Praktis Topikal Matematik Tambahan T5

Praktis Topikal SPM: Matematik Tambahan Tingkatan 5 – Jawapan J2 J3 Luas kawasan rumput yang boleh diragut oleh kambing Area that goat can access = 38.70 + 7.937 = 46.64 m2 BAB 2 Pembezaan KERTAS 1 BAHAGIAN A 1 (a) x = 8 + 2k3 dx dk = 6k2 (b) dy dx = dy dk × dk dx = 12k5 × 1 6k2 = 2k3 = 2 x 2 – 42 x = 8 + 2k3 2k3 = x – 8 k3 = x – 8 2 = x 2 – 4 = x – 8 2 (a) y = x(3x + 1)(2x – 3) = x(6x2 – 7x – 3) = 6x3 – 7x2 – 3x dy dx = 18x2 – 14x – 3 (b) y = 3 2 u5 , u = 2x – 4 dy du = 5 3 2 u4 , du dx = 2 dy dx = dy du × du dx = 5 3 2 u4 × 2 = 5 3 2 (2x – 4)4 × 2 = 15(2x – 4)4 3 (a) f(x) = 5x2 + 3x + 4 f’(x) = 10x + 3 f’’(x) = 10 (b) f(x) = x2 – 2 x f(x) = x2 – 2x–1 f’(x) = 2x + 2x–2 f’’(x) = 2 – 4x–3 = 2 – 4 x3 4 (a) y = x2 – 27 x2 dy dx = 2x + 54x–3 d2 y dx2 = 2 – 162x–4 Apabila/When d2 y dx2 = 0, 2 – 162x–4 = 0 2 = 162 x4 x4 = 81 x = 3 atau/or x = –3 (b) dy dx = 4px + 3 Apabila/When x = 2, dy dx = 11 4p(2) + 3 = 11 8p = 8 p = 1 5 (a) y = 3x(x – 2) = 3x2 – 6x dy dx = 6x – 6 (b) Apabila y minimum, dy dx = 0 When y is minimum, dy dx = 0 6x – 6 = 0 6x = 6 x = 1 (c) Nilai minimum y, Minimum value of y, Daripada/From y = 3x(x – 2) y = 3x2 – 6x = 3(1)2 – 6(1) = –3 6 (a) y = 2x2 – 5x dy dx = 4x – 5 Pada titik/At point (2, –5), dy dx = 4(2) – 5 = 8 – 5 = 3 (b) mnormal = – 1 3 y – (–5) = – 1 3 (x – 2) y + 5 = – 1 3 x + 2 3 3y + 15 = –x + 2 3y = –x – 13 7 (a) y = 2x2 – 7x + 3 dy dx = 4x – 7 Pada titik/At point (1, 3), dy dx = 4(1) – 7 = 4 – 7 = –3 (b) y – 3 = –3(x – 1) y – 3 = –3x + 3 y = –3x + 6 8 (a) y = 8 x2 = 8x–2 dy dx = –16x–3 = – 16 x3 δx = (4 + h) – 4 = h δy = dy dx × δx = – 16 (4)3 × h = – 1 4 h (b) y = 5 – 8 x = 5 – 8x–1 dy dx = 8x–2 = 8 x2 Apabila/ When y = 3, 3 = 5 – 8 x 8 x = 2 x = 4 Maka,/Therefore, dy dx = 8 (4)2 = 0.5 Apabila y menokok daripada 3 kepada 3 + q When y increases from 3 to 3 + q, δy = (3 + q) – 3 = q δy δx = dy dx δx = dx dy × δy = 1 0.5 × q = q 0.5 9 (a) dv dt = dv dr × dr dt 90p = 4pk2 × 2.5 k2 = 9 k = 3 (b) V = 4 3 pr3 dV dr = 4pr2 Diberi/Given r = 1 cm dV dr = 4p(1)2 = 4p dV dt = dV dr × dr dt – p 2 = (4p) × dr dt dr dt = – 1 8 Kadar perubahan jejari belon apabila jejari 1 cm ialah – 1 8 cm s–1. The rate of change of the radius of the balloon when the radius is 1 cm is 1 8 cm s–1. 10 y = 2x2 – 4x + 6 dy dx = 4x – 4 Apabila/When x = 2 , y = 2(2)2 – 4(2) + 6 = 6 dy dx = 4(2) – 4 = 4 dx = 3 100 × 2 = 0.06 dy = dy dx × dx = 4 × 0.06 = 0.24 dy y × 100 = 0.24 6 × 100 = 4% BAHAGIAN B 11 (a) dV dt = –3.5p V = 4 3 pr 3 dV dr = 4pr 2 Apabila/When r = 5, dV dr = 4p(5)2 = 100p dV dt = dV dr × dr dt dr dt = dV dt × dr dV = –3.5p × 1 100p = –0.035 cm s–1 (b) y = x3 dy dx = 3x2 Apabila/When x = 5, y = 53 = 125 dy dx = 3(5)2 = 75 δy = dy dx × δx = 75 × (5.003 – 5) = 0.225 Maka, nilai hampir bagi y ialah Therefore, the approximate value of y is y + δy = (5)3 + 0.225 = 125 + 0.225 = 125.23 10_PT SPM Add Math F5_2023.indd 3 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J4 J5 12 (a) V = x3 dV dx = 3x2 Diberi/Given dV dt = 75.6 dV dt = dV dx × dx dt dx dt = dV dt × dx dV = 75.6 × 1 3(5)2 = 1.008 cm s–1 (b) dV dt = 6.4p , dr dt = 0.4 V = 4 3 pr 3 dV dr = 4pr 2 dV dt = dV dr × dr dt 6.4p = 4pr2 × 0.4 4pr2 = 16p 4r2 = 16 r2 = 4 r = 2 cm KERTAS 2 BAHAGIAN A 1 (a) y = 3x(1 – x)4 u = 3x , v = (1 – x)4 du dx = 3 dv dx = 4(1 – x)3 (–1) = –4(1 – x)3 dy dx = 3x[–4(1 – x)3 ] + 3[(1 – x)4 ] = –12x(1 – x)3 + 3(1 – x)4 = 3(1 – x)3 [–4x + (1 – x)] = 3(1 – x)3 (1 – 5x) Pada titik/At point (2, 6), dy dx = 3(1 – 2)3 [1 – 5(2)] = 3(–1)(–9) = 27 Kecerunan tangen pada titik A ialah 27. Maka, kecerunan normal pada titik A ialah – 1 27 . Gradient of tangent at point A is 27. Thus, the gradient of normal at point A is – 1 27 . (b) Persamaan normal Equation of the normal, y – 6 = – 1 27 (x – 2) 27y – 162= –x + 2 27y + x = 164 2 (a) y = x2 (2 – x) + 3 = 2x2 – x3 + 3 dy dx = 4x – 3x2 = 4(2) – 3(2)2 = –4 (b) Titik pusingan/Turning point, dy dx = 0 4x – 3x2 = 0 x(4 – 3x) = 0 x = 0, x = 4 3 Apabila/When x = 0, y = 2(0)2 – (0)3 + 3 = 3 Apabila/When x = 4 3 , y = 2 4 3  2 –  4 3  2 + 3 = 113 27 \ Koordinat titik pusingan ialah (0, 3) dan  4 3 , 113 27  \ The coordinates of the turning points are (0, 3) and  4 3 , 113 27  (c) d2 y dx2 = 4 – 6x Apabila/When x = 0, d2 y dx2 = 4 – 6(0) = 4 > 0 Maka, (0, 3) ialah titik minimum. Thus, (0, 3) is a minimum point. Apabila/When x = 4 3 , d2 y dx2 = 4 – 6 4 3  = –4 < 0 Maka,  4 3 , 113 27  ialah titik maksimum. Thus,  4 3 , 113 27  is a maximum point. 3 (a) y = 12 x3 = 12x–3 dy dx = –36x–4 = – 36 x4 Apabila/When x = 2, dy dx = – 36 (2)4 = – 36 16 = – 9 4 (b) Apabila/When x = 2, y = 12 (2)3 = 12 8 = 3 2 δx = 2.01 – 2 = 0.01 12 (x + δx)3 ≈ y + dy dx × δx 12 (2 + 0.01)3 = 3 2 + – 9 4 (0.01) = 1.478 ZON KBAT 1 (a) dy dx = mx2 – nx Pada titik pusingan (2, –8), dy dx = 0 At turning point (2, –8), dy dx = 0 m(2)2 – n(2) = 0 4m – 2n = 0 2m – n = 0 ------- ➀ dy dx = 8 apabila/when x = 1 m(–1)2 – n(–1) = 8 m + n = 8 ------- ➁ Tambahkan ➀ dan ➁, Adding ➀ and ➁, 3m = 8 m = 8 3 Gantikan m = 8 3 ke dalam ➁, Substitute m = 8 3 into ➁, 8 3 + n= 8 n= 16 3 \ m = 8 3 , n = 16 3 (b) dy dx = 8 3 x2 – 16 3 x = 8 3 × 32 – 16 3 × 3 = 24 – 16 = 8 mnormal = – 1 8 y – 4 = – 1 8 (x – 3) 8y – 32 = –x + 3 8y + x = 35 BAB 3 Pengamiran KERTAS 1 BAHAGIAN A 1 (a) ∫ g(x) dx = ∫ d dx (7x) dx = 7x + c (b) ∫ 4(8 – x)2 dx = 4(8 – x)3 3(–1) = 4(8 – x)3 –3 \ p = – 4 3 , q = 3 2 (a) Pada titik pusingan (3, 1), dy dx = 0 At turning point (3, 1), dy dx = 0 hx – 6 = 0 h(3) – 6 = 0 3h = 6 h = 2 (b) Daripada/From dy dx = 2x – 6 y = ∫(2x – 6) dx = 2x2 2 – 6x + c = x2 – 6x + c Pada titik/At point (3, 1) 1 = (3)2 – 6(3) + c 1 = 9 – 18 + c c = 10 \ y = x2 – 6x + 10 3 (a) ∫(3x2 + 2) dx = 3x3 3 + 2x + c = x3 + 2x + c Bandingkan/Compare hx3 + 2x + c = x3 + 2x + c h = 1 (b) Apabila/When x = 1, ∫(3x2 + 2) dx = 10 13 + 2(1) + c = 10 3 + c = 10 c = 7 4 (a) ∫ 1 5 m(x) dx = – ∫ 5 1 m(x) dx = – 9 (b) ∫ 5 1 [kx – m(x)] dx = 23 ∫ 5 1 kx dx – ∫ 5 1 m(x) dx = 23  kx2 2  5 1 – 9 = 23 k(5)2 2 – k(1)2 2 = 32 25k 2 – k 2 = 32 24k = 64 k = 64 24 = 8 3 10_PT SPM Add Math F5_2023.indd 4 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Praktis Topikal SPM: Matematik Tambahan Tingkatan 5 – Jawapan J4 J5 5 (a) ∫ 4 6 q(x) dx = –∫ 6 4 q(x) dx = – 7 (b) ∫ 6 4 [mx – q(x)] dx = 21 ∫ 6 4 mx dx – ∫ 6 4 q(x) dx = 21  mx2 2  6 4 – 7 = 21 m(6)2 2 – m(4)2 2 = 28 36m 2 – 16m 2 = 28 10m = 28 m = 14 5 6 (a) ∫ 2 0 f(x) dx =  5x 7 – x  2 0 =  5(2) 7 – 2  –  5(0) 7 – 0  = 10 5 – 0 = 2 (b) ∫ 1 0 2g(x) dx = 2∫ 1 0 g(x) dx = 2 2x x2 + 1  1 0 = 2 2(1) (1)2 + 1 – 0 = 2 7 (a) ∫ q p f(x) dx = – 12 ∫ q p 4f(x) dx = 4∫ q p f(x) dx = 4( – 12) = – 48 (b) ∫ k 1 (x + 2) dx =  x2 2 + 2x k 1 =  k2 2 + 2k –  (1)2 2 + 2(1) = k2 2 + 2k – 1 2 – 2 = k2 2 + 2k – 5 2 8 (a) ∫ 2 7 m(x) dx = –∫ 7 2 m(x) dx = – 4 (b) ∫ 7 2 [3 – m(x)] dx = ∫ 7 2 3 dx – ∫ 7 2 m(x) dx = [3x] 7 2 – 4 = 3(7) – 3(2) – 4 = 21 – 6 – 4 = 11 9 (a) ∫ 8 3 [3f(x) – px] dx = 28 ∫ 8 3 3f(x) dx – ∫ 8 3 px dx = 28 3 ∫ 8 3 f(x) dx – 3 px2 2 4 8 3 = 28 3(10) – 1 82 p 2 – 32 p 2 2 = 28 30 – 55p 2 = 28 55p 2 = 2 p = 4 55 (b) ∫ 8 3 1 2 [f(x) + 6] dx = 1 2 ∫ 8 3 f(x) dx + ∫ 8 3 3 dx = 1 2 (10) + 3x4 8 3 = 5 + [3(8) – 3(3)] = 20 10 (a) y = x2 + 3 ............... y – 5 = –x ..................... Gantikan  ke dalam , Substitute  into , (x2 + 3) – 5 = –x x2 + x – 2 = 0 (x + 2)(x – 1) = 0 x = –2 atau/or x = 1 Maka, p = –2 dan q = 1. Thus, p = –2 and q = 1 (b) Luas rantau berlorek Area of the shaded region = ∫ 1 –2 [(5 – x) – (x2 + 3)] dx = ∫ 1 –2 (–x2 – x + 2) dx = – x3 3 – x2 2 + 2x4 1 –2 = – (1)3 3 – (1)2 2 + 2(1)4 – – (–2)3 3 – (–2)2 3 + 2(–2)4 = 7 6 – 1– 10 3 2 = 9 2 unit2 /units2 11 (a) ∫ p 2 [f(x) + 2] dx = 10 ∫ p 2 f(x) dx + ∫ p 2 2 dx = 10 6 + [2x] p 2 = 10 6 + [2p – 2(2)] = 10 2p = 10 – 6 + 4 2p = 8 p = 4 (b) x = 4 y x2 = 16 y2 Isi padu janaan/Generated volume = ∫ 3 1 px2 dy = ∫ 3 1 p1 16 y2 2 dy = ∫ 3 1 p(16y–2) dy = p3 16y–1 –1 4 3 1 = –p1 16 3 – 16 1 2 = –p1– 32 3 2 = 32 3 p unit3 /units3 BAHAGIAN B 12 (a) dy dx = 4x + 9 y = ∫(4x + 9) dx = 4x2 2 + 9x + c = 2x2 + 9x + c Pada titik A(2, -3) At point A(2, – 3), –3 = 2(2)2 + 9(2) + c –3 = 8 + 18 + c c = – 29 \ Persamaan lengkung ialah The equation of the curve is y = 2x2 + 9x – 29 (b) dy dx = 3x2 – 3x + 2 y = ∫(3x2 – 3x + 2) dx = 3x3 3 – 3x2 2 + 2x + c Pada titik (1, –3), At point (1, –3), –3 = 3(1)3 3 – 3(1)2 2 + 2(1) + c –3 = 1 – 3 2 + 2 + c c = – 9 2 \ Persamaan lengkung ialah Equation of the curve is y = x3 – 3x2 2 + 2x – 9 2 13 (a) dy dx = 4x – 4 x2 Pada titik pusingan, dy dx = 0 At turning point, dy dx = 0 4x – 4 x2 = 0 4x3 – 4 = 0 4(x3 – 1) = 0 x3 – 1 = 0 x = 1 \ h = 1 dy dx = 4x – 4 x2 y = ∫(4x – 4x–2) dx = 2x2 + 4x–1 + c Pada titik (1, 12)/At point (1, 12), 12 = 2(1)2 + 4 (1) + c c = 6 Persamaan bagi lengkung ialah Equation of the curve is y = 2x2 + 4 x + 6 (b) dy dx = px2 – qx Pada titik (2, –8)/At point (2, –8), p(2)2 – q(2) = 0 2p – q = 0................. dy dx = px2 – qx 6 = p(–1)2 – q(–1) 6 = p + q q = 6 – p....................... Gantikan  ke dalam , Substitute  into , 2p – (6 – p) = 0 p = 6 Gantikan p = 6 ke dalam , Substitute p = 6 into , q = 6 – 2 = 4 KERTAS 2 BAHAGIAN A 1 (a) (i) ∫ 4 1 f(x) dx = – 10 (luas rantau berlorek di bawah paksi-x) (area of the shaded region below the x-axis) (ii) ∫ 4 1 [x – 4f(x)] dx = ∫ 4 1 x dx – 4∫ 4 1 f(x) dx =  x2 2  4 1 – 4( –10) 10_PT SPM Add Math F5_2023.indd 5 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J6 J7 =  (4)2 2 – (1)2 2  + 40 = 16 2 – 1 2 + 40 = 47 1 2 (iii)∫ 1 4 [x + 3f(x)] dx = ∫ 1 4 x dx + 3∫ 1 4 f(x) dx =  x2 2  1 4 + 3(10) =  (1)2 2 – (4)2 2  + 30 = 1 2 – 16 2 + 30 = 22 1 2 (b) f’(x) = x – 4 f(x) = ∫(x – 4) dx = x2 2 – 4x + c Pada/At x = 1, f(x) = 0 0 = (1)2 2 – 4(1) + c c = 7 2 \ f(x) = x2 2 – 4x + 7 2 BAHAGIAN B 2 (a) x + 4 = (x – 2)2 x + 4 = x2 – 4x + 4 x2 – 4x + 4 – x – 4 = 0 x2 – 5x = 0 x(x – 5) = 0 x = 0 atau/or x = 5 Pada titik L/At point L, x = 5, y = 5 + 4 = 9 \ Koordinat titik L = (5, 9) Coordinates of point L = (5, 9) (b) (i) ∫ 5 0 [(x + 4) – (x – 2)2 ] dx = ∫ 5 0 (x + 4) dx – ∫ 5 0 (x2 – 4x + 4) dx =  x2 2 + 4x 5 0 –  x3 3 – 4x2 2 + 4x 5 0 =  52 2 + 4(5) –  53 3 – 2(5)2 + 4(5) =  25 2 + 20 –  125 3 – 50 + 20 = 65 2 – 35 3 = 125 6 unit2 /units2 (b) (ii) y = (x – 2)2 Apabila/When y = 0, x = 2 V = p ∫ 2 0 [(x – 2)2 ] 2 dx = p ∫ 2 0 (x – 2)4 dx = p (x – 2)5 5  2 0 = p0 – (0 – 2)5 5  = p 32 5  = 32 5 p unit3 /units3 3 (a) Fungsi kecerunan, Gradient function, dy dx = 4x y = ∫ 4x dx = 4x2 2 + c = 2x2 + c Pada titik L(1, 6), At point L(1, 6), 6 = 2(1)2 + c 6 = 2 + c c = 4 Maka, persamaan lengkung ialah y = 2x2 + 4. Hence, the equation of the curve is y = 2x2 + 4. (b) O y x K y = 2x2 + 4 L = (1, 6) M(1, 0) 4 Luas rantau berlorek Area of the shaded region = Luas segi empat tepat OKLM Area of rectangle OKLM – ∫ 1 0 y dx = (1  6) – ∫ 1 0 (2x2 + 4) dx = 6 –  2x3 3 + 4x 1 0 = 6 –  2(1)3 3 + 4(1) – 0 = 6 – 2 3 – 4 = 1 1 3 unit2 /units2 (c) y = 2x2 + 4 2x2 = y – 4 x2 = y – 4 2 = y 2 – 2 V = ∫ 6 4 p x2 dy = p∫ 6 4  y 2 – 2dy = p y2 4 – 2y 6 4 = p 62 4 – 2(6) –  42 4 – 2(4) = p[(9 – 12) – (4 – 8)] = p[–3 – (–4)] = p(–3 + 4) = p unit3 /units3 ZON KBAT 1 Titik persilangan/Points of intersection (x – 3)(x – 6) = 2x x2 – 6x – 3x + 18 = 2x x2 – 6x – 3x – 2x + 18 = 0 x2 – 11x + 18 = 0 (x – 2)(x – 9) = 0 x = 2 atau/or x = 9 Apabila/When x = 2, y = 2(2) = 4 0 2 3 4 A B Luas rantau berlorek A: Area of the shaded region A: A = 1 2 (2)(4) = 4 unit2 /units2 Luas rantau berlorek B: Area of the shaded region B: ∫ 3 2 (x – 3)(x – 6) dx = ∫ 3 2 (x2 – 9x + 18) dx =  x3 3 – 9x2 2 + 18x 3 2 =  33 3 – 9(3)2 2 + 18(3) –  23 3 – 9(2)2 2 + 18(2) = 9 – 81 2 + 54 –  8 3 – 18 + 36 = 15 6 unit2 /units2 Jumlah luas kawasan berlorek Total area of the shaded region = 4 + 1 5 6 = 5 5 6 unit2 /units2 2 (a) y = x2 + 1 dy dx = 2x Pada titik/At point (1, 3), dy dx = 2(1) = 2 ∴ Kecerunan garis KL ialah – 1 2 . The gradient of line KL is – 1 2 . 3 – 0 1 – p = – 1 2 (3)(2) = (–1)(1 – p) 6 = p – 1 p = 7 (b) Luas rantau berlorek Area of the shaded region = ∫ 1 0 (x2 + 1) dx + 1 2 (7 – 1)(3) =  x3 3 + x 1 0 + 9 = 1 13 3 + 12 + 9 = 4 3 + 9 = 10 1 3 unit2 /units2 BAB 4 Pilih Atur dan Gabungan KERTAS 1 BAHAGIAN A 1 (a) (4 P1 × 3 P1 × 2) + (4 P1 × 3 P1 × 1) = 24 + 12 = 36 (b) 8 P3 = 336 2 (a) 6 P2 × 7 P2 = 1 260 (b) 3 × 7 P2 × 6 = 756 3 (a) 5! = 120 (b) P P P C C 3! × 3! = 36 4 (a) 5! = 120 (b) 2! × 2! × 1 = 4 5 (a) 7! = 5 040 (b) 3 × 2! × 5! = 720 6 (a) 8 P8 = 40 320 (b) 6 P6 × 3 P2 = 4 320 10_PT SPM Add Math F5_2023.indd 6 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Praktis Topikal SPM: Matematik Tambahan Tingkatan 5 – Jawapan J6 J7 7 (a) (5 C2 × 3 C1 ) + (5 C1 × 3 C2 ) = 30 + 15 = 45 (b) 2 C2 × 3 C1 = 3 (c) (4 C1 × 3 C1 ) + 3 C2 = 12 + 3 = 15 8 (a) 7 C4 × 5 C3 = 350 (b) 3! × 5! = 720 9 (a) 6 C2 × 5 C3 = 15 + 10 = 25 (b) (6 C4 × 5 C1 ) + (6 C3 × 5 C2 ) = 75 + 200 = 275 10 (a) 8 C4 × 5 C2 = 700 (b) (8 C4 × 5 C2 ) + (8 C5 × 5 C1 ) + (8 C6 × 5 C0 ) = 1 008 11 (a) 13C7 = 1 716 (b) (7 C6 × 6 C1 ) + (7 C7 × 6 C0 ) = 43 12 (a) 6 P3 = 120 (b) 6 C4 + 6 C5 + 6 C6 = 22 ZON KBAT 1 Kes 1: Jika bermula dengan 3 atau 5 Case 1: If starts with 3 or 5 2 × 3 × 2 × 1 × 2 = 24 Kes 2: Jika bermula dengan 4 Case 2: If starts with 4 1 × 3 × 2 × 1 × 3 = 18 \ 24 + 18 = 42 cara/ways 2 (a) 2 P1 × 4 P3 = 48 (b) 4 P3 × 3 P1 = 72 BAB 5 Taburan Kebarangkalian KERTAS 1 BAHAGIAN A 1 (a) P(X  2) = P(X = 2 ) + P(X = 3 ) = 27 64 + 27 64 = 27 32 (b) 1 – 27 64 – 27 64 – 1 64 = 9 64 2 (a) np = 6, npq = 4 6q = 4 q = 4 6 \ p = 1 – 4 6 p = 1 3 (b) np = 6 n 1 3 = 6 n = 18 3 n = 10, p = 3 7 , q = 4 7 Katakan X ialah bilangan calon yang lulus ujian. Let X be the number of candidates that passed the test. (a) P(X = 0) = 10C0 3 7 0  4 7 10 = 0.0037 (b) P(X , 9) = 1 – P(X > 9) = 1 – P(X = 9) – P(X = 10) = 1 – 3 10C9 3 7 9  4 7 1 4 – 3 10C10 3 7 10 4 7 0 4 = 1 – 0.00279 – 0.0002090 = 0.9970 (c) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0037 + 3 10C1 3 7 1  4 7 9 4 + 3 10C2 3 7 2  4 7 8 4 + 3 10C3 3 7 3  4 7 7 4 + 3 10C4 3 7 4  4 7 6 4 = 0.0037 + 0.0278 + 0.09396 + 0.1879 + 0.2467 = 0.5601 4 (a) p = 0.98, q = 0.02 P(X = n) = 0.6099 nCn(0.98)n(0.02)0 = 0.6099 0.98n = 0.6099 n log 0.98 = log 0.6099 n = log 0.6099 log 0.98 n = 24.47 n = 24 (b) Min/Mean = np = 24(0.98) = 23.52 Varians/Variance = npq = 23.52(0.02) = 0.4704 5 (a) Z = 69 – 64 3 = 1.667 (b) P 60 – 64 3 , Z , 69 – 64 3  P(– 1.333 , Z , 1.667) = 1 – P(Z  1.333) – P(Z  1.667) = 1 – 0.0913 – 0.0478 = 0.8609 6 (a) X – 1.65 0.35 = – 0.22 X – 1.65 = – 0.077 X = 1.573 (b) Z = 1.6 – 1.65 0.35 = – 0.1429 P(Z  – 0.1429) = 1 – P(Z  0.1429) = 1 – 0.4431 = 0.5569 × 100 % = 55.69 % 7 (a) 0.45 = X – 43 8 X = 46.6 kg (b) PX  40 – 43 8  = P(Z  – 0.375) = 1 – P(Z  0.375) = 1 – 0.3538 = 0.6462 × 100 % = 64.62% 8 (a) P(0.38 < Z < 1.65) = P(Z  0.38) – P(Z  1.65) = 0.3520 – 0.0495 = 0.3025 (b) P(–1.30 < Z < –0.65) = P(Z  0.65 ) – P(Z  1.30) = 0.2578 – 0.0968 = 0.161 9 (a) P(X  162) = PZ  162 – 155 6  = P(Z  1.167) = 0.1216 (b) P(145 , X , 167) = P 145 – 155 6 , Z , 167 – 155 6  = P( –1.667 , Z , 2) = 1 – P(Z  1.667) – P(Z  2) = 1 – 0.0478 – 0.0228 = 0.9294 10 (a) μ = 54 (b) Z = X – μ σ 1.8 = k – 54 7 12.6 = k – 54 k = 66.6 BAHAGIAN B 11 (a) P( 3 , X , 7 ) = P 3 – 6 3 , Z , 7 – 6 3  = P(– 1 , Z , 0.3333) = 1 – P(Z  1) – P(Z  0.333) = 1 – 0.1587 – 0.3696 = 0.4717 (b) P(Z  0) = 0.5 P(Z  –m) = 0.5 + 0.3336 = 0.8336 KERTAS 2 BAHAGIAN A 1 (a) p = 0.25, q = 0.75 P(X  5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 8 C5 (0.25)5 (0.75)3 + 8 C6 (0.25)6 (0.75)2 + 8 C7 (0.25)7 (0.75)1 + 8 C8 (0.25)8 (0.75)0 = 0.0231 + 0.00385 + 0.000366 + 0.0000153 = 0.02733 (b) Min/Mean = np = (450) (0.25) = 112.5 Sisihan piawai/Standard deviation =  (450)(0.25)(0.75) = 9.186 BAHAGIAN B 2 (a) (i) μ = 0, σ = 1 (ii) P(–2.1 < Z < 0) = P(Z  0) – P(Z  2.1) = 0.5 – 0.0179 = 0.4821 (b) (i) n = 8, p = 1 3 , q = 2 3 P(X  1) = 1 – P(X = 0) = 1 – 8 C0 1 3 0  2 3 8 = 1 – 0.0390 = 0.961 (ii) n 1 32 3 = 52 n = 234 Maka, bilangan murid Tingkatan 5 ialah 234 orang. Thus, the number of Form 5 students is 234. 3 (a) (i) Z = 56.4 – 44 10 = 1.24 (ii) P(Z , k) = 0.9107 P(Z  k) = 1 – 0.9107 P(Z  k) = 0.0893 k = 1.345 (b) p = 0.83, q = 1 – 0.83 = 0.17 n = 23 + 17 = 40 (i) Min/Mean, μ = np = (40)(0.83) = 33.2 (ii) Varians/Variance = npq = (40)(0.83)(0.17) = 5.644 10_PT SPM Add Math F5_2023.indd 7 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J8 J9 (iii) Sisihan piawai/Standard deviation =  5.644 = 2.376 4 (a) (i) P(X  3 ) = 1 – P(X = 0) + P(X = 1) + P(X = 2) = 1 – [10C0 (0.3)0 (0.7)10 + 10C1 (0.3)1 (0.7)9 + 10C2 (0.3)2 (0.7)8 ] = 1 – 0.028 – 0.121 – 0.233 = 0.618 (ii) npq = 189 n(0.3)(0.7) = 189 n = 900 (b) P(X  62.2) = 0.0853 P(X , 41.3) = 0.0281 62.2 – μ σ = 1.37 62.2 – μ = 1.37σ ……. ① 41.3 – μ σ = –1.91 41.3 – μ = –1.91σ …… ② ① – ②: 20.9 = 3.28σ σ = 6.372 62.2 – μ = 1.37(6.372) μ = 53.47 5 (a) (i) p = 0.18, q = 0.82, n = 8 P(X  2) = 1 – P(X = 0) – P(X = 1) = 1 – 8 C0 (0.18)0 (0.82)8 – 8 C1 (0.18)1 (0.82)7 = 1 – 0.2044 – 0.3590 = 0.4366 (ii) σ2 = npq 130 = n(0.18)(0.82) n = 880 (b) μ = 66.56, σ2 = 43.56 kg2 , σ = 6.6 P(50 < X < 70) = P 50 – 66.56 6.6 < Z < 70 – 66.56 6.6  = P(–2.509 < Z < 0.5212) = 1 – P(Z  2.509) – P(Z  0.5212) = 1 – 0.00481 – 0.3012 = 0.6940 P(50 < X < 70) = 0.6940 180 n = 0.6940 n = 259 ZON KBAT 1 (a) n = 8, p = 0.7 P(X = r) = 8 Cr (0.7)r (0.3)8 – r P(X = 4) = 8 C4 (0.7)4 (0.3)4 = 0.1361 (b) P(X  1)  0.90 1 – P(X = 0)  0.90 P(X = 0) , 0.10 nC0 (0.7)0 (0.3)n , 0.10 log(0.3)n , log 0.10 nlog 0.3 , log 0.10 n  log 0.10 log 0.3 n  1.912 n = 2 2 (a) Min/Mean= 12 np = 12 .................. ① Varians/Variance = 4 np(1 – p) = 4 ………… ② Gantikan ① ke dalam ②, Substitute ① into ②, 12(1 – p) = 4 1 – p = 4 12 p = 1 – 1 3 p = 2 3 n × 2 3 = 12 n = 18 \ p = 2 3 dan bilangan percubaan ialah 18. p = 2 3 and the number of trials is 18. (b) n = 6, p = 2 3 P(X  1) = 1 – P(X = 0) = 1 – 6 C0 2 3 0  1 3 6 = 0.9986 BAB 6 Fungsi Trigonometri KERTAS 1 BAHAGIAN A 1 (a) y k θ 1 – x 1 – k2 kosek θ/cosec θ = 1 sin q = 1 k (b) sin 2θ = 2 sin θ kos θ/cos θ = 2(k)(– 1 – k2 ) = –2k 1 – k2 2 y A 13 –12 5 x 0 y 3 4 5 x 0 B (a) kot A/cot A = 1 tan A = kos A/cos A sin A = – 12 13 5 13 = – 12 5 (b) kos (A + B) = kos A kos B – sin A sin B cos (A + B) = cos A cos B – sin A sin B = – 12 13  4 5  –  5 13  3 5  = – 48 65 – 15 65 = – 63 65 3 y 3 4 5 A x y –24 –7 25 x B (a) kos A/cos A = 4 5 (b) sin (A – B) = sin A kos B – kos A sin B sin (A – B) = sin A cos B – cos A sin B =  3 5 – 24 25  –  4 5 – 7 25  = – 72 125 + 28 125 = – 44 125 4 y m 1 x θ 1 – m2 (a) kos θ/cos θ = 1 – m2 tan θ = sin θ kos q/cos q = m 1 – m2 (b) sin 2θ = 2 sin θ kos θ 2 sin θ cos θ = 2(m)( 1 – m2 ) = 2m 1 – m2 5 y 13 –12 5 x A y 5 3 4 x B (a) tan A = – 5 12 (b) kos (A – B) = kos A kos B + sin A sin B cos(A – B) = cos A cos B + sin A sin B = – 12 13  4 5  + 5 13  3 5  = – 48 65 + 15 65 = – 33 65 6 (a) Diberi kos θ = m Given cos θ = m 1 + tan2 θ = sek2 θ/sec2 θ tan2 θ = sek2 θ/sec2 θ – 1 = 1 kos2 θ/cos2 θ – 1 = 1 m2 – 1 = 1 – m2 m2 (b) kot x + 2 sek x = 0 cot x + 2 sec x = 0 kos x sin x + 2 kos x = 0 cos x sin x + 2 cos x = 0 kos x 1 sin x + 2 = 0 cos x 1 sin x + 2 = 0 kos/cos x = 0 x = 90°, 270° atau/or 1 sin x + 2= 0 sin x = – 1 2 x = 210°, 330° ∴ x = 90°, 210°, 270°, 330° 10_PT SPM Add Math F5_2023.indd 8 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Praktis Topikal SPM: Matematik Tambahan Tingkatan 5 – Jawapan J8 J9 BAHAGIAN B 7 (a) 3 sin x sek x – kos x = 0 3 sin x sec x – cos x = 0 3 sin x kos x – kos x = 0 3 sin x cos x – cos x = 0 kos x(3 sin x – 1) = 0 cos x(3 sin x – 1) = 0 kos/cos x = 0 x = 90°, 270° atau/or 3 sin x – 1= 0 sin x = 1 3 x = 19°28’, 160°32’ \ x = 19°28’, 90°, 160°32’, 270° (b) 15 – sin θ – 4 sin 30° = 15 kos2 θ 15 – sin θ – 4 sin 30° = 15 cos2 θ 15 – 15 kos2 θ = sin θ + 4 sin 30° 15 – 15 cos2 θ = sin θ + 4 sin 30° 15(1 – kos2 θ) = sin θ + 4 sin 30° 15(1 – cos2 θ) = sin θ + 4 sin 30° 15 sin2 θ = sin θ + 4 sin 30° 15 sin2 θ = sin θ + 4(0.5) 15 sin2 θ – sin θ – 2 = 0 (5 sin θ – 2)(3 sin θ + 1) = 0 5 sin θ – 2 = 0 sin θ = 2 5 θ = 23°35’, 156°25’ atau/or 3 sin θ + 1 = 0 sin θ = – 1 3 θ = 199°28’, 340°32’ \θ = 23°35’, 156°25’, 199°28’, 340°32’ 8 (a) 3 kos 2x = 8 sin x – 5 3 cos 2x = 8 sin x – 5 3(1 – 2 sin2 x) = 8 sin x – 5 3 – 6 sin2 x = 8 sin x – 5 6 sin2 x + 8 sin x – 8 = 0 3 sin2 x + 4 sin x – 4 = 0 (3 sin x – 2)(sin x + 2) = 0 sin x = 2 3 atau/or sin x = –2 x = 41°49’, 138°11’ (b) 1 tan x + 2 kos x = 0 1 tan x + 2 cos x = 0 kos x sin x + 2 kos x = 0 cos x sin x + 2 cos x = 0 kos x + 2 sin x kos x = 0 cos x + 2 sin x cos x = 0 kos x (1 + 2 sin x) = 0 cos x (1 + 2 sin x) = 0 kos x/cos x = 0 x = 90°, 270° atau/or 1 + 2 sin x = 0 2 sin x = –1 sin x = –0.5 x = 210°, 330° \ x = 90°, 210°, 270°, 330° 9 (a) 2 tan x kos x – 1 kot x = 0 2 tan x cos x – 1 cot x = 0 2 tan x kos x – tan x = 0 2 tan x cos x – tan x = 0 tan x(2 kos x – 1) = 0 tan x(2 cos x – 1) = 0 tan x = 0 x = 0°, 180°, 360° atau/or 2 kos x – 1 = 0 2 cos x – 1 = 0 kos/cos x = 1 2 x = 60°, 300° ∴ x = 0°, 60°, 180°, 300°, 360° (b) 2 tan2 α – 5 tan α + 2 = 0 Katakan/Let tan a = x 2x2 – 5x + 2 = 0 (2x – 1)(x – 2) = 0 2x – 1 = 0 x = 1 2 tan a = 1 2 a = 26°34’, 206°34’ atau/or x – 2 = 0 x = 2 tan a = 2 a = 63°26’, 243°26’ \ a = 26°34’, 63°26’, 206°34’, 243°26’ KERTAS 2 BAHAGIAN A 1 (a) Sebelah kiri/LHS = kosek2 x – 2 sin2 x – kot2 x cosec2 x – 2 sin2 x – cot2 x = 1 + kot2 x – 2 sin2 x – kot2 x 1 + cot2 x – 2 sin2 x – cot2 x = 1 – 2 sin2 x = kos/cos 2x (Sebelah kanan/RHS) (b) (i) (ii) π 2π y 1 –1 0 x 2 π 2 3π y = – 3π x 1 3 y = kos/cos 2x – 1 3 3(kosek2 x – 2 sin2 x – kot2 x) = x p – 1 3(cosec2 x – 2 sin2 x – cot2 x) = x p – 1 3 kos 2x = x p – 1 3 cos 2x = x p – 1 kos 2x/cos 2x= x 3p – 1 3 y = x 3p – 1 3 ∴ Garis lurus/The straight line: y = x 3p – 1 3 Apabila/When x = 0, y = – 1 3 x = p, y = 0 Bilangan penyelesaian = 4 Number of solutions = 4 2 (a) π 2π y x 2 π x π y = 3π 2 y = –2 kos/cos x 2 0 –2 (b) p x + 2 kos x = 0 p x + 2 cos x = 0 –2 kos/cos x = p x y = p x x p 2 p 2p y 2 1 0.5 Bilangan penyelesaian = 2 Number of solutions = 2 3 (a) π 2π x 2 π 3π 2 3 0 2 y y = |3 kos/cos 2x| y = 2 – x 2π (b) 2 – | 3 kos/cos 2x | = x 2p 2 – y = x 2p y = 2 – x 2p Apabila/When x = 0, y = 2 x = 2p, y = 1 ∴ Bilangan penyelesaian = 8 Number of solutions = 8 4 (a) π 4 π 2 π 3π 4 5π 4 3π 2 y = kos/cos 2x 3 2 x y 3 2 3 4 0 3 – 4 3 – 2 9 – 4 y = 2x π 9 4 – (b) 4x 3p – kos/cos 2x = 3 2 kos/cos 2x = 4x 3p – 3 2 3 2 kos/cos 2x = 3 2  4x 3p – 3 2  y = 2x p – 9 4 x 0 3p 2 y – 9 4 3 4 Bilangan penyelesaian = 3 Number of solutions = 3 10_PT SPM Add Math F5_2023.indd 9 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J10 J11 5 (a) y x –2 1 3 0 4 π 2 π 2 2π y = 1 + 3 kos/cos x 3π y = 3 – 3x 2π (b) 6p kos/cos x = 4p – 3x 3 kos/cos x = 2 – 3x 2p 1 + 3 kos/cos x = 2 – 3x 2p + 1 1 + 3 kos/cos x = 3 – 3x 2p y = 3 – 3x 2p x 0 2p y 3 0 Bilangan penyelesaian = 2 Number of solutions = 2 6 (a) π 3 π y = –3 sin 3x 2 y x 3 2π 3 4π 3 5π 2π y = x π 3 –3 0 (b) p x + 3 sin 3x 2 = 0 p x = –3 sin 3x 2 y = p x x 0 p 2p y ∞ 1 1 2 Bilangan penyelesaian = 2 Number of solutions = 2 7 (a) Sebelah kiri/LHS: sin2 θ 1 – kos/cos θ = 1 – kos2 /cos2 θ 1 – kos/cos θ = (1 – kos/cos θ)(1+ kos/cos θ) 1 – kos/cos θ = 1 + kos/cos θ (Sebelah kanan/RHS) (b) sin2 θ 1 – kos/cos θ = 3 2 1 + kos/cos θ = 3 2 kos/cos θ = 3 2 – 1 kos/cos θ = 0.5 θ = 60°, 300° 8 (a) Sebelah kanan/RHS: kot/cot x(1 – kos/cos 2x) = kos/cos x sin x (2 sin2 x) = 2 sin x kos/cos x = sin 2x (Sebelah kiri/LHS) (b) (i) (ii) π 2 π 2 1 y = – sin 2x y x 2π 1 –1 2 1 – 0 y = – + x 2π 2 2 1 3π 2 kot/cot x(1 – kos/cos 2x) = x p – 1 2 sin 2x = x p – 1 sin 2x = x 2p – 1 2 –sin 2x = 1 2 – x 2p y = 1 2 – x 2p x 0 p y 1 2 0 Bilangan penyelesaian = 3 Number of solutions = 3 ZON KBAT 1 sin (m + n) sin (m – n) = (sin m kos/cos n + kos/cos m sin n) × (sin m kos/cos n – kos/cos m sin n) = (sin m kos/cos n)2 – (kos/cos m sin n)2 = r2 –  1 r  2 = r2 – 1 r2 = r4 – 1 r2 2 Sebelah kiri/LHS: 2 kos/cos (p + 45°) kos/cos (p – 45°) = 2(kos/cos p kos/cos 45° – sin p sin 45°) × (kos/cos p kos/cos 45° + sin p sin 45°) = 2(kos2 /cos2 p kos/cos 45° – sin2 p sin2 45°) = 2  kos2 /cos2 p  1 2  2 – sin2 p  1 2  2  = 2  kos2 /cos2 p 1 2  – sin2 p 1 2  = kos2 /cos2 p – sin2 p = kos/cos 2p(Sebelah kanan/RHS) BAB 7 Pengaturcaraan Linear KERTAS 2 BAHAGIAN C 1 (a) I : 5x + 6y > 1 200 II : 0.4x + 0.5y < 400 III : y – x < 150 (b) 900 800 700 600 500 400 300 200 100 0 y x 200 400 600 800 1 000 1 200 1 400 0.4x + 0.5y = 400 (360, 510) 5x + 6y = 1 200 y – x = 150 R (c) (i) Bilangan pokok bunga ros putih Number of white roses plants = 450 (ii) Titik maksimum Maximum point = (360, 510) Keuntungan maksimum Maximum profit = 4.5(360) + 4(510) = RM3 660 2 (a) I : 25x + 20y < 2 500 5x + 4y < 500 II : x > 20 III : y – x  10 (b) 160 140 120 100 80 60 40 20 0 y x 20 40 60 80 100 120 y – x = 10 (20, 30) 5x + 4y = 500 x = 20 R (c) (i) Bilangan buku rujukan Biologi Number of Biology reference books = 100 (ii) Titik minimum Minimum point = (20, 30) Jumlah kos minimum Total minimum cost = 25(20) + 20(30) = RM1 100 3 (a) I : x + y < 80 II : y < 3x III : y – x < 10 (b) 90 80 70 60 50 40 30 20 10 0 y x 10 20 30 40 50 60 70 80 x + y = 80 (20, 60) y – x = 10 y = 3x R (c) (i) 30 < y < 60 (ii) Titik maksimum Maximum point = (20, 60) Jumlah yuran maksimum Total maximum fees = 700(20) + 800(60) = RM62 000 10_PT SPM Add Math F5_2023.indd 10 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Praktis Topikal SPM: Matematik Tambahan Tingkatan 5 – Jawapan J10 J11 4 (a) I : y – x < 150 II : y  2 3 x III : x + y < 400 (b) 450 400 350 300 250 200 150 100 50 0 y x 50 100 150 200 250 300 350 400 x + y = 400 (125, 275) y – x = 150 y = R 2 3 x (c) (i) Titik maksimum Maximum point = (125, 275) Jumlah yuran maksimum Total maximum fees = 20(125) + 30(275) = RM10 750 (ii) Bilangan minimum murid Tingkatan 3 Minimum number of Form 3 students = 100 Bilangan maksimum murid Tingkatan 3 Maximum number of Form 3 students = 250 5 (a) I : x + y < 70 II : x < 2y III : y – x < 40 (b) 80 70 60 50 40 30 20 10 0 y x 10 20 30 40 50 60 70 x + y = 70 (15, 55) y – x = 40 x = 2y R (c) (i) Bilangan pokok rambutan Number of rambutan trees = 10 (ii) Titik maksimum Maximum point = (15, 55) Jumlah kos maksimum Maximum total cost = 3(15) + 5(55) = RM320 6 (a) I : 40x + 40y < 20(22) 40x + 40y < 440 x + y < 11 II : 300x + 600y < 4 200 x + 2y < 14 III : x > 4 (b) 16 14 12 10 8 6 4 2 0 y x 2 4 6 8 10 12 14 16 x + y = 11 (8, 3) x = 4 x + 2y = 14 R (c) (i) 5 buah kerusi/chairs (ii) Titik maksimum Maximum point = (8, 3) Jumlah keuntungan maksimum Total maximum profit = 300(8) + 500(3) = RM3 900 BAB 8 Kinematik Gerakan Linear KERTAS 2 BAHAGIAN C 1 (a) vA = 6 + 4t – 2t 2 Halaju maksimum apabila a = 0 Maximum velocity when a = 0 dv dt = 4 – 4t = 0 4t = 4 t = 1 Apabila/When t = 1, vA = 6 + 4(1) – 2(1)2 = 10 – 2 = 8 m s–1 (b) Pada/At R, vA = 0 sA = (6 + 4t – 2t 2 ) dt = 6t + 4t 2 2 – 2t 3 3 = 6t + 2t 2 – 2t 3 3 vA = 0 6 + 4t – 2t 2 = 0 2t 2 – 4t – 6 = 0 t 2 – 2t – 3 = 0 (t – 3)(t + 1) = 0 t = 3 atau/or t = –1 t  0, maka/thus t = 3 Apabila/When t = 3, sA = 6(3) + 2(3)2 – 2(3)3 3 = 18 + 18 – 18 = 18 m Maka, jarak di antara R dengan P ialah 18 m. Therefore, the distance between R and P is 18 m. (c) Apabila A berada di R, t = 3 When A is at R, t = 3 sB = (–3)(3) = –9 m ∴ Jarak di antara A dengan B Distance between A and B = 30 – 18 – 9 = 3 m P 18 m 3 m –9 m A B 30 m Q 2 (a) v = 5t 2 – 13t – 6 a = 10t – 13 Pada K/at K, v = 0 5t 2 – 13t – 6 = 0 (5t + 2)(t – 3) = 0 t = – 2 5 atau/or t = 3 t > 0, maka/thus t = 3 a = 10(3) – 13 = 30 – 13 = 17 m s–2 (b) vmin apabila/when a = 0, 10t – 13 = 0 t = 13 10 vmin = 5 13 10  2 – 13 13 10  – 6 = 5 169 100 – 169 10 – 6 = –14.45 m s–1 (c) Jumlah jarak/Total distance =  13 10 0 (5t 2 – 13t – 6) dt +  3 13 10 (5t 2 – 13t – 6) dt =  5t 3 3 – 13t 2 2 – 6t 13 10 0 +  5t 3 3 – 13t 2 2 – 6t 3 13 10 = |–15.12|+|–16.38| = 31.50 m 3 (a) v = 15 + 4t – 3t 2 Apabila/When t = 0, v = 15 + 4(0) – 3(0)2 = 15 cm s–1 (b) a = dv dt = 4 – 6t Apabila/When t = 0, a = 4 – 6(0) = 4 cm s–2 (c) Halaju maksimum, a = 0 Maximum velocity, a = 0 4 – 6t= 0 t = 2 3 vmaksimum/maximum = 15 + 4 2 3  – 3 2 3  2 = 16 1 3 cm s–1 (d) 15 + 4t – 3t 2 = 0 (5 + 3t)(3 – t) = 0 t = – 5 3 atau/or t = 3 t  0, maka/thus t = 3 s =  (15 + 4t – 3t 2 ) dt = 15t + 2t 2 – t 3 + c Apabila/When t = 0, s = 0, c = 0 ∴ s = 15t + 2t 2 – t 3 10_PT SPM Add Math F5_2023.indd 11 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J12 J13 Apabila/When t = 3, s = 15(3) + 2(3)2 – (3)3 = 45 + 18 + 27 = 36 cm 4 (a) v =  a dt = (6 – 2t) dt = 6t – t 2 + c Apabila/When t = 0, v = –8 m s–1, c = –8 ∴ v = 6t – t 2 – 8 Pada halaju maksimum, a = 0 At maximum velocity, a = 0, 6 – 2t = 0 t = 3 v = 6(3) – (3)2 – 8 = 18 – 9 – 8 = 1 m s–1 (b) Apabila halaju positif, v > 0 When the velocity is positive, v > 0 6t – t 2 – 8 > 0 t 2 – 6t + 8 < 0 (t – 4)(t – 2) < 0 2 4 t 2 < t < 4 (c)  2 0 v dt +  4 2 v dt =  2 0 (6t – t 2 – 8) dt +  4 2 (6t – t 2 – 8) dt =  6t 2 2 – t 3 3 – 8t 2 0 +  6t 2 2 – t 3 3 – 8t 4 2 = 6 2 3 + –5 1 3 + 6 2 3  = 20 3 – 16 3 + 20 3 = 24 3 = 8 m 5 (a) Apabila/When a = 0, 2 – 4t = 0 4t = 2 t = 1 2 (b) v = (2 – 4t) dt = 2t – 4t 2 2 + c Apabila/When t = 0, v = 4, c = 4 ∴ v = 2t – 2t 2 + 4 Pada halaju maksimum, a = 0 At maximum velocity, a = 0 t = 1 2 vmaksimum/maximum = 2 1 2  – 2 1 2  2 + 4 = 1 – 1 2 + 4 = 4 1 2 m s–1 (c) Apabila zarah berehat seketika, v = 0 When the particle is instantaneously at rest, v = 0 2t – 2t 2 + 4 = 0 (2t + 2)(t – 2) = 0 t = –1 atau/or t = 2 t > 0, ∴ t = 2 (d) Jumlah jarak dalam 5 saat yang pertama Total distance in the first 5 seconds =  2 0 (2t – 2t 2 + 4) dt +  5 2 (2t – 2t 2 + 4) dt = t 2 – 2t 3 3 + 4t 2 0 + t 2 – 2t 3 3 + 4t 5 2 = 6 2 3 + |–45| = 51 2 3 m 6 (a) (i) Apabila/When t = 0, v = (0)2 – 9(0) + 18 = 18 m s–1 (ii) Sesaran maksimum, v = 0 Maximum displacement, v = 0 t 2 – 9t + 18 = 0 (t – 3)(t – 6) = 0 t = 3 atau/or t = 6 s = ∫(t 2 – 9t + 18) dt = t 3 3 – 9t 2 2 + 18t + c Apabila/When t = 0, s = 0, c = 0 \ s = t 3 3 – 9t 2 2 + 18t Apabila/When t = 3, s = (3)3 3 – 9(3)2 2 + 18(3) = 221 2 m Apabila/When t = 6, s = (6)3 3 – 9(6)2 2 + 18(6) = 18 m Maka, sesaran maksimum ialah 221 2 m. Thus, the maximum displacement is 22 1 2 m. (b) Apabila zarah berhenti seketika, v = 0 When the particle stops instantaneously, v = 0 Apabila/When t = 3, s = 221 2 m Apabila/When t = 6, s = 18 m \ Jarak di antara A dengan B Distance between A and B = 221 2 m – 18 m = 41 2 m (c) Halaju awal ialah apabila t = 0, The initial velocity is when t = 0, v = 02 – 9(0) + 18 = 18 m s–1 Apabila/When t = 0, v = 18 m s–1 Apabila/When t = 3 atau/or t = 6, v = 0 m s–1 3 6 0 t v 18 Jumlah jarak/Total distance = ∫ 3 0 v dt + ∫ 6 3 v dt =  t 3 3 – 9t 2 2 + 18t 3 0 +  t 3 3 – 9t 2 2 + 18t 6 3 = 221 2 + 18 – 221 2 = 221 2 + 9 2 = 27 m 7 (a) v = ds dt = 6 – 2t Apabila/When t = 0, v = 6 – 2(0) = 6 m s–1 (b) v = 6 – 2t Apabila zarah bergerak ke kiri, When the particle moves to the left, v < 0 6 – 2t < 0 2t > 6 t > 3 (c) Jumlah jarak/Total distance = ∫ 4 0 (6 – 2t) dt = [6t - t 2 ] 4 0 = 24 – 16 – 0 = 8 m ZON KBAT 1 (a) Daripada/From s = 4t 2 – 1 3 t 3 , v = ds dt v = 8t – t 2 Apabila zarah berhenti seketika, v = 0. When the particle is instantaneously at rest, v = 0. 8t – t 2 = 0 t(8 – t) = 0 t = 8 s = 4(8)2 – 1 3 (8)3 = 256 – 1 3 (512) = 851 3 m (b) Apabila zarah melalui O sekali lagi, s = 0. When the particle passes through point O again, s = 0. 4t 2 – 1 3 t 3 = 0 t 2 4 – 1 3 t = 0 4 – 1 3 t = 0 1 3 t = 4 t = 12 v = 8t – t 2 Apabila/When t = 12 s, v = 8(12) – (12)2 = 96 – 144 = –48 m s–1 (ke sebelah kiri/to the left) (c) Pada halaju seragam, a = 0 At uniform velocity, a = 0 Daripada/From v = 8t – t 2 a = 8 – 2t Apabila/When a = 0 8 – 2t = 0 2t = 8 t = 4 2 (a) Apabila zarah berhenti seketika, v = 0 When the particle is at rest, v = 0 10_PT SPM Add Math F5_2023.indd 12 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Praktis Topikal SPM: Matematik Tambahan Tingkatan 5 – Jawapan J12 J13 a = 2t + 1 v = ∫ a dt = 2t 2 2 + t + c = t 2 + t + c Apabila/When t = 1 s, v = 5 m s–1 5 = 12 + 1 + c c = 3 \v = t 2 + t + 3 Apabila/When t = 5 s, v = 52 + 5 + 3 = 25 + 5 + 3 = 33 m s–1 (b) s = ∫ v dt = ∫(t 2 + t + 3) dt = t 3 3 + t 2 2 + 3t + c Apabila/When t = 0, s = 0, maka/then c = 0 \ s = t 3 3 + t 2 2 + 3t Apabila/When t = 5 s, s = 53 3 + 52 2 + 3(5) = 415 6 = 691 6 m 3 (a) v = mt2 + nt – 6 a = 2mt + n Apabila/When t = 2, v = 0 4m + 2n – 6 = 0 4m + 2n = 6 ..............① Apabila/When t = 1 2 , a = 2 m s–2, m + n = 2 ..............② 2 × ②, 2m + 2n = 4 ..............③ ① – ③, 2m = 2 m = 1 Gantikan m = 1 ke dalam ②, Substitute m = 1 into ②, 1 + n = 2 n = 1 (b) v = t 2 + t – 6 Apabila zarah bergerak ke kanan, v > 0. When the particle moves to the right, v > 0 t 2 + t – 6 > 0 (t – 2)(t + 3) > 0 t = 2 atau/or t = –3 –3 2 t Oleh sebab/Since t  0, \ t > 2 (c) Jumlah jarak yang dilalui Total distance travelled = ∫ 5 4 (t 2 + t – 6) dt =  t 3 3 + t 2 2 – 6t 5 4 =  125 3 + 25 2 – 30 –  64 3 + 16 2 – 24 = 145 6 – 16 3 = 113 6 m KERTAS MODEL SPM KERTAS 1 BAHAGIAN A 1 (a) f 3 (x) = f 2 f(x) f 2 (x) = f(ax + b) = a(ax + b) + b = a2 x + ab + b f 2 f(x) = f 2 (ax + b) = a2 (ax + b) + ab + b = a3 x + a2 b + ab + b Diberi/Given f 3 (x) = 8x + 7 a3 = 8 a = 2 a2 b + ab + b = 7 (2)2 b + 2b + b = 7 7b = 7 b = 1 \ a = 2, b = 1 (b) f(x) = 2x + 1 f 2 (x) = (2)2 x + (2)(1) + 1 = 4x + 3 f 4 (x) = f 2 f 2 (x) = f 2 (4x + 3) = 4(4x + 3) + 3 = 16x + 12 + 3 = 16x + 15 2 Katakan/Let y = x + 2 x – 2 y(x – 2) = x + 2 yx – 2y = x + 2 x(y – 1) = 2 + 2y x = 2 + 2y y – 1 f –1(x) = 2 + 2x x – 1 , x ≠ 1 Katakan/Let y = mx + c x = y – c m g–1(x) = x – c m , x ≠ m g–1(3) = f(3) 3 – c m = 3 + 2 3 – 2 3 – c = 5m c = 3 – 5m ................① gf –1(2) = 7 m 2 + 2(2) (2) – 1  + c = 7 6m + c = 7 c = 7 – 6 m................② Gantikan ① ke dalam ②, Substitute ① into ②, 3 – 5m = 7 – 6m m = 4 Gantikan m = 4 ke dalam ①, Substitute m = 4 into ①, c = 3 – 5(4) c = –17 \ m = 4, c = –17 3 J = p(1 + k)n = 800 0001 + 7 100  20 = 800 000(1 + 0.07)20 = 800 000 (3.86968) = 3 095 747.57 Maka, jumlah pelaburannya selepas 20 tahun ialah RM3 095 747.57. Hence, his total investment after 20 years is RM3 095 747.57. 4 (a) 1 4 3 = 1 4 3 × 4 3 4 3 = 4 3 4 × 4 ×  3 ×  3 = 4 3 48 =  3 12 (b) 1 3 2 + 4 3 = 1 3 2 + 4 3 × 3 2 – 4 3 3 2 – 4 3 = 3 2 – 4 3 (3 2 + 4 3 )(3 2 – 4 3 ) = 3 2 – 4 3 (3 2 )2 – (4 3 )2 = 3 2 – 4 3 –30 5 (a) ln (4x – 3) = 5 loge (4x – 3) = 5 e5 = 4x – 3 148.41 = 4x – 3 4x = 151.41 x = 37.85 (b) 10e3x = 48 e3x = 4.8 ln e3x = ln 4.8 3x ln e = ln 4.8 3x = ln 4.8 x = ln 4.8 3 = 0.5229 6 (a) k – 5h = (h + 4) – k k + k = 5h + h + 4 2k = 6h + 4 k = 3h + 2 (b) 5h, 3h + 2, h + 4 a = 5h, d = (3h + 2) – 5h = 2 – 2h T10 = 5h + 9(2 – 2h) = 5h + 18 – 18h = 18 – 13h 7 (a) (i) v = 0.0048, w = 0.000048 (ii) r = 0.0048 0.48 = 0.01 (b) a = 4, r = 8 4 = 2 Tn = 2 048 4(2)n – 1 = 2 048 2n – 1 = 512 (n – 1) log 2 = log 512 n – 1 = log 512 log 2 n – 1 = 9 n = 10 S10 = 4(210 – 1) 2 – 1 = 4 092 8 (a) a = 1, r = 4 S10 = 1(410 – 1) 4 – 1 = 349 525 (b) Sn = 5 461 1(4n – 1) 4 – 1 = 5 461 4n – 1 = 16 383 4n = 16 384 10_PT SPM Add Math F5_2023.indd 13 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J14 J15 nlog 4 = log 16 384 n = log 16 384 log 4 n = 7 9 (a) v = ∫(6t – 4) dt = 3t 2 – 4t + c Apabila/When v = 30, t = 2 30 = 3(2)2 – 4(2) + c c = 26 \ v = 3t 2 – 4t + 26 (b) d dx[f(x)] = 3g(x) ∫ 3g(x) dx = f(x) (÷3) ∫ g(x) dx = 1 3 f(x) 10 (a) 9 P7 = 181 440 (b) 8 P6 × 6 P1 = 120 960 (c) 3 P1 × 7 P1 × 6 P1 = 45 360 11 (a) X ~ (20, 102 ) P(X  28) = PZ  28 – 20 10  = P(Z  0.8) = 0.2119 (b) P(X , t) = 8 50 = 0.16 PZ , t – 20 10  = 0.16 t – 20 10 = –0.994 t = 10.06 minit/ minutes 12 (a) 2 tan2 q – tan q – 3 = 0 (2 tan q – 3)(tan q + 1) = 0 2 tan q – 3 = 0 tan q = 3 2 q = 56.31°, 236.31° atau/or tan q + 1 = 0 tan q = –1 q = 135°, 315° \ q = 56.31°, 135°, 236.31°, 315° (b) kos/cos (2q + 30°) = –0.1793 Sudut asas/Basic angle = 79.67° 0° < q < 360° 0° < 2q < 720° 30° < 2q + 30° < 720° 2q + 30° = 100.33°, 259.67°, 460.33°, 619.67° 2q = 70.33°, 229.67°, 430.33°, 589.67° q = 35.17°, 114.84°, 215.17°, 294.84° BAHAGIAN B 13 (a) (i) P →R = P →O + O →R P →R = –5a ∼ + 15b ∼ (ii) O →Q = O →P + P →Q O →Q = 5a ∼ + 1 3 (15b ∼) = 5a ∼ + 5b ∼ (b) O →T = mO→Q = m(5a ∼ + 5b ∼) = 5ma∼ + 5mb∼ P →T = P →O + O →T = –5a ∼ + 5ma∼ + 5mb∼ = (5m – 5)a ∼ + 5mb∼ P →T = kP →R (5m – 5)a ∼ + 5mb∼ = k(15b ∼ – 5a ∼) Bandingkan pekali a ∼ dan b ∼ , Compare the coefficients of a ∼ and b ∼ , 5m – 5 = –5k , 5m = 15k 5m + 5k = 5 m = 3k .........② m + k = 1 .........① Gantikan ② ke dalam ①, Substitute ② into ①, 3k + k = 1 4k = 1 k = 1 4 m = 3 1 4 = 3 4 \ k = 1 4 , m = 3 4 14 (a) y = x3 – 6x2 + 9x + 2 dy dx = 3x2 – 12x + 9 Pada/At A(2, 4), dy dx = 3(2)2 – 12(2) + 9 = –3 (b) m1 m2 = –1 –3 × m2 = –1 m2 = 1 3 Persamaan normal, Equation of the normal, y – 4 = 1 3 (x – 2) 3y – 12 = x – 2 3y – x = 10 (c) dy dx = 0 3x2 – 12x + 9 = 0 x2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 x = 3 atau/or x = 1 Apabila/When x = 1, y = (1)3 – 6(1)2 + 9(1) + 2 y = 6 \ Q(1, 6) d2 y dx2 = 6x – 12 Apabila/When x = 1, d2 y dx2 = 6(1) – 12 = –6 < 0 \ (1, 6) ialah titik maksimum. (1, 6) is a maximum point. 15 (a) 122 = 62 + 162 – 2(6)(16) kos/cos q kos/cos q = 0.7708 q = 39.57° = 39.57° × p 180° = 0.6907 rad (b) Panjang lengkok AB/ Arc length AB = 16(0.6907) = 11.05 cm Panjang CB/ Length of CB = 16 – 6 = 10 cm Perimeter kawasan berlorek Perimeter of the shaded region = 10 cm + 12 cm + 11.05 cm = 33.05 cm (c) Luas sektor AOB Area of sector AOB = 1 2 (16)2 (0.6907) = 88.41 cm2 Luas ΔAOC / Area of ∆AOC = 1 2 (6)(16) sin 39.57° = 30.58 cm2 Luas kawasan berlorek Area of the shaded region = 88.41 cm2 – 30.58 cm2 = 57.83 cm2 KERTAS 2 BAHAGIAN A 1 (a) y = 2x(3 – x)4 dy dx = (2x) d dx(3 – x)4 + (3 – x)4 d dx(2x) = 2(x)(4)(3 – x)3 (–1) + (3 – x)4 (2) = –8x(3 – x)3 + 2(3 – x)4 Apabila/When x = 2, dy dx = –8(2)[3 – (2)]3 + 2[3 – (2)]4 = –16(1)3 + 2(1)3 = –14 (b) Kecerunan tangen ialah –14, maka kecerunan normal ialah 1 14 . The gradient of the tangent is –14, therefore, the gradient of the normal is 1 14 . Persamaan normal: Equation of the normal: y – 4 = 1 14 (x – 2) y – 4 = 1 14 x – 1 7 y = 1 14 x + 27 7 2 (a) (i) gf(x) = g(4 – x) = m(4 – x)2 – n = mx2 – 8mx + 16m – n Bandingkan dengan 3x2 – 24x + 3: Compare with 3x2 – 24x + 3: m = 3, 16m – n = 3 16(3) – n = 3 n = 45 (ii) g(5) = 3(5)2 – 45 = 30 (iii)fg(4) = f[3(4)2 – 45] = f(3) = –3 + 4 = 1 (b) 63 45 30 –1 4 10 0 3 y x Julat/Range: 0 < y < 63 3 (a) Luas ΔQOR/Area of ∆QOR = 1 4  0 6 3 0 0 5 –4 0 = 1 2 × |0 – 24 + 0 – 0 – 15 + 0| = 1 2 × |–39| = 19.5 unit2 /units2 (b) T =  2(6) + 1(3) 2 + 1 , 2(5) + 1(–4) 2 + 1  = (5, 2) (c) Lokus P(x, y) dan PT = 4 unit Locus of P(x, y) and PT = 4 units  (x – 5)2 + (y – 2)2 = 4 (x – 5)2 + (y – 2)2 = 16 4 sin q = 5 10 q = 30° PM =  102 + 52 = 8.66 cm RPS = 60° = 1.047 rad P R M 10 cm 5 cm q 10_PT SPM Add Math F5_2023.indd 14 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Praktis Topikal SPM: Matematik Tambahan Tingkatan 5 – Jawapan J14 J15 Luas kawasan RTSU Area of region RTSU = 2 × (Luas sektor PRUS – Luas segi tiga PRS) 2 × (Area of sector PRUS – Area of triangle PRS) = 2 ×  1 2 (10)2 (1.047) – 1 2 × 10 × 8.66 = 2 × (52.35 – 43.30) = 2 × 9.05 = 18.1 cm2 Luas dua bulatan kecil Area of two small circles = 2(3.142 × 0.82 ) = 4.02 cm2 Jumlah luas kawasan berlorek Total area of the shaded region = 18.1 cm2 + 4.02 cm2 = 22.12 cm2 5 (a) Sn = 641 –  3 4  n  Sn – 1 = 641 –  3 4  n – 1 Tn = Sn – Sn – 1 = 641 –  3 4  n  – 641 –  3 4  n – 1 = 643 4  n – 1 –  3 4  n  = 64 3 4  n – 1 – 11 – 3 4 2 = 16 3 4  n – 1 (b) Tn , 4 16 3 4  n – 1 , 4  3 4  n – 1 , 1 4 (n – 1)log10 3 4 , log10 1 4 (n – 1)(–0.1249) , –0.6021 n  5.8206 Sebutan pertama yang kurang daripada 4 ialah T6 . The first term that is less than 4 is T6 . 6 (a) Sebelah kanan/RHS = tan x + tan x kos/cos 2x = tan x (1 + kos/cos 2x) = sin x kos/cos x (1 + 2 kos2 /cos2 x – 1) = sin x kos/cos x (2 kos2 /cos2 x) = 2 sin x kos/cos x = sin 2x(Sebelah kiri/LHS) (b) (i) y = 3|kos/cos x| y = x 2p 3p 2 p 2 p 2p x y 3 1 O (ii) 6p|kos/cos x| – x = 0 3|kos/cos x| = x 2p y = x 2p Apabila/When x = 2p, y = 1 Apabila/When x = 0, y = 0 Bilangan penyelesaian = 4 Number of solutions = 4 7 (a) dy dx = x2 + 5 y = ∫ (x2 + 5) dx y = 1 3 x 3 + 5x + c Apabila/When x = 3 dan/and y = 8 8 = 1 3 (3)3 + 5(3) + c c = –16 y = 1 3 x3 + 5x – 16 (b) Kecerunan tangen pada titik (3, 8) The gradient of the tangent at point (3, 8) dy dx = 32 + 5 = 14 Persamaan tangen, y = 14x + c Equation of tangent, y = 14x + c At point/Pada titik (3, 8) y = 14(3) + c c = –34 \ y = 14x – 34 (c) dy dx = 14, dy dt = 3.5 dx dt = dx dy × dy dt = 1 14 × 3.5 = 0.25 unit per saat 0.25 unit per second BAHAGIAN B 8 py = x2 + qx y x = 1 p x + q p (a) x 2 3 4 5 6 7 y x 2.50 3.00 3.50 4.00 4.50 5.00 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0 1 2 3 4 5 6 7 x x y (b) (i) 1 p = 5 – 3.5 7 – 4 1 p = 1.5 3 1 p = 0.5 p = 2 (ii) q p = 1.5 q 2 = 1.5 q = 3 9 (a) (i) O →P = O →A + A →P = 6a ∼ + 2b ∼ (ii) B →A = B →O + O →A = –8b ∼ + 6a ∼ (b) (i) O →S = hO→P = h(6a ∼ + 2b ∼) = 6ha∼ + 2hb∼ (ii) B →S = kB →A = k(6a ∼ – 8b ∼) = 6ka∼ – 8kb∼ O →S = O →B + B →S = 8b ∼ + 6ka∼ – 8kb∼ = 8b ∼ – 8kb∼ + 6ka∼ = 6ka∼ + 8(1 – k)b ∼ (c) 6ha∼ + 2hb∼ = 6ka∼ + 8(1 – k)b ∼ Bandingkan pekali a ∼ , Compare the coefficient of a∼ , 6h = 6k h = k Bandingkan pekali b ∼ , Compare the coefficient of b∼ , 2h = 8(1 – k) = 8 – 8k, diberi/given h = k, 2k = 8 – 8k 10k = 8 k = 8 10 = 4 5 h = 4 5 \ h = 4 5 , k = 4 5 10 (a) y = x3 dy dx = 3x2 Kecerunan tangen di A Gradient of the tangent at A = 3(1)2 = 3 Kecerunan normal kepada lengkung di A = – 1 3 Gradient of the normal to the curve at A = – 1 3 Persamaan normal kepada lengkung di A ialah y = – 1 3 x + c Equation of the normal to the curve at A is y = – 1 3 x + c Gantikan koordinat A(1, 1). Substitute the coordinates of A(1, 1). 1 = – 1 3 (1) + c c = 4 3 \ Persamaan normal kepada lengkung di A ialah y = – 1 3 x + 4 3 . Equation of the normal to the curve at A is y = – 1 3 x + 4 3 . (b) y = – 1 3 x + 4 3 y x y = x3 A(1, 1) 0 1 B Pada titik B, y = 0 At point B, y = 0 0 = – 1 3 x + 4 3 x = 4 Luas rantau berlorek Area of the shaded region = ∫ 1 0 x3 dx + 1 2 × (4 – 1) × 1 =  x4 4  1 0 + 3 2 10_PT SPM Add Math F5_2023.indd 15 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J16 JPB =  1 4 – 0 + 3 2 = 1 3 4 unit2 /units2 (c) Isi padu janaan Generated volume = ∫ 1 0 p(x3 )2 dx + 1 3 × p(1)2 × (4 – 1) = p x7 7  1 0 + p = p 1 7 – 0 + p = 1 1 7 p unit3 /units3 11 (a) p = 1 5 , q = 4 5 , n = 7 (i) P(X = 3) = 7 C3 1 5  3  4 5  4 = 0.1147 (ii)  = np = 560 × 1 5 = 112 σ = npq = 112 × 4 5 = 9.4657 (b) (i) P(X , 170) = 0.75 PZ , 170 – 155 σ  = 0.75 PZ , 15 σ  = 0.75 PZ  15 σ  = 0.25 15 σ = 0.674 σ = 22.26 (ii) P(150 , X , 160) = P 150 – 155 22.26 , Z , 160 – 155 22.26  = P(–0.2246 < Z < 0.2246) = 1 – P(Z  0.2246) – P(Z  0.2246) = 1 – 2(0.411) = 0.178 Bilangan murid/Number of students = 0.178  1 200 = 213.6 = 213 orang murid/students z f(z) –0.2247 0 0.2247 BAHAGIAN C 12 (a) s = 6t 2 – 5t – t 3 0 = 6t 2 – 5t – t 3 t 3 – 6t 2 + 5t = 0 t(t 2 – 6t + 5) = 0 t(t – 1)(t – 5) = 0 t = 0, t = 1 atau/or t = 5 ∴ Zarah melalui titik O sekali lagi apabila t = 1 dan t = 5. The particle passes through O again when t = 1 and t = 5. (b) s = 6t 2 – 5t – t 3 v = ds dt = 12t – 5 – 3t 2 dv dt = 12 – 6t 12 – 6t = 0 t = 2 Apabila/When t = 2, v = 12(2) – 5 – 3(2)2 = 24 – 5 – 12 = 7 m s–1 (c) a = dv dt = 12 – 6t Apabila/When t = 3, a= 12 – 6(3) = 12 – 18 = –6 m s–2 13 (a) 24 x × 100 = 130 x = 18.46 y 12 × 100 = 150 y = 18 z = 20 14 × 100 = 142.9 (b) – I = 150(70) + 130(40) + 142.9(100) + 120(150) 360 = 47 990 360 = 133.31 300 P2019  100 = 133.31 P2019 = 225.04 Kos penghasilan sabun pencuci pada tahun 2019 ialah RM225.04. The cost of making the detergent in the year 2019 was RM225.04. (c) (i) 133.31 100 = x 130 x = 173.30 (ii) P2023 P2019 × 100 = 173.30 P2023 225.04 × 100 = 173.30 P2023 = 389.99 Kos penghasilan sabun pencuci pada tahun 2023 ialah RM390. The price of making the detergent in the year 2023 is RM390. 14 (a) sin 49° 3 = sin PRQ 5.4 sin PRQ = 0.6792 PRQ = 42.78° RQP = 180° – 49° – 42.78° = 88.22° PR sin 88.22° = 6.0 sin 49° PR = 7.95 cm (b) PS2 = PR2 + SR2 – 2(PR)(SR) kos/cos 130° = 7.952 + 4.82 – 2(7.95)(4.8) kos/cos 130° = 135.30 PS = 11.63 cm (c) sin PSR 7.95 = sin 130° 11.63 sin PSR = 0.5237 PSR = 31.58° RPS = 180° – 130° – 31.58° = 18.42° PRT = 31.58° – 18.42° = 13.16° Luas ΔPRT/Area of ∆PRT = 1 2 × 4.8 × 7.95 × sin 13.16° = 4.344 cm2 S T R P 18.42° 31.58° 13.16° 7.95 cm 4.8 cm 15 (a) I : 200x + 100y < 1 200 2x + y < 12 II : y < 4x III : y  3 (b) 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 x y = 4x y = 3 (2, 8) 2x + y = 12 R y (c) Kos pengangkutan, Transportation cost, k = 50x + 30y 50x + 30y = 150 5x + 3y = 15 Titik maksimum = (2, 8) Maximum point = (2, 8) Jumlah kos pengangkutan maksimum Maximum total transportation cost = 2(50) + 8(30) = RM340 10_PT SPM Add Math F5_2023.indd 16 28/02/2023 2:20 PM PENERBIT ILMU BAKTI SDN. BHD.