MODULE AND MORE Matematik Tambahan Tingkatan 5

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 198 8 (b) V = 100 – 2t – t 2 (i) Apabila t = 0, V = 100 cm3 (ii) dv dt = –2 – 2t Apabila t = 4, dv dt = –2 –2(4) = –10 cm3 s–1 10. (a) Rajah 2 / Diagram 2 Dalam Rajah 2, diberi luas rantau P ialah 4 unit2 dan ∫ 4 –2 f(x) = 6. Cari In Diagram 2, it is given that the area of P is 4 unit2 and ∫ 4 –2 f(x) = 6. Find (i) ∫ 4 0 f(x)dx (ii) ∫ 0 –2 f(x)dx (iii)luas Q. the area of Q. [3 markah / marks] (a) (i) ∫ 4 0 f(x)dx = 6 + 4 = 10 (ii) ∫ 0 –2 f(x)dx = –4 (iii)luas Q/ Area Q = 6 × 4 – 10 = 14 unit2 (b) Satu lengkung mempunyai terbitan kedua d2 y dx2 = 1 − x2 . Persamaan tangen kepada lengkung pada (1, 4) ialah y = 5 − x. Tentukan persamaan lengkung itu. A curve has a second derivation d2 y dx2 = 1 − x2. The equation of tangent to the curve at (1, 4) is y = 5 − x. Determine the equation of the curve. [3 markah / marks] d2 y dx2 = 1 – x2 dy dx = x – 1 3 x3 + c Apabila x = 1, dy dx = –1 1 – 1 3 + c = –1 c = –1 – 2 3 = –5 3 dy dx = x – 1 3 x3 – 5 3 y = x2 2 – 1 12x4 – 5 3 x + c1 (1, 4); 4 = 1 2 – 1 12 – 5 3 + c c1 = 21 4 ∴ y = x2 2 – 1 12x4 – 5 3 x + 21 4 11. (a) Terdapat n pasu bunga di hadapan rumah Raju. Dia ingin menyusun 2 daripada n pasu bunga itu. Didapati bahawa bilangan cara menyusunnya ialah 56. Cari nilai n. There are n flower pots in front of Raju’s house. He wants to arrange 2 of the n flower pots. It is found that the number of ways to arrange it is 56. Find the value of n. [2 markah / marks] n p2 = 56 n! (n – 2)! = 56 n(n – 1) = 56 n2 – n – 56 = 0 (n – 8)(n + 7) = 0 n = 8 (b) 10 orang dijemput duduk pada suatu meja bulat. Cari bilangan cara menyusun kesemuanya jika 10 people are invited to sit on a round table. Find the number of ways to arrange all of them if (i) tidak ada halangan. there is no restrictions. 0 y Q P x 6 –2 4

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 199 (ii) terdapat dua pasang suami isteri mesti duduk bersama. there are two couples who need to sit next to each other. [2 markah / marks] (i) 9! = 362 880 (ii) 7! 2!2! = 1 260 12. Suatu pemboleh ubah rawak diskret X mempunyai taburan P(X = x) = x 2k untuk x = 1, 2, 3, 4 dan 5. A random discrete variable X has a distribution P(X = x) = x 2k for x = 1, 2, 3, 4 and 5. (a) Cari nilai k. Find the value of k. (b) Bina jadual taburan kebarangkalian bagi X. Construct a probability distribution table for X. (c) Cari/Find P(2 < X ≤ 5) [5 markah / marks] (a) P[X = 1] + P[X = 2] + P[X = 3] + P[X = 4] + P[X = 5] = 1 1 2k + 2 2k + 3 2k + 4 2k + 5 2k = 1 15 2k = 1 k = 15 2 (b) x 1 2 3 4 5 P(X = x) 1 15 2 15 3 15 4 15 5 15 (c) P[2 < x ≤ 5] = 1 5 + 4 15 + 1 3 = 12 15 = 4 5 Bahagian B / Section B [16 markah / marks] Jawab mana-mana dua soalan daripada bahagian ini. Answer any two questions from this section. 13. Rajah 3 menunjukkan segi tiga ABC. Diagram 3 shows a triangle ABC. Rajah 3 / Diagram 3 (a) Hitung sudut cakah ABC. Calculate the obtuse angle ABC. [2 markah / marks] (b) Lakar dan labelkan sebuah segi tiga berlainan APC daripada ABC, dengan keadaan ∠ACB, AB dan AC tidak berubah. Sketch and label another different triangle APC from ABC, such that ∠ACB, AB and AC remain unchanged. [2 markah / marks] (c) Hitung panjang CP. Calculate the length of CP. [2 markah / marks] (d) Cari luas ABC. Find the area of ABC. [2 markah / marks] A B C 30° 7 cm 10 cm

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 200 8 (a) sin B 10 = sin 30° 7 B = 134° 25’ (b) (c) ∠APC = 45°35’ ; ∠PAC = 104°25’ PC2 = 72 + 102 – 2(7)(10) kos/ cos 104°25’ PC = 13.56 cm (d) Luas/ Area ABC = 1 2 (10)(7) sin(150° – 134°35’) = 9.4 cm2 14. (a) Jika tan A = 3 dan tan (A + B) = 1, cari tan B dalam bentuk surd. If tan A = 3 and tan (A + B) = 1, find tan B in surd form [2 markah / marks] (b) (i) Rajah 4 menunjukkan graf y = a sin bx + c. Diagram 4 shows a graph of y = a sin bx + c. Rajah 4 / Diagram 4 Nyatakan nilai a, b dan c. State the value of a, b and c. [3 markah / marks] (ii) Pada graf yang sama, lakar satu graf yang sesuai untuk menyelesaikan persamaan x(a sin bx + c) = π. Nyatakan bilangan penyelesaian. On the same axes, sketch a suitable graph to solve the equation x(a sin bx + c) = π. State the number of solutions. [3 markah / marks] (a) tan A = 3 tan(A + B) = tan A + tan B 1 – tan A tan B = 1 3 + tan B = 1 – 3 tan B (1 + 3)tan B = 1 – 3 tan B = (1 – 3)(1 – 3) (1 + 3)(1 – 3) = 1 – 2 3 + 3 1 – 3 = 3 – 2 10 cm 7 cm 7 cm 30° A P B C y 1 0 π 2π 2 3 –1 x

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 201 (i) y = a sin bx + c a = 3 , b = 1 2 , c = −1 (ii) x(a sin bx + c) = π a sin bx + c = π x y = π x Apabila/When x = π, y = 1 Apabila/When x = π 2 , y = 2 2 penyelesaian/ solutions 15. Dua duit syiling yang masing-masing berjejari 1.5 cm dan 1 cm menyentuh antara satu sama lain di luar dan berdiri tegak di atas garis lurus PQ seperti ditunjukkan dalam Rajah 5. The two coins with a radius of 1.5 cm and 1 cm respectively touch to each other and stand on a straight line as shown in Diagram 5. Rajah 5 / Diagram 5 Cari / Find (a) perimeter bagi rantau berlorek. the perimeter of the shaded region. [4 markah / marks] (b) luas bagi rantau berlorek. the area of the shaded region. [4 markah / marks] (a) Kos/ cos θ = 0.5 2.5 θ = 78°27’ Panjang lengkok CD = 78°27’ 360° × 2p(1.5) = 2.05 cm Length of arc CD Panjang lengkok BD = 90° + 11.33’ 360° × 2p(1) = 1.77 cm Length of arc BD CB = 2.52 – 0.52 = 2.45 cm Perimeter = 2.05 + 1.77 + 2.45 = 6.27 cm (b) luas berlorek = luas trapezium OABC – sektor ODC – sektor ADB Shaded area = area of trapezium OABC – sector ODC – sector ADB = 1 2 [1 + 1.5]2.45 – 78°27’ 360° × p(1.5)2 – 101°33’ 360° × p(1)2 = 0.64 cm2 y 1 0 2 π π 2π 2 3 –1 x P 1.5 cm 1 cm Q P 1.5 cm 1 cm O D C B 0.5 cm A 2.5 cm θ Q

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 202 8 Kertas 2 / Paper 2 [2 jam 30 minit / 2 hours 30 minutes] Bahagian A / Section A [50 markah / marks] Jawab semua soalan. Answer all questions. 1. Selesaikan sistem persamaan linear berikut. Solve the following system of linear equations. x + 2y − z = 4 2x + y + z = −2 x + 2y + z = 2 [6 markah / marks] 2. Satu fungsi ditakrifkan oleh f : x → 3 − 2x A function is defined by f : x → 3 − 2x. (a) Lakar graf bagi f untuk domain −1 ≤ x ≤ 3. Sketch the graph for f for the domain −1 ≤ x ≤ 3. [2 markah / marks] (b) Pada paksi yang sama, lakar graf yang sepadan bagi f −1. On the same axes, sketch the corresponding graph of f −1. [3 markah / marks] (c) Nyatakan domain bagi f −1. State the domain for f −1. [1 markah / mark] (a) & (b) x + 2y – z = 4 ...➀ 2x + y + z = –2 ...➁ ➀ + ➁ 3x + 3y = 2 ...➃ x + 2y + z = 2 ...➂ ➀ + ➂ 2x + 4y = 6 x + 2y = 3 ...➄ 3[3 – 2y] + 3y = 2 9 – 6y + 3y = 2 7 = 3y y = 7 3 x = 3 – 21 7 3 2 = –5 3 –5 3 + 14 3 – 4 = z z = –1 0 1 1 2 3 (–3, 3) (3,–3) (5,–1) (–1, 5) 4 5 –1 –3 –2 –1 –2 –3 2 3 4 5 f(x) f –1(x) f(x) = 3 – 2x x (c) Domain bagi f−1(x) Domain for f −1(x) −3 ≤ x ≤ 5

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 203 3. (a) Cari nilai p, dengan keadaan y = 6 ialah satu tangen kepada lengkung y = x2 + (1 − p)x + 2p. Find the value of p, such that y = 6 is a tangent to the curve y = x2 + (1 − p)x + 2p. [3 markah / marks] (b) Rajah 1 menunjukkan sebuah segi empat sama ABCD dengan setiap sisi berukuran 4 cm. BEFG ialah sebuah segi empat tepat. Diagram 1 shows a square ABCD with each side of 4 cm. BEFG is a rectangle. A E D F C B G 4 cm 2x cm Rajah 1 / Diagram 1 Jika 2AE = BG = 2x, If 2AE = BG = 2x, (i) tunjukkan bahawa luas BEFG diberi oleh L(x) = 8x − 2x2 . show that the area of BEFG is given by L(x) = 8x − 2x2. [2 markah / marks] (ii) cari nilai x supaya luas BEFG ialah maksimum. find the value of x so that the area of BEFG is maximum. [2 markah / marks] (iii) cari luas maksimum. find the maximum area. [1 markah / mark] 4. (a) Rajah 2 menunjukkan garis lurus y = x + 4 menyilang lengkung y = 4 + 3x − x2 pada A dan B. Diagram 2 shows a straight line y = x + 4 intersects the curve y = 4 + 3x − x2 at A and B. A 0 B y x Rajah 2 / Diagram 2 (a) y = x2 + (1 – p)x + 2p y = (1 – p)2 4 + 2p – x + 1 1 – p 2 2 2 2p – (1 – p)2 4 = 6 8p – (1 – 2p + p2 ) = 24 p2 – 10p + 25 = 0 (p – 5)(p – 5) = 0 p = 5 (b) (i) L(x) = 2x(4 – x) = 8x – 2x2 (ii) dL(x) dx = 8 – 4x = 0 x = 2 Apabila x = 2 L(2) = 8(2) – 2(2)2 = 16 – 8 = 8 cm2 A E D C B F G 4 4 2x x

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 204 8 (i) Cari koordinat A dan B. Find the coordinates of A and B [2 markah / marks] (ii) Seterusnya, cari luas yang dibatasi oleh lengkung dan garis lurus itu. Then, find the area enclosed by the curve and the line. [3 markah / marks] (b) Luas yang dibatasi oleh lengkung, x = 0 dan x = 1 diputarkan 180° pada paksi-x, cari isi padu yang dijanakan. The area enclosed by the curve, x = 0 and x = 1 is rotated 180° about the x-axis, find the volume generated. [3 markah / marks] 5. (a) Kebarangkalian sebuah mesin menghasilkan plastik yang rosak ialah 0.01. The probability that a machine produces a spoilt plastic is 0.01. (i) Cari min dan sisihan piawai bagi taburan plastik yang rosak apabila jumlahnya ialah 1 000. Find the mean and the standard deviation for the plastic distribution when there are total of 1 000 plastics. (ii) Jika 5 plastik dipilih secara rawak sebagai sampel, hitung kebarangkalian sekurang-kurangnya satu yang rosak. If 5 plastics are chosen at random as a sample, calculate the probability that at least one of them is spolit. [4 markah / marks] (b) Pendapatan pekerja-pekerja di sebuah syarikat bertaburan normal dengan min RM200 seminggu dan sisihan piawai RM40. The income of the workers in a company is normally distributed with the mean of RM200 per week and the standard deviation of RM40. (i) Hitung kebarangkalian bahawa seorang pekerja dipilih secara rawak mendapat pendapatan di antara RM150 dan RM180 dalam seminggu. Calculate the probability that a worker is chosen at random has an income between RM150 and RM180 in a week. (ii) Jika 10 orang pekerja mendapat pendapatan lebih daripada RM230 seminggu, cari bilangan pekerja dalam syarikat itu. If 10 workers have incomes more than RM230, find the number of workers in the company. [4 markah / marks] (a) (i) y = x + 4 y = 4 + 3x – x2 x + 4 = 4 + 3x – x2 x2 – 2x = 0 x(x – 2) = 0 x = 0 ; 2 A(0, 4) B(2, 6) (ii) Luas/Area = ∫ 2 0 (4 + 3x – x2 )dx – 1 2 [4 + 6]2 = 4x + 3 2 x2 – x3 3  2 0 – 10 = 1 11 3 – 10 = 1 1 3 unit2 (b) Isipadu = ∫ 1 0 y2 dx Volume = ∫ 1 0 (4 + 3x – x2 )2 dx = ∫ 1 0 (16 + 24x + x2 – 6x3 + x4 )dx = 16x + 12x2 + 1 3 x3 – 3 2 x4 + 1 5 x5 4 1 0 = 1 27 30 unit3 ∴ isipadu sebenar = 31 13 60 unit3 (a) (i) p = 0.01, n = 1000 m = np = 1000(0.01) = 10 σ = npq = 1000(0.01)(0.99) = 3.15 (ii) P[x ≥ 1] = 1 – P[X = 0] = 1 – 5 C0 (0.01)0 (0.99)5 = 0.049 (b) (i) m = 200, σ = 40 P[ 150 ≤ z ≤ 180]

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 205 6. Rajah 3 menunjukkan sebuah segi tiga ADC. Diagram 3 shows a triangle ADC. A D C Rajah 3 / Diagram 3 Diberi → AD = i ~ + 2 j ~, → DC = 8 i ~ + j ~ dan → DB = 2 i ~ – j ~. Given → AD = i ~ + 2 j ~ , → DC = 8 i ~ + j ~ and → DB = 2 i ~ – j ~ . (a) Cari vektor-vektor → AB dan → AC dalam sebutan i dan j. Find the vector → AB and → AC in terms of i ~ and j ~ . [2 markah / marks] (b) Tunjukkan bahawa B terletak pada garis AC. Seterusnya, cari AB : BC. Show that B lies on the line AC. Then, find AB : BC. [3 markah / marks] (c) Jika luas ADC = 24 unit2 , cari luas segi tiga ADB. If the area of ADC = 24 unit2, find the area of triangle ADB. [2 markah / marks] = P 150 – 200 40 < z < 180 – 200 40 4 = P 5 – 4 < z < 1 – 2 4 = 0.2029 (ii) P[X > 230]= P[Z > 230 – 200 40 ] = P[Z > 3 4 ] = 0.2266 Bilangan pekerja = 8 0.2266 Number of wokers  35 (a) → AB = → AD + → DB = i ~ + 2 j ~ + 2 i ~ – j ~ = 3 i ~ + j ~ → AC = → AD + → DC = i ~ + 2 j ~ + 8 i ~ + j ~ = 9 i ~ + 3 j ~ (b) → AC = 3 → AB A, B dan C segaris A, B and C collinear AB : BC = 2 : 1 (c) ∆ADC = 1 2 ACh = 24 ∴∆ADB = 2 3 × 24 = 16 unit2 A h B D C 2 1

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 206 8 7. Jadual 1 menunjukkan nilai-nilai eksperimen bagi x dan y yang diperolehi daripada suatu eksperimen. Table 1 shows the values of x and y obtained from an experiment. x 1 2 3 4 5 y 5.9 5,6 5.8 6.0 6.2 Jadual 1 / Table 1 Diketahui x dan y dihubungkan oleh persamaan y = a x + b x . It is known that x and y are related by the equation y = a x + b x . (a) Ungkapkan persamaan ini dalam bentuk Y = mX + c Express the equation in the form of Y = mX + c [1 markah / mark] (b) Seterusnya, lukis satu graf penyuaian terbaik. Then, draw a best fit graph. [4 markah / marks] (c) Daripada graf, anggarkan From the graph, estimate (i) nilai a dan b. the value of a and b. (ii) nilai y apabila x = 2.5. the value of y when x = 2.5 [2 markah / marks] (a) y = a x + b x y x = ax + b (b) x 1 2 3 4 5 y x 5.9 7.9 10.0 12 13.9 2 1 0 2 3 4 5 4 6 8 10 12 14 y x (c) (i) b = 4 a = 14 – 4 5 = 2 (ii) Apabila/When x = 2.5 y x = 9 y = 9 2.5 = 5.69

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 207 Bahagian B / Section B [30 markah / marks] Jawab mana-mana tiga soalan daripada bahagian ini. Answer any three questions from this section. 8. (a) Selesaikan 7 tan x = 4 tan (45° – x) untuk 0° ≤ x ≤ 360°. Solve 7 tan x = 4 tan (45° – x) for 0° ≤ x ≤ 360°. [4 markah / marks] (b) Lakarkan pada paksi yang sama bagi lengkung y = kos x dan y = sin 2x untuk 0 ≤ x ≤ 2π. Nyatakan bilangan penyelesaian bagi persamaan-persamaan yang berikut. Sketch on the same axes for the curve y = cos x and y = sin 2x for 0 ≤ x ≤ 2π. State the number of solutions for the following equations. (i) sin 2x = 0.5 (ii) sin 2x = kos/ cos x (iii)sin 2x – kos/ cos x = 1 [6 markah / marks] (a) 7 tan x = 4 tan (45° – x) = 4[1 – tan x] 1 + tan x 7 tan x + 7 tan2 x = 4 – 4 tan x 7 tan2 x + 11 tan x – 4 = 0 tan x = –11 ± 112 – 4(7)(–4) 14 tan x = –11 + 233 14 x = 16° 56’, 196°56’ tan x = –11 – 233 14 x = 118°4’, 298°4’ (b) (i) 4 penyelesaian/solutions (ii) 4 penyelesaian/solutions (iii)sin 2x = 1 + kos/ cos x 2 penyelesaian/solutions y x 16°56′ 16°56′ y x 61°56′ 61°56′ y y = 1 + kos x y = sin 2x y = kos x y = 0.5 x 2 1 0 π 2π –1

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 208 8 9. (a) Rajah 4 menunjukkan sebuah segi empat selari PQRS. Diagram 4 shows a parallelogram PQRS. Rajah 4 / Diagram 4 (i) Jika QS selari dengan persamaan 5x = 2 − y, cari persamaan QS. If QS is parallel to the equation 5x = 2 − y, find the equation of QS. (ii) Jika PR berserenjang dengan QR, cari persamaan QR. If PR is perpendicular to QR, find the equation of QR. (iii) Seterusnya, cari Then, find (a) koordinat Q dan S. the coordinates of Q and S. (b) luas segi empat selari PQRS. the area of the parallelogram PQRS. [6 markah / marks] (b) Satu titik A(x, y) bergerak supaya sudut PAQ sentiasa 90°. Tunjukkan bahawa lokus A melalui titik R. A point A(x, y) moves such that the angle PAQ is always 90°. Show that the locus of A passes through the point R. [4 markah / marks] y S Q P(–1, 5) 0 R(5, 1) x (a) (i) 5x = 2 – y y = –5x + 2 Titik tengah PR = (2, 3) ∴Persamaan QS / Equation QS y – 3 = –5(x – 2) y = –5x + 13 ...➀ (ii) mPR = 5 – 1 –1 – 5 = 4 –6 = 2 –3 Persamaan QR,/ Equation QS, y – 1 = 3 2 (x – 5) y = 3 2 x – 15 2 + 1 y = 3 2 x – 13 2 ...➁ (iii)(a) ➀ = ➁ –5x + 13 = 3 2 x – 13 2 –10x + 26 = 3x – 13 39 = 13x x = 3 y = –2 ∴Q(3, –2) 3 + x 2 = 2 –2 + y 2 = 3 x = 1 y = 8 S = (1, 8) (b) luas = 1 2  –1 3 5 1 –1 5 –2 1 8 5  area = 1 2 |2 + 3 + 40 + 5 – 15 + 10 –1 + 8| = 1 2 |52| = 26 unit2 (b) PAQ = 90° mPA = y – 5 x + 1 , mQA = y + 2 x – 3 ( y – 5 x + 1 )( y + 2 x – 3 ) = –1 y2 – 3y – 10 = –(x2 – 2x – 3) y2 + x2 – 2x – 3y – 13 = 0 Jika/If x = 5 y2 + 25 – 10 – 3y – 13 = 0 y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y = 1 ; 2 (5, 1) ialah R

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 209 10. (a) Rajah 5 menunjukkan susunan luas segi tiga dalam satu baris mengikut janjang aritmetik. Diagram 5 shows the arrangement of the area of triangles in a row that follow an arithmetic progression. Rajah 5 / Diagram 5 Didapati bahawa hasil tambah luas bagi sebutan 2n pertama bilangan segi tiga adalah sama dengan hasil tambah luas sebutan n segi tiga yang berikutnya. Jika segi tiga pertama mempunyai luas 12 cm2 dan beza sepunya ialah 3 cm2 , It is found that the sum of the first 2n terms of the number of triangles is the same as the sum of the next n terms. If the first triangle has an area of 12 cm2 and the common difference is 3 cm2, (i) cari nilai n. find the value of n. [4 markah / marks] (ii) yang manakah satu segi tiga mempunyai luas 48 cm2 ? which triangle has an area of 48 cm2? [2 markah / marks] (b) Dalam suatu janjang geometri, sebutan kedua melebihi sebutan pertama sebanyak 20 dan sebutan ke empat melebihi sebutan kedua sebanyak 15. Cari nilai yang mungkin bagi sebutan pertama janjang tersebut. In a geometric progression, the second term exceeds the first term by 20 and the fourth term exceeds the second term by 15. Find the possible values for the first term of the progression. [4 markah / marks] (a) (i) S2n = S3n – S2n 2S2n = S3n 2[2n 2 (2a + (2n – 1)d] = 3n 2 [2a + (3n – 1) d] a = 12, d = 3 4n(24 + 6n – 3) = 3n[24 + 9n – 3] 84 + 24n = 63 + 27n 21 = 3n n = 7 (ii) Tn = a + (n – 1)d = 48 12 + 3(n – 1) = 48 (n – 1) = 36 3 = 12 n = 13 (b) ar – a = 20 ar3 – ar = 15 a(r – 1) = 20 ...➀ ar(r2 – 1) = 15 ...➁ ➁ ÷ ➀ r(r + 1) = 15 20 4r2 + 4r = 3 4r2 + 4r – 3 = 0 (2r + 3)(2r – 1) = 0 r = – 3 2 ; 1 2 ∴a[– 3 2 –1] = 20 atau a( 1 2 – 1) = 20 a = –8 a = –40

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 210 8 11. Rajah 6 menunjukkan sebuah sektor major bulatan AOBC dengan pusat O dan jejari 8 cm, dan sektor OAC berpusat A. Diagram 6 shows a major sector of a circle AOBC with centre O and the radius of 8 cm, and sector OAC with centre A. A O B C Rajah 6 / Diagram 6 Jika sudut minor AOB ialah 0.8π, cari If the minor angle AOB is 0.8π, find (a) panjang perentas BC. the length of the chord BC. [2 markah / marks] (b) perimeter rantau yang tidak berlorek. the perimeter of the region that is not shaded. [4 markah / marks] (c) luas rantau yang tidak berlorek dalam sebutan π. the area of the region that is not shaded in terms of π. [4 markah / marks] Bahagian C / Section C [20 markah / marks] Jawab mana-mana dua soalan daripada bahagian ini. Answer any two questions from this section. 12. Suatu zarah bergerak pada satu garis lurus dan melalui satu titik tetap O. Halajunya, v m s−1, diberi oleh v = 2(3t 2 − 2t − 1), dengan keadaan t ialah masa, dalam saat, selepas melalui O. Cari A particle moves along a straight line and passes through a fixed point O. Its velocity, v m s−1, is given by v = 2(3t 2 − 2t − 1), where t is the time, in seconds, after passing through O. Find (a) pecutan awal, dalam m s−2. the initial acceleration, in m s−2. [1 markah / mark] (b) halaju maksimum, dalam m s−1. the maximum velocity, in m s−1. [2 markah / marks] (a) COB = 2π – 0.8π – π 3 = 13π 15 = 156° BC2 = 82 + 82 – 2(8)(8) kos 156° BC = 15.65 cm (b) Perimeter = 6 5 π(8) + (8) + 8 + π 3 (8) = 54.54 cm (c) Luas = 1 2 (8)2 1 13π 15 2 – 1 2 (8)2 1 π 3 2 Area = 323 13π 15 – π 3 4 = 256π 15 = 17 π 15 cm2 A 60° 0.8π O B C

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 211 (c) nilai t apabila zarah berhenti seketika. the value of t when the particle stops instantaneously. [3 markah / marks] (d) jumlah jarak, dalam m, yang dilalui oleh zarah itu dalam 2 saat pertama dengan melakar graf halaju-masa. the total distance, in m, travelled by the particle in the first 2 seconds by sketching a velocity-time graph. [4 markah / marks] (a) v = 2(3t 2 – 2t – 1) dv dt = a = 12t – 4 Apabila/When t = 0 a = –4 m s–2 (b) Apabila/When dv dt = 0 t = 1 3 ∴v = 2[31 1 3 2 2 – 21 1 3 2 – 1] = 2 –2 3 m s–1 (c) v = 0 3t2 – 2t – 1 = 0 (3t + 1)(t – 1) = 0 t = 1 (d) jumlah jarak/total distance = ∫ 1 0 (6t 2 – 4t – 2)dt + ∫ 2 1 (6t2 – 4t – 2)dt =  (2t 3 – 2t2 – 2t] 1 0 + 32t 3 – 2t 2 – 2t 4 2 1 = |–2| + 6 = 8 m 13. Seorang peniaga mempunyai 50 kg kacang tanah dan 75 kg kacang walnut. Dia ingin membungkus dengan mencampurkannya kepada dua gred, A dan B seperti ditunjukkan dalam Jadual 2 A businessman has 50 kg of groundnuts and 75 kg of walnuts. He wants to pack by mixing them into two grades, A and B as shown in Table 2. Campuran / Mixture Kacang tanah / Groundnuts (g) Kacang walnut / Walnuts (g) Gred A / Grade A 25 150 Gred B / Grade B 50 50 Jadual 2 / Table 2 Jika x mewakili bilangan bungkusan gred A dan y mewakili bilangan bungkusan gred B, If x represents the number of packages of grade A and y represents the number of packages of grade B, (a) tulis semua ketaksamaan yang memuaskan kekangan yang diberi, selain daripada x ≥ 0 dan y ≥ 0. write all the inequalities that satisfy the constraints given, other than x ≥ 0 and y ≥ 0. [2 markah / marks] (b) Menggunakan skala 4 cm kepada 500 unit pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua kekangan di atas. Using a scale of 4 cm to 500 units on both axes, construct and shade the region R which satisfies all the above constraints. [3 markah / marks]

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 212 8 (c) Gunakan graf yang dibina di (b) untuk menjawab soalan-soalan yang berikut Use the graph constructed in (b) to answer the following questions. (i) Cari bilangan maksimum bungkusan gred A dan B jika peniaga ingin membuat bilangan bungkusan gred B dua kali lebih banyak daripada bilangan bungkusan gred A. Find the maximum number of packages of grade A and B if the businessman wants to make the number of grade B packages two times more than the grade A packages. [3 markah / marks] (ii) Harga asal setiap bungkusan gred A dan gred B masing-masing ialah RM3.50 dan RM2.50. Cari keuntungan maksimum jika dia ingin menjual setiap bungkusan gred A dan gred B dengan harga masing-masing ialah RM7.50 dan RM4.50. The initial cost price for grade A and grade B are RM3.50 and RM2.50 respectively. Find the maximum profit if he wants to sell each package of grade A and grade B with the price of RM7.50 and RM4.50 respectively. [2 markah / marks] (a) 25x + 50y ≤ 50 000 x + 2y ≤ 2 000 ... ➀ 150x + 50y ≤ 75 000 6x + 2y ≤ 3 000 ... ➁ 3x + y ≤ 1 500 y = 2x (b) (c) (i) Gred/ Grade A = 300 , Gred/ Grade B = 600 (ii) p = 4x + 2y 2y = −4x + p y = −2x + p 2 p = 4(200) + 2(900) = 800 + 1 800 = RM2 600 1000 y x 0 R 1000 3x + y = 1500 2000 2000 y = 2x (200, 900) (300, 600) x + 2y = 2000

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 213 14. Jadual 3 menunjukkan indeks harga dan pemberat bagi empat jenis tiket pengangkutan. Table 3 shows the price indices and weightages of four types of transportation tickets. Harga tiket Cost of ticket Indeks harga pada tahun 2020 berasaskan tahun 2018 Price indices for year 2020 based on year 2018 Pemberat Weightages Bas / Bus n 3 Keretapi / Train 125 4 Kapal terbang / Aeroplane 130 m Kapal laut / Ship 110 1 Jadual 3 / Table 3 (a) Diberi tiket bas pada tahun 2018 dan 2020 masing-masing ialah RM20 dan RM24. Cari nilai n. Given that the bus ticket in year 2018 and 2020 is RM20 and RM24 respectively. Find the value of n. [1 markah / mark] (b) Diberi indeks gubahan pada tahun 2020 berasaskan tahun 2018 ialah 125, cari nilai m. Given that the composite index for year 2020 based on the year 2018 is 125, find the value of m. [3 markah / marks] (c) Dijangka harga tiket bas dan kapal terbang akan meningkat sebanyak 10% pada tahun 2021. Cari indeks gubahan pada tahun 2021 berasaskan tahun 2018. It is expected that the bus ticket and the plane ticket will increase by 10% in the year 2021. Find the composite index for the year 2021 based on the year 2018. [3 markah / marks] (d) Jika satu tiket kapal terbang pada tahun 2021 berharga RM650, cari harganya pada tahun 2018. If a plane ticket in the year 2021 costs RM650, find the cost in the year 2018. [3 markah / marks] (a) P20 P18 × 100 = n 24 20 × 100 = n = 120 (b) 3(120) + 4(125) + 130 m + 110 (8 + m) = 125 970 + 130 m= 1000 + 125 m 5 m = 30 m = 6 (c) 110 100 × 120 = 132 110 100 × 130 = 143 I = 132 × 3 + 125 × 4 + 143 × 6 + 110 14 = 133.14 (d) P21 P18 × 100 = 143 P18 = 650 × 100 143 = RM454.55

Matematik Tambahan Tingkatan 5 Kertas Pra-SPM 214 8 15. Rajah 7 menunjukkan sebuah bongkah, dengan permukaan ABCD, PQRS, PADS dan QBCR adalah segi empat tepat. Diagram 7 shows a block, where the surfaces ABCD, PQRS, PADS and QBCR are rectangles. Rajah 7 / Diagram 7 Diberi PQ = 2AB dan M ialah titik tengah BC, cari Given that PQ = 2AB and M is the midpoint of BC, find (a) AM [2 markah / marks] (b) AR [2 markah / marks] (c) ∠ARM [3 markah / marks] (d) luas segi tiga PMR. the area of the triangle PMR. [3 markah / marks] (a) AM2 = 52 + 42 AM = 6.4 cm (b) AR2 = 62 + PR2 = 36 + 102 + 82 AR = 14.14 cm (c) MR2 = 42 + 62 + 52 MR = 8.77 cm 6.42 = 77 + 200 – 2 77 × 200 kos ARM kos ARM = 77 + 200 – 41 cos ARM 2 77 × 200 ARM = 18°2’ (d) MP2 = 36 + 41 MP = 77 = 8.77 PR = 100 + 64 = 12.81 S = 12.81 + 8.77 + 8.77 2 = 15.17 ∆PMR = 15.17(15.17 − 8.77)2 (15.17 − 12.81) = 38.29 cm2 A M 6 cm 5 cm 10 cm 8 cm P Q R S θ D C B A M 6 cm 10 cm 8 cm P Q R S D C B

J1 JAWAPAN BAB Sukatan Membulat 1 Circular Measure 1. (a) Gunakan/Use 43°32’ 360° = θ rad 2π rad θ = 43°32’ × 2π 360° = 0.76 rad (b) Gunakan/Use 157°16’ 360° = θ rad 2π rad θ = 157°16’ × 2π 360° = 2.745 rad (c) Gunakan/Use 247°54’ 360° = θ rad 2π rad θ = 247°54’ × 2π 360° = 4.327 rad (d) Gunakan/Use 303.68° 360° = θ rad 2π rad θ = 303.68° × 2π 360° = 5.3 rad (e) Gunakan/Use 354.74° 360° = θ rad 2π rad θ = 354.74° × 2π 360° = 6.191 rad 2. (a) θ° 360° = 2 3 π rad 2π rad θ° = 360° × 2 3 π 2π = 120° (b) θ° 360° = 4 9 rad 2π rad θ° = 360° × 4 9 2π = 25.46° atau/or 25°28’ (c) θ° 360° = 4.65 rad 2π rad θ° = 360° × 4.65 2π = 266.43° atau/or 266°26’ (d) θ° 360° = 1.67π rad 2π rad θ° = 360° × 1.67π 2π = 300.6° atau/or 300°36’ (e) θ° 360° = 0.72π rad 2π rad θ° = 360° × 0.72π 2π = 129.6° atau/or 129°36’ 3. (a) 0.8 rad 2π rad = panjang lengkok, AB Arc length AB 2πj Panjang lengkok, s Arc length, s = 0.8 rad × 2π(7.4) 2π rad = 5.92 cm (b) 0.5π rad 2π rad = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = 0.5π rad × 2π(4.5) 2π rad = 7.07 cm (c) Sudut yang dicangkum = 360° – 125° = 235° Angle subtended θ 360° = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = 235° × 2π(10.2) 360° = 41.84 cm (d) Sudut yang dicangkum = (2π − 1.3) rad Angle subtended (2π – 1.3) rad 2π rad = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = (2π – 1.3) rad × 2π(20.5) 2π rad = 102.16 cm (e) Sudut yang dicangkum = (2π − 4.5) rad Angle subtended (2π – 4.5) rad 2π rad = panjang lengkok, s 2πj Panjang lengkok, s Arc length, s = (2π – 4.5) rad × 2π(8.4) 2π rad = 15 cm

J2 4. (a) 204° 360° = panjang lengkok, s 2πj j = 20.3 × 360° 204° × 2π = 5.7 cm (b) j = 14.5 2π – 1.6 = 3.1 cm (c) j = 11.5 2π – 0.8π = 3.05 cm (d) 360° – 105° 360° = panjang lengkok, s 2πj j = 14.2 × 360° 255° × 2π = 3.19 cm (e) j = 3 2π – 3 4 π = 0.76 cm 5. (a) (2π – θ) = 7.2 6.1 θ = 5.1 rad (b) (2π – θ) = 4.8 3.2 θ = 4.78 rad (c) (2π – 2θ)= 2.8 1.8 θ = 2.36 rad (d) (2π – θ) 3 = 3.2 2.5 θ = 2.44 rad (e) 2πj = 6.3 + 8.2 + 11.4 j = 4.12 cm θ = 11.4 4.12 = 2.77 rad 6. (a) 0.4π rad = 0.4π × 360° 2π = 72° Maka/Hence, PQ = 9.62 + 9.62 − 2(9.6)2 kos72° = 11.28 cm Panjang lengkok PQ= jθ Arc length PQ = 9.6(0.4π) = 12.06 cm Perimeter = (12.06 + 11.28) cm = 23.34 cm (b) Panjang lengkok minor PQ = 2π(8) – 30 Minor arc length PQ = 20.27 cm ∠POQ = 20.27 8 = 2.53 rad = 145.14° Maka, PQ = 82 + 82 −2(8)2 kos145.14° = 15.27 cm Perimeter = 15.27 + 20.27 Hence = 35.54 cm (c) Panjang lengkok minor PQ = 2π(3) – 4.5(3) Minor arc length PQ = 5.35 cm ∠POQ = 2π − 4.5 = 1.78 rad = 102.17° Maka, PQ = 32 + 32 − 2(3)2 kos102.17° Hence = 4.67 cm Perimeter = 5.35 cm + 4.67 cm = 10.02 cm (d) ∠POQ = 150° Panjang jejari/Arc length, j = (7.3 × 360°) (210 × 2π) = 1.99 cm Maka/Hence, PQ = 1.992 + 1.992 − 2(1.99)2 kos150° = 3.84 cm Panjang lengkok = 150° 360° × 2π(1.99) Arc length = 5.2 cm Perimeter = 5.2 cm + 3.84 cm = 9.04 cm (e) Panjang lengkok PQR = 2π(5.3) − 5.3(1.4π) Arc length PQR = 10 cm ∠POQ = (2π − 1.4π) ÷ 2 = 0.3π = 54° Panjang perentas PQ Length of chord PQ = 5.32 + 5.32 − 2(5.3)2 kos 54° = 4.81 cm Perimeter = (4.81 × 2 + 10) cm = 19.62 cm 7. (a) 0.7 rad = 0.7 × 360° 2π = 40.1° sin 40.1° = AD 9.2 AD = 5.93 cm OD = 9.22 − 5.932 = 7.03cm Panjang lengkok = 9.2 × 0.7 Arc length = 6.44 cm Perimeter = 5.93 + 6.44 + (9.2 – 7.03) = 14.54 cm (b) Katakan OM = j cm Let Maka, jθ + 2j + 2jθ = 2j + jα Hence, α = 3θ (c) (i) Diberi AP = 2 cm, maka OA = 6 cm Given hence 6a + 2θ = 10 θ = 5 − 3a (ii) a = 30° = π 6 rad θ = 5 − 3 π 6  = 3.43 rad

J3 (d) π 3 = 60° Maka, AP = AB = j. Hence OA = j 2 = 35π 6 = πj 3 + πj 2 j = 7 cm (e) Perimeter = OM + ON + panjang lengkok/ arc length MN + MK + panjang lengkok/arc length KL + ML = 5 + (5 2 − 5) + 5 π 4  + (5 2 − 5) + 5 2 π 4  + 5 = 23.62 cm 8. (a) Sudut yang tercangkum = 5 6 π rad Angle subtended Maka, luas = 5 6 π 2 (3.5)2 = 16.04 cm2 Hence, the area (b) Sudut yang tercangkum = 2 3 π rad Angle subtended Maka, luas =  2 3 π 2 ( 5.2)2 Hence, the area = 28.32 cm2 (c) Sudut yang tercangkum = (360 – 215)° Angle subtended = 145° Maka, luas = 145° 360° × π × ( 7.8)2 Hence, the area = 76.98 cm2 (d) Sudut yang tercangkum = 6 2.5 = 2.4 rad Angle subtended Maka, luas = 2.4 2 × (2.5)2 = 7.50 cm2 Hence, the area (e) Jejari sektor = 10 3π 5 Radius of the sector = 50 3π cm Maka, luas =  3 5 π 2  50 3π  2 Hence, the area = 26.53 cm2 9. (a) Gunakan /Use luas sektor luas bulatan = θ 2p area of sector area of circle = θ 2p 10.21 πj 2 = 2π – 3.5 2π j 2 = 10.21 × 2 2p – 3.5 j = 2.71 cm (b) Gunakan/Use panjang lengkok 2πj = luas sektor luas bulatan arc length 2πj = area of sector area of circle 12 2πj = 42 πj 2 j = 42 × 2 12 = 7 cm (c) Gunakan/Use 38 2πj = πj2 – 85 πj 2 j = 2(πj2 – 85) 38 2(πj2 – 85) = 38j πj2 – 19j – 85 = 0 j = 19 ± (–19)2 – 4π(–85) 2π j = 9.04 cm (d) Gunakan/ Use luas sektor/area of sector luas bulatan/area of circle = θ 360° 7.5 πj 2 = 88 360° j 2 = 7.5 × 360° 88π j = 3.13 cm (e) Gunakan/ Use 30 πj 2 = 12.5 2πj j = 30 × 2 12.5 = 4.8 cm 10. (a) Gunakan/ Use luas sektor luas bulatan = θ 2p area of sector area of circle = θ 2p 33 π(4.8)2 = 2π – θ 2π 66π 23.04π = 2π – θ 2.865 = 2π – θ θ = 2p − 2.865 = 3.42 radian (b) Gunakan/Use luas sektor luas bulatan = panjang lengkok 2πj area of sector area of circle = arc length 2πj 46 πj 2 = 12 2πj j = 46 × 2 12 = 7.67 cm s = jθ θ = 12 7.67 = 1.56 radian

J4 (c) 2j + jθ = 15 …… ➀ 1 2 j 2 θ = 30 …… ➁ ➁ ÷ ➀ 1 2 j 2 q 2j + jq = 2 jq 2(2 + q) = 2 jθ = 8 + 4θ θ = 8 j – 4 (d) 1 2 × 82 × θ = 43 θ = 1.34 radian (e) Gunakan/Use luas sektor luas bulatan = panjang lengkok 2πj = θ 2π area of sector area of circle = arc length 2πj = θ 2π 54 πj 2 = 21 2πj = θ 2π j = 54 × 2 21 = 5.14 cm 2p – θ = 21 5.14 θ = 2.2 radian 11. (a) Luas berlorek/Shaded area = luas sektor POQ – luas segi tiga POQ area of sector POQ – area of triangle POQ Luas sektor = 1 2 j 2 2π − 4π 3  Area of sector = 1 2 (8.1)2  2π 3 2 = 68.71 cm2 2π 3 rad = 120° Luas segi tiga = 1 2 × (8.1)2 × sin 120° Area of triangle = 28.41 cm2 Maka, luas tembereng = 68.71 − 28.41 Hence, the area of segment = 40.3 cm2 (b) Luas sektor POS = luas sektor QOR area of sector POS = area of sector QOR 1 2 j2 θ = 1 2 (3.5)2 (π − 1.4) = 10.67 cm2 1.4 rad = 1.4 2π × 360° = 80.21° Luas segi tiga = 1 2 × (3.5)2 × sin80.21° Area of triangle = 6.04 cm2 Maka, luas tembereng = 2(10.67 − 6.04) Hence, the area of segment = 9.26 cm2 (c) Perentas AB = 5.52 + 5.52 − 2(5.5)2 kos50° Chord AB = 4.65 cm Kos/ cos ∠AOB = 4.32 + 4.32 − 4.652 2(4.3)2 ∠AOB = 65.46° = 1.14 rad 50° = 0.87 rad Luas tembereng = 1 2 (5.5)2 (0.87) − 1 2 (5.5)2 Area of segment sin50° + 1 2 (4.3)2 (1.14)− 1 2 (4.3)2 sin65.46° = 3.7 cm2 12. (a) OP = 8 2 cm Luas sektor major = 270° 360° × π(8 2)2 Area of major sector = 96π cm2 Luas segi tiga = 1 2 × (8 2)2 Area of triangle = 64 cm2 Jumlah luas = 96π + 64 Total area Luas semibulatan = π(8)2 = 64π Area of semiclrcle Luas rantau berlorek = 96π + 64 − 64π Area of shaded region = 32π + 64 = 32(π + 2) cm2 (b) Luas kain Q Area of cloth Q = 1 2 × (20)2 × 1.5 − 1 2 × (15)2 × 1.5 = 300 cm2 – 168.75 cm2 = 131.25 cm2 Luas kain P Area of cloth P = 1 2 × (15)2 × 1.5 − 1 2 × (10)2 × 1.5 = 168.75 cm2 – 75 cm2 = 93.75 cm2 Beza luas Difference of area = 131.25 cm2 − 93.75 cm2 = 37.5 cm2 (c) (i) Lilitan/ Circumference OPQ = 9π Lilitan/ Circumference QSR = 5π Lilitan/ Circumference OTR = 4π Perimeter = 2(9π + 5π + 4π) = 36π cm (ii) Luas semibulatan OPQ = 1 2 × (9)2 × π Area of semicircle OPQ Luas semibulatan QSR = 1 2 × (5)2 × π Area of semicircle QSR Luas semibulatan OTR = 1 2 × (4)2 × π Area of semicircle OTR Luas keseluruhan = 2 × 1 2 π × (81 + 25 – 16) Total area = 90π cm2

J5 Praktis SPM 1 Kertas 1 1. 42 = 52 + 52 – 2(5)(5) kos θ kos θ = 52 + 52 – 42 2(5)(5) θ = 47.16° ∠APB = ∠CPD = 180° − 47.16° 2 = 66.42° Luas sektor APB = 66.42° 360° × π(5)2 = 14.49 cm2 Area of sector APB Luas ∆BCP = 1 2 (5)(5)sin 47.16° = 9.17 cm2 Luas rantau berlorek = 10 × 8 − 9.17 – 2(14.49) Area of shaded region = 41.85 m2 2. (a) 82 = 152 + 152 – 2(15)2 kos θ kos θ = 152 + 152 – 82 2(15)(15) θ = 30.93° 30.93° 360° × 2π = 0.54 rad (b) Luas tembereng Area of segments = 1 2 (15)2 (0.54)− 1 2 (15)2 sin30.93° = 2.93 cm2 3. (a) 10 = 7(2π − θ) 7θ = 14p − 10 θ = 4.85 rad (b) Luas = 1 2 j 2 θ = 1 2 (7)2 (2p − θ) Area = 1 2 (7)2  10 7  = 35 cm2 Kertas 2 1. (a) kos θ = 6 12 = 1 cos θ 2 θ = 60° = π 3 rad (b) kos 30° = 18 AB cos 30° AB = 20.78 cm Panjang lengkok BD = 20.78 π 6  Arc length BD = 10.88 cm (c) Luas sektor BAD = 1 2 (20.78)2  π 6  Area of sector BAD = 113.05 cm2 Luas ∆ABE = 1 2 (18)(20.78) sin30° Area = 93.51 cm2 Luas rantau berlorek = 19.54 cm2 Area of shaded region A P D B C F E 4 m 10 m 8 m 5 m fi 2. (a) PQ2 = 152 + 52 = 250 PQ = 5 10 cm Panjang lengkok AB = 2π(7.5) Arc length AB = 15π Panjang lengkok CD = 2π(12.5) Arc length CD = 25π Maka, 15π = jq ….. ➀ 25π = (j + 5 10 )q ….. ➁ ➁ ÷ ➀ 5 3 = j + 5 10 j 2j = 15 10 j = 15 10 2 cm q = 15π × 2 15 10 = 1.99 rad = 113.84° kos ∠EOC = BO OC  15 10 2 + 5 10  kos(180° − 113.84°) = EO EO = 15.98 cm Panjang kad = 15.98 + 15 10 2 + 5 10 Length of card = 55.5 cm ≈ 56 cm Lebar kad / Breadth of card = j + 5 10 = 15 10 2 + 5 10 = 39.53 cm ≈ 40 cm (b) Luas kad tidak digunakan Area of unused card = 56 × 40 −  39.532 2 × 1.99 + 1 2  15 10 2  2 (1.99) = 1244.88 cm2 3. (a) 60° = π 3 rad (b) AC = 2 3 tan 30°  = 10.39 cm (c) Luas satu tembereng Area of a segment = 1 2 (10.39)2  π 3  − 1 2 (10.39)2 sin60° = 9.78 cm2 A B C O P Q 60fi 30fi B A 12 cm O E D C 6 cm fi fi 2 15 cm 12.5 cm 7.5 cm P Q O fi E j B D A C 5 10 √

J6 Luas bulatan = π(3)2 Area of circle Jumlah luas = 9π + 3(9.78) Total area = 57.61 cm2 4. (a) 5 = 11q q = 5 11 rad = 0.45 rad (b) Luas sektor = 1 2 (11)2 (0.45) Area of sector = 27.2 cm2 tan 50° = TS 11 TS = 13.11 cm Luas/Area ∆TSR = 1 2 (11)(13.11) = 72.1 cm2 50° = 50° 360° × 2p = 0.87 rad Luas sektor SRP = 1 2 (11)2 (0.87) Area of sector SRP = 52.6 cm2 Luas berlorek = 72.1 − 52.6 + 27.2 Shaded area = 46.7cm2 Sudut KBAT (a) PQ2 = 102 + 102 − 2(10)2 kos POQ kos POQ = 102 + 102 − 102 2(100) ∠POQ = 60° ∠SOP = 120°−60° 2 = 30° = π 2 rad (b) Luas tembereng PQ /Area of segment PQ = 1 2 (10)2  π 3  − 1 2 (10)2 sin 60° = 9.06 cm2 Luas tembereng SR / Area of segment SR = 1 2 (10)2  2π 3  − 1 2 (10)2 sin 120° = 61.42 cm2 Maka, luas berlorek Hence, the shaded area = 61.42 – 9.06 = 52.36 cm2 (c) Panjang lengkok SP = QR Arc length SP = QR 10 π 6  = 5π 3 SR2 = 102 + 102 − 2(10)2 kos 120° SR = 10 3 Perimeter = 10 + 2 5π 3  + 10 3 = 101+ π 3 + 3 S R Q P T 50fi 11 cm 5 cm ff P S Q R O 10 cm 120fi 10 cm

J7 BAB Pembezaan 2 Differentiation 1. (a) had k x → 0 2k + x limx → 0 = 1 2 (b) had 4 – 2x + x2 x → 0 limx → 0 = 4 = 2 2. (a) 6 (b) 0 (c) 2 3 (d) had 2(x2 – 4) x → 2 x – 2 limx → 2 = had 2(x + 2)(x – 2) x → 2 x – 2 limx → 2 = 8 (e) had 2(x2 – 3x – 4) x → 2 x + 1 limx → 2 = had 2(x – 4)(x + 1) x → 2 x + 1 limx → 2 = –4 (f) had 1 x + 2x x 4 x + x x x → ∞ limx → ∞ = 0 + 2 0 + 1 = 2 3. (a) δy = f(x + δx) = 4 − x2 = 4 – (x + δx)2 – 4 + x2 = 4 – x2 – 2xδx – (δx)2 – 4 + x2 = −(2x + δx)δx δy δx = –(2x + δx) had δy δx = dy dx = –2x δx → 0 (b) x2 – x – 2 δy = (x + δx)2 – (x + δx) – 2 – x2 + x + 2 = x2 + 2x·δx + (δx)2 – x − δx – x2 + x = 2xδx – δx + (δx)2 δy = [(2x – 1) + δx]δx δy δx = (2x – 1) + δx had δy δx = dy dx = 2x – 1 δx → 0 (c) δy = 1 (x + δx)2 – 1 x2 = x2 – (x + δx)2 x2 (x + δx)2 = –2xδx – (δx)2 x2 (x + δx)2 δy δx = –2x – δx x2 (x + δx)2 had δy δx = dy dx = –2x x4 = –2 x3 δx → 0 (d) δy = 1 (x + δx + 1) – 1 (x + 1) = (x + 1) – x – 1 – δx (x + 1)(x + 1 + δx) = –δx (x + 1)(x + 1 + δx) δy δx = –1 (x + 1)(x + 1 + δx) had δy δx = –1 (x + 1)(x + 1) = – 1 (x + 1)2 δx → 0 4. (a) d dx  1 2  = 0 (b) d dx ( x ) = d dx x 1 2  = 1 2 x (c) d dx  8x4 5  = 32 5 x3 (d) d dx  –6 (2x)3  = d dx  –6 8x3  = d dx  –3 4 x–3 = 9 4x4 (e) d dx  –5x 3x4  = d dx  –5 3 x –3 = 5 x 4 (f) d dx  2x x  = d dx ( 2x – 1 2 ) = – 2 2 x – 3 2 = – 2 2x x 5. (a) f(x) = 3 x = 3x 1 – 2 f’(x) = 3 – 2 x 3 – 2 = –3 2x x (b) f(x) = 10 3x –3 = 10 3 x3 f’(x) = 10x2 (c) f(x) = –5 4x–2 = –5 4 x2 f’(x) = –5 2 x 6. (a) y = 3x 3 2 dy dx = 9 2 x 1 2 = 9 2 x

J8 Apabila x = 4, dy dx = 92 (2) When = 9 (b) y = 7x 2 x = 72 x 12 dy dx = 74 x 1 – 2 = 7 4 x Apabila x = 9, dy dx = 7 When 4 9 = 7 12 (c) y = –8 5x–2 = 8 – 5 x2 dy dx = 16 – 5 x Apabila x = 1 – 8 , dy dx = 16 – 5  1 – 8  When = 25 7. (a) y = x5 – 12 x3 + x –1 dy dx = 5x4 – 32 x2 – 1x2 (b) y = –2x – 5 – x – 2 + 2 dy dx = 10x6 + 2x3 (c) y = 2x2 + 3x dy dx = 4x – 3 (d) y = x + 4x 52 dy dx = 1 + 10x 32 = 1 + 10x x (e) y = 2x – 2x4 + 3 – 3x3 3x2 = 23 x –1 – 23 x2 + x –2 – x dy dx = –23x2 – 43 x – 1x3 – 1 (f) y = 2(x2 + 2x + 1) 3x = 23 (x + 2 + x –1) dy dx = 23 1 – 1x2  (g) y = x39 – 3x + 14 x2 = 9x3 – 3x4 + 14 x5 dy dx = 27x2 – 12x3 + 54 x4 = x227 – 12x + 54 x2 (h) y = 47 x2 – 3x  = 87 x – 127 dy dx = 87 8. (a) y = 83 x–1 – x2 + x 12 dy dx = –83x2 – 2x + 1 2 x Apabila/When x = 4, dy dx = –8 3(4)2 – 8 + 14 = –711 12 (b) y = 5x – x–1 4x2 = 54 x–1 – 14 x–3 dy dx = – 54 x–2 + 34 x–4 Apabila/When x = –1, dy dx = – 54 + 34 = – 12 9. (a) y = (1 – 4x2)2 3 = 13 (1 – 4x2)2 dy dx = 23 (1 – 4x2 )· d dx (1 – 4x2) = 23 (1 – 4x2 )(–8x) = – 163 x(1 – 4x2) (b) y = 2 – 1x 5 dy dx = 52 – 1x 4 · d dx (2 – x–1) = 52 – 1x 4 · 1x2  = 5x2 2 – 1x 4 (c) y = 12 3x3 – 25 x24 dy dx = 12 (4)3x3 – 25 x23· d dx 3x3 – 25 x2 = 23x3 – 25 x23·9x2 – 45 x (d) y = 2  5x2 + x3 = 2 5x2 + x –3 dy dx = 2(–3) 5x2 + x –4– 10x3 + 1 = –61 – 10x3   5x2 + x4 = 6(10 – x3) x3 5x2 + x4

J9 10. (a) y = 5 4 (2 + x2 ) 1 2 dy dx = 5 4  1 2 (2 + x2 ) 1 – 2 (2x) = 5x 4 2 + x2 Apabila x = 4, dy dx = 5(4) 4 18 When = 5 3 2 = 5 2 6 (b) y = (x4 – 3x2 – 2)3 dy dx = 3(x4 – 3x2 – 2)2 (4x3 – 6x) = 6(2x3 – 3x)(x4 – 3x2 – 2)2 Apabila x = –1, dy dx = 6(–2 + 3)(1 – 3 – 2)2 When = 96 (c) y = 6(x + x2 )–1 dy dx = –6(x + x2 )–2(1 + 2x) = –6(1 + 2x) (x + x2 )2 Apabila x = –2, dy dx = –6(–3) (–2 + 4)2 When = 9 2 11. (a) y = x3 (1 – 2x)2 u = x3 v = (1 – 2x)2 du dx = 3x2 dv dx = 2(1 – 2x)(–2) dy dx = x3 (2)(–2)(1 – 2x) + (1 – 2x)2 (3x2 ) = x2 (1 – 2x)[–4x + 3(1 – 2x)] = x2 (1 – 2x)(3 – 10x) (b) y = (x + 4)2 (1 – 3x)2 dy dx = (x + 4)2 (2)(–3)(1 – 3x) + (1 – 3x)2 (2)(x + 4) = 2(1 – 3x)(x + 4)[–3(x + 4) + 1 – 3x] = 2(1 – 3x)(x + 4)[–6x – 11] (c) y = (2x –1)(1 + x2 )4 dy dx = (2x – 1)(4)(2x)(1 + x2 )3 + (2)(1 + x2 )4 = 2(1 + x2 )3 [4(2x – 1)(x) + 1 + x2 ] = 2(1 – x2 )3 [9x2 – 4x + 1] (d) y = x2 (x – 1) 1 2 dy dx = x2  1 2 (x – 1) 1 – 2 + x – 1(2x) = x2 2 x – 1 + 2x x – 1 = x2 + 4x(x – 1) 2 x – 1 = 5x2 – 4x 2 x – 1 (e) y = 2(x – 2)(1–3x)3 dy dx = 2(x – 2)(3)(–3)(1 – 3x)2 + 2(1 – 3x)3 = 2(1 – 3x)2 [–9(x – 2) + 1 – 3x] = 2(1 – 3x)2 [19 – 12x] 12. (a) y = (x – 3)3 (1 + 2x)2 dy dx = (1 + 2x)2 (3)(x – 3)2 – (x – 3)3 (2)(2)(1 + 2x) (1 + 2x)4 = (1 + 2x)(x – 3)2 [3(1 + 2x) – 4(x – 3)] (1 + 2x)4 = (x – 3)2 [15 + 2x) (1 + 2x)3 (b) y = x2 (1 – 2x)3 dy dx = (1 – 2x)3 (2x) – x2 (3)(–2)(1 – 2x)2 (1 – 2x)6 = 2x(1 – 2x)2 [1 – 2x + 3x] (1 – 2x)6 = 2x(1 + x) (1 – 2x)4 (c) y = 1 + 2x2 2 + x3 dy dx = (2 + x 3 )(4x) – (1 + 2x2 )(3x2 ) (2 + x3 )2 = x[4(2 + x3 ) – 3(1 + 2x2 )x] (2 + x3 )2 = x(8 + 4x3 – 3x – 6x3 ) (2 + x3 )2 = x[8 – 3x – 2x3 ] (2 + x3 )2 (d) y = 2x3 – 3 (x – 1) dy dx = (x – 1)(6x2 ) – (2x3 – 3) (x – 1)2 = 4x3 – 6x2 + 3 (x – 1)2 (e) y = x x + 1 = x 1 2 x + 1 dy dx = (x + 1) 2 x – x (x + 1)2 = x + 1 – 2x 2 x (x + 1)2 = 1 – x 2 x (x + 1)2 13. (a) dy dx = (x – 3)2 (–2x) + (4 – x2 )(2)(x – 3) = 2(x – 3)[–x(x – 3) + 4 – x2 ] = 2(x – 3)(3x + 4 – 2x2 ] d2 y dx2 = (2x – 6)(–4x + 3) + (– 2x2 + 3x + 4)(2) = –8x2 + 6x + 24x – 18 – 4x2 + 6x + 8 = –12x2 + 36x – 10  dy dx  2 = 4(x – 3)2 (3x + 4 – 2x2 ]2 ∴ d2 y dx2 ≠  dy dx  2 3

J10 (b) y = x –1 – x–2 + 2x2 dy dx = – x–2 + 2x –3 + 4x d2 y dx2 = 2x–3 – 6x–4 + 4 = 2 x3 – 6 x4 + 4  dy dx  2 = 4x – 1 x2 + 2 x3  2 ∴ d2 y dx2 ≠  dy dx  2 (c) dy dx = (2x + 1) – 2x (2x + 1)2 = 1 (2x + 1)2 d2 y dx2 = –4 (2x + 1)3  dy dx  2 = 1 (2x + 1)4 ∴ d2 y dx2 ≠  dy dx  2 14. (a) y = x2 + 2x y’ = 2x + 2 y’’ = 2 Sebelah kiri = 2x2 [2] – (2x + 2)2 + 4(2x + 2) Left side = 4x2 – [4x2 + 8x + 4] + 8x + 8 = 4 sebelah kanan right side (b) y = x3 – 2x2 y’ = 3x2 – 4x y’’ = 6x – 4 Sebelah kiri = x2 [6x – 4] – 6[x3 – 2x2 ] Left side = –4x2 + 12x2 = 8x2 sebelah kanan right side (c) y = x + 2x –1 y’ = 1 – 2x –2 y’’ = 4x –3 Sebelah kiri = 4 x3 + 2 x 1 – 2 x2  + x Left side = 2 x + x sebelah kanan/right side 15. (a) y = 5(–2) – 3(–2)2 x = –10 – 12 = –22 y’ = 5 – 6x Pada/At x = –2 dy dx = 5 – 6(–2) = 17 (b) f(–1) = 4(–1)2 – 6(–1) + 1 = 11 f’(x) = 8x – 6 f’(–1) = 8(–1) – 6 = –14 ∴ Pada (–1, 11), f’(x) = –14 At (c) f(1) = –2 + 4 = 2 f’(x) = –2 + 8x f’(1) = –2 + 8(1) = 6 ∴ Pada (1, 2), f’(x) = 6 At 16. (a) y = px–1 + qx 4 = p 2 + 2q y’ = –p x2 + q Apabila x = 2 –p 4 + q = –8 When –p + 4q = –32 ...➁ ➁ + ➂ p + 4q = 8 ...➂ 8q = –24 q = –3 p = (4 + 6)2 = 20 (b) y = px3 + qx (–1, 1) 1 = –p – q ...➀ y’ = 3px2 + q Apabila x = –1 y’ = 3p + q = 5 ...➁ When ➁ + ➀ 2p = 6 p = 3, q = –4 (c) y = px2 + qx + 1 (2, 5) 5 = 4p + 2q + 1 2 = 2p + q ...➀ y’ = 2px + q, Apabila x = 2 y’ = 4p + q = 8 ...➁ When ➁ – ➀ 2p = 6 p = 3, q = –4 17. (a) y = 1 x x = 2, y = 1 2 dy dx = – 1 x2 Pada x = 2, dy dx = – 1 4 Persamaan tangen/Equation of tangent y – 1 2 = – 1 4 (x – 2) y = – 1 4 x + 1 Persamaan normal//Equation of normal y – 1 2 = 4(x – 2) y = 4x – 15 2 (b) y = x3 – 4x x = –1 y = –1 – 4(–1) = 3 dy dx = 3x2 – 4 pada x = –1 dy dx = 3(–1)2 – 4 = –1 Persamaan tangen/Equation of tangent y – 3 = –(x + 1) y = –x + 2 Persamaan normal/Equation of normal y – 3 = x + 1 y = x + 4 (c) y = x = x 1 2 x = 4, y = 2 dy dx = 1 2 x

J11 Pada x = 4, dy dx = 1 4 Persamaan tangen/Equation of tangent y – 2 = 1 4 (x – 4) y = 1 4 x + 1 Persamaan normal//Equation of normal y – 2 = –4(x – 4) y = –4x + 18 18. (a) (i) y = 3x2 – 6x dy dx = 6x – 6 = 12 6x = 18 x = 3 y = 3(3)2 – 6(3) = 9 ∴P[3, 9] (ii) Persamaan normal pada (3, 9) Equation of the normal at (3, 9) y – 9 = – 1 12(x – 3) y = – 1 12x + 37 4 (b) y = 2x + 8 x dy dx = 2 – 8 x2 = 0 2x2 = 8 x = ±2 y = 2(2) + 8 2 , 2(–2) – 8 2 = 8, –8 ∴(2, 8), (–2, –8) (c) y = ax2 + bx + 2 dy dx = 2ax + b pada 1, 1 2 , dy dx = 2a + b y = x2 + 6x + 4 dy dx = 2x + 6 pada (–2, –4), dy dx = 2(–2) + 6 = 2 ∴2a + b = – 1 2 ...➀ 1 2 = a + b + 2 a + b = – 3 2 ...➁ ➀ – ➁ a = 1, b = –1 = – 5 2 19. (a) dy dx = 4x − 4 Untuk titik pusingan, kecerunan dy dx = 0 For the turning point, gradient dy dx = 0 Maka, 4x − 4 = 0 Then x = 1 dan y = 2 − 4 − 1 = −3 Titik pusingan ialah (1, –3) The turning point is (1, –3) Dari jadual berikut. From the following table. x Sedikit kurang 1 1 Sedikit lebih 1 dy dx − 0 + Lakaran tangen Kecerunan dy dx menukar tanda dari negaif ke positif semasa melalui x = 1, jadi titik pusingan itu ialah minimum. The gradient dy dx changes from negative to positive as it passes through x = 1, hence the turning point is minimum. (b) dy dx = 3x2 − 6x Untuk titik pusingan, kecerunan dy dx = 0 For the turning point, gradient dy dx = 0 Maka 3x2 − 6x = 0 x = 0 dan 2 y = 0 dan −4 Titik pusingan ialah (0, 0) dan (2, −4) The turning point is (0, 0) and (2, −4) Dari jadual berikut./From the following table. x Sedikit kurang 0 0 Sedikit lebih 0 dy dx − 0 + Lakaran tangen x Sedikit kurang 2 2 Sedikit lebih 2 dy dx + 0 – Lakaran tangen (0, 0) ialah titik minimum dan (2, −4) ialah titik maksimum. (0, 0) is a minimum point and (2, –4) is a maximum point. (c) dy dx = x2 − 2x – 3 Untuk titik pusingan, kecerunan dy dx = 0 Maka x2 − 2x – 3 = 0 (x – 3)(x + 1) = 0 x = 3 dan –1 y = –8 dan 8 3

J12 Titik pusingan ialah (3, –8) dan −1, 8 3  Dari jadual berikut. x Sedikit kurang 3 0 Sedikit lebih 3 dy dx − 0 + Lakaran tangen x Sedikit kurang –1 –1 Sedikit lebih –1 dy dx + 0 – Lakaran tangen (3, −8) ialah titik minimum dan −1, 8 3  ialah titik maksimum. (3, –8) is a minimum point and –1, 8 3  is a maximum point. 20. (a) Perimeter = 26 = 2y + 4x y = 13 – 2x Luas A = xy +  1 2 x2sin 60°2 Area A = x[13 – 2x] + x2sin60° Untuk A maksimum, dA dx = 0 For a maximum A dA dx = 13 – 4x + 2xsin60° = 0 13 = x4 – 2 3 2  ∴x = 13 4 – 3 (b) (i) 4t + 8x + 4x = 360 t + 3x = 90 I = 2x2 t = 2x2 [90 – 3x] = 180x2 – 6x3 (ii) dI dx = 360x – 18x2 = 0 x[360 – 18x] = 0 x = 0 x = 20 Dimensi 20 × 40 × 30 cm Dimension (c) (i) x 10 = ED 20 ED = 2x ∴AE = 26 – 2x Luas AEFG = x[26 – 2x] Area AEFG (ii) dA dx = 26 – 4x = 0 x = 13 2 cm Luas maksimum = 13 2 26 – 2 13 2  Maximum area = 84 1 2 cm2 21. Diberi dx dt = 0.5 unit s−1 apabila x = 1. Given When dy dx = 6(3x − 6) Maka dy dt = dy dx × dx dt Hence = 6(3x − 6) dx dt  Apabila x = 1 dan dx dt = 0.5 When and Maka dy dt = 6(−3)(0.5) Hence = −9 unit s−1 Kadar perubahan y ialah −9 unit s−1. y menyusut. The rate of change of y is –9 unit s–1. y decreases. (b) Diberi dx dt = 0.5 unit s−1 apabila x = 4. Given when dy dx = (x – 1) – x (x – 1)2 = –1 (x – 1)2 Maka dy dt = dy dx × dx dt Hence = –1 (x – 1)2  dx dt  Apabila x = 4 dan dx dt = 0.5 When and Maka dy dt = –1 (4 – 1)2(0.5) Hence = – 1 18 unit s−1 Kadar penyusutan y ialah 1 18 unit s−1 The rate of y decreasing is 1 18 unit s–1 22. (a) Diberi dx dt = 8 unit s−1 apabila x = 3 Given when dy dx = (x – 1)(8x) – 4x2 (x – 1)2 Maka dy dt = dy dx × dx dt Hence = (x – 1)(8x) – 4x2 (x – 1)2  dx dt  Apabila x = 3 dan dy dt = 8 When and Maka 8 = (3 – 1)(8)(3) – 4(3)2 (3 – 1)2  dx dt  Hence dx dt = 8 3 unit s−1 2x x t

J13 Kadar perubahan x ialah 8 3 unit s−1. x bertambah. The rate of change of x is 8 3 unit s–1. x increases. (b) Diberi dy dt = 8 unit s−1 apabila x = 2 Given when dy dx = 1 2x – 3 Maka dy dt = dy dx × dx dt Hence = 1 2x – 3  dx dt  Apabila x = 2 dan dx dt = 8 When and Maka 8 = 1 dx dt  Hence dx dt = 8 unit s−1 Kadar perubahan x ialah 8 unit s−1. x bertambah. The rate of change of x is 8 unit s–1. x increases. 23. (a) tan 30° = r h = 1 3 r = h 3 Isipadu V = 1 3 πr2 h Volume = 1 3 π h 3 2 h V = 1 9 πh3 dV dh = 1 3 πh2 dV dt = 1 3 πh2 dh dt Jika h = 4 cm, dV dt = –4 cm3 s–1 –4 = 1 3 π(4)2 dh dt 3 – 4π = dh dt ∴ dh dt = –3 4π cm s–1 h menyusut h decreases (b) L = πr2 dL dr = 2πr dL dt = 2πr dr dt π = 2πr dr dt  1 = 2(3.5) dr dt dr dt = 1 7 cm s–1 Jejari bertambah The radius increases (c) h = 2r V = πr2 h = πr2 (2r) V = 2πr3 dV dr = 6πr2 dV dt = 6πr2 dr dt –1.2= 6π(7)2 dr dt dr dt = –1.2 6π(7)2 = –0.0013 cm s–1 Jejari menyusut The radius decreases 24. (a) Diberi y = 4x − 3 dy dx = 4 x berubah daripada 5 kepada 5.01 x changes from 5 to 5.01 Jadi/So δx = 5.01 – 5 = 0.01 δy δx ≈ dy dx δy ≈ dy dx · δx ≈ 4 δx ≈ 4 (0.01) ≈ 0.04 Jadi, y bertambah sebanyak 0.04. So, y increases 0.04. Apabila x = 5, y = 17 Maka, nilai baharu y Hence, the new value of y = y + δy = 17 + (0.04) = 17.04 (b) Diberi y = 9 x2 dy dx = 18 – x3 x berubah daripada 3 kepada 2.98. x changes from 3 to 2.98. Jadi, δx = 2.98 − 3 = −0.02 δy δx ≈ dy dx δy ≈ dy dx · δx ≈ 18 – x3 · δx ≈ 18 – (3)3 (−0.02) ≈ 0.013 Jadi, y bertambah sebanyak 0.013. So, y increases 0.013. Apabila x = 3, y = 1 Maka, nilai baharu y = y + δy Hence, the new value of y = y + δy = 1 + 0.013 = 1.013 h 30° r h r

J14 (c) Diberi y = x2 (x + 2) dy dx = 3x2 + 4x x berubah daripada 3 ke 2.99. x changes from 3 to 2.99. Jadi, δx = 2.99 – 3 = −0.01 δy δx ≈ dy dx δy ≈ dy dx · δx ≈ (3x2 + 4x)δx ≈ [3(3)2 + 4(3)](−0.01) ≈ −0.39 Jadi, y menyusut sebanyak 0.39. so, y decreases 0.39. 25. (a) (i) L = 2πj × 8 + πj 2 = π[16j + j 2 ] (ii) δj = 3.98 – 4 = –0.02 cm dL dj = 16π + 2πj Apabila j = 4 cm dan δj = –0.02 δL = dL dj . δj = [16π + 2π(4)])(–0.02) = 24π(–0.02) = –0.48π cm2 Luas menyusut sebanyak 0.48π cm2 The area decreases 0.48π cm2 (b) V = 6x2 – 4x dV dx = 12x – 4 δV = dV dx · δx δx = 0.01, x = 10 = (12x – 4)δx = (120 – 4)(0.01) = 1.16 cm3 (c) T = 2π L 10 = 2π 10 L 1 2 dT dL = π 10 1 L δT = 0.025 δT = dT dL . δL 0.025 = π 10(0.8) δL 0.025 8 π = δL δL = 0.0225 cm Praktis SPM 2 Kertas 1 1. x = 3t 3 – 2, dy dt = 15t 2 (a) dx dt = 9t 2 (b) dy dx = dy dt · dt dx = 15t 2 × 1 9t 2 = 5 3 2. (a) y = x – 3 x + 1 dy dx = (x + 1) – (x – 3) (x + 1)2 = 4 (x + 1)2 Apabila y = 0, x = 3 P(3, 0) When Pada P(3, 0), dy dx = 4 (3 + 1)2 = 1 4 At q = 1 4 3. (a) dy dx = 0 pada titik B dan D. at point B and D. (b) dy dx  0 pada titik C dan E. at point C and E. (c) dy dx  0 pada titik A. at point A. 4. y = x(x – 2)2 dy dx = 2x(x –2) + (x – 2)2 = (x – 2)(2x + x – 2) = (x – 2)(3x – 2) dy dx = (x – 2)(3x – 2) = 0 x = 2 ; x = 2 3 y = 0 2 3  2 3 – 2 2 = 32 27 d2 y dx2 = (x – 2)(3)+ (3x – 2) = 6x – 8 Apabila x = 2 d2 y dx2 = 12 – 8 = 4 > 0 When ∴(2, 0) ialah titik minimum is a minimum point Apabila x = 2 3 , d2 y dx2 = 6 2 3  – 8 When = –4 < 0  2 3 , 32 27 ialah titik maksimum is a maximum point Kertas 2 1. (a) dy dx = 2(1 + x2 ) – (2x – 4)(2x) (1 + x2 )2 = 2[1 + x2 – 2x2 + 4x] (1 + x2 )2 Apabila x = 2, dy dx = 2[1 + 4 – 8 + 8] 25 dy dx = 2 5 (b) persamaan garis normal Equation of the normal line y = – 5 2 x + c 0 = – 5 2 (2) + c c = 5 ∴y = – 5 2 x + 5 8 cm j

J15 2. (a) Perimeter = 120 = 2πj + 2x 60 = πj + x Luas, L = x(2j) Area = 2j[60 – πj] = 120j – 2πj 2 (b) dL dj = 120 – 4πj = 0 j = 120 4π = 30 π cm d2 L dj2 = –4π < 0 ∴ Luas ialah maksimum The area is maximum Apabila j = 30 π cm L = 120 30 π  – 2π 30 π  2 = 1800 π Sudut KBAT 1. (a) dV dt = –6 cm3 s–1 r 10 = h 15 r = 2 3 h V = 1 3 πr2 h = 1 3 π 2 3 h 2 h V = 4 27πh3 dV dh = 4 9 πh2 dV dt = 4 9 πh2 · dh dt –6 = 4 9 π[9]2 dh dt –6 36π = dh dt dh dt = – 1 6π cm s–1 (b) L = πr2 = π 2 3 h4 2 L = 4 9 πh2 dL dh = 8 9 πh dL dt = 8 9 πh dh dt = 8 9 π[9]– 1 6π = – 4 3 cm2 s–1 2. (a) dp dt = 10 unit s–1 A = 1 2 [p + 1]p2 A = 1 2 p3 + 1 2 p2 (b) dA dp = 3 2 p2 + p dA dt =  3 2 p + p · dp dt p = 2, dA dt =  3 2 (2)2 + 210 = 80 unit2 s–1 (c) p2 = PQ d(PQ) dp = 2p d(PQ) dt = 2p · dp dt p = 1, d(PQ) dt = 2(1)(10) = 20 unit s–1 A F D E C B x x j 15 cm 10 cm r h p2 y x y = x2 R(–1, 0) 0 P(p, 0)

J16 BAB Pengamiran 3 Integration 1. (a) d dx  23 x6 = 4x5 ∫d 23 x6 = ∫4x5 dx ∫4x5 dx = 23 x6 + c (b) 3∫g(x)dx = 3[4f(x)] + c = 12f(x) + c (c) ∫ 12 h(x)dx = (2x – 5) (3 – 4x)3 ∫ 1a h(x)dx = 2a  (2x – 5) (3 – 4x)3  (d) y = (x − 1)(x − 3)2 dy dx = 2(x − 1)(x − 3) + (x − 3)2 = (x − 3)(2x − 2 + x − 3] = (x − 3)(3x – 5) ∫(x − 3)(3x − 5)dx = (x − 1)(x − 3)2 (e) y = 2x x + 1 = 2(x + 1) – 2x (x + 1)2 = 2 (x + 1)2 ∫ 2 (x + 1)2 dx = 2x (x + 1) ∫ 6 (x + 1)2 dx = 3∫ 2 (x + 1)2 dx = 3(2x) (x + 1) = 6x (x + 1) (f) ∫f(x)dx = x + 1 x2 2π ∫f(x)dx = 2π  x + 1 x2  2. (a) ∫ 34 x dx = 34 x22 + c = 38 x2 + c (b) ∫ 23x3 dx = ∫ 23 x–3 dx = 23 x–2 –2 + c = 1 – 3x2 + c (c) ∫ x (16x4) dx = ∫ 1 16 x–3dx = 1 16 1 (–2)x2 + c = 1 – 32x2 + c (d) ∫ 5x 12 2x2 dx = ∫ 52 x – 32 dx = 52 × (–2)x + c = 5 – x + c (e) ∫ x 3 x dx = ∫ 13 x 12 dx = 13 × 23 x x + c = 29 x x + c (f) ∫ 8x3 4x5 dx = ∫2x–2dx = –2x + c (g) 23 ∫ x2 x 12 dx = 23 ∫x 32 dx = 23 × 25 x2 x + c = 4 15x2 x + c 3. (a) ∫2x2 dx – ∫4xdx + ∫x–2dx = 23 x3 – 2x2 – 1x + c (b) ∫5x2 dx – ∫ 10x2 dx = 53 x3 + 10x + c (c) ∫ 3x2 x 23 dx – ∫ 4x 23 dx = ∫3x 43 dx – ∫4x 2 – 3 dx = 37 3x 73  – 34x 13  = 97 x 73 – 12x 13 + c (d) = ∫2t–3dt – ∫6 dt + ∫3t –2dt = –t–2 – 6t – 3t –1 = 1 – t2 – 6t – 3t + c (e) ∫(8x – 4x2 + 12 – 6x)dx = ∫(2x – 4x2 + 12)dx = x2 – 43 x3 + 12x + c (f) 85 ∫(x2 – 3x)dx = 85  x33 – 3x2 2  + c (g) ∫ x(x – 5)(x + 1) (x – 5) dx = ∫(x2 + x)dx = 1 3 x3 + x22 + c 4. (a) Katakan u = x – 5 du dx = 1, dx = du1

J17 = ∫ 34 u3 du1 = 3(x – 5)4 16 + c (b) Katakan u = 3 – 4x du dx = –4, dx = du –4 = ∫2u–2 du –4 = u– 2 + 1 –2(–1) + c = 1 2(3 – 4x) + c (c) Katakan u = x – 2 du dx = 1, dx = du = ∫6u 12 du = 12u 32 3 + c = 4(x – 2) x – 2 + c (d) Katakan u = 3 – x du dx = –1, dx = –du = ∫ –23 u–3 du = –2u–3 + 1 3(–2) + c = 1 3(3 – x)2 + c (e) Katakan u = 1 – 2x du dx = –2, dx = du –2 = ∫ 3 –2 u –12 du = 3u 1 – 2 + 1 –2 12  + c = –3 1 – 2x + c 5. (a) dy dx = 3 – x2 – 4x ∫dy = ∫(3 – x2 – 4x)dx y = 3x – 13 x3 – 2x2 + c Apabila x = 0, y = 4 c = 4 Persamaan lengkung ialah The equation of the curve is y = 3x – 13 x3 – 2x2 + 4 (b) dy dx = (x – 2) 12 ∫dy= ∫(x – 2) 12 dx y = 23 (x – 2) 32 + c Apabila x = 6, y = 1 1 = 163 + c c = – 133 Persamaan lengkung ialah The equation of the curve is y = 23 (x – 2) 32 – 133 (c) dy dx = 2x3 + x2 ∫dy = ∫(2x3 + x2)dx y = 12 x 4 + 13 x3 + c Apabila x = 1, y = –1 –1 = 12 + 13 + c c = – 116 Persamaan lengkung ialah The equation of the curve is y = 12 x4 + 13 x3 – 116 (d) dy dx = ax – 3 ∫dy = ∫(ax – 3)dx y = a2 x2 – 3x + c Apabila x = –1, y = 8 8 = 12 a + 3 + c 5 = 12 a + c ...➀ Apabila x = 3, y = 4 4 = 92 a – 9 + c 13 = 92 a + c ...➁ ➁ – ➀: 8 = 4a a = 2, c = 4 Persamaan lengkung ialah The equation of the curve is y = x2 – 3x + 4 (e) dy dx = 3x2 + b ∫dy = ∫(3x2 + b)dx y = x3 + bx+ c Apabila x = 1, y = 3 3 = 1 + b + c 2 = b + c ...➀ Apabila x = –1, y = –3 –3 = –1 – b + c –2 = –b + c ...➁ ➁ – ➀: –4 = –2b b = 2, c = 0 Persamaan lengkung ialah The equation of the curve is y = x3 + 2x 6. (a) ∫0 –2 x(2x – 1)dx = ∫0 –2(2x2 – x)dx =  23 x 3 – 12 x2 0 –2 = 0 –  –163 – 2 = 223

J18 (b) ∫3 1 4x3 – 3 x2 dx = ∫31 (4x – 3x–2)dx = 2x 2 + 3x 31 = (18 + 1) – (2 + 3) = 14 (c) ∫1 –2(x + 2x–3)dx =  12 x 2 – 1x2 1 –2 =  12 – 1 – 2 – 14  = 34 – 3 = 1 –2 4 (d) ∫ 40 2(2 – x)–3dx =  2(2 – x)–2 (–2)(–1) 40 = 122 – 122 = 0 (e) ∫2 –2 (2x + 3)(2x – 3) (2x – 3) dx = [x2 + 3x]2 –2 = (4 + 6) – (4 – 6) = 12 7. (a) (i) ∫ –3 3 f(x)dx = ∫ –1 –3 f(x)dx + ∫ 3 –1 f(x)dx = –1 + 6 = 5 (ii) ∫ –1 –3 g(x)dx – 2∫ –1 –3 f(x)dx = 4 – 2(–1) = 6 (iii)∫ 0 –3 f(x)dx + ∫ 30 f(x)dx = ∫ 3 –3 f(x)dx = –1 + 6 = 5 (iv) 12 ∫ 2 –1 f(x)dx + ∫ 32 f(x)dx = 12 [6] = 3 (v) ∫ –1 –3 2g(x)dx – ∫ –1 –3 kx dx = 20 2(4) –  kx2 2  –1 –3 = 20 8 –  k2 – 9k2  = 20 4k = 12 k = 3 8. (a) ∫k –2(x – 4)dx = –10  x22 – 4x k –2 = –10 k22 – 4k – (2 + 8) = –10 k(k – 8) = 0 k = 0 atau k = 8 (b) ∫2k 2x–3dx = 34  1 – x2  2k = 34  1 – 4  –  1 – k2  = 34 1k2 – 14 = 34 4 – k2 = 3k2 4k2 = 4 k = ±1 (c) ∫4 –12xdx – ∫4 –1 g(x)dx = [x2]4 –1 –  1 (x – 2)2  4 –1 =(16 – 1) – 3 14 – 19 4 –1 = 15 – 5 36 = 31 14 36 (d) d2y dx2 = 1 – 6x2 dy dx = x – 2x3 + c x = –1, dy dx = –1 –1 + 2 + c = –1 c = –2 dy dx = x – 2x3 – 2 y = x22 – 12 x4 – 2x + c1, x = –1 y = 2 2 = 12 – 12 + 2 + c1, c1 = 0 ∴y = 12 x2 – 12 x4 – 2x (e) dy dx = 43 x3 – 2x + c x = 0, dy dx = 4, c = 4 y = 13 x4 – x2 + 4x + c1 x = 0, y = 1, c1 = 1 ∴y = 13 x4 – x2 + 4x + 1 9. (a) (i) ∫73 f(x)dx = 8 + 4.5 = 12.5 unit2 (ii) ∫3 –2 f(x)dx = −4.5 unit2 (iii) Jumlah luas = 12.5 + 4.5 Total area = 17 unit2 (b) (i) Luas H = luas segiempat Area H = Area of rectangle –∫0 –4 f(x)dx = 12 – 6 = 6 unit2 (ii) Luas K = ∫20 f(x)dx = 5

J19 (iii)∫ 2 –4 f(x)dx = ∫ 0 –4 f(x)dx + ∫ 2 0 f(x)dx = 6 + 5 = 11 unit2 (c) (i) Luas P/ Area P = 25 − Q = 25 − 18 = 7 unit2 (ii) Luas Q = ∫ 5 1xdy Area Q 10. (a) Apabila y = 0 2 + x – x2 = 0 (2 – x)(1 + x) = 0 x = 2; –1 B(2, 0) A(0, 2) luas = ∫ 2 0 ydx = ∫ 2 0 (2 + x – x2 )dx area = 2x + x2 2 – x3 3  2 0 = 4 + 2 – 8 3 4 = 10 3 unit2 (b) luas = ∫ 3 1 ydx area = ∫ 3 1 1 x2 dx = ∫ 3 1 x –2dx = – 1 x 4 3 1 = – 1 3 4 – [–1] = 1 – 1 3 = 2 3 unit2 (c) y = (x2 + x)(2 – x) = –x3 + 2x2 + 2x – x2 y = –x3 + x2 + 2x luas = ∫ 0 –1(–x3 + x2 + 2x)dx + ∫ 2 0 (–x3 + x2 + 2x)dx area =  –x4 4 + 1 3 x3 + x2  0 –1 +  –x4 4 + x3 3 + x2  2 0 =  5 – 12 + [–4 + 8 3 + 4] = 5 12 + 8 3 = 1 3 12 unit2 (d) Apabila y = 0 (x + 2)(3 – x) = 0 x = 3; –2 P[3,0] Luas di atas paksi-x/Area above the x-axis = ∫ 3 1 (x + 2)(3 – x)dx = ∫ 3 1 (6 + x – x2 )dx = 6x + 1 2 x2 – x3 3 4 3 1 = 318 + 9 2 – 94 – 6 + 1 2 – 1 3 4 = 1 7 3 unit2 Luas dari x = 3 ke x = 4 ∫4 3 (6 + x – x2 )dx = 6x + 1 2 x2 – x3 3 4 4 3 = 24 + 8 – 64 3 4 – 9 + 9 2  = 5 –2 6 ∴Jumlah luas = 1 7 3 +  5 –2 6  = 1 10 6 unit2 Total area (e) Apabila y = 0 4 – x2 = 0 x = ±2 Luas dari x = –3 ke –2 = ∫ –2 –3(4 – x2 )dx = 34x – x3 3 4 –2 –3 = 31–8 + 8 3 2 – (–12 + 9)4 = 1 –2 3 unit2 luas dari x = 2 ke x = 4 = ∫ 4 2 (4 – x2 )dx = 34x – x3 3 4 4 2 = – 2 10 3 unit2 Jumlah luas =  1 –2 3  +  – 2 10 3  Total area = 13 unit2 11. (a) x = y2 – 4y luas sebelah kiri/area on the left = ∫ 4 0 x dy = ∫ 4 0 (y2 – 4y)dy =  y3 3 – 2y2  4 0 =  64 3 – 324 – 0 = – 2 10 3 ∫0 –1(y2 – 4y)dy =  y3 3 – 2y2 4 0 –1 = 0 – 1 1 – 3 – 22 = 1 2 3 Jumlah luas =  – 2 10 3  + 1 2 3 Total area = 13 unit2 (b) x = y 1 2 luas = 2∫ 4 1xdy area = 2∫ 4 1 y 1 2 dy = 23 2 3 y 3 2 4 4 1 = 231 2 3 (8) – 2 3 (1) = 23 14 3 4 = 28 3 unit2 (c) x2 = 4 y x = 2 y = 2y 1 – 2 luas = ∫ 2 1 2 2y 1 – 2 dy area = 4 y 4 2 1 2 = 4 2 – 4 2 = 8 – 4 2 = 4 2 = 2 2 unit2

J20 12. (a) x2 + 1 = –x + 3 x2 + x – 2 = 0 (x – 1)(x + 2) = 0 x = 1; –2 y = 2 A[1, 2] luas = ∫ 1 0 ydx + ∆ABC area = ∫ 1 0 (x2 + 1)dx + 1 2 (2)(2) = 3 x3 3 + x4 1 0 + 2 = 1 1 3 + 12 + 2 = 1 3 3 unit2 (b) Titik persilangan Intersection point x2 = x x4 = x x(x3 – 1) = 0 x = 0; x = 1 y = 0 y = 1 luas berlorek = ∫ 1 0 x dx – ∫ 1 0 x2 dx shaded area =  2 3 x 3 2 4 1 0 –  1 3 x3 4 1 0 = 2 3 – 1 3 = 1 3 unit2 (c) Titik persilangan/Intersection point (1 – x)(x + 4) = x + 4 x + 4 – x2 – 4x = x + 4 x2 + 4x = 0 x(x + 4) = 0 x = 0; x = –4 y = 4 ; y = 0 luas berlorek = luas segi tiga + ∫ 1 0 f(x)dx shaded area = area of triangle + ∫ 1 0 f(x)dx = 1 2 (4)(4) + ∫ 1 0(4 – 3x – x2 )dx = 8 + 4x – 3 2 x2 – x3 3 4 1 0 = 8 + 34 – 3 2 – 1 3 4 = 1 10 6 unit2 13. (a) Isi padu janaan The generated volume = π∫ 3 1y2 dx = π∫ 3 1 1 x2 dx = π∫ 3 1x –2 dx = π 1 – x  3 1 = π 1 – 3 –(–1)4 = 2 3 π unit3 (b) Titik persilangan dengan paksi-x (3, 0) ,(0, 0) The intersection point at the x-axis of (3, 0), (0, 0) 1 1 3 y O x y = x2 + 1 y = –x + 3 A B dx C Isipadu dijanakan The generated volume = π∫ 3 1y2 dx = π∫ 3 1x2 (x – 3)2 dx = π∫ 3 1x4 – 6x3 + 9x2 )dx = π 1 5 x5 – 3 2 x4 + 3x3  3 1 = π( 35 5 – 3 2 (3)4 + 3(3)3 ) – 1 1 5 – 3 2 + 324 = 2 6 5 π unit3 (c) Apabila y = 0, x = –4 Isipadu dijanakan The generated volume = π∫ 0 –4 y2 dx = π∫ 0 –41 x + 4 2 2dx = π 1 4 x2 + 2x 0 –4 = π0 – (4 – 8)4 = 4π unit3 14. (a) Apabila/When x = 0, y = 4 Isipadu/Volume = 10 3 π = π∫ 4 k x2 dy = π∫ 4 k1 16 – y2 4 2dy = π4y – 1 12 y3  4 k = 10 3 π 116 – 64 12 2 – 4k + k3 12 = 10 3 k3 12 – 4k + 22 3 = 0 k3 – 48k + 88 = 0 Jika/If k = 2, 8 – 48(2) + 88 = 0 ∴k = 2 (b) Apabila/When x = 0, y = k Isipadu/Volume = 2π 2π = π∫ 4 k (y – k)dy ∫ 4 k (y – k)dy = 2  y2 2 – ky 4 k = 2 (8 – 4k) – 1 k2 2 – k2 2 = 2 k2 2 – 4k + 6 = 0 k2 – 8k + 12 = 0 (k – 2)(k – 6) = 0 ∴k = 2 k = 6 (c) Isipadu = 152 3 π Volume = π∫ k 1 (y + 3)2 dy π 3 (y + 3)3 3 4 k 1 = 152 3 π (k + 3)3 3 – 43 3 = 152 3

J21 (k + 3)3 = 216 k + 3 = 6 k = 3 15. (a) Isipadu dijanakan oleh lengkung Volume generated by the curve = π∫ 2 0 (6 + 7x – 3x2 )2 dx = π∫ 2 0 (36 + 84x + 13x2 – 42x3 + 9x4 )dx = 164.27π atau 4 164 15π Isipadu kon Volume of cone = 1 3 π(8)2 (2) = 42.67 π atau 2 42 3 π Isipadu janaan/Generated volume = 121.6 π unit3 atau 3 121 5 π unit3 (b) Koordinat A/Coordinates of A 4 x = x x = 2, y = 2 Isipadu/Volume = π∫ 3 2 x2 dy + kon = π∫ 3 2 16 y2 dy + 1 3 π(2)2 (2) = π –16 y  3 2 + 8 3 π = π –16 3 –  –16 2 24 + 8 3 π = 1 5 3 π unit3 (c) y = 9 – x2 dy dx = –2x x = 2, dy dx = –4 persamaan tangen pada (2, 5) The equation of tangent at (2, 5) y – 5 = –4(x – 2) y = –4x + 13 Tangen menyilang paksi-x pada 1 13 4 , 02. Tangent intersects the x-axis at 1 13 4 , 02 Isipadu kon = 1 3 π(5)2 1 5 4 2 Volume of cone = 125 12 π unit3 Isipadu dijanakan oleh lengkung Volume generated by the curve = π∫ 3 2 y2 dx = π∫ 3 2 (9 – x2 )2 dx = π∫ 3 2 (81 – 18x2 + x4 )dx = π381x – 6x3 + 1 5 x5 4 3 2 = 1 9 5 π unit3 Isipadu diperlu = 73 60 π unit3 Volume needed 16. (a) (i) Dengan teorem Pythagoras, x2 + y2 = r2 (ii) Isi padu janaan/Generated volume = 2π∫ r 0x2 dy = 2π∫ r 0(r2 – y2 )dy = 2πr2 y – y3 3 4 r 0 = 2π3r3 – r3 3 4 = 2π 3 2 3 r3 4 = 4 3 πr3 (b) Persamaan garis lurus The equation of straight line y = h r x + h Isipadu dijana/Generated volume = π∫ h 0x2 dy = π∫ h 0 r(y – h) h 4 2 dy = r2 π h2 ∫ h 0(y – h)2 dy = r2 π h2  (y – h)3 3 4 h 0 = r2 π h2 30 – 1 –h3 3 24 = r2 π h2 3 h3 3 4 = 1 3 πr2 h unit3 (c) y = 14 – 4 9 x2 π∫ 14 10 x2 dy = 9 4 π∫ 14 10(14 – y)dy = 9 4 π314y – y2 2 4 14 10 = 9 4 π[(196 – 98) – (140 – 50)] = 9 4 π[8] = 18π unit3 Isipadu kon/Volume of cone = 1 3 π(3)2 (10) = 30π Jumlah isipadu/Total volume = 18π + 30π = 48π unit3 Satu bekas isipadu = 20 × 30 × 40 = 24000 cm3 Volume of a container Dua bekas/Two containers = 48000 cm3 Bilangan kon = 48000 48π = 318 kon Number of cone Jumlah jualan/Total sales = 318 × RM3.50 = RM1113 P(x, y) y y x O x r

J22 Kos/cost = RM345 × 2 = RM690 Keuntungan/Profit = RM423 Praktis SPM 3 Kertas 1 1. (a) ∫ 4 –2 f(x)dx = 5 ∫ –2 4 f(x)dx = –5 (b) ∫ 4 –22f(x)dx – ∫ 4 –23x dx = 10 –  32 x24 4 –2 = 10 – [24 – 6] = –8 2. (a) dy dx = 2x – 4 Apabila x = 3, dy dx = 2(3) – 4 = 2 (b) y = ∫(2x – 4)dx = x2 – 4x + c Pada/At (3, –2), –2 = 9 – 12 + c c = 1 ∴y = x2 – 4x + 1 3. (a) –h (h, 6k) (0, 3k) h x y O (b) ∫h 0 f(x)dx = 8 + ∫h 0 g(x)dx = 8 + 12 [3k + 6k]h = 8 + 9hk2 4. ∫h –1(3 – 2x)dx = 6 [3x – x2]h –1 = 6 3h – h2 – (–3 – 1) = 6 3h – h2 – 2 = 0 h2 – 3h + 2 = 0 (h – 1)(h – 2) = 0 h = 1; 2 5. (a) ∫ b a f(x)dx = 8 a = 1, b = 3 (b) ∫ 3 –2 f(x)dx = ∫ 1 –2 f(x)dx + ∫ 3 1 f(x)dx = –5 + 8 = 3 6. f’(x) = 4x – 4 f’(x) = 4x – 4 = 0 x = 1 ∴(1, –4) ialah titik minimum (1, –4) is a minimum point f(x) = ∫(4x – 4)dx = 2x2 – 4x + c Pada/At (1, –4) –4 = 2 – 4 + c c = –2 f(x) = 2x2 – 4x – 2 7. (a) ∫ 2 h p(x)dx = – 34 (b) ∫h 2 p(x)dx – ∫h 2 2xdx = –4 14 34 + 4 14 = [x2]h 2 = h2 – 4 5 + 4 = h2 h = 3 Kertas 2 1. (a) y = 3x2 – 27 Apabila y = 0 x = ±3 dy dx = 6x Apabila dy dx = 6x = 6 x = 1 y = –24 ∴P[1, –24] (b) Persamaan tangen y + 24 = 6(x – 1) Equation of tangent y = 6x – 30 luas berlorek shaded area = 12 [4][24] – ∫3 1 ydx = 48 – ∫31 (3x2 – 27)dx = 48 – [x3 – 27]31 ∴48 – [0 + 26] = 22 unit2 (c) Isipadu = π∫0k x2 dy = 21 1103 2π Volume ∫ 0k1 13 y + 92dx = 2203  16 y2 + 9y40k = 2203 0 – [k26 + 9k] = 2203 k2 + 54k + 440 = 0 (k + 44)(k + 10) = 0 k = –44 ; –10 ∴k = –10 2. (a) 4y = x + 3 ; x = 4y2 – 2 4y – 3 = 4y2 – 2 4y2 – 4y + 1 = 0 (2y – 1)2 = 0 y = 12 , x = 2 – 3 = –1 A–1, 12 2 (b) Apabila/When y = 0, x = –3 luas berlorek/shaded area = 12 (2) 12 2 – ∫ –1 –2 ydx = 12 – ∫ –1 –2 12 (x + 2) 12 dx = 12 – 3 12 . 23 (x + 2) 32 4 –1 –2 = 12 – 3 13 – 04 = 16 unit2 –3 –2 –1 x y O –1, 12 fi ff

J23 (c) Isipadu/Volume = π∫ 1 –2 y2 dx = π∫ 1 –2 1 4 (x + 2)dx = π3 (x + 2)2 8 4 1 –2 = π3 9 8 – 04 = 9π 8 Isipadu sebenar Actual volume = 9π 16 unit3 3. (a) y = 6x – x2 dy dx = 6 – 2x = 0 x = 3 Katakan persamaan tangen Let the equation of tangent y = mx + c 10 = 3m + c ...➀ 6x – x2 = mx + c x2 + (m – 6)x + c = 0 (m – 6)2 – 4c = 0 (m – 6)2 = 4(10 – 3m) m2 – 12m + 36 = 40 – 12m m2 = 4 m = ±2 Apabila m = –2, c = 10 – 3(–2) = 16 ∴6 – 2x = –2 8 = 2x x = 4; y = 6(4) – 16 = 8 ∴A(4, 8) ; a = 4 (b) luas berlorek = luas trapezium – ∫ 4 3 ydx shaded area = area of trapezium – ∫ 4 3 ydx = 1 2 [10 + 8][1] – ∫ 4 3 (6x – x2 )dx = 9 – 33x2 – 1 3 x3 4 4 3 = 9 – 326 2 3 – 184 = 1 3 unit2 Sudut KBAT (a) y = x3 + 1 Apabila y = 0 ; 0 = x3 + 1 x3 = –1 x = –1 P[–1, 0] dy dx = 3x2 Apabila x = –1, dy dx = 3 When y = 3(x + 1) Jika y = 3x + 3 menyilang lengkap If y = 3x + 3 intersects completely 3x + 3 = x3 + 1 x3 – 3x – 2 = 0 Apabila x = 2 23 – 3(2) – 2 = 8 – 8 = 0 ∴y = 23 + 1 = 9 Q(2, 9) (b) luas A = ∫ 2 –1 ydx area A = ∫ 2 –1(x3 + 1)dx = 3 x4 4 + x4 2 –1 = (4 + 2) – 1 1 4 – 12 = 6 + 3 4 = 3 6 4 unit2 luas/area PQR = 1 2 (3)9 = 27 2 unit2 ∴luas/area B = 27 2 – 3 6 4 = 3 6 4 unit2 Nisbah A : B = 3 6 4 : 3 6 4 Ratio = 1 : 1 (c) Isipadu = π 2 ∫ 2 –1 y2 dx = π 2 ∫ 2 –1(x3 + 1)2dx Volume = π 2 3 1 7 x7 + 1 2 x4 + x4 2 –1 = π 2 3 13 28 14 4 = 13 14 28 π unit3 O 4 x y 3 (3, 10) (4, 8) –1O 2 P x y Q R (2, 9)

J24 BAB Pilih Atur dan Gabungan 4 Permutation and Combination 1. (a) (i) n! (n – 0)! = 1 (ii) n! (n – n + 1)! = n! (b) (i) 10! (10 – 3)! = 720 (ii) 5! (1)! × 6! (4)! = 5 × 6! = 3 6000 (c) (i) n! (n – n + 1)! × (n – 1)! (n – 1 – 2)! = n!(n – 1)(n – 2) (ii) m! (m – 2)! ÷ (m + 1)! (m + 1 – 2)! = m(m – 1) ÷ (m + 1)! (m – 1)! = m – 1 m + 1 2. (a) Bilangan cara Number of ways = 4 × 3 = 12 (b) Bilangan cara Number of ways = 5 × 6 = 30 (c) Bilangan cara Number of ways = 6 × 4 × 3 = 72 3. (a) Bilangan cara = 7! = 5 040 Number of ways (b) Bilangan cara = 8! = 40 320 Number of ways (c) Bilangan cara = 6! = 720 Number of ways 4. (a) Bilangan cara = 10P4 = 10 × 9 × 8 × 7 Number of ways = 5 040 (b) Bilangan cara = 9 P3 Number of ways = 504 (c) Bilangan cara = 6 P3 Number of ways = 120 5. (a) Bilangan cara susunan Number of ways of arrangement = 9! 4!3!2! = 1 260 (b) Bilangan cara susunan Number of ways of arrangement = 15! 5!6!4! = 630 630 (c) Bilangan cara susunan Number of ways of arrangement = 9! 3!3!2! = 5 040 6. (a) Bilangan cara = 7! Number of ways = 5 040 (b) Bilangan cara = 5! Number of ways = 120 (c) Bilangan cara = (10 − 1)! Number of ways = 362 880 7. (a) (i) Bilangan cara = 2 × 4! = 48 Number of ways (ii) Bilangan cara = 4! × 2! = 48 Number of ways (iii) Susunan dalam bentuk VKVKV Arrangement in the form of VKVKV Bilangan cara = 2! × 3! = 12 Number of ways (b) (i) Bilangan cara = 1 × 4! = 24 Number of ways (ii) Bilangan cara = 1! × 2! × 2! × 2! = 8 Number of ways (c) (i) 5! × 2 = 240 (ii) 4! × 4! = 576 5! × 6 P2 = 3 600 (d) (i) 6 × 6 × 6 x 6 × 3 = 3 888 (ii) 3! × 3 × 2 = 36 8. (a) (i) Bilangan cara menyusun = (10 − 1)! Number of ways to arrange = 362 880 (ii) Bilangan cara menyusun Number of ways to arrange = (5 − 1)! × 6! = 17 280 (iii) Bilangan cara menyusun Number of ways to arrange = (10 − 1)! − (7 − 1)! × 4! = 345 600 (b) (i) Bilangan cara = (8 −1)! × 2! Number of ways = 10 080 (ii) Bilangan cara menyusun = (7 − 1)! × 3! Number of ways to arrange = 4 320 (c) (i) Bilangan cara menyusun = (5 − 1)! × 4! Number of ways to arrange = 576 (ii) Bilangan cara menyusun = (4 − 1)! × 4! Number of ways to arrange = 144 9. (a) (i) Sebelah kiri = r! n Cr Let side = r!n! (n − r)!r! = n! (n − r)! = n Pr Sebelah kanan Right side Maka/Hence, r! n Cr = n Pr (ii) n C0 = n! (n − 0)!0! = n! (n)! = 1 = sebelah kanan Right side (b) (i) 8 C2 = 8! (8 – 2)!2! = 8! (6)!2! = 28 Sudut Kalkulator Tekan 8 SHIFT 2 = 28 n Cr

J25 (ii) 4 C2 × 9 C4 = 4! (4 – 2)!2! × 9! (9 – 4)!4! = 756 (c) (i) 10C7 ÷ 6 C3 = 10! (10 – 7)!7! ÷ 6! (6 – 3)!3! = 6 (ii) n Cn – 2 = n! (n − n + 2)!(n – 2)! = n! (2)!(n – 2)! = n(n – 1)(n – 2)! (2)!(n – 2)! = n(n – 1) (2) 10. (a) Bilangan cara pilihan = 8 C4 Number of ways = 70 (b) Bilangan cara pilihan = 12C5 Number of ways = 792 (c) Bilangan cara pilihan = 7 C4 Number of ways = 35 11. (a) (i) Untuk melukis satu garisan, kita memerlukan mana-mana dua titik sahaja. To draw a straight line, we neeed any two points only. Maka, bilangan cara = 6 C2 = 15. So, number of ways (ii) Untuk melukis satu segi tiga, kita memerlukan mana-mana tiga titik sahaja. To draw a triangle, we need any three points only. Maka, bilangan cara = 6 C3 = 20. So, number of ways (iii) Untuk melukis satu segi empat, kita memerlukan mana-mana empat titik sahaja. To draw a rectangle, we need any four points only. Maka, bilangan cara = 6 C4 = 15. So, number of ways (b) (i) Bilangan cara = 4 C1 × 5 C4 = 20 Number of ways (ii) Kes 1: 3 perempuan, 2 lelaki Bilangan cara = 5 C3 × 4 C2 = 60 Number of ways Kes 2: 4 perempuan, 1 lelaki Bilangan cara = 5 C4 × 4 C1 = 20 Number of ways Kes 3: 5 perempuan, 0 lelaki Bilangan cara = 5 C5 = 1 Number of ways Jumlah cara = 81 (iii) Bilangan cara = 6 C2 = 15 Number of ways (c) Bilangan cara = 2 [3 C1 × 2 C1 × 6 C3 × 3 C3 ] Number of ways = 240 (d) (i) Bilangan cara = 23C5 = 33 649 Number of ways (ii) Bilangan cara = 10C1 × 7 C1 × 6 C1 × 20C2 Number of ways = 79 800 (iii) Bilangan cara = 17C5 = 6 188 Number of ways (e) (i) Jika memilih baris tiga kerusi. If choose the row of three chairs Bilangan cara = 5 C1 × 4 C4 = 5 Number of ways Jika memilih baris empat kerusi If choose the row of four chairs Bilangan cara = 5 C2 × 3 C3 = 10 Jumlah = 15 (ii) 1 cara sahaja Only 1 way (f) (i) Bilangan cara = 7 C3 = 35 Number of ways (ii) Bilangan cara = 5 C4 = 5 Number of ways (iii) Bilangan cara = 7 C3 × 2 = 70 Number of ways Praktis SPM 4 Kertas 1 1. (a) 1 ≤ n < 7 dengan n ialah integer (b) x Cm = x Cn x! (x − m)!m! = x! (x − n)!n! n + m = x 2. 4! 2! – 3! = 6 3. Kes 1: Baju biru Case 1: Blue blouses 3 C1 × 2 C1 = 6 cara Kes 2: Seluar biru Case 2: Blue pants 2 C1 × 2 C1 = 4 cara Kes 3: Bukan baju biru dan seluar biru Case 3: Not blue blouses and blue pants 2 C1 × 2 C1 = 4 cara Jumlah cara = 14 cara Total ways Sudut KBAT (a) Kes 1: nombor berdigit 4 = 3 × 5 P3 = 180 Kes 2: nombor berdigit 5 = 5 × 5 P5 = 600 Kes 3: nombor berdigit 6 = 5 × 5 P5 = 600 Jumlah = 600 + 600 + 180 = 1380 (b) 2 × 4 × 3 + 2 × 5 × 4 × 3 = 144 (c) 5 P4 × 1 P1 = 120

J26 BAB Taburan Kebarangkalian 5 Probability Distribution 1. (a) Pemboleh ubah ialah menang, tewas atau seri The variable is win, lose or draw (b) Pemboleh ubah ialah tinggi pelajar itu. The variable is the height of a student. (c) Pemboleh ubah ialah {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. The variable is 2. (a) X = {0, 1, 2, 3, 4,… 20} adalah diskret kerana dapat dikira. X = {0, 1, 2, 3, 4,… 20} is discrete because they can be counted. (b) Y = {y: 1 1 2 ≤ y ≤ 3} adalah selanjar kerana sudah menentukan dengan tepat masa yang sewajar. Y = {y: 1 1 2 ≤ y ≤ 3} is continuous because it has determined by the time accordingly. (c) Z = {z: 3 ≤ z ≤ 8} adalah selanjar kerana sudah menentukan dengan tepat masa yang sewajarnya. Z = {z: 3 ≤ z ≤ 8} is continuous because it has determind by the time accordingly. (d) X = {0, 1, 2, 3, ... , n} adalah diskret kerana dapat dikira. X = {0, 1, 2, 3, ... , n} is discrete because they can be counted. 3. X = {0, 1, 2} P[X = 0] = P[M, M] = 1 3 × 1 3 = 1 9 P[X = 1] = P[H, M] + P[M, H] = 2 3 × 1 3 + 1 3 × 2 3 = 4 9 P[X = 2] = P[H, H] = 2 3 × 2 3 = 4 9 X = x 0 1 2 P[X = x] 1 9 4 9 4 9 (b) X = {0, 1, 2, 3} B B B B B B B B’ B’ B’ B’ B’ B’ B’ 7 10 7 10 7 10 7 10 7 10 7 10 7 10 3 10 3 10 3 10 3 10 3 10 3 10 3 10 P[X = 0] = P[B’, B’, B’] =  7 10 3 = 343 1000 P[X = 1] = 3 C1 3 10 7 10 2 = 441 1000 P[X = 2] = 3 C21 3 10 2  7 10 = 189 1000 P[X = 3] = 1 3 10 3 = 27 1000 X = x 0 1 2 3 P[X = x] 343 1000 441 1000 189 1000 27 1000 4. (a) (i) P(X = 0) = 0 6 = 0 P(X = 1) = 1 6 P(X = 2) = 2 6 = 1 3 P(X = 3) = 3 6 = 1 2 X = x 0 1 2 3 P(X = x) 0 1 6 1 3 1 2 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 1 6 + 1 3 + 1 2 = 1 (Terbukti)/(Proved) H H H M M M 2 3 2 3 2 1 3 3 1 3 1 3

J27 (ii) x 1 3 1 2 1 6 0 0 1 2 3 P(X = x) (b) (i) P(X = 1) = (1 + 1)2 54 = 2 27 P(X = 2) = (2 + 1)2 54 = 1 6 P(X = 3) = (3 + 1)2 54 = 8 27 P(X = 4) = (4 + 1)2 54 = 25 54 X = x 1 2 3 4 P(X = x) 2 27 1 6 8 27 25 54 P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 2 27 + 1 6 + 8 27 + 25 54 = 1 (Terbukti)/(Proved) (ii) x 9 54 16 54 25 54 4 54 0 1 2 3 4 P(X = x) 5. (a) (i) P(X ≤ 6) = 1 − P(X = 8) = 1 − 0.27 = 0.73 (ii) P(2 ≤ X ≤ 6) = 1 − P(X = 8) = 0.73 (b) Y 1 2 3 4 5 6 7 P(Y = y) 0.1 0.2 0.1 0.2 0.1 0.2 0.1 ∑7 0 P(Y = y) = 0.1 × 4 + 0.2 × 3 = 0.4 + 0.6 = 1 Maka, Y ialah suatu pemboleh ubah rawak diskret. Hence, y is a random discrete variable. (i) P(Y < 4) = P(Y = 1) + P(Y = 2) + P(Y = 3) = 0.1 + 0.2 + 01 = 0.4 (ii) P(3 < Y < 5) = P(Y = 4) = 0.2 (c) (i) P(Z > 2) = P(Z = 3) + P(Z = 4) = 4 9 q + q = 4 9 q = 2 9 P(Z = 0) + P(Z = 1) + P(Z = 2) = 5 9 p + p + p + q = 5 9 3p + 2 9 = 5 9 3p = 3 9 = 1 9 (ii) z 0 1 2 3 4 P(Z = z) 1 9 1 9 1 3 2 9 2 9 (iii) 0 0 1 2 3 4 z P(Z = z) 1 9 2 9 3 9 6. (a) Ini ialah percubaan Bernoulli, kerana kebarangkalian mendapat bola merah setiap kali ialah tak bersandar, iaitu tetap = 5 8 dan percubaan ini berlaku 4 kali, n = 4. This is Bernoulli’s trial, because the probability of getting a red ball every time is not dependent, that is fixed = 5 8 and the trial occurs 4 times, n = 4. (b) Ini ialah percubaan Bernoulli, kerana terdapat dua kesudahan sahaja, dengan p = 0.85 dan percubaan ini berlaku 10 kali, n = 10. This is a Bernoulli trial because there are two outcomes, where p = 0.85 and this trial occurs 10 times, n = 10.

J28 (c) Ini bukan percubaan Bernoulli, kerana kebarangkalian mendapat A pertama kali, p = 3 6 , kedua kali p = 2 5 dan ketiga kali p = 1 4 dengan setiap kali p berubah, iaitu bersandar. This is not Bernoulli trial, because the probability to get A the first time, p = 3 6 , the second time p = 2 5 and the third time p = 1 4 , where each time p changes, that is dependent. 7. (a) n = 14, p = 0.05, q = 0.95 (i) P(X = 4) = 14C4 (0.05)4 (0.95)10 = 0.0037 (ii) P(X < 12) = 1 − P[X = 12] − P(X = 13) + P(X = 14) = 1 − (14C12(0.05)12(0.95)2 + 14C13(0.05)13(0.95)1 + 14C14(0.05)14) = 1 (iii) P(5 ≤ X ≤ 7) = P(X = 5) + P[X = 6) + P(X = 7) = [14C5 (0.05)5 (0.95)9 + 14C6 (0.05)6 (0.95)8 + 14C7 (0.57)(0.95)7 ] = 0.00043 (b) n = 8, p = 0.9, q = 0.1 (i) P(X = 8) = 8 C8 (0.9)8 = 0.4305 Maka, 43.05% dijangkiti Covid-19. Hence, 43.05% contracted Covid-19. (ii) P(X > 6) = P(X = 7) + P(X = 8) = 8 C7 (0.9)7 (0.1)1 + 8 C8 (0.9)8 = 0.8131 Maka, 81.31% dijangkiti Covid-19. Hence, 81.31% contracted Covid-19. (c) n = 10, p = 0.95, q = 0.05 (i) P(X > 8) = P(X = 9] + P(X = 10) = 10C9 (0.95)9 (0.05)1 + 10C10(0.95)10 = 0.9139 (ii) P(2 ≤ X ≤ 8) = 1 − P(X = 0) − P(X = 1) − P(X = 9) − P(X = 10) = 0.0861 (d) p = 2 3 , q = 1 3 (i) P(X = 0) = 0.0014 = n C0 2 3 2 0  1 3 2 n = 0.0014 n log  1 3 2 = log 0.0014 n = log0.0014 log 1 3 = 6 (ii) P(X > 4) = P(X = 5) + P(X = 6) = 6 C5 2 3 2 5  1 3 2 + 6 C6 2 3 2 6 = 0.3512 8. (a) 0.20 + 0.25 + p + p + 0.15 + 0.05= 1 0.65 + 2p = 1 2p = 0.35 p = 0.175 P (X = x) x 0 0 1 2 3 4 5 0.05 0.10 0.15 0.20 0.25 0.30 p (b) n = 4, p = 1 2 , q = 1 2 P(X = 0) = 4 C0 1 2 2 0  1 2 2 4 = 0.0625 P(X = 1) = 4 C1 1 2 2 1  1 2 2 3 = 0.2500 P(X = 2) = 4 C2 1 2 2 2  1 2 2 2 = 0.3750 P(X = 3) = 4 C3 1 2 2 3  1 2 2 1 = 0.2500 P(X = 4) = 4 C4 1 2 2 4  1 2 2 0 = 0.0625 P (X = x) 0 x 0 1 2 3 4 0.1 0.2 0.3 0.4 9. (a) (i) n = 3, p = 1 6 P(mendapat dua ‘5’) = 3 C2 1 6 2 2  5 6 2 = 5 72 (ii) E(X) = np = 216 × 5 72 = 15 kali (b) (i) E(X) = np = 3 n 3 14 2 = 3 n = 14

J29 (ii) Varians = npq = 14 × 3 14 × 11 14 = 2 5 14 Sisihan piawai = 1.54 Standard deviation (b) (i) q = 0.35, p = 0.65 Sisihan piawai = 1.5 = n(0.65)(0.35) Standard deviation n = 1.52 (0.65)(0.35) = 10 (ii) Min/Mean = np = 10(0.65) = 6.5 10. (a) p = 0.02, q = 0.98 P(X ≥ 1) > 0.95 1 – P(X = 0) > 0.95 1 − n C0 (0.02)0 (0.98)n > 0.95 0.98n < 0.05 n log 0.98 < log 0.05 n > log0.05 log0.98 n > 148.28 n = 149 (b) (i) P(X = 3) = 5 C3 1 6 2 3  5 6 2 2 = 0.0322 (ii) E(X) = np = 550 × 0.0322 = 17.68 (c) n = 5, p = 0.8, q = 0.2 (i) P(X = 3) = 5 C3 (0.8)3 (0.2)2 = 0.2048 (ii) P(X < 2) = P(X = 0) + P(X = 1) = 0.00672 (iii) E(X) = np = 50 × 0.8 = 40 11. (a) (i) µ = 32 (ii) P(X > 32) = 0.5 (iii) P(23 < X < 32) = 0.5 – 0.16 = 0.34 (b) (i) P(X > 54) = 1 – 0.26 = 0.74 (ii) P(54 ≤ X ≤ 76) = 1 − 2(0.26) = 0.48 (c) x f(x) 0 2.5 2.8 3.0 12. Pemboleh ubah X Variable X Min, µ Mean, µ Sisihan piawai, σ Standard deviation, σ Skor-z z-score (a) 0.8 = X − 67 X = 15(0.8) + 67 = 79 67 15 0.8 (b) 4.6 −2.1 = 4.6 – µ 0.9 µ = 4.6 + 2.1(0.9) = 6.49 0.9 −2.1 (c) 130 143 –1.85 = 130 − 143 σ σ = –13 –1.85 = 7.03 –1.85 13. (a) P(Z < –0.4) = P(Z > 0.4) = 0.3446 z –0.4 0 f(z) Graf ialah simetri pada f(z) = 0. Graph is symmetrical about f(z) = 0. (b) P(Z > –1.2) = 1 – P(Z ≥ 1.2) = 1 – 0.1151 = 0.8849 Jumlah luas di bawah graf ialah 1. Total area under the graph is 1. z –1.2 0 f(z) (c) 1 – P(Z > 1.5) – P(Z > 2.1) = 1 – 0.0668 – 0.0179 = 0.9153 z –1.5 0 2.1 f(z)

J30 14. (a) skor-z = –1.78 z 0 z f(z) (b) 1 – P(Z ≤ z) = 0.9126 P(Z ≤ z) = 0.0874 skor-z = –1.357 Luas adalah lebih daripada 0.5, maka skor-z ialah negatif. Area is more than 0.5, so z-score is negative. z 0 z f(z) (c) 1 – P(Z > z) = 0.8251 P(Z > z) = 0.1749 skor-z = 0.935 Luas lebih daripada 0.5, maka skor-z ialah positif. Area is more than 0.5, so z-score is positive. z 0 z f(z) 15. (a) P(Z > k − 15 22 ) = 0.4152 k − 15 22 = 0.214 k = 19.71 z 0 0.214 f(z) 0.4152 (b) PZ < k − 250 16 2 = 0.32 k − 250 16 = –0.468 k = 242.51 z –0.468 0 f(z) 0.32 (c) P 35 – 40.2 12.5 < Z < k – 40.2 12.5  = 0.458 P–0.416 < Z < k – 40.2 12.5 2 = 0.458 1 – P(Z < –0.416) – P1Z > k – 40.2 12.5 2 = 0.458 P1Z < k – 40.2 12.5 2 = 0.2033 ∴ k – 40.2 12.5 = 0.83 k = 50.58 z –0.416 0 0.83 f(z) 0.2033 0.3387 (d) P 30 – 35 k < Z < 40 – 35 k  = 0.578 P– 5 k < Z < 5 k 2 = 0.578 1 – 2P(Z ≥ 5 k ) = 0.578 P1Z ≥ 5 k 2 = 0.211 5 k = 0.803 k = 6.23 z – 0 f(z) 0.211 5 k 5 k 0.578 16. (a) (i) µ = 53, σ = 9 P[42 < X < 58] = P 42 – 53 9 < Z < 58 – 53 9 4 = P– 11 9 < Z < 5 9 4 = 0.5999 (ii) P(X < 40] = PZ < 40 – 53 9 4 = 0.0743 0.0743 × 100% = 7.43% (iii) P[X > m] = 0.74 P3Z > m – 53 9 4 = 0.74 m – 53 9 = –0.643 m = 53 – 9(0.643) = 47.21 (b) P(X < 60) = 0.8 P1Z < 60 – µ σ 2 = 0.8 60 – µ σ = 0.842 60 – µ = 0.842σ ...➀ P(X < 70) = 0.9 P1Z < 70 – µ σ 2 = 0.9 70 – µ σ = 1.282 70 – µ = 1.282σ ...➁ ➁ – ➀ 10 = 0.44 σ σ = 22.73 varians/variance = 516.53 µ = 60 – 0.842(22.73) = 40.86

J31 Praktis SPM 5 Kertas 1 1. (a) min = 0 dan sisihan piawai = 1 Mean = 0 and standard deviation = 1 (b) P(0 < Z < 1) = 0.3413 2. (a) 1.8 = 38 −32 σ , σ = 3.33 (b) P(X < k) = P(Z < k −32 3.33 ) = 0.4213 k −32 3.33 = −0.199 k = 31.34 3. (a) P(X = 0) + P(X > 2) = 1 – h – k (b) P(X = 0)= 4 C0 p0 (1 − p)4 = 4 27 1 − p = 0.037 p = 0.963 Kertas 2 1. (a) n = 7, p = 0.22 P(X = 4) = 7 C4 (0.22)4 (0.78)3 = 0.0389 (b) (i) Min 0.46 kg dan sisihan piawai m kg. Mean 0.46 kg and the standard deviation of m kg. P(X > 0.52) = 0.18 PZ > 0.52 – 0.46 m 2 = 0.18 0.52 – 0.46 m = 0.916 m = 0.52 – 0.46 0.916 = 0.066 kg (ii) P(0.4 < X < 0.5) = P1 0.4 – 0.46 0.066 < Z < 0.5 – 0.46 0.066 2 = 0.5461 0.5461 × 375 = 205 bungkus 2. (a) (i) P(X > 1.2) = PZ > 1.2 – 0.75 0.3 2 = 0.0668 (ii) P(X < m) = PZ < m – 0.75 0.3 2 = 0.15 m – 0.75 0.3 = −1.036 m = 0.4392 kg (b) (i) p = 0.4 P(X = 1) = 8 × P(X = 0) n C1 (0.4)(0.6)n − 1 = 8 × n C0 (0.6)n n × 0.4 0.6 = 8 n = 8 × 0.6 0.4 = 12 (ii) Var(X) = npq = 12(0.4)(0.6) Sisihan piawai = 1.7 Standard deviation 3. (a) (i) Min/Mean = 90 P(X > 140) = 0.12 P(X > 140) = PZ > 140 – 90 σ 2 = 0.12 140 – 90 σ = 1.175 σ = 42.55 (ii) P(50 < X < 100) = P 50 – 90 42.55 < Z < 100 – 90 42.55 2 = 0.4193 0.4193 × 45 = 18.87 = 19 (b) P(X < y)= PZ < y – 90 42.55 2 = 0.23 y – 90 42.55 = −0.739 y = RM58.56 4. (a) RM24 boleh bermain 6 set permainan RM24 can play 6 sets of game. n = 6 P(menjatuhkan semua tin dalam tiga lontaran berturut-turut) P(topple all the tins in three consecutive throws) = (0.7)3 = 0.343 P(X ≥ 1) = 1 − P(X = 0) = 1 − 6 C0 (0.343)0 (0.657)6 = 0.9196 Peratus menang ialah 91.96%. Awang patut bermain. Percentage to win is 91.96%. Awang should play (b) E(x) = np 3 = n(0.3430) n = 8.75 Dia perlu bermain 9 set. He needs to play 9 sets. Sudut KBAT (a) min/mean = 16.4 P[X > 17] = 0.16 P3Z > 17 – 16.4 σ 4 = 0.16 17 – 16.4 σ = 0.994 σ = 0.6036 m (b) P[15.2 < X < 17] = 13.5 + 34 + 34 = 81.5% (c) P[X < 15.2] = 0.025 P3Z < 15.2 – 16.4 0.6036 4 = 0.01563 0.01563 × 1000 = 15.63 pokok

J32 BAB Fungsi Trigonometri 6 Trigonometric Functions 1. (a) (d) fi 121ff y x II 0 α = 180° − 121° α = 412° − 360° = 59° = 52° (b) (e) 2 – 3 π = –2 × 180° 3 = –120° α = 360° − 316° = 44° α = 60° (c) (f) 9 8 p = 9 × 180° 8 = 202.5° α = 180° − 160° = 20° α = 9 8 p – p = π 8 2. (a) θ = 360° + (180° – 42°) = 498° (b) θ = –3π – π 6  = – 17 6 π kerana arah ikut jam. because follows clockwise direction. 3. (a) sin 249° = –sin 69° (b) sek (–200°) = 1 kos(–200°) 1 –kos 20° = –sek 20° (c) kot/ cot 219° = 1 tan 219° = kot/ cot 39° (d) kos/ cos (−152°) = kos/ cos 28° (e) –kosek(–32°) = – 1 sin(–32°) = – 1 –sin 32° = kosek/ cosec 32° (f) –tan 3 4 π2 = – tan  3 × 180° 4 2 = –tan 135° = −(−tan 45°) = tan 45° (g) –sin 8 – 3 π2 = –sin(–480°) = –(–sin 60°) = sin 60° (h) –kot( 7 4 π) = – 1 tan 315° = – 1 –tan 45° = kot/ cot 45° fi 316ff y x 0 I fi 520ff y x II 0 fi –412ff y x 0 IV –120fi 60fi y x 0 III fi 9ff 8 y x 0 y x 69fi 0 y x 20fi 0 y x 39fi 0 y x 28fi 152fi 0 y x 0 –32fi y x 45fi 135fi 0 y x 60fi 0 y x 0 45fi

J33 4. (a) y x –5 –12 13 fi 0 (i) sin θ = –5 13 (ii) sek/ sec θ = 1 kos/ cos θ = 1 –12 13 = 13 – 12 (iii) kot/ cot θ = 1 tan θ = 12 5 (b) sek/sec θ = – 2 = 1 cos θ kos/cos θ = –1 2 (i) kot/cot θ = –1 (ii) kosek/cosec θ = 1 sin θ = 2 (iii)sin θ = 1 2 (c) kot/cot θ = – 3 tan θ = –1 3 (i) kos/cos θ = 3 2 (ii) kosek/cosec θ = 1 sin θ = 1 1 – 2 = –2 (iii)sek/sec θ = 1 kos/cos θ = 1 3 2 = 2 3 5. (a) (i) kos/cos θ = –1 1 + p2 (ii) kosek/cosec θ = 1 sin θ = 1 + p2 p (iii)sek/sec θ = 1 kos/cos θ = – 1 + p2 (b) (i) kot/cot θ = p2 – 4 2 (ii) sek/sec θ = 1 kos θ = –p p2 – 4 (iii) kos/cos θ = – p2 – 4 2 (c) (i) sin θ = 1 2 + 2p + p2 (ii) kosek/cosec θ = 1 sin θ = 2 + 2p + p2 (iii) kos/cos θ = 1 + p 2 + 2p + p2 6. (a) kos/cos θ = sin 13° 42’ = kos 76° 18’ θ = 76° 18’ (b) kot/cot θ = tan 82° 15’ = kot 7° 45’ ∴θ = 7° 45’ (c) kosek/ cosec θ = sek/ sec p° 1 sin θ = 1 kos/ cos p° sin θ = kos/ cos p° = sin 90° − p° θ = 90° − p (d) sin θ = kos/ cos 34° 54’ = sin 55° 6’ θ = 55° 6’ (e) tan (90° − θ) = kot/ cot 68° 34’ = tan 21° 26’ 90° − θ = 21° 26’ 90° − 21° 26’ = θ θ = 68° 34’ y x √2 –1 1 fi 0 y x √3 –1 2 0 fi x y –2 p fi –√p2 –4 0 y x –1 √1+p2 p fi 0 x y 1 (1+p) fi √2+2p+p2 0 y x 13fi42ff 76fi18ff 0 y x 82fi15ff 7fi45ff 0 y x pfi 90fi–pfi 0 y x 55fi6ff 34fi54ff 0 y x 21fi26ff 68fi34ff 0