Module & More Matematik Tambahan Tg4

PELANGI BESTSELLER

MoMduOlRe E&

Pembelajaran BERPANDU dan SISTEMATIK

MATEMATIK
TAMBAHAN

Additional Mathematics

Tee Hock Tian
Penerbitan Pelangi Sdn Bhd All Rights Reserved

EDISI GURU

4TINGKATAN DSKP &
FORMAT BAHARU
KSSM
SPM 2021
Pakej PdPR

Pengajaran dan Pembelajaran di Rumah
Rekod Pencapaian Lembaran PBD Nota
KBAT Ekstra Online Quick Quiz Praktis SPM

KANDUNGAN

BABPenerbitan Pelangi Sdn Bhd All Rights ReservedFungsi 1 BAB Indeks, Surd dan Logaritma 58

1 Functions 4 Indices, Surds and Logarithms

1.1 Fungsi................................................................................... 1 4.1 Hukum Indeks...................................................................58
Functions Laws of Indices
1.2 Fungsi Gubahan.................................................................. 5 4.2 Hukum Surd......................................................................61
Composite Functions Laws of Surds
1.3 Fungsi Songsang...............................................................11 4.3 Hukum Logaritma............................................................67
Inverse Functions Laws of Logarithms
4.4 Aplikasi Indeks, Surd dan Logaritma.............................74
Praktis SPM 1.............................................................................14 Applications of Indices, Surds and Logarithms

Sudut KBAT...............................................................................17 Praktis SPM 4.............................................................................75

BAB Sudut KBAT...............................................................................78

2 Fungsi Kuadratik 19

Quadratic Functions BAB

2.1 Persamaan dan Ketaksamaan Kuadratik.......................19 5 Janjang 79
Quadratic Equations and Inequalities
2.2 Jenis-jenis Punca Persamaan Kuadratik........................24 Progressions
Types of Roots of Quadratic Equations
2.3 Fungsi Kuadratik...............................................................27 5.1 Janjang Aritmetik..............................................................79
Quadratic Functions Arithmetic Progressions
5.2 Janjang Geometri..............................................................87
Praktis SPM 2.............................................................................35 Geometric Progressions

Sudut KBAT...............................................................................42 Praktis SPM 5.............................................................................96

Sudut KBAT.............................................................................102

BAB Sistem Persamaan 43 BAB Hukum Linear 103

3 Systems of Equations 6 Linear Law

3.1 Sistem Persamaan Linear dalam Tiga 6.1 Hubungan Linear dan Tak Linear................................103
Pemboleh Ubah ................................................................43 Linear and Non-Linear Relations
6.2 Hukum Linear dan Hubungan Tak Linear..................107
Systems of Linear Equations in Three Variables Linear Law and Non-Linear Relations
3.2 Persamaan Serentak yang melibatkan Satu Persamaan 6.3 Aplikasi Hukum Linear..................................................113
Application of Linear Law
Linear dan Satu Persamaan Tak Linear.........................48
Simultaneous Equations involving One Linear Equation and Praktis SPM 6...........................................................................117

One Non-Linear Equation Sudut KBAT.............................................................................122

Praktis SPM 3.............................................................................54

Sudut KBAT...............................................................................57

iii

BAB Geometri Koordinat 123 9.3 Luas Segi Tiga..................................................................171
Area of a Triangle
7 Coordinate Geometry 9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas

7.1 Pembahagi Tembereng Garis........................................123 Segi Tiga...........................................................................173
Application of Sine Rule, Cosine Rule and Area of a Triangle
Divisor of a Line Segment
7.2 Garis Lurus Selari dan Garis Lurus Serenjang............126 Praktis SPM 9...........................................................................174

Parallel Lines and Perpendicular Lines Sudut KBAT.............................................................................179
7.3 Luas Poligon....................................................................130

Areas of Polygons
7.4 Persamaan Lokus............................................................133

Equations of Loci

Praktis SPM 7...........................................................................135

Sudut KBAT.............................................................................140
Penerbitan Pelangi Sdn Bhd All Rights Reserved BAB Nombor Indeks 180

10 Index Numbers

BAB Vektor 142 10.1 Nombor Indeks...............................................................180
Index Numbers
8 Vectors 10.2 Indeks Gubahan..............................................................187
Composite Index

Praktis SPM 10........................................................................191

Sudut KBAT.............................................................................195

8.1 Vektor...............................................................................142 Kertas Pra-SPM............................................................... 196
Vectors
8.2 Penambahan dan Penolakan Vektor............................146 Jawapan
Addition and Subtraction of Vectors Lembaran PBD
8.3 Vektor dalam Satah Cartes............................................150
Vectors in a Cartesian Plane http://www.epelangi.com/Module&More2021/
MatematikTambahan/T4/LPBD.pdf
Praktis SPM 8...........................................................................156
Jawapan Lembaran PBD
Sudut KBAT.............................................................................163

BAB Penyelesaian Segi Tiga 154

9 Solution of Triangles

9.1 Petua Sinus.......................................................................164

Sine Rule
9.2 Petua Kosinus..................................................................168

Cosine Rule

BONUS Lembaran PBD dengan Jawapan

untuk Guru http://www.epelangi.com/Module&More2021/MatematikTambahan/T4/
BonusLPBD.

iv

BAB

2 Fungsi Kuadratik
Quadratic Functions
NOTA IMBASAN
Penerbitan Pelangi Sdn Bhd All Rights Reserved
2.1 Persamaan dan Ketaksamaan Kuadratik

Quadratic Equations and Inequalities

NOTA IMBASAN

Persamaan dan ketaksamaan kuadratik
Quadratic equations and inequalities

Penyelesaian persamaan Hasil tambah punca (HTP) dan hasil Penyelesaian ketaksamaan
Solving equation darab punca (HDP). kuadratik
Sum of roots (SOR) and product of roots Solving quadratic inequalities
(a) Penyempurnaan kuasa dua (POR)
Completing the square (a) Lakaran graf
(b) Rumus kuadratik (a) Punca-punca ialah dan Graph sketching
Quadratic formula Roots are and (b) Garis nombor
(b) HTP/SOR Number line
(c) HDP/SOR (c) Jadual / Table
(d) Persamaan / Equation

1. Selesaikan setiap persamaan berikut menggunakan kaedah penyempurnaan kuasa dua. TP 3

Solve each of the following equations using completing the square method. 

CONTOH (a) x2 – 8x + 10 = 0

−2x2 + 8x + 13 = 0 x 2 – 8x = –10
Penyelesaian:
−2x2 + 8x = −13 Pindah sebutan pemalar ke kanan. x2 – 8x + 1– 8 22 = –10 + 1– 8 22
x2 – 4x = ­123 Move the constant to the right. 2 2
Pekali x2 = +1
Coefficient of x2 = +1 (x – 4)2 = –10 + 16

√(x – 4)2 = ±√6

x2 – 4x + 1– 4 22 = 13 + 1– 4 22 1 2Tambah sebutanp2bers2admi aan. x – 4 = √6    atau x – 4 = –√6
2 2 2 = 6.449 = 1.551
kedua-dua belah

(x – 2)2 = 13 + 4 1 2Add the term b 2 on both
2
sides of equatio2n.

√(x – 2)2 = ±√10.5 Masukkan ± apabila
mengambil punca kuasa
x – 2 = √10.5 x – 2 = –√10.5 dua.
Put in ± when taking
x = 5.240 x = –1.240 square roots.

25

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

(b) –2x2 + 13x – 8 = 0 (c) –3x2 – 18x + 7 = 0

2x2 – 13x + 8 = 0 –3x2 – 18x = –7

x2 – 13 x = – 4 x2 + 6x = 7
4 3
BAB 2
1 2 1 2 x2 – 143x + – 13 2 = –4 + – 13 2 x2 + 6x + (3)2 = 7 + (3)2
Penerbitan Pelangi Sdn Bhd All Rights Reserved44 3

1 2x– 13 2 = – 4 + 169 (x + 3)2 = 34
4 16 3

1 2x – 13 2 ± 105 √(x + 3)2 = ± 34
4 16 3
=

x – 13 = 105    x – 143 = – 105 x+3= 34   x + 3 = – 34
4 16 16 3 3

x = 5.812 = 0.6883 x = 0.3665 x = –6.3665

2. Selesaikan setiap persamaan berikut menggunakan persamaan kuadratik.  TP 3

Solve each of the following equations using quadratic formula. 

CONTOH (a) –3x2 + 16x – 9 = 0

4x2 − 17x + 5 = 0 a = –3,  b = 16,  c = –9

Penyelesaian: Bandingkan dengan bentuk am. x = –16 ± √(16)2 – 4(–3)(–9)
Compare with the general form. 2(–3)
  ax2 + bx + c = 0
a = 4,  b = –17,  c = 5 = –16 + √148
–16
x = –b ± √b2 – 4ac Guna rumus
2a Use the formula –16 + √148 –16 – √148
x = –6   atau x= –6
–(–17) ± √(–17)2 – 4(4)(5)
= 2(4) = 0.6391 x = 4.6943

= 17 ± √209
8

x = 17 + √209   x= 17 – √209
8 8

= 3.9321 = 0.3179

(b) 5x2 – 19x + 8 = 0 (c) –7x2 + 22x – 13 = 0

a = 5,  b = –19,  c = 8 a = –7,  b = 22,  c = –13

x = –b ± √b2 – 4ac x = –b ± √b2 – 4ac
2a 2a

x = –(–19) ± √(–19)2 – 4(5)(8) x = –(22) ± √(22)2 – 4(–7)(–13)
2(5) 2(–7)

= 19 ± √201 = –22 ± √120
10 –14

x = 19 + √201   x = 19 – √41 x = –22 + √120   x = –22 – √120  
10 10 –14 –14

= 3.3177 = 1.260 = 0.7890 = 2.3539

26

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

3. Tentukan hasil tambah dan hasil darab punca bagi setiap persamaan kuadratik yang berikut.  TP 2

Determine the sum and the product of roots for each of the following quadratic equations.

CONTOH (a) 6x2 + 14x – 7 = 0

5x2 – 12x – 9 = 0
Penerbitan Pelangi Sdn Bhd All Rights Reserved a = 6,  b = 14,  c = –7

Penyelesaian: BAB 2Bandingkan dengan bentuk am. HTP/SOR = –b=–14=–7
ax2 + bx + c = 0 Compare with the general form. a 5 3
a = 5,  b = –12,  c = –9
HDP/POR = c = – 7
a 6
HTP/SOR = – b = – –12 = 12
a 5 5

HDP/POR = c = – 9
a 5

(b) –4x2 + 12x – 3 = 0 (c) 9x2 – 6x – 11 = 0

a = –4,  b = 12,  c = –3 a = 9,  b = –6,  c = –11

HTP/SOR = – b = – 12 = 3 HTP/SOR = – b =– –6 = 2
a –4 a 6 3

HDP/POR = c = –3 = 3 HDP/POR = c = –11
a –4 4 a 9

4. Bentukkan satu persamaan kuadratik 3x2 – 6x – 8 = 0 dengan punca-punca yang diberi. TP 4

Form a quadratic equation 3x2 – 6x – 8 = 0 with the given roots.

CONTOH (a) a + 5, b + 5

a – 4,  b – 4

HTP/SOR:

Penyelesaian: Tentukan HTP dan HDP bagi (a + 5) + (b + 5) = (a + b) + 10
persamaan yang diberi.
a = 3,  b = –6,  c = –8 Determine SOR and POR for = 2 + 10 = 12
the equation given.
b –6 HDP/POR:
a 3
a + b = – =– =2 (a + 5)(b + 5) = ab + 5a + 5b + 25

ab = c =– 8 = 1– 8 2 + 5(2) + 25
a 3 3

HTP/SOR: (a – 4) + (b – 4) = a + b – 8 = 97
= 2 – 8 = –6 3

HDP/POR: (a – 4)(b – 4) = ab – 4a – 4b + 16 Persamaan kuadratik / Quadratic equation
x2 – (HTP)x + HDP = 0
= 1– 8 2 – 4(a + b) + 16
3 x2 – (12)x + 19372 = 0

= 1– 8 2 – 4(2) + 16 = 16 3x2 – 36x + 97 = 0
3 3

Persamaan kuadratik / Quadratic equation

x2 – (HTP)x + (HDP) = 0

x2 – (–6)x + 136 = 0
3x2 + 18x + 16 = 0

27

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

(b) 3a, 3b (c) a2, b2



HTP/SOR: HTP/SOR:

3a + 3b = 3(a + b)  a2 + b2 = (a + b)2 – 2ab
BAB 2
= 3(2) = 6 = 22 – 21– 8 2 = 28
Penerbitan Pelangi Sdn Bhd All Rights Reserved 3 3
HDP/POR:

(3a)(3b) = 9ab HDP/POR:

= 91– 8 2 (a2)(b2) = (ab)2
3
1– 8 22 64
= –24 = 3 = 9

Persamaan kuadratik / Quadratic equation Persamaan kuadratik / Quadratic equation
x2 – (HTP)x + HDP = 0
x2 – (HTP)x + HDP = 0
x2 – 12382x + 16942 = 0
x2 – 6x + (–24) = 0
9x2 – 84x + 64 = 0
x2 – 6x – 24 = 0

5. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

Contoh

Jika a dan b ialah punca-punca persamaan x2 – 4x + 7 = 0, bentukkan persamaan kuadratik dengan
punca-punca 3a dan 3b.

If α and β are roots of quadratic equation x2 – 4x + 7 = 0, form a quadratic equation with roots 3α and 3b.

Penyelesaian:

a = 1,  b = –4,  c = 7

HTP/SOR =– b HDP/POR = c
a a

a + b = – 1 –4 2 = 4 ab = 7 =7
1 1

HTP baharu/ New SOR = 3a + 3b HDP baharu/ New POR = 3a × 3b
= 3(a + b) = 3(4) = 12 = 9ab = 9(7) = 63

Persamaan baharu ialah/ New equation is
x2 – (HTP)x + HDP = 0
x2 – 12x + 63 = 0

Jika a dan b ialah punca-punca persamaan 3x2 – 9x + 5 = 0, bentukkan persamaan kuadratik dengan
punca-punca 2a dan 2b.

If α and β are roots of quadratic equation 3x2 – 9x + 5 = 0, form a quadratic equation with roots 2α and 2b.

a = 3,  b = –9,  c = 5

HTP = – b HDP = c Persamaan baharu ialah
a a
x2 – (HTP)x + HDP = 0

a + b = – 1 –9 2 = 3 ab = 5 x2 – 6x + 20 = 0
3 3 3

HTP baharu = 2a + 2b HDP baharu = 2a × 2b 3x2 – 18x + 20 = 0
= 2(a + b) = 2(3) = 6
= 4ab = 41 5 2 = 20
3 3

28

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

6. Selesaikan setiap yang berikut.   TP 5

Solve each of the following.

CONTOH

Salah satu daripada punca bagi persamaan 2x2 – 8x + p = 0 ialah tiga kali punca yang satu lagi. Cari
punca-punca itu dan nilai p.

One of the roots of quadratic equation 2x2 – 8x + p = 0 is three times the other. Find the roots and the value of p.
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Penyelesaian:
BAB 2
a = 2,  b = –8,  c = p ;  Katakan punca-punca ialah a dan 3a./ Let roots are a and 3a.

HTP/ SOR = – b HDP/ POR = c
a a

a + 3a = – 1 –8 2 = 4 a × 3a = p
2 2
p
4a = 4   →   a = 1 3a2 = 2

3a = 3(1) = 3 p = 6a2 = 6(1)2 = 6

\  Punca-punca ialah 1 dan 3. Manakala, nilai p ialah 6.
Roots are 1 and 3. While, value of p is 6.

Salah satu punca bagi persamaan x2 – 15x + m = 0 ialah dua kali punca yang satu lagi. Cari punca-punca
itu dan nilai m.

One of the roots of quadratic equation x2 – 15x + m = 0 is two times the other. Find the roots and the value of m.

a = 1,  b = –15,  c = m ;  Katakan punca-punca ialah a dan 2a.

HTP = – b HDP = c
a a

a + 2a = – 1 –15 2 a × 2a = m
1 1

3a = 15 m = 2a2
= 2(5)2
a = 5
= 50
2a = 2(5) = 10

Punca-punca ialah 5 dan 10. Nilai m ialah 50.

7. Tentukan ketaksamaan berikut.  TP 5 CONTOH 1 

Solve the following inequalities.

CONTOH 1 

x2 – 7x + 10 < 0 –x2 + 7x – 12 < 0

Penyelesaian: Penyelesaian:

(i) Kaedah 1:  lakaran graf/ Method 1: graph sketching (i) Kaedah 1:  lakaran graf/ Method 1: graph sketching

a . 0 bentuk graf ialah/ the shape of the graph is a , 0 bentuk graf ialah/ the shape of the graph is

apabila/ when x 2 – 7x + 10 < 0 apabila/ when –x2 + 7x – 12 < 0

(x – 2)(x – 5) < 0 (–x + 4)(x – 3) < 0

x – 2 = 0   atau/ or x – 5 = 0 –x + 4 = 0 atau/ or   x – 3 = 0

x = 2 x=5 x = 4 x=3

2 5 34

Maka/ Thus, 2 < x < 5 Maka/ Thus, x < 3  atau/ or  x > 4
29

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

(ii) Kaedah 2:  garis nombor/ Method 2: number line (ii) Kaedah 2:  garis nombor/ Method 2: number line

x2 – 7x + 10 < 0 –x2 + 7x – 12 < 0

(x – 2)(x – 5) < 0 (–x + 4)(x – 3) < 0

Pertimbangkan x – 2 > 0 dan/ and   x – 5 > 0 Pertimbangkan –x + 4 > 0 dan/ and  x – 3 > 0
BAB 2
Consider x > 2 x>5 Consider x < 4 x>3
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–– + –+ +

–+ + ++ –

+2 – 5+ –3 + 4 –

Maka/ Thus, 2 < x < 5 Maka/ Thus, x < 3  atau/ or  x > 4

(iii) Kaedah 3:  jadual / Method 3: table (iii) Kaedah 3:  jadual/ Method 3: table

x2 – 7x + 10 < 0 –x2 + 7x – 12 < 0

(x – 2)(x – 5) < 0 (–x + 4)(x – 3) < 0

Pertimbangkan x – 2 = 0 dan/ and  x – 5 = 0 Pertimbangkan –x + 4 = 0 dan/ and  x – 3 = 0

Consider x = 2 x=5 Consider x = 4 x=3

(x – 2): – + + (x – 3) : – + +

(x – 5): – – + (–x + 4) : + + –

(x – 2)(x – 5): + – + (x – 3)(–x + 4) : – + –
25 34

Maka/ Thus, 2 < x < 5 Maka/ Thus, x < 3  atau/ or  x > 4

(a) x2 – 6x + 8 > 0 (b) 4x2 + 8x – 45 , 0

Apabila x2 – 6x + 8 = 0 Apabila 4x2 + 8x – 45 = 0

(x – 2)(x – 4) = 0 (2x – 5)(2x + 9) = 0
5 9
x – 2 = 0 atau   x – 4 = 0 x = 2 atau   x =– 2

x = 2 x=4

24 – –29 –52

Maka, x < 2  atau  x > 4 Maka, – 9 ,x , 5
2 2

(c) –3x2 + 17x – 10 . 0 (d) 7x2 – 24x – 16 > 0

Apabila –3x2 + 17x – 10 = 0 Apabila 7x2 – 24x – 16 = 0

(–x + 5)(3x – 2) = 0 (7x + 4)(x – 4) = 0
2 4
x=5  atau   x= 3 x=– 7   atau   x=4

–23 5 – –47 4

Maka, 2 , x ,5 Maka, x<– 4  atau x > 4
3 7

30

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

2.2 Jenis-jenis Punca Persamaan Kuadratik

Types of Roots of Quadratic Equations

Penerbitan Pelangi Sdn Bhd All Rights ReservedNOTA IMBASAN

BAB 2
1. Jenis punca persamaan kuadratik ditentukan oleh nilai pembeza layan, b2 – 4ac.
The types of roots of a quadratic equation are determined by the value of discriminant, b2 – 4ac.

2. b2 – 4ac . 0 b2 – 4ac = 0 b2 – 4ac , 0

Dua punca yang berbeza. Dua punca yang sama. Tiada punca.
Two different roots. Two equal roots. No roots.

8. Tentukan jenis punca bagi setiap persamaan kuadratik yang berikut. TP 4

Determine the type of roots for each of the following quadratic equations.

CONTOH

(i) 9x2 – 12x + 4 = 0 (ii) 4x2 – 13x + 3 = 0 (iii) 6x2 + 7x + 5 = 0

Penyelesaian:

(i) a = 9, b = –12, c = 4 (ii) a = 4, b = –13, c = 3 (ii) a = 6, b = 7, c = 5

b2 – 4ac = (–12)2 – 4(9)(4) b2 – 4ac = (–13)2 – 4(4)(3) b2 – 4ac = 72 – 4(6)(5)

= 144 – 144 = 169 – 48 = 49 – 120

=0 = 121 . 0 = –71 , 0

Persamaan itu mempunyai Persamaan itu mempunyai Persamaan itu tidak
mempunyai punca.
dua punca yang sama. dua punca yang berbeza.
The equation has no roots.
The equation has two equal roots. The equation has two distinct

roots.

(a) 2x2 – 5x – 4 = 0 (b) 3x2 + 7x + 8 = 0

a = 2,  b = –5,  c = –4 a = 3,  b = 7,  c = 8

b2 – 4ac = (–5)2 – 4(2)(–4) b2 – 4ac = 72 – 4(3)(8)

= 25 + 32 = 49 – 96

= 57 . 0 = –47 , 0

Persamaan itu mempunyai dua punca yang Persamaan itu tidak mempunyai punca.
berbeza.

(c) 4x2 – 28x + 49 = 0 (d) 6x2 – 9x + 2 = 0

a = 4,  b = –28,  c = 49 a = 6,  b = –9,  c = 2

b2 – 4ac = (–28)2 – 4(4)(49) b2 – 4ac = (–9)2 – 4(6)(2)

= 784 – 784 = 81 – 48

=0 = 33 . 0

Persamaan itu mempunyai dua punca yang Persamaan itu mempunyai dua punca yang
sama. berbeza.

31

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

9. Cari nilai-nilai p jika setiap persamaan kuadratik berikut mempunyai dua punca yang sama. TP 4

Find the values of p if each of the following quadratic equations has two equal roots.

CONTOH (a) x2 + 2px + 3p + 4 = 0

x2 – 2px + 6p + 16 = 0

Penyelesaian:
a = 1,  b = –2p,  c = 6p + 16
b2 – 4ac = 0
(–2p)2 – 4(1)(6p + 16) = 0
4p2 – 24p – 64 = 0
p2 – 6p – 16 = 0
(p + 2)(p – 8) = 0
BAB 2 a = 1, b = 2p, c = 3p + 4

Penerbitan Pelangi Sdn Bhd All Rights Reserved b2 – 4ac = 0

Syarat untuk dua (2p)2 – 4(1)(3p + 4) = 0
punca yang sama.
Condition for two 4p2 – 12p – 16 = 0
equal roots.
p2 – 3p – 4 = 0

(p + 1)(p – 4) = 0

p + 1 = 0   atau p – 4 = 0

p = –1 p=4

p + 2 = 0   atau p – 8 = 0

p = –2 p=8

(b) 3x2 + px + 12 = 0 (c) 4x2 – 4px + 8p + 9 = 0

a = 3, b = p, c = 12 a = 4, b = –4p, c = 8p + 9

b2 – 4ac = 0 b2 – 4ac = 0

p2 – 4(3)(12) = 0 (–4p)2 – 4(4)(8p + 9) = 0

p2 – 144 = 0 16p2 – 128p – 144 = 0

p2 = 144 p2 – 8p – 9 = 0

p = ±12 (p + 1)(p – 9) = 0

p + 1 = 0   atau p – 9 = 0

p = –1 p=9

10. Cari julat nilai p jika setiap persamaan kuadratik berikut mempunyai dua punca yang berbeza. TP 4

Find the range of values of p if each of the following quadratic equations has two different roots.

CONTOH (a) x2 + 2x + p – 3 = 0

x2 – 4x – 3 + p = 0 b2 – 4ac . 0

Penyelesaian: 22 – 4(1)(p – 3) . 0
b2 – 4ac . 0
(–4)2 – 4(1)(–3 + p) . 0 Syarat untuk dua 4 – 4p + 12 . 0
16 + 12 – 4p . 0 punca yang berbeza.
4p , 28 Condition for two 16 – 4p . 0
p , 7 distinct roots.
4p , 16

p , 4

(b) 2x2 – 7x + p = 0 (c) (p + 1)x2 + 4x – 9 = 0

b2 – 4ac . 0 b2 – 4ac . 0

(–7)2 – 4(2)(p) . 0 42 – 4(p + 1)(–9) . 0

49 – 8p . 0 16 + 36p + 36 . 0

8p , 49 36p . –52

p , 49 p . – 13
8 9

32

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

11. Cari julat nilai p jika setiap persamaan kuadratik berikut tidak mempunyai punca. TP 4

Find the range of values of p if each of the following quadratic equations has no roots.

CONTOH (a) x2 – 2x + p – 6 = 0

x2 + 6x + p – 4 = 0 Penerbitan Pelangi Sdn Bhd All Rights Reserved b2 – 4ac , 0

Penyelesaian: BAB 2Syarat untuk tiada punca. (–2)2 – 4(1)(p – 6) , 0
b2 – 4ac , 0 Condition for no roots.
62 – 4(1)(p – 4) , 0 4 – 4p + 24 , 0
36 – 4p + 16 , 0 Kesalahan Lazim
52 – 4p , 0 28 – 4p , 0
4p . 52 Tidak songsangkan simbol
p . 13 ketaksamaan apabila 4p . 28
darabkan ketaksamaan
dengan satu nombor negatif. p . 7
Does not change the
inequality sign when multiply
with negative number.

(b) 4x2 + 3x + p = 0 (c) (2p – 1)x2 – 6x + 8 = 0

b2 – 4ac , 0 b2 – 4ac , 0

32 – 4(4)(p) , 0 (–6)2 – 4(2p – 1)(8) , 0

9 – 16p , 0 36 – 64p + 32 , 0

–16p , –9 68 , 64p

p . 9 64p . 68
16
p . 17
16

12. Selesaikan setiap yang berikut. TP 5

Solve of the following.

CONTOH

Cari julat nilai p jika persamaan kuadratik x2 – (p + 5)x + 4 = 0 mempunyai dua punca yang berbeza.

Find the range of values of p if the quadratic equation x2 – (p + 5)x + 4 = 0 has two different roots.

Penyelesaian:

b2 – 4ac . 0 p + 1 = 0 p
[ –(p + 5)]2 – 4(1)(4) . 0 p = –1 –9 –1
p2 + 10p + 25 – 16 . 0
p2 + 10p + 9 . 0 p + 9 = 0
(p + 1)(p + 9) . 0 p = –9

Maka, p , –9  atau  p . –1

Cari julat nilai p jika persamaan kuadratik x2 – 2px + 4p – 3 = 0 tidak mempunyai punca.

Find the range of values of p if the quadratic equation x2 – 2px + 4p – 3 = 0 has no roots.

b2 – 4ac , 0 p – 1 = 0 p
(–2p)2 – 4(1)(4p – 3) , 0 p = 1 13
4p2 – 16p + 12 , 0
p2 – 4p + 3 , 0 p – 3 = 0 Maka, 1 , p , 3
(p – 1)(p – 3) , 0 p = 3

33

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

2.3 Fungsi Kuadratik

Quadratic Functions

BAB 2 NOTA IMBASAN

Penerbitan Pelangi Sdn Bhd All Rights Reserved
1. Bentuk am bagi fungsi kuadratik ialah f(x) = ax2 + bx + c, dengan keadaan a, b dan c adalah pemalar dan a ≠ 0.
The general form of a quadratic function is f(x) = ax2 + bx + c, where a, b and c are constants and a ≠ 0.

2. (a) Jika a . 0, bentuk graf ialah . (b) Jika a , 0, bentuk graf ialah .
If a . 0, the shape of the graph is . If a , 0, the shape of the graph is .

3. Kedudukan graf / Position of the graph

Nilai a b2 – 4ac . 0 b2 – 4ac = 0 b2 – 4ac , 0
Value of a

a.0 x

xx

xx

a,0 x

Persamaan f(x) = 0 mempunyai Persamaan f(x) = 0 mempunyai Persamaan f(x) = 0 tidak

dua punca yang berbeza. dua punca yang sama. mempunyai punca.

The equation f(x) = 0 has two different The equation f(x) = 0 has two equal The equation f(x) = 0 has no roots.
roots. roots.
NOTA IMBASAN
4. Dengan kaedah penyempurnaan kuasa dua f(x) = ax2 + bx + c boleh diungkapkan dalam bentuk f(x) = a(x – h)2 + k di mana

a, h dan k adalah pemalar.

By completing the square, f(x) = ax2 + bx + c can be expressed in the form f(x) = a(x – h)2 + k where a, h and k are constants.

(a) Jika a . 0, fungsi kuadratik mempunyai nilai minimum k apabila x = h dan titik minimum (h, k).

If a . 0, the quadratic function has a minimum value k when x = h and minimum point (h, k).

(b) Jika a , 0, fungsi kuadratik mempunyai nilai maksimum k apabila x = h dan titik maksimum (h, k).

If a , 0, the quadratic function has maximum value k when x = h and maximum point (h, k).

(c) Paksi simetri ialah satu garis menegak yang melalui titik maksimum atau titik minimum.

x = h adalah persamaan paksi simetri.

Axis of symmetry is a vertical line passing through the maximum point or minimum point.

x = h is the equation of axis of symmetry. b
2a
(d) Paksi simetri boleh ditentukan dengan menggunakan x = – .
Axis of symmetry can be determined
by using x = – b .
2a

5. Langkah-langkah untuk melakar graf fungsi kuadratik:

Steps for sketching the graph of quadratic function:

(a) Kenal pasti nilai a dan lakarkan bentuk graf itu.

Identify the value of a and sketch the shape of the graph.

(b) Cari nilai b2 – 4ac untuk menentukan kedudukan graf.

Find the value of b2 – 4ac to determine the position of the graph.

(c) Ungkapkan f(x) = ax2 + bx + c dalam bentuk f(x) = a(x – h)2 + k dengan kaedah penyempurnaan kuasa dua untuk

menentukan titik minimum atau titik maksimum (h, k).

Expressed f(x) = ax2 + bx + c in the form of f(x) = a(x – h)2 + k by completing the square to determine the minimum or maximum point (h,k).

(d) Cari titik persilangan antara graf dengan paksi-y dengan menggantikan x = 0.

Find the point of intersection of the graph with the y-axis by substituting x = 0.

(e) Cari titik persilangan antara graf dengan paksi-x dengan menyelesaikan f(x) = 0.

Find the point of intersection of the graph with the x-axis by solving f(x) = 0.

(f ) Lakarkan graf dengan menyambungkan semua titik diperoleh daripada langkah di atas.

Sketch the graph by joining all the points obtained in the steps above.

34

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

13. Tentukan sama ada setiap fungsi yang berikut ialah fungsi kuadratik atau bukan. TP 1

Determine whether each of the following functions is a quadratic function.

CONTOH (a) f(x) = 3 – 4x – 2x2

f(x) = 5x2 – x + 9

Penyelesaian:
Kuasa tertinggi bagi x ialah 2/Highest power of x is 2
Maka f(x) ialah fungsi kuadratik.

Thus, f(x) is a quadratic function.
Penerbitan Pelangi Sdn Bhd All Rights Reserved Kuasa tertinggi bagi x ialah 2
Maka f(x) ialah fungsi kuadratik.
BAB 2
(b) f(x) = 7x3 – x2 + 8 (c) f(x) = –x – 3x2

Kuasa tertinggi bagi x ialah 3 Kuasa tertinggi bagi x ialah 2
Maka f(x) bukan fungsi kuadratik. Maka f(x) ialah fungsi kuadratik.

14. Bagi setiap fungsi kuadratik berikut, tentukan bentuk graf itu dan tentukan juga jenis punca apabila
f(x) = 0. TP 3

For each of the following quadratic functions, determine the shape of the graph and determine also the type of roots when
f(x) = 0.

CONTOH

(i) f(x) = x2 – 6x + 3 (ii) f(x) = –4x2 + 8x – 4 (iii) f(x) = 3x2 – 9x + 7

a = 1 . 0, bentuk graf ialah/ a = –4 , 0, bentuk graf ialah/ a = 3 . 0, bentuk graf ialah/

shape of graph is shape of graph is shape of graph is

b2 – 4ac = (–6)2 – 4(1)(3) b2 – 4ac = (8)2 – 4(–4)(–4) b2 – 4ac = (–9)2 – 4(3)(7)

= 36 – 12 = 64 – 64 = 81 – 84

= 24 . 0 =0 = –3 , 0

Maka, f(x) = 0 mempunyai Maka, f(x) = 0 mempunyai Maka, f(x) = 0 tidak

dua punca yang berbeza. dua punca yang sama. mempunyai punca.
Thus, f(x) = 0 has two distinct roots.
Thus, f(x) = 0 has two equal roots. Thus, f(x) = 0 has no roots.



(a) f(x) = 2x2 – 7x + 5 (b) f(x) = –4 – 6x – 3x2

a = 2 . 0, bentuk graf ialah a = –3 , 0, bentuk graf ialah

b2 – 4ac = (–7)2 – 4(2)(5) b2 – 4ac = (–6)2 – 4(–3)(–4)

= 49 – 40 = 36 – 48

=9.0 = –12 , 0

Maka, f(x) = 0 mempunyai dua punca yang Maka, f(x) = 0 tidak mempunyai punca.

berbeza.

(c) f(x) = 9x2 – 12x + 4 (d) f(x) = 5x2 + 3x + 1

a = 9 . 0, bentuk graf ialah a = 5 . 0, bentuk graf ialah

b2 – 4ac = (–12)2 – 4(9)(4) b2 – 4ac = (3)2 – 4(5)(1)

= 144 – 144 = 9 – 20

=0 = –11 , 0

Maka, f(x) = 0 mempunyai dua punca yang Maka, f(x) = 0 tidak mempunyai punca.

sama.

35

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

15. Cari julat nilai bagi k jika setiap graf bagi fungsi kuadratik berikut menyilangi paksi-x pada dua titik berlainan.

Find the range of values of k if each of the following graphs of quadratic function intersects the x-axis at two different
points.  TP 4

CONTOH (a) f(x) = (2k – 3)x2 – 4x – 8
BAB 2
f(x) = 3x2 – 8x + k – 6 a = 2k – 3,  b = –4,  c = –8
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Penyelesaian: f(x) mempunyai dua punca yang berlainan

a = 3,  b = –8,  c = k – 6 apabila b2 – 4ac . 0

f(x) mempunyai dua punca yang berlainan apabila (–4)2 – 4(2k – 3)(–8) . 0

f(x) has two distinct roots when 16 + 64k – 96 . 0

b2 – 4ac . 0  64k . 80

(–8)2 – 4(3)(k – 6) . 0 k . 5
4
64 – 12k + 72 . 0

12k , 136

k , 34
3

16. Cari nilai-nilai m jika setiap graf bagi fungsi kuadratik berikut menyilangi paksi-x pada satu titik. TP 4

Find the values of m if each of the following graphs of quadratic function intersects the x-axis at one point.

CONTOH (a) f(x) = x2 + 2mx + m + 6

f(x) = mx2 – 6x + 9 a = 1,  b = 2m,  c = m + 6

Penyelesaian: f(x) mempunyai dua punca yang sama apabila
a = m,  b = –6,  c = 9
f(x) mempunyai dua punca yang sama apabila b2 – 4ac = 0
f(x) has two equal roots when
(2m)2 – 4(1)(m + 6) = 0
b2 – 4ac = 0 
4m2 – 4m – 24 = 0
(–6)2 – 4(m)(9) = 0
36 – 36m = 0 m2 – m – 6 = 0
36m = 36
m = 1 (m – 3)(m + 2) = 0

m – 3 = 0 atau   m + 2 = 0

m = 3 m = –2

17. Cari julat nilai p jika setiap graf fungsi kuadratik berikut tidak menyilangi paksi-x. TP 4

Find the range of values of p if each of the following graphs of quadratic function does not intersects the x-axis.

CONTOH (a) f(x) = x2 + 2(p + 1)x + p2 – 1

f(x) = (2p + 5)x2 – 6x + 9 a = 1,  b = 2(p + 1),  c = p2 – 1

Penyelesaian: f(x) tidak mempunyai punca apabila
a = 2p + 5,  b = –6,  c = 9
f(x) tidak mempunyai punca apabila b2 – 4ac , 0
f(x) has no roots when
[2(p + 1)]2 – 4(1)(p2 – 1) , 0
b2 – 4ac , 0 
4(p + 1)2 – 4p2 + 4 , 0
(–6)2 – 4(2p + 5)(9) , 0
36 – 72p – 180 , 0 4p2 + 8p + 4 – 4p2 + 4 , 0
–72p , 144
p . –2 8p , –8

p , –1

36

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

18. Ungkapkan setiap fungsi kuadratik yang berikut dalam bentuk a(x – h)2 + k. Nyatakan nilai maksimum atau
minimum dan nilai sepadan bagi x. TP 4

Express each of the following quadratic functions in the form a(x – h)2 + k. State the maximum or minimum value and the
corresponding value of x.

CONTOH
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(i) f(x) = 2x2 + 8x – 1 (ii) f(x) = 2 + 6x – 3x2
BAB 2
Penyelesaian: Penyelesaian:

f(x) = 2x2 + 8x – 1 f(x) = –3x2 + 6x + 2

= 2(x2 + 4x) – 1 = –3(x2 – 2x) + 2

= 32 x2 + 4x + 1 4 22 – 1 4 224 – 1 = 3–3 x2 – 2x + 1 –2 22 – 1 –2 224 + 2
2 2 2 2

= 2(x + 2)2 – 8 – 1 = –3(x – 1)2 + 3 + 2

= 2(x + 2)2 – 9 = –3(x – 1)2 + 5

Oleh sebab a . 0, f(x) mempunyai nilai minimum Oleh sebab a , 0, f(x) mempunyai nilai
–9 apabila (x + 2) = 0 iaitu x = –2. maksimum 5 apabila (x – 1) = 0 iaitu x = 1.

Since a . 0, f(x) has minimum value of –9 when Since a , 0, f(x) has maximum value of 5 when
(x + 2) = 0 which is x = –2. (x – 1) = 0 which is x = 1.

(a) f(x) = x2 + 4x – 3 (b) f(x) = –x2 + 5x – 11

f(x) = –(x2 – 5x) – 11

1 2 1 2 4 2 4 2 = 3– x2 – 5x + 1 –5 22 – 1 –5 224 – 11
f(x) = x2 + 4x + 2 2 2 2
– –3

= (x + 2)2 – 4 – 3 1 2 = – x – 5 2 25 – 11
2 4
= (x + 2)2 – 7 +

1 2 5 2 19
Oleh sebab a . 0, f(x) mempunyai nilai minimum = – x – 2 4
–7 apabila (x + 2) = 0 iaitu x = –2. –

Oleh sebab a , 0, f(x) mempunyai nilai maksimum

191 2– x– 5 = 0 iaitu x= 5 .
4 apabila 2 2

(c) f(x) = 2x2 – 8x + 15 (d) f(x) = –2x2 – 12x + 9

f(x) = 2(x2 – 4x) + 15 f(x) = 3–2 x2 + 6x + 1 6 22 – 1 6 224 + 9
2 2
= 32 x2 – 4x + 1 –4 22 – 1 –4 224 + 15
2 2 = –2(x + 3)2 + 18 + 9

= 2(x – 2)2 – 8 + 15 = –2(x + 3)2 + 27

= 2(x – 2)2 + 7 Oleh sebab a , 0, f(x) mempunyai nilai maksimum
27 apabila (x + 3) = 0 iaitu x = –3.
Oleh sebab a . 0, f(x) mempunyai nilai minimum

7 apabila (x – 2) = 0 iaitu x = 2.

(e) f(x) = 4x2 – 8x + 17 (f) f(x) = –5 – 21x – 3x2

f(x) = 4(x2 – 2x) + 17 f(x) = –3x2 – 21x – 5

= 34 x2 – 2x + 1 –2 22 – 1 –2 224 + 17 = –3(x2 + 7x) – 5
2 2
= 3–3 x2 + 7x + 1 7 22 – 1 7 224 – 5
= 4(x – 1)2 – 4 + 17 2 2

= 4(x – 1)2 + 13 7 2 147
2 4
1 2 Oleh sebab a . 0, f(x) mempunyai nilai minimum
= –3 x + + – 5

13 apabila (x – 1) = 0 iaitu x = 1. 1 2 = –3 x + 7 2 + 127
2 4

Oleh sebab a , 0, f(x) mempunyai nilai maksimum

1 2127 x+ 7 = 0 iaitu x = –  27 .
4 2
apabila

37

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

19. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH

y Rajah menunjukkan graf bagi fungsi f(x) = –(x – k)2 – 9 dengan keadaan
O
–11 k ialah pemalar. Cari

The diagram shows the graph of the function f(x) = –(x – k)2 – 9, where k is a
constant. Find

(a) nilai k.

the value of k.

(b) persamaan paksi simetri.

the equation of axis of symmetry.

(c) koordinat titik maksimum.

the coordinates of the maximum point.
BAB 2 x

Penerbitan Pelangi Sdn Bhd All Rights Reservedy = f(x)
(6, –11)

Penyelesaian:

(a) Titik tengah bagi (0, –11) dan (6, –11)/ Midpoint of (0, –11) and (6, –1)

0 + 6 –11 – 11
2 2
1 2 = , = (3, –11)

Pada titik maksimum/ At the maximum point, x = 3

3–k=0
k = 3

(b) Persamaan paksi simetri ialah/ Equation of axis of symmetry is x = 3

(c) f(x) = –(x – 3)2 – 9
Maka, titik maksimum ialah/ Thus, maximum point is (3, –9).

Rajah menunjukkan bentuk bagi graf fungsi kuadratik f(x) = a(x + m)2 + n. Tentukan nilai-nilai a, m dan n.

The diagram shows the shapes of the graph of quadratic function f(x) = a(x + m)2 + n. Determine the values of a, m and n.

(a) f(x) (b) f(x)
–4 3
6 x

O4 x –22
–2

x + m = 0 x + m = 0

4 + m = 0 3 + m = 0

m = –4 m = –3

n = nilai minimum n = nilai maksimum

= –2 = –4

f(x) = a(x – 4)2 – 2 f(x) = a(x – 3)2 – 4

Pada titik (0, 6), 6 = a(0 – 4)2 – 2 Pada titik (0, –22), –22 = a(0 – 3)2 – 4

16a = 8 9a = –18
8
a = 16 a = –2

= 1
2

38

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

20. Lakarkan graf bagi setiap fungsi kuadratik yang berikut. Nyatakan persamaan paksi simetri bagi setiap
graf. TP 5

Sketch the graph of each of the following quadratic functions. State the equation of the axis of symmetry for each graph.

CONTOH 1
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f(x) = x2 + 8x + 12
BAB 2
Penyelesaian: 1 Tentukan bentuk graf. the graph. Apabila/When x = 0 4 Tentukan pintasan-y.
a=1.0 Determine the shape of Determine y-intercept.
b2 – 4ac = (8)2 – 4(1)(12) f(x) = (0)2 + 8(0) + 12 5 Lakar graf.
Sketch the graph.
= 16 . 0 = 12

Maka, graf f(x) berbentuk dengan titik minimum

dan menyilangi paksi-x pada dua titik yang berbeza. f(x)

Thus, graph f(x) has shape with minimum point and

intersect the x-axis at two distinct points.

f(x) = x2 + 8x + 12 12

1 2 1 2 = x2 + 8x + 8 2 8 2 2 Tentukan titik minimum
2 2 atau maksimum.
– + 12
Determine the minimum
= (x + 4)2 – 16 + 12 or maximum point. –6(–4, –4)–2 0 x

= (x + 4)2 – 4

Titik minimum ialah/ Minimum point is (–4, –4).

Apabila/When f(x) = 0, Persamaan paksi simetri ialah/Equation of axis of
x2 + 8x + 12 = 0 symmetry is x = –4.
(x + 2)(x + 6) = 0 3 Tentukan pintasan-x jika ada.
x = –2   atau/or   x = –6 Determine x-intercept if exist.

CONTOH 2

f(x) = –2x2 + 6x – 5

Penyelesaian: Apabila/When x = 0
a = –2 , 0 f(x) = –2(0)2 + 6(0) – 5
= –5
b2 – 4ac = (6)2 – 4(–2)(–5)
f(x)
= –4 , 0
x
Maka, graf f(x) berbentuk dengan titik maksimum 0

dan tidak menyilang paksi-x.

Thus, graph f(x) has shape with maximum point and
does not intersect the x-axis.

f(x) = –2x2 + 6x – 5 ΂–23 , – –12 ΃

= 3–2 x2 – 3x + 1 –3 22 – 1 –3 224 – 5
2 2

1 2 = –2 x– 3 2 9 –5 –5
2 2
+

1 2 = –2 x– 3 2 1
2 2
–

Persamaan paksi simetri ialah/ Equation of axis of
Titik maksimum ialah / Maximum point is 3
symmetry is x = 2 .
1 3 1 2.
2 , – 2

39

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

(a) f(x) = x2 – 6x – 7

a=1.0 Apabila x = 0
f(x) = (0)2 – 6(0) – 7
b2 – 4ac = (–6)2 – 4(1)(–7) = –7

f(x)
BAB 2 = 64 . 0

Penerbitan Pelangi Sdn Bhd All Rights Reserved Maka, graf f(x) berbentuk dengan titik
minimum dan menyilangi paksi-x pada dua
titik yang berbeza.

f(x) = x2 – 6x – 7 –10 x
7
1 2 1 2 = x2 – 6x + –6 2 – –6 2 – 7
2 2
–7

= (x – 3)2 – 9 – 7 (3, –16)

= (x – 3)2 – 16

Titik minimum ialah (3, –16).

Apabila f(x) = 0, Persamaan paksi simetri ialah x = 3.
x2 – 6x – 7 = 0
(x + 1)(x – 7) = 0
x = –1   atau   x = 7

(b) f(x) = –x2 + 6x – 5

a = –1 , 0 Apabila x = 0
f(x) = –(0)2 + 6(0) – 5
b2 – 4ac = (6)2 – 4(–1)(–5) = –5

= 16 . 0

Maka, graf f(x) berbentuk dengan titik f(x)
maksimum dan menyilang paksi-x pada dua titik (3, 4)
yang berbeza.

f(x) = –x2 + 6x – 5 x
5
= 3– x2 – 6x + 1 –6 22 – 1 –6 224 – 5 01
2 2 –5

= –(x – 3)2 + 9 – 5

= –(x – 3)2 + 4

Titik maksimum ialah (3, 4).

Apabila f(x) = 0, Persamaan paksi simetri ialah x = 3.
–x2 + 6x – 5 = 0
x2 – 6x + 5 = 0
(x – 1)(x – 5) = 0
x = 1   atau   x = 5

40

(c) f(x) = 2x2 – 5x + 4   Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

a=2.0 Apabila x = 0
f(x) = 2(0)2 – 5(0) + 4
b2 – 4ac = (–5)2 – 4(2)(4) = 4

= –7 , 0 f(x)

4
Penerbitan Pelangi Sdn Bhd All Rights Reserved Maka, graf f(x) berbentuk dengan titik
minimum dan tidak menyilang paksi-x.
BAB 2
f(x) = 2x2 – 5x + 4

= 32 x2 – 5 x + 1 –5 22 – 1 –5 224 + 4
2 4 4

1 2 = 2 x– 5 2 25 +4 ΂–54 , –78΃
4 8
–

5 2 7 0 x
4 8
1 2 = 2 x– +

1 2 5 7 . Persamaan paksi simetri ialah x = 5 .
Titik minimum ialah 4 , 8 4

PRAKTIS SPM 12 QR code

Kertas 1 2. Cari julat nilai x dengan keadaan fungsi kuadratik
SPM f(x) = 4 + 3x – x2 ialah negatif.
1. Diberi −3 ialah salah satu punca persamaan
SPM kuadratik (x − p)2 = 25, dengan keadaan p ialah 2017 Find the range of value of x such that the quadratic
2015 pemalar. Cari nilai-nilai p. function f(x) = 4 + 3x – x2 is negative.

Given −3 is one of the roots of the quadratic equation f(x) = 4 + 3x – x2
(x − p)2 = 25, where p is a constant. Find the values of p.
4 + 3x – x2 , 0

x2 – 3x – 4 . 0

(x − p)2 = 25 (x – 4)(x + 1) . 0
x − p = ±5

Apabila x = −3,

−3 − p = 5 , −3 − p = −5 –1 4

p = −8 , p = 2 x , –1  atau  x . 4


41

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

3. (a) Diberi bahawa satu daripada punca-punca (a) f(x) = x2 + 12x + h

SPM bagi persamaan kuadratik x2 + (p + 8)x – p2 = 0 = x2 + 12x +  12 2 –  12 2 + h
2017 dengan keadaan p ialah pemalar, adalah 2 2

negatif kepada yang satu lagi. Cari nilai bagi = (x + 6)2 – 36 + h

hasil darab punca.

It is given that one of the roots of the quadratic
equation x2 + (p + 8)x – p2 = 0 where p is a constant,
is negative of the other. Find the value of the product
of roots.

(b) Diberi bahawa persamaan kuadratik

mx2 – 2nx + 4m = 0, dengan keadaan m dan

n ialah pemalar, mempunyai dua punca yang

sama. Cari m : n.

It is given that the quadratic equation
mx2 – 2nx + 4m = 0, where m and n are constants,
has two equal roots. Find m : n.
BAB 2 (b) –36 + h = –20
h = 36 – 20
Penerbitan Pelangi Sdn Bhd All Rights Reserved h = 16

(a) x2 + (p + 8)x – p2 = 0 5. Persamaan kuadratik 2x2 + hx − 6k = 0 mempunyai
BUKAN punca-punca a dan b. Persamaan kuadratik
HTP = a + (–a) = –(p + 8) RUTIN 3x2 + 1 = h + 12x juga mempunyai punca-punca a
KBAT dan b. Cari nilai h dan nilai k.
0 = –p – 8
The quadratic equation 2x2 + hx − 6k = 0 has roots a and
p = –8 b. The quadratic equation 3x2 + 1 = h + 12x also has roots
a and b. Find the value of h and of k.
HDP = –p2
= –(–8)2 Untuk 2x2 + hx − 6k = 0
= –64

(b) mx2 – 2nx + 4m = 0 x2 + hx − 3k = 0
2
b2 – 4ac = 0
h
(–2n)2 – 4(m)(4m) = 0 HTP: − 2 = a + b …… 1

4n2 – 16m2 = 0 HDP: −3k = ab …… 2

16m2 = 4n2

 m 2 = 4 Untuk 3x2 + 1 = h + 12x
n 16
3x2 – 12x + 1 – h = 0
 m 2 =  1  =  1 2
n 4 2 x2 − 4x + 1 1 – h 2 = 0
3
\  m : n = 1 : 2
HTP:  4 = a + b …… 3

Gantikan 3 ke 1: − h = 4
2

h = −8

4. Fungsi kuadratik f ditakrifkan oleh x2 + 12x + h, HDP: 1 – h = ab …… 4
3
SPM dengan keadaan h ialah pemalar.
Gantikan 4 ke 2 : −3k = 1 – h
2017 The quadratic function is defined by x2 + 12x + h, where 3
h is a constant.
−9k = 1 − (−8) h = –8
(a) Ungkapkan f(x) dalam bentuk (x + m)2 + n
–9k = 9
dengan keadaan m dan n ialah pemalar.
k = −1
Express f(x) in the form (x + m)2 + n where m and
n are constants.

(b) Diberi nilai minimum bagi f(x) ialah –20, cari

nilai h.

Given the minimum value of f(x) is –20, find the
value of h.

42

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

6. Rajah 1 menunjukkan pandangan hadapan bagi 7. Fungsi kuadratik f(x) = –x2 + 7x – 10 boleh
SPM empat keping kayu dengan lebar yang sama.
2016 Jumlah luas permukaan hadapan keempat-empat diungkapkan dalam bentuk f(x) = –1x – 7 22 + p,
2
kayu itu ialah 84 cm2. Keempat-empat keping dengan keadaan p ialah pemalar.
kayu itu digunakan untuk menghasilkan sebuah
bingkai lukisan berbentuk segi empat tepat seperti The quadratic function f(x) = –x2 + 7x – 10 can be
yang ditunjukkan dalam Rajah 2. Hitung lebar,
dalam cm, kepingan kayu itu.

Diagram 1 shows the front view of the four pieces of
wood with the same width. The total front area of the
four pieces of wood is 84 cm2. The four pieces of wood are
used to produce a rectangular painting frame as shown
in Diagram 2. Calculate the width, in cm, of the wood.
Penerbitan Pelangi Sdn Bhd All Rights Reserved 1 2expressed in the form of f(x) = – x– 7 2 + p, where p is
2
BAB 2a constant.

(a) Cari nilai p.

Find the value of p.

(b) Lakar graf f(x).

Sketch the graph of f(x).

(a) f(x) = –x2 + 7x – 10

= –(x2 – 7x) – 10

= – 3x2 – 7x + 1– 7 22 – 1– 7 224 – 10
2 2

60 cm 60 cm = – 31x – 7 22 – 49 4 – 10
2 4

= – 1x – 7 22 + 49 – 10
2 4

Rajah 1 / 25 cm = – 1x – 7 22 + 9
Diagram 1 2 4
Rajah 2 /
Diagram 2 Maka, p = 9
4

(b) Titik maksimum ialah 1 7 , 9 2
2 4

60 cm Apabila f(x) = 0

–x2 + 7x – 10 = 0

(25 – 2x) cm (–x + 2)(x – 5) = 0
x cm
x cm x = 2 atau x = 5

Jumlah luas = 84 Apabila x = 0, f(x) = –10

2(x × 60) + 2(x × (25 – 2x)) = 84 ΂ ΃f(x) 7 , 9
120x + 50x – 4x2 – 84 = 0 2 4
–4x2 + 170x – 84 = 0
4x2 – 170x + 84 = 0 O2 5 x
2x2 – 85x + 42 = 0
(2x – 1)(x – 42) = 0 –10

2x – 1 = 0 atau x – 42 = 0

2x = 1 x = 42 (ditolak)

x = 1
2

Lebar ialah 1 cm
2

43

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

8. Diberi fungsi kuadratik f(x) = (2m + 1)x2 – 3mx + (a) Nilai x bagi titik tengah = –2 + 8 =3
2
SPM 2(m – 2), dengan keadaan m ialah pemalar, adalah
2016 h = 3,  m = –5
sentiasa positif apabila m > p atau m < q. Cari nilai
\  Q(3, –5)
p dan nilai q.

Given the quadratic function f(x) = (2m + 1)x2 – 3mxBAB 2 (b) y = a(x – 3)2 – 5

Penerbitan Pelangi Sdn Bhd All Rights Reserved+ 2(m – 2), where m is a constant, is always positive Gantikan x = 8, y = 0

when m . p or m , q. Find the value of p and of q.

0 = a(8 – 3)2 – 5

a = 2m + 1, b = –3m, c = 2m – 4 0 = 25a – 5

b2 – 4ac < 0 25a = 5

(–3m)2 – 4(2m + 1)(2m – 4) < 0 a = 1
5
9m2 – 4(4m2 – 8m + 2m – 4) < 0

9m2 – 16m2 + 24m + 16 < 0

–7m2 + 24m + 16 < 0

7m2 – 24m – 16 > 0 10. Diberi bahawa lengkung y = (k – 3)x2 – 6x + 1,
SPM dengan keadaan k ialah pemalar, bersilang dengan
(7m + 4)(m – 4) > 0 2018 garis lurus y = 2x + 5 pada dua titik. Cari julat

7m + 4 = 0 atau m – 4 = 0 nilai k.

(7m + 4) : – + + It is given that the curve y = (k – 3)x2 – 6x + 1, where k
(m – 4) : – – + is a constant, intersects with the straight line y = 2x + 5
at two points. Find the range of values of k.
+–+
4 4 y = (k – 3)x2 – 6x + 1 …… 1
– 7 y = 2x + 5 …… 2
Gantikan 1 ke dalam 2,
m < – 4 m > 4 2x + 5 = (k – 3)x2 – 6x + 1
7 (k – 3)x2 – 6x – 2x – 5 + 1 = 0
(k – 3)x2 – 8x – 4 = 0
\ p = 4 dan q = – 4 b2 – 4ac . 0
7 (–8)2 – 4(k – 3)(–4) . 0
64 + 16k – 48 . 0
9. Rajah menunjukkan graf y = a(x – h)2 + m, dengan 16k . –16
k . –1
SPM keadaan a, h dan m ialah pemalar. Garis lurus y = –5
2018 ialah tangen kepada lengkung pada titik Q.

Diagram shows the graph y = a(x – h)2 + m, where a,
h and m are constants. The straight line y = – 5 is the

tangent to the curve at point Q.

y 11. Graf fungsi kuadratik g(x) = hx2 + (k − 1)x + 4,

–2 0 x SPM dengan keadaan h dan k ialah pemalar, mempunyai
8 2015 satu titik minimum.

Q The graph of a quadratic function g(x) = hx2 + (k − 1)x + 4,
where h and k are constants, has a minimum point.
(a) Nyatakan koordinat Q.
(a) Nyatakan nilai h jika h ialah suatu integer
State the coordinates of Q.
dengan keadaan −1 < h < 1.
(b) Cari nilai a.
State the value of h if h is an integer such that
Find the value of a. −1 < h < 1.

(b) Dengan menggunakan jawapan di (a), cari

julat nilai k jika graf itu tidak menyilang

paksi-x.

Using the answer from (a), find the range of values
of k if the graph does not intersect the x-axis.

44

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

(a) Diberi –1 < h < 1. Iaitu, h = –1, 0, 1 13. Rajah menunjukkan graf bagi fungsi kuadratik
Suatu graf fungsi f(x) = ax2 + bx + c mempunyai p
SPM f(x) = xn + qx + r, dengan keadaan p, q, r, m dan n
titik minimum (bentuk graf ) jika a . 0. 2015
Jadi, h = 1.
ialah pemalar.

Diagram shows the graph of a quadratic function
Penerbitan Pelangi Sdn Bhd All Rights Reserved 7
(b) Graf fungsi g(x) = x2 + (k − 1)x + 4 tidak f(x) = 2 + qx, where p, q, r, m and n are constants.
BAB 2
menyilang paksi-x. Jadi, graf itu tidak

mempunyai punca nyata. N2001c

b2 − 4ac , 0

(k − 1)2 − 4(1)(4) , 0

k2 − 2k + 1 − 16 , 0

k2 − 2k − 15 , 0

(k − 5)(k + 3) , 0 –3 k
k − 5 = 0 , k + 3 = 0 5
k = 5 , k = −3

\ −3 , k , 5

12. Persamaan kuadratik (px)2 + 7qx + 9 = 0 (a) Nyatakan nilai n.
SPM mempunyai dua punca yang sama. Manakala,
2019 persamaan kuadratik ℎx2 − 2x + 2p = 0 tiada State the value of n.

punca, dengan keadaan ℎ, p dan q ialah pemalar. (b) Jika f(x) = 0 dan hasil darab punca ialah −r,
Ungkapkan julat q dalam sebutan ℎ. nyatakan nilai

The quadratic equation (px)2 + 7qx + 9 = 0 has two equal If f(x) = 0 and the product of roots is −r, state the
roots. Meanwhile, the quadratic equation ℎx2 − 2x + 2p = value of
0 has no roots, where ℎ, p and q are constants. Express
the range of q in terms of ℎ. (i) q
(ii) p

p2x2 + 7qx + 9 = 0 (a) x–n = x2
–n = 2
a = p2,  b = 7q,  c = 9  n = –2

b2 – 4ac = 0 (b) f(x) = 0

(7q)2 – 4(p2)(9) = 0 px2 + qx + r = 0

49q2 – 362 = 0 a = p,  b = q,  c = r

49q2 = 36p2 (i) Hasil tambah punca / Sum of roots

√p2 = 49q2 = m + (–m) = 0
36
b
p = ± 76q …… (1) – a = 0
hx2 – 2x + 2p = 0
– ­qp = 0
a = h,  b = –2,  c = 2p q = 0

b2 – 4ac , 0

(–2)2 – 4(h)(2p) , 0 (ii) Hasil darab punca / Product of roots = –r

4 – 8hp , 0 c = –r
a
4 , 8hp

1 , 2hp r = –r
( 1) → (2) p . 21h …… (2) p
r = –pr

7q . 1 – 7q , 1 r + pr = 0
6 2h 6 2h
r(1 + p) = 0

q . 1 × 6 q , 1 × (– 6 ) r = 0,  p = –1
2h 7 2h 7


q . 3 q , – 3
7h 7h

45

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

14. Graf bagi fungsi kuadratik f(x) = 4[3ℎ − (x − 2)2], (a) x2 – 6(2x – h) = 0

SPM dengan keadaan ℎ ialah pemalar mempunyai titik x2 – 12x + 6h = 0
2019 maksimum (2, ℎ − 11).

The graph of a quadratic function f(x)= 4[3ℎ − (x − 2)2], a = 1, b = –12, c = 6h
where ℎ is a constant, has maximum point (2, ℎ − 11).
BAB 2 m + 3m = – b
(a) Nyatakan nilai h. a
Penerbitan Pelangi Sdn Bhd All Rights Reserved
State the value of ℎ 4m = – 1 –12 2
1
(b) Nyatakan jenis punca bagi f(x) = 0.
4m = 12
Justifikasikan jawapan anda.
m = 3
State the type of roots for f(x) = 0. Justify your
answer.

(a) f(x) = 12h – 4(x – 2)2 m × 3m = c
a
Nilai maksimum / Maximum value = 12h

h – 11 = 12h 3m2 = 6h
1
11h = –11

h = –1 3 × 32 = 6h

(b) f(x) = 0, –12 – 4(x – 2)3 = 0 6h = 27

–12 – 4(x2 – 4x + 4) = 0 h = 27
6
–12 – 4x2 + 16x – 16 = 0
9
–4x2 + 16x – 28 = 0 = 2

–x2 + 4x – 7 = 0

a = –1,  b = 4,  c = –7 (b) Punca-punca baharu ialah 6 dan –1.

b2 – 4ac = 42 – 4(–1)(–7) HTP baharu = 6 + (–1)

= 16 – 18 = –12 =5

b2 – 4ac , 0 HDP baharu = 6 × (–1)

Tidak ada punca nyata / No real roots = –6

Persamaan baharu ialah

x2 – (5)x + (–6) = 0

x2 – 5x – 6 = 0

Kertas 2 2. Persamaan kuadratik k − 4x = x2 − x + 1, dengan

1. Persamaan kuadratik x2 – 6(2x – h) = 0, dengan SPM keadaan k ialah pemalar, mempunyai punca-
keadaan h ialah pemalar mempunyai punca- 2015 punca a dan b.
punca m dan 3m, m ≠ 0.
The quadratic equation k − 4x = x2 − x + 1, where k is a
A quadratic equation x2 – 6(2x – h) = 0, where h is a
constant, has roots a and b.
constant has roots m and 3m, m ≠ 0.
(a) Cari julat nilai k jika a ≠ b.
(a) Cari nilai m dan nilai h.
Find the range of values of k if a ≠ b.
Find the value of m and of h.
(b) Diberi a + 1 dan b + 1 adalah punca-punca
(b) Seterusnya, bentukkan persamaan kuadratik
yang mempunyai punca-punca m + 3 dan bagi satu lagi persamaan kuadratik
m – 4.
2x2 − hx + 4 = 0, dengan keadaan h ialah
Hence, form the quadratic equation with the roots
m + 3 and m – 4. pemalar. Cari nilai k dan nilai h.
Given a + 1 and b + 1 are the roots of another
quadratic equation 2x2 − hx + 4 = 0, where h is a

constant. Find the value of k and of h.

46

(a) x2 − x + 1 = k − 4x   Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

x2 − x + 4x + 1 − k = 0 3. Fungi kuadratik f(x) = 2x2 – 8x + k mempunyai
nilai minimum 5 apabila x = h.
x2 + 3x + 1 − k = 0
The quadratic function f(x) = 2x2 – 8x + k has a minimum
value of 5 when x = h.

(a) Cari nilai h dan nilai k.

Find the value of h and of k.

(b) Seterusnya, dengan menggunakan nilai h dan
nilai k di (a), lakarkan graf f(x) = 2x2 – 8x + k.

Hence, by using the value of h and of k in (a), sketch
the graph of f(x) = 2x2 – 8x + k.
a ≠ b bermaksud dua punca berbezaPenerbitan Pelangi Sdn Bhd All Rights Reserved

BAB 2b2 − 4ac . 0

32 − 4(1)(1 − k) . 0

9 − 4 + 4k . 0

5 + 4k . 0

4k . −5

k . – 5 (a) f(x) = 2x2 – 8x + k
4 = 2(x2 – 4x) + k

(b) Untuk 2x2 − hx + 4 = 0 = 23x2 – 4x + 1 –4 22 – 1 –4 224 + k
2 2

x2 − h x + 2 = 0 = 2[(x – 2)2 – 4] + k
2
= 2(x – 2)2 – 8 + k

HTP: h2 = (a + 1) + (b + 1) –8 + k = 5

h k = 13
2
− 2 = a + b …… 1 x – 2 = 0

x = 2

HDP: 2 = (a + 1)(b + 1) \  h = 2

2 = ab + a + b + 1 …… 2

(b) f(x) = 2(x – 2)2 + 5

Untuk x2 + 3x + 1 − k = 0,   a = 2 > 0
HTP: −3 = a + b …… 3
HDP: 1 − k = ab …… 4 f(x) mempunyai nilai minimum

Titik minimum ialah (2, 5)

f(x) = 2x2 – 8x + 13

Gantikan 3 ke 1: h − 2 = −3 b2 – 4ac = (–8)2 – 4(2)(13)
2
h = –40 < 0
2
= −1 Graf f(x) tidak menyilang paksi-x.

h = −2 Apabila x = 0, f(x) = 2(0)2 – 8(0) + 13

= 13

 Gantikan 4 dan 3 ke 2: 2 = (1 − k) + (−3) + 1 f(x)

2 = −1 − k

k = −3 13

(2, 5) x
0

47

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 5. Lengkung fungsi kuadratik f(x) = –2(x + h)2 + 2k

4. Diberi persamaan kuadratik p(x2 + 16) = –4qx SPM menyilang paksi-x pada titik-titik (2, 0) dan
BUKAN mempunyai dua punca yang sama, cari nisbah 2016 (6, 0). Garis lurus y = 8 menyentuh titik maksimum
RUTIN p : q. Seterusnya, selesaikan persamaan itu.
lengkung itu.
Given that the quadratic equation p(x2 + 16) = – 4qx
has two equal roots, find the ratio p : q. Hence, solve the
equation.
BAB 2 The curve of a quadratic function f(x) = –2(x + h)2 + 2k
intersects the x-axis at points (2, 0) and (6, 0). The
Penerbitan Pelangi Sdn Bhd All Rights Reserved straight line y = 8 touches the maximum point of the

p(x2 + 16) = –4qx curve.

px2 + 4qx + 16p = 0 (a) Tentukan nilai h dan nilai k.

b2 – 4ac = 0 Determine the value of h and of k.

(b) Seterusnya, lakar graf f(x) untuk 0 < x < 7.

(4q)2 – 4p(16p) = 0 Hence, sketch the graph of f(x) for 0 < x < 7.

16q2 – 64p2 = 0 (c) Nyatakan persamaan bagi lengkung itu jika

64p2 = 16q2 graf itu dipantulkan pada paksi-x.

p2 16 State the equation of the curve if the graph is
q2 64
= reflected in the x-axis.

= 1 (a) Paksi simetri, x = 2+6
4 2
=4

1 p 22 = 1 1 22 x + h = 0
q 2
4 + h = 0

p : q = 1 : 2 h = – 4

2k = 8

Apabila q = 2p, k = 4
p(x2 + 16) = –4(2p)x
x2 + 16 = –8x (b) f(x) = –2(x – 4)2 + 8
x2 + 8x + 16 = 0 = –2(x2 – 8x + 16) + 8
(x + 4)(x + 4) = 0 = –2x2 + 16x – 32 + 8
x + 4 = 0 = –2x2 + 16x – 24
x = –4 \ Apabila x = 0, f(x) = –24

x = 7, f(x) = –2(7)2 + 16(7) – 24

= –10

f(x)
8

0 2 67 x
–10

–24

(c) Persamaan baru ialah
f(x) = –[–2(x – 4)2 + 8]
= 2(x – 4)2 – 8

48

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

Sudut KBAT

1. ABCD ialah sebuah segi empat tepat dengan panjang 5x cm dan lebar (4 – x) cm. KBAT Ekstra

ABCD is a rectangle with a length of 5x cm and a width of (4 – x) cm.
Penerbitan Pelangi Sdn Bhd All Rights Reserved
5x
BAB 2 4–x

Cari perimeter, dalam cm, segi empat ABCD jika luas ABCD adalah maksimum.
Seterusnya, nyatakan nilai luas yang maksimum, dalam cm2, bagi segi empat ABCD.

Find the perimeter, in cm, of the rectangle ABCD if the area of ABCD is a maximum.
Hence, state the maximum value of the area, in cm2, of the rectangle ABCD.

Katakan luas segi empat tepat = f(x)

f(x) = 5x(4 – x)

= –5x2 + 20x

= –5(x2 – 4x) 4 4
2 2
= –53x2 – 4x + 1– 22 – 1– 224

= –5(x – 2)2 + 20

a = –5 < 0, maka, f(x) mempunyai nilai maksimum

x–2=0

\  x = 2
Perimeter = 2(5x) + 2(4 – x)

= 2(5 × 2) + 2(4 – 2)
= 24 cm

Luas maksimum = 5(2)(4 – 2)

= 20 cm2

2. (a) Cari julat nilai-nilai m dengan keadaan fungsi f(x) = 2x2 – 7x + m adalah sentiasa positif bagi semua nilai x.

Find the range of values of m such that the function f(x) = 2x2 – 7x + m is always positive for all values of x.

(b) Tunjukkan fungsi g(x) = 3x – 8 – 4x2 adalah sentiasa negatif bagi semua nilai x.
Show that the function g(x) = 3x – 8 – 4x2 is always negative for all values of x.

(a) f(x) = 2x2 – 7x + m

a = 2, b = –7, c = m

b2 – 4ac < 0

(–7)2 – 4(2)(m) < 0

49 – 8m < 0

8m > 49
49
m > 8

(b) g(x) = – 4x2 + 3x – 8 y x
a = –4, a < 0, graf maksimum O
b2 – 4ac = 32 – 4(–4)(–8)
= 9 – 128 Kuiz 1
= –119 < 0, tidak mempunyai punca.
g(x) adalah sentiasa negatif.

49

MoMduOlRe E& 4TINGKATAN RC184131S Module & more MATEMATIK TAMBAHAN TINGKATAN 4

KSSM

MATEMATIK TAMBAHAN Additional Mathematics
Penerbitan Pelangi Sdn Bhd All Rights Reserved
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NOTA IMBASAN CONTOH TERKERJA KESALAHAN LAZIM

Diselitkan dalam teks untuk Menunjukkan langkah-langkah Menunjukkan kesilapan yang biasa
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PETA KONSEP secara sistematik SOALAN KLON SPM

Mengintegrasikan konsep-konsep PRAKTIS SPM
sesuatu bab dan mengukuhkan
Menyediakan latihan berorientasikan Mendedahkan murid kepada soalan
kefahaman peperiksaan di akhir setiap bab berpiawai SPM

KBAT/i-THINK JAWAPAN LENGKAP DWIBAHASA

Menerapkan keperluan terkini Membantu murid menyemak jawapan Meningkatkan pemahaman teks
melalui peta i-THINK dan KBAT untuk membina keyakinan diri melalui penggunaan bahasa Melayu

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