Module & More Add Math Tg 5

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

(b) V = 100 – 2t – t2 d2y = 1 – x2
dx2
(i) Apabila t = 0, V = 100 cm3

(ii) ddvt = –2 – 2t dy =x– 1 x3 + c
Apabila t = 4, dx 3
8 dv
dt = –2 –2(4) Apabila x = 1, dy = –1
dx
= –10 cm3 s–1
1
1– 3 + c = –1

y c = –1 – 2 = –5
6 3 3
10. (a)
Q dy 1 5
dx =x– 3 x3 – 3

y= x2 – 1 x 4 – 5 x + c1
2 12 3

(1, 4); 4 = 1 – 1 – 5 +c
2 12 3
x
–2 P 0 4 c1 = 21
4

Rajah 2 / Diagram 2 ∴y= x2 – 1 x 4 – 5 x + 21
2 12 3 4
Dalam Rajah 2, diberi luas rantau P ialah

∫ 4 unit2 dan 4 f(x) = 6. Cari
–2
In Diagram 2, it is given that the area of P is 4 unit2

∫and 4 f(x) = 6. Find
–2

∫(i) 4 f(x)dx 11. (a) Terdapat n pasu bunga di hadapan rumah
0 Raju. Dia ingin menyusun 2 daripada n pasu
bunga itu. Didapati bahawa bilangan cara
∫(ii) 0 f(x)dx
–2 menyusunnya ialah 56. Cari nilai n.

(iii) luas Q. [3 markah / marks] There are n flower pots in front of Raju’s house. He
wants to arrange 2 of the n flower pots. It is found
the area of Q. that the number of ways to arrange it is 56. Find the
value of n.

[2 markah / marks]
∫(a) (i) 4 f(x)dx
0

= 6 + 4 = 10

∫ (ii) 0 f(x)dx np2 = 56
–2
(n n! = 56
= –4 – 2)!

(iii) luas Q/ Area Q = 6 × 4 – 10 n(n – 1) = 56

= 14 unit2 n2 – n – 56 = 0

(n – 8)(n + 7) = 0

n = 8

(b) Satu lengkung mempunyai terbitan kedua

d2y = 1 − x2. Persamaan tangen kepada
dx2
(b) 10 orang dijemput duduk pada suatu
lengkung pada (1, 4) ialah y = 5 − x. Tentukan
meja bulat. Cari bilangan cara menyusun
persamaan lengkung itu.
kesemuanya jika
A curve has a second derivation d2y = 1 − x2. The
dx2 10 people are invited to sit on a round table. Find
the number of ways to arrange all of them if
equation of tangent to the curve at (1, 4) is y = 5 − x.
Determine the equation of the curve. (i) tidak ada halangan.

[3 markah / marks] there is no restrictions.

198

(ii) terdapat dua pasang suami isteri mesti   Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 
duduk bersama.
(b) Bina jadual taburan kebarangkalian bagi X.
there are two couples who need to sit next to
each other. Construct a probability distribution table for X.

[2 markah / marks] (c) Cari/Find P(2 < X ≤ 5)
[5 markah / marks]

(i) 9! = 362 880 (a) P[X = 1] + P[X = 2] + P[X = 3] +

(ii) 27!2! ! = 1 260 P[X = 4] + P[X = 5] = 1

1 + 2 + 3 + 4 + 5 = 1
2k 2k 2k 2k 2k

15 = 1 k = 15
2k 2

(b)

12. Suatu pemboleh ubah rawak diskret X mempunyai x 12345
x
taburan P(X = x) = 2k untuk x = 1, 2, 3, 4 dan 5. 12345
15 15 15 15 15
A random discrete variable X has a distribution P(X = x) P(X = x)

= x for x = 1, 2, 3, 4 and 5.
2k
(a) Cari nilai k. (c) P[2 < x ≤ 5]

Find the value of k. = 1 + 4 + 1 = 12 = 4
5 15 3 15 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.

B

7 cm

A 30° C
10 cm

Rajah 3 / Diagram 3

(a) Hitung sudut cakah ABC. [2 markah / marks]

Calculate the obtuse angle ABC.

(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. [2 markah / marks]
[2 markah / marks]
Calculate the length of CP.

(d) Cari luas ABC.

Find the area of ABC.

199

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

(a) sin B = sin 30°
10 7

8 B = 134° 25’
(b)
B
P

7 cm

7 cm
A 10 cm 30° C

(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 (10)(7) sin(150° – 134°35’)
2

= 9.4 cm2

14. (a) Jika tan A = 3 dan tan (A + B) = 1, cari tan B dalam bentuk surd. [2 markah / marks]

If tan A = 3 and tan (A + B) = 1, find tan B in surd form


(b) (i) Rajah 4 menunjukkan graf y = a sin bx + c.

Diagram 4 shows a graph of y = a sin bx + c.

y

3

2

1
x

0 π 2π
–1

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
1 – tan A tan B

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 = 3–2
1–3

200

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 

(i) y = a sin bx + c x(a sin bx + c) = π

a = 3 , b = 1 , c = −1 a sin bx + c= π
2 x

(ii) y y = π
x

3 Apabila/When x = π, y = 1

2 Apabila/When x = π , y = 2
2

1 2 penyelesaian/ solutions
x

0 π π 2π
–1 2

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.

1.5 cm

1 cm

P Q

Cari / Find Rajah 5 / Diagram 5
(a) perimeter bagi rantau berlorek.
[4 markah / marks]
the perimeter of the shaded region. [4 markah / marks]

(b) luas bagi rantau berlorek.

the area of the shaded region.

(a) Kos/ cos θ = 0.5
2.5

θ = 78°27’

Panjang lengkok CD = 78°27’ × 2p(1.5) = 2.05 cm O D A
360° 0.5 cm θ 2.5 cm 1 cm
Length of arc CD 1.5 cm
B
Panjang lengkok BD = 90° + 11.33’ × 2p(1) = 1.77 cm P C Q
360°
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 [1 + 1.5]2.45 – 78°27’ × p(1.5)2 – 101°33’ × p(1)2
2 360° 360°

= 0.64 cm2

201

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

Kertas 2 / Paper 2
[2 jam 30 minit / 2 hours 30 minutes]

8 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]

x + 2y – z = 4 ...➀ 3[3 – 2y] + 3y = 2 –5 + 14 – 4 = z
2x + y + z = –2 ...➁ 3 3
➀ + ➁ 3x + 3y = 2 ...➃ 9 – 6y + 3y = 2 z = –1
x + 2y + z = 2 ...➂
➀ + ➂ 2x + 4y = 6 7 = 3y
x + 2y = 3 ...➄
y = 7
3
7
x = 3 – 2 3 2

= –5
3

2. Satu fungsi ditakrifkan oleh f : x → 3 − 2x [2 markah / marks]
[3 markah / marks]
A function is defined by f : x → 3 − 2x. [1 markah / mark]

(a) Lakar graf bagi f untuk domain −1 ≤ x ≤ 3.

Sketch the graph for f for the domain −1 ≤ x ≤ 3.



(b) Pada paksi yang sama, lakar graf yang sepadan bagi f −1.

On the same axes, sketch the corresponding graph of f −1.

(c) Nyatakan domain bagi f −1.

State the domain for f −1.

(a) & (b) f (x) (c) Domain bagi f−1(x)

(–1, 5) 5 f(x) = 3 – 2x Domain for f −1(x)
(–3, 3) 4 f –1(x)
3 −3 ≤ x ≤ 5
2
1

–3 –2 –1–1 0 1 2 3 4 5 x
–2 (5,–1)
–3 (3,–3)

202

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 

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.

AE B

4 cm 2x cm
G
F

DC [2 markah / marks]
Rajah 1 / Diagram 1 [2 markah / marks]
[1 markah / mark]
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.

(ii) cari nilai x supaya luas BEFG ialah maksimum.

find the value of x so that the area of BEFG is maximum.

(iii) cari luas maksimum.

find the maximum area.

(a) y = x2 + (1 – p)x + 2p (b) (i) L(x) = 2x(4 – x) Ax E B
= 8x – 2x2 2x
1–p 2 G
2
 2 y = (21p–4–p()12+2p–x+ (ii) dLd(xx) = 8 – 4x = 0 C
p)2 = 6 x = 2 4
– F
4

8p – (1 – 2p + p2) = 24 Apabila x = 2 L(2) = 8(2) – 2(2)2 D 4
= 16 – 8
p2 – 10p + 25 = 0

(p – 5)(p – 5) = 0 = 8 cm2

p = 5

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.

y
B

A

x
0
Rajah 2 / Diagram 2

203

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

(i) Cari koordinat A dan B.

Find the coordinates of A and B

[2 markah / marks]

8 (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]

(a) (i) y = x + 4 ∫(b) Isipadu = 1 y2dx0
y = 4 + 3x – x2 Volume
x + 4 = 4 + 3x – x2 ∫ = 1(4 + 3x – x2)2 dx
x2 – 2x = 0 0
x(x – 2) = 0
x = 0 ; 2 ∫ = 1(16 + 24x + x2 – 6x3 + x4)dx
A(0, 4) B(2, 6) 0

 4 = 1 – 3 + 1 1
16x + 12x2 + 3 x3 2 x4 5 x5 0

(ii) Luas/Area = 2(4 + 3x – x2)dx – 1 [4 + 6]2 = 2 7 310 unit3
∫ 0 2
  = 4x + ∴ isipadu sebenar = 1 3 3601 unit3
3 x2 – x3 2– 10
2 3
0

= 11 13 – 10 = 1 31 unit2

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) p = 0.01, n = 1000 (ii) P[x ≥ 1] = 1 – P[X = 0]

m = np = 1000(0.01) = 1 – 5C0(0.01)0(0.99)5
= 0.049
= 10

σ = npq = 1000(0.01)(0.99) (b) (i) m = 200, σ = 40

= 3.15 P[ 150 ≤ z ≤ 180]

204

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 

= P 150 – 200 < z < 180 – 200 4 (ii) P[X > 230] = P[Z > 230 – 200 ]
40 40 40
3
= P – 45 < z < – 21 4 = P[Z > 4 ]

= 0.2029 = 0.2266

Bilangan pweokkeerrjsa = 0.28266
Number of

 35

6. Rajah 3 menunjukkan sebuah segi tiga ADC.

Diagram 3 shows a triangle ADC.

D

A C

Rajah 3 / Diagram 3

Diberi A→D = ~i 2+~j ,2D~→jC, D→C = 8~i +~j dan D→B = 2~i – ~j .

Given A→D = ~i + = 8~i + ~j and D→B = 2~i – ~j .

(a) Cari vektor-vektor A→B dan A→C dalam sebutan i dan j.
Find the vector A→B and A→C in terms of ~i and ~j .
[2 markah / marks]
(b) Tunjukkan bahawa B terletak pada garis AC. Seterusnya, cari AB : BC. [3 markah / marks]
[2 markah / marks]
Show that B lies on the line AC. Then, find AB : BC.

(c) Jika luas ADC = 24 unit2, cari luas segi tiga ADB.

If the area of ADC = 24 unit2, find the area of triangle ADB.

(a) A→B = A→D + D→B (c) D

==A→C3~i ~i=++A2→~~Djj + 2~i – ~j h
+ D→C
A B1 C
2

= 9~i ~i++23~j ~j+ 8~i + ~j ∆ADC = 1 ACh = 24
= 2

(b) A→C = 3A→B ∴∆ADB = 2 × 24 = 16 unit2
3

A, B dan C segaris
A, B and C collinear

AB : BC = 2 : 1

205

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

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.

8 x12345
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

(b) Seterusnya, lukis satu graf penyuaian terbaik. [1 markah / mark]
[4 markah / marks]
Then, draw a best fit graph.
[2 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

(a) y = a x + b (c) (i) b = 4
x
a = 14 – 4 =2
y x = ax + b 5
(b)
(ii) Apabila/When x = 2.5
x 12 3 4 5
y x = 9
y x 5.9 7.9 10.0 12 13.9 y = 9 = 5.69
2.5

y

14
12
10

8
6
4
2
0 12345 x

206

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 

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 tan2x + 11 tan x – 4 = 0

tan x = –11 ± 112 – 4(7)(–4)
14

tan x = –11 + 233
14
y y
16°56′
x = 16° 56’, 196°56’ x

–11 – 233 16°56′ 61°56′ x
tan x 14 61°56′
=

x = 118°4’, 298°4’

(b) y y = 1 + kos x

2

1 y = kos x

y = 0.5
x
0 π 2π

–1 y = sin 2x

(i) 4 penyelesaian/solutions
(ii) 4 penyelesaian/solutions
(iii) sin 2x = 1 + kos/ cos x
2 penyelesaian/solutions

207

8   Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

9. (a) Rajah 4 menunjukkan sebuah segi empat selari PQRS.
Diagram 4 shows a parallelogram PQRS.

yS

P(–1, 5)

0 R(5, 1)
Q x

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]

(a) (i) 5x = 2 – y 3 + x = 2 –2 + y = 3
2 2
y = –5x + 2
x = 1 y = 8
Titik tengah PR = (2, 3)
S = (1, 8)
∴Persamaan QS / Equation QS

  (b) luas =

area
=
y – 3 = –5(x – 2) 1 –1 3 5 1 –1
2 5 –2 1 8 5
y = –5x + 13 ...➀ 1
2 |2 + 3 + 40 + 5 – 15
(ii) mPR = 5–1 = 4 = 2 + 10 –1 + 8|
–1 – 5 –6 –3

Persamaan QR,/ Equation QS, = 1 |52| = 26 unit2
2
y – 1 = 3 (x – 5)
2 (b) PAQ = 90°
y – 5 y+2
y = 3 x – 15 + 1 mPA = x + 1 , mQA = x–3
2 2
y – 5 y + 2
y = 3 x – 13 ...➁ ( x + 1 )( x – 3 ) = –1
2 2
3 13 y2 – 3y – 10 = –(x2 – 2x – 3)
(iii) (a) ➀ = ➁ –5x + 13 = 2 x – 2
y2 + x2 – 2x – 3y – 13 = 0

–10x + 26 = 3x – 13 Jika/If x = 5

39 = 13x y2 + 25 – 10 – 3y – 13 = 0

x = 3 y2 – 3y + 2 = 0

y = –2 (y – 1)(y – 2) = 0

∴Q(3, –2) y = 1 ; 2

(5, 1) ialah R

208

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 

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 (b) ar – a = 20

2S2n = S3n ar3 – ar = 15
– 1)d] =
2[ 2n (2a + (2n 3n [2a + (3n – 1) a(r – 1) = 20 ...➀
2 2
d] a r(r2 – 1) = 15 ...➁

➁÷➀

a = 12, d = 3 r(r + 1) = 15
20
4n(24 + 6n – 3) = 3n[24 + 9n – 3]

84 + 24n = 63 + 27n 4r2 + 4r = 3

21 = 3n 4r2 + 4r – 3 = 0

n = 7 (2r + 3)(2r – 1) = 0

(ii) Tn = a + (n – 1)d = 48 r = – 3 ; 1
12 + 3(n – 1) = 48 2 2

(n – 1) = 36 = 12 ∴a[– 3 –1] = 20 atau a( 1 – 1) = 20
3 2 2

n = 13 a = –8 a = –40

209

8   Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

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.

B

AO

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.

(b) perimeter rantau yang tidak berlorek. [2 markah / marks]
[4 markah / marks]
the perimeter of the region that is not shaded.

(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]

(a) COB π (c) Luas = 1 (8)2 13π 2 – 1 (8)2 π 2 B
= 2π – 3 2 15 2 3
0.8π – Area

= 13π = 156° = 323 13π – π 4 A 0.8 π
15 15 3 60° O

BC2 = 82 + 82 – 2(8)(8) kos 156° = 256π = 17 π cm2
15 15
BC = 15.65 cm

(b) Perimeter = 6 π(8) + (8) + 8 + π (8) C
5 3

= 54.54 cm

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(3t2 − 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(3t2 − 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]

210

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 

(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(3t2 – 2t – 1)
dv =a= 12t – 4
dt

Apabila/When t = 0 a = –4 m s–2

(b) Apabila/When dv = 0 t= 1
dt 3

∴v = 2[31 1 2 – 21 1  – 1]
3 3
= – 2 32 m s–1

(c) v = 0 3t2 – 2t – 1 = 0

(3t + 1)(t – 1) = 0

t = 1

(d) jumlah jarak/total distance

∫  ∫ = 1(6t2 – 4t – 2)dt + 2(6t2 – 4t – 2)dt
01

= (2t3 – 2t2 – 2t]1 + 2t3 – 2t2 – 2t 2

  3 40 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]

211

8   Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

(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
... ➁
150x + 50y ≤ 75 000

6x + 2y ≤ 3 000

3x + y ≤ 1 500

y = 2x

(b) y

2000 y = 2x

3x + y = 1500

1000 (200, 900)

R (300, 600) x + 2y = 2000

0 1000 2000 x

(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

212

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM 

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 Indeks harga pada tahun 2020 Pemberat
berasaskan tahun 2018
Cost of ticket Weightages
Price indices for year 2020 based on
year 2018

Bas / Bus n3

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 × 100 = n
P18

24 × 100 = n = 120
20

(b) 3(120) + 4(125) + 130 m + 110 = 125
(8 + m)

970 + 130 m= 1000 + 125 m

5 m = 30

m = 6

(c) 110 × 120 = 132
100

110 × 130 = 143
100

I = 132 × 3 + 125 × 4 + 143 × 6 + 110
14
= 133.14

(d) P21 × 100 = 143
P18
650 × 100
P18 = 143 = RM454.55

213

  Matematik Tambahan  Tingkatan 5  Kertas Pra-SPM

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.

DC

8 cm M

A BS R

6 cm

P 10 cm Q

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]
[2 markah / marks]
(b) AR [3 markah / marks]

(c) ∠ARM [3 markah / marks]

(d) luas segi tiga PMR.
the area of the triangle PMR.

(a) AM2 = 52 + 42 DC
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 8 cm M
2 77 × 200
cos ARM A BS θ R
5 cm
ARM = 18°2’
6 cm
(d) MP2 = 36 + 41

MP = 77 = 8.77 P 10 cm Q

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

214

JAWAPAN

BAB Sukatan Membulat (d) θ° = 1.67π rad
360° 2π rad
1 Circular Measure
θ° = 360° × 1.67π
2π

1. (a) Gunakan/Use = 300.6° atau/or 300°36’

43°32’ = θ rad (e) θ° = 0.72π rad
360° 2π rad 360° 2π rad

θ = 43°32’ × 2π θ° = 360° × 0.72π
360° 2π

= 0.76 rad = 129.6° atau/or 129°36’

(b) Gunakan/Use panjang lengkok, AB

157°16’ = θ rad 3. (a) 0.8 rad = Arc length AB
360° 2π rad 2π rad
2πj

θ = 157°16’ × 2π Panjang lengkok, s
360°
Arc length, s

= 2.745 rad = 0.8 rad × 2π(7.4)
2π rad
(c) Gunakan/Use

247°54’ = θ rad = 5.92 cm
360° 2π rad
(b) 0.5π rad = panjang lengkok, s
θ = 247°54’ × 2π 2π rad 2πj
360°
Panjang lengkok, s
= 4.327 rad
(d) Gunakan/Use Arc length, s

303.68° θ rad = 0.5π rad × 2π(4.5)
360° 2π rad 2π rad
=

θ = 303.68° × 2π = 7.07 cm
360°
(c) Sudut yang dicangkum = 360° – 125° = 235°
= 5.3 rad
Angle subtended

(e) Gunakan/Use θ = panjang lengkok, s
360° 2πj
354.74° θ rad
360° = 2π rad Panjang lengkok, s

354.74° × 2π Arc length, s
360°
θ = =235° × 2π(10.2)
360°
= 6.191 rad
= 41.84 cm
2 rad
θ° 3 π (d) Sudut yang dicangkum = (2π − 1.3) rad
360°
2. (a) = 2π rad Angle subtended

360° × 2 π (2π – 1.3) rad = panjang lengkok, s
3 2π rad 2πj
θ° =
2π Panjang lengkok, s

= 120° Arc length, s

4 rad =(2π – 1.3) rad × 2π(20.5)
9 2π rad
(b) θ° =
360° 2π rad = 102.16 cm

360° × 4 (e) Sudut yang dicangkum = (2π − 4.5) rad
9
θ° = Angle subtended

2π (2π – 4.5) rad = panjang lengkok, s
2π rad 2πj
= 25.46° atau/or 25°28’

(c) θ° = 4.65 rad Panjang lengkok, s
360° 2π rad
Arc length, s

θ° = 360° × 4.65 = (2π – 4.5) rad × 2π(8.4)
2π 2π rad

= 266.43° atau/or 266°26’ = 15 cm

J1

4. (a) 326004°° = panjang lengkok, s Perimeter = 15.27 + 20.27
2πj
Hence = 35.54 cm

j = 20.3 × 360° (c) Panjang lengkok minor PQ = 2π(3) – 4.5(3)
204° × 2π
Minor arc length PQ = 5.35 cm

= 5.7 cm ∠POQ = 2π − 4.5

14.5 = 1.78 rad
2π – 1.6
(b) j = = 102.17°

= 3.1 cm Maka, PQ = 32 + 32 − 2(3)2 kos102.17°

Hence = 4.67 cm

(c) j = 11.5 Perimeter = 5.35 cm + 4.67 cm
2π – 0.8π
= 10.02 cm

= 3.05 cm (d) ∠POQ = 150°
Panjang jejari/Arc length, j
360° – 105° panjang lengkok, s
(d) 360° = 2πj = (7.3 × 360°) = 1.99 cm
(210 × 2π)
14.2 × 360°
j = 255° × 2π Maka/Hence,

= 3.19 cm PQ = 1.992 + 1.992 − 2(1.99)2 kos150°
= 3.84 cm
3
(e) j = – 3 Panjang lengkok = 150° × 2π(1.99)
4 360°
2π π Arc length

= 0.76 cm   = 5.2 cm

Perimeter = 5.2 cm + 3.84 cm

5. (a) (2π – θ) = 7.2 = 9.04 cm
6.1
(e) Panjang lengkok PQR = 2π(5.3) − 5.3(1.4π)

θ = 5.1 rad Arc length PQR = 10 cm

(b) (2π – θ) = 4.8 ∠POQ = (2π − 1.4π) ÷ 2
3.2
= 0.3π
θ = 4.78 rad = 54°

(c) (2π – 2θ)= 2.8 Panjang perentas PQ
1.8
Length of chord PQ

θ = 2.36 rad = 5.32 + 5.32 − 2(5.3)2 kos 54°
= 4.81 cm
(2π – θ) = 3.2
(d) 3 2.5 Perimeter = (4.81 × 2 + 10) cm

θ = 2.44 rad = 19.62 cm

(e) 2πj = 6.3 + 8.2 + 11.4 7. (a) 0.7 rad = 0.7 × 360° = 40.1°
2π
j = 4.12 cm

θ = 11.4 sin 40.1° = AD
4.12 9.2

= 2.77 rad AD = 5.93 cm

6. (a) 0.4π rad = 0.4π × 360° OD = 9.22 − 5.932
2π = 7.03cm

= 72° Panjang lengkok = 9.2 × 0.7

Maka/Hence, Arc length = 6.44 cm

PQ = 9.62 + 9.62 − 2(9.6)2 kos72° Perimeter = 5.93 + 6.44 + (9.2 – 7.03)

= 11.28 cm = 14.54 cm

Panjang lengkok PQ = jθ (b) Katakan OM = j cm
Let
Arc length PQ = 9.6(0.4π)
Maka, jθ + 2j + 2jθ = 2j + jα
= 12.06 cm
Hence, α = 3θ
Perimeter = (12.06 + 11.28) cm
(c) (i) Diberi AP = 2 cm, maka OA = 6 cm
= 23.34 cm
Given hence
(b) Panjang lengkok minor PQ = 2π(8) – 30
6a + 2θ = 10
Minor arc length PQ = 20.27 cm
θ = 5 − 3a
20.27
∠POQ = 8 (ii) a = 30°

= 2.53 rad = π rad
6

= 145.14° θ = 5 − 3 π 
6
Maka, PQ = 82 + 82−2(8)2 kos145.14°
= 15.27 cm = 3.43 rad

J2

(d) π = 60° (b) Gunakan/Use
3
panjan2gπljengkok = luas sektor
Maka, AP = AB = j. luas bulatan
Hence
arc length = area of sector
OA = j 2πj area of circle
2
12 = 42
= 35π = πj + πj 2πj πj2
6 3 2
j = 42 × 2
j = 7 cm 12

(e) Perimeter = OM + ON + panjang lengkok/ = 7 cm

arc length MN + MK + panjang (c) Gunakan/Use

lengkok/arc length KL + ML 38 = πj2 – 85
2πj πj2
= 5 + (5 2 − 5) + 5 π  + (5 2 − 5) +
5 4
2 π +5 2(πj2 – 85)
4 j = 38

= 23.62 cm 2(πj2 – 85) = 38j

8. (a) Sudut yang tercangkum = 5 π rad πj2 – 19j – 85 = 0
6
Angle subtended j = 19 ± (–19)2 – 4π(–85)
2π
5
6 π j = 9.04 cm

Maka, luas = 2 (3.5)2 = 16.04 cm2 (d) Gunakan/ Use
=
Hence, the area 2 luas sektor/area of sector θ
3 luas bulatan/area of circle 360°
(b) Sudut yang tercangkum π rad =

Angle subtended 7.5 88
πj2 360°
  2 π =
3

Maka, luas = 2 ( 5.2)2 j2 = 7.5 × 360°
88π
Hence, the area

= 28.32 cm2 j = 3.13 cm

(c) Sudut yang tercangkum = (360 – 215)° (e) Gunakan/ Use

Angle subtended = 145° π3j02 = 12.5
2πj
Maka, luas = 145° × π × ( 7.8)2
360°
Hence, the area 30 × 2
j = 12.5
= 76.98 cm2
6
(d) Sudut yang tercangkum = 2.5 = 2.4 rad = 4.8 cm

Angle subtended 10. (a) Gunakan/ Use

Maka, luas = 2.4 × (2.5)2 = 7.50 cm2 luluaassbsuelkattoarn = θ
2 2p
Hence, the area
aarreeaa of sector θ
(e) Jejari sektor = 10 of circle = 2p
3π
Radius of the sector 33 = 2π – θ
5 π(4.8)2 2π

= 50 cm 236.60π4π = 2π – θ
3π

  3 π 2.865 = 2π – θ
5
  50 2 θ = 2p − 2.865
Maka, luas = 3π
2 = 3.42 radian
Hence, the area
(b) Gunakan/Use
= 26.53 cm2
luluaassbsuelkattoarn panjang lengkok
9. (a) Gunakan /Use = 2πj

luluaassbsuelkattoarn = θ aarreeaa of sector = arc length
2p of circle 2πj

aarreeaa of sector = θ 46 = 12
of circle 2p πj2 2πj

10.21 = 2π – 3.5 j = 46 × 2 = 7.67 cm
πj2 2π 12

j2 = 10.21 × 2 s = jθ
2p – 3.5
θ = 12 = 1.56 radian
j = 2.71 cm 7.67

J3

(c) 2j + jθ  = 15 …… ➀ Kos/ cos ∠AOB = 4.32 + 4.32 − 4.652
2(4.3)2
1 j2θ  = 30 …… ➁
2 ∠AOB = 65.46°
= 1.14 rad
➁÷➀ 50° = 0.87 rad

1 j2q LAureaas otfesmegbmereen nt g = s1212in((5540..°53))+22(s012in.8(6745)..34−)62(°121.(154.)5−)2
2
=2
2j + jq

jq q) = 2
2(2 +

jθ = 8 + 4θ = 3.7 cm2

θ= 8 12. (a) OP = 8 2 cm
j–4

(d) 1 × 82 × θ  = 43 Luas sektor major = 270° × π(8 2)2
2 Area of major sector 360°

θ = 1.34 radian
(e) Gunakan/Use = 96π cm2

luas sektor = panjang lengkok = θ Luas segi tiga = 1 × (8 2)2
luas bulatan 2πj 2π 2

area of sector arc length θ Area of triangle = 64 cm2
area of circle 2πj 2π
= = Jumlah luas = 96π + 64

Total area

54 = 21 = θ Luas semibulatan = π(8)2 = 64π
πj2 2πj 2π
Area of semiclrcle

j = 54 × 2 = 5.14 cm Luas rantau berlorek = 96π + 64 − 64π
21
Area of shaded region = 32π + 64

21 = 32(π + 2) cm2
5.14
2p – θ = (b) Luas kain Q

Area of cloth Q

θ = 2.2 radian = 1 × (20)2 × 1.5 − 1 × (15)2 × 1.5
2 2
11. (a) Luas berlorek/Shaded area

= luas sektor POQ – luas segi tiga POQ = 300 cm2 – 168.75 cm2

area of sector POQ – area of triangle POQ = 131.25 cm2

1 4π
2 3
 
Luas sektor = j2 2π − Luas kain P

Area of sector Area of cloth P

= 1 (8.1)2 23π2 = 1 × (15)2 × 1.5 − 1 × (10)2 × 1.5
2 2 2

= 68.71 cm2 = 168.75 cm2 – 75 cm2

2π rad = 120° = 93.75 cm2
3
1 Beza luas
2
Luas segi tiga = × (8.1)2 × sin 120° Difference of area
Area of triangle =
28.41 cm2 = 131.25 cm2 − 93.75 cm2

Maka, luas tembereng = 68.71 − 28.41 = 37.5 cm2
Hence, the area of segment = 40.3 cm2
(c) (i) Lilitan/ Circumference OPQ = 9π

(b) Luas sektor POS = luas sektor QOR Lilitan/ Circumference QSR = 5π

area of sector POS = area of sector QOR Lilitan/ Circumference OTR = 4π

1 j2θ = 1 (3.5)2(π − 1.4) Perimeter = 2(9π + 5π + 4π)
2 2
= 36π cm

= 10.67 cm2 1
(ii) Luas semibulatan OPQ = 2 × (9)2 × π
1.4
1.4 rad = 2π × 360° Area of semicircle OPQ

= 80.21° 1 Luas semibulatan QSR = 1 × (5)2 × π
2 Area of semicircle QSR 2
Luas segi tiga = × (3.5)2 × sin80.21°

Area of triangle 1
= 6.04 cm2 Luas semibulatan OTR = 2 × (4)2 × π
Area of semicircle OTR
Maka, luas tembereng = 2(10.67 − 6.04)

Hence, the area of segment = 9.26 cm2 LToutaasl keseluruhan = 2× 1 π × (81 + 25 – 16)
90π 2
(c) Perentas AB = 5.52 + 5.52 − 2(5.5)2 kos50° area
Chord AB = 4.65 cm = cm2

J4

Praktis SPM 1 2. (a) PQ2 = 152 + 52 7.5 cm
P
Kertas 1 = 250
15 cm
1. 42 = 52 + 52 – 2(5)(5) kos θ PQ = 5 10 cm
Q
52 + 52 – 42 Panjang lengkok AB = 2π(7.5) 12.5 cm
2(5)(5)
kos θ = F 10 m E Arc length AB = 15π
B 4m C 8m
θ = 47.16° D Panjang lengkok CD = 2π(12.5)
∠APB = ∠CPD ␪
A 5m P Arc length CD = 25π

Maka, 15π = jq ….. ➀

= 180° − 47.16° 25π = (j + 5 10 )q ….. ➁ E Oj B D
2 ➁÷➀ ␪ 5√10

= 66.42° 35 = j+5 10 A
j
Luas sektor APB = 66.42° × π(5)2 = 14.49 cm2 C
360°
Area of sector APB 2j = 15 10

Luas ∆BCP = 1 (5)(5)sin 47.16° j = 15 10 cm
2 2

= 9.17 cm2 q = 15π ×2 = 1.99 rad
15 10
Luas rantau berlorek = 10 × 8 − 9.17 – 2(14.49)

Area of shaded region = 41.85 m2 = 113.84°

2. (a) 82 = 152 + 152 – 2(15)2 kos θ BO
kos θ = OC
152 + 152 – 82 kos ∠EOC =
2(15)(15)
 15 10 10  kos(180° − 113.84°) = EO
θ = 30.93° 2 +5

30.93°
360° × 2π = 0.54 rad EO = 15.98 cm

(b) Luas tembereng Panjang kad = 15.98 + 15 10 +5 10
2
Area of segments Length of card

= 1 (15)2(0.54)− 1 (15)2 sin30.93° = 55.5 cm
2 2
≈ 56 cm

= 2.93 cm2 Lebar kad / Breadth of card
= j + 5 10
3. (a) 10 = 7(2π − θ)

7θ = 14p − 10 = 15 10 +5 10
2
θ = 4.85 rad

 (b) Luas = 1 j2θ = 1 (7)2(2p − θ) = 39.53 cm
2 2
Area ≈ 40 cm
=
1 (7)2 10 (b) Luas kad tidak digunakan
2 7
Area of unused card

= 35 cm2 = 56 × 40 −  39.532 × 1.99 +
2

Kertas 2 6 1 21  15 10 2(1.99)
12 2 2
1. ( a) ckoossθθ = =
= 1244.88 cm2
π B
θ = 60° = 3 rad ␪ 3. (a) 60° = π rad
2␪ 3
(b) kos 30° = 18 A 12 cm O E D C
cos 30° AB A
6 cm

AB = 20.78 cm P Q
O 30Њ
π
6
 
Panjang lengkok BD = 20.78

Arc length BD B 60Њ C
= 10.88 cm

(c) 1 π 2 3 
2 6 tan 30°
 
Luas sektor BAD = (20.78)2 (b) AC =

Area of sector BAD
= 113.05 cm2
= 10.39 cm

Luas ∆ABE = 1 (18)(20.78) sin30° (c) Luas satu tembereng
Area 2
Area of a segment

= 93.51 cm2 = 1 (10.39)2 π  − 1 (10.39)2 sin60°
2 3 2
Luas rantau berlorek = 19.54 cm2
= 9.78 cm2
Area of shaded region

J5

Luas bulatan = π(3)2 ∠SOP = 120°−60°
2
Area of circle
= 30°
Jumlah luas = 9π + 3(9.78)
= π rad
Total area = 57.61 cm2 2

4. (a) 5 = 11q

q = 5 rad 10 cm
11 PQ
SR
= 0.45 rad
120Њ
T O 10 cm
P 5 cm Q

␪ 11 cm

50Њ (b) Luas tembereng PQ /Area of segment PQ
SR
1 π 1
(b) Luas sektor = 1 (11)2(0.45) = 2 (10)2 3  − 2 (10)2 sin 60°
2
Area of sector = 9.06 cm2

= 27.2 cm2 Luas tembereng SR / Area of segment SR
tan 50° =
TS = 1 (10)2 2π  − 1 (10)2 sin 120°
11 2 3 2

TS = 13.11 cm = 61.42 cm2

Luas/Area ∆TSR = 1 (11)(13.11) Maka, luas berlorek
2
Hence, the shaded area
= 61.42 – 9.06
50° = 72.1 cm2
50° = 360° × 2p = 0.87 rad = 52.36 cm2
(c) Panjang lengkok SP = QR
1
Luas sektor SRP = 2 (11)2(0.87) Arc length SP = QR

Area of sector SRP 10 π  = 5π
6 3
= 52.6 cm2

Luas berlorek = 72.1 − 52.6 + 27.2 SR2 = 102 + 102 − 2(10)2 kos 120°

Shaded area = 46.7cm2 SR = 10 3

Sudut KBAT Perimeter = 10 + 2 5π  + 10 3
(a) PQ2 = 102 + 102 − 3
2(10)2 kos POQ
= 101+ π + 3
3
102 + 102 − 102
kos POQ = 2(100)

∠POQ = 60°

J6

BAB Pembezaan (d) δy = (x + 1 – 1
δx + 1) (x + 1)
2 Differentiation
= ((xx + 1) – x – 1 – δx
+ 1)(x + 1 + δx)

1. (a) had k δδyx == ((xx + –δx 1 + δx)
2k + x + 1)(x + 1 + δx)
x → 0
lim –1
x→0 1 1)(x +
= 2

δy –1 1
δx (x + 1)(x + 1) + 1)2
(b) had 4 – 2x + x2 had = = – (x

x → 0 δx → 0

lim = 4 d 1
dx 2
x→0 4. (a)   = 0

2. ( a) 6 = 2 (b) 0 (c) 32 (b) d ( x ) = d x 1 
2(x2 – 4) dx dx 2
(d) had x–2
1
x→2 2x
lim
x→2 =
= had
2(x + 2)(x – 2) (c) d  8x4 
x → 2 x–2 dx 5
lxi→m2
= 32
=8 5 x3

(e) had 2(x2 – 3x – 4) (d) d  –6  = d  –6 
x+1 dx (2x)3 dx 8x3
x → 2
lim
x→2 d –3
2(x – 4)(x + 1) dx  4 x–3
= had x+1 =

xli→m2 = 9
4x4
x→2

= –4 1 + 2x (e) d  –5x  = d  –5 x –3
x + x dx 3x4 dx 3
(f) had 4 x
xlxi→→m∞∞ x x = 5
x4


= 0 + 2 = 2  (f) d 2x = d ( – 1 )
0 + 1 dx x dx 2
2x

3. (a) δy = f(x + δx) = 4 − x2 2 – 3
2 2
= 4 – (x + δx)2 – 4 + x2 =– x

= 4 – x2 – 2xδx – (δx)2 – 4 + x2 2
x
= −(2x + δx)δx = –
δδyx = –(2x + δx) 2x

had δy = dy = –2x 5. (a) f(x) = 3 – 1
δx dx
δx → 0 x = 3x 2

(b) x2 – x – 2 – 32 – 3

δy = (x + δx)2 – (x + δx) – 2 – x2 + x + 2 f’(x) = x 2

= x2 + 2x·δx + (δx)2 – x − δx – x2 + x = –3
2x x
= 2xδx – δx + (δx)2

δy = [(2x – 1) + δx]δx

δδxy = (2x – 1) + δx (b) f(x) = 10 = 10 x3
3x –3 3
δy dy
had δx = dx = 2x – 1 f’(x) = 10x2

δx → 0 1 1 –5 –5
(x + δx)2 x2 4x –2 4
(c) δy = – (c) f(x) = = x 2

= x2 – (x + δx)2 f’(x) = –5 x
x2(x + δx)2 2

= –2xδx – (δx)2 3
δδxy = x2(x + δx)2
–2x – δx 6. (a) y = 3x 2 1

x2(x + δx)2 dy = 9 x 2
dx 2

had δy = dy = –2x = –2 = 92 x
δx dx x4 x3
δx → 0

J7

= dy = 9 (2) 4 x2 – 3 
dx 2 7 x

Apabila x 4, (h) y =

When 8 12
= 9 7 7
= x –
1
7x 7
(b) y = 2x = 2 x 2 dy 8
dx 7
=

dy = 7 – 1 8. (a) y = 8 x–1 – x2 + 1
dx 4 3
x 2 x2

= 7 ddxy = –8 – 2x + 1
4x 3x2 2x
dy 7
dx = 49


Apabila x = 9, Apabila/When x = 4,

When dy –8 1
dx 3(4)2 4
= 7 = –8+
12
–8 = – 85 x2 11
(c) y = 5x –2 = –7 12

dy = – 156 x (b) y = 5x – x–1 = 5 x–1 – 1 x –3
dx 4x2 4 4

AWpheanb ila x = – 81 , ddyx == – 156  – 18  ddxy = – 5 x–2 + 3 x–4
2 4 4
5
Apabila/When x = –1,

7. (a) y = x5 – 1 x3 + x –1 dy = – 5 + 3
2 dx 4 4

ddxy = 5x4 – 3 x2 – 1 = – 1
2 x2 2

(b) y = –2x – 5 – x – 2 + 2 9. (a) y = (1 – 4x2)2
3
ddyx = 10 2
x6 + x3 = 13 (1 – 4x2)2

(c) y = 2x2 + 3x ddxy = 2 (1 – 4x2)· d (1 – 4x2)
3 dx
ddxy = 4x – 3
5 = 32 (1 – 4x2)(–8x)

(d) y = x + 4x 2 = – 16 x(1 – 4x2)
3
dy 3
dx
= 1 + 10x 2 (b) y = 2 – 1 5
x
= 1 + 10x x
dy 1 d
(e) y = 2x – 2x4 + 3 – 3x3   dx = 5 2 – x 4· dx (2 – x –1)
3x2
= 52 – 1 4·  1 
= 2 x –1 – 2 x2 + x –2 – x x x2
3 3
= 5 2 – 1 4
ddxy = –2 4 1 x2 x
3x2 – 3 x – x3 –1
1 2
  (c) y = 2 5 4
3x3 – x2
2(x2 + 2x + 1)
(f) y = 3x
    dy 1 2 d 2
2 dx = 2 (4) 3x3 – 5 x2 3· dx 3x3 – 5 x 2
3
= (x + 2 + x –1)  2 4
5
ddxy = 2 1 1  = 23x3 – 5 x2 3· 9x2 – x
3 x2
– 2 5
x2
    (d) y = =2 +x –3

(g) y = x39 – 3x + 1 x2 5 +x 3
4 x2

= 9x3 – 3x4 + 1 x5 dy = 2(–3) 5 + x–4– 10 + 1
4 dx x2 x3

dy = 27x2 – 12x3 + 5 x4 –61 – 10  6(10 – x3)
dx 4 x3 +x4
=  =
= x227 – 12x + 5 x2  5 x4 5
4 x2 + x3 x2

J8

10. (a) y = 5 (2 + 1 12. (a) y = (x – 3)3
4 (1 + 2x)2
x2) 2

  dy = 5 1 – 1 (1 + 2x)2(3)(x – 3)2 –
dx 4 2
(2 + x2) 2 (2x) dy = (x – 3)3(2)(2)(1 + 2x)
dx (1 + 2x)4
= 5x
4 2 + x2 = (1 + 2x)(x – 3)2[3(1 + 2x) – 4(x – 3)]
(1 + 2x)4 3
Apabila x = 4, dy = 5(4)
dx (x – 3)2[15 + 2x)
When 4 18 = (1 + 2x)3

= 5 = 52 (b) y = x2
32 6 – 2x)3
(1
(b) y = (x4 – 3x2 – 2)3

dy = 3(x4 – 3x2 – 2)2(4x3 – 6x) ddyx = (1 – 2x)3(2x) – x2 (3)(–2)(1 – 2x)2
dx (1 – 2x)6

= 6(2x3 – 3x)(x4 – 3x2 – 2)2 2x(1 – 2x)2[1 – 2x + 3x]
(1 – 2x)6
AWphaenb ila x = – 1, ddxy == 6(–2 + 3)(1 – 3 – 2)2 =
96
= 2(1x(–1 + x)
(c) y = 6(x + x2)–1 2x)4

ddyx = –6(x + x2)–2(1 + 2x) (c) y = 1 + 2x2
2 + x3
–6(1 + 2x)
= (x + x2)2 ddyx = (2 + x 3)(4x) – (1 + 2x2)(3x2)
(2 + x3)2
Apabila x = –2, dy = –6(–3)
When dx (–2 + 4)2
x[4(2 + x3) – 3(1 + 2x2)x]
9 = (2 + x3)2
= 2

= x(8 + 4x3 – 3x – 6x3)
(2 + x3)2
11. (a) y = x3(1 – 2x)2

u = x3 v = (1 – 2x)2 = x[8(–2 3x – 2x 3]
+ x3)2
du = 3x2 dv = 2(1 – 2x)(–2)
dx dx 2x3 – 3
(d) y = (x – 1)
dd xy == x3(2)(–2)(1 – 2x) + (1 – 2x)2(3x2)
x2(1 – 2x)[–4x + 3(1 – 2x)] ddyx = (x – 1)(6x2) – (2x3 – 3)
(x – 1)2
= x2(1 – 2x)(3 – 10x)

(b) y = (x + 4)2 (1 – 3x)2 = 4x3(x– 6x2 + 3
– 1)2
ddxy = (x + 4)2(2)(–3)(1 – 3x) + (1 – 3x)2(2)(x + 4)
= 2(1 – 3x)(x + 4)[–3(x + 4) + 1 – 3x] 1
= 2(1 – 3x)(x + 4)[–6x – 11]
(e) y = x = x2
x+1 x+1

(c) y = (2x –1)(1 + x2)4 (x + 1) –
2x
ddyx = (2x – 1)(4)(2x)(1 + x2)3 + (2)(1 + x2)4 ddyx = (x + 1)2 x
= 2(1 + x2)3[4(2x – 1)(x) + 1 + x2]

= 2(1 – x2)3[9x2 – 4x + 1] = x + 1 – 2x
2 x (x + 1)2
1

(d) y = x2(x – 1) 2 = 1 – x
2 x (x + 1)2
  ddxy = x2 1 – 1 + x – 1(2x)
2
(x – 1) 2

= x2 + 2x x – 1 13. (a) dy = (x – 3)2(–2x) + (4 – x2)(2)(x – 3)
2 x–1 dx
= 2(x – 3)[–x(x – 3) + 4 – x2]
= x2 + 4x(x – 1)
2 x–1 = 2(x – 3)(3x + 4 – 2x2]

= 5x2 – 4x d2y = (2x – 6)(–4x + 3) + (– 2x2 + 3x + 4)(2)
2 x–1 dx2
= –8x2 + 6x + 24x – 18 – 4x2 + 6x + 8

(e) y = 2(x – 2)(1–3x)3 = –12x2 + 36x – 10

ddyx = 2(x – 2)(3)(–3)(1 – 3x)2 + 2(1 – 3x)3   dy 2 = 4(x – 3)2(3x + 4 – 2x2]2
= 2(1 – 3x)2[–9(x – 2) + 1 – 3x] dx
= 2(1 – 3x)2[19 – 12x]
∴ dd2xy2 ≠ dy
  dx 2

J9

(b) y = x–1 – x–2 + 2x2 16. (a) y = px–1 + qx

ddxy = – x–2 + 2x –3 + 4x 4 = p + 2q
2
–p
dd2xy2 = 2x–3 – 6x–4 + 4 y’ = x2 +q

= 2 – 6 +4 Apabila x = 2 –p + q = –8
x3 x4 When 4
–p + 4q = –32 ...➁
 ddyx 2 = 4x – 1 2 2
x2 + ➁ + ➂ p + 4q = 8 ...➂
x3
8q = –24

∴ dd2xy2 ≠ dy q = –3
  dx 2
p = (4 + 6)2

(c) dy = (2x + 1) – 2x = 1 = 20
dx (2x + 1)2 (2x + 1)2
(b) y = px3 + qx (–1, 1)

dd2xy2 = –4 1 = –p – q ...➀
(2x + 1)3
y’ = 3px2 + q

 ddxy 2 = 1 1)4 Apabila x = –1 y’ = 3p + q = 5 ...➁
(2x +
When ➁ + ➀ 2p = 6

∴ dd2xy2 ≠ dy p = 3, q = –4
  dx 2
(c) y = px2 + qx + 1 (2, 5)

14. (a) y = x2 + 2x 5 = 4p + 2q + 1

y’ = 2x + 2 2 = 2p + q ...➀

y’’ = 2 y’ = 2px + q,

Sebelah kiri = 2x2[2] – (2x + 2)2 + 4(2x + 2) Apabila x = 2 y’ = 4p + q = 8 ...➁

Left side = 4x2 – [4x2 + 8x + 4] + 8x + 8 When ➁ – ➀ 2p = 6

= 4 sebelah kanan p = 3, q = –4

right side 1 1
x 2
(b) y = x3 – 2x2 17. (a) y = x = 2, y =

y’ = 3x2 – 4x dy 1
dx x2
y’’ = 6x – 4 = –

Sebelah kiri = x2[6x – 4] – 6[x3 – 2x2] dy 1
dx 4
Left side = –4x2 + 12x2 Pada x = 2, =–

= 8x2 sebelah kanan Persamaan tangen/Equation of tangent

right side 1 1
2 4
(c) y = x + 2x –1 y – = – (x – 2)

y’ = 1 – 2x –2 1
4
y’’ = 4x–3 y = – x + 1

  Left side 4 + 2 2 +x Persamaan normal//Equation of normal
Sebelah kiri = x3 x 1– x2

= 2 +x y – 1 = 4(x – 2)
x 2
15
sebelah kanan/right side y = 4x – 2

1 5. (a) y = 5(–2) – 3(–2)2 (b) y = x3 – 4x x = –1
x = –10 – 12 = –22
y’ = 5 – 6x y = –1 – 4(–1)

= 3

Pa da/At x = – 2 dd xy == 517– 6(–2) ddxy = 3x2 – 4 pada x = –1
(b) f(–1) = 4(–1)2 – 6(–1) + 1
= 11 ddyx == –31(–1)2 – 4
f’(x) = 8x – 6
f ’(–1) = 8(–1) – 6 = –14 Persamaan tangen/Equation of tangent
∴ Pada (–1, 11), f’(x) = –14
y – 3 = –(x + 1)
At
y = –x + 2
(c) f(1) = –2 + 4 = 2
f’(x) = –2 + 8x Persamaan normal/Equation of normal
f’(1) = –2 + 8(1) = 6
∴ Pada (1, 2), f’(x) = 6 y – 3 = x + 1
At
y = x + 4

1

(c) y = x = x 2 x = 4, y = 2

ddyx = 1
2x

J10

Pada x = 4, dy = 1 Titik pusingan ialah (1, –3)
dx 4
The turning point is (1, –3)
Persamaan tangen/Equation of tangent
Dari jadual berikut.
y – 2 = 1 (x – 4)
4 From the following table.

y = 1 x + 1 x Sedikit 1 Sedikit
4 kurang 1 lebih 1
dy
Persamaan normal//Equation of normal dx −

y – 2 = –4(x – 4) Lakaran 0+
tangen
y = –4x + 18

18. (a) (i) y = 3x2 – 6x

ddxy = 6x – 6 = 12 Kecerunan dy menukar tanda dari negaif ke
dx
6x = 18 x = 3

y = 3(3)2 – 6(3) positif semasa melalui x = 1, jadi titik pusingan

= 9 itu ialah minimum.

∴P[3, 9] The gradient dy changes from negative to positive
dx
(ii) Persamaan normal pada (3, 9) as it passes through x = 1, hence the turning point is
Equation of the normal at (3, 9)
minimum.
y – 9 = – 1 (x – 3)
12 (b) dy = 3x2 − 6x
dx
y = – 1 x + 37
12 4 Untuk titik pusingan, kecerunan dy =0
dx
(b) y = 2x + 8
x For the turning point, gradient dy =0
dx
ddyx = 2 – 8
x2 =0 Maka 3x2 − 6x = 0

2x2 = 8 x = ±2 x = 0 dan 2

8 8 y = 0 dan −4
2 2
y = 2(2) + , 2(–2) – Titik pusingan ialah (0, 0) dan (2, −4)

= 8, –8 The turning point is (0, 0) and (2, −4)

∴(2, 8), (–2, –8) Dari jadual berikut./From the following table.

(c) y = ax2 + bx + 2 x Sedikit 0 Sedikit
kurang 0 lebih 0
ddyx = 2ax + b dy
dx − 0+
pada 1, 1 , dy = 2a + b Lakaran
2 dx tangen Sedikit
kurang 2
y = x2 + 6x + 4 x
+
ddyx = 2x + 6 dy Sedikit
pada (–2, –4 ), ddyx dx lebih 2
= 2(–2) + 6 Lakaran 2
= 2 tangen
1
∴2a + b = – 2 ...➀ 0–

1 = a + b + 2
2

a + b = – 3 ...➁
2
(0, 0) ialah titik minimum dan (2, −4) ialah
➀ – ➁ a = 1, b = –1
titik maksimum.
= – 5
2 (0, 0) is a minimum point and (2, –4) is a maximum
point.

19. (a) dy = 4x − 4 (c) dy = x2 − 2x – 3
dx dx
dy dy
Untuk titik pusingan, kecerunan dx =0 Untuk titik pusingan, kecerunan dx =0

For the turning point, gradient dy =0 Maka x2 − 2x – 3 = 0
dx
(x – 3)(x + 1) = 0
Maka, 4x − 4 = 0
x = 3 dan –1
Then
8
x = 1 dan y = 2 − 4 − 1 y = –8 dan 3

= −3

J11

Titik pusingan ialah (3, –8) dan −1, 8  Luas maksimum = 13 26 – 2 13 
3 2 2
Dari jadual berikut. Maximum area
1
= 84 2 cm2

x Sedikit 0 Sedikit 21. Diberi dx = 0.5 unit s−1 apabila x = 1.
kurang 3 lebih 3 dt
dy Given When
dx − 0+
Lakaran ddyx = 6(3x − 6)
tangen Sedikit
kurang –1 Maka dy = dy × dx
x dt dx dt
+ Hence
dy –1 Sedikit = 6(3x − 6) dx 
dx lebih –1 dt
Lakaran
tangen 0– Apabila x = 1 dan dx = 0.5
dt
When and

Maka dy = 6(−3)(0.5)
Hence dt = −9 unit s−1

(3, −8) ialah titik minimum dan −1, 8  ialah Kadar perubahan y ialah −9 unit s−1.
3
y menyusut.
The rate of change of y is –9 unit s–1.
titik maksimum.
y decreases.
 (3, –8) is a minimum point and –1, 8 is a maximum
3 ( b) DGiivbeenr i ddxt = 0.5 unit s−1 apabila x = 4.
point.
when

20. (a) Perimeter = 26 = 2y + 4x dy = (x – 1) – x
dx (x – 1)2
y = 13 – 2x

Luas A = xy + 1 = –1
2 (x – 1)2
  Area A
x2sin 60° 2

= x[13 – 2x] + x2sin60°   Maka dy = dy × dx
dt dx dt
Untuk A maksimum, dA = 0 Hence
dx = –1 dx
For a maximum A dt
(x – 1)2
dA
dx = 13 – 4x + 2x sin60° = 0 Apabila x = 4 dan dx = 0.5
dt
13 = x4 – 23  When and
2
Maka dy = –1 (0.5)
∴x = 13 dt (4 – 1)2
4– 3 Hence
= – 1 unit s−1
18

(b) (i) 4t + 8x + 4x = 360 t Kadar penyusutan y ialah 1 unit s−1
t + 3x = 90 x 18
I = 2x2t
= 2x2[90 – 3x] The rate of y decreasing is 1 unit s–1
= 180x2 – 6x3 18

(ii) ddxI = 360x – 18x2 = 0 2x 2 2. (a) Diberi dx = 8 unit s−1 apabila x = 3
x[360 – 18x] = 0 dt
Given when

x = 0 x = 20 dy = (x – 1)(8x) – 4x2
dx (x – 1)2
Dimensi 20 × 40 × 30 cm
dy dy dx
Dimension dt = dx dt
  Maka ×

x ED Hence
(c) (i) 10 = 20 = (x – 1)(8x) – 4x2 dx
dt
ED = 2x (x – 1)2
∴AE = 26 – 2x
Luas AEFG = x[26 – 2x] Apabila x = 3 dan dy =8
Area AEFG dt
When and

(3 – 1)(8)(3) – 4(3)2 dx
(3 – 1)2 dt
(ii) ddAx = 26 – 4x = 0   Maka 8 =

Hence

x = 13 cm dx = 8 unit s−1
2 dt 3

J12

Kadar perubahan x ialah 8 unit s−1. Jejari bertambah
3
The radius increases
x bertambah. 8 r
3 (c) h = 2r
The rate of change of x is unit s–1. V = πr2h
= πr2(2r)
x increases. V = 2πr3 h

(b) DGiivbeenr i ddyt = 8 unit s−1 apabila x = 2 ddVr = 6πr2

when

dy = 1 ddVt = 6πr2 dr
dx 2x – 3 dt

Maka dy = dy × dx –1.2= 6π(7)2 dr
Hence dt dx dt dt

= 1 3  dx  ddrt = –1.2 = –0.0013 cm s–1
2x – dt 6π(7)2

Apabila x = 2 dan dx =8 Jejari menyusut
dt
When and The radius decreases

1 dx  24. (a) Diberi y = 4x − 3
dt
Maka 8 = dy =4
dx
Hence

dx = 8 unit s−1 x berubah daripada 5 kepada 5.01
dt
x changes from 5 to 5.01
Kadar perubahan x ialah 8 unit s−1.
Jadi/So δx = 5.01 – 5 = 0.01

x bertambah. δδyx ≈ dy
The rate of change of x is 8 unit s–1. dx

x increases. dy
dx
23. (a) tan 30° = r = 1 δy ≈ · δx
h 3
≈ 4 δx
r = h ≈ 4 (0.01)
3 ≈ 0.04
1 Jadi, y bertambah sebanyak 0.04.
  Isipadu V = 3 πr 2h
So, y increases 0.04.
Volume 1 π h 2h
3 3 Apabila x = 5, y = 17
= Maka, nilai baharu y

V = 1 πh3 Hence, the new value of y
9
dV 1 = y + δy
dh = 3 πh2 = 17 + (0.04)

= 17.04

dV = 1 πh2 dh (b) Diberi y = 9
dt 3 dt x2
= – 1x8 3
Jika h = 4 cm, dV = –4 cm3 s–1 dy
dt dx

–4 = 1 π(4)2 dh x berubah daripada 3 kepada 2.98.
3 dt
x changes from 3 to 2.98.
– 43π = dh
dt Jadi, δx = 2.98 − 3 = −0.02

∴ dh = –3 cm s–1 r δδyx ≈ dy
dt 4π 30° dx

h δy ≈ dy · δx
dx
h menyusut – – 1(x138 38 )3·(δ−x0.02)
h decreases
≈
(b) L = πr2 ≈

ddLr = 2πr

ddLt = 2πr dr ≈ 0.013
dt Jadi, y bertambah sebanyak 0.013.

π = 2πr dr  So, y increases 0.013.
dt
Apabila x = 3, y = 1
1 = 2(3.5) dr Maka, nilai baharu y = y + δy
dt
dr 1 Hence, the new value of y = y + δy
dt = 7 cm s–1
= 1 + 0.013

= 1.013

J13

(c) Diberi y = x2(x + 2)   2. (a) y = x –3
x +1
dy = 3x2 + 4x
dx dy (x + 1) – (x – 3) 4
dx = (x + 1)2 = + 1)2
x berubah daripada 3 ke 2.99. (x
x changes from 3 to 2.99.
Apabila y = 0, x = 3 P(3, 0)
Jadi, δx = 2.99 – 3 = −0.01 When
dy 4 1
δδxy ≈ dy Pada P(3, 0), dx = (3 + 1)2 = 4
dx
At

δy ≈ dy · δx q = 1
dx 4

≈ (3x2 + 4x)δx 3. ( a ) ddyx = 0 pada titik B dan D.

≈ [3(3)2 + 4(3)](−0.01) at point B and D.

≈ −0.39 (b ) ddxy  0 pada titik C dan E.

Jadi, y menyusut sebanyak 0.39. at point C and E.
so, y decreases 0.39.
(c ) ddyx  0 pada titik A.
25. (a) (i) L = 2πj × 8 + πj2
at point A.
= π[16j + j2]
  4. y = x(x – 2)2
(ii) δj = 3.98 – 4 = –0.02 cm
8 cm ddxy = 2x(x –2) + (x – 2)2
dL = 16π + 2πj j = (x – 2)(2x + x – 2) = (x – 2)(3x – 2)
dj

Apabila j = 4 cm dan δj = –0.02 ddyx = (x – 2)(3x – 2) = 0

δL = dL . δj x = 2 ; x= 2
dj 3
y = 0
= [16π + 2π(4)])(–0.02) 2  2 – 22
3 3
= 24π(–0.02)

= –0.48π cm2 = 32
27
Luas menyusut sebanyak 0.48π cm2 d2y
dx2 = (x – 2)(3)+ (3x – 2)
The area decreases 0.48π cm2

(b) V = 6x2 – 4x = 6x – 8
ddVx = 12x – 4
Apabila x = 2 d2y = 12 – 8 = 4 > 0
δV = dV · δx δx = 0.01, x = 10 dx2
dx When

= (12x – 4)δx ∴(2, 0) ialah titik minimum
= (120 – 4)(0.01)
= 1.16 cm3 is a minimum point

 Apabila x= 2 , d2y = 6 2 –8
3 dx2 3
When
1
(c) T = 2π L = 2π = –4 < 0
10 L2
10   23 , 2372
ialah titik maksimum
dT = π 1 δT = 0.025
dL 10 L is a maximum point

δT = dT . δL Kertas 2 dy 2(1 + x2) – (2x – 4)(2x)
dL dx (1 + x2)2
1. (a) =
0.025 = π δL
10(0.8)
= 2[1 + x2 – 2x2 + 4x]
0.025 8 = δL (1 + x2)2

π δL = 0.0225 cm Apabila x = 2, dy = 2[1 + 4 – 8 + 8]
dx 25

Praktis SPM 2 dy = 2
dx 5

Kertas 1 dy (b) persamaan garis normal
dt
  1. x = 3t3 – 2, = 15t2 Equation of the normal line

dx y = – 5 x + c
dt 2
(a) = 9t2
5
dy dy dt 1 0 = – 2 (2) + c
dx dt dx 9t2
(b) = · = 15t2 × c = 5

= 5 ∴y = – 5 x + 5
3 2

J14

2. (a) Perimeter = 120 = 2πj + 2x (b) L = πr2 = π 2 h4 2
3
60 = πj + x 4
9
Luas, L = x(2j) A x F L = πh2
Area = 2j[60 – πj] E j
ddhL = 8
= 120j – 2πj2 B 9 πh

(b) dL = 120 – 4πj = 0 C xD ddLt = 8 πh dh
dj 9 dt

j = 120 = 30 cm   = 8 π[9] – 1
4π π 9 6π

d2L = –4π < 0 = – 4
dj2 3 cm2 s–1

∴ Luas ialah maksimum

The area is maximum 2. (a) dp = 10 unit s–1
dt
Apabila j = 30 cm y
π 1 y = x2
A= 2 [p + 1]p2
L = 120 30  – 2π 30 2 = 1800
π π π 1 1
A= 2 + 2 p2
Sudut KBAT p3 p2 P(p, 0) x

dV (b) dA = 3 p2 + p R(–1, 0) 0
dt dp 2
1. (a) = –6 cm3 s–1 10 cm

r = h r= 2 h r dA =  3 p + p · dp
10 15 3 h 15 cm dt 2 dt

V = 1 πr2h   dA 3 80 unit2
3 p = 2, dt = 2 (2)2 + 2 10 = s–1

= 1 π  2 h2h (c) p2 = PQ
3 3
d(PQ) = 2p
V = 4 π h3 dp
27
d(PQ) dp
ddVh = 4 π h2 dt = 2p · dt
9
d(PQ)
ddVt = 4 π h2· dh p = 1, dt = 2(1)(10)
9 dt

–6 = 4 π [9]2 dh = 20 unit s–1
9 dt

–6 = dh dh = – 1 cm s–1
36π dt dt 6π

J15

BAB 1 3

3 Pengamiran (d) ∫ 5x 2 dx = ∫ 5 x– 2 dx
2x2 2
Integration

= 5 × (–2) +c
2 x
(a) ddx  2 x6 =
1. 3 4x5 = – 5x + c

∫d 2 x6 = ∫4x5 dx (e) ∫ x dx = ∫ 1 1
3 3x 3
2 x 2 dx
3
∫4x5 dx = x6 + c 1 2
3 3
(b) 3∫g(x)dx = 3[4f(x)] + c = × x x +c

= 12f(x) + c = 2 x x +c
9
1 (2x – 5) 8x3
(c) ∫ 2 h(x)dx = (3 – 4x)3 (f) ∫ 4x5 dx = ∫2x–2dx

∫ 1 h(x)dx = 2  (2x – 5)  = –2 +c
a a (3 – 4x)3 x
3
(d) y = (x − 1)(x − 3)2 (g) 2 ∫ x2 dx = 2
3 3 ∫x 2 dx
ddxy = 2(x − 1)(x − 3) + (x − 3)2 1
= (x − 3)(2x − 2 + x − 3]
x2 = 2 × 2 x2 x +c
3 5


= (x − 3)(3x – 5) = 4 x2 x +c
15
∫(x − 3)(3x − 5)dx
3. (a) ∫2x2dx – ∫4xdx + ∫x–2dx
= (x − 1)(x − 3)2

(e) y = 2x = 2 x3 – 2x2 – 1 +c
x+1 3 x

= 2(x + 1) – 2x (b) ∫5x2dx – ∫ 10 dx
(x + 1)2 x2

= 2 = 5 x3 + 10 + c
(x + 1)2 3 x

∫ 2 dx = 2x (c) ∫ 3x2 dx – ∫ 4 dx
+ 1)2 (x + 1)
(x 2 2

∫ (x 6 dx = 3∫ (x 2 1)2 dx x3 x3
+ 1)2 + 4 – 2

= ∫3x 3 dx – ∫4x 3 dx
7 1
= 3(2x) = 3 3x 3  – 34x 3 
(x + 1) 7
7 1
= 6x = 9 x 3 – + c
(x + 1) 7 12x 3

(f) ∫f(x)dx = x+1 (d) = ∫2t–3dt – ∫6 dt + ∫3t –2dt
x2
= –t–2 – 6t – 3t–1

2π ∫ f(x)dx = 2  x + 1  = – t12 – 6t – 3 +c
π x2 t

2. (a) ∫ 3 x dx = 3 x2 +c (e) ∫(8x – 4x2 + 12 – 6x) dx
4 4 2
= ∫(2x – 4x2 + 12) dx
3 4
= 8 x2 + c = x2 – 3 x3 + 12x + c

(b) ∫ 2 dx = ∫ 2 x–3 dx (f) 8 ∫(x2 – 3x)dx
3x3 3 5

= 2 x –2 +c ( g) ∫=x85(x–(x3x35–)–(5x3)2+x21 )+dcx
3 –2

= – 31x2 + c

(c) ∫ x dx = ∫ 1 x –3 dx = ∫(x2 + x)dx
(16x4) 16 1 x2
= 3 x3 + 2 +c
1 1
= 16 (–2)x2 +c 4. (a) Katakan u = x – 5

= – 321x2 + c du = 1, dx = du
dx 1

J16

= ∫ 3 u3 du (c) ddxy = 2x3 + x2
4 1

= 3(x – 5)4 + c ∫dy = ∫(2x3 + x2)dx
16
1 1
(b) Katakan u = 3 – 4x y = 2 x4 + 3 x3 + c

du = –4, dx = du Apabila x = 1, y = –1
dx –4
–1 = 1 + 1 +c
= ∫2u–2 du = u– 2 + 1 +c 2 3
–4 –2(–1)
11
= 1 + c c = – 6
–
2(3 4x) Persamaan lengkung ialah

(c) Katakan u = x – 2 The equation of the curve is

du = 1, dx = du y = 1 x4 + 1 x3 – 11
dx 2 3 6
3
1 (d) ddxy = ax – 3
= du = 12u 2 +c
∫6u 2 3

= 4(x – 2) x – 2 + c ∫dy = ∫(ax – 3)dx
(d) Katakan u = 3 – x
y = a x2 – 3x + c
du 2
dx = –1, dx = –du
Apabila x = –1, y = 8

= ∫ –2 u–3du = –2u–3 + 1 + c 8 = 1 a+ 3+ c
3 3(–2) 2

= 1 x)2 + c 5 = 1 a+ c ...➀
3(3 – 2

(e) Katakan u = 1 – 2x Apabila x = 3, y = 4

du = –2, dx = du 4 = 9 a– 9+ c
dx –2 2
– 1 + 1
3 –1 13 = 9 a + c ...➁
–2 3u 2 2
= ∫ u 2 du = +c

= –3 1 – 2x + c –2 1  ➁ – ➀: 8 = 4a
2 a = 2, c = 4
Persamaan lengkung ialah
5. (a) dy = 3 – x2 – 4x
dx The equation of the curve is

∫dy = ∫(3 – x2 – 4x)dx y = x2 – 3x + 4
1
y = 3x – 3 x3 – 2x2 + c (e) ddxy = 3x2 + b

Apabila x = 0, y = 4 ∫dy = ∫(3x2 + b)dx

c = 4 y = x3 + bx+ c

Persamaan lengkung ialah Apabila x = 1, y = 3

The equation of the curve is 3=1+b+c

y = 3x – 1 x3 – 2x2 + 4 2 = b + c ...➀
3
Apabila x = –1, y = –3

(b) ddyx 1 –3 = –1 – b + c

= (x – 2) 2 –2 = –b + c ...➁
1
➁ – ➀: –4 = –2b
∫dy = ∫(x – 2) 2 dx
3 b = 2, c = 0

y = 2 (x – 2) 2 + c Persamaan lengkung ialah
3
The equation of the curve is

Apabila x = 6, y = 1 y = x3 + 2x

1= 16 +c 6. (a) 0 x(2x – 1)dx = ∫0–2(2x2 – x)dx
3
∫ –2

c = – 13  = 2 x 3 – 1 x2 0
3 3 2 –2

Persamaan lengkung ialah = 0 – –136 – 2

The equation of the curve is

2 3 13 22
3 3 3
y = (x – 2) 2 – =

J17

(b) 3 4x3 – 3 dx = ∫13(4x – 3x–2)dx (b) 2 2x–3dx = 3
x2 4
∫1 ∫k

 = 3 3   – x12 2 = 3
2x 2 + x 1 4
k
= (18 + 1) – (2 + 3)
= 14  – 14  –  – k12  = 3
4

(c) ∫1–2(x + 2x–3)dx 1 – 1 = 3
k2 4 4
  = 1 1
2 x 2 – x2 1 4 – k2 = 3k2
–2
4k2 = 4
1 1
=  2 – 1 – 2 – 4  k = ±1

= 3 –3 (c) ∫–412xdx – ∫4 g(x)dx
4 –1

14
(x – 2)2 –1
 
= –2 41 = [x2]4–1 –

(d) 4 2(2 – x)–3dx =(16 – 1) – 3 1 – 41 –1
4
∫0 9

2(2 – x)–2 4 5 14 3316
(–2)(–1) 0 36
  = = 15 – =

= 1 – 1 =0 (d) d2y = 1 – 6x2
22 22 dx2

(e) 2 (2x + 3)(2x – 3) dx ddyx = x – 2x3 + c
(2x – 3)
∫–2 dy
dx
= [x2 + 3x]–22 x = –1, = –1
= (4 + 6) – (4 – 6)
= 12 –1 + 2 + c = –1

c = –2

7. (a) (i) ∫ –33f(x)dx = ∫ –1 f(x)dx + ∫ 3 f(x)dx ddxy = x – 2x3 – 2
–3 –1
x2 1
= –1 + 6 = 5 y = 2 – 2 x4 – 2x + c1,

(ii) ∫ ––13 g(x)dx – 2∫ –1 f(x)dx x = –1 y = 2
–3

= 4 – 2(–1) 2= 1 – 1 + 2 + c1,
2 2
= 6

(iii) ∫ 0–3 f(x)dx + ∫ 3 f(x)dx c1 = 0
0
∴y = 1 x2 – 1 x4 – 2x
= ∫ 3 f(x)dx = –1 + 6 = 5 2 2
–3

  (iv) 12 ∫ 2 3 (e) dy = 4 x3 – 2x + c
–1 f(x)dx + ∫ 2 f(x)dx dx 3

= 1 [6] = 3 x = 0, dy = 4, c = 4
2 dx

  (v) ∫ 2––(3142)g–(x)kd2xx2 ∫– –1 kx dx = 20 y = 1 x4 – x2 + 4x + c1
–3 3

–1 = 20 x = 0, y = 1, c1 = 1

–3 ∴y = 1 x4 – x2 + 4x + 1
3
8 –  k – 9k  = 20
2 2 9. (a) (i) 7 f(x)dx = 8 + 4.5

4k = 12 ∫3 = 12.5 unit2
k = 3
(ii) ∫3–2 f(x)dx = −4.5 unit2
∫k–2(x – 4)dx = –10 (iii) Jumlah luas = 12.5 + 4.5
8. (a)
Total area = 17 unit2
 x2 k =
– 4x –10 (b) (i) Luas H = luas segiempat
2 –2 Area H = Area of rectangle

 k22 – 4k – (2 + 8) = –10 –∫ 0 f(x)dx
–4

k(k – 8) = 0 = 12 – 6 = 6 unit2

k = 0 atau k = 8 (ii) Luas K = 2 f(x)dx = 5

∫0

J18

(iii) ∫2–4 f(x)dx = 0 f(x)dx + 2 f(x)dx 7 31 – 2 65 = 1 0 16 unit2
  =
∫ –4 ∫0

= 6 + 5 = 11 unit2 ∴Jumlah luas +

(c) (i) Luas P/ Area P = 25 − Q Total area

= 25 − 18 (e) Apabila y = 0

= 7 unit2 4 – x2 = 0

5 x = ±2
1
(ii) Luas Q = ∫ x dy Luas dari x = –3 ke –2

Area Q 3 4∫ = x3
––23(4 – x2)dx = 4x – 3 –2
10. (a) Apabila –3

y = 0 = 31–8 + 8 2 – (–12 + 9)4
3
2 + x – x2 = 0

(2 – x)(1 + x) = 0 = – 2 31 unit2

x = 2; –1

B(2, 0) A(0, 2) luas dari x = 2 ke x = 4

  luas = 2 y dx = ∫20(2 + x – x2)dx 3 4= ∫42(4 – x2)dx = x3 4
∫ 0 – 4x – 3 2
area x3 2
= 2x + x2 30 = – 1 0 23 unit2
2
– 2 13  – 10 23 
= 4 + 2 – 8 4 Jumlah luas = +
3
Total area

= 10 unit2 = 13 unit2
3
1 1. (a) x = y2 – 4y

(b) luas = 3 ydx luas sebelah kiri/area on the left

area ∫1 = 4 x dy

1 ∫0
x2 = ∫40(y2 – 4y)dy
= 3 dx

∫1  y3 – 4  64 324
3 3
∫= 3 x –2dx = 2y2 0 = – – 0
1
= – 1 0 23
= – 1 43 = – 1 4 – [–1]
x 1 3
 4 y3
=1– 1 = 2 unit2 ∫0–1(y2 – 4y)dy = 3 – 2y2 0
3 3 –1

(c) y = (x2 + x)(2 – x) = 0 – 1 – 31 – 22 = 2 13

= –x3 + 2x2 + 2x – x2 – 1 0 32 + 2 13
 


y = –x3 + x2 + 2x Jumlah luas =

  luas = Total area

area
∫0–1(–x3 + x2 + 2x)dx + ∫02(–x3 + x2 + 2x)dx = 13 unit2

    = –x4 + 1 x3 + x2 0+ –x4 + x3 + x2 2 1
4 3 4 3 0
–1 (b) x = y 2

  = – 152 8 luas = 2∫ 4 x dy
3 1

+ [–4 + + 4] area 4 1
1
5 8 = 3 112 = ∫2 y2 dy
12 3
= + unit2 3 4 = 2 2 3 4
3 1
(d) Apabila y = 0 y2

(x + 2)(3 – x) = 0 = 231 2 (8) – 2 (1)
3 3
x = 3; –2
3 4 14 28
P[3,0] =2 3 = 3 unit2

Luas di atas paksi-x/Area above the x-axis 4
y
= 3 (x + 2)(3 – x)dx = 3 (6 + x – x2)dx (c) x2 =

∫1 1 4 3x3 ∫1 9 1 1
2 2 2 3
= 6x + x2 – 3 3= 18 + – 94 – 6 + – 4 x = 2 – 1

1 y = 2y 2

= 7 13 unit2 laureaas = ∫ 212 2y – 21 dy

Luas dari x = 3 ke x = 4

 4 1 x3 = 4 4y 2
∫43(6 + x – x2)dx = 6x + 2 x2 – 3 4 1
3
2
64 9
= 24 + 8 – 3 4 – 9 + 2  = 4 2 – 4 = 8 – 4 = 4
22 2
= –2 65
= 2 2 unit2

J19

12. (a) x2 + 1 = –x + 3 y Isipadu dijanakan
x2 + x – 2 = 0 y = x2 + 1
(x – 1)(x + 2) = 0 A The generated volume
x = 1; –2
y = 2 1 y = –x + 3 = π∫ 3 y 2dx
A[1, 2] 1

= π∫ 3 x2(x – 3)2dx
1

= ∫π 3 x 4 – 6x3 + 9x2)dx
1

luas = 1 ydx + ∆ABC O dx B C x  = π 1 x5 – 3 x4 + 3x3 3
1 3 5 2 1
area ∫0

= 1 (x 2 + 1)dx + 1 (2)(2) = π( 35 13 + – 1 – 3 + 324
2 5 5 2
∫0 2
– (3)4 3(3)3)

3 4 = x3 +x 1 + 2 = 6 52 π unit3
3 0

= 1 1 + 12 + 2 (c) Apabila y = 0, x = –4
3
Isipadu dijanakan
= 3 31 unit2
The generated volume

(b) Titik persilangan ∫= π 0 y2dx
–4

Intersection point = 1∫π 0 x + 4 2 dx
–4 2
x2 = x
 = π 1 x2 + 2x 0
x4 = x 4 –4

x(x3 – 1) = 0 x = 0; x = 1 = π0 – (4 – 8)4

y=0y=1

luas berlorek = 1 x dx – ∫1 x2dx = 4π unit3
– 0
shaded area ∫0 3 14. (a) Apabila/When x = 0, y = 4
1
 4  4 = 2 x2 1 3 x3 1 Isipadu/Volume = 10 π
3 0 0 3

= 2 – 1 = 1 unit2 = π∫ 4 x2dy
3 3 3 k

(c) Titik persilangan/Intersection point = 1π∫ 4 16 – y2 2 dy
k 4
(1 – x)(x + 4) = x + 4
1 10
x + 4 – x2 – 4x = x + 4   = 12 4= 3
π 4y – y3 π
x2 + 4x = 0 k

x(x + 4) = 0 116 – 64 2 – 4k + k3 = 10
12 12 3
x = 0; x = –4
k3 22
y = 4 ; y = 0 12 – 4k + 3 = 0

luas berlorek = luas segi tiga + 1 f(x)dx k3 – 48k + 88 = 0

∫0

shaded area = area of triangle + 1 f(x)dx Jika/If k = 2, 8 – 48(2) + 88 = 0

∫0 ∴k = 2
1
= 2 (4)(4) + ∫10(4 – 3x – x2)dx (b) Apabila/When x = 0, y = k

 4= 8 + 3 x3 1 Isipadu/Volume = 2π
4x – 2 x2 – 3 0
2π = π∫k4 (y – k)dy
= 8+ 34 – 3 – 1 4
2 3 ∫4 (y – k)dy = 2
k
= 1 0 61 unit2
 y2 – ky 4 = 2
k
2
13. (a) Isi padu janaan k2
(8 – 4k) – 1 2 – k22 = 2
The generated volume

= π∫ 3 y2dx k2 – 4k + 6 = 0
1 2

= π∫ 3 1 dx k2 – 8k + 12 = 0
1 x2
(k – 2)(k – 6) = 0
∫= π 3 x –2dx
1 ∴k = 2 k = 6

= π – 1x 3 (c) Isipadu = 1352π
1
Volume
= π – 31 –(–1)4
= π∫1k (y + 3)2dy

2 (y + 3)3 1352π
3 3
= π unit3 3 4 π k =
1

(b) Titik persilangan dengan paksi-x (3, 0) ,(0, 0) (k + 3)3 – 43 = 152
The intersection point at the x-axis of (3, 0), (0, 0) 3 3 3


J20

(k + 3)3 = 216 Isipadu diperlu = 73 π unit3
60
k + 3 = 6 Volume needed

k = 3 16. (a) (i) Dengan teorem Pythagoras,

15. (a) Isipadu dijanakan oleh lengkung x2 + y2 = r2

Volume generated by the curve y

= π∫20(6 + 7x – 3x2)2 dx P(x, y)

= π∫20(36 + 84x + 13x2 – 42x3 + 9x4)dx ry

= 164.27π atau 164 145π Ox x

Isipadu kon

Volume of cone (ii) Isi padu janaan/Generated volume

= 1 π(8)2(2) = 2π∫ r0x2dy
3
= 42.67 π atau 4 2 32 π
2π∫ r0(r2 – y2)dy

2π r2y – y3 r
Isipadu janaan/Generated volume  4 = 30

= 121.6 π unit3 atau 1 21 53 π unit3
(b) Koordinat A/Coordinates of A =

= 2π 3r3 – r3 4
3

4 =x = 2π 3 2 r34 = 4 πr 3
x 3 3

x = 2, y = 2 (b) Persamaan garis lurus

Isipadu/Volume The equation of straight line

= π∫ 3 x2dy + kon y = h x + h
2 r

= π∫ 3 16 dy + 1 π(2)2(2) Isipadu dijana/Generated volume
2 y2 3
= π∫ h0x2dy
 –16  3 8  4
= π y 2 + 3 π = π∫ h r(y – h) 2dy
0 h


= π–136 – –12624 + 8 π = r2π ∫ h (y – h)2dy
3 h2 0

= 5 31 π unit3 r2π (y – h)3 h
h2 30
 4 =

(c) y = 9 – x2 = r2π 30 – 1 –h3 24
h2 3
dy = –2x
dx r2π h3 1
dy 3 4 = h2 3 = 3 πr2h unit3
x = 2, dx = –4

persamaan tangen pada (2, 5) (c) y = 14 – 4 x 2
9
The equation of tangent at (2, 5) 14
∫π 10 x 2dy
y – 5 = –4(x – 2)
9
y = –4x + 13 13 = 4 ∫π 1140(14 – y)dy
4
Tangen menyilang paksi-x pada 1 , 02. 3 4 9 y2 14
4 2 10
1 2 13 , 0 = π 14y –
Tangent intersects the x-axis at 4
9
1 2 Volume of cone 1 5 = 4 π[(196 – 98) – (140 – 50)]
Isipadu kon = 3 π(5)2 4
9
= 11225π unit3 = 4 π[8] = 18π unit3

Isipadu dijanakan oleh lengkung Isipadu kon/Volume of cone = 1 π(3)2(10) = 30π
3
Volume generated by the curve
Jumlah isipadu/Total volume = 18π + 30π =
= 3
π∫ 2 y2dx 48π unit3

= π∫32(9 – x2)2dx Satu bekas isipadu = 20 × 30 × 40 = 24000 cm3

Volume of a container

= π∫32(81 – 18x2 + x4)dx Dua bekas/Two containers = 48000 cm3

3 4 = π 1 3 Bilangan kon = 48000 = 318 kon
81x – 6x3 + 5 x5 2 48π
Number of cone

= 9 51 π unit3 Jumlah jualan/Total sales = 318 × RM3.50

= RM1113

J21

Kos/cost = RM345 × 2 = RM690 7. (a) ∫ 2 p(x)dx = – 3
Keuntungan/Profit = RM423 h 4

Praktis SPM 3 (b) ∫ h p(x)dx – ∫ h 2xdx = –4 1
2 2 4

Kertas 1 3 +4 1 = [x2]h2 = h2 – 4
4 4

1. (a) ∫ 4 f(x)dx = 5 5 + 4 = h2
–2
h = 3
–2 f(x)dx = –5
∫ 4
Kertas 2
 4(b) ∫ 4 2f(x)dx –∫ 4 3x dx
–2 –2 1. (a) y = 3x2 – 27
x2
3 4 Apabila y = 0
= 10 – 2
–2 x = ±3

= 10 – [24 – 6] = –8 dy = 6x
dx
2. (a) dy = 2x – 4 Apabila ddyx = 6x = 6
dx
dy
Apabila x = 3, dx = 2(3) – 4 x = 1

=2 y = –24

(b) y = ∫(2x – 4)dx ∴P[1, –24]

= x2 – 4x + c (b) Persamaan tangen y + 24 = 6(x – 1)

Pada/At (3, –2), –2 = 9 – 12 + c Equation of tangent

c = 1 y = 6x – 30

∴y = x2 – 4x + 1 luas berlorek

3. (a) y shaded area
1
= 2 [4][24] – ∫3 ydx
1

(h, 6k) = 48 – ∫31(3x2 – 27)dx

(0, 3k) = 48 – [x3 – 27]31
∴48 – [0 + 26] = 22 unit2
–h O h x

1 2(c) = 0 = 2 110
k 3

h h Isipadu π∫ x2 dy π

(b) ∫0 f(x)dx = 8 + ∫0 g(x)dx Volume

=8+ 1 [3k + 6k]h ∫ 01 1 y + 92dx = 220
2 3 3
k

= 8 + 9hk  4 1 y2 + 9y 0 = 220
2 6 3
k

4. ∫h–1(3 – 2x)dx = 6 0 – [ k2 + 9k] = 220
6 3

[3x – x2]h–1 = 6 k2 + 54k + 440 = 0

3h – h2 – (–3 – 1) = 6 (k + 44)(k + 10) = 0

3h – h2 – 2 = 0 k = –44 ; –10

h2 – 3h + 2 = 0 ∴k = –10

(h – 1)(h – 2) = 0 2. (a) 4y = x + 3 ; x = 4y2 – 2

h = 1; 2 4y – 3 = 4y2 – 2

b 4y2 – 4y + 1 = 0

5. (a) ∫a f(x)dx = 8 (2y – 1)2 = 0

a = 1, b = 3 y= 1 , x = 2 – 3 = –1
2
( b) 3 f(x)dx = 1 f(x)dx + ∫ 3 f(x)dx 1
1 A–1, 2 2
∫ –2 ∫ –2

= –5 + 8 = 3

6. f’(x) = 4x – 4 (b) Apabila/When y = 0, x = –3

f’(x) = 4x – 4 = 0 luas berlorek/shaded area y

x = 1  2 = 1 (2) 1 ∫– –1 y dx
2 2 –2
∴(1, –4) ialah titik minimum ΂–1, 12΃
1 1 1
(1, –4) is a minimum point 2 2 –3 –2 –1 O
∫– –1 2
f(x) = ∫(4x – 4)dx = –2 (x + 2) dx
x
= 2x2 – 4x + c
3 4 = 3 –1
Pada/At (1, –4) 1 – 1 . 2 (x + 2) 2 –2
2 2 3
–4 = 2 – 4 + c

c = –2 = 1 – 3 1 – 04 = 1 unit2
2 3 6
f(x) = 2x2 – 4x – 2

J22

(c) Isipadu/Volume = ∫π 1 y2dx Sudut KBAT
–2
1
= ∫π 1 4 (x + 2)dx (a) y = x3 + 1
–2
Apabila y = 0 ; 0 = x3 + 1
(x + 2)2
8 x3 = –1
3 4 = π 1
–2 x = –1

= π 3 9 – 04 = 9π Pdd[xy–1=, 0]
8 8 3x2

Isipadu sebenar

Actual volume Apabila x = –1, dy = 3
9π When dx
= 16 unit3

3. (a) y = 6x – x2 y y = 3(x + 1)

ddyx = 6 – 2x = 0 (3, 10) Jika y = 3x + 3 menyilang lengkap
x = 3 (4, 8) If y = 3x + 3 intersects completely

x 3x + 3 = x3 + 1
34
x3 – 3x – 2 = 0

O Apabila x = 2 23 – 3(2) – 2

=8–8=0

Katakan persamaan tangen ∴y = 23 + 1 = 9

Let the equation of tangent Q(2, 9) ∫2 y
–1
y = mx + c (b) luas A = ydx Q
(2, 9)
10 = 3m + c ...➀ area A ∫2–1(x3 + 1)dx
Rx
6x – x2 = mx + c = 2

x2 + (m – 6)x + c = 0 3 4= x4 +x2
4 –1
(m – 6)2 – 4c = 0 P
1 1 12 –1O
(m – 6)2 = 4(10 – 3m) = (4 + 2) – 4 –

m2 – 12m + 36 = 40 – 12m 3
4
m2 = 4 =6+

m = ±2 = 6 34 unit2

Apabila m = –2, c = 10 – 3(–2)

= 16 1 27
2 2
∴6 – 2x = –2 luas/area PQR = (3)9 = unit2

8 = 2x 27 6 34 = 6 43
2
x = 4; y = 6(4) – 16 ∴luas/area B = – unit2

= 8

∴A(4, 8) ; a = 4 Nisbah A : B = 6 34 : 6 34
Ratio
(b) luas berlorek = luas trapezium – 4 ydx
=1:1
∫3

shaded area = area of trapezium – 4 ydx
∫ 3 π 2 = π ∫2–1(x3 + 1)2dx
3 4(c) Isipadu = 2 y2dx 2
1 ∫ –1
2 ∫4 Volume
[10 + 8][1] – 3 (6x – x2)dx =
3 4 = π 1 x7 + 1 x4 + x 2
1 4 2 7 2 –1
9– 3 3
=
3x2 – x3 = π 3 28 1143 4
2
= 9 – 326 2 – 184 = 1 unit2
3 3 = 1 4 1238 π unit3

J23

BAB (c) Bilangan cara = (10 − 1)!

4 Pilih Atur dan Gabungan Number of ways = 362 880

Permutation and Combination 7. (a) (i) Bilangan cara = 2 × 4! = 48

Number of ways

1. (a) (i) n! = 1 (ii) Bilangan cara = 4! × 2! = 48
– 0)!
(n Number of ways

(ii) (n – n! = n! (iii) Susunan dalam bentuk VKVKV
n+
1)! Arrangement in the form of VKVKV

(b) (i) 10! = 720 Bilangan cara = 2! × 3! = 12
(10 – 3)!
Number of ways

(ii ) (51)!! 6! (b) (i) Bilangan cara = 1 × 4! = 24
(4)!
× = 5 × 6! Number of ways
= 3 6000
(ii) Bilangan cara = 1! × 2! × 2! × 2! = 8

(c) (i) n! (n – 1)! Number of ways
(n – n + 1)! (n – 1 – 2)!
× (c) (i) 5! × 2 = 240

= n!(n – 1)(n – 2) (ii) 4! × 4! = 576

(ii) (mm–!2)! ÷ (m + 1)! 5! × 6P2 = 3 600
(m + 1 – 2)! (d) (i) 6 × 6 × 6 x 6 × 3 = 3 888

(m + 1)! (ii) 3! × 3 × 2 = 36
(m – 1)!
= m(m – 1) ÷ 8. (a) (i) Bilangan cara menyusun = (10 − 1)!

m–1 Number of ways to arrange = 362 880
m+1
= (ii) Bilangan cara menyusun

2. (a) Bilangan cara Number of ways to arrange

Number of ways = (5 − 1)! × 6!

=4×3 = 17 280

= 12 (iii) Bilangan cara menyusun

(b) Bilangan cara Number of ways to arrange

Number of ways = (10 − 1)! − (7 − 1)! × 4!

=5×6 = 345 600

= 30 (b) (i) Bilangan cara = (8 −1)! × 2!

(c) Bilangan cara Number of ways = 10 080

Number of ways (ii) Bilangan cara menyusun = (7 − 1)! × 3!

=6×4×3 Number of ways to arrange = 4 320

= 72 (c) (i) Bilangan cara menyusun = (5 − 1)! × 4!

3. (a) Bilangan cara = 7! = 5 040 Number of ways to arrange = 576

Number of ways (ii) Bilangan cara menyusun = (4 − 1)! × 4!

(b) Bilangan cara = 8! = 40 320 Number of ways to arrange = 144

Number of ways 9. (a) (i) Sebelah kiri = r! nCr
Let side =
(c) Bilangan cara = 6! = 720 r!n!
(n − r)!r!
Number of ways

4. (a) Bilangan cara = 10P4 = 10 × 9 × 8 × 7 n!
Number of ways = 5 040 = (n − r)!

(b) Bilangan cara = 9P3
Number of ways = 504 = nPr Sebelah kanan

(c) Bilangan cara = 6P3 Right side
Number of ways = 120
Maka/Hence, r! nCr = nPr

5. (a) Bilangan cara susunan (ii) nC0 = (n n!
− 0)!0!
Number of ways of arrangement
9!
= 4!3!2! = 1 260 = n! = 1
(n)!
(b) Bilangan cara susunan
= sebelah kanan
Number of ways of arrangement
Right side
15!
= 5!6!4! = 630 630 (b) (i) 8C2 = 8!
– 2)!2!
(c) Bilangan cara susunan (8

Number of ways of arrangement = 8!
(6)!2!
9!
= 3!3!2! = 5 040 = 28

6. (a) Bilangan cara = 7! Sudut Kalkulator

Number of ways = 5 040 Tekan
(b) Bilangan cara = 5! 8 SHIFT nCr 2 = 28

Number of ways = 120

J24

(ii) 4C2 × 9C4 (c) Bilangan cara = 2 [3C1 × 2C1 × 6C3 × 3C3]
Number of ways = 240
= 4! × 9!
(4 – 2)!2! (9 – 4)!4! (d) (i) Bilangan cara = 23C5 = 33 649

= 756 Number of ways

(c) (i) 10C7 ÷ 6C3 = 10! ÷ 6! (ii) Bilangan cara = 10C1 × 7C1 × 6C1 × 20C2
(10 – 7)!7! – 3)!3! Number of ways = 79 800

(6 (iii) Bilangan cara = 17C5 = 6 188

= 6 Number of ways

(ii) nCn – 2 (e) (i) Jika memilih baris tiga kerusi.

= n! If choose the row of three chairs
2)!(n
(n − n + – 2)! Bilangan cara = 5C1 × 4C4 = 5
Number of ways
= n!
(2)!(n – 2)! Jika memilih baris empat kerusi

n(n – 1)(n – 2)! If choose the row of four chairs
(2)!(n – 2)!
= Bilangan cara = 5C2 × 3C3 = 10
Jumlah = 15
n(n – 1)
= (2) (ii) 1 cara sahaja

Only 1 way

10. (a) Bilangan cara pilihan = 8C4 (f) (i) Bilangan cara = 7C3 = 35

Number of ways = 70 Number of ways

(b) Bilangan cara pilihan = 12C5 (ii) Bilangan cara = 5C4 = 5

Number of ways = 792 Number of ways

(c) Bilangan cara pilihan = 7C4 (iii) Bilangan cara = 7C3 × 2 = 70

Number of ways = 35 Number of ways

11. (a) (i) Untuk melukis satu garisan, kita Praktis SPM 4

memerlukan mana-mana dua titik sahaja.

To draw a straight line, we neeed any two points Kertas 1

only. 1. (a) 1 ≤ n < 7 dengan n ialah integer

Maka, bilangan cara = 6C2 = 15. (b) xCm = xCn

So, number of ways x! x!
m)!m! − n)!n!
(ii) Untuk melukis satu segi tiga, kita (x − = (x

memerlukan mana-mana tiga titik sahaja. n + m = x

To draw a triangle, we need any three points

only. 2. 4! – 3! = 6
2!
Maka, bilangan cara = 6C3 = 20.

So, number of ways 3. Kes 1: Baju biru

(iii) Untuk melukis satu segi empat, kita Case 1: Blue blouses

memerlukan mana-mana empat titik 3C1 × 2C1 = 6 cara
Kes 2: Seluar biru
sahaja.

To draw a rectangle, we need any four points Case 2: Blue pants

only. 2C1 × 2C1 = 4 cara
Kes 3: Bukan baju biru dan seluar biru
Maka, bilangan cara = 6C4 = 15.

So, number of ways Case 3: Not blue blouses and blue pants

(b) (i) Bilangan cara = 4C1 × 5C4 = 20 2C1 × 2C1 = 4 cara
Jumlah cara = 14 cara
Number of ways

(ii) Kes 1: 3 perempuan, 2 lelaki Total ways

Bilangan cara = 5C3 × 4C2 = 60

Number of ways Sudut KBAT

Kes 2: 4 perempuan, 1 lelaki (a) Kes 1: nombor berdigit 4 = 3 × 5P3 = 180
Kes 2: nombor berdigit 5 = 5 × 5P5 = 600
Bilangan cara = 5C4 × 4C1 = 20 Kes 3: nombor berdigit 6 = 5 × 5P5 = 600
Jumlah = 600 + 600 + 180 = 1380
Number of ways (b) 2 × 4 × 3 + 2 × 5 × 4 × 3 = 144
(c) 5P4 × 1P1 = 120
Kes 3: 5 perempuan, 0 lelaki

Bilangan cara = 5C5 = 1

Number of ways

Jumlah cara = 81

(iii) Bilangan cara = 6C2 = 15

Number of ways

J25

BAB (b) X = {0, 1, 2, 3}

5 Taburan Kebarangkalian 3

Probability Distribution 10 B

1. (a) Pemboleh ubah ialah menang, tewas atau seri 3B
10 7 B’
The variable is win, lose or draw
10
(b) Pemboleh ubah ialah tinggi pelajar itu. B3

The variable is the height of a student. 3 7 10 B
B’
(c) Pemboleh ubah ialah {2, 3, 4, 5, 6, 7, 8, 9, 10, 10 10 7 B’

11, 12}. 10

The variable is 3

2. (a) X = {0, 1, 2, 3, 4,… 20} adalah diskret kerana 10 B

dapat dikira. 7 3 B
10 10 7 B’
X = {0, 1, 2, 3, 4,… 20} is discrete because they can 10
be counted. B’ 3

(b) Y = {y: 1 21 ≤ y ≤ 3} adalah selanjar kerana 7 10 B
sudah menentukan dengan tepat masa yang B’
10 7 B’
sewajar.
10
Y = {y: 1 12 ≤ y ≤ 3} is continuous because it has
determined by the time accordingly. P[X = 0] = P[B’, B’, B’] = 1703

(c) Z = {z: 3 ≤ z ≤ 8} adalah selanjar kerana = 343
1000
sudah menentukan dengan tepat masa yang
P[X = 1] = 3C11301702
sewajarnya.
= 441
Z = {z: 3 ≤ z ≤ 8} is continuous because it has 1000
determind by the time accordingly.
1 3C23 2 170
(d) X = {0, 1, 2, 3, ... , n} adalah diskret kerana 10

dapat dikira.

X = {0, 1, 2, 3, ... , n} is discrete because they can be
counted.

3. X = {0, 1, 2}

2 P[X = 2] =

2 3H = 189
3 1000
H1 M
P[X = 3] = 11303
3

2 = 27
1 3H 1000
M
3 1 M

3 X=x 0 1 2 3
P[X = x]
P[X = 0] = P[M, M] 343 441 189 27
1000 1000 1000 1000

= 1 × 1 = 1 4. (a) (i) P(X = 0) = 0 =0
3 3 9 6

P[X = 1] = P[H, M] + P[M, H] P(X = 1) = 1
6
= 2 1 + 1 2
3 × 3 3 × 3 P(X = 2) = 2 = 1
6 3
4
= 9 P(X = 3) = 3 = 1
6 2
P[X = 2] = P[H, H]

= 2 2 = 4 X=x 0123
3 3 9
× P(X = x) 0 1 1 1
632
X=x 0 1 2

1 4 4 P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0 + 1
P[X = x] 999 1 + 1 + 2
6 3

= 1 (Terbukti)/(Proved)

J26

(ii) P(X = x) ∑ 7 P(Y = y) = 0.1 × 4 + 0.2 × 3
0
= 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

1 (ii) P(3 < Y < 5) = P(Y = 4)
2
= 0.2
4
1 (c) (i) P(Z > 2) = P(Z = 3) + P(Z = 4) = 9
3
4
1 q + q = 9
6
q = 2
x 9
5
00 1 2 3 P (Z = 0) + P(Z = 1) + P(Z = 2) = 9

p + p + p + q = 5
9
(b) (i) P(X = 1) = (1 + 1)2 = 2
54 27 3p + 2 = 5
9 9
P(X = 2) = (2 + 1)2 = 1
54 6 3p = 3
9
P(X = 3) = (3 + 1)2 = 8
54 27 = 19

P(X = 4) = (4 + 1)2 = 25
54 54
(ii) z 0 1234
X=x 1234 P(Z = z)
1 1122
P(X = x) 2 1 8 25 9 9399
27 6 27 54
(iii)
P(Z = z)

P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 2 + 1 + 8 + 25
27 6 27 54
3
= 1 (Terbukti)/(Proved) 9

(ii) P(X = x)

2
9

25 1
54 9

16 z
54 0 01234

9
54
6. (a) Ini ialah percubaan Bernoulli, kerana
4
54 kebarangkalian mendapat bola merah setiap
5
x kali ialah tak bersandar, iaitu tetap = 8 dan
0 1234

percubaan ini berlaku 4 kali, n = 4.
This is Bernoulli’s trial, because the probability of
5. (a) (i) P(X ≤ 6) = 1 − P(X = 8)
getting a red ball every time is not dependent, that is
5
= 1 − 0.27 = 0.73 fixed = 8 and the trial occurs 4 times, n = 4.

(ii) P(2 ≤ X ≤ 6) = 1 − P(X = 8) (b) Ini ialah percubaan Bernoulli, kerana terdapat

= 0.73 dua kesudahan sahaja, dengan p = 0.85 dan

(b) percubaan ini berlaku 10 kali, n = 10.

Y 1234567 This is a Bernoulli trial because there are two

P(Y = y) 0.1 0.2 0.1 0.2 0.1 0.2 0.1 outcomes, where p = 0.85 and this trial occurs
10 times, n = 10.

J27

(c) Ini bukan percubaan Bernoulli, kerana P (X = x)

kebarangkalian mendapat A pertama kali,

p= 3 , kedua kali p = 2 dan ketiga kali p = 1
6 5 4

dengan setiap kali p berubah, iaitu bersandar. 0.30
This is not Bernoulli trial, because the probability to

get A the first time, p = 3 , the second time p = 2 and 0.25
6 5
1
the third time p = 4 , where each time p changes,
0.20
that is dependent. p

7. (a) n = 14, p = 0.05, q = 0.95 0.15

(i) P(X = 4) 0.10

= 14C4(0.05)4(0.95)10 = 0.0037 0.05
(ii) P(X < 12) = 1 − P[X = 12] − P(X = 13) + 0

P(X = 14) x

= 1 − (14C12(0.05)12(0.95)2 + 01 2345
14C13(0.05)13(0.95)1 + 14C14(0.05)14)
1 1
= 1 2 2

(iii) P(5 ≤ X ≤ 7) = P(X = 5) + P[X = 6) + P(X = 7) (b) n = 4, p = , q =

= [14C5(0.05)5(0.95)9 + 14C6(0.05)6(0.95)8 + P(X = 0) = 4C0 1 20 1 24 = 0.0625
14C7(0.57)(0.95)7] = 0.00043 2 2
(b) n = 8, p = 0.9, q = 0.1
1 1
(i) P(X = 8) = 8C8(0.9)8 = 0.4305 P(X = 1) = 4C1 2 21 2 23 = 0.2500
Maka, 43.05% dijangkiti Covid-19.
Hence, 43.05% contracted Covid-19. 1 1
P(X = 2) = 4C2 2 22 2 22 = 0.3750
(ii) P(X > 6) = P(X = 7) + P(X = 8)

= 8C7(0.9)7(0.1)1 + 8C8(0.9)8 P(X = 3) = 4C3 1 23 1 21 = 0.2500
= 0.8131 2 2

Maka, 81.31% dijangkiti Covid-19. P(X = 4) = 4C4 1 24 1 20 = 0.0625
Hence, 81.31% contracted Covid-19. 2 2

(c) n = 10, p = 0.95, q = 0.05 P (X = x)

(i) P(X > 8) = P(X = 9] + P(X = 10)

= 10C9(0.95)9(0.05)1 + 10C10(0.95)10 0.4
= 0.9139

(ii) P(2 ≤ X ≤ 8)

= 1 − P(X = 0) − P(X = 1) − P(X = 9) −

P(X = 10) 0.3

= 0.0861

(d) p = 2 , q = 1 0.2
3 3

(i) P(X = 0) = 0.0014

 = nC0 2 20 1 2n = 0.0014 0.1
3 3
1
n log  3 2 = log 0.0014

n = log0.0014 = 6 0 x

log 1 01 234
3

(ii) P(X > 4) = P(X = 5) + P(X = 6) 9. (a) (i) n = 3, p = 1
6
= 6C5 2 25 1 2 + 6C6 2 26 1 5
3 3 3 P(mendapat dua ‘5’) = 3C2 6 22 6 2

= 0.3512 = 5
72
8. (a) 0.20 + 0.25 + p + p + 0.15 + 0.05 = 1

0.65 + 2p = 1 (ii) E(X) = np = 216 5 = 15 kali
72
2p = 0.35 ×

p = 0.175 (b) (i) E(X) = np = 3

n 134 2 = 3

n = 14

J28

(ii) Varians = npq = 14 × 3 × 11 = 2 5 (ii) E(X) = np
14 14 14
= 550 × 0.0322

Sisihan piawai = 1.54 = 17.68

Standard deviation (c) n = 5, p = 0.8, q = 0.2

(b) (i) q = 0.35, p = 0.65 (i) P(X = 3) = 5C3(0.8)3(0.2)2
= 0.2048
Sisihan piawai = 1.5 = n(0.65)(0.35)
(ii) P(X < 2) = P(X = 0) + P(X = 1)
Standard deviation

n = 1.52   = 0.00672
(0.65)(0.35)
(iii) E(X) = np
= 10
= 50 × 0.8
(ii) Min/Mean = np
= 40
= 10(0.65)
11. (a) (i) µ = 32
= 6.5
(ii) P(X > 32) = 0.5
10. (a) p = 0.02, q = 0.98
(iii) P(23 < X < 32) = 0.5 – 0.16
P(X ≥ 1) > 0.95
= 0.34
1 – P(X = 0) > 0.95
(b) (i) P(X > 54) = 1 – 0.26
1 − nC0(0.02)0(0.98)n > 0.95
0.98n < 0.05 = 0.74

n log 0.98 < log 0.05 (ii) P(54 ≤ X ≤ 76) = 1 − 2(0.26)

= 0.48

n > log0.05 (c) f(x)
log0.98

n > 148.28
n = 149

(b) (i) P(X = 3) = 5C3 1 23 5 22 0 x
6 6
2.5 2.8 3.0

= 0.0322

12.

Pemboleh ubah X Min, µ Sisihan piawai, σ Skor-z

Variable X Mean, µ Standard deviation, σ z-score

(a) 0.8 = X − 67 67 15 0.8

X = 15(0.8) + 67
= 79

(b) 4.6 −2.1 = 4.6 – µ 0.9 −2.1
0.9

µ = 4.6 + 2.1(0.9)

= 6.49

(c) 130 143 –1.85 = 130 − 143 –1.85
σ
–13
σ = –1.85

= 7.03

13. (a) P(Z < –0.4) = P(Z > 0.4) Jumlah luas di bawah graf ialah 1.

= 0.3446 Total area under the graph is 1.

f(z) f(z)

–0.4 0 z z

Graf ialah simetri pada f(z) = 0. –1.2 0 f(z)

Graph is symmetrical about f(z) = 0. (c) 1 – P(Z > 1.5) – P(Z > 2.1)
= 1 – 0.0668 – 0.0179
(b) P(Z > –1.2) = 0.9153
= 1 – P(Z ≥ 1.2)
= 1 – 0.1151 –1.5 0 2.1 z
= 0.8849

J29

14. (a) skor-z = –1.78 ∴ k – 40.2 = 0.83
12.5
f(z)
k = 50.58

f(z)
0.3387

z0 z 0.2033

(b) 1 – P(Z ≤ z) = 0.9126 z
P(Z ≤ z) = 0.0874 –0.416 0 0.83
skor-z = –1.357
Luas adalah lebih daripada 0.5, maka skor-z (d) P 30 – 35 <Z< 40 – 35  = 0.578
k k
ialah negatif.
P– 5 < Z < 5 2 = 0.578
Area is more than 0.5, so z-score is negative. k k

f(z) 1 – 2P(Z ≥ 5 ) = 0.578
k

P1Z ≥ 5 2 = 0.211
k

z0 z 5 = 0.803
k

(c) 1 – P(Z > z) = 0.8251 k = 6.23
P(Z > z) = 0.1749
skor-z = 0.935 0.578 f(z)
0.211

Luas lebih daripada 0.5, maka skor-z ialah 5 0 5 z
positif. k k
–
Area is more than 0.5, so z-score is positive.
16. (a) (i) µ = 53, σ = 9
f(z)

P[42 < X < 58]

= P 42 – 53 < Z < 58 – 53 4
9 9
z
0z P– 11 5 4
= 9 < Z < 9
k − 15
15. (a) P(Z > 22 ) = 0.4152 = 0.5999
(ii) P(X < 40] = 40 – 53
PZ < 9 4
k − 15 = 0.214
22 = 0.0743

k = 19.71 0.0743 × 100% = 7.43%

f(z) (iii) P[X > m] = 0.74

0.4152 P3Z > m – 53 4 = 0.74
9
m – 53
z 9 m = –0.643
0 0.214 = 53 – 9(0.643)

(b) PZ < k − 250 2 = 0.32 = 47.21
16
(b) P(X < 60) = 0.8

k − 250 = –0.468 P1Z < 60 – µ 2 = 0.8
16 σ

k = 242.51 60 – µ = 0.842
σ
f(z)
60 – µ = 0.842σ ...➀

0.32 P(X < 70) = 0.9

z P1Z < 70 – µ 2 = 0.9
σ
–0.468 0
70 – µ = 1.282
σ
(c) P 35 – 40.2 < Z < k – 40.2  = 0.458
12.5 12.5
70 – µ = 1.282σ ...➁

P–0.416 < Z < k – 40.2 2 = 0.458 ➁–➀
12.5
10 = 0.44 σ

1 – P(Z < –0.416) – P1Z > k – 40.2 2 = 0.458 σ = 22.73
12.5
varians/variance = 516.53

P1Z < k – 40.2 2 = 0.2033 µ = 60 – 0.842(22.73)
12.5
= 40.86

J30

Praktis SPM 5 3. (a) (i) Min/Mean = 90
P(X > 140) = 0.12
Kertas 1
140 – 90 2
1. (a) min = 0 dan sisihan piawai = 1 P(X > 140) = PZ > σ

Mean = 0 and standard deviation = 1 = 0.12

(b) P(0 < Z < 1) = 0.3413 140 – 90 = 1.175
σ
2. (a) 1.8 = 38 −32 , σ = 3.33
(b) P(X < k) σ <
k −32 σ = 42.55
= P(Z 3.33 ) = 0.4213
(ii) P(50 < X < 100) =

k3−.3332 = −0.199 P  5402–.5950 < Z < 10402.–55902 = 0.4193
k = 31.34
0.4193 × 45 = 18.87
3. (a) P(X = 0) + P(X > 2) = 1 – h – k
( b) P(X < y)= PZ < y42–.5950 2 = 19
(b) P(X = 0)= 4C0 p0 (1 − p)4 = 4
27
1 − p = 0.037 = 0.23

p = 0.963 y – 90 = −0.739
42.55
Kertas 2
y = RM58.56

1. (a) n = 7, p = 0.22 4. (a) RM24 boleh bermain 6 set permainan

P(X = 4) = 7C4(0.22)4(0.78)3 RM24 can play 6 sets of game.

= 0.0389 n=6

(b) (i) Min 0.46 kg dan sisihan piawai m kg. P(menjatuhkan semua tin dalam tiga lontaran
Mean 0.46 kg and the standard deviation of
berturut-turut)
m kg.
P(topple all the tins in three consecutive throws)
P(X > 0.52) = 0.18
= (0.7)3 = 0.343
0.52 – 0.462 =
PZ > m 0.18 P(X ≥ 1) = 1 − P(X = 0)

0.52 – 0.46 = 0.916 = 1 − 6C0(0.343)0(0.657)6
m m = = 0.9196
0.52 – 0.46
0.916 Peratus menang ialah 91.96%. Awang patut

bermain.
Percentage to win is 91.96%. Awang should play
= 0.066 kg
(b) E(x) = np
( ii) P=(P01.40.<40X.–0<606.04.65)< Z <
0.5 – 0.46 3 = n(0.3430)
0.066
2 n = 8.75

= 0.5461 Dia perlu bermain 9 set.

He needs to play 9 sets.

0.5461 × 375 = 205 bungkus

2. (a) (i) P(X > 1.2) = PZ > 1.2 – 0.75 2 = 0.0668 Sudut KBAT
0.3
(a) min/mean = 16.4
(ii) P(X < m) = PZ < m – 0.75 2 = 0.15 P[X > 17] = 0.16
0.3
P3Z 17 – 16.4 4 = 0.16
m – 0.75 > σ
0.3 = −1.036

m = 0.4392 kg 17 – 16.4 = 0.994
σ
(b) (i) p = 0.4
σ = 0.6036 m
P(X = 1) = 8 × P(X = 0)
(b) P[15.2 < X < 17] = 13.5 + 34 + 34
n C1(0.4)(0.6)n − 1 = 8 × nC0(0.6)n
n × 00..64 = 8 = 81.5%

(c) P[X < 15.2] = 0.025

n = 8 × 0.6 = 12 P3Z < 15.2 – 16.4 4 = 0.01563
0.4 0.6036

(ii) Var(X) = npq = 12(0.4)(0.6) 0.01563 × 1000 = 15.63 pokok

Sisihan piawai = 1.7

Standard deviation

J31

BAB Fungsi Trigonometri (c) y x

6 Trigonometric Functions 39Њ 0

1. (a) y (d) y 1
tan 219°
II kot/ cot 219° =
121Њ
x 0 x = kot/ cot 39°
␣
–412Њ ␣ (d) y
0

IV

α = 180° − 121° α = 412° − 360° 0x
28Њ
= 59° = 52°
152Њ

(b) y (e) – 23 π = –2 × 180° kos/ cos (−152°) = kos/ cos 28°
3 (e) y

I = –120°

␣ x y
0
x
316Њ 0 –32Њ

0 x
60Њ
1
α = 360° − 316° –120Њ –kosek(–32°) = – sin(–32°)
III

= 44° 1
–sin 32°
α = 60° = –

(c) y (f) 9 p = 9 × 180° = kosek/ cosec 32°
8 8 (f)
y
II = 202.5°

␣x y 135Њ

0 520Њ 9␲ 45Њ x
8x 0
␣0

α = 180° − 160° –tan 3 π2 = – tan 3 × 4180°2
4
= 20° α = 98 p – p = π
8 = –tan 135°

2. (a) θ = 360° + (180° – 42°) = −(−tan 45°)

= 498° = tan 45°

–3π – π  (g) y
6
(b) θ =

= – 17 π 0 x
6 60Њ

kerana arah ikut jam.

because follows clockwise direction. –sin – 38 π2 = –sin(–480°)

3. (a) y

x = –(–sin 60°)

69Њ 0 = sin 60°

(h) y

sin 249° = –sin 69°

(b) y x
0 45Њ

20Њ 0 x

1 –kot( 7 π) = – 1
kos(–200°) 4 tan 315°
sek (–200°) =
1
–kos1 20° = –sek 20° = – –tan 45°

= kot/ cot 45°

J32

4. (a) y (c) (i) sin θ = 1

–12 x 2 + 2p + p2 y
T0
(ii) kosek/cosec θ = 1 √2+2p+p 2
–5 sin θ
13

1

–5 = 2 + 2p + p2 ␪ x
13 (iii) kos/cos θ = (1+p)
(i) sin θ = 1+p 0

(ii) sek/ sec θ = 1 = 1 = – 1123 2 + 2p + p2
kos/ cos θ –12
6. (a) y
13
76Њ18Ј

(iii) kot/ cot θ = 1 θ = 12
tan 5
y 13Њ42Ј x
1 √2 0
(b) sek/sec θ = – 2 = cos
θ
kos/cos θ = –1
2 1 kos/cos θ = sin 13° 42’
␪
x = kos 76° 18’
–1 0
(i) kot/cot θ = –1 θ = 76° 18’

(ii) kosek/cosec θ = 1 = 2 (b) y
sin θ
7Њ45Ј
(iii) sin θ = 1

2 82Њ15Ј x
0
(c) kot/cot θ = – 3 y

tan θ = –1 3 √3 x kot/cot θ = tan 82° 15’
3 2 0␪ –1 = kot 7° 45’
∴θ = 7° 45’
(i) kos/cos θ = 2
(c) y
(ii) kosek/cosec θ = 1
sin θ 90Њ–pЊ

= 1 pЊ x
– 12 0

= –2 kosek/ cosec θ = sek/ sec p°

(iii) sek/sec θ = 1 sin1 θ = 1
kos/cos θ kos/ cos p°

=1=2 sin θ = kos/ cos p°
33 = sin 90° − p°
θ = 90° − p
2

5. (a) (i) kos/cos θ = –1 (d) y
1 + p2
55Њ6Ј

(ii) kosek/cosec θ = 1 y
sin θ √1+p2
34Њ54Ј x
p ␪ 0
1 + p2 –1 0
= p x sin θ = kos/ cos 34° 54’
= sin 55° 6’
(iii) sek/sec θ = 1 θ = 55° 6’
kos/cos θ
(e) y
= – 1 + p2
21Њ26Ј
p2 – 4
(b) (i) kot/cot θ = 2 y

(ii) sek/sec θ = 1 –√p2–4 x 68Њ34Ј x
kos θ ␪ 0
0

= –p –2 p tan (90° − θ) = kot/ cot 68° 34’
p2 – 4
= tan 21° 26’

90° − θ = 21° 26’

(iii) kos/cos θ = – p2 – 4 90° − 21° 26’ = θ
2
θ = 68° 34’

J33