Strategi A+ Ting 4 - Matematik Tambahan

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ii Bab 1 Fungsi Revisi Pantas 1 Praktis PBD 1.1 Fungsi 2 1.2 Fungsi Gubahan 7 1.3 Fungsi Songsang 13 Praktis Berformat SPM 16 Zon KBAT 17 Bab 2 Fungsi Kuadratik Revisi Pantas 18 Praktis PBD 2.1 Persamaan dan Ketaksamaan Kuadratik 19 2.2 Jenis-jenis Punca Persamaan Kuadratik 26 2.3 Fungsi Kuadratik 29 Praktis Berformat SPM 37 Zon KBAT 38 Bab 3 Sistem Persamaan Revisi Pantas 39 Praktis PBD 3.1 Sistem Persamaan Linear dalam Tiga Pemboleh Ubah 40 3.2 Persamaan Serentak yang Melibatkan Satu Persamaan Linear dan Satu Persamaan Tak Linear 46 Praktis Berformat SPM 53 Zon KBAT 53 Bab 4 Indeks, Surd dan Logaritma Revisi Pantas 54 Praktis PBD 4.1 Hukum Indeks 55 4.2 Hukum Surd 57 4.3 Hukum Logaritma 62 4.4 Aplikasi Indeks, Surd dan Logaritma 65 Praktis Berformat SPM 66 Zon KBAT 67 Bab 5 Janjang Revisi Pantas 68 Praktis PBD 5.1 Janjang Aritmetik 69 5.2 Janjang Geometri 77 Praktis Berformat SPM 87 Zon KBAT 88 Bab 6 Hukum Linear Revisi Pantas 89 Praktis PBD 6.1 Hubungan Linear dan Tak Linear 90 6.2 Hukum Linear dan Hubungan Tak Linear 96 6.3 Aplikasi Hukum Linear 99 Praktis Berformat SPM 102 Zon KBAT 104 Bab 7 Geometri Koordinat Revisi Pantas 105 Praktis PBD 7.1 Pembahagi Tembereng Garis 106 7.2 Garis Lurus Selari dan Garis Lurus Serenjang 108 7.3 Luas Poligon 115 7.4 Persamaan Lokus 119 Praktis Berformat SPM 123 Zon KBAT 126 Bab 8 Vektor Revisi Pantas 127 Praktis PBD 8.1 Vektor 128 8.2 Penambahan dan Penolakan Vektor 132 8.3 Vektor dalam Satah Cartes 135 Praktis Berformat SPM 138 Zon KBAT 140 Bab 9 Penyelesaian Segi Tiga Revisi Pantas 141 Praktis PBD 9.1 Petua Sinus 142 9.2 Petua Kosinus 147 9.3 Luas Segi Tiga 151 9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas Segi Tiga 156 Praktis Berformat SPM 159 Zon KBAT 160 Bab 10 Nombor Indeks Revisi Pantas 161 Praktis PBD 10.1 Nombor Indeks 161 10.2 Indeks Gubahan 168 Praktis Berformat SPM 175 Zon KBAT 176 Kandungan Rekod Prestasi Murid R1 Jawapan 177 00a Strategi A+ SPM Mate Tam Tg4_Kandungan_Final1.indd 2 6/11/2023 2:49:13 PM MU BAKTI SDN. BHD. PENERBIT ILMU BAKTI SDLMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BAKTBIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

1 1.1 Fungsi Functions 1 Fungsi ialah suatu hubungan yang memetakan setiap unsur x dalam set X kepada hanya satu unsur y dalam set Y. Perhatikan gambar rajah anak panah di bawah. A function is a relation that maps each element x in set X to only one element y in set Y. Refer the arrow diagram below. 1 2 3 4 8 12 Set X Set Y 2 Unsur di dalam set X, iaitu 1, 2, dan 3 dikenali sebagai objek, manakala unsur di dalam set Y, iaitu 4, 8 dan 12 dikenali sebagai imej. The elements in set X, which are 1, 2 and 3 are known as objects, while the elements in set Y, which are 4, 8 and 12 are known as images. 3 Suatu graf boleh ditentukan sama ada graf tersebut ialah fungsi atau bukan dengan menggunakan ujian garis mencancang. A graph can be determined whether the graph is a function or not by using the vertical line test. i-THINK Peta Pokok Ujian garis mencancang Vertical line test Fungsi Function Bukan fungsi Not a function Garis mencancang memotong graf hanya pada satu titik. The vertical line cuts the graph at only one point. Garis mencancang tidak memotong manamana titik pada graf atau memotong lebih daripada satu titik. The vertical line does not cut the graph at any point or cuts more than one point. y x O y x O 4 Jika fungsi f memetakan set X kepada set Y, If function f maps set X to set Y, (a) set X dikenali sebagai domain set X is known as domain (b) set Y dikenali sebagai kodomain set Y is known as codomain (c) unsur dalam set Y yang dipetakan dari X dikenali sebagai julat the elements in set Y that are mapped from X is known as range 5 Imej f(x) boleh ditentukan dengan menggantikan nilai objek x ke dalam suatu fungsi./The image f(x) can be determined by substituting the value of object x in a function. 6 Objek x juga boleh ditentukan dengan menggantikan nilai f(x). The object x can also be determined by substituting the value of f(x). 1.2 Fungsi Gubahan Composite Functions 1 Jika fungsi f memetakan set A kepada set B dan fungsi g memetakan set B kepada set C, maka fungsi gf memetakan set A kepada set C./If function f maps set A to set B and function g maps set B to set C, then function gf maps set A to set C. x x f gf g f(x) gf(x) g(x) Set A Set B Set C 2 Fungsi gf dikenali sebagai fungsi gubahan. The function gf is known as composite function. 1.3 Fungsi Songsang Inverse Functions 1 Fungsi songsang bagi suatu fungsi f ditulis sebagai f –1. The inverse function of a function f is written as f –1. f f –1 x y 2 Jika fungsi f : x → y, maka f –1 : y → x. If function f : x → y, then f –1 : y → x. 3 Ujian garis mengufuk boleh digunakan untuk menentukan sama ada graf bagi suatu fungsi itu mempunyai fungsi songsang atau tidak. The horizontal line test can be used to determine whether the graph of a function has an inverse function or not. 4 Fungsi songsang hanya wujud jika garis mengufuk memotong graf itu hanya pada satu titik. The inverse function exists only if the horizontal line cuts the graph at only one point. Fungsi songsang wujud. Inverse function exist. Fungsi songsang tidak wujud. Inverse function does not exist. y x O y = f(x) Ujian garis mengufuk Horizontal line test y x O y = f(x) Ujian garis mengufuk Horizontal line test Revisi Pantas Fungsi Functions Bidang Pembelajaran: Algebra Bab 1 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 1 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

2 Contoh 1 2 4 6 8 1 2 3 4 Penyelesaian Hubungan ini ialah suatu fungsi kerana setiap objek hanya mempunyai satu imej sahaja walaupun unsur 3 tidak mempunyai objek. This relation is a function because each object has only one image even though element 3 has no object. 1 1 2 3 3 4 5 9 Hubungan ini ialah suatu fungsi kerana setiap objek hanya mempunyai satu imej walaupun unsur 9 tidak mempunyai objek. This relation is a function because each object has only one image even though element 9 has no object. 2 a b c 2 4 6 7 Hubungan ini bukan fungsi kerana objek b mempunyai lebih daripada satu imej. This relation is not a function because object b has more than one image. 3 1 2 3 9 3 4 5 Hubungan ini bukan fungsi kerana objek 9 tidak mempunyai imej. This relation is not a function because the object 9 has no image. 4 a b c d 1 6 7 Hubungan ini ialah suatu fungsi kerana setiap objek hanya mempunyai satu imej. This relation is a function because each object has only one image. 5 3 5 7 13 15 25 26 35 49 Hubungan ini bukan fungsi kerana objek 5 dan 7 mempunyai lebih daripada satu imej. This relation is not a function because objects 5 and 7 have more than one image. 1.1 Fungsi/ Functions Latihan 1 Tentukan sama ada setiap hubungan yang berikut ialah fungsi atau bukan. Beri alasan anda. Determine whether each of the following relations is a function or not. Give your reason. TP 1 TP 1 Mempamerkan pengetahuan asas tentang fungsi. Latihan 2 Tentukan sama ada setiap graf yang berikut ialah fungsi atau bukan. Beri alasan anda. Determine whether each of the following graphs is a function or not. Give your reason. TP 1 TP 1 Mempamerkan pengetahuan asas tentang fungsi Praktis PBD Contoh 2 y O x Penyelesaian Graf ini ialah suatu fungsi kerana apabila diuji dengan garis mencancang, hanya satu titik memotong graf itu. The graph is a function because when tested with the vertical line, there is only one point that cuts the graph. 1 y O x Graf ini ialah suatu fungsi kerana apabila diuji dengan garis mencancang, hanya satu titik memotong graf itu. The graph is a function because when tested with the vertical line, there is only one point that cuts the graph. 2 y O x Graf ini bukan fungsi kerana apabila diuji dengan garis mencancang, terdapat lebih daripada satu titik yang memotong graf itu. The graph is not a function because when tested with the vertical line, there is more than one point that cuts the graph. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 2 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

3 1 5 2 6 3 7 4 8 Domain = {2, 3, 4} Kodomain/Codomain = {5, 6, 7, 8} Julat/Range = {5, 6, 7} 2 1 2 3 5 3 4 5 7 9 Domain = {1, 2, 3, 5} Kodomain/Codomain = {3, 4, 5, 7, 9} Julat/Range = {3, 4, 7} 3 {(a, 3), (b, 5), (c, 7), (d, 8)} Domain = {a, b, c, d} Kodomain/Codomain = {3, 5, 7, 8} Julat/Range = {3, 5, 7, 8} Latihan 3 Nyatakan domain, kodomain dan julat bagi fungsi berikut. TP 1 State the domain, codomain and the range of the following functions. TP 1 Mempamerkan pengetahuan asas tentang fungsi. 3 y O x 1 2 Graf ini ialah suatu fungsi kerana apabila diuji dengan garis mencancang, hanya satu titik memotong graf itu. The graph is a function because when tested with the vertical line, there is only one point that cuts the graph. 4 y O x Graf ini bukan suatu fungsi kerana apabila diuji dengan garis mencancang, terdapat lebih daripada satu titik yang memotong graf itu. The graph is not a function because when tested with the vertical line, there is more than one point that cuts the graph. 5 y O x Graf ini ialah suatu fungsi kerana apabila diuji dengan garis mencancang, hanya satu titik memotong graf itu. The graph is a function because when tested with the vertical line, there is only one point that cuts the graph. Contoh 3 1 2 3 7 2 6 11 14 Penyelesaian Domain = {1, 2, 3, 7} Kodomain/Codomain = {2, 6, 11, 14} Julat/Range = {2, 6, 14} Tip Bestari 1 Domain mengandungi objek sahaja. Domain contains the objects only. 2 Kodomain mengandungi imej dan bukan imej. Codomain contains images and non-images. 3 Julat mengandungi imej sahaja. Range contains the images only. Kesilapan Umum Julat/Range = {2, 6, 11, 14} Julat adalah sama dengan kodomain hanya jika setiap unsur dalam kodomain ialah imej bagi fungsi itu. The range is equal to the codomain only if each element in the codomain is the image of the function. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final - Tajul Updated.indd 3 6/11/2023 3:15:12 PM MU BAKTI SDN. BHD. PENERBIT ILMU BAKTI SDLMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BAKTBIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

4 4 {(2, p), (4, q), (6, p), (8, r)} Domain = {2, 4, 6, 8} Kodomain/Codomain = {p, q, r} Julat/Range = {p, q, r} 5 a b c d e 1 2 3 4 5 6 y x Domain = {a, b, c, d, e} Kodomain/Codomain = {1, 2, 3, 4, 5, 6} Julat/Range = {1, 3, 4, 5, 6} 6 1 2 4 5 6 1 2 3 4 5 6 3 y x Domain = {1, 2, 3, 4, 5, 6} Kodomain/Codomain = {1, 2, 3, 4, 5, 6} Julat/Range = {2, 4, 6} Contoh 4 f(x) = |2x – 3|; –1  x  3 Penyelesaian x –1 3 2 3 f(x) 5 0 3 f(x) 3 –1 3 x 5 0 3 2 Julat Range Julat/Range: 0  f(x)  5 1 f(x) = |x2 – 4|; – 2  x  3 x –2 0 2 3 f(x) 0 4 0 5 f(x) 5 x –3 –2 –1 0 1 3 2 4 Julat/Range: 0  f(x)  5 2 f(x) = |x + 1|; –2  x  2 x –2 –1 2 f(x) 1 0 3 f(x) –1 x 3 –2 0 1 2 1 Julat/Range: 0  f(x)  3 3 h(x) = |2x + 3|; –2  x  1 x –2 – 3 2 0 1 h(x) 1 0 3 5 h(x) –1–2 x 5 0 3 1 1 3 2 – Julat/Range: 0  h(x)  5 4 h(x) = |x2 – 1|; –2  x  2 x –2 –1 0 1 2 h(x) 3 0 1 0 3 h(x) 3 x 0 1 –2 –1 1 2 Julat/Range: 0  h(x)  3 5 f(x) = |3 – x|; –2  x  5 x –2 0 3 5 f(x) 5 3 0 2 f(x) x –2 0 2 5 3 3 5 Julat/Range: 0  f(x)  5 Latihan 4 Lengkapkan jadual dan lakar graf bagi domain yang diberi. Kemudian, nyatakan julat yang sepadan untuk domain itu. TP 3 Complete the table and sketch the graph for the given domain. Then, state the corresponding range for the given domain. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 4 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

5 Contoh 5 f : x → 3x 4x – 2; x = 5 Penyelesaian f(5) = 3(5) 4(5) – 2 = 15 18 = 5 6 Imej bagi 5 ialah 5 6 . The image of 5 is 5 6 . 1 f : x → 4x – 5; x = 3 f(3) = 4(3) − 5 = 7 Imej bagi 3 ialah 7. The image of 3 is 7. 2 f : x → x2 – 3x − 6; x = −2 f(−2) = (−2)2 – 3(−2) − 6 = 4 Imej bagi −2 ialah 4. The image of −2 is 4. 3 g : x → 2x3 – 5x; x = 2 g(2) = 2(2)3 – 5(2) = 16 − 10 = 6 Imej bagi 2 ialah 6. The image of 2 is 6. 4 h : x → 5x + 4 x – 1 ; x = 4 h(4) = 5(4) + 4 (4) – 1 = 24 3 = 8 Imej bagi 4 ialah 8. The image of 4 is 8. 5 f : x → x2 – 1 3x + 2 ; x = –4 f(–4) = (–4)2 – 1 3(–4) + 2 = 15 –10 = – 3 2 Imej bagi –4 ialah – 3 2 . The image of –4 is – 3 2 . Latihan 5 Cari imej bagi setiap fungsi berikut. TP 2 Find the image for each of the following functions. TP 2 Mempamerkan kefahaman tentang fungsi. Contoh 6 Diberi f : x → 3x − 12, cari objek apabila imej ialah 9. Given that f : x → 3x – 12, find the object when the image is 9. Penyelesaian f(x) = 3x – 12 9 = 3x – 12 21 = 3x x = 7 Objek bagi 9 ialah 7. The object of 9 is 7. 1 Diberi f : x → 4x + 7, cari objek apabila imej ialah 15. Given that f : x → 4x + 7, find the object when the image is 15. f(x) = 4x + 7 15 = 4x + 7 8 = 4x x = 2 Objek bagi 15 ialah 2. The object of 15 is 2. 2 Diberi f : x → 6x – 2, cari objek apabila imej ialah 1. Given that f : x → 6x – 2, find the object when the image is 1. f(x) = 6x – 2 1 = 6x – 2 3 = 6x x = 3 6 = 1 2 Objek bagi 1 ialah 1 2 . The object of 1 is 1 2 . Latihan 6 Selesaikan setiap yang berikut. TP 2 Solve each of the following. TP 2 Mempamerkan kefahaman tentang fungsi. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 5 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

6 Contoh 7 Diberi bahawa f(x) = |2x – 3|, cari objek apabila f(x) = 7. It is given that f(x) = |2x – 3|, find the object when f(x) = 7. Penyelesaian |2x – 3| = 7 2x – 3 = –7 atau/or 2x – 3 = 7 2x = –7 + 3 atau/or 2x = 7 + 3 2x = –4 atau/or 2x = 10 x = –2 atau/or x = 5 1 Diberi bahawa f(x) = |3x|, cari objek apabila f(x) = 12. It is given that f(x) = |3x|, find the object when f(x) = 12. |3x| = 12 3x = –12 atau/or 3x = 12 x = –4 atau/or x = 4 Latihan 7 Cari nilai-nilai x bagi setiap fungsi berikut. TP 3 Find the values of x for each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 3 Diberi f : x → 3x + 2 2 , cari objek apabila imej ialah 10. Given that f : x → 3x + 2 2 , find the object when the image is 10. f(x) = 3x + 2 2 10 = 3x + 2 2 20 = 3x + 2 18 = 3x x = 6 Objek bagi 10 ialah 6. The object of 10 is 6. 4 Diberi f : x → 5x – 6 2x , cari objek apabila imej ialah 4. Given that f : x → 5x – 6 2x , find the object when the image is 4. f(x) = 5x – 6 2x 4 = 5x – 6 2x 8x = 5x – 6 3x = –6 x = –2 Objek bagi 4 ialah –2. The object of 4 is –2. 5 Diberi f : x → 4x – 2 x – 2 , cari objek apabila imej ialah 6. Given that f : x → 4x – 2 x – 2 , find the object when the image is 6. f(x) = 4x – 2 x – 2 6 = 4x – 2 x – 2 6x − 12 = 4x – 2 2x = 10 x = 5 Objek bagi 6 ialah 5. The object of 6 is 5. 6 Rajah di bawah menunjukkan hubungan antara set P dengan set Q. The diagram below shows the relation between set P and set Q. 1 • 5 • h • 20 • • 1 • 0.2 • 0.1 • 0.05 Set P Set Q f(x) (a) Menggunakan tatatanda fungsi, ungkapkan f dalam sebutan x. Using the function notation, express f in terms of x. (b) Cari nilai h. Find the value of h. (c) Jika set P mewakili semua nombor bulat, adakah setiap objek mempunyai imejnya? KBAT Mengaplikasi If set P represents all the whole numbers, does every object has its image? (a) x = 1; 1 (1) = 1 x = 5; 1 (5) = 0.2 x = 20; 1 (20) = 0.05 ∴ f(x) = 1 x , x ≠ 0 (b) f(h) = 1 h 0.1 = 1 h h = 10 (c) Tidak, kerana imej bagi 0 adalah tidak tertakrif. No, because the image of 0 is undefined. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 6 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

7 1 g(x) = x – 8 h(x) = 2x + 1 (a) gh(x) = g(2x + 1) = (2x + 1) – 8 = 2x + 1 – 8 = 2x – 7 (b) hg(x) = h(x – 8) = 2(x – 8) + 1 = 2x – 16 + 1 = 2x – 15 (c) gh(5) = 2(5) – 7 = 10 – 7 = 3 2 g(x) = x + 2 h(x) = 3x – 7 (a) g2 (x) = g(x + 2) = (x + 2) + 2 = x + 4 (b) hg(x) = h(x + 2) = 3(x + 2) – 7 = 3x + 6 – 7 = 3x – 1 (c) hg(2) = 3(2) – 1 = 6 – 1 = 5 2 Diberi bahawa f(x) = |x – 2|, cari objek apabila f(x) = 3. It is given that f(x) = |x – 2|, find the object when f(x) = 3. |x – 2| = 3 x – 2 = –3 atau/or x – 2 = 3 x = –3 + 2 atau/or x = 3 + 2 x = –1 atau/or x = 5 3 Diberi bahawa f(x) = |2x + 1|, cari objek apabila f(x) = 5. It is given that f(x) = |2x + 1|, find the object when f(x) = 5. |2x + 1|= 5 2x + 1 = –5 atau/or 2x + 1 = 5 2x = –5 – 1 atau/or 2x = 5 – 1 2x = –6 atau/or 2x = 4 x = –3 atau/or x = 2 4 Diberi bahawa f(x) = |2x – 5|, cari objek apabila f(x) = 9. It is given that f(x) = |2x – 5|, find the object when f(x) = 9. |2x – 5| = 9 2x – 5 = –9 atau/or 2x – 5 = 9 2x = –9 + 5 atau/or 2x = 9 + 5 2x = –4 atau/or 2x = 14 x = –2 atau/or x = 7 5 Diberi bahawa f(x) = 3x – 1 2 , cari objek apabila f(x) = 6. It is given that f(x) = 3x – 1 2 , find the object when f(x) = 6. 3x – 1 2 = 6 3x – 1 2 = –6 atau/or 3x – 1 2 = 6 3x – 1 = –12 atau/or 3x – 1 = 12 3x = –12 + 1 atau/or 3x = 12 + 1 3x = –11 atau/or 3x = 13 x = – 11 3 atau/or x = 13 3 1.2 Fungsi Gubahan/ Composite Functions Latihan 8 Selesaikan setiap yang berikut. TP 3 Solve each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. Contoh 8 g(x) = 5x – 9 h(x) = x + 3 Cari/Find (a) gh(x), (b) hg(x), (c) gh(–2). Penyelesaian (a) gh(x) = g(x + 3) = 5(x + 3) – 9 = 5x + 15 – 9 = 5x + 6 (b) hg(x) = h(5x – 9) = (5x – 9) + 3 = 5x – 9 + 3 = 5x – 6 (c) gh(–2) = 5(–2) + 6 = –10 + 6 = –4 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 7 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

8 Contoh 9 Diberi f(x) = 3x dan g(x) = x + 5. Cari fg(4). Given f(x) = 3x and g(x) = x + 5. Find fg(4). Penyelesaian g(4) = (4) + 5 = 9 fg(4) = f(9) = 3(9) = 27 1 Diberi f(x) = 2x dan g(x) = x – 3. Cari fg(7). Given f(x) = 2x and g(x) = x – 3. Find fg(7). g(7) = (7) – 3 = 4 fg(7) = f(4) = 2(4) = 8 2 Diberi f(x) = 6x dan g(x) = x – 2. Cari gf(5). Given f(x) = 6x and g(x) = x – 2. Find gf(5). f(5) = 6(5) = 30 gf(5) = g(30) = 30 – 2 = 28 3 Diberi f(x) = x + 4 dan g(x) = 2x – 5. Cari gf(3). Given f(x) = x + 4 and g(x) = 2x – 5. Find gf(3). f(3) = (3) + 4 = 7 gf(3) = g(7) = 2(7) – 5 = 9 4 Diberi g(x) = x2 dan h(x) = x + 3. Cari hg(3). Given g(x) = x2 and h(x) = x + 3. Find hg(3). g(3) = (3)2 = 9 hg(3) = h(9) = (9) + 3 = 12 5 Diberi g(x) = 2x2 dan h(x) = 3x + 1. Cari gh(4). Given g(x) = 2x2 and h(x) = 3x + 1. Find gh(4). h(4) = 3(4) + 1 = 13 gh(4) = g(13) = 2(13)2 = 338 3 g(x) = 3x h(x) = 6 x + 6 (a) gh(x) = g( 6 x + 6 –) = 3( 6 x + 6 ) = 18 x + 6 , x ≠ –6 (b) h2 (x) = h( 6 x + 6 ) = 6 ( 6 x + 6 ) + 6 = 6 ( 6 + 6x + 36 x + 6 ) = 6(x + 6) 6x + 42 = 6(x + 6) 6(x + 7) = x + 6 x + 7 , x ≠ –7 (c) h2 (3) = (3) + 6 (3) + 7 = 9 10 4 g(x) = x2 + 1 h(x) = x – 3 (a) gh(x) = g(x – 3) = (x – 3)2 + 1 = (x2 – 6x + 9) + 1 = x2 – 6x + 10 (b) hg(x) = h(x2 + 1) = (x2 + 1) – 3 = x2 – 2 (c) hg(–4) = (–4)2 – 2 = 16 – 2 = 14 5 g( ) = x + 4 h(x) = 2x2 – 9 (a) g2 (x) = g(x + 4) = (x + 4) + 4 = x + 8 (b) hg(x) = h(x + 4) = 2(x + 4)2 – 9 = 2(x2 + 8x + 16) – 9 = 2x2 + 16x + 32 – 9 = 2x2 + 16x + 23 (c) hg(–2) = 2(–2)2 + 16(–2) + 23 = 2(4) – 32 + 23 = –1 Latihan 9 Cari imej bagi setiap fungsi gubahan berikut. TP 3 Find the image for each of the following composite functions. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 8 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

9 Contoh 10 f(x) = 3x + 1 g(x) = 2x – 1 fg(x) = 1 Penyelesaian fg(x) = f(2x – 1) = 3(2x – 1) + 1 = 6x – 3 + 1 = 6x – 2 Diberi/Given fg(x) = 1 6x – 2 = 1 6x = 3 x = 3 6 = 1 2 1 g(x) = x + 3 h(x) = 2x + 4 gh(x) = 5 gh(x) = g(2x + 4) = (2x + 4) + 3 = 2x + 7 Diberi/Given gh(x) = 5 2x + 7 = 5 2x = 5 – 7 2x = –2 x = –1 2 f(x) = 2x g(x) = x – 5 fg(x) = 6 fg(x) = f(x – 5) = 2(x – 5) = 2x – 10 Diberi/Given fg(x) = 6 2x – 10 = 6 2x = 6 + 10 2x = 16 x = 8 3 f(x) = 4 x , x ≠ 0 g(x) = 2x + 1 fg(x) = 15 fg(x) = f(2x + 1) = 4 (2x + 1) Diberi/Given fg(x) = 15 4 (2x + 1) = 15 4 = 15(2x + 1) 4 = 30x + 15 –15 + 4 = 30x –11 = 30x x = – 11 30 4 f(x) = 12 – x 2 f 2 (x) = 4 f 2 (x) = f ( 12 – x 2 ) = 12 – ( ) 12 – x 2 2 = ( ) 24 – 12 + x 2 2 = 12 + x 2 × 1 2 = 12 + x 4 Diberi/Given f 2 (x) = 4 12 + x 4 = 4 12 + x = 16 x = 4 5 h(x)= 6x – 7 h2 (x) = 23 h2 (x) = h(6x – 7) = 6(6x – 7) – 7 = 36x – 42 – 7 = 36x – 49 Diberi/Given h2 (x) = 23 36x – 49 = 23 36x = 72 x = 2 Contoh 11 g(x) = x + 6 fg(x) = 2x + 16 Penyelesaian f(x + 6) = 2x + 16 Katakan/Let y = x + 6 x = y – 6 f(y) = 2x + 16 = 2(y – 6) + 16 = 2y – 12 + 16 = 2y + 4 f(x) = 2x + 4 1 g(x)= x – 5 fg(x) = x – 2 f(x – 5) = x – 2 Katakan/Let y = x – 5 x = y + 5 f(y) = x – 2 = (y + 5) – 2 = y + 3 f(x) = x + 3 2 g(x)= x + 8 fg(x) = x – 6 f(x + 8) = x – 6 Katakan/Let y = x + 8 x = y – 8 f(y) = x – 6 = (y – 8) – 6 = y – 14 f(x) = x – 14 Latihan 10 Cari nilai x bagi setiap fungsi gubahan berikut. TP 3 Find the value of x for each of the following composite functions. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. Latihan 11 Cari fungsi f(x) bagi setiap yang berikut. TP 3 Find the function f(x) for each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 9 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

10 3 g(x)= 2x – 4 fg(x) = 2x + 1 f(2x – 4) = 2x + 1 Katakan/Let y = 2x – 4 2x = y + 4 x = y + 4 2 f(y) = 2x + 1 = 2( y + 4 2 ) + 1 = y + 5 f(x) = x + 5 4 g(x)= x + 6 fg(x) = 3x + 17 f(x + 6) = 3x + 17 Katakan/Let y = x + 6 x = y – 6 f(y) = 3x + 17 = 3(y – 6) + 17 = 3y – 1 f(x) = 3x – 1 5 g(x)= 3x – 5 fg(x) = 6x – 9 f(3x – 5) = 6x – 9 Katakan/Let y = 3x – 5 3x = y + 5 x = y + 5 3 f(y) = 6x – 9 = 6( y + 5 3 ) – 9 = 2y + 1 f(x) = 2x + 1 Contoh 12 f(x) = 3x + 2 fg(x) = 3x – 19 Penyelesaian f(x) = 3x + 2 maka/then fg(x) = 3g(x) + 2 Diberi/Given fg(x) = 3x – 19 3g(x) + 2 = 3x – 19 3g(x) = 3x – 19 – 2 3g(x) = 3x – 21 3g(x) = 3(x – 7) g(x) = x – 7 1 f(x) = x + 3 fg(x) = x – 8 f(x) = x + 3 maka/then fg(x) = g(x) + 3 Diberi/Given fg(x) = x – 8 g(x) + 3 = x – 8 g(x) = x – 8 – 3 g(x) = x – 11 2 f(x) = x – 6 fg(x) = x + 9 f(x) = x – 6 maka/then fg(x) = g(x) – 6 Diberi/Given fg(x) = x + 9 g(x) – 6 = x + 9 g(x) = x + 9 + 6 g(x) = x + 15 3 f(x) = x – 12 fg(x) = 3x – 4 f(x) = x – 12 maka/then fg(x) = g(x) – 12 Diberi/Given fg(x) = 3x – 4 g(x) – 12 = 3x – 4 g(x) = 3x – 4 + 12 g(x) = 3x + 8 4 f(x) = 3x – 7 fg(x) = 3x + 20 f(x) = 3x – 7 maka/then fg(x) = 3g(x) – 7 Diberi/Given fg(x) = 3x + 20 3g(x) – 7 = 3x + 20 3g(x) = 3x + 20 + 7 3g(x) = 3x + 27 g(x) = 3(x + 9) 3 g(x) = x + 9 5 f(x) = 2x – 3 fg(x) = 10x + 5 f(x) = 2x – 3 maka/then fg(x) = 2g(x) – 3 Diberi/Given fg(x) = 10x + 5 2g(x) – 3 = 10x + 5 2g(x) = 10x + 5 + 3 2g(x) = 10x + 8 g(x) = 2(5x + 4) 2 g(x) = 5x + 4 Latihan 12 Cari fungsi g(x) bagi setiap yang berikut. TP 3 Find the function g(x) for each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 10 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

11 f : x → 2x + k g : x → x – 4 fg : x → mx + 6 Cari nilai k dan nilai m. Find the value of k and of m. Penyelesaian fg(x) = f(x – 4) = 2(x – 4) + k = 2x – 8 + k Bandingkan/Compare fg(x) = 2x – 8 + k fg(x) = mx + 6 –8 + k = 6 , m = 2 k = 14 ∴ m = 2, k = 14 Contoh 13 1 f : x → 14 – 3x g : x → 2 – 4x fg : x → hx + k Cari nilai h dan nilai k. Find the value of h and of k. fg(x) = f(2 – 4x) = 14 – 3(2 – 4x) = 14 – 6 + 12x = 8 + 12x = 12x + 8 Bandingkan/Compare fg(x) = hx + k fg(x) = 12x + 8 ∴ h = 12, k = 8 2 f : x → 2x – 7 g : x → 3x gf : x → ax + b Cari nilai a dan nilai b. Find the value of a and of b. gf(x) = g(2x – 7) = 3(2x – 7) = 6x – 21 Bandingkan/Compare gf(x) = ax + b gf(x) = 6x – 21 ∴ a = 6, b = –21 3 f : x → 8x + k g : x → x – 4 fg : x → mx – 18 Cari nilai k dan nilai m. Find the value of k and of m. fg(x) = f(x – 4) = 8(x – 4) + k = 8x – 32 + k Bandingkan/Compare fg(x) = mx – 18 fg(x) = 8x – 32 + k m = 8; –32 + k = –18 k = –18 + 32 = 14 ∴ m = 8, k = 14 4 f : x → 6 – 4x g : x → ax + b gf : x → 2 – 12x Cari nilai a dan nilai b. Find the value of a and of b. gf(x) = g(6 – 4x) = a(6 – 4x) + b = 6a – 4ax + b = 6a + b – 4ax Bandingkan/Compare gf(x) = 2 – 12x gf(x) = 6a + b – 4ax –4a = –12; 6a + b = 2 a = 3 6(3) + b = 2 18 + b = 2 b = 2 – 18 = –16 ∴ a = 3, b = –16 5 f : x → 2x + h g : x → 3x – 9 gf : x → kx + 12 Cari nilai h dan nilai k. Find the value of h and of k. gf(x) = g(2x + h) = 3(2x + h) – 9 = 6x + 3h – 9 Bandingkan/Compare gf(x) = kx + 12 gf(x) = 6x + 3h – 9 k = 6; 3h – 9 = 12 3h = 12 + 9 3h = 21 h = 7 ∴ h = 7, k = 6 Latihan 13 Selesaikan setiap yang berikut. TP 3 Solve each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 11 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

12 Latihan 14 Selesaikan setiap yang berikut. TP 3 TP 4 KBAT Menilai Solve each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah rutin yang mudah. 1 Diberi h(x) = – 6 x , cari Given that h(x) = – 6 x , find (a) h2 (x), (a) h2 (x) = – 6 (– 6 x ) = 6 × x 6 = x (b) h8 (x), (b) h3 (x) = (– 6 x ) = – 6 x h4 (x) = – 6 (– 6 x ) = x h8 (x) = h4 (h4 (x)) = x (c) h17(x). (c) h17(x) = h16[h(x)] = (– 6 x ) = – 6 x 2 Rokiah ialah seorang penjual nasi lemak. Keuntungan harian yang diperoleh, dalam RM, diberi oleh w : x → 3x – 118 4 , dengan keadaan x ialah bilangan peket nasi lemak yang dijual dalam sehari. Rokiah is a seller of nasi lemak. The daily profit obtained, in RM, is given by w : x → 3x – 118 4 , where x is the number of packets of nasi lemak sold in a day. (a) Hitung keuntungan harian yang diperoleh Rokiah jika dia menjual 1 197 peket nasi lemak dalam seminggu. Calculate the average daily profit obtained by Rokiah if she sold 1 197 packets of nasi lemak in a week. (b) Cari bilangan minimum peket nasi lemak yang perlu dijual dalam sehari supaya Rokiah tidak mengalami sebarang kerugian. Find the minimum number of packets of nasi lemak that must be sold in a day so that Rokiah does not experience any loss. (a) 1 minggu/week : 1 197 peket nasi lemak/packets of nasi lemak 1 hari/day : 171 peket nasi lemak/packets of nasi lemak w(x) = 3x – 118 4 = 3(171) – 118 4 = 98.75 ∴ Keuntungan harian/The daily profit = RM98.75 (b) 3x – 118 4 > 0 3x – 118 > 0 3x > 118 x > 39.33 x = 40 Bilangan minimum peket nasi lemak yang perlu dijual dalam sehari ialah 40 peket. The minimum number of packets of nasi lemak that must be sold in a day is 40 packets. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 12 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

13 1.3 Fungsi Songsang/ Inverse Functions Latihan 15 Tentukan sama ada setiap fungsi f berikut mempunyai fungsi songsang atau tidak. Beri alasan anda. TP 3 Determine whether each of the following functions f has an inverse or not. Give your reason. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. O x y f (–1, –18) (5, –18) Domain: –1  x  5 Penyelesaian Apabila ujian garis mengufuk dilakukan, garis mengufuk memotong graf f pada dua titik. Ini bermaksud fungsi f ini bukan fungsi satu dengan satu. Maka, fungsi f tidak mempunyai fungsi songsang. When the horizontal line test is carried out, the horizontal line cuts the graph of function f at two points. This means that the function f is not a one-to-one function. Thus, the function f has no inverse function. Contoh 15 1 O x y f (–4, –16) (6, 8) Domain: –4  x  6 Apabila ujian garis mengufuk dilakukan, garis mengufuk memotong graf f hanya pada satu titik. Ini bermaksud jenis fungsi f ini ialah fungsi satu dengan satu. Maka, fungsi f mempunyai fungsi songsang. When the horizontal line test is carried out, the horizontal line cuts the graph of function f at only one point. This means that the function f is a one-to-one function. Thus, the function f has an inverse function. 2 O x y f (6, −3) (−2, 9) Domain: –2  x  6 Apabila ujian garis mengufuk dilakukan, garis mengufuk memotong graf f hanya pada satu titik. Ini bermaksud jenis fungsi f ini ialah fungsi satu dengan satu. Maka, fungsi f mempunyai fungsi songsang. When the horizontal line test is carried out, the horizontal line cuts the graph of function f at only one point. This means that the function f is a one-to-one function. Thus, the function f has an inverse function. 3 O x y f (−3, 0) (3, 0) Domain: –3  x  3 Apabila ujian garis mengufuk dilakukan, garis mengufuk memotong graf f pada tiga titik. Ini bermaksud fungsi f ini bukan fungsi satu dengan satu. Maka, fungsi f tidak mempunyai fungsi songsang. When the horizontal line test is carried out, the horizontal line cuts the graph of function f at three points. This means that the function f is not a one-to-one function. Thus, the function f has no inverse function. 4 O x y f (1, −2) (−1, 4) Domain: –1  x  1 Apabila ujian garis mengufuk dilakukan, garis mengufuk memotong graf f hanya pada satu titik. Ini bermaksud jenis fungsi f ini ialah fungsi satu dengan satu. Maka, fungsi f mempunyai fungsi songsang. When the horizontal line test is carried out, the horizontal line cuts the graph of function f at only one point. This means that the function f is a one-to-one function. Thus, the function f has an inverse function. 5 O x y f (0, 2) (3, 2) Domain: 0  x  3 Apabila ujian garis mengufuk dilakukan, garis mengufuk memotong graf f pada empat titik. Ini bermaksud fungsi f ini bukan fungsi satu dengan satu. Maka, fungsi f tidak mempunyai fungsi songsang. When the horizontal line test is carried out, the horizontal line cuts the graph of function f at four points. This means that the function f is not a one-to-one function. Thus, the function f has no inverse function. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 13 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

14 Latihan 16 Cari fungsi songsang bagi setiap fungsi berikut. TP 3 Find the inverse function for each of the following functions. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 1 f(x) = 3x + 2 f(x) = 3x + 2 x = f –1(3x + 2) Katakan/Let y = 3x + 2 y – 2 = 3x y – 2 3 = x Maka/Hence, f –1(y) = y – 2 3 f –1(x) = x – 2 3 2 f(x) = 3x – 5 f(x) = 3x – 5 x = f –1(3x – 5) Katakan/Let y = 3x – 5 y + 5 = 3x y + 5 3 = x Maka/Hence, f –1(y) = y + 5 3 f –1(x) = x + 5 3 3 f(x) = 6 – 4x f(x) = 6 – 4x x = f –1(6 – 4x) Katakan/Let y = 6 – 4x 4x = 6 – y x = 6 – y 4 Maka/Hence, f –1(y) = 6 – y 4 f –1(x) = 6 – x 4 4 f(x) = 5 – x 3 f(x) = 5 – x 3 x = f –1( 5 – x 3 ) Katakan/Let y = 5 – x 3 3y = 5 – x x = 5 – 3y Maka/Hence, f –1(y) = 5 – 3y f –1(x) = 5 – 3x 5 f(x) = x – 6 x , x ≠ 0 f(x) = x – 6 x x = f –1( x – 6 x ) Katakan/Let y = x – 6 x xy = x – 6 6 = x – xy 6 = x(1 – y) 6 1 – y = x Maka/Hence, f –1(y) = 6 1 – y f –1(x) = 6 1 – x , x ≠ 1 6 f(x) = 7 + 7x f(x) = 7 + 7x x = f –1(7 + 7x) Katakan/Let y = 7 + 7x y – 7 = 7x y – 7 7 = x Maka/Hence, f –1(y) = y – 7 7 f –1(x) = x – 7 7 f(x) = 2x + 3 x , x ≠ 0 Penyelesaian x = f –1 ( ) 2x + 3 x Katakan/Let y = 2x + 3 x xy = 2x + 3 xy – 2x = 3 x(y – 2) = 3 x = 3 y – 2 Daripada/From , f –1(y) = 3 y – 2 ∴ f –1(x) = 3 x – 2 , x ≠ 2 Contoh 16 Kesilapan Umum f –1(x) = 3 x – 2 adalah tidak lengkap./is incomplete. Syarat x ≠ 2 mesti ditulis dalam jawapan. The condition of x ≠ 2 must be written in the answer. 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 14 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

15 1 f(x) = 2x – 14 f –1(x) = px + q f(x) = 2x – 14 x = f –1(2x – 14) Katakan/Let y = 2x – 14 Maka/Then x = y + 14 2 f –1(y) = y + 14 2 f –1(x) = x + 14 2 = x 2 + 7 Bandingkan dengan Compare with f –1(x) = px + q ∴ p = 1 2 dan/and q = 7 2 f(x) = 6x + q f –1(x) = px – 12 f(x) = 6x + q x = f –1(6x + q) Katakan/Let y = 6x + q Maka/Then x = y – q 6 f –1(y) = y – q 6 f –1(x) = x – q 6 = x 6 – q 6 Bandingkan dengan Compare with f –1(x) = px – 12 ∴ p = 1 6 dan/and q 6 = 12 q = 72 3 f(x) = 2p – 3x f –1(x) = 1 – q 3 x f(x) = 2p – 3x x = f –1(2p – 3x) Katakan/Let y = 2p – 3x Maka/Then x = 2p – y 3 f –1(y) = 2p – y 3 f –1(x) = 2p – x 3 Bandingkan dengan Compare with f –1(x) = 1 – q 3 x = 3 – qx 3 ∴2p = 3 p = 3 2 dan/and q = 1 Latihan 17 Cari nilai p dan nilai q bagi setiap yang berikut. TP 3 KBAT Mengaplikasi Find the value of p and of q for each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. f(x) = 4x + q f –1(x) = px + 2 Penyelesaian f(x) = 4x + q x = f –1(4x + q) Katakan/Let y = 4x + q Maka/Then x = y – q 4 f –1(y) = y – q 4 f –1(x) = x – q 4 = x 4 – q 4 Bandingkan/Compare f –1(x) = x 4 – q 4 f –1(x) = px + 2 ∴ 1 4 = p dan/and – q 4 = 2 p = 1 4 q = –8 Contoh 17 Untuk tujuan pembelajaran Imbas kod QR atau layari https://www. coolmath. com/ algebra/16- inversefunctions/05- how-tofind-theinverse-of-afunction-01 untuk nota tambahan tentang cara untuk mencari fungsi songsang. Laman Web Laman Web 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 15 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

16 (a) f −1, (b) f −1g(2). [5 markah/marks] 4 Diberi fungsi f(x) = 3 − x dan g(x) = px² + q. Jika fungsi gubahan gf(x) = 3x² − 8x + 15, cari Given that the functions f(x) = 3 − x and g(x) = px² + q. If the composite function gf(x) = 3x² − 8x + 15, find (a) nilai p dan nilai q, the value of p and of q, [3 markah/marks] (b) nilai bagi g²(0). the value of g²(0). [2 markah/marks] 5 Tiga fungsi ditakrifkan oleh f : x → 3x − 4, fg : x → 6x + 11 dan hf : x → 9x² − 24x + 10. Cari Three functions are defined by f : x → 3x − 4, fg : x → 6x + 11 and hf : x → 9x² − 24x + 10. Find (a) g(x), (b) h(x). [6 markah/marks] 6 (a) Diberi bahawa g(x) = x + 4 2x – 5 , x ≠ 5 2 . Cari nilai g–1(–3). It is given that g(x) = x + 4 2x – 5 , x ≠ 5 2 . Find the value of g–1(–3). [2 markah/marks] (b) Diberi fungsi f(x) = –ax + b, dengan keadaan a dan b ialah pemalar. Cari nilai a dan nilai b dengan keadaan f –1(7) = 4 dan f –1(–3) = 9. Given that the function f(x) = –ax + b, where a and b are constants. Find the value of a and of b such that f –1(7) = 4 and f –1(–3) = 9. [3 markah/marks] 7 (a) Harga seunit bagi suatu barang ialah RMx. Setiap unit barang yang dibeli secara dalam talian dikenakan caj 2%. Diberi kos penghantaran ialah RM15 dan jumlah perbelanjaan untuk membeli seunit barangan itu ialah f(x). Jika t(x) mewakili fungsi bagi bayaran berserta caj, ungkapkan f(x) dalam sebutan x. The price per unit of an item is RMx. Each item purchased online is to be charged by 2%. Given the delivery fee is RM15 and the total expenditure to buy a unit of the item is f(x). If t(x) represents the function for the payment with charge, express f(x) in terms of x. [4 markah/marks] (b) Diberi fungsi g : x → 6x + d dan g2 : x → cx – 28, dengan keadaan c dan d ialah pemalar dan c > 0. Cari Given that the function g : x →6x + d and g2 : x → cx – 28, where c and d are constants and c > 0. Find (i) nilai c dan nilai d, the value of c and of d, (ii) (g−1) 2 (x). [4 markah/marks] Praktis Berformat SPM Kertas 1 Bahagian A 1 (a) Rajah 1 menunjukkan fungsi gubahan kh yang memetakan set u kepada set w. Diagram 1 shows the composite function kh that maps set u to set w. u v kh w Rajah 1/Diagram 1 Nyatakan/State (i) fungsi yang memetakan set u kepada set v, the function that maps set u to set v, (ii) k−1(w). [2 markah/marks] (b) Diberi fungsi f : x → 6x + 4 dan g : x → 3x − 8, cari g−1f(x). Given that the function f : x → 6x + 4 and g : x → 3x − 8, find g−1f(x). [3 markah/marks] 2 Diberi bahawa fungsi f : x → p + qx. It is given that the function f : x → p + qx. (a) Cari f −1(x) dalam sebutan p dan q. Find f −1(x) in terms of p and q. [2 markah/marks] (b) Jika f −1(11) = −1 dan f(4) = −4, cari nilai p dan q. If f −1(11) = −1 and f(4) = −4, find the value of p and of q. [4 markah/marks] 3 Fungsi f dan g ditakrifkan oleh f : x → 8 − 6x dan g : x → x² − 6. Cari The functions f and g are defined by f : x → 8 − 6x and g : x → x² − 6. Find Bahagian B 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 16 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

17 1 Diberi bahawa g : x → 2x – 3 dan h : x → x 3 + 2, cari It is given that g : x → 2x – 3 and h : x → x 3 + 2, find (a) g−1(x), [2 markah/marks] (b) g−1h(x), [2 markah/marks] (c) fungsi k(x) dengan keadaan kh(x) = 2x + 8. [3 markah/marks] the function k(x) such that kh(x) = 2x + 8. 2 Diberi bahawa g : x → 3x + 15 dan h : x → 2x − 12. It is given that g : x → 3x + 15 and h : x → 2x − 12. (a) Cari/Find (i) h(8), (ii) nilai p jika g(p − 1) = 1 2 h(8) + 1, the value of p if g(p − 1) = 1 2 h(8) + 1, (iii) hg(x). [5 markah/marks] (b) (i) Lakar graf bagi y = |hg(x)| untuk –6  x  0. Sketch the graph of y = |hg(x)| for –6  x  0. (ii) Cari nilai q dengan keadaan hg(q) = 2gh(q). [5 markah/marks] Find the value of q such that hg(q) = 2gh(q). Kertas 2 Bahagian A Bahagian B 1 Tadika Desa mula beroperasi pada tahun 2016 dan bilangan murid bagi 12 tahun yang pertama diberi oleh f : t → 16 + 7t, dengan keadaan t ialah bilangan tahun selepas tahun 2016. KBAT Mengaplikasi Desa Kindergarten started operating in 2016 and the number of students for the first 12 years is given by f : t → 16 + 7t, such that t is the number of years after 2016. (a) Cari bilangan murid selepas 5 tahun. Find the number of students after 5 years. (b) Pada tahun berapakah bilangan murid akan menjadi 79 orang? In which year will the number of students be 79? Zon KBAT 01 Strategi A+ SPM Mate Tam Tg4_C1_1-17_Final.indd 17 10/17/23 2:37 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

18 2.1 Persamaan dan Ketaksamaan Kuadratik Quadratic Equations and Inequalities 1 Bentuk am persamaan kuadratik ialah ax2 + bx + c = 0, dengan keadaan a, b dan c ialah pemalar dan a ≠ 0. The general form of a quadratic equation is ax2 + bx + c = 0, where a, b and c are constants and a ≠ 0. 2 Ciri-ciri persamaan kuadratik: Characteristics of a quadratic equation: (a) Melibatkan hanya satu pemboleh ubah. Involves only one variable. (b) Mempunyai tanda ‘=’ dan boleh dinyatakan dalam bentuk ax2 + bx + c = 0. Has an equal sign ‘=’ and can be expressed in the form ax2 + bx + c = 0. (c) Kuasa tertinggi bagi pemboleh ubah ialah 2. The highest power of the variable is 2. 3 Terdapat dua kaedah yang boleh digunakan untuk menyelesaikan persamaan kuadratik: There are two methods that can be used to solve quadratic equations: (a) Kaedah penyempurnaan kuasa dua Completing the square method (b) Kaedah rumus Formula method x = –b ± b2 – 4ac 2a 4 Persamaan kuadratik dengan punca-punca yang diberi boleh ditulis sebagai A quadratic equation with given roots can be expressed as x2 – (hasil tambah punca)x + (hasil darab punca) = 0 x2 – (sum of roots)x + (product of roots) = 0 5 Kaedah yang boleh digunakan untuk menyelesaikan ketaksamaan kuadratik: The methods that can be used to solve a quadratic inequality: (a) Kaedah lakaran graf/Graph sketching method (b) Kaedah garis nombor/Number line method (c) Kaedah jadual/Table method 2.2 Jenis-jenis Punca Persamaan Kuadratik Types of Roots of Quadratic Equations 1 Jenis-jenis punca bagi suatu persamaan kuadratik boleh ditentukan dengan mencari nilai b2 – 4ac. Nilai ini dikenali sebagai pembezalayan. Types of roots of a quadratic equation can be determined by finding the value of b2 – 4ac. This value is known as discriminant. 2 Terdapat tiga jenis punca persamaan kuadratik: There are three types of roots of quadratic equations: (a) Dua punca nyata yang sama, b2 – 4ac = 0 Two real and equal roots, b2 – 4ac = 0 (b) Dua punca nyata dan berbeza, b2 – 4ac > 0 Two real and different roots, b2 – 4ac > 0 (c) Tidak mempunyai punca nyata, b2 – 4ac < 0 Has no real roots, b2 – 4ac < 0 2.3 Fungsi Kuadratik Quadratic Functions 1 Diberi fungsi kuadratik f(x) = ax2 + bx + c, Given that a quadratic function f(x) = ax2 + bx + c, (a) jika a adalah positif (a > 0), maka graf fungsi kuadratik itu mempunyai satu titik minimum, if a is positive (a > 0), then the graph of the quadratic function has a minimum point, (b) jika a adalah negatif (a < 0), maka graf fungsi kuadratik itu mempunyai satu titik maksimum. if a is negative (a < 0), then the graph of the quadratic function has a maximum point. 2 Perubahan nilai a, b dan c memberi kesan terhadap bentuk dan kedudukan graf. The changes in the values of a, b and c affect the shape and position of the graph. 3 Bentuk dan kedudukan graf tersebut adalah seperti berikut: The shape and position of the graph is as follows: Bab 2 Fungsi Kuadratik Quadratic Functions Bidang Pembelajaran: Algebra Revisi Pantas Pembezalayan Discriminant Jenis punca dan kedudukan graf Types of roots and position of the graph f(x) = ax2 + bx + c a > 0 a < 0 b2 – 4ac > 0 • Dua punca nyata dan berbeza/Two real and different roots • Graf menyilang paksi-x pada dua titik yang berbeza The graph intersects the x-axis at two different points x x b2 – 4ac = 0 • Dua punca nyata yang sama/Two real and equal roots • Graf menyentuh paksi-x hanya pada satu titik The graph touches the x-axis at one point only x x b2 – 4ac < 0 • Tidak mempunyai punca nyata/No real roots • Graf tidak menyilang paksi-x The graph does not intersect the x-axis x x 4 Fungsi kuadratik boleh ditulis dalam bentuk am, bentuk verteks dan bentuk pintasan. A quadratic function can be written in general form, vertex form and intercept form. Bentuk am/General form f(x) = ax2 + bx + c Bentuk verteks/Vertex form f(x) = a(x – h)2 + k Bentuk pintasan/Intercept form f(x) = a(x – p)(x – q) Verteks Vertex 1– b 2a , f1– b 2a 22 (h, k) 1 p + q 2 , f1 p + q 2 22 Paksi simetri Axis of symmetry x = – b 2a x = h x = p + q 2 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 18 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

19 x² – 5x – 7 = 0 Penyelesaian x² – 5x – 7 = 0 x² – 5x = 7 x² – 5x + 1 –5 2 2 2 = 7 + 1 –5 2 2 2 1x – 5 2 2 ² = 7 + 25 4 1x – 5 2 2 ² = 53 4 x – 5 2 = ± 53 4 x = 5 2 ± 53 2 x = 5 – 53 2 , 5 + 53 2 x = –1.140, 6.140 Contoh 1 1 x² – 3x – 1 = 0 x² – 3x – 1 = 0 x² – 3x = 1 x² – 3x + 1 –3 2 2 2 = 1 + 1 –3 2 2 2 1x – 3 22 2 = 1 + 9 4 1x – 3 22 2 = 13 4 x – 3 2 = ± 13 4 x = 3 2 ± 13 2 x = 3 – 13 2 , 3 + 13 2 x = –0.303, 3.303 2 x² – 7x + 4 = 0 x² – 7x + 4 = 0 x² – 7x = –4 x² – 7x + 1 –7 2 2 2 = –4 + 1 –7 2 2 2 1x – 7 2 2 2 = –4 + 49 4 1x – 7 2 2 2 = 33 4 x – 7 2 = ± 33 4 x = 7 2 ± 33 2 x = 7 – 33 2 , 7 + 33 2 x = 0.628, 6.372 2.1 Persamaan dan Ketaksamaan Kuadratik/ Quadratic Equations and Inequalities Latihan 1 Selesaikan persamaan kuadratik berikut dengan menggunakan kaedah penyempurnaan kuasa dua. Beri jawapan betul kepada tiga tempat perpuluhan. TP 2 Solve the following quadratic equations by using completing the square method. Give the answers correct to three decimal places. TP 2 Mempamerkan kefahaman tentang fungsi kuadratik. Praktis PBD Latihan 2 Selesaikan persamaan kuadratik berikut dengan menggunakan kaedah penyempurnaan kuasa dua. Beri jawapan betul kepada tiga tempat perpuluhan. TP 2 Solve the following quadratic equations by using completing the square method. Give the answers correct to three decimal places. TP 2 Mempamerkan kefahaman tentang fungsi kuadratik. 5x2 – 3x – 9 = 0 Penyelesaian 5x2 – 3x – 9 = 0 x2 – 3 5 x – 9 5 = 0 x2 – 3 5 x = 9 5 x2 – 3 5 x + (– 3 10) 2 = 9 5 + 9 100 (x – 3 10) 2 = 189 100 x – 3 10 = ± 189 10 x = 3 10 ± 189 10 x= 3 10 – 189 10 , 3 10 + 189 10 x= –1.075, 1.675 Contoh 2 1 2x2 + 6x – 5 = 0 2x2 + 6x – 5 = 0 × 1 2 x2 + 3x – 5 2 = 0 x2 + 3x = 5 2 x2 + 3x + ( 3 2 ) 2 = 5 2 + ( 3 2 ) 2 (x + 3 2 ) 2 = 5 2 + 9 4 (x + 3 2 ) 2 = 19 4 x + 3 2 = ±19  2 x = – 3 2 ± 19  2 x = – 3 2 – 19  2 , – 3 2 + 19  2 x = –3.679, 0.679 2 7x2 – 6x – 36 = 0 7x2 – 6x – 36 = 0 × 1 7 x2 – 6 7 x – 36 7 = 0 x2 – 6 7 x = 36 7 x2 – 6 7 x + (– 3 7 ) 2 = 36 7 + (– 3 7 ) 2 (x – 3 7 ) 2 = 36 7 + 9 49 (x – 3 7 ) 2 = 261 49 x – 3 7 = ± 261  7 x = 3 7 ± 261  7 x = 3 7 – 261  7 , 3 7 + 261  7 x = –1.879, 2.736 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 19 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

20 3x2 – 4x – 6 = 0 Penyelesaian 3x2 – 4x – 6 = 0 a = 3, b = –4, c = –6 x = –(–4) ±  (–4)2 – 4(3)(–6) 2(3) = 4 ± 16 + 72 6 = 4 ± 88 6 x = 4 – 88 6 atau/or x = 4 + 88 6 x = –0.8968 atau/or x = 2.230 Tip Bestari Jika/If ax2 + bx + c = 0 maka/then x = –b ±  b2 – 4ac 2a Contoh 3 1 x2 – 3x = 5 x2 – 3x = 5 x2 – 3x – 5 = 0 a = 1, b = –3, c = –5 x = –(–3) ±  (–3)2 – 4(1)(–5) 2(1) x = 3 ± 9 + 20 2 x = 3 – 29 2 atau/or x = 3 + 29 2 x = –1.193 atau/or x = 4.193 2 3x2 – 5x – 7 = 0 3x2 – 5x – 7 = 0 a = 3, b = –5, c = –7 x = –(–5) ±  (–5)2 – 4(3)(–7) 2(3) = 5 ± 25 + 84 6 = 5 ± 109 6 x = 5 – 109 6 atau/orx = 5 + 109 6 x = –0.9067 atau/or x = 2.573 3 x2 + 4x = 6 x2 + 4x = 6 x2 + 4x – 6 = 0 a = 1, b = 4, c = –6 x = –4 ± 42 – 4(1)(–6) 2(1) x = –4 ± 16 + 24 2 x = –4 ± 40 2 x = –2 + 10 atau/or x = –2 –10 x = 1.162 atau/or x = –5.162 4 2x2 – 3x = 8 2x2 – 3x = 8 2x2 – 3x – 8 = 0 a = 2, b = –3, c = –8 x = –(–3) ±  (–3)2 – 4(2)(–8) 2(2) x = 3 ± 9 + 64 4 x = 3 ± 73 4 x = 3 – 73 4 atau/or x = 3 + 73 4 x = –1.386 atau/or x = 2.886 5 Rajah di bawah menunjukkan sebuah kotak tisu berbentuk kuboid. The diagram below shows a cuboid-shaped tissue box. 4 cm x cm (2x + 1) cm Diberi bahawa isi padu kotak itu ialah 92 cm3 . Cari nilai x. It is given that the volume of the box is 92 cm3 . Find the value of x. Isi padu/Volume = 92 4x(2x + 1) = 92 2x2 + x = 23 2x2 + x – 23 = 0 a = 2, b = 1, c = –23 x = –(1) ± (1)2 – 4(2)(–23) 2(2) x = –3.650 atau/or x = 3.150 Nilai negatif bagi panjang adalah tidak sah. The negative value for length is not valid. ∴ x = 3.150 6 Rajah di bawah menunjukkan sebuah trapezium. The diagram below shows a trapezium. (x + 3) cm (x + 4) cm (2x) cm (2x + 1) cm Cari nilai x. Find the value of x. x + 3 x + 4 2x x + 3 x – 2 x + 4 Menggunakan teorem Pythagoras, Using the Phythagoras’ theorem, (x + 4)2 + (x – 2)2 = (2x) 2 x2 + 8x + 16 + x2 – 4x + 4 = 4x2 2x2 – 4x – 20 = 0 x2 – 2x – 10 = 0 a = 1, b = –2, c = –10 x = –(–2) ± (–2)2 – 4(1)(–10) 2(1) x = –2.317 atau/or x = 4.317 Oleh sebab/Since x > 0, maka/thus x = 4.317 Latihan 3 Selesaikan setiap yang berikut dengan menggunakan rumus. Beri jawapan betul kepada empat angka bererti. TP 2 Solve each of the following by using formula. Give the answers correct to four significiant figures. TP 2 Mempamerkan kefahaman tentang fungsi kuadratik. 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 20 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

21 Punca-punca/Roots: 3, – 1 2 Penyelesaian Hasil tambah punca Sum of roots = 3 + fi– 1 2 ff = 5 2 Hasil darab punca Product of roots = (3)fi– 1 2 ff = – 3 2 Persamaan kuadratik: The quadratic equation: x2 – fi 5 2 ffx + fi– 3 2 ff = 0 2x2 – 5x – 3 = 0 Kaedah Alternatif x = 3, x = – 1 2 x – 3 = 0, 2x = –1 x – 3 = 0, 2x + 1 = 0 (x – 3)(2x + 1) = 0 2x2 + x – 6x – 3 = 0 2x2 – 5x – 3 = 0 Contoh 4 1 Punca-punca/Roots: 3, 2 Hasil tambah punca Sum of roots = 3 + 2 = 5 Hasil darab punca Product of roots = (3)(2) = 6 Persamaan kuadratik: The quadratic equation: x2 – (5)x + (6) = 0 x2 – 5x + 6 = 0 x = 3, x = 2 x – 3 = 0, x – 2 = 0 (x – 3)(x – 2) = 0 x2 – 2x – 3x + 6 = 0 x2 – 5x + 6 = 0 Kaedah Alternatif 2 Punca-punca/Roots: 5, –3 Hasil tambah punca Sum of roots = 5 + (–3) = 2 Hasil darab punca Product of roots = (5)(–3) = –15 Persamaan kuadratik: The quadratic equation: x2 – (2)x + (–15) = 0 x2 – 2x – 15 = 0 x = 5, x = –3 x – 5 = 0, x + 3 = 0 (x – 5)(x + 3) = 0 x2 + 3x – 5x – 15 = 0 x2 – 2x – 15 = 0 Kaedah Alternatif 3 Punca/Root: 5 7 sahaja/only Hasil tambah punca Sum of roots = fi 5 7 ff + fi 5 7 ff = 10 7 Hasil darab punca Product of roots = fi 5 7 ff fi 5 7 ff = 25 49 Persamaan kuadratik: The quadratic equation: x2 – fi 10 7 ffx + fi 25 49ff = 0 49x2 – 70x + 25 = 0 x = 5 7 7x – 5 = 0 (7x – 5)2 = 0 (7x – 5)(7x – 5) = 0 49x2 – 35x – 35x + 25 = 0 49x2 – 70x + 25 = 0 Kaedah Alternatif 4 Punca-punca/Roots: 3, – 2 3 Hasil tambah punca Sum of roots = 3 + fi– 2 3 ff = 7 3 Hasil darab punca Product of roots = (3)fi– 2 3 ff = –2 Persamaan kuadratik: The quadratic equation: x2 – fi 7 3ffx + (–2) = 0 3x2 – 7x – 6 = 0 x = 3, x = – 2 3 x – 3 = 0, 3x + 2 = 0 (x – 3)(3x + 2) = 0 3x2 + 2x – 9x – 6 = 0 3x2 – 7x – 6 = 0 Kaedah Alternatif 5 Punca-punca/Roots: 3 2 , – 1 3 Hasil tambah punca Sum of roots = 3 2 + fi– 1 3 ff = 7 6 Hasil darab punca Product of roots = fi 3 2 fffi– 1 3 ff = – 1 2 Persamaan kuadratik: The quadratic equation: x2 – fi 7 6 ff x + fi– 1 2 ff = 0 6x2 – 7x – 3 = 0 x = 3 2 , x = – 1 3 2x – 3 = 0, 3x + 1 = 0 (2x – 3)(3x + 1) = 0 6x2 + 2x – 9x – 3 = 0 6x2 – 7x – 3 = 0 Kaedah Alternatif Latihan 4 Bentukkan persamaan kuadratik dengan punca-punca yang diberi. TP 2 Form a quadratic equation with the given roots. TP 2 Mempamerkan kefahaman tentang fungsi kuadratik. 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final - Tajul Updated.indd 21 6/11/2023 3:04:58 PM MU BAKTI SDN. BHD. PENERBIT ILMU BAKTI SDLMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BAKTBIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

22 1 Punca-punca bagi persamaan kuadratik 2x2 – 8x + p = 5 ialah 1 dan 3. Cari nilai p. The roots of the quadratic equation 2x2 – 8x + p = 5 are 1 and 3. Find the value of p. Hasil tambah punca/Sum of roots = 1 + 3 = 4 Hasil darab punca/Product of roots = (1)(3) = 3 2x2 – 8x + p = 5 2x2 – 8x + p – 5 = 0 x2 – 4x + p – 5 2 = 0 Hasil darab punca Product of roots p – 5 2 = 3 p – 5 = 6 p = 5 + 6 = 11 2 Punca-punca bagi persamaan kuadratik 2x2 + px – 16 = 0 ialah 2 dan –4. Cari nilai p. The roots of the quadratic equation 2x2 + px – 16 = 0 are 2 and –4. Find the value of p. Hasil tambah punca/Sum of roots = 2 + (–4) = –2 Hasil darab punca/Product of roots = (2)(–4) = –8 2x2 + px – 16 = 0 x2 + p 2 x – 8 = 0 Hasil tambah punca Sum of roots – p 2 = –2 p = 4 3 Punca-punca bagi persamaan kuadratik 3x2 + px + 6 = q ialah –2 dan 6. Cari nilai p dan nilai q. The roots of the quadratic equation 3x2 + px + 6 = q are –2 and 6. Find the value of p and of q. Hasil tambah punca/Sum of roots = –2 + 6 = 4 Hasil darab punca/Product of roots = (–2)(6) = –12 3x2 + px + 6 = q 3x2 + px + 6 – q = 0 x2 + p 3 x + 6 – q 3 = 0 Hasil tambah punca Sum of roots – p 3 = 4 p = –12 Hasil darab punca Product of roots 6 – q 3 = –12 6 – q = –36 q = 42 4 Diberi bahawa p dan 5 ialah punca-punca bagi 2x2 + 4x + q = 0. Cari nilai p dan nilai q. Given p and 5 are the roots of 2x2 + 4x + q = 0. Find the value of p and of q. Hasil tambah punca/Sum of roots = p + 5 Hasil darab punca/Product of roots = 5p 2x2 + 4x + q = 0 x2 + 2x + q 2 = 0 Hasil tambah punca Sum of roots p + 5 = –2 p = –2 – 5 p = –7 Hasil darab punca Product of roots 5p = q 2 5(–7) = q 2 –35 = q 2 q = –70 Latihan 5 Selesaikan setiap yang berikut. TP 3 Solve each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. Punca-punca bagi persamaan kuadratik 2x2 + px + 6q = 12 ialah 3 dan –5. Cari nilai p dan nilai q. The roots of the quadratic equation 2x2 + px + 6q = 12 are 3 and –5. Find the value of p and of q. Penyelesaian Hasil tambah punca Sum of roots 3 + (–5) = –2 Hasil darab punca Product of roots (3)(–5) = –15 Hasil tambah punca Sum of roots – p 2 = –2 p = 4 Hasil darab punca Product of roots 3q – 6 = –15 3q = –9 q = –3 2x2 + px + 6q = 12 2x2 + px + 6q – 12 = 0 x2 + p 2 x + 6q – 12 2 = 0 x2 + p 2 x + 3q – 6 = 0 Bandingkan dengan Compare with x2 – (HTP) x + (HDP) = 0 Contoh 5 Tip Bestari Untuk menyelesaikan soalan ini, persamaan perlu diungkapkan sebagai x2 – (HTP)x + (HDP) = 0. Maka, pekali bagi x2 mestilah ditukarkan kepada 1. To solve this question, the equation must be expressed as x2 – (HTP)x + (HDP) = 0. Therefore, the coefficient of x2 must be changed to 1. Kaedah Alternatif x = 3, x = –5 (x – 3)(x + 5) = 0 x2 + 5x – 3x – 15 = 0 x2 + 2x – 15 = 0 (32) 2x2 + 4x – 30 = 0 Bandingkan/Compare 2x2 + px + 6q – 12 = 0 p = 4 , 6q – 12 = –30 6q = –18 q = –3 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 22 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

23 Latihan 6 Selesaikan. TP 3 Solve. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. (a) a dan β ialah punca-punca bagi persamaan kuadratik 2x2 + 4x – 9 = 0. Bentukkan persamaan kuadratik dengan punca-punca a + 2 and β + 2.  and  are the roots of the quadratic equation 2x2 + 4x – 9 = 0. Form a quadratic equation with roots  + 2 and β + 2. Penyelesaian 2x2 + 4x – 9 = 0 a = 2, b = 4, c = –9 Punca-punca/Roots: a, β a + β = – b a = – 4 2 = –2 aβ = c a = – 9 2 Punca-punca baharu: a + 2, β + 2 New roots: a + 2, β + 2 Hasil tambah punca baharu Sum of new roots (a + 2) + (β + 2) = (a + β) + 2 + 2 = (–2) + 4 = 2 Hasil darab punca baharu Product of new roots (a + 2)(b + 2) = aβ + 2a + 2β + 4 = aβ + 2(a + β) + 4 = 1 – 9 2 2 + 2(–2) + 4 = – 9 2 Persamaan kuadratik yang baharu New quadratic equation x2 – (HTP)x + (HDP) = 0 x2 – (2)x + 1 – 9 2 2 = 0 x2 – 2x – 9 2 = 0 2x2 – 4x – 9 = 0 (b) Diberi bahawa 3 4 dan β ialah punca-punca bagi persamaan kuadratik 20x2 – 7x – 6 = 0, cari nilai β. Given that 3 4 and β are the roots of the quadratic equation 20x2 – 7x – 6 = 0, find the value of β. Penyelesaian 20x2 – 7x – 6 = 0 a = 20, b = –7, c = –6 Punca-punca/Roots: 3 4 , β 3 4 + β = – b a 3 4 + β = – (–7) 20 β = 7 20 – 3 4 = – 2 5 Contoh 6 Kaedah Alternatif 3 4 (β) = c a 3 4 β = –6 20 β = – 2 5 Atau/Or 20x2 – 7x – 6 = 0 (4x – 3)(5x + 2) = 0 4x – 3 = 0 atau/or 5x + 2 = 0 4x = 3 5x = –2 x = 3 4 x = – 2 5 Punca-punca/Roots: 3 4 , β ∴ β = – 2 5 1 Murid-murid dibahagikan kepada kumpulan yang terdiri daripada empat orang. Students are divided into groups of four. 2 Setiap kumpulan dikehendaki menyediakan satu soalan tentang membentuk persamaan kuadratik daripada puncapunca. Kemudian, tukar soalan dengan kumpulan lain. Each group is required to construct a question of forming quadratic equations from the given roots. Then, trade the problem with other groups. 3 Guru mengadakan perbincangan bersama-sama murid. Teacher holds a discussion with students. Aktiviti PAK-21 Trade a Problem PdPc Aktiviti PAK-21 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 23 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

24 1 Diberi α dan β ialah punca-punca bagi persamaan kuadratik x2 – 7x + 14 = 0, bentukkan persamaan kuadratik dengan punca-punca α 3 dan β 3 . Given that a and β are the roots of the quadratic equation x2 – 7x + 14 = 0, form a quadratic equation with roots α 3 and β 3 . x2 – 7x + 14 = 0 a = 1, b = –7, c = 14 Punca-punca/Roots: α, β α + β = – b a = – (–7) 1 = 7 αβ = c a = 14 1 = 14 Punca-punca baharu/New roots: α 3 , β 3 Hasil tambah punca baharu Sum of new roots α 3 + β 3 = α + β 3 = 7 3 Hasil darab punca baharu Product of new roots 1 α 321β 3 2 = αβ 9 = 14 9 Persamaan kuadratik baharu New quadratic equation x2 – 1 7 3 2 x + 1 14 9 2 = 0 9x2 – 21x + 14 = 0 2 Diberi α dan b ialah punca-punca bagi persamaan kuadratik 3x2 – x – 15 = 0, bentukkan persamaan kuadratik dengan punca-punca 5 – α dan 5 – b. Given that a and β are the roots of the quadratic equation 3x2 – x – 15 = 0, form a quadratic equation with roots 5 – α and 5 – β. 3x2 – x – 15 = 0 a = 3, b = –1, c = –15 Punca-punca/Roots: α, β α + β = – b a = – (–1) 3 = 1 3 αβ = c a = –15 3 = –5 Punca-punca baharu/New roots: 5 – α, 5 – β Hasil tambah punca baharu Sum of new roots (5 – α) + (5 – β) = 10 – α – β = 10 – (α + β) = 10 – 1 1 3 2 = 29 3 Hasil darab punca baharu Product of new roots (5 – α)(5 – β) = 25 – 5β – 5α + αβ = 25 – 5(β + α) + αβ = 25 – 51 1 3 2 + (–5) = 55 3 Persamaan kuadratik baharu New quadratic equation x2 – 1 29 3 2 x + 1 55 3 2 = 0 3x2 – 29x + 55 = 0 3 Diberi α dan –1 ialah punca-punca bagi persamaan kuadratik 2x2 – 6x + k = 0, cari nilai α dan nilai k. Given that α and –1 are the roots of the quadratic equation 2x2 – 6x + k = 0, find the value of α and of k. 2x2 – 6x + k = 0 a = 2, b = –6, c = k Punca-punca/Roots: α, –1 Hasil tambah punca Sum of roots α + (–1) = – b a α – 1 = – (–6) 2 α – 1 = 3 α = 4 Hasil darab punca/Product of roots α(–1) = c a –α = k 2 –(4) = k 2 –8 = k k = –8 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 24 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

25 4 Satu daripada punca bagi persamaan kuadratik 2x2 – mx + m = 0, dengan keadaan m ≠ 0 ialah dua kali punca yang satu lagi. Cari nilai m dan tentukan nilai punca-punca itu. KBAT Mengaplikasi One of the roots of the quadratic equation 2x2 – mx + m = 0, where m ≠ 0 is twice the other root. Find the value of m and determine the values of the roots. 2x2 – mx + m = 0 a = 2, b = –m, c = m Katakan punca-punca ialah α dan 2α. Let the roots be α and 2α. Hasil tambah punca: Sum of roots: α + 2α = – b a 3α = – (–m) 2 α = m 6 1 Hasil darab punca: Product of roots: α(2α) = c a 2α2 = m 2 α2 = m 4 2 Gantikan 1 ke dalam 2 . Substitute 1 into 2 . 1 m 6 2 2 = m 4 m2 36 = m 4 m2 – 9m = 0 m(m – 9) = 0 m = 0 atau/or m = 9 m ≠ 0, maka/thus m = 9 Gantikan m = 9 ke dalam persamaan kuadratik. Substitute m = 9 into the quadratic equation. 2x2 – 9x + 9 = 0 (2x – 3)(x – 3) = 0 2x – 3 = 0 atau/or x – 3 = 0 x = 3 2 x = 3 ∴ Punca-punca ialah 3 2 dan 3. The roots are 3 2 and 3. x –8 1 x2 + 7x > 8 Penyelesaian x2 + 7x > 8 x2 + 7x – 8 > 0 (x – 1)(x + 8) > 0 Punca-punca/Roots: x = –8, 1 ∴ x < –8 atau/or x > 1 Contoh 7 1 x2 < 6 + x x2 < 6 + x x2 – x – 6 < 0 (x + 2)(x – 3) < 0 Punca-punca/Roots: x = –2, 3 x –2 3 ∴ –2 < x < 3 2 x2 + 2x > 15 x2 + 2x > 15 x2 + 2x – 15 > 0 (x + 5)(x – 3) > 0 Punca-punca/Roots: x = –5, 3 x –5 3 ∴ x < –5 atau/or x > 3 3 2x2 – 6x  8 2x2 – 6x  8 2x2 – 6x – 8  0 2(x2 – 3x – 4)  0 2(x + 1)(x – 4)  0 Punca-punca/Roots: x = –1, 4 x –1 4 ∴ –1  x  4 4 –x2 + 2x + 8  0 –x2 + 2x + 8  0 x2 – 2x – 8  0 (x – 4)(x + 2)  0 Punca-punca/Roots: x = –2, 4 x –2 4 ∴ –2  x  4 5 2 > 3x – x2 2 > 3x – x2 x2 – 3x + 2 > 0 (x – 1)(x – 2) > 0 Punca-punca/Roots: x = 1, 2 x 1 2 ∴ x < 1 atau/or x > 2 Latihan 7 Cari julat nilai x bagi setiap ketaksamaan kuadratik yang berikut. TP 2 Find the range of values of x for each of the following quadratic inequalities. TP 2 Mempamerkan kefahaman tentang fungsi kuadratik. Tip Bestari (a) (x – m)(x – n) < 0 atau/or (x – m)(x – n)  0 m < x < n m  x  n x m n (b) (x – m)(x – n) > 0 atau/or (x – m)(x – n)  0 x < m atau/or x > n x  m atau/or x  n x m n 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 25 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

26 3x2 – 2x + 5 = 0 Penyelesaian a = 3, b = –2, c = 5 b2 – 4ac = (–2)2 – 4(3)(5) = 4 – 60 = –56 (< 0) Maka, persamaan kuadratik 3x2 – 2x + 5 = 0 tidak mempunyai punca nyata. So, the quadratic equation 3x2 – 2x + 5 = 0 has no real roots. Contoh 8 1 x2 – 2x – 3 = 0 a = 1, b = –2, c = –3 b2 – 4ac = (–2)2 – 4(1)(–3) = 4 + 12 = 16 (> 0) Maka, persamaan kuadratik x2 – 2x – 3 = 0 mempunyai dua punca nyata dan berbeza. So, the quadratic equation x2 – 2x – 3 = 0 has two real and different roots. 2 4x2 – 4x + 1 = 0 a = 4, b = –4, c = 1 b2 – 4ac = (–4)2 – 4(4)(1) = 16 – 16 = 0 Maka, persamaan kuadratik 4x2 – 4x + 1 = 0 mempunyai dua punca nyata yang sama. So, the quadratic equation 4x2 – 4x + 1 = 0 has two real and equal roots. 3 2x2 + 3x – 5 = 0 a = 2, b = 3, c = –5 b2 – 4ac = (3)2 – 4(2)(–5) = 9 + 40 = 49 (> 0) Maka, persamaan kuadratik 2x2 + 3x – 5 = 0 mempunyai dua punca nyata dan berbeza. So, the quadratic equation 2x2 + 3x – 5 = 0 has two real and different roots. 4 2x2 – x + 6 = 0 a = 2, b = –1, c = 6 b2 – 4ac = (–1)2 – 4(2)(6) = 1 – 48 = –47 (< 0) Maka, persamaan kuadratik 2x2 – x + 6 = 0 tidak mempunyai punca nyata. So, the quadratic equation 2x2 – x + 6 = 0 has no real roots. 5 9x2 – 12x + 4 = 0 a = 9, b = –12, c = 4 b2 – 4ac = (–12)2 – 4(9)(4) = 144 – 144 = 0 Maka, persamaan kuadratik 9x2 – 12x + 4 = 0 mempunyai dua punca nyata yang sama. So, the quadratic equation 9x2 – 12x + 4 = 0 has two real and equal roots. 2.2 Jenis-jenis Punca Persamaan Kuadratik/ Types of Roots of Quadratic Equations Latihan 8 Tentukan jenis punca bagi setiap persamaan kuadratik berikut. TP 2 Determine the types of roots for each of the following quadratic equations. TP 2 Mempamerkan kefahaman tentang fungsi kuadratik. 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 26 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

27 2x2 + 6x + 5 = k Penyelesaian 2x2 + 6x + 5 = k 2x2 + 6x + 5 – k = 0 a = 2, b = 6, c = 5 – k Mempunyai dua punca nyata dan berbeza, Has two real and different roots, b2 – 4ac > 0 (6)2 – 4(2)(5 – k) > 0 36 – 40 + 8k > 0 –4 + 8k > 0 8k > 4 k > 1 2 Contoh 9 1 x2 – 5x + 3 = k x2 – 5x + 3 = k x2 – 5x + 3 – k = 0 a = 1, b = –5, c = 3 – k Mempunyai dua punca nyata dan berbeza, Has two real and different roots, b2 – 4ac > 0 (–5)2 – 4(1)(3 – k) > 0 25 – 12 + 4k > 0 13 + 4k > 0 4k > –13 k > – 13 4 2 3x2 + 2x + k = 5 3x2 + 2x + k = 5 3x2 + 2x + k – 5 = 0 a = 3, b = 2, c = k – 5 Mempunyai dua punca nyata dan berbeza, Has two real and different roots, b2 – 4ac > 0 (2)2 – 4(3)(k – 5) > 0 4 – 12k + 60 > 0 –12k + 64 > 0 –12k > –64 12k < 64 k < 64 12 k < 16 3 3 kx2 + 6x = –1 kx2 + 6x = –1 kx2 + 6x + 1 = 0 a = k, b = 6, c = 1 Mempunyai dua punca nyata dan berbeza, Has two real and different roots, b2 – 4ac > 0 (6)2 – 4(k)(1) > 0 36 – 4k > 0 –4k > –36 4k < 36 k < 9 4 kx2 + 2x = 1 – x2 kx2 + 2x = 1 – x2 kx2 + x2 + 2x – 1 = 0 (k + 1)x2 + 2x – 1 = 0 a = k + 1, b = 2, c = –1 Mempunyai dua punca nyata dan berbeza, Has two real and different roots, b2 – 4ac > 0 (2)2 – 4(k + 1)(–1) > 0 4 + 4k + 4 > 0 8 + 4k > 0 4k > –8 k > –2 5 2x2 + 4x + k = 1 2x2 + 4x + k = 1 2x2 + 4x + k – 1 = 0 a = 2, b = 4, c = k – 1 Mempunyai dua punca nyata dan berbeza, Has two real and different roots, b2 – 4ac > 0 (4)2 – 4(2)(k – 1) > 0 16 – 8k + 8 > 0 24 – 8k > 0 –8k > –24 8k < 24 k < 3 Latihan 9 Cari julat bagi nilai k jika persamaan kuadratik berikut mempunyai dua punca nyata dan berbeza. TP 3 Find the range of values of k if the following quadratic equations have two real and different roots. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 27 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

28 2x2 + 3x + 6 = k Penyelesaian 2x2 + 3x + 6 = k 2x2 + 3x + 6 – k = 0 a = 2, b = 3, c = 6 – k Mempunyai dua punca nyata yang sama, Has two real and equal roots, b2 – 4ac = 0 (3)2 – 4(2)(6 – k)= 0 9 – 48 + 8k = 0 –39 + 8k = 0 8k = 39 k = 39 8 Contoh 10 1 x2 – 6x + 1 = k x2 – 6x + 1 = k x2 – 6x + 1 – k = 0 a = 1, b = –6, c = 1 – k Mempunyai dua punca nyata yang sama, Has two real and equal roots, b2 – 4ac = 0 (–6)2 – 4(1)(1 – k) = 0 36 – 4 + 4k = 0 32 + 4k = 0 4k = –32 k = –8 2 3x2 + 4x + k = 2 3x2 + 4x + k = 2 3x2 + 4x + k – 2 = 0 a = 3, b = 4, c = k – 2 Mempunyai dua punca nyata yang sama, Has two real and equal roots, b2 – 4ac = 0 (4)2 – 4(3)(k – 2) = 0 16 – 12k + 24 = 0 40 – 12k = 0 –12k = –40 k = 10 3 3 kx2 – 2x = 6 kx2 – 2x = 6 kx2 – 2x – 6 = 0 a = k, b = –2, c = –6 Mempunyai dua punca nyata yang sama, Has two real and equal roots, b2 – 4ac = 0 (–2)2 – 4(k)(–6) = 0 4 + 24k = 0 24k = –4 k = – 1 6 4 x2 + kx + 1 = –x x2 + kx + 1 = –x x2 + kx + x + 1 = 0 x2 + (k + 1)x + 1 = 0 a = 1, b = k + 1, c = 1 Mempunyai dua punca nyata yang sama, Has two real and equal roots, b2 – 4ac = 0 (k + 1)2 – 4(1)(1) = 0 (k + 1)2 – 4 = 0 (k + 1 + 2)(k + 1 – 2) = 0 (k + 3)(k – 1) = 0 k = –3 atau/or k = 1 5 x2 + 2x = kx2 + 3 x2 + 2x = kx2 + 3 x2 – kx2 + 2x – 3 = 0 (1 – k)x2 + 2x – 3 = 0 a = 1 – k, b = 2, c = –3 Mempunyai dua punca nyata yang sama, Has two real and equal roots, b2 – 4ac = 0 (2)2 – 4(1 – k)(–3)= 0 4 + 12 – 12k = 0 16 – 12k = 0 12k = 16 k = 4 3 3x2 + 2x + 1 = k Penyelesaian 3x2 + 2x + 1 = k 3x2 + 2x + 1 – k = 0 a = 3, b = 2, c = 1 – k Tidak mempunyai punca nyata, Has no real roots b2 – 4ac < 0 (2)2 – 4(3)(1 – k) < 0 4 – 12 + 12k < 0 –8 + 12k < 0 12k < 8 k < 2 3 Contoh 11 1 x2 – 5x + 3 = k x2 – 5x + 3 = k x2 – 5x + 3 – k = 0 a = 1, b = –5, c = 3 – k Tidak mempunyai punca nyata, Has no real roots, b2 – 4ac < 0 (–5)2 – 4(1)(3 – k) < 0 25 – 12 + 4k < 0 13 + 4k < 0 4k < –13 k < – 13 4 2 2x2 + 4x + k = 1 2x2 + 4x + k = 1 2x2 + 4x + k – 1 = 0 a = 2, b = 4, c = k – 1 Tidak mempunyai punca nyata, Has no real roots, b2 – 4ac < 0 (4)2 – 4(2)(k – 1) < 0 16 – 8k + 8 < 0 –8k < –24 8k > 24 k > 3 Latihan 10 Cari nilai bagi k jika persamaan kuadratik berikut mempunyai dua punca nyata yang sama. Find the value of k if the following quadratic equations have two real and equal roots. TP 3 TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. Latihan 11 Cari julat bagi nilai k jika persamaan kuadratik berikut tidak mempunyai punca nyata. Find the range of values of k if the following quadratic equations have no real roots. TP 3 TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 28 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

29 3 kx2 + 3x = 7 kx2 + 3x = 7 kx2 + 3x – 7 = 0 a = k, b = 3, c = –7 Tidak mempunyai punca nyata, Has no real roots, b2 – 4ac < 0 (3)2 – 4(k)(–7) < 0 9 + 28k < 0 28k < –9 k < – 9 28 4 kx2 + x = x2 – 2 kx2 + x = x2 – 2 kx2 – x2 + x + 2 = 0 (k – 1)x2 + x + 2 = 0 a = k – 1, b = 1, c = 2 Tidak mempunyai punca nyata, Has no real roots, b2 – 4ac < 0 (1)2 – 4(k – 1)(2) < 0 1 – 8k + 8 < 0 –8k < –9 8k > 9 k > 9 8 5 2kx2 + 3x = x2 + 2 2kx2 + 3x = x2 + 2 2kx2 – x2 + 3x – 2 = 0 (2k – 1)x2 + 3x – 2 = 0 a = 2k – 1, b = 3, c = –2 Tidak mempunyai punca nyata, Has no real roots, b2 – 4ac < 0 (3)2 – 4(2k – 1)(–2) < 0 9 + 16k – 8 < 0 1 + 16k < 0 16k < –1 k < – 1 16 2.3 Fungsi Kuadratik/ Quadratic Functions Latihan 12 Selesaikan masalah yang berikut. TP 2 Solve the following problems. TP 2 Mempamarkan kefahaman tentang fungsi kuadratik. 1 Rajah di sebelah menunjukkan graf fungsi f(x) = –4x2 + 2x – 3. Buat generalisasi dan lakar graf f(x) yang baharu jika The diagram on the right shows the graph function f(x) = –4x2 + 2x – 3. Make a generalisation and sketch the new graph of f(x) if (a) nilai a berubah kepada –6, the value of a changes to –6, (b) nilai b berubah kepada –2, the value of b changes to –2, (c) nilai c berubah kepada 3. the value of c changes to 3. x y 0 –3 f(x) = –4x2 + 2x – 3 (a) Nilai a semakin kecil, maka kelebaran graf semakin berkurang. Bentuk graf dan pintasan-y tidak berubah. The value of a is smaller, the width of the graph decreases. The shape of the graph and the y-intercept remains unchanged. x y 0 –3 f(x) = –6x2 + 2x – 3 (b) Nilai b  0, maka verteks berada di sebelah kiri paksi-y. Bentuk graf dan pintasan-y tidak berubah. The value of b  0, thus the vertex is on the left side of the y-axis. The shape of the graph and the y-intercept remains unchanged. x y 0 –3 f(x) = –4x2 – 2x – 3 (c) Graf bergerak 6 unit ke atas. Bentuk graf tidak berubah. The graph moves 6 units upwards. The shape of the graph remains unchanged. x y 0 3 f(x) = –4x2 + 2x + 3 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 29 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

30 f(x) = 3x2 – 2x + k – 5 Penyelesaian a = 3, b = –2, c = k – 5 Jika lengkung menyilang paksi-x pada dua titik yang berbeza, maka f(x) mempunyai dua punca nyata dan berbeza. If the curve intersects the x-axis at two different points, then f(x) has two real and different roots. b2 – 4ac > 0 (–2)2 – 4(3)(k – 5) > 0 4 – 12k + 60 > 0 –12k + 64 > 0 –12k > –64 k < 16 3 Contoh 13 Kesilapan Umum –12k > –64 k > –64 –12 k > 16 3 Apabila mendarab atau membahagi suatu ketaksamaan dengan suatu nombor negatif, songsangkan arah tatatanda ketaksamaan itu. When multiplying or dividing an inequality by a negative number, reverse the direction of the inequality notation. 1 f(x) = x2 + 6x + k a = 1, b = 6, c = k b2 – 4ac > 0 (6)2 – 4(1)(k) > 0 36 – 4k > 0 –4k > –36 4k < 36 k < 9 Latihan 14 Cari julat nilai k jika lengkung bagi fungsi kuadratik berikut menyilang paksi-x pada dua titik yang berbeza. TP 3 Find the range of values of k if the curve of the following quadratic functions intersect the x-axis at two different points. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. Latihan 13 Tentukan jenis punca bagi setiap fungsi kuadratik berikut. Kemudian, lakar graf dan buat generalisasi tentang kedudukan graf pada paksi-x. TP 3 Determine the types of roots for each of the following quadratic functions. Then, sketch the graph and make a generalisation on the position of the graph at the x-axis. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. f(x) = x2 + 2x + 3 Penyelesaian a = 1, b = 2, c = 3 b2 – 4ac = (2)2 – 4(1)(3) = –8 (0) Oleh sebab a  0 dan b2 – 4ac  0, graf itu berbentuk pararabola minimum dan tidak menyilang pada mana-mana titik pada paksi-x. Since a  0 and b2 – 4ac  0, the graph is a parabola minimum and does not intersect at any point on the x-axis. x Contoh 12 1 f(x) = 2x2 + 4x – 6 a = 2, b = 4, c = –6 b2 – 4ac = (4)2 – 4(2)(–6) = 64 (0) Oleh sebab a > 0 dan b2 – 4ac > 0, graf itu berbentuk pararabola minimum dan menyilang paksi-x pada dua titik yang berbeza. Since a  0 and b2 – 4ac  0, the graph is a parabola minimum and intersects the x-axis at two different points. x 2 f(x) = –3x2 + x – 4 a = –3, b = 1, c = –4 b2 – 4ac = (1)2 – 4(–3)(–4) = –47 (0) Oleh sebab a  0 dan b2 – 4ac  0, graf itu berbentuk pararabola maksimum dan tidak menyilang pada mana-mana titik pada paksi-x. Since a  0 and b2 – 4ac  0, the graph is a parabola maximum and does not intersect at any point on the x-axis. x 3 f(x) = x2 – 6x + 9 a = 1, b = –6, c = 9 b2 – 4ac = (–6)2 – 4(1)(9) = 0 Oleh sebab a  0 dan b2 – 4ac = 0, graf itu berbentuk pararabola minimum dan menyentuh paksi-x pada satu titik sahaja. Since a  0 and b2 – 4ac = 0, the graph is a parabola minimum and touches the x-axis at one point only. x 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 30 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

31 2 f(x) = –x2 + 2x – k + 4 a = –1, b = 2, c = –k + 4 b2 – 4ac > 0 (2)2 – 4(–1)(–k + 4) > 0 4 + 4(–k + 4) > 0 4 – 4k + 16 > 0 –4k + 20 > 0 –4k > –20 4k < 20 k < 5 3 f(x) = 4x2 + kx + 4 a = 4, b = k, c = 4 b2 – 4ac > 0 k2 – 4(4)(4) > 0 k2 – 64 > 0 (k + 8)(k – 8) > 0 k –8 8 k < –8 atau/or k > 8 4 f(x) = (k – 3)x2 – 2x – 4 a = k – 3, b = –2, c = –4 b2 – 4ac > 0 (–2)2 – 4(k – 3)(–4)> 0 4 + 16k – 48 > 0 16k – 44 > 0 16k > 44 k > 11 4 f(x) = mx2 + 3x + 6 Penyelesaian a = m, b = 3, c = 6 Jika lengkung menyilang paksi-x pada satu titik sahaja, maka f(x) mempunyai dua punca nyata yang sama. If the curve intersects the x-axis at one point only, then f(x) has two real and equal roots. b2 – 4ac = 0 (3)2 – 4(m)(6) = 0 9 – 24m = 0 24m = 9 m = 3 8 Contoh 14 1 f(x) = –2x2 + x + m a = –2, b = 1, c = m b2 – 4ac = 0 (1)2 – 4(–2)(m) = 0 1 + 8m = 0 8m = –1 m = – 1 8 2 f(x) = 4x2 – 6x – m – 2 a = 4, b = –6, c = –m – 2 b2 – 4ac = 0 (–6)2 – 4(4)(–m – 2) = 0 36 + 16m + 32 = 0 16m = –68 m = – 17 4 3 f(x) = 1 – 8x – mx2 a = –m, b = –8, c = 1 b2 – 4ac = 0 (–8)2 – 4(–m)(1)= 0 64 + 4m = 0 4m = –64 m = –16 4 f(x) = x2 – mx + m + 8 a = 1, b = –m, c = m + 8 b2 – 4ac = 0 (–m) 2 – 4(1)(m + 8) = 0 m2 – 4m – 32 = 0 (m + 4)(m – 8) = 0 m + 4 = 0 atau/or m – 8= 0 m = –4 m = 8 Latihan 15 Cari nilai (atau nilai-nilai) m jika lengkung bagi fungsi kuadratik berikut menyentuh paksi-x pada satu titik sahaja. TP 3 Find the value (or values) of m if the curve of the following quadratic functions touch the x-axis at one point only. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. f(x) = 2x2 – 6x + 1 – n Penyelesaian a = 2, b = –6, c = 1 – n Jika lengkung itu tidak menyilang paksi-x, maka f(x) tidak mempunyai punca nyata. If the curve does not intersect the x-axis, then f(x) does not have real roots. b2 – 4ac < 0 (–6)2 – 4(2)(1 – n) < 0 36 – 8 + 8n < 0 8n < –28 n < – 7 2 Contoh 15 1 f(x) = –2x2 – 4x – n + 1 a = –2, b = –4, c = –n + 1 b2 – 4ac < 0 (–4)2 – 4(–2)(–n + 1) < 0 16 – 8n + 8 < 0 –8n < –24 8n > 24 n > 3 Latihan 16 Cari julat nilai n jika lengkung bagi fungsi kuadratik berikut tidak menyilang paksi-x. TP 3 Find the range of values of n if the curve of the following quadratic functions do not intersect the x-axis. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 31 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

32 2 f(x) = nx2 – 2x – 8 a = n, b = –2, c = –8 b2 – 4ac < 0 (–2)2 – 4(n)(–8) < 0 4 + 32n < 0 32n < –4 n < – 1 8 3 f(x) = nx2 – 12x + 6 a = n, b = –12, c = 6 b2 – 4ac < 0 (–12)2 – 4(n)(6) < 0 144 – 24n < 0 –24n < –144 24n > 144 n > 6 4 f(x) = 6 – 2x + (n – 1)x2 a = n – 1, b = –2, c = 6 b2 – 4ac < 0 (–2)2 – 4(n – 1)(6) < 0 4 – 24n + 24 < 0 –24n < –28 24n > 28 n > 7 6 f(x) = 2x2 – 7x – 4 Penyelesaian f(x) = 2x2 – 7x – 4 = 21x2 – 7 2 x – 22 = 2x2 – 7 2 x + 1– 7 4 2 2 – 1– 7 4 2 2 – 24 = 21x – 7 4 2 2 – 49 16 – 24 = 21x – 7 4 2 2 – 81 164 = 21x – 7 4 2 2 – 81 8 a > 0, f(x) mempunyai titik minimum. a > 0, f(x) has a minimum point. Verteks minimum = 1 7 4 , – 81 8 2 Minimum vertex Contoh 16 1 f(x) = x2 – 6x – 8 f(x) = x2 – 6x – 8 = x2 – 6x + 1– 6 2 2 2 – 1– 6 2 2 2 – 8 = (x – 3)2 – 9 – 8 = (x – 3)2 – 17 a > 0, f(x) mempunyai titik minimum./f(x) has a minimum point. Verteks minimum = (3 , –17) Minimum vertex 2 f(x) = 3x2 – 12x + 15 f(x) = 3x2 – 12x + 15 = 3(x2 – 4x + 5) = 33x2 – 4x + 1– 4 2 2 2 – 1– 4 2 2 2 + 54 = 3[(x – 2)2 – 4 + 5] = 3[(x – 2)2 + 1] = 3(x – 2)2 + 3 a > 0, f(x) mempunyai titik minimum./f(x) has a minimum point. Verteks minimum = (2 , 3) Minimum vertex 3 f(x) = –2(x + 3)(x – 1) f(x) = –2(x + 3)(x – 1) = –2(x2 + 2x – 3) = –2 x2 + 2x + 1 2 2 2 2 – 1 2 2 2 2 – 34 = –2[(x + 1)2 – 4] = –2(x + 1)2 + 8 a < 0, f(x) mempunyai titik maksimum./f(x) has a maximum point. Verteks maksimum/Maximum vertex = (–1, 8) 4 f(x) = (3x + 6)(x + 1) f(x) = (3x + 6)(x + 1) = 3x2 + 9x + 6 = 3(x2 + 3x + 2) = 3 x2 + 3x + 1 3 2 2 2 – 1 3 2 2 2 + 24 = 31 x + 3 2 2 2 – 1 4 4 = 31 x + 3 2 2 2 – 3 4 a > 0, f(x) mempunyai titik minimum./f(x) has a minimum point. Verteks minimum/Minimum vertex = 1 – 3 2 , – 3 4 2 Latihan 18 Buat generalisasi terhadap bentuk dan kedudukan graf f(x) = a(x + h) 2 + k apabila nilai a, h atau k berubah. TP 3 Make a generalisation about the shape and the position of the graph f(x) = a(x + h)2 + k when the value of a, h or k is changed. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. Rajah di sebelah menunjukkan graf f(x) = (x + 1)2 + 2, dengan keadaan a = 1, h = –1 dan k = 2. The diagram on the right shows the graph of f(x) = (x + 1) 2 + 2, where a = 1, h = –1 and k = 2. Latihan 17 Ungkapkan fungsi kuadratik berikut dalam bentuk verteks, f(x) = a(x – h) 2 + k. Seterusnya, nyatakan koordinat verteks bagi fungsi itu. TP 3 Express the following quadratic functions in the vertex form, f(x) = a(x – h)2 + k. Hence, state the coordinates of the vertex of the function. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. x 3 (–1, 2) 0 f(x) 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 32 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

33 1 f(x) = (x – 1)2 + 2 Apabila nilai h berubah daripada –1 kepada 1, graf dengan bentuk yang sama bergerak 2 unit ke kanan. Paksi simetri ialah x = 1 dan nilai minimum tidak berubah, iaitu 2. When the value of h changes from –1 to 1, the graph with the same shape moves 2 units to the right. The axis of symmetry is x = 1 and the minimum value remains unchanged, that is 2. Apabila/When x = 0, f(x) = (0 – 1)2 + 2 = 3 x 3 (1, 2) 0 f(x) 2 1 2 f(x) = (x + 1)2 – 2 Apabila nilai k berubah daripada 2 kepada –2, graf dengan bentuk yang sama bergerak 4 unit ke bawah. Nilai minimum menjadi –2 dan paksi simetri tidak berubah, iaitu x = –1. When the value of k changes from 2 to –2, the graph with the same shape moves 4 units downward. The minimum value becomes –2 and the axis of symmetry remains unchanged, that is x = –1. Apabila/When x = 0, f(x) = (0 + 1)2 – 2 = –1 x –1 (–1, –2) –1 0 f(x) –2 Latihan 19 Lakarkan graf bagi setiap fungsi kuadratik berikut. TP 3 KBAT Menganalisis Sketch the graph for each of the following quadratic functions. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. x –2 12 6 f(x) 0 (2, 16) 16 2 x 2 (–1, 2) –1 0 f(x) 11 f(x) = 9(x + 1)2 + 2 Penyelesaian Apabila nilai a berubah daripada 1 kepada 9, kelebaran graf berkurang. Paksi simetri dan nilai minimum graf tidak berubah. Apabila x = 0, f(x) = 9(0 + 1)2 + 2 = 11 When the value of a changes from 1 to 9, the width of the graph decreases. The axis of symmetry and the minimum value of the graph remain unchanged. When x = 0, f(x) = 9(0 + 1)2 + 2 = 11 Contoh 17 f(x) = –x2 + 4x + 12 Penyelesaian a = –1 < 0, graf berbentuk ∩ a = –1 < 0, graph has a shape of ∩ f(x) = –x2 + 4x + 12 = –(x2 – 4x – 12) = –x2 – 4x + 1– 4 2 2 2 – 1– 4 2 2 2 – 124 = –[(x – 2)2 – 16] = –(x – 2)2 + 16 Titik maksimum = (2, 16) Maximum point = (2, 16) Untuk mencari pintasan-x, gantikan f(x) = 0. To find the x-intercept, substitute f(x) = 0. –x2 + 4x + 12 = 0 x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0 x = –2 atau/or x = 6 Untuk mencari pintasan-f(x), gantikan x = 0. To find the f(x)-intercept, substitute x = 0. f(0) = –(0)2 + 4(0) + 12 = 12 Contoh 18 Kaedah Alternatif Titik maksimum juga boleh diperoleh dengan mencari imej bagi 2. The maximum point can also be obtained by finding the image of 2. f(2) = –(2)2 + 4(2) + 12 = 16 ∴ Titik maksimum/Maximum point = (2, 16) 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 33 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

34 1 f(x) = –x2 + 6x + 16 a = –1 < 0, graf berbentuk ∩ a = –1 < 0, graph has a shape of ∩ f(x) = –x2 + 6x + 16 = –(x2 – 6x – 16) = –[x2 – 6x + 1– 6 2 2 2 – 1– 6 2 2 2 – 164 = –[x2 – 6x + (–3)2 – (–3)2 – 16] = –[(x – 3)2 – 25] = –(x – 3)2 + 25 Titik maksimum/Maximum point = (3, 25) Apabila/When f(x) = 0, –x2 + 6x + 16 = 0 (–x – 2)(x – 8) = 0 –x – 2 = 0 atau/or x – 8 = 0 x = –2 x = 8 Apabila/When x = 0, f(x) = –(0)2 + 6(0) + 16 = 16 x –2 16 8 (3, 25) f(x) 25 0 3 2 f(x) = x2 – 4x – 5 a = 1 > 0, graf berbentuk ∪ a = 1 > 0, graph has a shape of ∪ f(x) = x2 – 4x – 5 = x2 – 4x + 1– 4 2 2 2 – 1– 4 2 2 2 – 5 = x2 – 4x + (–2)2 – (–2)2 – 5 = (x – 2)2 – 9 Titik minimum/Minimum point = (2, –9) Apabila/When f(x) = 0, x2 – 4x – 5 = 0 (x + 1)(x – 5) = 0 x + 1 = 0 atau/or x – 5 = 0 x = –1 x = 5 Apabila/When x = 0, f(x) = (0)2 – 4(0) – 5 = –5 x –1 5 f(x) (2, –9) –5 2 –9 0 3 f(x) = 2x2 – 4x – 16 a = 2 > 0, graf berbentuk ∪ a = 2 > 0, graph has a shape of ∪ f(x) = 2x2 – 4x – 16 = 2(x2 – 2x – 8) = 2x2 – 2x + 1– 2 2 2 2 – 1– 2 2 2 2 – 84 = 2[x2 – 2x + (–1)2 – (–1)2 – 8] = 2[(x – 1)2 – 9] = 2(x – 1)2 – 18 Titik minimum/Minimum point = (1, –18) Apabila/When f(x) = 0, 2x2 – 4x – 16 = 0 2(x2 – 2x – 8) = 0 2(x + 2)(x – 4) = 0 x + 2 = 0 atau/or x – 4 = 0 x = –2 x = 4 Apabila/When x = 0, f(x) = 2(0)2 + 4(0) – 16 = –16 x –2 4 f(x) (1, –18) –16 0 1 –18 4 f(x) = –3x2 + 12x + 15 a = –3 < 0, graf berbentuk ∩ a = –3 < 0, graph has a shape of ∩ f(x) = –3x2 + 12x + 15 = –3(x2 – 4x – 5) = –3x2 – 4x + 1– 4 2 2 2 – 1– 4 2 2 2 – 54 = –3[x2 – 4x + (–2)2 – (–2)2 – 5] = –3[(x – 2)2 – 9] = –3(x – 2)2 + 27 Titik maksimum/Maximum point = (2, 27) Apabila/When f(x) = 0, –3x2 + 12x + 15 = 0 –3(x2 – 4x – 5) = 0 –3(x + 1)(x – 5) = 0 x + 1 = 0 atau/or x – 5 = 0 x = –1 x = 5 Apabila/When x = 0, f(x) = –3(0)2 + 12(0) + 15 = 15 x –1 0 15 5 (2, 27) f(x) 27 2 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 34 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

35 1 Bilangan pelajar yang tamat pengajian di sebuah kolej diberi oleh fungsi f(x) = x2 – 6x + 84. Pada tahun ke berapakah bilangan pelajar yang tamat pengajian sekurang-kurangnya 100 orang? The number of students who graduated from a college is given by the function f(x) = x2 – 6x + 84. In which year at least 100 students graduated from the college? f(x)  100 x2 – 6x + 84  100 x2 – 6x – 16  0 (x + 2)(x – 8)  0 2 Sebiji bola dilambung ke atas dari suatu titik. Ketinggian, h dalam meter, bola itu pada masa t saat diberi oleh fungsi h(t) = –6t2 + 24t + 30. A ball is thrown vertically upwards from a point. The height, h in metre, of the ball at time t seconds is given by the function h(t) = –6t2 + 24t + 30. (a) Berapakah ketinggian bola itu apabila t = 1? What is the height of the ball when t = 1? (b) Hitung masa, dalam saat, apabila bola itu mencapai ketinggian maksimum. Calculate the time, in seconds, when the ball reach the maximum height. (c) Cari ketinggian maksimum, dalam m, bola itu. Find the maximum height, in m, of the ball. (a) Apabila/When t = 1, h(t) = –6t 2 + 24t + 30 = –6(1)2 + 24(1) + 30 = –6 + 24 + 30 = 48 m (b) h(t) = –6t 2 + 24t + 30 = –6(t 2 – 4t – 5) = –6[t 2 – 4t + (–2)2 – (–2)2 – 5] = –6[(t – 2)2 – 9] = –6(t – 2)2 + 54 Bola itu mencapai ketinggian maksimum apabila t = 2. The ball reaches the maximum height when t = 2. x –2 8 Daripada graf, x < –2 atau x  8. Maka, sekurang-kurangnya 100 orang pelajar tamat pengajian pada tahun ke-8. From the graph, x < –2 or x  8. So, at least 100 students graduated from the college in the 8th year. (c) Titik maksimum/Maximum point = (2, 54) ∴ Ketinggian maksimum ialah 54 m. The maximum height is 54 m. Latihan 20 Selesaikan. TP 4 TP 5 Solve. TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah rutin yang mudah. TP 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi kuadratik dalam konteks penyelesaian masalah rutin yang kompleks. Suatu objek dilambung ke atas dari suatu kedudukan. Ketinggian, h dalam meter, objek itu pada masa t saat diberi oleh fungsi h(t) = –4t2 + 8t + 12. An object is thrown vertically upwards from a position. The height, h in metre, of the object at time t seconds is given by the function h(t) = –4t2 + 8t + 12. (a) Cari ketinggian objek itu apabila t = 1 2 . Find the height of the object when t = 1 2 . (b) Hitung masa, dalam saat, apabila objek itu mencapai ketinggian maksimum. Calculate the time, in seconds, when the object reach the maximum height. (c) Cari ketinggian maksimum, dalam m, objek itu. Find the maximum height, in m, of the object. Penyelesaian (a) Apabila/When t = 1 2 , h(t) = –41 1 2 2 2 + 81 1 2 2 + 12 = –1 + 4 + 12 = 15 m (b) h(t) = –4t 2 + 8t + 12 = –4(t 2 – 2t – 3) = –4[t 2 – 2t + (–1)2 – (–1)2 – 3] = –4[(t – 1)2 – 4] = –4(t – 1)2 + 16 Objek itu mencapai ketinggian maksimum apabila t = 1. The object reaches the maximum height when t = 1. (c) Titik maksimum/Maximum point = (1, 16) ∴ Ketinggian maksimum ialah 16 m. The maximum height is 16 m. Contoh 19 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 35 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

36 3 Rajah di sebelah menunjukkan sebuah pintu gerbang berbentuk parabola. Ketinggian, dalam m, pintu gerbang itu diwakili oleh fungsi f(x) = – 1 50 x2 + 8. Jarak di antara dua hujung lengkung, PQ ialah 70 m dan tinggi PQ daripada permukaan tanah mengufuk, EF ialah 12 m. Cari ketinggian maksimum, dalam m, pintu gerbang itu dari permukaan tanah mengufuk. The diagram on the right shows a parabolic gateway. The height, in m, of the gateway is represented by the function f(x) = – 1 50 x2 + 8. The distance between the two ends of the curve, PQ is 70 m and the height of PQ from the horizontal ground, EF is 12 m. Find the maximum height, in m, of the gateway from the horizontal ground. T E P F Q Apabila/When x = 0, f(0) = – 1 50(0)2 + 8 = 8 Apabila/When x = 70 2 = 35, f(35) = – 1 50(35)2 + 8 = –16.5 T E P F Q x y 70 m 8 m 16.5 m 12 m 0 (35, –16.5) Ketinggian maksimum/The maximum height = 8 m + 16.5 m + 12 m = 36.5 m 4 Dalam satu eksperimen, suatu objek dilancarkan pada suatu sudut tirus daripada permukaan tanah. Ketinggian, dalam m, objek itu diwakili oleh graf y = – 1 36 x2 + 2x, dengan keadaan x ialah jarak mengufuk dari titik pelancaran. In an experiment, an object is launched at an acute angle from the ground. The height, in m, of the object is represented by the graph y = – 1 36 x2 + 2x, where x is the horizontal distance from the point of the launching. (a) Berapakah jarak mengufuk, dalam m, objek itu dari titik pelancaran apabila ia mencapai ketinggian maksimum? What is the horizontal distance, in m, of the object from the point of launching when it reaches the maximum height? (b) Cari ketinggian maksimum, dalam m, objek itu. Find the maximum height, in m, of the object. (c) Hitung jarak mengufuk, dalam m, apabila objek itu menyentuh permukaan tanah semula. Calculate the horizontal distance, in m, when the object touches the ground again. (a) y = – 1 36 x2 + 2x = – 1 36 (x2 – 72x) = – 1 36 3x2 – 72x + 1– 72 2 2 2 – 1– 72 2 2 2 4 = – 1 36 3x2 – 72x + (–36)2 – 362 4 = – 1 36 [(x – 36)2 – 362 ] = – 1 36 (x – 36)2 + 36 Titik maksimum/Maximum point = (36, 36) Apabila objek itu mencapai ketinggian maksimum, jarak mengufuk ialah 36 m. When the object reaches the maximum height, the horizontal distance is 36 m. (b) 36 m (c) Apabila/When y = 0 – 1 36 x2 + 2x = 0 x1– 1 36 x + 22 = 0 x = 0 atau/or – 1 36 x + 2 = 0 – 1 36 x = –2 x = 72 Maka, jarak mengufuk ialah 72 m. Thus, the horizontal distance is 72 m. 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 36 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

37 1 Ungkapkan 2x(x − 3) = (2 − x)(x + 3) dalam bentuk am. Seterusnya, selesaikan persamaan kuadratik itu. Beri jawapan betul kepada tiga tempat perpuluhan. Express 2x(x − 3) = (2 − x)(x + 3) in general form. Hence, solve the quadratic equation. Give the answers correct to three decimal places. [4 markah/marks] 2 (a) Diberi 2 ialah salah satu punca bagi persamaan kuadratik 3(x + p)² = 42 − p, dengan keadaan p ialah pemalar. Cari nilai-nilai p. Given that 2 is one of the roots of the quadratic equation 3(x + p)² = 42 − p, where p is a constant. Find the values of p. [3 markah/marks] (b) Jika α dan β ialah punca-punca bagi persamaan kuadratik 2x² − 6x − 7 = 0, bentukkan persamaan kuadratik dengan punca-punca 1 α dan 1 β . If α and β are the roots of the quadratic equation 2x² − 6x − 7 = 0, form a quadratic equation with the roots 1 α and 1 β . [3 markah/marks] 3 (a) Diberi bahawa persamaan kuadratik mx² + 6x + n = 1 mempunyai dua punca nyata yang sama, dengan keadaan m dan n ialah pemalar. Ungkapkan m dalam sebutan n. It is given that the quadratic equation mx² + 6x + n = 1 has two real and equal roots, such that m and n are constants. Express m in terms of n. [2 markah/marks] (b) Persamaan kuadratik 3x² + 2x + 5 = p mempunyai dua punca nyata dan berbeza. Cari julat nilai p. The quadratic equation 3x² + 2x + 5 = p has two real and different roots. Find the range of values of p. [3 markah/marks] 6 (a) Selesaikan |2x – 6| = 8. Seterusnya, lakar graf bagi f(x) = |2x – 6|untuk 0  f(x)  8. Solve |2x – 6| = 8. Hence, sketch the graph of f(x) = |2x – 6| for 0  f(x)  8. [4 markah/marks] (b) Daud menanam pokok buah-buahan pada tanah berbentuk segi empat tepat. Dia merancang untuk memasang pagar di sekeliling tanah yang berdimensi 8p meter × (60 – p) meter. Tunjukkan bahawa luas tanah itu, A = a(p + m)2 + n, dengan keadaan a, m 4 Diberi graf bagi fungsi kuadratik g(x) = q – 2(x + p)², dengan keadaan p dan q ialah pemalar, mempunyai titik pusingan maksimum (1, 3p + 8). Given that the graph of a quadratic function g(x) = q – 2(x + p)², such that p and q are constants, has a maximum turning point (1, 3p + 8). (a) Nyatakan nilai p dan q. State the value of p and of q. (b) Nyatakan jenis punca bagi g(x) = 0. Beri alasan anda. State the type of roots for g(x) = 0. Give your reason. [4 markah/marks] 5 Rajah 1 menunjukkan graf bagi fungsi kuadratik f(x) = px + q – 1 xn , dengan keadaan p, q dan n ialah pemalar. Diagram 1 shows the graph of a quadratic function f(x) = px + q – 1 xn, such that p, q and n are constants. x O β f(x) = px + q – 1 xn f(x) β + 4 Rajah 1/Diagram 1 (a) Nyatakan nilai n. State the value of n. (b) Jika β dan β + 4 ialah punca-punca bagi f(x) dan hasil darab punca itu ialah 12, cari nilai p dan nilai q. If β and β + 4 are the roots of f(x) and the product of roots is 12, find the value of p and of q. [4 markah/marks] dan n ialah pemalar. Seterusnya, cari jumlah panjang, dalam m, pagar yang Daud perlu beli apabila luas tanah itu adalah maksimum. Daud has planted some fruit trees on a rectangular land. He plans to fence up the land which has a dimension of 8p metres × (60 – p) metres. Show that the area of the land, A = a(p + m)2 + n, where a, m and n are constants. Hence, find the total length, in m, of the fence that Daud has to buy when the area of the land is maximum. [4 markah/marks] Praktis Berformat SPM Kertas 1 Bahagian A Bahagian B 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 37 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

38 1 (a) Persamaan kuadratik x2 – 8x + 12 = 0 mempunyai punca-punca p dan q, dengan keadaan p > q. Cari The quadratic equation x2 – 8x + 12 = 0 has roots p and q, where p > q. Find (i) nilai p dan nilai q, the value of p and of q, (ii) julat nilai x jika x2 – 8x + 12 > 0. the range of values of x if x2 – 8x + 12 > 0. [4 markah/marks] (b) Menggunakan nilai p dan q daripada (a)(i), bentukkan persamaan kuadratik dengan punca-punca p + 3 dan 2q + 3. Using the values of p and q from (a)(i), form a quadratic equation with roots p + 3 and 2q + 3. [3 markah/marks] 2 (a) Jika a dan b ialah punca-punca bagi persamaan kuadratik 2x2 + 5x – 1 = 0, bentukkan persamaan kuadratik dengan punca-punca 7a − 2 dan 7b − 2. If a and b are the roots of the quadratic equation 2x2 + 5x – 1 = 0, form a quadratic equation with roots 7a − 2 and 7b − 2. [4 markah/marks] (b) Cari julat nilai x dengan keadaan (3x + 2)(x – 1) > (x + 2)(x + 3). Find the range of values of x such that (3x + 2)(x – 1) > (x + 2)(x + 3). [3 markah/marks] 3 Lengkung bagi fungsi kuadratik f(x) = 3x2 – 12x – 36 bersilang dengan paksi-x pada titik P dan titik Q. Diberi bahawa f(x) boleh ditulis sebagai f(x) = a(x + h)2 + k, dengan keadaan a, h dan k ialah pemalar. The curve of a quadratic function f(x) = 3x2 – 12x – 36 intersects the x-axis at point P and point Q. It is given that f(x) can be written as f(x) = a(x + h)2 + k, where a, h and k are constants. (a) Cari/Find (i) koordinat titik P dan titik Q, the coordinates of point P and point Q, (ii) nilai-nilai a, h dan k, the values of a, h and k, (iii) koordinat titik pusingan. the coordinates of turning point. [6 markah/marks] (b) (i) Lakar graf bagi f(x) = a(x + h) 2 + k untuk –2 < x < 6. Sketch the graph of f(x) = a(x + h)2 + k for –2 < x < 6. (ii) Diberi r(x) = a1 (x + h1 )2 + k1 mewakili pantulan bagi f(x), nyatakan nilai-nilai a1 , h1 dan k1 . Given r(x) = a1 (x + h1 )2 + k1 represents the reflection of f(x), state the values of a1 , h1 and k1 . [4 markah/marks] Kertas 2 Bahagian A Bahagian B 1 Azmi memotong bentuk segi empat sama bersisi 5 cm daripada setiap bucu sekeping kadbod yang berbentuk segi empat sama seperti ditunjukkan dalam Rajah 1. Dia kemudian melipat kadbod tersebut menjadi sebuah kotak seperti ditunjukkan dalam Rajah 2. Azmi cuts the square shapes with sides measuring 5 cm from each corner of a square cardboard as shown in Diagram 1. He then folds the cardboard to become a box as shown in Diagram 2. 5 cm 5 cm 5 cm Rajah 1/ Diagram 1 Rajah 2/ Diagram 2 (a) Jika isi padu kotak itu ialah 1 280 cm3 , bentukkan satu persamaan kuadratik untuk mewakili isi padu tersebut. If the volume of the box is 1 280 cm3 , form a quadratic equation to represent the volume. KBAT Mencipta (b) Seterusnya, cari panjang asal, dalam cm, sisi kadbod itu. KBAT Mengaplikasi Hence, find the original length, in cm, of the sides of the cardboard. Zon KBAT 02 Strategi A+ SPM Mate Tam Tg4_C2_18-38_Final.indd 38 10/17/23 11:38 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

39 3.1 Sistem Persamaan Linear dalam Tiga Pemboleh Ubah Systems of Linear Equations in Three Variables 1 Sistem persamaan linear terdiri daripada dua atau lebih persamaan linear yang melibatkan set pemboleh ubah yang sama. A system of linear equations consists of two or more linear equations that contain the same set of variables. 2 Bentuk am bagi suatu sistem persamaan linear dalam tiga pemboleh ubah ialah ax + by + cz = d, dengan keadaan a, b dan c ≠ 0. The general form of a linear equation in three variables is ax + by + cz = d, where a, b and c ≠ 0. 3 Terdapat dua kaedah yang boleh digunakan untuk menyelesaikan sistem persamaan linear dalam tiga pemboleh ubah: There are two methods that can be used to solve systems of linear equations in three variables: (a) Kaedah penghapusan/ Elimination method (b) Kaedah penggantian/ Substitution method 4 Sistem persamaan linear dalam tiga pemboleh ubah digunakan untuk menyelesaikan masalah dalam kehidupan seharian. Masalah yang diberi diungkapkan sebagai satu sistem persamaan linear dan kemudian diselesaikan untuk menentukan nilai bagi setiap pemboleh ubah. Kadang-kadang, sistem persamaan itu terdiri daripada tiga persamaan linear tetapi tidak semestinya setiap persamaan linear melibatkan tiga pemboleh ubah. Systems of linear equations in three variables are used to solve problems in daily life. The given problem is expressed as a system of linear equations and then solved to determine the value of each variable. Sometimes, the systems of equations consists of three linear equations but not every linear equation involves three variables. 3.2 Persamaan Serentak yang Melibatkan Satu Persamaan Linear dan Satu Persamaan Tak Linear Simultaneous Equations Involving One Linear Equation and One Non-Linear Equation 1 Langkah-langkah untuk menyelesaikan persamaan serentak: The steps to solve simultaneous equations: Peta Alir i-THINK Selesaikan persamaan kuadratik itu dengan menggunakan pemfaktoran atau rumus x = –b ± b2 – 4ac 2a . Solve the quadratic equation by using factorisation or formula x = –b ± b2 – 4ac 2a . Gantikan penyelesaian satu demi satu ke dalam persamaan linear untuk mendapatkan nilai-nilai pemboleh ubah. Substitute the solution one by one into the linear equation to obtain the value of the variables. Jadikan salah satu pemboleh ubah sebagai subjek persamaan. Make one of the variables as the subject of the equation. Kenal pasti persamaan linear. Identify the linear equation. Gantikan pemboleh ubah itu ke dalam persamaan tak linear, menjadikan satu persamaan kuadratik dengan satu pemboleh ubah. Substitute the variable into the non-linear equation, giving a quadratic equation in one variable. Bab 3 Sistem Persamaan Systems of Equations Bidang Pembelajaran: Algebra Revisi Pantas 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 39 10/17/23 11:49 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

40 Sistem Persamaan Linear dalam Tiga Pemboleh Ubah Systems of Linear Equations in Three Variables 3.1 Latihan 1 Bentukkan tiga persamaan linear dalam tiga pemboleh ubah bagi setiap situasi yang berikut. Form three linear equations in three variables for each of the following situations. TP 1 TP 1 Mempamerkan pengetahuan asas tentang sistem persamaan. Praktis PBD Suatu konsert menawarkan tiga jenis tiket. Harga tiket bagi seorang dewasa, seorang warga emas dan seorang kanak-kanak masing-masing ialah RM50, RM25 dan RM10. Penganjur itu berjaya menjual 240 keping tiket dengan harga RM8 100. Bilangan tiket warga emas yang dijual ialah dua kali bilangan tiket dewasa. A concert offers three types of tickets. The ticket prices for an adult, a senior citizen and a child are RM50, RM25 and RM10 respectively. The organiser managed to sell 240 tickets with a total value of RM8 100. The number of senior citizen tickets sold is twice the number of adult tickets. Penyelesaian Katakan/Let x = Bilangan tiket dewasa yang dijual Number of adult tickets sold y = Bilangan tiket warga emas yang dijual Number of senior citizen tickets sold z = Bilangan tiket kanak-kanak yang dijual Number of children tickets sold Maka/Hence x + y + z = 240 y = 2x 50x + 25y + 10z = 8 100 Contoh 1 1 Esah, Fatimah dan Jamilah masing-masing membawa sejumlah wang ke sekolah. Jumlah wang mereka ialah RM26. Jumlah wang Esah ialah RM4 kurang daripada jumlah wang Fatimah. Jumlah wang Jamilah ialah tiga kali jumlah wang Fatimah. Esah, Fatimah and Jamilah each brought a certain amount of money to school. The total money is RM26. Esah’s total money is RM4 less than Fatimah’s total money. Jamilah’s total money is three times Fatimah’s total money. Katakan/Let x = Jumlah wang Esah, dalam RM Esah’s total money, in RM y = Jumlah wang Fatimah, dalam RM Fatimah’s total money, in RM z = Jumlah wang Jamilah, dalam RM Jamilah’s total money, in RM Maka/Hence x + y + z = 26 y – x = 4 z = 3y 2 Rajah di bawah menunjukkan tiga pakej jualan yang ditawarkan oleh sebuah kedai. The diagram below shows three sales packages offered by a shop. RM18 RM20 RM17.50 Katakan/Let x = Harga sebiji donat, dalam RM The price of a doughnut, in RM y = Harga sepeket kentang jejari, dalam RM The price of a packet of french fries, in RM z = Harga sekotak susu, dalam RM The price of a carton of milk, in RM Maka/Hence x + 2y + z = 18 x + y + 2z = 20 2x + y + z = 17.5 3 Jumlah umur Aizan, Kamil dan Hamdan ialah 95 tahun. Umur Kamil ialah dua kali umur Hamdan. Umur Aizan ialah 7 tahun lebih daripada jumlah umur Kamil dan Hamdan. The total age of Aizan, Kamil and Hamdan is 95 years old. Kamil’s age is twice Hamdan’s age. Aizan’s age is 7 years more than the total of Kamil’s and Hamdan’s age. Katakan/Let x = Umur Aizan/Aizan’s age y = Umur Kamil/Kamil’s age z = Umur Hamdan/Hamdan’s age Maka/Hence x + y + z = 95 y = 2z x – (y + z) = 7 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 40 10/17/23 11:49 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

41 Latihan 2 Tentukan sama ada persamaan-persamaan berikut ialah sistem persamaan linear dalam tiga pemboleh ubah atau bukan. Beri justifikasi anda. TP 2 Determine whether the following equations are systems of linear equations in three variables or not. Give your justification. TP 2 Mempamerkan kefahaman tentang penyelesaian sistem persamaan. 2x + 3y + 5z = 6 x − 2y + 7z = −5 3y − 4z + 8x = 12 Penyelesaian Ya, kerana ketiga-tiga persamaan mempunyai tiga pemboleh ubah dengan kuasa pemboleh ubah bernilai 1. Yes, because all three equations have three variables with the power of 1. Contoh 2 1 4p − 7q = 15r q + 9p − 4r = 12 6r + 2p + q = 6 Ya, kerana ketiga-tiga persamaan mempunyai tiga pemboleh ubah dengan kuasa pemboleh ubah bernilai 1. Yes, because all three equations have three variables with the power of 1. 2 3m + 5n + p = −8 m2 + 2p + n = 0 4m + 5n + 6p = 10 Bukan, kerana persamaan kedua mempunyai kuasa pemboleh ubah bernilai 2. No, because the second equation has the variable with the power of 2. 3 e + 3(f + 2g) = 6 2e + 4f = 9g − 8 f 2 + 3e + g = 7 Ya, kerana ketiga-tiga persamaan mempunyai tiga pemboleh ubah dengan kuasa pemboleh ubah bernilai 1. Yes, because all three equations have three variables with the power of 1. Latihan 3 Selesaikan setiap yang berikut. TP 3 Solve each of the following. TP 3 Mengaplikasikan kefahaman tentang sistem persamaan untuk melaksanakan tugasan mudah. 2x + y + 3z = 5 x + y – z = 6 5x – 2y – z = 12 Penyelesaian 2x + y + 3z = 5 ——— 1 x + y – z = 6 ——— 2 5x – 2y – z = 12 ——— 3 Hapuskan z daripada 1 dan 2 , Eliminate z from 1 and 2 , 1 : 2x + y + 3z = 5 3 × 2 : 3x + 3y – 3z = 18 (+) 5x + 4y = 23 ——— 4 Hapuskan z daripada 2 dan 3 , Eliminate z from 2 and 3 , 2 : x + y – z = 6 3 : 5x – 2y – z = 12 (–) –4x + 3y = –6 ——— 5 Selesaikan 4 dan 5 untuk mencari nilai x dan y, Solve 4 and 5 to find the values of x and y, 3 × 4 : 15x + 12y = 69 4 × 5 : –16x + 12y = –24 (–) 31x = 93 x = 3 Daripada/From 4 , 5(3) + 4y = 23 4y = 8 y = 2 Daripada/From 1 , 2(3) + (2) + 3z = 5 3z = –3 z = –1 Maka/Hence, x = 3, y = 2, z = –1 Contoh 3 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 41 10/17/23 11:49 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

42 1 x + y + 2z = 5 x – 3y + 4z = 14 3x + y + 2z = 17 x + y + 2z = 5 ——— 1 x – 3y + 4z = 14 ——— 2 3x + y + 2z = 17 ——— 3 Hapuskan x daripada 1 dan 2 , Eliminate x from 1 and 2 , 1 : x + y + 2z = 5 2 : x – 3y + 4z = 14 (–) 4y – 2z = –9 ——— 4 Hapuskan x daripada 1 dan 3 , Eliminate x from 1 and 3 , 3 × 1 : 3x + 3y + 6z= 15 3 : 3x + y – 2z = 17 (–) 2y + 4z = –2 ——— 5 Selesaikan 4 dan 5 untuk mencari nilai y dan z, Solve 4 and 5 to find the values of y and z, 4 : 4y – 2z = –9 2 × 5 : 4y + 8z = –4 (–) –10z = –5 z = 1 2 Daripada/From 4 , 4y – 2 1 2 2 = –9 4y = –8 y = –2 Daripada/From 1 , x + (–2) + 2 1 2 2 = 5 x = 6 Maka/Hence, x = 6, y = –2, z = 1 2 2 2x + y + 5z = 13.5 x + 3y – 2z = 6 3x – 2y + 4z = 11.5 2x + y + 5z = 13.5 ——— 1 x + 3y – 2z = 6 ——— 2 3x – 2y + 4z = 11.5 ——— 3 Hapuskan x daripada 1 dan 2 , Eliminate x from 1 and 2 , 1 : 2x + y + 5z = 13.5 2 × 2 : 2x + 6y – 4z = 12 (–) –5y + 9z = 1.5 ——— 4 Hapuskan x daripada 2 dan 3 , Eliminate x from 2 and 3 , 3 × 2 : 3x + 9y – 6z = 18 3 : 3x – 2y + 4z = 11.5 (–) 11y – 10z = 6.5 ——— 5 Selesaikan 4 dan 5 untuk mencari nilai y dan z, Solve 4 and 5 to find the values of y and z, 10 × 4 : –50y + 90z = 15 9 × 5 : 99y – 90z = 58.5 (+) 49y = 73.5 y = 1.5 Daripada/From 4 , –5(1.5) + 9z = 1.5 9z = 9 z = 1 Daripada/From 2 , x + 3(1.5) – 2(1) = 6 x = 3.5 Maka/Hence, x = 3.5, y = 1.5, z = 1 Sudut Kalkulator Selesaikan sistem persamaan linear berikut. Solve the following system of linear equation. x + 2y + z = 3 2x – y + 3z = 13 2x + 3y + 4z = 11 Langkah 1: Pilih ‘Equation/Function’ pada Menu. Pilih Simul Equation dan pilih 3 pemboleh ubah. Step 1: Choose ‘Equation/Function’ in Menu. Choose Simul Equation and select 3 unknowns. MENU ALPHA (–) 1 3 Langkah 2: Masukkan semua nilai pekali bagi pemboleh ubah dan tekan ‘=’ untuk mendapatkan nilai-nilai x, y dan z. Step 2: Key in all the coefficients of the variables and press ‘=’ to get the values of x, y and z. = 1 1 = 3 2 = = 2 1 = = 3 3 = = 1 (–) 4 2 1 1 = = = = = 3 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 42 10/17/23 11:49 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

43 1 Jamil, Minah dan Chandran pergi ke sebuah kedai untuk membeli beberapa barang untuk menghias kelas mereka. Jamil membeli dua keping kad manila, tiga kotak pen penanda dan empat botol gam dengan harga RM16.00. Minah membeli enam keping kad manila, empat kotak pen penanda dan dua botol gam dengan harga RM23.00. Chandran membeli dua keping kad manila, lima kotak pen penanda dan tiga botol gam dengan harga RM21.00. Cari harga seunit bagi setiap jenis barang tersebut, dalam RM. Jamil, Minah and Chandran went to a shop to purchase some items to decorate their classroom. Jamil bought two manila cards, three boxes of marker pens and four glue sticks for RM16.00. Minah bought six manila cards, four boxes of marker pens and two glue sticks for RM23.00. Chandran bought two manila cards, five boxes of marker pens and three glue sticks for RM21.00. Find the unit price of each item, in RM. Katakan harga bagi sekeping kad manila = x harga bagi sekotak pen penanda = y harga bagi sebotol gam = z Let the price of a manila card = x the price of a box of marker pens = y the price of a glue stick = z 2x + 3y + 4z = 16 ——— 1 6x + 4y + 2z = 23 ——— 2 2x + 5y + 3z = 21 ——— 3 Hapuskan z daripada 1 dan 2 , Eliminate z from 1 and 2 , 1 : 2x + 3y + 4z = 16 2 × 2 : 12x + 8y + 4z = 46 (–) 10x + 5y = 30 ——— 4 Hapuskan z daripada 1 dan 3 , Eliminate z from 1 and 3 , 1 × 3 : 6x + 9y + 12z = 48 3 × 4 : 8x + 20y + 12z = 84 (–) 2x + 11y = 36 ——— 5 Selesaikan 4 dan 5 untuk mencari nilai x dan y, Solve 4 and 5 to find the values of x and y, 4 : 10x + 5y = 30 5 × 5 : 10x + 55y = 180 (–) –50y = –150 y = 3 Daripada/From 4 , 10x + 5(3) = 30 10x = 15 x = 1.5 Daripada/From 1 , 2(1.5) + 3(3) + 4z = 16 4z = 4 z = 1 Maka, harga bagi sekeping kad manila ialah RM1.50, harga bagi sekotak pen penanda ialah RM3 dan harga bagi sebotol gam ialah RM1. Hence, the price of a manila card is RM1.50, the price of a box of marker pens is RM3 and the price of a glue stick is RM1. Latihan 4 Selesaikan setiap yang berikut. TP 4 Solve each of the following. TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sistem persamaan dalam konteks penyelesaian masalah rutin yang mudah. Siti membeli 2 kg lobak merah, 1 kg kubis dan 3 kg ubi kentang dengan harga RM21.50, manakala Salmah membeli 4 kg lobak merah, 1 kg kubis dan 2 kg ubi kentang dengan harga RM27.50. Diberi jumlah harga bagi 1 kg lobak merah dan 1 kg kubis adalah melebihi harga bagi 1 kg ubi kentang sebanyak RM5. Cari harga untuk sekilogram bagi setiap jenis sayur, dalam RM. Siti buys 2 kg of carrots, 1 kg of cabbages and 3 kg of potatoes for RM21.50, while Salmah buys 4 kg of carrots, 1 kg of cabbages and 2 kg of potatoes for RM27.50. Given the total price of 1 kg of carrots and 1 kg of cabbages exceeds the price of 1 kg of potatoes by RM5. Find the price for a kilogram of each type of vegetable, in RM. Contoh 4 Katakan harga bagi 1 kg lobak merah = x harga bagi 1 kg kubis = y harga bagi 1 kg ubi kentang = z Let the price of 1 kg of carrots = x the price of 1 kg of cabbages = y the price of 1 kg of potatoes = z 2x + y + 3z = 21.5 ——— 1 4x + y + 2z = 27.5 ——— 2 x + y – z = 5 ——— 3 Hapuskan y daripada 1 dan 2 , Eliminate y from 1 and 2 , 1 : 2x + y + 3z = 21.5 2 : 4x + y + 2z = 27.5 (–) –2x + z = –6 ——— 4 Hapuskan y daripada 2 dan 3 , Eliminate y from 2 and 3 , 2 : 4x + y + 2z = 27.5 3 : x + y – z = 5 (–) 3x + 3z = 22.5 x + z = 7.5 ——— 5 Selesaikan 4 dan 5 untuk mencari nilai x dan z, Solve 4 and 5 to find the values of x and z, 4 : –2x + z = –6 5 : x + z = 7.5 (–) –3x = –13.5 x = 4.5 Daripada/From 5 , (4.5) + z = 7.5 z = 3 Daripada/From 1 , 2(4.5) + y + 3(3) = 21.5 y = 3.5 Maka, harga bagi 1 kg lobak merah ialah RM4.50, harga bagi 1 kg kubis ialah RM3.50 dan harga bagi 1 kg ubi kentang ialah RM3. Hence, the price of 1 kg of carrots is RM4.50, the price of 1 kg of cabbages is RM3.50 and the price of 1 kg of potatoes is RM3. 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 43 10/17/23 11:49 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

44 2 Jadual di bawah menunjukkan bilangan jualan bagi tiga jenis telefon bimbit oleh tiga buah kedai dalam seminggu. The table below shows the number of sales for three types of handphones by three stores in a week. Kedai Store Bilangan jualan/Number of sales Jumlah (RM) Telefon bimbit A Total (RM) Handphone A Telefon bimbit B Handphone B Telefon bimbit C Handphone C P 8 5 4 18 400 Q 2 6 9 22 300 R 4 6 8 22 400 Hitung harga seunit bagi setiap jenis telefon bimbit, dalam RM. Calculate the unit price of each type of handphone, in RM. Katakan harga bagi seunit telefon bimbit A, dalam RM = x harga bagi seunit telefon bimbit B, dalam RM = y harga bagi seunit telefon bimbit C, dalam RM = z Let the price of 1 unit handphone A, in RM = x the price of 1 unit handphone B, in RM = y the price of 1 unit handphone C, in RM = z 8x + 5y + 4z = 18 400 ——— 1 2x + 6y + 9z = 22 300 ——— 2 4x + 6y + 8z = 22 400 ——— 3 Hapuskan x daripada 1 dan 2 , Eliminate x from 1 and 2 , 1 : 8x + 5y + 4z = 18 400 4 × 2 : 8x + 24y + 36z = 89 200 (–) –19y – 32z = –70 800 ——— 4 Hapuskan x daripada 2 dan 3 , Eliminate x from 2 and 3 , 2 × 2 : 4x + 12y + 18z = 44 600 3 : 4x + 6y + 8z = 22 400 (–) 6y + 10z = 22 200 ——— 5 Selesaikan 4 dan 5 untuk mencari nilai y dan z, Solve 4 and 5 to find the values of y and z, 4 : –19y – 32z = –70 800 3.2 × 5 : 19.2y + 32z = 71 040 (+) 0.2y = 240 y = 1 200 Daripada/From 5 , 6(1 200) + 10z = 22 200 10z = 15 000 z = 1 500 Daripada/From 1 , 8x + 5(1 200) + 4(1 500) = 18 400 8x = 6 400 x = 800 Maka, harga bagi seunit telefon bimbit A ialah RM800, harga bagi seunit telefon bimbit B ialah RM1 200 dan harga bagi seunit telefon bimbit C ialah RM1 500. Hence the price of 1 unit of handphone A is RM800, the price of 1 unit of handphone B is RM1 200 and the price of 1 unit of handphone C is RM1 500. 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 44 10/17/23 11:49 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

45 3 Sebuah kedai cenderamata menawarkan tiga pakej cenderamata kepada pelanggan seperti ditunjukkan dalam jadual di bawah. A souvenir shop offers three souvenir packages to customers as shown in the table below. RM92 RM101 RM111 Hitung harga seunit bagi setiap barang tersebut, dalam RM. Calculate the unit price of each of the item, in RM. Katakan harga bagi sebuah topi, dalam RM = x harga bagi sepasang selipar, dalam RM = y harga bagi sehelai baju, dalam RM = z Let the price of a cap, in RM = x the price of a pair of slippers, in RM = y the price of a shirt, in RM = z 2x + y + z = 92 ——— 1 x + 2y + z = 101 ——— 2 x + y + 2z = 111 ——— 3 Hapuskan z daripada 1 dan 2 , Eliminate z from 1 and 2 , 1 : 2x + y + z = 92 2 : x + 2y + z = 101 (–) x – y = – 9 ——— 4 Hapuskan z daripada 2 dan 3 , Eliminate z from 2 and 3 , 2 × 2 : 2x + 4y + 2z = 202 3 : x + y + 2z = 111 (–) x + 3y = 91 ——— 5 Selesaikan 4 dan 5 untuk mencari nilai x dan y, Solve 4 and 5 to find the values of x and y, 4 : x – y = –9 5 : x + 3y = 91 (–) –4y = –100 y = 25 Daripada/From 5 , x + 3(25) = 91 x = 16 Daripada/From 1 , 2(16) + (25) + z = 92 z = 35 Maka, harga bagi sebuah topi ialah RM16, harga bagi sepasang selipar ialah RM25 dan harga bagi sehelai baju ialah RM35. Hence, the price of a cap is RM16, the price of a pair of slippers is RM25 and the price of a shirt is RM35. 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 45 10/17/23 11:50 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

46 Persamaan Serentak yang Melibatkan Satu Persamaan Linear dan Satu Persamaan Tak Linear Simultaneous Equations Involving One Linear Equation and One Non-Linear Equation 3.2 Latihan 5 Selesaikan setiap persamaan serentak yang berikut. TP 3 Solve each of the following simultaneous equations. TP 3 Mengaplikasikan kefahaman tentang sistem persamaan untuk melaksanakan tugasan mudah. 1 2x + y = 5 xy 4 = 1 2 2x + y = 5 ——— 1 xy 4 = 1 2 ——— 2 Daripada/From 1 , y = 5 – 2x Gantikan y = 5 – 2x ke dalam 2 , Substitute y = 5 – 2x into 2 , x(5 – 2x) 4 = 1 2 5x – 2x2 = 2 2x2 – 5x + 2 = 0 (2x – 1)(x – 2) = 0 2x – 1 = 0 atau/or x – 2 = 0 x = 1 2 x = 2 Apabila/When x = 1 2 , y = 5 – 2 1 2 2 = 4 Apabila/When x = 2, y = 5 – 2(2) = 1 Oleh itu/Hence, x = 1 2 , y = 4 atau/or x = 2, y = 1 2 x + y = 5 xy – 2y = 2 x + y = 5 ——— 1 xy – 2y = 2 ——— 2 Daripada/From 1 , x = 5 – y Gantikan x = 5 – y ke dalam 2 , Substitute x = 5 – y into 2 , (5 – y)y – 2y = 2 5y – y2 – 2y = 2 –y2 + 3y – 2 = 0 y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y – 1 = 0 atau/or y – 2 = 0 y = 1 y = 2 Apabila/When y = 1, x = 5 – (1) = 4 Apabila/When y = 2, x = 5 – (2) = 3 Oleh itu/Hence, x = 4, y = 1 atau/or x = 3, y = 2 x + 3y = 5 x2 + 2y2 = 6 Penyelesaian Langkah 1/Step 1 x + 3y = 5 ——— 1 x2 + 2y2 = 6 ——— 2 Langkah 2/Step 2 Daripada/From 1 , x = 5 – 3y Langkah 3/Step 3 Gantikan x = 5 – 3y ke dalam 2 , Substitute x = 5 – 3y into 2 , (5 – 3y)2 + 2y2 = 6 25 – 30y + 9y2 + 2y2 = 6 11y2 – 30y + 19 = 0 ——— 3 Langkah 4/Step 4 (11y – 19)(y – 1) = 0 11y – 19 = 0 atau/or y – 1 = 0 y = 19 11 y = 1 Langkah 5/Step 5 Apabila/When y = 19 11, x = 5 – 3 19 112 = – 2 11 Apabila/When y = 1, x = 5 – 3(1) = 2 Oleh itu,/Hence, x = – 2 11, y = 19 11 atau/or x = 2, y = 1 Contoh 5 Tip Bestari Gantikan jawapan ke dalam persamaan tak linear untuk menyemak. Substitute the answers back to the non-linear equation for checking. x2 + 2y2 = 6 (tak linear/non-linear) Apabila/When x = – 2 11 , y = 19 11 – 2 112 2 + 2 19 112 2 = 6 Apabila/When x = 2, y = 1 (2)2 + 2(1)2 = 6 Maka, penyelesaian adalah betul. Thus, the solutions are correct. 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 46 10/17/23 11:50 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

47 3 3x + y = 5 x2 – y2 = –3 3x + y = 5 ——— 1 x2 – y2 = –3 ——— 2 Daripada/From 1 , y = 5 – 3x Gantikan y = 5 – 3x ke dalam 2 , Substitute y = 5 – 3x into 2 , x2 – (5 – 3x)2 = –3 x2 – (25 – 30x + 9x2 ) = –3 x2 – 25 + 30x – 9x2 = –3 –8x2 + 30x – 22 = 0 4x2 – 15x + 11 = 0 (4x – 11)(x – 1) = 0 4x – 11 = 0 atau/or x – 1 = 0 x = 11 4 x = 1 Apabila/When x = 11 4 , y = 5 – 3 11 4 2 = – 13 4 Apabila/When x = 1, y = 5 – 3(1) = 2 Oleh itu/Hence, x = 11 4 , y = – 13 4 atau/or x = 1, y = 2 4 2x – y = 3 x2 – 3xy + y2 = 5 2x – y = 3 ——— 1 x2 – 3xy + y2 = 5 ——— 2 Daripada/From 1 , y = 2x – 3 Gantikan y = 2x – 3 ke dalam 2 , Substitute y = 2x – 3 into 2 , x2 – 3x(2x – 3) + (2x – 3)2 = 5 x2 – 6x2 + 9x + 4x2 – 12x + 9 = 5 –x2 – 3x + 4 = 0 x2 + 3x – 4 = 0 (x – 1)(x + 4) = 0 x – 1 = 0 atau/or x + 4 = 0 x = 1 x = –4 Apabila/When x = 1, y = 2(1) – 3 = –1 Apabila/When x = –4, y = 2(–4) – 3 = –11 Oleh itu/Hence, x = 1, y = –1 atau/or x = –4, y = –11 Latihan 6 Selesaikan setiap yang berikut. TP 3 Solve each of the following. TP 3 Mengaplikasikan kefahaman tentang fungsi kuadratik untuk melaksanakan tugasan mudah. 2x – y = 3 4 x + 3 y = 5 Penyelesaian 2x – y = 3 ——— 1 4 x + 3 y = 5 ——— 2 2 × (xy), 4y + 3x = 5xy ——— 3 Daripada/From 1 , y = 2x – 3 Gantikan y = 2x – 3 ke dalam 3 , Substitute y = 2x – 3 into 3 , 4(2x – 3) + 3x = 5x(2x – 3) 8x – 12 + 3x = 10x2 – 15x 10x2 – 15x – 8x + 12 – 3x = 0 10x2 – 26x + 12 = 0 5x2 – 13x + 6 = 0 (5x – 3)(x – 2) = 0 5x – 3 = 0 atau/or x – 2 = 0 x = 3 5 x = 2 Apabila/When x = 3 5 , y = 2 3 5 2 – 3 = – 9 5 Apabila/When x = 2, y = 2(2) – 3 = 1 Oleh itu/Hence, x = 3 5 , y = – 9 5 atau/or x = 2, y = 1 Contoh 6 1 x – y = 2 3 x + 2y = 3 x – y = 2 ——— 1 3 x + 2y = 3 ——— 2 Daripada/From 2 , 3 + 2xy = 3x ——— 3 Daripada/From 1 , y = x – 2 Gantikan y = x – 2 ke dalam 3 , Substitute y = x – 2 into 3 , 3 + 2x(x – 2) = 3x 3 + 2x2 – 4x – 3x = 0 2x2 – 7x + 3 = 0 (2x – 1)(x – 3) = 0 2x – 1 = 0 atau/or x – 3 = 0 x = 1 2 x = 3 Apabila/When x = 1 2 , y =  1 2 2 – 2 = – 3 2 Apabila/When x = 3, y = (3) – 2 = 1 Oleh itu/Hence, x = 1 2 , y = – 3 2 atau/or x = 3, y = 1 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 47 10/17/23 11:50 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B

48 2 x + y = 1 x + 3 y = 3 x + y = 1 ——— 1 x + 3 y = 3 ——— 2 Daripada/From 2 , xy + 3 = 3y ——— 3 Daripada/From 1 , x = 1 – y Gantikan x = 1 – y ke dalam 3 , Substitute x = 1 – y into 3 , (1 – y)y + 3 = 3y y – y2 + 3 – 3y = 0 –y2 – 2y + 3 = 0 y2 + 2y – 3 = 0 (y – 1)(y + 3) = 0 y – 1 = 0 atau/or y + 3 = 0 y = 1 y = –3 Apabila/When y = 1, x = 1 – (1) = 0 Apabila/When y = –3, x = 1 – (–3) = 4 Oleh itu/Hence, x = 0, y = 1 atau/or x = 4, y = –3 3 x + 3y = 5 2 x + 1 y = 2 x + 3y = 5 ——— 1 2 x + 1 y = 2 ——— 2 2 × (xy), 2y + x = 2xy ——— 3 Daripada/From 1 , x = 5 – 3y Gantikan x = 5 – 3y ke dalam 3 , Substitute x = 5 – 3y into 3 , 2y + (5 – 3y) = 2(5 – 3y)y 2y + 5 – 3y = 10y – 6y2 6y2 – 11y + 5 = 0 (6y – 5)(y – 1) = 0 6y – 5 = 0 atau/or y – 1 = 0 y = 5 6 y = 1 Apabila/When y = 5 6 , x = 5 – 3 5 6 2 = 5 2 Apabila/When y = 1, x = 5 – 3(1) = 2 Oleh itu/Hence, x = 5 2 , y = 5 6 atau/or x = 2, y = 1 4 x + 2y = 7 2 x – 3 y = 1 x + 2y = 7 ——— 1 2 x – 3 y = 1 ——— 2 2 × (xy), 2y – 3x = xy ——— 3 Daripada/From 1 , x = 7 – 2y Gantikan x = 7 – 2y ke dalam 3 , Substitute x = 7 – 2y into 3 , 2y – 3(7 – 2y) = (7 – 2y)y 2y – 21 + 6y = 7y – 2y2 2y – 21 + 6y – 7y + 2y2 = 0 2y2 + y – 21 = 0 (2y + 7)(y – 3) = 0 2y + 7 = 0 atau/or y – 3 = 0 y = – 7 2 y = 3 Apabila/When y = – 7 2 , x = 7 – 2– 7 2 2 = 14 Apabila/When y = 3, x = 7 – 2(3) = 1 Oleh itu/Hence, x = 14, y = – 7 2 atau/or x = 1, y = 3 5 2x + y = 2 3 2x – 2 y = 1 2x + y = 2 ——— 1 3 2x – 2 y = 1 ——— 2 2 × (2xy), 3y – 4x = 2xy ——— 3 Daripada/From 1 , y = 2 – 2x Gantikan y = 2 – 2x ke dalam 3 , Substitute y = 2 – 2x into 3 , 3(2 – 2x) – 4x = 2x(2 – 2x) 6 – 6x – 4x = 4x – 4x2 6 – 6x – 4x – 4x + 4x2 = 0 4x2 – 14x + 6 = 0 2x2 – 7x + 3 = 0 (2x – 1)(x – 3) = 0 2x – 1 = 0 atau/or x – 3 = 0 x = 1 2 x = 3 Apabila/When x = 1 2 , y = 2 – 2 1 2 2 = 1 Apabila/When x = 3, y = 2 – 2(3) = –4 Oleh itu/Hence, x = 1 2 , y = 1 atau/or x = 3, y = –4 03 Strategi A+ SPM Mate Tam Tg4_C3_39-53_Final.indd 48 10/17/23 11:50 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTILMU BAKTI SDN. BHD. PENERBIT ILMU BAKT ILMU BAKTI SDN. BHD. PENERBIT ILMU B