Modul A+1 Matematik Tambahan Tingkatan 4 KSSM

4 DWIBAHASA Tingkatan Matematik Tambahan Additional Mathematics Dr. M. K. Wong (Penulis Buku Teks) S. S. Law Pan Asia Publications Sdn. Bhd. 199101016590 (226902-X) No. 2-16, Jalan SU 8, Taman Perindustrian Subang Utama, Seksyen 22, 40300 Shah Alam, Selangor Darul Ehsan, Malaysia. Tel: +603-5614 4168 Faks: +603-5614 4268 E-mel: enquiry@panasiapub.com Laman web: www.panasiapub.com © Pan Asia Publications Sdn. Bhd. Hak cipta terpelihara. Sebarang bahagian dalam buku ini tidak boleh diterbitkan semula, disimpan dalam cara yang boleh dipergunakan lagi ataupun dipindahkan dalam sebarang bentuk atau cara, baik secara elektronik, mekanik, gambar, rakaman atau sebagainya, tanpa kebenaran daripada penerbit. Cetakan Pertama 2022 MODUL A+1 MATEMATIK TAMBAHAN Tingkatan 4 ISBN 978-967-466-628-6 Dicetak oleh Metrobay Industry Sdn. Bhd. (430102-W) • PDF Manual Guru • Rancangan Pengajaran Tahunan Bonus Guru

ii BAB 1 Fungsi Functions 1.1 Fungsi Functions...................................................................... 2 1.2 Fungsi Gubahan Composite Functions..................................................... 6 1.3 Fungsi Songsang Inverse Functions........................................................ 10 Soalan Berformat SPM.................................................... 13 BAB 2 Fungsi Kuadratik Quadratic Functions 2.1 Persamaan dan Ketaksamaan Kuadratik Quadratic Equations and Inequalities ........................... 21 2.2 Jenis-jenis Punca Persamaan Kuadratik Types of Roots of Quadratic Equations.......................... 27 2.3 Fungsi Kuadratik Quadratic Functions.................................................... 28 Soalan Berformat SPM.................................................... 37 BAB 3 Sistem Persamaan Systems of Equations 3.1 Sistem Persamaan Linear dalam Tiga Pemboleh Ubah Systems of Linear Equations in Three Variables ............. 43 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...................................... 49 Soalan Berformat SPM.................................................... 54 BAB 4 Indeks, Surd dan Logaritma Indices, Surds and Logarithms 4.1 Hukum Indeks Laws of Indices ........................................................... 57 4.2 Hukum Surd Laws of Surds.............................................................. 60 4.3 Hukum Logaritma Laws of Logarithms..................................................... 64 4.4 Aplikasi Indeks, Surd dan Logaritma Applications of Indices, Surds and Logarithms............... 70 Soalan Berformat SPM.................................................... 73 BAB 5 Janjang Progressions 5.1 Janjang Aritmetik Arithmetic Progressions............................................... 76 5.2 Janjang Geometri Geometric Progressions............................................... 81 Soalan Berformat SPM.................................................... 86 BAB 6 Hukum Linear Linear Law 6.1 Hubungan Linear dan Tak Linear Linear and Non-Linear Relations.................................. 92 6.2 Hukum Linear dan Hubungan Tak Linear Linear Law and Non-Linear Relations........................... 97 6.3 Aplikasi Hukum Linear Applications of Linear Law ........................................ 101 Soalan Berformat SPM.................................................. 108 BAB 7 Geometri Koordinat Coordinate Geometry 7.1 Pembahagi Tembereng Garis Divisor of a Line Segment .......................................... 114 7.2 Garis Lurus Selari dan Garis Lurus Serenjang Parallel Lines and Perpendicular Lines....................... 118 7.3 Luas Poligon Areas of Polygons...................................................... 123 7.4 Persamaan Lokus Equations of Loci ...................................................... 128 Soalan Berformat SPM.................................................. 131 BAB 8 Vektor Vectors 8.1 Vektor Vectors ..................................................................... 137 8.2 Penambahan dan Penolakan Vektor Addition and Subtraction of Vectors ............................ 141 8.3 Vektor dalam Satah Cartes Vectors in a Cartesian Plane....................................... 148 Soalan Berformat SPM.................................................. 152 BAB 9 Penyelesaian Segi Tiga Solution of Triangles 9.1 Petua Sinus Sine Rule .................................................................. 157 9.2 Petua Kosinus Cosine Rule .............................................................. 161 9.3 Luas Segi Tiga Area of a Triangle...................................................... 166 9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas Segi Tiga Application of Sine Rule, Cosine Rule and Area of a Triangle.................................................................... 169 Soalan Berformat SPM.................................................. 173 BAB 10 Nombor Indeks Index Numbers 10.1 Nombor Indeks Index Numbers.......................................................... 175 10.2 Indeks Gubahan Composite Index........................................................ 178 Soalan Berformat SPM.................................................. 184 Jawapan ......................................................................MG-1 Lembaran PBD......................................................... MG-11 00_Kand_Modul A+ MateTam Tg4.indd 2 08/10/2021 11:12 AM

1 Nota 1.1 Fungsi/Functions 1. Hubungan melibatkan dua set; domain dan kodomain. Unsur dalam domain disebut sebagai objek dan unsur dalam kodomain disebut sebagai imej. A relation involves two sets; a domain and a codomain. The elements in a domain are called objects and the elements in a codomain are called images. a b c 1 2 3 Domain = {a, b, c} Kodomain/Codomain = {1, 2, 3} Objek/Object = a, b, c Imej/Image = 1, 2 Julat/Range = {1, 2} 2. Terdapat empat jenis hubungan. There are four types of relations. (a) (b) Satu kepada satu One to one Satu kepada banyak One to many (c) (d) Banyak kepada satu Many to one Banyak kepada banyak Many to many 3. Fungsi ialah hubungan khas dengan setiap objek dalam domain mempunyai satu imej sahaja. A function is a special relation where each object in the domain has only one image. 4. Fungsi ialah hubungan satu dengan satu atau hubungan banyak dengan satu. Function is a one to one relation or a many to one relation. 5. Fungsi boleh ditulis dengan tatatanda f : x → y atau f(x) = y The functions can be written in the notation f : x → y or f(x) = y 1.2 Fungsi Gubahan/Composite Functions 1. x y X Y Z g z gf f 2. gf(x) ≠ fg(x) 3. f 2 (x) = ff(x), f 3 (x) = ff 2 (x) = f 2 f(x) = fff(x) 1.3 Fungsi Songsang/Inverse Functions 1. Fungsi songsang memetakan imej kepada objeknya. An inverse function maps an image to its object. Jika fungsi f memetakan x kepada y dan fungsi g memetakan y kepada z, maka fungsi gubahan gf memetakan x kepada z. If the function f maps x to y and the function g maps y to z, then the composite function gf maps x to z. f : x → y atau/or f(x) = y g : y → z atau/or g(y) = z gf : x → z atau/or gf(x) = g(y) = z f X Y f –1 x y Fungsi Functions BAB 1 01_Modul A+ MateTam Tg4.indd 1 08/10/2021 11:18 AM

2 BAB 1 1. Nyatakan sama ada hubungan yang berikut ialah fungsi atau bukan. Berikan justifikasi anda. State whether the following relations is a function or not. Give your justification. TP 1 2. Tentukan sama ada graf yang berikut ialah fungsi atau bukan dengan menggunakan ujian garis mencancang. Determine whether the following graphs is a function or not by using the vertical line test. TP 1 0 y x Fungsi Function Contoh/Example (a) 0 y x Bukan fungsi (b) y x 0 Fungsi (a) x f 12 —x 2 3 4 6 12 4 3 1 Fungsi. Satu objek mempunyai satu imej. (b) p M N q r a b c d Bukan fungsi. Satu objek mempunyai lebih daripada satu imej. (c) {(w, 3), (x, 4), (y, 7), (w, 8), (z, 7), (x, 3)} Bukan fungsi. Hubungan banyak dengan banyak. (d) Set Q Set P 5 10 15 20 a b c d Fungsi. Hubungan banyak dengan satu. 1.1 Fungsi Functions Buku Teks m.s. 2-11 PAUTAN INTERAKTIF PAUTAN INTERAKTIF Apakah itu fungsi? What are functions? Contoh/Example 5 7 8 16 25 49 64 Fungsi. Satu objek mempunyai satu imej. Function. One object has one image. 01_Modul A+ MateTam Tg4.indd 2 08/10/2021 11:18 AM

3 BAB 1 3. Dengan menggunakan tatatanda fungsi, ungkapkan hubungan berikut dalam sebutan x. By using the function notation, express the following relations in terms of x. TP 2 –3 3 4 4 11 x f f(x) f(x) = x2 – 5 Contoh/Example (a) 8 27 64 4 5 3 x g g(x) g(x) = 3 x + 1 (b) –1 1 3 14 19 11 x m m(x) m(x) = x2 + 10 (c) 3 6 9 2 3 1 x h h(x) h(x) = x 3 4. Tentukan domain, kodomain dan julat bagi setiap fungsi yang berikut. Determine the domain, codomain and range for each of the following functions. TP 2 –2 2 4 6 13 1 Set A Set B Domain = {–2, 2, 4} Kodomain/Codomain = {1, 6, 13} Julat/Range = {1, 13} Contoh/Example (a) {(5, a), (6, b), (8, c), (9, a)} Domain = {5, 6, 8, 9} Kodomain = {a, b, c} Julat = {a, b, c} (b) 1 2 3 4 5 6 P Q R S T Set B Set A Domain = {1, 2, 3, 4, 5, 6} Kodomain = {P, Q, R, S, T} Julat = {Q, R, S, T} 5. Lakarkan graf bagi fungsi yang berikut untuk domain –1  x  4. Seterusnya, nyatakan julat yang sepadan dengan domain yang diberi. Sketch the graph for the following functions in the domain –1  x  4. Hence, state the corresponding range for the given domain. TP 3 f(x) = x – 3 f(–1) = –1 – 3 = –4 f(4) = 4 – 3 = 1 0 4 –4 1 –1 y x Julat sepadan/Corresponding range: –4  f(x)  1 Contoh/Example (a) f(x) = |x – 3| f(–1) = |–1 – 3| = |– 4| = 4 f(4) = 4 – 3 = 1 4 3 1 –4 –3 4 y x 0 –1 3 Julat sepadan: 0  f(x)  4 (b) f(x) = |5 − 3x| f(–1) = |5 – 3(–1)| = 8 f(4) = |5 – 3(4)| = |–7| = 7 –7 8 7 5 4 – 3 y x 0 5 –1 Julat sepadan: 0  f(x)  8 01_Modul A+ MateTam Tg4.indd 3 08/10/2021 11:18 AM

4 BAB 1 6. Selesaikan setiap yang berikut. Solve each of the following. TP 3 Fungsi h ditakrifkan oleh h : x → 4x2 – 12. Function h is defined by h : x → 4x2 – 12. (i) Cari imej bagi –1. Find the image of –1. (ii) Diberi imej bagi p ialah p + 2, cari nilai-nilai yang mungkin bagi p. Given the image of p is p + 2, find the possible values of p. (i) h(–1) = 4(–1)2 – 12 = –8 (ii) h(p) = p + 2 4p2 – 12 = p + 2 4p2 – p – 14 = 0 p = 2 atau/or p = –1.75 Contoh/Example (a) Fungsi f ditakrifkan oleh f(x) = |9m + 2x|. Cari nilai-nilai m dengan keadaan Function f is defined by f(x) = |9m + 2x|. Find the values of m such that (i) f(3) = 7 (ii) f(m) = 2 (iii) f(m) = m + 5 (i) |9m + 2(3)| = 7 9m + 6 = 7 atau 9m + 6 = –7 9m = 1 9m = –13 9m = 1 9 9 m = – 13 9 (ii) |9m + 2m| = 2 11m = 2 atau 11m = –2 11m = 2 11 11m = – 2 11 (iii) |9m + 2m| = m + 5 11m = m + 5 atau 11m = –m – 5 11m – m = 5 11m + m = –5 10m = 5 12m = –5 1 1m = 1 2 m = – 5 12 (b) Diberi fungsi f(x) = |x + 8|, cari Given the function f(x) = |x + 8|, find (i) f(–10), (ii) nilai-nilai x dengan keadaan f(x) = 5. the values of x such that f(x) = 5. (i) |–10 + 8| = |–2| = 2 (ii) |x + 8| = 5 x + 8 = 5 atau x + 8 = –5 x = –3 x = –13 (c) Diberi fungsi f(x) = |1 + 2x|. Nyatakan Given the function f(x) = |1 + 2x|. State (i) objek-objek bagi 15, the objects of 15, (ii) domain bagi 0  f(x)  15. the domain for 0  f(x)  15. (i) |1 + 2x| = 15 1 + 2x = –15 atau 1 + 2x = 15 x = –8 x = 7 (ii) –8  x  7 (d) Rajah di bawah menunjukkan gambar rajah anak panah yang mewakili fungsi h(x) = s – tx, dengan keadaan s dan t ialah pemalar. Cari nilai s dan nilai t. The diagram below shows an arrow diagram which represents the function h(x) = s – tx, where s and t are constants. Find the values of s and t. –13 –16 6 5 h s – t(5) = –13 s – 5t = –13 … s – t(6) = –16 s – 6t = –16 … Persamaan serentak  – : –t = –3 –t = 3 s – 5(3) = –13 s = –13 + 15 s = 2 (e) Diberi satu fungsi f : x → 6 1 – 2x , x ≠ m. Given a function f : x → 6 1 – 2x , x ≠ m. (i) Nyatakan nilai m. State the value of m. (ii) Cari nilai-nilai k dengan keadaan f(k) = –k. Find the values of k such that f(k) = –k. (i) Oleh sebab x ≠ 1 2 , maka m = 1 2 (ii) 6 1 – 2k = –k 6 = –k(1 – 2k) 6 = –k + 2k2 0 = 2k2 – k – 6 k = –1.5 atau k = 2 01_Modul A+ MateTam Tg4.indd 4 08/10/2021 11:18 AM

5 BAB 1 Uji Kendiri 1.1 1. Fungsi f ditakrifkan oleh f : x → |3x – x2 |. Function f is defined by f : x → |3x – x2 |. KBAT Mengaplikasi (a) Cari imej bagi –2, 5 dan 1 3 . Find the images of –2, 5 and 1 3 . (b) Diberi imej bagi k ialah 5k, cari nilai-nilai yang mungkin bagi k. Given the image of k is 5k, find the possible values of k. (a) f(–2) = |3(–2) – (–2)2 | = |–10| = 10 f(5) = |3(5) – 52 | = |–10| = 10 f  1 3  = |3 1 3  –  1 3  2 | = |1 – 1 9 | = 8 9 (b) f(k) = 5k |3(k) – k2 | = 5k 3k – k2 = 5k atau 3k – k2 = –5k –k2 – 2k = 0 –k2 + 8k = 0 k = 0, –2 k = 0, 8 ∴ k = –2, 0, 8 2. Diberi g(x) = 5 x + mx, x ≠ 0. Cari nilai m jika g(5) = 16. Given g(x) = 5 x + mx, x ≠ 0. Find the value of m if g(5) = 16. g(5) = 16 5 5 + m(5) = 16 5m = 15 m = 3 3. Fungsi h ditakrifkan oleh h : x → |x + 2|. Cari domain bagi h(x) > 3. The function of h is defined by h : x → |x + 2|. Find the domain for h(x) > 3. |x + 2| > 3 x + 2 > 3 atau x + 2 < –3 x > 1 x < –5 4. Diberi m(x) = 3x – 5 x – 3 , x ≠ 3. Suatu nilai x memetakan kepada dirinya sendiri di bawah fungsi m. Cari nilainilai x. Given m(x) = 3x – 5 x – 3 , x ≠ 3. A value of x maps to itself under the function m. Find the values of x. 3x – 5 x – 3 = x 3x – 5 = x(x – 3) 3x – 5 – x2 + 3x = 0 –x2 + 6x – 5 = 0 x = 1 atau x = 5 5. Fungsi k ditakrifkan oleh k : x → ax + b. Diberi k(3) = 2 dan k(1) = –5, cari nilai a dan nilai b. Function k is defined by k : x → ax + b. Given k(3) = 2 and k(1) = –5, find the values of a and b. 3a + b = 2 …… a + b = –5 ……  – : 2a = 7 a = 7 2 7 2 + b = –5 b = –8 1 2 6. Rajah di bawah menunjukkan hubungan antara set A dengan set B. The diagram below shows the relation between set A and set B. –1 5 6 7 18 4 Set A Set B (a) Nyatakan jenis hubungan tersebut. State the type of relation. (b) Nyatakan julat bagi hubungan tersebut. State the range for the relation. (a) Hubungan banyak kepada satu (b) {4, 18} 7. Gambar rajah anak panah di bawah menunjukkan fungsi g. The arrow diagram below shows function g. –4 4 6 –8 5 6 t 4 x g(x) (a) Nyatakan nilai t. State the value of t. (b) Dengan menggunakan tatatanda fungsi, tulis hubungan bagi g(x). By using the function notation, write the relation of g(x). (c) Cari imej bagi –5 di bawah pemetaan ini. Find the image of –5 under this mapping. (a) 8 (b) g(x) = |x| (c) 5 01_Modul A+ MateTam Tg4.indd 5 08/10/2021 11:18 AM

6 BAB 1 1. Selesaikan setiap yang berikut. Solve each of the following. TP 3 Dalam gambar rajah anak panah di bawah, fungsi f memetakan set M kepada set N manakala fungsi g memetakan set N kepada set P. In the arrow diagram below, function f maps set M to set N whereas function g maps set N to set P. x x + 2 3x2 Set M Set N Set P f g Dengan menggunakan tatatanda fungsi, nyatakan By usingn the function notation, state (i) fungsi f, (ii) fungsi g, the function f, the function g, (iii) fungsi f 2 , (iv) fungsi gf 2 . the function f 2 , the function gf 2 . (i) f(x) = x + 2 (ii) gf(x) = 3x2 g(x + 2) = 3x2 Katakan/Let x + 2 = y x = y – 2 g(y) = 3(y – 2)2 ∴ g(x) = 3(x – 2)2 (iii) f 2 (x) = f(x + 2) = (x + 2) + 2 = x + 4 (iv) gf 2 (x) = g(x + 4) = 3(x + 4 – 2)2 = 3(x + 2)2 Contoh/Example (a) Fungsi f dan fungsi g masing-masing ditakrifkan oleh f : x → 2x2 – 5 dan g : x → –11 + 8x. Cari ungkapan bagi fg dan gf. Seterusnya, cari nilai-nilai x apabila Functions f and g are defined by f : x → 2x2 – 5 and g : x → –11 + 8x respectively. Find the expression of fg and gf. Hence, find the values of x when (i) f = g (ii) fg = 3gf + 6 fg(x) = f(–11 + 8x) = 2(–11 + 8x)2 – 5 gf(x) = g(2x2 – 5) = –11 + 8(2x2 – 5) = 16x2 – 51 (i) 2x2 – 5 = –11 + 8x 2x2 – 8x + 6 = 0 x = 3 atau x = 1 (ii) 2(–11 + 8x)2 – 5 = 3(16x2 – 51) + 6 2(64x2 – 176x + 121) – 5 – 3(16x2 ) + 3(51) – 6 = 0 128x2 – 352x + 242 – 5 – 48x2 + 153 – 6 = 0 80x2 – 352x + 384 = 0 x = 2.4 atau x = 2 (b) Diberi h : x → px + q dan h2 : x → 9x + 8. Cari nilai bagi p + 2q jika p > 0. Given h : x → px + q and h2 : x → 9x + 8. Find the value of p + 2q if p > 0. hh(x) = p(px + q) + q hh(x) = p2 x + pq + q Bandingkan dengan h2 (x) = 9x2 + 8 p2 = 9 2 p = 3 (sebab p > 0) 3q + q = 8 4q = 8 q = 2 p + 2q = 3 + 2(2) = 7 (c) Diberi f(x) = 3m – 7x dan g(x) = 2n + 3x dengan keadaan f 2 = gf. Ungkapkan x dalam sebutan m dan n. Given f(x) = 3m – 7x and g(x) = 2n + 3x such that f 2 = gf. Express x in terms of m and n. ff(x) = 3m – 7(3m – 7x) = 3m – 21m + 49x = –18m + 49x gf(x) = 2n + 3(3m – 7x) = 2n + 9m – 21x –18m + 49x = 2n + 9m – 21x 49x + 21x = 2n + 9m + 18m 70x = 2n + 27m x = 2n + 27m 70 1.2 Fungsi Gubahan Composite Functions Buku Teks m.s. 12-19 01_Modul A+ MateTam Tg4.indd 6 08/10/2021 11:18 AM

7 BAB 1 (d) Dalam rajah di bawah, suatu fungsi v memetakan x kepada y dan fungsi w memetakan y kepada z. In the diagram below, a function v maps x to y and a function w maps y to z. 8 k 9 x y v w z Diberi v(x) = 1 + x dan w(x) = x2 – 1, cari Given v(x) = 1 + x and w(x) = x2 – 1, find (i) nilai k, the value of k, (ii) fungsi yang memetakan x kepada z. the function that maps x to z. (i) k = 92 – 1 = 80 (ii) wv(x) = w(1 + x) = (1 + x)2 – 1 = 1 + 2x + x2 – 1 = 2x + x2 (e) Diberi f(x) = x – 2 x – 3, x ≠ 3 dan g(x) = 3 x – 2, x ≠ 2, cari f 2 (6) dan gf 2 (6). Given f(x) = x – 2 x – 3, x ≠ 3 and g(x) = 3 x – 2 , x ≠ 2, find f 2 (6) and gf 2 (6). f 2 (6) = ff(6) = f  6 – 2 6 – 3 = f  4 3  = 4 3 – 2 4 3 – 3 = – 2 3 – 5 3 = 2 5 gf 2 (6) = g 2 5  = 3 2 5 – 2 = –1 7 8 2. Selesaikan setiap yang berikut. Solve each of the following. TP 3 Diberi f : x → x + 1 dan fg : x → 2x2 + x – 1, cari fungsi g(x). Given f : x → x + 1 and fg : x → 2x2 + x – 1, find the function g(x). f [g(x)] = 2x2 + x – 1 g(x) + 1 = 2x2 + x – 1 g(x) = 2x2 + x – 1 – 1 g(x) = 2x2 + x – 2 Contoh/Example (a) Diberi f : x → x + 6 dan gf : x → x2 – x + 1, cari fungsi g(x). Given f : x → x + 6 and gf : x → x2 – x + 1, find the function g(x). g[f(x)] = x2 – x + 1 g[x + 6] = x2 – x + 1 Katakan x + 6 = y x = y – 6 Gantikan g(y) = (y – 6)2 – (y – 6) + 1 g(y) = y2 – 12y + 36 – y + 6 + 1 g(y) = y2 – 13y + 43 g(x) = x2 – 13x + 43 (b) Diberi fungsi m : x → 5 x , x ≠ 0 dan nm : x → 10x, cari Given the function m : x → 5 x , x ≠ 0 and nm : x → 10x, find (i) n(x), (ii) nilai x apabila mn(x) = 12. the value of x when mn(x) = 12. (i) n[m(x)] = 10x n 5 x  = 10x Katakan 5 x = y x = 5 y Gantikan n(y) = 10 5 y  n[y] = 50 y n(x) = 50 x (ii) mn(x) = 12 5  50 x  = 12 5 ÷ 50 x = 12 5 × x 50 = 12 x = 120 01_Modul A+ MateTam Tg4.indd 7 08/10/2021 11:18 AM

8 BAB 1 Uji Kendiri 1.2 1. Tuliskan fungsi berulang f 2 , f 3 , f 4 , f 25 dan f 50 bagi fungsi f(x) = 4 x , x ≠ 0. Write the iterated function f 2 , f 3 , f 4 , f 25 and f 50 for the function f(x) = 4 x , x ≠ 0. KBAT Mengaplikasi f 2 (x) = f f(x) = 4  4 x  = x f 3 (x) = f 2 f(x) = 4 x f 4 (x) = f 2 f 2 (x) = x f 25(x) = 4 x f 50(x) = x 2. Luas permukaan riak air di permukaan kolam, L, dalam cm2 , diberi oleh fungsi L(j) = πj 2 dengan j ialah jejari riak air, dalam cm. Jejari riak air bertambah mengikut fungsi j(t) = 1 4 t 2 , t > 0, dengan t ialah masa, dalam saat. Cari luas permukaan riak air di permukaan kolam selepas 4 saat. The surface area of water ripple on a pool, L, in cm2 , is given by the function L(j) = πj 2 , where j is the radius of the water ripple, in cm. The radius of the water ripple increases according to the function j(t) = 1 4 t 2 , t > 0, where t is the time, in seconds. Find the surface area of water ripple on the pool after 4 seconds. Lj(t) = π 1 4 t 2  2 = 1 16πt 4 Lj(4) = 1 16π(4)4 = 16π cm2 3. Diberi f(x) = 5 x , cari setiap yang berikut. Given f(x) = 5 x , find each of the following. TP 3 f 2 (x) f 2 (x) = f  5 x  = 5  5 x  = 5 ÷ 5 x = 5 × x 5 = x Contoh/Example (a) f 5 (x) f 5 (x) = f 2 f 2 f 1 = f 4  5 x  = 5 x (b) f 3 (x) f 3 (x) = f 2  5 x  = 5 x (c) f 4 (x) f 4 (x) = f 2 f 2 = f 2 (x) = x (d) f 8 (x) f 8 (x) = f 4 f 4 = f 4 (x) = x (e) f 13(x) f 13(x) = f 4 f 4 f 4 f 1 = f 12 5 x  = 5 x 4. Seorang penjaja sate memperoleh keuntungan harian, dalam RM, mengikut fungsi f : x → 2x – 100 4 dengan x ialah bilangan sate yang dijual setiap hari. A satay hawker earns a daily profit, in RM, according to the function f : x → 2x – 100 4 , where x is the number of satay sold per day. TP 4 (a) Pada satu minggu, penjaja sate tersebut menjual 3 580 sate dalam 5 hari. Kira purata keuntungan harian bagi minggu itu. In a week, the satay hawker sold 3 580 satay in 5 days. Calculate the average daily profit for the week. 3 580 ÷ 5 = 716 Purata keuntungan harian = 2(716) – 100 4 = RM333 (b) Berapakah bilangan minimum sate yang perlu dijual supaya penjaja sate itu tidak rugi? What is the minimum number of satay that needs to be sold so that the satay hawker does not suffer loss? 2x – 100 4  0 2x  100 x  50 Bilangan minimum sate yang perlu dijual ialah 50 cucuk supaya dia tidak rugi. 01_Modul A+ MateTam Tg4.indd 8 08/10/2021 11:18 AM

9 BAB 1 3. Fungsi f ditakrifkan oleh f : x → 5 – x. Cari fungsi g jika fg : x → 5x2 + 1. Function f is defined by f : x → 5 – x. Find the function g if fg : x → 5x2 + 1. f[g(x)] = 5x2 + 1 5 – g(x) = 5x2 + 1 – g(x) = 5x2 + 1 – 5 g(x) = –5x2 + 4 5. Fungsi q ditakrifkan oleh q : x → 5 – x 2 . Cari fungsi p jika pq : x → x – 3. Function q is defined by q : x → 5 – x 2 . Find the function p if pq : x → x – 3. p[q(x)] = x – 3 p 5 – x 2  = x – 3 Katakan 5 – x 2 = y 5 – x = 2y x = 5 – 2y ∴ p(y) = (5 – 2y) – 3 = 2 – 2y Maka, p(x) = 2 – 2x 4. Diberi f(x) = 5x – 9 dan fg(x) = 2x2 + 3. Cari fungsi g(x). Given f(x) = 5x – 9 and fg(x) = 2x2 + 3. Find the function g(x). fg(x) = 2x2 + 3 5g(x) – 9 = 2x2 + 3 5g(x) = 2x2 + 3 + 9 g(x) = 2x2 + 12 5 6. Diberi k(x) = mx – n dan k3 (x) = 27x + 12. Cari nilai m dan nilai n. Given k(x) = mx – n and k3 (x) = 27x + 12. Find the values of m and n. KBAT Menganalisis k[k(x)] = m(mx – n) – n = m2 x – mn – n k3 (x) = k2 k(x) = m2 (mx – n) – mn – n k3 (x) = m3 x – m2 n – mn – n Bandingkan dengan k3 (x) = 27x + 12 m3 = 27 m = 3 –m2 n – mn – n = 12 –32 n – 3n – n = 12 –9n – 3n – n = 12 n = – 12 13 7. Diberi u(x) = 5 7 – x , x ≠ 7 dan vu(x) = 3 + x. Cari fungsi v. Given u(x) = 5 7 – x , x ≠ 7 and vu(x) = 3 + x. Find the function v. vu(x) = 3 + x v 5 7 – x = 3 + x Katakan 5 7 – x = y 5 = 7y – xy xy = 7y – 5 x = 7y – 5 y v(y) = 3 + 7y – 5 y = 3y + 7y – 5 y = 10y – 5 y v(x) = 10x – 5 x 8. Suatu fungsi m ditakrifkan oleh m(x) = x 1 – x . Tunjukkan bahawa m3 (x) = x 1 – 3x . Seterusnya, cadangkan ungkapan yang sesuai bagi mn (x). Function m is defined by m(x) = x 1 – x . Show that m3 (x) = x 1 – 3x . Hence, suggest a suitable expression for mn (x). KBAT Menilai m2 (x) = x 1 – x 1 – x 1 – x = x 1 – x ÷ 1 – x 1 – x = x 1 – x ÷  1 – x 1 – x – x 1 – x = x 1 – x ×  1 – x 1 – 2x = x 1 – 2x m3 (x) = x 1 – x 1 – 2 x 1 – x = x 1 – x ÷ 1 – 2x 1 – x = x 1 – x ÷  1 – x 1 – x – 2x 1 – x = x 1 – x ×  1 – x 1 – 3x = x 1 – 3x mn (x) = x 1 – nx 01_Modul A+ MateTam Tg4.indd 9 08/10/2021 11:18 AM

10 BAB 1 1. Suatu fungsi ditakrifkan sebagai f(x) = x – 5 3 . Tentukan setiap yang berikut. A function is defined as f(x) = x – 5 3 . Determine each of the following. TP 2 f(3) f(3) = 3 – 5 3 = – 2 3 Contoh/Example (a) f –1(–1) Katakan a = f –1(–1) f(a) = –1 a = 2 a – 5 3 = −1 ∴ f –1(–1) = a = 2 a – 5 = –3 (b) Imej bagi 6 di bawah f. The image of 6 under f. f(6) = 6 – 5 3 = 1 3 (c) Objek bagi 1. The object of 1. x – 5 3 = 1 x – 5 = 3 x = 8 2. Tentukan sama ada setiap fungsi berikut mempunyai fungsi songsang atau tidak. Berikan justifikasi anda. Determine whether each of the following functions has an inverse function or not. Give your justification TP 2 –2 2 4 6 13 1 Set A Set B f x f(x) Tidak. f –1 mempunyai dua imej. No. f –1 has two images. Contoh/Example (a) f(x) f (x) = 3 – x x 3 3 0 Ya. f –1 mempunyai satu imej. (b) {(1, 3), (4, 2), (5, 7), (2, 8)} Ya. f –1 mempunyai satu imej. 3. Selesaikan setiap yang berikut. Solve each of the following. TP 5 1.3 Fungsi Songsang Inverse Functions Buku Teks m.s. 20-30 Contoh/Example Tentukan sama ada fungsi h : x → 4x – 3 dan g : x → x + 3 4 adalah songsang antara satu sama lain. Determine whether the functions h : x → 4x – 3 and g : x → x + 3 4 are inverse to each other. (Kaedah 2/Procedure 2) hg(x) = h x + 3 4  = 4 x + 3 4  – 3 = x + 3 – 3 = x Maka, h(x) dan g(x) adalah songsang antara satu sama lain. Hence, h(x) and g(x) are inverse to each other. gh(x) = g(4x – 3) = (4x – 3) + 3 4 = 4x 4 = x (Kaedah 1/Procedure 1) h : x → 4x – 3 Katakan/Let 4x – 3 = y x = y + 3 4 h–1(x) = x + 3 4 g : x → x + 3 4 Katakan/Let x + 3 4 = y x = 4y –3 g–1(x) = 4x – 3 Maka, h(x) dan g(x) adalah songsang antara satu sama lain. Hence, h(x) and g(x) are inverse to each other. 01_Modul A+ MateTam Tg4.indd 10 08/10/2021 11:18 AM

11 BAB 1 (a) Lakarkan graf bagi f : x → x2 untuk domain 0  x  4. Pada satah yang sama, lakarkan graf bagi f –1(x). Seterusnya, nyatakan domain dan julat bagi f –1(x). Sketch the graph of f : x → x2 for the domain 0  x  4. On the same plane, sketch the graph of f –1(x). Hence, state the domain and range of f –1(x). x –1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 1 2 3 4 f(x) f(x) = x2 f –1 (x) = √ —x Domain bagi f –1(x) ialah 0  x  4 dan julat bagi f –1(x) ialah 0  f –1(x)  2. (b) Fungsi songsang f –1(x) ditakrifkan oleh f –1 : x→ 6x 5 + x , x ≠ a. The inverse function f –1(x) is defined by f –1 : x → 6x 5 + x , x ≠ a. (i) Nyatakan nilai a. State the value of a. (ii) Cari f(3). Find f(3). (iii) Cari nilai ff –1(x). Find the value of ff –1(x). (i) 5 + x ≠ 0 x ≠ –5 ∴ a = –5 (ii) f(3) = x ⇒ f –1(x) = 3 6x 5 + x = 3 6x = 15 + 3x 3x = 15 x = 5 ∴ f(3) = 5 (iii) Katakan 6x 5 + x = y 6x = 5y + xy 6x – xy = 5y x(6 – y) = 5y x = 5y 6 – y f(x) = 5x 6 – x (c) Suatu fungsi f ditakrifkan sebagai f : x → 3 – 2x 5x , x ≠ 0, cari A function f is defined as f : x → 3 – 2x 5x , x ≠ 0, find (i) f –1(2), (ii) nilai-nilai x dengan keadaan f(x) = f –1(x). the values of x such that f(x) = f –1(x). (i) f –1(2) ⇒ f(x) = 2 3 – 2x 5x = 2 3 – 2x = 10x 3 = 12x x = 1 4 f –1(2) = 1 4 (ii) Katakan 3 – 2x 5x = y 3 – 2x = 5xy –2x – 5xy = –3 2x + 5xy = 3 x(2 + 5y) = 3 x = 3 2 + 5y Maka, f –1(x) = 3 2 + 5x f(x) = f –1(x) 3 – 2x 5x = 3 2 + 5x (3 – 2x)(2 + 5x) = 15x 6 + 15x – 4x – 10x2 – 15x = 0 –10x2 – 4x + 6 = 0 x = –1 atau x = 0.6 ff –1(x) = 5 6x 5 + x 6 – 6x 5 + x =  30x 5 + x 6(5 + x) 5 + x – 6x 5 + x = 30x 5 + x ÷ 30 5 + x = x 01_Modul A+ MateTam Tg4.indd 11 08/10/2021 11:18 AM

12 BAB 1 Uji Kendiri 1.3 (d) Diberi fungsi s(x) = x 4 , x ≠ 0. Cari nilai x dengan keadaan s(x) = s–1(x). Seterusnya, lakarkan graf bagi s(x) dan s–1(x). Given the function s(x) = x 4 , x ≠ 0. Find the value of x such that s(x) = s–1(x). Hence, sketch the graph of s(x) and s–1(x). Katakan x 4 = y x = 4y s–1(x) = 4x s(x) = s–1(x) x 4 = 4x x = 16x 0 = 16x – x 0 = 15x x = 0 (e) Diberi g(x) = 2x + 4 x – 3 , x ≠ 3. Cari nilai bagi 2 7 g–1(–6). Given g(x) = 2x + 4 x – 3 , x ≠ 3. Find the value of 2 7 g–1(–6). Katakan 2x + 4 x – 3 = y 2x + 4 = xy – 3y 2x – xy = –3y – 4 x(2 – y) = –3y – 4 x = –3y – 4 2 – y g–1(x) = –3x – 4 2 – x g–1(–6) = –3(–6) – 4 2 – (–6) = 7 4 2 7 g–1(–6) = 2 7 × 7 4 = 1 2 1. Cari fungsi f bagi f –1(x) = 2 + x x . Seterusnya, cari nilai m apabila f(m) = 3. Find the function of f for f –1(x) = 2 + x x . Hence, find the value of m when f(m) = 3. Katakan 2 + x x = y 2 + x = xy x – xy = –2 x(1– y) = –2 x = –2 1 – y x = 2 y – 1 f(x) = 2 x – 1 2. Diberi fungsi g(x) = 5x – a dan g–1(x) = bx + 2. Cari nilai a dan nilai b. Given the functions g(x) = 5x – a and g–1(x) = bx + 2. Find the values of a and b. Katakan 5x – a = y 5x = y + a x = y + a 5 g–1(x) = x 5 + a 5 … Bandingkan  dengan g–1(x) = bx + 2: b = 1 5 a 5 = 2 a = 10 3. Diberi dua fungsi m : x → ax + b dan n : x → x + 1 x – 2, x ≠ 2. Diberi m–1(11) = n(3) dan mn–1(4) = 9, cari nilai a dan nilai b. Given two functions m : x → ax + b and n : x → x + 1 x – 2, x ≠ 2. Given m–1(11) = n(3) and mn–1(4) = 9, find the values of a and b. KBAT Menganalisis Katakan ax + b = y ax = y – b ax = y – b a m–1(x) = x – b a m–1(11) = n(3) 11 – b a = 3 + 1 3 – 2 11 – b a = 4 11 – b = 4a 4a + b = 11 … Katakan x + 1 x – 2 = y x + 1 = xy – 2y x – xy = –2y – 1 x(1 – y) = –2y – 1 x = –2y – 1 1 – y n–1(x) = –2x – 1 1 – x n–1(4) = –2(4) – 1 1 – 4 = 3 m(3) = 9 3a + b = 9 …  – : a = 2 4(2) + b = 11 b = 11 – 8 b = 3 2 m – 1 = 3 2 = 3m – 3 3m = 2 + 3 m = 5 3 –1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 1 2 3 4 x s(x) s(x) = x — 4 s–1 (x) = 4x 01_Modul A+ MateTam Tg4.indd 12 08/10/2021 11:18 AM

13 BAB 1 Bahagian A/Section A 1. (a) Rajah di bawah menunjukkan fungsi gubahan hk yang memetakan p kepada r. The diagram below shows the composite function hk which maps p to r. KBAT Mengaplikasi p q hk r Nyatakan State (i) fungsi yang memetakan q kepada r. the function that maps q to r. (ii) k –1h–1(r) [2 markah/marks] (b) Diberi fungsi f(x) = 2mx dan g(x) = 3x + n, dengan keadaan m dan n ialah pemalar. Ungkapkan n dalam sebutan m supaya fungsi gubahan gf memetakan 2 kepada dirinya sendiri. Given the functions f(x) = 2mx and g(x) = 3x + n, where m and n are constants. Express n in terms of m such that the composite function gf maps 2 to itself. KBAT Mengaplikasi [3 markah/marks] (a) (i) h (ii) p (b) gf(2) = 2 3(2m(2)) + n = 2 12m + n = 2 n = 2 – 12m 2. (a) Diberi fungsi k : x → 2x + 9, cari Given the function k : x → 2x + 9, find KBAT Mengaplikasi (i) nilai x apabila k(x) = 3x, the value of x when k(x) = 3x, (ii) nilai a dengan keadaan k(a + 9) = a. the value of a such that k(a + 9) = a [4 markah/marks] (b) Diberi bahawa q(x) = x – m n ialah fungsi songsang bagi fungsi k(x) = 2x + 9. Tentukan nilai bagi 2m – n. Given that q(x) = x – m n is the inverse function of the function k(x) = 2x + 9. Determine the value of 2m – n. [2 markah/marks] (a) (i) 2x + 9 = 3x –x = –9 x = 9 (ii) 2(a + 9) + 9 = a 2a + 27 = a a = –27 (b) k : x ˜ 2x + 9 2x + 9 = y x = y – 9 2 k–1(x) = x – 9 2 Jadi, m = 9 dan n = 2. 2m – n = 2(9) – 2 = 16 3. (a) Diberi f : x → 1 – 2mx, g : x → 5x – n dan fg : x → m2 x + n. Cari nilai-nilai m dan n. Given f : x → 1 – 2mx, g : x → 5x – n and fg : x → m2 x + n. Find the values of m and n. KBAT Mengaplikasi [4 markah/marks] (b) Seterusnya, dengan menggunakan nilai m yang negatif, ungkapkan gf(x) menggunakan tatatanda fungsi. Hence, by using the negative value of m, express gf(x) by using function notation. [2 markah/marks] (a) fg(x) = 1 – 2m(5x – n) = 1 – 10mx + 2mn m2 x + n = –10mx + 2mn + 1 Bandingkan pekali x: m2 = –10m m2 + 10m = 0 m = 0 atau m = –10 Bandingkan pemalar: n = 2mn + 1 Jika m = 0 ⇒ n = 1 Jika m = –10 ⇒ n = 2(–10)n + 1 n = –20n + 1 21n = 1 n = 1 21 (b) m = –10 Ú n = 1 21 f(x) = 1 – 2(–10)x = 1 + 20x g(x) = 5x – 1 21 gf(x) = 5(1 + 20x) – 1 21 = 5 + 100x – 1 21 = 100x + 104 21 KLON SPM KLON SPM KLON SPM Soalan Berformat SPM Kertas 1 Paper 1 01_Modul A+ MateTam Tg4.indd 13 08/10/2021 11:18 AM

14 BAB 1 Bahagian B/Section B 4. (a) Rajah di bawah menunjukkan hubungan antara set A, set B dan set C. The diagram below shows the relation among set A, set B and set C. KBAT Mengaplikasi A C B pq(x) = x2 – 2x + 1 Diberi set A dipetakan kepada set B oleh fungsi 2 x + 1 dan dipetakan kepada set C oleh pq(x) = x2 – 2x + 1. Given that set A maps to set B by the function 2 x + 1 and maps to set C by pq(x) = x2 – 2x + 1. (i) Tulis fungsi yang memetakan set A kepada set B dengan menggunakan tatatanda fungsi. Write down the function that maps set A to set B by using the function notation. (ii) Cari fungsi yang memetakan set B kepada set C. Find the function which maps set B to set C. [3 markah/marks] (b) Diberi fungsi s(x) = x – 2 dan rs(x) = 3x2 + x + 1, cari Given the functions s(x) = x – 2 and rs(x) = 3x2 + x + 1, find KBAT Mengaplikasi (i) s –1(x) (ii) r(x) [3 markah/marks] (c) Diberi fungsi m(x) = x + 2, lakarkan graf bagi m–1(x). Given the function m(x) = x + 2, sketch the graph of m–1(x). [2 markah/marks] (a) (i) q : x → 2 x + 1 (ii) p 2 x + 1 = x2 – 2x + 1 Katakan 2 x + 1 = y, x = 2 y –1 p(y) =  2 y – 1 2 – 2 2 y – 1 + 1 = 4 y2 − 4 y + 1 − 4 y + 2 + 1 = 4 y2 – 8 y + 4 p : x → 4 x2 – 8 x + 4 (b) (i) Katakan x – 2 = y, x = y + 2 Maka, s –1(x) = x + 2 (ii) rs(x) = 3x2 + x + 1 r(x – 2) = 3x2 + x + 1 Katakan x – 2 = y, x = y + 2 r(y) = 3(y + 2)2 + (y + 2) + 1 = 3(y2 + 4y + 4) + y + 2 + 1 = 3y2 + 12y + 12 + y + 2 + 1 = 3y2 + 13y + 15 r(x) = 3x2 + 13x + 15 (c) x y = x m(x) m–1(x) y 0 2 –2 –2 2 KLON SPM 5. (a) Rajah di bawah menunjukkan graf bagi fungsi f : x → |2x + 1| untuk domain –3  x  4. The diagram below shows the graph of the function f : x → |2x + 1| for domain –3  x  4. KBAT Mengaplikasi f(x) x 9 (–3, 5) 4 0 Nyatakan/State (i) objek bagi 9, the object of 9, (ii) imej bagi 2, the image of 2, (iii) domain bagi 0  f(x)  5. the domain of 0  f(x)  5. [3 markah/marks] (b) Diberi fungsi f : x → 3x – 7, cari Given the function f : x→ 3x – 7, find KBAT Mengaplikasi (i) f –1(x), (i) nilai k dengan keadaan f 2  3 2k = –1. the value of k such that f 2  3 2k = –1. [3 markah/marks] (c) Diberi p(x) = 5x – 3 7 + x , x ≠ t, nyatakan nilai t. Seterusnya, tentukan p(t). Given that p(x) = 5x – 3 7 + x , x ≠ t, state the value of t. Hence, determine p(t). [2 markah/marks] (a) (i) 4 (ii) 5 (iii) |2x + 1| = 5 2x + 1 = 5 atau 2x + 1 = –5 x = 2 x = –3 Maka, –3  x  2. (b) (i) Katakan 3x – 7 = y x = y + 7 3 Maka, f –1(x) = x + 7 3 (ii) f 2  3 2k = –1 f 3 3 2k – 7 = –1 33 3 2k – 7 – 7= –1 27 2k − 21 − 7 = –1 27 2k = – 1 + 28 2k = 1 k = 1 2 (c) 7 + t = 0 t = –7 Nilai bagi p(t) adalah tidak tertakrif. KLON SPM 01_Modul A+ MateTam Tg4.indd 14 08/10/2021 11:18 AM

15 BAB 1 Bahagian A/Section A 1. (a) Gambar rajah anak panah di sebelah menunjukkan fungsi f. The arrow diagram on the right shows a function f. (i) Nyatakan hubungan yang diwakili oleh f(x) menggunakan tatatanda fungsi. State the relation represented by f(x) by using the function notation. [1 markah/mark] (ii) Tulis hubungan di atas dalam bentuk pasangan tertib. Write down the above relation in the form of ordered pair. [2 markah/marks] (b) Rajah di bawah menunjukkan graf fungsi f(x) = x2 + 6. The diagram below shows the graph of a function f(x) = x2 + 6. KBAT Mengaplikasi x f(x) = x2 + 6 y 0 6 Lakarkan graf songsangan bagi f(x). Seterusnya, nyatakan sama ada songsangan tersebut merupakan suatu fungsi atau bukan. Sketch the graph of the inverse f(x). Hence, state whether the inverse is a function or not. [3 markah/marks] Bahagian B/Section B 2. Diberi bahawa m : x → x + 3 dan n : x → 1 – x. Given that m : x → x + 3 and n : x → 1 – x. KBAT Mengaplikasi (a) Cari Find (i) n(2), (ii) nilai k jika m(k + 2) = 2 3 n(2), the value of k if m(k + 2) = 2 3 n(2), (iii) nm(x). [5 markah/marks] (b) Seterusnya, lakarkan graf y = |nm(x)| untuk –2  x  3. Nyatakan julat bagi y. Hence, sketch the graph y = |nm(x)| for –2  x  3. State the range of y. [3 markah/marks] (c) Tentukan sama ada (nm) –1(x) = nm(x). Determine whether (nm) –1(x) = nm(x). [2 markah/marks] 3. Rajah di bawah menunjukkan fungsi f yang memetakan set K kepada set M dan fungsi g yang memetakan set M kepada set N. The diagram below shows the function f which maps set K to set M and the function g which maps set M to set N. KBAT Mengaplikasi x f K M N g 8x – 5 24x + 15 (a) Cari Find (i) fungsi yang memetakan set M kepada set K, dalam sebutan x, the function which maps set M to set K, in terms of x, [1 markah/mark] (ii) g(x), [3 markah/marks] (iii) nilai x dengan keadaan fg(x) = 18x + 1. the value of x such that fg(x) = 18x + 1. [3 markah/marks] (b) Tentukan f –1g–1(x). Seterusnya, nyatakan hubungan antara f –1g–1(x) dengan (gf )–1(x). Determine f–1g–1(x). Hence, state the relation between f–1g–1(x) and (gf) –1(x). [3 markah/marks] Kertas 2 Paper 2 2 • –2 • –3 • • 5 • 10 x f(x) 01_Modul A+ MateTam Tg4.indd 15 08/10/2021 11:18 AM

21 BAB 2 1. Selesaikan persamaan kuadratik yang berikut dengan menggunakan kaedah penyempurnaan kuasa dua. Berikan jawapan betul kepada dua tempat perpuluhan. Solve the following quadratic equations by using completing the square method. Give answers correct to two decimal places. TP 2 (i) x2 – 6x + 3 = 0 x2 – 6x +  –6 2  2 = –3 +  –6 2  2 x2 –6x + (−3)2 = –3 + (−3)2 (x – 3)2 = –3 + 9 (x – 3)2 = 6 x – 3 = ±AB6 x = 3 ± AB6 x = 5.45 atau/or x = 0.55 Contoh/Example (ii) –2x2 – 4x + 1 = 0 x2 + 2x – 1 2 = 0 x2 + 2x +  2 2  2 = 1 2 +  2 2  2 (x + 1)2 = 1 2 + 1 (x + 1)2 = 3 2 x + 1 = ± 3 2 x = –1 ± 3 2 x = –2.22 atau/or x = 0.22 (a) x2 + 4x + 1 = 0 x2 + 4x +  4 2  2 = –1 +  4 2  2 (x + 2)2 = –1 + 4 (x + 2)2 = 3 x = –2 ± AB3 x = –0.27 atau x = –3.73 (b) x2 + 7x + 3 = 0 x2 + 7x +  7 2  2 = –3 +  7 2  2 x + 7 2  2 = –3 + 49 4 = 37 4 x = – 7 2 ± A37 4 x = –0.46 atau x = –6.54 (c) –x2 + 6x + 4 = 0 x2 – 6x – 4 = 0 x2 – 6x +  –6 2  2 = 4 +  –6 2  2 (x – 3)2 = 4 + 9 = 13 x = 3 ± AB13B x = 6.61 atau x = –0.61 (d) 2x2 – 5x + 2 = 0 x2 – 5 2 x +  –5 4  2 = –1 +  –5 4  2 x – 5 4  2 = –1 + 25 16 = 9 16 x = 5 4 ± A 9 16 x = 0.5 atau x = 2 (e) –2x2 + 6x + 1 = 0 x2 – 3x – 1 2 = 0 x2 – 3x +  –3 2  2 = 1 2 +  –3 2  2 x – 3 2  2 = 1 2 + 9 4 = 11 4 x = 3 2 ± A11 4 x = –0.16 atau x = 3.16 (f) 4x2 – 8x + 2 = 0 x2 – 2x + 1 2 = 0 x2 – 2x +  –2 2  2 = – 1 2 +  –2 2  2 (x – 1)2 = 1 2 x = 1 ± A 1 2 x = 1.71 atau x = 0.29 2.1 Persamaan dan Ketaksamaan Kuadratik Quadratic Equations and Inequalities Buku Teks m.s. 36-44 Tip SPM Perhatikan bahawa Note that (x + a) 2 = x2 + 2ax + a2 Tip SPM 1. Pindahkan sebutan pemalar ke sebelah kanan. Move the constant term to the right hand side. 2. Tambahkan sebutan  pekali x 2  2 di kedua-dua belah persamaan. Add the term  coefficient of x 2  2 to both sides of the equation. Tip SPM Bahagikan kedua-dua belah persamaan dengan –2 supaya pekali bagi x2 adalah 1. Divide both sides of the equation by –2 so that the coefficient of x2 is 1. 02_Modul A+ MateTam Tg4.indd 21 08/10/2021 11:31 AM

MG-1 BAB 1 Kertas 2/Paper 2 Bahagian A/Section A 1. (a) (i) f(x) = x2 + 1 (ii) (2, 5), (–2, 5), (–3, 10) (b) x y = x y 0 6 6 ∴ Bukan fungsi. Hubungan satu kepada banyak. Not a function. One to many relation. Bahagian B/Section B 2. (a) (i) n(2)= 1 – 2 = –1 (ii) m(k + 2) = 2 3 n(2) k + 2 + 3 = 2 3 (–1) k = – 17 3 (iii) nm(x) = 1 – (x + 3) = 1 – x – 3 = –2 – x (b) –1 1 2 3 4 5 6 –1 1 2 3 0 –2 y x 0  y  5 (c) nm(x) = –2 – x Katakan/Let –2 – x = y –2 – y = x Jadi/Hence, (nm) –1(x) = –2 – x ∴ (nm) –1(x) = nm(x) 3. (a) (i) Katakan/Let 8x – 5 = y x = y + 5 8 f –1(x) = x + 5 8 (ii) gf(x) = 24x + 15 g[8x – 5] = 24x + 15 g( y) = 24 y + 5 8  + 15 g( y) = 3(y + 5) + 15 g( y) = 3y + 30 \ g(x) = 3x + 30 (iii) fg(x) = 18x + 1 8[3x + 30] – 5 = 18x + 1 24x + 240 – 5 – 18x – 1 = 0 6x + 234 = 0 x = –39 (b) f(x) = 8x – 5 gf(x) = 24x + 15 g[8x – 5] = 24x + 15 Katakan/Let 8x – 5 = y x = y + 5 8 g(y) = 24[ y + 5 8 ] + 15 = 3y + 15 + 15 = 3y + 30 f –1(x) = x + 5 8 g–1(x) = x – 30 3 f –1g–1(x) = x – 30 3 + 5 8 = x – 30 + 15 24 = x – 15 24 ∴ f –1g–1(x) = (gf) –1(x) BAB 2 Kertas 2/Paper 2 Bahagian A/Section A 1. (a) (i) h + k = b a hk = – c a c b = c a × a b = –hk 1 h + k = –hk h + k (ii) a + b + c c = a c + b c + 1 = –1 hk + h + k –hk + 1 = –1 – h – k + hk hk = 1 –  1 + h + k hk  (b) 1 k + 1 h = h + k hk = b a ×  –a c  = – b c  1 k  1 h  = 1 hk = – a c x2 + b c x – a c = 0 cx2 + bx – a = 0 2. (a) 2x2 + px – 2x + 8 = 0 (p – 2)2 – 4(2)(8) > 0 p2 – 4p + 4 – 64 > 0 p2 – 4p – 60 > 0 ( p + 6)(p – 10) > 0 p < –6 atau/or p > 10 (b) α 2 + β 4 = –k 2 αβ 4 = 2 2 α + β = –k αβ = 4 –k = –( p – 2) 2 –2k = –p + 2 p = 2 + 2k 3. (a) f(x) = –(x – h) 2 + 2k –2 = –h2 + 2k … 1 2k = –1 k = – 1 2 h2 = 2 – 1 2  + 2 h2 = 1 h = ±1 (b) f(x) = –(x – 1)2 – 1 0 1 –1 –2 (1, –1) x f(x) (c) f(x) = (x – 1)2 + 1 4. (a) Katakan α dan β ialah puncapunca bagi x2 + px + q = 0. Let α and β are the roots of x2 + px + q = 0. α + β = –p ..... 1 α – β = 23 .....  αβ = q = 6 1 + : 2α = 23 – p ..... 3 1 – : 2β = –p – 23 ..... 4 3 × 4: 4ab = –(23 – p)(p + 23) 4(6) = –(12 – p2 ) 24 = –12 + p2 p2 = 36 p = ±6 q = 6 (b) Apabila/When p > 0, p = 6 Daripada/From 3, α = 23 – 6 2 = 3 – 3 Daripada/From 4, β = –6 – 23 2 = –3 – 3 12_JawG_Modul A+ MateTam Tg4.indd 1 27/10/2021 11:15 AM

MG-11 BAB 1 Fungsi Tahap Penguasaan Tafsiran 1 Mempamerkan pengetahuan asas tentang fungsi. 2 Mempamerkan kefahaman tentang fungsi. 3 Mengaplikasikan kefahaman tentang fungsi untuk melaksanakan tugasan mudah. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah rutin yang mudah. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. Gambar rajah Venn di bawah menunjukkan suatu fungsi f yang memetakan x kepada y. The Venn diagram below shows a function f that maps x to y. TP 1 a • b • c • • r • s • t • u x y f Nyatakan/State (a) domain/the domain, (b) kodomain/the codomain, (c) objek/the object, (d) imej/the image, (e) julat/the range, (f) jenis hubungan tersebut/the type of relation. (a) {a, b, c} (b) {r, s, t, u} (c) a, b, c (d) r, s, t, u (e) {r, t} (f) Banyak kepada satu/Many to one 2. Suatu fungsi h(x) = 2x – 3 memetakan x kepada dirinya sendiri. Cari nilai x. A function h(x) = 2x – 3 maps x to itself. Find the value of x. TP 2 2x – 3 = x 2x – x = 3 x = 3 3. Diberi fungsi f : x ˜ 1 x dan fungsi g : x ˜ x2 . Adakah fungsi gubahan fg sama dengan gf? Given the functions f : x ˜ 1 x and g : x ˜ x2 . Is the composite functions of fg the same as gf? TP 2 fg(x) = 1 x2 gf(x) = ( 1 x ) 2 Ya, fg sama dengan gf. Yes, fg is the same as gf. Lembaran Pentaksiran Bilik Darjah (PBD) Lemb PBD Modul A+ MateTam Tg4.indd 11 27/10/2021 11:15 AM

MG-24 Lembaran PBD BAB 6 Hukum Linear Tahap Penguasaan Tafsiran 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah rutin yang mudah. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang hukum linear dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. Satu eksperimen untuk mengkaji kekerasan spring dijalankan dengan menggunakan pemberat, M yang berlainan jisim. Pemanjangan spring, x m, diperoleh dengan menggantungkan pemberat-pemberat itu dan keputusan yang diperoleh ditunjukkan dalam jadual di bawah. An experiment to investigate the stiffness of the spring with different masses, M was carried out. The extension of the spring, x m, that is obtained with the weights and the results are shown in the table below. TP 5 M (kg) 0.1 0.2 0.3 0.4 0.5 x (m) 0.01 0.023 0.028 0.042 0.051 (a) Diberi M = kx, dengan keadaan k ialah pemalar spring, plot M melawan x. Given that M = kx, where k is the spring constant, plot M against x. (b) Menggunakan graf di (a), Using the graph in (a), (i) tentukan pemalar spring, determine the spring constant, (ii) jika spring telah diregangkan sebanyak 65 cm, berapakah jisim pemberat yang digantungkan itu? if the spring is extended by 65 cm, what is the mass of the weight that was hung on it? (iii) hitung tenaga yang diperlukan untuk meregangkan spring sebanyak 40 cm jika tenaga yang diberi ialah 1 2 kx2 joule. calculate the energy needed to extend the spring by 40 cm if the energy is given by the formula 1 2 kx2 joule. (a) M 0.5 0.4 0.3 0.2 0.1 0 0.01 0.02 0.03 0.04 0.05 0.06 x (b) (i) Kecerunan/Gradient, k = 0.5 0.051 – 0 = 9.8 kg m–1 (ii) Apabila/When x = 0.65 m, M = 9.8(0.65) = 6.37 kg (iii) Apabila/When x = 0.4 m, Tenaga/Energy = 1 2 (9.8)(0.4)2 = 0.784 J Lemb PBD Modul A+ MateTam Tg4.indd 24 27/10/2021 11:15 AM

Dapatkan sekarang! Ingin menjadi penulis kami? Sertai kami dengan menghantar e-mel ke alamat author@panasiapub.com. www.panasiapub.com 199101016590 (226902-X) ISBN 978-967-466-628-6 Sem. M’sia RM 11.90 Sabah/Sarawak RM 12.90 Mata Pelajaran / Tingkatan 4 5 Sejarah Matematik Matematik Tambahan Kimia Fizik Perniagaan Ekonomi English Judul-judul dalam siri Modul A+1 Siri MODUL A+1 ini menyediakan modul pembelajaran komprehensif yang dihasilkan berdasarkan Dokumen Standard Kurikulum dan Pentaksiran (DSKP). Modul ini telah dirancang dan ditulis oleh guru-guru yang berpengalaman dalam membantu proses PdPc dengan lebih efektif. Latihan yang disediakan mencakupi Tahap Penguasaan yang perlu dikuasai oleh murid untuk mengoptimumkan kefahaman mereka. AKSES DIGITAL ■ Jawapan Lengkap Soalan Berformat SPM ■ Lembaran Pentaksiran Bilik Darjah (PBD) 9 789674 666286