find the values
Bentukkan satu persamaan kuadratik yang of p and q. PL 4
1
mempunyai punca-punca — dan –4. Diberi (p + 1) dan 5 merupakan punca-punca
2 bagi persamaan kuadratik x + (q – 2)x – 10 = 0,
2
10. One of the roots for the quadratic equation dengan keadaan p dan q ialah pemalar. Cari
3x – kx + 4 = 0 is one third of another root. nilai p dan q.
2
Find the possible values of k. PL 4
Salah satu punca bagi persamaan kuadratik 15. y
3x – kx + 4 = 0 ialah satu per tiga daripada
2
punca yang satu lagi. Cari nilai-nilai k yang
mungkin.
x
11. Given the quadratic equation x(x + 5) = –6. PL 4 0 2 5
Diberi persamaan kuadratik x(x + 5) = –6.
(a) State the sum of roots of the quadratic The diagram above shows the graph for the
2
equation. quadratic equation y = x + bx + c. PL 4
Nyatakan hasil tambah punca bagi Rajah di atas menunjukkan graf bagi
2
persamaan kuadratik tersebut. persamaan kuadratik y = x + bx + c.
(b) State the product of roots of the quadratic (a) State the roots when y = 0.
equation. Nyatakan punca-punca apabila y = 0.
Nyatakan hasil darab punca bagi (b) Find the values of b and c.
persamaan kuadratik tersebut. Cari nilai b and c.
(c) If a and b are the roots of the quadratic 16. If a and b are the roots of the quadratic
equation, form a quadratic equation which equation 3x – 7x + 2 = 0, form a quadratic
2
has the roots (a + 4) and (b + 4). 4 4
Jika a dan b merupakan punca-punca equation which has the roots — and —. PL 4
a
b
bagi persamaan kuadratik tersebut, Jika a dan b ialah punca-punca bagi
bentukkan satu persamaan kuadratik persamaan kuadratik 3x – 7x + 2 = 0,
2
yang mempunyai punca-punca (a + 4) dan bentukkan persamaan kuadratik yang
(b + 4). 4 4
mempunyai punca-punca — dan —.
a b
12. Given the quadratic equation 3x – px + 5 = 0,
2
2
where p is a constant. Find the value of p if PL 4 17. If the roots of the equation 2x – 3x – 5 = 0 is
Diberi persamaan kuadratik 3x – px + 5 = 0, a and b, form a quadratic equation using the
2
dengan keadaan p ialah pemalar. Cari nilai p roots PL 4
2
jika Jika punca-punca persamaan 2x – 3x – 5 = 0
(a) one of the roots of the equation is –1. ialah a dan b, bentukkan persamaan kuadratik
salah satu punca bagi persamaan ialah –1. menggunakan punca-punca
(b) the sum of roots of the equation is 4. (a) (a – 4) and/ dan (b – 4)
1
1
hasil tambah punca bagi persamaan (b) —a and/ dan —b
ialah 4. 3 3
18. Given the quadratic equation 9x – 12x + 4 = 0
2
13. Given –6 is one of the roots of the quadratic has roots a and b. Form a quadratic equation
equation (x + p) = 25, such that p is a with the roots 3a and 3b. PL 4
2
constant. Find the values of p. PL 4 Diberi persamaan kuadratik 9x – 12x + 4 = 0
2
Diberi –6 ialah satu daripada punca mempunyai punca a dan b. Bentukkan
persamaan kuadratik (x + p) = 25, dengan
2
keadaan p ialah pemalar. Cari nilai-nilai p. persamaan kuadratik dengan punca-punca 3a
dan 3b.
9 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Practice 2 Quadratic Functions
x
19. Given the quadratic equation px – 3x + q = 0 26. It is given that the quadratic equation ——– = m,
2
2
SPM where p and q are constants has roots a and x – 2
2018
2a. Express p in terms of q. PL 4 where m is a constant, has two equal real roots.
Diberi persamaan kuadratik px – 3x + q = 0 Find the possible values of m. PL 3
2
x
2
dengan keadaan p dan q ialah pemalar Diberi bahawa persamaan kuadratik ——– = m,
mempunyai punca-punca a dan 2a. x – 2
Ungkapkan p dalam sebutan q. dengan keadaan m ialah pemalar, mempunyai
dua punca nyata yang sama. Cari nilai-nilai m
20. By using the graph sketching method, find the yang mungkin.
range of values of x for 2x – x + 3 > x(x – 5).
2
Dengan menggunakan kaedah lakaran graf, 27. Determine the range of values of k if the
2
cari julat nilai x bagi 2x – x + 3 > x(x – 5). quadratic equation 2x = 5(3x – 1) + 4k has no
2
PL 3 real roots. PL 3
Tentukan julat nilai k jika persamaan kuadratik
21. By using the table, find the range of values of 2x = 5(3x – 1) + 4k tidak mempunyai punca
2
x for x – x < 12. PL 3 nyata.
2
Dengan menggunakan jadual, cari julat nilai x
2
bagi x – x < 12. 28. The quadratic equation (q – 1)x – 4x + 5 = 0,
2
where q is a constant, has no real roots. Find
22. By using the number lines, find the range of the range of values of q. PL 3
values of x for 5x + 3 , 4x(x + 4). PL 3 Persamaan kuadratik (q – 1)x – 4x + 5 = 0,
2
Dengan menggunakan garis nombor, cari julat dengan keadaan q ialah pemalar, tidak
nilai x bagi 5x + 3 , 4x(x + 4). mempunyai punca nyata. Cari julat nilai q.
23. Find the range of values of x for 2x + 11x < 6. 29. Find the range of values of p for the quadratic
2
2
Cari julat nilai x bagi 2x + 11x < 6. PL 3 equation 2x – 7x + 3p = 0 if the quadratic
2
equation PL 3
Cari julat nilai p bagi persamaan kuadratik
Types of Roots of Quadratic 2x – 7x + 3p = 0 jika persamaan kuadratik itu
2
Equations
Jenis-jenis Punca Persamaan (a) has two different real roots
2.2 Kuadratik TEXTBOOK mempunyai dua punca nyata yang
pp. 45 – 48
berbeza
(b) has two equal real roots
24. Determine the types of roots of the following mempunyai dua punca nyata yang sama
quadratic equations by finding the values of (c) has no real roots
discriminant of the quadratic equations. PL 3 tiada punca nyata
Tentukan jenis punca bagi persamaan
kuadratik yang berikut dengan mencari nilai 30. Find the value of h if the straight line y = hx – 9
pembezalayan bagi persamaan kuadratik is the tangent to the curve y = x + 5x – 9.
2
tersebut. Cari nilai h jika garis lurus y = hx – 9 ialah
(a) x + 5x – 6 = 0 tangen kepada lengkung y = x + 5x – 9.
2
2
(b) 3x(3x + 2) = –1 PL5
25. Given the quadratic equation 2x – 3 = 8x + 5px , 31. Given that a curve x(x + y) – 4 = 0 does not
2
2
where p is a constant, has two different real intersect the straight line 2x + y = p, where p is
roots. Find the range of values of p. PL 3 a constant. Find the range of values of p.
Diberi persamaan kuadratik 2x – 3 = 8x + 5px , Diberi suatu lengkung x(x + y) – 4 = 0 tidak
2
2
dengan keadaan p ialah pemalar, mempunyai menyilang garis lurus 2x + y = p, dengan
dua punca nyata yang berbeza. Cari julat keadaan p ialah pemalar. Cari julat nilai p.
nilai p. PL5



© Penerbitan Pelangi Sdn. Bhd. 10

Additional Mathematics Form 4 Practice 2 Quadratic Functions
32. (a) It is given that one of the roots of the 35. The quadratic function f(x) = q – 6x + x ,
2
2
2
2
SPM quadratic equation x + (p + 2)x – p = 0 where q is a constant, has the minimum value
2017
where p is a constant, is negative of the 7. Find the values of q. PL 3
other. Find the value of the product of Fungsi kuadratik f(x) = q – 6x + x , dengan
2
2
roots. PL 5 keadaan q ialah pemalar, mempunyai nilai
Diberi bahawa salah satu punca bagi minimum 7. Cari nilai-nilai q.
persamaan kuadratik x + (p + 2)x – p = 0
2
2
dengan keadaan p ialah pemalar, adalah 36. y
negatif kepada yang satu lagi. Cari nilai
bagi hasil darab punca. y = f(x)
4
(b) The quadratic equation mx – 3nx + — m = 0 (0, 16)
2
9
where m and n are constants has two x
equal real roots. Find m : n. PL 5 (–4, 0) 0
4
Persamaan kuadratik mx – 3nx + — m = 0
2
9
dengan keadaan m dan n ialah pemalar The diagram above shows the graph of a
mempunyai dua punca nyata yang sama. quadratic function for f(x) = (x + m) + n, such
2
Cari m : n. that m and n are constants. State PL 3
Rajah di atas menunjukkan graf fungsi
33. It is given that the curve y = (p – 1)x – x + 5 kuadratik f(x) = (x + m) + n, dengan keadaan
2
2
SPM where p is a constant, intersects with the m dan n ialah pemalar. Nyatakan
2018
straight line y = 2x + 3 at two points. Find the (a) the values of m and n.
range of values of p. PL 5 nilai m dan n.
Diberi bahawa lengkung y = (p – 1)x – x + 5 (b) the equation of the axis of symmetry.
2
dengan keadaan p ialah pemalar, menyilang persamaan paksi simetri.
garis lurus y = 2x + 3 pada dua titik. Cari
julat nilai p.
37. y
Quadratic Functions x
2.3 Fungsi Kuadratik TEXTBOOK –3 0 5
pp. 49 – 63
34. –15
y
(m, 12) The diagram above shows the graph of a
10
2
quadratic function for f(x) = (x – p) – 16
where p is a constant. Find PL 3
x
–2 0 4 Rajah di atas menunjukkan graf fungsi
kuadratik f(x) = (x – p) – 16 dengan keadaan
2
The diagram above shows the graph of the p ialah pemalar. Cari
function y = –(x – m) + 12, where m is a (a) the equation of the axis of symmetry.
2
constant. Find PL 3 persamaan paksi simetri.
Rajah di atas menunjukkan graf bagi fungsi (b) the value of p.
y = –(x – m) + 12, dengan keadaan m ialah nilai p.
2
pemalar. Cari (c) the coordinates of the minimum point.
(a) the value of m. koordinat bagi titik minimum.
nilai m.
(b) the equation of the axis of symmetry.
persamaan paksi simetri.



11 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Practice 2 Quadratic Functions
38. f(x) 45. Determine the maximum value and the
(m, n) corresponding value at x for the quadratic
2
function f(x) = –x + 4x – 8. PL 3
Tentukan nilai maksimum dan nilai x yang
sepadan bagi fungsi kuadratik f(x) = –x + 4x – 8.
2
x
0 p 6
46. Given the quadratic function f(x) = x – 5x + 4
2
has the minimum point at (p, q). Find PL 3
The diagram above shows the graph of a Diberi fungsi kuadratik f(x) = x – 5x + 4
2
quadratic function for f(x) = 4 – (x – 4) . Find mempunyai titik minimum pada (p, q). Cari
2
the values of PL 3 (a) the value of p and of q.
Rajah di atas menunjukkan graf bagi fungsi nilai p and nilai q.
kuadratik f(x) = 4 – (x – 4) . Cari nilai bagi
2
(a) m (b) n (b) the equation of the axis of symmetry.
(c) p persamaan paksi simetri.
47. Given f(x) = x – kx + 2k – 3, such that k is a
2
39. Given the quadratic function f(x) = –x + 6x – 13. constant, is always positive when m , k , n.
2
Without drawing the graph, find the maximum Find the value of m and of n. PL 4
point of f(x). PL 3 Diberi f(x) = x – kx + 2k – 3, dengan keadaan
2
Diberi fungsi kuadratik f(x) = –x + 6x – 13. k ialah pemalar, adalah sentiasa positif
2
Tanpa melukis graf, cari titik maksimum bagi apabila m , k , n. Cari nilai m dan nilai n.
f(x).
48. The quadratic function f(x) = (x + 4) – 20 + k,
2
40. Determine the type of roots for the following where k is a constant, has minimum point
quadratic function f(x) = 0. PL 3 (–4, 5k). PL 3
Tentukan jenis punca bagi fungsi kuadratik Fungsi kuadratik f(x) = (x + 4) – 20 + k,
2
f(x) = 0 yang berikut. dengan keadaan k ialah pemalar, mempunyai
(a) f(x) = x – 4x – 18 titik minimum (–4, 5k).
2
(b) f(x) = –5x + 11x – 7 (a) Find the value of k.
2
Cari nilai k.
41. Find the range of values of x such that the (b) State the type of roots for f(x) = 0. Justify
2
SPM quadratic function f(x) = 3x – 14x – 5 is
2017 your answer.
negative. PL 4 Nyatakan jenis punca bagi f(x) = 0.
Cari julat nilai x dengan keadaan fungsi Justifikasikan jawapan anda.
kuadratik f(x) = 3x – 14x – 5 ialah negatif.
2
49. Given that the graph of a quadratic function
42. Find the range of values of m if the quadratic f(x) = mx – 6x + n, where m and n are
2
function f(x) = x – 4x + m intersects the x-axis constants, has a minimum point. PL 4
2
at two different points. PL 4 Diberi bahawa graf bagi fungsi kuadratik
Cari julat nilai m jika graf bagi fungsi f(x) = mx – 6x + n, dengan keadaan m dan n
2
kuadratik f(x) = x – 4x + m menyilang paksi-x ialah pemalar, mempunyai titik minimum.
2
pada dua titik yang berlainan. (a) Given m is an integer such that –2 , m , 2,
state the value of m.
43. Find the value of p if f(x) = px + 5x – 6 Diberi m ialah integer dengan keadaan
2
touches the x-axis at one point only. PL 4 –2 , m , 2, nyatakan nilai m.
Cari nilai p jika f(x) = px + 5x – 6 menyentuh
2
paksi-x pada satu titik sahaja. (b) Using the answer from (a), find the value
of n when the graph touches the x-axis at
44. Find the minimum value and the equation of the one point only.
2
axis of symmetry for f(x) = x – 3x + 4. PL 3 Dengan menggunakan jawapan daripada
Cari nilai minimum dan persamaan paksi (a), cari nilai n apabila graf tersebut
simetri untuk f(x) = x – 3x + 4. menyentuh paksi-x pada satu titik sahaja.
2
© Penerbitan Pelangi Sdn. Bhd. 12

Additional Mathematics Form 4 Practice 2 Quadratic Functions
50. Given the quadratic function f(x) = –2x + 6x – p , 51. A quadratic function is defined by
2
2
such that p is a constant. Find PL 4 SPM f(x) = x + 8x + h where h is a constant. PL 4
2
2017
Diberi fungsi kuadratik f(x) = –2x + 6x – p , Suatu fungsi kuadratik adalah ditakrifkan
2
2
dengan keadaan p ialah pemalar. Cari sebagai f(x) = x + 8x + h dengan keadaan h
2
(a) the equation of the axis of symmetry. ialah pemalar.
persamaan paksi simetri. (a) Express f(x) in the form (x + p) + q
2
(b) the possible values of p if f(x) has the where p and q are constants.
1
2
maximum value –20—. Ungkapkan f(x) dalam bentuk (x + p) + q
2 dengan keadaan p dan q ialah pemalar.
nilai-nilai p yang mungkin jika f(x) (b) Given the minimum value of f(x) is –10,
1
mempunyai nilai maksimum –20—. find the value of h.
2 Diberi nilai minimum bagi f(x) ialah –10,
cari nilai bagi h.
PAPER 2

1. (a) Given the quadratic equation (n + 3)x – 12x = n – 10, such that n is a constant. Find the values of n
2
TEXTBOOK or ranges of values of n if the quadratic equation has PL3 Subtopic 2.1 & 2.2
pp. 36 – 48
Diberi persamaan kuadratik (n + 3)x – 12x = n – 10, dengan keadaan n ialah pemalar. Cari nilai n
2
atau julat nilai n jika persamaan kuadratik tersebut mempunyai
(i) two different real roots/ dua punca nyata yang berbeza
(ii) two equal real roots/ dua punca nyata yang sama
(iii) no real roots/ tiada punca nyata

(b) Using the values of n from (a)(ii), solve the quadratic inequality (n + 3)x – 12x – n + 10 < 3x.
2
Gunakan nilai-nilai n daripada (a)(ii), selesaikan ketaksamaan kuadratik (n + 3)x – 12x – n + 10 < 3x.
2
2
2. The roots of the quadratic equation x – 7x + k = 0 are a and b such that a . b. PL3 Subtopic 2.1 & 2.2
2
TEXTBOOK Punca-punca bagi persamaan kuadratik x – 7x + k = 0 ialah a dan b dengan keadaan a . b.
pp. 36 – 48
(a) Find the range of values of k.
Cari julat nilai k.
(b) If k = 10, find the values of a and b.
Jika k = 10, cari nilai a dan b.
(c) Form a new quadratic equation which has the following roots.
Bentukkan persamaan kuadratik baru yang mempunyai punca-punca berikut.
(i) (a – b) only/ sahaja
2
1
1
(ii) — and/ dan —
a b
3. The diagram shows a trapezium PQRS. Given that 5PS = 4QR and the area P (2x + 5) cm S
2
TEXTBOOK of the trapezium is 320 cm . PL5 Subtopic 2.1
pp. 36 – 44
Rajah di sebelah menunjukkan trapezium PQRS. Diberi 5PS = 4QR dan (3x – 4) cm
luas trapezium tersebut ialah 320 cm .
2
(a) Form a quadratic equation which represents the area of the trapezium. Q R
Bentukkan satu persamaan kuadratik yang mewakili luas trapezium tersebut.
(b) Find the value of x and give your answer in three decimal places.
Cari nilai x dan berikan jawapan anda betul kepada tiga tempat perpuluhan.




13 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Practice 2 Quadratic Functions

b
a
4. Given that — and — are the roots of the quadratic equation mx(2x – 3) = 10x – n, calculate the values of
3
3
TEXTBOOK 19 45
pp. 36 – 44 m and n if a + b = ––– and ab = –––. PL5 Subtopic 2.1
2 2
b
a
Diberi — dan — ialah punca-punca persamaan kuadratik mx(2x – 3) = 10x – n, hitung nilai m dan nilai
3 3
45
19
n jika a + b = —– dan ab = —–.
2 2
2
5. The diagram shows the graph of y = 2x + px + q. PL4 Subtopic 2.1 y
Rajah di sebelah menunjukkan graf bagi y = 2x + px + q.
2
TEXTBOOK
pp. 36 – 44
(a) If a and b are the roots of the quadratic equation 2x + px + q = 0, q
2
state the value of
Jika a dan b ialah punca-punca persamaan kuadratik 2x + px + q = 0,
2
nyatakan nilai 0 2 5 x
(i) a + b,
(ii) ab.
(b) Find the values of p and q.
Cari nilai p dan nilai q.
2
6. The diagram shows the curve of y = –x + 4x + p. PL4 Subtopic 2.1 & 2.2 y
Rajah di sebelah menunjukkan graf bagi lengkung y = –x + 4x + p.
2
TEXTBOOK
pp. 36 – 48
(a) Show that the quadratic equation –x + 4x + p = 0 has real roots if
2
p + 4 > 0. m 0 n x
Tunjukkan bahawa persamaan kuadratik –x + 4x + p = 0 mempunyai y = f(x)
2
punca nyata jika p + 4 > 0.
(b) Find the roots, m and n, when p = 12.
Cari punca-punca, m dan n, apabila p = 12.
7. The diagram shows two curves, y = x + 2x + hx + k and y
2
1
TEXTBOOK y = 3(x – 1) + k – 1, that meet at a point on the x-axis. Find y 2 y 1
2
pp. 36 – 63 2
Rajah di sebelah menunjukkan graf bagi dua lengkung,
y = x + 2x + hx + k dan y = 3(x – 1) + k – 1, yang bertemu pada
2
2
2
1
satu titik pada paksi-x. Cari PL5 Subtopic 2.1 & 2.3 0 x
(a) the values of h and k,
nilai h dan nilai k,
(b) the root of the equations if y = y = 0.
2
1
punca persamaan jika y = y = 0.
1 2
8. It is given that a and b are the roots of the quadratic equation x(x – 4) = 3h + 2 where h is a constant.
TEXTBOOK Diberi bahawa a dan b merupakan punca-punca bagi persamaan kuadratik x(x – 4) = 3h + 2 dengan
pp. 36 – 48
keadaan h ialah pemalar. PL5 Subtopic 2.1 & 2.2
(a) Find the range of values of h if a ≠ b.
Cari julat nilah h jika a ≠ b.
b
a
(b) Given — and — are the roots of another quadratic equation, 4x – kx – 5 = 0 where k is a constant.
2
2 2
Find the values of h and k.
a
b
Diberi — dan — merupakan punca-punca bagi persamaan kuadratik yang lain, iaitu 4x – kx – 5 = 0
2
2
2
dengan keadaan k ialah pemalar. Cari nilai h dan k.
© Penerbitan Pelangi Sdn. Bhd. 14

Additional Mathematics Form 4 Practice 2 Quadratic Functions
9. (a) Find the range of values of x for x + 4 < 3x – 4x + 6 < 5. PL3 Subtopic 2.1
2
2
TEXTBOOK Cari julat nilai x bagi x + 4 < 3x – 4x + 6 < 5.
pp. 36 – 63
(b) A quadratic function is defined by f(x) = x + 8x + p, such that p is a constant. PL3 Subtopic 2.1 & 2.3
2
Suatu fungsi kuadratik ditakrifkan oleh f(x) = x + 8x + p, dengan keadaan p ialah pemalar.
2
(i) Express f(x) in the form f(x) = (x – h) + k.
2
Ungkapkan f(x) dalam bentuk f(x) = (x – h) + k.
2
(ii) Given the minimum value is 0, find the value of p.
Diberi nilai minimum ialah 0, cari nilai p.
(iii) Solve this quadratic function if f(x) < –x + 11x + 36.
2
Selesaikan fungsi kuadratik ini jika f(x) < –x + 11x + 36.
2
10. It is given that a quadratic function has the maximum point (2, 5). The quadratic function can be written
2
TEXTBOOK as f(x) = –2(x – h) + k. PL3 Subtopic 2.1 & 2.3
pp. 36 – 63
Diberi bahawa suatu fungsi kuadratik mempunyai titik maksimum (2, 5). Fungsi kuadratik tersebut boleh
ditulis sebagai f(x) = –2(x – h) + k.
2
(a) State the quadratic function f(x) in the form f(x) = ax + bx + c.
2
Nyatakan fungsi kuadratik f(x) dalam bentuk f(x) = ax + bx + c.
2
(b) Find the roots of the quadratic function f(x). Give your answer in three decimal places.
Cari punca-punca bagi fungsi kuadratik f(x) tersebut. Berikan jawapan anda betul kepada tiga
tempat perpuluhan.
(c) Solve f(x) < –x + 1.
Selesaikan f(x) < –x + 1.
11. A toy rocket is released into the air. The height, in m, of the toy rocket from horizontal ground after t
2
TEXTBOOK seconds is given by the quadratic function h(t) = –t + 4t + 12. PL5 Subtopic 2.3
pp. 49 – 63
Sebuah roket mainan dilepaskan ke udara. Ketinggian, dalam m, roket mainan dari tanah mengufuk
selepas t saat diberi oleh fungsi kuadratik h(t) = –t + 4t + 12.
2
(a) When will the toy rocket reach the maximum height?
Bilakah roket mainan itu akan mencapai tinggi maksimum?
(b) Find the range of time when the toy rocket is in the air, in seconds.
Cari julat masa apabila roket mainan itu berada di udara, dalam saat.
(c) Find the time when the toy rocket hit the ground.
Cari masa apabila roket mainan itu menyentuh tanah.


12. The diagram shows the graph of f(x) = 2(x + 1) + 5 with a = 2, h = –1 y
2
TEXTBOOK and k = 5. PL3 Subtopic 2.3
2
pp. 49 – 63 f(x) = 2(x + 1) + 5
Rajah di sebelah menunjukkan graf bagi f(x) = 2(x + 1) + 5 dengan
2
a = 2, h = –1 dan k = 5. 7
(a) Determine the coordinates of the minimum point and the equation 5
of the axis of symmetry. x
Tentukan koordinat bagi titik minimum dan persamaan paksi –1 0
simetri.
(b) Make a generalisation on the changes of each of the following values on the shape and position of
the graph.
Buat generalisasi tentang kesan perubahan setiap nilai yang berikut terhadap bentuk dan kedudukan
graf.
(i) The value of a changes to 7/ Nilai a berubah kepada 7
(ii) The value of h changes to 3/ Nilai h berubah kepada 3
(iii) The value of k changes to 9/ Nilai k berubah kepada 9



15 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Practice 2 Quadratic Functions
1
13. Given that f(x) = –x – x + 3p has a maximum value of 12—. PL3 Subtopic 2.3
2
TEXTBOOK 4
pp. 49 – 63 1
Diberi f(x) = –x – x + 3p mempunyai nilai maksimum 12—.
2
4
(a) Using the method of completing the square, find the value of p.
Dengan menggunakan kaedah penyempurnaan kuasa dua, cari nilai p.
(b) Hence, sketch the graph for f(x) = –x – x + 3p for –5 < x < 4.
2
Seterusnya, lakarkan graf untuk f(x) = –x – x + 3p untuk –5 < x < 4.
2
14. (a) Find the range of values of x if  x – 1 , 3. PL3 Subtopic 2.3
TEXTBOOK Cari julat nilai x jika  x – 1 , 3.
pp. 49 – 63
(b) Sketch the graph of the function f : x → (x – 1) – 3 for –2 < x < 4. State the range.
2
Lakarkan graf fungsi f : x → (x – 1) – 3 bagi –2 < x < 4. Nyatakan julatnya.
2
15. Given a function y = 3 – 4x – 5x . PL3 Subtopic 2.3
2
2
TEXTBOOK Diberi fungsi y = 3 – 4x – 5x .
pp. 49 – 63
(a) By using completing the square, determine the turning point of the function y. State the type of
turning point obtained.
Dengan menggunakan kaedah penyempurnaan kuasa dua, tentukan titik pusingan bagi fungsi y.
Nyatakan jenis titik pusingan yang diperoleh.
(b) If x = 2 – 3r, where r is a constant, is the axis of symmetry, calculate the value of r.
Jika x = 2 – 3r, dengan keadaan r ialah pemalar, ialah paksi simetri, hitung nilai r.
(c) Hence, sketch the graph of y = 3 – 4x – 5x for –2 < x < 3.
2
Seterusnya, lakarkan graf y = 3 – 4x – 5x bagi –2 < x < 3.
2
2
2
16. Given that f (x) = m + nx + x = (x + p) + q. PL3 Subtopic 2.3
2
2
TEXTBOOK Diberi f (x) = m + nx + x = (x + p) + q.
pp. 49 – 63
(a) State p and q in terms of m and/or n.
Nyatakan p dan q dalam sebutan m dan/atau n.
(b) If n = 3, state the axis of symmetry of the curve f (x).
Jika n = 3, nyatakan paksi simetri bagi lengkung f (x).
17. Given that the curve y = (x + p) – q passes through the point (1, –2) and x = 2p – 15 is the axis of
2
2
TEXTBOOK symmetry. PL5 Subtopic 2.3
pp. 49 – 63
Diberi lengkung y = (x + p) – q melalui titik (1, –2) dan x = 2p – 15 ialah paksi simetri.
2
2
(a) Calculate the values of p and of q.
Hitung nilai p dan nilai q.
(b) If p , 0, sketch the graph of the curve y.
Jika p , 0, lakarkan graf bagi lengkung y.
18. The curve of a quadratic function f(x) = 2(x – p) – q intersects the x-axis at points (2, 0) and (6, 0). The
2
TEXTBOOK straight line y = –8 touches the minimum point of the curve. PL4 Subtopic 2.3
pp. 49 – 63
Lengkung bagi fungsi kuadratik f(x) = 2(x – p) – q menyilang paksi-x pada titik (2, 0) dan (6, 0). Garis
2
lurus y = –8 menyentuh titik minimum bagi lengkung tersebut.
(a) Find the values of p and q.
Cari nilai p dan q.
(b) Hence, sketch the graph of f(x) for 0 < x < 8.
Seterusnya, lakarkan graf f(x) untuk 0 < x < 8.
(c) If the graph is reflected about the x-axis, write the equation of the curve.
Jika graf tersebut dipantulkan pada paksi-x, tulis persamaan bagi lengkung tersebut.

© Penerbitan Pelangi Sdn. Bhd. 16

Additional Mathematics Form 4 Practice 2 Quadratic Functions
19. The diagram shows a side elevation of the inner surface of a bowl y
1
2
TEXTBOOK which has the function represented by f(x) = —x , such that x and
pp. 49 – 63 25
y = f(x) measured in metres. Find the width and depth of the bowl, (–5, 1) (5, 1)
in metres. PL4 Subtopic 2.3 y = 1 x 2
Rajah di sebelah menunjukkan pandangan sisi dalam bagi sebuah 25 x
1
mangkuk yang mempunyai fungsi yang diwakili oleh f(x) = —x 0
2
25
dengan keadaan x dan y = f(x) diukur dalam meter. Cari lebar dan
kedalaman mangkuk itu, dalam meter.

20. The profit, in RM, obtained by Shee Yun from x units of the handphone sold by her shop is given by the
2
TEXTBOOK quadratic function P(x) = –0.04x + 200x – 10 000. PL4 Subtopic 2.3
pp. 49 – 63
Keuntungan, dalam RM, yang diperoleh Shee Yun daripada x unit telefon bimbit yang dijual oleh
kedainya diberi oleh fungsi kuadratik P(x) = –0.04x + 200x – 10 000.
2
(a) Sketch the graph for P(x).
Lakarkan graf untuk P(x).
(b) Based on the graph in (a), determine the maximum profit obtained by Shee Yun from the sales of
handphones.
Berdasarkan graf di (a), tentukan keuntungan maksimum yang diperoleh Shee Yun daripada jualan
telefon bimbit.
(c) How many units of handphone that she needs to sell to get the maximum profit?
Berapa unitkah telefon bimbit yang perlu dijualnya untuk mendapat keuntungan maksimum?








HOTS PRACTICES PAK-21


1. Given h and k are the roots of the quadratic equation 3x + 8x – 3 = 0 where h . k. Form a quadratic
2
equation with the roots (4 – 3h) and (2 + k).
Diberi h dan k ialah punca-punca persamaan kuadratik 3x + 8x – 3 = 0 dengan h . k. Bentukkan
2
persamaan kuadratik yang mempunyai punca-punca (4 – 3h) dan (2 + k).

2. The diagram shows a parabolic bowl represented by the equation y = (x – 1) .
2
Water is poured into the bowl. Find the range of values of x when the height
of the water is more than 9 cm.
Rajah di sebelah menunjukkan sebuah mangkuk berbentuk parabola yang
diwakili dengan persamaan y = (x – 1) . Air dituang ke dalam mangkuk itu.
2
Cari julat nilai x apabila ketinggian air melebihi 9 cm.

3. It is given that a right cylinder has a height of 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 cm .
2
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 cm .
2





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Additional Mathematics Form 4 Answers







Pract ice Functions 21. a = 2, b = –5; a = –2, b = 15
spm 1 Fungsi 22. (a) g(x) = 4x (b) hg(x) = 8x – 3
3q + 1
23. p = —––––
2
PAPER 1 24. –1
1. (a) {4, 6, 8} 25. (a) g(x) = 4x 2 (b) g(x) = 3x – 4
(b) one-to-one relation/ hubungan satu kepada satu 26. p = 5 – 2q
2. (a) square of 27. a = –b + —
5
kuasa dua bagi 3 x – 1 x – 1
28. (a) g : x → —––– or/ atau g(x) = —–––
1 1 2 2
(b) f(x) = 8x – 3
2
2 4
x – 5
3 9 29. f (x) = —–––
–1
2
4 16 x + 4
30. (a) h (x) = —–––
-1
9
(b) –13
x + 6
–x – 3
(b) The type of the relation is one-to-one relation. It 31. (a) f (x) = —––– (b) g (x) = —––––
–1
–1
1 – x
4
is a function. 32. 3
Jenis hubungan itu ialah hubungan satu kepada 33. (a) f(x) = 3 – 4x (b) 3
satu. Hubungan itu ialah satu fungsi.
x + 5
3. This graph is not a function because when tested with 34. (a) g (x) = —–––
–1
4
the vertical line test, the line cuts two points on the 7
graph. (b) —
2
Graf ini bukan fungsi kerana apabila diuji dengan 35. Has no inverse function because when the horizontal
ujian garis mencancang, garis itu memotong dua titik line test is performed, the horizontal line cuts the
pada graf. graph f at two points.
4. (a) 6 Tidak mempunyai fungsi songsang kerana apabila
(b) {–1, 0, 1, 2, 3, 4} ujian garis mengufuk dilakukan, garis mengufuk
(c) {–1, 1, 2, 4} memotong graf f pada dua titik.
5. 2 36. (a) f
6. f : x → x or/ atau f(x) = x 2 (b) g (c) = b
2
–1
3
7. x = 4, x = – —
2 37. f(x)
8. (a) p = 3, q = –8
(b) 4 f(x) = 6x – 3
2
9. (a) –4 < x < 2
(b) 1 0 x
(c) 2 0.5
10. (a) –10 (b) 3 –1.5
11. (a) 26 (b) x = 0, x = –2
5
17
12. (a) — (b) —
3 3 f(x) has inverse function because when the horizontal
13. (a) 6 (b) 7 line test is performed, the horizontal line cuts the
(c) –2 < x < 5 graph f(x) at one point only.
14. (a) 1 500 (b) 28 f(x) mempunyai fungsi songsang kerana apabila ujian
2
2
15. (a) fg(x) = 3x – 21 (b) gf(x) = 9x – 7 garis mengufuk dilakukan, garis mengufuk memotong
16. 8 graf f(x) pada satu titik sahaja.
17. 5 x – 1
3
2
–1
18. – — 38. (a) g (x) = —––– (b) f(x) = x – 2
2
2 x + 4
–1
3
19. (a) fg(x) = 10x + 11 39. (a) h (x) = —––– (b) 2
gf(x) = 10x + 1
9
(b) — PAPER 2 10
10 1. (a) (i) –5 (ii) —
2
20. (a) gf(x) = 10 – 33x (b) — (iii) gf(x) = 37 – 21x 3
3
A1 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
(b) (b) Domain : –8 < x < 12
y
Range/ Julat : 0 < f (x) < 10
–1
(c) 8
58
8. m = –6, n = 20
37
26 HOTS PRACTICES PAK-21
4x – 16x – 29
2
1. (a) q(x) = —–––––––––––
x 9
–1 0 37 3 4x – 16x – 23
2
21 (b) pq(x) = —–––––––––––
3
0 < y < 58 –1
2. (a) (c) No, (pq) is not a function.
f(x) Tidak, (pq) bukan fungsi.
–1
9
Pract ice
5 spm 2 Quadratic Functions
Fungsi Kuadratik
1 x
–2 0 2.5 3 PAPER 1
1. x = 1.854, x = –4.854
(b) 0 < f(x) < 9 (c) x = 1, x = 4 2. x = 3.098, x = –2.098
3. (a) a = 3, b = –4; a = –3, b = 2 3. x = 3.225, x = 0.7753
4
16
(b) x = —, x = — 4. x = 0.719, x = 2.781
9 9 5. x – 3x – 4 = 0
2
29
4. (a) m = 3, n = –2 (b) f (x) = 81x – 80 6. (a) sum of roots/ hasil tambah punca = – —
4
6
(c) f (x) = 3 x – (3 – 1) product of roots/ hasil darab punca = – —
n
n
n
5
x – n
5. (a) f (x) = —––– (b) m = 3, n = 4 2 6
–1
m (b) 6x + 29x – 5 = 0
6. (a) y 7. x = –4.089, x = 1.589
8. –3
9. 2x + 7x – 4 = 0
2
10. k = 8, k = –8
x
0 11. (a) –5 (b) 6
–4
(c) x – 3x + 2 = 0
2
12. (a) –8 (b) 12
13. p = 1, p = 11
(b) This function has inverse function because when
tested with the horizontal line test, the line cuts 14. p = –3, q = –1
the graph at one point only. 15. (a) 2 and/ dan 5 (b) b = –7, c = 10
Fungsi ini mempunyai fungsi songsang kerana 16. x – 14x + 24 = 0
2
apabila diuji dengan ujian garis mengufuk, garis
itu memotong graf pada satu titik sahaja. 17. (a) 2x + 13x + 15 = 0 (b) 18x – 9x – 5 = 0
2
2
x
+
–1
g (x) = AB4 18. x – 4x + 4 = 0
3

2
(c) y 2
q
g 19. p = —
g –1 20. x < –3 or/ atau x > –1
x 21. –3 < x < 4
–4 0 1
–4 22. x , –3 or/ atau x . —
4
1
23. –6 < x < —
2
7. (a) y f(x) = 2x – 8 24. (a) two different real roots
x + 8 dua punca nyata yang berbeza
f (x) = 2 (b) two equal real roots
–1
4 dua punca nyata yang sama
22
x 25. p , —
–8 0 4 15
26. m = 0, m = 8
–8
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Additional Mathematics Form 4 Answers
185
49
27. k , – —– 2. (a) k , — (b) a = 5, b = 2
32 4 2 2
9
28. q . — (c) (i) x – 18x + 81 = 0 (ii) 10x – 7x + 1 = 0
2
5 3. (a) 54x + 63x – 2 740 = 0
49 49 (b) x = 6.564
29. (a) p , — (b) p = —
24 24 4. m = 3, n = 15
49
(c) p . — 5. (a) (i) 7 (ii) 10
24 (b) p = –14, q = 20
30. h = 5 6. (b) m = –2, n = 6
31. –4 , p , 4 7. (a) h = –4, k = 1 (b) x = 1
32. (a) –4 (b) 9 : 4 8. (a) h . –2 2 (b) h = 1, k = 8
1
17 9. (a) — < x < —
3
3
33. p , —
8 (b) (i) (x + 4) + p – 16 (ii) 16
2
5
34. (a) 1 (b) x = 1 (iii) – — < x < 4
35. q = 4, q = –4 2
10. (a) –2x + 8x – 3 (b) x = 3.581, x = 0.419
2
36. (a) m = 4, n = 0 (b) x = –4 1
37. (a) x = 1 (b) 1 (c) x < —, x > 4
2
(c) (1, –16) 11. (a) t = 2 (b) 0 , t , 6
38. (a) 4 (b) 4 (c) t = 6
(c) 2 12. (a) minimum point/ titik minimum: (–1, 5)
39. (3, –4) axis of symmetry/ paksi simetri: x = –1
40. (a) two different real roots/ dua punca nyata yang (b) (i) As the value of a increases to 7, the width of
berbeza the graph decreases. The axis of symmetry
(b) no real roots/ tiada punca nyata and the minimum value remains the same.
1
41. – — , x , 5 / Apabila nilai a bertambah kepada 7,
kelebaran graf berkurang. Paksi simetri dan
3
nilai minimum kekal sama.
42. m , 4 (ii) When the value of h changes to 3, the graph
25
43. – — moves to the right. The equation of the
24 axis of symmetry becomes x = 3 and the
7
44. minimum value/ nilai minimum: — minimum value remains the same. / Apabila
4 nilai h berubah kepada 3, graf akan bergerak
3
axis of symmetry/ paksi simetri: x = — ke kanan. Persamaan paksi simetri menjadi
2
x = 3 dan nilai minimum kekal sama.
45. maximum value/ nilai maksimum: –4 (iii) When the value of k changes to 9, the graph
corresponding value of x/ nilai x yang sepadan: 2 of the same shape moves vertically 4 units
5
9
46. (a) p = —, q = – — upwards. The minimum value becomes 9
2 4 and the axis of symmetry remains the same.
5
(b) x = — / Apabila nilai k berubah kepada 9, graf
2 dengan bentuk yang sama bergerak secara
47. m = 2, n = 6 menegak 4 unit ke atas. Nilai minimum
48. (a) –5 menjadi 9 dan paksi simetri kekal sama.
(b) Since b – 4ac = 100 . 0, the type of roots for 13. (a) p = 4
2
f(x) = 0 is two different real roots. / Oleh sebab (b) f(x)
1
b – 4ac = 100 . 0, jenis punca bagi f(x) = 0 – 1 , 12
2
ialah dua punca nyata yang berbeza. 2 4 12
49. (a) 1 (b) 9 f(x) = –x – x + 12
2
3
50. (a) x = — (b) p = 5, p = –5
2 x
2
51. (a) (x + 4) + h – 16 (b) 6 –4 0 3
(–5, –8) (4, –8)
PAPER 2 14. (a) –2 , x , 4
1. (a) (i) n , 1, n . 6 (b)
(ii) n = 1, n = 6 f(x)
(iii) 1 , n , 6 6
3
(b) When/ Apabila n = 1, — < x < 3
4 Range/ Julat :
1
4
When/ Apabila n = 6, — < x < — –2 0 4 x –3 < f(x) < 6
3 3 –2
(1, –3)
A3 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
19
2
2
15. (a) 1 – —, —– , maximum point / titik maksimum 2. Let/ Katakan
x = value of a calculator/ nilai satu kalkulator
5
5
4
(b) r = — y = value of a pen drive/ nilai satu pemacu pena
5 z = value of an earphone/ nilai satu fon telinga
(c) y x + y + z = 150
2 19
– –, – x + 2y + z = 210
5 5
3 x 2x + y + 3z = 280
–2 0 3 3. a = 3, b = –14, c = –5
–9 4. p = –3, q = –4
5. (a) m + n = 5, m – n = 5 (b) m = 3, n = 2
2
2
6. (a) p = 6q, 2q = p + 8 (b) p = 24
2
–54
PAPER 2
3
2
n
n
16. (a) p = —, q = m – —– (b) x = – — 1. x = –3, y = 5, z = 2
2 4 2 2. x = 3, y = 5, z = –2
5
57
17. (a) p = —, q = —–; p = –3, q = 6 14 7
2 4 3. x = —, y = —; x = 1, y = –2
2
3
(b) y 4. x = 0.5504, y = –0.7520; x = 0.1929, y = 1.0355
3 5. x = –4.622, y = –3.244; x = –2.178, y = 1.644
4
1
19
x 6. x = —, y = 9; x = – —, y = —
0 3 2 3 3
11
11
2
y = (x – 3) – 6 7. x = 3, y = 4; x = —, y = – —
–6 10 2
18. (a) p = 4, q = 8 8. shirt: RM15, trousers: RM23, necktie:RM9
(b) baju: RM15, seluar: RM23, tali leher: RM9
f(x) 9. y
80
24 (8, 24)
f(x) = 2(x – 4) – 8 70
2
x
0 2 6 60
(4, –8)
50
(c) f(x) = –2(x – 4) + 8
2
19. width/ lebar = 10 m, depth/ kedalaman = 1 m 40
20. (a) P(x)
30
(2 500, 240 000)
20
10
x
0 50.51 4 949.49
–10 000 x
–4 –3 –2 –1 0 1 2 3
(b) RM240 000 (2.4, 29.5), (–1.6, 13.5)
3
(c) 2 500 units/ unit 10. x = —, y = 14; x = 5, y = 7
2
11. (4, 2), (–2, 8)
HOTS PRACTICES PAK-21 12. 35 m, 28 m, 21 m
2
1. x – 2x – 3 = 0
HOTS PRACTICES PAK-21
2. x , –2, x . 4
3. r . 8 1. p = 10, q = 3
4
Pract ice Systems of Equations Pract ice Indices, Surds and Logarithms
spm 3 Sistem Persamaan spm Indeks, Surd dan Logaritma
PAPER 1
PAPER 1 5
1. has no solution / tiada penyelesaian 1. 3p – — 6
© Penerbitan Pelangi Sdn. Bhd. A4

Additional Mathematics Form 4 Answers
2. (a) 3m n (b) a b 40. (a) log — (b) log 3
–2 –1
–2 6
5
2 1 2
2
3. (a) n = — (b) n = –1 3 1
3 41. (a) 3 (b) – —
2
2
4. x = — 42. x = 4.344
3
6
36
2x
x
5. RHS = —–– = —– = 6 2x – 3 = LHS [shown] 43. x = 1.4305 × 10 -6
216 6 3 44. —––—
2n + 1
p
p
p
p
6. 3 + 3 p + 1 + 3 p + 2 = 3 + (3 )(3) + (3 )(9) 3m
2h + 3k
= 13(3 ) [shown] 45. —–––—
p
7. x = –1 6
8. m = 32, n = 16 46. m = 2.714
1
6 + b
5
9. a = —––– 47. x = —, x = 25
b 48. x = 8
3k – 7
10. h = —–––– 49. p = 3
2 50. n = 4
11. n = m + 1 51. 9
1
12. m = —, n = 5 52. n = 0.4838, n = –0.3260
3 53. x = 1
132
5
13. (a) — (b) 8 —– 54. x = 1.392
1
11 999 55. p = 81, p = —–
14. (a) Surd (b) Surd 56. x = 9 81
(c) Not a surd/ Bukan surd 57. LHS = log (x y ) = 4 log x + 2 log y
4 2
2
2
2
15. (a) Surd. Its value is a non-recurring decimal. = 4a + 2(2b) = 4(a + b) = RHS [shown]
Surd. Nilainya ialah perpuluhan tidak berulang. 58. (a) 3r + s (b) s – 4r
(b) Not a surd. Its value is an integer. 59. x = 4
Bukan surd. Nilainya ialah integer. 60. x = 11.1196
p + 2q
9
9
16. No/ Tidak. AB ≠ 2AB 61. (a) q – p (b) —––––
17. (a) AB (b) AB 62. x = 0.9076 q
6
42
log x
1
2 log y
21
(c) AB 63. ab = (log y )(log x) = — ——– — —– 2 = 2 [shown]
273
2
7
11
18. (a) AB (b) AB 4p + 2 x y log x log y
19. (a) bAB (b) AB 64. — ——–
5
30
q
2
3
2
20. (a) 5AB (b) 10AB 65. (a) 2y (b) 3 + —
y
21. (a) 4AB (b) 3AB2 66. p = 2 – — 5
2
AB 2 67. (a) c . 0, c ≠ 1 (b) y = x 4
7
22. (a) AB (b) —
6 68. after 9 years/ selepas 9 tahun
AB 3AB5 + 5 69. (7.17, 18)
6
23. (a) — (b) — ––– 70. a = 3.77, n = 2.15
15
6
24. (a) 9AB (b) 24AB 71. A = 12 000, k = 0.05
2
3
200
25. (a) —–– (b) 10 PAPER 2 y
3 1. (a) rs (b) 1 + x – —
3
26. (a) AB2 – 10 (b) 10AB15 – 11 2
(c) 24 + 16AB 2. (a) x = –1 is a negative number, therefore no
2
solution. / x = –1 ialah nombor negatif, maka
4(4 + AB)
3
27. (a) AB5 – AB (b) —––––-–– tiada penyelesaian.
2
13
(b) x = 5
10
34 + 3AB 7 – 2AB 3. (a) 4 years 7 months/ 4 tahun 7 bulan
35
28. (a) —––––––– (b) —–––––––
29 3 (b) x = 2 (c) 7
4. (a) log 4 + 2 log 7 – log 5
9
29. (a) log 243 = 5 (b) log 64 = 6 3 3 log 5
2
3
3
4
30. (a) 2 = 16 (b) 4 = 64 = log 4 + log 7 – — —— 2
2
3
log 3
3
3
31. (a) 0.9031 (b) –2.6021 3
5
2
(c) –0.3522 = log (4 × 7 ) – log AB
3
3
32. (a) 3 (b) 4 = log —— [shown]
196
33. (a) 1.7207 (b) 0.2389 3 AB
5
34. (a) x = 36 (b) x = 2 187 (b) x = 16
35. (a) 19 (b) 0.5 5. 7(3 n + 4 ) ; n = –2, n = 3
36. (a) x = 0.6021 (b) x = 1.2091 6. (a) 6AB2 + 5AB 3
37. (a) 0 (b) 1.0438 q
—
38. (a) x = 34 (b) x = 12 (b) 2 = 5 ⇒ 2 = 5 …… a
p
q
p
39. (a) x = 136.1429 (b) x = 0.5584 5 = 50 ⇒ 5 = 2 × 5 …… b
q
r
q
2r
r
A5 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
Substitute a into b / Gantikan a ke b: 675
n
qr — (c) m = —––
5 = 5 × 5 2r 35. 64
p
q
qr 36. (a) p = 0, 1, –1 (any one/ mana-mana satu)
q = —– + 2r
p 3
pq (b) —
5
r = —–––– [shown]
q + 2p 37. 16
7. (a) 59.32°C (b) 14 minutes/ minit 38. k = 9, r = 3; k = –9, r = –3
39. (a) 2 (b) 78 732
40. m = 8, n = 4; m = –24, n = 36
HOTS PRACTICES PAK-21
1. p = 4 41. 624 3
42. (a) — (b) 20
2
43. (a) 3 (b) 1 048 512
Pract ice Progressions 44. 2— or/ atau —
8
38
spm 5 Janjang 15 15
45. r = –2, r = 3
1
46. —
PAPER 1 3
1
1. (a) An arithmetic progression because d = d 2 47. (a) p = 45, q = 5 (b) 67—
2
1
Janjang aritmetik kerana d = d 2 9
1
(b) Not an arithmetic progression because d ≠ d 2 48. (a) first term/ sebutan pertama = —,
n
1
n
Bukan janjang aritmetik kerana d ≠ d 2 common ratio/ nisbah sepunya = —
1
4
27
2. (a) 2 (b) — (b) —––––– 3
3
3. 5, 3, 1, –1 3n – n 2
5
2
4. (a) 1 (b) 49, 53, 57 49. (a) —, – — (b) —, —
5
2
5. 53 5 5 12 28
6. 12 3 (b) —
1
7. a = 4, d = 5 50. (a) x 2
8. m = –2, n = 12 51. (a) p = 0.000148, q = 0.000000148
9. h = 2, k = 8 (b) 0.001
10. (a) –21 (b) 8
94
(c) 1 100 52. —
99
11. (a) 43.5 (b) –32.5 53. p = 10, q = 11
16 – 2x
12. (a) y = —––––– (b) 5
2
5 54. —
9
13. (a) 7p + 6 (b) — 55. 8 3
4
1
14. (a) – — (b) –2 56. —
2
2
15. 20 3
16. (a) 4 (b) 1 125 57. (a) m = 0.0048, n = 0.000048
17. (a) 132 (b) 30.5 (b) —
16
33
h
18. k = — 58. (a) 5 (b) 4
3
19. 54 27
20. –5 59. b = — 2
a
21. (a) 4 (b) 315 60. 8 hours/ jam
22. a = 10, d = –2 61. 11 year/ tahun ke-11
th
23. (a) T = –3, T = –1, T = 1
2
1
3
(b) 60
21 – 10n
24. —–––––– PAPER 2
2 1. (a) a = –9, d = 8 (b) 5 880
25. a = –10, d = 8 2. (a) –15 (b) 6
26. 18 onwards / 18 dan seterusnya (c) 75
1
27. 34, 41, 48, 55 3. (a) 60 (b) a = –22— , d = 5
28. 11 week/ minggu ke-11 (c) 6 2
th
29. 6 4. (a) 4 (b) 9
30. 30 (c) 8
31. (a) 4 (b) 2 5. (a) 28.8, –28.8 (b) 660.24, –188.64
1
32. 4, 2, 1, — 6. (a) 5 (b) 4
2 (c) 8 388 608 (d) 10
33. No because r ≠ r ≠ r / Tidak kerana r ≠ r ≠ r 3
1
2
3
2
1
34. (a) 3 (b) 5 7. (a) (i) 104
(ii) 40
© Penerbitan Pelangi Sdn. Bhd. A6

Additional Mathematics Form 4 Answers
12q – 3
(b) arithmetic progression: T = –280, 14. p = ——–––
6
8
geometric progression : T = ——; r 3
125
6
x
not the same for both progressions 15. y = —––––
2 – 4x
janjang aritmetik: T = –280,
1
1 2
6 16. (a) xy = pq — – p
8
janjang geometri: T = ——; x
125
6
10
10
10
tidak sama bagi kedua-dua janjang (b) log y = (log n)x + log m
3
2
(c) 250 17. y = – —x + 9
2
y
8. (a) (i) 149 cm (ii) 2 295 cm 18. — = –x + 9
(b) 25 rectangle/ segi empat tepat ke-25; P x 2
th
9. (a) RM900 (b) RM1 850 19. (a) log y = 2x (b) 10 000
10
(c) RM48 750 20. xy = –2x + 11 1
10. RM6 960 21. (a) 3 (b) ———
1 024
11. (a) 80(0.90) n - 1 , n = 1, 2, 3, 4, …
(b) 31 cm PAPER 2
12. 102 1. (a) Graph 2 because the graph obtained is a straight
line.
HOTS PRACTICES PAK-21 Graf 2 kerana graf yang diperoleh merupakan
9
8
1
1. a = 9, r = —; a = —, r = — satu garis lurus.
9 8 9 (b) Graph 1 because the graph obtained is a curve.
Graf 1 kerana graf yang diperoleh merupakan
satu lengkung.
Pract ice Linear Law 2. (a) For Table 1/ Bagi Jadual 1:
spm 6 Hukum Linear y
30
PAPER 1
1. (a) 2.7 25
(b) 17
3
2. y = – —x + 9 20
2
1
3. (a) y = – —x + 4 15
2
(b) 4
4. (a) 10 10
(b) 1
5
5. p = –4, q = 8
b
6. (a) a = — x
2 –4 –2 0 2 4 6
6 – y
(b) m = 2 ⇒ —––– = 2 ⇒ y = 0
3 – 0 –5
Hence, passing through the origin (0, 0).
Maka, melalui titik asalan (0, 0). For Table 2/ Bagi Jadual 2:
y y q 1 y
1 2
7. (a) — = 4x – 3 (b) — = — x + —
x x p p
(c) log y = (log n)(x + 3) + log m 8
10 10 10
8. h = –4, k = 2 6
m – 1
2
1
2
9. (a) xy = — ––– x + 5
2
(b) m = –5, n = 1 4
10. p = 18 – 3q 2
1
1
11. h = —, k = – — 0 x 2
6
4
2 4 6 8 10 12
12. (a) log y = (log w)x + 3 –2
10
10
1
(b) k = 1, w = ——
100 –4
13. y = –4x + x + 12
2
A7 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
(b) Graph for Table 2 because the graph obtained is (b) y
a straight line. x
Graf bagi Jadual 2 kerana graf yang diperoleh
merupakan satu garis lurus. 22
3. y 20

18
80
16
70
14
60
12
50
10
40
8
30
6
20
4
10
2
0 x
5 10 15 20 25 30
0 x
4. (a) 1 2 3 4 5 6
y
(c) (i) 2
(ii) 5
100 6. (a) log (x + 2) 0.48 0.60 0.70 0.78 0.85
10
90 log y -0.38 -0.67 -0.90 -1.08 -1.24
10
(b)
80
log y
10
70
0.8
60 0.73
0.6
50
0.4
40
0.2
30 log (x + 2)
10
0
20 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
–0.2
10
–0.4
0 x
5 10 15 20 25 30 –0.6
(b) y-intercept/ pintasan-y = 0,
gradient/ kecerunan = 3.6 –0.8
(c) y = 3.6x
(d) 80 –1.0
y
5. (a) — = px + pq –1.2
x

(c) m = 5.37, n = 2.32




© Penerbitan Pelangi Sdn. Bhd. A8

Additional Mathematics Form 4 Answers

7. (a) x 0.1 0.2 0.3 0.4 0.5 0.55 9. (a) — 0.50 0.33 0.25 0.20 0.17 0.14
1
1 x
— 4.57 3.68 2.78 1.89 0.99 0.55
1
y — 0.98 1.85 2.28 2.54 2.71 2.83
y
(b) 1
y (b) 1
y
5.5 5.45
5.0 3.5 3.55

4.5 3.0

4.0 2.5
3.5 2.0
3.2
3.0 1.5
2.5 1.0
2.0 0.5

1.5 0 1 x
0.1 0.2 0.3 0.4 0.5
1.0
(c) (i) 0.28
(ii) –1.44
0.5
10. (a)
0 x x 2 1 4 9 16 25 36
0.1 0.2 0.3 0.4 0.5 0.6
xy 4.37 5.48 7.35 9.96 13.30 17.40
(c) (i) 0.18
(ii) –1.61 (b)
(iii) 0.31 xy
8. (a)
y

16
9 14

8 12
7 10

6 8
5 6

4 4
3 2

2 x 2
0
5 10 15 20 25 30 35
1
0.6 (c) (i) 0.124
0 x (ii) 16
1 2 3 4 5 6
(b) p = 1.92, q = 0.41




A9 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
11. (a) xy (c) (i) 5.18
(ii) y = AB
x
36.04 – 4
40
13. (a) log y
35 10
30 1.1
25 1.0

20 0.9
15 0.8
0.74
10 0.7
5 0.6

0 x 0.5
1 2 3 4 5 6
–5 0.4
–6
0.3
(b) (i) g = 65.12, h = –0.09
(ii) 5 0.2 0.18
12. (a) 1 0.1
— 0.69 0.52 0.50 0.40 0.25 0.20
x
0 x
y 2 1 2 3 4 5 6 7
— 20.86 14.74 13.99 10.40 5.00 3.20
x
(b) (i) 1.43
(b) y 2 (ii) 0.66
x (iii) 5.50
14. (a) log P = log a + (log 36)kt
10
10
10
20
(b) log P
18 10
16 2.5
14 2.0
12.2
12 1.5
10 1.0

8 0.5
0.075
6 0 t
1 2 3 4 5 6
4
(c) a = 1.19, k = 0.28
2
1
0 x
0.1 0.2 0.3 0.4 0.5 0.6
–2

–4




© Penerbitan Pelangi Sdn. Bhd. A10

Additional Mathematics Form 4 Answers
(d) y = 44.67 (e) p = 2.24, q = 1.65
HOTS PRACTICES PAK-21 (f) 7.94 (g) 4.4
1. (a) xy

y Pract ice Coordinate Geometry
–20 –15 –10 –5 spm 7 Geometri Koordinat
–20

–40 PAPER 1
13
8
–60 1. 1 —, — 2
5
5
–80 2. (a) 10 units/ unit (b) (6, 8)
3. x = –1, y = 4
–100 4. (3, 4)
5. (a) (–1, 7) (b) 1 : 1
–120 22
6. 1 8, — 2
5
–141 –140 7. (a) Parallel because m = m = 3
2
1
Selari kerana m = m = 3
1
2
(b) p = –5.47, q = –141 (b) Not parallel because m ≠ m
1
(c) 1 Tidak selari kerana m ≠ m 2 2
1
y 1
8. m = m = – —
AB BC 2
x AB and BC are parallel. Hence, A, B and C are
0 1 2 3 4 5 6 collinear. / AB dan BC adalah selari. Maka, A, B dan
C adalah segaris.
–0.01
9
9. – —
–0.02 2
1
10. y = —x + 7
–0.03 2
11. (a) Perpendicular because m × m = –1
1
2
–0.04 –0.039 Berserenjang kerana m × m = –1
2
1
(b) Not perpendicular because m × m ≠ –1
1
2
–0.05 Tidak berserenjang kerana m × m ≠ –1
1 2
1
–0.06 12. –17—
2
13. y = 2x – 6
–0.07 1 15
14. y = – —x – —
–0.08 2 2
15. –1 1 – n
gradient/ kecerunan = –0.007, 16. m = ——
6
16 28
1
y-intercept/ pintasan-y = –0.039 17. (a) y = – —x + 4 (b) 1 – —, — 2
2. (a) log y = log p + (log q)x 2 5 5
10
10
10
18
3k
(b) 18. (a) p = – — (b) a = —
p
h
n
x 1 2 3 4 5 6 7 7.5 19. —
m
log y 0.56 0.78 0.99 1.21 1.43 1.77 1.87 1.97 20. (a) –4 (b) (–3, –4)
10
21. –5
(c) log y 22. 19 unit 2
10
23. n = –2, n = 22
24. 34 unit 2
2.0 1
25. —|3p – 40|
1.65 2
1.5 26. 1
27. 32 unit 2
1.0 28. n = 4, n = –5
0.9 5
29. n = 6, n = –15—
0.5 30. 34 unit 2 7
0.35
31. 71 unit 2
0 4.4 x
1 2 3 4 5 6 7
A11 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
1
2
32. n = –3 (c) (i) 10— (ii) 16— unit 2
33. x + y + 4x – 10y + 13 = 0 3 3
2
2
34. x + 3y – 2 = 0 1
35. 3x + 3y – 4x + 38y – 13 = 0 14. (a) y = – —x + 7 (b) (18, –2)
2
2
2
36. 3x + 3y + 16x + 34y – 1 = 0 22 10
2
2
2
37. 11x – 7y – 35 = 0 (c) 1 —, — (d) 4
3
3
38. Locus of point P/ Lokus bagi titik P: 2 2 2
65
x + y – 6x + 4y – 51 = 0 15. (a) (x + 2) + (y – 5) = (AB )
2
2
x + 4x + 4 + y – 10y + 25 = 65
2
2
When/ Apabila y = 0, x – 6x – 51 = 0 2 2
2
b – 4ac = (–6) – 4(1)(–51) = 240 . 0 x + y + 4x – 10y – 36 = 0 [shown]
2
2
\ intersects with the x-axis (b) 1 27 41 2
—, — , (2, –2)
menyilang paksi-x [shown] 5 5
39. When/ Apabila y = 0, x + 5x + 28 = 0 16. (a) (–10, –2) (b) 60 unit 2
2
2
b – 4ac = (5) – 4(1)(28) = –87 , 0
2
\ does not intersect with the x-axis
tidak menyilang paksi-x [shown] HOTS PRACTICES PAK-21
40. a = 6, b = 14, c = 22 1. (a) (4, 2) (b) y = x – 2
41. t = –1, t = –4 (c) 3
2
42. (m – 1)x + (m – 1)y – (6m + 8)x + 16y + 9m (d) PRT : 2 unit , RTQ : 4 unit , PTS : 2 unit ,
2
2
2
2
2
2
2
2
– 80 = 0 QTS : 4 unit ,
2
(e) 12 unit 2
PAPER 2 (f) x + y – 8x – 4y – 80 = 0
2
2
7
2
2
1. (a) 14 unit 2 (b) 1 0, — 2 2. (a) x + y – 16x – 16y – 497 = 0
3
2
2
(c) 3x + 3y – 72x + 34y + 251 = 0 (b) y = 3x – 16
(c) (15.9, 31.7), (0.1, –15.7)
1
2. (a) y = – —x – 10
2
9
(b) (i) (0, –5) (ii) 1 —, 4 2 Pract ice
3. (a) m = 1, n = 2 2 8 Vectors
(b) m × m = –1 × 1 = –1 spm Vektor
BC BD
\ /CBD = 90°, hence can form a right-angled
triangle. / maka dapat membentuk sebuah segi PAPER 1
tiga bersudut tegak. 1. (a) scalar quantity because does not have direction
4. (a) 5 kuantiti skalar kerana tidak mempunyai arah
(b) (i) 3 : 2 (b) vector quantity because has magnitude and
2
2
(ii) 5x + 5y – 2x – 46y + 74 = 0 direction
24
3
5. (a) (i) (–8, 0) (ii) y = – —x – — kuantiti vektor kerana mempunyai magnitud dan
5
5
(iii) 96 unit 2 arah
(b) 3x + 3y – 8x + 72y = 0 2. (a) (b)
2
2
6. (a) y = –3x + 26 (b) (6, 8) 40 km h –1 Q 50 N P
(c) –1 (d) 10 unit 2 v ~ 40 km j –1
7. (a) x + y – 12x – 12y + 67 = 0
2
2
(b) n = 5, n = 7 (c) (10, –2) 3. (a), (b)
8. (a) (i) 126 m 2 (ii) (–1, 2)
(b) x + y – 8x – 14y + 56 = 0 Q
2
2
9. (a) 9x + 5y + 2 = 0
(b) (i) (–3, 5) 1 →
→
(ii) Both the roads do not pass through Pantai P –2PQ –PQ
2
Indah. / Kedua-dua jalan raya itu tidak
melalui Pantai Indah.
2 31 9
2
10. (a) y = —x + — (b) 1 3, —
5 5 2 4. (a) parallel/ selari (b) parallel/ selari
(c) 24 unit 2
→
→
2
2
(d) x + y – 12x + 6y + 20 = 0 5. ST = –6RS
3 → →
11. (a) y = —x – 7 (b) (2, –4) RS and ST are parallel, S is a common point, then R,
2
(c) 36 unit (d) 8 S and T are collinear.
2
→
→
12. (a) 3x + 3y – 52x – 58y + 311 = 0 RS dan ST adalah selari, S ialah titik sepunya, maka
2
2
R, S dan T adalah segaris. [shown]
(b) does not intersect with the y-axis
tidak menyilang paksi-y 6. m = 2, n = –3
→
→
43
1
13. (a) y = 3x – 9 (b) y = – —x + — 7. PQ = 6MN
3 3 8. (a) m = 2, n = 4 (b) m = 5, n = –4
© Penerbitan Pelangi Sdn. Bhd. A12

Additional Mathematics Form 4 Answers
→
9. KL = 4u 36. ±0.9682
~
3(n – 6)
10. –7 37. m = ————
1
11. h = —, k = 2 n + 2
2 3m – 20
12. h = 1, k = 2 38. h = ———— –3i + 5j
14
~
~
→ → 39. (a) –3i + 5j (b) —–––—
13. (a) AD (b) DC ~ ~ ABB
34
→
(c) AD

14. (a) 3a + 5b (b) 6a – b PAPER 2

~
~
~
~
15. (a) –4a̰ + 8b ̰ (b) –a – 6b 1. (a) (i) 6x – 9y (ii) 2x + 6y

~
~
~
~
16 28 3 ~ 3 ~
16. —–x + —–y (b) m = , n =

5 ~ 5 ~ 8 4
(c) 3
3
1
17. (a) 3x – 2y (b) – —x – —y 2. (a) (i) a – 2b (ii) 1 (a – 6b)


~
2 ~
2 ~
~
18. (a) ABB 3ABB ~ ~ 4 ~ ~
13 units/ unit
37
(b) (i) r – q (ii) 3r – q (b) —––— units/ unit
~
4
~
~
~
→ 2 1
19. (a) OC = a + 2b (c) m = 5 , n = 10
~
~
(b) 3. (a) (i) 8ma + (5 – 5m)b (ii) (2 – 2n)a + 10nb ~
~
~
~
(b) m = 1 , n = 3 (c) 50 units/ unit
C 7 7 7
4. (a) (i) 3a + 4b (ii) 3a – 6b
A ~ ~ ~ ~
D a ~ (b) cannot be seen because S, R and Q are colinear
tidak dapat dilihat kerana S, R dan Q adalah
B segaris
b ~ O 5. (a) (i) q + 3 p (ii) p – q ~
~
2 ~
~
5 5 (b) (i) 3 mp + mq (ii) (n + 1)p – nq
20. (a) 3x + —y (b) —y – x ~ 2 ~ 2 ~ ~ ~
~
2 ~
2 ~
2
1 2 ~ ~
21. (a) 2 (b) 6i + 2j (c) m = 5 , n = – 5
8
→
→
6. (a) MN = (–4 + h)a + 5b , PQ = –2a – 12b ~
~
~
~
1 2 ~ ~ (b) 29
22. (a) 1 (b) 4i – j
–2
1 2 7. (a) – 3 ~ i – j ~ (b) —––— units/ unit
109
23. 12 6 10 ABBB
–4
3
–10i – 3j
24. 12i – 15 j (c) —–––––—
~
~
~
~
1 2
109
25. –13 ABBB
–35
~
~
~
26. 15 8. (a) (i) 9a + 8b (ii) 4a + 5b ~
1
27. (a) 4i + (1 + p) j (b) p = 2, p = –4 (b) 2
~
~
28. (a) (10, 4) (b) ABB 9. (a) 15 (b) x = 2, y = 4
82 units/ unit
5i + 3j –2i + 5j 10. (a) 8 y
~
~
~
~
29. (a) —–—— (b) —–––—
ABB ABB
29
34
4
–3i + 2j
~
~
30. —–––— c ~ a ~
ABB 2
13
→ 1
31. (a) SU (b) —(q – p) –2 O 2 4 6 x
5 ~ ~
32. (a) 17 units/ unit (b) –8 –2 b
→ → ~
5
–4
33. (a) OK = 1 2 , OL = 1 2 –4
3
4
(b) 9i – j ~ –7i + 12j –9i + 7j
~
~
~
2i + 3j
~
~
~
~
34. (a) 4i + 6j ~ (b) —–––— (b) (i) —–––––— (ii) —–––––—
ABBB
ABBB
130
193
~
ABB
13
20i – 37j
i – j
~
1 2 AB 2 ABBBB
~
~
~
35. (a) 14 (b) —––— (iii) —–––––—
1769
–14
A13 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
i + j
~
~
11. (a) (i) 3i + 3j ~ (ii) —–––– (b) (i) K / K’
~
(b) 18 AB 2 M / M’
12. (a) (i) q – p (ii) n p + 3 q 12 cm
~ ~ 3 + n ~ 3 + n ~ 12 cm L
(b) 4
13. (a) (i) 2x + 3y (ii) –2x + y ~ L’
~
~
~
(b) (i) 2mx + 3my (ii) (2 – 2n)x + ny ~ (ii) 64.85°
~
~
~
→ → → → 2. (a) (i) 12.07 cm (ii) 51.86°
(c) QB = 1 AC or / atau AC = 3QB (b) (i) 51.68 cm (ii) 8.56 cm
2
3
3. (a) 68.11° (b) 19.08 m
(c) 84.03 m 2
HOTS PRACTICES PAK-21 4. (a) 15.28 cm (b) 139.91°
1. (a) (i) –x + y ~ (ii) 1 2 ~ ~ (c) 62.48 cm 2
(x + y)
~
(iii) –x + 2y ~ 5. (a) 17.89 cm (b) 16.51 cm 2
~
(d) 93.87 cm
(c) 94.14°
→ 1 → → → 6. (a) (i) 18.29 cm (ii) 136.18°
(b) RW = OP or / atau OP = 2RW
2 (iii) 88.75 cm 2
(b) 5.92 cm
7. (a) (i) 15.74° (ii) 49.86°
Pract ice Solution of Triangles (iii) 164.15 cm 2
spm 9 Penyelesaian Segi Tiga (b) (i) J’
J K
PAPER 1 M
(ii) 5.30 cm

1. (a) sin K = sin L = sin M 8. (a) 22.61 cm (b) 38.22 cm
k l m 2
(c) 205.68 cm
(b) sin T = sin U = sin V 9. (a) 58.99° (d) 16.45 cm
(b) 26.13°
u
t
v
2. 47.69° (c) 10.80 cm (d) 107.38 cm 2
3. 9.44 cm 10. (a) 143.13° (b) 34.18 cm
4. 16.17 cm (c) 96 cm (d) 16.31°
2
5. 121.44° 11. (a) 18.15 cm (b) 113.65°
6. (a) Yes/ Ya (c) 63.15° (d) 148.95 cm 2
(b) No/ Tidak 12. (a) (i) 15.68° (ii) 29.71 cm
7. 34.53° (b) 76.52 cm (c) 174.77 cm 2
2
8. 15.02 cm 13. (a) (i) 125°
9. 76.64° (ii) 23.89 cm
10. 28.54° (iii) 21.74°
11. (a) 41.80° (b) 189.18 cm 2
(b) 88.59°
(c) 49.61°
12. the smallest angle/ sudut terkecil: 42.32° HOTS PRACTICES PAK-21
the largest angle/ sudut terbesar: 85.38° 1. (a) (i) a = 12, b = 18, c = 9.69
13. (a) 34.49 cm (ii) 111.73°
(b) 89.26° (b) (i)
14. 12.86 cm 2 P
15. 18.39 cm 2
16. 15.39 cm 2 9.69 cm 12 cm
17. 31.34 cm 2 9.69 cm
18. 19.14 cm 2 Q Q’ R
19. 57 cm 2
20. 42.79 cm 2 (ii) 8.36 cm 2
21. 120°
22. 3 107 cm 2
23. (a) 23.60° Pract ice
(b) 3 447.20 cm 2 10 Index Numbers
spm
PAPER 2 Nombor Indeks
1. (a) (i) 32 cm PAPER 1
(ii) 115.15° 1. 125
(iii) 356.38 cm 2 2. 150
© Penerbitan Pelangi Sdn. Bhd. A14

Additional Mathematics Form 4 Answers
3. (a) 150 12. (a) RM3.51
(b) 125 (b) RM51 502.15
4. RM1 375 (c) Increased by / Meningkat 63.1%
5. Increased by / Meningkat 20%
6. I = 120 PAK-21
R
I = 90 HOTS PRACTICES
S
I = 120 1. 132
T
I = 120 2. 1 : 3 : 6
U
7.
Fuel / Bahan api Increased by/ Meningkat 16.67%
Utility / Utiliti Decreased by/ Menyusut 20% Assessment Paper
Assessment Paper
Assessment Paper
Salary/ Gaji Increased by/ Meningkat 28%
Rent / Sewa Increased by/ Meningkat 25% Paper 1
1. (a) (i) 16
8. 114.94 (ii) 3
9. 111.13 (b) f(x) = x or/ atau f : x → x 2
2
10. 132 2. (a) 5
11. 125 (b) x
12. 2 3. (a) b = –9, c = –5
13. (a) 125.7 (b) p = –3, p = 3
(b) RM3 142.50 11
14. 136.1 4. (a) (3, –49)
15. 121.4 (b) x = 3
(c) two different real roots
dua punca nyata yang berbeza
PAPER 2 (d) –4 , x , 10
1. (a) RM5 7
(b) 1 5. (a) 4
(c) RM35.70 — 1 5
(d) 125.93 (b) r = q
2. (a) RM36 6. x = 4, x = 9
(b) Increased by / Menokok 10% 7. p = –8, q = 3
(c) (i) 132.25 (ii) RM185.15
3. (a) RM1.50 (b) 125 8. (a) 2 1
(c) 102.4 (d) RM12.47 2
4. (a) RM44 000 (b) 116.25 (b) 22
(c) RM139 500 (d) 126.18 (c) 8 1
5. (a) x = 500, y = 1 612, z = 110 2
(b) 120.97 (c) RM604 850 9. (a) 5
(d) 157.26 (b) 22
6. (a) 112 (b) 130 10. (a) 9
(c) RM19.20 (d) 146 (b) 19 683
(e) RM140.00 (c) 29 520
7. (a) x = 0.25, y = 150, z = 0.40 11. (a) 8
(b) 129 (c) RM322.50 (b) (7, 0)
(d) 148.35 (c) x + y – 12x + 8y – 173 = 0
2
2
8. (a) I = 125, I = 150, I = 200, I = 80 12. (a) k = 4, k = –6
B
A
D
C
(b) 154.5 (c) RM3.09 (b) 30 unit 2
(d) 162.23 13. (a) (i) 6i + 8j

9. (a) (i) h = 132, k = 104 ~ ~

(ii) RM25 3i + 4j ~
~
(b) 4 (ii) 5
(c) RM144.90 (b) (i) 8i – 12j

~
~
10. (a) RM40 (ii) (18, –10)
(b) I = 180, I = 137.5, I = 140, I = 136 14. (a) 89.13°
C
B
A
D
(c) 154.83
(d) RM197.25 (b) 25.98 cm 2
(c) 594.20 cm
11. (a) (i) 120 (ii) 5.00 15. (a) 25%
(b) 125.71 (b) 117.25
(c) (i) 103.41 (c) RM1 961.62
(ii) 34 615 chicken nuggets / ketul nuget ayam
A15 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 4 Answers
Paper 2 (b) xy
1. x = 0.7595, y = 1.2785;
x = –0.1881, y = –1.5643 5.0
2. (a) (i) x – 5 (ii) 2x – 9
(b) 3 3 4.5
3. (a) (i) p = 5, q = –2 4.25
(ii) –2 , x , 5 4.0
(b) x – 5x + 4 = 0
2
4. (a) 4 3.5
(b) 6
(c) 8 3.0 2.95
5. a worker in factory / seorang pekerja di kilang:
RM150, a worker in office / seorang pekerja di 2.5
pejabat: RM220, a worker who deals with clients /
seorang pekerja yang berurusan dengan klien: RM270 2.0
6. (a) y = – 3 x – 1
2 1.5
(b) 3x + 3y – 16x – 74y + 339 = 0
2
2
15
12
7. (a) h = – —–, k = – —– 1.0
5 4
(b) y 0.5
1
y = 3x 2 – 12x – 15
21 0 0.05 0.10 0.15 0.20 0.25 0.30 x
(c) (i) 6.25
x (ii) 2.95
–2 –1 O 2 5 6 (iii) 0.89

–15 11. (a) (i) 15x – 3y
~
~

~
–27 (ii) 10x + 3y
~

(b) (i) 10mx + 3my
~
~
(c) y = –3x + 12x + 15 (ii) 15nx + (3 – 3n)y
2

8. (a) –9 – 4AB 5 3 ~ 2 ~
(b) log x = 3 ⇒ x = 1 000 (c) m = 5 , n = 5
10
log y = 2 ⇒ y = 100
10
2
LHS: x – 7xy = (1 000) – 7(1 000)(100) 12. (a) 25.13°
2
= 300 000 (b) 27.04 cm
RHS: 3000y = 3000(100) = 300 000 (c) 12.74 cm
LHS = RHS [shown] (d) 64.33° 2
(c) 5 (e) 292.82 cm
(d) x = 9.48, x = 0.52 13. (a) 112
9. (a) y = 2x – 3 (b) (i) 135
(ii) RM7.20

(b) y = – 1 x + 1 (c) 128
2 (d) RM60.70
(c) 1 8 5 , 1 2 14. (a) (i) 26.75 cm
5
(d) 21 3 unit 2 (ii) 24.10 cm 2
(iii) ∆PQT : 60.11 cm
10. (a) 5 ∆ PTS : 60.10 cm 2
∆ QRS : 88.47 cm 2
1 0.33 0.25 0.20 0.17 0.14 0.13 (b) 66.23 cm 2
x 15. (a) x = 150, y = 2.10, z = 4.50
xy 5.01 4.52 4.20 4.02 3.85 3.76 (b) 3
(c) RM1 485
(d) 137.5
© Penerbitan Pelangi Sdn. Bhd. A16

Praktis ! FC064130 Practice Makes Perfect!
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H E BA T Additional Praktis

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Matematik
Form SPM Mathematics H BA T
4
4 KSSM Tambahan Praktis HEBAT! SPM KSSM E
4
4





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