Matematik T5

Matematik Tingkatan 5 Bab 6 80 3 Rajah yang berikut menunjukkan satu bulatan unit dengan sudut θ. Tentukan nilai sin θ, kos θ dan tan θ. SP: 6.1.1 TP2 The following diagrams show a unit circle with angle θ. Determine the values of sin θ, cos θ and tan θ. Mudah (a) (–0.2588, 0.9659) θ x y O sin θ = 0.9659 kos/ cos θ = –0.2588 tan θ = 0.9659 –0.2588 = –3.7322 (b) θ x y O (–0.8192, –0.5736) sin θ = –0.5736 kos/ cos θ = –0.8192 tan θ = –0.5736 –0.8192 = 0.7 (c) (0.2588, –0.9659) θ x y O sin θ = –0.9659 kos/ cos θ = 0.2588 tan θ = –0.9659 0.2588 = –3.7322 (d) (–0.8660, 0.5) θ y x O sin θ = 0.5 kos/ cos θ = –0.866 tan θ = 0.5 –0.866 = –0.577 (e) (–0.6428, 0.7660) θ x y O sin θ = 0.766 kos/ cos θ = –0.6428 tan θ = 0.766 –0.6428 = –1.1917 θ (0.8660, 0.5) x y O sin θ = 0.5 kos/ cos θ = 0.866 tan θ = 0.5 0.866 = 0.577 Contoh 4 Cari nilai bagi setiap yang berikut dengan menggunakan sudut rujukan sepadan. Tulis jawapan anda dalam 3 tempat perpuluhan. SP: 6.1.2 TP3 Mudah Find the value for each of the following using the corresponding reference angle. Write your answer in 3 decimal places. (a) sin 359° = –sin (360° – 359°) = –sin 1° = –0.017 (b) tan 234° 21ʹ = tan (234° 21ʹ – 180°) = tan 54° 21ʹ = 1.394 (c) sin 115° = sin (180° – 115°) = sin 65° = 0.906 (d) tan 165° = –tan (180° – 165°) = –tan 15° = –0.268 (e) kos/ cos 325° = kos/ cos (360° – 325°) = kos/ cos 35° = 0.819 kos/ cos 145° = –kos/ cos (180° – 145°) = –kos/ cos 35° = –0.819 Contoh Tip Bestari tan θ = koordinat-y/ y-coordinate koordinat-x/ x-coordinate Penggunaan Kalkulator CONTOH

Matematik Tingkatan 5 Bab 6 81 5 Tentukan nilai bagi setiap yang berikut tanpa menggunakan kalkulator saintifik. SP: 6.1.2 TP3 Determine the value for each of the following without using scientific calculator. Sederhana (a) sin 135° = sin (180° – 135°) = sin 45° = 1 2 (b) kos/ cos 315° = kos/ cos (360° – 315°) = kos/ cos 45° = 1 2 (c) tan 120° = –tan(180° – 120°) = –tan 60° = – 3 (d) sin 240° = –sin (240° – 180°) = –sin 60° = – 3 2 (e) kos/ cos 300° = kos/ cos (360° – 300°) = kos/ cos 60° = 1 2 (f) tan 225° = tan (225° – 180°) = tan 45° = 1 (g) sin 330° = –sin (360° – 330°) = –sin 30° = –1 2 (h) kos/ cos 150° = –kos/ cos (180° – 150°) = –kos/ cos 30° = – 3 2 tan 210° = tan (210° – 180°) = tan 30° = 1 3 Contoh 6 Cari semua nilai yang mungkin bagi sudut θ dengan keadaan 0° ⩽ θ ⩽ 360°. Tulis jawapan anda dalam 1 tempat perpuluhan. SP: 6.1.3 TP3 Find all the possible values of angle θ where 0° ⩽ θ ⩽ 360°. Write your answers in 1 decimal place. Sederhana (a) tan θ = 5.6713 tan–1 5.6713 = 80° θ = 80° atau/ or (180° + 80°) = 80° atau/ or 260° (b) tan θ = –0.1228 tan–1 0.1228 = 7° θ = (180° – 7°) atau/ or (360° – 7°) = 173° atau/ or 353° (c) kos/ cos θ = –0.7193 kos–1/ cos–1 0.7193 = 44° θ = (180° – 44°) atau/ or (180° + 44°) = 136° atau/ or 224° (d) sin θ = 0.9063 sin–1 0.9063 = 65° θ = 65° atau/ or (180° – 65°) = 65° atau/ or 115° (e) kos/ cos θ = 0.9659 kos–1/ cos–1 0.9659 = 15° θ = 15° atau/ or (360° – 15°) = 15° atau/ or 345° (f) sin θ = –0.1736 sin–1 0.1736 = 10° θ = (180° + 10°) atau/ or (360° – 10°) = 190° atau/ or 350° (g) kos/ cos θ = 0 kos–1/ cos–1 0 = 90° θ = 90° atau/ or (360° – 90°) = 90° atau/ or 270° kos/ cos θ = –0.9976 kos–1/ cos–1 0.9976 = 3.97° θ = (180° – 3.97°) atau/ or (180° + 3.97°) = 176° atau/ or 184.0° Sudut berada di sukuan II dan III. The angles are in quadrants II and III. Cari sudut rujukan sepadan. Find the corresponding reference angle. Contoh Penggunaan Kalkulator CONTOH

Matematik Tingkatan 5 Bab 6 82 7 Selesaikan setiap masalah yang berikut. SP: 6.1.4 TP5 TP6 Solve each of the following problems. Sukar (a) Rajah di bawah menunjukkan sebuah segi tiga bersudut tegak ABD. ABC ialah satu garis lurus. The diagram below shows a right-angled triangle ABD. ABC is a straight line. Diberi AC = 20 cm dan AB : BC = 1 : 1.5. Cari Given that AC = 20 cm and AB : BC = 1 : 1.5. Find (i) sin z, (ii) kos/ cos x, (iii) tan y. AB AC = 1 1 + 1.5 AB 20 cm = 1 2.5 AB = 8 cm BD2 = AB2 + AD2 = 82 + 62 = 100 BD = 100 = 10 cm (i) sin z = 6 10 = 3 5 (ii) kos/ cos z = 8 10 = 4 5 kos/ cos x = –kos/ cos z = – 4 5 (iii) 270° < y < 360°, maka nilai tan y adalah negatif. 270° < y < 360°, thus the value of tan y is negative. tan y = – 8 6 = – 4 3 (b) PQR ialah sebuah segi tiga dengan keadaan PQ adalah berserenjang dengan QR. QR dipanjangkan ke S. Jika PQ = RS = 4 cm dan PR = 5 cm, cari PQR is a triangle where PQ is perpendicular to QR and QR is extended to S. If PQ = RS = 4 cm and PR = 5 cm, find (i) tan ∠PRQ, (ii) kos/ cos ∠QSP, seterusnya/ hence, (iii) ∠QSP. QR2 = PR2 – PQ2 = 52 – 42 = 9 QR = 9 = 3 cm 4 cm 4 cm P Q R S 5 cm (i) tan ∠PRQ = 4 3 (ii) PS2 = PQ2 + QS2 PS = 65 = 42 + (3 + 4)2 = 8.06 cm = 65 kos/ cos ∠QSP = 7 8.06 (iii) kos–1/ cos–1 7 8.06 = 29.72° atau/ or 29° 43ʹ (c) Diberi tan θ = 0.5774 dan sin θ = 0.5 dengan keadaan 0° ⩽ θ ⩽ 360°. Given that tan θ = 0.5774 and sin θ = 0.5 where 0° ⩽ θ ⩽ 360°. (i) Hitung nilai bagi kos θ. Nyatakan jawapan dalam empat tempat perpuluhan. Calculate the value of cos θ. State your answer in four decimal places. (ii) Wakilkan nilai-nilai trigonometri tersebut dalam bentuk bulatan unit. Represent the trigonometric values in the form of unit circle. (iii) Takrifkan nilai-nilai kos θ di tiga sukuan yang lain. Define the value of cos θ for the three other quadrant. (i) tan θ = sin θ kos/cos θ (ii) 0.5774 = 0.5 kos/cos θ kos/ cos θ = 0.5 0.5774 = 0.8660 (iii) Sukuan Quadrant Nilai kos θ The value of cos θ II –0.8660 III –0.8660 IV 0.8660 K B A T (0.8660, 0.5) y x O θ y D A B C x z 6 cm Soalan KBAT ekstra CONTOH

Matematik Tingkatan 5 Bab 6 83 6.2 Graf Fungsi Sinus, Kosinus dan Tangen The Graphs of Sine, Cosine and Tangent Functions Buku Teks m/s 184 – 191 1 Nyatakan jenis graf trigonometri bagi setiap yang berikut. TP1 State the type of trigonometric graph for each of the following. Mudah (a) x y 90° 180° 270° 1 0 –1 Graf kosinus/ Cosine graph (b) y x 180° 270° 360° 1 0 –1 Graf tangen/ Tangent graph (c) y x 90° 180° 270° 1 0 –1 Graf sinus/ Sine graph Contoh Graf sinus/ Sine graph x 180° 270° 360° 1 0 –1 y 2 Lakarkan graf bagi setiap yang berikut. SP: 6.2.1 TP2 Sketch the graph for each of the following. Mudah (a) y = kos/ cos x, 0° ⩽ x ⩽ 180° 0 1 –1 90° 180° y = kos/ cos x y x (b) y = tan x, 90° ⩽ x ⩽ 270° 90° 180° 270° y x 0 y = tan x (c) y = sin x, 30° ⩽ x ⩽ 330° 0 1 30° 90° –0.5 0.5 –1 180° 270° 330° y = sin x y x Contoh y = sin x, 0° ⩽ x ⩽ 180° y = sin x y 0 1 90° 180° x Info Digital 6.2 CONTOH

Matematik Tingkatan 5 Bab 6 84 3 Diberi graf fungsi trigonometri dengan keadaan 0° ⩽ x ⩽ 360°. Lengkapkan jadual berikut dengan menyatakan pintasan-x, pintasan-y, nilai maksimum dan nilai minimum. SP: 6.2.1 TP2 Given the graphs of trigonometric function such that 0° ⩽ x ⩽ 360°. Complete the following table by stating the x-intercept, the y-intercept, the maximum value and the minimum value. Mudah Fungsi trigonometri Trigonometric function y = sin x y = kos x y = cos x y = tan x Pintasan-x x-intercept 0°, 180°, 360° (a) 90°, 270° (b) 0°, 180°, 360° Pintasan-y y-intercept (c) 0 (d) 1 0 Nilai maksimum Maximum value (e) 1 1 (f) ∞ Nilai minimum Minimum value –1 (g) –1 –∞ 4 Tentukan amplitud dan tempoh bagi setiap fungsi trigonometri yang berikut. SP: 6.2.2 TP3 Determine the amplitude and period for each of the following trigonometric functions. Sederhana Fungsi trigonometri Trigonometric function Amplitud Amplitude Tempoh Period (a) y = 2 kos/ cos x 2 360° (b) y = tan 1 3 x 1 540° (c) y = 2 kos/ cos 2x – 4 2 180° (d) y = 1 7 tan 3x + 1 1 7 60° (e) y = 4 sin x 4 360° (f) y = 2 sin 2x 3 2 540° (g) y = kos/ cos 2x + 3 1 180° y = 1 3 sin 2x 1 3 180° Contoh Tip Bestari Dengan/ Where A = Amplitud/ Amplitude P = Tempoh/ Period y = tan x 0° 180° 360° 1P 1P A A 0° 360° y = kos/ cos x 1P A A 0° 180° 360° y = sin x y y y x x x CONTOH

Matematik Tingkatan 5 Bab 6 85 5 Rajah menunjukkan sebahagian graf fungsi trigonometri dengan keadaan 0° ⩽ x ⩽ 360°. Cari nilai p dan q bagi setiap yang berikut. SP: 6.2.1 TP3 The diagrams show part of a graph of trigonometric function such that 0° ⩽ x ⩽ 360°. Find the values of p and q for each of the following. Sederhana (a) y = sin x 0° 20° + q p x y p = 180° 20° + q = 90° q = 70° (b) kos/ cos x p p + q x y p = 270° p + q = 360° 270° + q = 360° q = 90° (c) y = sin x q – p q x y q = 270° q – p = 180° 270° – p = 180° p = 90° Contoh p + 36° = 90° q – 90° = 270° p = 54° q = 360° y = tan x p + 36° q – 90° y x 6 Lakar setiap fungsi sinus yang berikut bagi 0° ⩽ x ⩽ 360° dan nyatakan perubahan graf daripada graf y = sin x. SP: 6.2.2 TP3 Sketch each of the following sine functions for 0° ⩽ x ⩽ 360° and state the changes of graph from the graph of y = sin x. Sederhana (a) y = 3 sin x y = 3 sin x y x –3 3 90° 180° 270° 360° Amplitud telah menjadi 3 The amplitude has become 3 Contoh y = sin 2x Tempoh fungsi telah menjadi 180° The period of the function has become 180° y = sin 2x 1 y –1 45° 90° 135° 180°225° 270° 315° 360° x Graf y = sin x Graph of y = sin x CONTOH

Matematik Tingkatan 5 Bab 6 86 (b) y = sin x + 2 y = sin x + 2 y x 3 2 1 0 90° 180° 270° 360° (c) y = 1 2 sin x + 3 y = 1 2 sin x + 3 y x 3.5 2.5 0 90° 180° 270° 360° 3 Graf telah anjak 3 unit ke atas dan amplitud telah menjadi 1 2 The graph has shifted up by 3 units and the amplitude has become 1 2 Graf telah anjak 2 unit ke atas The graph has shifted up by 2 units 7 Lakar setiap fungsi kosinus yang berikut bagi 0° ⩽ x ⩽ 360° dan nyatakan perubahan graf daripada graf y = kos x. SP: 6.2.2 TP3 Sketch each of the following cosine functions for 0° ⩽ x ⩽ 360° and state the changes of graph from the graph of y = cos x. Sederhana (a) y = kos/ cos x – 1 y = kos/ cos x – 1 0 180° 360° –2 y x Graf telah anjak 1 unit ke bawah The graph shifted down by 1 unit (b) y = kos/ cos 2x y = kos/ cos 2x y x 1 0 –1 45° 90° 135° 180° 225° 270° 315° 360° Tempoh fungsi telah menjadi 180° The period of the function has become 180° (c) y = 3 kos/ cos 2x y = 3 kos/ cos 2x y 3 0 45° 90° 135° 180° 225° 270° 315° 360° –3 x Amplitud telah menjadi 3 dan tempoh fungsi telah menjadi 180° The amplitude has become 3 and the period of the function has become 180° Contoh y = 2 kos/ cos x Amplitud telah menjadi 2 The amplitude has become 2 y = 2 kos/ cos x x y 2 0 90° 180° 270° 360° –2 Graf y = kos x Graph of y = cos x CONTOH

Matematik Tingkatan 5 Bab 6 87 8 Lakar setiap fungsi tangen yang berikut bagi 0° ⩽ x ⩽ 360° dan nyatakan perubahan graf daripada graf y = tan x. SP: 6.2.2 TP3 Sketch each of the following tangent functions for 0° ⩽ x ⩽ 360° and state the changes of graph from the graph of y = tan x. Sederhana (a) y = tan 2x y = tan 2x y 0° 45° 90° 135° 180° 225° 270° 315° 360° x Tempoh fungsi telah menjadi 90° The period of the function has become 90° (b) y = 4 tan x y = 4 tan x y x 0° 90° 180° 270° 360° Tiada perubahan kerana fungsi tangen tiada amplitud There is no change because tangent function does not have amplitude (c) y = 2 tan 2x – 2 y x y = 2 tan 2x – 2 –2 0° 45° 90° 135°180° 225° 270° 315° 360° Graf telah anjak 2 unit ke bawah dan tempoh fungsi telah menjadi 90° The graph has shifted down by 2 units and the period of the function has become 90° Contoh y = tan x + 1 Graf telah anjak 1 unit ke atas The graph has shifted up by 1 unit y = tan x + 1 y 1 0 90° 180° 270° 360° x 9 Selesaikan masalah yang berikut. SP: 6.2.3 TP4 TP5 Solve the following problems. Sukar (a) Graf di sebelah mewakili suatu fungsi trigonometri. The graph beside represents a trigonometric function. Tentukan amplitud dan tempoh fungsi itu dan seterusnya tuliskan satu fungsi yang diwakili oleh graf tersebut. Determine the amplitude and period of the function and hence write a function represented by the graph. Daripada graf/ From the graph, graf telah anjak 1 unit ke bawah, maka c = –1; the graph has moved down by 1 unit, hence c = –1; Amplitud fungsi ialah 2, maka a = 2 ∴ Bentuk graf merupakan graf kosinus, maka fungsi graf ialah y = 2 kos 2x – 1. The shape of the graph shows a cosine graph; hence the function of the graph is y = 2 cos 2x – 1. The amplitude of the function is 2, hence a = 2 tempoh fungsi ialah 180°, maka the period of the function is 180°, hence 360° b = 180° b = 2 y 1 0 45° 90° 135° 180° 225° 270° 315° 360° –1 –3 x Graf y = tan x Graph of y = tan x CONTOH

Matematik Tingkatan 5 Bab 6 88 (b) Graf di bawah menunjukkan aras air yang dicatatkan dari masa 0000 jam hingga 0800 jam di sebuah limbungan kapal. The graph below shows the water level recorded from 0000 hour to 0800 hour at a shipyard. (i) Tuliskan satu fungsi yang memodelkan aras air, h, dalam sebutan t. Write a function that models the water level, h, in terms of t. (ii) Nyatakan aras air pada pukul 3 pagi. State the water level at 3 a.m. (iii) Cari beza antara aras air yang paling tinggi dan paling rendah. Find the difference between the highest and the lowest water level. (i) Berdasarkan bentuk graf, graf mewakili fungsi sinus, maka h = a sin bt + c. Based on the shape of the graph, the graph represents a sine function, hence h = a sin bt + c. Daripada graf, amplitud adalah 1, graf menunjukkan dua tempoh (b = 2) dan graf telah anjak 4 unit ke atas (c = +4). From the graph, the amplitude is 1, the graph shows two periods (b = 2) and has moved up by 4 units (c = +4). ∴ Fungsi yang memodelkan aras air ialah h = sin 2t + 4. The function that models the water level is h = sin 2t + 4. (ii) Pada pukul 3 pagi, aras air adalah 3 m. At 3 a.m., the water level is 3 m. (iii) Beza/ Difference: 5 – 3 = 2 m Kertas 1 Jawab semua soalan./ Answer all questions. 1 Antara graf berikut, yang manakah mewakili y = 2 sin x, 0° ⩽ x ⩽ 360°? Which of the following graphs represents y = 2 sin x, 0° ⩽ x ⩽ 360°? A y 1 0 90° 180° 270° 360° x –1 B y 2 0 90° 180° 270° 360° x –2 C y 2 0 90° 180° x D x y 1 0 90° 180° 2 Tentukan tempoh bagi fungsi y = tan x, 0° ⩽ x ⩽ 360°. Determine the period for the function y = tan x, 0° ⩽ x ⩽ 360°. A 90° C 270° B 180° D 360° 3 Antara berikut, yang manakah tidak mempunyai sudut rujukan sepadan 50°? Which of the following does not have a corresponding reference angle of 50°? A 130° C 300° B 230° D 310° 4 Cari nilai bagi tan 320°. Find the value of tan 320°. A –0.8391 B –0.0040 C 0.0040 D 0.8391 5 Antara berikut, yang manakah bukan nilai bagi 60°? Which of the following is not a value of 60°? A sin θ = 0.8661 B kos/ cos θ = 0.5 C tan θ = 1.7321 D sin θ = 0.9848 Aras air, h (m) Water level, h (m) 5 4 3 0000 0 0100 0300 0200 0400 0600 0800 0500 0700 Waktu, t (jam) Time, t (hour) Praktis Kendiri CONTOH

Matematik Tingkatan 5 Bab 6 89 6 Antara berikut, yang manakah pintasan-x bagi fungsi y = 2 kos x, 0° ⩽ x ⩽ 360°? Which of the following is the x-intercept of the function y = 2 cos x, 0° ⩽ x ⩽ 360°? A 90° dan/ and 270° B 0° dan/ and 180° C 0° dan/ and 360° D 90° dan/ and 180° 7 Rajah berikut menunjukkan satu bulatan unit dengan sudut θ. The following diagram shows a unit circle with angle θ. (0.4391, 0.8984) θ O x y Hitung nilai bagi tan θ. Calculate the value of tan θ. A 2.046 B 0.671 C –0.328 D –0.671 8 Diberi sudut rujukan sepadan ialah 45°, apakah nilai sudut yang berada di sukuan IV? Given that the corresponding reference angle is 45°, what is the value of the angle in quadrant IV? A 135° C 315° B 225° D 405° 9 Rajah berikut menunjukkan satu bulatan unit dengan sudut θ. The following diagram shows a unit circle with angle θ. (–0.9457, p) y x O θ Cari nilai bagi p. Find the value of p. A 0.945 C –0.325 B 0.325 D –0.945 10 Apakah amplitud bagi graf y = 2 kos 3x + 1? What is the amplitude for graph y = 2 cos 3x + 1? A 1 C 3 B 2 D 4 Kertas 2 Jawab semua soalan./ Answer all questions. Bahagian A/ Section A 1 Rajah menunjukkan dua buah segi tiga bersudut tegak yang kongruen, PQR dan TSR. The diagram shows two congruent right-angled triangles, PQR and TSR. P T S R Q 3 mm 53.13° Diberi PR = RT, PQ = 3 mm dan sudut RTS = 53.13°. Given PR = RT, PQ = 3 mm and angle RTS = 53.13°. (a) Hitung panjang, dalam cm, garis QS. Calculate the length, in cm, of line QS. [2 markah/ marks] (b) Cari sudut PRS. Find the angle of PRS. [2 markah/ marks] Jawapan/ Answer: (a) PQ = TS = 3 mm tan ÐRTS = RS TS tan 53.13° = RS 3 RS = 4 mm QR = RS = 4 mm ∴ QS = 8 mm (b) ÐPRS = ÐQRT tan ÐQRT = – 3 4 ÐQRT = –tan–1 3 4 = 180° – 36.87° = 143.13° CONTOH

Matematik Tingkatan 5 Bab 6 90 Bahagian B/ Section B 2 (a) (i) Lakar graf bagi y = sin 2x untuk 0° ⩽ x ⩽ 360°. Draw the graph of y = sin 2x for 0° ⩽ x ⩽ 360°. (ii) Nyatakan amplitud dan tempoh bagi graf y = sin 2x. State the amplitud and period of the graph y = sin 2x. [4 markah/ marks] (b) (i) Lakar graf bagi y = sin 2x yang bergerak secara mencancang 1 unit ke atas pada paksi yang sama. Sketch the graph for y = sin 2x which moves up vertically by 1 unit on the same axes. (ii) Seterusnya, nyatakan fungsi baharu itu. Hence, state the new function. [4 markah/ marks] Jawapan/ Answer: (a) (i), (ii) (b) (i) Tempoh/ Period: 360° 2 = 180° ∴ Amplitud bagi y = sin 2x ialah 1 dan tempoh fungsi ialah 180°. The amplitude for y = sin 2x is 1 and the period of the function is 180°. y 2 1 0 –1 y = sin 2x + 1 x 45° 90° 135° 180° 225° 270° 315° 360° y = sin 2x (ii) c = +1, maka fungsi baharu ialah y = sin 2x + 1. c = +1, hence the new function is y = sin 2x + 1. 3 (a) Lakar graf y = tan 2x dengan keadaan 0° ⩽ x ⩽ 360° dan seterusnya, nyatakan nilai amplitud dan tempoh bagi graf fungsi itu. Sketch the graph y = tan 2x where 0° ⩽ x ⩽ 360° and hence, state the value of the amplitude and period of the graph of the function. [3 markah/ marks] (b) (i) Lakar graf bagi y = tan 2x yang anjak 2 unit ke bawah pada paksi yang sama. Sketch the graph for y = tan 2x which moves down by 2 units on the same axes. (ii) Seterusnya, nyatakan fungsi bagi graf baharu itu. Hence, state the function of the new graph. [4 markah/ marks] Jawapan/ Answer: (a), (b)(i) Tempoh graf ialah/ The period of the graph is 180° 2 = 90°. Daripada graf, amplitud bagi y = tan 2x adalah tidak tertakrif dan tempoh fungsi ialah 90°. From the graph, the amplitude of y = tan 2x is undefined and the period of the function is 90°. y 0 –2 45° 90° 135° 180° 225° 270° 315° 360° x y = tan 2x y = tan 2x – 2 (b) (ii) c = –2 apabila graf anjak 2 unit ke bawah, maka fungsi bagi graf baharu ialah y = tan 2x – 2. c = –2 when the graph moves down by 2 units, hence the function of the new graph is y = tan 2x – 2. CONTOH

91 Praktis Intensif 7.1 Serakan Dispersion Buku Teks m/s 198 – 210 1 Tentukan saiz selang kelas bagi setiap set data yang berikut berdasarkan bilangan kelas yang diberi. Seterusnya, bina satu jadual kekerapan yang lengkap dengan selang kelas, had bawah, had atas, titik tengah, sempadan bawah dan sempadan atas. SP: 7.1.1 TP1 TP2 Mudah Determine the size of class interval for each of the following sets of data based on the given number of classes. Hence, construct a frequency table complete with class intervals, lower limit, upper limit, midpoint, lower boundary and upper boundary. Contoh Rajah di bawah menunjukkan markah yang diperoleh sekumpulan murid dalam satu kuiz Matematik. The diagram below shows the marks obtained by a group of students in a Mathematics quiz. 80 60 45 72 56 82 90 99 53 18 89 62 61 99 72 11 67 91 79 69 79 31 50 70 35 55 37 68 Bilangan kelas/ Number of classes = 5 Saiz selang kelas/ Size of class interval: 99 – 11 5 = 17.6 (≈ 20) Markah Marks Kekerapan Frequency Had bawah Lower limit Had atas Upper limit Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 1 – 20 2 1 20 10.5 0.5 20.5 21 – 40 3 21 40 30.5 20.5 40.5 41 – 60 6 41 60 50.5 40.5 60.5 61 – 80 11 61 80 70.5 60.5 80.5 81 – 100 6 81 100 90.5 80.5 100.5 Rajah di bawah menunjukkan jisim, dalam g, epal di dalam sebuah kotak. The diagram below shows the mass, in g, of apples in a carton. 100 89 93 88 94 100 102 97 91 104 81 93 99 101 99 105 98 99 102 100 Bilangan kelas/ Number of classes = 5 Saiz selang kelas/ Size of class intervals: 105 – 81 5 = 4.8 (≈ 5) Jisim (g) Mass (g) Kekerapan Frequency Had bawah Lower limit Had atas Upper limit Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 81 – 85 1 81 85 83 80.5 85.5 86 – 90 2 86 90 88 85.5 90.5 91 – 95 4 91 95 93 90.5 95.5 96 – 100 8 96 100 98 95.5 100.5 101 – 105 5 101 105 103 100.5 105.5 Sukatan Serakan Data Terkumpul 7 Measures of Dispersion for Grouped Data Bab Info Digital 7.1 CONTOH

Matematik Tingkatan 5 Bab 7 92 2 Lengkapkan setiap jadual kekerapan longgokan yang berikut. SP: 7.1.1 TP2 Complete each of the following cumulative frequency tables. Mudah (a) (b) (c) Tinggi (cm) Height (cm) Bilangan murid Number of students Kekerapan longgokan Cumulative frequency 150 – 154 3 3 155 – 159 4 7 160 – 164 12 19 165 – 169 12 31 170 – 174 1 32 175 – 179 1 33 Jisim (g) Mass (g) Bilangan murid Number of students Kekerapan longgokan Cumulative frequency 46 – 50 13 13 51 – 55 10 23 56 – 60 10 33 61 – 65 4 37 66 – 70 2 39 71 – 75 1 40 Wang saku (RM) Pocket money (RM) Bilangan murid Number of students Kekerapan longgokan Cumulative frequency 1.00 – 2.90 1 1 3.00 – 4.90 10 11 5.00 – 6.90 8 19 7.00 – 8.90 7 26 9.00 – 10.90 3 29 11.00 – 12.90 6 35 Contoh Pendapatan bulanan (RM) Monthly income (RM) Bilangan pekerja Number of workers Kekerapan longgokan Cumulative frequency 500 – 999 5 5 1 000 – 1 499 4 9 1 500 – 1 999 8 17 2 000 – 2 499 12 29 2 500 – 2 999 4 33 3 Lengkapkan jadual kekerapan bagi setiap set data yang berikut dan seterusnya, wakilkan data dengan graf taburan frekuensi yang dinyatakan. SP: 7.1.1 TP3 Sederhana Complete the frequency table for each of the following sets of data and hence, represent the data with the stated frequency distribution graph. Contoh Laju (km j‒1) Speed (km h–1) Bilangan motosikal Number of motorcycles Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 1 – 20 1 10.5 0.5 20.5 21 – 40 8 30.5 20.5 40.5 41 – 60 15 50.5 40.5 60.5 61 – 80 13 70.5 60.5 80.5 81 – 100 3 90.5 80.5 100.5 Jadual kekerapan di bawah menunjukkan laju, dalam km j–1, motosikal yang melalui Jalan Besar pada suatu petang. The frequency table below shows the speed, in km h–1, of motorcycles which passes Jalan Besar on one evening. Wakilkan data dengan histogram dan poligon kekerapan dengan menggunakan skala yang sama dan sesuai. Represent the data with a histogram and a frequency polygon using the same, suitable scales. 0 5 10 15 0.5 20.5 40.5 60.5 80.5 100.5 Laju (km j–1)/ Speed, (km h–1) Bilangan motosikal Number of motercycles CONTOH

93 Matematik Tingkatan 5 Bab 7 (a) Jadual kekerapan di bawah menunjukkan jisim, dalam kg, murid-murid di Kelas 5 Angsana. The frequency table below shows the mass, in kg, of students in Class 5 Angsana. Jisim (g) Mass (g) Bilangan murid Number of students Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 40 – 49 3 44.5 39.5 49.5 50 – 59 13 54.5 49.5 59.5 60 – 69 16 64.5 59.5 69.5 70 – 79 2 74.5 69.5 79.5 80 – 89 1 84.5 79.5 89.5 Wakilkan data dengan sebuah poligon kekerapan dengan menggunakan skala yang sesuai. Represent the data with a frequency polygon using suitable scales. 0 5 10 15 34.5 44.5 54.5 64.5 74.5 84.5 94.5 Bilangan murid/ Number of students Jisim (g) Mass (g) (b) Jadual kekerapan di bawah menunjukkan tinggi, dalam cm, murid-murid di Kelas 5 Ixora. The frequency table below shows the height, in cm, of students in Class 5 Ixora. Tinggi (cm) Height (cm) Bilangan murid Number of students Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 146 – 150 4 148 145.5 150.5 151 – 155 14 153 150.5 155.5 156 – 160 8 158 155.5 160.5 161 – 165 3 163 160.5 165.5 166 – 170 1 168 165.5 170.5 Wakilkan data dengan sebuah histogram dengan menggunakan skala yang sesuai. Represent the data with a histogram using suitable scales. 0 5 10 15 145.5 150.5 155.5 160.5 165.5 170.5 Bilangan murid/ Number of students Tinggi (cm) Height (cm) CONTOH

Matematik Tingkatan 5 Bab 7 94 4 Banding dan tafsirkan serakan kedua-dua set data terkumpul dan, seterusnya jawab setiap soalan yang berikut. SP: 7.1.2 TP4 Compare and interpret the dispersion of both sets of grouped data and, hence answer each of the following questions. Sukar Contoh Rajah di bawah menunjukkan dua histogram yang mewakili masa yang diambil untuk berlari 100 m dan 200 m larian oleh 20 orang atlet. The diagram below shows two histograms representing the time taken to run 100 m and 200 m by 20 athletes. 0 2 4 6 10.5 15.5 20.5 25.5 30.5 Bilangan atlet/ Number of athletes Masa (s)/ Time (s) 8 35.5 Larian 100 m/ 100 m run 0 2 4 6 10.5 15.5 20.5 25.5 30.5 Bilangan atlet/ Number of athletes Masa (s)/ Time (s) 8 35.5 Larian 200 m/ 200 m run (i) Nyatakan bentuk taburan histogram bagi kedua-dua acara tersebut. State the distribution shape of the histogram for both events. (ii) Acara manakah yang mempunyai serakan masa rekod yang lebih luas? Berikan sebab anda. Which event has a wider dispersion of the recorded time? Given your reason. (i) Bentuk taburan histogram bagi acara larian 100 m ialah pencong ke kiri manakala bagi acara larian 200 m ialah seragam. The distribution shape of histogram for the 100 m run event is skewed to the left whereas the 200 m run event shows a uniform distribution. (ii) Acara larian 100 m mempunyai serakan yang lebih luas kerana beza masa yang diambil adalah lebih besar, iaitu 15 s (28 s – 13 s). The 100 m run event has a wider dispersion because the difference in the time taken is larger, that is 15 s (28 s – 13 s). CONTOH

95 Matematik Tingkatan 5 Bab 7 (a) Rajah di bawah menunjukkan dua histogram bagi markah ujian Matematik yang diperoleh murid-murid di Kelas 5S1 dan Kelas 5P1. The diagram below shows two histograms for the Mathematics test marks obtained by the students in Class 5S1 and Class 5P1. 0 4 8 12 0.5 20.5 40.5 60.5 80.5 Bilangan murid/ Number of students Markah ujian/ Test marks 16 100.5 Kelas 5P1/ Class 5P1 0 4 8 12 0.5 20.5 40.5 60.5 80.5 Bilangan murid/ Number of students Markah ujian/ Test marks 16 100.5 Kelas 5S1/ Class 5S1 (i) Nyatakan bentuk taburan histogram bagi kedua-dua kelas. State the distribution shape of the histograms of both classes. (ii) Banding dan huraikan taburan markah ujian antara kedua-dua kelas. Compare and describe the dispersion of the test marks between both classes. (iii) Kelas manakah menunjukkan keputusan yang lebih cemerlang? Berikan sebab anda. Which class shows a more excellent result? Give your reason. (i) Bentuk taburan histogram bagi markah ujian Kelas 5P1 ialah bentuk loceng manakala bagi markah ujian Kelas 5S1 ialah pencong ke kiri. The histogram for the test marks by Class 5P1 shows a bell-shaped distribution whereas the distribution shape of the histogram for the test marks by Class 5S1 is skewed to the left. (ii) Markah ujian yang diperoleh murid-murid di Kelas 5P1 mempunyai serakan yang lebih luas kerana beza markah yang diperoleh adalah lebih besar, iaitu 80 markah (90.5 – 10.5). The test marks obtained by the students in Class 5P1 has a wider dispersion because the difference in the marks obtained is larger, which is 80 marks (90.5 – 10.5). (iii) Kelas 5S1 menunjukkan keputusan yang lebih cemerlang kerana kebanyakan markah murid adalah lebih tinggi. Class 5S1 shows a more excellent result because most of the student’s marks are higher. CONTOH

Matematik Tingkatan 5 Bab 7 96 (b) Rajah di bawah menunjukkan poligon kekerapan bagi harga jualan 20 buah rumah di kawasan A dan kawasan B pada bulan Mei. The diagram below shows the frequency polygons of the selling price of 20 houses in area A and area B in May. 0 5 6 7 99 999.5 Unit rumah House units 8 149 999.5 199 999.5 249 999.5 299 999.5 349 999.5 399 999.5 449 999.5 4 3 2 1 Harga jualan (RM) Selling price (RM) Kawasan A Area A Kawasan B Area B (i) Nyatakan bentuk taburan harga jualan bagi rumah di kedua-dua kawasan. State the distribution shape of selling price of the houses in both areas. (ii) Bandingkan serakan harga jualan bagi rumah di kedua-dua kawasan. Compare the dispersion of selling price for the houses in both areas. (iii) Pada pendapat anda, kawasan yang manakah merupakan kawasan bandar dan yang manakah merupakan kawasan luar bandar? In your opinion, which area is an urban area and which area is a rural area? (i) Bentuk taburan poligon kekerapan bagi harga jualan rumah di kawasan A ialah pencong ke kanan manakala bagi harga jualan rumah di kawasan B ialah pencong ke kiri. The distribution shape of frequency polygon for the house selling price in area A is skewed to the right whereas for the house selling price in area B is skewed to the left. (ii) Harga jualan bagi rumah di kawasan A dan kawasan B mempunyai serakan yang lebih kurang sama. The selling price of the houses in area A and area B has approximately the same dispersion. (iii) Kawasan A merupakan kawasan luar bandar kerana harga jualan rumah adalah lebih rendah manakala kawasan B merupakan kawasan bandar kerana harga jualan rumah adalah lebih tinggi. Area A is a rural area because the selling price of houses is lower whereas area B is an urban area because the selling price of houses is higher. CONTOH

97 Matematik Tingkatan 5 Bab 7 5 Selesaikan setiap yang berikut. SP: 7.1.3 TP4 Solve each of the following. Sederhana Contoh Jadual kekerapan berikut menunjukkan berat bagi 40 orang murid di sebuah kelas. The following frequency table shows the weight of 40 students in a class. Berat/ Weight (kg) 41 – 45 46 – 50 51 – 55 56 – 60 61 – 65 66 – 70 71 – 75 Kekerapan/ Frequency 2 5 10 12 6 4 1 (a) Bina satu ogif untuk mewakili data tersebut. Construct an ogive to represent the data. (b) Seterusnya, dengan menggunakan ogif yang dibina, cari Hence, using the constructed ogive, find (i) kuartil pertama, (ii) median, (iii) kuartil ketiga. the first quartile, median, the third quartile. (a) Berat (kg) Weight (kg) Kekerapan Frequency Sempadan atas Upper boundary Kekerapan longgokan Cumulative frequency 36 – 40 0 40.5 0 41 – 45 2 45.5 2 46 – 50 5 50.5 7 51 – 55 10 55.5 17 56 – 60 12 60.5 29 61 – 65 6 65.5 35 66 – 70 4 70.5 39 71 – 75 1 75.5 40 (b) (i) Kuartil pertama = 52.25 First quartile (ii) Median = 56.5 (iii) Kuartil ketiga = 61.25 Third quartile 0 25 30 35 Kekerapan longgokan Cumulative frequency 40 20 15 10 5 Berat (kg) Weight (kg) 40.5 45.5 50.5 55.5 60.5 65.5 70.5 75.5 52.25 56.5 61.25 CONTOH

Matematik Tingkatan 5 Bab 7 98 (a) Jadual kekerapan berikut menunjukkan pengambilan gula harian, dalam g, bagi 20 orang dewasa. The following frequency table shows the daily sugar intake, in g, of 20 adults. Pengambilan gula harian (g) Daily sugar intake (g) 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60 61 – 70 71 – 80 Kekerapan/ Frequency 1 1 5 7 3 2 1 (i) Bina satu ogif untuk mewakili data tersebut. Construct an ogive to represent the data. (ii) Seterusnya, dengan menggunakan ogif yang dibina, cari Hence, using the constructed ogive, find (a) kuartil pertama, (b) median, (c) kuartil ketiga. the first quartile, median, the third quartile. (i) Pengambilan gula harian (g) Daily sugar intake (g) 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60 61 – 70 71 – 80 Kekerapan/ Frequency 0 1 1 5 7 3 2 1 Sempadan atas Upper boundary 10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 Kekerapan longgokan Cumulative frequency 0 1 2 7 14 17 19 20 (ii) (a) kuartil pertama = 37 (b) median = 45 (c) kuartil ketiga = 53 first quartile median third quartile 0 14 16 18 Kekerapan longgokan Cumulative frequency 20 12 10 8 6 Pengambilan gula harian (g)/ Daily sugar intake (g) 10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 4 2 Q1 Q2 Q3 CONTOH

99 Matematik Tingkatan 5 Bab 7 (b) Jadual kekerapan berikut menunjukkan jisim bagi 60 buah betik yang dijual oleh seorang penjaja. The following frequency table shows the mass of 60 papayas which are sold by a vendor. Jisim (g) Mass (g) 501 – 600 601 – 700 701 – 800 801 – 900 901 – 1 000 Bilangan betik Number of papayas 7 12 17 14 10 (i) Bina satu ogif untuk mewakili data tersebut. Construct an ogive to represent the data. (ii) Seterusnya, dengan menggunakan ogif yang dibina, cari Hence, using the constructed ogive, find (a) kuartil pertama, (b) median, (c) kuartil ketiga. the first quartile, median, the third quartile. (i) Jisim betik (g) Mass of papaya(g) Kekerapan Frequency Sempadan atas Upper boundary Kekerapan longgokan Cumulative frequency 401 – 500 0 500.5 0 501 – 600 7 600.5 7 601 – 700 12 700.5 19 701 – 800 17 800.5 36 801 – 900 14 900.5 50 901 – 1 000 10 1 000.5 60 (ii) (a) kuartil pertama = 670.5 (b) median = 765.5 (c) kuartil ketiga = 860.5 first quartile median third quartile 0 30 40 50 Kekerapan longgokan Cumulative frequency 60 20 10 500.5 600.5 700.5 800.5 900.5 1 000.5 Q1 Q2 Q3 Jisim (g) Mass (g) CONTOH

Matematik Tingkatan 5 Bab 7 100 (c) Jadual berikut menunjukkan kandungan karbon dalam 50 sampel bahan kimia. The following table shows the carbon content in 50 samples of chemical substances. Kandungan karbon/ Carbon content (%) 3.1 – 4.0 4.1 – 5.0 5.1 – 6.0 6.1 – 7.0 7.1 – 8.0 Bilangan sampel/ Number of samples 8 12 23 5 2 (i) Bina satu histogram longgokan untuk mewakili data tersebut. Construct a cumulative histogram to represent the data. (ii) Seterusnya, plot satu ogif pada kertas graf yang sama. Hence, plot an ogive on the same graph paper. (iii) Berdasarkan ogif yang dibina, tentukan Based on the constructed ogive, determine (a) kuartil pertama, (b) median, (c) kuartil ketiga. the first quartile, median, the third quartile. (i) Kandungan karbon/ Carbon content (%) 3.1 – 4.0 4.1 – 5.0 5.1 – 6.0 6.1 – 7.0 7.1 – 8.0 Kekerapan/ Frequency 8 12 23 5 2 Sempadan bawah/ Lower boundary 3.05 4.05 5.05 6.05 7.05 Sempadan atas/ Upper boundary 4.05 5.05 6.05 7.05 8.05 Kekerapan longgokan/ Cumulative frequency 8 20 43 48 50 (ii) (a) kuartil pertama = 4.5 (b) median = 5.25 (c) kuartil ketiga = 5.75 first quartile median third quartile 0 30 40 50 Kekerapan longgokan Cumulative frequency 20 10 3.05 4.05 5.05 6.05 7.05 8.05 Q1 Q2 Q3 Kandungan karbon (%) Carbon content (%) CONTOH

101 Matematik Tingkatan 5 Bab 7 6 Berdasarkan ogif yang diberi, jawab semua soalan yang berikut. SP: 7.1.3 TP5 Based on the given ogives, answer all the following questions. Sukar (a) Ogif berikut menunjukkan skor yang diperoleh 40 orang murid dalam suatu ujian Matematik. The following ogive shows the scores obtained by 40 students in a Mathematics test. 0 50 Kekerapan longgokan Cumulative frequency 40 30 20 10 Skor Scores 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.5 31.5 59 74.5 (i) Cari persentil ke-15, P15 dan persentil ke-50, P50. Find the 15th percentile, P15 and the 50th percentile, P50. (ii) Murid yang berada pada persentil ke-85 dan ke atas akan mendapat sebuah hadiah daripada guru. Apakah skor minimum untuk mendapat sebuah hadiah? Students who are at the 85th percentile and above will get a present from the teacher. What is the minimum score to obtain a present? (iii) Berapakah peratusan murid yang memperoleh skor sekurang-kurangnya 70? What is the percentage of students who score at least 70? (i) 15% daripada jumlah kekerapan/ 15% of the total frequency = 15 100 � 40 = 6 Daripada ogif/ From the ogive, P15 = 40.5 50% daripada jumlah kekerapan/ 50% of the total frequency = 50 100 � 40 = 20 Daripada ogif/ From the ogive, P50 = 59 (ii) 85% daripada jumlah kekerapan/ 85% of the total frequency = 85 100 � 40 = 34 Daripada ogif/ From the ogive, P85 = 74.5 ∴ Skor minimum untuk mendapat sebuah hadiah ialah 74.5. The minimum scores to get a present is 74.5. (iii) Peratusan murid yang memperoleh skor sekurang-kurangnya 70: The percentage of students who scored at least 70: 40 – 31.5 40 � 100% = 21.25% CONTOH

Matematik Tingkatan 5 Bab 7 102 (b) Ogif berikut menunjukkan gaji bagi 36 orang pekerja dalam Syarikat A. The following ogive shows the salary of 36 workers in Company A. (i) Cari persentil ke-25, P25 dan persentil ke-50, P50. Find the 25th percentile, P25 and the 50th percentile, P50. (ii) Berapakah peratusan pekerja yang mendapat gaji kurang daripada RM2 500? What is the percentage of workers who have salary below RM2 500? (i) 25% daripada jumlah kekerapan/ 25% of the total frequency = 25 100 � 36 = 9 Daripada ogif/ From the ogive, P25 = RM1 925.50 50% daripada jumlah kekerapan/ 50% of the total frequency = 50 100 � 36 = 18 Daripada ogif/ From the ogive, P50 = RM2 575.50 (ii) Peratusan pekerja yang mendapat gaji kurang daripada RM2 500: The percentage of workers who have salary below RM2 500: 17 36 � 100% = 47.22% 0 Kekerapan longgokan Cumulative frequency 40 30 20 10 Gaji (RM) Salary (RM) 1 000.5 1 500.5 2 000.5 2 500.5 3 000.5 3 500.5 4 000.5 4 500.5 17 1 925.5 2 575.5 CONTOH

103 Matematik Tingkatan 5 Bab 7 7.2 Sukatan Serakan Measures of Dispersion Buku Teks m/s 198 – 210 1 Cari julat dan julat antara kuartil bagi setiap data terkumpul yang berikut. SP: 7.2.1 TP1 TP2 Find the range and interquartile range for each of the following grouped data. Mudah (a) (b) (c) Contoh Tinggi anak benih/ Height of seedlings Julat/ Range = 79.5 + 89.5 2 – 29.5 + 39.5 2 = 84.5 – 34.5 = 50 Julat antara kuartil/ Interquartile range: 66.5 – 50.5 = 16 0 10 20 30 40 50 Kekerapan longgokan Cumulative frequency 29.5 39.5 49.5 59.5 69.5 79.5 89.5 Tinggi/ Height (cm) 50.5 66.5 Q1 Q3 Jisim bagi 40 orang murid The mass of 40 students Julat/ Range = 55.5 + 60.5 2 – 35.5 + 40.5 2 = 58 – 38 = 20 Julat antara kuartil/ Interquartile range: 50.5 – 44.5 = 6 0 10 20 30 40 Kekerapan longgokan Cumulative frequency 35.5 40.5 45.5 50.5 55.5 60.5 Jisim/ Mass (kg) 44.5 50.5 Q1 Q3 Bilangan buku yang dibaca oleh 80 murid dalam 6 bulan Number of books read by 80 students in 6 months Julat/ Range = 12.5 + 15.5 2 – 0.5 + 3.5 2 = 14 – 2 = 12 Julat antara kuartil/ Interquartile range: 9.35 – 5.75 = 3.6 0 20 40 60 80 Kekerapan longgokan Cumulative frequency 0.5 3.5 6.5 9.5 12.5 15.5 Bilangan buku/ Number of books 5.75 9.35 Q1 Q3 Julat/ Range = 34.5 + 39.5 2 – 4.5 + 9.5 2 = 37 – 7 = 30 Julat antara kuartil/ Interquartile range: 28.0 – 20.5 = 7.5 0 20 40 60 80 Kekerapan longgokan Cumulative frequency 4.5 9.5 14.5 19.5 24.5 29.5 Jisim tin aluminium yang dikumpulkan (kg) The mass of aluminium cans collected (kg) 20.5 28.0 Q1 Q3 100 34.5 39.5 Info Digital 7.2 CONTOH

Matematik Tingkatan 5 Bab 7 104 2 Lengkapkan setiap jadual yang berikut dan seterusnya cari varians dan sisihan piawai bagi data tersebut. SP: 7.2.1 Complete each of the following tables and hence, find the variance and standard deviation for the data. TP3 TP4 Sederhana Contoh Isi padu air yang diminum (m) The volume of water drank (m) Bilangan murid, f The number of students, f Titik tengah, x Midpoint, x fx x 2 fx 2 1 – 500 5 250.5 1 252.5 62 750.25 313 751.25 501 – 1 000 14 750.5 10 507 563 250.25 7 885 503.5 1 001 – 1 500 8 1 250.5 10 004 1 563 750.25 12 510 002 1 501 – 2 000 6 1 750.5 10 503 3 064 250.25 18 385 501.5 2 001 – 2 500 7 2 250.5 15 753.5 5 064 750.25 35 453 251.75 ∑f = 40 ∑fx = 48 020 ∑fx2 = 74 548 010 Min/ Mean, x :̅ Varians/ Variance: Sisihan piawai: 48 020 40 = 1 200.5 m 74 548 010 40 – (1 200.5)2 Standard deviation: = 1 863 700.25 – 1 441 200.25 422 500 = 650 m = 422 500 m (a) (b) Bilangan buku dibaca Number of books read Bilangan murid, f The number of students, f Titik tengah, x Midpoint, x fx x 2 fx 2 1 – 10 5 5.5 27.5 30.25 151.25 11 – 20 14 15.5 217 240.25 3 363.5 21 – 30 8 25.5 204 650.25 5 202 31 – 40 6 35.5 213 1 260.25 7 561.5 41 – 50 7 45.5 318.5 2 070.25 14 491.75 ∑f = 40 ∑fx = 980 ∑fx2 = 30 770 Min/ Mean, x :̅ Varians/ Variance: Sisihan piawai: 980 40 = 24.5 30 770 40 – (24.5)2 = 769.25 – 600.25 Standard deviation: = 169 169 = 13 Wang saku (RM) Pocket money (RM) Bilangan murid, f The number of students, f Titik tengah, x Midpoint, x fx x 2 fx 2 11 – 15 3 13 39 169 507 16 – 20 5 18 90 324 1 620 21 – 25 6 23 138 529 3 174 26 – 30 8 28 224 784 6 272 31 – 35 8 33 264 1 089 8 712 ∑f = 30 ∑fx = 755 ∑fx2 = 20 285 Min/ Mean, x :̅ Varians/ Variance: Sisihan piawai: 755 30 = RM25.17 20 285 30 – 755 30 2 = 676.17 – 633.36 Standard deviation: = RM42.81 42.81 = RM6.54 Penggunaan Kalkulator CONTOH

105 Matematik Tingkatan 5 Bab 7 3 Bina satu plot kotak berdasarkan ogif yang diberi dan seterusnya nyatakan bentuk taburan bagi data tersebut. Construct a box plot based on the given ogive and hence state the distribution shape of the data. SP: 7.2.2 TP4 Sederhana Contoh Nilai minimum/ Minimum value: 40 Nilai maksimum/ Maximum value: 75 Kedudukan/ Position of Q1 : 1 4 � 100 = 25 Kedudukan/ Position of Q2 : 1 2 � 100 = 50 Kedudukan/ Position of Q3 : 3 4 � 100 = 75 Data ini mempunyai taburan pencong ke kiri kerana bahagian kiri plot kotak lebih besar daripada bahagian kanan plot kotak. The distribution of the data is skewed to the left because the left half of the box plot is longer than the right half of the box plot. Nilai minimum/ Minimum value: 2 Nilai maksimum/ Maximum value: 8 Kedudukan/ Position of Q1 : 1 4 � 40 = 10 Kedudukan/ Position of Q2 : 1 2 � 40 = 20 Kedudukan/ Position of Q3 : 3 4 � 40 = 30 Data ini mempunyai taburan pencong ke kanan kerana bahagian kanan plot kotak lebih besar daripada bahagian kiri plot kotak. The distribution of the data is skewed to the right because the right half of the box plot is longer than the left half of the box plot. 0 40 60 80 100 Kekerapan longgokan Cumulative frequency 20 40 45 50 55 60 65 70 75 Jisim (g) Mass (g) Q1 Q2 Q3 54.5 64.5 0 20 30 40 Kekerapan longgokan Cumulative frequency 10 2 4 6 8 Masa (min) Time (min) Q1 Q2 Q3 5.8 4.9 40 45 50 55 60 65 70 75 2 4 6 8 Jisim (g)/ Mass (g) Masa (min)/ Time (min) CONTOH

Matematik Tingkatan 5 Bab 7 106 4 Cari min, varians dan sisihan piawai bagi setiap yang berikut dan seterusnya jawab soalan yang diberi. SP: 7.2.3 Find the mean, variance and standard deviation for each of the following and hence answer the given question. TP4 TP5 Sukar Contoh Jadual di bawah menunjukkan jangka hayat bagi dua jenama mentol. The table below shows the lifespan of two brands of light bulbs. Jangka hayat (tahun) Lifespan (years) 0 – 0.9 1.0 – 1.9 2.0 – 2.9 3.0 – 3.9 4.0 – 4.9 Jenama A/ Brand A 5 7 16 15 7 Jenama B/ Brand B 7 12 12 14 5 Dengan menggunakan sukatan yang sesuai, tentukan mentol jenama yang manakah lebih baik dan tahan lama. By using suitable measures, determine which brand of light bulb is better and long-lasting. Bagi mentol A/ For bulb A, Jangka hayat (tahun) Lifespan (years) Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 0 – 0.9 5 0.45 2.25 0.2025 1.0125 1.0 – 1.9 7 1.45 10.15 2.1025 14.7175 2.0 – 2.9 16 2.45 39.2 6.0025 96.04 3.0 – 3.9 15 3.45 51.75 11.9025 178.5375 4.0 – 4.9 7 4.45 31.15 19.8025 138.6175 ∑f = 50 ∑fx = 134.5 ∑fx2 = 428.925 Min/ Mean: Varians/ Variance: Sisihan piawai: 134.5 50 = 2.69 428.925 50 – (2.69)2 = 1.3424 Standard deviation: 1.3424 = 1.1586 Bagi mentol B/ For bulb B, Jangka hayat (tahun) Lifespan (years) Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 0 – 0.9 7 0.45 3.15 0.2025 1.4175 1.0 – 1.9 12 1.45 17.4 2.1025 25.23 2.0 – 2.9 12 2.45 29.4 6.0025 72.03 3.0 – 3.9 14 3.45 48.3 11.9025 166.635 4.0 – 4.9 5 4.45 22.25 19.8025 99.0125 ∑f = 50 ∑fx = 120.5 ∑fx2 = 364.325 Min/ Mean: Varians/ Variance: Sisihan piawai: 120.5 50 = 2.41 364.325 50 – (2.41)2 = 1.4784 Standard deviation: 1.4784 = 1.2159 Mentol A lebih baik dan tahan lama kerana min jangka hayat mentol A adalah lebih tinggi daripada mentol B (2.69 > 2.41) dan sisihan piawainya yang lebih kecil (1.1586 < 1.2159) menunjukkan jangka hayat mentol A adalah lebih konsisten. Bulb A is better and long-lasting because the mean of lifespan of bulb A is higher than bulb B (2.69 > 2.41) and the smaller standard deviation (1.1586 < 1.2159) shows that the lifespan of bulb A is more consistent. CONTOH

107 Matematik Tingkatan 5 Bab 7 Jadual kekerapan di bawah menunjukkan bilangan soalan yang diselesaikan oleh murid-murid daripada dua kategori dalam Kuiz Matematik Antarabangsa dalam masa 20 minit. The frequency table below shows the number of questions solved by the students from two categories in the International Mathematics Quiz in 20 minutes. Bilangan soalan Number of questions 1 – 2 3 – 4 5 – 6 7 – 8 9 – 10 Kategori A/ Category A 5 7 5 4 9 Kategori B/ Category B 2 3 5 12 8 Tentukan murid kategori yang manakah mempunyai prestasi yang lebih baik. Justifikasikan jawapan anda dengan sukatan yang sesuai. Determine which category of students has a better performance. Justify your answer using suitable measures. Bagi kategori A/ For category A, Bilangan soalan Number of questions Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 1 – 2 5 1.5 7.5 2.25 11.25 3 – 4 7 3.5 24.5 12.25 85.75 5 – 6 5 5.5 27.5 30.25 151.25 7 – 8 4 7.5 30 56.25 225 9 – 10 9 9.5 85.5 90.25 812.25 ∑f = 30 ∑fx = 175 ∑fx2 = 1 285.5 Min/ Mean: Varians/ Variance: Sisihan piawai: 175 30 = 5.8333 1 285.5 30 – ( 175 30 ) 2 = 8.8222 Standard deviation: 8.8222 = 2.9702 Bagi kategori B/ For category B, Bilangan soalan Number of questions Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 1 – 2 2 1.5 3 2.25 4.5 3 – 4 3 3.5 10.5 12.25 36.75 5 – 6 5 5.5 27.5 30.25 151.25 7 – 8 12 7.5 90 56.25 675 9 – 10 8 9.5 76 90.25 722 ∑f = 30 ∑fx = 207 ∑fx2 = 1 589.5 Min/ Mean: Varians/ Variance: Sisihan piawai: 207 30 = 6.9 1 589.5 30 – (6.9)2 = 5.3733 Standard deviation: 5.3733 = 2.3180 Murid daripada kategori B mempunyai prestasi yang lebih baik kerana minnya adalah lebih tinggi daripada kategori A (6.9 > 5.8333). Students from category B has a better performance because the mean is higher than category A (6.9 > 5.8333). CONTOH

Matematik Tingkatan 5 Bab 7 108 5 Selesaikan setiap masalah yang berikut. SP: 7.2.4 TP5 TP6 Solve each of the following problems. Sederhana Sukar (a) Ogif di bawah menunjukkan masa yang diambil untuk mengulang kaji dalam seminggu oleh 100 orang murid. The ogive below shows the time taken to do revision in a week by 100 students. (i) Berdasarkan ogif, dengan menggunakan selang kelas yang sesuai, bina satu jadual kekerapan bagi masa yang diambil untuk mengulang kaji dalam seminggu. Based on the ogive, using a suitable class interval, construct a frequency table for the time taken to do revision in a week. (ii) Seterusnya, hitung min dan sisihan piawai bagi data tersebut. Hence, calculate the mean and standard deviation for the data. (i) Masa yang diambil untuk mengulang kaji (minit) Time taken to do revision (minutes) Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 60 – 119 18 89.5 1 611 8 010.25 144 184.5 120 – 179 28 149.5 4 186 22 350.25 625 807 180 – 239 34 209.5 7 123 43 890.25 1 492 268.5 240 – 299 20 269.5 5 390 72 630.25 1 452 605 ∑f = 100 ∑fx = 18 310 ∑fx2 = 3 714 865 (ii) Min/ Mean: Varians/ Variance: Sisihan piawai: 18 310 100 = 183.1 3 714 865 100 – (183.1)2 = 3 623.04 Standard deviation: 3 623.04 = 60.1917 0 100 Bilangan murid Number of students 80 60 40 20 Masa yang diambil (minit) 59.5 119.5 179.5 239.5 299.5 Time taken (minutes) CONTOH

109 Matematik Tingkatan 5 Bab 7 (b) Histogram di bawah menunjukkan jisim, dalam kg, bagi sekumpulan pekerja di suatu syarikat. The histogram below shows the mass, in kg, of a group of workers in a company. (i) Berdasarkan histogram, bina satu jadual kekerapan untuk menunjukkan jisim bagi kumpulan pekerja tersebut dengan selang kelas yang sesuai. Based on the histogram, construct a frequency table to show the mass of the group of workers with suitable class interval. (ii) Seterusnya, hitung min dan sisihan piawai bagi kumpulan pekerja tersebut. Hence, calculate the mean and standard deviation for the group of workers. (i) Jisim (kg) Mass (kg) Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 40 – 49 5 44.5 222.5 1 980.25 9 901.25 50 – 59 11 54.5 599.5 2 970.25 32 672.75 60 – 69 22 64.5 1 419 4 160.25 91 525.5 70 – 79 18 74.5 1 341 5 550.25 99 904.5 80 – 89 4 84.5 338 7 140.25 28 561 ∑f = 60 ∑fx = 3 920 ∑fx2 = 262 565 (ii) Min/ Mean: Varians/ Variance: Sisihan piawai: 3 920 60 = 65.3333 262 565 60 – (3 920 60 ) 2 = 107.6389 Standard deviation: 107.6389= 10.37 0 25 Bilangan pekerja Number of workers 20 15 10 5 Jisim (kg) 39.5 49.5 59.5 69.5 79.5 89.5 Mass (kg) CONTOH

Matematik Tingkatan 5 Bab 7 110 (c) Jadual kekerapan berikut menunjukkan masa yang diambil oleh murid-murid di Kelas 5 Cempaka untuk bermain permainan komputer dan mengulang kaji dalam seminggu. The following frequency table shows the time taken by the students in Class 5 Cempaka to play computer games and do revision in a week. Masa (jam) Time (hours) 4 – 8 9 – 13 14 – 18 19 – 23 24 – 28 29 – 33 Bermain permainan komputer Play computer games 3 7 15 10 4 1 Mengulang kaji Do revision – 6 24 8 2 – (i) Lengkapkan jadual berikut. Complete the following table. Bermain permainan komputer Play computer games Mengulang kaji Do revision Masa (jam) Time (hours) Sempadan atas Upper limit Kekerapan Frequency Kekerapan longgokan Cumulative frequency Kekerapan Frequency Kekerapan longgokan Cumulative frequency 0 – 3 3.5 0 0 – – 4 – 8 8.5 3 3 0 0 9 – 13 13.5 7 10 6 6 14 – 18 18.5 15 25 24 30 19 – 23 23.5 10 35 8 38 24 – 28 28.5 4 39 2 40 29 – 33 33.5 1 40 (ii) Berdasarkan jadual di (i), lukiskan dua ogif pada paksi yang sama dengan menggunakan skala yang sesuai. Seterusnya, bandingkan penggunaan masa murid-murid berdasarkan kuartil ketiga. Based on the table in (i), draw two ogives on the same axes using a suitable scale. Hence, compare the usage of time of the students based on the third quartile. 0 Bilangan murid Number of students 40 30 Masa (jam) 3.5 8.5 13.5 18.5 23.5 Time (hours) 20.75 28.5 33.5 20 10 Mengulang kaji Do revision Bermain permainan komputer Playing computer games Murid-murid di Kelas 5 Cempaka lebih cenderung bermain permainan komputer daripada mengulang kaji. The students in Class 5 Cempaka is more inclined on playing computer games than do revision. K B A T CONTOH

111 Matematik Tingkatan 5 Bab 7 Kertas 1 Jawab semua soalan./ Answer all questions. 1 Jadual kekerapan berikut menunjukkan tinggi, dalam cm, bagi murid-murid di Kelas 5 Petunia. The following frequency table shows the height, in cm, of the students in Class 5 Petunia. Tinggi (cm) Height (cm) Kekerapan Frequency 140 – 144 4 145 – 149 5 150 – 154 6 155 – 159 9 160 – 164 2 Cari kekerapan longgokan bagi selang kelas 155 – 159. Find the cumulative frequency for the class interval 155 – 159. A 9 C 24 B 15 D 26 2 Diberi selang kelas 130 – 134, 135 – 139, 140 – 144, 145 – 149 dan 150 – 154. Hitung saiz selang kelas bagi data terkumpul ini. Given the class intervals 130 – 134, 135 – 139, 140 – 144, 145 – 149 and 150 – 154. Calculate the size of the class intervals. A 4 C 6 B 5 D 24 3 Rajah berikut menunjukkan sebuah histogram. The following diagram shows a histogram. Kekerapan/ Frequency Pemboleh ubah/ Variable Apakah bentuk taburan bagi data yang diwakili oleh histogram? What is the distribution shape of the data represented by the histogram? A Pencong ke kanan C Bentuk loceng Skew to the right Bell-shaped B Pencong ke kiri D Seragam Skew to the left Uniform 4 Serakan bagi data terkumpul boleh diperhatikan dengan The dispersion of grouped data can be observed using I Histogram/ Histogram II Poligon kekerapan / Frequency polygon III Ogif/ Ogive IV Parabola/ Parabola A I dan II sahaja C I dan III sahaja I and II only I and III only B II dan IV sahaja D I, II dan III sahaja II and IV only I, II and III only 5 Rajah berikut menunjukkan sebuah ogif bagi jisim ikan yang dijual. The following diagram shows an ogive of the mass of fish sold. Kekerapan longgokan Cumulative frequency 60 45 30 15 0 1 3.5 4.2 5.8 10 Jisim (kg) Mass (kg) Cari julat antara kuartil bagi jualan ikan. Find the interquartile range for the sales of fish. A 9.0 kg C 2.5 kg B 4.2 kg D 2.3 kg 6 Jadual kekerapan berikut menunjukkan isi padu air yang diminum dalam sehari oleh murid-murid. The following frequency table shows the volume of water drank in one day by the students. Isi padu air () Volume of water () Bilangan murid Number of students 0.6 – 1.0 4 1.1 – 1.5 13 1.6 – 2.0 5 2.1 – 2.5 2 2.6 – 3.0 1 Cari min bagi data tersebut. Find the mean for the data. A 2.366 B 1.62 C 1.46 D 0.2344 7 Diberi jumlah kekerapan ialah 30, ∑fx2 = 1 491.5 dan sisihan piawai ialah 1.2828. Cari nilai min. Given the total frequency is 30, ∑fx2 = 1 491.5 and the standard deviation is 1.2828. Find the value of mean. A 48.07 B 7.17 C 6.93 D ‒6.93 Praktis Kendiri CONTOH

Matematik Tingkatan 5 Bab 7 112 Bahagian B/ Section B 2 Rajah berikut menunjukkan markah yang diperoleh 32 orang murid dalam suatu kuiz Matematik. The following diagram shows the marks obtained by 32 students in a Mathematics quiz. 45 47 57 56 79 56 80 88 54 52 52 67 68 76 92 89 59 67 78 76 82 90 41 63 98 60 69 82 71 91 85 70 (a) Lengkapkan jadual kekerapan yang disediakan di ruang jawapan. Complete the frequency table provided in the answer space. [3 markah/ marks] (b) Hitung varians dan sisihan piawai bagi data tersebut. Calculate the variance and standard deviation of the data. [3 markah/ marks] (c) (i) Wakilkan data tersebut dalam sebuah ogif. Represent the data in an ogive. (ii) Seterusnya, cari peratusan murid yang memperoleh lebih daripada 75 markah. Hence, find the percentage of students who obtained more than 75 marks. [4 markah/ marks] Kertas 2 Jawab semua soalan./ Answer all questions. Bahagian A/ Section A 1 Jadual kekerapan berikut menunjukkan jangka hayat bagi dua jenis bateri. The following frequency table shows the lifespans of two types of batteries. Jangka hayat (tahun) Lifespan (years) Bateri A Battery A Bateri B Battery B 0 – 0.4 4 10 0.5 – 0.9 6 15 1.0 – 1.4 8 13 1.5 – 1.9 15 3 2.0 – 2.4 8 6 2.5 – 2.9 9 3 (a) Cari min bagi bateri A dan bateri B. Find the mean for battery A and battery B. [2 markah/ marks] (b) Seterusnya, hitung beza min antara kedua-dua jenis bateri. Hence, calculate the difference in mean between both types of batteries. [1 markah/ mark] (c) Jenis bateri manakah yang boleh tahan lebih lama? Berikan sebab anda. Which type of battery can last longer? Give your reason. [1 markah/ mark] Jawapan/ Answer: (a) Titik tengah Midpoint fA fAx fB fBx 0.2 4 0.8 10 2 0.7 6 4.2 15 10.5 1.2 8 9.6 13 15.6 1.7 15 25.5 3 5.1 2.2 8 17.6 6 13.2 2.7 9 24.3 3 8.1 ∑ 50 82 50 54.5 x̅ A = 82 50 x̅ B = 54.5 50 = 1.64 tahun/ years = 1.09 tahun/ years (b) Beza/ Difference: 1.64 – 1.09 = 0.55 tahun/ years (c) Bateri A boleh tahan lebih lama kerana min jangka hayat adalah lebih panjang daripada bateri B. Battery A can last longer because the mean of lifespan is longer than battery B. CONTOH

113 Matematik Tingkatan 5 Bab 7 Jawapan/ Answer: (a) Markah Marks Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 Kekerapan longgokan Cumulative frequency 41 – 50 3 45.5 136.5 2 070.25 6 210.75 3 51 – 60 8 55.5 444 3 080.25 24 642 11 61 – 70 6 65.5 393 4 290.25 25 741.5 17 71 – 80 6 75.5 453 5 700.25 34 201.5 23 81 – 90 6 85.5 513 7 310.25 43 861.5 29 91 – 100 3 95.5 286.5 9 120.25 27 360.75 32 ∑f = 32 ∑fx = 2 226 ∑fx2 = 162 018 (b) Min/ Mean: Varians/ Variance: Sisihan piawai: 2 226 32 = 69.5625 162 018 32 – (69.5625)2 = 224.1211 Standard deviation: 224.1211 = 14.9707 (c) (i) 0 35 Kekerapan longgokan Cumulative frequency 30 25 20 15 Markah/ Marks 40.5 50.5 60.5 70.5 80.5 90.5 100.5 10 5 (ii) Daripada ogif, bilangan murid yang memperoleh lebih daripada 75 markah ialah 12 orang (32 – 20). From the ogive, the number of students who obtained more than 75 marks is 12 (32 – 20). Peratusan/ Percentage: 12 32 � 100% = 37.5% ≈ 20 CONTOH

Matematik Tingkatan 5 Bab 7 114 3 Rajah berikut menunjukkan bilangan telur yang dimakan oleh 20 orang dalam seminggu. The following diagram shows the number of eggs eaten by 20 people in a week. 4 7 15 18 7 10 5 14 8 9 11 6 4 13 10 4 11 8 5 5 (a) Dengan menggunakan data tersebut, lengkapkan jadual kekerapan di ruang jawapan. By using the data, complete the frequency table in the answer space. [3 markah/ marks] (b) Plot satu ogif untuk mewakili data tersebut. Plot an ogive to represent the data. [3 markah/ marks] (c) Berdasarkan ogif, bina satu plot kotak dan seterusnya, nyatakan bentuk taburan data tersebut. Based on the ogive, construct a box plot and hence, state the distribution shape of the data. [4 markah/ marks] Jawapan/ Answer: (a) Bilangan telur Number of eggs Bilangan orang, f Number of people, f Titik tengah, x Midpoint, x fx x 2 fx 2 Kekerapan longgokan Cumulative frequency 4 – 6 7 5 35 25 175 7 7 – 9 5 8 40 64 320 12 10 – 12 4 11 44 121 484 16 13 – 15 3 14 42 196 588 19 16 – 18 1 17 17 289 289 20 ∑f = 20 ∑fx = 178 ∑fx2 = 1 856 (b), (c) 0 20 Kekerapan longgokan Cumulative frequency 15 10 5 Bilangan telur (biji) Number of eggs 3.5 6.5 9.5 12.5 15.5 18.5 Q1 Q2 Q3 Data ini mempunyai taburan pencong ke kanan kerana bahagian kanan plot kotak lebih besar daripada bahagian kiri plot kotak. The distribution of the data is skewed to the right because the right half of the box plot is longer than the left half of the box plot. CONTOH

Pemodelan Matematik 8 Mathematical Modelling Bab 115 Praktis Intensif 8.1 Pemodelan Matematik Mathematical Modelling Buku Teks m/s 228 – 240 1 Kenal pasti dan definisikan setiap masalah yang berikut. Seterusnya, buat andaian dan tentukan pemboleh ubah untuk menyelesaikan masalah tersebut. SP: 8.1.1 TP1 TP2 TP3 Identify and define each of the following problems. Then, make the assumptions and determine the variables to solve the problems. Mudah Sederhana Contoh Seorang perenang mengambil masa 20 saat untuk berenang dari A ke B di sepanjang sungai dan mengambil masa 24 saat kembali ke A. Jika kadar aliran air sungai ialah 2 km j–1, apakah kelajuan air tenang? A swimmer took 20 seconds to swim from A to B along a river and 24 seconds to return to A. If the flow rate of the river water is 2 km h–1, what is the speed of calm water? Masalah/ Problem: Kita mengetahui kadar aliran air sungai dan tempoh berenang antara A dan B. Kita perlu mencari kelajuan air tenang. We know about the flow rate of the river water and the travel time between A and B. We need to find the speed of calm water. Andaian/ Assumptions: • Kelajuan perenang tidak berubah sepanjang masa. The speed of the swimmer does not change all the time. • Kesan geseran antara perenang dengan air sungai dan rintangan angin diabaikan. The effect of friction between the swimmer and river water as well as the wind resistance are neglected. Pemboleh ubah: v untuk kelajuan perenang, t untuk tempoh perjalanan dan d untuk jarak antara A dengan B Variables: v for the speed of the swimmer, t for travel time and d for the distance between A and B (a) Jason menyimpan RM50 000 dalam Bank A dengan kadar faedah sebanyak 1.6% setahun. Berapa lamakah Jason perlu menyimpan sekiranya dia bercadang untuk membayar wang pendahuluan sebuah rumah yang berharga RM55 000 dengan wang simpanannya? Jason saved RM50 000 in Bank A with an interest rate of 1.6% per annum. How long does Jason need to save if he plans to pay for the down payment of a house which cost RM55 000 with his saving? Masalah/ Problem: Kita mengetahui amaun prinsipal dan kadar faedah. Faedah mesti dimasukkan ke dalam prinsipal Jason untuk membayar wang pendahuluan sebuah rumah. Kita perlu mencari tempoh simpanan Jason di dalam bank. We know about the principal’s amount and the interest rate. The interest must be included into Jason’s principal to pay for the down payment of a house. We need to determine how long Jason needs to keep his saving in the bank. Andaian/ Assumptions: • Kadar faedah tidak berubah sepanjang tempoh simpanan. The interest rate does not change during the period of saving. • Harga rumah tidak berubah sehingga Jason mempunyai wang yang mencukupi untuk membayar wang pendahuluan rumah. The price of the house does not change until Jason has enough money to pay for the down payment of the house. Pemboleh ubah: Faedah, prinsipal sebanyak RM50 000, kadar faedah sebanyak 1.6% dan masa Variables: Interest, principal of RM50 000, interest rate of 1.6% and time Info Digital 81 CONTOH

Matematik Tingkatan 5 Bab 8 116 (b) Perjalanan sebuah sampan dari H ke K di sepanjang sungai mengambil masa selama 4 jam. Dalam perjalanan balik dari K ke H, sampan tersebut mengambil masa 4.5 jam. Jika kelajuan arus sungai ialah 2.5 km j–1, berapakah kelajuan sampan di atas air tenang? The journey of a canoe travelling from H to K along the river takes 4 hours. In the return journey from K to H, the canoe takes 4.5 hours. If the speed of the river current is 2.5 km h–1, what is the speed of the canoe on calm water? Masalah/ Problem: Kita mengetahui kelajuan arus sungai dan tempoh perjalanan antara H dan K. Kita perlu mencari kelajuan sampan di atas air tenang. We know about the speed of the river current and the travel time between H and K. We need to find the speed of the canoe on calm water. Andaian/ Assumptions: • Kelajuan arus sungai dan kelajuan sampan tidak berubah sepanjang masa. The speed of the river current and the speed of the canoe do not change throughout the time. • Kesan geseran antara permukaan sampan dengan air sungai dan rintangan angin melawan sampan diabaikan. The effect of friction between the surface of the canoe and the river water as well as the wind resistance against the canoe are neglected. Pemboleh ubah: v untuk kelajuan sampan, t untuk tempoh perjalanan dan d untuk jarak antara H dengan K Variables: v for the speed of the canoe, t for travel time and d for the distance between H and K (c) Bapa Jamie memberinya RM3 000 sebagai ganjaran mendapat keputusan yang cemerlang dalam peperiksaan SPM. Jamie menyimpan wangnya di dalam Bank XYZ dengan kadar faedah sebanyak 1.3% setahun. Dia ingin membeli sebuah telefon baharu yang berharga RM5 000 dengan wang tabungannya. Berapa lamakah Jamie perlu menyimpan wang tersebut? Jamie’s father gave her RM3 000 as a reward for getting excellent results in the SPM examination. Jamie saves all her money in Bank XYZ with an interest rate of 1.3% per annum. She wants to buy a new phone which costs RM5 000 with her saving. How long does Jamie need to save? Masalah/ Problem: Kita mengetahui prinsipal dan kadar faedah. Harga telefon baharu ialah prinsipal simpanan termasuk faedah. Kita perlu mencari tempoh simpanan Jamie di bank. We know about the principal and the interest rate. The price of the new phone is the saving’s principal including interest. We need to determine the duration of Jamie’s saving in the bank. Andaian/ Assumptions: • Kadar faedah tidak akan berubah sepanjang tempoh simpanan. The interest rate does not change during the period of saving. • Harga telefon tidak berubah sehingga Jamie berjaya mengumpul amaun wang yang diperlukan. The price of the phone does not change until Jamie has enough money to buy it. Pemboleh ubah: Faedah, prinsipal sebanyak RM3 000, kadar faedah sebanyak 1.3% dan masa Variables: Interest, principal of RM3 000, interest rate of 1.3% and time CONTOH

Matematik Tingkatan 5 Bab 8 117 2 Selesaikan setiap masalah yang berikut melalui pemodelan matematik. SP: 8.1.2 TP4 Solve each of the following problems through mathematical modelling. Sederhana Sukar Contoh Kenny ditawarkan kerja daripada Syarikat A dan Syarikat B. Syarikat A terletak di luar bandar manakala Syarikat B terletak di bandar. Kenny tidak perlu menyewa rumah sekiranya dia bekerja di Syarikat A. Namun, gaji pendahuluan dan kenaikan gaji tahunan Syarikat A adalah lebih rendah daripada Syarikat B. Syarikat yang manakah patut dipilih oleh Kenny untuk bekerja sekurang-kurangnya 5 tahun? Kenny received job offers from Company A and Company B. Company A is located in the rural area whereas Company B is located in the city. Kenny does not need to rent a house if he works for Company A. However, the starting salary and annual salary increment in Company A is lower than Company B. Which company should Kenny choose to work for at least 5 years? Masalah/ Problem: • Gaji dan kenaikan gaji tahunan di Syarikat B adalah lebih daripada Syarikat A The salary and annual salary increment in Company B is more than in Company A • Kenny mempunyai perbelanjaan berlebihan, iaitu sewa rumah sekiranya dia bekerja di Syarikat B Kenny has extra expense, which is house rent if he works in Company B • Tentukan beza pendapatan Kenny selepas bekerja 5 tahun di antara Syarikat A dan Syarikat B Determine the difference in Kenny’s income after working for 5 years between Company A and Company B Andaian/ Assumptions: • Sewa rumah di bandar adalah ditetapkan pada RM650 sebulan The house rent in the city is fixed at RM650 monthly • Keadaan kerja antara kedua-dua syarikat adalah sama The working conditions of both companies are the same • Kos hidup Kenny di luar bandar dan dalam bandar adalah sama, iaitu RM1 000 sebulan dan dia tiada perbelanjaan lain Kenny’s living costs in the rural area and in the city are the same, which is RM1 000 per month and he has no other expenses • Gaji daripada Syarikat A ialah RM2 500 dan gaji daripada Syarikat B ialah RM3 000 The salary from Company A is RM2 500 and the salary from Company B is RM3 000 • Kenaikan tahunan di Syarikat A ialah 3% manakala di Syarikat B ialah 3.5% The annual increment in Company A is 3% whereas in Company B is 3.5% Pemboleh ubah/ Variables: Gaji tahunan, RMx dan jumlah perbelanjaan tahunan, RMy dan jumlah pendapatan tahunan, RMz dalam 5 tahun Annual salary, RMx and total annual expenses, RMy and the total annual income, RMz in 5 years Mengaplikasi matematik/ Applying mathematics: Untuk mengira jumlah pendapatan Kenny bagi 5 tahun kerja: To calculate Kenny’s total income for working 5 years: Tahun bekerja Year of working Syarikat A Company A Syarikat B Company B 1 RM2 500 × 12 = RM30 000 RM3 000 × 12 = RM36 000 2 RM30 000 × 1.03 = RM30 900 RM36 000 × 1.035 = RM37 260 3 RM30 000 × (1.03)2 = RM31 827 RM36 000 × (1.035)2 = RM38 564.10 4 RM30 000 × (1.03)3 = RM32 781.81 RM36 000 × (1.035)3 = RM39 913.84 5 RM30 000 × (1.03)4 = RM33 765.26 RM36 000 × (1.035)4 = RM41 310.83 Jumlah, x Total, x RM159 274.07 RM193 048.77 Menolak perbelanjaan, y/ Deduct expenses, y Kos hidup/ Cost of living RM1 000 × 12 × 5 = RM60 000 RM1 000 × 12 × 5 = RM60 000 Sewa rumah/ House rent – RM650 × 12 × 5 = RM39 000 Jumlah pendapatan, z Total income, z RM99 274.07 RM94 048.77 CONTOH

Matematik Tingkatan 5 Bab 8 118 (a) Charles telah menyelesaikan 24 soalan Matematik dalam masa 2 jam. Hitung bilangan soalan yang dapat diselesaikan oleh Charles dalam masa 235 minit. Charles solved 24 Mathematics questions in 2 hours. Calculate the number of questions which can be solved by Charles within 235 minutes. Masalah/ Problem: • Tentukan bilangan soalan yang boleh diselesaikan dalam masa 235 minit. Determine the number of questions which can be solved in 235 minutes. • Diketahui bahawa semakin banyak masa diambil, semakin banyak soalan Matematik boleh diselesaikan. Oleh itu, bilangan soalan berubah secara langsung dengan masa. We know that the more time taken, the more Mathematics questions can be solved. Thus, the number of questions varies directly as the time. Andaian/ Assumptions: • Andaikan tahap kesukaran bagi semua soalan Matematik adalah sama Assume that the difficulty level for all Mathematics questions is the same • Charles tidak mempunyai masalah dalam penyelesaian soalan Matematik Charles does not have any problems in solving Mathematics questions Pemboleh ubah/ Variables: • Katakan t ialah masa yang diambil, dalam minit dan n ialah bilangan soalan Matematik Let t be the time taken, in minutes and n be the number of Mathematics questions • n berubah secara langsung dengan t, maka n = kt dengan keadaan k ialah pemalar n varies directly as t, hence n = kt where k is a constant Mengaplikasi matematik/ Applying mathematics: Gantikan t = 120 dan n = 24 ke dalam n = kt, 24 = k(120) k = 24 120 = 1 5 Substitute t = 120 and n = 24 into n = kt, Maka/ Hence, n = 1 5 t Persamaan ini menghuraikan hubungan antara bilangan soalan Matematik yang dapat diselesaikan dengan masa. This equation describes the relationship between the number of Mathematics questions which can be solved with the time. Apabila/ When t = 235, n = 1 5 × 235 = 47 soalan/ questions Maka, 47 soalan Matematik boleh diselesaikan oleh Charles dalam 235 minit. Hence, 47 Mathematics questions can be solved by Charles in 235 minutes. • Kenny boleh memilih bekerja di Syarikat A kerana dia dapat menyimpan lebih banyak pendapatan dan tidak perlu membayar sewa rumah. Kenny can choose to work in Company A because he can save more income and cut down on house rent. Menentusahkan dan mentafsir penyelesaian dalam konteks masalah berkenaan Verifying and interpreting solutions in the context of the problem • Berdasarkan pengiraan pada jadual, pendapatan akhir dapat diperoleh melalui penolakan perbelanjaan daripada gaji. Based on the calculations in the table, the final income can be obtained through the deduction of expenses from the salary. • Model matematik yang digeneralisasikan, iaitu z = x – y (jumlah pendapatan = gaji – perbelanjaan) boleh digunakan dalam semua situasi kehidupan. The generalised mathematical model, which is z = x – y (total income = salary – expenses) can be used in all life situations. Memurnikan model matematik: Refining the mathematical model: • Dalam model ini, kita andaikan Kenny hanya mempunyai perbelanjaan yang tetap dan tiada simpanan. Hal ini mungkin tidak benar sekiranya Kenny tinggal bersama dengan keluarganya di luar bandar dan perbelanjaannya termasuk keperluan keluarganya. In this model, we assume that Kenny has only a fixed expense and no saving. This may not be true if Kenny lives with his family in the rural area and his expenses include his family’s needs. • Model matematik ini boleh diperbaik dengan mengumpulkan lebih data supaya kejituan jawapan boleh ditambah baik. This mathematical model can be improvised by collecting more data to improve the accuracy of the answer. CONTOH

Matematik Tingkatan 5 Bab 8 119 Menentusahkan dan mentafsir penyelesaian dalam konteks masalah berkenaan Verifying and interpreting solutions in the context of the problem • Model fungsi linear n = 1 5 t yang diperoleh mungkin tidak dapat digunakan dalam semua situasi penyelesaian soalan kerana tahap kesukaran soalan Matematik adalah tidak sama. The linear function model n = 1 5 t obtained may not be used in all question solving situations because the difficulty levels of Mathematics questions are not the same. • Charles mungkin perlu mengambil lebih masa untuk memahami dan menyelesaikan soalan Matematik yang lebih sukar dan dia mungkin mengambil masa yang lebih singkat untuk menyelesaikan soalan Matematik yang lebih mudah. Charles may need to take more time to understand and solving a more difficult Mathematics question and he may take a shorter time to solve an easier Mathematics question. • Apabila diterjemah kembali ke dunia sebenar, model fungsi linear yang diperoleh tidak sesuai digunakan untuk menangani masalah berkenaan. When this is reflected to the real-world situation, the linear function model obtained is not suitable to solve this problem. Memurnikan model matematik: Refining the mathematical model: • Dalam masalah ini, kita tidak dapat memurnikan model memandangkan maklumat yang diberi adalah terhad. In this problem, we are not able to refine the model due to limited information given. (b) Rajah berikut menunjukkan keratan rentas sebuah kolam manakala jadual berikut menunjukkan data yang dikumpulkan oleh sekumpulan ahli sains. The following diagram shows the cross section of a pond while the following table shows the data collected by a group of scientists. Jarak dari tengah kolam, x (m) Distance from the centre of the pond, x (m) 0 5 10 15 20 Kedalaman kolam, y (m) Depth of the pond, y (m) 1.20 1.09 0.88 0.60 0.10 Jarak dari tengah kolam, x (m) Distance from the centre of the pond, x (m) Kedalaman kolam, y (m) Depth of the pond, y (m) Tentukan kedalaman kolam apabila jarak dari tengah kolam ialah 7.5 m dengan menggunakan pemodelan matematik. Determine the depth of the pond when the distance from the centre of the pond is 7.5 m through mathematical modelling Masalah/ Problem: • Bagaimana menentukan kedalaman kolam pada 7.5 m dari tengah kolam How to determine the depth of the pond at 7.5 m away from the centre of the pond Andaian dan pemboleh ubah/ Assumptions and variables: • Andaikan bentuk kolam ialah hemisfera. Assume that the shape of the pond is hemispherical. • Kolam adalah paling dalam di bahagian tengah. The pond is the deepest in the centre. • Pemboleh ubah yang terlibat dalam kajian ini ialah kedalaman kolam, y dan jarak dari tengah kolam, x. The variables involved in this study is the depth of the pond, y and the distance from the centre of the pond, x. Mengaplikasi matematik/ Applying mathematics: • Tulis jarak dari tengah kolam dan kedalaman kolam sebagai set pasangan tertib (x, y) dan lukis satu graf bagi data tersebut. Write the distance from the centre of the pond and the depth of the pond as a set of ordered pairs (x, y) and draw a graph for the data. • Graf yang dilukis menunjukkan lengkung penyuaian terbaik dan menyerupai graf fungsi kuadratik. The graph drawn shows the curve of best fit and resembles the graph of a quadratic function. • Nilai anggaran digunakan dalam pemodelan matematik ini untuk mewakili situasi sebenar. Daripada graf, kedalaman kolam ialah 1.0 m. An approximate value is used in this mathematical modelling to represent the actual situation. From the graph, the depth of the pond is 1.0 m. (7.5, 1.0) y x 10 20 30 1.4 1.2 1 0.8 0.6 0.4 0.2 0 K B A T n O t n = 1 5 t CONTOH

Matematik Tingkatan 5 Bab 8 120 Kertas 1 Jawab semua soalan./ Answer all questions. 1 Antara berikut, yang manakah masalah dunia sebenar yang dapat diterjemah sebagai masalah matematik? Which of the following real-world problems can be translated into a mathematical problem? I Adakah populasi negeri Pulau Pinang akan melebihi 1 000 000 orang? Will the population of Penang state exceed 1 000 000 people? II Berapakah kos bagi Ijazah Sarjana Muda Perniagaan selepas 10 tahun? What is the cost for a Bachelor’s Degree in Business after 10 years? III Berapa lamakah masa yang akan diambil oleh seseorang untuk memakan dua mangkuk mi? How much time will be taken by a person to eat two bowls of noodles? A I dan II sahaja C II dan III sahaja I and II only II and III only B I dan III sahaja D I, II dan III I and III only I, II and III 2 Maklumat berikut adalah tentang suatu situasi. The following information is about a situation. Jisim karbon-14 berkurang sebanyak setengah daripada jisim asalnya dalam setiap 5 730 tahun. Berapakah jisim karbon-14 selepas 20 ribu tahun jika jisim asalnya ialah 10.5 kg? The mass of carbon-14 reduces by half from its original mass in every 5 730 years. What is the mass of the carbon-14 after 20 thousand years if its original mass is 10.5 kg? Apakah jenis graf fungsi yang mungkin digunakan dalam pemodelan matematik bagi situasi di atas? What is the possible type of graph of function used in the mathematical modelling for the situation above? A Linear C Eksponen Linear Exponential B Kuadratik D Punca kuasa dua Quadratic Square root Menentusahkan dan mentafsir penyelesaian: Verifying and interpreting solutions: Tentukan pemalar fungsi kuadratik yang mempunyai bentuk y = ax2 + bx + c dengan menggantikan sebarang tiga data [misalnya (0, 1.2), (15, 0.6) dan (20, 0.1)]. Determine the constants of the quadratic function in the form of y = ax2 + bx + c by substituting any three data [for example (0, 1.2), (15, 0.6) and (20, 0.1)]. 1.2 = a(0)2 + b(0) + c → 1.2 = c 0.6 = a(15)2 + b(15) + c → 0.6 = 225a + 15b + c 0.1 = a(20)2 + b(20) + c → 0.1 = 400a + 20b + c Oleh sebab c ialah pemalar, sistem bagi dua persamaan linear dalam dua pemboleh ubah ialah Since c is a constant, the system for the two linear equations in two variables is 0.6 = 225a + 15b + 1.2 ② – ①: –0.015 = 5a –0.04 = 15a + b ………① a = –0.003 0.1 = 400a + 20b + 1.2 –0.055 = 20a + b ………② Gantikan/ Substitute a = –0.003 ke dalam/ into ①, –0.04 = 15(–0.003) + b b = 0.005 Maka fungsi kuadratik yang mungkin ialah y = –0.003x2 + 0.005x + 1.2 Hence, the possible quadratic function is y = –0.003x2 + 0.005x + 1.2 Apabila/ When x = 7.5, y = –0.003(7.5)2 + 0.005(7.5) + 1.2 = 1.07 m (hampir dengan jawapan yang diperoleh daripada graf/ approximate to the answer obtained from the graph) Memurnikan model matematik: Refining the mathematical model: • Dalam model ini, kita andaikan kedalaman kolam adalah paling dalam di bahagian tengah dan bentuk kolam ialah hemisfera. Model baharu diperlukan untuk andaian baharu. In this model, we assume the depth of the pond is the deepest at the centre and the shape of the pond is hemispherical. A new model is required for new assumptions. • Kejituan jawapan boleh diperbaik jika lebih banyak data dikumpulkan. The accuracy of the answer can be improved if more data are collected. Praktis Kendiri CONTOH

Matematik Tingkatan 5 Bab 8 121 3 Antara berikut, yang manakah bukan komponen penting dalam pemodelan matematik? Which of the following is not an important component in mathematical modelling? A Memurnikan model matematik Refining the mathematical model B Mentafsir penyelesaian tanpa penentusahan Interpreting the solutions without verification C Mengenal pasti dan mendefinisikan masalah Identify and defining the problems D Mengaplikasikan matematik untuk menyelesaikan masalah Applying mathematics to solve the problems 4 Maklumat berikut adalah tentang suatu situasi. The following information is about a situation. Parvina telah menyimpan RM1 000 dalam satu akaun simpanan dengan kadar faedah 1.3% setahun. Dia akan mengeluarkan wang untuk membeli sebuah basikal apabila simpanannya mencapai RM1 250. Parvina has deposited RM1 000 in a saving account with an interest rate of 1.3% per annum. She will withdraw the money to buy a bicycle when her saving reach RM1 250. Apakah andaian boleh dibuat daripada situasi di atas? What assumptions can be made from the situation above? I Kadar faedah tidak berubah sepanjang tempoh simpanan The interest rate does not change throughout the period of saving II Prinsipal tidak berubah sepanjang tempoh simpanan The principal does not change throughout the period of saving III Harga basikal tidak berubah sehingga Parvina mampu membeli basikal tersebut The price of the bicycle does not change until Parvina affords to buy the bicycle A I, II dan III C I dan III sahaja I, II and III I and III only B II dan III sahaja D I dan II sahaja II and III only I and II only 5 Apakah perlu dilakukan sebelum mengaplikasikan matematik untuk menyelesaikan masalah? What is needed to be done before applying mathematics to solve the problems? A Melaporkan dapatan Reporting the findings B Memurnikan model matematik Refining the mathematical model C Mengenal pasti dan mendefinisikan masalah Identifying and defining the problems D Membuat andaian dan mengenal pasti pemboleh ubah Making assumptions and identifying the variables 6 Maklumat berikut adalah tentang suatu situasi. The following information is about a situation. Makcik Falilah hendak membeli 10 kg beras. Harga beras di pasar raya A adalah lebih mahal daripada pasar raya B. Tetapi pasar raya B terletak lebih jauh daripada rumah Makcik Falilah berbanding dengan pasar raya A. Aunt Falilah wants to buy 10 kg of rice. The price of rice at supermarket A is more expensive than supermarket B. However, supermarket B is located further from Aunt Falilah’s house compared to supermarket A. Antara berikut, yang manakah merupakan masalah bagi situasi di atas? Which of the following is the problem for the situation above? A Keinginan Makcik Falilah pergi ke pasar raya B Aunt Falilah’s desire to go to supermarket B B Jisim beras dan harga penghantaran The mass of rice and the delivery charge C Harga beras dan jarak pasar raya The price of rice and the distance of supermarket D Jisim beras dan jarak pasar raya The mass of rice and the distance of supermarket 7 Antara berikut, yang manakah bukan merupakan pemboleh ubah dalam pemodelan matematik? Which of the following is not a variable in mathematical modelling? A Masa yang diambil The time taken B Pilihan warna Choices of colours C Bilangan kereta The number of cars D Isi padu minyak The volume of oil 8 Maklumat berikut adalah tentang suatu situasi. The following information is about a situation. Lisa telah menyimpan RM50 000 dalam suatu akaun simpanan dengan kadar faedah 3% setahun. Berapakah faedah Lisa akan dapat dalam masa 4 tahun? Lisa has deposited RM50 000 into a saving account with an interest rate of 3% per annum. How much interest will Lisa get in 4 years? Tentukan pemboleh ubah dalam penyelesaian masalah ini. Determine the variables in the solution of this problem. A Prinsipal dan kadar faedah Principal and interest rate B Prinsipal dan tempoh simpanan Principal and period of saving C Kadar faedah dan tempoh simpanan Interest rate and period of saving D Prinsipal, kadar faedah dan tempoh simpanan Principal, interest rate and period of saving CONTOH

Matematik Tingkatan 5 Bab 8 122 Kertas 2 Jawab semua soalan./ Answer all questions. Bahagian A/ Section A 1 Rajah menunjukkan dua jenis kerepek kentang dan harga masing-masing. The diagram shows two types of potato chips and their respective prices. RM3.80 RM2.80 Mak Cik Minah bercadang membeli kerepek kentang untuk kelas anaknya sempena Hari Kanak-kanak. Bajet anggaran Mak Cik Minah ialah RM75 dan bilangan murid adalah tidak melebihi 30 orang. Aunt Minah plans to buy potato chips for her child’s class during Children’s Day. Aunt Minah’s estimated budget is RM75 and the number of students is not more than 30. (a) Nyatakan dua masalah yang mungkin dihadapi oleh Mak Cik Minah. State two possible problems which will be faced by Aunt Minah. [2 markah/ marks] (b) Apakah andaian yang perlu dibuat oleh Mak Cik Minah? What assumptions should be made by Aunt Minah? [2 markah/ marks] Jawapan/ Answer: (a) Masalah 1: Sama ada bajet Mak Cik Minah cukup membeli kerepek kentang yang mencukupi untuk satu kelas murid. Problem 1: Whether Aunt Minah’s budget is enough to buy enough potato chips for one class of students. Masalah 2: Bilangan kerepek kentang yang perlu dibeli oleh Mak Cik Minah. Problem 2: The number of potato chips which need to be bought by Aunt Minah. (b) Andaian/ Assumptions: • Bilangan maksimum murid ialah 30 orang. The maximum number of students is 30. • Mak Cik Minah hanya mempunyai RM75 untuk membeli kerepek kentang. Aunt Minah only has RM75 to buy the potato chips. 2 Encik Zafri ialah seorang penghantar dan dia bekerja 7 jam sehari. Berdasarkan pengalamannya, dia mengambil masa purata 4 jam dalam perjalanan penghantaran setiap hari. Hari ini, dia mesti menghantar 45 buah bungkusan kepada 30 orang pelanggan. Mr Zafri is a delivery man and he works 7 hours a day. Based on his experience, he spent an average of 4 hours on his delivery trip every day. Today, he must send 45 parcels to 30 customers. (a) Apakah masalah yang dihadapi oleh Encik Zafri semasa penghantarannya? What problem does Mr Zafri face during his deliveries? [1 markah/ mark] (b) Berikan tiga andaian yang perlu dibuat oleh Encik Zafri untuk mengatur masa penghantarannya. Give three assumptions that need to be made by Mr Zafri to arrange his delivery schedule. [2 markah/ marks] Jawapan/ Answer: (a) Had masa untuk menghantar semua bungkusan kepada semua pelanggan. The time limit to send all the parcels to all the customers. (b) Andaian/ Assumptions: • Perjalanan penghantarannya tidak mengalami sesak jalan His delivery trip is not caught in traffic jam • Jarak antara setiap pasang pelanggannya adalah sama The distance between each pair of customers is the same • Pelanggannya menerima bungkusan sebaik sahaja dia sampai His customers receive the parcel right when he arrives CONTOH

Matematik Tingkatan 5 Bab 8 123 3 Jadual menunjukkan jualan nasi lemak dan mi goreng di sebuah gerai dalam tiga bulan. The table shows the sales of nasi lemak and fried noodles at a stall in three months. Bulan Month Mei May Jun June Julai July Jualan (bungkus) Sales (packets) Nasi lemak Nasi lemak 25 36 49 Mi goreng Fried noodles 20 30 40 (a) Nyatakan jenis graf fungsi yang boleh digunakan untuk meramalkan jualan masa depan bagi kedua-dua jenis makanan. State the types of graphs of function which can be used to predict future sales for both types of food. [2 markah/ marks] (b) Tentukan dua andaian yang perlu dibuat dalam pemodelan matematik ini. Determine two assumptions that need to be made in this mathematical modelling. [2 markah/ marks] Jawapan/ Answer: (a) Jualan nasi lemak/ Sales of nasi lemak: 25, 36, 49, …, n2 Fungsi kuadratik/ Quadratic function Jualan mi goreng/ Sales of fried noodles: 20, 30, 40, …, 10n Fungsi linear/ Linear function (b) Andaian/ Assumptions: • Penjualan makanan tidak mengalami masalah kekurangan pelanggan The selling of food does not face the lack of customers • Kualiti makanan kekal sama dan tidak menyebabkan bilangan pelanggan berkurang The food quality remains the same and does not cause a decrease in the number of customers 4 Encik Harun mulai memandu ke Ipoh dari Kuala Lumpur pada pukul 10 pagi. Dia menganggarkan bahawa dia akan tiba di Ipoh sebelum pukul 12 tengah hari. Mr Harun starts to drive to Ipoh from Kuala Lumpur at 10.00 a.m. He estimates that he will reach Ipoh before 12 p.m. (a) Diberi jarak antara Kuala Lumpur dan Ipoh ialah 205.2 km. Tentukan laju, dalam km/j, yang Encik Harun perlu pandu supaya anggarannya adalah betul. Given the distance between Kuala Lumpur and Ipoh is 205.2 km. Determine the speed, in km/h, which Mr Harun needs to drive so that his estimation is correct. [2 markah/ marks] (b) Nyatakan satu perkara yang membenarkan Encik Harun membuat anggaran itu dan adakah perkara itu dapat menentusahkan penyelesaian masalahnya. State one matter which allows Mr Harun to make that estimation and does that matter able to verify the solution to his problem. [2 markah/ marks] Jawapan/ Answer: (a) 1200 – 1000 = 2 jam/ hours 205.2 km ÷ 2 jam/ hours = 102.6 km/ j/ km/h (b) Encik Harun andaikan laju pemanduannya adalah kekal sama sepanjang perjalanannya. Perkara ini tidak dapat menentusahkan penyelesaian masalahnya sepenuhnya kerana apabila diterjemah kembali ke dunia sebenar, sesetengah perjalanan mempunyai had laju yang tetap. Mr Harun assumes his driving speed remains the same along his journey. This matter cannot fully verify the solution to his problem when translated to the real-world situation, some routes have fixed speed limits. CONTOH

Matematik Tingkatan 5 Bab 8 124 Bahagian B/ Section B 5 Kereta Danella menggunakan 34 liter petrol untuk bergerak sejauh 306 km. Berapakah petrol, dalam liter, yang akan digunakan jika Danella menghadiri satu mesyuarat di Syarikat A yang terletak 234 km daripada rumahnya? Selesaikan masalah ini melalui pemodelan matematik. Danella’s car consumed 34 litres of petrol to move 306 km. How much petrol, in litre, will be consumed if Danella attends a meeting in Company A which is located 234 km away from her house? Solve this problem through mathematical modelling. [12 markah/ marks] Jawapan/ Answer: Masalah/ Problem: • Tentukan isi padu petrol yang diperlukan untuk suatu jarak sejauh 234 km Determine the volume of petrol required for a distance of 234 km • Diketahui bahawa semakin jauh jarak, semakin banyak petrol digunakan. Oleh itu, isi padu petrol berubah secara langsung dengan jarak perjalanan. We know that the further the distance, the more petrol is consumed. Thus, the volume of petrol varies directly as the distance travelled. Andaian dan pemboleh ubah: Assumptions and variables: • Andaikan kelajuan memandu bagi kedua-dua perjalanan adalah sama. Assume that the driving speed for both routes are the same. • Danella tidak menghadapi sebarang masalah sepanjang perjalanannya. Danella does not face any problems throughout her travel. • Katakan d mewakili jarak perjalanan dan V mewakili isi padu petrol yang digunakan. Let d represents the distance travelled and V represents the volume of petrol consumed. • V berubah secara langsung dengan d, maka V = kd dengan keadaan k ialah pemalar. V varies directly as d, hence V = kd where k is a constant. Mengaplikasi matematik/ Applying mathematics: Gantikan V = 34 dan d = 306 ke dalam V = kd, Substitute V = 34 and d = 306 into V = kd, Maka/ Therefore, V = 1 9 d Persamaan ini menghuraikan hubungan antara isi padu petrol yang digunakan dengan jarak perjalanan. This equation describes the relationship between the volume of petrol consumed with the distance travelled. Apabila/ When d = 234, V = 1 9 (234) = 26 liter/ litres ∴ 26 liter petrol akan digunakan untuk suatu jarak sejauh 234 km. 26 litres of petrol will be consumed for a distance of 234 km. Menentusahkan dan mentafsir penyelesaian: Verifying and interpreting solutions: • Model fungsi linear V = 1 9 d yang diperoleh mungkin tidak dapat digunakan untuk semua situasi perjalanan. The linear function model V = 1 9 d obtained may not be used in all situations. • Hal ini kerana bukan semua perjalanan berada dalam keadaan yang sama. Misalnya, penggunaan petrol akan dipengaruhi sekiranya perjalanan ke Syarikat A sering menghadapi masalah kesesakan lalu lintas. This is because not all routes are in the same conditions. For example, the consumption of petrol will be affected if the route to Company A is always facing traffic jam problems. • Oleh itu, model fungsi linear yang diperoleh tidak sesuai digunakan untuk menangani masalah berkenaan di dunia sebenar. Therefore, the linear function model obtained is not suitable to solve this type of problem in the real-world situation. Memurnikan model matematik: Refining the mathematical model: Dalam masalah ini, kita tidak dapat memurnikan model memandangkan maklumat yang diberi adalah terhad. In this problem, we are not able to refine the model due to limited information given. 34 = k(306) k = 34 306 = 1 9 V O V = 1 9 d CONTOH d

Matematik Tingkatan 5 Bab 8 125 6 Pada awal tahun, Encik Chang menyimpan sebanyak RM100 000 dalam akaun simpanan tetap di Bank PB dengan kadar faedah 4.5% setahun dan pengkompaunan setiap tahun. Terbitkan satu model matematik bagi jumlah simpanan Encik Chang selepas t tahun penyimpanan. Selesaikan masalah ini melalui pemodelan matematik. At the beginning of a year, Mr Chang saves RM100 000 in his savings account at Bank PB with an interest rate of 4.5% per annum and compounded every year. Derive a mathematical model for Mr Chang’s total savings after t years of saving. Solve this problem through mathematical modelling. [12 markah/ marks] Jawapan/ Answer: Masalah/ Problem: • Encik Chang diberikan faedah kompaun dengan pengkompaunan sekali setahun. Mr Chang is given compound interest which compounds once per annum. • Prinsipal Encik Chang ialah RM100 000. Mr Chang’s principal is RM100 000. • Kadar faedah tahunan ialah 4.5%. The interest rate is 4.5% per annum. • Tertibkan satu model matematik bagi jumlah simpanan Encik Chang pada akhir tahun ke-t. Derive a mathematical model for Mr Chang’s total savings at the end of tth year. Andaian dan pemboleh ubah: Assumptions and variables: • Andaikan Encik Chang tidak mengeluarkan atau menambah wang simpanannya sepanjang tempoh penyimpanannya. Assumes Mr Chang did not withdraw or raise his savings throughout the period of saving. • Pemboleh ubah yang terlibat ialah prinsipal, RMP, kadar faedah tahunan, r = 0.045, bilangan kali faedah dikompaun, n dan masa, t tahun. The variables involved are the principal, RMP, the annual interest rate, r = 0.045, the number of times the interest is compounded, n and the time, t years. Mengaplikasi matematik untuk menyelesaikan masalah: Applying mathematics to solve the problem: Faedah kompaun ialah faedah yang dihitung berdasarkan hasil tambah prinsipal asal dan faedah yang terkumpul daripada penyimpanan sebelumnya, maka hasil simpanan boleh dihitungkan seperti jadual berikut: Compound interest is an interest which is calculated based on the sum of original principal and the accumulated interest from previous periods of savings, thus the amount of savings can be calculated as the following table: Tahun Year Prinsipal (RM) Principal (RM) Faedah yang diterima (RM) Collected interest (RM) Prinsipal + Faedah (RM) Principal + Interest (RM) Hasil simpanan (RM) Amount of savings (RM) 1 100 000 100 000 × 0.045 100 000 + (100 000 × 0.045) = 100 000(1 + 0.045) 100 000(1.045) 2 100 000(1.045) 100 000(1.045) × 0.045 100 000(1.045) + [100 000(1.045) × 0.045] = 100 000(1.045)(1 + 0.045) 100 000(1.045)2 3 100 000(1.045)2 100 000(1.045)2 × 0.045 100 000(1.045)2 + [100 000(1.045)2 × 0.045] = 100 000(1.045)2 (1 + 0.045) 100 000(1.045)3 4 100 000(1.045)3 100 000(1.045)3 × 0.045 100 000(1.045)3 + [100 000(1.045)3 × 0.045] = 100 000(1.045)3 (1 + 0.045) 100 000(1.045)4 5 100 000(1.045)4 100 000(1.045)4 × 0.045 100 000(1.045)4 + [100 000(1.045)4 × 0.045] =100 000(1.045)4 (1 + 0.045) 100 000(1.045)5 Berdasarkan jadual, hasil simpanan ialah suatu kuasa 1.045 didarab dengan prinsipal simpanan sebanyak RM100 000 dan kuasa adalah sepadan dengan bilangan tahun. Based on the table, the amount of savings is a power of 1.045 times the principal, RM100 000 and the power corresponds to the number of years. CONTOH

Matematik Tingkatan 5 Bab 8 126 Oleh sebab 1.045 = 1 + r, dengan menggunakan pemboleh ubah yang ditetapkan, model matematik bagi masalah ini boleh ditulis sebagai A(t) = P(1 + r)t dengan keadaan A(t) sebagai hasil simpanan dan t sebagai bilangan tahun. As 1.045 = 1 + r, using the set variables, the mathematical model for this problem can be written as A(t) = P(1 + r)t where A(t) as the amount of savings and t as the number of years. Hasil simpanan, A(t) (RM) Amount of saving, A(t) (RM) Masa, t (tahun) Time, t (years) 0 10 20 30 40 50 200 000 400 000 600 000 800 000 1 000 000 A(t) = 100 000(1.045)t Graf bagi model matematik ini menunjukkan suatu fungsi eksponen dengan pertumbuhan eksponen bagi hasil simpanan, A(t), apabila masa, t, bertambah. The graph for this mathematical model shows an exponent function with an exponential growth on the amount of savings, A(t), as time, t, increases. Memurnikan model matematik: Refining the mathematical model: Model matematik ini, A(t) = P(1 + r)t , digunakan untuk menghitung faedah kompaun dengan kekerapan pengkompaunan sekali setahun. Sekiranya pengkompaunan berlaku beberapa kali, n, setahun, maka model matematik ini boleh ditambah baik seperti berikut: This mathematical model, A(t) = P(1 + r)t , is used to calculate the annual compounding. If the compounding occurs few times, n, annually, thus this mathematical model can be improved as follows: A(t) = P(1 + r n ) t CONTOH

127 Tulis jawapan anda pada ruang jawapan yang disediakan. Anda boleh menggunakan kalkulator saintifik yang tidak boleh diprogramkan. 1 Arifi merupakan seorang atlet berbasikal yang akan menyertai acara berbasikal peringkat negeri. Oleh itu, dia sering berbasikal setiap hari di kawasan persekitaran rumahnya sebagai langkah persediaan. Arifi is a cycling athlete that will participate in a cycling event at state level. Therefore, he always cycles every day in his housing area as a preparation. (a) Diberi bahawa laluan latihan itu merupakan cantuman dua sektor bulatan yang berpusat O seperti dalam Rajah 1. Bagi setiap pusingan, Arifi akan bermula di titik A, kemudian melalui B, C, D dan kembali semula ke titik A dengan melalui titik O. AOD ialah garis lurus. Given that the training route is a combination of two circular sectors with centre O as shown in Diagram 1. For each round, Arifi will start at point A, than passes through points B, C , D and returns to point A by passing through point O. AOD is a straight line. A B C D O 120° 7 km 4 km Rajah 1/ Diagram 1 (i) Hitung jarak, dalam km, setiap laluan berbasikal yang dilalui oleh Arifi. Beri jawapan kepada dua tempat perpuluhan. Calculate the distance, in km, of each complete route of cycling travelled by Arifi. Give the answer to two decimal places. [Guna/ Use π = 22 7 ] [3 markah/ marks] (ii) Target Arifi ialah berbasikal dengan jarak minimum 65 km sehari. Hitung bilangan pusingan lengkap yang perlu dilalui oleh Arifi untuk mencapai targetnya. Arifi’s target is to cycle a minimum distance of 65 km per day. Calculate the number of complete routes that Arifi needs to travel to achieve his target. [2 markah/ marks] (b) Arifi berbasikal dengan laju seragam 8 m s–1 dalam 42 saat yang pertama. Kemudian, dia meningkatkan kelajuan sehingga mencapai 14 m s–1 pada saat ke-70 dan seterusnya menyahpecut sehingga berhenti pada minit ke-2. Arifi cycles at a uniform speed of 8 m s–1 in the first 42 seconds. Then, he increases the speed to 14 m s–1 at 70th second and hence decelerates constantly until stop at the 2nd minute. (i) Lakarkan graf laju-masa di ruang jawapan berdasarkan maklumat yang diberikan. Sketch the speed-time graph in the answer space based on the information given. [2 markah/ marks] (ii) Daripada graf di (b)(i), hitung nyahpecutan, dalam m s–2, basikal itu dalam tempoh 50 saat terakhir. From the graph in (b)(i), calculate the deceleration, in m s–2, of the bicycle in the last 50 seconds. [2 markah/ marks] (iii) Hitung jumlah jarak, dalam m, yang dilalui dalam tempoh 2 minit itu. Calculate the total distance, in m, travelled in the period of 2 minutes. [3 markah/ marks] (c) Keesokan harinya, Arifi terlibat dalam kemalangan ketika menjalani latihan. Oleh itu, dia menggunakan polisi insurans perubatan utama dengan peruntukan deduktibel sebanyak RM400 dan fasal penyertaan peratusan ko-insurans 80/20 dalam polisinya. Diberi kos perubatan yang dilindungi polisinya berjumlah RM15 300. On the next day, Arifi was involved in an accident while training. Hence, he used his major medical insurance policy with a deductible provision of RM400 and 80/20 co-insurance percentage clause in his policy. Given the medical cost covered by his policy is RM15 300. (i) Hitung kos perubatan selepas deduktibel. Calculate the medical cost after deductible. [1 markah/ mark] (ii) Hitung bayaran kos yang perlu ditanggung oleh Arifi sendiri. Calculate the cost to be borne by Arifi himself. [2 markah/ marks] Soalan SPM Kertas 2 Bahagian C CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 128 Jawapan/ Answers: (a) (i) Perimeter laluan Perimeter of the route = AB + BC + CD + DO + OA = ( 60 360 × 2 × 22 7 × 7) + 3 + ( 120 360 × 2 × 22 7 × 4) + 4 + 7 = 29.71 km (ii) Target jarak/ Distance target Jarak laluan/ Distance route = 65 29.71 = 2.19 � 3 ⸫ 3 pusingan lengkap/ complete routes (b) (i) Laju (m s–1) Speed (m s–1) O 42 8 14 70 120 Masa (saat) Time (second) (ii) Laju/ Speed Masa/ Time = 0 – 14 120 – 70 = – 0.28 m s−2 (iii) Jumlah jarak yang dilalui/ Total distance travelled: = Luas di bawah graf/ Area under the graph = (42 × 8) + [ 1 2 × (8 + 14) × 28] + ( 1 2 × 50 × 14) = 336 + 308 + 350 = 994 m (c) (i) Kos perubatan selepas deduktibel: Medical cost after deductible: = RM15 300 – RM400 = RM14 900 2 SMK Sulaiman telah mengadakan Minggu Sains dan Matematik pada minggu lepas. Sebahagian murid tingkatan lima telah menyertai cabaran matematik. Rajah 2 menunjukkan salah satu soalan matematik dalam cabaran matematik itu. SMK Sulaiman has organised Science and Mathematics Week last week. Some of the form five students had participated the mathematics challenge. Diagram 2 shows one of the mathematics questions in the mathematics challenge. r s p 21124 12206 q 13005 11117 11x Rajah 2/ Diagram 2 (a) (i) Berdasarkan Rajah 2, tulis dua persamaan matriks. Based on Diagram 2, write two matrix equations. (ii) Hitung nilai-nilai bagi p, q, r dan s. Calculate the values of p, q, r and s. (iii) Seterusnya, cari nilai x jika 11x ialah suatu nombor dalam asas sepuluh. Hence, find the value of x if 11x is a number in base ten. [7 markah/ marks] (ii) Kos ditanggung oleh Arifi: Cost borne by Arifi: = ( 20 100 × RM14 900) + RM400 = RM3 380 CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 129 (b) Jadual 1 menunjukkan skor yang diperoleh peserta cabaran matematik. Table 1 shows the scores obtained by the participants of the mathematics challenge. Skor/ Scores 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 Bilangan peserta Number of participants 4 6 13 18 9 Jadual 1/ Table 1 Tentukan min dan sisihan piawai bagi skor peserta dalam cabaran matematik itu. Seterusnya, komen tentang prestasi murid-murid tingkatan 5 dalam cabaran matematik ini. Determine the mean and standard deviation of the scores of the participants. Hence, comment on the performance of the form 5 students in this mathematics challenge. [5 markah/ marks] (c) SMK Sulaiman mempunyai 86 orang murid tingkatan 5 dan antara 60 orang murid telah menyertai cabaran sains. SMK Sulaiman has 86 form 5 students and 60 of them has participated in the science challenge. (i) Hitung bilangan murid yang menyertai kedua-dua cabaran. Calculate the number of students who has participated in both challenges. (ii) Diberi S mewakili bilangan murid yang menyertai cabaran sains dan M mewakili bilangan murid yang menyertai cabaran matematik. Lukis satu gambar rajah Venn untuk menunjukkan penyertaan cabaran oleh murid-murid tingkatan 5. Given that S represents the number of students who participated the science challenge and M represents the number of students who participated the mathematics challenge. Draw a Venn diagram to show the participation in the challenges by the form 5 students. [3 markah/ marks] Jawapan/ Answers: (a) (i) p q [r s] = 21124 13005 12206 11117 [p q] r s = [11x] (ii) p q [r s] = pr qr ps qs = 21124 13005 12206 11117 = 150 200 300 400 pr qr = 150 200 p q = 3 4 Apabila/ When p = 3, 3r = 150 r = 50 Apabila/ When p = 3, 3s = 300 s = 100 ⸫ p = 3, q = 4, r = 50, s = 100 (iii) [p q] r s = [pr + qs] 11x = 550 = [3(50) + 4(100)] x = 50 = [150 + 400] = [550] = [11x] CONTOH