MODULE AND MORE Matematik Tambahan Tingkatan 5

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 48 BAB 2 23. Selesaikan masalah yang melibatkan kadar perubahan bagi kuantiti yang terhubung dan mentafsir penyelesaian tersebut. Solve the problems involving the rate of change of related quantities and interpret the answers. TP 5 (a) Sebuah kon tegak yang terbalik mempunyai sudut separa 30°. Jika kon itu diisi dengan minyak hingga penuh dan minyak kemudiannya dibiar mengalir keluar dari muncung kon dengan kadar 4 cm3 s−1. Cari kadar perubahan aras minyak di dalam kon yang menurun ketika tinggi aras minyak itu ialah 4 cm. An inverted cone has a semi angle of 30°. If the cone is filled fully with oil and then the oil is allowed to drip out from the tip of the cone at a rate of 4 cm3 s−1. Find the rate of change of the height of oil in the cone that is decreasing when the height of the oil is 4 cm. CONTOH Setelah dipanaskan, panjang sisi x cm bagi sebuah kubus bertambah dengan kadar malar 0.002 cm s−1. Cari Once heated, the length of the side of x cm of a cube increases with a constant rate of 0.002 cm s−1. Find (i) kadar perubahan isi padunya jika x = 5 cm, the rate of change in the volume if x = 5 cm, (ii) kadar perubahan dalam luas permukaan kubus jika x = 6 cm. the rate of change in the surface area of cube if x = 6 cm. Penyelesaian: (i) Katakan V ialah isi padu kubus. Let V is the volume of cube. Maka/Hence V = x3 dV dx = 3x2 dV dt = dV dx × dx dt Diberi/Given dx dt = 0.002 cm s−1, x = 5 cm dV dt = 3x2  dx dt  = 3(5)2 (0.002) = 0.15 cm3 s−1 Isi padu bertambah dengan kadar 0.15 cm3 s−1. The volume increases with a rate of 0.15 cm3 s−1. (ii) Katakan luas permukaan ialah L. Let the surface area is L. L = 6x2 dL dx = 12x dL dt = dL dx × dx dt Diberi/Given x = 6 cm dan/and dx dt = 0.002 dL dt = 12(6)(0.002) = 0.144 cm2 s−1 Luas permukaan bertambah dengan kadar 0.144 cm2 s−1. The surface area increases with a rate of 0.144 cm2 s−1. tan 30° = r h = 1 3 r = h 3 Isipadu V = 1 3 πr2 h Volume = 1 3 π h 3 2 h V = 1 9 πh3 dV dh = 1 3 πh2 dV dt = 1 3 πh2 dh dt h 30fi r x x x Jika h = 4 cm, dV dt = –4 cm3 s–1 –4 = 1 3 π(4)2 dh dt 3 – 4π = dh dt ∴ dh dt = –3 4π cm s–1 h menyusut h decreases

BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 49 2 (b) Luas sebuah bulatan bertambah dengan kadar π cm2 s−1. Tentukan kadar perubahan jejarinya apabila jejarinya ialah 3.5 cm. The area of a circle increases at a rate of π cm2 s−1. Determine the rate of change of its radius when the radius is 3.5 cm. (c) Isi padu, V cm3 sebuah silinder menyusut dengan kadar 1.2 cm3 s−1 dan tingginya h cm sentiasa dua kali panjang jejari, r cm tapak silinder. Cari kadar perubahan jejari apabila r ialah 7 cm. The volume, V cm3, of a cylinder decreases at a rate of 1.2 cm3 s−1 and its height of h cm is always twice of the radius, r cm of the base of the cylinder. Find the rate of change of the radius when the r is 7 cm. 24. Tentukan perubahan kecil dan penghampiran suatu kuantiti bagi setiap yang berikut. Determine the small change and the approximated quantity for each of the following. TP 4 h r L = πr2 dL dr = 2πr dL dt = 2πr dr dt π = 2πr dr dt  1 = 2(3.5) dr dt dr dt = 1 7 cm s–1 Jejari bertambah The redius increases h = 2r V = πr2 h = πr2 (2r) V = 2πr3 dV dr = 6πr2 dV dt = 6πr2 dr dt –1.2 = 6π(7)2 dr dt dr dt = –1.2 6π(7)2 = –0.0013 cm s–1 Jejari menyusut The radius decreases CONTOH Diberi y = x2 + x, cari tokokan kecil dalam y apabila x berubah daripada 2 kepada 1.98. Seterusnya, cari nilai baharu y. Given that y = x2 + x, find the small change in y when x changes from 2 to 1.98. Then, find the new value of y. Penyelesaian: Diberi/Given y = x2 + x dy dx = 2x + 1 x berubah daripada 2 kepada 1.98. x changes from 2 to 1.98. Jadi/So δx = 1.98 – 2 = −0.02 δy δx ≈ dy dx δy ≈ dy dx ·δx ≈ (2x + 1)δx ≈ (2(2) + 1)(−0.02) ≈ −0.1 Jadi, y menyusut sebanyak 0.1. So, y decreases 0.1. Apabila/When x = 2, y = 22 + 2 = 6 Maka, nilai baharu y Hence, the new value of y = y + δy = 6 + (−0.1) = 5.9

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 50 BAB 2 (a) Diberi y = 4x − 3, cari tokokan kecil dalam y apabila x berubah daripada 5 kepada 5.01. Seterusnya, cari nilai baharu y. Given that y = 4x − 3, find the small change in y when x changes from 5 to 5.01. Then, find the new value of y. (b) Diberi y = 9 x2 , cari dy dx . Seterusnya, cari nilai hampir bagi 4 (2.98)2 . Beri jawapan betul kepada dua angka bererti. Given that y = 9 x2 , find dy dx . Then, find the approximate value for 4 (2.98)2 . Give your answer correct to two significant figures. (c) Diberi y = x2 (x + 2), cari tokokan kecil dalam y apabila x berubah daripada 3 kepada 2.99. Given that y = x2(x + 2), find the small change in y when x changes from 3 to 2.99. ≈ 4 δx ≈ 4 (0.01) ≈ 0.04 Jadi, y bertambah sebanyak 0.04. So, y increases 0.04. Apabila x = 5, y = 17 Maka, nilai baharu y Hence, the new value of y = y + δy = 17 + (0.04) = 17.04 Diberi y = 4x − 3 dy dx = 4 x berubah daripada 5 kepada 5.01 x changes from 5 to 5.01 Jadi/So δx = 5.01 – 5 = 0.01 δy δx ≈ dy dx δy ≈ dy dx · δx Diberi y = 9 x2 dy dx = 18 – x3 x berubah daripada 3 kepada 2.98. x changes from 3 to 2.98. Jadi, δx = 2.98 − 3 = −0.02 δy δx ≈ dy dx δy ≈ dy dx · δx ≈ 18 – x3 · δx ≈ 18 – (3)3 (−0.02) ≈ 0.013 Jadi, y bertambah sebanyak 0.013. So, y increases 0.013. Apabila x = 3, y = 1 Maka, nilai baharu y = y + δy Hence, the new value of y = y + δy = 1 + 0.013 = 1.013 Diberi y = x2 (x + 2) dy dx = 3x2 + 4x x berubah daripada 3 ke 2.99. x changes from 3 to 2.99. Jadi, δx = 2.99 – 3 = −0.01 δy δx ≈ dy dx δy ≈ dy dx · δx ≈ (3x2 + 4x)δx ≈ [3(3)2 + 4(3)](−0.01) ≈ −0.39 Jadi, y menyusut sebanyak 0.39. so, y decreases 0.39.

BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 51 2 25. Selesaikan setiap yang berikut. Solve each of the following. TP 4 (a) Sebuah silinder tanpa penutup mempunyai tinggi 8 cm dan jejari j cm. Jika luas permukaan silinder itu ialah L cm2 , A cylinder without a cover has a height of 8 cm and a radius of j cm. If the surface area of the cylinder is L cm2, (i) tunjukkan bahawa L = π(j 2 + 16j). show that L = π(j2 + 16j). (ii) Seterusnya, cari perubahan kecil bagi luasnya jika jejari silinder itu menyusut daripada 4 cm kepada 3.98 cm. Then, find the approximate change in the area if the radius of the cylinder decreases from 4 cm to 3.98 cm. (b) Isi padu, V cm3 , bagi suatu bekas diberi oleh V = x(6x − 4), dengan keadaan x cm ialah kedalaman cecair itu. Tunjukkan bahawa perubahan hampir dalam isi padu cecair apabila kedalaman cecair berubah daripada 10 cm kepada 10.01 cm ialah 1.16 cm3 The volume of a container, V cm3 , is given by V = x(6x − 4), where x cm is the depth of the liquid. Show that the approximate change in the volume of the liquid when the height changes from 10 cm to 10.01 cm is 1.16 cm3. (c) Tempoh ayunan, T s, satu bandul yang mempunyai panjang L cm diberi oleh T = 2π L 10 . Cari dT dL dan seterusnya hitung perubahan dalam L jika T berubah daripada 0.8 s kepada 0.825 s. The period of oscillation, T s, of a pendulum with a length of L cm is given by T = 2π L 10 . Find dT dL and then calculate the change in L if T changes from 0.8 s to 0.825 s. CONTOH Jejari sebuah sfera bertambah daripada 3 cm kepada 3.01 cm. Cari perubahan kecil bagi isi padunya secara penghampiran The radius of a sphere increases from 3 cm to 3.01 cm. Find the approximate change in its volume. Penyelesaian: Isi padu sfera, V = 4 3 πr3 , dengan r ialah jejari sfera. Volume of sphere, V = 4 3 πr3, where r is the radius of sphere. dV dr = 4πr 2 Diberi jejari sfera bertambah daripada 3 ke 3.01 cm. Given that the radius of sphere increases from 3 to 3.01 cm. Maka/Hence δr = 3.01 – 3 = 0.01 cm δV δr ≈ dV dr δV ≈ dV dr · δr ≈ (4πr2 )δr ≈ 4π(3)2 (0.01) = 0.36π cm3 Jadi, isi padu sfera bertambah sebanyak 0.36π cm3 . So, the volume of sphere increases 0.36π cm3. 8 cm j V = 6x2 – 4x dV dx = 12x – 4 δV = dV dx · δx δx = 0.01, x = 10 = (12x – 4)δx = (120 – 4)(0.01) = 1.16 cm3 (i) L = 2πj × 8 + πj 2 = π[16j + j 2 ] (ii) δj = 3.98 – 4 = –0.02 cm dL dj = 16π + 2πj Apabila j = 4 cm dan δj = –0.02 δL = dL dj . δj = [16π + 2π(4)])(–0.02) = 24π(–0.02) = –0.48π cm2 Luas menyusut sebanyak 0.48π cm2 The area decreases 0.48π cm2 T = 2π L 10 = 2π 10 L 1 2 dT dL = π 10 1 L δT = 0.025 δT = dT dL . δL 0.025 = π 10(0.8) δL 0.025 8 π = δL δL = 0.0225 cm

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 52 BAB 2 PRAKTIS PRAKTIS SPM SPM 12 Kertas 1 1. Diberi x = 3t 3 − 2 dan dy dt = 15t 2 , cari Given x = 3t3 − 2 and dy dt = 15t2, find (a) dx dt dalam sebutan t. dx dt in terms of t. (b) dy dx x = 3t 3 – 2, dy dt = 15t 2 (a) dx dt = 9t 2 (b) dy dx = dy dt · dt dx = 15t 2 × 1 9t 2 = 5 3 2. Rajah menunjukkan sebahagian daripada lengkung y = x – 3 x + 1 dan satu garis lurus y = qx + 2. The diagram shows a part of the curve y = x – 3 x + 1 and the straight line y = qx + 2. y x P O y = qx+2 y = x–3 x+1 Diberi bahawa garis lurus itu selari dengan tangen kepada titik P pada lengkung itu. Cari nilai q. Given that the straight line is parallel to the tangent at point P on the curve. Find the value of q. y = x – 3 x + 1 dy dx = (x + 1) – (x – 3) (x + 1)2 = 4 (x + 1)2 Apabila y = 0, x = 3 P(3, 0) When Pada P(3, 0), dy dx = 4 (3 + 1)2 = 1 4 At q = 1 4 3. Rajah menunjukkan suatu graf kubik y = f(x) dan graf linear y = g(x). Titik-titik A, B, C dan D terletak pada lengkung manakala E terletak pada garis lurus. Tangen pada B dan D adalah selari dengan paksi-x. The diagram shows a cubic graph y = f(x) and a linear graph y = g(x). The points A, B, C and D lie on the curve while E is on the line. The tangents at B and D are parallel to the x-axis. y x O A B C D E Nyatakan titik yang manakah memenuhi syaratsyarat berikut: State the point(s) which satisfy the following conditions: (a) dy dx = 0 (b) dy dx  0 (c) dy dx  0 (a) dy dx = 0 pada titik B dan D. at point B and D. (b) dy dx  0 pada titik C dan E. at point C and E. (c) dy dx  0 pada titik A. at point A. 2014 2016 2017

BAB Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 53 2 4. Carikan koordinat titik pegun bagi lengkung y = x(x − 2)2 dan tentukan sifat bagi titik-titik itu. Find the coordinates of the stationary points of the curve y = x(x − 2)2 and determine the characteristics of these points. y = x(x – 2)2 dy dx = 2x(x –2) + (x – 2)2 = (x – 2)(2x + x – 2) = (x – 2)(3x – 2) dy dx = (x – 2)(3x – 2) = 0 x = 2 ; x = 2 3 y = 0 2 3  2 3 – 2 2 = 32 27 d2 y dx2 = (x – 2)(3)+ (3x – 2) = 6x – 8 Apabila x = 2 d2 y dx2 = 12 – 8 = 4  0 When ∴(2, 0) ialah titik minimum is a minimum point Apabila x = 2 3 , d2 y dx2 = 6 2 3  – 8 When = –4  0  2 3 , 32 27 ialah titik maksimum is a minimum point Kertas 2 1. Diberi bahawa persamaan lengkung y = 2x – 4 1 + x2 dan lengkung ini melalui titik P(2, 0). Given the equation of the curve y = 2x – 4 1 + x2 and the curve passes through the point P(2, 0). (a) Cari kecerunan pada titik P. Find the gradient at the point P. (b) Hitung persamaan garis normal kepada lengkung pada titik P. Calculate the equation of the normal line to the curve at point P. (a) dy dx = 2(1 + x2 ) – (2x – 4)(2x) (1 + x2 )2 = 2[1 + x2 – 2x2 + 4x] (1 + x2 )2 Apabila x = 2, dy dx = 2[1 + 4 – 8 + 8] 25 dy dx = 2 5 (b) persamaan garis normal Equation of the normal line y = – 5 2 x + c 0 = – 5 2 (2) + c c = 5 ∴y = – 5 2 x + 5 2. Rajah menunjukkan sebuah rantau tertutup yang dibina daripada seutas dawai dengan panjang 120 cm. The diagram shows an enclosed area that is built by a wire of the length 120 cm. A F D E C B Diberi AF = CD dan ABC dan DEF adalah dua semibulatan dengan jejari j cm. Given that AF = CD and ABC and DEF are two semicircles with radius j cm. (a) Tunjukkan bahawa luas rantau tertutup, L ialah L = (120j − 2πj 2 ) cm2 Show that the enclosed area L is L = (120j − 2πj2) cm2 (b) Cari nilai j supaya luas rantau tertutup L ialah maksimum. Cari nilai L tersebut. Find the value of j so that the enclosed area L is maximum. Hence find the value of L. (a) Perimeter = 120 = 2πj + 2x 60 = πj + x Luas, L = x(2j) Area = 2j[60 – πj] = 120j – 2πj 2 (b) dL dj = 120 – 4πj = 0 j = 120 4π = 30 π cm d2 L dj2 = –4π  0 ∴ Luas ialah maksimum The area is maximum Apabila j = 30 π cm L = 120 30 π  – 2π 30 π  2 = 1800 π 2015 2016 A F D E C B x x j Praktis SPM Ekstra

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan 54 BAB 2 Sudut Sudut KBAT KBAT 1. Air mengalir keluar daripada sebuah kon yang tertonggeng dengan kadar 6 cm3 s−1. Jejari tapak dan tinggi kon masing-masing ialah 10 cm dan 15 cm. Hitung Water flows out of an inverted cone at a rate of 6 cm3 s−1. The radius and the height of the cone are 10 cm and 15 cm respectively. Calculate (a) kadar perubahan ketinggian paras air yang berkurang the rate of change of the height of the water that is decreases (b) kadar perubahan luas permukaan air yang berkurang the rate of change of the surface area of the water that is decreases apabila paras air adalah setinggi 9 cm. when the height of the water is 9 cm. (a) dV dt = –6 cm3 s–1 r 10 = h 15 r = 2 3 h V = 1 3 πr2 h = 1 3 π 2 3 h 2 h V = 4 27πh3 dV dh = 4 9 πh2 dV dt = 4 9 πh2 · dh dt –6 = 4 9 π[9]2 dh dt –6 36π = dh dt dh dt = – 1 6π cm s–1 (b) L = πr2 = π[ 2 3 h] 2 L = 4 9 πh2 dL dh = 8 9 πh dL dt = 8 9 πh dh dt = 8 9 π[9]– 1 6π = – 4 3 cm2 s–1 2. Rajah menunjukkan lengkung y = x2 dan titik R(−1, 0). Titik P(p, 0) adalah satu titik yang berubah pada paksi-x dan p bertambah dengan kadar 10 unit s−1. PQ adalah selari dengan paksi-y. The diagram shows a curve y = x2 and point R(−1, 0). Point P(p, 0) is a variable point on the x-axis and p increases at a rate of 10 unit s−1. PQ is parallel to the y-axis. Q O y x y = x2 R(–1, 0) P(p, 0) (a) Nyatakan luas, A unit2 , bagi PQR dalam sebutan p. State the area, of A unit2, for PQR in terms of p. (b) Hitung kadar perubahan A yang bertambah pada ketika p = 2. Calculate the rate of change of A that is increasing when p = 2. (c) Hitung kadar perubahan bagi panjang PQ pada ketika p = 1. Calculate the rate of change of the length of PQ when p = 1. (a) dp dt = 10 unit s–1 A = 1 2 [p + 1]p2 A = 1 2 p3 + 1 2 p2 (b) dA dp = 3 2 p2 + p dA dt =  3 2 p + p · dp dt p = 2, dA dt =  3 2 (2)2 + 210 = 80 unit2 s–1 (c) p2 = PQ d(PQ) dp = 2p 15 cm 10 cm r h p2 y x y = x2 R(–1, 0) 0 P(p, 0) KBAT Ekstra d(PQ) dt = 2p · dp dt p = 1, d(PQ) dt = 2(1)(10) = 20 unit s–1 Quiz 2

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 55 3 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN BAB 3 Pengamiran Integration 3.1 Pengamiran Sebagai Songsangan Pembezaan Integration as the Inverse of Differentiation 1. Dalam proses pembezaan, jika y = 1 2 x2 + c, dengan c ialah pemalar, maka fungsi kecerunan dy dx = x. Secara songsang, jika diberikan fungsi kecerunan dy dx = x, maka persamaan lengkung ialah y = 1 2 x2 + c. In the process of differentiation, if y = 1 2 x2 + c, where c is a constant, then the gradient function is dy dx = x. Conversely, if the gradient function is given by dy dx = x, then the equation of the curve is y = 1 2 x2 + c. 2. Proses ini adalah songsangan bagi pembezaan dan dikenali sebagai pengamiran. This process is the inverse of differentiation and is called integration. 3. Tata tanda bagi pengamiran ialah ∫ f(x)dx. The notation for integration is ∫f(x) dx. Sebagai contoh, dy dx = x For example, dy = x dx ∫dy = ∫ x dx y = 1 2 x2 + c. CONTOH 1 1. Cari kamiran melalui songsangan pembezaan terhadap x. Find the integral by inversing the differentiation with respect to x. TP 2 (a) Diberi/Given d dx ( 2 3 x6 ) = 4x5 , cari/find ∫4x5 dx. (b) Diberi/Given dy dx 4f(x) = g(x), cari/find 3∫ Diberi/Given g(x) dx. d dx (3x3 ) = 9x2 , Cari/Find ∫9x2 dx. Penyelesaian: d dx (3x3 ) = 9x2 d(3x3 ) = 9x2 ·dx ∫d(3x3 ) = ∫9x2 dx ∫9x2 dx = 3x3 + c Tip ∫9x2dx boleh ditulis sebagai 9∫x2dx. ∫9x2dx can be written as 9∫x2 dx. d dx  2 3 x6  = 4x5 d 2 3 x6  = 4x5 ·dx ∫d 2 3 x6  = ∫4x5 dx ∫4x5 dx = 2 3 x6 + c 3∫g(x)dx = 3[4f(x)] + c = 12f(x) + c

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 56 BAB 3 (c) Diberi d dx (2x – 5) (3 – 4x)3 = 1 2 h(x), cari ∫ 1 a h(x)dx, dengan a ialah pemalar. Given d dx (2x – 5) (3 – 4x)3 = 1 2 h(x), find ∫ 1 a h(x)dx, such that a is a constant. (d) Diberi y = (x − 1)(x − 3)2 , cari dy dx . Seterusnya, cari ∫(x − 3)(3x − 5) dx. Given y = (x − 1)(x − 3)2, find dy dx. Then, find ∫(x − 3)(3x − 5)dx. (e) Diberi y = 2x x + 1 , cari dy dx . Seterusnya, cari ∫ 6 (x + 1)2 dx. Given y = 2x x + 1 , find dy dx . Then , find ∫ 6 (x + 1)2 dx. (f) Diberi y = x + 1 x2 dan dy dx = f(x), dengan π ialah pemalar. Cari ∫ 2 π f(x)dx. Given y = x + 1 x2 and dy dx = f(x), such that π is a constant. Find ∫ 2 π f(x) dx. CONTOH 2 CONTOH 3 Diberi y = x2 (2 − x)2 , cari dy dx . Seterusnya, cari ∫ x(2 − x)(4 − 3x)dx. Given y = x2(2 − x)2, find dy dx . Then, find ∫ x(2 − x)(4 − 3x)dx. Penyelesaian: y = x2 (2 − x)2 dy dx = −x2 (2 − x) + 2x(2 − x)2 = x(2 − x)[−x + 2(2 − x)] = x(2 − x)(4 −3x) Maka/Hence ∫ x(2 − x)(4 − 3x)dx = x2 (2 − x)2 y = (x − 1)(x − 3)2 dy dx = 2(x − 1)(x − 3) + (x − 3)2 = (x − 3)(2x − 2 + x − 3] = (x − 3)(3x – 5) ∫(x − 3)(3x − 5)dx = (x − 1)(x − 3)2 ∫f(x)dx = x + 1 x2 + c 2 π ∫f(x)dx = 2 π  x + 1 x2  + c y = 2x x + 1 = 2(x + 1) – 2x (x + 1)2 = 2 (x + 1)2 ∫ 2 (x + 1)2 dx = 2x (x + 1) + c ∫ 6 (x + 1)2 dx = 3∫ 2 (x + 1)2 dx = 3(2x) (x + 1) + c = 6x (x + 1) + c ∫ 1 2 h(x)dx = (2x – 5) (3 – 4x)3 + c ∫ 1 a h(x)dx = 2 a  (2x – 5) (3 – 4x)3  + c Diberi y = x 2 – x , cari dy dx . Seterusnya, cari ∫ 1 (2 – x)2 dx. Given y = x 2 – x , find dy dx . Then, find ∫ 1 (2 – x)2 dx. Penyelesaian: y = x 2 – x dy dx = (2 – x)(1) + x (2 – x)2 = 2 (2 – x)2 ∫ 2 (2 – x)2 dx = x 2 – x ∫ 1 (2 – x)2 dx = x 2(2 – x) 2∫ 1 (2 – x) 2 dx = ∫ 2 (2 – x) 2 dx

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 57 NOTA IMBASAN NOTA IMBASAN 3 3.2 Kamiran Tak Tentu Indefinite Integral 1. Kamiran axn terhadap x. Integrate axn with respect to x. ∫axndx = axn + 1 (n + 1) + c, n ≠ −1, a dan c ialah pemalar. ∫axndx = axn + 1 (n + 1) + c, n ≠ −1, where a and c are constants. 2. Kamiran bagi ungkapan algebra Integration for algebraic expressions ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫ g(x) dx 3. Kamiran (ax + b) n terhadap x. Integrate (ax + b)n with respect to x. Katakan u = ax + b, maka Let u = ax + b, then du dx = adx = du a ∫(ax + b) n dx = ∫un du a = un + 1 a(n + 1) + c = (ax + b)n + 1 a(n + 1) , n ≠ −1, a, b dan n ialah pemalar. a, b, and n are constants. 2. Cari kamiran yang berikut terhadap x. Find the integral with respect to x of the following. TP 2 (a) ∫ 3 4 x dx (b) ∫ 2 3x3 dx (c) ∫ x (2x)4 dx (d) ∫ 5 x 2x2 dx CONTOH (i) ∫ 1 2 dx (ii) ∫ x dx Penyelesaian: (i) ∫ 1 2 dx = 1x0 + 1 2 + c = 1 2 x + c (ii) ∫ x dx = x 1 2 dx = x 1 2 + 1 1 2 + 1 + c = 2 3 x x + c mesti tukar kepada bentuk axn . must convert to axn . Tip x 1 2 + 1 = (x)(x) 1 2 = x x ∫ 3 4 x dx = 3 4 x2 2 + c = 3 8 x2 + c ∫ 2 3x3 dx = ∫ 2 3 x–3 dx = 2 3 x–2 –2 + c = 1 – 3x2 + c ∫ x (16x4 ) dx = ∫ 1 16 x–3dx = 1 16 1 (–2)x2 + c = 1 – 32x2 + c ∫ 5x 1 2 2x2 dx = ∫ 5 2 x – 3 2 dx = 5 2 × (–2) x + c = 5 – x + c

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 58 BAB 3 (e) ∫ x 3 x dx (f) ∫ (2x)3 4x5 dx (g) 2 3 ∫ x2 x dx 3. Tentukan kamiran tak tentu bagi setiap fungsi algebra yang berikut. Determine the indefinite integral for each of the following algebraic expression. TP 2 (a) ∫(2x2 – 4x + 1 x2 )dx (b) ∫5(x2 – 2 x2 )dx (c) ∫ 3x2 – 4 3 x2 dx (d) ∫  2 t 3 – 6 + 3 t 2  dt (e) ∫(2x + 3)(4 – 2x)dx (f) ∫ 8x(x2 – 9) 5(x + 3) dx (g) ∫ x(x2 – 4x – 5) (x – 5) dx ∫ x 3 x dx = ∫ 1 3 x 1 2 dx = 1 3 × 2 3 x x + c = 2 9 x x + c 2 3 ∫ x2 x 1 2 dx = 2 3 ∫x 3 2 dx = 2 3 × 2 5 x2 x + c = 4 15x2 x + c ∫2x2 dx – ∫4xdx + ∫x–2dx = 2 3 x3 – 2x2 – 1 x + c = ∫2t–3dt – ∫6 dt + ∫3t –2dt = –t–2 – 6t – 3t –1 = 1 – t 2 – 6t – 3 t + c ∫ 3x2 x 2 3 dx – ∫ 4 x 2 3 dx = ∫3x 4 3 dx – ∫4x 2 – 3 dx = 3 7 3x 7 3  – 34x 1 3  = 9 7 x 7 3 – 12x 1 3 + c ∫(8x – 4x2 + 12 – 6x)dx = ∫(2x – 4x2 + 12)dx = x2 – 4 3 x3 + 12x + c 8 5 ∫(x2 – 3x)dx = 8 5  x3 3 – 3x2 2  + c ∫ x(x – 5)(x + 1) (x – 5) dx = ∫(x2 + x)dx = 1 3 x3 + x2 2 + c ∫ 8x3 4x5 dx = ∫2x–2dx = –2 x + c CONTOH (i) ∫ x + 1 4 dx (ii) ∫ 3x(x2 – 4) 5 dx Penyelesaian: (i) ∫ x + 1 4 dx = ∫ x 4 dx + ∫ 1 4 dx = 1 4  x2 2  + 1 4 x + c = 1 8 x2 + 1 4 x + c (ii) ∫ 3x(x2 – 4) 5 dx = ∫ 3x3 5 dx – ∫ 12x 5 dx = 3x4 5(4) – 12x2 5(2) + c = 3x4 20 – 6x2 5 + c ∫5x2 dx – ∫ 10 x2 dx = 5 3 x3 + 10 x + c

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 59 3 4. Tentukan kamiran tak tentu bagi fungsi berikut yang berbentuk (ax + b)n , dengan keadaan a dan b ialah pemalar, n ialah integer dan n ≠ –1. Determine the indefinite integral for the following functions in the form of (ax + b)n, where a and b are constants, n is an integer and n ≠ –1. TP 3 (a) ∫ 3 4 (x – 5)3 dx (b) ∫ 2 (3 – 4x)2 dx (c) ∫6 x – 2 dx (d) ∫ 2 3(3 – x)3 dx (e) ∫ 3 1 – 2x dx 5. Tentukan persamaan lengkung, y daripada fungsi kecerunan yang berikut. Determine the equation of the curve, y from the following gradient functions. TP 3 (a) Diberi dy dx = 3 − x2 – 4x dan y = 4 apabila x = 0. Given that dy dx = 3 − x2 – 4x and y = 4 when x = 0. CONTOH ∫(2x + 3)4 dx Penyelesaian: Katakan/Let u = 2x + 3 du dx = 2, dx = du 2 = ∫u4 du 2 = u4 + 1 2(4 + 1) + c = (2x + 3)5 10 + c Katakan u = 3 – 4x du dx = –4, dx = du –4 = ∫2u–2 du –4 = u– 2 + 1 –2(–1) + c = 1 2(3 – 4x) + c Katakan u = 1 – 2x du dx = –2, dx = du –2 = ∫ 3 –2 u –1 2 du = 3u 1 – 2 + 1 –2 1 2  + c = –3 1 – 2x + c Katakan u = x – 5 du dx = 1, dx = du 1 = ∫ 3 4 u3 du 1 = 3(x – 5)4 16 + c dy dx = 3 – x2 – 4x ∫dy = ∫(3 – x2 – 4x)dx y = 3x – 1 3 x3 – 2x2 + c Apabila x = 0, y = 4 c = 4 Persamaan lengkung ialah The equation of the curve is y = 3x – 1 3 x3 – 2x2 + 4 Katakan u = 3 – x du dx = –1, dx = –du = ∫ –2 3 u–3 du = –2u–3 + 1 3(–2) + c = 1 3(3 – x)2 + c Katakan u = x – 2 du dx = 1, dx = du = ∫6u 1 2 du = 12u 3 2 3 + c = 4(x – 2) x – 2 + c CONTOH Diberi dy dx = 3x2 – 5 dan y = 1 apabila x = −1. Given that dy dx = 3x2 – 5 and y = 1 when x = −1. Penyelesaian: dy dx = 3x2 − 5 ∫dy = ∫(3x2 − 5)dx y = x3 − 5x + c Apabila/When x = −1, y = 1 1 = −1 + 5 + c c = −3 Persamaan lengkung ialah y = x3 − 5x – 3. The equation of the curve is y = x3 − 5x – 3. Persamaan lengkung ialah fungsi y dalam sebutan x Gunakan x dan y untuk mendapat nilai c. The equation of the curve is function y in terms of x. Use x and y to obtain a value of c.

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 60 BAB 3 (b) Diberi dy dx = x – 2 dan y = 1 apabila x = 6. Given that dy dx = x – 2 and y = 1 when x = 6. (c) Diberi dy dx = x2 (2x + 1) dan graf melalui titik (1, −1). Given that dy dx = x2(2x + 1) and the graph passes through the point (1, −1). (d) Diberi dy dx = (ax − 3) dan graf melalui titik (−1, 8) dan (3, 4), dengan a ialah pemalar. Given that dy dx = (ax − 3) and the graph passes through the points of (−1, 8) and (3, 4), where a is a constant. (e) Diberi dy dx = 3x2 + b dan graf melalui titik (1, 3) dan (−1, −3), dengan b ialah pemalar. Given that dy dx = 3x2 + b and the graph passes through the points (1, 3) and (−1, −3), where b is a constant. dy dx = (x – 2) 1 2 ∫dy = ∫(x – 2) 1 2 dx y = 2 3 (x – 2) 3 2 + c Apabila x = 6, y = 1 1 = 16 3 + c c = – 13 3 Persamaan lengkung ialah The equation of the curve is y = 2 3 (x – 2) 3 2 – 13 3 dy dx = 2x3 + x2 ∫dy = ∫(2x3 + x2 )dx y = 1 2 x 4 + 1 3 x3 + c Apabila x = 1, y = –1 –1 = 1 2 + 1 3 + c c = – 11 6 Persamaan lengkung ialah The equation of the curve is y = 1 2 x4 + 1 3 x3 – 11 6 dy dx = 3x2 + b ∫dy = ∫(3x2 + b)dx y = x3 + bx+ c Apabila x = 1, y = 3 3 = 1 + b + c 2 = b + c ...➀ Apabila x = –1, y = –3 –3 = –1 – b + c –2 = –b + c ...➁ ➁ – ➀: –4 = –2b b = 2, c = 0 Persamaan lengkung ialah The equation of the curve is y = x3 + 2x dy dx = ax – 3 ∫dy = ∫(ax – 3)dx y = a 2 x2 – 3x + c Apabila x = –1, y = 8 8 = 1 2 a + 3 + c 5 = 1 2 a + c ...➀ Apabila x = 3, y = 4 4 = 9 2 a – 9 + c 13 = 9 2 a + c ...➁ ➁ – ➀: 8 = 4a a = 2, c = 4 Persamaan lengkung ialah The equation of the curve is y = x2 – 3x + 4

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 61 3 3.3 Kamiran Tentu Definite Integral NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 1. Nilai ∫f(x) dx dapat ditentukan apabila x = a digantikan ke dalam persamaan ∫f(x)dx = F(x) + c, iaitu apabila x = a, ∫f(x)dx = F(a) + c. The value of ∫f(x) dx can be obtained when x = a is substituted into the equation ∫f(x) dx = F(x) + c, that is when x = a, ∫f(x)dx = F(a) + c. 2. Pengamiran fungsi f(x) terhadap x dari x = a hingga x = b ialah ditulis seperti: The integration of the function f(x) with respect to x from x = a to x = b is written as: ∫ b a f(x)dx = [F(x) + c] b a = [F(b) + c] – [F(a) + c] = F(b) – F(a). 3. Beberapa petua bagi kamiran tentu. Rules for definite integral (i) ∫ b a [f(x) ± g(x)]dx = ∫ b a f(x)dx ± ∫ b ag(x)dx (ii) Jika/If d dx [f(x)] = f(x), maka/then ∫ b a f(x)dx = F(b) – F(a) (iii) ∫ a a f(x)dx = 0 (iv) ∫ c a f(x)dx + ∫ b c f(x)dx = ∫ b a f(x)dx (v) ∫ b a f(x)dx = –∫ a bf(x)dx 4. Luas di bawah suatu lengkung Area under the curve (i) luas rantau di antara lengkung y = f(x) dan paksi –x dari x = a ke x = b diberi oleh: Area of the region between the curve y = f(x) and the x-axis from x = a to x = b is given by: ∫ b a y dx = ∫ b a f(x)dx (ii) luas rantau di antara lengkung x = g(y) dan paksi-y dari y = a ke y = b diberi oleh: Area of the region between the curve x = g(y) and the y-axis from y = a to y = b is given by: A = ∫ b a x dy = ∫ b a g(y)dy (i) (ii) 5. (i) Luas dibatasi oleh lengkung y = f(x) dan satu graf lagi y = g(x) dari x = a ke x = b diberi oleh: Area of the region bounded by the curve y = f(x) and another graph y = g(x) from x = a to x = b is given by: (ii) Luas dibatasi oleh lengkung x = f(y) dan satu graf lagi x = g(y) dari y = a ke y = b. Area of the region between the curve x = f(y) and another curve x = g(y) from y = a to y = b is given by: 6. Isi padu janaan Generated Volume (i) Apabila kawasan berlorek dalam rajah diputarkan melalui 360° pada paksi-x, isi padu janaan ialah I = π∫ b a y2dx. When the shaded region in the diagram is rotated through 360° about the x-axis, the generated volume is I = π∫ b a y2dx. x y x y O a dx b y = f(x) x y a b dy O x x = g(y) O a O = – O y b x a y x b y = f(x) a y b x y = g(x) b dx y O x y a y = f(x) y x b a O = – O O y x b a x = g(y) dy y x = f(y) x b dy a

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 62 BAB 3 6. Nilaikan yang berikut. Evaluate the following. TP 3 (a) ∫ 0 –2 x(2x – 1)dx (b) ∫ 3 1 4x3 – 3 x2 dx (c) ∫ 1 –2x + 2 x3 dx (d) ∫ 4 0 2 (2 – x)3 dx (e) ∫ 2 –2 4x2 – 9 (2x – 3) dx a O b dy x y x x = f(y) CONTOH ∫ 3 –1(x – 1)(x + 3)dx Penyelesaian: ∫ 3 –1(x – 1)(x + 3)dx = ∫ 3 –1(x2 + 2x – 3)dx =  1 3 x3 + x2 – 3x 3 –1 = [9 + 9 – 9] –  1 – 3 + 1 + 3 = 1 5 3 ∫ 0 –2 x(2x – 1)dx = ∫ 0 –2(2x2 – x)dx =  2 3 x 3 – 1 2 x2  0 –2 = 0 –  –16 3 – 2 = 22 3 ∫ 3 1 4x3 – 3 x2 dx = ∫ 3 1 (4x – 3x–2)dx = 2x 2 + 3 x  3 1 = (18 + 1) – (2 + 3) = 14 ∫ 2 –2 (2x + 3)(2x – 3) (2x – 3) dx = [x2 + 3x] 2 –2 = (4 + 6) – (4 – 6) = 12 ∫ 4 0 2(2 – x)–3dx =  2(2 – x)–2 (–2)(–1)  4 0 = 1 22 – 1 22 = 0 ∫ 1 –2(x + 2x–3)dx =  1 2 x 2 – 1 x2  1 –2 =  1 2 – 1 – 2 – 1 4  = 3 4 – 3 = 1 –2 4 (ii) Apabila kawasan berlorek dalam rajah diputarkan melalui 360° pada paksi-y, isi padu janaan ialah I = π∫ b a x2dy. When the shaded region in the diagram is rotated through 360° about the y-axis, the generated volume is I = π∫ b a x2dy.

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 63 3 CONTOH Diberi ∫ 4 –2 f(x)dx = 4, ∫ 4 0 f(x)dx = 2 dan ∫ 4 –2 g(x)dx = 5, cari nilai bagi It is given that ∫ 4 –2 f(x)dx = 4, ∫ 4 0 f(x)dx = 2 and ∫ 4 –2 g(x)dx = 5, find the value of (i) ∫ –2 4 2f(x)dx (iv) ∫ 2 –2 f(x)dx + ∫ 4 2 f(x)dx (ii) ∫ 4 –2[f(x) – g(x)]dx (v) k jika/if ∫ 4 –2[k − f(x)]dx = 8 (iii)∫ 0 –2 f(x)dx Penyelesaian: (iv) ∫ 2 –2 f(x)dx + ∫ 4 2 f(x)dx = ∫ 4 –2 f(x)dx = 4 (v) ∫ 4 –2[k − f(x)]dx = 8 ∫ 4 –2kdx – ∫ 4 –2 f(x)dx = 8 [kx]4 −2 = 8 + 4 4k + 2k = 12 k = 2 (i) ∫ –2 4 2f(x)dx = −8 (ii) ∫ 4 –2 [f(x)dx – g(x)]dx = ∫ 4 –2 f(x)dx − ∫ 4 –2 g(x)dx = 4 – (5) = −1 (iii)∫ 0 –2 f(x)dx = ∫ 4 –2 f(x)dx – ∫ 4 0 f(x)dx = 4 – 2 = 2 7. Cari nilai bagi yang berikut. Find the value of the following. TP 4 (a) Diberi ∫ 3 –1 f(x) = 6, ∫ –1 –3 f(x)dx = −1 dan ∫ –1 –3 g(x)dx = 4, cari nilai bagi Given that ∫ 3 –1 f(x) = 6, ∫ –1 –3 f(x)dx = −1 and ∫ –1 –3 g(x)dx = 4, find the value of (i) ∫ 3 –3 f(x)dx (iv) ∫ 2 –1 1 2 f(x)dx + ∫ 3 2 1 2 f(x)dx (ii) ∫ –1 –3 [g(x) – 2f(x)]dx (v) k jika/if ∫ –1 –3[2g(x) − kx]dx = 20 (iii)∫ 0 –3 f(x) + ∫ 3 0 f(x)dx (i) ∫ –3 3 f(x)dx = ∫ –1 –3 f(x)dx + ∫ 3 –1 f(x)dx = –1 + 6 = 5 (ii) ∫ –1 –3 g(x)dx – 2∫ –1 –3 f(x)dx = 4 – 2(–1) = 6 (iii)∫ 0 –3 f(x)dx + ∫ 3 0 f(x)dx = ∫ 3 –3 f(x)dx = –1 + 6 = 5 (iv) 1 2 ∫ 2 –1 f(x)dx + ∫ 3 2 f(x)dx = 1 2 [6] = 3 (v) ∫ –1 –3 2g(x)dx – ∫ –1 –3 kx dx = 20 2(4) –  kx2 2  –1 –3 = 20 8 –  k 2 – 9k 2  = 20 4k = 12 k = 3

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 64 BAB 3 8. Selesaikan yang berikut. Solve the following. TP 4 (a) Diberi ∫ k –2(x − 4)dx = −10, cari nilai k, dengan keadaan k ≥ 0. Given that ∫ k –2(x − 4)dx = −10, find the value of k, where k ≥ 0. (b) Diberi ∫ 2 k ( 2 x3 )dx = 3 4 , cari nilai k. Given that ∫ 2 k( 2 x3 )dx = 3 4 , find the value of k. (c) Diberi d dx  1 (x – 2)2 4 = g(x), cari nilai bagi ∫ 4 –1[2x – g(x)]dx. Given that d dx 3 1 (x – 2)2  = g(x), find the value of ∫ 4 –1[2x – g(x)]dx. CONTOH Diberi ∫ k 1(3x − 2)dx = 5 2 , cari nilai k, dengan keadaan k > 0. Given that ∫ k 1(3x − 2)dx = 5 2 , find the value of k, where k > 0. Penyelesaian: ∫ k 1 (3x − 2)dx = 5 2  3x2 2 – 2x k 1 = 5 2 3k2 2 – 2k –  3 2 – 2 = 5 2 3k2 – 4k + 1 = 5 3k2 – 4k – 4 = 0 (3k + 2)(k – 2) = 0 k = 2 ∫k –2(x – 4)dx = –10  x2 2 – 4x k –2 = –10  k2 2 – 4k – (2 + 8) = –10 k(k – 8) = 0 k = 0 atau k = 8 ∫ 4 –12xdx – ∫ 4 –1 g(x)dx = [x2 ] 4 –1 –  1 (x – 2)2  4 –1 =(16 – 1) – 3 1 4 – 1 9 4 –1 = 15 – 5 36 = 31 14 36 ∫2 k 2x–3dx = 3 4  1 – x2  2 k = 3 4  1 – 4  –  1 – k2  = 3 4 1 k2 – 1 4 = 3 4 4 – k2 = 3k2 4k2 = 4 k = ±1

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 65 3 (d) Diberi d2 y dx2 = 1 − 6x2 apabila dy dx = −1, x = −1 dan y = 2, cari nilai y dalam sebutan x. Given that d2y dx2 = 1 − 6x2 when dy dx = −1, x = −1 and y = 2, find the value of y in terms of x. (e) Diberi d2 y dx2 = 4x2 − 2 apabila dy dx = 4, x = 0 dan y = 1, cari nilai y dalam sebutan x. Given that d2y dx2 = 4x2 − 2 when dy dx = 4, x = 0 and y = 1, find the value of y in terms of x. 9. Cari kamiran tentu atau luas berdasarkan rajah yang diberi. Find the definite integral or area based on the diagram given. TP 3 (a) Diberi luas P ialah 4.5 unit2 dan ∫ 7 −2 f(x)dx = 8 unit2 , cari Given that the area of P is 4.5 unit2 and ∫ 7 −2 f(x)dx = 8 unit2, find (i) ∫ 7 3 f(x)dx (ii) ∫ 3 −2 f(x)dx (iii)jumlah luas P dan Q. the total area of P and Q. d2 y dx2 = 1 – 6x2 dy dx = x – 2x3 + c x = –1, dy dx = –1 –1 + 2 + c = –1 c = –2 dy dx = 4 3 x3 – 2x + c x = 0, dy dx = 4, c = 4 y = 1 3 x4 – x2 + 4x + c x = 0, y = 1, c1 = 1 ∴y = 1 3 x4 – x2 + 4x + 1 dy dx = x – 2x3 – 2 y = x2 2 – 1 2 x4 – 2x + c, x = –1 y = 2 2 = 1 2 – 1 2 + 2 + c, c1 = 0 ∴y = 1 2 x2 – 1 2 x4 – 2x CONTOH y x A a O b B c y = f(x) (i) ∫ 7 3 f(x)dx = 8 + 4.5 = 12.5 unit2 (ii) ∫ 3 –2 f(x)dx = −4.5 unit2 (iii) Jumlah luas = 12.5 + 4.5 Total area = 17 unit2 Diberi luas rantau A ialah 10 unit2 dan luas B ialah 6 unit2 , cari Given that the area of the region A is 10 unit2 and the area of B is 6 unit2, find (i) ∫ c b f(x)dx (ii) ∫ c a f(x)dx (iii)∫ b a f(x)dx + |∫ c b f(x)dx| Penyelesaian: (i) ∫ c b f(x)dx = Luas di bawah paksi-x = Area under the x-axis = 6 (ii) ∫ c a f(x)dx = ∫ b a f(x)dx + ∫ c b f(x)dx = 10 – 6 = 4 (iii)∫ b a f(x)dx + |∫ c b f(x)dx| = 10 + |–6| = 16 P Q O –2 3 7 x y y = f(x) Nilai ialah –6. Value is –6.

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 66 BAB 3 (b) Diberi ∫ 0 –4 f(x)dx = 6 dan ∫ 2 0 f(x)dx = 5. Given that ∫ 0 –4 f(x)dx = 6 and ∫ 2 0 f(x)dx = 5. Cari / Find (i) luas H / the area of H (ii) luas K / the area of K (iii)∫ 2 –4 f(x)dx (c) Diberi luas Q ialah 18 unit2 , Given that the area of Q is 18 unit2, (i) cari luas P, find the area of P, (ii) ungkapkan luas Q dalam tatatanda kamiran. express the area of Q in integration notation. 10. Cari luas rantau berlorek bagi setiap rajah yang berikut. Find the area of the shaded region for each of the following diagrams. TP 4 (a) y = 2 + x – x2 y H K x –4 2 3 O y = f(x) O 5 5 1 y x Q P x = f(y) (i) Luas P/ Area P = 25 − Q = 25 − 18 = 7 unit2 (ii) Luas Q = ∫ 5 1xdy Area Q Apabila x = 0 y = 2 y = 0 2 + x – x2 = 0 (2 – x)(1 + x) = 0 x = 2; –1 B(2, 0) A(0, 2) luas = ∫ 2 0 ydx = ∫ 2 0 (2 + x – x2 )dx area = 2x + x2 2 – x3 3  2 0 = 4 + 2 – 8 3 4 = 10 3 unit2 CONTOH y = x2 – 2x Penyelesaian y = x2 – 2x = x[x – 2] luas berlorek/shaded area = |∫ 2 0 ydx| + ∫ 3 2 ydx = |∫ 2 0 (x2 – 2x)dx| + ∫ 3 2 (x2 – 2x)dx = | x3 3 – x2  2 0| +  x3 3 – x2  3 2 = |1 8 3 – 42| + 1 27 3 – 92 – 1 8 3 – 42 = | –4 3 | + 0 + 4 3 4 = 2 2 3 unit2 y y = x2 – 2x O 3 x y = 2 + x – x2 A y x O B Tip Perlu buat kamiran berasingan kerana luas di bawah paksi-x ialah negatif. Jika ∫ 3 0 ydx, kita dapat Need to perform the integration separately because the area under x-axis is negative. ∫ 3 0 (x2 – 2x)dx =  x3 3 – x2  3 0 = 0 Luas tidak mungkin sifar. Area cannot be zero. Guna/Use ∫ b a ydx (i) Luas H = luas segiempat Area H = Area of rectangle –∫ 0 –4 f(x)dx = 12 – 6 = 6 unit2 (ii) Luas K = ∫ 2 0 f(x)dx = 5 (iii)∫ 2 –4 f(x)dx = ∫ 0 –4 f(x)dx + ∫ 2 0 f(x)dx = 6 + 5 = 11 unit2

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 67 3 (b) y = 1 x2 (c) y = x(x + 1)(2 – x) (d) y = (x + 2)(3 – x) (e) y = 4 – x2 (1, 1) 3, 1 9 fi fi y = 1 x2 y x O –1 2 y x O y = x(x + 1)(2 – x) 1 4 y x O P y = (x + 2)(3 – x) –3 O 4 y x y = 4 – x2 y = (x2 + x)(2 – x) = –x3 + 2x2 + 2x – x2 y = –x3 + x2 + 2x luas = ∫ 0 –1(–x3 + x2 + 2x)dx + ∫ 2 0 (–x3 + x2 + 2x)dx area =  –x4 4 + 1 3 x3 + x2  0 –1 +  –x4 4 + x3 3 + x2  2 0 =  5 – 12 + [–4 + 8 3 + 4] = 5 12 + 8 3 = 1 3 12 unit2 Apabila y = 0 4 – x2 = 0 x = ±2 Luas dari x = –3 ke –2 = ∫ –2 –3(4 – x2 )dx = 34x – x3 3 4 –2 –3 = 31–8 + 8 3 2 – (–12 + 9)4 = 1 –2 3 unit2 luas dari x = 2 ke x = 4 = ∫ 4 2 (4 – x2 )dx = 34x – x3 3 4 4 2 = – 2 10 3 unit2 Jumlah luas =  1 –2 3  +  – 2 10 3  Total area = 13 unit2 luas = ∫ 3 1 ydx area = ∫ 3 1 1 x2 dx = ∫ 3 1 x –2dx = – 1 x 4 3 1 = – 1 3 4 – [–1] = 1 – 1 3 = 2 3 unit2 Apabila y = 0 (x + 2)(3 – x) = 0 x = 3; –2 P[3,0] Luas di atas paksi-x/Area above the x-axis = ∫ 3 1 (x + 2)(3 – x)dx = ∫ 3 1 (6 + x – x2 )dx = 6x + 1 2 x2 – x3 3 4 3 1 = 318 + 9 2 – 94 – 6 + 1 2 – 1 3 4 = 1 7 3 unit2 Luas dari x = 3 ke x = 4 ∫ 4 3 (6 + x – x2 )dx = 6x + 1 2 x2 – x3 3 4 4 3 = 24 + 8 – 64 3 4 – 9 + 9 2  = 5 –2 6 ∴Jumlah luas = 1 7 3 +  5 –2 6  = 1 10 6 unit2 Total area

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 68 BAB 3 11. Cari luas rantau berlorek bagi setiap rajah yang berikut. Find the area of the shaded region for each of the following diagrams. TP 4 (a) x = y2 – 4y (b) y = x2 (c) y = 4 x2 CONTOH x = (y – 3)2 + 2 x = (y – 3)2 + 2 = y2 – 6y + 11 luas berlorek Shaded area = ∫ 4 1 (y2 – 6y + 11)dy = 3 y3 3 – 3y2 + 11y4 4 1 = 1 64 3 – 48 + 442 – 1 1 3 – 3 + 112 = 9 unit2 x = y2 – 4y luas sebelah kiri/area on the left = ∫ 4 0 x dy = ∫ 4 0 (y2 – 4y)dy =  y3 3 – 2y2  4 0 =  64 3 – 324 – 0 = – 2 10 3 ∫ 0 –1(y2 – 4y)dy =  y3 3 – 2y2 4 0 –1 = 0 – 1 1 – 3 – 22 = 1 2 3 Jumlah luas =  – 2 10 3  + 1 2 3 Total area = 13 unit2 x2 = 4 y x = 2 y = 2y 1 – 2 luas = ∫ 2 1 2 2y 1 – 2 dy area = 4 y 4 2 1 2 = 4 2 – 4 2 = 8 – 4 2 = 4 2 = 2 2 unit2 x = y 1 2 luas = 2∫ 4 1xdy area = 2∫ 4 1 y 1 2 dy = 23 2 3 y 3 2  4 1 = 231 2 3 (8) – 2 3 (1) = 23 14 3 4 = 28 3 unit2 1 4 y x O x = (y – 3)2 + 2 –1 O y x x = y2 – 4y Guna/Use ∫ 4 1 x dy 4 y O x 1 y = x2 2 O 1 2 y x y = 4 x2

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 69 3 12. Cari luas berlorek yang dibatasi oleh lengkung, garis lurus atau satu lagi lengkung yang diberi bagi setiap yang berikut. Find the shaded area bounded by the curve, the straight line or another curve that is given for each of the following. TP 4 (a) Diberi persamaan lengkung y = x2 + 1 dan garis lurus y = −x + 3. Given the equation of the curve y = x2 + 1 and the straight line y = −x + 3. CONTOH C A B O D x y = x + 3 y y = x2 – 3x + 6 1 y O x y = x2 + 1 y = –x + 3 Diberi persamaan lengkung y = x2 − 3x + 6 dan garis lurus y = x + 3. Given the equation of the curve y = x2 − 3x + 6 and the straight line y = x + 3. Penyelesaian Untuk mencari titik persilangan x2 – 3x + 6 = x + 3 x2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 x = 3 ; 1 Maka/Hence y = 6; 4 A(1, 4) B(3, 6) Luas berlorek = luas ABCD – luas di bawah graf lengkung dari x = 1 ke x = 3 Luas ABCD = ∫ 3 1 ydx = ∫ 3 1 (x + 3)dx =  x2 2 + 3x 3 1 =  9 2 + 9 –  1 2 + 3 = 1 13 2 – 1 3 2 = 10 unit2 Luas di bawah graf lengkung Area under the curve = ∫ 3 1 (x2 – 3x + 6)dx =  x3 3 – 3 2 x2 + 6x4 3 1 = 9 – 27 2 + 184 –  1 3 – 3 2 + 64 = 2 8 3 unit2 ∴luas berlorek shaded area = 10 – 2 8 3 = 1 1 3 unit2 x2 + 1 = –x + 3 x2 + x – 2 = 0 (x – 1)(x + 2) = 0 x = 1; –2 y = 2 A[1, 2] luas = ∫ 1 0 ydx + ∆ABC area = ∫ 1 0 (x2 + 1)dx + 1 2 (2)(2) = 3 x3 3 + x4 1 0 + 2 = 1 1 3 + 12 + 2 = 1 3 3 unit2 Tip • Perlu cari titik persilangan. Need to find the intersection point. • Dari titik persilangan, lukis garis lurus mencancang. From the intersection point, draw a vertical straight line. • Cari luas yang lebih besar di bawah garis. Find the biggest area under the line. • Tolak luas yang lebih kecil. Minus the smaller area 1 1 3 y O x y = x2 + 1 y = –x + 3 A B dx C Kaedah Alternatif Untuk mencari luas trapezium: To find the area of trapezium: 1 2 × (4 + 6) × 2 = 10 unit2

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 70 BAB 3 (b) Diberi dua persamaan lengkung, y = x2 dan x = y2 . Given the two equations of the curves, y = x2 and x = y2. (c) Diberi persamaan lengkung y = (1 − x)(x + 4) dan garis lurus y = x + 4. Given the equation of the curve y = (1 − x)(x + 4) and the straight line y = x + 4. 13. Tentukan isi padu janaan dalam sebutan π apabila luas berlorek itu diputarkan melalui 360° pada paksi-x. Determine the generated volume in terms of π when the shaded region is rotated through 360° about the x-axis. TP 3 y x O y = x2 x = y2 y x (0, 4) (1, 0) (–4, 0) O CONTOH Diberi persamaan y = 4 − x2 . Given the equation y = 4 − x2. Apabila/When x = 0 y = 4 Isipadu yang dijanakan The generated volume = π∫ 2 –2y2 dx = π∫ 2 –2(4 – x2 )2 dx = π∫ 2 –2(16 – 8x2 + x4 )dx = π16x – 8 3 x3 + 1 5 x5 4 2 –2 Tip Guna/Use ∫ b a πy2 dx –2 O 2 x y y = 4 – x2 Titik persilangan Intersection point x2 = x x4 = x x(x3 – 1) = 0 x = 0; x = 1 y = 0 y = 1 luas berlorek = ∫ 1 0 x dx – ∫ 1 0 x2 dx shaded area =  2 3 x 3 2 4 1 0 –  1 3 x3 4 1 0 = 2 3 – 1 3 = 1 3 unit2 Titik persilangan/Intersection point (1 – x)(x + 4) = x + 4 x + 4 – x2 – 4x = x + 4 x2 + 4x = 0 x(x + 4) = 0 x = 0; x = –4 y = 4 ; y = 0 luas berlorek = luas segi tiga + ∫ 1 0 f(x)dx shaded area = area of triangle + ∫ 1 0 f(x)dx = 1 2 (4)(4) + ∫ 1 0(4 – 3x – x2 )dx = 8 + 4x – 3 2 x2 – x3 3 4 1 0 = 8 + 34 – 3 2 – 1 3 4 = 1 10 6 unit2 = π132 – 64 3 + 32 5 2 – 1–32 + 64 3 – 32 5 24 = 2 34 15 π unit3 Kaedah Alternatif Oleh sebab graf ialah bersimetri pada x = 0. Isipadu dijanakan ialah Since the graph is symmetry at x = 0, the generated volume is = 2∫ 2 0πy2 dx = 2π∫ 2 0 (4 – x2 )2dx = 2π16x – 8 3 x3 + 1 5 x5 4 2 0 = 34 2 15 π unit3

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 71 3 (a) Diberi persamaan y = 1 x . Given the equation y = 1 x . (b) Diberi persamaan y = x(x − 3). Given the equation y = x(x − 3). (c) Diberi persamaan x = 2y2 – 4. Given the equation x = 2y2 – 4. 14. Tentukan nilai k diberikan isipadu dalam sebutan π yang dijanakan apabila luas berlorek itu diputarkan melalui 360° pada paksi-y. Determine the value of k with the given volume in terms of π that is generated when the shaded region is rotated through 360° about the y-axis. TP 4 O 1 3 x y y = 1 x –1 O y x y = x(x – 3) O y x x = 2y2 – 4 Isi padu janaan The generated volume = π∫ 3 1y2 dx = π∫ 3 1 1 x2 dx = π∫ 3 1x –2 dx = π 1 – x  3 1 = π 1 – 3 –(–1)4 = 2 3 π unit3 Titik persilangan dengan paksi-x (3, 0) ,(0, 0) The intersection point at the x-axis of (3, 0), (0, 0) Isipadu dijanakan The generated volume = π∫ 3 1y2 dx = π∫ 3 1x2 (x – 3)2 dx = π∫ 3 1x4 – 6x3 + 9x2 )dx = π 1 5 x5 – 3 2 x4 + 3x3  3 1 = π( 35 5 – 3 2 (3)4 + 3(3)3 ) – 1 1 5 – 3 2 + 324 = 2 6 5 π unit3 Apabila y = 0, x = –4 Isipadu dijanakan The generated volume = π∫ 0 –4 y2 dx = π∫ 0 –41 x + 4 2 2dx = π 1 4 x2 + 2x 0 –4 = π0 – (4 – 8)4 = 4π unit3 CONTOH Diberi isi padu janaan ialah 65 8 π unit3 oleh lengkung y = 4x2 dari y = 4 ke y = k. Given the generated volume is 65 8 π unit3 by the curve y = 4x2 from y = 4 to y = k. Isipadu dijanakan The generated volume = π∫ k 4x2 dy = 65 8 π ∫ k 4 1 4 ydy = 65 8  y2 2 4 k 4 = 65 2 k2 2 – 8 = 65 2 k2 2 = 81 2 k = 9 4 k y x O y = 4x2

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 72 BAB 3 (a) Diberi isi padu yang dijanakan oleh rantau berlorek ialah 10 3 π unit3 dengan graf y2 + 4x2 = 16 Given the volume generated by the shaded region is 10 3 π unit3 with the graph y2 + 4x2 = 16. (b) Diberi isi padu yang dijanakan oleh rantau berlorek ialah 2π unit3 dengan lengkung y = x2 + k. Given the volume generated by the shaded region is 2π unit3 with the curve y = x2 + k. (c) Diberi isi padu yang dijanakan oleh rantau berlorek ialah 152 3 π unit3 dengan graf y = x − 3. Given the volume generated by the shaded region is 152 3 π unit3 with the graph y = x − 3. Isipadu = 152 3 π Volume = π∫ k 1 (y + 3)2 dy π 3 (y + 3)3 3 4 k 1 = 152 3 π (k + 3)3 3 – 43 3 = 152 3 (k + 3)3 = 216 k + 3 = 6 k = 3 k O x y 4 y O x y = x2 + k 1 O k y x y = x – 3 Apabila/When x = 0, y = 4 Isipadu/Volume = 10 3 π = π∫ 4 k x2 dy = π∫ 4 k1 16 – y2 4 2dy = π4y – 1 12 y3  4 k = 10 3 π 116 – 64 12 2 – 4k + k3 12 = 10 3 k3 12 – 4k + 22 3 = 0 k3 – 48k + 88 = 0 Jika k = 2, 8 – 48(2) + 88 = 0 If ∴k = 2 Apabila/When x = 0, y = k Isipadu/Volume = 2π 2π = π∫ 4 k (y – k)dy ∫ 4 k (y – k)dy = 2  y2 2 – ky 4 k = 2 (8 – 4k) – ( k2 2 – k2 ) = 2 k2 2 – 4k + 6 = 0 k2 – 8k + 12 = 0 (k – 2)(k – 6) = 0 ∴k = 2 k = 6

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 73 3 15. Tentukan isi padu yang dijanakan bagi setiap yang berikut. Determine the volume generated for each of the following. TP 5 (a) Luas yang dibatasi oleh lengkung, y = 6 + 7x − 3x2 dan pakis-y diputarkan melalui 360° pada paksi-x. The area bounded by the curve y = 6 + 7x − 3x2 and y-axis is rotated through 360° about the x-axis. O (2, 8) x y y = 6 + 7x – 3x2 CONTOH Luas yang dibatasi oleh garis lurus y = x + 4 dan y = (x − 2)2 diputarkan melalui 360° pada paksi-x. The area bounded by the straight line y = x + 4 and y = (x − 2)2 is rotated through 360° about the x-axis. Langkah 1/Step 1 Cari titik persilangan/Find the intersection point (x – 2)2 = x + 4 x2 – 4x + 4 = x + 4 x2 – 5x = 0 x(x – 5) = 0 x = 0, x = 5 Langkah 2/Step 2 Apabila/When x = 5, y = 9 Langkah 3/Step 3 Cari isi padu janaan lengkung. Find the generated volume of the curve π∫ 5 0 y2 dx = π∫ 5 0 (x – 2)4 dx = π (x – 2)5 5  5 0 = 243 5 π + 32 5 π = 275 5 π Langkah 4/Step 4 Cari isi padu janaan garis lurus dari x = 0 ke x = 5. Find the generated volume of the straight line from x = 0 to x = 5. π∫ 5 0 (x + 4)2 dx = π (x + 4)3 3  5 0 = π1 93 3 – 43 3 2 = 665 3 π Isi padu janaan/Generated volume = 665 3 π − 275 5 π = 166 2 3 π unit3 y x O y = (x – 2)2 y = x + 4 y x O (5, 9) Isipadu dijanakan oleh lengkung Volume generated by the curve = π∫ 2 0 (6 + 7x – 3x2 )2 dx = π∫ 2 0 (36 + 84x + 13x2 – 42x3 + 9x4 )dx = 164.27π atau 4 164 15π Isipadu kon Volume of cone = 1 3 π(8)2 (2) = 42.67 π atau 2 42 3 π Isipadu janaan/Generated volume = 121.6 π unit3

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 74 BAB 3 (b) Luas yang dibatasi oleh paksi-y, lengkung y = 4 x dan y = x diputarkan melalui 360° pada paksi-y. The area bounded by the y-axis, the curve y = 4 x and y = x is rotated through 360° about the y-axis. (c) Luas yang dibatasi oleh lengkung y = 9 − x2 , tangen pada x = 2 dan paksi-x yang diputarkan melalui 360° pada paksi-x. The area bounded by the curve y = 9 − x2, the tangent at x = 2 and the x-axis is rotated through 360° about the x-axis. O 3 x y A y = x y = 4 x 2 y x y = 9 – x2 Koordinat A/Coordinates of A 4 x = x x = 2, y = 2 Isipadu/Volume = π∫ 3 2 x2 dy + kon = π∫ 3 2 16 y2 dy + 1 3 π(2)2 (2) = π –16 y  3 2 + 8 3 π = π –16 3 –  –16 2 24 + 8 3 π = 1 5 3 π unit3 y = 9 – x2 dy dx = –2x x = 2, dy dx = –4 persamaan tangen pada (2, 5) The equation of tangent at (2, 5) y – 5 = –4(x – 2) y = –4x + 13 Tangen menyilang paksi-x pada 1 13 4 , 02. Tangent intersects the x-axis at 1 13 4 , 02 Isipadu kon = 1 3 π(5)2 1 5 4 2 Volume of cone = 125 12 π unit3 Isipadu dijanakan oleh lengkung Volume generated by the curve = π∫ 3 2 y2 dx = π∫ 3 2 (9 – x2 )2 dx = π∫ 3 2 (81 – 18x2 + x4 )dx = π381x – 6x3 + 1 5 x5 4 3 2 = 1 9 5 π unit3 Isipadu diperlu = 73 60 π unit3 Volume needed

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 75 3 CONTOH 3.4 Aplikasi Pengamiran Application of Integration 16. Selesaikan masalah berikut yang melibatkan pengamiran. Solve the following problems involving integration. Rajah menunjukkan permukaan sebuah pintu. ABC ialah suatu lengkok simetri daripada sebahagian graf y = 10 − x2 3 . B ialah titik tertinggi dari DE. The diagram shows the front view of a door. ABC is a symmetrical arch from a part of the graph y = 10 − x2 3 . B is the highest point from DE. (i) Cari lebar DE. Find the width of DE. (ii) Cari luas permukaan pintu. Find the surface area of the door. (iii) Cari luas permukaan pintu baharu apabila AD dan CE ditambah 1 meter, manakala bentuk lengkok ABC dan lebar pintu dikekalkan. Find the new surface area of the door when AD and CE are lengthened by 1 metre, while the arch ABC and the width of the door remain the same. Penyelesaian (i) Diberi/Given y = 10 – x2 3 Apabila/When y = 7 7 = 10 – x2 3 x2 3 = 3 x = ±3 Maka lebar pintu Hence, the width of the door = (3 + 3)m = 6 m (ii) Luas pintu/Area of the door = ∫ 3 –310 – x2 3 2dx = 310x – 1 9 x3 4 3 –3 = 54 m2 (iii) Luas tambahan/Additional area = 1 × 6 = 6 m2 Luas pintu baharu/Area of new door = (54 + 6) m2 = 60 m2 A D E C B 7 m 7 m 8 m 1 6 Tip y = – 1 3 (x2 ) + 10 merupakan paksi-y yang melalui tengah-tengah pintu. Jika x = 0, y = 10 ialah titik maksimum y = – 1 3 (x2 ) + 10 is a y-axis that passes the middle of the door. If x = 0, y = 10 is a maximum point. 10 m y O x

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 76 BAB 3 (a) Satu titik P(x, y) bergerak supaya jaraknya dari titik asalan sentiasa r unit. A point P (x, y) moves such that it is always r units from the origin. (i) Tunjukkan bahawa persamaan bulatan yang dibina ialah x2 + y2 = r2 . Show that the equation of the circle formed is x2 + y2 = r2. (ii) Seterusnya, tunjukkan bahawa isi padu sfera berjejari r unit diberi oleh 4 3 πr3 . Then, show that the volume of the sphere with radius of r unit is given by 4 3 πr3. (b) Dengan cara yang sesuai, buktikan bahawa isi padu kon ialah 1 3 πr2 h, dengan keadaan r ialah jejarinya dan h ialah tinggi kon itu. With a suitable method, prove that the volume of a cone is 1 3 πr 2h, where r is the base radius and h is the height of the cone. (c) Satu kon aiskrim wafer dengan tinggi 10 cm dan jejari tapak ialah 3 cm akan dipenuhi oleh aiskrim sehingga bahagian atasnya berbentuk parabola dengan persamaan y = 14 − 4 9 x2 seperti ditunjukkan dalam rajah. Cari A wafer ice cream cone is 10 cm tall and a base radius of 3 cm will be filled with ice cream until the top is part of a parabola with the equation y = 14 − 4 9 x2 as shown in the diagram. Find (i) isi padu aiskrim bagi setiap kon. the volume of ice cream for each cone. (ii) keuntungan seorang penjual jika setiap kon boleh dijual dengan harga RM3.50 dan dia boleh menjual dua bekas aiskrim dengan dimensi (30 × 20 × 40) cm. Kos setiap bekas aiskrim ialah RM345. the profit of the seller if each cone is sold at RM3.50 and he can sell two containers of the ice cream whose dimensions are (30 × 20 × 40) cm. The cost of each container of the ice cream is RM 345. (i) Dengan teorem Pythagoras, x2 + y2 = r2 P(x, y) y y x O x r (ii) Isi padu janaan/Generated volume = 2π∫ r 0x2 dy = 2π∫ r 0(r2 – y2 )dy = 2πr2 y – y3 3 4 r 0 Persamaan garis lurus The equation of straight line y = h r x + h Isipadu dijana/Generated volume = π∫ h 0x2 dy 10 cm 3 cm Aiskrim Ice cream y = 14 – 4 9 x2 π∫ 14 10 x2 dy = 9 4 π∫ 14 10(14 – y)dy = 9 4 π314y – y2 2 4 14 10 = 9 4 π[(196 – 98) – (140 – 50)] = 9 4 π[8] = 18π unit3 = π∫ h 0 r(y – h) h 4 2 dy = r2 π h2 ∫ h 0(y – h)2 dy = r2 π h2  (y – h)3 3 4 h 0 = 2π3r3 – r3 3 4 = 2π 3 2 3 r3 4 = 4 3 πr3 = r2 π h2 30 – 1 –h3 3 24 = r2 π h2 3 h3 3 4 = 1 3 πr2 h unit3 Isipadu kon/Volume of cone = 1 3 π(3)2 (10) = 30π Jumlah isipadu/Total volume = 18π + 30π = 48π unit3 Satu bekas isipadu = 20 × 30 × 40 = 24000 cm3 Volume of a container Dua bekas/Two containers = 48000 cm3 Bilangan kon = 48000 48π = 318 kon Number of cone Jumlah jualan/Total sales = 318 × RM3.50 = RM1113 Kos/cost = RM345 × 2 = RM690 Keuntungan/Profit = RM423

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 77 3 PRAKTIS PRAKTIS SPM SPM 13 Kertas 1 1. Diberi ∫ 4 –2 f(x)dx = 5, cari Given that ∫ 4 –2 f(x)dx, find (a) ∫ –2 4 f(x)dx (b) ∫ 4 –2[2f(x) – 3x]dx (a) ∫ 4 –2 f(x)dx = 5 ∫ –2 4 f(x)dx = –5 (b) ∫ 4 –22f(x)dx – ∫ 4 –23x dx = 10 –  3 2 x2 4 4 –2 = 10 – [24 – 6] = –8 2. Fungsi kecerunan suatu lengkung ialah (2x – 4). Diberi P(3, –2) terletak pada lengkung itu, cari The gradient function of a curve is (2x – 4). Given that P(3, –2) lies on the curve, find (a) kecerunan tangen pada titik P. the gradient of the tangent at P. (b) persamaan lengkung itu. the equation of the curve. (a) dy dx = 2x – 4 Apabila x = 3, dy dx = 2(3) – 4 = 2 (b) y = ∫(2x – 4)dx = x2 – 4x + c Pada/At (3, –2), –2 = 9 – 12 + c c = 1 ∴y = x2 – 4x + 1 3. Rajah menunjukkan graf suatu lengkung y = f(x) dan garis lurus y = g(x). The diagram shows a graph for the curve y = f(x) and the line y = g(x). Diberi ∫ h 0 f(x)dx – ∫ h 0 g(x)dx = 8. Given that ∫ h 0 f(x)dx – ∫ h 0 g(x)dx = 8. (a) Pada rajah, lorekkan rantau yang diwakili oleh ∫ h 0 f(x)dx – ∫ h 0 g(x)dx. On the diagram, shade the region represented by ∫ h 0 f(x)dx – ∫ h 0 g(x)dx. (b) Cari ∫ h 0 f(x)dx dalam sebutan h dan k. Find ∫ h 0 f(x)dx in terms of h and k. (a) –h (h, 6k) (0, 3k) h x y O (b) ∫ h 0 f(x)dx = 8 + ∫ h 0 g(x)dx = 8 + 1 2 [3k + 6k]h = 8 + 9hk 2 2014 2015 –h (0, 3k) y = g(x) y = f(x) y x O h

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 78 BAB 3 4. Diberi ∫ h –1(3 – 2x)dx = 6, cari nilai yang mungkin bagi h. Given that ∫ h –1(3 – 2x)dx = 6, find the possible values of h. ∫ h –1(3 – 2x)dx = 6 [3x – x2 ] h –1 = 6 3h – h2 – (–3 – 1) = 6 3h – h2 – 2 = 0 h2 – 3h + 2 = 0 (h – 1)(h – 2) = 0 h = 1; 2 5. Rajah menunjukkan sebahagian lengkung y = f(x). Rantau berlorek ditakrifkan sebagai ∫ b a f(x)dx = 8. The diagram shows a part of the curve y = f(x). The shaded region is defined by ∫ b a f(x)dx = 8. O 1 3 5 y x –2 y = f(x) (a) Nyatakan nilai a dan nilai b. State the value of a and of b. (b) Diberi bahawa luas dibatasi oleh lengkung, paksi-x dan x = –2 ke x = 3 ialah 13 unit2 . Cari nilai ∫ 3 –2 f(x)dx. Given that the area bounded by the curve, the x-axis and x = –2 to x = 3 is 13 unit2. Find the value of ∫ 3 –2 f(x)dx. (a) ∫ b a f(x)dx = 8 a = 1, b = 3 (b) ∫ 3 –2 f(x)dx = ∫ 1 –2 f(x)dx + ∫ 3 1 f(x)dx = –5 + 8 = 3 6. Rajah di atas menunjukkan lengkung y = f(x). Garis lurus y = –4 ialah tangen kepada lengkung. Diberi f'(x) = 4x – 4, cari persamaan lengkung itu. Diagram above shows a curve y = f(x). The straight line y = –4 is a tangent to the curve. Given that f'(x) = 4x – 4, find the equation of the curve. f'(x) = 4x – 4 f'(x) = 4x – 4 = 0 x = 1 ∴(1, –4) ialah titik minimum (1, –4) is a minimum point f(x) = ∫(4x – 4)dx = 2x2 – 4x + c Pada/At (1, –4) –4 = 2 – 4 + c c = –2 f(x) = 2x2 – 4x – 2 7. Diberi ∫ h 2 p(x)dx = 3 4, cari Given that ∫ h 2 p(x)dx = 3 4 , find (a) ∫ 2 h p(x)dx (b) ∫ h 2 [p(x) – 2x]dx = –4 1 4 (a) ∫ 2 h p(x)dx = – 3 4 (b) ∫ h 2 p(x)dx – ∫ h 2 2xdx = –4 1 4 3 4 + 4 1 4 = [x2 ] h 2 = h2 – 4 5 + 4 = h2 h = 3 2016 2017 2018 2019 y x O y = –4

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 79 3 Kertas 2 1. Rajah menunjukkan lengkung y = 3x2 – 27 dan AB ialah tangen kepada lengkung itu pada titik P. The diagram shows the curve y = 3x2 – 27 and AB is a tangent to the curve at point P. x y O P B A Diberi bahawa kecerunan tangen ialah 6. Given that the gradient of the tangent is 6. (a) Cari koordinat P. Find the coordinate of P. (b) Hitung luas rantau berlorek itu. Calculate the area of the shaded region. (c) Apabila rantau yang dibatasi oleh lengkung, paksi-x dan y = k diputarkan melalui 180° pada paksi-y menghasilkan isi padu 2 36 3 π unit3 , cari nilai k. When the area bounded by the curve, the x-axis and y = k is rotated through 180° about the y-axis produces a volume of 2 36 3 π unit3, find the value of k. (a) y = 3x2 – 27 Apabila y = 0 x = ±3 dy dx = 6x Apabila dy dx = 6x = 6 x = 1 y = –24 ∴P[1, –24] (b) Persamaan tangen y + 24 = 6(x – 1) Equation of tangent y = 6x – 30 luas berlorek shaded area = 1 2 [4][24] – ∫ 3 1 ydx = 48 – ∫ 3 1 (3x2 – 27)dx = 48 – [x3 – 27]3 1 ∴48 – [0 + 26] = 22 unit2 (c) Isipadu = π∫ 0 k x2 dy = 21 110 3 2π Volume ∫ 0 k 1 1 3 y + 92dx = 220 3  1 6 y2 + 9y4 0 k = 220 3 0 – [k2 6 + 9k] = 220 3 k2 + 54k + 440 = 0 (k + 44)(k + 10) = 0 k = –44 ; –10 ∴k = –10 2. Rajah menunjukkan garis lurus 4y = x + 3 menyentuh lengkung x = 4y2 – 2 pada titik A. The diagram shows a straight line 4y = x + 3 touches the curve x = 4y2 – 2 at point A. A x = 4y2 – 2 x O y Cari / Find (a) koordinat A. the coordinates of A. (b) luas rantau berlorek. the area of the shaded region. (c) isi padu kisaran dalam sebutan π apabila rantau yang dibatasi oleh lengkung dan x = 1 diputarkan melalui 180° pada paksi-x. the generated volume in terms of π when the area bounded by the curve and x = 1 is rotated through 180° about the x-axis. (a) 4y = x + 3 ; x = 4y2 – 2 4y – 3 = 4y2 – 2 4y2 – 4y + 1 = 0 (2y – 1)2 = 0 y = 1 2 , x = 2 – 3 = –1 A–1, 1 2 2 2017 2018

Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 80 BAB 3 (b) Apabila/When y = 0, x = –3 luas berlorek/shaded area = 1 2 (2) 1 2 2 – ∫ –1 –2 ydx = 1 2 – ∫ –1 –2 1 2 (x + 2) 1 2 dx = 1 2 – 3 1 2 . 2 3 (x + 2) 3 2 4 –1 –2 = 1 2 – 3 1 3 – 04 = 1 6 unit2 (c) Isipadu/Volume = π∫ 1 –2 y2 dx = π∫ 1 –2 1 4 (x + 2)dx = π3 (x + 2)2 8 4 1 –2 = π3 9 8 – 04 = 9π 8 Isipadu sebenar Actual volume = 9π 16 unit3 3. Rajah menunjukkan lengkung y = 6x – x2 dan tangen kepada lengkung pada titik A melalui titik B(3, 10). The diagram shows a curve y = 6x – x2 and the tangent to the curve at point A passes through the point B(3, 10). y x O 3 a B(3, 10) A (a) Tunjukkan bahawa a = 4. Show that a = 4. (b) Hitung luas berlorek itu. Calculate the shaded area. (a) y = 6x – x2 dy dx = 6 – 2x = 0 x = 3 Katakan persamaan tangen Let the equation of tangent y = mx + c 10 = 3m + c ...➀ 6x – x2 = mx + c x2 + (m – 6)x + c = 0 (m – 6)2 – 4c = 0 (m – 6)2 = 4(10 – 3m) m2 – 12m + 36 = 40 – 12m m2 = 4 m = ±2 Apabila m = –2, c = 10 – 3(–2) = 16 ∴6 – 2x = –2 8 = 2x x = 4; y = 6(4) – 16 = 8 ∴A(4, 8) ; a = 4 (b) luas berlorek = luas trapezium – ∫ 4 3 ydx shaded area = area of trapezium – ∫ 4 3 ydx = 1 2 [10 + 8][1] – ∫ 4 3 (6x – x2 )dx = 9 – 33x2 – 1 3 x3 4 4 3 = 9 – 326 2 3 – 184 = 1 3 unit2 2019 O 4 x y 3 (3, 10) (4, 8) –3 –2 –1 x y O –1, 1 2 fi ff Praktis SPM Ekstra

BAB Matematik Tambahan Tingkatan 5 Bab 3 Pengamiran 81 3 Sudut Sudut KBAT KBAT –1O 2 P x y Q R (2, 9) Rajah menunjukkan sebahagian graf y = x3 + 1. P ialah titik persilangan lengkung pada paksi-x. Satu tangen kepada lengkung di P memotong lengkung sekali lagi di Q. QR adalah berserenjang dengan paksi-x. The diagram shows part of a graph y = x3 + 1. P is the intersection point of the curve at the x-axis. A tangent to the curve at P cuts the curve again at Q. QR is perpendicular to the x-axis. P y x R Q B A O y = x3 + 1 KBAT Ekstra Tentukan / Determine (a) koordinat P dan Q. the coordinates of P and Q. (b) nisbah luas A kepada luas B. the ratio of the area of A to B. (c) isi padu janaan apabila luas berlorek A diputarkan melalui 180° pada paksi-x. the generated volume when the shaded area A is rotated through 180° about the x-axis. = (4 + 2) – 1 1 4 – 12 = 6 + 3 4 = 3 6 4 unit2 luas/area PQR = 1 2 (3)9 = 27 2 unit2 ∴luas/area B = 27 2 – 3 6 4 = 3 6 4 unit2 Nisbah A : B = 3 6 4 : 3 6 4 Ratio = 1 : 1 (c) Isipadu = π 2 ∫ 2 –1 y2 dx = π 2 ∫ 2 –1(x3 + 1)2dx Volume = π 2 3 1 7 x7 + 1 2 x4 + x4 2 –1 = π 2 3 13 28 14 4 = 13 14 28 π unit3 (a) y = x3 + 1 Apabila y = 0 ; 0 = x3 + 1 x3 = –1 x = –1 P[–1, 0] dy dx = 3x2 Apabila x = –1, dy dx = 3 When y = 3(x + 1) Jika y = 3x + 3 menyilang lengkap If y = 3x + 3 intersects completely 3x + 3 = x3 + 1 x3 – 3x – 2 = 0 Apabila x = 2 23 – 3(2) – 2 = 8 – 8 = 0 ∴y = 23 + 1 = 9 Q(2, 9) (b) luas A = ∫ 2 –1 ydx area A = ∫ 2 –1(x3 + 1)dx = 3 x4 4 + x4 2 –1 Quiz 3

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 82 BAB 4 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN BAB 4 Pilih Atur dan Gabungan Permutation and Combination 4.1 Pilih Atur Permutation 1. Petua pendaraban: Multiplication rule: Jika suatu peristiwa boleh berlaku dalam m cara dan suatu peristiwa kedua boleh berlaku dalam n cara, maka kedua-dua peristiwa boleh berlaku dalam m × n cara. If an event can happen in m ways and another event can happen in n ways, then the two events can happen successively in m × n ways. 2. Jika n benda yang berlainan disusun dalam satu baris mengikut tertib, bilangan pilih atur If n different objects are arranged in a line according to order, the number of permutations = n! = n(n − 1)(n − 2) …. 3 . 2 . 1 Tatatanda n! dibaca ‘n faktorial’. The notation n! is read as ‘factorial n’. 3. Bilangan pilih atur untuk menyusun 0 objek ditandakan 0! = 1. The number of permutations to arrange 0 object is labelled as 0! = 1. 4. Jika r objek dipilih daripada n objek yang berlainan dan disusun mengikut tertib, bilangan pilih atar ialah n Pr = n! (n – r)! . If r object is chosen from n different objects and are arranged in order, the number of permutations is n Pr = n! (n – r)! . 5. Pilih atur bagi n objek yang berlainan dengan syarat tertentu. Permutations of n different objects according to some conditions. Bagi masalah yang melibatkan pilih atur objek dengan syarat tertentu, pastikan syarat tertentu dipenuhi terlebih dahulu sebelum menyusun yang lain. For problems involving permutations of the objects for given conditions, make sure that the certain given conditions need to be addressed first before permuting the rest. 6. Jika terdapat n benda, antaranya ada p benda yang serupa, q benda yang lain yang serupa juga, r benda lain lagi yang serupa lagi dan seterusnya, maka jumlah pilihatur ialah n! p!q!r!... If there are n objects, among which are p objects that are similar, another group of q objects that are similar to each other, another group of r objects that are similar to each other, then the total permutation is n! p!q!r!... 7. Jika terdapat n objek yang berlainan disusun dalam satu bulatan, maka bilangan pilih atur ialah (n − 1)! jika posisi tidak ditetapkan pada bulatan itu. If there are n different objects to be arranged in a circle, then the permutation is given by (n − 1)! if the positions are not marked in the circle. 1. Nilaikan yang berikut. Evaluate the following. TP 1 CONTOH (i) n Pn (ii) 6 P2 (iii) 7 P0 Penyelesaian: (i) n Pn = n (n – n)! = n! (0)! = n! (ii) 6 P2 = 6! (6 – 2)! = 6! 4! = 6.5.4.3.2.1 4.3.2.1 = 30 Tip (n – r)! ≠ n! – r! Guna /Use n Pr = n! (n – r)!

BAB Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 83 4 (a) (i) n P0 (ii) n Pn − 1 (b) (i) 10P3 (ii) 5 P4 × 6 P2 (c) (i) n Pn − 1 × n − 1P2 (ii) mP2 m + 1P2 2. Selesaikan masalah yang berikut. Solve the following problems. TP 2 (a) Terdapat 4 jenis sayur dan 3 jenis buah-buahan yang dijual di sebuah gerai yang berdekatan dengan rumah John. Cari bilangan cara John boleh membeli 1 jenis sayur dan 1 jenis buahbuahan. There are 4 types of vegetables and 3 types of fruits being sold at a store near to John’s house. Find the number of ways of John to buy 1 type of vegetable and 1 type of fruit. Bilangan cara Number of ways = 4 × 3 = 12 (i) n! (n – 0)! = 1 (ii) n! (n – n + 1)! = n! (i) 10! (10 – 3)! = 720 (ii) 5! (1)! × 6! (4)! = 5 × 6! = 3 6000 (iii) 7 P0 = 7! (7 – 0)! = 7! 7! = 7.6.5.4.3.2.1 7.6.5.4.3.2.1 = 1 Sudut Kalkulator (i) 7 P4 Tekan Press (ii) 5! Tekan Press 7 5 X! = 4 = 840 120 SHIFT SHIFT n Pr (i) n! (n – n + 1)! × (n – 1)! (n – 1 – 2)! = n!(n – 1)(n – 2) (ii) m! (m – 2)! ÷ (m + 1)! (m + 1 – 2)! = m(m – 1) ÷ (m + 1)! (m – 1)! = m – 1 m + 1 CONTOH Gopal ingin memilih sebuah buku dan sebuah majalah dari rak yang mengandungi 3 buah buku dan 2 buah majalah yang berlainan. Berapa cara berlainan yang boleh dilakukan oleh Gopal? Gopal wants to choose a book and a magazine from a shelf which has 3 different books and 2 different magazines. How many ways can this be done by Gopal. Penyelesaian: Gopal mempunyai 3 cara untuk memilih sebuah buku dan 2 cara untuk memilih sebuah majalah dari rak itu. Mengikut petua pendaraban, jumlah cara ialah 3 × 2 = 6. Gopal has 3 ways to choose a book and 2 ways to choose a magazine from the shelf. By using multiplication rule, the total number of ways is 3 × 2 = 6.

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 84 BAB 4 (b) Seorang murid ternampak 5 batang krayon dan 6 batang pensel berwarna di atas lantai. Guru kelas menyuruh murid itu memilih sebatang krayon dan pensel berwarna, cari berapa cara yang boleh dilakukan oleh murid itu A student saw 5 sticks of crayons and 6 colour pencils on the floor. The class teacher wants the student to pick up a crayon and a colour pencil, find the number of ways that can be done by the student. Bilangan cara Number of ways = 5 × 6 = 30 (c) Dalam satu pameran perabot, ada 6 jenis kerusi plastik, 4 jenis meja kecil dan 3 jenis kusyen. Puan Wan ingin memilih satu barang daripada setiap jenis barang itu. Cari bilangan cara berlainan yang boleh dilakukan oleh Puan Wan? In a furniture exhibition, there are 6 types of plastic chairs and 4 types of small tables and 3 types of cushions. Miss Wan wants to select one from each group of items. Find the number of ways this can be done by Miss Wan. Bilangan cara Number of ways = 6 × 4 × 3 = 72 3. Selesaikan masalah pilih atur n objek yang berbeza. Solve the permutation problems of n different objects. TP 2 (a) Seorang pekerja diminta menyusun 7 bahan berlainan penghasilan kilang dalam satu baris untuk disemak. Berapa cara boleh dilakukan? A worker is asked to arrange 7 different materials produced by a factory in a line to be checked. In how many ways can this be done? Bilangan cara = 7! =5 040 Number of ways (b) Terdapat 8 keping gambar perlu disimpan dalam sebuah buku album. Berapa cara ini boleh dilakukan? There are 8 photos to be arranged in a photo album. In how many ways can this be done? Bilangan cara = 8! = 40 320 Number of ways CONTOH Berapa cara berlainan digit 0, 1, 2, 3 dan 4 boleh disusun dalam satu baris? How many different ways of the digits 0, 1, 2, 3 and 4 can be arranged in a line? Penyelesaian: Terdapat lima tempat kosong perlu diisi oleh nombor 0, 1, 2, 3 atau 4. Tempat pertama ada 5 cara, tempat kedua ada 4 cara kerana satu digit telah digunakan, tempat ketiga ada 3 cara kerana dua digit telah digunakan, tempat keempat ada 2 cara kerana tiga digit telah digunakan. Tempat terakhir hanya ada 1 cara kerana satu digit yang tertinggal sahaja. Maka dengan petua pendaraban, bilangan cara = 5 × 4 × 3 × 2 × 1 = 120 atau 5! atau 5 P5 . There are five ways that need to be filled by the numbers 0, 1, 2, 3 or 4. The first place has 5 ways, the second place has 4 ways because one digit has been used, the third place has 3 ways because two digits have been used, the fourth place has 2 ways because three digits have been used. The last place has only 1 way because only one digit is left. So, with multiplication tips, the number of ways = 5 × 4 × 3 × 2 × 1 = 120 or 5! or 5P5.

BAB Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 85 4 CONTOH (c) 6 buah beg akan digantung pada 6 penyangkut dalam satu baris. Cari bilangan cara ini boleh dilakukan. 6 bags are to be hung on 6 hangers in a row. Find the number of ways this can be done. Bilangan cara = 6! = 720 Number of ways 4. Selesaikan masalah pilih atur n objek yang berbeza diambil r objek pada satu masa. Solve the problems of permutations of n different objects taken r objects at a time. TP 2 (a) Terdapat 4 buah kerusi sahaja di ruang menunggu tetapi ada 10 orang mencari tempat duduk. Berapakah bilangan cara tempat duduk ini dapat diduduki? There are 4 chairs in a waiting area but there are 10 people looking for the seats. How many ways can the people be seated? Bilangan cara = 10P4 = 10 × 9 × 8 × 7 Number of ways = 5 040 (b) Seorang pelukis mempamerkan 9 lukisan di papan pamerannya. Encik Tan ingin membeli tiga lukisan untuk rumah baharunya. Cari bilangan cara yang boleh dilakukan. An artist exhibits 9 drawings on an exhibition board. Mr.Tan wants to buy three paintings for his new house. Find the number of ways that can be done. Bilangan cara = 9 P3 Number of ways = 504 Satu kata laluan terdiri daripada 6 digit dengan menggunakan digit 0, 1, 2, 3, 4, 5, 6, 7, 8 dan 9 tanpa ulangan. Cari bilangan cara berlainan kata laluan ini boleh dilakukan. A password is made up of 6 digits taken from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 without repetition. Find the number of different passwords that can be formed. Penyelesaian: Terdapat lima tempat kosong perlu diisi oleh 10 nombor. There are five empty spaces to be filled by 10 numbers. Tempat pertama ada 10 pilihan. The first space has 10 choices. Tempat kedua ada 9 pilihan kerana satu digit telah digunakan. The second space has 9 choices since one digit has been used. Tempat ketiga ada 8 pilihan kerana dua digit telah digunakan. The third space has 8 choices since two digits have been used. Tempat keempat ada 7 pilihan kerana tiga digit telah digunakan. The fourth space has 7 choices since three digits have been used. Tempat kelima ada 6 pilihan kerana empat digit yang tertinggal sahaja. The fifth space has 6 choices since four digits have been used. Tempat terakhir masih ada 5 pilihan. The final space has 5 choices. Maka, dengan petua pendaraban, bilangan cara So, by multiplication rule, the number of ways = 10 × 9 × 8 × 7 × 6 × 5 = 151 200 atau/or 10P6 = 10! (10−6)! = 151 200

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 86 BAB 4 (c) Daripada 6 pakej yang ditawarkan oleh Astro, bapa Hassan ingin membeli 3 pakej itu. Cari bilangan cara yang boleh dilakukan. From the 6 packages offered by Astro, Hassan’s father wants to purchase 3 packages. Find the number of ways that can be done. Bilangan cara = 6 P3 Number of ways = 120 5. Selesaikan setiap soalan yang berikut yang melibat objek secaman. Solve each of the following questions involving similar objects. TP 3 (a) Terdapat 4 batang pen biru yang sama, 3 batang pen merah yang sama dan 2 pen hitam yang sama disusun dalam satu baris. Cari bilangan cara untuk menyusun semua pen itu. There are 4 blue pens of the same type, 3 red pens that are the same and 2 black pens that are also the same to be arranged in a row. Find the number of ways to arrange all the pens. Bilangan cara susunan Number of ways of arrangement = 9! 4!3!2! = 1 260 (b) Dalam sebuah kedai yang menjual telefon bimbit, terdapat 5 model A, 6 model B dan 4 model C. Cari bilangan cara pekerja dapat mempamerkan ketiga-tiga model itu di dalam kedai. In a shop that is selling smart phones, there are 5 model A, 6 Model B and 4 model C. Find the number of ways of the staff to exhibit all the three models in a shop. Bilangan cara susunan Number of ways of arrangement = 15! 5!6!4! = 630 630 (c) Rajah di bawah menunjukkan 9 kad yang bernombor. Diagram below shows 9 numbered cards. 2 2 2 5 6 6 6 8 8 Cari bilangan cara menyusun semua kad ini dalam satu baris. Find the number of ways to arrange all the cards in a row. Bilangan cara susunan Number of ways of arrangement = 9! 3!3!2! = 5 040 CONTOH Cari bilangan cara untuk menyusun semua huruf daripada perkataan KALKULATOR. Find the number of ways to arrange all the letters in the word KALKULATOR. Penyelesaian: Jika semua huruf adalah berlainan, maka terdapat 10! cara untuk menyusun semua huruf itu. Tetapi terdapat 2 huruf K yang sama, 2 huruf A yang sama dan 2 huruf L yang sama. Maka bilangan cara akan berkurang kerana tidak dapat mengasingkan huruf yang sama. Jadi, bilangan cara susunan If all the letters are different, then there are 10! ways to arrange all the letters. But there are 2 similar letters of K, 2 similar letters of A and 2 similar letters of L. Hence the number of ways will decrease because cannot separate the same letters. So, the number of ways of arrangement = 10! 2!2!2! = 453 600

BAB Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 87 4 CONTOH Berapa banyak nombor 4 digit dapat dibentuk daripada digit-digit 1, 2, 3, 4, 5, 6, 7 dan 8 jika ulangan tidak dibenarkan dan setiap nombor How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7 and 8 if repetition is not allowed and each number (i) ialah nombor ganjil. is an odd number. (ii) bermula dengan 5. starts with 5. (iii) berakhir dengan nombor genap. ends with an even number. (iv) bernilai lebih daripada 3 000 dan adalah nombor genap. greater than 3 000 and are even numbers. 6. Selesaikan soalan berikut yang melibatkan susunan objek dalam bulatan. Solve the following questions involving arrangement of objects in a circle. TP 3 (a) Dalam suatu permainan, seramai 8 orang kanak-kanak perlu duduk di atas lantai dalam suatu bulatan. Cari bilangan cara menyusun semua kanak-kanak itu. In a game, 8 children are to be seated in a circle on the floor. Find the number of ways to arrange the children. Bilangan cara = 7! Number of ways = 5 040 (b) Terdapat sebuah pokok di taman. Pekebun ingin menyusun 6 pasu bunga mengelilingi pokok itu. Cari bilangan cara dia dapat menyusun pasu bunga tersebut. There is a tree in a garden. A gardener wants to put 6 pots of flowers around the tree. Find the number of ways of the gardener to arrange the flower pots. Bilangan cara = 5! Number of ways = 120 (c) Dalam satu jamuan makan malam, terdapat 10 orang akan duduk pada satu meja bulat. Cari bilangan cara menyusun semua 10 orang tersebut. In a dinner, there are 10 people to be seated around a round table. Find the number of ways to arrange all the 10 people. Bilangan cara = (10 − 1)! Number of ways = 362 880 7. Selesaikan setiap yang berikut. Solve each of the following. TP 4 CONTOH Dengan berapa cara 7 orang boleh duduk mengelilingi sebuah meja bulat? In how many ways can 7 people be seated around a round table? Penyelesaian: Oleh sebab meja bulat, bilangan cara ialah (7 − 1)! = 720 Since it is a round table, the number of ways is (7 − 1)! = 720 Tip Jika objek di sebelah kiri dan kanan adalah sama, susunan itu adalah sama. Oleh itu, dalam kes ini, 7 mungkin telah hilang, iaitu 7! 7 = 6! If the objects on the left and on the right are the same, it is the same arrangement. Hence, in this case, 7 may have been lost, that is 7! 7 = 6!

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 88 BAB 4 Penyelesaian: (i) 1, 3, 5, atau 7 Nombor ganjil mesti berakhir dengan digit nombor ganjil An odd number must end with an odd digit number. Bilangan cara mengisi tempat terakhir ialah 4. Number of ways to fill the last place is 4. Selepas itu, masih tertinggal 7 nombor yang belum digunakan untuk mengisi 3 tempat kosong itu. After that, there are still 7 numbers that have not been used to fill the 3 empty spaces. Cara mengisi ialah 7 P3 . The way to fill is 7P3. Jumlah bilangan cara = 4 × 7 P3 = 840 Total number of ways = 4 × 7P3 = 840 (ii) 5 Nombor 5 mesti diletakkan pada tempat pertama. Bilangan cara mengisinya ialah 1. The number of 5 must be put at the first place. The number of ways to fill it is 1. Selepas itu, masih tertinggal 7 nombor yang belum digunakan untuk mengisi 3 tempat kosong itu. After that, there are still left 7 numbers that have not been used to fill the 3 empty spaces. Cara mengisi ialah 7 P3 . The ways to fill is 7P3 . Jumlah bilangan cara = 1 × 7 P3 = 210 Total number of ways = 1 × 7P3 = 210 (iii) 2,4,6 atau 8 Nombor genap mesti berakhir dengan digit nombor genap. An even number must be ended with a digit of even number. Bilangan cara mengisi tempat terakhir ialah 4. Selepas itu, masih tertinggal 7 nombor yang belum digunakan untuk mengisi 3 tempat kosong itu. The number of ways to fill the last place is 4. After that, there are left 7 numbers that have not been used to fill the 3 empty spaces. Cara mengisi ialah 7 P3 The ways to fill is 7P3. Jumlah cara ialah = 4 × 7 P3 = 840 Total number of ways = 4 × 7P3 = 840 (iv) Kotak di atas menunjukkan semua nombor yang mungkin bagi tempat pertama dan terakhir. The boxes above shows all the possible digits to be used for the first and the last spaces. Kes I / Case 1 Jika tempat pertama menggunakan 3, 5 atau 7, maka tempat terakhir mempunyai 4 pilihan dan terdapat 6 P2 untuk dua tempat di tengahtengah. If the first place using 3, 5 or 7, then the last place has 4 choices and 6P2 for the two middle places. Bilangan cara ialah 3 × 6 P2 × 4 = 360 Number of ways is 3 × 6P2 x 4 = 360 Kes 2 / Case 2 Jika tempat pertama menggunakan 4, 6 atau 8, maka tempat terakhir mempunyai 3 pilihan dan terdapat 6 P2 untuk dua tempat di tengahtengah. If the first place uses 4, 6 or 8, then the last place has 3 choices and there is 6P2 choices for the middle two places. Bilangan cara ialah 3 × 6 P2 × 3 = 270 Number of ways is 3 × 6 P2 × 3 = 270 Jumlah cara ialah = 360 +270 = 630 Total number of ways = 360 + 270 = 630 Tip Syarat perlu dipertimbangkan dahulu. Selepas itu gunakan petua pendaraban. The conditons need to be attended first. Then, use the multiplication rule. Boleh di isi oleh 3, 4, 5, 6, 7 atau 8 2, 4, 6 atau 8

BAB Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 89 4 (a) Seorang pelajar dijemput menyusun semua huruf daripada perkataan KETUA dalam satu baris. Cari bilangan cara untuk menyusun huruf jika A student is asked to arrange all the letters from the word KETUA in a row. Find the number of ways to arrange the letters if (i) susunan mesti bermula dengan konsonan. the arrangement must start with a consonant. (ii) konsonan mesti berada sebelahmenyebelah antara satu sama lain. the consonants must be side-by-side with each other. (iii)semua vokal mesti diasingkan. all the vowels must be separated. (i) Bilangan cara = 2 × 4! = 48 Number of ways (ii) Bilangan cara = 4! × 2! = 48 Number of ways (iii)Susunan dalam bentuk VKVKV Arrangement in the form of VKVKV Bilangan cara = 2! × 3! = 12 Number of ways (b) Ketua pengawas dan 4 orang pelajar lain yang terdiri daripada 2 lelaki dan 2 perempuan perlu disusun dalam satu baris dalam sesi bergambar, Cari bilangan cara menyusun semua 5 pelajar itu jika A head prefect and 4 other students consisting of 2 boys and 2 girls are to be arranged in a row in a photo session. Find the number of ways to arrange all the 5 students if (i) ketua pengawas mesti berada di tengahtengah baris. the head prefect must be in the middle of the row. (ii) ketua pengawas berada di tengah-tengah dan 2 orang lelaki mesti bersama. the head prefect must be in the middle and the 2 boys must be together. (i) Bilangan cara = 1 × 4! = 24 Number of ways (ii) Bilangan cara = 1! × 2! × 2! × 2! = 8 Number of ways (c) 4 orang lelaki, 2 orang perempuan dan seorang kanak-kanak diminta duduk dalam satu baris. Cari cara menyusun mereka jika 4 boys, 2 girls and a child are asked to sit in a row. Find the number of ways to arrange them if (i) kanak-kanak duduk di antara dua perempuan itu. the child sits between the two girls. (ii) 4 orang lelaki mesti duduk bersama. the 4 boys must sit together. (iii) 2 perempuan tidak boleh duduk bersama. the 2 girls cannot sit together. (i) 5! × 2 = 240 (ii) 4! × 4! = 576 5! × 6 P2 = 3 600 (d) Diberi digit-digit 1, 2, 3, 4, 5 dan 6 untuk membentuk 5 digit nombor. Jika semua digit boleh diulangi, cari bilangan cara membentuk Given the digits 1, 2, 3, 4, 5 and 6 to make a 5 digits number. If all the digits can be repeated, find the number of ways to form (i) nombor ganjil. odd numbers. (ii) suatu nombor dengan digit genap mesti berada di tempat ganjil. a number with the even digits must be at the odd places. (i) 6 × 6 × 6 x 6 × 3 = 3 888 (ii) 3! × 3 × 2 = 36

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 90 BAB 4 CONTOH 8. Selesaikan setiap yang berikut. Solve each of the following. TP 4 (a) 10 orang pengakap terdiri daripada 6 orang lelaki dan 4 orang perempuan duduk mengelilingi sebuah unggun api dalam satu bulatan. Cari bilangan cara menyusun mereka jika 10 scouts which is consisted of 6 boys and 4 girls sit around a camp fire. Find the number of ways to arrange them if (i) tidak ada halangan. there is no restrictions. (ii) semua lelaki mesti duduk sebelahmenyebelah. all the boys must sit side by side. (iii)semua perempuan mesti diasingkan all the girls must be separated. (i) Bilangan cara menyusun = (10 − 1)! Number of ways to arrange = 362 880 (ii) Bilangan cara menyusun Number of ways to arrange = (5 − 1)! × 6! = 17 280 (iii)Bilangan cara menyusun Number of ways to arrange = (10 − 1)! − (7 − 1)! × 4! = 345 600 (b) 5 orang wanita dan 4 orang lelaki duduk dalam satu bulatan untuk bermain satu permainan. Cari bilangan cara menyusun semua jika 5 women and 4 men sit in a circle to play a game. Find the number of ways to arrange all of them if (i) satu pasang suami isteri mesti duduk bersama. a couple of husband and wife must sit together. (ii) tiga orang perempuan tertentu mesti duduk bersama. three specific women must sit together. (i) Bilangan cara = (8 −1)! × 2! Number of ways= 10 080 (ii) Bilangan cara menyusun = (7 − 1)! × 3! Number of ways to arrange = 4 320 (c) Digit-digit 1, 2, 3, 4, 5, 6, 7 dan 8 disusun dalam satu bulatan. Cari bilangan cara menyusun jika The digits 1, 2, 3, 4, 5, 6, 7 and 8 are arranged in a circle. Find the number of ways to arrange them if (i) semua digit ganjil mesti bersama. all the odd digits must be together. (ii) digit nombor ganjil berselang-seli dengan digit nombor genap. the odd digits are alternate to the even digits. (i) Bilangan cara menyusun = (5 − 1)! × 4! Number of ways to arrange = 576 (ii) Bilangan cara menyusun = (4 − 1)! × 4! Number of ways to arrange = 144 Dengan berapa cara boleh menyusun 6 orang murid, dengan keadaan dua orang ialah anak kembar di satu meja bulat? In how many ways can arrange the 6 students with an identical twins on a round table? Penyelesaian: Oleh sebab meja ialah bulat, cara susunan ialah (6 − 1)! = 120 jika semua orang adalah berbeza. Tetapi ada sepasang anak kembar yang sama, maka cara susunan ialah (6−1)! 2! = 60. Since the table is round, the number of ways to arrange is (6 − 1)! = 120 if all the people are different. But there is an identical twins, hence the number of ways is (6−1)! 2! = 60.

BAB Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 91 4 NOTA IMBASAN NOTA IMBASAN 4.2 Gabungan Combination 1. Jika r objek dipilih daripada n objek tanpa mempertimbangkan tertib susunannya, bilangan gabungan yang berlainan adalah diberi oleh If r objects are chosen from n objects without arranging them, the number of different combinations is given by n Cr = n! (n − r)!r! . = n(n − 2)….3.2.1 (n − r)(n − r −1)…3.2.1.r(r − 1)(r − 2)…3.2.1 = n(n − 1)...(n – r + 1) r(r −1)(r − 2)…3.2.1 2. Perhatikan bahawa Note that (i) n Cn = 1 (ii) n Cn − 1 = n (iii) n C0 = 1 (iv) n Cr = n Cn − r (v) r! n Cr = n Pr 9. Selesaikan setiap yang berikut. Solve each of the following. TP 2 (a) Buktikan / Prove that (i) r! n Cr = n Pr (ii) n C0 = 1 CONTOH Buktikan / Prove that (i) n Cn = 1 (ii) n Cr = n Cn − r Penyelesaian: (i) n Cn = n! (n − n)!n! = n! (0)!n! = 1 (ii) Sebelah kiri / Left side = n Cr = n! (n − r)!r! Sebelah kanan / Right side = n Cn − r = n! (n − n + r)!(n – r)! = n! r!(n − r)! Maka / Hence n Cr = n Cn − r (i) Sebelah kiri = r! n Cr Let side = r!n! (n − r)!r! = n! (n − r)! = n Pr Sebelah kanan Right side Maka/Hence, r! n Cr = n Pr (ii) n C0 = n! (n − 0)!0! = n! (n)! = 1 = sebelah kanan Right side

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 92 BAB 4 (b) Nilaikan / Evaluate (i) 8 C2 (ii) 4 C2 × 9 C4 (c) Carikan / Find (i) 10C7 ÷ 6 C3 (ii) n Cn − 2 10. Selesaikan setiap yang berikut. Solve each of the following. TP 2 (a) Daripada sekumpulan 8 orang pekerja, 4 orang pekerja akan dipilih secara rawak untuk menghadiri satu seminar peningkatan kemahiran. Cari bilangan cara boleh dilakukan. From a group of 8 workers, 4 workers are chosen randomly to attend a skill improvement seminar. Find the number of ways to do this. Bilangan cara pilihan = 8 C4 Number of ways = 70 (b) Seorang guru ingin memilih 5 orang pelajar secara rawak daripada 12 orang pelajar di dalam kelasnya untuk menyertai suatu pertandingan perbahasan. Cari bilangan cara yang boleh dilakukan . A teacher wants to choose 5 students randomly from the class of 12 students to take part in a debate. Find the number of ways to do this. Bilangan cara pilihan = 12C5 Number of ways = 792 (c) Di sebuah gerai makan, terdapat 7 jenis laukpauk untuk dipilih. Awang ingin memilih 4 jenis makanan, Cari bilangan cara dia boleh memilih jika semua lauk-pauk adalah sedap. At a food stall, there are 7 types of side dishes to choose. Awang wants to choose 4 types of food, Find the number of ways he can choose if all the side dishes are delicious. Bilangan cara pilihan = 7 C4 Number of ways = 35 (i) 10C7 ÷ 6 C3 = 10! (10 – 7)!7! ÷ 6! (6 – 3)!3! = 6 (ii) n Cn – 2 = n! (n − n + 2)!(n – 2)! = n! (2)!(n – 2)! = n(n – 1)(n – 2)! (2)!(n – 2)! = n(n – 1) (2) (i) 8 C2 = 8! (8 – 2)!2! = 8! (6)!2! = 28 (ii) 4 C2 × 9 C4 = 4! (4 – 2)!2! × 9! (9 – 4)!4! = 756 CONTOH Cari bilangan cara memilih 3 orang sukarelawan daripada 12 orang sukarelawan. Find the number of ways to choose 3 volunteers from 12 volunteers. Penyelesaian: Oleh sebab pilihan dan susunan ini tidak penting, ini adalah masalah gabungan. Since this selection and arrangement is not important, this is a combination problem. Bilangan cara / Number of ways = 12C3 = 12! (12−3)!(3)! = 220 Sudut Kalkulator Tekan 8 SHIFT 2 = 28 n Cr

BAB Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 93 4 CONTOH Satu jawatan kuasa kebersihan terdiri daripada 5 orang yang dipilih daripada 5 orang guru lelaki, 6 orang guru perempuan dan seorang pengetua dari suatu sekolah. Cari bilangan cara memilih jawatan kuasa itu jika A cleanliness committee is formed by 5 people chosen from 5 male teachers, 6 female teachers and a principal of a school. Find the number of ways to select the committee if (i) tiada sebarang syarat dikenakan there is no given conditions. (ii) jawatan kuasa itu mesti ada pengetua. the committee must have principal. (iii)jawatan kuasa itu mesti ada pengetua dan dua orang guru lelaki. the committee must have the principal and two men teachers. Penyelesaian: (i) Jumlah orang = 12 dan memerlukan 5 orang. Total people = 12 and we need 5 people only. Maka, bilangan cara = 12C5 = 792 Hence, the number of ways = 12C5 = 792 (ii) Pengetua mesti ada di dalam jawatan kuasa, bilangan cara memilih pengetua = 1 C1 . The principal must be in the committee, the number of ways to select the principal = 1C1. Masih ada 4 tempat untuk diisi oleh 11 orang guru, maka bilangan cara memilih guru = 11C4 . There are 4 more places to be filled up by 11 available teachers, hence the number of ways to select teachers = 11C4. Dengan menggunakan petua pendaraban, jumlah bilangan cara = 11C4 × 1 C1 = 330 By using multiplication rule, the total number of ways = 11C4 × 1C1 = 330 (iii) Jawatan kuasa memerlukan 1 pengetua, 2 guru lelaki dan 2 guru perempuan. Maka, jumlah bilangan cara The committee needs 1 principal, 2 male and 2 female teachers. Thus, the total number of ways = 1 C1 × 5 C2 × 6 C2 = 150 11. Selesaikan setiap soalan bersyarat yang berikut. Solve each of the following questions with the given conditions. TP 5 (a) 6 titik telah dilukis pada suatu kertas dengan keadaan tiada 3 titik adalah segaris. Cari bilangan 6 points are marked on a paper such that there is no 3 points on a straight line. Find the number of (i) garis lurus boleh dibina. straight line that can be formed. (ii) segi tiga boleh dibina. triangles that can be formed. (iii)segi empat boleh dibentuk. rectangles that can be formed. (i) Untuk melukis satu garisan, kita memerlukan mana-mana dua titik sahaja. To draw a straight line, we neeed any two points only. Maka, bilangan cara = 6 C2 = 15. So, number of ways (ii) Untuk melukis satu segi tiga, kita memerlukan mana-mana tiga titik sahaja. To draw a triangle, we need any three points only. Maka, bilangan cara = 6 C3 = 20. So, number of ways (iii)Untuk melukis satu segi empat, kita memerlukan mana-mana empat titik sahaja. To draw a rectangle, we need any four points only. Maka, bilangan cara = 6 C4 = 15. So, number of ways

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 94 BAB 4 (b) Suatu pasukan tenis terdiri daripada 5 orang yang dipilih daripada 4 orang lelaki dan 5 orang perempuan. Cari bilangan cara boleh dilakukan jika pasukan itu A tennis team is formed by 5 people, which is chosen from 4 men and 5 women. Find the number of ways to form the team if the team (i) mesti mengandungi seorang lelaki sahaja. must have only one man. (ii) mesti mengandungi lebih perempuan daripada lelaki. must have more women than men. (iii) mesti mengandungi satu pasang lelaki dan perempuan yang tertentu. must have a pair of males and a specific female. (c) Terdapat 8 orang ingin pergi ke bandar menggunakan dua buah kereta yang masing-masing hanya boleh membawa 4 orang sahaja. Terdapat hanya tiga daripada 8 orang itu ada lesen memandu kereta. Cari bilangan cara mereka boleh pergi ke bandar. There are 8 people who want to go to town in two cars where each car can only take 4 people. There are only three of the 8 people have car driving license. Find the number of ways of them to go to town. Bilangan cara = 2 [3 C1 × 2 C1 × 6 C3 × 3 C3 ] Number of ways = 240 (d) Suatu hadiah yang mengandungi 5 bahan akan dipilih daripada 10 pensel berwarna-warni, 7 batang pen dan 6 batang pembaris. Cari bilangan cara untuk membungkus hadiah jika A prize which consists of 5 items will be chosen from 10 colour pencils, 7 pens and 6 rulers. Find the number of ways to package the prize if (i) tiada halangan. there is no restriction. (ii) hadiah mesti mengandungi sekurang-kurangnya satu bahan daripada tiga kategori itu. it must contain at least one item from the three categories. (iii) hadiah tidak mengandungi pembaris. it does not contain rulers. (i) Bilangan cara = 23C5 = 33 649 Number of ways (ii) Bilangan cara = 10C1 × 7 C1 × 6 C1 × 20C2 Number of ways = 79 800 (iii)Bilangan cara = 17C5 = 6 188 Number of ways (i) Bilangan cara = 4 C1 × 5 C4 = 20 Number of ways (ii) Kes 1: 3 perempuan, 2 lelaki Bilangan cara = 5 C3 × 4 C2 = 60 Number of ways Kes 2: 4 perempuan, 1 lelaki Bilangan cara = 5 C4 × 4 C1 = 20 Number of ways Kes 3: 5 perempuan, 0 lelaki Bilangan cara = 5 C5 = 1 Number of ways Jumlah cara = 81 (iii)Bilangan cara = 6 C2 = 15 Number of ways

BAB Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 95 4 (e) Rajah menunjukkan dua baris kerusi, satu baris berdekatan dengan pintu keluar. Cari bilangan pilihan cara tempat duduk bagi 7 orang yang terdiri daripada 2 kanak-kanak, 1 pasang suami-isteri dan 4 orang dewasa jika The diagram shows two rows of chairs, one row is near to the exit door. Find the number of ways of seats for 7 people that is formed by 2 children, 1 couple of husband-wife and 4 adults if (i) 2 kanak-kanak mesti duduk dalam baris yang sama. 2 children must sit in the same row. (ii) 1 pasang suami-isteri dan 2 kanak-kanak mesti duduk pada baris yang sama. 1 couple of husband-wife and 2 children must sit in the same row. (i) Jika memilih baris tiga kerusi. If choose the row of three chairs Bilangan cara = 5 C1 × 4 C4 = 5 Number of ways Jika memilih baris empat kerusi If choose the row of four chairs Bilangan cara = 5 C2 × 3 C3 = 10 Jumlah = 15 (ii) 1 cara sahaja Only 1 way (f) 4 huruf dipilih daripada perkataan PEMBARIS. Cari bilangan cara jika pilihan itu 4 letters are chosen from the word PEMBARIS. Find the number of ways to choose the letters if it (i) mengandungi huruf B. contains the letter B. (ii) tidak mengandungi vokal. does not contain a vowel. (iii) mengandungi huruf E atau R. it contains letter E or R. (i) Bilangan cara = 7 C3 = 35 Number of ways (ii) Bilangan cara = 5 C4 = 5 Number of ways (iii)Bilangan cara = 7 C3 × 2 = 70 Number of ways

Matematik Tambahan Tingkatan 5 Bab 4 Pilih Atur dan Gabungan 96 BAB 4 PRAKTIS PRAKTIS SPM SPM 14 3. Dalam almari Nur, 3 daripada 5 helai baju ialah biru, dan 2 daripada 4 pasang seluar ialah biru. Pada suatu hari, dia memilih 1 baju dan 1 pasang seluar untuk temuduga . Dicadangkan bahawa dia tidak boleh memakai warna sama bagi baju dan seluar. Cari berapa cara dia dapat memilih pakaiannya. In Nur’s wardrobe, 3 out of 5 blouses are blue, and 2 out of 4 pants are blue. On a certain day, she picks 1 blouse and 1 pair of pants for her interview. She is advised not to wear the same colour for her blouse and pants. Find the number of ways she can choose her clothes. Kes 1: Baju biru Case 1: Blue blouses 3 C1 × 2 C1 = 6 cara Kes 2: Seluar biru Case 2: Blue pants 2 C1 × 2 C1 = 4 cara Kes 3: Bukan baju biru dan seluar biru Case 3: Not blue blouses and blue pants 2 C1 × 2 C1 = 4 cara Jumlah cara = 14 cara Total ways 2018 2019 2019 Sudut Sudut KBAT KBAT (c) adalah nombor dengan lima digit dan berakhir dengan nombor ‘0’. is a five digit number and ends with a number ‘0’. (a) Kes 1: nombor berdigit 4 = 3 × 5 P3 = 180 Kes 2: nombor berdigit 5 = 5 × 5 P5 = 600 Kes 3: nombor berdigit 6 = 5 × 5 P5 = 600 Jumlah = 600 + 600 + 180 = 1380 (b) 2 × 4 × 3 + 2 × 5 × 4 × 3 = 144 (c) 5 P4 × 1 P1 = 120 Dengan menggunakan digit-digit 0, 1, 2, 4, 6 atau 7, cari bilangan nombor yang dibentuk jika nombor itu By using the digits 0, 1, 2, 4, 6 or 7, find the number of ways that can be formed if the number (a) lebih daripada 3 000 tanpa mengulangi digit. is more than 3 000 without repeating the digits. (b) di antara 2 500 dan 7 000 tanpa mengulangi digit. is between 2 500 and 7 000 without repeating the digits. KBAT Ekstra Kertas 1 1. (a) Diberi 7 Cn > 1, senaraikan semua nilai yang mungkin bagi n. Given 7Cn > 1, list all the possible values of n. (b) Diberi bahawa x Cm = x Cn , ungkapkan x dalam sebutan m dan n. Given that x Cm = x Cn, express x in terms of m and n. (a) 1 ≤ n < 7 dengan n ialah integer (b) x Cm = x Cn x! (x − m)!m! = x! (x − n)!n! n + m = x 2. Rajah menunjukkan kata laluan bagi mangga Yong. The diagram shows a password for Yong’s padlock. 0 2 4 2 Yong hendak menetapkan semula kata laluan, dengan keadaan kod baharu itu tidak boleh mengandungi digit 2 diikuti oleh digit 4. Berapa bilangan kod laluan berbeza yang dapat dibentuk? Yong wants to reset the password, where the new code cannot consist of digit 2 followed by digit 4. How many different passwords can be formed? 4! 2! – 3! = 6 Quiz 4 Praktis SPM Ekstra

BAB Matematik Tambahan Tingkatan 5 Bab 5 Taburan Kebarangkalian 97 NOTA IMBASAN NOTA IMBASAN 5 NOTA IMBASAN NOTA IMBASAN BAB 5 Taburan Kebarangkalian Probability Distribution 5.1 Pemboleh Ubah Rawak Random Variable 1. Pemboleh ubah rawak ialah suatu pemboleh ubah dengan nilainya ialah kesudahan numerik yang dapat ditentukan daripada suatu fenomena rawak. A random variable is a variable whose value is an outcome is numeric from a random phenomena. 2. Pemboleh ubah rawak diskret ialah suatu pemboleh ubah yang dapat dikira. Kesudahan yang mungkin, X dapat ditulis dalam tata tanda set, X = {x: x = 0, 1, 2, 3, ….}. A discrete random variable is a variable which can be counted. The possible outcomes, X is written in set notation, X = {x: x = 0, 1, 2, 3, ….}. 3. Pemboleh ubah rawak selanjar ialah suatu pemboleh ubah Y yang nilainya tidak dapat dikira dan ditulis dalam bentuk Y = {y : a < y ≤ b}, dengan keadaan a dan b adalah pemalar. A continuous random variable is a variable Y whose value cannot be counted and is written in the form Y = {y : a < y ≤ b} , where a and b are constants. 4. Taburan kebarangkalian pemboleh ubah rawak diskret ialah suatu taburan menunjukkan kebarangkalian semua kesudahan yang mungkin dalam bentuk graf, jadual atau rajah pokok. The probability distribution of a random discrete variable is a distribution to show the probability of all the possible outcomes in the form of graph, table or a tree diagram. 5. Nilai setiap kebarangkalian P(X = x) berada di antara 0 dan 1, iaitu 0 ≤ P(X = x) ≤ 1. The value of each probability P(X = x) is between 0 and 1, that is 0 ≤ P(X = x) ≤ 1. 6. Hasil tambah kebarangkalian bagi semua kesudahan yang mungkin ialah 1. ∑n 0 P[X= x] = 1, x = 0, 1, 2, . . . , n The sum of probability of all the possible outcomes is 1. ∑n 0 P[X = x] = 1, x = 0,1, 2, . . . , n 1. Nyatakan pemboleh ubah rawak bagi setiap situasi yang berikut. State the random variable for each of the following situations. TP 1 CONTOH (i) Satu duit sylling yang adil dilambungkan sekali. A fair coin is tossed once. (ii) Jisim sebiji nanas di dalam bakul adalah di antara 0.5 kg dan 2.8 kg. The mass of a pineapple in the basket is between 0.5 kg and 2.8 kg. Penyelesaian: (i) Pemboleh ubah ialah gambar di permukaan atas duit syiling, iaitu {kepala, bunga}. The variable is the picture on the upper surface of coin, that is {head, flower}. (ii) Pemboleh ubah ialah jisim nanas The variable is the mass of the pineapple. (a) Keputusan permainan catur. The results of a chess game. Pemboleh ubah ialah menang, tewas atau seri The variable is win, lose or draw