J7 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan (b) 6 2 5 – 2 = 6 2 5 – 2 × (2 5 + 2 ) ( 2 5 + 2 ) = 6(2 5 + 2 ) (2 5 )2 – ( 2 )2 = 6(2 5 + 2 ) 20 – 2 = 6(2 5 + 2 ) 18 = 2 5 + 2 3 8 (a) loga 27a 16 = loga 27 + loga a – loga 16 = loga 33 + 1 – loga 42 = 3 loga 3 + 1 – 2 loga 4 = 3q + 1 – 2p (b) log81 u – log3 w = 1 2 log3 u log3 81 – log3 w = 1 2 log3 u 4 – log3 w = 1 2 log3 u – 4 log3 w = 2 log3 u – log3 w4 = 2 log3( u w4 ) = 2 u w4 = 32 u = 9w4 9 (a) 1 + log5 x = log5 (x + 7) 1 = log5 (x + 7) – log5 x 1 = log5 (x + 7) x 51 = x + 7 x 5x = x + 7 4x = 7 x = 7 4 (b) log2 √ x – log4 3 = 3 2 log2 √ x – log2 3 log2 22 = 3 2 log2 √ x – log2 3 2 = 3 2 2 log2 √ x – log2 3 = 3 log2 (√ x ) 2 – log2 3 = 3 log2 x – log2 3 = 3 log2 x 3 = 3 x 3 = 23 x = 24 10 (a) p = 3m ⇒ log3 p = m q = 3n ⇒ log3 q = n log3 ( pq5 81 ) = log3 p + log3 q5 – log3 81 = log3 p + log3 q5 – log3 34 = m + 5 log3 q – 4 log3 3 = m + 5n – 4(1) = m + 5n – 4 (b) log27 p – logq 9 = log3 p log3 27 – log3 9 log3 q = m log3 33 – log3 32 n = m 3 log3 3 – 2 log3 3 n = m 3 – 2 n 11 (a) log2 7.5 = log2 ( 15 2 ) = log2 15 – log2 2 = log2 (3 × 5) – log2 2 = log2 3 + log2 5 – log2 2 = 1.585 + 2.322 – 1 = 2.907 (b) log4 54 = log2 54 log2 4 = log2 (2 × 27) log2 22 = log2 2 + log2 33 2 log2 2 = 1 + 3 log2 3 2 = 1 + 3(1.585) 2 = 2.878 12 (a) log10 ( x2 + 21 x ) = 1 x2 + 21 x = 101 x2 + 21 = 10x x2 – 10x + 21 = 0 (x – 3)(x – 7) = 0 x – 3 = 0 atau/or x – 7 = 0 x = 3 x = 7 (b) 4log2 x = 6 log2 (4log2 x ) = log2 6 (log2 x)log2 4 = log2 6 log2 x (log2 22 ) = log2 6 (log2 x)(2) = log2 6 log2 x2 = log2 6 x2 = 6 x = 6 = 2.449 13 (a) v = 28(1.16)t = 28(1.16)12.5 = 179.01°C (b) 1 200 = 28(1.16)t 1.16t = 42.857 log 1.16t = log 42.857 t log 1.16 = log 42.857 t = log 42.857 log 1.16 = 25.32 saat/seconds BAHAGIAN B 14 (a) 1 logm mn + 1 logn mn = 1 ( logmn mn logmn m ) + 1 ( logmn mn logmn n ) = 1 ( 1 logmn m ) + 1 ( 1 logmn n ) = logmn m + logmn n = logmn mn = 1 (Tertunjuk/Shown) log6 (2x – 1) = 1 logm mn + 1 logn mn log6 (2x – 1) = 1 2x – 1 = 61 2x = 7 x = 7 2 (b) 52x – 5x + 1 = 50 52x – (5x )(51 ) – 50 = 0 52x – 5(5x ) – 50 = 0 (5x + 5)(5x – 10) = 0 5x + 5 = 0 atau/or 5x – 10 = 0 5x = –5 5x = 10 Apabila 5x = –5, x tidak tertakrif. When 5x = –5, x is not defi ned. Apabila/When 5x = 10, x = log 10 log 5 = 1.431 KERTAS 2 BAHAGIAN A 1 log3 (3x + 4) – 7 log9 x2 + 6 log3 x = log3 (3x + 4) – 14 log9 x + 6 log3 x = log3 (3x + 4) – 14( log3 x log3 9 )+ 6 log3 x = log3 (3x + 4) – 14( log3 x 2 ) + 6 log3 x = log3 (3x + 4) – 7 log3 x + 6 log3 x = log3 (3x + 4) – log3 x = log3 3x + 4 x Seterusnya/Hence, log3 (3x + 4) – 7 log9 x2 + 6 log3 x = 2 log3 3x + 4 x = 2 3x + 4 x = 32 3x + 4 x = 9 3x + 4 = 9x 4 = 6x x = 2 3 2 Diberi/Given log2 m3 p2 = 21 log2 m3 + log2 p2 = 21 3 log2 m + 2 log2 p = 21 1 Diberi/Given log2 m4 p = 17 log2 m4 – log2 p = 17 4 log2 m – log2 p = 17 2 Selesaikan 1 dan 2 , Solve 1 and 2 , 3 log2 m + 2 log2 p = 21 1 (+) 8 log2 m – 2 log2 p = 34 2 × 2 11 log2 m = 55 log2 m = 5 m = 25 m = 32 Daripada/From 1 , 3(5) + 2 log2 p = 21 15 + 2 log2 p = 21 2 log2 p = 6 log2 p = 3 p = 23 p = 8 Maka/Hence, m = 32, p = 8 BAHAGIAN B 3 (a) (9x )(3y ) = 1 1 8x 4y = 4 2 PT SPM Add Math Tkt 4-Jawapan.indd 7 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J8 Daripada/From 1 , (9x )(3y ) = 1 ((32 )x )(3y ) = 1 (32x )(3y ) = 1 32x + y = 30 2x + y = 0 3 Daripada/From 2 , 8x 4y = 4 (23 )x (22 )y = 22 23x 22y = 22 23x – 2y = 22 3x – 2y = 2 4 Daripada/From 3 , y = –2x Gantikan y ke dalam 4 , Substitute y into 4 , 3x – 2(–2x) = 2 3x + 4x = 2 7x = 2 x = 2 7 Daripada/From 3 , y = –2( 2 7 ) = – 4 7 ∴ x = 2 7 , y = – 4 7 (b) log2 x + log2 y = 3 1 log3 (3y – 1) – 2 log9 x = 1 2 Daripada/From 1 , log2 xy = 3 xy = 23 xy = 8 3 Daripada/From 2 , log3 (3y – 1) – 2 ( log3 x log3 9 ) = 1 log3 (3y – 1) – 2 ( log3 x 2 ) = 1 log3 (3y – 1) – log3 x = 1 log3 3y – 1 x = 1 3y – 1 x = 31 3y – 1 = 3x 4 Daripada/From 3 , x = 8 y Gantikan x ke dalam 4 , Substitute x into 4 , 3y – 1 = 3( 8 y ) 3y2 – y – 24 = 0 (3y + 8)(y – 3) = 0 3y + 8 = 0 atau/or y – 3 = 0 y = – 8 3 y = 3 Oleh sebab/Since y > 0, y = 3 Apabila/When y = 3, x = 8 (3) = 8 3 ∴ x = 8 3 , y = 3 ZON KBAT 1 log2 (13 + log3 X) = 4 13 + log3 X = 24 13 + log3 X = 16 log3 X = 3 X = 33 = 27 log5 (4X – 87 + log2 Y) = 2 4X – 87 + log2 Y = 52 4(27) – 87 + log2 Y = 25 log2 Y = 4 Y = 24 = 16 Maka/Then, X + Y = 27 + 16 = 43 BAB 5 Janjang KERTAS 1 BAHAGIAN A 1 (a) T2 – T1 = T3 – T2 18 – (12 – x) = 4x – 18 18 – 12 + x = 4x – 18 18 – 12 + 18 = 4x – x 24 = 3x x = 8 d = 4(8) – 18 = 14 (b) T1 = a = 12 – (8) = 4 T10 + T11 + T12 + ….. + T19 + T20 = S20 – S9 = 20 2 [2(4) + (20 – 1)(14)] – 9 2 [2(4) + (9 – 1)(14)] = 2 740 – 540 = 2 200 2 (a) S7 = 7 2 [5(7) + 3] = 7 2 (38) = 133 (b) T7 = S7 – S6 = 133 – 6 2 [5(6) + 3] = 133 – 99 = 34 3 (a) p = q – 3 q = p + 3 (b) q – 3 ⩾ 1 q ⩾ 4 4 (a) d = T5 – T4 = 20 – 14 = 6 (b) T4 = 14 T3 = 14 – 6 = 8 T2 = 8 – 6 = 2 T1 = a = 2 – 6 = –4 S35 = 35 2 [2(–4) + (35 – 1)(6)] = 3 430 5 (a) T3 = ar2 = 32 1 T3 + T4 = 16 (32) + ar3 = 16 ar3 = –16 (ar2 )r = –16 (32)r = –16 r = – 1 2 Gantikan r = – 1 2 ke dalam 1 , Substitute r = – 1 2 into 1 , a(– 1 2 ) 2 = 32 a 4 = 32 a = 128 (b) S∞ = a 1 – r = 128 1 – (– 1 2) = 128 3 2 = 128 × 2 3 = 256 3 6 (a) 5w – 1 3w + 1 = 7w + 1 5w – 1 (5w – 1)2 = (3w + 1)(7w + 1) 25w2 – 10w + 1 = 21w2 + 10w + 1 4w2 – 20w = 0 4w(w – 5) = 0 w = 0 atau/or w = 5 Oleh sebab w > 0, maka w = 5. Since w > 0, thus w = 5. (b) 36 + h + k = 28 h + k = –8 1 h 36 = k h k = h2 36 2 Gantikan 2 ke dalam 1 , Substitute 2 into 1 , h + ( h2 36 ) = –8 36h + h2 = –288 h2 + 36h + 288 = 0 (h + 12)(h + 24) = 0 h + 12 = 0 , h + 24 = 0 h = –12 h = –24 Apabila/When h = –12, k = (–12)2 36 = 4 Apabila/When h = –24, k = (–24)2 36 = 16 BAHAGIAN B 7 (a) (i) Isi padu tiga buah kon yang pertama ialah Volume of the fi rst three cones are V1 = 1 3 × π × j2 × 5 = 5 3 πj 2 V2 = 1 3 × π × j 2 × 8 = 8 3 πj 2 V3 = 1 3 × π × j 2 × 11 = 11 3 πj 2 V2 – V1 = 8 3 πj 2 – 5 3 πj 2 = πj 2 V3 – V2 = 11 3 πj 2 – 8 3 πj 2 = πj 2 ∴ Beza sepunya/Common difference = πj 2 (ii) a = V1 = 5 3 πj 2 , d = πj 2 n 2 [2a + (n –1)d] = Sn n 2 [2( 5 3 πj 2 ) + (n – 1)( πj 2 )] = 124 3 πj 2 πj 2 × n 2 ( 10 3 + n – 1) = 124 3 πj 2 n 2 ( 10 + 3n – 3 3 ) = 124 3 PT SPM Add Math Tkt 4-Jawapan.indd 8 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
J9 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan n( 3n + 7 3 ) = 248 3 3n2 + 7n – 248 = 0 (3n + 31)(n – 8) = 0 3n + 31 = 0 atau/or n – 8 = 0 n = – 31 3 n = 8 Oleh sebab/Since n > 0, n = 8 (b) (i) Sebutan pertama/First term r = –2 S7 = a[(–2)7 – 1] (–2) – 1 129 = a(–129) –3 129 = 43a a = 3 (ii) T9 = (3)(–2)9 – 1 = (3)(–2)8 = 768 KERTAS 2 BAHAGIAN A 1 (a) Isi padu tiga buah silinder yang pertama Volume of the fi rst three cylinders = π × 42 × 5, π × 42 × 8, π × 42 × 11 = 80π, 128π, 176π d = 128π – 80π = 48π Isi padu silinder mengikut janjang aritmetik. The volume of the cylinder follows arithmetic progression. a = 80π, d = 48π Tn = a + (n – 1)d T15 = 80π + (15 – 1)(48π) = 80π + 672π = 752π (b) Sn = 13 984π n 2 [2(80π) + (n – 1)(48π)] = 13 984π nπ[80 + 24(n – 1)] = 13 984π n(80 + 24n – 24) = 13 984 n(24n + 56) = 13 984 24n2 + 56n – 13 984 = 0 3n2 + 7n – 1 748 = 0 (3n + 76)(n – 23) = 0 3n + 76 = 0 atau/or n – 23 = 0 n = – 76 3 n = 23 Oleh sebab/Since n > 0, n = 23 2 (a) a = 24 000 r = 100 + 8 100 = 1.08 Dari tahun 2016 ke 2021 = 6 tahun From year 2016 to 2021 = 6 years Tn = ar5 T6 = (24 000)(1.08)5 = 35 263.87 Gaji tahunannya pada tahun 2021 ialah RM35 264. His annual salary in 2021 is RM35 264. (b) a = 24 000, r = 1.08 Tn > 60 000 (24 000)(1.08)n – 1 > 60 000 1.08n – 1 > 2.5 log10 1.08n – 1 > log10 2.5 (n – 1) log10 1.08 > log10 2.5 n – 1 > log10 2.5 log10 1.08 n > 11.906 + 1 n > 12.906 n = 13 (c) S6 = 24 000(1.086 – 1) 1.08 –1 = 24 000(0.5869) 0.08 = 176 070 3 (a) T7 = 210, a = x, d = y x + (7 – 1)y = 210 x + 6y = 210 x = 210 – 6y 1 S9 = 1 440 9 2 [2x + (9 – 1)y] = 1 440 2x + 8y = 320 x + 4y = 160 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , (210 – 6y) + 4y = 160 –2y = –50 y = 25 Daripada/From 1 , x = 210 – 6(25) = 60 (b) Untuk Kedai Bistari/For Bistari Shop, Tn = x + (n – 1)y = 60 + (n – 1)25 = 60 + 25n – 25 = 25n + 35 1 Untuk kedai Cemerlang, For Cemerlang shop, Tn = 95 + (n – 1)20 = 95 + 20n – 20 = 20n + 75 2 1 = 2 : 25n + 35 = 20n + 75 5n = 40 n = 8 4 (a) T6 = 9 × T4 ar 6 – 1 = 9ar 4 – 1 r 5 = 9r 3 r 2 = 9 r = 3 (b) (i) Sn = 5 465 5(3n – 1) 3 – 1 = 5 465 5(3n – 1) = 10 930 3n – 1 = 2 186 3n = 2 187 log10 3n = log10 2 187 n log10 3 = log10 2 187 n = log10 2 187 log10 3 n = 7 (ii) T7 = 5(3)7 – 1 = 3 645 Panjang bahagian terakhir dawai itu ialah 3 645 mm. The length of the last part of the wire is 3 645 mm. BAHAGIAN B 5 (a) (i) Janjang aritmetik: Arithmetic progression: a = 140, n = 28, S28 = 2 030 28 2 [2(140) + (28 – 1)d] = 2 030 14[280 + (27)d] = 2 030 280 + 27d = 145 27d = –135 d = –5 (ii) Sn = n 2 [2(140) + (n – 1)(–5)] = n 2 (280 – 5n + 5) = n 2 (285 – 5n) (iii) Tn = 140 + (n – 1)(–5) = 140 – 5n + 5 = 145 – 5n (b) (i) πj 2 = 1 600π j = 40 Katakan Cn = lilitan bulatan bagi bulatan ke-n Let Cn = circumference of the nth circle C1 = 2π(40) = 80π C2 = 2π(40 × 3 5) = 48π C3 = 2π [40 × ( 3 5) 2 ] = 144 5 π C4 = 2π [40 × ( 3 5) 3 ] = 432 25 π C5 = 2π [40 × ( 3 5) 4 ] = 1 296 125 π (ii) a = 80π, r = 3 5 S∞ = 80π 1 – 3 5 = 80π ( 2 5 ) = 200π ZON KBAT 1 Katakan/Let S = 1 5 + 2 52 + 3 53 + 4 54 + 5 55 + 6 56 + ... 1 × 5 5S = 1 + 2 5 + 3 52 + 4 53 + 5 54 + 6 55 + ... 2 2 – 1 : 4S = 1 + 1 5 + 1 52 + 1 53 + 1 54 + 1 55 + 1 56 + ... 4S = 1 1 – 1 5 4S = 5 4 S = 5 16 Maka/Thus, 1 5 + 2 52 + 3 53 + 4 54 + 5 55 + 6 56 + ... = 5 16 BAB 6 Hukum Linear KERTAS 1 BAHAGIAN B 1 (a) Kecerunan/Gradient, m = – 18 9 = –2 Y = mX + c xy = –2x2 + 18 y = –2x + 18 x (b) y3 = 3x(7 – x) y3 x = 3(7 – x) y3 x = 21 – 3x y3 x = –3x + 21 PT SPM Add Math Tkt 4-Jawapan.indd 9 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J10 Untuk/For (5, u), u = –3(5) + 21 = 6 Untuk/For (w, 0), 0 = –3(w) + 21 3w = 21 w = 7 Maka/Thus, u = 6, w = 7 2 (a) xy = 12x – 4x3 (÷ x), y = 12 – 4x2 y = – 4x2 + 12 (b) Y = y, m = –4, X = x2 , c = 12 (i) m = –4 (ii) (0, 12) 3 (a) y = mx3 log10 y = log10 mx3 log10 y = log10 m + log10 x3 log10 y = log10 m + 3 log10 x log10 y = 3 log10 x + log10 m (b) (i) log10 m = 2 (ii) Kecerunan/Gradient = 3 p – 2 1 – 0 = 3 p – 2 = 3 p = 5 4 (a) 5y = (h – 2)x2 + 20 x 5xy = (h – 2)x3 + 20 xy = h – 2 5 x3 + 4 (b) Kecerunan/Gradient = h – 2 5 pintasan-xy/xy-intercept = 4 p 2 = 4 p = 8 Kecerunan/Gradient = – 3 10 h – 2 5 = – 3 10 h – 2 = – 3 2 h = 1 2 5 (a) y = v 6x log10 y = log10 v 6x log10 y = log10 v – log10 6x log10 y = log10 v – x log10 6 log10 y = – (log10 6)x + log10 v (b) log10 v = –4 v = 10–4 = 1 10 000 BAHAGIAN B 6 (a) (i) 1 y = g( 1 x ) + h g = 10 – 0 6 – 4 = 10 2 = 5 Katakan/Let K = (0, h) 10 – h 6 – 0 = 5 10 – h = 30 h = –20 (ii) 1 y = 5 x + (–20) 1 y = 5 – 20x x y = x 5 – 20x (b) (i) pintasan-Y/Y-intercept = 1 ∴ p = 1 log y = n log x + 1 n = 7 – 1 4 – 0 = 6 4 = 3 2 (ii) log y = 3 2 log x + 1 log y = log x 3 2 + 1 log y – log x 3 2 = 1 log y x 3 2 = 1 y x 3 2 = 101 y = 10x 3 2 KERTAS 2 BAHAGIAN A 1 (a) x 1 2 3 4 6 √ y 5.5 7.3 9.5 11.7 15.5 8 10 12 14 16 4 3.4 2 6 0 123456 x y (b) q = 3.4 p = 15.5 – 3.4 6 – 0 = 2.017 BAHAGIAN B 2 (a) x 2 3 4 5 6 7 y x 4.8 5.4 5.95 6.6 7.1 7.7 y = (2p – 3)x2 + u p x y x = (2p – 3)x + u p Kecerunan/Gradient = 2p – 3 pintasan-Y/Y-intercept = u p 4 5 6 7 8 3.65 y x 2 1 3 0 123456 7 x (b) Kecerunan/Gradient, 2p – 3 = 7.7 – 3.65 7 – 0 2p – 3 = 0.5786 2p = 3.5786 p = 1.7893 pintasan-Y/Y-intercept = 3.65 u 1.7893 = 3.65 u = 6.531 3 (a) R = β(2.8)– α H log10 R = log10 β(2.8)– α H log10 R = log10 β + log10 (2.8)– α H log10 R = log10 β + (– α H log10 2.8) log10 R = –(α log10 2.8) 1 H + log10 β Kecerunan/Gradient = –α log10 2.8 pintasan-Y/Y-intercept = log10 β (b) 1 H 0.010 0.026 0.040 0.057 0.069 log10 R 1.465 1.220 1.004 0.742 0.558 0.8 1.0 1.2 1.4 1.6 1.64 0.4 0.2 0.6 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 1.8 0.08 log10 R 1 H Kecerunan/Gradient = –αlog10 2.8 0.558 – 1.64 0.069 – 0 = –α(0.4472) –15.68 = –0.4472α α = 35.06 pintasan-Y/Y-intercept = log10 β 1.64 = log10 β β = 43.65 PT SPM Add Math Tkt 4-Jawapan.indd 10 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
J11 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan 4 (a) y = uwx – 2 log10 y = log10 uwx – 2 log10 y = log10 u + log10 wx – 2 log10 y = log10 u + (x – 2)log10 w log10 y = (log10 w) (x – 2) + log10 u Kecerunan/Gradient = log10 w pintasan-Y/Y-intercept = log10 u x – 2 1 2 3 4 6 log10 y 1.54 1.28 1.02 0.76 0.24 0.8 1.0 1.2 1.4 1.6 1.82 0.4 0.2 0.6 0 123456 1.8 log10 y x – 2 (b) Kecerunan/Gradient = log10 w 0.24 – 1.54 6 – 1 = log10 w log10 w = – 0.26 w = 0.5495 pintasan-Y/Y-intercept = log10 u 1.82 = log10 u u = 66.069 ZON KBAT 1 (a) t2 0.25 1.0 2.25 4 6.25 d 0.18 0.90 1.90 3.34 5.2 4 5 6 6.5 7 2 1 3 0 123456 7 t 2 d (b) Apabila/When t = 2.8 t 2 = 2.82 = 7.84 Daripada graf/From the graph, d = 6.5 m sin β = 1.35 6.5 sin β = 0.2077 β = 11.99º 6.5 1.35 Q R P Kaedah alternatif: Alternative method: Diberi/Given d ∝ t 2 d = mt2 , m ialah pemalar/is a constant Kecerunan/Gradient, 5.2 – 0 6.25 – 0 = 0.832 Maka/Thus, d = 0.832t 2 Apabila/When t = 2.8 d = 0.832(2.8)2 = 6.5230 m sin β = 1.35 6.5230 sin β = 0.2070 β = 11.95º BAB 7 Geometri Koordinat KERTAS 1 BAHAGIAN A 1 (a) (n)(–5)+ (m)(7) m + n = 3 –5n + 7m = 3(m + n) –5n + 7m = 3m + 3n 7m – 3m = 3n + 5n 4m = 8n m n = 8 4 m n = 2 1 m : n = 2 : 1 (b) Bagi koordinat-y/For y-coordinate, (1)(u) + (2)(9) 2 + 1 = –2 u + 18 = –6 u = –24 2 (a) Katakan/Let C = (h, k) 2(–3) + 3(h) 3 + 2 = 3 –6 + 3h = 15 3h = 21 h = 7 2(1) + 3(k) 3 + 2 = 4 2 + 3k = 20 3k = 18 k = 6 ∴ C = (7, 6) (b) Luas/Area = 1 2 | 0 –3 7 0 9 1 6 9 | = 1 2 |[(0)(1) + (–3)(6) + (7)(9)] – [(–3)(9) + (7)(1) + (0)(6)]| = 1 2 |45 – (–20)| = 1 2 |65| = 32.5 unit2 /units2 3 (a) Titik tengah GH/Midpoint of GH = (10 + 0 2 , 0 + (–4) 2 ) = (5, –2) mGH = –4 – 0 0 – 10 = 2 5 ∴ mQR = – 5 2 Persamaan QR ialah Equation of QR is y – (–2) = (– 5 2 ) (x – 5) 2y + 4 = –5x + 25 2y = –5x + 21 (b) Apabila/When y = 0, 2(0) = –5x + 21 x = 21 5 Apabila/When x = 0, 2y = –5(0) + 21 y = 21 2 Katakan Q berada pada paksi-x. Let Q is on the x-axis. H(0, –4) G(10, 0) O x y R Q Luas/Area ΔQOR = 1 2 × 21 5 × 21 2 = 441 20 = 22.05 unit2 /units2 4 (a) Bagi garis lurus/For straight line QS, x f + y g = 1, S = (0, g) Bagi garis lurus SR, For straight line SR, S = (0, – 8 e ) Maka/Hence, g = – 8 e e = – 8 g (b) Bagi garis lurus QS, For straight line QS, x f + y g = 1, Q = (f, 0) mPQ = – h f mSR = 2 e PQ//SR ⇒ mPQ = mSR – h f = 2 e – h f = 2 (– 8 g ) – h f = – g 4 h = fg 4 5 (a) Apabila/When x = 0; (0)2 + y2 + 3(0) – 3y – 8 = 0 y2 – 3y – 8 = 0 a = 1, b = –3, c = –8 b2 – 4ac = (–3)2 – 4(1)(–8) = 41 ( > 0) y2 – 3y – 8 = 0 mempunyai dua punca nyata dan berbeza. y2 – 3y – 8 = 0 has two real and different roots. ∴ Lokus L bersilang dengan paksi-y pada dua titik. ∴ Locus of L intersects the y-axis at two points. (b) y = x – 2 1 x2 + y2 + 3x – 3y – 8 = 0 2 PT SPM Add Math Tkt 4-Jawapan.indd 11 01/03/2023 9:18 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J12 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , x2 + (x – 2)2 + 3x – 3(x – 2) – 8 = 0 x2 + x2 – 4x + 4 + 3x – 3x + 6 – 8 = 0 2x2 – 4x + 2 = 0 a = 2, b = –4, c = 2 (–4)2 – 4(2)(2) = 0 2x2 – 4x + 2 = 0 mempunyai dua punca yang sama. 2x2 – 4x + 2 = 0 has two equal real roots. Maka, y = x – 2 bersilang dengan x2 + y2 + 3x – 3y – 8 = 0 hanya pada satu titik. Thus, y = x – 2 intersects x2 + y2 + 3x – 3y – 8 = 0 only at one point. ∴ y = x – 2 ialah tangen kepada lokus L. ∴ y = x – 2 is a tangent to the locus of L. BAHAGIAN B 6 (a) 1(p) + 4(8) 4 + 1 = 6 p + 32 = 30 p = –2 1(5) + 4(0) 4 + 1 = q 5q = 5 q = 1 (b) Katakan kedudukan lampu isyarat = F Let the position of the traffi c light = F x y O C A B F P(6, 5) y = –4x + 12 Diberi persamaan AB ialah Given the equation of AB is y = –4x + 12 1 mAB = –4 Maka/Thus, mCP = 1 4 Persamaan CP ialah/Equation of CP is y – 5 = 1 4 (x – 6) 4y – 20 = x – 6 4y = x + 14 x = 4y – 14 2 Gantikan 2 ke dalam 1 , Substitute 2 into 1 , y = –4(4y – 14) + 12 y = –16y + 56 + 12 17y = 68 y = 4 Daripada/From 2 , x = 4(4) – 14 x = 2 F = (2, 4) Maka, kedudukan lampu isyarat ialah (2, 4). Thus, the position of the traffi c light is (2, 4). KERTAS 2 BAHAGIAN A 1 (a) (i) PQ: 2x + y = 9 Apabila y = 1 dan x = h When y = 1 and x = h 2h + 1 = 9 2h = 8 h = 4 (ii) PQ: 2x + y = 9 y = –2x + 9 mPQ = –2, maka/thus mQR = 1 2 Persamaan QR ialah Equation of QR is y = 1 2 x – 1 (iii) Luas/Area ΔPQR = 1 2 × PQ × QR = 1 2 × √(4 – 2)2 + (1 – 5)2 × √(4 – 0)2 + (1 – (–1))2 = 1 2 × √22 + (–4)2 × √(4)2 + (2)2 = 1 2 × √20 × √20 = 1 2 × 20 = 10 unit2 /units2 (b) Katakan/Let S = (h, k) y x 2x + y = 9 Q(4, 1) (2) (3) R(0, –1) S(h, k) P(2, 5) O 2(h) + 3(4) 3 + 2 = 0 2h + 12 = 0 2h = –12 h = – 6 2(k) + 3(1) 3 + 2 = –1 2k + 3 = –5 2k = – 8 k = – 4 Maka/Thus, S = (–6, –4) BAHAGIAN B 2 (a) mHG = mEF = 2 Persamaan HG ialah/Equation of HG is y – 13 = 2(x – 8) y – 13 = 2x – 16 y = 2x – 3 (b) mEH = – 1 2 oleh sebab/since EH HG. Persamaan EH ialah/Equation of EH is y – 3 = – 1 2 (x – (–2)) y – 3 = – 1 2 x – 1 y = – 1 2 x + 2 (c) Selesaikan y = 2x – 3 dan y = – 1 2 x + 2. Solving y = 2x – 3 and y = – 1 2 x + 2. 2x – 3 = – 1 2 x + 2 5 2 x = 5 x = 2 y = 2(2) – 3 = 1 Maka/Thus, H = (2, 1) (d) Luas segi empat tepat EFGH Area of rectangle EFGH = 2 × ΔEGH = 2 × 1 2 × –2 8 2 –2 3 13 1 3 = |[–2)(13) + (8)(1) + (2)(3)] – [(–2)(1) + (2)(13) + (8)(3]| = |–12 – 48| = 60 Luas segi empat tepat EFGH ialah 60 unit2 . The area of rectangle EFGH is 60 units2 . 3 (a) (i) Gantikan y = 2x + 8 ke dalam 5y = x + 22. Substitute y = 2x + 8 into 5y = x + 22. 5(2x + 8) = x + 22 10x + 40 = x + 22 9x = –18 x = –2 y = 2(–2) + 8 = 4 Maka/Thus, G = (–2, 4) (ii) Untuk/For y = 2x + 8 Apabila/When y = 0, 0 = 2x + 8 x = –4 F = (–4, 0) Luas/Area of ΔEFG = 1 2 8 –4 –2 8 6 0 4 6 = 1 2 × |[(8)(0) + (–4)(4) + (–2)(6)] – [(6)(–4) + (0)(–2) + (4)(8)]| = 1 2 × |–28 – 8| = 1 2 × |–36| = 1 2 × 36 = 18 Luas/Area of ΔEFG = 18 unit2 /units2 (b) Katakan/Let H = (p, q) 2(p) + 1(8) 1 + 2 = –4 2p + 8 = –12 2p = –20 p = –10 2(q) + 1(6) 1 + 2 = 0 2q + 6 = 0 2q = –6 q = –3 Thus, H = (–10, –3). PT SPM Add Math Tkt 4-Jawapan.indd 12 01/03/2023 9:18 AM PENERBIT ILMU BAKTI SDN. BHD.
J13 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan (c) y x y = 2x + 8 5y = x + 22 E(8, 6) H(–10, –3) L(x, y) F O G mLH = y – (–3) x – (–10) = y + 3 x + 10 mLG = y – 4 x – (–2) = y – 4 x + 2 mLH × mLG = –1 oleh sebab LH berseranjang dengan LG. mLH × mLG = –1 since LH is perpendicular to LG. Lokus L ialah/The locus of L is y + 3 x + 10 × y – 4 x + 2 = –1 (y + 3)(y – 4) = –(x + 10)(x + 2) y2 – 4y + 3y – 12 = –(x2 + 2x + 10x + 20) y2 – y – 12 = –x2 – 12x – 20 x2 + y2 + 12x – y + 8 = 0 4 (a) (i) EH: 2y = 3x + 32 y = 3 2 x + 16 mEH = 3 2 , maka/thus mEF = – 2 3 Persamaan EF ialah Equation of EF is y – 1 = – 2 3 (x – 3) y – 1 = – 2 3 x + 2 y = – 2 3 x + 3 (ii) 3 2 x + 16 = – 2 3 x + 3 13 6 x = –13 x = –6 y = 3 2 (–6) + 16 = 7 Maka/Thus, E = (–6, 7) (iii) mFG = 1 – (–3) 3 – 0 = 4 3 Persamaan FG ialah Equation of FG is y – 1 = 4 3 (x – 3) y – 1 = 4 3 x – 4 y = 4 3 x – 3 (b) Bagi titik/For point H, y = –5 2(–5) = 3x + 32 3x = –42 x = –14 H = (–14, –5) Diberi/Given PH = 6 √(x – (–14))2 + (y – (–5))2 = 6 (x + 14)2 + (y + 5)2 = 62 x2 + 28x + 196 + y2 + 10y + 25 = 36 x2 + y2 + 28x + 10y + 185 = 0 ZON KBAT 1 Persamaan bulatan/Equation of the circle: (x – 1)2 + (y – 1)2 = 52 (x2 – 2x + 1) + (y2 – 2y + 1) = 25 x2 + y2 – 2x – 2y – 23 = 0 Gantikan y = 2x – 3 ke dalam persamaan bulatan: Substitute y = 2x – 3 into the equation of the circle: x2 + (2x – 3)2 – 2x – 2(2x – 3) – 23 = 0 x2 + (4x2 – 12x + 9) – 2x – 4x + 6 – 23 = 0 5x2 – 18x – 8 = 0 (5x + 2)(x – 4) = 0 5x + 2 = 0 atau/or x – 4 = 0 x = – 2 5 x = 4 Apabila/When x = – 2 5 , y = 2(– 2 5) – 3 = – 19 5 Apabila/When x = 4, y = 2(4) – 3 = 5 Maka, y = 2x – 3 bersilang dengan bulatan x2 + y2 – 2x – 2y – 23 = 0 di titik (– 2 5 , – 19 5 ) dan (4, 5). Thus, y = 2x – 3 intersects the circle x2 + y2 – 2x – 2y – 23 = 0 at points (– 2 5 , – 19 5 ) and (4, 5). BAB 8 Vektor KERTAS 1 BAHAGIAN A 1 (a) FG →//GH → Maka, FG → = λGH →, dengan keadaan λ ialah pemalar. Thus, FG → = λGH →, where λ is a constant 6p ~ – 4q ~ = λ[4p ~ + (2u – 1)q ~ ] 6p ~ – 4q ~ = 4λp ~ + λ(2u – 1)q ~ Bandingkan pekali p ~ , Compare the coeffi cient of p~ , 6 = 4λ λ = 3 2 Bandingkan pekali q ~ , Compare the coeffi cient of q~ , –4 = λ(2u – 1) –4 = 3 2 (2u – 1) –4( 2 3) = 2u – 1 – 8 3 + 1 = 2u u = – 5 6 (b) FG GH = λ = 3 2 Maka/Thus, FG : GH = 3 : 2 2 (a) FH → = 5k ~ – 6h ~ (b) GH → = 3 5 (5k ~ – 6h ~ ) = 3k ~ – 18 5 h ~ Dalam/In ΔEGH, EG → + GH → = EH → EG → + (3k ~ – 18 5 h ~) = 5k ~ EG → = 18 5 h ~ + 5k ~ – 3k ~ EG → = 18 5 h ~ + 2k ~ 3 (a) u ~ + w ~ = ( m + 1 –9 ) + ( m 1 ) = ( m + 1 + m –9 + 1 ) = ( 2m + 1 –8 ) (b) |u ~ + w ~ | = 10 √(2m + 1)2 + (–8)2 = 10 (2m + 1)2 + (64) = 102 4m2 + 4m + 1 + 64 = 100 4m2 + 4m – 35 = 0 (2m + 7)(2m – 5) = 0 2m + 7 = 0 , 2m – 5 = 0 m = – 7 2 , m = 5 2 4 (a) α + 5 = 0 , 6β – 2 = 0 α = –5 6β = 2 β = 1 3 (b) MK → + KL → = ML → MK → + (7i ~ + 3j ~ ) = –3i ~ + 4j ~ MK → = (–3i ~ + 4j ~ ) – (7i ~ + 3j ~ ) MK → = –3i ~ + 4j ~ – 7i ~ – 3j ~ MK → = –10i ~ + j ~ 5 (a) OP → + PQ → = OQ → (–6i ~ + 2j ~ ) + PQ → = 3i ~ + 7j ~ PQ → = (3i ~ + 7j ~ ) – (–6i ~ + 2j ~ ) PQ → = 9i ~ + 5j ~ (b) Vektor unit dalam arah PQ → Unit vector in the direction of PQ→ = 1 |PQ →| × PQ → = 1 √ 92 + 52 × (9i ~ + 5j ~ ) = 1 √ 106 (9i ~ + 5j ~ ) 6 (a) p ~ – q ~ = (9i ~ + j ~ ) – (6i ~ – nj ~ ) = 9i ~ + j ~ – 6i ~ + nj ~ = 3i ~ + (1 + n)j ~ (b) |p ~ – q ~ | = √34 |3i ~ + (1 + n)j ~ | = √34 √(3)2 + (1 + n)2 = √34 (3)2 + (1 + n)2 = 34 9 + (1 + 2n + n2 ) = 34 n2 + 2n – 24 = 0 (n – 4)(n + 6) = 0 n = 4 atau/or n = –6 BAHAGIAN B 7 (a) QS → = PS → – PQ→ = (4a ~ + 9b ~) – (–3a ~ + 5b ~) = 4a ~ + 9b ~ + 3a ~ + 5b ~ = 7a ~ + 4b ~ RS → = 3 4QS → = 3 4(7a ~ + 4b ~) = 21 4 a ~ + 3b ~ PT SPM Add Math Tkt 4-Jawapan.indd 13 01/03/2023 9:18 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J14 PR → = PS → – RS→ = (4a ~ + 9b ~) – ( 21 4 a ~ + 3b ~) = 4a ~ + 9b ~ – 21 4 a ~ – 3b ~ = – 5 4 a ~ + 6b ~ (b) AB → – BC→ = ( 6 q ) (OB → – OA→) – (OC → – OB→) = ( 6 q ) (( p 2 ) – ( 4 p )) – (( q 8 ) – ( p 2 )) = ( 6 q ) ( p 2 ) – ( 4 p ) – ( q 8 ) + ( p 2 ) = ( 6 q ) p – 4 – q + p = 6 2p – q = 10 q = 2p – 10 1 dan/and 2 – p – 8 + 2 = q p + q = – 4 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , p + (2p – 10) = – 4 3p = 6 p = 2 q = 2(2) – 10 = –6 ∴ p = 2, q = – 6 KERTAS 2 BAHAGIAN A 1 (a) EG → = EF → + FG → = 4d ~ + 3 4 (12e ~ ) = 4d ~ + 9e ~ (b) (i) EP → = 2 3 (12e ~ ) = 8e ~ PQ → = (h – 1)EF → = (h – 1)(4d ~ ) = 4(h – 1)d ~ EQ → = EP → + PQ → = 8e ~ + 4(h – 1)d ~ = 4(h – 1)d ~ + 8e ~ , (ii) Diberi E, Q dan G adalah segaris. Given E, Q and G are collinear. Maka/Thus, EQ → = λEG → = λ(4d ~ + 9e ~ ) = 4λd ~ + 9λe ~ Bandingkan pekali d ~ dan e ~ , Compare the coeffi cient of d~ and e~ , 9λ = 8 , 4(h – 1) = 4λ λ = 8 9 4(h – 1) = 4( 8 9 ) h – 1 = 8 9 h = 17 9 2 (a) OS → = 2 5 OQ → OR → = 3 5OP → = 2 5 (10y ~ ) = 3 5 (15x ~ ) = 4y ~ = 9x ~ (i) Dalam/In ΔOPT, OT → = OP → + PT → = 15x ~ + αPS → = 15x ~ + α(OS → – OP →) = 15x ~ + α(4y ~ – 15x ~ ) = 15x ~ + 4αy ~ – 15αx ~ = (15 – 15α)x ~ + 4αy ~ (ii) Dalam/In ΔOQT, OT → = OQ → + QT → = 10y ~ + βQR → = 10y ~ + β(OR → – OQ →) = 10y ~ + β(9x ~ – 10y ~ ) = 10y ~ + 9βx ~ – 10βy ~ = 9βx ~ + (10 – 10β)y ~ (b) Bandingkan pekali x ~ , Compare the coeffi cient of x ~ , 15 – 15α = 9β 15α + 9β = 15 1 Bandingkan pekali y ~ , Compare the coeffi cient of y~ , 4α = 10 – 10β α = 10 – 10β 4 α = 5 – 5β 2 2 Gantikan 2 ke dalam 1 , Substitute 2 into 1 , 15( 5 – 5β 2 ) + 9β = 15 75 – 75β + 18β = 30 β = 15 19 Gantikan β = 15 19 ke dalam 2 , Substitute β = 15 19 into 2 , α = 5 – 5( 15 19 ) 2 = 10 19 BAHAGIAN B 3 (a) (i) DG → = 4DH → = 4(16n ~ ) = 64n ~ EG → = DG → – DE → EG → = 64n ~ – 40m ~ (ii) HG → = 3DH → = 3(16n ~ ) = 48n ~ HF → = HG → + GF → = 48n ~ + (50m ~ – 48n ~ ) = 50m ~ (b) JG → = HG → – HJ → = 48n ~ – 3 5 HF → = 48n ~ – 3 5 (50m ~ ) = 48n ~ – 30m ~ EG → = 64n ~ – 40m ~ = 4(16n ~ – 10m ~ ) = 4 × 1 3 (3 × 16n ~ – 3 × 10m ~ ) = 4 3 (48n ~ – 30m ~ ) = 4 3 JG → Maka, E, J dan G adalah segaris. Hence, E, J and G are collinear. (c) |EG →| = √[40(4)]2 + [64(3)]2 = √62 464 = 249.93 unit/units 4 (a) (i) F →C = D →C – D→F = 12q ~ – (1 3 × 12p ~) = 12q ~ – 4p ~ D →B = D →A + A →B = 12p ~ + 1 2 (12q ~ – 4p ~ ) = 12p ~ + 6q ~ – 2p ~ = 10p ~ + 6q ~ D →T = h D→B = h(10p ~ + 6q ~ ) = 10hp ~ + 6hq ~ (ii) Dalam/In ΔDFC, D →T = D →F + F→T = D →F + k F→C = 4p ~ + k(12q ~ – 4p ~ ) = 4p ~ + 12kq ~ – 4kp ~ = (4 – 4k)p ~ + 12kq ~ (b) 10hp ~ + 6hq ~ = (4 – 4k)p ~ + 12kq ~ Bandingkan pekali q ~ , Compare the coeffi cient of q ~ , 6h = 12k h = 2k Bandingkan pekali p ~ , Compare the coeffi cient of p ~ , 10h = 4 – 4k 10(2k) = 4 – 4k 20k = 4 – 4k 24k = 4 k = 1 6 Maka/Hence, h = 2( 1 6 ) = 1 3 (c) A →B = 1 2 F →C = 1 2 (12q ~ – 4p ~ ) = 6q ~ – 2p ~ F →T = kF→C = 1 6 (12q ~ – 4p ~ ) = 1 3 (6q ~ – 2p ~ ) = 1 3 A →B ∴ FT : AB = 1 : 3 5 (a) J →F = D →F – D→J J →F = 48p ~ – 36q ~ J →H = 1 3 J →F = 1 3 (48p ~ – 36q ~ ) = 16p ~ – 12q ~ J →E = D →E – D→J = 12p ~ – 36q ~ D →G = D →E – G→E = D →E – 1 3 J →E = 12p ~ – 1 3 (12p ~ – 36q ~ ) = 12p ~ – 4p ~ + 12q ~ = 8p ~ + 12q ~ (b) D →H = D →J + J→H = D →J + 1 3 J →F = 36q ~ + 1 3 (48p ~ – 36q ~ ) = 36q ~ + 16p ~ – 12q ~ = 16p ~ + 24q ~ PT SPM Add Math Tkt 4-Jawapan.indd 14 01/03/2023 9:18 AM PENERBIT ILMU BAKTI SDN. BHD.
J15 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan = 2(8p ~ + 12q ~ ) = 2 D →G Maka, D, G dan H adalah segaris. Hence, D, G and H are collinear. (c) D →G = 8p ~ + 12q ~ |D →G|2 = |8p ~ |2 + |12q ~ |2 |D →G|2 = |8 × 2|2 + |12 × 3|2 = 1 552 |D →G| = 39.40 Jarak/Distance of DG = 39.40 m ZON KBAT 1 (a) BD → = AD → – AB → = (12i ~ + hj ~ ) – (4i ~ + 8j ~ ) = 12i ~ + hj ~ – 4i ~ – 8j ~ = 8i ~ + (h – 8)j ~ BC → = m m + n [8i ~ + (h – 8)j ~ ] Dalam/In ΔABC, AC → = AB → + BC → 10i ~ + 5j ~ = (4i ~ + 8j ~ ) + m m + n [8i ~ + (h – 8)j ~ ] 10i ~ + 5j ~ = 4i ~ + 8j ~ + 8m m + n i ~ + m m + n (h – 8)j ~ 10i ~ + 5j ~ = (4 + 8m m + n )i ~ + [8 + m m + n (h – 8)]j ~ Bandingkan pekali i ~ , Compare the coeffi cient of i ~ , 10 = 4 + 8m m + n 6 = 8m m + n 6m + 6n = 8m 6n = 2m 3 1 = m n Maka/Thus, m : n = 3 : 1 (b) Bandingkan pekali j ~ , Compare the coeffi cient of j~ , 5 = 8 + m m + n (h – 8) –3 = (3) (3) + (1)(h – 8) –3 = 3 4 (h – 8) –12 = 3(h – 8) –12 = 3h – 24 3h = 12 h = 4 BAB 9 Penyelesaian Segi Tiga KERTAS 2 BAHAGIAN C 1 (a) DF2 = 282 + 242 – 2(28)(24) kos/cos 68° DF2 = 856.529 DF = 29.267 cm (b) D G G’ E F 24 cm 28 cm 26 cm 68º 29.267 cm 43º sin ∠DGF 29.267 = sin 43° 26 sin ∠DGF = 29.267 × sin 43° 26 sin ∠DGF = 0.7677 ∠DGF = 50°9’ atau/or 180° – 50°9’ = 50°9’ atau/or 129°51’ Maka/Thus, ∠DGF = 50°9’ dan/and ∠DG’F = 129°51’ (c) (i) ∠FDG = 180° – 50°9’ – 43° = 86°51’ FG sin 86° 51' = 26 sin 43° FG = 38.066 cm (ii) Luas sisi empat DEFG Area of quadrilateral DEFG = ΔDEF + ΔDFG = 1 2 × DE × EF × sin 68° + 1 2 × DF × FG × sin 43° = 1 2 × 28 × 24 × sin 68° + 1 2 × 29.267 × 38.066 × sin 43° = 311.534 + 379.90 = 691.43 cm2 2 (a) Luas/Area ΔEFG = 29.5 1 2 × 9 × 7 × sin ∠EFG = 29.5 sin ∠EFG = 0.9365 ∠EFG = 69°28’ (b) EG2 = 92 + 72 – 2(9)(7) kos/cos 69°28’ EG2 = 85.805 EG = 9.263 cm (c) sin ∠EHG 9.263 = sin 50° 13 sin ∠EHG = 9.263 × sin 50° 13 = 0.5458 ∠EHG = 33°5’ ∠GEH = 180° – 50° – 33°5’ = 96°55’ (d) E H F G 13 cm 9 cm 7 cm 50º 69º28’ 96º55’ 9.263 cm Luas sisi empat EFGH Area of quadrilateral EFGH = Luas ΔEFG + Luas ΔEGH Area of ΔEFG + Area of ΔEGH = 29.5 + 1 2 × 13 × 9.263 × sin 96°55’ = 29.5 + 59.771 = 89.27 cm2 3 (a) (i) PR2 = 19.52 + 92 – 2(19.5)(9) kos/cos 48° PR2 = 226.385 PR = 15.046 cm (ii) sin ∠PRS 5 = sin 116° 15.046 sin ∠PRS = 5 sin 116° 15.046 = 0.2987 ∠PRS = 17°23’ (b) (i) 9 cm 5 cm 19.5 cm 48º Q R S P’ P 116º 17º23’ (ii) ∠RPS = 180° – 116° – 17°23’ = 46°37’ ∠PSP’ = 180° – 2(46°37’) = 86°46’ Luas/Area of ΔP’SR = Luas ΔPRS – Luas ΔPSP’ Area of ΔPRS – Area of ΔPSP’ = 1 2 × 15.046 × 5 × sin 46°37’ – 1 2 × 5 × 5 × sin 86°46’ = 27.338 – 12.48 = 14.858 cm2 4 (a) (i) sin ∠EGF 8 = sin 54° 12 sin ∠EGF = 8 × sin 54° 12 = 0.5393 ∠EGF = 32°38’ ∠EGH = 180° – 32°38’ = 147°22’ (ii) ∠EHG = 180° – 147°22’ – 17° = 15°38’ EH sin 147°22’ = 12 sin 15°38’ EH = 12 × sin 147°22’ sin 15°38’ = 24.0 cm (b) AN = √ 62 + 52 = √61 AC = √ 132 + 62 = √205 NC = √ 132 + 52 = √194 (i) kos/cos ∠NAC = (√61 ) 2 + (√205) 2 – (√194 ) 2 2(√61 )(√205) = 0.3219 ∠NAC = 71°13’ (ii) Luas/Area ΔNAC = 1 2 × √61 × √205 × sin 71°13’ = 52.94 cm2 (iii) Katakan jarak terdekat dari A ke NC = h cm Let the shortest distance from A to NC = h cm 1 2 × h × NC = Luas/Area of ΔNAC 1 2 × h × √194 = 52.94 h = 2 × 52.94 √194 = 7.602 cm ZON KBAT 1 (a) (i) EG2 = (8)2 + (17.4)2 – 2(8)(17.4) kos/cos 48° = 180.474 EG = 13.43 cm PT SPM Add Math Tkt 4-Jawapan.indd 15 01/03/2023 9:18 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J16 (ii) sin ∠EGF 6.4 = sin 108° 13.43 sin ∠EGF = 6.4 sin 108° 13.43 = 0.4532 ∠EGF = 26.95° (b) (i) E F’ F G 72o 108o 13.43 cm 26.95o ) ) ) (ii) ∠F’EG = 180° – 72° – 26.95° = 81.05° Luas/Area of ΔEF’G = 1 2 (EF’)(EG) sin ∠F’EG = 1 2 (6.4)(13.43) sin 81.05° = 42.45 cm2 2 (a) sin ∠EGF 12.8 = sin 28° 10.4 sin ∠EGF = 12.8 sin 28° 10.4 = 0.5778 ∠EGF = 35.30° ∠EGF ialah sudut cakah. ∠EGF is an obtuse angle. ∴ ∠EGF = 180° – 35.30° = 144.7° (b) 12.8 cm 28o F G’ E = =10.4 cm G BAB 10 Nombor Indeks KERTAS 2 BAHAGIAN C 1 (a) q RM3.00 × 100 = 115 q = RM3.00 × 115 100 = RM3.45 (b) s r × 100 = 140 s = 1.4r 1 s – r = 1.60 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , (1.4r) – r = 1.60 0.4r = 1.6 r = 4 s = 1.4(4) = 5.6 Maka/Thus, r = 4.00, s = 5.60 (c) (i) RM61.50 P2019 = 123 P2019 = RM61.50 123 × 100 = RM50.00 (ii) Bahan Ingredient Harga per kilogram (RM) Price per kilogram (RM) I2022/2019 Pemberat Weightage Tahun (w) Year 2019 Tahun Year 2022 A 6.00 7.20 120 3 B 3.00 3.45 115 5 C 6.00 7.50 125 n D 4.00 5.60 140 2 E 2.50 3.25 130 1 Indeks gubahan/Composite index = 123 3(120) + 5(115) + n(125) + 2(140) + 1(130) 3 + 5 + n + 2 + 1 = 123 125n + 1 345 n + 11 = 123 125n + 1 345 = 123n + 1 353 2n = 8 n = 4 2 (a) RM1 800 P2016 × 100 = 160 P2016 = RM1 800 160 × 100 = RM1 125 (b) Sudut sektor P/Angle of sector P = 360° – 200° – 60° = 100° Sudut sektor/Angle of sector Q = 200° Sudut sektor/Angle of sector R = 60° Telefon pintar Smartphone I2018/2016 w Iw P 160 100 16 000 Q 130 200 26 000 R 120 60 7 200 Jumlah/Total 360 49 200 I ¯ = 49 200 360 = 136.67 (c) Q2108 RM540 000 × 100 = 136.67 Q2018 = RM540 000 × 136.67 100 = RM738 018 Jumlah jualan telefon pintar pada tahun 2018 ialah RM738 018. The total sales of the smartphones in the year 2018 is RM738 018. (d) Bagi/For P, I2023/2016 = 160 × 115 100 = 184 Bagi/For Q, I 2023/2016 = 130 × 130 100 = 169 Telefon pintar Smartphone I2018/2016 I2023/2016 w Iw P 160 184 100 18 400 Q 130 169 200 33 800 R 120 120 60 7 200 Jumlah/Total 360 59 400 I ¯ = 59 400 360 = 165 3 (a) x = P2021 P2018 × 100 = 1.75 1.40 × 100 .= 125 PT SPM Add Math Tkt 4-Jawapan.indd 16 01/03/2023 9:18 AM PENERBIT ILMU BAKTI SDN. BHD.
J17 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan 4.50 y × 100 = 180 y = 4.50 × 100 180 = 2.5 z 2.00 × 100 = 160 z = 160 × 2.00 100 = 3.20 (b) Item Harga (RM) Price (RM) Tahun P2021/2018 w Iw Year 2018 Tahun Year 2021 Nasi Rice 1.80 2.70 150 90 13 500 Tepung gandum Wheat fl our 1.40 1.75 125 60 7 500 Minyak masak Cooking oil 2.50 4.50 180 40 7 200 Gula Sugar 2.00 3.20 160 30 4 800 Garam Salt 0.50 0.60 120 20 2 400 Jumlah/Total 240 35 400 Indeks gubahan/Composite index, I – = 35 400 240 = 147.5 (c) Katakan J = Jumlah jualan pada tahun 2018 Let J = total sales in the year 2018 RM637 200 J × 100 = 147.5 J = RM637 200 147.5 × 100 = RM432 000 4 (a) P2018 P2014 × 100 = 125 1 P2021 P2014 × 100 = 160 2 2 ÷ 1 : P2021 P2018 = 160 125 P2021/2018 = 160 125 × 100 = 128 (b) (i) m = 24.00 16.00 × 100 = 150 (ii) P2018 RM16.00 × 100 = 130 P2018 = RM16.00 × 130 100 = RM20.80 (c) I– 2018/2014 = 3(125) + 2(130) + 4(n) 3 + 2 + 4 123 = 4n + 635 9 4n = 9(123) – 635 4n = 472 n = 118 (d) RM177 P2014 × 100 = 123 P2014 = RM177 123 × 100 = RM143.90 5 (a) (i) RM175 p × 100 = 140 p = RM175 140 × 100 = RM125 (ii) q RM180 × 100 = 165 q = RM180 × 165 100 = RM297 (iii) r = RM140 RM80 × 100 = 175 (b) Perbelanjaan Expenditure Perbelanjaan bulanan (RM) Monthly expenditure (RM) I2021/2019 w Iw Tahun Year 2019 Tahun Year 2021 Sewa Rental 750 1 500 200 150 30 000 Pengangkutan Transport 125 175 140 60 8 400 Makanan Food 320 576 180 90 16 200 Hiburan Entertainment 180 297 165 40 6 600 Bil utiliti Utility bills 80 140 175 20 3 500 Jumlah/Total 360 64 700 Indeks gubahan/Composite index, I – = 64 700 360 = 179.72 (c) RM3 400 Q2019 × 100 = 179.72 Q2019 = RM3 400 179.72 × 100 = RM1 891.83 ZON KBAT 1 P2019 P2017 × 100 = 108 P2023 P2019 × 100 = 115 Indeks gubahan/Composite index, = P2023 P2019 × P2019 P2017 × 100 = 115 100 × 108 100 × 100 = 115 100 × 108 = 124.2 PT SPM Add Math Tkt 4-Jawapan.indd 17 01/03/2023 9:18 AM PENERBIT ILMU BAKTI SDN. BHD.
RUMUS –b ± b2 – 4ac 1 x = ––––––––––––– 2a 2 am × an = am + n 3 am ÷ an = am – n 4 (am)n = amn 5 loga mn = loga m + loga n m 6 loga –– = loga m – loga n n 7 loga mn = n loga m logc b 8 loga b = –––––– logc a 9 Tn = a + (n – 1)d n 10 Sn = ––[2a + (n – 1)d] 2 11 Tn = arn – 1 a(rn – 1) a(1 – rn ) 12 Sn = –––––––– = ––––––––, r ≠ 1 r – 1 1 – r a 13 S∞ = –––––, |r| < 1 1 – r ALGEBRA dy dv du 1 y = uv, ––– = u––– + v––– dx dx dx du dv v––– – u––– u dy dx dx 2 y = ––, ––– = –––––––––––– v dx v2 dy dy du 3 ––– = ––– × ––– dx du dx 4 Luas di bawah lengkung = ∫ b a y dx atau = ∫ b a x dy 5 Isi padu kisaran = ∫ b a πy2 dx atau = ∫ b a πx2 dy ∑x 1 x – = ––– N ∑fx 2 x – = –––– ∑f 3 σ = ∑(x – x –) 2 √ N = √∑x2 N – x –2 4 σ = ∑f(x – x –) 2 ∑f √ = √∑fx2 ∑f – x –2 1 ––N – F 2 5 m = L + (––––––––– )C fm Q1 6 I = ––– × 100 Q0 ∑Wi Ii 7 I – = ––––––– ∑Wi n! 8 n Pr = ––––––– (n – r)! n! 9 n Cr = ––––––––– (n – r)! r! 10 P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 11 P(X = r) = n Cr pr qn – r , dengan keadaan p + q = 1 12 Min, μ = np 13 σ = npq X – μ 14 Z = –––––– σ KALKULUS STATISTIK PT SPM Add Math Tkt 4-PAT.indd 94 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
1 Jarak = √(x2 – x1 )2 + (y2 – y1 )2 2 Titik tengah x1 + x2 y1 + y2 (x, y) = (–––––––, –––––––) 2 2 3 Titik yang membahagi suatu tembereng garis nx1 + mx2 ny1 + my2 (x, y) = (–––––––––, –––––––––) m + n m + n 4 Luas segi tiga 1 = ––|(x1 y2 + x2 y3 + x3 y1 ) – (x2 y1 + x3 y2 + x1 y3 )| 2 5 |r|~ = √x2 + y2 ~ xi + ~ yj 6 r ∧ = ––––––– √x2 + y2 1 Panjang lengkok, s = jq 1 2 Luas sektor, A = ––j 2 q 2 3 sin2 A + kos2 A = 1 4 sek2 A = 1 + tan2 A 5 kosek2 A = 1 + kot2 A 6 sin 2A = 2 sin A kos A 7 kos 2A = kos2 A – sin2 A = 2 kos2 A – 1 = 1 – 2 sin2 A 8 tan 2A = 2 tan A 1 – tan2 A 9 sin (A ± B) = sin A kos B ± kos A sin B 10 kos (A ± B) = kos A kos B ∓ sin A sin B tan A ± tan B 11 tan (A ± B) = –––––––––––––– 1 ∓ tan A tan B a b c 12 ––––– = ––––– = ––––– sin A sin B sin C 13 a2 = b2 + c2 – 2bc kos A 14 Luas segi tiga 1 = ––ab sin C 2 GEOMETRI TRIGONOMETRI PT SPM Add Math Tkt 4-PAT.indd 95 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
Bil Perkara Kertas 1 (3472/1) Kertas 2 (3472/2) 1 Jenis Instrumen Ujian Bertulis 2 Jenis Item • Subjektif Respons Terhad • Subjektif Respons Terhad Berstruktur 3 Bilangan Soalan Bahagian A 12 soalan (64 markah) (Jawab semua soalan) Bahagian B 3 soalan (16 markah) (Jawab dua soalan) Bahagian A 7 soalan (50 markah) (Jawab semua soalan) Bahagian B 4 soalan (30 markah) (Jawab tiga soalan) Bahagian C 4 soalan (20 markah) (Jawab dua soalan) 4 Jumlah Markah 80 100 5 Konstruk • Mengingat & Memahami • Mengaplikasi • Menganalisis • Menilai • Mencipta • Mengingat & Memahami • Mengaplikasi • Menganalisis • Menilai • Mencipta 6 Tempoh Ujian 2 jam 2 jam 30 minit 7 Cakupan Konstruk Standard kandungan dan standard pembelajaran dalam Dokumen Standard Kurikulum dan Pentaksiran (DSKP) KSSM (Tingkatan 4 dan Tingkatan 5) 8 Aras Kesukaran Rendah : Sederhana : Tinggi 5 : 3 : 2 9 Kaedah Penskoran Analitik 10 Alat Tambahan Kalkulator saintifik yang tidak boleh diprogram Format Instrumen Peperiksaan SPM Mulai Tahun 2021 Mata Pelajaran Matematik Tambahan (3472) Format Peperiksaan SPM Mata Pelajaran Matematik Tambahan (3472) PENERBIT ILMU BAKTI SDN. BHD.