(a) Diberi –1 < h < 1. Iaitu, h = –1, 0, 1 Suatu graf fungsi f(x) = ax2 + bx + c mempunyai titik minimum (bentuk graf ) jika a . 0. Jadi, h = 1. Given that –1 < h < 1, that is h = –1, 0, 1 A function graph f(x) = ax2 + bx + c has minimum point (graph ) if a . 0. Then, h = 1. (b) Graf fungsi g(x) = x2 + (k − 1)x + 4 tidak menyilang paksi-x. Jadi, graf itu tidak mempunyai punca nyata. The function graph g(x) = x2 + (k – 1)x + 4 does not intersect the x-axis. Then, the graph does not have real root. b2 − 4ac , 0 (k − 1)2 − 4(1)(4) , 0 k2 − 2k + 1 − 16 , 0 k2 − 2k − 15 , 0 (k − 5)(k + 3) , 0 k − 5 = 0 , k + 3 = 0 k = 5 , k = −3 \ −3 , k , 5 7. Persamaan kuadratik (px)2 + 7qx + 9 = 0 mempunyai dua punca yang sama. Manakala, persamaan kuadratik hx2 − 2x + 2p = 0 tiada punca, dengan keadaan h, p dan q ialah pemalar. Ungkapkan julat q dalam sebutan h. The quadratic equation (px)2 + 7qx + 9 = 0 has two equal roots. Meanwhile, the quadratic equation hx2 − 2x + 2p = 0 has no roots, where h, p and q are constants. Express the range of q in terms of h. p2 x2 + 7qx + 9 = 0 a = p2 , b = 7q, c = 9 b2 – 4ac = 0 (7q)2 – 4(p2 )(9) = 0 49q2 – 362 = 0 49q2 = 36p2 √p2 = 49q2 36 p = ± 7q 6 …… 1 hx2 – 2x + 2p = 0 a = h, b = –2, c = 2p b2 – 4ac , 0 (–2)2 – 4(h)(2p) , 0 4 – 8hp , 0 4 , 8hp 1 , 2hp p . 1 2h …… 2 –3 5 k SPM 2019 1 → 2 7q 6 . 1 2h – 7q 6 , 1 2h q . 1 2h × 6 7 q , 1 2h × (– 6 7 ) q . 3 7h q , – 3 7h 8. Rajah menunjukkan graf bagi fungsi kuadratik f(x) = p xn + qx + r, dengan keadaan p, q, r, m dan n ialah pemalar. Diagram shows the graph of a quadratic function f(x) = p xn + qx + r, where p, q, r, m and n are constants. 0 x –m m f(x) (a) Nyatakan nilai n. State the value of n. (b) Jika f(x) = 0 dan hasil darab punca ialah −r, nyatakan nilai If f(x) = 0 and the product of roots is −r, state the value of (i) q (ii) p (a) x–n = x2 –n = 2 n = –2 (b) f(x) = 0 px2 + qx + r = 0 a = p, b = q, c = r (i) Hasil tambah punca / Sum of roots = m + (–m) = 0 – b a = 0 – q p = 0 q = 0 SPM 2015 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 40 BAB 2
(ii) Hasil darab punca / Product of roots = –r c a = –r r p = –r r = –pr r + pr = 0 r(1 + p) = 0 r = 0, p = –1 9. Graf bagi fungsi kuadratik f(x) = 4[3h − (x − 2)2 ], dengan keadaan h ialah pemalar mempunyai titik maksimum (2, h − 11). The graph of a quadratic function f(x)= 4[3h − (x − 2)2], where h is a constant, has maximum point (2, h − 11). (a) Nyatakan nilai h. State the value of h. (b) Nyatakan jenis punca bagi f(x) = 0. Justifikasikan jawapan anda. State the type of roots for f(x) = 0. Justify your answer. (a) f(x) = 12h – 4(x – 2)2 Nilai maksimum / Maximum value = 12h h – 11 = 12h 11h = –11 h = –1 (b) f(x) = 0, –12 – 4(x – 2)2 = 0 –12 – 4(x2 – 4x + 4) = 0 –12 – 4x2 + 16x – 16 = 0 –4x2 + 16x – 28 = 0 –x2 + 4x – 7 = 0 a = –1, b = 4, c = –7 b2 – 4ac = 42 – 4(–1)(–7) = 16 – 28 = –12 b2 – 4ac , 0 Tidak ada punca nyata / No real roots Kertas 2 1. Persamaan kuadratik x2 – 6(2x – h) = 0, dengan keadaan h ialah pemalar mempunyai puncapunca m dan 3m, m ≠ 0. A quadratic equation x2 – 6(2x – h) = 0, where h is a constant has roots m and 3m, m ≠ 0. (a) Cari nilai m dan nilai h. Find the value of m and of h. SPM 2019 (b) Seterusnya, bentukkan persamaan kuadratik yang mempunyai punca-punca m + 3 dan m – 4. Hence, form the quadratic equation with the roots m + 3 and m – 4. (a) x2 – 6(2x – h) = 0 x2 – 12x + 6h = 0 a = 1, b = –12, c = 6h m + 3m = – b a 4m = – 1 –12 1 2 4m = 12 m = 3 m × 3m = c a 3m2 = 6h 1 3 × 32 = 6h 6h = 27 h = 27 6 = 9 2 (b) Punca-punca baharu ialah 6 dan –1. The new roots are 6 and –1. HTP baharu/new SOP = 6 + (–1) = 5 HDP baharu/new POR = 6 × (–1) = –6 Persamaan baharu ialah The new equation is x2 – (5)x + (–6) = 0 x2 – 5x – 6 = 0 2. Persamaan kuadratik k − 4x = x2 − x + 1, dengan keadaan k ialah pemalar, mempunyai puncapunca a dan b. The quadratic equation k − 4x = x2 − x + 1, where k is a constant, has roots a and b. (a) Cari julat nilai k jika a ≠ b. Find the range of values of k if a ≠ b. (b) Diberi a + 1 dan b + 1 adalah punca-punca bagi satu lagi persamaan kuadratik 2x2 − hx + 4 = 0, dengan keadaan h ialah pemalar. Cari nilai k dan nilai h. Given a + 1 and b + 1 are the roots of another quadratic equation 2x2 − hx + 4 = 0, where h is a constant. Find the value of k and of h. SPM 2015 Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 41 BAB 2
(a) x2 − x + 1 = k − 4x x2 − x + 4x + 1 − k = 0 x2 + 3x + 1 − k = 0 a ≠ b bermaksud dua punca berbeza a ≠ b means two different roots b2 − 4ac . 0 32 − 4(1)(1 − k) . 0 9 − 4 + 4k . 0 5 + 4k . 0 4k . −5 k . – 5 4 (b) Untuk/For 2x2 − hx + 4 = 0 x2 − h 2 x + 2 = 0 HTP/SOR: h 2 = (a + 1) + (b + 1) h 2 − 2 = a + b …… 1 HDP/POR: 2 = (a + 1)(b + 1) 2 = ab + a + b + 1 …… 2 Untuk/For x2 + 3x + 1 − k = 0, HTP/SOR: −3 = a + b …… 3 HDP/POR: 1 − k = ab …… 4 Gantikan 3 ke 1: h 2 − 2 = −3 Replace 3 into 1 h 2 = −1 h = −2 Gantikan 4 dan 3 ke 2: 2 = (1 − k) + (−3) + 1 Replace 4 and 3 into 2 2 = −1 − k k = −3 3. Fungi kuadratik f(x) = 2x2 – 8x + k mempunyai nilai minimum 5 apabila x = h. The quadratic function f(x) = 2x2 – 8x + k has a minimum value of 5 when x = h. (a) Cari nilai h dan nilai k. Find the value of h and of k. (b) Seterusnya, dengan menggunakan nilai h dan nilai k di (a), lakarkan graf f(x) = 2x2 – 8x + k. Hence, by using the value of h and of k in (a), sketch the graph of f(x) = 2x2 – 8x + k. (a) f(x) = 2x2 – 8x + k = 2(x2 – 4x) + k = 23x2 – 4x + 1 –4 2 2 2 – 1 –4 2 2 2 4 + k = 2[(x – 2)2 – 4] + k = 2(x – 2)2 – 8 + k –8 + k = 5 k = 13 x – 2 = 0 x = 2 \ h = 2 (b) f(x) = 2(x – 2)2 + 5 a = 2 > 0 f(x) mempunyai nilai minimum f(x) has minimum value Titik minimum ialah (2, 5) The minimum point is (2, 5) f(x) = 2x2 – 8x + 13 b2 – 4ac = (–8)2 – 4(2)(13) = –40 < 0 Graf f(x) tidak menyilang paksi-x. Graph f(x) does not intersect the x-axis. Apabila/When x = 0, f(x) = 2(0)2 – 8(0) + 13 = 13 x 0 f(x) (2, 5) 13 Praktis SPM Ekstra Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 42 BAB 2
Sudut Sudut KBAT KBAT Kuiz 2 KBAT Ekstra 1. ABCD ialah sebuah segi empat tepat dengan panjang 5x cm dan lebar (4 – x) cm. ABCD is a rectangle with a length of 5x cm and a width of (4 – x) cm. 5x 4 – x Cari perimeter, dalam cm, segi empat ABCD jika luas ABCD adalah maksimum. Seterusnya, nyatakan nilai luas yang maksimum, dalam cm2 , bagi segi empat ABCD. Find the perimeter, in cm, of the rectangle ABCD if the area of ABCD is a maximum. Hence, state the maximum value of the area, in cm2, of the rectangle ABCD. Katakan luas segi empat tepat/Let the the area of rectangle = f(x) f(x) = 5x(4 – x) = –5x2 + 20x = –5(x2 – 4x) = –53x2 – 4x + 1– 4 2 2 2 – 1– 4 2 2 2 4 = –5(x – 2)2 + 20 a = –5 < 0, maka, f(x) mempunyai nilai maksimum a = –5 < 0, hence, f(x) has maximum value x – 2 = 0 \ x = 2 Perimeter = 2(5x) + 2(4 – x) = 2(5 × 2) + 2(4 – 2) = 24 cm Luas maksimum = 5(2)(4 – 2) Maximum area = 20 cm2 2. (a) Cari julat nilai-nilai m dengan keadaan fungsi f(x) = 2x2 – 7x + m adalah sentiasa positif bagi semua nilai x. Find the range of values of m such that the function f(x) = 2x2 – 7x + m is always positive for all values of x. (b) Tunjukkan fungsi g(x) = 3x – 8 – 4x2 adalah sentiasa negatif bagi semua nilai x. Show that the function g(x) = 3x – 8 – 4x2 is always negative for all values of x. (a) f(x) = 2x2 – 7x + m a = 2, b = –7, c = m b2 – 4ac < 0 (–7)2 – 4(2)(m) < 0 49 – 8m < 0 8m > 49 m > 49 8 (b) g(x) = – 4x2 + 3x – 8 a = –4, a < 0, graf maksimum/ maximum graph b2 – 4ac = 32 – 4(–4)(–8) = 9 – 128 = –119 < 0, tidak mempunyai punca. does not have roots g(x) adalah sentiasa negatif. g(x) is always negative. O y x Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik 43 BAB 2
44 BAB 3 Sistem Persamaan Systems of Equations NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 3.1 Sistem Persamaan Linear dalam Tiga Pemboleh Ubah Systems of Linear Equations in Three Variables Sistem persamaan dalam tiga pemboleh ubah Systems of Linear Equations in Three Variables
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 45 BAB 3 1. Tentukan sama ada persamaan berikut adalah sistem persamaan linear dalam tiga pemboleh ubah atau bukan. Determine whether the following equations are systems of linear equations in three variables or not. TP 1 CONTOH 2p + 3q – r = 17 5p + 2q – r2 = 0 6p – q + 2r = 1 Bukan. Kuasa bagi r ialah 2. Ini bukan linear. No. The power of r is 2. This is not linear. (a) a + b + c = 6 2a – b + 3c = 21 a – 3b + c = 2 Ya. Semua tiga persamaan mempunyai tiga pemboleh ubah a, b, dan c dengan kuasa 1. Yes. All three equations have variables a, b, and c, of power 1. (b) 2m + n – 3p = –12 3m + 2n – p = –1 5m – n + 6p = 3 (c) 2x + y = z 3x – 2y – 4z = –2 5x – y2 + 2x = 21 Ya. Semua tiga persamaan mempunyai tiga pemboleh ubah m, n dan p dengan kuasa 1. Yes. All three equations have variables m, n and p, of power 1. Bukan. Kuasa bagi y ialah 2. Ini bukan linear.. No. The power of y is 2. This is not linear. 2. Selesaikan sistem persamaan linear berikut menggunakan kaedah penghapusan. TP 4 Solve the following systems of linear equations using the elimination method. CONTOH 1 4p – q + 2r = –3 2p + 3q = 7 – 4r 3p – 2q + r = –7 Penyelesaian: 4p − q + 2r = −3… 1 2p + 3q + 4r = 7… 2 3p − 2q + r = −7… 3 Susun semula supaya urutan susunan pemboleh ubah itu sama. Rearrange so that the arrangement of variables is the same. 1 × 3: 12p − 3q + 6r = −9 … 4 2p + 3q + 4r = 7 … 2 4 + 2: 14p + 10r = −2 … 5 Pilih 1 dan 2 untuk menghapuskan q dengan penambahan. Choose 1 and 2 to eliminate q by addition. 1 × 2: 8p − 2q + 4r = −6 … 6 3p − 2q + r = −7 … 3 6 − 3: 5p + 3r = 1 … 7 Pilih satu pasang persamaan lain untuk menghapuskan q. Choose another pair of equations to eliminate q. 5 × 3: 42p + 30r = −6 … 8 7 × 10: 50p + 30r = 10 … 9 9 − 8: 8p = 16 p = 2 Hapuskan r dalam 5 dan 7 Eliminate r in 5 and 7 5(2)+3r = 1 3r = −9 r = −3 Gantikan p = 2 ke dalam 7 Substitute p = 2 into 7 4(2) − q + 2(−3) = −3 8 − q − 6 = −3 8 − 6 + 3 = q q = 5 \ p = 2, q = 5, r = −3 Gantikan p = 2 dan r = –3 ke dalam 1 Substitute p = 2 and r = –3 into 1
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 46 BAB 3 Selesaikan sistem persamaan berikut menggunakan kaedah penggantian. TP 4 Solve the following systems of linear equations using the substitution method. CONTOH 2 3x – y + 2z = 25 …… 1 2x + 2y + z = 5 …… 2 4x + 3y – 2z = 2 …… 3 Penyelesaian: Dari 1 / From 1: 3x + 2z – 25 = y y = 3x + 2z – 25 …… 4 Ungkapkan y dalam sebutan x dan z dari 1 Express y in terms of x and z from 1 Dari 2 / From 2: 2x + 2(3x + 2z – 25) + z = 5 2x + 6x + 4z – 50 + z = 5 8x + 5z = 55 …… 5 Dari 3 / From 3: 4x + 3(3x + 2x – 25) – 2z = 2 4x + 9x + 6x – 75 – 2z = 2 13x + 4z = 77 …… 6 Gantikan 4 ke dalam 2 dan 3 Substitute 4 into 2 and 3 5 × 4: 32x + 20z = 220 …… 7 6 × 5: 65x + 20z = 385 …… 8 Jadikan pekali z dalam 5 dan 6 sama. Make the coefficient of z in 5 and 6 the same. 8 − 7: 33x = 165 x = 5 Dari 5 / From 5: 8(5) + 5z = 55 5z = 15 z = 3 Dari 4 / From 4: y = 3(5) + 2(3) – 25 = 15 + 6 – 25 = –4 \ x = 5, y = –4, z = 3 Tentukan nilai pekali x, y dan z Determine the value of coefficient of x, y and z
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 47 BAB 3 Selesaikan sistem persamaan berikut menggunakan kaedah penghapusan atau kaedah penggantian. TP 4 Solve the following system of linear equations using the elimination method or substitution method. (a) 5x – 2y – 4z = 3 3x + 3y + 2z = –3 –2x + 5y + 3z = 3 (b) x + 4y – z = 20 3x + 2y + z = 8 2x – 3y + 2z = –16 5x – 2y – 4z = 3 …… 1 3x + 3y + 2z = –3 …… 2 –2x + 5y + 3z = 3 …… 3 2 × 2: 6x + 6y + 4z = –6 1: 5x – 2y – 4z = 3 Tambah/Add: 11x + 4y = –3 …… 4 2 × 3: 9x + 9y + 6z = –9 3 × 2: –4x + 10y + 6z = 6 Tolak/Subtract: 13x – y = –15 …… 5 5 × 4: 52x – 4y = –60 4: 11x + 4y = –3 Tambah/Add: 63x = –63 x = –1 Ganti/Replace x = –1 ke dalam/into 5: 13(–1) – y = –15 y = 2 Ganti/Replace x = –1, y = 2 ke dalam/into 2: 3(–1) + 3(2) + 2z = –3 2z = –6 z = –3 Maka/Thus, x = –1, y = 2, z = –3 x + 4y – z = 20 …… 1 3x + 2y + z = 8 …… 2 2x – 3y + 2z = –16 …… 3 1: x + 4y – z = 20 2: 3x + 2y + z = 8 Tambah/Add: 4x + 6y = 28 …… 4 2 × 2: 6x + 4y + 2z = 16 3: 2x – 3y + 2z = –16 Tolak/Subtract: 4x + 7y = 32 …… 5 4: 4x + 6y = 28 5: 4x + 7y = 32 Tolak/Subtract: –y = –4 y = 4 Ganti/Replace y = 4 ke dalam/into 4: 4x + 6(4) = 28 4x = 4 x = 1 Ganti/Replace x = 1, y = 4 ke dalam/into 1: 1 + 4(4) – z = 20 –z = 3 z = –3 Maka/Thus, x = 1, y = 4, z = –3 (c) 2y – z = 7 x + 2y + z = 17 2x – 3y + 2z = –1 (d) 2x + y = 2 x + y – z = 4 3x + 2y + z = 0 2y – z = 7 …… 1 x + 2y + z = 17 …… 2 2x – 3y + 2z = –1 …… 3 2 × 2: 2x + 4y + 2z = 34 3: 2x – 3y + 2z = –1 Tolak/Subtract: 7y = 35 y = 5 Ganti/Replace y = 5 ke dalam/into 1: 2(5) – z = 7 z = 3 Ganti/Replace y = 5, z = 3 ke dalam/into 2: x + 2(5) + 3 = 17 x = 4 Maka/Thus, x = 4, y = 5, z = 3 2x + y = 2 …… 1 x + y – z = 4 …… 2 3x + 2y + z = 0 …… 3 2: x + y – z = 4 3: 3x + 2y + z = 0 Tambah/Add: 4x + 3y = 4 …… 4 1 × 2: 4x + 2y = 4 4: 4x + 3y = 4 Tolak/Subtract: –y = 0 y = 0 Ganti/Replace y = 0 ke dalam/into 1:2x + 0 = 2 x = 1 Ganti/Replace x = 1, y = 0 ke dalam/into 2: 1 + 0 – z = 4 z = –3 Maka/Thus, x = 1, y = 0, z = –3
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 48 BAB 3 (e) x – y + 3z = 8 3x + y – 2z = –2 2x + 4y + z = 0 x – y + 3z = 8 …… 1 3x + y – 2z = –2 …… 2 2x + 4y + z = 0 …… 3 1: x – y + 3z = 8 2: 3x + y – 2z = –2 Tambah/Add: 4x + z = 6 …… 4 2 × 4: 12x + 4y – 8z = –8 3: 2x + 4y + z = 0 Tolak/Subtract: 10x – 9z = –8 …… 5 4 × 9: 36x + 9z = 54 5: 10x – 9z = –8 Tambah/Add: 46x = 46 x = 1 Ganti/Replace x = 1 ke dalam/into 4: 4(1) + z = 6 z = 2 Ganti/Replace x = 1, z = 2 ke dalam/into 1: 1 – y + 3(2) = 8 y = –1 Maka/Thus, x = 1, y = –1, z = 2 (f) 2x – 4y + 3z = 17 x + 2y – z = 0 4x – y – z = 6 2x – 4y + 3z = 17 …… 1 x + 2y – z = 0 …… 2 4x – y – z = 6 …… 3 2 × 2: 2x + 4y – 2z = 0 1: 2x – 4y + 3z = 17 Tolak/Subtract: 8y – 5z = –17 …… 4 2 × 4: 4x + 8y – 4z = 0 3: 4x – y – z = 6 Tolak/Subtract: 9y – 3z = –6 …… 5 4 × 3: 24y – 15z = –51 5 × 5: 45y – 15z = –30 Tolak: –21y = –21 y = 1 Ganti/Replace y = 1 ke dalam/into 5: 9(1) – 3z = –6 z = 5 Ganti/Replace y = 1, z = 5 ke dalam/into 2: x + 2(1) – 5 = 0 x = 3 Maka/Thus, x = 3, y = 1, z = 5 (g) 2x + y – 2z = –1 3x – 3y – z = 5 x – 2y + 3z = 6 2x + y – 2z = –1 …… 1 3x – 3y – z = 5 …… 2 x – 2y + 3z = 6 …… 3 2 × 2: 6x – 6y – 2z = 10 1: 2x + y – 2z = –1 Tolak/Subtract: 4x – 7y = 11 …… 4 2 × 3: 9x – 9y – 3z = 15 3: x – 2y + 3z = 6 Tambah/Add: 10x – 11y = 21 …… 5 4 × 10: 40x – 70y = 110 5 × 4: 40x – 44y = 84 Tolak/Subtract: –26y = 26 y = –1 Ganti/Replace y = –1 ke dalam/into 4: 4x – 7(–1) = 11 x = 1 Ganti/Replace x = 1, y = –1 ke dalam/into 2: 3(1) – 3(–1) – z = 5 z = 1 Maka/Thus, x = 1, y = –1, z = 1 (h) x + 3y + 5z = 20 y – 4z = –16 3x – 2y + 9z = 36 x + 3y + 5z = 20 …… 1 y – 4z = –16 …… 2 3x – 2y + 9z = 36 …… 3 1 × 3: 3x + 9y + 15z = 60 3: 3x – 2y + 9z = 36 Tolak/Subtract: 11y + 6z = 24 …… 4 2 × 11: 11y – 44z = –176 4: 11y + 6z = 24 Tolak/Subtract: –50z = –200 z = 4 Ganti/Replace z = 4 ke dalam/into 2: y – 4(4) = –16 y = 0 Ganti/Replace y = 0, z = 4 ke dalam/into 1: x + 3(0) + 5(4) = 20 x = 0 Maka/Thus, x = 0, y = 0, z = 4
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 49 BAB 3 3. Selesaikan setiap yang berikut. TP 5 Solve each of the following. CONTOH Wahidah, Atikah dan Faruq memesan tiga gabungan makanan yang berlainan di sebuah restoran. Wahidah memesan dua bahagian nasi goreng dan satu bahagian ayam goreng. Atikah memesan satu bahagian nasi goreng, satu bahagian ayam goreng dan satu bahagian sup cendawan. Faruq memesan satu bahagian sup cendawan dan dua bahagian ayam goreng. Harga makanan yang dipesan oleh Wahidah, Atikah dan Faruq masing-masing ialah RM20, RM21 dan RM26. Cari harga bagi satu bahagian ayam goreng. Wahidah, Atikah and Faruq ordered three combinations of food in a restaurant. Wahidah ordered two parts of fried rice and one part of fried chicken. Atikah ordered one part of fried rice, one part of fried chicken and one part of mushroom soup. Faruq ordered one part of mushroom soup and two parts of fried chicken. The cost of the food ordered by Wahidah, Atikah and Faruq are RM20, RM21 and RM26 respectively. Find the price of one part of fried chicken. Penyelesaian: Katakan x = harga satu bahagian nasi goreng Let price of one part of fried rice y = harga satu bahagian ayam goreng price of one part of fried chicken z = harga satu bahagian sup cendawan price of one part of mushroom soup 2x + y = 20 …… 1 x + y + z = 21 …… 2 2y + z = 26 …… 3 1: 2x + y = 20 2: x + y + z = 21 Tolak/Subtraction: x – z = –1 …… 4 1 × 2: 4x + 2y = 40 3: 2y + z = 26 Tolak/Subtraction:4x – z = 14 …… 5 4: x – z = –1 5: 4x – z = 14 Tolak: –3x = –15 Subtraction x = 5 Gantikan x = 5 ke dalam 4/Substitute x = 5 into 4: 5 – z = –1 z = 6 Gantikan x = 5 ke dalam 1/Substitute x = 5 into 1: 2(5) + y = 20 y = 10 Maka, harga bagi satu bahagian ayam goreng ialah RM10. Thus, the price of one part of fried chicken is RM10. 1 Selesaikan ketiga-tiga persamaan. List all three equations. 2 Turunkan kepada dua pemboleh ubah. Reduce to two variables. Sebanyak 360 kupon telah dijual semasa karnival di sebuah sekolah. Harga kupon ialah RM8, RM10 dan RM12 dan jumlah pendapatan daripada jualan kupon ialah RM3 500. Gabungan jualan kupon berharga RM8 dan RM10 adalah lima kali bilangan kupon RM12. Cari bilangan kupon bagi setiap jenis yang dijual. There were 360 coupons sold during a school carnival. The coupon prices were RM8, RM10 and RM12 and the total income from the coupon sales was RM3 500. The combined number of RM8 coupons and RM10 coupons sold was five times the number of RM12 coupons sold. Find the number of coupons of each type sold. Katakan x = bilangan kupon/number of coupons RM8 Let say y = bilangan kupon/number of coupons RM10 z = bilangan kupon/number of coupons RM12 x + y + z = 360 …… 1 8x + 10y + 12z = 3 500 …… 2 x + y = 5z …… 3 1: x + y + z = 360 3: x + y – 5z = 0 Tolak/Subtract: 6z = 360 z = 60 Gantikan/Replace z = 60 ke dalam/into 1: x + y + 60 = 360 x + y = 300 …… 4 Gantikan/Replace z = 60 ke dalam/into 2: 8x + 10y + 12(60) = 3 500 8x + 10y = 2 780 …… 5 4 × 8: 8x + 8y = 2 400 5: 8x + 10y = 2 780 Tolak/Subtract: –2y = –380 y = 190 Gantikan/Replace y = 190, z = 60 ke dalam/into 4: x + 190 = 300 x = 110 Maka/Thus, x = 110, y = 190, z = 60
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 50 BAB 3 Persamaan Serentak yang melibatkan Satu Persamaan Linear dan Satu Persamaan Tak Linear Simultaneous Equations involving One Linear Equation and One Non-Linear Equation 3.2 NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 4. Selesaikan persamaan serentak berikut menggunakan kaedah penggantian. TP 4 Solve the following simultaneous equations using the substitution method. CONTOH 1 3x + y = 5 …… 1 8y2 – 6x = –7 …… 2 Penyelesaian: Dari 1 / From 1: y = 5 − 3x …… 3 3 → 2: 8(5 – 3x)2 – 6x + 7 = 0 8(25 – 15x – 15x + 9x2 ) – 6x + 7 = 0 200 – 240x + 72x2 – 6x + 7 = 0 72x2 − 246x + 207 = 0 24x2 − 82x + 69 = 0 (2x – 3)(12x – 23) = 0 x = 3 2 x = 23 12 y = 5 – 31 3 2 2 y = 5 – 31 23 122 = 1 2 = – 3 4 \ x = 3 2 , y = 1 2 dan/ and x = 23 12, y = – 3 4 Jadikan x sebagai perkara rumus. Make x as the subject of formula.. Gantikan 3 dalam 2. Substitute 3 into 2. Pemfaktoran. Factorisation. Gantikan nilai x ke dalam 3. Substitute the values of x into 3.
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 51 BAB 3 Selesaikan persamaan serentak berikut menggunakan kaedah penghapusan. TP 4 Solve the following simultaneous equations using the elimination method. CONTOH 2 4x – y = 6 …… 1 x2 + 3xy = 5 …… 2 Penyelesaian: 1 × 3x: 12x2 – 3xy = 18x …… 3 Hapuskan 3xy Eliminate 3xy 2 × 3: 13x2 = 5 + 18x 13x2 – 18x – 5 = 0 x = –b ± √b2 – 4ac 2a Guna rumus kuadratik Use quadratic formula = –(–18) ± √(–18)2 – 4(13)(–5) 2(13) = 18 ± √584 26 x = 18 + √584 26 x = 18 – √584 26 = 1.622 = −0.237 y = 4x − 6 y = 4(−0.237) − 6 = 4(1.622) − 6 = −6.948 = 0.488 \ x = 1.62, y = 0.488 dan/ and x = −0.237, y = −6.948 Selesaikan persamaan serentak berikut menggunakan kaedah penghapusan atau kaedah penggantian. TP 4 Solve the following simultaneous equations using elimination method or substitution method. (a) x + y – 12 = 0 y2 – 8x = 9 x + y – 12 = 0 …… 1 y2 – 8x = 9 …… 2 Daripada 1: x = 12 – y …… 3 From Gantikan/Replace 3 ke dalam/into 2: y2 – 8(12 – y) = 9 y2 – 96 + 8y – 9 = 0 y2 + 8y – 105 = 0 (y – 7)(y + 15) = 0 y – 7 = 0 atau y + 15 = 0 y = 7 or y = –15 Gantikan nilai-nilai y ke dalam 3: Replace the values of y into 3: Apabila/When y = 7 x = 12 – 7 = 5 Apabila/When y = –15 x = 12 – (–15) = 27 Maka, x = 5, y = 7 dan x = 27, y = –15 Thus and
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 52 BAB 3 (b) x – 2y + 1 = 0 x2 – xy – 3 = 0 x – 2y + 1 = 0 …… 1 x2 – xy – 3 = 0 …… 2 Daripada/From 1: x = 2y – 1 …… 3 Gantikan/Replace 3 ke dalam/into 2: (2y – 1)2 – (2y – 1)y – 3 = 0 4y2 – 4y + 1 – 2y2 + y – 3 = 0 2y2 – 3y – 2 = 0 (2y + 1)(y – 2) = 0 2y + 1 = 0 atau/or y – 2 = 0 y = – 1 2 y = 2 Gantikan nilai-nilai y ke dalam 3: Replace the values of y into 3: Apabila/When y = – 1 2 x = 21– 1 2 2 – 1 = –2 Apabila/When y = 2 x = 2(2) – 1 = 3 Maka/Thus, x = –2, y = – 1 2 dan/and x = 3, y = 2 (c) 3 + 2y – x = 0 3x2 + 4y2 = 5 – 3xy 3 + 2y – x = 0 …… 1 3x2 + 4y2 = 5 – 3xy …… 2 Daripada/From 1: x = 3 + 2y …… 3 Gantikan/Replace 3 ke dalam/into 2: 3(3 + 2y)2 + 4y2 = 5 – 3(3 + 2y)y 3(9 + 12y + 4y2 ) + 4y2 = 5 – 9y – 6y2 27 + 36y + 12y2 + 4y2 – 5 + 9y + 6y2 = 0 22y2 + 45y + 22 = 0 y = –45 ± √(45)2 – 4(22)(22) 2(22) y = –45 + √89 44 atau/or y = –45 – √89 44 y = –0.8083 atau y = –1.2371 Gantikan nilai-nilai y ke dalam 3: Replace the values of y into 3: Apabila/When y = –0.8083 x = 3 + 2(–0.8083) = 1.383 Apabila/When y = –1.2371 x = 3 + 2(–1.2371) = 0.5258 Maka/Thus, x = 1.383, y = –0.8083 dan/and x = 0.5258, y = –1.2371 (d) 4x – y – 5 = 0 2x2 – 3y2 + 19 = 0 4x – y – 5 = 0 …… 1 2x2 – 3y2 + 19 = 0 …… 2 Daripada/From 1: y = 4x – 5 …… 3 Gantikan/Replace 3 ke dalam/into 2: 2x2 – 3(4x – 5)2 + 19 = 0 2x2 – 3(16x2 – 40x + 25) + 19 = 0 2x2 – 48x2 + 120x – 75 + 19 = 0 –46x2 + 120x – 56 = 0 23x2 – 60x + 28 = 0 (23x – 14)(x – 2) = 0 23x – 14 = 0 atau/or x – 2 = 2 x = 14 23 x = 2 Gantikan nilai-nilai x ke dalam 3: Replace the values of x into 3: Apabila/When x = 14 23 y = 41 14 232 – 5 = – 59 23 Apabila/When x = 2 y = 4(2) – 5 = 3 Maka/Thus, x = 14 23, y = – 59 23 dan/and x = 2, y = 3
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 53 BAB 3 (e) x – 3y = 5 x2 – xy + y2 – 6 = 0 x – 3y = 5 …… 1 x2 – xy + y2 – 6 = 0 …… 2 Daripada/From 1: x = 3y + 5 …… 3 Gantikan/Replace 3 ke dalam/into 2: (3y + 5)2 – (3y + 5)y + y2 – 6 = 0 9y2 + 30y + 25 – 3y2 – 5y + y2 – 6 = 0 7y2 + 25y + 19 = 0 y = –25 ± √(25)2 – 4(7)(19) 2(7) y = –25 + √93 14 atau/or y = –25 – √93 14 y = –1.0969 atau y = –2.4745 Gantikan nilai-nilai y ke dalam 3: Replace the values of y into 3: Apabila/When y = –1.0969 x = 3(–1.0969) + 5 = 1.7093 Apabila/When y = –2.4745 x = 3(–2.4745) + 5 = –2.4235 Maka/Thus, x = 1.7093, y = –1.0969 dan/and x = –2.4235, y = –2.4745 (f) 3x + 2y = 10 3 x + 2 y = 5 3x + 2y = 10 …… 1 3y + 2x = 5xy …… 2 Daripada/From 1: y = 10 – 3x 2 …… 3 Gantikan/Replace 3 ke dalam/into 2: 31 10 – 3x 2 2 + 2x = 5x1 10 – 3x 2 2 30 – 9x + 4x = 50x – 15x2 15x2 – 55x + 30 = 0 3x2 – 11x + 6 = 0 (3x – 2)(x – 3) = 0 x = 2 3 atau/or x = 3 Gantikan nilai-nilai y ke dalam 3: Replace the values of y into 3: Apabila/When x = 2 3 y = 10 – 31 2 3 2 2 = 4 Apabila/When x = 3 y = 10 – 3(3) 2 = 1 2 Maka/Thus, x = 2 3 , y = 4 dan/and x = 3, y = 1 2 (g) x + 3y = 1 4y x + 3x y = –13 x + 3y = 1 …… 1 4y2 + 3x2 = –13xy …… 2 Daripada/From 1: x = 1 – 3y …… 3 Gantikan/Replace 3 ke dalam/into 2: 4y2 + 3(1 – 3y)2 = –13(1 – 3y)y 4y2 + 3(1 – 6y + 9y2 ) = –13y + 39y2 4y2 + 3 – 18y + 27y2 + 13y – 39y2 = 0 –8y2 – 5y + 3 = 0 8y2 + 5y – 3 = 0 (8y – 3)(y + 1) = 0 8y – 3 = 0 atau/or y + 1 = 0 y = 3 8 y = –1 Gantikan nilai-nilai y ke dalam 3: Replace the values of y into 3: Apabila/When y = 3 8 x = 1 – 31 3 8 2 = – 1 8 Apabila/When y = –1 x = 1 – 3(–1) = 4 Maka/Thus, x = – 1 8 , y = 3 8 dan/and x = 4, y = –1
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 54 BAB 3 Selesaikan persamaan serentak berikut menggunakan kaedah perwakilan graf. TP 5 Solve the following simultaneous equations using the graphical representation method. CONTOH 3 y + 2x = 7 y2 = 4x + 1 Penyelesaian: Bina satu jadual nilai / Construct a table of values x 0 1 2 3 4 5 6 7 y = 7 – 2x 7 5 3 1 –1 –3 –5 –7 y = ±√4x + 1 ±1 ±2.24 ±3 ±3.61 ±4.12 ±4.58 ±5 ±5.39 Bina satu graf / Construct a graph 1 2 3 (2, 3) (6, –5) –1 4 5 6 7 8 –1 1 2 3 4 5 6 7 8 –2 –3 –4 –5 –6 –7 0 y x Titik persilangan mewakili penyelesaian persamaan serentak: Point of intersection represents the solution of simultaneous equations: (2,3) dan /and (6,−5).
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 55 BAB 3 Selesaikan persamaan serentak berikut menggunakan kaedah perwakilan graf. 5 Solve the following simultaneous equations using the graphical representation method. (h) x 0 1 2 3 4 5 6 7 y = 7 – 2x 7 5 3 1 –1 –3 –5 –7 y = 10 – x2 x – 9 3 0.33 –1.5 –3 –4.33 –5.57 1 2 3 (2, 3) (5, –3) –1 4 5 6 7 8 –1 1 2 3 4 5 6 7 8 –2 –3 –4 –5 –6 –7 0 y x Penyelesaian / Solutions: (2,3),(5,−3) (i) x –3 –2 –1 0 1 2 3 y = 2 – x 5 4 3 2 1 0 –1 y = ±√ 9– x2 0 ±2.24 ±2.83 ±3 ±2.83 ±2.24 0 1 2 3 (2.871, –0.871) (–0.871, 2.871) –1 –1 1 2 3 –2 –3 –3 –2 0 Penyelesaian / Solutions: (−0.8,2.85),(2.8,−0.80) y + 2x = 7 x2 + xy = 10 0 < x < 7 x + y = 2 x2 + y2 = 9 –3 < x < 3
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 56 BAB 3 5. Selesaikan setiap yang berikut: TP 5 Solve each of the following: (a) Cari koordinat titik-titik persilangan antara garis lurus y = –3x – 4 dan lengkung xy + 40 = y2 . Find the coordinates of the points of intersection between the line y = –3x – 4 and the curve xy + 40 = y2. y = –3x – 4 …… 1 xy + 40 = y2 …… 2 Gantikan 1 ke dalam 2: Replace 1 into 2: x(–3x – 4) + 40 = (–3x – 4)2 –3x2 – 4x + 40 = 9x2 + 24x + 16 12x2 + 28x – 24 = 0 3x2 + 7x – 6 = 0 (3x – 2)(x + 3) = 0 3x – 2 = 0 atau x + 3 = 0 x = 2 3 or x = –3 Gantikan nilai-nilai x ke dalam 1: Replace the values of x into 1: Apabila/When x = 2 3 , y = –31 2 3 2 – 4 = – 6 Apabila/When x = –3, y = –3(–3) – 4 = 5 Maka, titik-titik persilangan ialah 1 2 3 , –62 dan (–3, 5). Thus, the points of intersections are 1 2 3 , –62 and (–3, 5). (b) Diberi panjang hipotenus sebuah segi tiga tepat ialah 35 cm dan perimeter segi tiga tepat itu ialah 84 cm. Cari panjang yang mungkin bagi dua sisi yang lain bagi segi tiga tepat itu. Given the length of the hypotenuse of a right-angled triangle is 35 cm and the perimeter of the right- angled triangle is 84 cm. Find the possible length of the other two sides of the triangle. x + y + 35 = 84 x + y = 49 …… 1 x2 + y2 = 352 x2 + y2 = 1 225 …… 2 Daripada/From 1: y = 49 – x …… 3 Gantikan 3 ke dalam 2: Replace 3 into 2: x2 + (49 – x)2 = 1 225 x2 + (2 401 – 98x + x2 ) = 1 225 2x2 – 98x + 1 176 = 0 x2 – 49x + 588 = 0 (x – 28)(x – 21) = 0 x – 28 = 0 atau x – 21 = 0 x = 28 x = 21 Gantikan nilai-nilai x ke dalam 3: Replace the values of x into 3: Apabila/When x = 28, y = 49 – 28 = 21 Apabila/When x = 21, y = 49 – 21 = 28 Maka, panjang sisi-sisi segi tiga ialah 21 cm dan 28 cm. Thus, the length of the sides of the triangle are 21 cm and 28 cm.
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 57 BAB 3 Kertas 2 1. Selesaikan persamaan serentak berikut. Solve the following simultaneous equations. 5 + y – 3x = 0, x2 + y2 – 2y – 25 = 0 Beri jawapan anda betul kepada dua tempat perpuluhan. Give your answers correct to two decimal places. 5 + y – 3x = 0 ………… 1 x2 + y2 – 2y – 25 = 0 ………… 2 Daripada/From 1, y = 3x – 5 …… 3 Gantikan 3 ke dalam 2. Replace 3 into 2: x2 + (3x – 5)2 – 2(3x – 5) – 25 = 0 x2 + 9x2 – 30x + 25 – 6x + 10 – 25 = 0 10x2 – 36x + 10 = 0 5x2 – 18x + 5 = 0 x = –(–18) ± (–18)2 – 4(5)(5) 2(5) = 0.30 atau 3.30 Gantikan nilai-nilai x ke dalam 3. Replace the values of x into 3. Apabila/When x = 0.30, y = 3(0.30) – 5 = –4.10 Apabila/When x = 3.30, y = 3(3.30) – 5 = 4.90 Maka, x = 0.30, y = –4.10 dan x = 3.30, y = 4.90. Thus, x = 0.30, y = –4.10 and x = 3.30, y = 4.90. 2. Rajah menunjukkan pelan bagi laman belakang sebuah rumah banglo berbentuk segi empat tepat PQRS. Laman itu terdiri daripada sebuah kolam renang berbentuk sukuan bulatan dan kawasan berumput PQRUT. Diagram shows the plan of a rectangular backyard of a bungalow PQRS. The backyard consists of a swimming pool in a shape of a quadrant of a circle and a grassy area PQRUT. T U R S Q P SPM 2014 SPM 2018 Diberi bahawa SR = x m dan PT = y m. Luas laman berbentuk segi empat tepat PQRS ialah 77 m2 dan perimeter kawasan berumput ialah 33 m. Kolam renang dengan kedalaman seragam mengandungi 46.2 m3 air. Dengan menggunakan π = 22 7 , cari kedalaman, dalam m, air dalam kolam itu. It is given that SR = x m and PT = y m. The area of a rectangular backyard is 77 m2 and the perimeter of the grassy area is 33 m. The swimming pool with uniform depth contains 46.2 m3 of water. By using π = 22 7 , find the depth, in m, of water in the pool. Luas laman/backyard area = 77 (y + x)x = 77 xy + x2 = 77 …… 1 Perimeter kawasan berumput = 33 Perimeter of grassy area = 33 y + (y + x) + x + 1 2 πj = 33 2y + 2x + 1 2 1 22 7 2x = 33 2y + 2x + 11 7 x = 33 14y + 14x + 11x = 231 14y + 25x = 231 …… 2 Daripada/From 2: y = 231 – 25x 14 …… 3 Gantikan 3 ke dalam 1: Replace 3 into 1: x1 231 – 25x 14 2 + x2 = 77 231x – 25x2 + 14x2 = 1 078 –11x2 + 231x – 1 078 = 0 x2 – 21x + 98 = 0 (x – 14)(x – 7) = 0 x – 14 = 0 atau x – 7 = 0 x = 14 or x = 7 Gantikan nilai-nilai x ke dalam 3: Replace the values of x into 3. Apabila/When x = 14, y = 231 – 25(14) 14 = – 17 2 (tidak diterima) Apabila/When x = 7, y = 231 – 25(7) 14 = 4 Maka/Thus, x = 7, y = 4 PRAKTIS SPM 3
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 58 BAB 3 Praktis SPM Ekstra Katakan kedalaman air = d Let the depth of water = d Isi padu/Volume = luas/area × d 46.2 = 1 4 pj 2 d 46.2 = 1 4 1 22 7 2(7)2 d 46.2 = 154 4 d d = 1.2 Maka/Thus, d = 1.2 meter 3. Diberi persamaan berikut: Given the following equations: P = x – 2y Q = 3x – y + 1 R = x2 + 4y2 Cari nilai-nilai x dan y jika Q = R = 2P. Find the values of x and y if Q = R = 2P. Q = R = 2P 3x – y + 1 = x2 + 4y2 = 2(x – 2y) 3x – y + 1 = x2 + 4y2 …… 1 3x – y + 1 = 2(x – 2y) …… 2 Daripada 2, 3x – y + 1 = 2x – 4y From 3x – 2x = y – 4y – 1 x = –3y – 1 …… 3 Gantikan 3 ke dalam 1. Replace 3 into 1: 3(–3y – 1) – y + 1 = (–3y – 1)2 + 4y2 –9y – 3 – y + 1 = 9y2 + 6y + 1 + 4y2 13y2 + 16y + 3 = 0 (13y + 3)(y + 1) = 0 y = – 3 13 atau y = –1 Gantikan nilai-nilai y ke dalam 3. Replace the values of y into 3: Apabila/When y = – 3 13 , Apabila/When y = –1, x = –3(–1) – 1 = 2 x = –31– 3 132 – 1 = – 4 13 Maka/Thus, x = – 4 13 , y = – 3 13 dan/and x = 2, y = –1. SPM 2017 BUKAN RUTIN 4. Selesaikan persamaan serentak berikut. Solve the following simultaneous equations. 3 + y – 2x = 0, 3x2 + 2y2 – 4xy = 9 3x2 + 2y2 – 4xy = 9 …… 1 3 + y – 2x = 0 ⇒ y = 2x – 3 …… 2 Gantikan/Replace 2 ke/into 1. 3x2 + 2(2x − 3)2 – 4x(2x − 3) = 9 3x2 + 2(4x2 – 12x + 9) – 8x2 + 12x – 9 = 0 3x2 + 8x2 – 24x + 18 – 8x2 + 12x – 9 = 0 3x2 – 12x + 9 = 0 ÷ 3 x2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x – 1 = 0 , x – 3 = 0 x = 1 , x = 3 Apabila/When x = 1, y = 2(1) – 3 = –1 Apabila/When x = 3, y = 2(3) – 3 = 3 Maka/Thus, x = 1, y = 1 dan/and x = 3, y = 3. 5. Selesaikan persamaan serentak x + 3y = 2 dan 4 x – 3 y = 8. Beri jawapan anda betul kepada tiga tempat perpuluhan. Solve the simultaneous equations x + 3y = 2 and 4 x – 3 y = 8. Give you answers correct to three decimal places. x + 3y = 2 x = 2 – 3y …… 1 4 x – 3 y = 8 × xy → 4y – 3x = 8xy …… 2 1→2: 4y – 3(2 – 3y) = 8y(2 – 3y) 4y – 6 + 9y = 16y – 24y2 24y2 – 3y – 6 = 0 y = –b ± b2 – 4ac 2a = –(–1) ± (–1)2 – 4(8)(–2) 2(8) = 1 ± 65 16 y = 1 + 65 16 y = 1 – 65 16 = 0.566 = –0.441 x = 2 – 3(0.566) x = 2 – 3(–0.441) = 0.302 = 3.323 \ x = 0.302, y = 0.566 dan/and x = 3.323, y = –0.441 SPM 2015 SPM 2019
Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan 59 BAB 3 1. Diberi (2m, 3n) adalah penyelesaian kepada persamaan serentak y + 2x = –4 dan 4 y – 3 2x = – 13 6 , cari nilai-nilai m dan n. Given that (2m, 3n) is the solution to the simultaneous equations y + 2x = –4 and 4 y – 3 2x = – 13 6 , find the values of m and of n. x = 2m, y = 3n y + 2x = –4 3n + 4m = –4 3n = –4 – 4m n = –4 – 4m 3 … 4 y – 3 2x = – 13 6 8x – 3y 2xy = – 13 6 6(8x – 3y) = –13(2xy) 48x – 18y = –26xy 48(2m) – 18(3n) = –26(2m)(3n) 96m – 54n + 156mn = 0 … Gantikan kepada : Replace into : 96m – 541 –4 – 4m 3 2 + 156m 1 –4 – 4m 3 2 = 0 96m – 18(–4 – 4m) + 52m (–4 – 4m) = 0 96m + 72 + 72m – 208m – 208m2 = 0 208m2 + 40m – 72 = 0 52m2 + 10m – 18 = 0 26m2 + 5m – 9 = 0 (13m + 9)(2m – 1) = 0 13m + 9 = 0 , 2m – 1 = 0 m = – 9 13 m = 1 2 n = 1 3–4 – 41– 9 132 n = 1 3–4 – 41 1 22 = – 16 39 = –2 2. Suatu jenama minuman tin tertentu dikeluarkan secara pek dengan 6, 12 dan 24 tin setiap jenis dan masing-masing berharga RM10, RM18 dan RM36 setiap pek. Sebuah stor telah menjual 14 pek dengan jumlah 162 tin dan menerima bayaran RM248. Cari bilangan pek setiap jenis yang telahdijual. A certain brand of canned drinks comes in packages of 6, 12 and 24 cans costing RM10, RM18 and RM36 per package respectively. A store sold 14 packages containing a total of 162 cans and received RM248. Find the number of packages of each type sold. Katakan x = bilangan pek 6 tin Let y = bilangan pek 12 tin z = bilangan pek 24 tin x + y + z = 14 …… 1 10x + 18y + 36z = 248 …… 2 6x + 12y + 24z = 162 …… 3 3 ÷ 6: x + 2y + 4z = 27 1: x + y + z = 14 Tolak/Subtract: y + 3z = 13 …… 4 1 × 5: 5x + 5y + 5z = 70 2 ÷ 2: 5x + 9y + 18z = 124 Tolak/Subtract: –4y – 13z = –54 …… 5 4 × 4: 4y + 12x = 52 5: –4y – 13z = –54 Tambah/Add: –z = –2 z = 2 Ganti z = 2 ke dalam 4: Replace z = 2 into 4: y + 3(2) = 13 y = 7 Ganti y = 7, z = 2 ke dalam 1: Replace y = 7, z = 2 into 1: x + 7 + 2 = 14 x = 5 Maka/Thus, x = 5, y = 7, z = 2 Sudut Sudut KBAT KBAT KBAT Ekstra Kuiz 3
1. Permudahkan setiap ungkapan algebra yang berikut. TP 1 Simplify each of the following algebraic expressions. CONTOH 52n + 5 × 53 – n × 25n + 5 25(2n – 1) = 52n + 5 × 53 – n × 52(n + 5) 52(2n – 1) = 52n + 5 + 3 – n + 2n + 10 54n – 2 = 53n + 18 − (4n − 2) = 52 – n (a) m3 × (n–2)4 × (5m)4 (3m2 n)3 = m3 × n–8 × 625m4 27m6 n3 = 625m7 n–8 27m6 n3 = 625 27 m7 – 6n–8 – 3 = 625 27 mn−11 NOTA IMBASAN NOTA IMBASAN Hukum indeks Laws of indices 60 4.1 BAB 4 Indeks, Surd dan Logaritma Indices, Surds and Logarithms 4.1 Hukum Indeks Laws of Indices
(b) x2m + 2 × (yn − 1)3 ÷ (xyn )4 = x2m + 2 × y3n − 3 ÷ x4 y4n = x2m + 2 − 4 y3n − 3 − 4n = x2m − 2 y −3 − n (c) 31 3x2 4y3 2 4 × 1 16y2 9x3 2 3 4 2 = 1 81x8 256y12 × 4096y6 729x9 2 2 = 1 16 9xy6 2 2 = 256 81x2 y12 2. Permudahkan setiap ungkapan algebra yang berikut. TP 2 Simplify each of the following algebraic expressions. CONTOH 7(2n ) − 8 1 3 n − 1 + 4 1 2 n + 3 = 7(2n )−(23 ) 1 3 n − 1 + (22 ) 1 2 n + 3 = 7(2n )−2n − 3 + 2n + 6 = 7(2n )−2n (2−3) + 2n (26 ) = 2n 17 – 1 8 + 642 = 70 7 8 (2n ) = 567 8 (2n ) = 567(2n – 3) (a) 8 2 3 x + 43 + x −16 1 2 x + 1 = (23 ) 2 3 x + (22 )3 + x −(24 ) 1 2 x + 1 = 22x + 26 + 2x − 22x + 4 = 22x + 26 (22x ) − 24 (22x ) = 22x (1 + 26 − 24 ) = 4x (1 + 64 − 16) = 49(4x ) (b) a3 + n × (a2 )n − 1 a4 − n = a3 + n × a2n − 2 a4 − n = a3 + n + 2n − 2 − (4 − n) = a3n + 1 − 4 + n = a4n − 3 (c) 5n + 3 +125 1 3 (n – 1) − 25 1 2 n = 5n (53 ) + (53 ) 1 3 (n – 1) – (52 ) 1 2 n = 5n (53 ) + 5n − 1 − 5n = 5n 153 + 1 5 – 12 = 124 1 5 (5n ) = 621 5 (5n ) = 621(5n − 1) 3. Buktikan. TP 3 Prove. CONTOH 23x – 4 = 8x 16 23x – 4 = 23x ÷ 24 = 23x 24 = 8x 16 (a) 34 – 2x = 81 9x 34 − 2x = 34 ÷ 32x = 34 32x = 81 9x Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 61 BAB 4
(b) 7n – 2m = 7n 49m 7n − 2m = 7n ÷ 72m = 7n 72m = 7n 49m (c) 43m – 2 = 64m 16 43m − 2 = 43m ÷ 42 = 43m 42 = 64m 16 4. Tunjukkan. TP 3 Show. CONTOH Tunjukkan bahawa 7n + 7n + 2 + 7n + 3 boleh dibahagi tepat oleh 3 bagi semua integer positif n. Show that 7n + 7n + 2 + 7n + 3 is divisible by 3 for all positive integers n. 7n + 7n + 2 + 7n + 3 =7n (1) + 7n (72 ) + 7n (73 ) =7n (1 + 49 + 343) =7n (393) =7n (3 × 131) Boleh dibahagi tepat oleh 3 kerana 3 ialah satu faktor. Divisble by 3 since 3 is a factor. (a) Tunjukkan bahawa 9n + 9n + 3 − 9n + 2 boleh dibahagi tepat oleh 11 bagi semua integer positif n. Show that 9n + 9n + 3 − 9n + 2 is divisble by 11 for all positive integers n. 9n + 9n + 3 − 9n + 2 = 9n (1) + 9n (93 ) − 9n (92 ) = 9n (1 + 729 − 81) = 9n (649) = 9n (11 × 59) Boleh dibahagi tepat oleh 11 kerana 11 ialah satu faktor. Divisible by 11 since 11 is a factor. (b) Tunjukkan bahawa 5n + 2 – 5n + 1 + 5n + 3 boleh dibahagi tepat oleh 29 bagi semua integer positif n. Show that 5n + 2 – 5n + 1 + 5n + 3 is divisble by 29 for all positive integers n. 5n + 2 – 5n + 1 + 5n + 3 = 5n (52 ) – 5n (51 ) + 5n (53 ) = 5n (25 – 5 + 125) = 5n (145) = 5n (29 × 5) Boleh dibahagi tepat oleh 29 kerana 29 ialah satu faktor. Divisible by 29 since 29 is a factor. (c) Tunjukkan bahawa 6n + 3 + 6n + 2 − 6n + 1 boleh dibahagi tepat oleh 41 bagi semua integer positif n. Show that 6n + 3 + 6n + 2 − 6n + 1 is divisble by 41 for all positive integers n. 6n + 3 + 6n + 2 − 6n + 1 = 6n (63 ) + 6n (62 ) − 6n (61 ) = 6n (216 + 36 − 6) = 6n (246) = 6n (41 × 6) Boleh dibahagi tepat oleh 41 kerana 41 ialah satu faktor. Divisible by 41 since 41 is a factor. Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 62 BAB 4
5. Selesaikan setiap persamaan yang berikut. TP 4 Solve each of the following equations. CONTOH 3x + 2 – 3x – 1 = 26 27 3x (32 ) – 3x (3–1) = 26 27 3x 19 – 1 3 2 = 26 27 3x 1 26 3 2 = 26 27 26(3x – 1) = 261 1 272 3x – 1 = 3–3 x – 1 = –3 x = –2 (a) 4x × 22x − 1 = 128 22x × 22x − 1 = 27 22x + 2x − 1 = 27 2x + 2x − 1 = 7 4x = 8 x = 2 (b) 53(x – 1) = 25x – 1 53x + 3 = 52(x – 1) 3x + 3 = 2x – 2 x = –5 (c) 4x + 2 16x = 64x + 1 4x + 2 16x = (43 )x + 1 4x + 2 – 2x = 43x + 3 2 − x = 3x + 3 4x = –1 x = – 1 4 (d) 8x + 2 = 1 512x + 1 8x + 2 × 512x + 1 = 1 8x + 2 × 83(x + 1) = 80 x + 2 + 3x + 3 = 0 4x = –5 x = – 5 4 (e) 52x + 1 + 52x − 1 = 26 52x (51 ) + 52x (5−1) = 26 52x 15 + 1 5 2 = 26 52x 1 26 5 2 = 26 52x = 5 2x = 1 x = 1 2 (f) 62x + 1 + 62x − 1 = 1332 62x (61 )+62x (6−1) = 1332 62x 16 + 1 6 2 = 1332 62x 1 37 6 2 = 1332 62x = 1332 × 6 37 62x = 216 62x = 63 2x = 3 x = 3 2 (g) 4x + 2 − 4x − 1 = 252 4x (42 ) – 4x (4−1) = 252 4x 116 – 1 4 2 = 252 4x 1 63 4 2 = 252 4x = 252 × 4 63 4x = 16 4x = 42 x = 2 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 63 BAB 4
6. Selesaikan masalah yang berikut. TP 4 Solve the following problems. (a) Suhu, T, bagi sebuah aluminium, dalam °C, selepas dipanaskan selama t minit diberikan oleh T = 28.5(1.1)t . Cari suhunya selepas setengah jam. The temperature, T, of a piece of aluminium, in °C, after its heated for t minutes is given by T = 28.5(1.1)t . Find its temperature after heated for half an hour. T = 28.5(1.1)t t = 30 minit / minutes T = 28.5(1.1)30 = 497.31°C (b) (b) Nilai sebuah kereta menyusut sebanyak 8% daripada nilainya pada awal setiap tahun. Jika harga asal kereta itu ialah RM145 000, nilainya p, dalam RM, selepas t tahun diberi oleh p = 145 000(0.92)t . Carikan harga kereta itu selepas 7 tahun. The value of a car decreases by 8% from its value at the beginning of the year. If its original price was RM145 000, its value p, in RM, after t years is given by p = 145 000(0.92)t . Find the price of the car after 7 years. p = 145000(0.92)t t = 7 p = 145000(0.92)7 = 80 887.76 4.2 Hukum Surd Laws of Surds NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN Surd Surd Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 64 BAB 4
7. Tukar perpuluhan berulang berikut kepada pecahan. TP 3 Convert the following recurring decimals to fractions. CONTOH 0.353535… Penyelesaian: Biar / Let N = 0.353535 …… 1 100N = 35.3535 …… 2 2 – 1: 99N = 35 N = 35 99 (a) 0.7777… Biar / Let N = 0.7777 …… 1 10N = 7.7777 …… 2 2 – 1: 9N = 7 N = 7 9 (b) 0.818181… Biar / Let N = 0.818181 …… 1 100N = 81.8181 …… 2 2 – 1: 99N = 81 N = 81 99 = 9 11 (c) 0.717171… Biar / Let N = 0.717717717 …… 1 1000N = 717.717717 …… 2 2 – 1: 999N = 717 N = 717 999 = 239 333 8. Tentukan sama ada setiap yang berikut ialah surd. Nyatakan sebab anda. TP 3 Determine whether each of the foillowing is a surd. State your reason. CONTOH 4 √80 = 2.990697562… (Gunakan kalkulator / Use calculator) 4 √80 ialah sebuah surd kerana bukan perpuluhan berulang. 4 √80 is a surd because it is a non-recurring decimal. (a) 5 √243 = 5 √243 = 3 Nilainya ialah satu integer. 5 √243 bukan sebuah surd. The value is an integer. 5√243 is not a surd. (b) 4 16 625 = 4 16 625 = 2 5 Nilainya ialah satu pecahan. 4 16 625 bukan sebuah surd. The value is a fraction. 4 16 625 is not a surd. (c) 6 √88 = 6 √88 = 2.109018772… 6 √88 ialah sebuah surd kerana bukan perpuluhan berulang. 6√88 is a surd because it is a non-recurrung decimal. Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 65 BAB 4
9. Tulis setiap yang berikut sebagai satu surd tunggal. TP 3 Write each of the following as a single surd. CONTOH √5x × √7x = √5x × 7x = √35x2 = x√35 (a) √3 × √11 = √3 × 11 = √33 (b) √78 √6 = 78 6 = √13 (c) √13m × √2m = √13m × 2m = √26m2 = m√26 (d) √20n √32n = 20n 32n = 5 8 (e) √6 × √10 √5 = 6 × 10 5 = √12 10. Permudahkan setiap yang berikut: TP 3 Simplify each of the following: CONTOH √12 Penyelesaian: √12 = √4 × 3 = √4 × √3 = 2√3 (a) √32 √32 = √16 × 2 = √16 × √2 = 4√2 (b) √45 √45 = √9 × 5 = √9 × √5 = 3√5 (c) √125 √125 = √25 × 5 = √25 × √5 = 5√5 (d) 3 √24 3 √24 = 3 √8 × 3 = 3 √8 × 3 √3 = 23 √3 (e) 3 √135 3 √135 = 3 √27 × 5 = 3 √27 × 3 √5 = 33 √5 11. Permudahkan setiap yang berikut: TP 3 Simplify each of the following: CONTOH 1 3√8 + 2√8 – 4√8 Penyelesaian: 3√8 + 2√8 – 4√8 = (3 + 2 – 4)√8 = √8 (a) 3√2 + 5√2 – √2 3√2 + 5√2 – √2 = (3 + 5 – 1)√2 = 7√2 (b) 4√3 – 7√3 4√3 – 7√3 = (4 – 7)√3 = –3√3 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 66 BAB 4
CONTOH 2 √18 – √8 Penyelesaian: = √9 × 2 – √4 × 2 = 3√2 – 2√2 = (3 – 2)√2 = √2 Tukar ke sebutan surd yang serupa. Change to equal surd term. (c) 13√28 – 7√7 = 13√4 × 7 – 7√7 = 13√4√7 – 7√7 = 26√7 – 7√7 = (26 – 7)√7 = 19√7 (d) 2√12 + 5√48 – 7√3 = 2√4 × 3 + 5√16 × 3 – 7√3 = 2√4√3 + 5√16√3 – 7√3 = 4√3 + 20√3 – 7√3 = (4 + 20 – 7)√3 = 17√3 (e) 3√20 – √5 – 2√45 3√20 – √5 – 2√45 = 3√4 × 5 – √5 – 2√9 × 5 = 3√4√5 – √5 – 2√9√5 = 6√5 – √5 – 6√5 = (6 – 1 – 6)√5 = –√5 (f) 2√6 + √150 – 3√54 2√6 + √150 – 3√54 = 2√6 + √25 × 6 – 3√9 × 6 = 2√6 + 5√6 – 9√6 = (2 + 5 – 9)√6 = –2√6 (g) √2 × √3 + 4√6 √2 × √3 + 4√6 = √2 × 3 + 4√6 = √6 + 4√6 = (1 + 4)√6 = 5√6 CONTOH 3 28 9 Penyelesaian: 28 9 = √28 3 = √4 × 7 3 = √4 × √7 3 = 2√7 3 (h) 8 25 8 25 = √8 5 = √4 × 2 5 = √4 × √2 5 = 2√2 5 (i) 20 9 20 9 = √20 3 = √4 × 5 3 = √4 × √5 3 = 2√5 3 (j) 18 75 18 75 = 6 25 = √6 √25 = √6 5 (k) 3 24 27 3 24 27 = 3 √24 3 √27 = 3 √8 × 3 3 = 3 √8 × 3 √3 3 = 23 √3 3 (l) 3 8 16 3 8 16 = 3 1 2 = 1 3 √2 × 3 √4 3 √4 = 3 √4 3 √8 = 3 √4 2 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 67 BAB 4
12. Permudahkan setiap yang berikut: TP 3 Simplify each of the following: CONTOH 1 2 √3 Penyelesaian: 2 √3 = 2 √3 × √3 √3 = 2√3 3 (a) 5 √2 5 √2 = 5 √2 × √2 √2 = 5√2 2 (b) 2 3 2 3 = √2 √3 × √3 √3 = √6 3 (c) √5 √8 √5 √8 = √5 √4 × 2 = √5 2√2 × √2 √2 = √10 4 (d) √7 √12 √7 √12 = √7 √4 × 3 = √7 2√3 × √3 √3 = √21 6 (e) 6 √5 6 √5 = 6 √5 × √5 √5 = 6√5 5 CONTOH 2 3√5 4√3 Penyelesaian: 3√5 4√3 = 3√5 4√3 × √3 √3 = 3√15 12 = √15 4 (f) 3√2 5√3 3√2 5√3 = 3√2 5√3 × √3 √3 = 3√6 15 = √6 5 (g) 6√7 5√27 6√7 5√27 = 6√7 5√9 × 3 = 6√7 5√9√3 = 6√7 15√3 × √3 √3 = 6√21 45 = 2√21 15 (h) 5√5 3√2 5√5 3√2 = 5√5 3√2 × √2 √2 = 5√10 6 (i) 8√5 3√12 8√5 3√12 = 8√5 3√4 × 3 = 8√5 6√3 = 8√5 6√3 × √3 √3 = 8√15 18 = 4√15 9 (j) 4√45 –6√20 4√45 –6√20 = 4√9 × 5 –6√4 × 5 = 4√9√5 –6√4√5 = 12√5 –12√5 = –1 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 68 BAB 4
13. Permudahkan setiap yang berikut: TP 3 Simplify each of the following: CONTOH 1 3 √2 + √5 Penyelesaian: 3 (√2 + √5) (√2 – √5) (√2 – √5) Darabkan dengan konjugat. Multiply with conjugate. = 3√2 – 3√5 2 – 5 = 3√2 – 3√5 –3 = √5 – √2 CONTOH 2 4 √3 – √2 Penyelesaian: 4 (√3 – √2) (√3 + √2) (√3 + √2) = 4√3 + 4√2 3 – 2 = 4√3 + 4√2 (a) 3 √7 – √2 = 3 (√7 – √2) (√7 + √2) (√7 + √2) = 3√7 + 3√2 7 – 2 = 3√7 + 3√2 5 (b) √3 √8 + √5 = √3 (√8 + √5) (√8 – √5) (√8 – √5) = √3√8 – √3√5 8 – 5 = √3√4 × 2 – √15 3 = √3√4√2 – √15 3 = 2√6 – √15 3 (c) 4 √7 – 1 = 4 (√7 – 1) (√7 + 1) (√7 + 1) = 4√7 + 4 7 – 1 = 4√7 + 4 6 = 2(√7 + 1) 3 (d) 3 2√3 + 3√5 = 3 (2√3 + 3√5) (2√3 – 3√5) (2√3 – 3√5) = 3(2√3 – 3√5) 4(3) – 9(5) = 3(2√3 – 3√5) –33 = 2√3 – 3√5 –11 = 3√5 – 2√3 11 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 69 BAB 4
(e) 3√2 2√6 – √10 = 3√2 (2√6 – √10) (2√6 + √10) (2√6 + √10) = 3√2(2√6 + √10) 4(6) – 10 = 6√12 + 3√20 14 = 6√4 × 3 + 3√4 × 5 14 = 12√3 + 6√5 14 = 6√3 + 3√5 7 (f) 2√20 1 – √5 = 2√20 (1 – √5) (1 + √5) (1 + √5) = 2√20 + 2√100 1 – 5 = 2√4 × 5 + 2(10) –4 = 4√5 + 20 –4 = –√5 – 5 (g) √y – 4 √y + 2 = √y – 4 (√y + 2) (√y – 2) (√y – 2) = y – 2√y – 4√y + 8 y – 4 = y – 6√y + 8 y – 4 (h) 8√m + 5√n √m – √n = (8√m + 5√n) (√m – √n) (√m + √n) (√m + √n) = 8m + 8√mn + 5√mn + 5n m – n = 8m + 13√mn + 5n m – n 14. Selesaikan setiap yang berikut: TP 4 Solve each of the following: CONTOH √2x + 1 = 3 Penyelesaian: (√2x + 1)2 = 32 2x + 1 = 9 2x = 8 x = 4 (a) √7x – 3 = 4 (√7x – 3)2 = 42 7x – 3 = 16 7x = 19 x = 19 7 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 70 BAB 4
(b) √2y + 1 = √5y – 11 (√2y + 1)2 = (√5y – 11)2 2y + 1 = 5y – 11 5y – 2y = 12 3y = 12 y = 4 (c) 3√2x – 1 = 3 3√2x – 1 = 3 √2x – 1 = 3 3 (√2x – 1)2 = 12 2x – 1 = 1 2x = 2 x = 1 (d) 5√m + 1 = 6 (5√m + 1)2 = 62 25(m + 1) = 36 25m + 25 = 36 25m = 11 m = 11 25 (e) √p + 6 = p √p + 6 = p (√p)2 = (p – 6)2 p = p2 – 12p + 36 p2 – 13p + 36 = 0 (p – 4)(p – 9) = 0 (p – 4) = 0 atau p – 9 = 0 p = 4 p = 9 4.3 Hukum Logaritma Laws of Logarithms NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN Logaritma/ Logarithms Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 71 BAB 4
15. Tukar setiap yang berikut dalam bentuk logaritma. TP 2 Change each of the following in logarithmic form. CONTOH 81 = 34 N = ax log3 81 = 4 loga N = x (a) 35 = 243 243 = 35 log3 243 = 5 (b) 53 = 125 125 = 53 log5 125 = 3 (c) 7–2 = 1 49 log7 1 49 = 7–2 log7 1 49 = –2 (d) 102 = 100 100 = 102 log10 100 = 2 (e) 9 —1 2 = 3 3 = 9 1 2 log9 3 = 1 2 16. Tukar setiap yang berikut dalam bentuk indeks. TP 2 Change each of the following in index form. CONTOH log2 8 = 3 23 = 8 Tip Jika / If loga N = x, maka / then ax = N (a) log2 32 = 5 25 = 32 (b) log4 64 = 3 43 = 64 (c) log7 √7 = 1 2 7 —1 2 = √7 (d) log10 0.001 = –3 10–3 = 0.001 17. Cari nilai bagi setiap yang berikut. TP 3 Find the value of each of the following. CONTOH (i) log5 √5 Penyelesaian: (ii) log3 1 27 (i) Katakan log5 √5 = x Let 5x = √5 Tukar kepada bentuk indeks. Change to index form. 5x = 5—1 2 Bandingkan indeksnya. Compare index. x = 1 2 (ii) Katakan log3 1 27 = y Let 3y = 1 27 = 1 33 = 3–3 y = –3 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 72 BAB 4
(a) log4 256 Katakan log4 256 = x Let 4x = 256 4x = 44 x = 4 (b) log6 1 36 Katakan log6 1 36 = x Let 6x = 1 36 6x = 6–2 x = –2 (c) log 3 81 Katakan log 3 81 = x Let ( 3)x = 81 3 —1 2 x = 34 1 2 x = 4 x = 8 (d) log10 0.01 Katakan log10 0.01 = x Let 10x = 0.01 10x = 10–2 x = –2 (e) logm 1 Katakan logm 1 = x Let mx = 1 mx = m0 x = 0 (f) log—1 2 0.125 Katakan log—1 2 0.125 = x Let 1 1 2 2 x = 0.125 1 1 2 2 x = 1 8 1 1 2 2 x = 1 1 2 2 3 x = 3 18. Cari nilai x dalam setiap persamaan yang berikut. TP 3 Find the value of x in each of the following equations. CONTOH (i) log3 x = –4 x = 3–4 x = 1 34 x = 1 81 (ii) logx 6 = 1 3 x —1 3 = 6 (x —1 3 )3 = 63 x = 216 (a) log2 x = 6 x = 26 x = 64 (b) log3 x = 5 x = 35 x = 243 (c) log4 x = 1 2 4 —1 2 = x x = (22 ) —1 2 x = 2 (d) logx 5 = 1 3 x —1 3 = 5 1x —1 3 2 3 = 53 x = 125 (e) log2x 1 8 = 1 2 (2x) —1 2 = 1 8 12x —1 2 2 2 = 1 1 8 2 2 2x = 1 64 x = 1 128 (f) log—1 4 x = 2 1 1 4 2 2 = x x = 1 16 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 73 BAB 4
19. Ungkapkan setiap yang berikut sebagai satu logaritma tunggal. TP 4 Express each of the following as a single logarithm. CONTOH 3 loga x + 4 loga y – loga x2 y = loga x3 + loga y 4 – loga x2 y n loga x = loga x n = loga 1 x3 × y4 x2 y 2 loga x + loga y = loga xy loga x – loga y = loga x y (a) loga m + 3 loga n – loga m2 n = loga m + loga n3 – loga m2 n = loga 1 m × n3 m2 n 2 = loga n2 m (b) loga p – 3 loga q + 2 loga r = loga p – loga q3 + loga r2 = loga 1 pr2 q3 2 (c) 1 2 loga m – 3 loga n + 2 loga q = loga m—1 2 – loga n3 + loga q2 = loga 1 q2 m n3 2 20. Cari nilai bagi setiap yang berikut. TP 4 Find the value of each of the following. CONTOH 2 log3 5 + log3 12 – 2 log3 10 = log3 52 + log3 12 – log3 102 n loga x = loga x n = log3 1 25 × 12 100 2 loga x + loga y = loga xy loga x – loga y = loga x = log y 3 3 = 1 (a) log2 48 + log2 3 – log2 9 = log2 1 48 × 3 9 2 = log2 16 = log2 24 = 4 (b) log4 24 – log4 3 4 + log4 2 = log4 1 24 × 2 3 4 2 = log4 64 = log4 43 = 3 (c) log3 18 + log3 4 – 3 log3 2 = log3 18 + log3 4 – log3 23 = log3 1 18 × 4 23 2 = log3 9 = log3 32 = 2 (d) log7 1 7 402 + log7 80 – log7 6 + log7 21 = log7 1 7 40 × 80 × 21 6 2 = log7 49 = log7 72 = 2 (e) 2 log2 6 – 1 2 log2 16 – 2 log2 3 = log2 1 62 (16)—1 2 × 32 2 = log2 1 36 4 × 9 2 = log2 1 = 0 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 74 BAB 4
21. Selesaikan setiap yang berikut. TP 5 Solve each of the following. CONTOH Diberi log3 2 = 0.631 dan log3 5 = 1.465, cari nilai bagi setiap yang berikut. Given log3 2 = 0.631 and log3 5 = 1.465, find the value of each of the following. (i) log3 2 1 2 = log3 5 2 = log3 5 – log3 2 loga x y = loga x – loga y = 1.465 – 0.631 = 0.834 (ii) log3 200 = log3 (52 × 23 ) = log3 52 + log3 23 loga xy = loga x + loga y = 2 log3 5 + 3 log3 2 loga x n = n loga x = 2(1.465) + 3(0.631) = 4.823 Diberi log4 3 = 0.792 dan log4 5 = 1.161, cari nilai bagi setiap yang berikut. Given log4 3 = 0.792 and log4 5 = 1.161, find the value of each of the following. (a) log4 1 2 3 = log4 5 3 = log4 5 – log4 3 = 1.161 – 0.792 = 0.369 (b) log4 20 = log4 20—1 2 = 1 2 log4 20 = 1 2 log4 (4 × 5) = 1 2 (log4 4 + log4 5) = 1 2 (1 + 1.161) = 1.0805 (c) log4 3.75 = log4 3 3 4 = log4 15 4 = log4 1 3 × 5 4 2 = log4 3 + log4 5 – log4 4 = 0.792 + 1.161 – 1 = 0.953 (d) log4 45 = log4 (9 × 5) = log4 (32 × 5) = log4 32 + log4 5 = 2 log4 3 + log4 5 = 2(0.792) + 1.161 = 2.745 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 75 BAB 4
22. Cari nilai bagi setiap yang berikut. TP 3 Find the value of each of the following. CONTOH log5 21 = log10 21 log10 5 = 1.892 Tukarkan asas 5 kepada asas 10. Change base 5 to base 10. (a) log4 7 = log10 7 log10 4 = 1.404 (b) log14 69 = log10 69 log10 14 = 1.604 (c) log5 3 4 = log10 3 4 log10 5 = –0.1787 (d) log4 0.48 = log10 0.48 log10 8 = –0.3530 23. Diberi log3 p = x, ungkapkan setiap yang berikut dalam sebutan x. TP 4 Given log3 p = x, express each of the following in terms of x. CONTOH logp 3 = log3 3 log3 p = 1 log3 p loga a = 1 = 1 x (a) log9 p = log3 p log3 9 = x log3 32 = x 2 log3 3 = x 2 (b) log9 27p = log3 27p log3 9 = log3 (33 × p) log3 32 = 3 log3 3 + log3 p 2 log3 3 = 3 + x 2 (c) log9p 81 = log3 81 log3 9p = log3 34 log3 (32 × p) = 4 log3 3 2 log3 3 + log3 p = 4 2 + x Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 76 BAB 4
24. Selesaikan setiap persamaan yang berikut. TP 4 Solve each of the following equations. CONTOH 5x + 2 = 95 log10 5x + 2 = log10 95 (x + 2) log10 5 = log10 95 x + 2 = log10 95 log10 5 x = 2.8295 – 2 x = 0.8295 Ambil log10 pada keduadua belah persamaan. Log10 on both sides. (a) 8x – 1 = 46 log10 8x – 1 = log10 46 (x – 1) log10 8 = log10 46 x – 1 = log10 46 log10 8 x = 1.8412 + 1 x = 2.8412 (b) 5x – 2 = 4x + 1 log10 5x – 2 = log10 4x + 1 (x – 2) log10 5 = (x + 1) log10 4 x log10 5 – 2 log10 5 = x log10 4 + log10 4 x log10 5 – x log10 4 = log10 4 + 2 log10 5 x(log10 5 – log10 4) = log10 4 + 2 log10 5 x = log10 4 + 2 log10 5 log10 5 – log10 4 x = 20.64 (c) 2x 3x = 5x + 1 6x = 5x + 1 x log10 6 = (x + 1) log10 5 x log10 6 = x log10 5 + log10 5 x log10 6 – x log10 5 = log10 5 x (log10 6 – x log10 5) = log10 5 x = log10 5 log10 6 – log10 5 x = 8.827 25. Selesaikan setiap persamaan yang berikut. TP 4 Solve each of the following equations. CONTOH log3 4x = 3 log3 8 4x = 83 4x = 512 x = 512 4 x = 128 (a) logx 750 – 3 = logx 6 logx 750 – logx 6 = 3 logx 1 750 6 2 = 3 logx 125 = 3 x3 = 125 x3 = 53 x = 5 (b) 2 log7 (x – 2) = log7 25 log7 (x – 2)2 = log7 25 (x – 2)2 = 25 (x – 2)2 = 52 x – 2 = 5 x = 7 (c) log2 4x = 6 – log2 (5x – 2) log2 4x + log2 (5x – 2) = 6 log2 4x(5x – 2) = 6 4x(5x – 2) = 26 20x2 – 8x – 64 = 0 5x2 – 2x – 16 = 0 (x – 2)(5x + 8) = 0 x = 2 atau x = – 8 5 Jawapan tidak diterima kerana logaritma bagi nombor negatif tidak tertakrif. Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 77 BAB 4
(d) log5 (8x – 4) = 2 log5 3 + log5 4 log5 (8x – 4) = log5 32 + log5 4 log5 (8x – 4) = log5 (32 × 4) log5 (8x – 4) = log5 36 8x – 4 = 36 x = 5 (e) 2 log4 3 + log4 (1 – x) – log4 3x = 1 2 log4 32 + log4 (1 – x) – log4 3x = 1 2 log4 1 9 × (1 – x) 3x 2 = 1 2 9 × (1 – x) 3x = 4—1 2 9 – 9x 3x = 2 9 – 9x = 6x 15x = 9 x = 3 5 4.4 Aplikasi Indeks, Surd dan Logaritma Applications of Indices, Surds and Logarithms 26. Selesaikan setiap masalah yang berikut. TP 5 Solve each of the following problem. CONTOH Diberi T = 2p M 10 , di mana T mewakili tempoh ayunan bandul dalam saat dan M ialah jisim bagi bandul dalam kg. Cari jisim bandul apabila tempoh ayunan bandul ialah 1.14 saat. Given T = 2p M 10 , where T represents the period of the pendulum expressed in seconds and M represents the mass of the pendulum in kg. Find the mass of the pendulum when the period of the pendulum is 1.14 seconds. Penyelesaian: T = 2p M 10 M 10 = T 2p 1 M 10 2 2 = 1 T 2p2 2 M 10 = T2 4p2 M = 10T2 4p2 Apabila T = 1.14 s, When Maka, M = 10(1.14)2 41 22 7 2 2 Thus, = 0.33 kg Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 78 BAB 4
(a) Diberi bahawa dua pemboleh ubah x dan y dihubungkan oleh persamaan Given that two variables x and y are related by the equation y = 9 4 (x – 0.4)2 Cari nilai x apabila y = 1.9. Find the value of x when y = 1.9. y = 9 4 (x – 0.4)2 (x – 0.4)2 = 4y 9 x – 0.4 = 4y 9 x = 2√y 3 + 0.4 Apabila /When y = 1.9, Maka/ Thus, x = 2√1.9 3 + 0.4 = 1.319 (b) Rajah menunjukkan sebuah segi empat tepat PQRS dengan panjang sisi ialah (1 + √5) cm dan luas segi empat ialah √80 cm2 . Hitungkan lebar, dalam cm, segi empat tepat itu. Ungkapkan jawapan anda dalam bentuk m + n√5 di mana m dan n ialah integer. The diagram shows a rectangle PQRS with the length of (1 + √5) cm and the area of the rectangle is √80 cm2. Calculate the width, in cm, of the rectangle. Express your answer in the form of m + n√5 where m and n are integers. Luas = Panjang × Lebar Area = Length × Width √80 = (1 + √5) × Lebar Lebar = √80 Width (1 + √5) = √80 (1 + √5) 1 – √5 (1 – √5) = √16 × 5 (1 + √5) 1 – √5 (1 – √5) = 4√5 (1 + √5) 1 – √5 (1 – √5) = 4√5 – 4√5√5 1 – 5 = 4√5 – 20 –4 = (5 – √5) cm (1 + 5) cm fiff R Q S P Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 79 BAB 4
Kertas 1 1. Diberi 27k – 2 243p + 1 = 1, ungkapkan p dalam sebutan k. Given 27k – 2 243p + 1 = 1, express p in terms of k. 27k – 2 243p + 1 = 1 27k – 2 = 243p + 1 (33 )k – 2 = (35 )p + 1 33k – 6 = 35p + 5 3k – 6 = 5p + 5 5p = 3k – 11 p = 3k – 11 5 2. Selesaikan persamaan: Solve the equation: logh 864 – log√h 2h = 1 logh 864 – log√h 2h = 1 logh 864 – logh 2h logh √h = 1 logh 864 – logh 2h logh h 1 2 = 1 logh 864 – logh 2h 1 1 2 2 = 1 logh 864 – 2 logh 2h = 1 logh 864 – logh (2h)2 = 1 logh 864 4h2 = 1 864 4h2 = h 216 = h3 63 = h3 h = 6 SPM 2017 SPM 2017 3. Selesaikan persamaan: Solve the equation: log4 3 + log4 (2x – 1) = 2 log4 3 + log4 (2x – 1) = 2 log4 3(2x – 1) = 2 3(2x – 1) = 42 6x – 3 = 16 6x = 19 x = 19 6 4. Diberi log3 x = p dan log3 y = q, ungkapkan log3 81x2 y dalam sebutan p dan q. Given log3 x = p and log3 y = q, express log3 81x2 y in terms of p and q. log3 81x2 y = log3 1 81x2 y 2 1 2 = log3 1 34 x2 y 2 1 2 = log3 1 32 x y 1 2 2 = log3 32 + log3 x – log3 y 1 2 = 2 log3 3 + log3 x – 1 2 log3 y = 2 + p – q 2 5. Maklumat berikut adalah berkaitan dengan hukum indeks: The following information is on the laws of indices. (am)10 = 3 √a × 3 √a × 3 √a × … × 3 √a p kali / p times dengan keadaan m dan p ialah pemalar. where m and p are constants. Nyatakan nilai m dan nilai p. State the value of m and of p. (am)10 = (a 1 3 )p \ m = 1 3 , p = 10 SPM 2013 SPM 2019 PRAKTIS PRAKTIS SPM SPM 4 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 80 BAB 4
6. Diberi logn K = 3x, ungkapkan dalam sebutan x, Given logn K = 3x, express in terms of x, (a) logn 1 K (b) log√nKn4 (a) logn 1 K = logn K–1 = –logn K = –3x (b) log√nKn4 = log√nK + log√n n4 = logn K logn n 1 2 + 4 logn n logn n 1 2 = 3x 1 2 + 4 1 2 = 6x + 8 Kertas 2 1. Diberi bahawa h = 5x dan k = 5y . It is given that h = 5x and k = 5y . (a) Ungkapkan 25x + y 125y dalam sebutan h dan k. Express 25x + y 125y in terms of h and k. (b) Cari log25 125h k2 dalam sebutan x dan y. Find log25 125h k2 in terms of x and y. (a) 25x + y 125y = 52(x + y) 53y = 52x + 2y – 3y = 52x – y = (5x )2 5y = h2 k (b) log25 125h k2 = log25 125 + log25 h – log25 k2 = log5 53 log5 52 + log5 h log5 52 – log5 k2 log5 52 = log5 53 + log5 5x – log5 52y log5 52 = 3 + x – 2y 2 SPM 2019 SPM 2014 2. Selesaikan persamaan: Solve the equation: log4 (7x + 3) – log4 x2 + log16 x2 = 3 log4 (7x + 3) – log4 x2 + log4 x2 log4 16 = 3 log4 (7x + 3) – log4 x2 + log4 x2 log4 42 = 3 2 log4 (7x + 3) – 2 log4 x2 + log4 x2 = 6 log4 (7x + 3)2 x2 x4 = 6 (7x + 3)2 x2 x4 = 46 (7x + 3)2 x2 = 4 096 (7x + 3)2 = 4 096x2 (7x + 3)2 = (64x)2 7x + 3 = 64x x = 1 19 3. Diberi log9 x = m dan log3 y = n, Given log9 x = m and log3 y = n, (a) ungkapkan xy dan x y dalam bentuk indeks dengan asas 3. express xy and x y in index form with base 3. (b) Seterusnya, hitung nilai m dan nilai n diberikan xy = 243 dan x y = 27. Hence, calculate the value of m and of n given xy = 243 and x y = 27. (a) log9 x = m x = 9m = 32m log3 y = n y = 3n (b) xy = 243 32m + n = 35 2m + n = 5 …… 1 x y = 27 32m – n = 33 2m – n = 3 ……… 2 1 + 2, 4m = 8 m = 2 Daripada 1, 2(2) + n = 5 n = 1 KBAT xy = 32m × 3n = 32m + n x y = 32m 3n = 32m – n Praktis SPM Ekstra Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 81 BAB 4
1. Farid telah berkhidmat di sebuah syarikat dengan gaji tahunannya diberi oleh persamaan G = 18 000(1.05)t – 1, dengan keadaan G ialah gaji tahunan dan t ialah bilangan tahun dia berkhidmat.Cari nilai minimum bagi t supaya gaji tahunannya akan melebihi RM60 000. Farid started working for a company with an annual salary given by an equation G = 18 000(1.05)t – 1 such that G is the annual salary and t are the number of years he is working in the company. Find the minimum value of t such that his annual salary will exceed RM60 000. 18 000(1.05)t – 1 . 60 000 (1.05)t – 1 . 60 000 18 000 (1.05)t – 1 . 10 3 (t – 1) log10 1.05 . log10 1 10 3 2 t – 1 . log10 1 10 3 2 log10 1.05 t – 1 . 24.68 t . 25.68 t = 26 2. Suatu syarikat pelaburan menawarkan suatu pulangan r% setahun. Sejumlah wang RMp dilaburkan. Jumlah pelaburan selepas suatu tempoh n tahun diberikan oleh RMp11 + r 100 2 n tertakluk kepada syarat iaitu pelaburan itu tidak dikeluarkan. Swee Hoe melaburkan RM30 000 dan pulangan ialah 6% setahun. An investment company offers a return of r% per annum. A sum of money RMp is invested. The total investment after a period of n years is given by RMp 11 + r 100 2 n subject to the condition that the investment is not withdrawn. Swee Hoe invested RM30 000 and the return is 6% per annum. (a) Berapakah jumlah pelaburan selepas 5 tahun? How much is the investment after 5 years? (b) Selepas berapa tahunkah jumlah pelaburan akan melebihi RM100 000 untuk kali pertama? After how many years will the total investment exceed RM100 000 for the first time? (a) Jumlah pelaburan = p11 + r 100 2 rt = 30 00011 + 6 100 2 5 = 30 000 (1.06)5 = RM40 146.77 (b) Jumlah pelaburan > 100 000 30 000 (1.06)n > 100 000 n log10 1.06 > log101 10 3 2 n > log10 1 10 3 2 log10 1.06 n > 20.66 n = 21 tahun Sudut Sudut KBAT KBAT KBAT Ekstra Kuiz 4 Matematik Tambahan Tingkatan 4 Bab 4 Indeks, Surd dan Logaritma 82 BAB 4
NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 1. Tentukan sama ada setiap yang berikut ialah janjang aritmetik atau bukan. TP 1 Determine whether each of the following is an arithmetic progression. CONTOH 1 19,15,11,7,… Penyelesaian: T2 − T1 = 15 − 19 = –4 T3 − T2 = 11 − 15 = –4 T4 − T3 = 7 − 11 = –4 d = –4 ialah pemalar/is a constant. \ Ini ialah janjang aritmetik. \ This is an arithmetic progression. CONTOH 2 log10 5, log10 10, log10 20, log10 80 Penyelesaian: T2 − T1 = log10 10 − log10 5 = log10 10 5 = log10 2 T3 − T2 = log10 20 − log10 10 = log10 20 10 = log10 2 T4 − T3 = log10 80 − log10 20 = log10 80 20 = log10 4 d bukan satu pemalar/d is not a constant. Bukan suatu janjang aritmetik. Not an arithmetic progression. 4.1 Janjang aritmetik Arithmetic progressions 83 BAB 5 Janjang Progressions 5.1 Janjang Aritmetik Arithmetic Progressions
(a) 3 4 , 1 1 2 , 9 4 , 3, … T2 − T1 = 1 1 2 − 3 4 = 3 4 T3 − T2 = 9 4 − 1 1 2 = 3 4 T4 − T3 = 3 − 9 4 = 3 4 d = 3 4 (pemalar/ constant) Suatu janjang aritmetik An arithmetic progression (b) log5 4, log5 12, log5 36, log5 108, … T2 − T1 = log5 12 4 = log5 3 T3 − T2 = log5 36 12 = log5 3 T4 − T3 = log5 108 36 = log5 3 d = log5 3 (pemalar/ constant) Suatu janjang aritmetik An arithmetic progression (c) 3p – 2, 2p – 1, p, 5 + p, … T2 − T1 = 2p − 1 – (3p – 2) = 1 – p T3 − T2 = p − (2p – 1) = 1 – p T4 − T3 = 5 + p − p = 5 d bukan satu pemalar Bukan suatu janjang aritmetik d is not a constant Not an arithmetic progression 2. Nyatakan sebutan pertama dan beza sepunya bagi setiap janjang aritmetik yang berikut. TP 2 State the first term and the common difference for each of the following arithmetic progression. CONTOH 1 −3, 5, 13, 21, … Penyelesaian: a = −3 d = 5 − (−3) = 8 CONTOH 2 17, 12, 7, 2, … Penyelesaian: a = 17 d = 12 − 17 = –5 CONTOH 3 x + 3, 2x + 5, 3x + 7, 4x + 9, … Penyelesaian: a = x + 3 d = 2x + 5 – (x + 3) = x + 2 (a) –6, 3, 12, 21, … a = –6 d = 3 − (–6) = 9 (b) 11, 8 1 2 , 6, 3 1 2 , … a = 11 d = 6 − 8 1 2 = –2 1 2 (c) 3x + 7, 4x + 5, 5x + 3, 6x + 1, … a = 3x + 7 d = 5x + 3 – (4x + 5) = x – 2 3. Tentukan nilai sebutan tertentu bagi janjang aritmetik yang berikut. TP 3 Determine the value of the specific term for the arithmetic progression. CONTOH 1 11, 7, 3, … Cari/ Find T9 . Penyelesaian: a = T1 = 11 d = T2 – T1 = 7 – 11 = –4 Tn = a + (n – 1)d T9 = 11 + (9 – 1)(–4) = 11 – 32 = –21 CONTOH 2 log 3, log 6, log 12,…, … Cari/ Find T7 . Penyelesaian: a = T1 = log 3 d = T3 – T2 = log 12 – log 6 = log 12 6 = log 2 Tn = a + (n – 1)d T7 = log 3 + (7 – 1) log 2 = log 3(26 ) = log (3 × 64) = log 192 Matematik Tambahan Tingkatan 4 Bab 5 Janjang 84 BAB 5
(a) −8, −2, 4, 10, … Cari / Find T12 a = −8 d = 4 − (−2) = 6 T12 = –8 + 11(6) = 58 (b) log 10, log 100, log 1000, … Cari / Find T8 a = log 10 d = log 100 – log 10 = log 100 10 = log 10 T8 = log10 10 + 7(log 10) = log 10(107 ) = 8 log 10 4. Hitungkan bilangan sebutan bagi setiap janjang aritmetik yang berikut. TP 3 Calculate the number of terms in each of the following arithmetic progressions. CONTOH 1 4.1, 4.7, 5.3, 5.9, …, 11.3 Penyelesaian: a = 4.1, d = 4.7 − 4.1 = 0.6 Tn = a + (n − 1)d = 4.1 + (n − 1)(0.6) = 0.6n + 3.5 0.6n + 3.5 = 11.3 0.6n = 7.8 n = 13 CONTOH 2 3y, –y, –5y, …, –65y Penyelesaian: a = 3y, d = –y − 3y = –4y Tn = a + (n − 1)d = 3y + (n − 1)(–4y) = 7y − 4ny –4ny + 7y = –65y –4ny = –72y n = 18 (a) −11, −7, −3, …, 125 a = −11 , d = −7 − (−11) = 4 Tn = a + (n − 1)d = –11 + (n – 1)(4) = 4n − 15 4n – 15 = 125 4n = 140 n = 35 (b) 17, 13, 9, …, –31 a = 17, d = 13 − 17 = −4 Tn = a + (n − 1)d = 17 + (n − 1)( −4) = −4n + 21 –4n + 21 = –31 4n = 52 n = 13 (c) 1 5 , 2 5 , 3 5 , …, 14 5 a = 1 5 , d = 2 5 − 1 5 = 1 5 Tn = a + (n − 1)d = 1 5 + (n – 1)( 1 5 ) = 1 5 n 1 5 n = 14 5 n = 14 (d) 3m, m, –m, –3m, …, −13m a = 3m , d = m − 3m = −2m Tn = a + (n − 1)d = 3m + (n − 1)( −2m) = 5m – 2nm 5m – 2mn = –13m 2nm = 18m 2n = 18 n = 9 Matematik Tambahan Tingkatan 4 Bab 5 Janjang 85 BAB 5
5. Tentukan hasil tambah n sebutan pertama bagi suatu janjang aritmetik. TP 3 Determine the sum of the first n terms in an arithmetic progression. CONTOH 1 3, 5, 7, 9, … (12 sebutan yang pertama) (first 12 terms) Penyelesaian: a = 3, d = 5 − 3 = 2 S12 = n 2 [2a + (n – 1)d] = 12 2 [2(3) + 11(2)] = 6(6 + 22) = 168 CONTOH 2 5, 2, –1, –4, … (10 sebutan yang pertama) (first 10 terms) Penyelesaian: a = 5, d = 2 − 5 = −3 Sn = n 2 [2a + (n – 1)d] S10 = 10 2 [2(5) + 9(–3)] = 5(10 – 27) = –85 (a) −15, −11, −7, −3, … (16 sebutan yang pertama) (first 16 terms) a = −15, d = −11 − (−15) = 4 S16 = n 2 [2a + (n – 1)d] = 16 2 [2(–15) + 15(4)] = 8(–30 + 60) = 240 (b) 19, 14, 9, 4, … (14 sebutan yang pertama) (first 14 terms) a = 19, d = 14 − 19 = –5 S14 = n 2 [2a + (n – 1)d] = 14 2 [2(19) + 13(–5)] = 7(38 – 65) = –189 6. Tentukan hasil tambah sebutan tertentu yang berturutan bagi suatu janjang aritmetik. TP 3 Determine the sum of specific number of consecutive terms in an arithmetic progression. CONTOH 1 23, 17, 11, …., –67 Penyelesaian: a = 23, d = 17 − 23 = −6, l = –67 Tn = a + (n – 1)d = 23 + (n – 1)(–6) = 29 – 6n 29 – 6n = –67 6n = 96 n = 16 S16 = n 2 [a + l] = 16 2 [23 + (–67)] = 8(–44) = –352 CONTOH 2 –6, –4, –2, 0, …., 16 Penyelesaian: a = –6, d = –4 − (–6) = 2, l = 16 Tn = a + (n – 1)d = –6 + (n – 1)(2) = 2n – 8 2n – 8 = 16 2n = 24 n = 12 S12 = n 2 [a + l] = 12 2 [–6 + 16] = 6(10) = 60 Matematik Tambahan Tingkatan 4 Bab 5 Janjang 86 BAB 5
(a) −15, −12, −9, …, 30 a = −15, d = −12 − (−15) = 3, l = 30 Tn = a + (n – 1)d = –15 + (n – 1)(3) = 3n – 18 3n – 18 = 30 3n = 48 n = 16 S16 = n 2 [a + l] = 16 2 [–15 + 30] = 120 (b) 9, 4, –1, …, –86 a = 9, d = 4 − 9 = –5, l = –86 Tn = a + (n – 1)d = 9 + (n – 1)(–5) = 14 – 5n 14 – 5n = –86 5n = 100 n = 20 S20 = n 2 [a + l] = 20 2 [9 – 86] = –770 7. Hitung hasil tambah sebutan tertentu yang berturutan bagi suatu janjang aritmetik. TP 4 Calculate the sum of specific number of consecutive terms in an arithmetic progression. CONTOH 1 3, 10, 17, …. (sebutan ke-7 hingga sebutan ke-12) (7th term to 12th term) Penyelesaian: Kaedah 1/Method 1 a = 3, d = 10 − 3 = 7 Hasil tambah T7 hingga T12/Sum of T7 to T12 = S12 – S6 = 12 2 [2(3) + 11(7)] – 6 2 [2(3) + 5(7)] = 6(83) – 3(41) = 375 Kaedah 2/Method 2 T7 = 3 + 6(7) = 45 T12 = 3 + 11(7) = 80 45, …, 80 (6 sebutan/terms) Hasil tambah T7 hingga T12/Sum of T7 to T12 S6 = n 2 [a + l] = 6 2 [45 + 80] = 375 Tip S12 = T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12 S6 = T1 + T2 + T3 + T4 + T5 + T6 \ S12 – S6 = T7 + T8 + T9 + T10 + T11 + T12 {Hasil tambah T7 hingga T12} {Sum of T7 to T12} (a) −22, −16, −10, … (sebutan ke-5 hingga sebutan ke-14) (5th term to 14th term) a = −22, d = −16 − (−22) = 6 Hasil tambah T5 hingga T14 Sum of T5 to T14 = S14 – S4 = 14 2 [2(–22) + 13(6)] – 4 2 [2(–22) + 3(6)] = 7(34) – 2(–26) = 290 (b) –1, 4, 9, … (sebutan ke-4 hingga sebutan ke-12) (4th term to 12th term) a = –1, d = 4 − (–1) = 5 Hasil tambah T4 hingga T12 Sum of T4 to T12 = S12 – S3 = 12 2 [2(–1) + 11(5)] – 3 2 [2(–1) + 2(5)] = 6(53) – 12 = 306 Matematik Tambahan Tingkatan 4 Bab 5 Janjang 87 BAB 5
(c) 6.7, 5.3, 3.9, … (sebutan ke-6 hingga sebutan ke-10) (6th term to 10th term) a = 6.7, d = 5.3 − 6.7 = −1.4 Hasil tambah T6 hingga T10 Sum of T6 to T10 = S10 – S5 = 10 2 [2(6.7) + 9(–1.4)] – 5 2 [2(6.7) + 4(–1.4)] = 5(0.8) – 19.5 = –15.5 (d) 5, 9 2 , 4, … (sebutan ke-9 hingga sebutan ke-16) (9th term to 16th term) a = 5, d = 9 2 − 5 = –0.5 Hasil tambah T9 hingga T16 Sum of T9 to T16 = S16 – S8 = 16 2 [2(5) + 15(–0.5)] – 8 2 [2(5) + 7(–0.5)] = 8(2.5) – 4(6.5) = –6 8. Selesaikan masalah berikut yang melibatkan janjang aritmetik. TP 5 Solve the following problems involving arithmetic progressions. CONTOH 1 Diberi x + 3, 2x – 3 dan x – 5 adalah tiga sebutan berturut-turut bagi janjang aritmetik. Cari nilai bagi x. Given that x + 3, 2x – 3 and x – 5 are three consecutive terms of an arithmetic progressions. Find the value of x. Penyelesaian: 2x – 3 – (x + 3) = x – 5 – (2x – 3) 2x – 3 – x – 3 = x – 5 – 2x + 3 x – 6 = –x – 2 2x = 4 x = 2 (a) Diberi p, 3p – 1 dan 4p adalah tiga sebutan berturut-turut bagi janjang aritmetik. Cari nilai bagi p. Given that p, 3p – 1 and 4p are three consecutive terms of an arithmetic progression. Find the value of p. 4p – (3p – 1) = 3p – 1 – p p + 1 = 2p – 1 p = 2 CONTOH 2 Sebutan ke-n, Tn suatu janjang aritmetik diberi sebagai Tn = 4n + 7. Cari hasil tambah 5 sebutan pertama janjang itu. The nth term, Tn of an arithmetic progression is given as Tn = 4n + 7. Find the sum of the first 5 terms of the progression. Penyelesaian: Tn = 4n + 7 a = T1 = 4(1) + 7 = 11 T2 = 4(2) + 7 = 15 d = T2 – T1 = 15 – 11 = 4 S5 = 5 2 [2(11) + 4(4)] = 95 (b) Sebutan ke-n, Tn suatu janjang aritmetik diberi sebagai Tn = 5n – 2. Cari hasil tambah 7 sebutan pertama janjang itu. The nth term, Tn of an arithmetic progression is given as Tn = 5n – 2. Find the sum of the first 7 terms of the progression. Tn = 5n – 2 a = T1 = 5(1) – 2 = 3 T2 = 5(2) – 2 = 8 d = T2 – T1 = 8 – 3 = 5 S7 = 7 2 [2(3) + 6(5)] = 126 Tip d = Tn – Tn – 1 Matematik Tambahan Tingkatan 4 Bab 5 Janjang 88 BAB 5
CONTOH 3 Diberi hasil tambah n sebutan pertama bagi janjang aritmetik 25, 18, 11, … ialah 4. Cari nilai bagi n. Given the sum of the first n term of the arithmetic progression 25, 18, 11, … is 4. Find the value of n. Penyelesaian: a = 25, d = 18 – 25 = –7 Sn = n 2 [2(25) + (n – 1)(–7)] n 2 [50 + 7 – 7n]= 4 7n2 – 57n + 8 = 0 (7n – 1)(n – 8) = 0 n = 1 7 (tidak diterima/not accepted), n = 8 (c) Diberi hasil tambah n sebutan pertama bagi janjang aritmetik –22, –16, –10, … ialah 132. Cari nilai bagi n. Given the sum of the first n term of the arithmetic progression –22, –16, –10, … is 132. Find the value of n. a = –22, d = –16 – (–22) = 6 Sn = n 2 [2(–22) + (n – 1)(6)] n 2 [–44 – 6 + 6n] = 132 6n2 – 50n – 264 = 0 3n2 – 25n – 132 = 0 (3n + 11)(n – 12) = 0 n = – 11 3 (tidak diterima/not accepted), n = 12 CONTOH 4 Dalam suatu janjang arimetrik, sebutan ke-6 ialah 28 dan sebutan ke-10 ialah 16. Cari sebutan pertama dan beza sepunya. In an arithmetic progression, the 6th term is 28 and the 10th term is 16. Find the first term and the common difference. Penyelesaian: T6 = 28 = a + 5d …… (i) T10 = 16 = a + 9d …… (ii) (ii) – (i): 4d = –12 d = –3 28 = a + 5(–3) a = 43 Tip Tn = Sn – Sn – 1 (d) Dalam suatu janjang arimetrik, sebutan ke-5 ialah 13 dan sebutan ke-12 ialah 27. Cari sebutan pertama dan beza sepunya. In an arithmetic progression, the 5th term is 13 and the 12th term is 27. Find the first term and the common difference. T5 = 13 = a + 4d …… (i) T12 = 27 = a + 11d ……(ii) (ii) –(i): 7d = 14 d = 2 13 = a + 4(2) a = 5 CONTOH 5 Hasil tambah n sebutan pertama suatu janjang aritmetik diberi oleh Sn = 3n2 + n. Hitung sebutan ke-6. The sum of the first n term of an arithmetic progression is given by Sn = 3n2 + n. Calculate the 6th term. Penyelesaian: T6 = S6 – S5 = [3(6)2 + 6] – [3(5)2 + 5] = 114 – 80 = 34 (e) Hasil tambah n sebutan pertama suatu janjang aritmetik diberi oleh Sn = 5 2 n2 – 9n. Hitung sebutan ke-5. The sum of the first n term of an arithmetic progression is given by Sn = 5 2 n2 – 9n. Calculate the 5th term. T5 = S5 – S4 = 3 5 2 (5)2 – 9(5)4 – 3 5 2 (4)2 – 9(4)4 = 3 125 2 – 454 – [40 – 36] = 35 2 – 4 = 13.5 Matematik Tambahan Tingkatan 4 Bab 5 Janjang 89 BAB 5