Practice paper, not an official KPM paper

Percubaan Sijil Pelajaran MalaysiaTrial Sijil Pelajaran Malaysia

Matematik TambahanAdditional Mathematics

Tingkatan 5, 2 jam / 2 hours, 100 marks

Nama / Name

Kelas / Class

Tarikh / Date

Markah / Marks: / 100

Instructions

Kertas soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C.This question paper consists of three sections: Section A, Section B and Section C.
Jawab semua soalan.Answer all questions.
Tunjukkan langkah-langkah penting dalam kerja mengira anda. Markah diberikan untuk kaedah.Show your working. Marks are given for method.
Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.The diagrams in this paper are not drawn to scale unless stated.
Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram.You may use a non-programmable scientific calculator.

Answer schemes are in English (DLP).

Practice shows answers as you go. Exam times you and hides answers until you submit.

Bahagian ASection A

40 marks

Jawab semua soalan.Answer all questions.

Question 1

Sebuah pemegang pintu kaca berbentuk sektor bulatan berjejari 9 cm. Gunakan π = 3.142.A glass door handle is shaped like a sector of a circle of radius 9 cm. Use π = 3.142.

[1 mark] (a)Tukarkan 108° kepada radian, dalam sebutan π.Convert 108° to radians, in terms of π.

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108 × π180 = 3π5 rad [1]1

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[2 marks] (b)Sektor itu mencangkum sudut 0.7 rad di pusat. Hitung panjang lengkok dan luas sektor itu.The sector subtends an angle of 0.7 rad at the centre. Calculate the arc length and the area of the sector.

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Arc length = 9 × 0.7 = 6.3 cm [1]1
Area = 12 × 9² × 0.7 = 28.35 cm² [1]1

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For the teacher: Applying, low difficulty, Chapter 1, standard 1.2.1

Textbook p. 6

Question 2

Diberi y = (5 − 2x)⁴.Given y = (5 − 2x)⁴.

[2 marks] (a)Cari dydx.Find dydx.

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Chain rule: 4(5 − 2x)³ × (−2) [1]1
= −8(5 − 2x)³ [1]1

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[1 mark] (b)Cari nilai dydx apabila x = 1.Find the value of dydx when x = 1.

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−8(3)³ = −216 [1]1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.2.3

Textbook p. 43

Question 3

Fungsi kecerunan suatu lengkung ialah dydx = 4x³ − 6x + 1. Lengkung itu melalui titik (1, 4).The gradient function of a curve is dydx = 4x³ − 6x + 1. The curve passes through the point (1, 4).

[3 marks] (a)Cari persamaan lengkung itu.Find the equation of the curve.

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y = x⁴ − 3x² + x + c [1]1
4 = 1 − 3 + 1 + c, c = 5 [1]1
y = x⁴ − 3x² + x + 5 [1]1

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[1 mark] (b)Cari nilai y apabila x = 2.Find the value of y when x = 2.

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16 − 12 + 2 + 5 = 11 [1]1

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For the teacher: Applying, medium difficulty, Chapter 3, standard 3.2.4

Textbook p. 90

Question 4

Selesaikan setiap soalan berikut.Answer each of the following.

[3 marks] (a)Nilaikan ∫₁⁴ (3x² − 2x) dx.Evaluate ∫₁⁴ (3x² − 2x) dx.

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[x³ − x²] from 1 to 4 [1]1
(64 − 16) − (1 − 1) [1]1
= 48 [1]1

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[2 marks] (b)Diberi ∫₁⁴ g(x) dx = 7. Cari nilai ∫₄¹ [g(x) + 2] dx.Given ∫₁⁴ g(x) dx = 7. Find the value of ∫₄¹ [g(x) + 2] dx.

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∫₁⁴ [g(x) + 2] dx = 7 + 2(3) = 13 [1]1
Reversing the limits changes the sign: −13 [1]1

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For the teacher: Applying, low difficulty, Chapter 3, standard 3.3.1

Textbook p. 94

Question 5

Selesaikan setiap soalan berikut.Answer each of the following.

[1 mark] (a)Hitung nilai ⁷P₃.Calculate the value of ⁷P₃.

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7 × 6 × 5 = 210 [1]1

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[2 marks] (b)Berapakah bilangan susunan berbeza semua huruf dalam perkataan TOMATO?How many different arrangements can be made from all the letters of the word TOMATO?

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6! / (2! × 2!) [1]1
= 180 [1]1

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[2 marks] (c)Seramai 3 orang murid dipilih daripada 4 murid lelaki dan 5 murid perempuan. Cari bilangan cara pemilihan jika sekurang-kurangnya seorang murid lelaki dipilih.3 pupils are chosen from 4 boys and 5 girls. Find the number of ways if at least one boy is chosen.

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Total − no boys: ⁹C₃ − ⁵C₃ [1]1
= 84 − 10 = 74 [1]1

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For the teacher: Applying, low difficulty, Chapter 4, standard 4.1.2

Textbook p. 127

Question 6

Kebarangkalian sebiji benih cili bercambah ialah 0.8. Enam biji benih ditanam dan X ialah bilangan benih yang bercambah.The probability that a chilli seed germinates is 0.8. Six seeds are planted and X is the number of seeds that germinate.

[1 mark] (a)Nyatakan taburan bagi X.State the distribution of X.

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X ~ B(6, 0.8) [1]1

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[3 marks] (b)Cari kebarangkalian bahawa sekurang-kurangnya 5 biji benih bercambah.Find the probability that at least 5 seeds germinate.

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P(X = 5) = ⁶C₅(0.8)⁵(0.2) = 0.3932 [1]1
P(X = 6) = (0.8)⁶ = 0.2621 [1]1
P(X ≥ 5) = 0.6554 [1]1

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For the teacher: Applying, medium difficulty, Chapter 5, standard 5.2.2

Textbook p. 155

Question 7

Selesaikan persamaan sin 2x = cos x bagi 0° ≤ x ≤ 180°.Solve the equation sin 2x = cos x for 0° ≤ x ≤ 180°.

[4 marks] (a)Tunjukkan langkah kerja anda.Show your working.

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2 sin x cos x = cos x [1]1
cos x (2 sin x − 1) = 0 [1]1
cos x = 0, x = 90° [1]1
sin x = ½, x = 30°, 150° [1]1

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For the teacher: Applying, medium difficulty, Chapter 6, standard 6.5.2

Textbook p. 218

Question 8

Sebuah kedai roti membuat x kelompok roti dan y kelompok kek setiap hari. Satu kelompok roti menggunakan 1 kg tepung dan 2 jam masa membakar. Satu kelompok kek menggunakan 2 kg tepung dan 1 jam masa membakar. Setiap hari, tepung yang ada ialah 16 kg dan masa membakar ialah 20 jam. Sekurang-kurangnya 2 kelompok setiap jenis dibuat. Keuntungan ialah RM30 bagi satu kelompok roti dan RM40 bagi satu kelompok kek.A bakery makes x batches of bread and y batches of cake each day. One batch of bread uses 1 kg of flour and 2 hours of baking time. One batch of cake uses 2 kg of flour and 1 hour of baking time. Each day there are 16 kg of flour and 20 hours of baking time available. At least 2 batches of each kind are made. The profit is RM30 per batch of bread and RM40 per batch of cake.

[2 marks] (a)Tulis empat ketaksamaan linear yang memenuhi semua syarat itu.Write four linear inequalities that satisfy all the conditions.

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x + 2y ≤ 16 and 2x + y ≤ 20 [1]1
x ≥ 2 and y ≥ 2 [1]1

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[2 marks] (b)Cari keuntungan harian maksimum, dalam RM.Find the maximum daily profit, in RM.

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Vertices of the region: (2, 2), (2, 7), (8, 4), (9, 2); (8, 4) from solving x + 2y = 16 and 2x + y = 20 [1]1
Profit 30x + 40y: 140, 340, 400, 350; maximum RM400 at (8, 4) [1]1

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For the teacher: Analysing, high difficulty, Chapter 7, standard 7.1.1

Textbook p. 238

Question 9

Satu zarah bergerak di sepanjang suatu garis lurus melalui titik tetap O. Pecutannya ialah a = (2t − 8) m s⁻², dengan keadaan t saat selepas melalui O. Halaju awalnya ialah 12 m s⁻¹.A particle moves along a straight line and passes through the fixed point O. Its acceleration is a = (2t − 8) m s⁻², t seconds after passing O. Its initial velocity is 12 m s⁻¹.

[2 marks] (a)Cari masa, dalam saat, zarah itu berada dalam keadaan rehat seketika.Find the times, in seconds, at which the particle is instantaneously at rest.

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v = t² − 8t + 12 [1]1
(t − 2)(t − 6) = 0, t = 2 and t = 6 [1]1

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[2 marks] (b)Cari jumlah jarak, dalam m, yang dilalui zarah itu dalam 4 saat pertama.Find the total distance, in m, travelled by the particle in the first 4 seconds.

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s = t³3 − 4t² + 12t; s(2) = 323, s(4) = 163 [1]1
Distance = 323 + (323 − 163) = 16 m [1]1

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For the teacher: Analysing, high difficulty, Chapter 8, standard 8.2.3

Textbook p. 265

Question 10

Diberi sin θ = −9/41 dengan 180° < θ < 270°. Tanpa menggunakan kalkulator, cariGiven sin θ = −9/41 with 180° < θ < 270°. Without using a calculator, find

[2 marks] (a)nilai kos θ,the value of cos θ,

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Adjacent = √(41² − 9²) = 40 [1]1
cos θ = −40/41 (θ in the third quadrant) [1]1

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[2 marks] (b)nilai tan θ.the value of tan θ.

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tan θ = sin θ / cos θ [1]1
= (−9/41) ÷ (−40/41) = 9/40 [1]1

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For the teacher: Applying, low difficulty, Chapter 6, standard 6.2.2

Textbook p. 197

Bahagian BSection B

40 marks

Jawab semua soalan.Answer all questions.

Question 11

Rajah 1 menunjukkan sebuah bulatan berpusat O dan berjejari 9 cm. Perentas AB mencangkum sudut 2π/3 rad di O. Kawasan berlorek ialah tembereng minor. [Guna π = 3.142 dan √3 = 1.732 jika perlu]Diagram 1 shows a circle with centre O and radius 9 cm. The chord AB subtends an angle of 2π/3 rad at O. The shaded region is the minor segment. [Use π = 3.142 and √3 = 1.732 where needed]

Circle with centre O and radius 9 cm, chord AB and the minor segment shaded — Rajah 1Diagram 1

[1 mark] (a)Tukarkan 2π/3 rad kepada darjah.Convert 2π/3 rad to degrees.

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(2π/3) × 180/π = 120° [1]1

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[1 mark] (b)Hitung panjang lengkok minor AB.Calculate the length of the minor arc AB.

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s = rθ = 9 × 2π/3 = 6π = 18.85 cm [1]1

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[2 marks] (c)Hitung luas sektor OAB.Calculate the area of the sector OAB.

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A = 12 r²θ = 12 × 81 × 2π3 [1]1
= 27π = 84.83 cm² [1]1

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[3 marks] (d)Hitung luas kawasan berlorek.Calculate the area of the shaded region.

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Area of triangle OAB = 12 × 9² × sin 120° [1]1
= 35.07 cm² [1]1
Segment = 84.83 − 35.07 = 49.76 cm² [1]1

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[3 marks] (e)Hitung perimeter kawasan berlorek.Calculate the perimeter of the shaded region.

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Chord AB = 2 × 9 × sin 60° [1]1
= 9√3 = 15.59 cm [1]1
Perimeter = 18.85 + 15.59 = 34.44 cm [1]1

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For the teacher: Applying, low difficulty, Chapter 1, standard 1.4.1

Textbook p. 20

Question 12

Huruf-huruf dalam perkataan PENGAWAS disusun dalam satu baris.The letters of the word PENGAWAS are arranged in a row.

[2 marks] (a)Cari bilangan susunan berbeza semua huruf itu.Find the number of different arrangements of all the letters.

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8! / 2! [1]1
= 20 160 [1]1

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[2 marks] (b)Cari bilangan susunan yang bermula dengan P dan berakhir dengan S.Find the number of arrangements that begin with P and end with S.

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Remaining 6 letters include two A: 6! / 2! [1]1
= 360 [1]1

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[3 marks] (c)Cari bilangan susunan jika semua huruf vokal (E, A, A) berada bersebelahan.Find the number of arrangements in which all the vowels (E, A, A) are next to each other.

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Vowels as one block: 3!/2! = 3 ways inside the block [1]1
Block and 5 consonants: 6! = 720 [1]1
3 × 720 = 2160 [1]1

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[3 marks] (d)Satu pasukan 5 orang dipilih daripada 6 murid lelaki dan 5 murid perempuan. Cari bilangan cara pemilihan jika bilangan murid lelaki melebihi bilangan murid perempuan.A team of 5 is chosen from 6 boys and 5 girls. Find the number of ways if the boys outnumber the girls.

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3 boys + 2 girls: ⁶C₃ × ⁵C₂ = 200 [1]1
4 boys + 1 girl: ⁶C₄ × ⁵C₁ = 75; 5 boys: ⁶C₅ = 6 [1]1
Total = 281 [1]1

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For the teacher: Applying, low difficulty, Chapter 4, standard 4.1.3

Textbook p. 129

Question 13

Kebarangkalian sebuah bas sekolah tiba mengikut masa pada suatu hari ialah 0.7. Dalam 10 hari yang dipilih secara rawak, X ialah bilangan hari bas itu tiba mengikut masa. Ketibaan pada hari yang berbeza adalah tidak bersandar.The probability that a school bus arrives on time on a day is 0.7. In 10 days chosen at random, X is the number of days the bus arrives on time. Arrivals on different days are independent.

[2 marks] (a)Cari min dan varians bagi X.Find the mean and the variance of X.

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Mean = np = 10 × 0.7 = 7 [1]1
Variance = npq = 10 × 0.7 × 0.3 = 2.1 [1]1

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[2 marks] (b)Cari P(X = 8).Find P(X = 8).

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¹⁰C₈(0.7)⁸(0.3)² [1]1
= 0.2335 [1]1

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[3 marks] (c)Cari P(X ≥ 9).Find P(X ≥ 9).

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P(X = 9) = 10(0.7)⁹(0.3) = 0.1211 [1]1
P(X = 10) = (0.7)¹⁰ = 0.0282 [1]1
P(X ≥ 9) = 0.1493 [1]1

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[3 marks] (d)Cari kebarangkalian bahawa bilangan hari bas itu tiba mengikut masa terletak dalam lingkungan satu sisihan piawai daripada min.Find the probability that the number of days the bus arrives on time lies within one standard deviation of the mean.

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σ = √2.1 = 1.449, so the interval is 5.55 < X < 8.45 [1]1
X = 6, 7, 8 [1]1
P = 0.2001 + 0.2668 + 0.2335 = 0.7004 [1]1

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For the teacher: Applying, medium difficulty, Chapter 5, standard 5.2.4

Textbook p. 162

Question 14

Sebuah kereta api mainan bergerak di sepanjang landasan lurus. Sesarannya daripada titik tetap O ialah s = (6t² − t³) m, dengan keadaan t saat selepas ia melalui O.A toy train moves along a straight track. Its displacement from the fixed point O is s = (6t² − t³) m, t seconds after it passes O.

[2 marks] (a)Cari halajunya apabila t = 1.Find its velocity when t = 1.

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v = ds/dt = 12t − 3t² [1]1
v(1) = 12 − 3 = 9 m s⁻¹ [1]1

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[2 marks] (b)Cari pecutannya apabila t = 3.Find its acceleration when t = 3.

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a = dv/dt = 12 − 6t [1]1
a(3) = −6 m s⁻² [1]1

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[3 marks] (c)Cari halaju maksimum kereta api itu.Find the maximum velocity of the train.

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a = 0, 12 − 6t = 0 [1]1
t = 2 [1]1
v = 24 − 12 = 12 m s⁻¹ [1]1

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[3 marks] (d)Cari halajunya apabila ia kembali ke O dan nyatakan arah gerakannya.Find its velocity when it returns to O and state its direction of motion.

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s = 0: t²(6 − t) = 0, t = 6 [1]1
v(6) = 72 − 108 = −36 m s⁻¹ [1]1
Negative velocity: it moves in the negative direction (back towards the starting side of O) [1]1

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

Textbook p. 263

Bahagian CSection C

20 marks

Jawab semua soalan.Answer all questions.

Question 15

Rajah 2 menunjukkan keratan rentas sebuah pendakap logam. Ia dibatasi oleh lengkung y = 6x − x² dan garis lurus y = 2x yang bersilang di asalan O dan di titik A. Semua ukuran dalam cm.Diagram 2 shows the cross-section of a metal bracket. It is bounded by the curve y = 6x − x² and the straight line y = 2x, which meet at the origin O and at the point A. All measurements are in cm.

Curve y = 6x minus x squared and line y = 2x with the region between them shaded — Rajah 2Diagram 2

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

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6x − x² = 2x, x(4 − x) = 0 [1]1
A = (4, 8) [1]1

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[3 marks] (b)Cari luas kawasan berlorek.Find the area of the shaded region.

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Area = ∫₀⁴ (4x − x²) dx [1]1
[2x² − x³3] from 0 to 4 [1]1
32 − 643 = 323 cm² [1]1

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[5 marks] (c)Kawasan berlorek diputarkan melalui 360° pada paksi-x. Cari isi padu janaan, dalam sebutan π.The shaded region is revolved through 360° about the x-axis. Find the volume generated, in terms of π.

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V = π∫₀⁴ [(6x − x²)² − (2x)²] dx [1]1
= π∫₀⁴ (32x² − 12x³ + x⁴) dx [1 for expansion]1
π[32x³3 − 3x⁴ + x⁵5] from 0 to 4 [1 for integration]1
π(20483 − 768 + 10245) [1]1
= 1792π15 cm³ [1]1

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For the teacher: Analysing, medium difficulty, Chapter 3, standard 3.3.5

Textbook p. 107

Question 16

Kedalaman air, h meter, di sebuah pelabuhan A ialah h = 6 + 3 sin (30t)°, dengan keadaan t ialah bilangan jam selepas jam 0600 dan 0 ≤ t ≤ 12.The depth of water, h metres, at harbour A is h = 6 + 3 sin (30t)°, where t is the number of hours after 0600 hours and 0 ≤ t ≤ 12.

[3 marks] (a)Nyatakan kedalaman maksimum, kedalaman minimum dan tempoh bagi model itu.State the maximum depth, the minimum depth and the period of the model.

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Maximum depth = 6 + 3 = 9 m [1]1
Minimum depth = 6 − 3 = 3 m [1]1
Period = 360° / 30° = 12 hours [1]1

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[3 marks] (b)Lakarkan graf h melawan t bagi 0 ≤ t ≤ 12.Sketch the graph of h against t for 0 ≤ t ≤ 12.

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Sine shape for one full cycle [1]1
Starts at h = 6, maximum 9 at t = 3, minimum 3 at t = 9 [1]1
Crosses h = 6 again at t = 6 and ends at (12, 6) [1]1

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[4 marks] (c)Di sebuah pelabuhan B, kedalaman air, d meter, ialah d = 4 + 6 sin (15t)° kos (15t)°. (i) Tunjukkan bahawa d = 4 + 3 sin (30t)°. (ii) Cari nilai t apabila d = 5.5 bagi 0 ≤ t ≤ 12. (iii) Pelabuhan manakah mempunyai kedalaman maksimum yang lebih besar?At harbour B, the depth of water, d metres, is d = 4 + 6 sin (15t)° cos (15t)°. (i) Show that d = 4 + 3 sin (30t)°. (ii) Find the values of t when d = 5.5 for 0 ≤ t ≤ 12. (iii) Which harbour has the greater maximum depth?

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(i) 6 sin A cos A = 3(2 sin A cos A) = 3 sin 2A, with A = (15t)°, so d = 4 + 3 sin (30t)° [1]1
(ii) 3 sin (30t)° = 1.5, sin (30t)° = 0.5 [1]1
30t = 30° or 150°, so t = 1 or t = 5 [1]1
(iii) Harbour A: 9 m, harbour B: 4 + 3 = 7 m, so harbour A [1]1

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For the teacher: Evaluating, high difficulty, Chapter 6, standard 6.5.2

Textbook p. 218

Test specification (JSU)

Chapter	Remembering	Understanding	Applying	Analysing	Evaluating	Creating	Items	marks
1. Circular Measure	–	–	2 (13)	–	–	–	2	13
2. Differentiation	–	–	1 (3)	–	–	–	1	3
3. Integration	–	–	2 (9)	1 (10)	–	–	3	19
4. Permutation and Combination	–	–	2 (15)	–	–	–	2	15
5. Probability Distribution	–	–	2 (14)	–	–	–	2	14
6. Trigonometric Functions	–	–	2 (8)	–	1 (10)	–	3	18
7. Linear Programming	–	–	–	1 (4)	–	–	1	4
8. Kinematics of Linear Motion	–	–	1 (10)	1 (4)	–	–	2	14
Total	0 (0)	0 (0)	12 (72)	3 (18)	1 (10)	0 (0)	16	100

Each cell shows items (marks).