TINGKATAN KSSM MoRE MoRE ModulE ModulE && Eksklusif Eksklusif PELANGI ONLINE TEST https://qr.pelangibooks.com/?u=POTMT4 Bonus Digital Lembaran PBD dengan Jawapan Terkini SPM Format PELANGI BESTSELLER Pembelajaran BERPANDU dan SISTEMATIK Praktis Ekstra SPM Kod QR Lembaran PBD Kod QR Jawapan Kod QR i-THINK/KBAT Klu Soalan Praktis SPM Studi 1 Minit Latihan modular MATEMATIK TAMBAHAN Additional Mathematics DWIBAHASA Tee Hock Tian Wong Kiat Hi
iii Bab 1 Fungsi 1 Functions 1.1 Fungsi..................................................................................1 Functions 1.2 Fungsi Gubahan.................................................................7 Composite Functions 1.3 Fungsi Songsang ..............................................................14 Inverse Functions Praktis SPM 1............................................................................19 Praktis Ekstra SPM 1 Kod QR ............................................19 Sudut KBAT ..............................................................................21 ........................................................................................ 21 Bab 2 Fungsi Kuadratik 22 Quadratic Functions 2.1 Persamaan dan Ketaksamaan Kuadratik ......................22 Quadratic Equations and Inequalities 2.2 Jenis-jenis Punca Persamaan Kuadratik .......................28 Types of Roots of Quadratic Equations 2.3 Fungsi Kuadratik..............................................................31 Quadratic Functions Praktis SPM 2............................................................................38 Praktis Ekstra SPM 2 Kod QR ............................................38 Sudut KBAT ..............................................................................43 ........................................................................................ 43 Bab 3 Sistem Persamaan 44 Systems of Equations 3.1 Sistem Persamaan Linear dalam Tiga Pemboleh Ubah ...............................................................44 Systems of Linear Equations in Three Variables 3.2 Persamaan Serentak yang melibatkan Satu Persamaan Linear dan Satu Persamaan Tak Linear.........................50 Simultaneous Equations involving One Linear Equation and One Non-Linear Equation Praktis SPM 3............................................................................57 Praktis Ekstra SPM 3 Kod QR ............................................57 Sudut KBAT ..............................................................................59 ........................................................................................ 59 Bab 4 Indeks, Surd dan Logaritma 60 Indices, Surds and Logarithms 4.1 Hukum Indeks..................................................................60 Laws of Indices 4.2 Hukum Surd.....................................................................64 Laws of Surds 4.3 Hukum Logaritma ...........................................................71 Laws of Logarithms 4.4 Aplikasi Indeks, Surd dan Logaritma............................78 Applications of Indices, Surds and Logarithms Praktis SPM 4............................................................................80 Praktis Ekstra SPM 4 Kod QR ............................................80 Sudut KBAT ..............................................................................82 ........................................................................................ 82 Bab 5 Janjang 83 Progressions 5.1 Janjang Aritmetik.............................................................83 Arithmetic Progressions 5.2 Janjang Geometri.............................................................91 Geometric Progressions Praktis SPM 5..........................................................................101 Praktis Ekstra SPM 5 Kod QR ..........................................101 Sudut KBAT ............................................................................105 ...................................................................................... 105 Kandungan Penerbitan Pelangi Sdn. Bhd.
iv Bab 6 Hukum Linear 106 Linear Law 6.1 Hubungan Linear dan Tak Linear ...............................106 Linear and Non-Linear Relations 6.2 Hukum Linear dan Hubungan Tak Linear.................112 Linear Law and Non-Linear Relations 6.3 Aplikasi Hukum Linear.................................................118 Application of Linear Law Praktis SPM 6..........................................................................122 Praktis Ekstra SPM 6 Kod QR ..........................................122 Sudut KBAT ............................................................................125 ...................................................................................... 125 Bab 7 Geometri Koordinat 126 Coordinate Geometry 7.1 Pembahagi Tembereng Garis .......................................126 Divisor of a Line Segment 7.2 Garis Lurus Selari dan Garis Lurus Serenjang...........130 Parallel Lines and Perpendicular Lines 7.3 Luas Poligon ...................................................................135 Areas of Polygons 7.4 Persamaan Lokus...........................................................137 Equations of Loci Praktis SPM 7..........................................................................140 Praktis Ekstra SPM 7 Kod QR ..........................................140 Sudut KBAT ............................................................................143 ...................................................................................... 143 Bab 8 Vektor 144 Vectors 8.1 Vektor..............................................................................144 Vectors 8.2 Penambahan dan Penolakan Vektor ...........................148 Addition and Subtraction of Vectors 8.3 Vektor dalam Satah Cartes ...........................................152 Vectors in a Cartesian Plane Praktis SPM 8..........................................................................159 Praktis Ekstra SPM 8 Kod QR ..........................................159 Sudut KBAT ............................................................................164 ...................................................................................... 164 Bab 9 Penyelesaian Segi Tiga 165 Solution of Triangles 9.1 Petua Sinus......................................................................165 Sine Rule 9.2 Petua Kosinus.................................................................169 Cosine Rule 9.3 Luas Segi Tiga.................................................................172 Area of a Triangle 9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas Segi Tiga..........................................................................174 Application of Sine Rule, Cosine Rule and Area of a Triangle Praktis SPM 9..........................................................................175 Praktis Ekstra SPM 9 Kod QR ..........................................175 Sudut KBAT ............................................................................178 ...................................................................................... 178 Bab 10 Nombor Indeks 179 Index Numbers 10.1 Nombor Indeks..............................................................179 Index Numbers 10.2 Indeks Gubahan.............................................................185 Composite Index Praktis SPM 10 .......................................................................190 Praktis Ekstra SPM 10 Kod QR ........................................190 Sudut KBAT ............................................................................193 ...................................................................................... 193 Kertas Pra-SPM......................................................................195 Jawapan http://www.epelangi.com/Module&More2021/ MatematikTambahan/T4/JawapanKeseluruhan.pdf Lembaran PBD dan Jawapan http://www.epelangi.com/Module&More2021/ MatematikTambahan/T4/LPBD.pdf Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 1 Fungsi Functions 1.1 STUDI 1 Minit Buku Teks: 2 – 11 Bab 1 Fungsi Functions Analisis Soalan SPM 2021 Kertas 1 Kertas 2 S9 – Tiga Cara Perwakilan Hubungan / Three Ways to Represent Relations Suatu hubungan memadankan unsur-unsur satu set dengan unsur-unsur set lain. A relation matches the elements of one set with the elements of another set. 1 Gambar rajah anak panah Arrow diagram 1 Set A separuh daripada is half of Set B 2 3 4 2 4 6 8 1 Satu dengan satu One-to-one A B 2 Banyak dengan satu Many-to-one C D 3 Satu dengan banyak One-to-many P Q 4 Banyak dengan banyak Many-to-many M N 2 Graf / Graph Set A 8 6 4 2 1 0 2 3 4 Set B 3 Pasangan bertertib Ordered pairs {(1, 2), (2, 4), (3, 6), (4, 8)} 7 4 3 16 9 64 49 x f y Empat Jenis Hubungan / Four Types of Relations Tatatanda fungsi/Function notation: f(x) = x2 atau / or f : x → x2 Domain / Domain = {3, 4, 7} Kodomain / Codomain = (9, 16, 49, 64} Objek / Objects = 3, 4, 7 Imej / Images = 9, 16, 49 Julat / Range = {9, 16, 49} Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 2 CONTOH 1. Tentukan sama ada setiap hubungan yang berikut ialah satu fungsi atau bukan. Beri sebab untuk jawapan anda. TP1 Determine whether each of the following relations is a function. Give the reason for your answer. c b a q p s r Suatu fungsi. Setiap objek hanya mempunyai satu imej. A function. Each object has only one image. (a) r q p 8 4 12 Suatu fungsi. Setiap objek mempunyai hanya satu imej. A function. Each object has only one image. (b) 3 m 11 7 5 n p Bukan fungsi. 11 tidak mempunyai imej. Not a function. 11 does not have any image. (c) 2 4 6 8 3 5 7 Bukan fungsi. Objek 8 tidak mempunyai imej. Not a function. The object 8 does not have any image. (d) 0 2 x y Satu fungsi. Garis mencancang memotong graf itu pada satu titik sahaja. A function. The vertical line cuts the graph at only one point. (e) y x 0 Bukan fungsi. Garis mencancang menyilang graf di dua titik berlainan. Not a function. The vertical line cuts the graph at two different points. Tip Penting Gunakan ujian garis mencancang. Use the vertical line test. Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 3 CONTOH (h) {(6 , p), (6, q) , (9, q) , (13, r)} Bukan fungsi. Objek 6 mempunyai lebih daripada satu imej, iaitu p dan q. Not a function. The object 6 has more than one image, that is, p and q. (i) {(a , 2) , (b , 2) , (c, 4) , (d, 8), (f, 2)} Fungsi. Setiap objek mempunyai hanya satu imej. A function. Every object has only one image. 2. Tulis setiap fungsi yang berikut menggunakan tatatanda fungsi. TP2 Write each of the following by using the function notation. f 1 2 3 5 6 7 g : x → x + 4 g(x) = x + 4 (a) g 2 3 4 4 6 8 10 g : x → 2x g(x) = 2x (b) {(3, 1 9 ), (5, 1 25), (6, 1 36)}. f : x → 1 x2 , x ≠ 0 f(x) = 1 x2 , x ≠ 0 (c) 3 0 (2, 5) (4, 7) x h(x) h : x → x + 3 h(x) = 2x Tip Penting Fungsi sering diwakili oleh huruf kecil seperti f, g, h dan k. Functions are often represented by small letters such as f, g, h and k. Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 4 CONTOH 3. Lengkapkan jadual yang berikut. TP1 Complete the following table. Fungsi Function Domain Domain Kodomain Codomain Objek Object Imej Image Julat Range 0 13 –3 1 x f(x) 2 6 {0 < x < 6} {−3 <f(x) < 10} 0 < x < 6 –3 < f(x) < 13 {0 < f(x) < 13} (a) 0 10 6 2 x f(x) –2 3 4 {−2 < x < 4} {0 <f(x) < 10} −2 < x < 4 0 < f(x) < 10 {0 < f(x) < 10} (b) 0 6 5 2 –2 x f(x) 31 4 3 1 –1 –1 2 {−1, 0, 1, 2, 3} {−2, 0, 2, 4, 6} −1, 0, 1, 2, 3 −2, 0, 2, 4, 6 {−2, 0, 2, 4, 6} (c) 4 2 18 10 5 9 11 {4, 10, 18} {2, 5, 9, 11} 4, 10, 18 2, 5, 9 {2, 5, 9} (d) {(3, 10), (5, 26), (6, 37),(9, 82)} {3, 5, 6, 9} {10, 26, 37, 82} 3, 5, 6, 9 10, 26, 37, 82 {10, 26, 37, 82} Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 5 CONTOH 4. Lakar graf bagi setiap fungsi berikut. Seterusnya, nyatakan julat sepadan bagi domain yang diberikan. TP4 Sketch the graph of the following functions. Hence, state the corresponding range for the given domain. f(x) = |1 − 2x|, −1 < x < 3 x –1 0 1 2 1 2 3 f(x) 3 1 0 1 3 5 0 5 2 x f(x) 1 3 4 3 1 –1 2 Julat/ Range: 0 < f(x) < 5 (a) y = |x2 − 1|, −3 < x < 3 0 2 x f(x) 3 (–3, 8) (3, 8) 1 4 3 –3 –2 –1 2 5 6 7 8 1 Julat/ Range: 0 < f(x) < 8 5. Selesaikan setiap yang berikut. TP5 Solve each of the following. CONTOH Fungsi f ditakrifkan sebagai f : x → 4x + 3 x , x ≠ 0. Cari Function f is defined as f : x → 4x + 3 x , x ≠ 0. Find (i) f(−2) (ii) imej bagi 1 4 di bawah f. the image of 1 4 under f. (iii) nilai-nilai f apabila imej ialah 13. the values of f when the image is 13. Penyelesaian: (i) f(x) = 4x + 3 x f(–2) = –8 – 3 2 = –9 1 2 (ii) f(x) = 4x + 3 x f 1 1 4 2 = 41 1 4 2 + 3 1 4 = 1 + 12 = 13 (iii) f(x) = 13 4x + 3 x = 13 4x2 + 3 = 13x 4x2 – 13x + 3 = 0 (4x – 1)(x – 3) = 0 x = 1 4 , x = 3 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 6 CONTOH Fungsi g ditakrifkan sebagai g : x → |2 – 8x|, cari Function g is defined as g : x → |2 – 8x|, find (i) g1 1 2 2 (ii) nilai-nilai x yang memetakan hubungan dirinya. the values of x which maps to itself. Penyelesaian: (i) g(x) = |2 – 8x| g1 1 2 2 = |2 − 4| = |−2| = 2 (ii) g(x) = x |2 – 8x| = x 2 – 8x = x 2 – 8x = –x 2 = x + 8x 2 = 8x – x 9x = 2 7x = 2 x = 2 9 x = 2 7 (a) Fungsi f ditakrifkan sebagai f : x → 12 x – 2 , x ≠ 2. Cari The function f is defined as f : x → 12 – 2 , x ≠ 2. Find (i) nilai bagi f(−4) dan f(10). the value of f(−4) and f(10). (ii) nilai-nilai x dengan keadaan f(x) = 3 4 . the values of x for which f(x)= . (i) f(x) = 12 x – 2 f(–4) = 12 –6 = –2 f(10) = = 3 2 (ii) f(x) = 3 4 12 x – 2 = 3 4 3(x – 2) = 48 x – 2 = 16 x = 18 (b) Diberi g(x) = u3x – 7u, cari Given g(x) = u3x – 7u, (i) nilai bagi g(−2) dan g(5). the value of g(−2) and g(5). (ii) objek dengan keadaan imej ialah 14. the objects for which the image is 14. (i) g(x) = u3x – 7u g(–2) = u–6 – 7u = 13 g(5) = u15 – 7u = 8 (ii) g(x) = 14 u3x – 7u = 14 3x – 7 = 14 3x – 7 = –14 3x = 21 3x = –7 x = 7 x = – 7 3 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 7 CONTOH (c) Diberi h(x) = 2x2 + mx − 30 dengan keadaan m ialah satu pemalar dan h(2) = −34. Cari Given that h(x) = 2x2 + mx − 30 for which m is a constant and h(2) = −34. Find (i) nilai m. the value of m. (ii) nilai-nilai x yang memetakan kepada dirinya. the values of x which map to itself. (i) h(x) = 2x2 + mx − 30 h(2) = 8 + 2m − 30 = −34 2m = −34 + 22 m = −6 (ii) h(x) = x 2x2 − 6x − 30 = x 2x2 − 7x − 30 = 0 (x + 5)(x − 6) = 0 x = − 5 2 , x = 6 (d) Fungsi f ditakrifkan sebagai f(x) = 7 + |3x|. Cari The function f is defined as f(x) = 7 + |3x|. Find (i) nilai bagi f(−1) dan f(6). the values of f(−1) and f(6). (ii) objek-objek dengan keadaan imejnya ialah 19. the objects for which the image is 19. (i) f(x) = 7 + |3x| f(−1) = 7 + |−3| = 10 f(6) = 7 + |18| = 25 (ii) f(x) = 19 7 + |3x| = 19 |3x| = 12 3x = 12 3x = –12 x = 4 x = –4 Fungsi Gubahan Composite Functions 1.2 STUDI 1 Minit Buku Teks: 12 – 19 Fungsi Gubahan Composite Functions x y z A f g B gf C • Fungsi f memetakan x kepada y dan fungsi g memetakan y kepada z. Function f maps x onto y, and function g maps y onto z. • Fungsi gubahan gf memetakan x terus kepada z. Composite function gf maps x directly onto z. gf(x) = g[f(x)] 6. Huraikan fungsi gubahan untuk setiap kes yang berikut. TP2 Describe the composite function for each of the following case. f u v g h p q r s u = gf k = hg (a) p a b c d e q h k r s h = qp k = rs Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 8 CONTOH (b) f w x y z g g h k h = gf k = gg / g2 (c) f p q r s f g u v u = ff / f 2 v = fg 7. Bagi setiap pasangan fungsi yang berikut, cari (i) fg(x) dan (ii) gf(x). TP3 For each of the following pairs of functions, find (i) fg(x) and (ii) gf(x). f(x) = x2 g(x) = 3x + 1 Penyelesaian: (i) fg(x) = f(3x + 1) = (3x + 1)2 = 9x2 + 6x + 1 = 5 – 2x (ii) gf(x) = g(x2 ) = 3x2 + 1 (a) f(x) = 5 – 3x g(x) = 2x + 1 (i) fg(x) = f(2x + 1) = 5 – 3(2x + 1) = 2 – 6x (ii) gf(x) = g(5 – 3x) = 2(5 – 3x) + 1 = 11 – 6x (b) f(x) = x + 3 g(x) = 5x2 – 2 (i) fg(x) = f(5x2 – 2) = 5x2 – 2 + 3 = 5x2 + 1 (ii) gf(x) = g(x + 3) = 5(x + 3)2 – 2 = 5(x2 + 6x + 9) – 2 = 5x2 + 30x + 43 (c) f(x) = 1 – x g(x) = 4 x (i) fg(x) = f1 4 x 2 = 1 – 4 x , x ≠ 0 (ii) gf(x) = g(1 – x) = 4 1 – x , x ≠ 1 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 9 CONTOH 8. Bagi setiap pasangan fungsi yang berikut, cari TP3 For each pair of the following functions, find (i) fg(3) (ii) gf(4) f : x → 5x, g : x → 4 – 2x Penyelesaian: (i) g(3) = 4 – 2(3) = –2 fg(3) = f(–2) = 5(–2) = –10 (ii) f(4) = 5(4) = 20 gf(4) = g(20) = 4 – 2(20) = –36 (a) f : x → 3x + 2, g : x → 2 – x2 (i) g(3) = 2 – (3)2 = –7 fg(3) = f(–7) = 3(–7) + 2 = –19 (ii) f(4) = 3(4) + 2 = 14 gf(4) = g(14) = 2 – (14)2 = –194 (b) f : x → x – 3, g : x → 3 x , x ≠ 0 (i) g(3) = = 1 fg(3)= f(1) = 1 – 3 = –2 (ii) f(4) = 4 – 3 = 1 gf(4)= g(1) = 3 1 = 3 (c) f : x → 2 – 8x, g : x → x x + 5 , x ≠ –5 (i) g(3) = 3 3 + 5 = 3 8 fg(3)= f 1 3 8 2 = 2 – 81 3 8 2 = –1 (ii) f(4) = 2 – 8(4) = –30 gf(4)= g(–30) = –30 –30 +5 = 6 5 1 Gantikan 4 ke dalam f, anda akan dapat nilai 20. Insert 4 into f, you will get 20. 2 Gantikan 20 ke dalam g. Insert 20 into g. Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 10 CONTOH 9. Bagi setiap pasangan fungsi yang berikut, cari TP3 For each pair of the following functions, find (i) f 2 (x) (ii) g2 (x) (iii) f 2 (2) (iv) g2 (3) f(x) = 2x, g(x) = x – 5 Penyelesaian: (i) f 2 (x) = f(2x) = 2(2x) = 4x (ii) g2 (x) = g(x – 5) = (x – 5) – 5 = x – 10 (iii)f 2 (2) = 4(2) = 8 (iv) g2 (3) = 3 – 10 = –7 (a) f(x) = 3x – 1, g(x) = 4x + 2 (i) f 2 (x) = f(3x – 1) = 3(3x – 1) – 1 = 9x – 4 (ii) g2 (x) = g(4x + 2) = 4(4x + 2) + 2 = 16x + 10 (iii)f 2 (2) = 9(2) – 4 = 18 – 4 = 14 (iv) g2 (3) = 16(3) + 10 = 48 + 10 = 58 (b) f(x) = x – 6, g(x) = x2 + 1 (i) f 2 (x) = f(x – 6) = (x – 6) – 6 = x – 12 (ii) g2 (x) = g(x2 + 1) = (x2 + 1)2 + 1 = x2 + 2x2 + 1 (iii)f 2 (2)= 2 – 12 = –10 (iv) g2 (3)= 32 + 2(3)2 + 3 = 81 + 18 + 1 = 100 (c) f : x → 2 + 5x, g : x → x x – 1 , x ≠ 1 (i) f2 (x) = f(2 + 5x) = 2 + 5(2 + 5x) = 12 + 25x (ii) g2 (x) = x x – 1 ( x x – 1 ) – 1 = x x – 1 x – (x – 1) x – 1 = x (iii)f2 (2) = 12 + 25(2) = 62 (iv) g2 (3) = 3 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 11 CONTOH CONTOH 10. Selesaikan setiap yang berikut. TP4 Solve each of the following. Diberi f : x → 2x + 4, g : x → x – 2 dan fg(x) = 2. Given f : x → 2x + 4, g : x → x – 2 and fg(x) = 2. Penyelesaian: fg(x) = f(x – 2) 1 Cari fungsi gubahan fg(x). Find the composite function fg(x). = 2(x – 2) + 4 = 2x fg(x) = 2x = 2 2 Samakan fg(x) dengan 2. Equate fg(x) with 2. x = 1 (a) Diberi f : x → x + 3, g : x → 9 – 2x dan fg(x) = 6. Given f : x → x + 3, g : x → 9 – 2x and fg(x) = 6. fg(x) = f(9 – 2x) = 9 – 2x + 3 = 12 – 2x fg(x) = 12 – 2x = 6 2x = 6 x = 3 (b) Diberi f : x → x – 3, g : x → 3 – 5x dan gf(x) = –2. Given f : x → x – 3, g : x → 3 – 5x and gf(x) = –2. gf(x) = g(x – 3) = 3 – 5(x – 3) = 18 – 5x gf(x) = 18 – 5x = –2 5x = 20 x = 4 (c) Diberi f : x → 3 x , g : x → 2x + 1 dan gf(x) = –1. Given f : x → 3 x , g : x → 2x + 1 and gf(x) = –1. gf(x) = g1 3 x 2 = 21 3 x 2 + 1 = 6 + x x , x ≠ 0 gf(x) = 6 + x x = –1 6 + x = –x 2x = –6 x = –3 11. Cari fungsi g. TP4 Find the function g. f(x) = x + 1, fg(x) = x + 7 Penyelesaian: f(x) = x + 1 fg(x) = g(x) + 1 x + 7 = g(x) + 1 2 Samakan kedua-dua fg(x). Equate both fg(x). g(x) = x + 7 – 1 = x + 6 3 Selesaikan untuk g(x). Solve for g(x). g : x → x + 6 (a) f(x) → x + 2, fg(x) = 7 – 3x f(x) = x + 2 fg(x) = g(x) + 2 7 – 3x = g(x) + 2 g(x) = 7 – 3x – 2 = 5 – 3x g : x → 5 – 3x 1 Gantikan g(x) ke dalam f(x). Insert g(x) into f(x). Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 12 CONTOH (b) f(x) → 2x – 1, fg(x) = 9 – 4x f(x) = 2x – 1 fg(x) = 2g(x) – 1 9 – 4x = 2g(x) – 1 2g(x) = 9 – 4x + 1 2g(x) = 10 – 4x g(x) = 10 – 4x 2 = 5 – 2x g : x → 5 – 2x (c) f(x) = 4 – x, fg(x) = 16 – x 8 f(x) = 4 – x fg(x) = 4 – g(x) 16 – x 8 = 4 – g(x) g(x) = 4 – 1 16 – x 8 2 = 32 – (16 – x) 8 = 16 + x 8 g : x → 16 + x 8 f(x) = x + 2, gf(x) = 7 – 3x Penyelesaian: gf(x) = 7 – 3x g(x + 2) = 7 – 3x Katakan/ Let y = x + 2 x = y – 2 g(y) = 7 – 3(y – 2) = 7 – 3y + 6 = 13 – 3y g(x) = 13 – 3x g : x → 13 – 3x (d) f(x) = 5 – x, gf(x) = 18 – x gf(x) = 18 – x g(5 – x) = 18 – x Katakan y = 5 – x x = 5 – y g(y) = 18 – (5 – y) = 13 + y g(x) = 13 + x g : x → 13 + x (e) f : x → 6 x , x ≠ 0, gf : x → 2x + 3 gf(x) = 2x + 3 g1 6 x 2 = 2x + 3 Katakan y = 6 x x = 6 g(y) = 21 6 y 2 + 3 = 12 y + 3 g(x) = 12 x + 3 g : x → 12 x + 3, x ≠ 0 (f) f(x) = x + 2, gf(x) = 8 – 3x2 gf(x) = 8 – 3x2 g(x + 2) = 8 – 3x2 Katakan y = x + 2 x = y – 2 g(y) = 8 – 3(y – 2)2 = 8 – 3(y2 – 4y + 4) = 8 – 3y2 + 12y – 12 = –3y2 + 12y – 4 g(x) = –3x2 + 12x – 4 g : x → –3x2 + 12x – 4 1 Gantikan f(x) = x + 2 ke dalam gf(x). Insert f(x) = x + 2 into gf(x). 2 Gantikan y = x + 2 dan x = y – 2 ke dalam g(x + 2). Insert y = x + 2 and x = y – 2 into g(x + 2). 3 Selesaikan untuk g(y) ≈ g(x). Solve for g(y) ≈ g(x). Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 13 CONTOH 12. Selesaikan setiap yang berikut. TP5 Solve each of the following. Diberi fungsi f : x → 9 – 2x, g : x → ax + b dan fg : x → 1 – 6x. Cari nilai bagi a dan b. Given the function f : x → 9 – 2x, g : x → ax + b and fg : x → 1 – 6x. Find the value of a and of b. Penyelesaian: fg(x) = f(ax + b) 1 Cari fungsi gubahan fg(x). Find composite function fg(x). = 9 – 2(ax + b) = 9 – 2b – 2ax 9 – 2b – 2ax = 1 – 6x 2 Samakan kedua-dua fg(x). Equate both fg(x). 3 Bandingkan pemalar. Compare the constant. Bandingkan pemalar: 9 – 2b = 1, –2a = –6 Compare the constants: 2b = 8 a = 3 b = 4 (a) Diberi fungsi f : x → 4x + k, g : x → x – 3 dan fg : x → mx – 8. Cari nilai bagi k dan m. Given the function f : x → 4x + k, g : x → x – 3 and fg : x → mx – 8. Find the value of k and of m. fg(x) = f(x – 3) = 4(x – 3) + k = 4x – 12 + k 4x – 12 + k = mx – 8 Bandingkan pemalar: Compare the constants: 4x = mx, –12 + k = –8 m = 4 k = 4 (b) Diberi fungsi f : x → 5x – 1, g : x → 2x dan gf : x → ax + b. Cari nilai bagi a dan b. Given the function f : x → 5x – 1, g : x → 2x and gf : x → ax + b. Find the value of a and of b. gf(x) = g(5x – 1) = 2(5x – 1) = 10x – 2 10x – 2 = ax + b Bandingkan pemalar: Compare the constants: 10x = ax, b = –2 a = 10 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 14 CONTOH Fungsi Songsang Inverse Functions 1.3 STUDI 1 Minit Buku Teks: 20 – 30 Suatu fungsi f yang berhubungan satu dengan satu mempunyai fungsi songsangan f –1. A function f with a one-to-one relation has an inverse function f–1. x y f f –1 Langkah-langkah menentukan f –1: Steps to determine f–1: 1 Ungkapkan fungsi f(x) = y sebagai x dalam sebutan y. Rewrite the function f(x) = y as x in terms of y. 2 Tulis x sebagai f –1(y). Write x as f –1(y). 3 Gantikan pemboleh ubah y dengan x. Replace variable y with x. CONTOH/EXAMPLE: f(x) = 2x + 6 Katakan/Let f(x) = y 2x + 6 = y x = y – 6 2 Oleh kerana / Since f –1(y) = x, f –1(y) = y Jadi / So, f –1(x) = x – 6 2 13. Cari nilai bagi setiap yang berikut. TP3 Find the values for each of the following. Dalam gambar rajah anak panah yang berikut, fungsi f memetakan x kepada y. Cari In the following arrow diagram, the function f maps x to y. Find 5 4 f x y –1 2 –2 –8 (i) f(2) (ii) f −1(−8) (iii)f −1(−2) (iv) f(4) Penyelesaian: (i) f(2) = 4 (ii) f(−1) = −8, f −1(−8) = −1 (iii)f(5) = −2, f −1(−2) = 5 (iv) f(2) = 4, f −1(4) = 2 (a) Dalam gambar rajah anak panah yang berikut, fungsi f memetakan x kepada y. Cari In the following arrow diagram, the function f maps x to y. Find 4 2 f x y –6 1 3 – 4 –1 (i) f(1) (ii) f −1(−1) (iii)f −1(2) (iv) f −11 3 4 2 (i) f(1) = −1 (ii) f(1) = −1, f −1(−1) = 1 (iii)f(4) = 2, f −1(2) = 4 (iv) f(−6) = 3 4 , f −11 3 4 2 = −6 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 15 CONTOH CONTOH Fungsi f ditakrifkan oleh f(x) = 8 x – 1 , x ≠ 1. Cari Function f is defined by f(x) = 8 x – 1 , x ≠ 1. Find (i) f(3) (ii) f(5) (iii) f –1(2) (iv) f –1(4) Penyelesaian: (i) f(3) = 8 3 – 1 = 4 (ii) f(5) = 8 5 – 1 = 2 (iii)f –1(2) = k f(k) = 2 8 k – 1 = 2 k = 5 (iv) f –1(4) = x f(x) = 4 8 x – 1 = 4 x = 3 (b) Diberi g(x) = 3x x – 2 , x ≠ 2. Cari Given that g(x) = 3x x – 2 , x ≠ 2. Find (i) f(4) (ii) f(3) (iii) f −1(9) (iv) f −1(10) (i) f(4) = 3(4) 4 – 2 = 6 (ii) f(3) = = 9 (iii) f –1(9) = k f(k) = 9 3k k – 2 = 9 3k = 9k – 18 6k = 18 k = 3 (iv) f –1(10) = k f(k) = 10 3k k – 2 = 10 3k = 10k – 20 7k = 20 k = 20 7 14. Tentukan sama ada setiap fungsi f berikut mempunyai fungsi songsang atau tidak. Beri satu sebab untuk jawapan anda. TP3 Determine whether each of the following function f has an inverse function or not. Give a reason for your answer. 0 x f(x) f bukan fungsi satu dengan satu kerana garis mengufuk memotong graf itu pada dua titik. f tidak ada fungsi songsang. f is not a one-to-one function because the horizontal line cuts the graph at two points. f has no inverse function. (a) f x y p –3 r q 5 7 f ialah fungsi satu dengan satu. f mempunyai fungsi songsang. f is a one-to-one function. f has an inverse function. Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 16 CONTOH (b) 0 x f(x) f ialah fungsi satu dengan satu. f mempunyai fungsi songsang. f is a one-to-one function. f has an inverse function. (c) 0 x f(x) f bukan fungsi satu dengan satu kerana garis mengufuk memotong graf itu pada dua titik. f tidak ada fungsi songsang. f is not a one-to-one function because the horizontal line cuts the graph at two points.. f has no inverse function. 15. Rajah berikut menunjukkan graf bagi fungsi satu dengan satu, f. Dalam setiap kes, lakar graf bagi f −1. TP4 The following diagrams show the graph of one-to-one function, f. In each case, sketch the graph of f −1. 0 (0, 1) (1, 0) (3, 5) (5, 3) x f(x) y = x f (a) 0 5 –3 –3 5 x f(x) f –1 y = x f (b) 0 4 4 x f(x) f –1 y = x f (c) 0 3 3 x f(x) f –1 y = x f 8 8 Tip Penting (1, 0) → (0, 1) (5, 3) → (3, 5) Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 17 CONTOH CONTOH 16. Cari fungsi songsang bagi setiap fungsi yang berikut. TP3 Find the inverse function for each of the following functions. f : x → x – 8 Penyelesaian: Katakan/ Let f –1(x) = y f(y) = x y – 8 = x y = x + 8 \ f –1(x) = x + 8 f –1 : x → x + 8 (a) f : x → 3 – 4x Katakan f –1(x) = y Let f(y) = x 3 – 4y = x y = 3 – x 4 f –1(x) = 3 – x 4 (b) g : x → 5x Katakan g –1(x) = y Let g(y) = x 5y = x y = x 5 g –1(x) = x 5 (c) h : x → – 2 x – 1 , x ≠ 1 Katakan h–1(x) = y Let h(y) = x 2 y – 1 = x y – 1 = 2 x y = 2 x + 1 h–1(x) = 2 x + 1, x ≠ 0 17. Tentukan fungsi gubahan berikut yang melibatkan fungsi sonsang. TP4 Determine the following composite function involving inverse function. Diberi dua fungsi f(x) → x – 3 dan g(x) = 2 x + 1, x ≠ –1. Cari Given two functions f(x) → x – 3 and g(x) = 2 x + 1 , x ≠ 1. Find (i) f –1g (ii) gf –1 Penyelesaian: Katakan/ Let f –1(x) = y f(y) = x y – 3 = x y = x + 3 \ f –1(x) = x + 3 (i) f –1g(x) = f –11 2 x + 12 = 2 x + 1 + 3 = 2 + 3(x + 1) x + 1 = 5 + 3x x + 1 , x ≠ –1 (ii) gf –1 = g(x + 3) = 2 x + 3 + 1 = 2 x + 4, x ≠ –4 1 Katakan f –1(x) = y Let f –1(x) = y 2 Jadikan y sebagai perkara rumus. Make y into subject of the equation. 3 Gantikan y ke dalam f –1(x) = y Insert y into f –1(x) = y Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 18 (a) Diberi dua fungsi f(x) = 6 – 3x dan g(x) = 1 2 – 1 , x ≠ 1 2 . Cari Given two function f(x) = 6 – 3x and g(x) = 1 2x – 1 , x ≠ 1 2 . Find (i) f –1g (ii) gf –1 Katakan f –1(x) = y f(y) = x 6 – 3y = x y = 6 – x 3 f –1(x) = 6 – x 3 (i) f –1g(x) = f –1 1 1 2 – 1 2 = 6 – 1 1 2x – 1 2 3 = 6(2x – 1) – 1 3(2x – 1) = 12 – 7 6 – 3 , x ≠ (ii) gf –1(x) = g1 6 – x 3 2 = 1 21 6 – x 3 2 – 1 = 1 1 12 – 2x – 3 3 2 = 3 9 – 2x , x ≠ 9 2 (b) Diberi fungsi f : x → 7 – 4x. Cari Given the function f : x → 7 – 4x. Find (i) ff –1 (ii) f –1f Katakan f –1(x) = y f(y) = x 7 – 4y = x y = (7 – x) 4 f –1(x) = 7 – x 4 (i) ff –1(x) = f1 7 – x 4 2 = 7 – 4 1 7 – x 4 2 = 7 – 7 + x = x (ii) f –1f(x) = f –1(7 – 4x) = 7 – (7 – 4x) 4 = 7 – 7 + 4x 4 = 4x 4 = x (c) Diberi f(x) = 3 2x , x ≠ 0 dan g(x) = 4 + x. Cari It is given that f(x) = 3 2x , x ≠ 0 and g(x) = 4 + x. Find (i) f g(x) (ii) f g(g–1) (i) f g(x) = 3 2g(x) = 3 2(4 + x) = (7 – x) 4 , x ≠ –4 (ii) Katakan / Let y = 4 + x x = y – 4 g–1(y) = y – 4 g–1(x) = x – 4 fg(g–1) = 3 8 + 2g–1(x) = 3 8 + 2(x – 4) = 3 2x , x ≠ 0 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 19 Kertas 1 1. (a) Dalam rajah di bawah, fungsi f memetakan set X kepada set Y dan fungsi g memetakan set Y kepada set Z. In the diagram below, function f maps set X to set Y and function g maps set Y to set Z. x X Y Z 3x – 7 g(y) f g Diberi g(y) = 4 y – 2 , y ≠ 2, cari fungsi yang memetakan set Z kepada set X dalam sebutan x. Given that g(y) = 4 y – 2 , y ≠ 2, find the function which maps set Z to set X in terms of x. (b) Rajah menunjukkan suatu graf fungsi y = h(x). The diagram shows the graph of the function y = h(x). y x y = h(x) (0, 3) 0 Pada rajah yang sama, lakarkan graf h–1(x). On the same diagram, sketch the graph of h–1(x). (a) Katakan / Let gf(x) = k(x) Maka / Then, k(x) = 4 f(x) – 2 k(x) = 4 3x – 7 – 2 k(x) = 4 3x – 9 , x ≠ 3 Katakan/ Let k(x) = y y = 4 3x – 9 Buku Teks ms.12-30 3x – 9 = 4 y x = 4 3y + 3 Maka/ Then, k–1(x) = 4 3y + 3, x ≠ 0. (b) y x y = h(x) 0 (0, 3) (1, 0) 2. Diberi fungsi g(x) = x 4 − 3 dan fg(x)= x 2 − 7, cari Given function g(x) = x 4 − 3 and fg(x) = x 2 − 7, find (a) g –1(x), (b) f(x), (c) fg –1(4). (a) Katakan / Let g(x) = y x 4 − 3 = y x = 4y + 12 g –1(x) = 4x + 12 (b) fg(g–1) = g–1 2 – 7 f(x) = 4x + 12 4 – 7 f(x) = 2x – 1 (c) g–1(4) = 4(4) + 12 g–1(4) = 28 fg–1(4) = f(28) = 2(28) – 1 = 55 Buku Teks ms.20-30 1 Praktis Ekstra SPM 1 Praktis SPM Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 20 3. Diberi fungsi p(x) = mx + 3, q(x) = 2x − 7 dan pq(x) = 2mx + n, ungkapkan m dalam sebutan n. Given the functions p(x) = mx + 3, q(x) = 2x − 7 and pq(x) = 2mx + n, express m in terms of n. pq(x) = m[q(x)] + 3 = m(2x – 7) + 3 = 2mx – 7m + 3 Membandingkan dengan pq(x) = 2mx + n, Comparing with pq(x) = 2mx + n, n = –7m + 3 m = 3 – n 7 4. Diberi fungsi f(x) = 3x − 7, cari Given the function f(x) = 3x − 7, find (a) f –1 (x), (b) nilai k jika f2 1 4s 3 2 = 20. the value of k if f2 1 4s 3 2 = 20. (a) Katakan / Let 3x – 7 = y x = y + 7 3 f −1(x) = x + 7 3 (b) f 2 (x) = 3(3x – 7) – 7 f 2 (x) = 9x – 28 f 21 4s 3 2 = 91 4s 3 2 – 28 20 = 12s – 28 s = 4 Kertas 2 Klu Soalan Pintasan-x iaitu nilai x apabila y = 0 perlu dicari sebagai titik di mana suatu graf linear dipantul menjadi bentuk-V. The x-intercept, which is value of x when y = 0, must be determined as it is the point at which the linear graph is reflected to form a V-shape. 1. Fungsi f dan g masing-masing ditakrifkan oleh f : x → 4 – 5x dan g : x → 3x – 5. The functions f and g are defined as f : x → 4 – 5x and g : x → 3x – 5. Buku Teks ms.12-19 Buku Teks ms.12-30 Buku Teks ms. 2-19 (a) (i) Cari/ Find f(4), (ii) Seterusnya, cari nilai p jika g(p + 3) = f(4). Hence, find the value of p if g(p + 3) = 1 8 f(4). (b) (i) Tentukan gf(x) Determine gf(x). (ii) Seterusnya lakarkan graf y = ugf(x)u untuk –1 < x < 2. Nyatakan julat bagi y. Hence, sketch the graph of y = ugf(x)u for –1 < x < 2. State the range of y. (a) (i) f(4) = 4 – 5(4) = –16 (ii) g(p + 3) = 1 8 f(4) 3(p + 3) – 5 = 1 8 (–16) 3p + 9 – 5 = –2 3p = –6 p = –2 (iii) gf(x) = g(4 – 5x) = 3(4 – 5x) – 5 = 12 – 15x – 5 = 7 – 15x (b) x –1 0 7 15 2 y 22 7 0 23 y 0 2 22 7 23 x –1 7 ––15 Julat/Range y : 0 < y < 23 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 21 Klu Soalan Titik pusingan (x, y) bagi graf kuadratik ditentukan dengan mengambil x = purata pintasan-x iaitu 2 + 4 2 = 3 dan y = |gf –1(3)|. The turning point (x, y) of a quadratic graph is determined by taking x = average of the x-intercepts which is 2 + 4 2 = 3 and y = |gf –1(3)|. 2. Fungsi f dan g diberi oleh f(x) = 4 – x dan g(x) = ax2 + bx, dengan keadaan a dan b ialah pemalar. The functions f and g are given by f(x) = 4 – x and g(x) = ax2 + bx where a and b are constants. (a) Diberi fungsi gubahan gf –1 → x2 – 6x + 8, tentukan nilai a dan nilai b. Given the composite function gf–1(x) = x2 – 6x + 8, determine the value of a and of b. (b) Lakar graf y = |gf −1(x)| untuk domain 0 < x < 7 dan nyatakan julat yang sepadan dengannya. Sketch the graph of y = |gf −1(x)| for the domain 0 < x < 7 and state the corresponding range. (a) f(x) = 4 – x Mengambil/ Taking y = f(x), 4 – x = y 4 – y = x f –1(x) = 4 – x Diberi / Given: gf –1(x) = x2 – 6x + 8 g(4 – x) = x2 – 6x + 8 Mengambil/ Taking, 4 – x = u Buku Teks ms. 2-30 x = 4 – u g(u) = (4 – u)2 – 6(4 – u) + 8 = 16 – 4u – 4u + u2 – 24 + 6u + 8 = u2 – 2u g(x) = x2 – 2x Membanding dengan g(x) = ax2 + bx Compairing with g(x)= ax2 + bx, a = 1, b = –2 (b) y = |gf –1(x)| = |x2 – 6x + 8| = |(x – 2)(x – 4)| Apabila/When y = 0: x = 2, x = 4 Apabila/When x = 0: y = 8 Apabila/When x = 7: y = |(7 – 2)(7 – 4)| = 15 0 (3, 1) 8 2 x f(x) 15 4 7 Julat/Range : 0 < f(x) < 15 Penerbitan Pelangi Sdn. Bhd.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi BAB 1 22 Rajah di atas menunjukkan pemetaan y kepada x di bawah fungsi f(y) = 5y – 3 dan pemetaan y kepada z di bawah fungsi g(y) = my , y ≠ 1 4 . Diagram above shows the mapping of y onto x by the function f(y) = 5y – 3 and the mapping of y onto z by the function of g(y) = m 4y – 1 , y ≠ 1 4 . (a) Cari nilai m. Find the value of m. (b) Cari fungsi yang memetakan x kepada y. Find the function which maps x onto y. (c) Cari fungsi yang memetakan x kepada z. Find the function which maps x onto z. (d) Lakar rajah untuk menunjukkan pemetaan fg(1). Sketch a diagram to show the mapping of fg(1). KBAT Sudut KBAT KBAT EKSTRA Jawapan Bab 1 Enrolment key m*mAtT4*M x 1 2 –2 y z (a) g(y) = m 4y – 1 Daripada rajah / From the diagram, g(1) = 2 m = –2 m = –6 (b) f(y) = 5y – 3 Katakan/ Let f –1(y) = x f(x) = y 5x – 3 = y x = + 3 5 f –1(y) = y + 3 5 Jadi/ So, f –1(x) = x + 3 5 (c) Fungsi yang memetakan x kepada z ialah gf –1(x). The function that mapes x onto z is gf –1(x). gf –1(x) = g1 x + 3 5 2 = –6 41 x + 3 5 2 – 1 = –6 1 4x + 12 – 5 5 2 gf –1(x) = – 30 4x + 7 , x ≠ – 7 4 (d) fg(1) = f (–2) = 5(–2) – 3 = –13 1020c Penerbitan Pelangi Sdn. Bhd.
TINGKATAN KSSM ModulE ModulE & MoRE MoRE & RC184133S PELANGI MATEMATIK TAMBAHAN Additional Mathematics Judul dalam Siri Module & More Imbas kod QR atau layari link bagi POT untuk Create new account. 1 2 Semak e-mel untuk mengaktifkan akaun. 3 Log in ke akaun anda. Masukkan Enrolment Key yang boleh dijumpai di halaman akhir setiap bab buku. 4 5 Mulakan ujian! Eksklusif Eksklusif PELANGI ONLINE TEST (Portal Ujian Soalan Objektif) Cara Mengakses POT Subjek/Tingkatan Biologi Biology Fizik Physics Kimia Chemistry Matematik Mathematics Matematik Tambahan Additional Mathematics Sejarah 4 5 CIRI-CIRI HEBAT Meningkatkan pemahaman teks melalui penggunaan bahasa Melayu dan bahasa Inggeris. LEMBARAN PBD DWIBAHASA Lembaran kerja PBD tambahan yang meliputi semua bab disediakan dalam kod QR. QR Kod JAWAPAN LENGKAP Jawapan boleh disemak terus di belakang setiap bab atau di halaman Kandungan melalui kod QR. QR Kod KLU SOALAN Tip membantu menjawab soalan-soalan Kertas 2 di bahagian Praktis SPM. STUDI 1 MINIT Fakta penting bagi setiap bab yang telah diringkaskan untuk mempercepat ulang kaji. PRAKTIS SPM & EKSTRA Latihan berorientasikan SPM disediakan di akhir setiap bab dan juga di dalam kod QR. QR Kod i-THINK/KBAT Penerapan soalan-soalan berformat i-THINK dan berbentuk aras tinggi.