Matematik TambahanAdditional Mathematics

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1

Question 1

Rajah 1 menunjukkan keratan rentas sebuah gerbang berbentuk parabola di pintu masuk sebuah sekolah. Ketinggian gerbang itu, h meter, pada jarak mengufuk x meter dari kaki kirinya diberikan oleh h(x) = 2x − x²/4, dengan 0 ≤ x ≤ 8.Diagram 1 shows the cross-section of a parabolic arch at the entrance of a school. The height of the arch, h metres, at a horizontal distance of x metres from its left foot is given by h(x) = 2x − x²/4, for 0 ≤ x ≤ 8.

[3 marks] (a)Ungkapkan h(x) dalam bentuk −¼(x − p)² + q, dengan p dan q ialah pemalar.Express h(x) in the form −¼(x − p)² + q, where p and q are constants.

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• h(x) = −¼(x² − 8x) [1 mark: factor out −¼]1
• = −¼[(x − 4)² − 16] [1 mark: complete the square]1
• = −¼(x − 4)² + 4 [1 mark]1

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[1 mark] (b)Nyatakan ketinggian maksimum gerbang itu.State the maximum height of the arch.

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• 4 m1

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[3 marks] (c)Sebuah sepanduk dipasang di antara dua titik pada gerbang itu yang berada 3 m di atas tanah. Hitung panjang sepanduk itu.A banner is hung between the two points on the arch which are 3 m above the ground. Calculate the length of the banner.

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• 2x − x²/4 = 3, so x² − 8x + 12 = 0 [1 mark: form the equation]1
• (x − 2)(x − 6) = 0, so x = 2 and x = 6 [1 mark]1
• Length = 6 − 2 = 4 m [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.3.6

Textbook p. 62

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Question 2

Diberi fungsi f : x → 2x − 7 dan g : x → x² − 3. Fungsi h ditakrifkan oleh h(x) = |2x − 6|.The functions f : x → 2x − 7 and g : x → x² − 3 are given. The function h is defined by h(x) = |2x − 6|.

[2 marks] (a)Cari nilai-nilai x supaya h(x) = 10.Find the values of x such that h(x) = 10.

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• 2x − 6 = 10 or 2x − 6 = −10 [1 mark]1
• x = 8 or x = −2 [1 mark]1

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[2 marks] (b)Cari f⁻¹(x) dan seterusnya nilai f⁻¹(5).Find f⁻¹(x) and hence the value of f⁻¹(5).

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• f⁻¹(x) = (x + 7)/2 [1 mark]1
• f⁻¹(5) = 6 [1 mark]1

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[2 marks] (c)Cari fg(x). Seterusnya, selesaikan fg(x) = 5.Find fg(x). Hence, solve fg(x) = 5.

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• fg(x) = 2(x² − 3) − 7 = 2x² − 13 [1 mark]1
• 2x² − 13 = 5, so x = 3 or x = −3 [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 1, standard 1.3.2

Textbook p. 24

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Question 3

Rajah 2 menunjukkan segi tiga OAB. Titik P terletak pada OA dengan keadaan OP : PA = 1 : 3 dan M ialah titik tengah AB. Diberi OA = 4x dan OB = 6y, iaitu vektor-vektor x dan y tidak selari.Diagram 2 shows triangle OAB. Point P lies on OA such that OP : PA = 1 : 3 and M is the midpoint of AB. It is given that OA = 4x and OB = 6y, where the vectors x and y are not parallel.

[1 mark] (a)Ungkapkan →AB dalam sebutan x dan y.Express →AB in terms of x and y.

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• →AB = −4x + 6y (or 6y − 4x)1

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[3 marks] (b)Seterusnya, cari →OM dan →PM.Hence, find →OM and →PM.

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• →OM = 4x + ½(6y − 4x) = 2x + 3y [2 marks]2
• →PM = (2x + 3y) − x = x + 3y [1 mark]1

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[2 marks] (c)Diberi x = 2i + j dan y = −i + 2j. Hitung |→PM|.Given x = 2i + j and y = −i + 2j. Calculate |→PM|.

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• →PM = (2 − 3)i + (1 + 6)j = −i + 7j [1 mark]1
• |→PM| = √(1 + 49) = √50 = 7.07 [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 8, standard 8.2.2

Textbook p. 224

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Question 4

Persamaan kuadratik 3x² − 7x + 1 = 0 dan x² − 6x + 2k = 0 diberikan, dengan k ialah pemalar.The quadratic equations 3x² − 7x + 1 = 0 and x² − 6x + 2k = 0 are given, where k is a constant.

[3 marks] (a)Selesaikan 3x² − 7x + 1 = 0. Beri jawapan betul kepada tiga tempat perpuluhan.Solve 3x² − 7x + 1 = 0. Give the answers correct to three decimal places.

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• x = [7 ± √(49 − 12)] / 6 [1 mark: correct substitution]1
• x = 2.180 [1 mark]1
• x = 0.153 [1 mark]1

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[2 marks] (b)Cari julat nilai k supaya x² − 6x + 2k = 0 mempunyai dua punca nyata yang berbeza.Find the range of values of k for which x² − 6x + 2k = 0 has two distinct real roots.

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• b² − 4ac > 0: 36 − 8k > 0 [1 mark]1
• k < 4.5 (or k < 9/2) [1 mark]1

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[2 marks] (c)Selesaikan ketaksamaan x² − 2x − 15 ≥ 0.Solve the inequality x² − 2x − 15 ≥ 0.

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• (x − 5)(x + 3) ≥ 0, critical values −3 and 5 [1 mark]1
• x ≤ −3 or x ≥ 5 [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.1.1

Textbook p. 38

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Question 5

Zulkifli berlatih larian untuk Hari Sukan sekolah. Jarak larian hariannya membentuk suatu janjang aritmetik: 2.0 km pada hari pertama, 2.4 km pada hari kedua, 2.8 km pada hari ketiga dan seterusnya.Zulkifli trains for his school Sports Day. His daily running distances form an arithmetic progression: 2.0 km on the first day, 2.4 km on the second day, 2.8 km on the third day, and so on.

[2 marks] (a)Cari jarak larian pada hari ke-15.Find the distance he runs on the 15th day.

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• T15 = 2.0 + (15 − 1)(0.4) [1 mark]1
• = 7.6 km [1 mark]1

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[2 marks] (b)Hitung jumlah jarak larian dalam 20 hari pertama.Calculate the total distance he runs in the first 20 days.

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• S20 = (20/2)[2(2.0) + 19(0.4)] [1 mark]1
• = 116 km [1 mark]1

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[4 marks] (c)Zulkifli berhenti berlatih apabila jumlah jarak larian melebihi 150 km. Cari bilangan hari yang paling sedikit sebelum dia berhenti.Zulkifli stops training when his total running distance exceeds 150 km. Find the least number of days after which he stops.

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• Sn = (n/2)[4 + 0.4(n − 1)] > 150, so n² + 9n − 750 > 0 [2 marks]2
• n > 23.26 [1 mark]1
• n = 24 (S23 = 147.2 km, S24 = 158.4 km) [1 mark]1

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For the teacher: Applying, medium difficulty, Chapter 5, standard 5.1.4

Textbook p. 138

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Question 6

Titik P(−3, 4) dan titik Q(9, −8) ialah dua tiang pada sebuah padang. Titik M terletak pada garis lurus PQ dengan keadaan PM : MQ = 1 : 3.Point P(−3, 4) and point Q(9, −8) are two posts on a field. Point M lies on the straight line PQ such that PM : MQ = 1 : 3.

[2 marks] (a)Cari koordinat M.Find the coordinates of M.

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• M = ((3(−3) + 1(9))/4, (3(4) + 1(−8))/4) [1 mark]1
• M(0, 1) [1 mark]1

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[3 marks] (b)Cari persamaan pembahagi dua sama serenjang bagi PQ.Find the equation of the perpendicular bisector of PQ.

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• Gradient of PQ = (−8 − 4)/(9 + 3) = −1, so perpendicular gradient = 1 [1 mark]1
• Midpoint of PQ = (3, −2) [1 mark]1
• y + 2 = 1(x − 3), so y = x − 5 [1 mark]1

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[3 marks] (c)Titik T(x, y) bergerak dengan keadaan TP : TQ = 1 : 2. Cari persamaan lokus T.Point T(x, y) moves such that TP : TQ = 1 : 2. Find the equation of the locus of T.

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• 2√[(x + 3)² + (y − 4)²] = √[(x − 9)² + (y + 8)²], square: 4[(x + 3)² + (y − 4)²] = (x − 9)² + (y + 8)² [1 mark]1
• Expand: 4x² + 24x + 36 + 4y² − 32y + 64 = x² − 18x + 81 + y² + 16y + 64 [1 mark]1
• x² + y² + 14x − 16y − 15 = 0 [1 mark]1

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For the teacher: Applying, medium difficulty, Chapter 7, standard 7.4.1

Textbook p. 201

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Question 7

Jawab semua bahagian soalan ini. Tunjukkan langkah-langkah penting dalam kerja mengira anda.Answer all parts of this question. Show the important steps in your working.

[2 marks] (a)Selesaikan 4^x+1 = 8^x−1.Solve 4^x+1 = 8^x−1.

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• 2^(2x+2) = 2^(3x−3) [1 mark]1
• 2x + 2 = 3x − 3, so x = 5 [1 mark]1

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[2 marks] (b)Diberi log₃ 2 = p, ungkapkan log₃ √12 dalam sebutan p.Given log₃ 2 = p, express log₃ √12 in terms of p.

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• log₃ √12 = ½ log₃ (4 × 3) = ½(2 log₃ 2 + log₃ 3) [1 mark]1
• = ½(2p + 1) = p + ½ [1 mark]1

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[2 marks] (c)Selesaikan log₂ (x + 10) − log₂ x = 2.Solve log₂ (x + 10) − log₂ x = 2.

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• log₂ [(x + 10)/x] = 2, so (x + 10)/x = 4 [1 mark]1
• x = 10/3 [1 mark]1

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[2 marks] (d)Jisim suatu ubat dalam badan pesakit susut mengikut formula M = 240(0.8)^t mg, dengan t ialah masa dalam jam selepas dos diambil. Cari bilangan jam penuh yang paling kecil supaya jisim ubat itu kurang daripada 60 mg.The mass of a medicine in a patient's body decreases according to the formula M = 240(0.8)^t mg, where t is the time in hours after the dose is taken. Find the smallest whole number of hours for which the mass of the medicine is less than 60 mg.

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• 240(0.8)^t < 60, so 0.8^t < 0.25, t > log 0.25 / log 0.8 = 6.21 [1 mark]1
• t = 7 hours [1 mark]1

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For the teacher: Analysing, high difficulty, Chapter 4, standard 4.4.1

Textbook p. 122

8Choose 3

Question 8

Program kitar semula sebuah sekolah mengumpul 8 kg kertas pada minggu pertama. Jisim kertas yang dikumpul pada setiap minggu ialah 1.5 kali jisim pada minggu sebelumnya.A school recycling programme collects 8 kg of paper in the first week. The mass of paper collected in each week is 1.5 times the mass collected in the previous week.

[1 mark] (a)Nyatakan nisbah sepunya bagi janjang geometri ini.State the common ratio of this geometric progression.

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• r = 1.51

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[2 marks] (b)Cari jisim kertas yang dikumpul pada minggu keenam.Find the mass of paper collected in the sixth week.

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• T6 = 8(1.5)^5 [1 mark]1
• = 60.75 kg [1 mark]1

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[3 marks] (c)Hitung jumlah jisim kertas yang dikumpul dalam lapan minggu pertama, betul kepada dua tempat perpuluhan.Calculate the total mass of paper collected in the first eight weeks, correct to two decimal places.

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• S8 = 8(1.5⁸ − 1)/(1.5 − 1) [1 mark: correct formula and values]1
• = 16(25.6289 − 1) [1 mark]1
• = 394.06 kg [1 mark]1

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[4 marks] (d)Cari bilangan minggu yang paling sedikit supaya jumlah kertas yang dikumpul melebihi 1 000 kg.Find the least number of weeks for the total mass of paper collected to exceed 1 000 kg.

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• 8(1.5ⁿ − 1)/0.5 > 1000, so 1.5ⁿ > 63.5 [2 marks]2
• n > log 63.5 / log 1.5 = 10.2 [1 mark]1
• n = 11 weeks (S10 = 906.64 kg, S11 = 1 367.96 kg) [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 5, standard 5.2.5

Textbook p. 151

9Choose 3

Question 9

Sebuah kantin sekolah menjual kek cawan. Keuntungan mingguan, P(x) ringgit, bergantung pada harga jualan x ringgit bagi satu kotak kek cawan, dengan P(x) = −2x² + 28x − 66.A school canteen sells cupcakes. The weekly profit, RM P(x), depends on the selling price of RM x for one box of cupcakes, where P(x) = −2x² + 28x − 66.

[2 marks] (a)Ungkapkan P(x) dalam bentuk a(x − h)² + k.Express P(x) in the form a(x − h)² + k.

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• P(x) = −2(x² − 14x) − 66 = −2[(x − 7)² − 49] − 66 [1 mark]1
• = −2(x − 7)² + 32 [1 mark]1

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[1 mark] (b)Nyatakan keuntungan maksimum dan harga jualan yang memberikan keuntungan itu.State the maximum profit and the selling price which gives that profit.

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• Maximum profit RM32 when x = RM71

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[3 marks] (c)Lakar graf y = P(x). Tandakan pintasan-x, pintasan-y dan titik maksimum.Sketch the graph of y = P(x). Mark the x-intercepts, the y-intercept and the maximum point.

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• Shape: an inverted parabola (a < 0) [1 mark]1
• Cuts the x-axis at (3, 0) and (11, 0) and the y-axis at (0, −66) [1 mark]1
• Maximum point (7, 32) [1 mark]1

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[2 marks] (d)Cari julat harga jualan x supaya keuntungan mingguan sekurang-kurangnya RM20.Find the range of the selling price x for which the weekly profit is at least RM20.

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• −2(x − 7)² + 32 ≥ 20, so (x − 7)² ≤ 6 [1 mark]1
• 7 − √6 ≤ x ≤ 7 + √6, i.e. 4.55 ≤ x ≤ 9.45 [1 mark]1

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[2 marks] (e)Pemilik kantin mendakwa keuntungan RM35 boleh dicapai. Gunakan pembezalayan untuk menunjukkan bahawa dakwaan itu tidak benar.The owner claims that a profit of RM35 can be achieved. Use the discriminant to show that the claim is not true.

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• −2x² + 28x − 66 = 35, so 2x² − 28x + 101 = 0 [1 mark]1
• b² − 4ac = 784 − 808 = −24 < 0, no real roots, so the claim is false [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.3.6

Textbook p. 62

10Choose 3

Question 10

Jawab kedua-dua bahagian soalan ini.Answer both parts of this question.

[5 marks] (a)Di koperasi sekolah, harga bagi sebatang pen, sebuah pembaris dan sebuah buku nota ialah RMx, RMy dan RMz masing-masing. Tiga orang murid membayar seperti berikut:
Aiman: 3 pen, 2 pembaris dan 1 buku nota, RM19
Bavani: 1 pen, 4 pembaris dan 2 buku nota, RM18
Chong: 2 pen, 1 pembaris dan 3 buku nota, RM19
Bentukkan tiga persamaan linear dan cari nilai x, y dan z.At the school co-operative shop, the prices of one pen, one ruler and one notebook are RM x, RM y and RM z respectively. Three pupils pay as follows:
Aiman: 3 pens, 2 rulers and 1 notebook, RM19
Bavani: 1 pen, 4 rulers and 2 notebooks, RM18
Chong: 2 pens, 1 ruler and 3 notebooks, RM19
Form three linear equations and find the values of x, y and z.

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• 3x + 2y + z = 19, x + 4y + 2z = 18, 2x + y + 3z = 19 [1 mark: any two correct; 2 marks: all three]2
• Eliminate one variable, e.g. put z = 19 − 3x − 2y into the other two equations: −5x = −20 and −7x − 5y = −38 [1 mark]1
• Solve the two equations: x = 4, then y = 2 [1 mark]1
• z = 19 − 12 − 4 = 3, so x = 4, y = 2, z = 3 [1 mark]1

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[5 marks] (b)Sebuah poster segi empat tepat mempunyai panjang yang melebihi lebarnya sebanyak 7 cm dan panjang pepenjurunya ialah 13 cm. Cari panjang dan lebar poster itu.A rectangular poster has a length that exceeds its width by 7 cm and the length of its diagonal is 13 cm. Find the length and the width of the poster.

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• l = w + 7 and l² + w² = 169 [1 mark]1
• w² + (w + 7)² = 169, so 2w² + 14w − 120 = 0, i.e. w² + 7w − 60 = 0 [2 marks]2
• (w − 5)(w + 12) = 0, so w = 5 (reject w = −12) [1 mark]1
• Width 5 cm, length 12 cm [1 mark]1

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For the teacher: Applying, medium difficulty, Chapter 3, standard 3.1.3

Textbook p. 77

11Choose 3

Question 11

Pemboleh ubah x dan y dihubungkan oleh persamaan y = px^q, dengan p dan q ialah pemalar. Jadual 1 menunjukkan nilai-nilai x dan y daripada satu eksperimen.

x	1	2	4	5	8
y	3.0	8.5	24.0	33.5	67.9

Rajah 3 menunjukkan garis lurus penyuaian terbaik bagi log₁₀ y melawan log₁₀ x. Garis itu melalui titik P(0.30, 0.93) dan Q(0.90, 1.83).The variables x and y are related by the equation y = px^q, where p and q are constants. Table 1 (above) shows values of x and y from an experiment.
Diagram 3 shows the line of best fit for log₁₀ y against log₁₀ x. The line passes through the points P(0.30, 0.93) and Q(0.90, 1.83).

[3 marks] (a)Tukarkan y = px^q kepada bentuk linear Y = mX + c. Nyatakan Y, X, m dan c.Reduce y = px^q to the linear form Y = mX + c. State Y, X, m and c.

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• log₁₀ y = q log₁₀ x + log₁₀ p [1 mark]1
• Y = log₁₀ y, X = log₁₀ x [1 mark]1
• m = q, c = log₁₀ p [1 mark]1

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[4 marks] (b)Daripada Rajah 3, hitung nilai q dan nilai p, betul kepada satu tempat perpuluhan bagi p.From Diagram 3, calculate the value of q and the value of p, correct to one decimal place for p.

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• q = gradient = (1.83 − 0.93)/(0.90 − 0.30) = 1.5 [2 marks]2
• Y-intercept: 0.93 = 1.5(0.30) + c, so c = 0.48 [1 mark]1
• p = 10^0.48 = 3.0 [1 mark]1

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[3 marks] (c)Anggarkan nilai y apabila x = 6, betul kepada nombor bulat terdekat.Estimate the value of y when x = 6, correct to the nearest whole number.

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• log₁₀ y = 0.48 + 1.5 log₁₀ 6 = 0.48 + 1.5(0.778) = 1.647 [2 marks]2
• y = 44 [1 mark]1

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For the teacher: Analysing, high difficulty, Chapter 6, standard 6.2.1

Textbook p. 164

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