MODULE AND MORE Matematik Tambahan Tingkatan 5

J3 0 4 18 y x –3 –4 6 x = 3 + y2 x = 2 – y2 2 1 -1 0 1 3 5 4 x y 2 2 1 2 0 4 6 x y x = (y – 2)2 y = – x + 2 3. (a) (i) ∫ 6 –3 g(x)dx = 14, ∫ 4 0 g(x)dx = –4 ∫ 0 –3 g(x)dx = 2∫ 6 4 g(x)dx ∴ ∫ 0 –3 g(x)dx = 12 ∫ 4 –3 g(x)dx = 12 – 4 = 8 (ii) ∫ 4 0 g(x)dx = ∫ 6 4 g(x)dx = 4 ∴∫ 0 –3 g(x)dx = 18 – 4 = 14 ∴2∫ 0 –3 g(x)dx = 28 (iii)∫ 6 0 g(x)dx = 2 ∴∫ 6 4 g(x)dx = 6 ∫ 0 –3 g(x)dx = 18 – 6 = 12 (b) (i) luas = ∫ 1 –1 [3 + y2 – (2 – y 2 )]dy area = ∫ 1 –1 (1 + 2y2 )dy = y + 2 3 y3 4 1 –1 = 1 + 2 3 4 – –1 – 2 3 4 = 3 1 3 unit2 (ii) Isipadu = π 2 ∫ 5 3 y2 dx Volume = π 2 ∫ 5 3 (x – 3)dx = π 2  x2 2 – 3x4 5 3 = π 2 1 25 2 – 152 – 1 9 2 – 924 = π 2 [8 – 6] = π unit3 (c) (i) luas  = 1 2 [2](4) = 4 unit2 area luas berlorek shaded area = 4 – ∫ 2 0 (y – 2)2dy = 4 –  (y – 2)3 3 4 2 0 = 4 – + 8 3 4 = 4 3 unit2 (ii) Isipadu = ∫ 2 0 πx2dy Volume = π∫ 2 0 (y – 2)4 dy = π  (y – 2)5 5 4 2 0 = π  32 5 4 = 32 5 π unit3 BAB 4: PILIH ATUR DAN GABUNGAN 1. (a) (i) 4 C1 × 5 C1 = 20 cara (ii) 4 baju 5 seluar 2 hitam 2 hitam Kes 1: Jika pakai baju hijam Case 1: It she wears black blouse Bilangan cara = 2 C1 × 3 C1 = 6 Kes 2: Jika pakai seluar hitam Case 2: It she wears black skirt Bilangan cara = 2 C1 × 2 C1 = 4 Kes 3: Jika bukan baju hitam dan seluar hitam Case 3: It not black blouse and black skirt. Bilangan cara = 2 C1 × 3 C1 = 6 Number of ways Jumlah cara = 6 + 4 + 6 = 16 Total of ways (b) (i) 6! 3! = 120 cara/ways (ii) Jika mesti bersama/If must together = 4! Mesti diasingkan/Must be separated 6! 3! − 4! = 96 (iii) M _ _ _ _ K 4! 3! = 4 cara/ways (iv) 1 cara/way 2. (a) (i) Bilangan cara = (8 – 1)!3! Number of ways = 30240 (ii) Jika duduk bersama = (9 – 1)!2! It sit next to each other Jika tidak ada halangan = (10 – 1)! If no restriction ∴Tidak duduk bersama = 9! – 8!2! Not sit next to each other = 282240 (iii) Bilangan cara = 10C8 (8 – 1)! Number of ways = 226800 (b) (i) 0, 1, 3, 4, 5, 7, 8 6 6 5 4 = 720 cara/ways (ii) 4 3 2 1 = 24 cara/ways

J4 0.1 3 0 5 7 9 r 0.2 0.3 P[X = r] 1 6 –1 0 0 1 2 x P(X = x) 2 6 3 6 4 6 (iii) 4 6 6 6 = 864 cara/ways (iv) 5 6 5 4 = 600 cara/ways 3, 4, 5, 7, 8 (c) (i) 4 merah/red, 3 biru/blue, 4 hijau/green, 3 kuning/yellow 4 C1 × 3 C1 × 4 C1 × 3 C1 = 144 (ii) Kes 1: 2 merah, 2 berlainan warna 2 red, 2 different colour 4 C2 × 10C2 = 270 Kes 2: 3 merah, 1 berlainan warna 3 red, 1 different colour 4 C3 × 10C1 = 40 Kes 3: 4 merah/red = 4 C4 = 1 Jumlah/Total = 270 + 40 + 1 = 311 (iii) Jika tiada guli hijau dan merah If no green and red marble 6 C4 = 15 Jika tiada guli merah sahaja If no red marble only 10C4 = 210 Sekurang-kurangnya 1 hijau At least 1 green 210 – 15 = 195 (d) (i) n C2 = 253 n! (n – 2)! = 253 n(n – 1) = 506 n2 – n – 506 = 0 (n + 22)(n – 23) = 0 n = 23 (ii) Bilangan salam antara 3 graduan. Number of handshakes among 3 graduates. = 3 C2 = 3 Bilangan salam/Number of handshakes = 253 – 3 = 250 (e) m + n = 26 ...➀ n C2 + mC2 = 160 n! (n – 2)!2! + m! (m – 2)!2! = 160 n(n – 1) + m(m – 1) = 320 n2 + m2 – (n + m) = 320 (n + m)2 – 2mn = 320 + 26 262 – 346 = 2mn mn = 165 m(26 – m) = 165 m2 – 26m + 165 = 0 (m – 11)(m – 15) = 0 n = 15, m = 11 m = 11 ; 15 BAB 5: TABURAN KEBARANGKALIAN 1. (a) (i) P[X = 3] = P[X = 7] = 0.3 = p 1 – 0.3 – 0.2 – 0.3 = q q = 0.2 (ii) P[X ≥ 5] = 0.2 + 0.3 + 0.2 = 0.7 (b) 2. (a) (i) p = 3 100 n = 10 P[X = 5] = 10C5 (0.03)5 (0.97)5 = 5.26 × 10–6 (ii) P[X ≥ 2] = 1 – P[X < 2] = 1 – P[X = 0] – P[X = 1] = 1 – 10C0 (0.03)0 (0.97)10 – 10C1 (0.03)(0.97)9 = 0.0345 (b) n = 2500, p = 0.03 µ = np = 2500 × 0.03 = 75 σ2 = npq = 75(0.97) σ = 8.53 3. (a) P[X = x] = k(1 – x)2 X –1 0 1 2 P[X = x] 4k k 0 k 4k + 2k = 1 k = 1 6 (b) P[X < 0] + P[X > 1] = 4 6 + 1 6 = 5 6 (c) 4. (a) (i) p = 0.6 n = 8 q = 0.4 P[X = 5] = 8 C5 (0.6)5 (0.4)3 = 0.2787 (ii) P[X ≥ 5] = P[X = 5] + P[X = 6] + P[X = 7] + P[X = 8] = 8 C5 (0.6)5 (0.4)3 + 8 C6 (0.6)6 (0.4)2 + 8 C7 (0.6)7 (0.4) + 8 C8 (0.6)8 = 0.5941

J5 (b) µ = 200 × 0.6 = 120 (c) P[X = n]= n Cn (0.6)n = 0.07776 n = log 0.07776 log 0.6 = 5 5. (a) (i) X ~ N(45, 8.32 ) Z = 1 = X – 45 8.3 X = 45 + 8.3 = 53.3 (ii) –2 = X – 45 8.3 X = 45 – 2(8.3) = 45 – 16.6 = 28.4 (b) P[X > 40] = P X – µ σ > 40 – 45 8.3 4 = PZ > –5 8.3 4 = 0.7266 (c) P[X > n] = 0.3589 P3 Z > n – 45 8.3 4 = 0.3589 n – 45 = 8.3(0.361) n = 48 6. (a) X ~ N(142, σ2 ) P[139 < X < 145] = 0.95 P 139 – 142 σ < Z < 145 – 142 σ 4 = 0.95 P– 3 σ < Z < 3 σ4 = 0.95 PZ > 3 σ4 = 0.05 2 = 0.025 3 σ = 1.96 σ = 3 1.96 = 1.53 (b) P[X < 141]= PZ < 141 – 142 1.53 4 = 0.2567 (c) P[X > 140.5] = PZ > 140.5 – 142 1.53 4 = 0.8366 Bilangan orang perempuan = 25 0.8366 Number of girls = 30 BAB 6: FUNGSI TRIGONOMETRI 1. (a) tan ( π 3 – A) = kot π 3 tan (60° – A) = kot 60° = 1 3 60° – A = 30° A = 30° = π 6 (b) kot A + tan A = sek A kosek A cot A + tan A = sec A cosec A Sebelah kiri LHS = 1 tan A + tan A = 1 + tan2 A tan A = sek2 A tan A = 1 kos2 A · kos A tan A = sek A kosek A = sec A cosec A = Sebelah kanan RHS 2. (a) cos A 1 − tan A + sin A 1 − cot A = sin A + cos A LHS = cos2 A cos A – sin A + sin2 A sin A − cos A (sebelah kiri) = cos2 A – sin2 A cos A − sin A = (cosA + sinA)(cos A – sin A) cos A − sin A = sin A + cos A = RHS (sebelah kanan) (b) (i) 4 sin2 x + 8 cos x – 7 = 0 4[1 – cos2 x] + 8 cos x – 7 = 0 4 cos2 x – 8 cos x + 3 = 0 (2 cosx – 1)(2 cosx – 3) = 0 cos x = 1 2 cos x = 3 2 ∴x = 60°, 300° (ii) Jika x = θ 2 θ 2 = 60°, 300° θ = 120°, 600° (Tidak diterima) (Rejected) 3. (a) 2 sin x = 4 cos x – 1 tan x 2 sin2 x cos x = 4 cos x – 1 2[1 – cos2 x] = 4 cos2 x – cos x 0 = 6 cos2 x – cos x – 2 6 cos2 x – cos x – 2 = 0 (Tertunjuk) (b) (3 cos x – 2)(2 cos x + 1) = 0 cos x = 2 3 , – 1 2 2 fiff3 1 60° y x 30° 30° y x 1 2 fiff3 60°

J6 y x 48.19° y x 1 2 60° y y = 4 cos x – 1 y = 4 kos x – 1 x 0 1 2 y y = y = tan 2x y = tan 2x x x 2 fi fi 1 0 x = 48.19°, 311.81° atau/or x = 120°, 240° ∴x = 48.19°, 120°, 240°, 311.81° (c) 2π sin x tan x = 2π – x 4 kos x – 1= 2 – x π = y 4 cos x – 1 = 2 – x π = y 2 penyelesaian/solution 4. (a) tan 2x = 5 sin 2x sin 2x kos 2x = 5 sin 2x sin 2x cos 2x = 5 sin 2x sin2x = 5 sin 2x · cos 2x sin 2x[1 – 5 cos 2x] = 0 (Tertunjuk/Proved) (b) sin 2x(1 – 5 cos 2x) = 0 sin 2x = 0 2x = 0°, 180°, 360° 540°, 720° x = 0°, 90°, 180°, 270°, 360° cos 2x = 1 5 2x = 78.46°, 281.54° 438.46°, 641.54° x = 39.23°, 140.77°, 219.23°, 320.77° (c) 5x sin 2x = 1 5 sin 2x = 1 x = tan 2x y = 1 x 2 penyelesaian/solutions 5. (a) 2 sin2 x + 2 = 7 cos x 2 sin2 x – 7 cos x + 2= 0 2[1 – cos2 x] – 7 cos x + 2 = 0 2 cos2 x + 7 cos x – 4 = 0 (2 cos x – 1)(cos x + 4) = 0 cos x = 1 2 atau/or cos x = –4 ∴x = 60°, 300° = π 3 , 5π 3 (b) (i) y = 3 sin bx a = 3 b → 2π 1 → π ∴b = 2 (ii) 3 sin 2x = 1.4 sin 2x = 1.4 3 x = 27.82°, 152.18° A(0.486, 1.4) B(2.656, 1.4) atau A(0.155π, 1.4) B(0.845π, 1.4) BAB 7: PENGATURCARAAN LINEAR 1. (a) x + y ≤ 5 3x + 2y ≤ 12 (b) (c) (i) P = 30x + 25y Titik optimal/Optimal point = (2, 3) Keuntungan maksimum/Maximum profit P = 30(2) + 25(3) = RM 135 (ii) Jika/If x = 1 , 0 ≤ y ≤ 4 2. (a) 6x + 3y ≤ 36 atau 2x + y ≤ 12 y ≤ 4x x ≤ 4y x ≤ 4 5 4 3 2 1 1 0 2 3 4 5 x R y 3x + 2y fi 12 x + y fi 5 6 7

J7 3. (a) vA = 3 + 2t – t 2 dvA dt = 2 – 2t = 0 t = 1 vmax = 3 + 2(1) – (1)2 = 4 m s–1 (b) Di R vA = 0 = 3 + 2t – t 2 0 = (3 – t)(1 + t) ∴t = 3 sA = 3t + t 2 – 1 3 t 3 t = 3, sA = 3(3) + 9 – 1 3 (27) = 9 m ∴RP = 9 m (c) vB = –5 sB = –5t + c Apabila t = 0 sB = 30 m ∴sB = –5t + 30 Apabila t = 3, sB = –5(3) + 30 = 15 m ∴Jarak di antara A dan B Distance between A and B = (15 – 9) m = 6 m (d) 3 + 2t – t 2 = –5 t 2 – 2t – 8 = 0 (t + 2)(t – 4) = 0 ∴ t = 4 4. (a) (i) a = –10 v = –10t + c t = 0 v = 30 ∴v = –10t + 30 Apabila tinggi maksimum When it is the maximum height v = 0 ∴10t = 30 t = 3 (ii) s = –5t2 + 30t Apabila s = 0 5t[–t + 6] = 0 t = 6 ∴v = –10(6) + 30 = –30 m s–1 (b) s = –5(9) + 90 = 45 m (c) Jumlah jarak = 1 2 (3)(30) × 2 Total distance = 90 m (b) (c) Untung/Profit P = 30x + 10y Titik optimum/ Optimum point = (4, 4) Untung/Profit P = 60(4) + 20(4) = RM320 BAB 8: KINEMATIK GERAKAN LINEAR 1. (a) v = 2t(6 – t) = 12t – 2t 2 dv dt = 12 – 4t = 0 t = 3 ∴vmak = 2(3)(6 – 3) = 18 m s–1 (b) v = 0, 2t(6 – t) = 0 t = 0, 6 ∴t = 6 (c) (d) Jumlah jarak/Total distance = ∫ 6 0 (12t – 2t 2 )dt + ∫ 7 6 (12t – 2t2 )dt = 36t 2 – 2 3 t3 4 6 0 + 36t2 – 2 3 t 3 4 7 6 = 72 + –6 2 3  = 78 2 3 m 2. (a) s = t(t – 4)2 ds dt = v = t(2)(t – 4) + (t – 4)2 = (t – 4)[2t + t – 4] = (t – 4)(3t – 4) Apabila t = 3 v = (3 – 4)(9 – 4) = –5 m s–1 (b) v = 0 = (t – 4)(3t – 4) ∴ t = 4 s dan/and 4 3 s (c) v = (t – 4)(3t – 4) dv dt = a = (t – 4)3 + (3t – 4) Apabila/When t = 4 a = (4 – 4)3 + (3(4) – 4) = 8 m s–2 8 y x R 6 4 2 0 2 y fi 4x x fi 4 6x + 3y fi 36 4 v 18 0 6 7 t s t 45 m t = 3 0 3 6 V t(s) 30 –30 0 3 6