Matematik TambahanAdditional Mathematics

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1

Question 1

Diberi fungsi kuadratik f(x) = x² − 8x + 13.A quadratic function is given by f(x) = x² − 8x + 13.

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

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• (x − 4)² [1 mark: half of −8 squared, 16, used]1
• (x − 4)² − 3 [1 mark]1

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[2 marks] (b)Nyatakan nilai minimum bagi f(x) dan nilai x apabila nilai minimum itu berlaku.State the minimum value of f(x) and the value of x at which it occurs.

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• Minimum value = −3 [1 mark]1
• x = 4 [1 mark]1

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[3 marks] (c)Cari julat nilai x bagi f(x) ≥ 13.Find the range of values of x for which f(x) ≥ 13.

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• x² − 8x ≥ 0, so x(x − 8) ≥ 0 [1 mark]1
• Critical values x = 0 and x = 8, with a sketch or sign test [1 mark]1
• x ≤ 0 or x ≥ 8 [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.2.1

Textbook p. 46

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Question 2

Diberi fungsi f(x) = 2x + 5 dan g(x) = x² − 1.Given the functions f(x) = 2x + 5 and g(x) = x² − 1.

[2 marks] (a)Cari f⁻¹(x).Find f⁻¹(x).

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• y = 2x + 5, so x = (y − 5)/2 [1 mark]1
• f⁻¹(x) = (x − 5)/2 [1 mark]1

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[2 marks] (b)Cari fg(x).Find fg(x).

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• f(x² − 1) = 2(x² − 1) + 5 [1 mark]1
• fg(x) = 2x² + 3 [1 mark]1

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[2 marks] (c)Cari nilai-nilai x dengan keadaan gf(x) = 24.Find the values of x such that gf(x) = 24.

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• gf(x) = (2x + 5)² − 1 = 24, so (2x + 5)² = 25, 2x + 5 = ±5 [1 mark]1
• x = 0 or x = −5 [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 1, standard 1.2.4

Textbook p. 16

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Question 3

Rajah 1 menunjukkan segi tiga OAB. Titik P terletak pada OA dengan keadaan OP = ¹/₃OA, titik Q terletak pada OB dengan keadaan OQ = ¹/₃OB dan M ialah titik tengah AB. Diberi OA = 6a dan OB = 8b.Diagram 1 shows triangle OAB. Point P lies on OA such that OP = ¹/₃OA, point Q lies on OB such that OQ = ¹/₃OB, and M is the midpoint of AB. Given OA = 6a and OB = 8b.

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

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• AB = −6a + 8b (or 8b − 6a) [1 mark]1

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[2 marks] (b)Ungkapkan OM dalam sebutan a dan b.Express OM in terms of a and b.

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• OM = OA + ½AB = 6a + ½(8b − 6a) [1 mark]1
• OM = 3a + 4b [1 mark]1

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[1 mark] (c)Ungkapkan PQ dalam sebutan a dan b.Express PQ in terms of a and b.

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• PQ = −2a + (8/3)b [1 mark]1

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[2 marks] (d)Tunjukkan bahawa PQ selari dengan AB. Seterusnya, nyatakan nisbah PQ : AB.Show that PQ is parallel to AB. Hence state the ratio PQ : AB.

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• PQ = (1/3)(8b − 6a) = (1/3)AB, so PQ is parallel to AB [1 mark]1
• PQ : AB = 1 : 3 [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

Rajah 2 menunjukkan segi tiga ABC pada satah Cartes dengan A(−3, 0), B(1, 2) dan C(3, −2).Diagram 2 shows triangle ABC on a Cartesian plane with A(−3, 0), B(1, 2) and C(3, −2).

[2 marks] (a)Tunjukkan bahawa ∠ABC = 90°.Show that ∠ABC = 90°.

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• Gradient of AB = (2 − 0)/(1 + 3) = 1/2 and gradient of BC = (−2 − 2)/(3 − 1) = −2 [1 mark]1
• (1/2) × (−2) = −1, so AB is perpendicular to BC and ∠ABC = 90° [1 mark]1

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[1 mark] (b)Cari koordinat titik tengah M bagi AC.Find the coordinates of the midpoint M of AC.

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• M = ((−3 + 3)/2, (0 − 2)/2) = (0, −1) [1 mark]1

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[2 marks] (c)Hitung luas segi tiga ABC.Calculate the area of triangle ABC.

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• Area = ½ |(−3)(2) + (1)(−2) + (3)(0) − [(0)(1) + (2)(3) + (−2)(−3)]| [1 mark]1
• = ½ |−8 − 12| = 10 square units [1 mark]1

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[2 marks] (d)Cari persamaan garis lurus yang melalui B dan selari dengan AC.Find the equation of the straight line that passes through B and is parallel to AC.

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• Gradient of AC = (−2 − 0)/(3 + 3) = −1/3 [1 mark]1
• y − 2 = −(1/3)(x − 1), i.e. x + 3y = 7 [1 mark]1

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For the teacher: Applying, medium difficulty, Chapter 7, standard 7.2.1

Textbook p. 187

5

Question 5

Hana membuka satu akaun simpanan. Pada bulan pertama dia menyimpan RM100. Setiap bulan berikutnya dia menyimpan 8% lebih banyak daripada bulan sebelumnya.Hana opens a savings account. In the first month she saves RM100. In every month after that she saves 8% more than in the month before.

[1 mark] (a)Tunjukkan bahawa simpanan bulanan itu membentuk suatu janjang geometri dan nyatakan nisbah sepunya.Show that the monthly savings form a geometric progression and state the common ratio.

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• Each amount is 108% of the previous one, so the ratio is constant: r = 1.08 [1 mark]1

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[2 marks] (b)Hitung simpanan Hana pada bulan ke-10, betul kepada sen terdekat.Calculate Hana's saving in the 10th month, correct to the nearest sen.

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• T₁₀ = 100(1.08)⁹ [1 mark]1
• RM199.90 [1 mark]1

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[2 marks] (c)Hitung jumlah simpanan Hana dalam 12 bulan pertama, betul kepada sen terdekat.Calculate Hana's total savings in the first 12 months, correct to the nearest sen.

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• S₁₂ = 100[(1.08)¹² − 1]/(1.08 − 1) [1 mark]1
• RM1 897.71 [1 mark]1

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[3 marks] (d)Pada bulan yang ke berapakah simpanan bulanan Hana melebihi RM500 buat kali pertama?In which month does Hana's monthly saving first exceed RM500?

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• 100(1.08)ⁿ⁻¹ > 500, so (1.08)ⁿ⁻¹ > 5 [1 mark]1
• n − 1 > log 5 / log 1.08 = 20.91 [1 mark]1
• n = 22, the 22nd month [1 mark] (check: T₂₁ = RM466.10, T₂₂ = RM503.38)1

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

Textbook p. 151

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Question 6

Titik A ialah (−4, 0) dan titik B ialah (2, 0). Titik P(x, y) bergerak dengan keadaan jaraknya dari A sentiasa dua kali jaraknya dari B.Point A is (−4, 0) and point B is (2, 0). Point P(x, y) moves such that its distance from A is always twice its distance from B.

[3 marks] (a)Cari persamaan lokus bagi P.Find the equation of the locus of P.

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• PA = 2PB, so PA² = 4PB² [1 mark]1
• (x + 4)² + y² = 4[(x − 2)² + y²] [1 mark]1
• x² + y² − 8x = 0 [1 mark]1

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[1 mark] (b)Tunjukkan bahawa titik (4, −4) terletak pada lokus itu.Show that the point (4, −4) lies on the locus.

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• 16 + 16 − 32 = 0, so (4, −4) satisfies the equation [1 mark]1

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[2 marks] (c)Garis lurus y = x bersilang dengan lokus itu di dua titik. Cari koordinat kedua-dua titik itu.The straight line y = x meets the locus at two points. Find the coordinates of the two points.

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• x² + x² − 8x = 0, so 2x(x − 4) = 0 [1 mark]1
• (0, 0) and (4, 4) [1 mark]1

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[2 marks] (d)Hitung luas segi tiga yang bucu-bucunya ialah A dan dua titik dalam (c).Calculate the area of the triangle whose vertices are A and the two points in (c).

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• Area = ½ |(−4)(0) + 0(4) + 4(0) − [0(0) + 0(4) + 4(−4)]| [1 mark]1
• = ½ × 16 = 8 square units [1 mark]1

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For the teacher: Analysing, high difficulty, Chapter 7, standard 7.3.3

Textbook p. 195

7

Question 7

Jawab semua bahagian soalan ini.Answer all parts of this question.

[2 marks] (a)Permudahkan (2ⁿ⁺² − 2ⁿ) ÷ 2ⁿ⁻¹ × ¹/₃.Simplify (2ⁿ⁺² − 2ⁿ) ÷ 2ⁿ⁻¹ × ¹/₃.

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• 2ⁿ(4 − 1) ÷ 2ⁿ⁻¹ × 1/3 = 3 × 2 × 1/3 [1 mark]1
• = 2 [1 mark]1

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[2 marks] (b)Nyatakan 12/(3 + √3) dalam bentuk a + b√3 dengan merasionalkan penyebut.Express 12/(3 + √3) in the form a + b√3 by rationalising the denominator.

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• 12(3 − √3)/[(3 + √3)(3 − √3)] = 12(3 − √3)/6 [1 mark]1
• = 6 − 2√3 [1 mark]1

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[2 marks] (c)Diberi log₃ 5 = k. Ungkapkan log₃ 45 dalam sebutan k.Given log₃ 5 = k. Express log₃ 45 in terms of k.

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• log₃ 45 = log₃ 9 + log₃ 5 [1 mark]1
• = 2 + k [1 mark]1

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[2 marks] (d)Selesaikan persamaan log₂ (x + 3) = 1 + log₂ x.Solve the equation log₂ (x + 3) = 1 + log₂ x.

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

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For the teacher: Applying, low difficulty, Chapter 4, standard 4.3.3

Textbook p. 116

8Choose 3

Question 8

Sebuah dewan sekolah mempunyai 20 baris tempat duduk. Baris pertama mempunyai 24 tempat duduk dan setiap baris seterusnya mempunyai 3 tempat duduk lebih daripada baris di hadapannya.A school hall has 20 rows of seats. The first row has 24 seats and each row after that has 3 more seats than the row in front of it.

[2 marks] (a)Cari bilangan tempat duduk dalam baris ke-20.Find the number of seats in the 20th row.

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• a = 24, d = 3; T₂₀ = 24 + 19(3) [1 mark]1
• 81 seats [1 mark]1

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[2 marks] (b)Hitung jumlah bilangan tempat duduk dalam dewan itu.Calculate the total number of seats in the hall.

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• S₂₀ = 20/2 (24 + 81) [1 mark]1
• 1 050 seats [1 mark]1

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[3 marks] (c)Untuk satu majlis, sekurang-kurangnya 600 orang hendak duduk bermula dari baris hadapan. Cari bilangan baris yang paling sedikit yang perlu digunakan.For an event, at least 600 people are to be seated starting from the front row. Find the least number of rows that must be used.

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• n/2 [48 + 3(n − 1)] ≥ 600, so n² + 15n − 400 ≥ 0 [1 mark]1
• Positive root n = 13.86 (or test S₁₃ = 546, S₁₄ = 609) [1 mark]1
• n = 14 rows [1 mark]1

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[3 marks] (d)Hitung jumlah tempat duduk dalam lima baris yang paling belakang.Calculate the total number of seats in the last five rows.

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• S₁₅ = 15/2 (24 + 66) = 675 [1 mark]1
• Last five rows = S₂₀ − S₁₅ [1 mark]1
• 1 050 − 675 = 375 seats [1 mark]1

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

Textbook p. 138

9Choose 3

Question 9

Diberi fungsi kuadratik f(x) = −2x² + 12x − 10.A quadratic function is given by f(x) = −2x² + 12x − 10.

[2 marks] (a)Ungkapkan f(x) dalam bentuk a(x + p)² + q.Express f(x) in the form a(x + p)² + q.

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• −2(x² − 6x) − 10 = −2[(x − 3)² − 9] − 10 [1 mark]1
• −2(x − 3)² + 8 [1 mark]1

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[2 marks] (b)Nyatakan nilai maksimum bagi f(x) dan persamaan paksi simetri.State the maximum value of f(x) and the equation of the axis of symmetry.

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• Maximum value = 8 [1 mark]1
• x = 3 [1 mark]1

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[3 marks] (c)Cari julat nilai x bagi f(x) > 0.Find the range of values of x for which f(x) > 0.

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• −2(x − 1)(x − 5) > 0, so (x − 1)(x − 5) < 0 [1 mark]1
• Critical values 1 and 5 with sketch of a graph or sign test [1 mark]1
• 1 < x < 5 [1 mark]1

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[3 marks] (d)Persamaan f(x) = k tidak mempunyai punca nyata. Cari julat nilai k.The equation f(x) = k has no real roots. Find the range of values of k.

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• −2(x − 3)² + 8 = k, so (x − 3)² = (8 − k)/2 [1 mark]1
• No real roots when (8 − k)/2 < 0 [1 mark] (or b² − 4ac < 0 for 2x² − 12x + 10 + k = 0)1
• k > 8 [1 mark]1

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For the teacher: Analysing, medium difficulty, Chapter 2, standard 2.3.3

Textbook p. 55

10Choose 3

Question 10

Perimeter sebuah papan tulis segi empat tepat ialah 28 cm dalam model skala yang dibuat oleh Kamal dan panjang pepenjurunya ialah 10 cm. Katakan panjang dan lebarnya ialah x cm dan y cm, dengan x > y.In a scale model made by Kamal, the perimeter of a rectangular whiteboard is 28 cm and the length of its diagonal is 10 cm. Let its length and width be x cm and y cm, where x > y.

[2 marks] (a)Tulis dua persamaan yang menghubungkan x dan y.Write two equations that connect x and y.

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• x + y = 14 [1 mark]1
• x² + y² = 100 [1 mark]1

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[4 marks] (b)Selesaikan persamaan serentak dalam (a) untuk mencari nilai x dan nilai y.Solve the simultaneous equations in (a) to find the values of x and y.

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• y = 14 − x substituted: x² + (14 − x)² = 100 [1 mark]1
• x² − 14x + 48 = 0 [1 mark]1
• (x − 6)(x − 8) = 0, so x = 6 or x = 8 [1 mark]1
• x = 8, y = 6 (since x > y) [1 mark]1

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[1 mark] (c)Hitung luas papan tulis itu.Calculate the area of the whiteboard.

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• 8 × 6 = 48 cm² [1 mark]1

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[3 marks] (d)Selesaikan persamaan serentak y = x − 1 dan y = x² − 3x + 2. Cari koordinat titik persilangannya.Solve the simultaneous equations y = x − 1 and y = x² − 3x + 2. Find the coordinates of the points of intersection.

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• x − 1 = x² − 3x + 2, so x² − 4x + 3 = 0 [1 mark]1
• (x − 1)(x − 3) = 0, so x = 1 or x = 3 [1 mark]1
• (1, 0) and (3, 2) [1 mark]1

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For the teacher: Applying, medium difficulty, Chapter 3, standard 3.2.2

Textbook p. 84

11Choose 3

Question 11

Pemboleh ubah x dan y dihubungkan oleh persamaan y = px² + qx, dengan p dan q ialah pemalar. Rajah 3 menunjukkan garis lurus yang diperoleh apabila y/x diplot melawan x.The variables x and y are related by the equation y = px² + qx, where p and q are constants. Diagram 3 shows the straight line obtained when y/x is plotted against x.

[2 marks] (a)Tunjukkan bahawa persamaan itu boleh ditukar kepada bentuk linear Y = mX + c. Nyatakan Y, X, m dan c dalam sebutan x, y, p dan q.Show that the equation can be reduced to the linear form Y = mX + c. State Y, X, m and c in terms of x, y, p and q.

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• Divide by x: y/x = px + q [1 mark]1
• Y = y/x, X = x, m = p, c = q [1 mark]1

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[3 marks] (b)Berdasarkan Rajah 3, cari nilai p dan nilai q.Based on Diagram 3, find the value of p and the value of q.

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• Gradient = (17 − 5)/(5 − 1) [1 mark]1
• p = 3 [1 mark]1
• Substitute (1, 5): 5 = 3(1) + q, so q = 2 [1 mark]1

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[2 marks] (c)Cari nilai y apabila x = 6.Find the value of y when x = 6.

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• y/x = 3(6) + 2 = 20 [1 mark]1
• y = 120 [1 mark]1

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[3 marks] (d)Cari nilai x dengan keadaan y = 14x.Find the value of x for which y = 14x.

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• y = 14x means y/x = 14 [1 mark]1
• 3x + 2 = 14 [1 mark]1
• x = 4 [1 mark]1

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For the teacher: Analysing, high difficulty, Chapter 6, standard 6.1.3

Textbook p. 157

12Choose 2

Question 12

Jadual 1 menunjukkan harga empat jenis makanan di sebuah kantin sekolah pada tahun 2022 dan 2024. Tahun asas ialah 2022.Table 1 shows the prices of four items at a school canteen in 2022 and 2024. The base year is 2022.

[2 marks] (a)Cari nilai x.Find the value of x.

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• x/1.50 × 100 = 120 [1 mark]1
• x = 1.80 [1 mark]1

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[2 marks] (b)Cari nilai y.Find the value of y.

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• 1.20/0.80 × 100 [1 mark]1
• y = 150 [1 mark]1

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[3 marks] (c)Hitung indeks gubahan bagi harga makanan itu pada tahun 2024 berasaskan tahun 2022.Calculate the composite index of the prices of these items in 2024 based on 2022.

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• Σ IW = 130(4) + 120(3) + 150(2) + 120(1) = 1 300 [1 mark]1
• ΣW = 10 [1 mark]1
• Composite index = 1 300/10 = 130 [1 mark]1

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[1 mark] (d)Sebuah set makanan berharga RM8.00 pada tahun 2022. Anggarkan harganya pada tahun 2024.A food set cost RM8.00 in 2022. Estimate its price in 2024.

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• 8.00 × 130/100 = RM10.40 [1 mark]1

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[2 marks] (e)Dari 2024 ke 2026, indeks gubahan bagi harga itu meningkat sebanyak 10%. Cari indeks gubahan pada tahun 2026 berasaskan tahun 2022.From 2024 to 2026 the composite index of these prices rises by 10%. Find the composite index in 2026 based on 2022.

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• Index 2026 / 2024 = 110 [1 mark]1
• 130 × 110/100 = 143 [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 10, standard 10.2.1

Textbook p. 280

13Choose 2

Question 13

Jadual 2 menunjukkan indeks harga bagi tiga jenis alat basikal pada tahun 2024 berasaskan tahun 2021 dan pemberatnya. Pemberat bagi kunci ialah k.Table 2 shows the price indices of three cycling items in 2024 based on 2021, and their weightages. The weightage for the lock is k.

[2 marks] (a)Harga sebuah topi keledar pada tahun 2024 ialah RM70. Cari harganya pada tahun 2021.A helmet costs RM70 in 2024. Find its price in 2021.

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• 70/P₀ × 100 = 140 [1 mark]1
• P₀ = RM50 [1 mark]1

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[3 marks] (b)Indeks gubahan bagi ketiga-tiga alat itu pada tahun 2024 ialah 124. Cari nilai k.The composite index of the three items in 2024 is 124. Find the value of k.

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• [140(5) + 120k + 100(3)]/(5 + k + 3) = 124 [1 mark]1
• 1 000 + 120k = 992 + 124k [1 mark]1
• k = 2 [1 mark]1

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[2 marks] (c)Indeks gubahan pada tahun 2026 berasaskan tahun 2024 ialah 110. Hitung indeks gubahan pada tahun 2026 berasaskan tahun 2021.The composite index in 2026 based on 2024 is 110. Calculate the composite index in 2026 based on 2021.

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• 124 × 110/100 [1 mark]1
• 136.4 [1 mark]1

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[3 marks] (d)Sebuah set alat basikal berharga RM250 pada tahun 2021. Anggarkan harganya pada tahun 2026.A set of cycling items cost RM250 in 2021. Estimate its price in 2026.

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• 250 × 136.4/100 [1 mark for method]1
• RM341 [1 mark]1
• Uses composite index based on 2021 (not 110 or 124) [1 mark]1

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For the teacher: Analysing, medium difficulty, Chapter 10, standard 10.2.2

Textbook p. 284

14Choose 2

Question 14

Rajah 4 menunjukkan tiga tiang bendera A, B dan C di atas sebuah padang. AB = 9 km, BC = 7 km dan ∠ABC = 110°.Diagram 4 shows three flagpoles A, B and C on a field. AB = 9 km, BC = 7 km and ∠ABC = 110°.

[3 marks] (a)Hitung panjang AC, betul kepada dua tempat perpuluhan.Calculate the length of AC, correct to two decimal places.

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• AC² = 9² + 7² − 2(9)(7) cos 110° [1 mark]1
• AC² = 173.1 [1 mark]1
• AC = 13.16 km [1 mark]1

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[3 marks] (b)Hitung ∠BAC, betul kepada satu tempat perpuluhan.Calculate ∠BAC, correct to one decimal place.

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• sin A / 7 = sin 110° / 13.16 [1 mark]1
• sin A = 0.4999 [1 mark]1
• ∠BAC = 30.0° [1 mark]1

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[2 marks] (c)Hitung luas segi tiga ABC.Calculate the area of triangle ABC.

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• ½ × 9 × 7 × sin 110° [1 mark]1
• 29.60 km² [1 mark]1

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[2 marks] (d)Hitung jarak terdekat dari B ke garis lurus AC.Calculate the shortest distance from B to the straight line AC.

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• ½ × AC × h = 29.60, so h = 2 × 29.60 / 13.16 [1 mark]1
• h = 4.50 km [1 mark]1

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For the teacher: Applying, low difficulty, Chapter 9, standard 9.2.1

Textbook p. 252

15Choose 2

Question 15

Rajah 5 menunjukkan sebuah sisi empat ABCD. Diberi AB = 10 cm, BC = 8 cm, ∠ABC = 60°, AD = 6.5 cm dan CD = 7.5 cm.Diagram 5 shows a quadrilateral ABCD. Given AB = 10 cm, BC = 8 cm, ∠ABC = 60°, AD = 6.5 cm and CD = 7.5 cm.

[2 marks] (a)Hitung panjang AC, betul kepada dua tempat perpuluhan.Calculate the length of AC, correct to two decimal places.

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• AC² = 10² + 8² − 2(10)(8) cos 60° = 84 [1 mark]1
• AC = 9.17 cm [1 mark]1

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[3 marks] (b)Hitung ∠ADC, betul kepada dua tempat perpuluhan.Calculate ∠ADC, correct to two decimal places.

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• cos D = (6.5² + 7.5² − 84) / (2 × 6.5 × 7.5) [1 mark]1
• cos D = 0.1487 [1 mark]1
• ∠ADC = 81.45° [1 mark]1

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[3 marks] (c)Hitung luas sisi empat ABCD.Calculate the area of quadrilateral ABCD.

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• Area of ABC = ½ × 10 × 8 × sin 60° = 34.64 cm² [1 mark]1
• Area of ACD = ½ × 6.5 × 7.5 × sin 81.45° = 24.10 cm² [1 mark]1
• Total area = 58.74 cm² [1 mark]1

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[2 marks] (d)Hitung jarak terdekat dari D ke AC.Calculate the shortest distance from D to AC.

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• ½ × AC × h = 24.10, so h = 2 × 24.10 / 9.165 [1 mark]1
• h = 5.26 cm [1 mark]1

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For the teacher: Analysing, high difficulty, Chapter 9, standard 9.3.3

Textbook p. 261

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