Module & MORE Pembelajaran BERPANDU dan SISTEMATIK Additional Mathematics MATEMATIK TAMBAHAN Tee Hock Tian 4 K S S M TINGKATAN TINGKATAN
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iii BAB Fungsi 1 Functions 1 1.1 Fungsi ..................................................................................1 Functions 1.2 Fungsi Gubahan.................................................................7 Composite Functions 1.3 Fungsi Songsang ..............................................................13 Inverse Functions Praktis SPM 1............................................................................19 Sudut KBAT ..............................................................................21 Online Quick Quiz QR code .......................................................21 BAB Fungsi Kuadratik 22 2 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 Sudut KBAT ..............................................................................43 Online Quick Quiz QR code .......................................................43 BAB Sistem Persamaan 44 3 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 Sudut KBAT ..............................................................................59 Online Quick Quiz QR code .......................................................59 BAB Indeks, Surd dan Logaritma 60 4 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 Sudut KBAT ..............................................................................82 Online Quick Quiz QR code .......................................................82 BAB Janjang 83 5 Progressions 5.1 Janjang Aritmetik.............................................................83 Arithmetic Progressions 5.2 Janjang Geometri.............................................................91 Geometric Progressions Praktis SPM 5..........................................................................101 Sudut KBAT ............................................................................105 Online Quick Quiz QR code .....................................................105 BAB Hukum Linear 106 6 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 Sudut KBAT ............................................................................125 Online Quick Quiz QR code .....................................................125 KANDUNGAN
iv BAB Geometri Koordinat 126 7 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 Sudut KBAT ............................................................................143 Online Quick Quiz QR code .....................................................143 BAB Vektor 144 8 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 Sudut KBAT ............................................................................164 Online Quick Quiz QR code .....................................................164 BAB Penyelesaian Segi Tiga 165 9 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 Sudut KBAT ............................................................................178 Online Quick Quiz QR code .....................................................178 BAB Nombor Indeks 179 Index Numbers 10 10.1 Nombor Indeks..............................................................179 Index Numbers 10.2 Indeks Gubahan.............................................................185 Composite Index Praktis SPM 10 .......................................................................190 Sudut KBAT ............................................................................193 Online Quick Quiz QR code .....................................................194 Kertas Pra-SPM.............................................................. 195 Jawapan http://www.epelangi.com/Module&More2021/ MatematikTambahan/T4/JawapanKeseluruhan.pdf Lembaran PBD http://www.epelangi.com/Module&More2021/ MatematikTambahan/T4/LPBD.pdf Jawapan Lembaran PBD http://www.epelangi.com/Module&More2021/ MatematikTambahan/T4/JawapanLPBD.pdf BONUS untuk Guru Lembaran PBD dengan Jawapan http://www.epelangi.com/Module&More2021/MatematikTambahan/T4/ BonusLPBD.
v © Penerbitan Pelangi Sdn. Bhd. Rekod Pencapaian Pentaksiran Murid Matematik Tambahan Tingkatan 4 / Additional Mathematics Form 4 Nama murid: Kelas: Student’s name Class Bab Chapter Tahap penguasaan Performance level Deskriptor Descriptor Penguasaan Achievement (3) Menguasai Achieve (7) Belum Menguasai Not yet achieve 1 FUNGSI FUNCTIONS 1 Mempamerkan pengetahuan asas tentang fungsi. Demonstrate the basic knowledge of functions. 2 Mempamerkan kefahaman tentang fungsi. Demonstrate the understanding of functions. 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. Apply the understanding of functions to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of functions in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of functions in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of functions in the context of non-routine problem solving in a creative manner. 2 FUNGSI KUADRATIK QUADRATIC FUNCTIONS 1 Mempamerkan pengetahuan asas tentang fungsi kuadratik. Demonstrate the basic knowledge of quadratic functions. 2 Mempamerkan kefahaman tentang fungsi kuadratik. Demonstrate the understanding of quadratic functions. 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. Apply the understanding of quadratic functions to perform simple tasks.
© Penerbitan Pelangi Sdn. Bhd. Matematik Tambahan Tingkatan 4 Rekod Pencapaian Pentaksiran Murid vi 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of quadratic functions in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of quadratic functions in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of quadratic functions in the context of non-routine problem solving in a creative manner. 3 SISTEM PERSAMAAN SYSTEMS OF EQUATIONS 1 Mempamerkan pengetahuan asas tentang sistem persamaan. Demonstrate the basic knowledge of systems of equations. 2 Mempamerkan kefahaman tentang penyelesaian sistem persamaan. Demonstrate the understanding of systems of equations. 3 Mengaplikasikan kefahaman tentang sistem persamaan untuk melaksanakan tugasan mudah. Apply the understanding of systems of equations to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sistem persamaan dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of systems of equations in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sistem persamaan dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of systems of equations in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sistem persamaan dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of systems of equations in the context of non-routine problem solving in a creative manner. 4 INDEKS, SURD DAN LOGARITMA INDICES, SURDS AND LOGARITHMS 1 Mempamerkan pengetahuan asas tentang indeks, surd dan logaritma. Demonstrate the basic knowledge of indices, surds and logarithms. 2 Mempamerkan kefahaman tentang indeks, surd dan logaritma. Demonstrate the understanding of indices, surds and logarithms.
© Penerbitan Pelangi Sdn. Bhd. Matematik Tambahan Tingkatan 4 Rekod Pencapaian Pentaksiran Murid vii 3 Mengaplikasikan kefahaman tentang indeks, surd dan logaritma untuk melaksanakan tugasan mudah. Apply the understanding of indices, surds and logarithms to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang indeks, surd dan logaritma dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of indices, surds and logarithms in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang indeks dan logaritma dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of indices and logarithms in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang indeks dan logaritma dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of indices and logarithms in the context of non-routine problem solving in a creative manner. 5 JANJANG PROGRESSIONS 1 Mempamerkan pengetahuan asas tentang janjang. Demonstrate the basic knowledge of progressions. 2 Mempamerkan kefahaman tentang janjang aritmetik dan janjang geometri. Demonstrate the understanding of arithmetic progressions and geometric progressions. 3 Mengaplikasikan kefahaman tentang janjang aritmetik dan janjang geometri untuk melaksanakan tugasan mudah. Apply the understanding of arithmetic progressions and geometric progressions to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang janjang aritmetik dan janjang geometri dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of arithmetic progressions and geometric progressions in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang janjang aritmetik dan janjang geometri dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of arithmetic progressions and geometric progressions in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang janjang aritmetik dan janjang geometri dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of arithmetic progressions and geometric progressions in the context of non-routine problem solving in a creative manner.
© Penerbitan Pelangi Sdn. Bhd. Matematik Tambahan Tingkatan 4 Rekod Pencapaian Pentaksiran Murid viii 6 HUKUM LINEAR LINEAR LAW 1 Mempamerkan pengetahuan asas tentang garis lurus penyuaian terbaik. Demonstrate the basic knowledge of lines of best fit. 2 Mempamerkan kefahaman tentang garis lurus penyuaian terbaik. Demonstrate the understanding of lines of best fit. 3 Mengaplikasikan kefahaman tentang hukum linear untuk melaksanakan tugasan mudah. Apply the understanding of linear law to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of linear law in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of linear law in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of linear law in the context of non-routine problem solving in a creative manner. 7 GEOMETRI KOORDINAT COORDINATE GEOMETRY 1 Mempamerkan pengetahuan asas tentang pembahagi tembereng garis. Demonstrate the basic knowledge of divisor of line segments. 2 Mempamerkan kefahaman tentang pembahagi tembereng garis. Demonstrate the understanding of divisor of line segments. 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah. Apply the understanding of coordinate geometry to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang geometri koordinat dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of coordinate geometry in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang geometri koordinat dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of coordinate geometry in the context of complex routine problem solving.
© Penerbitan Pelangi Sdn. Bhd. Matematik Tambahan Tingkatan 4 Rekod Pencapaian Pentaksiran Murid ix 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang geometri koordinat dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of coordinate geometry in the context of non-routine problem solving in a creative manner. 8 VEKTOR VECTORS 1 Mempamerkan pengetahuan asas tentang vektor. Demonstrate the basic knowledge of vectors. 2 Mempamerkan kefahaman tentang vektor. Demonstrate the understanding of vectors. 3 Mengaplikasikan kefahaman tentang vektor untuk melaksanakan tugasan mudah. Apply the understanding of vectors to perform simple tasks 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang vektor dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of vectors in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang vektor dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of vectors in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang vektor dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of vectors in the context of nonroutine problem solving in a creative manner. 9 PENYELESAIAN SEGI TIGA SOLUTION OF TRIANGLES 1 Mempamerkan pengetahuan asas tentang petua sinus dan petua kosinus. Demonstrate the basic knowledge of sine rule and cosine rule. 2 Mempamerkan kefahaman tentang petua sinus dan petua kosinus. Demonstrate the understanding of sine rule and cosine rule. 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah. Apply the understanding of sine rule, cosine rule and area of a triangle to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang penyelesaian segi tiga dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of solution of triangles in the context of simple routine problem solving.
© Penerbitan Pelangi Sdn. Bhd. Matematik Tambahan Tingkatan 4 Rekod Pencapaian Pentaksiran Murid x 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang penyelesaian segi tiga dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of solution of triangles in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang penyelesaian segi tiga dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of solution of triangles in the context of non-routine problem solving in a creative manner. 10 NOMBOR INDEKS INDEX NUMBERS 1 Mempamerkan pengetahuan asas tentang nombor indeks. Demonstrate the basic knowledge of index numbers. 2 Mempamerkan kefahaman tentang nombor indeks. Demonstrate the understanding of index numbers. 3 Mengaplikasikan kefahaman tentang nombor indeks untuk melaksanakan tugasan mudah. Apply the understanding of index numbers to perform simple tasks. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang nombor indeks dalam konteks penyelesaian masalah rutin yang mudah. Apply appropriate knowledge and skills of index numbers in the context of simple routine problem solving. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang nombor indeks dalam konteks penyelesaian masalah rutin yang kompleks. Apply appropriate knowledge and skills of index numbers in the context of complex routine problem solving. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang nombor indeks dalam konteks penyelesaian masalah bukan rutin secara kreatif. Apply appropriate knowledge and skills of index numbers in the context of non-routine problem solving in a creative manner.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 1 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA 1 BAB Fungsi Functions 1.1 Fungsi Functions Functions / Fungsi 7 4 3 16 9 64 49 Domain = {3, 4, 7} Kodomain / Codomain = {9, 16, 49, 64} Objek / Objects = 3, 4, 7 Julat / Range = {9,16,49} Fungsi mutlak, f(x) = |x| Absolute function, f(x) = |x| |x| = x, jika / if x > 0 –x, jika / if x > 0 0 x f(x) Fungsi / Function f(x) 5 3 2 x y 27 125 8 Tatatanda fungsi / Function notation f : x → x3 f(x) = x3 0 x f(x) Ujian garis mencancang / Vertical line test Bukan fungsi / Not a function f(x) z y x x y 4 2 7 6 0 x f(x) Garis mencancang memotong pada dua titik Vertical line cuts at 2 points
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 2 BAB 1 1. Tentukan sama ada setiap hubungan yang berikut ialah satu fungsi atau bukan. Beri sebab untuk jawapan anda. TP 1 Determine whether each of the following relations is a function. Give reason for your answer. CONTOH 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) 13 9 6 y x z Bukan fungsi. 6 mempunyai lebih daripada satu imej. Not a function. 6 has more than 1 image. 2. Menggunakan ujian garis mencancang untuk menentukan sama ada graf berikut ialah satu fungsi atau bukan. Using the vertical line test to determine whether each of the following graph is a function or not. TP 2 CONTOH 0 3 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. (a) 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.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 3 BAB 1 (b) 0 –4 4 –9 9 x y Bukan fungsi. Garis mencancang memotong graf itu pada lebih daripada satu titik. Not a function. The vertical line cuts the graph at more than one point. (c) 0 5 – 2 x y –2 Satu fungsi. Garis mencancang memotong graf itu pada satu titik sahaja. A function. The vertical line cuts the graph at only one point. 3. Tulis setiap fungsi yang berikut menggunakan tatatanda fungsi. TP 2 Write each of the following by using the function notation. CONTOH –3 –27 f 5 4 64 125 f : x → x3 f(x) = x3 (a) –5 5 g 4 –2 7 11 g : x → |2x – 1| g(x) = |2x – 1| (b) h 6 5 3 1 – 9 1 – 25 1 – 36 h : x → 1 x2 h(x) = 1 x2 (c) k 4 3 1 14 11 5 k : x → 3x + 2 k(x) = 3x + 2
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 4 BAB 1 4. Lengkapkan jadual yang berikut. TP 1 Complete the following table. Fungsi Function Domain Domain Kodomain Codomain Objek Object Imej Image Julat Range CONTOH 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}
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 5 BAB 1 5. Lakar graf bagi setiap fungsi berikut. Seterusnya, nyatakan julat sepadan bagi domain yang diberikan. Sketch the graph of the following functions. Hence, state the corresponding range for the given domain. TP 4 CONTOH 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 6. Selesaikan setiap yang berikut. TP 5 Solve each of the following. CONTOH 1 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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 6 BAB 1 CONTOH 2 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 x – 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 x . the values of x for which f(x)= 3 x . (i) f(x) = 12 x – 2 f(–4) = 12 –6 = –2 f(10) = 12 8 = 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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 7 BAB 1 (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 1.2 Fungsi Gubahan Composite Functions NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN x f(x) gf(x) A f g B gf C Dalam rajah, fungsi f memetakan set A kepada set B dan fungsi g memetakan set B kepada set C. Fungsi yang memetakan set A kepada set C ialah fungsi gubahan, gf. In the diagram, the function f maps set A to set B and the function g maps set B to set C. The function which maps set A to set C is the composite function, gf. 7. Bagi setiap pasangan fungsi yang berikut, cari For each pair of the following function, find TP 3 (i) fg(x), (ii) gf(x). CONTOH f(x) = 2x + 3 g(x) = 1 – x Penyelesaian: (i) fg(x) = f(1 – x) fg(x) = f(g(x)) = 2(1 – x) + 3 = 2 – 2x + 3 = 5 – 2x (ii) gf(x) = g(2x + 3) gf(x) = g(f(x)) = 1 – (2x + 3) = 1 – 2x – 3 = –2 – 2x (a) f(x) = 3x + 5 g(x) = 1 – 2x (i) fg(x) = f(1 – 2x) = 3(1 – 2x) + 5 = 3 – 6x + 5 = 8 – 6x (ii) gf(x) = g(3x + 5) = 1 – 2(3x + 5) = 1 – 6x – 10 = –9 – 6x
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 8 BAB 1 (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 8. Bagi setiap pasangan fungsi yang berikut, cari For each pair of the following functions, find TP 3 (i) fg(3) (ii) gf(4) CONTOH 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 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. (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) = 3 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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 9 BAB 1 9. Bagi setiap pasangan fungsi yang berikut, cari For each pair of the following functions, find TP 3 (i) fg(x) (ii) gf(x) (iii) fg(2) (iv) gf(3) CONTOH f(x) = 2x, g(x) = x + 5 Penyelesaian: (i) fg(x) = f(x + 5) = 2(x + 5) = 2x + 10 (ii) gf(x) = g(2x) = 2x + 5 (iii)fg(2) = 2(2) + 10 = 14 (iv) gf(3) = 2(3) + 5 = 11 (a) f(x) = 3x + 1, g(x) = 2 – x (i) fg(x) = f(2 – x) = 3(2 – x) + 1 = 7 – 3x (ii) gf(x) = g(3x + 1) = 2 – (3x + 1) = 1 – 3x (iii)fg(2) = 7 – 3(2) = 1 (iv) gf(3) = 1 – 3(3) = –8 (b) f(x) = x – 6, g(x) = x2 + 1 (i) fg(x) = f(x2 + 1) = x2 + 1 – 6 = x2 – 5 (ii) gf(x) = g(x – 6) = (x – 6)2 + 1 = x2 – 12x + 37 (iii)fg(2)= 22 – 5 = 4 – 5 = –1 (iv) gf(3)= (3)2 – 12(3) + 37 = 9 –36 + 37 = 10 (c) f : x → 2 + 5x, g : x → x x – 1 , x ≠ 1 (i) fg(x) = f1 x x – 1 2 = 2 + 51 x x – 1 2 = 2(x – 1) + 5x x – 1 = 7x – 2 x – 1 , x ≠ 1 (ii) gf(x) = g(2 + 5x) = 2 + 5x 2 + 5x – 1 = 2 + 5x 1 + 5x , x ≠ – 1 5 (iii)fg(2) = 7(2) – 2 2 – 1 = 12 (iv) gf(3) = 2 + 5(3) 1 + 5(3) = 17 16
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 10 BAB 1 10. Selesaikan setiap yang berikut. Solve each of the following. TP 4 CONTOH 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. Find the function g. TP 4 CONTOH 1 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 1 Gantikan g(x) ke dalam f(x). Insert g(x) into f(x). (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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 11 BAB 1 (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 CONTOH 2 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 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). (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 y 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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 12 BAB 1 12. Selesaikan setiap yang berikut. Solve each of the following. TP 5 CONTOH 1 Fungsi g ditakrifkan sebagai g : x → 4x – 1, cari Function g is defined as g : x → 4x – 1, find (a) g2 (b) g2 1 1 2 2 Penyelesaian: (a) g2 (x) = gg(x) = g(4x – 1) = 4(4x – 1) – 1 = 16x – 5 g2 : x → 16x – 5 (b) g2 1 1 2 2 = 161 1 2 2 – 5 = 3 Fungsi f ditakrifkan sebagai f : x → 4 x , x ≠ k. The function f is defined as f : x → 4 x , x ≠ k. (a) Tentukan nilai k. Determine the value of k. (b) Cari f 2 dan seterusnya cari nilai f 2 (3). Find f 2 and hence find the value of f 2 (3). f(x) = 4 x x ≠ 0 \ k = 0 (b) f 2 (x) = ff(x) = f 1 4 x 2 = 4 1 4 x 2 = x f 2 : x → x f 2 (3) = 3 CONTOH 2 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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 13 BAB 1 (c) 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 (d) 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 1.3 Fungsi Songsang Inverse Functions NOTA IMBASAN NOTA IMBASAN 1. 8 5 2 6 3 A B f 9 Fungsi f memetakan unsur dalam set A kepada unsur dalam set B. The function f maps the elements in set A onto the elements in set B. 8 5 2 6 3 A B f –1 9 Fungsi f –1 memetakan unsur dalam set B kepada unsur dalam set A. The function f –1 maps the elements in set B onto the elements in set A. 2. f –1 ialah fungsi songsang bagi f. f –1 is called the inverse function of f. 3. Hanya fungsi dengan hubungan satu kepada satu mempunyai fungsi songsang. Only a function with one-to-one relation has an inverse function. Tip f –1(x) ≠ 1 f(x)
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 14 BAB 1 NOTA IMBASAN NOTA IMBASAN 4. f dan g ialah fungsi songsang antara satu sama lain jika dan hanya jika f and g are inverse functions of one another if and only if (i) fg(x) = x, x dalam domain g, dan fg(x) = x, x in domain g, and (ii) gf(x) = x, x dalam domain f. gf(x) = x, x in domain f. 5. Jika f dan g ialah fungsi songsang antara satu sama lain, maka If f and g are inverse function of one another then (i) domain f = julat / range g, dan / and (ii) domain g = julat / range f. (iii) Graf g adalah pantulan graf f pada garis y = x. Graph g is a reflection of graph f in the line y = x. 6. Jika f dan g ialah fungsi songsang antara satu sama lain maka titik (a, b) berada pada graf f dan titik (b, a) berada pada graf g. If f and g are inverse function of one another then point (a, b) lies on the graph f and point (b, a) lies on the graph g. 7. Kaedah untuk mendapatkan fungsi songsang: Method to find inverse funtion: (i) Katakan/ Let: f –1(x) = y → f(y) = x. (ii) Jadikan y sebagai perkara rumus bagi f(y) = x. Make y into subject for f(y) = x. (iii) Gantikan y ke dalam f –1(x) = y. Insert y into f –1(x) = y. 13. Cari nilai bagi setiap yang berikut. TP 3 Find the values for each of the following. CONTOH 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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 15 BAB 1 (b) 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 –2 3 3 – 2 6 (i) f(3) (ii) f −1(4) (iii)f −1(6) (iv) f −11 3 2 2 (i) f(3) = 3 2 (ii) f(–2) = 4, f −1(4) = –2 (iii)f(5) = 6, f −1(6) = 5 (iv) f(3) = 3 2 , f −11 3 2 2 = 3 (c) 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 8 –5 f x y –2 3 1 –7 (i) f(–2) (ii) f −1(−7) (iii)f −1(1) (iv) f −1(−5) (i) f(–2) = 1 (ii) f(3) = −7, f −1(−7) = 3 (iii)f(–2) = 1, f −1(1) = –2 (iv) f(8) = –5, f −1(−5) = 8 14. Tentukan sama ada setiap fungsi f berikut mempunyai fungsi songsang atau tidak. Beri satu sebab untuk jawapan anda. TP 3 Determine whether each of the following function f has an inverse function or not. Give a reason for your answer. CONTOH 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.
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 16 BAB 1 (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. TP 4 The following diagrams show the graph of one-to-one function, f. In each case, sketch the graph of f −1. CONTOH 0 (0, 1) (1, 0) (3, 5) (5, 3) x f(x) y = x f Tip (1, 0) → (0, 1) (5, 3) → (3, 5) (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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 17 BAB 1 16. Cari fungsi songsang bagi setiap fungsi yang berikut. TP 3 Find the inverse function for each of the following functions. CONTOH 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 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 (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. Selesaikan setiap yang berikut. Solve each of the following. TP 4 CONTOH 1 Fungsi f ditakrifkan sebagai f : x → 2x – 7. Cari The function f is defined as f : x → 2x – 7. Find (i) f –1(x) (ii) f –1(3) Penyelesaian: (i) Katakan/ Let f –1(x) = y f(y) = x 2y – 7 = x y = x + 7 2 \ f –1(x) = x + 7 2 (ii) f –1(3) = 3 + 7 2 = 5 (a) Fungsi f ditakrifkan sebagai f : x → 2x – 5 x + 2 , x ≠ –2. Cari The function f is defined as f : x → 2x – 5 x + 2 , x ≠ –2. Find (i) f –1(x) (ii) f –1(5) (i) Katakan f –1(x) = y Let f(y) = x 2y – 5 y + 2 = x 2y – 5 = x(y + 2) 2y – 5 = xy + 2x 2y – xy = 2x + 5 y(2 – x) = 2x + 5 y = 2x + 5 2 – x f –1(x) = 2x + 5 2 – x , x ≠ 2 (ii) f –1(5) = 2(5) + 5 2 – 5 = 15 –3 = –5
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 18 BAB 1 CONTOH 2 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 (b) Diberi dua fungsi f(x) = 6 – 3x dan g(x) = 1 2x – 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 2x – 1 2 = 6 – 1 1 2x – 1 2 3 = 6(2x – 1) – 1 3(2x – 1) = 12x – 7 6x – 3 , x ≠ 1 2 (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 (c) 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
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 19 BAB Kertas 1 1 1. Rajah menunjukkan fungsi gubahan fg yang memetakan set A kepada set C. Diagram shows the composite function fg that maps set A to set C. x A 2x fg f B 2x + 1 g C Nyatakan State (a) fungsi yang memetakan set A kepada set C. the function that maps set A to set C. (b) f –1(x) (a) fg(x) = 2x + 1 (b) f –1(x) = x – 1 2. Rajah menunjukkan graf bagi fungsi f : x → |2x – 3| untuk domain –1 < x < 6. Diagram shows the graph of the function f : x → |2x – 3| for the domain –1 < x < 6. f(x) (–1, 5) 0 6 9 x Nyatakan State (a) objek bagi 9, the object of 9, (b) imej bagi 4, the image of 4, (c) domain bagi 0 < f(x) < 5. the domain of 0 < f(x) < 5. (a) 6 (b) f(4) = |2(4) – 3| = 5 (c) bila x = –1, f(x) = 5 bila x = 4, f(x) = 5 domain: –1 < x < 4 2015 2017 3. Diberi fungsi f : x → 4 – 3x, cari Given the function f : x → 4 – 3x, find (a) nilai x apabila f(x) memeta kepada diri sendiri, the value of x when f(x) maps onto itself, (b) nilai h dengan keadaan f(3 – h) = 2h. the value of h such that f(3 – h) = 2h. (a) f(x) = x 4 – 3x = x 4 = x + 3x x = 1 (b) f(3 – h) = 2h 4 – 3(3 – h)= 2h 4 – 9 + 3h = 2h –5 + 3h = 2h 3h – 2h = 5 h = 5 4. Diberi fungsi f : x → 2x – 7, g : x → px + 3 dan gf : x → 2px + 5q. Ungkapkan q dalam sebutan p. Given the functions f : x → 2x – 7, g : x → px + 3 and gf : x → 2px + 5q. Express q in terms of p. gf(x) = g(2x – 7) = p(2x – 7) + 3 2px + 5q = 2px – 7p + 3 5q = –7p + 3 q = 3 – 7p 5 2016 2016 PRAKTIS PRAKTIS SPM SPM 1
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 20 BAB 1 Kertas 2 1. Diberi bahawa f : x → 4 – 5x dan g : x → 3x – 5. It is given that f : x → 4 – 5x and g : x → 3x – 5. (a) Cari Find (i) f(4), (ii) nilai p jika g(p + 3) = 1 8 f(4). the value of p if g(p + 3) = 1 8 f(4). (iii) gf(x) (b) 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 2018 2. Diberi fungsi f : x → 4 – x dan fungsi g : x → ax2 + bx, dengan keadaan a dan b ialah pemalar. Fungsi gubahan gf –1 diberi oleh Given the function f : x → 4 – x and function g : x → ax2 + bx, such that a and b are constants. The composite function gf –1 is given by gf –1 → x2 – 6x + 8 (a) Cari nilai a dan nilai b. Find 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 4 – x = y 4 – y = x f –1(x) = 4 – x gf –1(x) = x2 – 6x + 8 g(4 – x) = x2 – 6x + 8 4 – x = u 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 g(x) = ax2 + bx \ a = 1, b = –2 (b) y = |gf –1(x)| = |x2 – 6x + 8| = |(x – 2)(x – 4)| \ y = 0 → x = 2, x = 4 \ x = 0 → y = 8 \ 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 2018 Praktis SPM Ekstra
Matematik Tambahan Tingkatan 4 Bab 1 Fungsi 21 BAB 1 x 1 2 –2 y z 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) = m 4y – 1 , 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. (a) g(y) = m 4y – 1 g(1) = m 4(1) – 1 = –2 m = –6 (b) f(y) = 5y – 3 Katakan f –1(y) = x Let f(x) = y 5x – 3 = y x = y + 3 5 f –1(y) = y + 3 5 \ f –1(x) = x + 3 5 (c) 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 Sudut Sudut KBAT KBAT KBAT Ekstra Kuiz 1
1. Selesaikan setiap persamaan berikut menggunakan kaedah penyempurnaan kuasa dua. TP 3 Solve each of the following equations using completing the square method. CONTOH −2x2 + 8x + 13 = 0 Penyelesaian: −2x2 + 8x = −13 x2 – 4x = 13 2 x2 – 4x + 1– 4 2 2 2 = 13 2 + 1– 4 2 2 2 (x – 2)2 = 13 2 + 4 √(x – 2)2 = ±√10.5 x – 2 = √10.5 x – 2 = –√10.5 x = 5.240 x = –1.240 Pindah sebutan pemalar ke kanan. Move the constant to the right. Pekali x2 = +1 Coefficient of x2 = +1 Tambah sebutan 1 b 2 2 2 di kedua-dua belah persamaan. Add the term 1 b 2 2 2 on both sides of equation. Masukkan ± apabila mengambil punca kuasa dua. Put in ± when taking square roots. (a) x2 – 8x + 10 = 0 x 2 – 8x = –10 x2 – 8x + 1– 8 2 2 2 = –10 + 1– 8 2 2 2 (x – 4)2 = –10 + 16 √(x – 4)2 = ±√6 x – 4 = √6 atau x – 4 = –√6 = 6.449 or = 1.551 BAB 2 Fungsi Kuadratik Quadratic Functions 2.1 Persamaan dan Ketaksamaan Kuadratik Quadratic Equations and Inequalities NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN Persamaan dan ketaksamaan kuadratik Quadratic equations and inequalities Penyelesaian persamaan Solving equation (a) Penyempurnaan kuasa dua Completing the square (b) Rumus kuadratik Quadratic formula x = –b ± b2 – 4ac 2a Penyelesaian ketaksamaan kuadratik Solving quadratic inequalities (a) Lakaran graf Graph sketching (b) Garis nombor Number line (c) Jadual / Table Hasil tambah punca (HTP) dan hasil darab punca (HDP). Sum of roots (SOR) and product of roots (POR) (a) Punca-punca ialah α dan β Roots are α and β (b) HTP/SOR = a + b = – b a (c) HDP/POR = ab = c a (d) Persamaan / Equation x2 – (HTP)x + HDP = 0 x2 – (SOR)x + POR = 0 22
(b) –2x2 + 13x – 8 = 0 2x2 – 13x + 8 = 0 x2 – 13 2 x = –4 x2 – 13 4 x + 1– 13 4 2 2 = –4 + 1– 13 4 2 2 1x – 13 4 2 2 = – 4 + 169 16 1x – 13 4 2 2 = ± 105 16 x – 13 4 = 105 16 x – 13 4 = – 105 16 x = 5.812 = 0.6883 (c) –3x2 – 18x + 7 = 0 –3x2 – 18x = –7 x2 + 6x = 7 3 x2 + 6x + (3)2 = 7 3 + (3)2 (x + 3)2 = 34 3 √(x + 3)2 = ± 34 3 x + 3 = 34 3 x + 3 = – 34 3 x = 0.3665 x = –6.3665 2. Selesaikan setiap persamaan berikut menggunakan persamaan kuadratik. TP 3 Solve each of the following equations using quadratic equation. CONTOH 4x2 − 17x + 5 = 0 Penyelesaian: ax2 + bx + c = 0 a = 4, b = –17, c = 5 x = –b ± √b2 – 4ac 2a = –(–17) ± √(–17)2 – 4(4)(5) 2(4) = 17 ± √209 8 x = 17 + √209 8 x = 17 – √209 8 = 3.9321 = 0.3179 Bandingkan dengan bentuk am. Compare with the general form. Guna rumus Use the formula (a) –3x2 + 16x – 9 = 0 a = –3, b = 16, c = –9 x = –16 ± √(16)2 – 4(–3)(–9) 2(–3) = –16 + √148 –6 x = –16 + √148 –6 atau x = –16 – √148 –6 = 0.6391 or x = 4.6943 (b) 5x2 – 19x + 8 = 0 a = 5, b = –19, c = 8 x = –b ± √b2 – 4ac 2a x = –(–19) ± √(–19)2 – 4(5)(8) 2(5) = 19 ± √201 10 x = 19 + √201 10 x = 19 – √201 10 = 3.3177 = 0.4823 (c) –7x2 + 22x – 13 = 0 a = –7, b = 22, c = –13 x = –b ± √b2 – 4ac 2a x = –(22) ± √(22)2 – 4(–7)(–13) 2(–7) = –22 ± √120 –14 x = –22 + √120 –14 x = –22 – √120 –14 = 0.7890 = 2.3539 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 23 BAB 2
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 5x2 – 12x – 9 = 0 Penyelesaian: ax2 + bx + c = 0 Bandingkan dengan bentuk am. Compare with the general form. a = 5, b = –12, c = –9 HTP/SOR = – b a = – –12 5 = 12 5 HDP/POR = c a = – 9 5 (a) 6x2 + 14x – 7 = 0 a = 6, b = 14, c = –7 HTP/SOR = – b a = – 14 6 = – 7 3 HDP/POR = c a = – 7 6 (b) –4x2 + 12x – 3 = 0 a = –4, b = 12, c = –3 HTP/SOR = – b a = – 12 –4 = 3 HDP/POR = c a = –3 –4 = 3 4 (c) 9x2 – 6x – 11 = 0 a = 9, b = –6, c = –11 HTP/SOR = – b a = – –6 9 = 2 3 HDP/POR = c a = –11 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 – 4, b – 4 Penyelesaian: a = 3, b = –6, c = –8 a + b = – b a = – –6 3 = 2 ab = c a = – 8 3 HTP/SOR: (a – 4) + (b – 4) = a + b – 8 = 2 – 8 = –6 HDP/POR: (a – 4)(b – 4) = ab – 4a – 4b + 16 = 1– 8 3 2 – 4(a + b) + 16 = 1– 8 3 2 – 4(2) + 16 = 16 3 Persamaan kuadratik / Quadratic equation x2 – (HTP)x + (HDP) = 0 x2 – (–6)x + 16 3 = 0 3x2 + 18x + 16 = 0 Tentukan HTP dan HDP bagi persamaan yang diberi. Determine SOR and POR for the equation given. (a) a + 5, b + 5 HTP/SOR: (a + 5) + (b + 5) = (a + b) + 10 = 2 + 10 = 12 HDP/POR: (a + 5)(b + 5) = ab + 5a + 5b + 25 = 1– 8 3 2 + 5(2) + 25 = 97 3 Persamaan kuadratik / Quadratic equation x2 – (HTP)x + HDP = 0 x2 – (12)x + 1 97 3 2 = 0 3x2 – 36x + 97 = 0 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 24 BAB 2
(b) 3a, 3b HTP/SOR: 3a + 3b = 3(a + b) = 3(2) = 6 HDP/POR: (3a)(3b) = 9ab = 91– 8 3 2 = –24 Persamaan kuadratik / Quadratic equation x2 – (HTP)x + HDP = 0 x2 – 6x + (–24) = 0 x2 – 6x – 24 = 0 (c) a2 , b2 HTP/SOR: a2 + b2 = (a + b)2 – 2ab = 22 – 21– 8 3 2 = 28 3 HDP/POR: (a2 )(b2 ) = (ab)2 = 1– 8 3 2 2 = 64 9 Persamaan kuadratik / Quadratic equation x2 – (HTP)x + HDP = 0 x2 – 1 28 3 2x + 1 64 9 2 = 0 9x2 – 84x + 64 = 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 the 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 a a + b = – 1 –4 1 2 = 4 HTP baharu/ New SOR = 3a + 3b = 3(a + b) = 3(4) = 12 HDP/POR = c a ab = 7 1 = 7 HDP baharu/ New POR = 3a × 3b = 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 a a + b = – 1 –9 3 2 = 3 HTP baharu/ New SOR = 2a + 2b = 2(a + b) = 2(3) = 6 HDP = c a ab = 5 3 HDP baharu/ New POR = 2a × 2b = 4ab = 41 5 3 2 = 20 3 Persamaan baharu ialah New equation is x2 – (HTP)x + HDP = 0 x2 – 6x + 20 3 = 0 3x2 – 18x + 20 = 0 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 25 BAB 2
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. Penyelesaian: a = 2, b = –8, c = p ; Katakan punca-punca ialah a dan 3a./ Let roots are a and 3a. HTP/ SOR = – b a a + 3a = – 1 –8 2 2 = 4 4a = 4 → a = 1 3a = 3(1) = 3 HDP/ POR = c a a × 3a = p 2 3a2 = p 2 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./ Let roots are a and 2a HTP = – b a a + 2a = – 1 –15 1 2 3a = 15 a = 5 2a = 2(5) = 10 HDP = c a a × 2a = m 1 m = 2a2 = 2(5)2 = 50 Punca-punca ialah 5 dan 10. Nilai m ialah 50. Roots are 5 and 10. The value of m is 50. 7. Tentukan ketaksamaan berikut. TP 5 Solve the following inequalities. CONTOH 1 x2 – 7x + 10 < 0 Penyelesaian: (i) Kaedah 1: lakaran graf/ Method 1: graph sketching a . 0 bentuk graf ialah/ the shape of the graph is apabila/ when x2 – 7x + 10 < 0 (x – 2)(x – 5) < 0 x – 2 = 0 atau/ or x – 5 = 0 x = 2 x = 5 2 5 Maka/ Thus, 2 < x < 5 CONTOH 2 –x2 + 7x – 12 < 0 Penyelesaian: (i) Kaedah 1: lakaran graf/ Method 1: graph sketching a , 0 bentuk graf ialah/the shape of the graph is apabila/when –x2 + 7x – 12 < 0 (–x + 4)(x – 3) < 0 –x + 4 = 0 atau/ or x – 3 = 0 x = 4 x = 3 3 4 Maka/ Thus, x < 3 atau/ or x > 4 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 26 BAB 2
(ii) Kaedah 2: garis nombor/ Method 2: number line x2 – 7x + 10 < 0 (x – 2)(x – 5) < 0 Pertimbangkan x – 2 > 0 dan/ and x – 5 > 0 Consider x > 2 x > 5 + 2 – + – + + – – + 5 Maka/ Thus, 2 < x < 5 (iii) Kaedah 3: jadual / Method 3: table x2 – 7x + 10 < 0 (x – 2)(x – 5) < 0 Pertimbangkan x – 2 = 0 dan/ and x – 5 = 0 Consider x = 2 x = 5 2 + – + – – + – + + 5 (x – 2): (x – 5): (x – 2)(x – 5): Maka/ Thus, 2 < x < 5 (ii) Kaedah 2: garis nombor/ Method 2: number line –x2 + 7x – 12 < 0 (–x + 4)(x – 3) < 0 Pertimbangkan –x + 4 > 0 dan/ and x – 3 > 0 Consider x < 4 x > 3 – 3 + – + + – – + + 4 Maka/ Thus, x < 3 atau/ or x > 4 (iii) Kaedah 3: jadual/ Method 3: table –x2 + 7x – 12 < 0 (–x + 4)(x – 3) < 0 Pertimbangkan –x + 4 = 0 dan/ and x – 3 = 0 Consider x = 4 x = 3 3 – + – + + – – + + 4 (x – 3): (–x + 4): (x – 3)(–x + 4): Maka/ Thus, x < 3 atau/ or x > 4 (a) x2 – 6x + 8 > 0 Apabila x2 – 6x + 8 = 0 When (x – 2)(x – 4) = 0 x – 2 = 0 atau/or x – 4 = 0 x = 2 x = 4 2 4 Maka/Thus, x < 2 atau x > 4 (b) 4x2 + 8x – 45 , 0 Apabila 4x2 + 8x – 45 = 0 When (2x – 5)(2x + 9) = 0 x = 5 2 atau/or x = – 9 2 5 – 2 9 – – 2 Maka/Thus, – 9 2 , x , 5 2 (c) –3x2 + 17x – 10 . 0 Apabila –3x2 + 17x – 10 = 0 When (–x + 5)(3x – 2) = 0 x = 5 atau/or x = 2 3 2 – 3 5 Maka/Thus, 2 3 , x , 5 (d) 7x2 – 24x – 16 > 0 Apabila 7x2 – 24x – 16 = 0 When (7x + 4)(x – 4) = 0 x = – 4 7 atau/or x = 4 4 4 – – 7 Maka/Thus, x < – 4 7 atau/or x > 4 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 27 BAB 2
2.2 Jenis-jenis Punca Persamaan Kuadratik Types of Roots of Quadratic Equations NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 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. Two different roots. Dua punca yang sama. Two equal roots. Tiada punca. 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 Penyelesaian: (ii) 4x2 – 13x + 3 = 0 (iii) 6x2 + 7x + 5 = 0 (i) a = 9, b = –12, c = 4 b2 – 4ac = (–12)2 – 4(9)(4) = 144 – 144 = 0 Persamaan itu mempunyai dua punca yang sama. The equation has two equal roots. (ii) a = 4, b = –13, c = 3 b2 – 4ac = (–13)2 – 4(4)(3) = 169 – 48 = 121 . 0 Persamaan itu mempunyai dua punca yang berbeza. The equation has two distinct roots. (ii) a = 6, b = 7, c = 5 b2 – 4ac = 72 – 4(6)(5) = 49 – 120 = –71 , 0 Persamaan itu tidak mempunyai punca. The equation has no roots. (a) 2x2 – 5x – 4 = 0 a = 2, b = –5, c = –4 b2 – 4ac = (–5)2 – 4(2)(–4) = 25 + 32 = 57 . 0 Persamaan itu mempunyai dua punca yang berbeza. The equation has two distanct roots. (b) 3x2 + 7x + 8 = 0 a = 3, b = 7, c = 8 b2 – 4ac = 72 – 4(3)(8) = 49 – 96 = –47 , 0 Persamaan itu tidak mempunyai punca. The equation has no roots. (c) 4x2 – 28x + 49 = 0 a = 4, b = –28, c = 49 b2 – 4ac = (–28)2 – 4(4)(49) = 784 – 784 = 0 Persamaan itu mempunyai dua punca yang sama. The equation has two equal roots. (d) 6x2 – 9x + 2 = 0 a = 6, b = –9, c = 2 b2 – 4ac = (–9)2 – 4(6)(2) = 81 – 48 = 33 . 0 Persamaan itu mempunyai dua punca yang berbeza. The equation has two distanct roots. Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 28 BAB 2
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 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 p + 2 = 0 atau/or p – 8 = 0 p = –2 p = 8 Syarat untuk dua punca yang sama. Condition for two equal roots. (a) x2 + 2px + 3p + 4 = 0 a = 1, b = 2p, c = 3p + 4 b2 – 4ac = 0 (2p)2 – 4(1)(3p + 4) = 0 4p2 – 12p – 16 = 0 p2 – 3p – 4 = 0 (p + 1)(p – 4) = 0 p + 1 = 0 atau/or p – 4 = 0 p = –1 p = 4 (b) 3x2 + px + 12 = 0 a = 3, b = p, c = 12 b2 – 4ac = 0 p2 – 4(3)(12) = 0 p2 – 144 = 0 p2 = 144 p = ±12 (c) 4x2 – 4px + 8p + 9 = 0 a = 4, b = –4p, c = 8p + 9 b2 – 4ac = 0 (–4p)2 – 4(4)(8p + 9) = 0 16p2 – 128p – 144 = 0 p2 – 8p – 9 = 0 (p + 1)(p – 9) = 0 p + 1 = 0 atau/or 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 x2 – 4x – 3 + p = 0 Penyelesaian: b2 – 4ac . 0 (–4)2 – 4(1)(–3 + p) . 0 16 + 12 – 4p . 0 4p , 28 p , 7 Syarat untuk dua punca yang berbeza. Condition for two distinct roots. (a) x2 + 2x + p – 3 = 0 b2 – 4ac . 0 22 – 4(1)(p – 3) . 0 4 – 4p + 12 . 0 16 – 4p . 0 4p , 16 p , 4 (b) 2x2 – 7x + p = 0 b2 – 4ac . 0 (–7)2 – 4(2)(p) . 0 49 – 8p . 0 8p , 49 p , 49 8 (c) (p + 1)x2 + 4x – 9 = 0 b2 – 4ac . 0 42 – 4(p + 1)(–9) . 0 16 + 36p + 36 . 0 36p . –52 p . – 13 9 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 29 BAB 2
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 x2 + 6x + p – 4 = 0 Penyelesaian: b2 – 4ac , 0 62 – 4(1)(p – 4) , 0 36 – 4p + 16 , 0 52 – 4p , 0 4p . 52 p . 13 Syarat untuk tiada punca. Condition for no roots. Kesalahan Lazim Tidak songsangkan simbol ketaksamaan apabila darabkan ketaksamaan dengan satu nombor negatif. Does not change the inequality sign when multiply with negative number. (a) x2 – 2x + p – 6 = 0 b2 – 4ac , 0 (–2)2 – 4(1)(p – 6) , 0 4 – 4p + 24 , 0 28 – 4p , 0 4p . 28 p . 7 (b) 4x2 + 3x + p = 0 b2 – 4ac , 0 32 – 4(4)(p) , 0 9 – 16p , 0 –16p , –9 p . 9 16 (c) (2p – 1)x2 – 6x + 8 = 0 b2 – 4ac , 0 (–6)2 – 4(2p – 1)(8) , 0 36 – 64p + 32 , 0 68 , 64p 64p . 68 p . 17 16 12. Selesaikan setiap yang berikut. TP 5 Solve each 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 + 5)]2 – 4(1)(4) . 0 p = –1 p2 + 10p + 25 – 16 . 0 p + 9 = 0 p2 + 10p + 9 . 0 p = –9 (p + 1)(p + 9) . 0 –1 p –9 Maka/ Thus, 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 (–2p)2 – 4(1)(4p – 3) , 0 p = 1 4p2 – 16p + 12 , 0 p – 3 = 0 p2 – 4p + 3 , 0 p = 3 (p – 1)(p – 3) , 0 1 3 p Maka/ Thus, 1 , p , 3 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 30 BAB 2
NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 2.3 Fungsi Kuadratik Quadratic Functions 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 Value of a b2 – 4ac . 0 b2 – 4ac = 0 b2 – 4ac , 0 a . 0 x x x a , 0 x x x Persamaan f(x) = 0 mempunyai dua punca yang berbeza. The equation f(x) = 0 has two different roots. Persamaan f(x) = 0 mempunyai dua punca yang sama. The equation f(x) = 0 has two equal roots. Persamaan f(x) = 0 tidak mempunyai punca. The equation f(x) = 0 has no roots. 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. (d) Paksi simetri boleh ditentukan dengan menggunakan x = – b 2a . 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. NOTA Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 31 BAB 2
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 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. (a) f(x) = 3 – 4x – 2x2 Kuasa tertinggi bagi x ialah 2 Maka f(x) ialah fungsi kuadratik. Highest power of x is 2. Thus, f(x) is a quadratic function. (b) f(x) = 7x3 – x2 + 8 Kuasa tertinggi bagi x ialah 3 Maka f(x) bukan fungsi kuadratik. Highest power of x is 3. Thus, f(x) is not a quadratic function. (c) f(x) = –x – 3x2 Kuasa tertinggi bagi x ialah 2 Maka f(x) ialah fungsi kuadratik. Highest power of x is 2. Thus, f(x) is a quadratic function. 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 a = 1 . 0, bentuk graf ialah/ shape of graph is b2 – 4ac = (–6)2 – 4(1)(3) = 36 – 12 = 24 . 0 Maka, f(x) = 0 mempunyai dua punca yang berbeza. Thus, f(x) = 0 has two distinct roots. (ii) f(x) = –4x2 + 8x – 4 a = –4 , 0, bentuk graf ialah/ shape of graph is b2 – 4ac = (8)2 – 4(–4)(–4) = 64 – 64 = 0 Maka, f(x) = 0 mempunyai dua punca yang sama. Thus, f(x) = 0 has two equal roots. (iii)f(x) = 3x2 – 9x + 7 a = 3 . 0, bentuk graf ialah/ shape of graph is b2 – 4ac = (–9)2 – 4(3)(7) = 81 – 84 = –3 , 0 Maka, f(x) = 0 tidak mempunyai punca. Thus, f(x) = 0 has no roots. (a) f(x) = 2x2 – 7x + 5 a = 2 . 0, bentuk graf ialah/shape of graph is b2 – 4ac = (–7)2 – 4(2)(5) = 49 – 40 = 9 . 0 Maka, f(x) = 0 mempunyai dua punca yang berbeza. Thus, f(x) = 0 has two distinct roots. (b) f(x) = –4 – 6x – 3x2 a = –3 , 0, bentuk graf ialah/shape of graph is b2 – 4ac = (–6)2 – 4(–3)(–4) = 36 – 48 = –12 , 0 Maka, f(x) = 0 tidak mempunyai punca. Thus, f(x) = 0 has no roots. (c) f(x) = 9x2 – 12x + 4 a = 9 . 0, bentuk graf ialah/shape of graph is b2 – 4ac = (–12)2 – 4(9)(4) = 144 – 144 = 0 Maka, f(x) = 0 mempunyai dua punca yang sama. Thus, f(x) = 0 has two equal roots. (d) f(x) = 5x2 + 3x + 1 a = 5 . 0, bentuk graf ialah/shape of graph is b2 – 4ac = (3)2 – 4(5)(1) = 9 – 20 = –11 , 0 Maka, f(x) = 0 tidak mempunyai punca. Thus, f(x) = 0 has no roots. Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 32 BAB 2
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 f(x) = 3x2 – 8x + k – 6 Penyelesaian: a = 3, b = –8, c = k – 6 f(x) mempunyai dua punca yang berlainan apabila f(x) has two distinct roots when b2 – 4ac . 0 (–8)2 – 4(3)(k – 6) . 0 64 – 12k + 72 . 0 12k , 136 k , 34 3 (a) f(x) = (2k – 3)x2 – 4x – 8 a = 2k – 3, b = –4, c = –8 f(x) mempunyai dua punca yang berlainan apabila f(x) has two distinct roots when b2 – 4ac . 0 (–4)2 – 4(2k – 3)(–8) . 0 16 + 64k – 96 . 0 64k . 80 k . 5 4 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 f(x) = mx2 – 6x + 9 Penyelesaian: a = m, b = –6, c = 9 f(x) mempunyai dua punca yang sama apabila f(x) has two equal roots when b2 – 4ac = 0 (–6)2 – 4(m)(9) = 0 36 – 36m = 0 36m = 36 m = 1 (a) f(x) = x2 + 2mx + m + 6 a = 1, b = 2m, c = m + 6 f(x) mempunyai dua punca yang sama apabila f(x) has two equal roots when b2 – 4ac = 0 (2m)2 – 4(1)(m + 6) = 0 4m2 – 4m – 24 = 0 m2 – m – 6 = 0 (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 f(x) = (2p + 5)x2 – 6x + 9 Penyelesaian: a = 2p + 5, b = –6, c = 9 f(x) tidak mempunyai punca apabila f(x) has no roots when b2 – 4ac , 0 (–6)2 – 4(2p + 5)(9) , 0 36 – 72p – 180 , 0 –72p , 144 p . –2 (a) f(x) = x2 + 2(p + 1)x + p2 – 1 a = 1, b = 2(p + 1), c = p2 – 1 f(x) tidak mempunyai punca apabila f(x) has no roots when b2 – 4ac , 0 [2(p + 1)]2 – 4(1)(p2 – 1) , 0 4(p + 1)2 – 4p2 + 4 , 0 4p2 + 8p + 4 – 4p2 + 4 , 0 8p , –8 p , –1 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 33 BAB 2
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 (i) f(x) = 2x2 + 8x – 1 Penyelesaian: f(x) = 2x2 + 8x – 1 = 2(x2 + 4x) – 1 = 23x2 + 4x + 1 4 2 2 2 – 1 4 2 2 2 4 – 1 = 2(x + 2)2 – 8 – 1 = 2(x + 2)2 – 9 Oleh sebab a . 0, f(x) mempunyai nilai minimum –9 apabila (x + 2) = 0 iaitu x = –2. Since a . 0, f(x) has minimum value of –9 when (x + 2) = 0 which is x = –2. (ii) f(x) = 2 + 6x – 3x2 Penyelesaian: f(x) = –3x2 + 6x + 2 = –3(x2 – 2x) + 2 = –33x2 – 2x + 1 –2 2 2 2 – 1 –2 2 2 2 4 + 2 = –3(x – 1)2 + 3 + 2 = –3(x – 1)2 + 5 Oleh sebab a , 0, f(x) mempunyai nilai maksimum 5 apabila (x – 1) = 0 iaitu x = 1. Since a , 0, f(x) has maximum value of 5 when (x – 1) = 0 which is x = 1. (a) f(x) = x2 + 4x – 3 f(x) = x2 + 4x + 1 4 2 2 2 – 1 4 2 2 2 – 3 = (x + 2)2 – 4 – 3 = (x + 2)2 – 7 Oleh sebab a . 0, f(x) mempunyai nilai minimum –7 apabila (x + 2) = 0 iaitu x = –2. Since a . 0, f(x) has minimum value of –7 when (x + 2) = 0 which is x = –2. (b) f(x) = –x2 + 5x – 11 f(x) = –(x2 – 5x) – 11 = –3x2 – 5x + 1 –5 2 2 2 – 1 –5 2 2 2 4 – 11 = –1x – 5 2 2 2 + 25 4 – 11 = – 1x – 5 2 2 2 – 19 4 Oleh sebab a , 0, f(x) mempunyai nilai maksimum – 19 4 apabila 1x – 5 2 2 = 0 iaitu x = 5 2 . Since a , 0, f(x) has maximum value of – 19 4 when 1x – 5 2 2 = 0 which is x = 5 2 . (c) f(x) = 2x2 – 8x + 15 f(x) = 2(x2 – 4x) + 15 = 23x2 – 4x + 1 –4 2 2 2 – 1 –4 2 2 2 4 + 15 = 2(x – 2)2 – 8 + 15 = 2(x – 2)2 + 7 Oleh sebab a . 0, f(x) mempunyai nilai minimum 7 apabila (x – 2) = 0 iaitu x = 2. Since a . 0, f(x) has minimum value of 7 when (x – 2) = 0 which is x = 2. (d) f(x) = –2x2 – 12x + 9 f(x) = –23x2 + 6x + 1 6 2 2 2 – 1 6 2 2 2 4 + 9 = –2(x + 3)2 + 18 + 9 = –2(x + 3)2 + 27 Oleh sebab a , 0, f(x) mempunyai nilai maksimum 27 apabila (x + 3) = 0 iaitu x = –3. Since a , 0, f(x) has maximum value of 27 when (x + 3) = 0 which is x = –3. (e) f(x) = 4x2 – 8x + 17 f(x) = 4(x2 – 2x) + 17 = 43x2 – 2x + 1 –2 2 2 2 – 1 –2 2 2 2 4 + 17 = 4(x – 1)2 – 4 + 17 = 4(x – 1)2 + 13 Oleh sebab a . 0, f(x) mempunyai nilai minimum 13 apabila (x – 1) = 0 iaitu x = 1. Since a . 0, f(x) has minimum value of 13 when (x – 1) = 0 which is x = 1. (f) f(x) = –5 – 21x – 3x2 f(x) = –3x2 – 21x – 5 = –3(x2 + 7x) – 5 = –33x2 + 7x + 1 7 2 2 2 – 1 7 2 2 2 4 – 5 = –31x + 7 2 2 2 + 147 4 – 5 = –31x + 7 2 2 2 + 127 4 Oleh sebab a , 0, f(x) mempunyai nilai maksimum 127 4 apabila 1x + 7 2 2 = 0 iaitu x = – 7 2 . Since a , 0, f(x) has maximum value of 127 4 when 1x + 7 2 2 = 0 which is x = – 7 2 . Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 34 BAB 2
19. Selesaikan setiap yang berikut. TP 5 Solve each of the following. CONTOH x y y = f(x) –11 (6, –11) O Rajah menunjukkan graf bagi fungsi f(x) = –(x – k)2 – 9 dengan keadaan 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. Penyelesaian: (a) Titik tengah bagi (0, –11) dan (6, –11)/ Midpoint of (0, –11) and (6, –1) = 1 0 + 6 2 , –11 – 11 2 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) x f(x) 4 6 O –2 (b) x f(x) 3 –4 –22 x + m = 0 4 + m = 0 m = –4 n = nilai minimum/minimum value = –2 f(x) = a(x – 4)2 – 2 Pada titik (0, 6), 6 = a(0 – 4)2 – 2 At point 16a = 8 a = 8 16 = 1 2 x + m = 0 3 + m = 0 m = –3 n = nilai maksimum/maximum value = –4 f(x) = a(x – 3)2 – 4 Pada titik (0, –22), –22 = a(0 – 3)2 – 4 At point 9a = –18 a = –2 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 35 BAB 2
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 f(x) = x2 + 8x + 12 Penyelesaian: a = 1 . 0 b2 – 4ac = (8)2 – 4(1)(12) = 16 . 0 Maka, graf f(x) berbentuk dengan titik minimum dan menyilangi paksi-x pada dua titik yang berbeza. Thus, graph f(x) has shape with minimum point and intersect the x-axis at two distinct points. f(x) = x2 + 8x + 12 = x2 + 8x + 1 8 2 2 2 – 1 8 2 2 2 + 12 = (x + 4)2 – 16 + 12 = (x + 4)2 – 4 Titik minimum ialah/ Minimum point is (–4, –4). Apabila/When f(x) = 0, x2 + 8x + 12 = 0 (x + 2)(x + 6) = 0 x = –2 atau/or x = –6 Apabila/When x = 0 f(x) = (0)2 + 8(0) + 12 = 12 x 0 f(x) –2 (–4, –4) 12 –6 Persamaan paksi simetri ialah/Equation of axis of symmetry is x = –4. 1 Tentukan bentuk graf. Determine the shape of the graph. 2 Tentukan titik minimum atau maksimum. Determine the minimum or maximum point. 3 Tentukan pintasan-x jika ada. Determine x-intercept if exist. 5 Lakar graf. Sketch the graph. 4 Tentukan pintasan-y. Determine y-intercept. CONTOH 2 f(x) = –2x2 + 6x – 5 Penyelesaian: a = –2 , 0 b2 – 4ac = (6)2 – 4(–2)(–5) = –4 , 0 Maka, graf f(x) berbentuk dengan titik maksimum 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 = –23x2 – 3x + 1 –3 2 2 2 – 1 –3 2 2 2 4 – 5 = –21x – 3 2 2 2 + 9 2 – 5 = –21x – 3 2 2 2 – 1 2 Titik maksimum ialah / Maximum point is 1 3 2 , – 1 2 2. Apabila/When x = 0 f(x) = –2(0)2 + 6(0) – 5 = –5 x 0 f(x) 1 2 3 2fi–, – –ff –5 Persamaan paksi simetri ialah/ Equation of axis of symmetry is x = 3 2 . Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 36 BAB 2
(a) f(x) = x2 – 6x – 7 a = 1 . 0 b2 – 4ac = (–6)2 – 4(1)(–7) = 64 . 0 Maka, graf f(x) berbentuk dengan titik minimum dan menyilangi paksi-x pada dua titik yang berbeza. Thus, graph f(x) has shape with minimum point and intersect the x-axis at two distinct points. f(x) = x2 – 6x – 7 = x2 – 6x + 1 –6 2 2 2 – 1 –6 2 2 2 – 7 = (x – 3)2 – 9 – 7 = (x – 3)2 – 16 Titik minimum ialah/Minimum point is (3, –16). Apabila f(x) = 0, x2 – 6x – 7 = 0 (x + 1)(x – 7) = 0 x = –1 atau/or x = 7 Apabila/When x = 0 f(x) = (0)2 – 6(0) – 7 = –7 x 0 7 (3, –16) f(x) –7 –1 Persamaan paksi simetri ialah x = 3. Equation of axis of symmetry is x = 3. (b) f(x) = –x2 + 6x – 5 a = –1 , 0 b2 – 4ac = (6)2 – 4(–1)(–5) = 16 . 0 Maka, graf f(x) berbentuk dengan titik maksimum dan menyilang paksi-x pada dua titik yang berbeza. Thus, graph f(x) has shape with maximum point and intersect the x-axis at two distinct points. f(x) = –x2 + 6x – 5 = –3x2 – 6x + 1 –6 2 2 2 – 1 –6 2 2 2 4 – 5 = –(x – 3)2 + 9 – 5 = –(x – 3)2 + 4 Titik maksimum ialah/Maximum point is (3, 4). Apabila f(x) = 0, –x2 + 6x – 5 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x = 1 atau x = 5 Apabila x = 0 f(x) = –(0)2 + 6(0) – 5 = –5 x 0 5 (3, 4) f(x) 1 –5 Persamaan paksi simetri ialah x = 3. Equation of axis of symmetry is x = 3. Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 37 BAB 2
(c) f(x) = 2x2 – 5x + 4 a = 2 . 0 b2 – 4ac = (–5)2 – 4(2)(4) = –7 , 0 Maka, graf f(x) berbentuk dengan titik minimum dan tidak menyilang paksi-x. Thus, graph f(x) has shape with minimum point and intersect the x-axis. f(x) = 2x2 – 5x + 4 = 23x2 – 5 2 x + 1 –5 4 2 2 – 1 –5 4 2 2 4 + 4 = 21x – 5 4 2 2 – 25 8 + 4 = 21x – 5 4 2 2 + 7 8 Titik minimum ialah 1 5 4 , 7 8 2. Minimum point is Apabila x = 0 f(x) = 2(0)2 – 5(0) + 4 = 4 x 0 f(x) 4 7 8 5 4fi–, –ff Persamaan paksi simetri ialah x = 5 4 . Equation of axis of symmetry is x = 5 4 . Kertas 1 1. Diberi −3 ialah salah satu punca persamaan kuadratik (x − p)2 = 25, dengan keadaan p ialah pemalar. Cari nilai-nilai p. Given −3 is one of the roots of the quadratic equation (x − p)2 = 25, where p is a constant. Find the values of p. (x − p)2 = 25 x − p = ±5 Apabila/When x = −3, −3 − p = 5 , −3 − p = −5 p = −8 , p = 2 SPM 2015 2. Cari julat nilai x dengan keadaan fungsi kuadratik f(x) = 4 + 3x – x2 ialah negatif. Find the range of value of x such that the quadratic function f(x) = 4 + 3x – x2 is negative. f(x) = 4 + 3x – x2 4 + 3x – x2 , 0 x2 – 3x – 4 . 0 (x – 4)(x + 1) . 0 –1 4 x , –1 atau/or x . 4 SPM 2017 PRAKTIS PRAKTIS SPM SPM 2 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 38 BAB 2
3. Diberi fungsi kuadratik f(x) = (2m + 1)x2 – 3mx + 2(m – 2), dengan keadaan m ialah pemalar, adalah sentiasa positif apabila m > p atau m < q. Cari nilai p dan nilai q. Given the quadratic function f(x) = (2m + 1)x2 – 3mx + 2(m – 2), where m is a constant, is always positive when m . p or m , q. Find the value of p and of q. a = 2m + 1, b = –3m, c = 2m – 4 b2 – 4ac < 0 (–3m)2 – 4(2m + 1)(2m – 4) < 0 9m2 – 4(4m2 – 8m + 2m – 4) < 0 9m2 – 16m2 + 24m + 16 < 0 –7m2 + 24m + 16 < 0 7m2 – 24m – 16 > 0 (7m + 4)(m – 4) > 0 7m + 4 = 0 atau/or m – 4 = 0 – + + – – + + – + (7m + 4): (m – 4): – 4 7 4 m < – 4 7 m > 4 \ p = 4 dan/and q = – 4 7 4. Rajah menunjukkan graf y = a(x – h)2 + m, dengan keadaan a, h dan m ialah pemalar. Garis lurus y = –5 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. x 0 y –2 8 Q (a) Nyatakan koordinat Q. State the coordinates of Q. (b) Cari nilai a. Find the value of a. SPM 2016 SPM 2018 (a) Nilai x bagi titik tengah = –2 + 8 2 = 3 The value of x of the midpoint h = 3, m = –5 \ Q(3, –5) (b) y = a(x – 3)2 – 5 Gantikan/Replace x = 8, y = 0 0 = a(8 – 3)2 – 5 0 = 25a – 5 25a = 5 a = 1 5 5. Diberi bahawa lengkung y = (k – 3)x2 – 6x + 1, dengan keadaan k ialah pemalar, bersilang dengan garis lurus y = 2x + 5 pada dua titik. Cari julat nilai k. It is given that the curve y = (k – 3)x2 – 6x + 1, where k is a constant, intersects with the straight line y = 2x + 5 at two points. Find the range of values of k. y = (k – 3)x2 – 6x + 1 …… 1 y = 2x + 5 …… 2 Gantikan 1 ke dalam 2, Replace 1 into 2, 2x + 5 = (k – 3)x2 – 6x + 1 (k – 3)x2 – 6x – 2x – 5 + 1 = 0 (k – 3)x2 – 8x – 4 = 0 b2 – 4ac . 0 (–8)2 – 4(k – 3)(–4) . 0 64 + 16k – 48 . 0 16k . –16 k . –1 6. Graf fungsi kuadratik g(x) = hx2 + (k − 1)x + 4, dengan keadaan h dan k ialah pemalar, mempunyai satu titik minimum. 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 nilai h jika h ialah suatu integer dengan keadaan −1 < h < 1. State the value of h if h is an integer such that −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. SPM 2018 SPM 2015 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 39 BAB 2