Modul A+1 Matematik Tingkatan 4 KSSM

4 DWIBAHASA Tingkatan Matematik Mathematics Chai Mun Pan Asia Publications Sdn. Bhd. 199101016590 (226902-X) No. 2-16, Jalan SU 8, Taman Perindustrian Subang Utama, Seksyen 22, 40300 Shah Alam, Selangor Darul Ehsan, Malaysia. Tel: +603-5614 4168 Faks: +603-5614 4268 E-mel: [email protected] Laman web: www.panasiapub.com © Pan Asia Publications Sdn. Bhd. Hak cipta terpelihara. Sebarang bahagian dalam buku ini tidak boleh diterbitkan semula, disimpan dalam cara yang boleh dipergunakan lagi ataupun dipindahkan dalam sebarang bentuk atau cara, baik secara elektronik, mekanik, gambar, rakaman atau sebagainya, tanpa kebenaran daripada penerbit. Cetakan Pertama 2022 MODUL A+1 MATEMATIK Tingkatan 4 ISBN 978-967-466-625-5 Dicetak oleh World Line Marketing Sdn. Bhd. (1115599-K) • PDF Manual Guru • Rancangan Pengajaran Tahunan Bonus Guru

ii BAB 1 Fungsi dan Persamaan Kuadratik dalam Satu Pemboleh Ubah Quadratic Functions and Equations in One Variable 1.1 Fungsi dan Persamaan Kuadratik Quadratic Functions and Equations...........................1 Soalan Berformat SPM.....................................................13 BAB 2 Asas Nombor Number Bases 2.1 Asas Nombor/Number Bases ....................................15 Soalan Berformat SPM.....................................................30 BAB 3 Penaakulan Logik Logical Reasoning 3.1 Pernyataan/Statements...............................................32 3.2 Hujah/Argument ........................................................42 Soalan Berformat SPM.....................................................53 BAB 4 Operasi Set Operations on Sets 4.1 Persilangan Set/Intersection of Sets ..........................58 4.2 Kesatuan Set/Union of Sets.......................................63 4.3 Gabungan Operasi Set Combined Operations on Sets...................................69 Soalan Berformat SPM.....................................................73 BAB 5 Rangkaian dalam Teori Graf Network in Graph Theory 5.1 Rangkaian/Network ...................................................77 Soalan Berformat SPM.....................................................87 BAB 6 Ketaksamaan Linear dalam Dua Pemboleh Ubah Linear Inequalities in Two Variables 6.1 Ketaksamaan Linear dalam Dua Pemboleh Ubah Linear Inequalities in Two Variables.........................91 6.2 Sistem Ketaksamaan Linear dalam Dua Pemboleh Ubah Systems of Linear Inequalities in Two Variables.......99 Soalan Berformat SPM...................................................104 BAB 7 Graf Gerakan Graphs of Motion 7.1 Graf Jarak-Masa/Distance-Time Graphs.................107 7.2 Graf Laju-Masa/Speed-Time Graphs.......................111 Soalan Berformat SPM...................................................120 BAB 8 Sukatan Serakan Data Tak Terkumpul Measures of Dispersion for Ungrouped Data 8.1 Serakan/Dispersion .................................................124 8.2 Sukatan Serakan/Measures of Dispersion...............127 Soalan Berformat SPM...................................................144 BAB 9 Kebarangkalian Peristiwa Bergabung Probability of Combined Events 9.1 Peristiwa Bergabung/Combined Events ..................147 9.2 Peristiwa Bersandar dan Peristiwa Tak Bersandar Dependent Events and Independent Events.............150 9.3 Peristiwa Saling Eksklusif dan Peristiwa Tidak Saling Eksklusif Mutually Exclusive Events and Non-Mutually Exclusive Events......................................................155 9.4 Aplikasi Kebarangkalian Peristiwa Bergabung Application of Probability of Combined Events......160 Soalan Berformat SPM...................................................164 BAB 10 Matematik Pengguna: Pengurusan Kewangan Consumer Mathematics: Financial Management 10.1 Perancangan dan Pengurusan Kewangan Financial Planning and Management.....................168 Soalan Berformat SPM...................................................173 Jawapan ...................................................................... MG–1 Lembaran Pentaksiran Bilik Darjah (PBD)............ MG–9 00_Modul A+ MM Tg4_Kand.indd 2 01/10/2021 3:16 PM

Nota 1. Ungkapan kuadratik dalam satu pemboleh ubah ialah suatu ungkapan yang berbentuk ax2 + bx + c dengan a, b, c sebagai pemalar, a ≠ 0 dan x ialah pemboleh ubah. A quadratic expression in one variable is an expression of the form ax2 + bx + c with a, b, c as constants, a ≠ 0 and x is a variable. Contoh/Example, 2x2 + 3x + 5, x2 – 7x + 4, 4x2 – 9, 8x – x2 2. Ciri-ciri ungkapan kuadratik dalam satu pemboleh ubah. Characteristics of quadratic expressions in one variable. • Ungkapan mempunyai hanya satu pemboleh ubah. The expression has only one variable. • Kuasa pemboleh ubah ialah suatu nombor bulat. The power of the variable is a whole number. • Kuasa tertinggi bagi pemboleh ubah ialah 2. The highest power of the variable is 2. Tip SPM Pemboleh ubah x dalam ungkapan kuadratik juga boleh diwakili oleh huruf-huruf abjab yang lain. The variable x in quadratic expressions can also be represented by other alphabet letters. 3. Fungsi kuadratik ialah suatu hubungan banyak-kepada-satu. A quadratic function is a many-to-one relation. Ujian garis mengufuk Horizontal line test Satu garis lurus mengufuk memotong graf fungsi pada satu titik. A horizontal line cuts the graph of the function at one point. Fungsi ialah hubungan satu-kepadasatu. Function is a one-to-one relation. Satu garis lurus mengufuk memotong graf fungsi pada dua titik. A horizontal line cuts the graph of the function at two points. Fungsi ialah hubungan banyakkepada-satu. Function is a many-to-one relation. 4. Ciri-ciri fungsi kuadratik. Characteristics of quadratic functions. • Graf berbentuk melengkung atau . Graph is curved shape or . • Satu titik maksimum atau satu titik minimum. One maximum point or one minimum point. • Paksi simetri graf adalah selari dengan paksi-y. Axis of symmetry of the graph is parallel to the y-axis. Fungsi dan Persamaan Kuadratik dalam Satu Pemboleh Ubah Quadratic Functions and Equations in One Variable 1.1 Fungsi dan Persamaan Kuadratik Quadratic Functions and Equations BAB 1 Buku Teks m.s. 2–29 1 01_Modul A+ MM Tg4.indd 1 12/10/2021 3:31 PM

Bentuk graf Shape of graph y = ax2 + bx + c Nilai-nilai a yang positif Positive values of a Nilai-nilai a yang negatif Negative values of a Titik minimum Minimum point Titik maksimum Maximum point 5. Punca suatu persamaan kuadratik ialah nilai bagi pemboleh ubah yang memuaskan persamaan kuadratik itu. A root of a quadratic equation is the value of the variable that satisfies the quadratic equation. Contoh/Example, Apabila/When x = 1, 3x2 – 4x + 1 = 3(1)2 – 4(1) + 1 = 3 – 4 + 1 = 0 ∴ x = 1 ialah satu punca bagi persamaan kuadratik 3x2 – 4x + 1 = 0. ∴ x = 1 is a root of the quadratic equation 3x2 – 4x + 1 = 0. 6. Punca-punca bagi persamaan kuadratik boleh ditentukan dengan kaedah pemfaktoran. The roots of quadratic equations can be determined by method of factorisation. Contoh/Example, 2x2 + 5x – 12 = 0 (2x – 3)(x + 4) = 0 2x – 3 = 0 atau/or x + 4 = 0 x = —3 2 atau/or x = –4 7. 2x –3 –3x x +4 +8x 2x2 –12 +5x Graf fungsi kuadratik Graphs of quadratic functions y = ax2 + c (i) a . 0 (ii) a , 0 y = ax2 + bx (i) a . 0 (ii) a , 0 y = a(px + m) 2 (i) a . 0 (ii) a , 0 y = a(px + m)(qx + n) (i) a . 0 (ii) a , 0 y x O c y x O c y x O b – — a y x O b – — a y x O m– — p y x O m– — p y x m O – — p n – — q y x O n – — q m– — p 2 BAB 1 01_Modul A+ MM Tg4.indd 2 12/10/2021 3:31 PM

1. Tandakan ✓ bagi ungkapan kuadratik atau ✗ bagi bukan ungkapan kuadratik dalam satu pemboleh ubah. Mark ✓ for quadratic expression or ✗ for not quadratic expression in one variable. TP 1 Contoh/Example 5t 2 – 7t – 2 ✓ (a) 4q – —v 2 + —1 v ✗ (b) –3w2 + w3 ✗ (c) x2 + y2 ✗ (d) 8n2 + 15 ✓ (e) 2m2 + mn – 4 ✗ 2. Padankan setiap yang berikut. Match each of the following. TP 1 Contoh/Example 4 – a2 6k + 32 8r2 + 10r 15 – 2v (i) Hubungan satu-kepada-satu One-to-one relation (ii) Hubungan banyak-kepada-satu Many-to-one relation (a) (b) (c) 3. Tandakan ✓ untuk menunjukkan ciri-ciri fungsi kuadratik f(x) = ax2 + bx + c. Mark ✓ to show the characteristics of quadratic function f(x) = ax2 + bx + c. TP 2 (a) Bentuk graf/Shape of graph (i) a . 0 (ii) a , 0 ✓ ✓ (b) Paksi simetri graf/Axis of symmetry of graph Selari dengan paksi-x/Parallel to the x-axis Selari dengan paksi-y/Parallel to the y-axis ✓ 4. (a) Tentukan nilai-nilai a atau c yang sepadan dengan graf I dan II berikut. Determine the values of a or c that correspond to the following graphs I and II. TP 2 (i) Graf/Graph I Graf/Graph II a . 1 0 , a , 1 y = ax2 x y O y = x2 I II (ii) Graf/Graph I Graf/Graph II c . 0 c , 0 y = x2 + c I II y x O y = x2 PAUTAN INTERAKTIF PAUTAN INTERAKTIF Persamaan kuadratik dalam satu pemboleh ubah https://bit.ly/3d1ZKeM 3 BAB 1 01_Modul A+ MM Tg4.indd 3 12/10/2021 3:31 PM

(b) Tandakan ✓ untuk menunjukkan kesan perubahan nilai a dan nilai c ke atas graf fungsi kuadratik di (a). Mark ✓ to show the effect of the change of values of a and c on the graph of the quadratic functions in (a). TP 2 (i) Apabila nilai a bertambah, lebar graf y = ax2 When the value of a increases, the width of the graph y = ax2 bertambah lebih sempit daripada graf y = x2 ✓ increases narrower than the graph y = x2 bertambah lebih luas daripada graf y = x2 increases wider than the graph y = x2 (ii) Apabila nilai c (. 0) bertambah, graf y = x2 + c When the value of c (. 0) increases, the graph y = x2 + c berada c unit ke atas graf y = x2 ✓ lies c units above the graph y = x2 berada c unit ke bawah graf y = x2 lies c units below the graph y = x2 (iii) Apabila nilai c (, 0) berkurang, graf y = x2 + c When the value of c (, 0) decreases, the graph y = x2 + c berada c unit ke bawah graf y = x2 ✓ lies c units below the graph y = x2 berada c unit ke atas graf y = x2 lies c units above from the graph y = x2 5. (a) Rajah menunjukkan graf fungsi y = x2 + bx. Tentukan nilai-nilai b yang sepadan dengan graf I dan II. Diagram shows the graph of the function y = x2 + bx. Determine the values of b that correspond to graphs I and II. TP 2 Graf/Graph I Graf/Graph II b , 0 b . 0 x O y y = x II 2 I (b) Tandakan ✓ untuk menunjukkan paksi simetri bagi Mark ✓ to show the axis of symmetry for (i) graf/graph I (ii) graf/graph II x = —b 2 x = —b 2 x = – —b 2 ✓ x = – —b 2 ✓ 6. Bentukkan satu ungkapan kuadratik bagi luas dalam setiap rajah yang berikut. Form a quadratic expression for the area in each of the following diagrams. TP 3 Contoh/Example (4x + 8) cm (x + 3) cm Luas/Area = —1 2 × (4x + 8) × (x + 3) = —1 2 (4x2 + 12x + 8x + 24) = —1 2 (4x2 + 20x + 24) = (2x2 + 10x + 12) cm2 (a) (18 – x) cm x cm Luas/Area = (18 – x) × x = (18x – x2 ) cm2 (b) (x + 5) cm x cm Luas/Area = x2 + 2x(x + 5) = x2 + 2x2 + 10x = (3x2 + 10x) cm2 4 BAB 1 01_Modul A+ MM Tg4.indd 4 12/10/2021 3:31 PM

7. Bentukkan persamaan kuadratik daripada maklumat yang diberikan. Form a quadratic equation from the given information. TP 3 Contoh/Example (x + 6) cm (2x – 3) cm A B D C Luas segi empat tepat ABCD = 50 cm2 Area of rectangle ABCD = 50 cm2 (x + 6)(2x – 3) = 50 2x2 – 3x + 12x – 18 = 50 2x2 + 9x – 68 = 0 (a) (x + 5) cm A (3x + 2) cm B C Luas segi tiga ABC = 85 cm2 Area of triangle ABC = 85 cm2 —1 2 × (x + 5) × (3x + 2) = 85 3x2 + 2x + 15x + 10 = 170 3x2 + 17x – 160 = 0 (b) (x + 7) cm (x + 2) cm x cm A B D C Luas trapezium ABCD = 162 cm2 Area of trapezium ABCD = 162 cm2 —1 2 × 3(x + 7) + x4 × (x + 2)= 162 (2x + 7)(x + 2) = 324 2x2 + 4x + 7x + 14 = 324 2x2 + 11x – 310 = 0 8. (a) Lengkapkan rajah berikut. Complete the following diagram. TP 3 x = 1 (i) x = –1 (ii) x = 2 = 12 – 1 – 2 = 1 – 1 – 2 = –2 = 22 – 2 – 2 = 4 – 2 – 2 = 0 = (–1)2 – (–1) – 2 = 1 + 1 – 2 = 0 Nilai bagi/Value of x2 – x – 2 Contoh/Example (b) Tandakan ✓ bagi nilai-nilai x yang merupakan punca persamaan kuadratik x2 – x – 2 = 0. Mark ✓ for the values of x that are the roots of the quadratic equation x2 – x – 2 = 0. x = 1 x = –1 ✓ x = 2 ✓ 9. Cari punca-punca bagi persamaan kuadratik yang berikut. Find the roots of the following quadratic equations. TP 3 Contoh/Example (x – 7)(3x + 1) = 0 x – 7 = 0 atau/or 3x + 1 = 0 x = 7 atau/or x = – —1 3 (a) x(2x + 3) = 0 x = 0 atau/or 2x + 3 = 0 x = 0 atau/or x = – — 3 2 (b) (x – 1)(x – 4) = 0 x – 1 = 0 atau/or x – 4 = 0 x = 1 atau/or x = 4 10. Selesaikan./Solve. TP 3 Contoh/Example (4k + 5)(2k – 3) = 0 4k + 5 = 0 atau/or 2k – 3 = 0 4k = –5 atau/or 2k = 3 k = – —5 4 atau/or k = —3 2 (a) (4f – 1)(3f + 2) = 0 4f – 1 = 0 atau/or 3f + 2 = 0 4f = 1 atau/or 3f = –2 f = — 1 4 atau/or f = – — 2 3 (b) (5 – 3p)(1 + 5p) = 0 5 – 3p = 0 atau/or 1 + 5p = 0 3p = 5 atau/or 5p = –1 p = —5 3 atau/or p = – —1 5 (c) (3 + 2m)(2 + 7m) = 0 3 + 2m = 0 atau/or 2 + 7m = 0 2m = –3 atau/or 7m = –2 m = – —3 2 atau/or m = – —2 7 5 BAB 1 01_Modul A+ MM Tg4.indd 5 12/10/2021 3:31 PM

11. Selesaikan dengan kaedah pemfaktoran. Solve by factorisation method. TP 3 Contoh/Example 2x2 – 5x – 12 = 0 (2x + 3)(x – 4) = 0 2x + 3 = 0 atau/or x – 4 = 0 x = – —3 2 atau/or x = 4 (a) x2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x – 1 = 0 atau/or x – 3 = 0 x = 1 atau/or x = 3 (b) 2k2 + 7k + 3 = 0 (2k + 1)(k + 3) = 0 2k + 1 = 0 atau/or k + 3 = 0 k = – — 1 2 atau/or k = –3 (c) 3p2 – 2p = 0 p(3p – 2) = 0 p = 0 atau/or 3p – 2 = 0 p = 0 atau/or p = — 2 3 12. Cari punca-punca bagi persamaan kuadratik yang berikut. Find the roots of the following quadratic equations. TP 3 Contoh/Example 5x(x – 3) = 2(x – 7) 5x2 – 15x = 2x – 14 5x2 – 17x + 14 = 0 (5x – 7)(x – 2) = 0 5x – 7 = 0 atau/or x – 2 = 0 x = — 7 5 atau/or x = 2 (a) 2x(2x + 1) = 2x + 1 4x2 + 2x = 2x + 1 4x2 – 1 = 0 (2x + 1)(2x – 1) = 0 2x + 1 = 0 atau/or 2x – 1 = 0 x = – — 1 2 atau/or x = — 1 2 (b) (7 – 2f)2 = 9 49 – 28f + 4f 2 = 9 4f 2 – 28f + 40 = 0 f 2 – 7f + 10 = 0 (f – 2)(f – 5) = 0 f – 2 = 0 atau/or f – 5 = 0 f = 2 atau/or f = 5 (c) (r + 3)(r – 3) = 7 r2 – 9 = 7 r2 – 16 = 0 (r + 4)(r – 4) = 0 r + 4 = 0 atau/or r – 4 = 0 r = –4 atau/or r = 4 13. Lakar graf bagi setiap fungsi kuadratik berikut. Sketch the graph for each of the following quadratic functions. TP 3 Contoh/Example y = 2x2 y x O (a) y = –3x2 y x O (b) y = 4x2 + 3 y x 3 O (c) y = – —1 2 x2 – 2 y x –2 O 6 BAB 1 01_Modul A+ MM Tg4.indd 6 12/10/2021 3:31 PM

14. Lakar graf bagi fungsi kuadratik berikut. Sketch the graph of the following quadratic functions. TP 3 Contoh/Example y = x2 – 9 y x O –9 –3 3 (a) y = 2x2 – 8 y x O –8 –2 2 (b) y = 16 – x2 y x O 16 –4 4 (c) y = —1 2 (25 – x2 ) y x O —– 25 2 –5 5 15. Lakar graf bagi setiap fungsi berikut. Sketch the graph for each of the following fucntions. TP 3 Contoh/Example y = (x – 1)(x + 3) 1 –3 –3 y x O (a) y = 2x – x2 2 y x O (b) y = 2x2 + 15x + 25 –5 5 – — 2 y x O (c) y = 24 – x – 3x2 –3 24 —8 3 y x O 16. Lakar graf fungsi berikut. Sketch the graph of the following functions. TP 3 Contoh/Example y = (x – 3)2 y x O 9 3 (a) y = 2(x + 2)2 y x O 8 –2 (b) y = –x2 – 8x – 16 y x –4 O –16 (c) y = –4x2 + 28x – 49 y x O —7 2 –49 Tip SPM Nilai c ialah pintasan-y bagi graf y = x2 + c. The value of c is the y-intercept of the graph y = x2 + c. 7 BAB 1 01_Modul A+ MM Tg4.indd 7 12/10/2021 3:31 PM

17. Rajah yang diberi menunjukkan sebuah segi empat tepat PQRS. Diberi MS = NP = x cm dan luas segi tiga MNQ ialah 38 cm2 . Hitung nilai-nilai x yang mungkin. The diagram given shows a rectangle PQRS. Given MS = NP = x cm and the area of triangle MNQ is 38 cm2 . Calculate the possible values of x. TP 4 Luas segi empat tepat PQRS/ Area of rectangle PQRS = 12 × 9 = 108 cm2 Luas ∆PQN/ Area of ∆PQN = — 1 2 x(12) = 6x cm2 Luas ∆MNS/ Area of ∆MNS = — 1 2 x(9 – x) = (— 9 2 x – — 1 2 x2 ) cm2 Luas ∆QRM/ Area of ∆QRM = —1 2 (9)(12 – x) = (54 – —9 2 x) cm2 108 – 6x – (—9 2 x – —1 2 x2 ) – (54 – —9 2 x) = 38 54 – 6x + —1 2 x2 = 38 —1 2 x2 – 6x + 16 = 0 x2 – 12x + 32 = 0 (x – 4)(x – 8) = 0 x = 4 atau/or x = 8 18. Sebidang lot banglo yang berbentuk segi empat tepat mempunyai perimeter 120 m dan luas 884 m2 . Cari panjang dan lebarnya. A rectangular bungalow lot has perimeter 120 m and area 884 m2 . Find its length and width. TP 5 2(x + y) = 120 x + y = 60 y = 60 – x xy = 884 x(60 – x) = 884 60x – x2 = 884 x2 – 60x + 884 = 0 (x – 34)(x – 26) = 0 x = 34 atau/or x = 26 Jika / If x = 34, y = 60 – 34 = 26 Jika / If x = 26, y = 60 – 26 = 34 ∴ Panjang ialah 34 m dan lebar ialah 26 m. ∴ Length is 34 m and width is 26 m. 19. Tentukan nilai positif x yang ditunjukkan dalam rajah berikut. Determine the positive value of x as shown in the following diagrams. TP 5 Contoh/Example 40 cm (x – 1)(x + 2) cm (x – 1)(x + 2) = 40 x2 + 2x – x – 2 = 40 x2 + x – 42 = 0 (x – 6)(x + 7) = 0 x – 6 = 0 atau/or x + 7 = 0 x = 6 atau/or x = –7 ∴ Nilai positif x ialah 6. ∴ The positive value of x is 6. (a) (2x + 3) cm x cm (x + 7) cm (2x + 3)2 = x2 + (x + 7)2 4x2 + 12x + 9 = x2 + x2 + 14x + 49 2x2 – 2x – 40 = 0 x2 – x – 20 = 0 (x – 5)(x + 4) = 0 x – 5 = 0 atau/or x + 4 = 0 x = 5 atau/or x = –4 ∴ Nilai positif x ialah 5. ∴ The positive value of x is 5. 12 cm 9 cm P Q S M R N x m y m KBAT Menganalisis 8 BAB 1 01_Modul A+ MM Tg4.indd 8 12/10/2021 3:31 PM

20. Rajah menunjukkan graf bagi suatu fungsi kuadratik y = ax2 + bx + c. Diagram shows the graph of a quadratic function y = ax2 + bx + c. TP 6 y x –1 O 5 (3, –8) (a) Tentukan nilai-nilai bagi a, b dan c. Determine the values of a, b and c. y = a(x + 1)(x – 5) x = 3, y = –8, –8 = a(3 + 1)(3 – 5) –8 = a(4)(–2) a = 1 y = (x + 1)(x – 5) = x2 – 4x – 5 b = –4, c = –5 (b) Seterusnya, lakar graf bagi fungsi kuadratik y = –ax2 – bx – c. Hence, sketch the graph of the quadratic function y = –ax2 – bx – c. y x –1O 5 21. Graf fungsi kuadratik f(x) = ax2 + b mempunyai pintasan-y, 3 dan memotong paksi-x di 1 dan k. The graph of the quadratic function f(x) = ax2 + b has the y-intercept, 3 and cuts the x-axis at 1 and k. TP 6 (a) Lakar graf f(x) = ax2 + b. Sketch the graph f(x) = ax2 + b. x y O 1 3 k (b) Cari nilai-nilai bagi a, b dan k. Seterusnya, lakar graf y = f(x) – 4. Find the values of a, b and k. Hence, sketch the graph y = f(x) – 4. b = 3 f(x) = ax2 + 3 x = 1, f(x) = 0, 0 = a(1)2 + 3 a = –3 k = –1 y = –3x2 + 3 – 4 = –3x2 – 1 y –1 y = –3x2 – 1 x O 9 BAB 1 01_Modul A+ MM Tg4.indd 9 12/10/2021 3:31 PM

Uji Kendiri 1.1 1. Rajah menunjukkan graf bagi fungsi kuadratik y = 2x2 + bx. Diagram shows the graph of the quadratic function y = 2x2 + bx. (a) Tentukan nilai b dan cari koordinat titik M. Determine the value of b and find the coordinates of point M. Paksi simetri / Axis of symmetry: x = – —– b 2a – ——b 2(2) = – —9 4 b = 9 y = 2x2 + 9x Apabila / When x = – —9 4 , y = 21– —9 4 2 2 + 91– —9 4 2 = —– 81 8 – —– 81 4 = – —– 81 8 M1– —9 4 , – —– 81 8 2 (b) Pada rajah yang sama, lakar graf y = 2x2 – bx. On the same diagram, sketch the graph y = 2x2 – bx. 2. Rajah menunjukkan sebuah kon tegak dengan jejari r cm dan jumlah luas permukaan 156π cm2 . Diagram shows a right cone with radius r cm and total surface area 156π cm2 . (a) Bentuk satu persamaan kuadratik dalam bentuk ar2 + br + c = 0. Form a quadratic equation in the form ar2 + br + c = 0. πr2 + 2πr(10) = 156π r2 + 20r = 156 r2 + 20r – 156 = 0 (b) Seterusnya, tentukan nilai r. Hence, determine the value of r. (r – 6)(r + 26) = 0 r = 6 atau / or r = –26 r . 0, ∴ r = 6 3. (a) Lengkapkan jadual yang berikut. Complete the following table. x –3 –2 —1 2 2 2x2 + 5x – 3 0 –5 0 15 (b) Daripada jadual di (a), nyatakan punca-punca bagi persamaan 2x2 + 5x = 3. From the table in (a), state the roots of the equation 2x2 + 5x = 3. x = –3, —1 2 r –6 –6r r +26 +26r r2 –156 +20r 10 cm y x y = 2x2 + bx y = 2x2 – bx O M 9 – — 2 9 — 2 10 BAB 1 01_Modul A+ MM Tg4.indd 10 26/10/2021 9:22 AM

4. (a) Tukarkan persamaan (x – 5)2 = 3x – 17 kepada bentuk ax2 + bx + c = 0. Change the equation (x – 5)2 = 3x – 17 to the form of ax2 + bx + c = 0. (x – 5)2 = 3x – 17 x2 – 10x + 25 = 3x – 17 x2 – 13x + 42 = 0 (b) Seterusnya, cari nilai-nilai x yang memuaskan persamaan (x – 5)2 = 3x – 17. Hence, find the values of x that satisfy the equation (x – 5)2 = 3x – 17. (x – 6)(x – 7) = 0 x = 6 atau / or x = 7 5. Selesaikan setiap persamaan yang berikut. Solve each of the following equations. (a) (p + 8)2 – 102 = 0 (p + 8)2 – 102 = 0 p2 + 16p + 64 – 100 = 0 p2 + 16p – 36 = 0 (p – 2)(p + 18) = 0 p = 2 atau / or p = –18 (b) w2 ———– – 12 4 = w + 5 w2 ———– – 12 4 = w + 5 w2 – 12 = 4w + 20 w2 – 4w – 32 = 0 (w – 8)(w + 4) = 0 w = 8 atau / or w= –4 6. Selesaikan setiap persamaan yang berikut. Solve each of the following equations. (a) (w + 5)2 + w2 = 13 (w + 5)2 + w2 = 13 w2 + 10w + 25 + w2 = 13 2w2 + 10w + 12 = 0 w2 + 5w + 6 = 0 (w + 2)(w + 3) = 0 w = –2 atau / or w = –3 (b) 2x + 3 = —– 14 x 2x + 3 = —– 14 x 2x2 + 3x – 14 = 0 (2x + 7)(x – 2) = 0 x = – — 7 2 atau / or x = 2 7. Pada satu gambar rajah, lakar graf y = a(x – 1)2 bagi a = 1, a = 3, a = –2 dan a = –4. On one diagram, sketch the graphs y = a(x – 1)2 for a = 1, a = 3, a = –2 and a = –4. y O x a = 3 a = 1 a = –2 a = –4 4 2 –2 –4 8. Lakar graf fungsi kuadratik yang berikut. Sketch the following graphs of quadratic functions. (a) y = 3x2 + 2 y 2 x O (b) y = 4 – x2 y –2 O 2 4 x (c) y = x2 + 2x –2 y x O (d) y = x2 – x – 2 x y O –2 –1 2 11 BAB 1 01_Modul A+ MM Tg4.indd 11 12/10/2021 3:31 PM

9. Titik maksimum/Maximum point 1—5 2 , —– 49 8 2 Melalui titik (0, 3) dan (6, 0) Passes through points (0, 3) and (6, 0) Suatu graf fungsi kuadratik f(x) = ax2 + bx + c mempunyai maklumat yang diberi. The graph of a quadratic function f(x) = ax2 + bx + c has the given information. (a) Lakar graf itu. Sketch the graph. x y 6 3 5 49 fi—, —–ff 2 8 O (b) Tentukan nilai-nilai bagi a, b dan c. Determine the values of a, b and c. c = 3 – —– b 2a = — 5 2 b = –5a ........................... y = ax2 + bx + 3 x = —5 2 , y = —– 49 8 , —– 49 8 = a1—5 2 2 2 + b1—5 2 2 + 3 —– 25 8 = —– 25 4 a + —5 2 b —5 8 = —5 4 a + —1 2 b 10. Sebuah padang rumput yang berbentuk segi empat tepat mempunyai perimeter 60 m dan luas 216 m2 . Jika panjang padang rumput itu diwakili oleh x m, A rectangular grass field has a perimeter of 60 m and an area of 216 m2 . If the length of the grass field is represented by x m, (a) bentukkan satu persamaan kuadratik dalam sebutan x. form a quadratic equation in terms of x. 2(x + y) = 60 x + y = 30 ................... xy = 216 x(30 – x) = 216 30x – x2 = 216 x2 – 30x + 216 = 0 (b) cari lebar padang rumput itu. find the width of the grass field. (x – 12)(x – 18) = 0 x = 12 atau / or x = 18 Apabila / When x = 12, y = 18 Apabila / When x = 18, y = 12 x . y, ∴ y = 12 Lebar / Width = 12 m 10a + 4b = 5 ............... 10a – 20a = 5 –10a = 5 a = – —1 2 b = —5 2 y m x m 12 BAB 1 01_Modul A+ MM Tg4.indd 12 12/10/2021 3:31 PM

11. Rajah menunjukkan sebidang tanah berbentuk segi empat tepat dengan panjang 32 m dan lebar 15 m. Bahagian berlorek dengan luas 342 m2 ditanami cili. Diagram shows a rectangular piece of land with length 32 m and width 15 m. The shaded region with an area of 342 m2 is planted with chillies. KBAT Mengaplikasi (a) Bentukkan satu persamaan kuadratik dalam sebutan x. Form a quadratic equation in terms of x. 32 × 15 – —1 2 x2 – —1 2 × (x + 10) × 15 = 342 480 – —1 2 x2 – —– 15 2 x – —–– 150 2 = 342 960 – x2 – 15x – 150 = 684 x2 + 15x – 126 = 0 (b) Cari nilai x. Find the value of x. (x – 6)(x + 21) = 0 x = 6 atau/or x = –24 x . 0, ∴ x = 6 12. x cm x cm (x + 3) cm 6 cm 6 cm 8 cm Dalam rajah, hasil tambah luas segi empat sama, segi empat tepat dan segi tiga ialah 933 cm2 . In the diagram, the sum of areas of the square, rectangle and triangle is 933 cm2 . KBAT Menganalisis (a) Persamaan kuadratik yang dibentuk adalah ax2 + bx + c = 0. Nyatakan nilai-nilai bagi a, b dan c. The quadratic equation formed is ax2 + bx + c = 0. State the values of a, b and c. x2 + 6(x + 3) + — 1 2 × 8 × 6 = 933 x2 + 6x + 18 + 24 = 933 x2 + 6x – 891 = 0 a = 1, b = 6, c = –891 (b) Cari jumlah perimeter bagi segi empat sama, segi empat tepat dan segi tiga itu. Find the total perimeter of the square, rectangle and triangle. (x – 27)(x + 33) = 0 x = 27 atau / or x = –33 x . 0, ∴ x = 27 Jumlah perimeter / Total perimeter = 4x + 2(x + 9) + 24 = 6 (27) + 42 = 204 cm 15 m 22 m x m x m x m Kertas 1 Paper 1 Soalan Berformat SPM 1. Persamaan kuadratik x2 – kx – 18 = 0 mempunyai satu punca –3. Cari nilai k. The quadratic equation x2 – kx – 18 = 0 has a root –3. Find the value of k. A –9 B –3 C 3 D 9 2. Selesaikan persamaan kuadratik 30m2 – 79m + 44 = 0. Solve the quadratic equation 30m2 – 79m + 44 = 0. A m = 6 11 atau/or 4 5 C m = 11 6 atau/or 4 5 B m = 6 11 atau/or 5 4 D m = 11 6 atau/or 5 4 3. Rajah di bawah menunjukkan sebuah segi empat tepat dengan luas 27 cm2 . The diagram below shows a rectangle with an area of 27 cm2 . x cm (4x – 3) cm Cari nilai x. Find the value of x. A 3 B 4 C 5 D 6 13 BAB 1 01_Modul A+ MM Tg4.indd 13 12/10/2021 3:31 PM

Bahagian A/Section A 1. Selesaikan persamaan kuadratik berikut. Solve the following quadratic equation. x2 + 3x – 12 = 2(x + 4) [4 markah] [4 marks] 2. Selesaikan persamaan kuadratik yang berikut. Solve the following quadratic equation. 8x(x + 3) = 13 + 19x [4 markah] [4 marks] 3. Dengan menggunakan pemfaktoran, selesaikan persamaan kuadratik berikut. By using factorisation, solve the following quadratic equation. 2x(2x – 3) = 14 – 5x [4 markah] [4 marks] Bahagian B/Section B 4. (a) Selesaikan persamaan kuadratik berikut. Solve the following quadratic equation. 6x2 + 7x = 3(8 – x) [4 markah] [4 marks] (b) Rajah menunjukkan satu laluan berbentuk segi empat tepat yang dibina atas sebuah kolam. Laluan itu dibina dengan 5 keping batu pemijak membulat yang kongruen. Diagram shows a rectangular path built on a pond. The path is constructed by 5 congruent circular stepping stones. x m (y + 6) m Diberi luas laluan itu ialah 45 m2 , cari diameter, dalam m, bagi sekeping batu pemijak itu. [5 markah] Given the area of the path is 45 m2 , find the diameter, in m, for one piece of the stepping stone. [5 marks] 4. Antara graf berikut, yang manakah menunjukkan fungsi y = –x2 + 3? Which of the following graphs shows the function y = –x2 + 3? A O y 3 x C O y –3 x B O y 3 x D O y 3 x 5. (a) Selesaikan persamaan kuadratik berikut. Solve the following quadratic equation. 3(x – 1)(x – 3) = (2x – 5)2 KBAT Mengaplikasi [5 markah] [5 marks] (b) Lakar graf bagi setiap fungsi berikut. Sketch the graph for each of the following functions. (i) y = x2 + 3x (ii) y = –x2 + 4 [4 markah] [4 marks] Bahagian C/Section C 6. (a) Selesaikan persamaan yang berikut. Solve the following equations. (i) (x + 3)2 = 2(x + 7) [4 markah] [4 marks] (ii) 3x ———– x – 14 = 1 ——— x – 4 [5 markah] [5 marks] (b) Rajah menunjukkan graf bagi fungsi y = ax2 – 5x + a. Diagram shows the graph of the function y = ax2 – 5x + a. 2 y x O k Tentukan nilai-nilai integer bagi a dan k. [6 markah] Determine the integer values of a and k. [6 marks] Kertas 2 Paper 2 5. Rajah berikut menunjukkan graf bagi fungsi y = x2 + 4x – 12. The following diagram shows the graph of the function y = x2 + 4x – 12. O L y –6 2 x L ialah titik minimum. Tentukan koordinat bagi titik L. L is a minimum point. Determine the coordinates of point L. A (–4, –12) B (–3, –15) C (–1, –15) D (–2, –16) KLON SPM KLON SPM 14 BAB 1 01_Modul A+ MM Tg4.indd 14 12/10/2021 3:31 PM

BAB BAB 1 1.1 1. (a) ✗ (b) ✗ (c) ✗ (d) ✓ (e) ✗ 2. (a) i (b) ii (c) i 3. (a) (i) (ii) (b) Selari dengan paksi–y Parallel to the y–axis 4. (a) (i) Graf/Graph II: 0 , a , 1 (ii) Graf/Graph I: c . 0 Graf/Graph II: c , 0 (b) (i) bertambah lebih sempit daripada graf y = x2 increases narrower than the graph y = x2 (ii) berada c unit ke atas graf y = x2 lies c units above the graph y = x2 (iii) berada c unit ke bawah graf y = x2 lies c units below the graph y = x2 5. (a) Graf/Graph I: b , 0 Graf/Graph II: b . 0 (b) (i) x = – —b 2 (ii) x = – —b 2 6. (a) (18x – x2 ) cm2 (b) (3x2 + 10x) cm2 7. (a) 3x2 + 17x – 160 = 0 (b) 2x2 + 11x – 310 = 0 8. (a) (i) 0 (ii) 0 (b) x = –1, x = 2 9. (a) 0, – —3 2 (b) 1, 4 10. (a) —1 4 , – —2 3 (b) —5 3 , – —1 5 (c) – —3 2 , – —2 7 11. (a) 1, 3 (b) – —1 2 , –3 (c) 0, —2 3 12. (a) – —1 2 , —1 2 (b) 2, 5 (c) –4, 4 13. (a) y x O (b) y x 3 O (c) y x –2 O 14. (a) y x O –8 –2 2 (b) y x O 16 –4 4 (c) y x O —– 25 2 –5 5 15. (a) 2 y x O (b) –5 5 – — 2 y x O (c) –3 24 — 8 3 y x O 16. (a) y x O 8 –2 (b) y x –4 O –16 (c) y x O —7 2 –49 17. x = 4 atau/or 8 18. Panjang/Length = 34 m Lebar/Width = 26 m 19. (a) x = 5 20. (a) a = 1, b = –4, c = –5 (b) y x –1O 5 21. (a) x y O 1 3 k (b) a = –3, b = 3, k = –1 y –1 y = –3x2 – 1 x O Uji Kendiri 1.1 1. (a) b = 9, M(– —9 4 , – —– 81 8 ) (b) y x y = 2x2 + bx y = 2x2 – bx O M 9 – — 2 9 — 2 2. (a) r2 + 20r – 156 = 0 (b) r = 6 3. (a) x –3 –2 —1 2 2 2x2 + 5x – 3 0 –5 0 15 (b) x = –3, —1 2 4. (a) x2 – 13x + 42 = 0 (b) x = 6 atau/or x = 7 5. (a) p = 2 atau/or p = –18 (b) w = 8 atau/or w = –4 6. (a) w = –2 atau/or w = –3 (b) x = – —7 2 atau/or x = 2 175 11_Modul A+ MM Tg4_JwpM.indd 175 12/10/2021 5:03 PM

7. y O x a = 3 a = 1 a = –2 a = –4 4 2 –2 –4 8. (a) y 2 x O (b) y –2 O 2 4 x (c) –2 y x O (d) x y O –2 –1 2 9. (a) x y 6 3 5 49 fi—, —–ff 2 8 O (b) a = – —1 2 , b = —5 2 , c = 3 10. (a) x2 – 30x + 216 = 0 (b) 12 m 11. (a) x2 + 15x – 126 = 0 (b) x = 6 12. (a) a = 1, b = 6, c = –891 (b) 204 cm Soalan Berformat SPM Kertas 1/Paper 1 1. C 2. C 3. A 4. B 5. D Kertas 2/Paper 2 1. x = 4 atau/or x = –5 2. x = – —– 13 8 atau/or x = 1 3. x = – —7 4 atau/or x = 2 4. (a) x = —4 3 atau/or x = –3 (b) Diameter/Diameter = 3 m 5. (a) x = 4 (b) (i) –3 y x O (ii) y –2 2 4 x O 6. (a) (i) x = –5 atau/or x = 1 (ii) x = — 7 3 atau/or x = 2 (b) a = 2, k = 2 BAB BAB 2 2.1 1. (a) 709 (b) 1305 (c) 10113 2. (a) 49 (b) 54 (c) 729 3. (a) 23 , 8 (b) 24 , 16 (c) 26 , 64 4. (a) 56 (b) 3072 5. (a) 2104 = 3610 (b) 3005 = 7510 (c) 2157 = 11010 6. (a) 2310 (b) 97610 7. (a) 2027 (b) 12104 8. (a) 10011012 (b) 10110112 (c) 100001112 9. (a) 5648 (b) 30108 10. (a) 2235 (b) 13245 11. (a) 100112 (b) 1001012 (c) 1100002 12. (a) 1138 (b) 4048 (c) 14608 13. (a) 2305 (b) 3315 (c) 10215 14. (a) 111003 (b) 3136 (c) 1409 15. (a) 328 (b) 5278 16. (a) 1110010002 (b) 111000100012 17. (a) 1100112 (b) 102023 (c) 12234 (d) 31204 (e) 20445 (f) 43235 (g) 23206 (h) 115106 (i) 40417 (j) 101207 (k) 63108 (l) 103018 (m) 30809 (n) 152319 18. (a) 111102 (b) 1113 (c) 334 (d) 31024 (e) 4345 (f) 23245 (g) 33456 (h) 2236 (i) 12667 (j) 25367 (k) 23778 (l) 26448 (m) 56549 (n) 22559 19. (a) 10112 (b) 2213 (c) 13324 (d) 31145 (e) 30536 (f) 41357 (g) 51347 (h) 23018 (i) 55429 20. x = 3423, y = 111101000, z = 1265 21. (a) 26010 (b) a = 8, b = 9 22. (a) (i) 4647 (ii) 50327 (b) t = 4235 Uji Kendiri 2.1 1. (a) 10112 (b) 215 (c) 147 (d) 129 2. (a) 73 72 71 70 (b) (i) 3 (ii) 7 (iii) 245 (iv) 686 3. (a) 11110102 , 11110112 , 11111002 , 11111012 , 11111102 (b) 3306 4. (a) 32 (b) 2017 5. (a) 100005 (b) x = 764 6. (a) 3584 (b) 1608 7. (a) 32 (b) Bilangan digit/Number of digits = 11 (c) w = 103330 (d) y = 5 524 8. (a) 10001112 (b) 2415 9. (a) 4328 (b) 1011103 10. (a) 16336 (b) x = 3 11. (a) (i) a = 4, b = 2, c = 5 (ii) a = 6, b = 5, c = 1 (b) (i) 101425 (ii) 1 7148 12. (a) 1508 (b) 11002 13. (a) 14356 (b) 50306 14. (a) 103014 (b) 27879 15. (a) x = 11201 (b) y = 1447 16. (a) (i) 20315 (ii) 4225 (b) a = 4130 Soalan Berformat SPM Kertas 1/Paper 1 1. C 2. B 3. B 4. B 5. C 6. C 7. B 8. D 9. C 10. C 11. B 12. C 13. B 14. A 15. B 16. B 17. B Kertas 2/Paper 2 1. (a) 243 (b) 20235 2. (a) (i) y = 110213 (ii) y = 11100112 (b) (i) m = 403 (ii) 1000000112 3. (a) p = 4, 1419 = 3146 (b) 2023 (c) (i) w = 5120 (ii) 14839 (iii) 11879 BAB BAB 3 3.1 1. (a) Bukan/Not (b) Bukan/Not (c) Ya/Yes 2. (a) Palsu/False (b) Benar/True (c) Palsu/False 3. (a) Benar/True (b) Palsu/False (c) Palsu/False 4. (a) Palsu False 240 minit tidak sama dengan 6 jam. 240 minutes is not equal to 6 hours. Benar True (b) Benar True 300° bukan sudut refleks. 300° is not a reflex angle. Palsu False 5. (a) 1 221 bukan nombor yang terbahagi dengan 7. 1 221 is not a number divisible by 7. (b) {1, 9, 11} tidak mempunyai 6 subset. {1, 9, 11} does not have 6 subsets. (c) — 3 7 adalah tidak kurang daripada —–5 14 ./ —3 7 is not less than —–5 14 . 176 11_Modul A+ MM Tg4_JwpM.indd 176 12/10/2021 5:03 PM

BAB BAB 1 Soalan Berformat SPM Kertas 2/Paper 2 1. x2 + 3x – 12 = 2(x + 4) x2 + 3x – 12 = 2x + 8 x2 + x – 20 = 0 (x – 4)(x + 5) = 0 x = 4 atau/or x = –5 2. 8x(x + 3) = 13 + 19x 8x2 + 24x = 13 + 19x 8x2 + 5x – 13 = 0 (8x + 13)(x – 1) = 0 x = – —– 13 8 atau/or x = 1 3. 2x(2x – 3) = 14 – 5x 4x2 – 6x = 14 – 5x 4x2 – x – 14 = 0 (4x + 7)(x – 2) = 0 x = – —7 4 atau/or x = 2 4. (a) 6x2 + 7x = 3(8 – x) 6x2 + 7x = 24 – 3x 6x2 + 10x – 24 = 0 3x2 + 5x – 12 = 0 (3x – 4)(x + 3) = 0 x = —4 3 atau/or x = –3 (b) y + 6 = 5x … (y + 6)x = 45 … 5x2 = 45 x2 = 9 x = ±3 x . 0, ∴ x = 3 Diameter/Diameter = 3 m 5. (a) 3(x – 1)(x – 3) = (2x – 5)2 3(x2 – 4x + 3) = 4x2 – 20x + 25 3x2 – 12x + 9 = 4x2 – 20x + 25 x2 – 8x + 16 = 0 (x – 4)2 = 0 x = 4 (b) (i) –3 y x O (ii) y –2 2 4 x O 6. (a) (i) (x + 3)2 = 2(x + 7) x2 + 6x + 9 = 2x + 14 x2 + 4x – 5 = 0 (x + 5)(x – 1) = 0 x = –5 atau/or x = 1 (ii) ——–– 3x x – 14 = ——– 1 x – 4 3x(x – 4) = x – 14 3x2 – 12x = x – 14 3x2 – 13x + 14 = 0 (3x – 7)(x – 2) = 0 x = —7 3 atau/or x = 2 (b) x = 0, y = 2 2 = 0 – 0 + a ∴ a = 2 y = 2x2 – 5x + 2 x = k, y = 0 0 = 2k2 – 5k + 2 (2k – 1)(k – 2) = 0 k = —1 2 atau/or k = 2 k ialah integer./k is an integer. ∴ k = 2 BAB BAB 2 Soalan Berformat SPM Kertas 2/Paper 2 1. (a) Nilai bagi digit 3/Value of digit 3 = 3 × 92 = 243 (b) 5247 = 5 × 72 + 2 × 71 + 4 × 70 = 245 + 14 + 4 = 26310 = 20235 2. (a) (i) y = 110213 (ii) y = 34 + 33 + 2 × 31 + 1 = 81 + 27 + 6 + 1 = 11510 = 11100112 (b) (i) 100034 = 1 × 44 + 0 × 43 + 0 × 42 + 0 × 41 + 3 × 40 = 256 + 0 + 0 + 0 + 3 = 25910 = 4038 ∴ m = 403 (ii) 100034 = 4 0 38 = 100 000 0112 3. (a) 1p19 = 11101102 1 × 92 + p × 91 + 1 × 90 = 1 × 26 + 1 × 25 + 1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 0 × 20 81 + 9p + 1 = 64 + 32 + 16 + 0 + 4 + 2 + 0 9p + 82 = 118 9p = 36 p = 4 1419 = 11810 = 3146 (b) 21103 – 12013 = 2023 (c) (i) 1 1 42036 + 5136 51206 ∴w = 5120 (ii) 51206 = 5 × 63 + 1 × 62 + 2 × 61 + 0 × 60 = 1080 + 36 + 12 + 0 = 112810 = 14839 Baki 5 263 Remainder 5 52 ……3 5 10 ……2 5 2 ……0 0 …… 2 Baki 2 115 Remainder 2 57 ……1 2 28 ……1 2 14 ……0 2 7 ……0 2 3 ……1 2 1 ……1 0 …… 1 Baki 6 118 Remainder 6 19 ……4 6 3 ……1 0 …… 3 1 4 0 3 21103 – 12013 2023 Baki 9 1128 Remainder 9 125 ……3 9 13 ……8 9 1 ……4 0 …… 1 Baki 8 259 Remainder 8 32 ……3 8 4 ……0 0 …… 4 MG-1

MG-9 BAB 1 Fungsi dan Persamaan Kuadratik dalam Satu Pemboleh Ubah Tahap Penguasaan Tafsiran 1 Mempamerkan pengetahuan asas tentang ungkapan, fungsi dan persamaan kuadratik dalam satu pemboleh ubah. 2 Mempamerkan kefahaman tentang ungkapan, fungsi dan persamaan kuadratik dalam satu pemboleh ubah. 3 Mengaplikasikan kefahaman tentang fungsi dan persamaan kuadratik dalam satu pemboleh ubah untuk melaksanakan tugasan mudah. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dan persamaan kuadratik dalam satu pemboleh ubah dalam konteks penyelesaian masalah rutin yang mudah. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dan persamaan kuadratik dalam satu pemboleh ubah dalam konteks penyelesaian masalah rutin yang kompleks. 1. k3 – 8 x2 + 3x – 6 4p2 + 9 w + 2 –y2 + 10y 2r2 – —7 r + 3 Berdasarkan ungkapan-ungkapan di atas, lengkapkan rajah berikut. Based on the expressions above, complete the following diagram. TP 1 Ungkapan / Expression (a) Ungkapan kuadratik Quadratic expression x2 + 3x – 6 4p2 + 9 –y2 +10y (b) Bukan ungkapan kuadratik Not quadratic expression k3 – 8 w + 2 2r2 – 7 r + 3 2. Tiga integer yang berturut-turut ialah n, n + 1 dan n + 2. Tulis satu ungkapan kuadratik, dalam bentuk an2 + bn + c, bagi The three consecutive integers are n, n + 1 and n + 2. Write a quadratic expression, in the form an2 + bn + c, for TP 2 (a) kuasa dua bagi hasil tambah tiga integer itu. the square for the sum of the three integers. (n + n + 1 + n + 2)2 = (3n + 3)2 = 9n2 + 18n + 9 (b) hasil tambah kuasa dua bagi setiap integer itu. the sum of squares for each of the integers. n2 + (n + 1)2 + (n + 2)2 = n2 + n2 + 2n + 1 + n2 + 4n + 4 = 3n2 + 6n + 5 Lembaran Pentaksiran Bilik Darjah (PBD)

MG-10 Lembaran PBD 3. Selesaikan persamaan kuadratik berikut. Solve the following quadratic equations. TP 3 (a) (x + 2)(x + 1) = 3(3x – 1) x2 + 3x + 2 = 9x – 3 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x – 1 = 0 atau/or x – 5 = 0 x = 1 atau/or x = 5 (b) x + 1 = (3x – 7)2 x + 1 = 9x2 – 42x + 49 9x2 – 43x + 48 = 0 (9x – 16)(x – 3) = 0 9x – 16 = 0 atau/or x – 3 = 0 x = 16 9 atau/or x = 3 4. (a) Selesaikan. Solve. TP 3 (i) 12t 2 – 31t – 15 = 0 (12t + 5)(t – 3) = 0 12t + 5 = 0 atau/or t – 3 = 0 t = – 5 12 atau/or t = 3 (ii) 5w2 + 27w + 8 = 0 (5w + 7)(w + 4) = 0 5w + 7 = 0 atau/or w + 4 = 0 w = – 7 5 atau/or w = –4 (b) Dalam rajah yang diberi, PQRS ialah sebuah segi empat tepat. Sebuah bulatan dan dua buah semi bulatan yang kongruen terterap dalam segi empat tepat itu. Jejari setiap semi bulatan itu ialah 1 cm. In the diagram given, PQRS is a rectangle. A circle and two congruent semicircles are inscribed in the rectangle. The radius of each semicircle is 1 cm. Ungkapkan Express TP 5 (i) panjang PS, dalam sebutan x, the length of PS, in terms of x, PS = 1 + 2x + 1 = (2 + 2x) cm (ii) luas kawasan berlorek, dalam bentuk ax2 + bx + c. the area of the shaded region, in the form ax2 + bx + c. Luas kawasan berlorek Area of the shaded region = (2 + 2x)(2x) – p(1)2 – px2 = 4x + 4x2 – p – px2 = [(4 – p)x2 + 4x – p] cm2 5. Rajah yang diberi menunjukkan suatu pepejal yang terdiri daripada gabungan sebuah kuboid dan sebuah piramid tegak. Tinggi pepejal itu ialah (x + 8) cm. Diagram given shows a solid consisting of the combination of a cuboid and a right pyramid. The height of the solid is (x + 8) cm. (a) Bentukkan satu ungkapan kuadratik, dalam bentuk ax2 + bx + c, bagi isi padu pepejal itu. Form a quadratic equation, in the form ax2 + bx + c, for the volume of the solid. TP 2 Isi padu pepejal Volume of solid = (4x – 5)(12)(3) + 1 3(4x – 5)(12)(x + 5) = 144x – 180 + 4(4x2 + 20x – 5x – 25) = 144x – 180 + 16x2 + 60x – 100 = (16x2 + 204x – 280) cm3 (b) Jika isi padu pepejal itu ialah 476 cm3 , hitung nilai x. If the volume of the solid is 476 cm3 , calculate the value of x. TP 4 16x2 + 204x – 280 = 476 16x2 + 204x – 756 = 0 4x2 + 51x – 189 = 0 (4x + 63)(x – 3) = 0 x = – 63 4 atau/or x = 3 x  5 4 , x = 3 Q R P S 2x cm (4x – 5) cm 3 cm 12 cm

Dapatkan sekarang! ISBN 978-967- 466-625-5 Sem. M’sia RM 11.90 Sabah/Sarawak RM 12.90 Mata Pelajaran / Tingkatan 4 5 Sejarah Matematik Matematik Tambahan Kimia Fizik Perniagaan Ekonomi English Judul-judul dalam siri Modul A+1 Siri MODUL A+1 ini menyediakan modul pembelajaran komprehensif yang dihasilkan berdasarkan Dokumen Standard Kurikulum dan Pentaksiran (DSKP). Modul ini telah dirancang dan ditulis oleh guru-guru yang berpengalaman dalam membantu proses PdPc dengan lebih efektif. Latihan yang disediakan mencakupi Tahap Penguasaan yang perlu dikuasai oleh murid untuk mengoptimumkan kefahaman mereka. AKSES DIGITAL ■ Aktiviti PAK-21 ■ Aktiviti Tarsia ■ Lembaran Pentaksiran Bilik Darjah (PBD) Ingin menjadi penulis kami? Sertai kami dengan menghantar e-mel ke alamat [email protected]. www.panasiapub.com 199101016590 (226902-X) 9 789674 666255