MODULE AND MORE Matematik Tambahan Tingkatan 5

J34 7. (a) kos/ cos 23° = sin(90° − 23°) = sin 67° = 0.9205 (b) tan 23° = kot/ cot = (90 – 23°) = kot/ cot 67° = 1 tan 67° = 1 2.3559 = 0.4245 (c) sek/ sec 23° = kosek/cosec (90° – 23°) = kosek/cosec 67° = 1 sin 67° = 1 0.9205 = 1.0864 8. (a) tan 225° = tan 45° = 1 (b) sek/ sec (−240°)= 1 kos/ cos(–240°) = 1 –kos/ cos 60° = 1 1 – 2 = –2 (c) – kosek/ cosec (330°) = – 1 sin 330° = – 1 –sin 30° = 1 1 2 = 2 (d) – sin (−300°) = – sin 60° = 3 – 2 (e) sin 315° + kos/ cos 135° = 1 – 2 +  1 – 2  = 2 – 2 y x –1 –1 √2 45fi 225fi 0 y x –1 60fi –240fi √3 2 0 y x –1 30fi √3 2 0 y x 1 60fi √3 2 0 (f) sek/sec 420° + kosek/cosec (−120°) = 1 kos/ cos 420° + 1 sin (–120°) = 1 kos/ cos 60° + 1 –sin 60° = 1 1 2 – 2 3 = 2 – 2 3 = 2 3 – 1 3 2 9. (a) x 0 π 4 π 2 3 4 π π 5 4 π 3 2 π 7 4 π 2π 0° 45° 90° 135° 180° 225° 270° 315° 360° kos/ cos x 1 0.71 0 –0.71 –1 –0.71 0 0.71 1 (b) x 0 π 4 π 2 3 4 π π 5 4 π 3 2 π 7 4 π 2π 0° 45° 90° 135° 180° 225° 270° 315° 360° tan x 0 1 ∞ –1 0 1 ∞ –1 0 y x 1 1 –1 –1 45fi 45fi √2 √2 y x 2 60fi 1 60fi √3 x kos/ cos x 1 –1 0 90fi 180fi 270fi 360fi x tan x 4 2 –4 –2 0 90fi 180fi 270fi 360fi

J35 10. (a) (b) (c) (d) (e) (f) 11. (a) y = a kos bx + c a = −2 1 1 2 kala dalam 2π 1 1 2 cycle in 2π b = 3 2 ; c = 0 (b) a = −2 ; b = 1 2 kala/cycle ; c = 0 (c) a = 2 ; b = 1 kala/cycle ; c = 1 12. (a) π(1 – tan x) = 2x 1 – tan x = 2x π 1 – 2x π = tan x Maka/Hence y = 1 – 2x π Apabila/When x = 0 y = 1 x = π y = –1 Terdapat 3 penyelesaian There are 3 solution (b) (i) (ii) 2|2 sin 2x| – 3 = –2x π |2 sin 2x| = 3 2 – x π ∴ y = 3 2 – x π Apabila/When x = 0, y = 3 2 x = 2π y = 3 2 – 2 = – 1 2 5 penyelesaian/solutions (c) (i) y = –2 sin 3 2 x 0 ≤ x ≤ 2π (ii) y = k = −2 atau/or 0 < k ≤ 2 x y – 0 3 4 3 4 fi 2fi 3 4 y = kos 2x 3 4 y = cos 2x x y –2 –1 0 2 1 fi y = 1+ kos 2x y = 1+ cos 2x x y –1 fi y = tan 2x–1 0 x y –3 –2 –1 0 1 2 3 fi 2fi 1 2 y = 3│kos x│ 1 2 y = 3│cos x│ x y –2 –1 0 1 fi 2fi y = –│sin x│–1 3 4 x y 0 2 fi 2fi y = 2 – │tan x│ y x 0 1 fi 2fi y = tan x y x 0 2 –2 fi 2fi y = │2 sin 2x│ y x 0 2 –2 2fi y = k y = k 3 2 y = –2 sin x

J36 (d) (i) (ii) π 2x = –2 kos 3 4 x ∴ y = π 2x = π 2  1 x  2 penyelesaian/solutions (e) (i) (ii) 2 sin 2x + 2 kos x + 1 = 0 2 sin 2x + 1 = –2 kos x ∴ y = –2 kos x 2 penyelesaian/solutions 13. (a) Sebelah kanan: sek2 A = 1 kos2 A = r2 x2 Right: sec2A = 1 cos2A = r2 x2 Sebelah kiri: 1 + tan2 A = 1 + y2 x2 = (x2 + y2 ) x2 Left: 1 + tan2 A = 1 + y2 x2 = (x2 + y2) x2 Tetapi/But x2 + y2 = r2 Maka/Hence 1 + tan2 A = r2 x2 = sebelah kanan 1 + tan2 A = sek2 A (b) Sebelah kanan: kosek2 A = 1 sin2 A = r2 y2 Right: cosec2A = 1 sin2A = r2 y2 Sebelah kiri: 1 + kot2 A = 1 + x2 y2 = (x2 + y2 ) y2 Left: 1 + cot2 A = 1 + x2 y2 = (x2 + y2) y2 Tetapi/But x2 + y2 = r2 Maka/Hence 1 + kot2 A = r2 y2 = sebelah kanan 1 + kot2 A = kosek2 A. 14. (a) Sebelah kiri: kos A 1 – sin2 A = kos A kos2 A = 1 kosA Left: cos A 1 – sin2A = cos A cos2A = 1 cos A = sek/ sec A = sebelah kanan/right side (b) Sebelah kiri / Left side = 1 − 2(1 − kos2 A) 1 − 2(1 − cos2 A) = 1 − 2 + 2 kos2 A 1 − 2 + 2 cos2 A = 2 kos2 A − 1 (sebelah kanan) 2 cos2 A − 1 (right side) y x 0 2 –2 2fi fi 2x y = y = –2 kos x 3 4 y = –2 cos x 3 4 (c) Sebelah kiri / Left side = sin A kos/ cos A + kos/ cos A sinA = sin2 A + kos2 A kos A sin A = 1 kos A sin A = sin2 A + cos2 A cos A sin A = 1 cos A sin A = sek A kosek A (sebelah kanan) = sec A cosec A (right side) (d) Sebelah kiri / Left side kot2 A + kos2 A = kosek2 A − 1 + kos2 A cot2 A + cos2 A = cosec2 A − 1 + cos2 A = kosek2 A – (1 − kos2 A) cosec2 A – (1 − cos2 A) = kosek2 A – sin2 A (sebelah kanan) cosec2 A – sin2 A (right side) (e) Sebelah kanan / Right side = (kosek A + kot A)2 (cosec A + cot A)2 = kosek2 A + 2 kosek A kot A + kot2 A cosec2 A + 2 cosec A cot A + cot2 A = 1 sin2 A + 2 kos A sin2 A + kos2 A sin2 A 1 sin2A + 2 cos A sin2A + cos2 A sin2A = 1 + 2 kos A + kos2 A sin2 A = (1 + kos A)2 sin2 A 1 + 2 cos A + cos2 A sin2A = (1 + cos A)2 sin2A = (1 + kos A)2 1 – kos2 A = (1 + kos A)2 (1 – kos A)(1 + kos A) (1 + cos A)2 1 – cos2A = (1 + cos A)2 (1 – cos A)(1 + cos A) = 1 + kos A 1 – kosA (sebelah kiri) 1 + cos A 1 – cos A (left side) 15. (a) kos (A + B) sin (A – B) = 1 – tan A + tan B tan A – tan B sebelah kanan/Right side = 1 – sin A sin B cos A cos B sin A cos A – sin B cos B = cos A cos B – sin A sin B cos A cos B sin A cos B – cos A sin B cos A cos B = cos (A + B) sin (A – B) = Sebelah kiri/Left side (b) sin (A – B) kos A kos B = tan A – tan B Sebelah kanan/Right side = sin A cos A – sin B cos B = sin A cos B – cos A sin B cos A cos B = sin (A – B) cos A cos B = Sebelah kiri/Left side y x 0 2 1 3 –2 –1 fi 2 3 2 fi fi y = 2 sin 2x+1

J37 (c) sin (A + B) + sin (A – B) sin (A + B) – sin (A – B) = tan A tan B Sebelah kiri/Left side = sin A cos B + cos A sin B + sin A cos B – cos A sin B sin A cos B + cos A sin B – sin A cos B + cos A sin B = 2 sin A cos B 2 cos A sin B = sin A cos A × cos B sin B = tan A × cot B = tan A tan B = Sebelah kanan/Right side 16. (a) Jika/If A = B Maka, kos (A + A) = kos A kos A − sin A sin A Hence, cos (A + A) = cos A cos A − sin A sin A kos 2A = kos2 A – sin2 A cos 2A = cos2 A – sin2 A Oleh kerana kos2 A = 1 − sin2 A Since cos2 A = 1 − sin2 A Maka/Hence kos/ cos 2A = 1 − sin2 A – sin2 A = 1 − 2 sin2 A Jika/If sin2 A = 1 − kos2 A sin2 A = 1 − cos2 A Maka/Hence kos 2A = kos2 A – (1 − kos2 A) cos 2A = cos2 A – (1 − cos2 A) = 2 kos2 A − 1 /2 cos2 A − 1 Jika/If A = θ 2 , kos θ = kos2 θ 2 – sin2 θ 2 cos θ = cos2 θ 2 – sin2 θ 2 = 2 kos2 θ 2 − 1 2 cos2 θ 2 − 1 = 1 − 2 sin2 θ 2 (b) Jika/If A = B tan (A + A) = sin (A + A) kos (A + A) = 2 sin A kos A kos2 A – sin2 A tan 2A = 2 sin A kos A kos2 A kos2 A – sin2 A kos2 A = 2 tan A 1 – tan2 A Jika A = θ 2 , maka tan θ = 2 tan θ 2 1 – tan2 θ 2 17. (a) Sebelah kanan / Right side (kos A + sin A)2 = kos2 A + 2 kos A sin A + sin2 A = 1 + 2 sin A kos A = 1 + sin 2A (sebelah kiri) (b) Sebelah kiri / Left side kos2 A + sin2 A kos2 A kos2 A – sin2 A kos2 A = 1 kos 2A = sek 2A (c) Sebelah kiri / Left side 1 – kos 2A 1 + kos 2A = 1 – (1 – 2 sin2 A) 1 + (2 kos2 A – 1) = 2 sin2 A 2 kos2 A = tan 2 A (d) Sebelah kanan / Right side sin θ 1 + kos θ = 2 sin θ 2 kos θ 2 1 + (2 kos2 θ 2 – 1) = 2 sin θ 2 2 kos θ 2 = tan θ 2 (e) Sebelah kiri / Left side sin A kos A + kos A sin A = sin2 A + kos2 A sinA kos A = 1 sin A kos A = 2 2 sin A kosA = 2 sin 2A = 2 kosek 2A 2 cosec 2A 18. (a) sin (132° − 72°) = sin 60° = 3 2 (b) tan (120° + 45°) = (tan 120° + tan 45°) 1 – tan 120° tan 45° = (– 3 + 1) 1 – (– 3 ) = 1 – 3 1 + 3 = 4 – 2 3 –2 = 3 – 2 (c) kos (45° − 30°) = kos 45° kos 30° + sin 45° sin 30° = 1 1 2 21 3 2 2 + 1 1 2 21 1 2 2 = ( 3 + 1) 2 2 19. (a) x y 51fi5ff 51fi5ff x = 180° – 51° 5’, 180° + 51.5’ = 128° 55’, 231° 5’ (b) x y 51fi34ff 51fi34ff

J38 2 tan x = 2.52 tan x = 1.26 x = 51° 34’, 180° + 51.34’ = 51° 34’, 231° 34’ (c) sin 2x = –0.357 2x = 200° 55’, 339° 5’ 560° 55’, 699° 5’ x = 100° 28’, 169° 33’, 280° 28’, 349° 33’ (d) kos 2x = 0.712 2x = 135° 24’, 224° 36’ 495° 24’, 584° 36 x = 67°42’. 112° 18’ 247° 42’, 292° 18’ (e) tan (x + 30) = 2.15 x + 30 = 65° 3’, 245° 3’ x = 35° 3’, 215° 3’ (f) sin(2x – 10°) = 0.681 (2x – 10°) = 42° 55’, 137° 5’ 402° 55’, 497° 5’ 2x = 52° 55’, 147° 5’ 412° 55’, 507° 5’ x = 26° 28’, 73° 32’, 206° 28’ 253° 32’ (g) sin 2x 3 = 0.782 = 51° 27’, 128° 33’ x = 77° 11’ , 192° 50’ (h) 2 kos(x – 25°) = 1.567 kos(x – 25°) = 0.7835 x – 25° = 38° 25’, 321° 35’ x = 63° 25’, 346° 35’ 20. (a) (i) kos (A + B) = kos A kos B – sin A sin B =  3 – 5 2 12 – 132 –  4 5 2 5 – 132 = 56 65 (ii) sin 2A = 2 sin A kos A = 2 4 5 2 3 – 5 2 = – 24 25 (iii)tan (A – B) = tan A – tan B 1 + tan A tan B = 4 – 3 –  5 12 2 1 +  4 – 3 2 5 12 2 = – 3 15 16 (b) (i) tan (A – B) = tan A – tan B 1 + tan A tan B = 5 12 – (–1) 1 +  5 12 2(–1) = 17 7 (ii) tan 2A = 2 tan A 1 – tan2 A = 2 5 12 2 1 – 25 144 = 1 1 119 = 120 119 (iii) kosek (A + B) = 1 sin (A + B) sin (A + B) = sin A kos B + kos A sin B =  5 13 2– 1 2 2 +  12 13 2 1 2 ) = 7 13 2 = kosek (A + B) = 13 2 7 x y 20fi55ff 20fi55ff x y 44fi36ff 44fi36ff x y 65fi3ff 65fi3ff x y 42fi55ff 42fi55ff x y 51fi27ff 51fi27ff x y 38fi25ff 38fi25ff x y –5 –3 13 4 5 A –12 y x –1 1 B A √2 12 5 13

J39 (c) (i) kos 2A= 2 kos2 A – 1 cos 2A = 2 cos2A – 1 = 2p2 – 1 (ii) tan A = 1 – p2 p (iii) kos 4A = kos 2(2A) cos 4A = cos 2(2A) = 2 kos2 2A –1 = 2[2p2 – 1]2 – 1 = 2[4p4 – 4p2 + 1] – 1 = 8p4 – 8p2 + 1 (d) (i) sin A = –1 1 + p2 (ii) sin 2A = 2 sin A kos A = 2( –1 1 + p2 )( –p 1 + p2 ) = 2p 1 + p2 (iii)tan A 2 tan A = 2tan A 2 1 – tan2 A 2 1 p = 2tan A 2 1 – tan2 A 2 1 – tan2A 2 = 2ptan A 2 tan2 A 2 + 2p tan A 2 – 1= 0 tan A 2 = –2p ± 4p2 – 4(–1) 2 = –2p ± 4p2 + 4 2 = –p ± p2 + 1 21. (a) kot x + kos x = 0 kos x sin x + kos x = 0 kos x[1 + sin x] = 0 kos x = 0 sin x = –1 x = 90°, 270° (b) 2 kos2 x – 1 = sin x 2[1 – sin2 x] – 1 = sin x 2 sin2 x + sin x – 1 = 0 (2 sin x – 1)(sin x + 1) = 0 sin x = 1 2 sin x = –1 x = 30°, 150°, 270° (c) 2 sek2 x = 5 tan x 2 kos2 x = 5 sin x kos x 2 kos x = 5 sin x kos2 x kos x[2 – 5 sin x kos x] = 0 kos x = 0 2 5 = sin x kos x sin 2x = 4 5 = 0.8 x = 90°, 270° 2x = 53° 8’, 126° 52’ 413° 8’, 486° 52’ x = 26° 34’, 63° 26’ 206° 34’, 243° 26’ (d) 4 tan 2x = 9 tan x 8 tan x 1 – tan2 x = 9 tan x 8 tan x = 9 tan x – 9 tan3 x 9 tan3 x = tan x tan x[9 tan2 x – 1] = 0 tan x = 0 tan x = ± 1 3 x = 0, 180°, 360° x = 18° 27’, 161° 34’, 198° 27’, 341° 34’ (e) 5 sin x kos x – 5 sin x – 2 kos x + 2 = 0 5 sin x[ kos x – 1] – 2[kos x – 1] = 0 (kos x – 1)(5 sin x – 2) = 0 kos x = 1 sin x = 2 5 x = 0°, 360°, 23° 35’, 156° 25’ (f) 3 kos x = 5 sin x 3 5 = tan x x = 30° 58’ 210° 58’ (g) 3 sin x + 3 kos x 1 + 1 sin x – kos x = 0 3[sin x + kos x][sin x – kos x] + 1 = 0 3[sin2 x – kos2 x] + 1 = 0 3[sin2 x – (1 – sin2 x)] + 1 = 0 3[2 sin2 x – 1] + 1 = 0 6 sin2 x – 2 = 0 sin2 x = 1 3 sin x = ± 1 3 x = 35° 16’, 144° 44’ = 215° 16’, 324° 44’ (h) 3 sin2 x = 8 sin x kos x + 3 kos2 x 3 sin2 x – 8 sin x kos x – 3 kos2 x = 0 (3 sin x + kos x)(sin x – 3 kos x) = 0 3 sin x = –kos x tan x = 3, tan x = 1 – 3 , x = 71° 34’, 251° 34’ x = 161° 34’, 341° 33’ (i) sin2 x = 1 – 2 kos2 x = 1 – 2[1 – sin2 x] = –1 + 2 sin2 x sin2 x = 1 sinx = ± 1 x = 90°, 270° 22. (a) (i) x2 + y2 = 16 y x A p 1 √1–p2 x A y –1 –p √1+p2 Q(–x, y) P(x, y) R S O x fi 4 y

J40 Maka OP = 4 luas berlorek = luas semibulatan – luas PQRS = 1 2 π(4)2 – (2x)(y) sin θ = y 4 , kos θ = x 4 ∴ luas berlorek = 8π – 2(4 sin q)(4 kos q) = 8π – 16 sin 2q (ii) Apabila/When q = 30°. Luas/Area = 8p − 16 sin 2(30°) = 8p − 16 3 2 = 8p − 8 3 (b) (i) P(x, y) OP = 5 OP2 = 52 x2 + y2 = 52 x2 + y2 = 25 QOP = 180° – 2θ (ii) luas berlorek = luas sektor – luas ∆QOP = (180° – 2θ) 360° × p(5)2 – 1 2 (2x)y = 25 (90° – θ) 180° π – xy x = 5 kos θ, y = 5 sin θ luas berlorek = 5 36(90° – θ)p – 25 sin θ kos θ (iii) Apabila θ = 15° luas berlorek = 5 36[90° – 15°]p – 25 2 sin 30° = 32.72 – 6.25 = 26.47 unit2 Praktis SPM 6 Kertas 1 1. sin 2x + sin x = 0 2 sin x kos x + sin x = 0 sin x[2 kos x + 1] = 0 sin x = 0 ; x = 0°, 180°, 360°, kos x = 1 – 2 x = 120°, 240° 2. sin (x – 45°) = sin x kos 45° – kos x sin 45° = sin x cos 45° – cos x sin 45° = 1 2 –5 13 2 – 1 2 –12 13 2 = 7 13 2 3. 3 kos x – kosek x + 2 = 0 3 kos x – 1 kos x + 2 = 0 3 kos2 x + 2 kos x – 1 = 0 [3 kos x – 1][kos x + 1] = 0 kos x = 1 3 ; –1 x = 70° 32’, 289° 28‘, 180° 4. (a) kos(180° – x) = kos 180° kos x + sin 180° sin x = –kos x = – 1 – k2 (b) kosek 2x = 1 sin 2x = 1 2k 1 – k2 Kertas 2 1. (a) kot2 x 1 – kos2 x = 1 2 kos 2x + 1 2 Sebelah kiri = kot2 x sin2 x Left side = kos2 x sin2 x ÷ 1 sin2 x = kos2 x Tetapi kos 2x = 2kos2 x – 1 kos2 x = kos 2x + 1 2 = 1 2 kos 2x + 1 2 (b) 1 2 kos 2x + 1 2 = 3 4 kos 2x = 1 2 2x = 60°, 300°, 420°, 660° x = 30°, 150°, 210°, 330° (c) (i) y = 1 2 kos2x + 1 2 y x 0 1 fi 2fi 1 2 y = kos 2x + 1 2 1 2 y = cos 2x + 1 2 1 2 y = 2x 3fi 3π( 1 2 kos 2x + 1 2 ) = 2x 1 2 kos 2x + 1 2 = 2x 3π y = 2x 3π x = 0, y = 0 x = π, y = 2 3 3 penyelesaian/solutions 2. (a) tan(x + 45°) + tan(x – 45°) = 2 tan 2x Sebelah kiri = tan x + 1 1 – tan x + tan x – 1 1 + tan x Left side = (1 + tan x)2 – (1 – tan x)2 1 – tan2 x = 1 + 2 tan x + tan2 x – 1 + 2 tan x – tan x 1 – tan2 x = 2(2 tan x) 1 – tan2 x = 2 tan 2x y x 60fi 2 1 y x –12 –5 x 13 y x √1–k2 k 1 x y x 60fi 2 0 1

J41 (b) 2 tan 2x = 1 2 tan 2x = 1 4 2x = 14° 2’, 194° 2’, 274° 2’, 554° 2’ x = 7° 1’, 97° 1’, 137° 1’, 277° 1’ (c) y = 1 2 [tan(x + 45°) + tan(x – 45°)] + 1 y = 1 2 [2 tan 2x] + 1 = tan 2x + 1 3. (a) (i) sek A – 1 tan A = tan A 2 Sebelah kiri Left side = 1 kos A – 1 sin A kos A = 1 – kos A sin A = 1 – [1 – 2 sin2 A 2 ] 2 sin A 2 kos A 2 = sin A 2 kos A 2 = tan A 2 (ii) tan 1 22 2 = sek 45° – 1 tan 45° = 1 kos 45° –1 = 2 – 1 (b) (i) & (ii) y x 0 fi 2fi y = tan x y = – x fi kot x 2 [sek x – 1) = – x π sek x – 1 tan x 2 = tan x = – x π ∴ y = – x π x = 0, , y = 0 x = π, y = –1 3 penyelesaian/solutions Sudut KBAT (a) sin(A – B) sin(A + B) = 3 5 5[sin A kos B – kos A sin B] = 3[sin A kos B + kos A sin B] 2 sin A kos B = 8 kos A sin B sin A kos B = 4 kos A sin B (b) (i) tan B = 1 2 sin A kos A = 4 tan B = 4 1 2 2 = 2 tan A = 2 (ii) tan 2A = 2 tan A 1 – tan2 A = 2(2) 1 – 4 = 4 –3 y x 0 1 fi

J42 BAB Pengaturcaraan Linear 7 Linear Programming 1. (a) x y 0 (b) x y 0 2. (a) y = 4 y > 4 (b) m = 5 − (−2) −4 − 4 = 7 – 8 y = 7 – 8 x + c c = 3 2 y = 7 – 8 x + 3 2 y ≤ 7 – 8 x + 3 2 3. (a) (b) (c) 4. (a) (b) (c) 5. (a) y ≤ 3x + 3 y ≥ 4 – 3 x + 4 y ≥ 1 2 x − 1 (b) x ≥ 0 x ≤ 2 y ≥ 2 3 x y ≤ 2x + 2 (c) y + x ≥ 4 3x + 2y ≤ 12 x + 2y ≤ 8 x y 0 R 2y fi 4x – 1 2 fi x + 2y x ff 4 x y 0 R 2y +3x + 2 fi 0 3x – 2y ff –5 x + y ffl 10 x y 0 R x ≤ y x ≥ –1 x + 2y ≥ 4 x y 0 R –x ≥ –4 x ≥ 1 – 2y –2 ≤ y x y 0 R x + 4y ≤ 100 x + y ≤ 60 x y 0 R y ≥ 4x – 12 y ≥ –3x – 15 2y + x ≥ –12 y ≤ 5

J43 6. (a) Pada (−2, 2) ialah titik minimum, maka At (−2, 2) is a minimum point, hence k = 2 + 2(−2) = −2 Pada (4, 5) ialah titik maksimum, maka At (4, 5) is a maximum point, hence k = 5 + 2(4) = 13 (b) Pada (3, 1) ialah titik minimum. Maka, k = 2(1) + 3 = 5. Tidak wujud titik maksimum kerana garis lurus boleh dianjak merentas rantau yang tidak ada had. At (3, 1) is a minimum point. Hence, k = 2(1) + 3 = 5. There is no maximum point because the straight lines can be moved across an unlimited region. 7. (a) I 4x + 5y ≤ 120 II 3x + 5y ≤ 150 III 600x + 1 500y ≤ 30 000 atau 2x + 5y ≤ 100 (b) (i) Jika perbelanjaan untuk penyuntingan dan penulisan ialah x, maka x ≤ 5 000. If the cost for editing and writing is x, then x ≤ 5 000. (ii) Jika perbelanjaan lain ialah y, maka y ≤ 1 000. If the other expenses is y, then y ≤ 1 000. (iii) Kos 150 ribu naskhah ialah= 150 × 2 000 20 = RM15 000. Kos untuk percetakan dan pengangkutan Z ≥ 15 000. The cost of 150 thousand copies is 150 × 2 000 20 = RM15 000. The cost for printing and transportation Z ≥ 15 000. 8. (a) (i) Ketaksamaan bagi setiap kekangan. The inequalities for each constraint. I y ≤ 80 II x − 3y ≤ 30 III 3x + 2y ≤ 300 (ii) (iii) (a) Daripada graf, julat bilangan B ialah From the graph, the range of the number of B is 10 ≤ y ≤ 60 (b) Keuntungan/Profit K = 3x + 6y Titik optimum ialah titik persilangan. The optimum point is the intersection point. y = 80 dan/and x − 3y = 300 Nilai integer x terdekat ialah 46. The value of the nearest integer x is 46. Maka, untung maksimum K Hence, the maximum profit K = 3(46) + 6(80) = RM618 (c) Untuk y = x, titik maksimum = (60, 60). For y = x, maximum point = (60, 60). Jenis A dan B masing-masing ialah 60 unit. Type A and type B are 60 units respectively. (b) (a) I x + y ≤ 100 II y ≥ 20 III y ≤ 2x (b) (c) (i) x maksimum / x maximum = 80 (ii) 20 ≤ y ≤ 50 (iii) Andaikan P = 1 200x + 600y Yuran maksimum / Maximum fees = 1 200(80) + 600(20) = RM108 000 Praktis SPM 7 Kertas 2 1. (a) I : 10x + 30y ≥ 30 x + 3y ≥ 3 II : y ≤ x III : x ≤ 9 (b) 40 20 –20 60 80 0 20 40 60 y x R 100 80 100 y = x 3x + 2y = 300 y = 80 x – 3y = 30 40 R 40 20 0 20 60 60 80 80 100 100 y x y = 2x (80, 20) y = 20 x + y = 100 R 4 4 2 0 2 6 6 8 8 10 10 12 x y y = x (3, 3) x = 9 x + 3y = 12

J44 (c) (i) Julat ialah 4 ≤ x ≤ 9. The range is 4 ≤ x ≤ 9. (ii) K = 120x + 180y Kos maksimum / Maximum cost K = 120(9) + 180(9) = RM2 700 2. (a) 50x + 60y ≤ 750 5x + 6y ≤ 75 75x + 45y ≤ 15 × 60 5x + 3y ≤ 60 (b) (c) (i) y = 2x, Dalam rantau, x maksimum = 4 dan y maksimum = 8 In the region, x maximum = 4 and y maximum = 8 (ii) U = 60x + 30y, apabila x = 9, y = 5 akan mendapat maksimum U = 60(9) + 30(5) = RM 690 3. (a) x + y ≥ 40 30x + 40y ≤ 4 000 3x + 4y ≤ 400 x − y ≤ 15 (b) (c) (i) Titik minimum = (20, 20) Minimum point Harga / Price = 30(20) + 40(20) = RM1 400 Baki = RM4000 − RM1 400 = RM2 600 (ii) Peruntukan yang tinggal = RM3 600 Allocation left = RM3 600 Maka/Hence 30x + 40y ≤ 3600 3x + 4y ≤ 360 Bilangan pen maksimum = 60 Maximum number of pens Sudut KBAT (a) x + y ≤ 20 80x + 50y ≤ 1 200 8x + 5y ≤ 120 (b) (c) (i) Julat / Range 0 ≤ x ≤ 11 (ii) Lukis x = 2y, maksimum kambing y = 5 10 5 15 20 –5 0 5 10 25 15 20 x + y = 20 8x + 5y = 120 x y 40 20 60 80 0 –20 –20 20 40 100 60 80 100 120 140 160 180 x + y = 40 x – y = 15 3x + 4y = 360 3x + 4y = 400 y x 4 2 12 6 -2 0 2 4 6 y x 8 10 8 10 12 14 16 18 20 5x + 3y = 60 5x + 6y = 75

J45 BAB Kinematik Gerakan Linear 8 Kinematics of Linear Motion 1. (a) (i) Jumlah jarak = jumlah semua perjalanan Total distance = total all the journey = (2 + 3 + 7) km = 12 km (ii) Titik asal Cik Nor ialah rumahnya dan titik akhir juga rumahnya. Maka sesaran = 0 km. Cik Nor’s original point is her house and the end point is also her house. Hence, displacement = 0 km. (b) (i) Jumlah jarak = jumlah semua perjalanan Total distance = total all the journey = (5 + 8) m = 13 m (ii) Titik asal zarah ialah di O dan titik akhirnya ialah 3 m ke kiri O. Maka sesaran = −3 m. The original point of the particle is at O and its end point is 3 m to the left of O. So, displacement = −3 m. 2. (a) (i) Apabila t = 0 s, s(0) = 4 – 2(0) − 02 = 4 m (zarah berada ke kanan O) the particle is at the right of O) (ii) Apabila t = 1 s, s(1) = 4 – 2(1) − 12 = 1 m (zarah berada ke kanan O) (the particle is at the right of O) (iii) Apabila t = 2 s, s(2)= 4 – 2(2) − 22 = −4m (zarah berada di titik O) (the particle is at O) (b) (i) Apabila t = 1 s, s(1) = 2 + 4(1) − 2(1)2 = 4 m (zarah berada ke kanan O) (the particle is at the right of O) (ii) Apabila t = 5 s, s(5) = 2 + 4(5) − 2(5)2 = −28 m (zarah berada ke kiri O) (the particle is at the left of O) (iii) Apabila t = 1 2 s, s( 1 2 ) = 2 + 4( 1 2 ) − 2( 1 2 )2 = 3 1 2 m (zarah berada ke kanan O) (the particle is at the right of O) (c) (i) Apabila t = 0 s, s(0) = (1 − 0)(3(0) + 4) = 4 m (zarah berada ke kanan O) (the particle is at the right of O) (ii) Apabila t = 1 3 s, s( 1 3 ) = (1 − 1 3 )(3( 1 3 )+ 4) = 3 1 3 m (zarah berada ke kanan O) (the particle is at the right of O) (iii) Apabila t = 1.5 s, s(1.5) = (1 − 1.5)(3(1.5) + 4) = −4 1 4 m (zarah berada ke kiri O) (the particle is at the left of O) 3. (a) (i) Apabila t = 0 s, v(0) = (2 − 0)(3(0) − 4) = −8 m s−1 (zarah bergerak ke arah kiri) (the particle moves to the left) (ii) Apabila t = 2 s, v(2) = (2 − 2)(3(2) − 4) = 0 m s−1 (zarah berhenti seketika) (the particle stops instantaneously) (iii) Apabila t = 5 s, v(5) = (2 − 5)(3(5) − 4) = −33 m s−1 (zarah bergerak ke arah kiri) (the particle moves to the left) (b) (i) Apabila t = 0 s, v(0) = 1 2 (0)2 – 2(0) = 0 m s−1 (zarah tidak bergerak) (the particle is not moving) (ii) Apabila t = 2 s, v(2) = 1 2 (2)2 – 2(2) = −2 m s−1 (zarah bergerak ke arah kiri) (the particle moves to the left) (iii) Apabila t = 5 s, v(5) = 1 2 (5)2 – 2(5) = 2 1 2 m s−1 (zarah bergerak ke arah kanan) (the particle moves to the right) (c) (i) Apabila t = 0 s, v(0) = 1 2 (2(0) − 3)2 − 4 = 1 2 m s−1 (zarah bergerak ke arah kanan) (the particle moves to the right) (ii) Apabila t = 1 s, v(1) = 1 2 (2(1) − 3)2 − 4 = 1 –3 2 m s−1 (zarah bergerak ke arah kiri) (the particle moves to the left) (iii) Apabila t = 2 s, v(2) = 1 2 (2(2) − 3)2 − 4 = 1 –3 2 m s−1 (zarah bergerak ke arah kiri) (the particle moves to the left) 4. (a) (i) Apabila t = 0 s, a(0) = 2 − 5(0) + 02 = 2 m s−2 (halaju zarah bertambah) (the velocity is increasing) (ii) Apabila t = 1 2 s, a 1 2  = 2 − 51 1 2  + 1 1 2  2 = − 1 4 m s−2 (halaju zarah berkurang) (the velocity is decreasing) (iii) Apabila t = 3 s, a(3) = 2 − 5(3 ) + (3)2 = −4 m s−2 (halaju zarah berkurang) (the velocity is decreasing) (b) (i) Apabila t = 1 s, a(1) = −2(1) + 6 = 4 m s−2 (halaju zarah bertambah) (the velocity is increasing) (ii) Apabila t = 3 s, a(3) = −2(3) + 6 = 0 m s−2 (halaju zarah adalah malar) (the velocity is constant) (iii) Apabila t = 4 s, a(4) = −2(4) + 6 = −2 m s−2 (halaju zarah berkurang) (the velocity is decreasing)

J46 (c) (i) Apabila t = 2 s, a(2) = −(2)2 + 4(2) = 4 m s−2 (halaju zarah bertambah) (the velocity is increasing) (ii) Apabila t = 4 s, a(4) = −(4)2 + 4(4) = 0 m s−2 (halaju zarah adalah malar) (the velocity is constant) (iii) Apabila t = 6 s, a(6) = −(6)2 + 4(6) = −12 m s−2 (halaju zarah berkurang) (the velocity is decreasing) 5. (a) Lakaran graf sesaranmasa Displacement-time graph Jumlah jarak dilalui Total distance travelled Jarak yang dilalui dalam saat ketiga Distance travelled in the third second (a) s(t) = 3t − 4, 0 ≤ t ≤ 3 s(t) = 3t − 4 Graf ini ialah graf garis lurus dengan kecerunan 3 dan titik pintasan-y ialah −4. This is a straight line graph with the gradient of 3 and the y-interception point is −4. 5 –4 0 3 t s(t) Dari graf, jumlah jarak From the graph, total distance = 4 + 5 = 9 m Jarak dalam saat ketiga Distance in the third second = s(3) − s(2) = 3(3) − 4 − [3(2) − 4)] = 3 m (b) s(t) = (t − 2)(t − 5), 0 ≤ t ≤ 6 s(t) = (t − 2)(t − 5) = t 2 − 7t + 10 Graf ini ialah graf minimum dengan puncapunca 2 dan 5. This is a minimum graph with the roots of 2 and 5. Titik minimum ds(t) dt = 2t − 7 = 0 t = 7 2 , s = 1 –2 4 10 –2 0 2 5 t s(t) 1 4 7 2 Dari graf, jumlah jarak From the graph, total distance = 10 + 1 2 4 + 1 2 4 + 4 = 1 18 2 m Jarak dalam saat ketiga Distance in the third second = s(3) − s(2) = (3 − 2)(3 − 5) – (2 − 2)(2 − 5) = −2 m = 2 m (c) s(t) = −t2 + 4t − 3, 0 ≤ t ≤ 3 s(t)= −t 2 + 4t − 3 = (1 − t)(t − 3) Graf ini ialah graf maksimum dengan punca-punca 1 dan 3 This is a maximum graph with the roots of 1 and 3. Titik maksimum ds(t) dt = −2t + 4 = 0 t = 2, s = 1 1 –3 0 123 t s(t) Dari graf, jumlah jarak From the graph, total distance = 3 + 1 + 1 = 5 m Jarak dalam saat ketiga Distance in the third second = s(3) − s(2) = (1 − 3)(3 − 3) − (1 − 2)(2 − 3) = 1 m

J47 6. (a) (i) Untuk sesaran positif, s > 0 For positive displacement s(t) = t2 − 3t > 0 t(t − 3) > 0 t > 3 (ii) Apabila s = 10 t2 − 3t = 10 t2 − 3t −10 = 0 (t − 5)(t + 2) = 0 t = 5 (b) (i) Untuk sesaran positif, s > 0. For positive displacement s(t) = t 2 − 3t – 4 > 0 (t − 4)(t + 1) > 0 t > 4 (ii) Apabila s = 25 – 4 4t 2 −12t – 16 = −25 4t 2 −12t + 9 = 0 (2t − 3)(2t − 3) = 0 t = 3 2 7. (a) ds dt = t 2 − 4t − 12 (i) Untuk sesaran maksimum atau minimum, For maximum or minimum displacement, ds dt = 0 t 2 − 4t −12 = 0 (t + 2)(t − 6) = 0 t = 6 Apabila t = 6, s = (6)3 3 − 2(6)2 − 12(6) + 4 = −68 m (ke kiri O) Untuk menentukan sesaran maksimum atau minimum. To determine maximum or minimum displacement. d2 s dt2 = 2t − 4 Apabila t = 6, d2 s dt2 = 8 > 0 Maka, s = −68 ialah minimum. Hence, s = −68 is minimum 0 3 t s(t) 4 s(t) –4 t (ii) Untuk halaju malar, pecutan mesti 0. For a uniform velocity, the acceleration must be 0. Jadi, a = d2 s dt2 = 2t − 4 = 0 t = 2 Apabila t = 2, halaju malar When t = 2, the velocity is uniform. v = 22 − 4(2) − 12 = −16 m s−1 (b) ds dt = −3t 2 + 6t + 9 (i) Untuk sesaran maksimum atau minimum, For maximum or minimum displacement, ds dt = 0 −3t 2 + 6t + 9 = 0 t 2 − 2t − 3 = 0 (t + 1)(t − 3) = 0 t = 3 Apabila t = 3, s = −33 + 3(3)2 + 9(3) = 27 m (ke kanan O) Untuk menentukan sesaran maksimum atau minimum. To determine maximum or minimum displacement. d2 s dt2 = −6t + 6 Apabila t = 6, d2 s dt2 = −30 < 0 Maka, s = 27 m ialah maksimum. Hence, s = 27 m is maximum (ii) Untuk halaju malar, pecutan mesti 0 For a uniform velocity, the acceleration must be 0. Jadi, a = d2 s dt2 = −6t + 6 =0 t = 1 Apabila t = 1, halaju malar. When t = 1, the velocity is uniform. v = −3(1)2 + 6(1) + 9 = 12 m s−1 8. (a) (i) ds dt = v = −6t 2 + 12t Apabila t = 0, v = 0 m s−1 (ii) dv dt = −12t + 12 Apabila t = 0, a = 12 m s−2 Apabila t = 2, a = −12(2) + 12 = −12 m s−2 (iii) Apabila halaju maksimum atau minimum, dv dt = 0. When velocity is maximum or minimum, dv dt = 0. −12t + 12 = 0 t = 1 Untuk menentukan sama ada maksimum atau minimum, kita guna To determine whether the velocity is maximum or minimum, we use 2 v 0 t

J48 d2 v dt2 = −12 (< 0 maksimum) Maka, apabila t = 1, halaju maksimum = −6(1)2 + 12 = 6 m s−1 (ke arah kanan / to the right) (b) (i) ds dt = 9t 2 − 15t + 6 Apabila t = 0, v = 6 m s−1 (ii) dv dt = 18t − 15 Apabila t = 0, a = −15 m s−2 Apabila t = 2, a = 18(2) − 15 = 21 m s−2 (iii) Apabila halaju maksimum atau minimum, dv dt = 0. 18t − 15 = 0 t = 5 6 Untuk menentukan sama ada maksimum atau minimum, kita guna To determine whether the velocity is maximum or minimum, we use d2 v dt2 = 18 ( > 0 minimum) Maka, apabila t = 5 6 , halaju mininum = 9 5 6 2 2 − 15 5 6 2 + 6 = 1 – 4 m s−1 ( ke arah kiri / to the left) 9. (a) (i) a = dv dt = 12 − 4t dv = (12 − 4t) dt ∫dv = ∫12t − 4t dt v = 12t − 2t2 + c Apabila t = 0, v = −2. Maka, c = −2 v = 12t − 2t 2 − 2 Apabila t = 2, v = 12(2) − 2(2)2 – 2 = 14 m s−1 (ii) Apabila pecutan ialah sifar, When the acceleration is zero 12 − 4t = 0 t = 3 Maka / Hence, v = 12(3) − 2(3)2 − 2 = 16 m s−1 (b) (i) a = dv dt = 2t − 3 dv = (2t − 3)dt ∫dv = ∫2t − 3 dt v = t 2 − 3t + c Apabila t = 0, v = 10. Maka, c = 10 v = t 2 − 3t + 10 Apabila t = 2, v = (2)2 − 3(2) + 10 = 8 m s−1 (ii) Apabila pecutan ialah sifar, When the acceleration is zero 2t − 3 = 0 t = 3 2 1 6 v t – 1 4 2 3 Maka / Hence, v = 1 3 2  2 − 31 3 2  + 10 = 3 7 4 m s–1 10. (a) (i) a = dv dt = 4 − t dv = (4 − t) dt ∫dv = ∫(4 − t) dt v = 4t – 1 2 t 2 + c Apabila/When t = 2, v = −5. Maka/Hence −5 = 4(2) – 1 2 (2)2 + c c = −11 v = 4t – 1 2 t 2 − 11 Apabila/When a = 2, 4 − t = 2 t = 2 v = 4(2) – 1 2 (2)2 − 11 v = −5 m s−1 (ii) s = ∫v dt =∫[ –1 2 t 2 + 4t −11] dt = –1 6 t 3 + 2t 2 − 11t + c Apabila/When t = 0, s = 0, c = 0 s = –1 6 t 3 + 2t 2 − 11t Apabila halaju maksimum atau minimum. When the velocity is maximum or minimum. dv dt = 0, 4 − t = 0 t = 4 s = –1 6 (4)3 + 2(4)2 − 11(4) = 2 –22 3 m (b) (i) a = dv dt = 6t dv = (6t) dt ∫dv= ∫(6t) dt v = 3t 2 + c Apabila/When t = 1, v = 4. Maka/Hence 4 = 3(1)2 + c c = 1 v = 3t 2 + 1 Apabila a = 2, 6t = 2 t = 1 3 v = 3 1 3 2 2 + 1 v = 1 1 3 m s−1 (ii) s = ∫v dt = ∫[3t 2 + 1] dt = t 3 + t + c Apabila t = 0, s = 0, c = 0

J49 s = t3 + t Apabila halaju maksimum atau minimum. When the velocity is maximum or minimum. dv dt = 0, 6t = 0 t = 0 Jadi/So s = 0 11. (a) v = 2t(6 − t) Luas dari 3 hingga 6 ialah The area from 3 to 6 is s1 = ∫ 63 (12t − 2t2)dt = [6t2 − 23 t3]63 = [6[6]2 − 23 (6)3 ] − [6(3)2 − 23 (3)3] = 36 m Luas dari t = 6 ke t = 8 The area from t = 6 to t = 8. s2 = ∫ 8 6 (12t − 2t2)dt = [6t2 − 23 t3]8 6 = [6[8]2 − 23 (8)3 ] − [6(6)2 − 23 (6)3 ] = 1 –29 3 Jumlah jarak/Total distance = 36 + 1 29 3 = 1 65 3 m (b) a = dv dt = (2 − t)2 dv = (2 − t)2 dt ∫dv = ∫(2 − t)2 dt v = (2 – t)3 –3 + c Apabila/When t = 2, maka/hence c = 0, v = (2 – t)3 –3 Jarak/Distance s = ∫ 42 (2 – t)3 –3 dt =[(2 – t)4 3(4) ]42 = 43 m 12. (a) (i) a = 8 – 2t v = 8t – t2 + c t = 0 v = 0 c = 0 v = 8t – t2 Apabila vmak, 8 – 2t = 0 t = 4 ∴ v = 8(4) – 42 = 16 m s–1 (ii) s = 4t2 – t33 + c t = 0, s = 0 c = 0 s = 4t2 – 13 t3 Untuk smak,8t – t2 = 0 [8 – t] = 0 t = 0 ; t = 8 Apabila t = 8, s = 4(8)2 – 13 (8)3 = 1 85 3 m (iii) Apabila balik ke O, s = 0 4t2 – 13 t3 = 0 t2 [4 – 13 t] = 0 t = 0 t = 12 (iv) s = ∫ 60 (8t – t2)dt = [4t2 – 13 t3]60 = 72 m (b) (i) t = 0 v = 32 ∫ = a + bt v = at + b2 t2 + c t = 0 v = 32 c = 32 v = at + b2 t2 + 32 Apabila t = 2 v = 72 72 = 2a + 2b + 32 a + b = 1 v = 0, t = 3 0 = 3a + 9b2 + 32 –3 = 6a + 9b –1 = 2a + 3b = 2(1 – b) + 3b –3 = b, a = 4 ∴ 4t – 32 t2 + 32 (ii) s = ∫ 214t – 32 t2 + 32 2dt = 32t2 – 12 t3 + 32 t4 21 = 3(2(2)2 – 4 + 3) – 2 – 12 + 32 24 = 4 m (c) (i) v = 6t – kt2 a = dv dt = 6 – 2kt Apabila t = 2, a = 2 2 = 6 – 2k(2) –4 = –4k k = 1 v = 6t – t2 (ii) dv dt = 6 – 2t = 0 t = 3 Apabila t = 3 v = 6(3) – 32 = 18 – 9 = 9 vmak = 9 m s–1 (iii)s = 3t2 – t33 + c t = 0; s = 0, c = 0 s = 3t2 – t33 O 6 8 t v (8, 108

J50 Apabila kembali ke O When return to 0. s = 0 = t23 – t3  ∴ t = 9 s (d) (i) s = 0 ; v = –10 a = p – 3t2 v = pt – t3 + c t = 0 v = –10 c = –10 v = pt – t3 – 10 s = pt2 2 – 14 t4 – 10t + c t = 1 s1 = p2 – 14 – 10 + c t = 2 s2 = 42 p – 4 – 20 + c s2 – s1 = 32 p – 3 13 4 = 1 4 4 32 p = 18 p = 12 (ii) v = 12t – t3 – 10 a = 12 – 3t2 = 0 t = 2 vmak = 12(2) – 23 – 10 = 6 m s–1 Praktis SPM 8 Kertas 1 1. (a) v = at2 – 8t a = dv dt = 2at – 8 Apabila t = 4, a = 12 12 = 8a – 8 a = 52 (b) 0 t v 85 165 v = 52 t2 – 8t v = t[ 52 t – 8] ∴ 85 < t < 165 (c) 52 t2 – 8t = 0 t[ 52 t – 8] = 0 t = 165 s (d) s = ∫ 30 ( 52 t2 – 8t)dt = [ 56 t3 – 4t2]30 = 13.5 m 2. (a) P O AQ 13 m vP = 6t2 + 12 SQ = 2t3 – t dsQ dt = 6t2 – 1 Apabila t = 0 dsQ dt = –1 ∴vQ = –1 m s–1 (b) s1 = ∫ 160 (6t2 – 1)dt = (2t3 – t) 16 0 = ( 2 6 6 – 16 ) =  –2 3 6  = 2 3 6 m s2 = ∫ 3 (6t2 – 1)dt 16 = 3(2t3 – t)4 3 16 = 51 – 3 2 6 6 – 16 4 = 51 – 3– 2 3 6 4 = 51 + 23 6 Jumlah jarak = 51 + 43 6 Total distance = 51.54 m (c) s p = 2t3 + 12t + c Apabila t = 0 s p = –13 s p = 2t3 + 12t – 13 Apabila berselisah sp = sQ 2t3 + 12t – 13 = 2t3 – t 13t = 13 t = 1 s p = 2 + 12 – 13 = 1 m sQ = 2 – 1 = 1 m dari A 3. (a) Sp = 6 + 6t – 6t2 Bila t = 0, Sp = 6 m SQ = 4t2 – 6t – 10 Bila t = 0, SQ = –10 m ∴ jarak AB = 10 + 6 = 16 m (b) Sp = SQ 6 + 6t – 6t2 = 4t2 – 6t – 10 10t2 – 12t – 16 = 0 0 3 t vQ 1 fiff6 Q B AP 0 6m

J51 0 1 2 t v 6 9 5t 2 – 6t – 8 = 0 (5t + 4)(t – 2)= 0 ∴ t = 2 (c) vp = 6 – 12t Jarak oleh P = 1 2 (6)( 1 2 ) + 1 2 ( 3 2 )(18) = 3 2 + 27 2 = 30 2 = 15 m 4. (a) v = t 3 – 5t 2 + 4t a = dv dt = 3t 2 – 10t + 4 Apabila t = 0, a = 4 m s–2 (b) 3t 2 – 10t + 4 < 12 3t 2 – 10t – 8 < 0 (3t + 2)(t – 4) < 0 0 ≤ t < 4 (c) v = 0 = t[t 2 – 5t + 4] t[(t – 4)(t – 1) = 0 t = 0 t = 1; 4 ∴ t = 1 atau 4 (d) s = ∫ 1 0 t 3 – 5t 2 + 4t dt + ∫ 4 1 t 3 – 5t 2 + 4t dt =  t 4 4 – 5 3 t 3 + 2t 2 4 1 0 +  t 4 4 – 5 3 t 3 + 2t 2 4 4 1 = 1 1 4 – 5 3 + 22 + 1 256 4 – 320 3 + 322 = 7 12 +  – 1 11 4  = 5 11 6 m v t 0 1 4 Sudut KBAT (a) v = 36t – 18t 2 Apabila berhenti v = 0 = 36t – 18t 2 0 = 18t[2 – t] t = 0 ; 2 ∴t = 2 s ke B (b) a = dv dt = 36 – 36t Apabila t = 2, a = 36 – 72 = –36 m s–2 (c) vmak apabila dv dt = 0, t = 1 v maks = 36(1) – 18(1)2 = 18 m s–1 (d) SAB = ∫ 2 0 (36t – 18t 2 )dt = [18t 2 – 6t 3 ] 2 0 = 24 m (e) Apabila s = 18t2 – 6t 3 = 0 6t2 [3 – t] = 0 t = 0 ; 3 ∴ t = 3 A B 0 2 t v (8, 1

J52 Kertas PRA SPM Kertas 1/Paper 1 Bahagian A/Section A 1. (a) (i) Merah / Red Putih / White Hitam / Black 5 4 ➤ 1 ➤ ➤ (ii) Satu kepada satu One to one (b) (i) k = 0 (ii) f(1) = 8 a + b = 13 6 ...➀ f(2) = 8 2a + b = 5 6 ...➁ ➀ – ➁ 4 a = 8 6 2a = 6 a = 3 b = 13 6 – 8 3 = 3 – 6 = 1 – 2 (iii) f(h) = 8 3h – 1 2 = 1 6 8 3h = 4 6 h = 4 2. (a) (i) x2 + px + 3q = 0 α + 2α = –p 3α = –p α(2α) = 3q 2 –p 3  2 = 3q 2p2 = 27q q = 2p2 27 (ii) x2 + px + 3q = 1x + p 2 2 2 – p2 4 + 3q = 1x + p 2 2 2 – p2 4 + 2p2 9 = 1x + p 2 2 2 – p2 36 ∴ Titik pusingan 1 –p 2 , –p2 36 2 Turning point (b) x(x – 2) > x + 4 x2 – 2x – x – 4 > 0 x2 – 3x – 4 > 0 (x + 1)(x – 4) > 0 Julat x /Range of x ∴x < –1, x > 4 3. (a) log2 8 + 2 log2 p = 5 3 + log2 p2 = 5 log2 p2 = 2 p2 = 22 p = 2 y A 7 1 3 B C x 1 0 – 2 (b) 32x + 3 + 6(3x ) – 1 = 0 27·32x + 6(3x ) – 1 = 0 (9·3x – 1)(3·3x + 1) = 0 3x = 1 9 3x = 1 – 3 = –(3)–1 3x = 3–2 Tidak diterima x = –2 Rejected 4. (a) 2x – 3y + 1 = 0 ...➀ x – 2y = 1 x = 1 + 2y ...➁ 2[1 + 2y] – 3y + 1 = 0 2 + 4y – 3y + 1 = 0 y = –3 x = 1 – 6 = –5 ∴Titik persilangan (–5, –3) Intersection point Persamaan J3 (y + 3)= 2(x + 5) Equation J3 y = 2x + 7 (b) y = 2x + 7 J2 : y = 2 3 x + 1 3 J1 : y = 1 2 x – 1 2 AB : BC 7 − 1 3 2 :  1 3 – − 1 2 22 20 3 : 5 6 = 8 : 1 5. (a) 2, 1.5, ... a = 2 , r = 3 4 S∞ = a 1 – r = 2 1 – 3 4 = 8 m Tidak sampai tempat makanan. Not reach the food. (b) x 2 = y x y = x2 2 a = 2 a + d = x ⇒ d = x – 2 a + 11d = y 2 + 11(x – 2) = x2 2 2 + 11x – 22 = x2 2 2(11x – 20) = x2 x2 – 22x + 40 = 0 (x – 20)(x – 2) = 0 x = 20 ; 2 y = 200, 2 2, 20, 200, x = 2, y = 2 ∴x = 20, y = 200 tidak diterima 6. (a) α, α + 4 α + α + 4 = 2k + 4 2α + 4 = 2k + 4 α = k α(α + 4) = 3k + 2 k(k + 4) = 3k + 2 k2 + k – 2 = 0 (k – 1)(k + 2) = 0 k = 1 ; –2

J53 (b) Jika k = 1 x2 – 6x + 5 = 0 (x –1)(x – 5) = 0 Jika k = –2 x2 – 4 = 0 x = ±2 7. (a) → PQ = –2 i ~ + j → ~ PR = 2 i ~ + 5 j → ~ PS = 4 i ~ + 7 j ~ → QR = → QP + → PR =  2 –1 +  2 5 2 =  4 4 2 → QS = → QP + → PS =  2 –12 +  4 7 2 =  6 6 2 → QR = 4(i ~ + j ~ ) → QS = 6(i ~ + j ~ ) = 6 4 → QR → QS = 3 2 → QR Oleh sebab titik Q ialah titik sepunya dan → QS = 3 2 → QR. Q, R dan S adalah segaris. Since point Q is a common point and → QS = 3 2 → QR. Q, R and S are collinear. (b) v ~q = 5 i ~ + 6 j ~ , v ~r = a i ~ + 3 j ~ 6 5 = 3 a 6a = 15 a = 5 2 8. (a) (i) y = 2x2 − 8 y x2 = – 8 x2 + 2 Y = y x2 , X = 1 x2 , m = −8 , c = 2 (ii) k = −8(2) + 2 = −14 (b) y = axn log y = log a + n log x n = 2 log a = 1.3 a ≈ 20 9. (a) y = 1 (x + 2)(x – 3) = 1 (x2 – x – 6) dy dx = –(x2 – x – 6)–2(2x – 1) = 1 – 2x (x2 – x – 6)2 dy dx + y2 (2x – 1) = 1 – 2x (x2 – x – 6)2 + (2x – 1) (x + 2)2 (x – 3)2 = 0 (x + 2)2 (x – 3)2 = 0 (b) V = 100 – 2t – t 2 (i) Apabila t = 0, V = 100 cm3 (ii) dv dt = –2 – 2t Apabila t = 4, dv dt = –2 –2(4) = –10 cm3 s–1 10. (a) (i) ∫ 4 0 f(x)dx = 6 + 4 = 10 (ii) ∫ 0 –2 f(x)dx = –4 (iii)luas Q/ Area Q = 6 × 4 – 10 = 14 unit2 (b) d2 y dx2 = 1 – x2 dy dx = x – 1 3 x3 + c Apabila x = 1, dy dx = –1 1 – 1 3 + c = –1 c = –1 – 2 3 = –5 3 dy dx = x – 1 3 x3 – 5 3 y = x2 2 – 1 12x4 – 5 3 x + c1 (1, 4); 4 = 1 2 – 1 12 – 5 3 + c c1 = 21 4 ∴ y = x2 2 – 1 12x4 – 5 3 x + 21 4 11. (a) n p2 = 56 n! (n – 2)! = 56 n(n – 1) = 56 n2 – n – 56 = 0 (n – 8)(n + 7) = 0 n = 8 (b) (i) 9! = 362 880 (ii) 7! 2!2! = 1 260 12. (a) P[X = 1] + P[X = 2] + P[X = 3] + P[X = 4] + P[X = 5] = 1 1 2k + 2 2k + 3 2k + 4 2k + 5 2k = 1 15 2k = 1 k = 15 2 (b) x 1 2 3 4 5 P(X = x) 1 15 2 15 3 15 4 15 5 15 (c) P[2 < x ≤ 5] = 1 5 + 4 15 + 1 3 = 12 15 = 4 5 Bahagian B/Section B 13. (a) sin B 10 = sin 30° 7 B = 134° 25’

J54 (b) (c) ∠APC = 45°35’ ; ∠PAC = 104°25’ PC2 = 72 + 102 – 2(7)(10) kos/ cos 104°25’ PC = 13.56 cm (d) Luas/ Area ABC = 1 2 (10)(7) sin(150° – 134°35’) = 9.4 cm2 14. (a) tan A = 3 tan(A + B) = tan A + tan B 1 – tan A tan B = 1 3 + tan B = 1 – 3 tan B (1 + 3)tan B = 1 – 3 tan B = (1 – 3)(1 – 3) (1 + 3)(1 – 3) = 1 – 2 3 + 3 1 – 3 = 3 – 2 (i) y = a sin bx + c a = 3 , b = 1 2 , c = −1 (ii) x(a sin bx + c) = π a sin bx + c = π x y = π x Apabila/When x = π, y = 1 Apabila/When x = π 2 , y = 2 2 penyelesaian/ solutions 15. (a) Kos/ cos θ = 0.5 2.5 θ = 78°27’ Panjang lengkok CD = 78°27’ 360° × 2p(1.5) Length of arc CD = 2.05 cm 10 cm 7 cm 7 cm 30° A P B C y 1 0 2 fi fi 2fi 2 3 –1 x Panjang lengkok BD = 90° + 11.33’ 360° × 2p(1) Length of arc BD = 1.77 cm CB = 2.52 – 0.52 = 2.45 cm Perimeter = 2.05 + 1.77 + 2.45 = 6.27 cm (b) luas berlorek = luas trapezium OABC – sektor ODC – sektor ADB Shaded area = area of trapezium OABC – sector ODC – sector ADB = 1 2 [1 + 1.5]2.45 – 78°27’ 360° × p(1.5)2 – 101°33’ 360° × p(1)2 = 0.64 cm2 Kertas 2/Paper 2 Bahagian A/Section A 1. x + 2y – z = 4 ...➀ 2x + y + z = –2 ...➁ ➀ + ➁ 3x + 3y = 2 ...➃ x + 2y + z = 2 ...➂ ➀ + ➂ 2x + 4y = 6 x + 2y = 3 ...➄ 3[3 – 2y] + 3y = 2 9 – 6y + 3y = 2 7 = 3y y = 7 3 x = 3 – 2 7 3 2 = –5 3 –5 3 + 14 3 – 4 = z z = –1 2. (a) & (b) (c) Domain bagi f−1(x) Domain for f −1(x) −3 ≤ x ≤ 5 3. (a) y = x2 + (1 – p)x + 2p y = (1 – p)2 4 + 2p – x +  1 – p 2 2 2 2p – (1 – p)2 4 = 6 8p – (1 – 2p + p2 ) = 24 p2 – 10p + 25 = 0 (p – 5)(p – 5) = 0 p = 5 P 1.5 cm 1 cm O D C B 0.5 cm A 2.5 cm fi Q 0 1 1 2 3 (–3, 3) (3,–3) (5,–1) (–1, 5) 4 5 –1 –3 –2 –1 –2 –3 2 3 4 5 f(x) f –1(x) f(x) = 3 – 2x x

J55 (b) (i) L(x) = 2x(4 – x) = 8x – 2x2 (ii) dL(x) dx = 8 – 4x = 0 x = 2 Apabila x = 2 L(2) = 8(2) – 2(2)2 = 16 – 8 = 8 cm2 4. (a) (i) y = x + 4 y = 4 + 3x – x2 x + 4 = 4 + 3x – x2 x2 – 2x = 0 x(x – 2) = 0 x = 0 ; 2 A(0, 4) B(2, 6) (ii) Luas/Area = ∫ 2 0 (4 + 3x – x2 )dx – 1 2 [4 + 6]2 = 4x + 3 2 x2 – x3 3  2 0 – 10 = 1 11 3 – 10 = 1 1 3 unit2 (b) Isipadu = ∫ 1 0 y2 dx Volume = ∫ 1 0 (4 + 3x – x2 )2 dx = ∫ 1 0 (16 + 24x + x2 – 6x3 + x4 )dx = 16x + 12x2 + 1 3 x3 – 3 2 x4 + 1 5 x5 4 1 0 = 1 27 30 unit3 ∴ isipadu sebenar = 31 13 60 unit3 5. (a) (i) p = 0.01, n = 1000 m = np = 1000(0.01) = 10 σ = npq = 1000(0.01)(0.99) = 3.15 (ii) P[x ≥ 1] = 1 – P[X = 0] = 1 – 5 C0 (0.01)0 (0.99)5 = 0.049 (b) (i) m = 200, σ = 40 P[ 150 ≤ z ≤ 180] = P 150 – 200 40 < z < 180 – 200 40 4 = P 5 – 4 < z < 1 – 2 4 = 0.2029 (ii) P[X > 230] = P[Z > 230 – 200 40 ] = P[Z > 3 4 ] = 0.2266 Bilangan pekerja = 8 0.2266 Number of wokers  35 A E D C B F G 4 4 2x x 6. (a) → AB = → AD + → DB = i ~ + 2 j ~ + 2 i ~ – j ~ = 3 i ~ + j ~ → AC = → AD + → DC = i ~ + 2 j ~ + 8 i ~ + j ~ = 9 i ~ + 3 j ~ (b) → AC = 3 → AB A, B dan C segaris A, B and C collinear AB : BC = 2 : 1 (c) ∆ADC = 1 2 ACh = 24 ∴∆ADB = 2 3 × 24 = 16 unit2 7. (a) y = a x + b x y x = ax + b (b) x 1 2 3 4 5 y x 5.9 7.9 10.0 12 13.9 (c) (i) b = 4 a = 14 – 4 5 = 2 (ii) Apabila/When x = 2.5 y x = 9 y = 9 2.5 = 5.69 Bahagian B/Section B 8. (a) 7 tan x = 4 tan (45° – x) = 4[1 – tan x] 1 + tan x A h B D C 2 1 2 1 0 2 3 4 5 4 6 8 10 12 14 y x

J56 7 tan x + 7 tan2 x = 4 – 4 tan x 7 tan2 x + 11 tan x – 4 = 0 tan x = –11 ± 112 – 4(7)(–4) 14 tan x = –11 + 233 14 x = 16° 56’, 196°56’ tan x = –11 – 233 14 x = 118°4’, 298°4’ (b) (i) 4 penyelesaian/solutions (ii) 4 penyelesaian/solutions (iii)sin 2x = 1 + kos/ cos x 2 penyelesaian/solutions 9. (a) (i) 5x = 2 – y y = –5x + 2 Titik tengah PR = (2, 3) ∴Persamaan QS / Equation QS y – 3= –5(x – 2) y= –5x + 13 ...➀ (ii) mPR = 5 – 1 –1 – 5 = 4 –6 = 2 –3 Persamaan QR,/ Equation QS, y – 1= 3 2 (x – 5) y= 3 2 x – 15 2 + 1 y= 3 2 x – 13 2 ...➁ (iii)(a) ➀ = ➁ –5x + 13 = 3 2 x – 13 2 –10x + 26 = 3x – 13 39 = 13x x = 3 y = –2 ∴Q(3, –2) 3 + x 2 = 2 –2 + y 2 = 3 x = 1 y = 8 S = (1, 8) (b) luas = 1 2  –1 3 5 1 –1 5 –2 1 8 5  area = 1 2 |2 + 3 + 40 + 5 – 15 + 10 –1 + 8| = 1 2 |52| = 26 unit2 y x 16°56′ 16°56′ y x 61°56′ 61°56′ y y = 1 + kos x y = sin 2x y = kos x y = 0.5 x 2 1 0 fi 2fi –1 (b) PAQ = 90° mPA = y – 5 x + 1 , mQA = y + 2 x – 3 ( y – 5 x + 1 )( y + 2 x – 3 ) = –1 y2 – 3y – 10 = –(x2 – 2x – 3) y2 + x2 – 2x – 3y – 13 = 0 Jika/If x = 5 y2 + 25 – 10 – 3y – 13= 0 y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y = 1 ; 2 (5, 1) ialah R 10. (a) (i) S2n = S3n – S2n 2S2n = S3n 2[2n 2 (2a + (2n – 1)d] = 3n 2 [2a + (3n – 1)d] a = 12, d = 3 4n(24 + 6n – 3) = 3n[24 + 9n – 3] 84 + 24n = 63 + 27n 21 = 3n n = 7 (ii) Tn = a + (n – 1)d = 48 12 + 3(n – 1) = 48 (n – 1) = 36 3 = 12 n = 13 (b) ar – a = 20 ar3 – ar = 15 a(r – 1) = 20...➀ ar(r2 – 1) = 15...➁ ➁ ÷ ➀ r(r + 1) = 15 20 4r2 + 4r = 3 4r2 + 4r – 3 = 0 (2r + 3)(2r – 1) = 0 r = – 3 2 ; 1 2 ∴a[– 3 2 –1] = 20 atau a( 1 2 – 1) = 20 a = –8 a = –40 11. (a) COB = 2π – 0.8π – π 3 = 13π 15 = 156° BC2 = 82 + 82 – 2(8)(8) kos 156° BC = 15.65 cm (b) Perimeter = 6 5 π(8) + (8) + 8 + π 3 (8) = 54.54 cm (c) Luas = 1 2 (8)2  13π 15 2 – 1 2 (8)2  π 3 2 Area = 323 13π 15 – π 3 4 = 256π 15 = 17 π 15 cm2 A 60° 0.8fi O B C

J57 Bahagian C/Section C 12. (a) v = 2(3t 2 – 2t – 1) dv dt = a = 12t – 4 Apabila/When t = 0 a = –4 m s–2 (b) Apabila/When dv dt = 0 t = 1 3 ∴v = 2[31 1 3  2 – 21 1 3  – 1] = 2 –2 3 m s–1 (c) v = 0 3t2 – 2t – 1 = 0 (3t + 1)(t – 1) = 0 t = 1 (d) jumlah jarak/total distance = ∫ 1 0 (6t 2 – 4t – 2)dt + ∫ 2 1 (6t2 – 4t – 2)dt =  (2t 3 – 2t2 – 2t] 1 0 + 32t 3 – 2t 2 – 2t 4 2 1 = |–2| + 6 = 8 m 13. (a) 25x + 50y ≤ 50 000 x + 2y ≤ 2 000 ... ➀ 150x + 50y ≤ 75 000 6x + 2y ≤ 3 000 ... ➁ 3x + y ≤ 1 500 y = 2x (b) (c) (i) Gred/ Grade A = 300 , Gred/ Grade B = 600 (ii) p = 4x + 2y 2y = −4x + p y = −2x + p 2 p = 4(200) + 2(900) = 800 + 1 800 = RM2 600 14. (a) P20 P18 × 100 = n 24 20 × 100 = n = 120 (b) 3(120) + 4(125) + 130 m + 110 (8 + m) = 125 970 + 130 m = 1000 + 125 m 5 m = 30 m = 6 (c) 110 100 × 120 = 132 110 100 × 130 = 143 I = 132 × 3 + 125 × 4 + 143 × 6 + 110 14 = 133.14 (d) P21 P18 × 100 = 143 P18 = 650 × 100 143 = RM454.55 15. (a)AM2 = 52 + 42 AM = 6.4 cm (b) AR2 = 62 + PR2 = 36 + 102 + 82 AR = 14.14 cm (c) MR2 = 42 + 62 + 52 MR = 8.77 cm 6.42 = 77 + 200 – 2 77 × 200 kos ARM kos ARM = 77 + 200 – 41 cos ARM 2 77 × 200 ARM = 18°2’ (d) MP2 = 36 + 41 MP = 77 = 8.77 PR = 100 + 64 = 12.81 S = 12.81 + 8.77 + 8.77 2 = 15.17 ∆PMR = 15.17(15.17 − 8.77)2 (15.17 − 12.81) = 38.29 cm2 A M 6 cm 5 cm 10 cm 8 cm P Q R S fi D C B 1000 y x 0 R 1000 3x + y = 1500 2000 2000 y = 2x (200, 900) (300, 600) x + 2y = 2000

Matematik Tambahan Tingkatan 5 Lembaran PBD B1 LEMBARAN PBD PBD 1 BAB 1: SUKATAN MEMBULAT Tahap Penguasaan Tafsiran 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian masalah rutin yang kompleks 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian masalah bukan rutin secara kreatif 1. Selesaikan setiap yang berikut. Solve each of the following. TP 5 Bentuk PQRSOP yang ditunjukkan dalam rajah terdiri daripada sebuah sektor bulatan ROS dengan pusat O menyambung dengan satu sektor POQ daripada satu bulatan berpusat O. Diberi bahawa panjang lengkok RS = 4.5 cm, ∠ROS = 0.55 radian dan POS ialah satu garis lurus dengan panjang 13 cm. Cari The shape of PQRSOP as shown in the diagram consists of a sector ROS of a circle with centre O joined to a sector POQ of a different circle also with centre O. Given that the arc length RS = 4.5 cm, ∠ROS = 0.55 radians and POS is a straight line of length 13 cm. Find (a) panjang OQ the length of OQ. (b) perimeter bagi PQRSOP. the perimeter of PQRSOP. (c) luas rantau berlorek itu. the area of the shaded region. P O S R Q

Matematik Tambahan Tingkatan 5 Lembaran PBD B2 LEMBARAN PBD PBD 2 2. Selesaikan setiap masalah yang berikut. Solve each of the following problems. TP 6 (a) Rajah menunjukkan dua sektor, POS daripada bulatan berpusat O dengan jejari 18 cm dan PRT dengan pusat R dan ORT ialah garis lurus. Diberi bahawa ∠POR ialah 1.5 radian dan OR = 8 cm. Cari The diagram shows two sectors, POS of a circle with centre O and radius 18 cm and PRT with centre R and ORT is a straight line. Given that ∠POR is 1.5 radians and OR = 8 cm. Find (i) jejari sektor PRS the radius of the sector PRS. (ii) perimeter rantau berlorek. the perimeter of the shaded region. (iii)luas rantau berlorek itu. the area of the shaded region. O S T P R

Matematik Tambahan Tingkatan 5 Lembaran PBD B3 LEMBARAN PBD PBD 3 (b) Rajah menunjukkan sebuah kotak kek yang digunakan oleh sebuah kedai. Kotak berbentuk prisma tegak dengan tinggi h cm. Luas keratan rentas adalah berbentuk sebuah sektor bulatan dengan jejari r cm bercangkum pada sudut 1.2 radian. Isi padu kotak ialah 216 cm3 . The diagram shows a cake box which is used by a shop. The box is a right prism of height h cm. The cross-sectional area is a sector of a circle with radius r cm and subtended at an angle of 1.2 radians. The volume of the box is 216 cm3. (i) Tunjukkan bahawa luas permukaan kotak, S cm2 diberi oleh 1.2r2 + 1152 r . Show that the surface area of the box, S cm2 is given by 1.2r2 + 1152 r . (ii) Sekeping kek dengan keratan rentas ialah satu sektor bulatan dengan panjang lengkok 6.5 cm dan jejari 5 cm akan dimasuk ke dalam kotak. Bolehkah kek ini dapat dimasukkan ke dalam kotak jika tinggi kotak ialah 10 cm? Tunjukkan langkah pengiraan anda. A piece of cake whose cross section is a sector of a circle with arc length 6.5 cm and radius 5 cm is to be put inside the box. Can this cake be put in the box if the height of the box is 10 cm? Show your working steps. r h 1.2

Matematik Tambahan Tingkatan 5 Lembaran PBD B4 LEMBARAN PBD PBD 4 BAB 2: PEMBEZAAN Tahap Penguasaan Tafsiran 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah rutin yang mudah. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. Selesaikan setiap yang berikut. Solve each of the following. TP 4 (a) Nilaikan had had x – 4 x – 16 . x → 16 Evaluate the limit lim x – 4 x – 16 . x → 16 (b) Cari pemalar a dengan keadaan had x2 – a2 x + a = −6 x → –4 Find the constant a such that lim x2 – a2 x + a = −6 x → –4 2. Selesaikan setiap masalah yang berikut. Solve each of the following problems. TP 5 (a) Sebuah tangga dengan panjang 12 m bersandar pada dinding mencancang. Hujung atas menggelongsor ke bawah dinding manakala hujung bawah tangga menggelongsor jauh dari dinding sepanjang tanah dengan kadar 0.5 m min−1. Apakah kadar hujung atas tangga menurun apabila jarak kaki tangga pada tanah mengufuk ialah 4 m dari dinding? A ladder of 12 m length leans against a vertical wall. Its top slides down the wall while the bottom moves away from the wall along the level ground at a rate of 0.5 m min−1. At what rate is the top of the ladder falling when the foot of the ladder on the ground is 4 m from the wall?

Matematik Tambahan Tingkatan 5 Lembaran PBD B5 LEMBARAN PBD PBD 5 (b) Dua buah keretapi meninggalkan stesen pada masa yang sama. Satu ke arah utara di atas landasan dengan halaju 60 km j−1. Satu lagi ke arah timur di atas landasan dengan halaju 45 km j−1. Berapakah laju kedua-dua keretapi itu bergerak antara satu sama lain, dalam km j−1 ketika keretapi ke utara itu ialah 120 km dari stesen. Two trains leave a station at the same time. One travels north on the track at a velocity of 60 km j−1. The other train travels east on the track at a velocity of 45 km h−1. How fast are they travelling away from each other, in km h−1 when the northbound train is 120 km from the station. (c) Jika f(x) = ax2 + bx + c. Diberi bahawa f(2) = 26, f’ (2) = 23 dan f’’(2) = 14. Cari If f(x) = ax2 + bx + c. Given that f(2) = 26, f' (2) = 23 and f ''(2) = 14. Find (i) a, b dan c. a, b and c. (ii) f(−1) 3. Selesaikan setiap masalah yang berikut. Solve each of the following problems. TP 6 (a) Sebuah bekas berbentuk silinder mempunyai isi padu 28π m3 . Bahan untuk membuat penutup bekas berharga RM5 meter persegi dan bahan digunakan membuat sekeliling dan tapak berharga RM2 setiap meter persegi. Apakah dimensi bekas supaya harga kos bekas adalah minimum? A storage container of a right circular cylinder has a volume of 28π m3. The material that is used to make the cover of the container costs RM5 per square and the material that is used to make the surrounding and the base costs RM2 for every square metre. What is the dimension of the container so that the cost of the container is minimum?

Matematik Tambahan Tingkatan 5 Lembaran PBD B6 LEMBARAN PBD PBD 6 (b) Diberi y = 4 x , cari Given y = 4 x , find (i) perubahan hampir dalam y apabila x menokok dari 4 ke 4.05. the approximate change in y when x increases from 4 to 4.05 (ii) peratus perubahan dalam y, dalam sebutan p jika x berubah dengan 2p%. the percentage change in y, in terms of p if x changes by 2p%. BAB 3: PENGAMIRAN Tahap Penguasaan Tafsiran 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengamiran dalam konteks penyelesaian masalah rutin yang mudah. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengamiran dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengamiran dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. Selesaikan setiap yang berikut. Solve each of the following. TP 4 Cari pengamiran tak tentu bagi setiap yang berikut. Find the indefinite integral for each of the following. (a) ∫x(2 − x)(x − 1)dx (b) ∫ 4 (2 − 2x)3dx

Matematik Tambahan Tingkatan 5 Lembaran PBD B7 LEMBARAN PBD PBD 7 2. Selesaikan setiap masalah yang berikut. Solve each of the following problems. TP 5 Fungsi kecerunan lengkung ialah (4x – 2) dan lengkung itu melalui titik A(−1, 3). Cari The gradient function of the curve is (4x − 2) and the curve passes through A(−1, 3). Find (a) persamaan normal pada A. the equation of the normal at A. (b) persamaan lengkung itu. the equation of the curve. 3. Selesaikan setiap masalah yang berikut. Solve each of the following problems. TP 6 (a) Diberi ∫ 6 –3 g(x)dx = 14 dan ∫ 4 0 g(x)dx = −4, cari Given ∫ 6 –3 g(x)dx = 14 and ∫ 4 0 g(x)dx = –4, find (i) ∫ 4 −3 g(x)dx jika/if ∫ 0 –3 g(x)dx = ∫ 6 4 2g(x)dx (ii) ∫ 0 −3 2g(x)dx jika/if ∫ 0 4 g(x)dx = ∫ 6 4 g(x)dx (iii)∫ 0 −3 g(x)dx jika/if ∫ 6 0 g(x)dx = 2

Matematik Tambahan Tingkatan 5 Lembaran PBD B8 LEMBARAN PBD PBD 8 (b) Rajah menunjukkan dua lengkung, x = 3 + y2 dan x = 2 − y2 . The diagram shows two curves, x = 3 + y2 and x = 2 − y2. Cari/ Find (i) luas dibatasi oleh dua lengkung, y = 1 dan y = −2. the area bounded by the two curves, y = 1 and y = −2. (ii) isi padu dijanakan apabila luas dibatasi oleh lengkung x = 3 − y2 dan garis x = 5 diputarkan melalui 180° pada paksi-x dalam sebutan π. the volume generated when area enclosed by the curve x = 3 − y2 and the line x = 5 is rotated through 180° about the x-axis in terms of π. (c) Rajah menunjukkan lengkung x = (y − 2)2 . Cari The diagram shows the curve x = (y − 2)2. Find (i) luas dibatasi oleh lengkung dan garis lurus y = 1 – 2 x + 2. the area bounded by the curve and the line y = 1 – 2 x + 2. (ii) Isi padu dijanakan oleh rantau yang dibatasi oleh lengkung, paksi-x dan paksi-y apabila diputarkan lengkap pada paksi-y dalam sebutan π. the volume generated by the region bounded by the curve , the x-axis and the y-axis when it is rotated completely around the y-axis in terms of π. x = 3 + y2 x = 2 – y2 2 1 -1 0 1 3 4 5 x y 1 2 3 0 2 4 6 x y

Matematik Tambahan Tingkatan 5 Lembaran PBD B9 LEMBARAN PBD PBD 9 BAB 4: PILIH ATUR DAN GABUNGAN Tahap Penguasaan Tafsiran 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. Selesaikan setiap yang berikut. Solve each of the following. TP 5 (a) Emma mempunyai 4 baju dan 5 seluar. Emma has 4 blouses and 5 skirts. (i) Berapa cara dia boleh memakai baju dan seluarnya jika tidak ada halangan. How many ways can she wears a blouse to a skirt if there is no restriction. (ii) Jika 2 daripada 4 baju ialah hitam dan 2 daripada 5 pasang seluarnya juga hitam. Berapa cara dia boleh memakai baju dan seluar itu jika dia tidak memakai baju hitam dengan seluar hitam pada masa yang sama. If 2 of the 4 blouses are black and 2 of the 5 skirts are black. How many ways can she wear blouse and skirt if she cannot wear her black blouse with her black skirt at the same time. (b) Rajah menunjukkan tiga biji guli biru yang sama dan tiga biji guli yang berlainan warna. Cari bilangan cara menyusun semua guli pada satu baris jika The diagram shows three identical blue marbles and three different coloured marbles. Find the number of ways to arrange all the marbles in a row if (i) tidak ada halangan./there is no restriction. (ii) tiga biji guli biru yang sama mest diasingkan. three identical blue marbles must be separated (iii) mesti bermula dengan guli merah dan berakhir dengan guli kuning. It must start with the red marble and ends with a yellow marble. (iv) Berapa bilangan cara untuk memilih empat guli dengan warna berlainan. How many ways to choose four different colour marbles.

Matematik Tambahan Tingkatan 5 Lembaran PBD B10 LEMBARAN PBD PBD 10 2. Selesaikan setiap yang berikut. Solve each of the following. TP 6 (a) 10 kawan dalam satu kumpulan dijemput ke suatu jamuan hari jadi dan duduk pada satu meja bulat. Berapa cara untuk menyusun semua jika 10 friends of a group are invited to a birthday party and to be seated at a round table. How many ways to arrange all of them if (i) tiga kawan tertentu mesti duduk sebelah menyebelah. three particular friends must sit next to each other. (ii) dua orang itu tidak boleh duduk bersama. two of them cannot sit next to each other. (iii) Berapa cara untuk menyusun mereka jika meja itu hanya boleh muat 8 orang sahaja. How many ways to arrange them if the table can only accommodate 8 of them. (b) Suatu kod empat digit akan dibentuk daripada digit-digit dari 0, 1, 3, 4, 5, 7 dan 8. Cari bilangan cara membentuk kod jika A four digit code is to be formed from the digits 0, 1, 3, 4, 5, 7 and 8. Find the number of ways to form the codes if (i) kod tidak boleh bermula dengan 0 dan ulangan tidak dibenarkan. the code cannot start with 0 and repetitions is not allowed. (ii) kod mengandungi digit ganjil sahaja dan ulangan tidak dibenarkan. the code contains only odd digits and repetitions is not allowed. (iii) 0 digunakan sekali sahaja manakala digit lain boleh diulang. 0 is used once only and the rest of the digits can be repeated. (iv) nilai kod lebih daripada 3 000 dan ulangan tidak dibenarkan. the value of the code is more than 3 000 and repetitions is not allowed.

Matematik Tambahan Tingkatan 5 Lembaran PBD B11 LEMBARAN PBD PBD 11 (c) Sebuah bakul mengandungi 4 biji merah, 3 biji biru, 4 biji hijau dan 3 biji kuning guli dengan setiap guli ditulis dengan nombor berlainan. Berapa cara memilih 4 biji guli jika A basket contains 4 red, 3 blue, 4 green and 3 yellow marbles each with a different number written on it. How many ways to choose 4 marbles if (i) semua mempunyai warna berlainan. all have different colours. (ii) sekurang-kurangnya 2 ialah merah. at least 2 are red. (iii)tiada guli merah dipilih tetapi sekurang-kurangnya satu guli ialah hijau. there is no red marble chosen but at least one marble is green. (d) (i) Pada suatu jamuan graduan, setiap graduan bersalam antara satu sama lain. Terdapat 253 salam. Berapa graduan yang hadir dalam jamuan itu? At a graduation party, every graduate shakes hands with each other. There are 253 handshakes. How many graduates are present in the party? (ii) Dari bahagian (i), jika 3 daripada graduan tidak bersalam antara satu sama lain tetapi bersalam dengan semua yang lain, berapa salam dibuat? From part (i), if 3 of them did not shake hands with each other but shake hands with the rest of them, how many handshakes are made?

Matematik Tambahan Tingkatan 5 Lembaran PBD B12 LEMBARAN PBD PBD 12 (e) Terdapat n orang pengakap lelaki dan m orang pengakap perempuan. Jika dua orang lelaki dan dua orang perempuan dipilih, bilangan gabungan ialah 160 dan jumlah orang pengakap ialah 26. Hitung nilai n dan m jika n > m. There are n boys scouts and m girls scouts. If two boy scouts and two girl scouts are chosen, the number of combinations is 160 and the total number of scouts is 26. Calculate the value of n and of m if n > m. BAB 5: TABURAN KEBARANGKALIAN Tahap Penguasaan Tafsiran 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang taburan kebarangkalian dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang taburan kebarangkalian dalam konteks penyelesaian masalah bukan rutin secara kreatif. Selesaikan setiap yang berikut. Solve each of the following. TP 5 1. (a) X ialah pemboleh ubah rawak diskret dengan taburan kebarangkalian berikut X is a discrete random variable with the following probability distribution. X = r 3 5 7 9 P(X = r) 0.3 0.2 p q Diberi bahawa P(X = 3) = P(X = 7), cari It is given that P(X = 3) = P(X = 7), cari (i) nilai p dan nilai q the value of p and of q. (ii) P(X ≥ 5). (b) Seterusnya, lukis graf taburan kebarangkalian itu. Then, draw a probability distribution graph

Matematik Tambahan Tingkatan 5 Lembaran PBD B13 LEMBARAN PBD PBD 13 2. Didapati bahawa 3 daripada 100 pesakit mengalami kesan sampingan daripada suatu ubat tertentu. Jika 10 pesakit dipilih secara rawak dari kumpulan pesakit itu, It is found that 3 out of 100 patients show side effects from a certain drug. If 10 patients are chosen at random from the group of patients, (a) cari kebarangkalian bahawa find the probability that (i) tepat separuhnya menunjukkan kesan sampingan daripada ubat tertentu itu. exactly half of them show the side effects of the drug. (ii) sekurang-kurangnya 2 menunjukkan kesan sampingan. at least 2 of them show the side effects. (b) Jika 2 500 pesakit menggunakan ubat tersebut, cari min dan sisihan piawai bilangan pesakit yang menunjukkan kesan sampingan. If there are 2 500 patients using the drug, find the mean and standard deviation of the number of patients who show side effects.

Matematik Tambahan Tingkatan 5 Lembaran PBD B14 LEMBARAN PBD PBD 14 Selesaikan setiap yang berikut. Solve each of the following. TP 6 3. Satu pemboleh ubah diskret X mempunyai fungi kebarangkalian P(X = x) = k(1 − x)2 untuk X = {−1, 0, 1, 2}. A discrete random variable X has the probability function P(X = x) = k(1 − x)2 for X = {−1, 0, 1, 2}. (a) Tunjukkan bahawa k = 1 6 . Show that k = 1 6 . (b) Cari P(X < 0) atau P(X > 1). Find P(X < 0) or P(X > 1). (c) Lukis graf taburan kebarangkalian itu. Draw the probability distribution graph. 4. Dalam sebuah sekolah, 60% daripada pelajar mempunyai akses internet di rumah mereka. In a school, 60% of the students have internet access at home. (a) Satu kumpulan 8 orang pelajar dipilih secara rawak. Cari kebarangkalian bahawa A group of 8 students are chosen at random. Find the probability that (i) 3 daripadanya tidak ada akses internet di rumah. 3 of them do not have internet access at home. (ii) sekurang-kurangnya 5 daripada mereka ada akses internet di rumah. at least 5 of them have internet access at home. (b) Cari bilangan min pelajar yang mempunyai akses internet jika 200 orang pelajar dipilih secara rawak. Find the mean of the number of students that have internet access if 200 students are chosen at random. (c) Cari bilangan pelajar yang dipilih jika kebarangkalian bahawa semua ada akses internet di rumah ialah 0.07776. Find the number of students chosen if the probability that all have internet access at home is 0.07776. 5

Matematik Tambahan Tingkatan 5 Lembaran PBD B15 LEMBARAN PBD PBD 15 5. Satu set data X mempunyai min 45 dan sisihan piawai 8.3 bertaburan normal. A set of data X has a mean of 45 and standard deviation of 8.3 is normally distributed. (a) Cari nilai X jika diberi jaraknya dari min ialah Find the value of X given its distance from the mean is (i) +1 sisihan piawai dari min. +1 standard deviation from the mean. (ii) −2 sisihan piawai dari min. −2 standard deviation from the mean. (b) Cari kebarangkalian bagi data yang dipilih secara rawak adalah lebih daripada 40. Find the probability of the data chosen at random is greater than 40. (c) Cari nilai n jika kebarangkalian data yang dipilih secara rawak lebih daripada n ialah 0.3589. Find the value of n if the probability of the data chosen at random greater than n is 0.3589. 6. Tinggi sekumpulan perempuan bertaburan normal dengan min 142 cm. Jika 95% orang perempuan mempunyai tinggi di antara 139 cm dan 145 cm, hitung The heights of a group of girls are normally distributed with mean 142 cm. If 95% of the girls have heights between 139 cm and 145 cm, calculate (a) sisihan piawai bagi kumpulan perempuan ini. the standard deviation of this group of girls. (b) kebarangkalian seorang perempuan dipilih secara rawak daripada kumpulan ini mempunyai tinggi kurang daripada 141 cm. the probability that a girl chosen at random from this group has a height less than 141 cm. (c) bilangan orang perempuan dalam kumpulan ini jika 25 orang mempunyai tinggi lebih daripada 140.5 cm. the number of girls in this group if 25 of the girls has heights more than 140.5 cm. 5

Matematik Tambahan Tingkatan 5 Lembaran PBD B16 LEMBARAN PBD PBD 16 BAB 6: FUNGSI TRIGONOMETRI Tahap Penguasaan Tafsiran 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi trigonometri dalam konteks penyelesaian masalah rutin yang mudah. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi trigonometri dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi trigonometri dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. Selesaikan setiap yang berikut. Solve each of the following. TP 4 (a) Jika tan ( π 3 − A) = kot π 3 , cari A untuk 0 ≤ A ≤ 2π. If tan ( π 3 − A) = cot π 3 , find A for 0 ≤ A ≤ 2π. (b) Buktikan bahawa kot A + tan A = sek A kosek A. Prove that cot A + tan A = sec A cosec A. Selesaikan setiap yang berikut. Solve each of the following. TP 5 2. (a) Buktikan bahawa kos A 1 – tan A + sin A 1 – kot A = sin A + kos A. Prove that cos A 1 – tan A + sin A 1 – cot A = sin A + cos A (b) (i) Selesaikan persamaan 4 sin2 x + 8 kos x − 7 = 0 untuk 0° ≤ x ≤ 360°. Solve the equation 4 sin2 x + 8 cos x − 7 = 0 for 0° ≤ x ≤ 360°. (ii) Seterusnya, cari penyelesaian bagi persamaan 4 sin2 θ 2 + 8 kos θ 2 − 7 = 0 untuk 0° ≤ x ≤ 360°. Hence, find the solution of the equation 4 sin2 θ 2 + 8 cos θ 2 − 7 = 0 for 0° ≤ x ≤ 360°. 6

Matematik Tambahan Tingkatan 5 Lembaran PBD B17 LEMBARAN PBD PBD 17 Selesaikan setiap yang berikut. Solve each of the following. TP 6 3. (a) Tunjukkan bahawa persamaan 2 sin x = 4 kos x − 1 tan x boleh diungkapkan dalam bentuk 6 kos2 x − kos x − 2 = 0. Show that the equation 2 sin x = 4 cos x − 1 tan x can be expressed in the form of 6 cos2 x − cos x − 2 = 0. (b) Seterusnya, selesaikan persamaan 2 sin x = 4 kos x − 1 tan x untuk 0° ≤ x ≤ 360°. Hence, solve the equation 2 sin x = 4 cos x − 1 tan x for 0°≤ x ≤ 360°. (c) Lakar graf y = 4 kos x − 1 untuk 0 ≤ x ≤ 2π. Seterusnya, cari bilangan penyelesaian bagi persamaan 2π sin x tan x + x = 2π untuk 0 ≤ x ≤ 2π. Sketch the graph y = 4 cos x − 1 for 0 ≤ x ≤ 2π. Hence, find the number of solution for the equation 2π sin x tan x + x = 2π for 0 ≤ x ≤ 2π.

Matematik Tambahan Tingkatan 5 Lembaran PBD B18 LEMBARAN PBD PBD 18 4. (a) Tunjukkan bahawa persamaan tan 2x = 5 sin 2x boleh ditulis dalam bentuk (1 − 5 kos 2x) sin 2x = 0. Show that the equation tan 2x = 5 sin 2x can be written in the form (1 − 5 cos 2x) sin 2x = 0. (b) Seterusnya, selesaikan persamaan tan 2x = 5 sin 2x untuk 0° ≤ x ≤ 360°. Hence, solve the equation tan 2x = 5 sin 2x for 0° ≤ x ≤ 360°. (c) Lakar graf y = tan 2x untuk 0 ≤ x ≤ π. Seterusnya, pada paksi sama, lakarkan satu graf sesuai untuk menyelesaikan persamaan 5x sin 2x = 1 dan nyatakan bilangan penyelesaian untuk 0 ≤ x ≤ π. Sketch the graph y = tan 2x for 0 ≤ x ≤ π. Hence, on the same axes, draw a suitable graph to solve the equation 5x sin 2x = 1 and state the number of solutions for 0 ≤ x ≤ π.

Matematik Tambahan Tingkatan 5 Lembaran PBD B19 LEMBARAN PBD PBD 19 5. (a) Selesaikan persamaan 2 sin2 x + 2 = 7 kos x untuk 0 ≤ x ≤ 2π. Beri jawapan dalam radian. Solve the equation 2 sin2 x + 2 = 7 cos x for 0 ≤ x ≤ 2π. Give your answer in radians. (b) Rajah menunjukkan sebahagian graf y = 3 sin bx. Garis lurus y = 1.4 menyilang lengkung pada A dan B. Cari The diagram shows part of the graph y = 3 sin bx. The line y = 1.4 cuts the curve at A and B. Find (i) nilai a dan nilai b. the value of a and of b. (ii) koordinat A dan B dalam radian. the coordinates of A and B in radians. y y = 1.4 a 0 –a x π

Matematik Tambahan Tingkatan 5 Lembaran PBD B20 LEMBARAN PBD PBD 20 BAB 7: PENGATURCARAAN LINEAR Tahap Penguasaan Tafsiran 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengaturcaraan linear dalam konteks penyelesaian masalah bukan rutin secara kreatif. Selesaikan setiap yang berikut. Solve each of the following problems. TP 6 1. Sebuah syarikat pengeluaran coklat menghasilkan dua jenis coklat, A dan B. Amaun tepung susu dan tepung koko digunakan dijadualkan di bawah. A chocolate manufacturing company produces two types of chocolates, A and B. The amount of milk and cocoa powder needed are shown in the table below. Jenis coklat Type of chocolate Tepung susu (unit) Milk powder (units) Tepung koko Cocoa powder A 1 3 B 1 2 Syarikat mempunyai sejumlah 5 unit tepung susu dan 12 unit tepung koko. The company has a total of 5 units of milk powder and 12 units of cocoa powder. (a) Tulis semua ketaksamaan yang memuaskan semua kekangan di atas, selain daripada x ≥ 0 dan y ≥ 0. Write all the inequalities that satisfy the above constraints, other than x ≥ 0 and y ≥ 0. (b) Pada skala yang sesuai, lukis dan lorek rantau R yang memuaskan semua kekangan tersebut. On a suitable scale, draw and shade the region R which satisfies the above constraints. (c) Gunakan graf / Using graph (i) cari keuntungan maksimum syarikat sehari jika setiap unit coklat A dan B memberi untung masingmasing sebanyak RM30 dan RM50. find the maximum profit per day of the company if each unit of chocolate A and B gives a profit of RM30 and RM50 respectively. (ii) julat bilangan coklat B dihasilkan jika syarikat ingin menghasilkan 1 unit coklat A. the range of the number of chocolate B that is produced if the company wants to produce 1 unit of chocolate A.

Matematik Tambahan Tingkatan 5 Lembaran PBD B21 LEMBARAN PBD PBD 21 2. Seorang tukang kayu membuat meja dan kerusi. Setiap meja boleh dijual dengan keuntungan RM30 manakala sebuah kerusi memberi untung RM10. Tukang kayu menggunakan 36 jam seminggu bekerja dan sebuah meja memerlukan 6 jam dan sebuah kerusi memerlukan 3 jam. Tukang kayu dikehendaki membuat sekurang-kurangnya empat kali lebih banyak kerusi daripada meja ke atas keperluan pelanggan. Meja memerlukan selebih-lebihnya empat kali lebih besar ruang simpanan daripada kerusi dan ruang yang ada hanya boleh muat selebih-lebihnya 4 buah meja seminggu. A carpenter makes tables and chairs. Each table can be sold for a profit of RM30 and each chair for a profit of RM10. The carpenter can afford to spend up to 36 hours per week working and takes 6 hours to make a table and 3 hours to make a chair. The carpenter must make at most four times as many chairs than tables from the demand of customers. Tables take up at most four times storage space as chairs and there is room for at most 4 tables a week. (a) Tulis semua ketaksamaan yang memuaskan semua kekangan di atas, selain daripada x ≥ 0 dan y ≥ 0. Write all inequalities that satisfy the above constraints, other than x ≥ 0 and y ≥ 0. (b) Pada skala yang sesuai, lukis dan lorek rantau R yang memuaskan semua kekangan tersebut. On a suitable scale, draw and shade the region R which satisfies the above constraints. (c) Gunakan graf, cari keuntungan maksimum syarikat seminggu. Use the graph, find the maximum profit of the company per week.

Matematik Tambahan Tingkatan 5 Lembaran PBD B22 LEMBARAN PBD PBD 22 BAB 8: KINEMATIK GERAKAN LINEAR Tahap Penguasaan Tafsiran 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang kinematik gerakan linear dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang kinematik gerakan linear dalam konteks penyelesaian masalah bukan rutin secara kreatif. Selesaikan setiap yang berikut. Solve each of the following problems. TP 5 1. Sebuah kereta bergerak pada satu jalan lurus dari lampu isyarat O. Halajunya, v m s−1, diberi oleh 2t(6 − t), dengan t ialah masa, dalam saat, selepas melalui lampu isyarat. (Anggap pergerakan ke kanan ialah positif). Cari A car moves along a straight road from a traffic lights O. Its velocity, v m s−1, is given by 2t(6 − t), where t is the time, in seconds, after passing the traffic lights. (Assume motion to the right is positive). Find (a) halaju maksimum kereta. the maximum velocity of the car. (b) masa apabila kereta berhenti seketika. the time when the car stops instantaneously. (c) Lakar graf halaju-masa bagi 0 ≤ t ≤ 7. Sketch the velocity-time graph for 0 ≤ t ≤ 7. (d) jumlah jarak yang dilalui dalam 7 saat pertama. the total distance travelled in the first 7 seconds. 2. Suatu zarah bergerak pada satu garis lurus dengan sesaran, s m selepas t saat dari titik tetap O, diberi oleh s = t(t − 4)2 . Cari A particle moves in a straight line so that its displacement, s m after t seconds from a fixed point O is given by s = t(t − 4)2. Find (a) halaju apabila t = 3. the velocity when t = 3. (b) nilai-nilai t apabila zarah berhenti seketika. the values of t when the particle stops instantaneously. (c) pecutan selepas 4 saat. the acceleration after 4 seconds.

Matematik Tambahan Tingkatan 5 Lembaran PBD B23 LEMBARAN PBD PBD 23 Selesaikan setiap masalah yang berikut. Solve each of the following problems. TP 6 3. Rajah menunjukkan kedudukan dan arah pergerakan dua objek, A dan B pada satu garis lurus dan melalui titik tetap, P dan Q pada masa yang sama. Jarak di antara P dan Q ialah 30 m. Halaju bagi A, vA = 3 + 2t − t 2 , dengan keadaan t ialah masa, dalam saat, selepas melalui P manakala B bergerak dengan halaju tetap −5 m s−1. Objek A berhenti seketika pada titik R. Anggapkan pergerakan ke kanan ialah positif, cari The diagram shows the position and direction of motion of two objects, A and B on a straight line passing two fixed points, P and Q at the same time. The distance between P and Q is 30 m. The velocity of A, vA = 3 + 2t − t2, where t is the time, in seconds, after passing P while B moves with a constant velocity of −5 m s−1. Object A stops instantaneously at R. Assume motion to the right is positive, find A P 30 m Q B (a) halaju maksimum A. the maximum velocity of A. (b) jarak R dari P. the distance of R from P (c) jarak, dalam m, di antara A dan B apabila A berada di titik R. the distance, in m, between A and B when A is at the point R. (d) masa apabila kedua-dua objek mempunyai halaju yang sama. the time when the two objects have the same velocity.

Matematik Tambahan Tingkatan 5 Lembaran PBD B24 LEMBARAN PBD PBD 24 4. Sebiji batu dilontar tegak ke atas dari tanah dengan halaju awal 30 m s−1. Anggapkan pecutan graviti ialah −10 m s−2. Cari A stone is thrown vertically upwards from the ground with an initial velocity of 30 m s−1. Assuming that the gravitational acceleration is −10 m s−2. Find (a) (i) masa yang diambil untuk mencapai tinggi maksimum. the time taken to reach the maximum height. (ii) halaju apabila batu menghentam tanah. the velocity when the stone hits the ground. (b) Lakar graf kedudukan-masa dan graf halaju-masa bagi pergerakan tersebut. Sketch the displacement-time graph and the velocity-time graph for the motion. (c) Cari jumlah jarak yang dilalui oleh batu sehingga batu kembali ke tanah lagi. Find the total distance travelled by the stone until it returns to the ground.

J1 JAWAPAN LEMBARAN PBD BAB 1: SUKATAN MEMBULAT 1. (a) OS(0.55) = 4.5 OS = 8.18 cm ∴OQ = 13 – 8.18 = 4.82 cm (b) Panjang lengkok PQ Arc length PQ = 4.82 (π – 0.55) = 12.49 cm Perimeter = 12.49 + (8.18 – 4.82) + 4.5 + 13 = 33.35 cm (c) luas berlorek = 1 2 (4.82)2 (π – 0.55) shaded area = 30.10 cm2 2. (a) (i) PR2 = 182 + 82 – 2(18)(8) kos 1.5 π × 180°2 PR = 19.17 cm (ii) panjang lengkok PS = 18(1.5) arc length PS = 27 cm ...➀ sin ORP 18 = sin  1.5 π × 180°2 19.17 ORP = 69.49° ∴PRT = 180° – 69.49° = 110.51° panjang lengkok PT = 110.51° 360° × 2π(19.17) arc length PT = 36.97 cm ...➁ Perimeter = 27 + 36.97 + 9.17 = 73.14 cm (iii)luas POR = 1 2 (18)(8) sin 1.5 π × 180°2 area POR = 71.82 cm2 luas sektor POS = 1 2 (18)2 (1.5) area of sector POS = 243 cm2 Luas sektor PRT = 110.51° 360° × π(19.17)2 area of sector PRT = 354.4 cm2 Luas rantau berlorek Shaded area 354.4 – [243 – 71.82] = 183.22 cm2 (b) (i) Isipadu = 216 Volume 1 2 r2 (1.2)h = 216 r2 h = 360 ...➀ Luas permukaan S Surface area S = 1 2 r2 (1.2) × 2 + 2rh + (1.2r)h = r2 (1.2) + 3.2rh = 1.2r2 + 3.2r 360 r2 2 = 1.2r2 + 1152 r (ii) Jika/If h = 10 cm r2 (10) = 360 r = 6 cm sudut θ = 6.5 5 angle = 1.3 rad Maka, CAO = (π – 1.3) rad Hence CO sin (π – 1.3)180° 5 4 = 5 sin 1.2 × 180° π 4 CO = 5.17 cm Oleh sebab OC < 6 cm Maka, kek dapat disimpan dalam kotak. Since OC < 6 cm Hence, the cake can be put in the box. BAB 2: PEMBEZAAN 1. (a) had x – 4 x → 16 x – 16 = had ( x – 4)( x + 4) (x – 16)( x + 4) x → 16 = had [x – 16] x → 16 (x – 16)( x + 4) = 1 8 (b) had x2 – a2 (x + a) = –6 x → –4 had (x + a)(x – a) (a + x) = –6 x → –4 –4 – a = –6 a = 2 2. (a) dx dt = 0.5 m/min x2 + y2 = 122 x2 = 144 – y2 x = (144 – y2 ) 1 2 dx dy = 1 2 [144 – y2 ] – 1 2 [–2y] dx dy = –y 144 – y2 ∴ dx dt = –y 144 – y2 · dy dt P O S R 4.5 cm 0.55 rad Q 13 cm r h 1.2 1.2 5 cm 6 cm Kek 5 cm 6.5 cm B A O fi O S T P R 18 cm 8 1.5 y 12 x

J2 r h Jika x = 4 m y2 = 144 – 16 If y = 128 = 8 2 0.5 = –8 2 16 · dy dt 1 2 = –2 2 · dy dt dy dt = –1 –4 2 m min–1 (b) VA = 60t VB = 45t AB2 = (60t)2 + (45t)2 AB = 5625t 2 = 75t Apabila A ialah 120 km dari stesen. When A is 120 km from the station. 60t = 120 t = 2 jam d(AB) dt = 75 km/j (c) (i) f(x) = ax2 + bx + c f(2) = 4a + 2b + c = 26 ...➀ f’(x) = 2ax + b f’(2) = 4a + b = 23 ...➁ f’’(x) = 2a f’’(2) = 14 = 2a a = 7 b = 23 – 28 = –5 4(7) + 2(–5) + c = 26 28 – 10 + c = 26 c = 8 (ii) f(x) = 7x2 – 5x + 8 f(–1) = 7 + 5 + 8 = 20 3. (a) Kos C/Cost C = 5πr2 + 2πr2 + 2πrh(2) = 7πr2 + 4πrh Isipadu = 28π = πr2 h Volume ∴ h = 28 r2 ∴ C = 7πr2 + 4πr( 28 r2 ) = 7πr2 + 112π r dC dr = 14πr – 112π r2 = 0 r3 = 112π 14π = 8 r = 2 d2 C dr2 = 14π + 224π r3 Apabila/When r = 2 d2 C dr2 = 14π + 28π = 42π > 0 ∴ Harga kos ialah minimum apabila r = 2 cm. The cost is minimum when r = 2 cm. h = 28 4 = 7 cm A B 60 km j –1 60 km h–1 45 km j –1 45 km h–1 (b) (i) y = 4x–1 dy dx = –4 x2 , x = 4 δx = 0.05 δy = dy dx · δx = –4 x2 (0.05) = –4 16 (0.05) = –0.0125 (ii) δx x × 100% = 2p% δx x × 100% = –4δx yx2 × 100% = –4δx 4 x · x2 × 100% = –δx x × 100% = –2p% BAB 3: PENGAMIRAN 1. (a) ∫x(2 – x)(x – 1)dx = ∫(2x – x2 )(x – 1)dx = ∫(3x2 – 2x – x3 )dx = x3 – x2 – 1 4 x4 + c (b) ∫ 4 (2 – 2x)3 dx = ∫ 4 8(1 – x)3 dx = 1 2 ∫(1 – x)–3dx = 1 2  (1 – x)–2 –2(–1) 4 + c = 1 4(1 – x)2 + c 2. (a) dy dx = 4x – 2 Apabila x = –1, dy dx = –6 When Persamaan normal/Equation of normal y = 1 6 x + c 3 = 1 6 (–1) + c 19 6 = c ∴y = 1 6 x + 19 6 (b) y = ∫(4x – 2)dx y = 2x2 – 2x + c Pada A(–1, 3) 3 = 2 + 2 + c c = –1 ∴y = 2x2 – 2x – 1