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 trek larian mempunyai bahagian melengkung berbentuk lengkok bulatan berjejari 24 m yang mencangkum sudut 105° di pusat. Ungkapkan 105° dalam radian dalam sebutan π, dan seterusnya cari panjang lengkok itu dalam sebutan π.A running track has a curved part in the shape of an arc of a circle of radius 24 m, subtending an angle of 105° at the centre. Express 105° in radians in terms of π, and hence find the length of the arc in terms of π.

[1 mark] (a)Tukarkan 105° kepada radian.Convert 105° to radians.

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105 × π/180 = 7π/12 rad [1]1

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[2 marks] (b)Cari panjang lengkok, dalam m.Find the arc length, in m.

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s = rθ = 24 × 7π/12 [1]1
= 14π m [1]1

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

Textbook p. 5

Question 2

Diberi y = (3x² − 5)⁴, cari nilai dy/dx apabila x = 1.Given y = (3x² − 5)⁴, find the value of dy/dx when x = 1.

[3 marks] Cari nilai dy/dx apabila x = 1.Find the value of dy/dx when x = 1.

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−192

Chain rule: dy/dx = 4(3x² − 5)³ × 6x [1]1
= 24x(3x² − 5)³ [1]1
x = 1: 24(−2)³ = −192 [1]1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.2.2

Textbook p. 41

Question 3

Sebiji bola getah ditiup supaya jejarinya, r cm, bertambah dengan kadar 0.2 cm s⁻¹. Isi padu bola ialah V = ⁴⁄₃πr³. Cari kadar perubahan isi padu apabila r = 6, betul kepada 2 tempat perpuluhan.A rubber ball is inflated so that its radius, r cm, increases at a rate of 0.2 cm s⁻¹. The volume of the ball is V = 4/3 πr³. Find the rate of change of the volume when r = 6, correct to 2 decimal places.

[4 marks] Cari dV/dt apabila r = 6.Find dV/dt when r = 6.

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90.48 cm³ s⁻¹

dV/dr = 4πr² [1]1
dV/dt = dV/dr × dr/dt = 4πr² × 0.2 [1]1
r = 6: 4π(36)(0.2) = 28.8π [1]1
= 90.48 cm³ s⁻¹ [1]1

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For the teacher: Applying, medium difficulty, Chapter 2, standard 2.3.1

Textbook p. 49

Question 4

Fungsi kecerunan suatu lengkung ialah dy/dx = 3x² − 8x. Lengkung itu melalui titik (2, 1).The gradient function of a curve is dy/dx = 3x² − 8x. The curve passes through the point (2, 1).

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

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y = x³ − 4x² + 9

y = ∫(3x² − 8x) dx = x³ − 4x² + c [1]1
Substitute (2, 1): 1 = 8 − 16 + c [1]1
c = 9, so y = x³ − 4x² + 9 [1]1

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[2 marks] (b)Hitung nilai bagi ∫₀² (3x² − 8x) dx.Calculate the value of ∫₀² (3x² − 8x) dx.

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−8

[x³ − 4x²] from 0 to 2 [1]1
= (8 − 16) − 0 = −8 [1]1

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

Textbook p. 90

Question 5

Sebuah pasukan olahraga 5 orang hendak dipilih daripada 7 pelajar lelaki dan 5 pelajar perempuan. Cari bilangan cara memilih pasukan itu jika pasukan mesti mempunyai sekurang-kurangnya 2 pelajar perempuan.An athletics team of 5 is to be chosen from 7 boys and 5 girls. Find the number of ways of choosing the team if it must contain at least 2 girls.

[4 marks] Cari bilangan cara.Find the number of ways.

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596

No condition: 12C5 = 792 [1]1
No girl: 7C5 = 21 [1]1
Exactly 1 girl: 5C1 × 7C4 = 5 × 35 = 175 [1]1
At least 2 girls: 792 − 21 − 175 = 596 [1]1

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For the teacher: Applying, medium difficulty, Chapter 4, standard 4.2.2

Textbook p. 133

Question 6

Seorang pemain bola keranjang menjaringkan lontaran bebas dengan kebarangkalian 0.7. Dia membuat 6 lontaran bebas. Biar X ialah bilangan jaringan yang berjaya.A basketball player scores a free throw with probability 0.7. She takes 6 free throws. Let X be the number of successful throws.

[3 marks] (a)Cari P(X = 4), betul kepada 4 tempat perpuluhan.Find P(X = 4), correct to 4 decimal places.

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0.3241

X ~ B(6, 0.7) [1]1
P(X = 4) = 6C4 (0.7)⁴(0.3)² [1]1
= 0.3241 [1]1

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

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4.2 and 1.26

Mean = np = 6 × 0.7 = 4.2 [1]1
Variance = npq = 6 × 0.7 × 0.3 = 1.26 [1]1

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

Textbook p. 155

Question 7

Selesaikan persamaan 2 kos²x − kos x − 1 = 0 bagi 0° ≤ x ≤ 360°.Solve the equation 2 cos²x − cos x − 1 = 0 for 0° ≤ x ≤ 360°.

[4 marks] Selesaikan persamaan itu.Solve the equation.

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x = 0°, 120°, 240°, 360°

(2 cos x + 1)(cos x − 1) = 0 [1]1
cos x = −1/2 or cos x = 1 [1]1
cos x = −1/2: x = 120°, 240° [1]1
cos x = 1: x = 0°, 360° [1]1

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For the teacher: Applying, medium difficulty, Chapter 6, standard 6.6.1

Textbook p. 223

Question 8

Diberi tan A = ¾ dengan A ialah sudut tirus, dan kos B = −5/13 dengan B ialah sudut cakah (90° < B < 180°). Cari nilai sin (A + B).Given tan A = 3/4 where A is an acute angle, and cos B = −5/13 where B is an obtuse angle (90° < B < 180°). Find the value of sin (A + B).

[4 marks] Cari sin (A + B).Find sin (A + B).

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33/65

sin A = 3/5, cos A = 4/5 (3-4-5 triangle) [1]1
sin B = 12/13 (B in the second quadrant, sin positive) [1]1
sin (A + B) = sin A cos B + cos A sin B = (3/5)(−5/13) + (4/5)(12/13) [1]1
= 33/65 [1]1

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For the teacher: Analysing, high difficulty, Chapter 6, standard 6.5.1

Textbook p. 216

Question 9

Sebuah kelab sukan membeli x biji bola sepak dan y biji bola jaring. Jumlah bola yang dibeli tidak melebihi 40. Bilangan bola sepak tidak melebihi dua kali bilangan bola jaring. Harga sebiji bola sepak ialah RM30 dan sebiji bola jaring RM25, dan bajet kelab ialah RM1 050.A sports club buys x footballs and y netballs. The total number of balls bought does not exceed 40. The number of footballs is at most twice the number of netballs. A football costs RM30 and a netball costs RM25, and the club's budget is RM1 050.

[3 marks] (a)Tulis tiga ketaksamaan linear, selain x ≥ 0 dan y ≥ 0, yang memenuhi semua kekangan.Write three linear inequalities, other than x ≥ 0 and y ≥ 0, that satisfy all the constraints.

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x + y ≤ 40 [1]1
x ≤ 2y [1]1
30x + 25y ≤ 1 050, i.e. 6x + 5y ≤ 210 [1]1

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[1 mark] (b)Adakah (20, 15) memenuhi ketiga-tiga ketaksamaan itu? Tunjukkan pengiraan anda.Does (20, 15) satisfy all three inequalities? Show your working.

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Yes

x + y = 35 ≤ 40; x = 20 ≤ 30; 6(20) + 5(15) = 195 ≤ 210, so yes [1]1

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

Textbook p. 237

Question 10

Sebutir zarah bergerak di sepanjang suatu garis lurus. Sesarannya daripada titik tetap O ialah s = t³ − 5t² + 3t meter, t saat selepas bermula.A particle moves along a straight line. Its displacement from a fixed point O is s = t³ − 5t² + 3t metres, t seconds after it starts.

[2 marks] (a)Cari halaju zarah apabila t = 4.Find the velocity of the particle when t = 4.

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11 m s⁻¹

v = ds/dt = 3t² − 10t + 3 [1]1
t = 4: 48 − 40 + 3 = 11 m s⁻¹ [1]1

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[2 marks] (b)Cari pecutan zarah apabila t = 4.Find the acceleration of the particle when t = 4.

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14 m s⁻²

a = dv/dt = 6t − 10 [1]1
t = 4: 14 m s⁻² [1]1

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

Textbook p. 263

Bahagian BSection B

40 marks

Jawab semua soalan.Answer all questions.

Question 11

Lengkung y = 2x³ − 9x² + 12x − 3 mempunyai dua titik pegun.The curve y = 2x³ − 9x² + 12x − 3 has two stationary points.

[2 marks] (a)Cari dy/dx, dan seterusnya cari kecerunan lengkung itu apabila x = 3.Find dy/dx, and hence the gradient of the curve at x = 3.

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dy/dx = 6x² − 18x + 12 [1]1
x = 3: 54 − 54 + 12 = 12 [1]1

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[4 marks] (b)Cari koordinat kedua-dua titik pegun itu dan tentukan sama ada setiap satu ialah titik maksimum atau minimum.Find the coordinates of the two stationary points and determine whether each is a maximum or a minimum.

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6(x − 1)(x − 2) = 0, so x = 1 or x = 2 [1]1
(1, 2) and (2, 1) [1]1
d²y/dx² = 12x − 18 [1]1
x = 1: −6 < 0, maximum; x = 2: +6 > 0, minimum [1]1

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[2 marks] (c)Cari perubahan kecil dalam y apabila x bertambah daripada 3 kepada 3.02.Find the small change in y when x increases from 3 to 3.02.

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0.24

δy ≈ (dy/dx) δx = 12 × 0.02 [1]1
= 0.24 [1]1

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[2 marks] (d)Pada suatu ketika x = 3 dan x bertambah dengan kadar 0.5 unit s⁻¹. Cari kadar perubahan y pada ketika itu.At a certain instant x = 3 and x is increasing at 0.5 units s⁻¹. Find the rate of change of y at that instant.

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6 units s⁻¹

dy/dt = dy/dx × dx/dt = 12 × 0.5 [1]1
= 6 units s⁻¹ [1]1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.4.2

Textbook p. 53

Question 12

Lengkung y = 12x − 3x² bersilang dengan paksi-x di titik O dan titik P.The curve y = 12x − 3x² meets the x-axis at the point O and the point P.

[2 marks] (a)Cari koordinat-x bagi O dan P.Find the x-coordinates of O and P.

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0 and 4

12x − 3x² = 0, 3x(4 − x) = 0 [1]1
x = 0 (O) and x = 4 (P) [1]1

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[3 marks] (b)Cari luas rantau yang dibatasi oleh lengkung itu dan paksi-x.Find the area of the region bounded by the curve and the x-axis.

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32 units²

Area = ∫₀⁴ (12x − 3x²) dx [1]1
= [6x² − x³] from 0 to 4 [1]1
= 96 − 64 = 32 units² [1]1

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[3 marks] (c)Cari luas rantau yang dibatasi oleh lengkung itu, paksi-x, garis x = 1 dan garis x = 3.Find the area of the region bounded by the curve, the x-axis, the line x = 1 and the line x = 3.

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22 units²

∫₁³ (12x − 3x²) dx = [6x² − x³] from 1 to 3 [1]1
(54 − 27) − (6 − 1) [1]1
= 27 − 5 = 22 units² [1]1

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[2 marks] (d)Garis x = k membahagikan rantau dalam (b) kepada dua bahagian yang sama luas, dengan 0 < k < 4. Cari nilai k.The line x = k divides the region in (b) into two parts of equal area, where 0 < k < 4. Find the value of k.

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k = 2

Each part has area 16; the curve is symmetrical about the line through the midpoint of O and P [1]1
k = 2 (check: 6(2)² − 2³ = 16) [1]1

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For the teacher: Applying, low difficulty, Chapter 3, standard 3.3.3

Textbook p. 98

Question 13

Dalam suatu pertandingan memanah, Zarif mengena sasaran dengan kebarangkalian 0.7 bagi setiap anak panah. Dia melepaskan 8 anak panah. Biar X ialah bilangan anak panah yang mengena sasaran.In an archery competition, Zarif hits the target with probability 0.7 for each arrow. He shoots 8 arrows. Let X be the number of arrows that hit the target.

[2 marks] (a)Cari P(X = 6), betul kepada 4 tempat perpuluhan.Find P(X = 6), correct to 4 decimal places.

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0.2965

P(X = 6) = 8C6 (0.7)⁶(0.3)² [1]1
= 0.2965 [1]1

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[2 marks] (b)Cari P(X ≥ 7), betul kepada 4 tempat perpuluhan.Find P(X ≥ 7), correct to 4 decimal places.

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0.2553

P(X = 7) + P(X = 8) = 8C7 (0.7)⁷(0.3) + (0.7)⁸ [1]1
= 0.1977 + 0.0576 = 0.2553 [1]1

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[2 marks] (c)Cari min dan sisihan piawai bagi X, betul kepada 2 tempat perpuluhan.Find the mean and the standard deviation of X, correct to 2 decimal places.

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5.6 and 1.30

Mean = 8 × 0.7 = 5.6 [1]1
σ = √(8 × 0.7 × 0.3) = √1.68 = 1.30 [1]1

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[2 marks] (d)Masa yang diambil oleh peserta untuk menamatkan sebuah maraton tertabur secara normal dengan min 210 minit dan sisihan piawai 20 minit. Seorang peserta dipilih secara rawak. Cari kebarangkalian bahawa masanya kurang daripada 180 minit.The time taken by participants to finish a marathon is normally distributed with mean 210 minutes and standard deviation 20 minutes. A participant is chosen at random. Find the probability that the time is less than 180 minutes.

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0.0668

Z = (180 − 210)/20 = −1.5 [1]1
P(Z < −1.5) = P(Z > 1.5) = 0.0668 [1]1

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[2 marks] (e)10% peserta terpantas layak untuk acara elit. Cari masa maksimum, dalam minit, untuk layak. (Gunakan z = 1.28.)The fastest 10% of the participants qualify for the elite event. Find the maximum time, in minutes, to qualify. (Use z = 1.28.)

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184.4 minutes

P(Z < −1.28) = 0.10, so (x − 210)/20 = −1.28 [1]1
x = 210 − 25.6 = 184.4 minutes [1]1

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For the teacher: Applying, medium difficulty, Chapter 5, standard 5.3.3

Textbook p. 172

Question 14

Soalan ini berkaitan dengan fungsi trigonometri bagi x dan 2x.This question is about trigonometric functions of x and 2x.

[3 marks] (a)Buktikan bahawa (1 − kos 2x)/sin 2x = tan x.Prove that (1 − cos 2x)/sin 2x = tan x.

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1 − cos 2x = 2 sin²x [1]1
sin 2x = 2 sin x cos x [1]1
2 sin²x / (2 sin x cos x) = sin x / cos x = tan x [1]1

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[4 marks] (b)Selesaikan persamaan 4 sin x kos x = 3 kos x bagi 0° ≤ x ≤ 360°.Solve the equation 4 sin x cos x = 3 cos x for 0° ≤ x ≤ 360°.

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x = 48.59°, 90°, 131.41°, 270°

cos x (4 sin x − 3) = 0 [1]1
cos x = 0: x = 90°, 270° [1]1
sin x = 3/4: basic angle 48.59° [1]1
x = 48.59°, 131.41° [1]1

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[3 marks] (c)Lakarkan graf y = 2 sin 2x bagi 0° ≤ x ≤ 360°.Sketch the graph of y = 2 sin 2x for 0° ≤ x ≤ 360°.

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Sine shape starting at the origin and finishing at (360°, 0) [1]1
Maximum value 2 and minimum value −2 (amplitude 2) [1]1
Two complete cycles (period 180°): zeros at 0°, 90°, 180°, 270°, 360°; maxima at 45° and 225°; minima at 135° and 315° [1]1

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For the teacher: Analysing, medium difficulty, Chapter 6, standard 6.3.1

Textbook p. 205

Bahagian CSection C

20 marks

Jawab semua soalan.Answer all questions.

Question 15

Sebutir zarah bergerak di sepanjang suatu garis lurus dan melalui titik tetap O apabila t = 0. Halajanya, v m s⁻¹, diberi oleh v = t² − 7t + 12, dengan t ialah masa dalam saat selepas melalui O. [Anggap gerakan ke kanan sebagai arah positif.]A particle moves along a straight line and passes through a fixed point O when t = 0. Its velocity, v m s⁻¹, is given by v = t² − 7t + 12, where t is the time in seconds after passing O. [Take motion to the right as positive.]

[1 mark] (a)Nyatakan halaju awal zarah.State the initial velocity of the particle.

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12 m s⁻¹

12 m s⁻¹ [1]1

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[2 marks] (b)Cari nilai-nilai t apabila zarah berada dalam keadaan rehat seketika.Find the values of t when the particle is instantaneously at rest.

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t = 3 and t = 4

t² − 7t + 12 = 0, (t − 3)(t − 4) = 0 [1]1
t = 3 and t = 4 [1]1

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[2 marks] (c)Cari pecutan zarah apabila t = 5, dan nilai t apabila pecutan zarah ialah sifar.Find the acceleration of the particle when t = 5, and the value of t when the acceleration is zero.

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3 m s⁻²; t = 3.5

a = dv/dt = 2t − 7; t = 5: 3 m s⁻² [1]1
2t − 7 = 0, t = 3.5 [1]1

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[2 marks] (d)Cari sesaran zarah daripada O pada masa t, iaitu s dalam sebutan t.Find the displacement of the particle from O at time t, that is, s in terms of t.

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s = t³/3 − 7t²/2 + 12t

s = ∫(t² − 7t + 12) dt = t³/3 − 7t²/2 + 12t + c [1]1
s = 0 when t = 0, so c = 0 [1]1

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[3 marks] (e)Cari jumlah jarak yang dilalui oleh zarah dalam 5 saat pertama.Find the total distance travelled by the particle in the first 5 seconds.

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14.5 m

Direction changes at t = 3 and t = 4 [1]1
s(3) = 13.5, s(4) = 13 1/3, s(5) = 14 1/6 [1]1
13.5 + (13.5 − 13 1/3) + (14 1/6 − 13 1/3) = 14.5 m [1]1

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For the teacher: Applying, low difficulty, Chapter 8, standard 8.3.1

Textbook p. 267

Question 16

Sebuah bengkel menghasilkan x buah basikal biasa dan y buah basikal lumba setiap minggu. Pengeluaran dihadkan oleh kekangan berikut. Jumlah basikal yang dihasilkan tidak melebihi 30. Satu basikal biasa memerlukan 1 jam di bilik cat dan satu basikal lumba memerlukan 2 jam, dan bilik cat tersedia selama 44 jam seminggu. Bilangan basikal biasa tidak melebihi dua kali bilangan basikal lumba. Keuntungan ialah RM70 bagi sebuah basikal biasa dan RM100 bagi sebuah basikal lumba.A workshop makes x standard bicycles and y racing bicycles each week. Production is limited by the following constraints. The total number of bicycles made does not exceed 30. A standard bicycle needs 1 hour in the paint room and a racing bicycle needs 2 hours, and the paint room is available for 44 hours a week. The number of standard bicycles is at most twice the number of racing bicycles. The profit is RM70 for a standard bicycle and RM100 for a racing bicycle.

[3 marks] (a)Tulis tiga ketaksamaan linear, selain x ≥ 0 dan y ≥ 0, yang memenuhi semua kekangan di atas.Write three linear inequalities, other than x ≥ 0 and y ≥ 0, that satisfy all the constraints above.

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x + y ≤ 30 [1]1
x + 2y ≤ 44 [1]1
x ≤ 2y [1]1

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[2 marks] (b)Dengan menggunakan skala 2 cm kepada 5 basikal pada kedua-dua paksi, lukis graf ketaksamaan itu dan lorekkan rantau R yang memenuhi semua kekangan.Using a scale of 2 cm to 5 bicycles on both axes, draw the graphs of the inequalities and shade the region R that satisfies all the constraints.

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Three boundary lines drawn correctly: x + y = 30 through (30, 0) and (0, 30); x + 2y = 44 through (44, 0) and (0, 22); x = 2y through (0, 0) and (20, 10) [1]1
Region R shaded: the quadrilateral with vertices (0, 0), (20, 10), (16, 14) and (0, 22) [1]1

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[3 marks] (c)Gunakan graf anda untuk mencari keuntungan maksimum mingguan, dan bilangan setiap jenis basikal yang dihasilkan untuk mencapainya.Use your graph to find the maximum weekly profit, and the number of each type of bicycle made to achieve it.

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RM2 520; 16 standard, 14 racing

Profit k = 70x + 100y [1]1
Test vertices: (0, 0) gives 0, (20, 10) gives 2 400, (16, 14) gives 2 520, (0, 22) gives 2 200 [1]1
Maximum profit RM2 520 with 16 standard and 14 racing bicycles [1]1

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[2 marks] (d)Keuntungan bagi sebuah basikal lumba meningkat kepada RM160. Cari keuntungan maksimum yang baharu dan bilangan setiap jenis basikal yang dihasilkan.The profit on a racing bicycle rises to RM160. Find the new maximum profit and the number of each type of bicycle made.

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RM3 520; 0 standard, 22 racing

New k = 70x + 160y: (16, 14) gives 3 360, (0, 22) gives 3 520, (20, 10) gives 3 000 [1]1
Maximum RM3 520 with 0 standard and 22 racing bicycles [1]1

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For the teacher: Analysing, high difficulty, Chapter 7, standard 7.2.1

Textbook p. 242

Test specification (JSU)

Chapter	Remembering	Understanding	Applying	Analysing	Evaluating	Creating	Items	marks
1. Circular Measure	–	–	1 (3)	–	–	–	1	3
2. Differentiation	–	–	3 (17)	–	–	–	3	17
3. Integration	–	–	2 (15)	–	–	–	2	15
4. Permutation and Combination	–	–	1 (4)	–	–	–	1	4
5. Probability Distribution	–	–	2 (15)	–	–	–	2	15
6. Trigonometric Functions	–	–	1 (4)	2 (14)	–	–	3	18
7. Linear Programming	–	–	1 (4)	1 (10)	–	–	2	14
8. Kinematics of Linear Motion	–	–	2 (14)	–	–	–	2	14
Total	0 (0)	0 (0)	13 (76)	3 (24)	0 (0)	0 (0)	16	100

Each cell shows items (marks).