Edisi Guru
Modul 360°
BonSuissipan 4
Jawapan
Tingkatan
• DSKP BerdaBsuakrukTaenks & KSSM
• KBAT
• KLON SPM Form at SPM
Terbah aru
Matematik
Tambahan
(Dwibahasa)
• Gan Bee Yeng Info KBAT
• Loh Chee Hoo Digital
• Noorwahida binti Daud @ Arifin
(Guru Cemerlang)
Kandungan
Rekod Pentaksiran Murid........................................................................................................ i
Bab 1 Fungsi Bab 5 Janjang
Functions Progressions
1.1 FFuunncgtsioi..n..s........................................................................... 1 5.1 Janjang Aritmetik.....................................................85
Arithmetic Progressions
1.2 Fungsi Gubahan.......................................................... 8 5.2 Janjang Geometri.....................................................92
Composite Functions Geometric Progressions
1.3 Fungsi Songsang........................................................14 Praktis ke Arah SPM .............................................................98
Inverse Functions
Praktis ke Arah SPM .............................................................18CONTOH Bab 6 Hukum Linear
Linear Law
Bab 2 Fungsi Kuadratik 6.1 Hubungan Linear dan Tak Linear............... 101
Quadratic Functions Linear and Non-Linear Relations
2.1 Persamaan dan Ketaksamaan Kuadratik.....22 6.2 Hukum Linear dan Hubungan Tak Linear.... 114
Quadratic Equations and Inequalities Linear Law and Non-Linear Relations
2.2 Jenis-jenis Punca Persamaan Kuadratik.....28 6.3 Aplikasi Hukum Linear...................................... 117
Types of Roots of Quadratic Equations Applications of Linear Law
2.3 Fungsi Kuadratik......................................................30 Praktis ke Arah SPM .......................................................... 121
Quadratic Functions
Praktis ke Arah SPM .............................................................52 Bab 7 Geometri Koordinat
Coordinate Geometry
Bab 3 Sistem Persamaan 7.1 Pembahagi Tembereng Garis.......................... 125
Systems of Equations Divisor of a Line Segment
3.1 Sistem Persamaan Linear dalam Tiga 7.2 Garis Lurus Selari dan Garis Lurus
Pemboleh Ubah......................................................... 56 Serenjang................................................................... 131
Systems of Linear Equations in Three Variables
Parallel Lines and Perpendicular Lines
3.2 Persamaan Serentak yang Melibatkan
7.3 Luas Poligon............................................................. 136
Satu Persamaan Linear dan Satu
Persamaan Tak Linear........................................... 61 Areas of Polygons
Simultaneous Equations involving One Linear
7.4 Persamaan Lokus................................................... 142
Equation and One Non-Linear Equation
Equations of Loci
Praktis ke Arah SPM .............................................................64
Praktis ke Arah SPM .......................................................... 146
Bab 4 Indeks, Surd dan Logaritma Bab 8 Vektor
Indices, Surds and Logarithms Vectors
4.1 Hukum Indeks............................................................67 8.1 VVeecktotorrs.......................................................................... 149
Laws of Indices 8.2 Penambahan dan Penolakan Vektor.......... 154
4.2 Hukum Surd................................................................70 Additions and Subtractions of Vectors
Laws of Surds 8.3 Vektor dalam Satah Cartes.............................. 156
4.3 Hukum Logaritma....................................................75 Vectors in Cartesian Plane
Laws of Logarithms Praktis ke Arah SPM .......................................................... 161
4.4 Aplikasi Indeks, Surd dan Logaritma...........81
Applications of Indices, Surds and Logarithms
Praktis ke Arah SPM .............................................................82
Bab 9 Penyelesaian Segi Tiga Bab 10 Nombor Indeks
Solution of Triangles Index Number
9.1 Petua Sinus............................................................... 166 10.1 Nombor Indeks....................................................... 178
Sine Rule Index Numbers
9.2 Petua Kosinus.......................................................... 170 10.2 Indeks Gubahan..................................................... 184
Cosine Rule Composite Index
9.3 Luas Segi Tiga......................................................... 172 Praktis ke Arah SPM .......................................................... 189
Area of a Triangle
9.4 Aplikasi Petua Sinus, Petua Kosinus Kertas Model Pra-SPM.......................... 191
dan Luas Segi Tiga............................................... 174
Application of Sine Rule, Cosine Rule
and Area of a Triangle
Praktis ke Arah SPM .......................................................... 176
CONTOH
Format Instrumen SPM
Matematik Tambahan (3472)
Bil. Perkara Kertas 1 (3472/1) Kertas 2 (3472/2)
1 Jenis Instrumen Ujian Bertulis
2 Jenis Item • Subjektif Respons Terhad
• Subjektif Respon Terhad Berstruktur
Bahagian A: Bahagian A:
12 soalan (64 markah) 7 soalan (50 markah)
(Jawab semua soalan) (Jawab semua soalan)
3 Bilangan Soalan
Bahagian B: Bahagian B:
3 soalan (16 markah) 4 soalan (30 markah)
(Jawab dua soalan) (Jawab tiga soalan)
Bahagian C
4 soalan (20 markah)
(Jawab dua soalan)
4 Jumlah Markah 80 markah 100 markah
• Mengingat dan Memahami • Mengingat dan Memahami
• Mengaplikasi
• Mengaplikasi • Menganalisis
• Menilai
5 Konstruk • Menganalisis • Mencipta
• Menilai
• Mencipta
6 Tempoh Ujian 2 jam 2 jam 30 minit
7 Cakupan Konstruk Standard kandungan dan standard pembelajaran
dalam Dokumen Standard Kurikulum dan Pentaksiran (DSKP) KSSM
(Tingkatan 4 dan Tingkatan 5)
8 Aras Kesukaran Rendah : Sederhana : Tinggi
5:3:2
9 Kaedah Penskoran Analitik
10 Alat Tambahan Kalkulator saintifik yang tidak boleh diprogram
Bab 1 Matematik Tambahan Tingkatan 4 Bab 1
Fungsi
Functions
MODUL PBD Praktis DSKP
1.1 Fungsi Buku Teks m/s 2 – 11 Info
Functions Digital 1.1
1 Wakilkan setiap hubungan berikut dengan menggunakan sebuah gambar rajah anak panah. SP: 1.1.1 TP1
Represent each of the following relations using an arrow diagram. Mudah
Contoh (a) Pasangan tertib {(1, –1), (2, –2), (3, –3), (5, –5)}
Hubungan antara set A dan set B ditakrifkan sebagai mentakrifkan hubungan yang memetakan unsur-
satu set pasangan tertib {(a, 1), (b, 2), (c, 3), (f, 6)}. unsur dari set A kepada unsur-unsur dalam set B.
The relation between set A and set B is defined by the set of
ordered pairs {(a, 1), (b, 2), (c, 3), (f, 6)}. The set of ordered pairs {(1, –1), (2, –2), (3, –3), (5, –5)}
defines the relation which maps the elements from set A
B to the elements in set B.
A
a⦁ ⦁1 AB
b⦁ ⦁2 1 –5
c⦁ ⦁3
f ⦁ ⦁6 2 –3
3 –2
CONTOH
5 –1
(b) Hubungan antara set A dan set B ditakrifkan (c) Hubungan ini memetakan x kepada 2x – 1.
sebagai satu set pasangan tertib {(0, –2), (2, 0), This relation maps x to 2x – 1.
(3, 2), (4, 0)}. x 2x – 1
The relation between set A and set B is defined by the set 7
of ordered pairs {(0, –2), (2, 0), (3, 2), (4, 0)}.
A 0 2 3 4 45
33
2
B
–2 0 2
1
Matematik Tambahan Tingkatan 4 Bab 1
2 Tentukan sama ada hubungan antara set A dan set B ialah suatu fungsi atau tidak. Justifikasikan sekiranya bukan.
Determine whether the relation between set A and set B is a function or not. Justify if it is not. SP: 1.1.1 TP1 Mudah
Contoh
A 1 2 3 Hubungan antara set A dan set B ialah hubungan satu kepada satu,
B maka hubungan ini ialah suatu fungsi.
The relation between set A and set B is one-to-one relation, so this relation
u v w is a function.
Tips Bestari
Hanya hubungan satu kepada satu dan hubungan banyak kepada satu merupakan fungsi.
Ony one-to-one relation and many-to-one relation are functions.
CONTOH B
(a)
Hubungan antara set A dan set B ialah hubungan banyak kepada
A a satu, maka hubungan ini ialah suatu fungsi.
2 The relation between set A and set B is many-to-one relation, so this
3 b relation is a function.
5
(b) B Hubungan antara set A dan set B ialah hubungan banyak
u kepada banyak, maka hubungan ini bukan suatu fungsi. Unsur
A v b mempunyai dua imej, iaitu v dan w.
a w The relation between set A and set B is many-to-many relation, so this
b relation is not a function. The element b has two images, which are
c v and w.
(c)
y
3 Hubungan antara set A dan set B bukan suatu fungsi kerana
unsur 3 tidak dipetakan kepada mana-mana imej.
2 The relation between set A and set B is not a function because the
element 3 is not mapped to any images.
(0, 1)
2
A0 x
12 3
B
Matematik Tambahan Tingkatan 4 Bab 1
3 Tentukan domain, kodomain dan julat bagi setiap perwakilan grafik fungsi yang berikut. SP: 1.1.2 TP1 Mudah
Determine the domain, codomain and range for each of the following graphical representations of function.
Perwakilan grafik fungsi Domain Kodomain Julat
Graphical representation of function Domain Codomain Range
Contoh
AB {1, 2, 3, 4} {3, 6, 11, 18} {3, 6, 11, 18}
13
26
3 11
4 18
CONTOH(a) Q {x, y, z} {3, 5, 7, 9} {3, 5, 9}
P {2, 3}
⦁3
x⦁ ⦁5
y⦁ ⦁7
z⦁ ⦁9
(b)
P 123456
Q 234 {1, 2, 3, 4, 5, 6} {2, 3, 4}
(c)
f (x)
30
‒5 N x N 20 ‒10 N f(x) N 30 ‒10 N f(x) N 30
–5 0 x
–10 20
(d)
f (x)
10
(4, 8)
(2, 2) {‒4, ‒1, 0, 2, 4} {‒4, ‒2, 2, 8, 10} {‒4, ‒2, 2, 8, 10}
0 x
(–4, –2)
(–1, –4)
3
Matematik Tambahan Tingkatan 4 Bab 1
4 Dengan menggunakan tatatanda fungsi, ungkapkan fungsi bagi setiap hubungan berikut dan seterusnya, selesaikan
masalah yang diberi. SP: 1.1.2 SP: 1.1.3 TP2 TP3
By using the function notation, express the function for each of the following relation and hence, solve the problems given.
Mudah Sederhana
Contoh
Fungsi f memetakan x kepada 1 x – 3. Tentukan julat (a) Fungsi f memetakan x kepada mx + 3 dan diberi
2 imej bagi 2 ialah ‒1. Cari
fungsi bagi domain = {0, 2, 3, 8}. The function f maps x onto mx + 3 and given the image
of 2 is –1. Find
The function f maps x onto 1 x – 3. Determine the range of
2 (i) julat bagi domain 2 < x < 5,
the function for the domain = {0, 2, 3, 8}. the range for domain 2 < x < 5,
(ii) domain dengan julat ‒1 < f(x) < 3.
f : x → 1 x – 3 atau/ or f (x) = 1 x – 3 the domain with the range of –1 < f(x) < 3.
2 2
f(2) = ‒1
Apabila/ When, 2m + 3 = ‒1
2m = ‒4
f(0) = 1 (0) –3 = ‒3 m = ‒2 → f (x) = ‒2x + 3
2
f(2) =CONTOH1 (2) –3 = ‒2
2
(i) Apabila domain/ When domain = {2, 5}
f(3) = 1 (3) –3 =‒ 3 f(2) = ‒2(2) + 3 = ‒1
2 2 f(5) = ‒2(5) + 3 = ‒7
f(8) = 1 (8) –3 =1
2
\ Julat fungsi/ The range function = {‒3, ‒2, ‒ 3 \ Julat/ Range = ‒1 < f (x) < ‒7
of 2 , 1 }
(ii) f(x) = ‒1 f(x) = 3
‒2x + 3 = ‒1 ‒2x + 3 = 3
2x = 4
2x = 0
x = 2
x=0
\ Domain = 0 < x < 2
(b) Fungsi h memetakan x kepada |2x – c| dengan (c) Fungsi f ditakrifkan oleh f : x → 4x – 7.
keadaan c ialah pemalar. Diberi imej bagi 3 ialah 0 The function f is defined by f : x → 4x – 7.
dalam fungsi h. (i) Lengkapkan jadual berikut.
The function h maps x onto |2x – c| where c is a constant. Complete the following table.
Given the image of 3 is 0 in function h. x23
(i) Cari objek bagi imej 4.
Find the object for image 4.
(ii) Tentukan julat fungsi h dengan melakarkan f(x) 1 5
graf fungsi h. (ii) Seterusnya, nyatakan domain bagi julat
1 N f (x) N5.
Determine the range of function h by sketching the
graph of function h. Hence, state the domain for the range of 1 N f (x) N 5.
h(3) = 0 f(x) = 4x – 7
|2(3) – c| = 0 (i) f(2) = 4(2) ‒ 7
= 1
6 – c = 0 → h(x) = |2x – 6|
c = 6
(i) h(x) = 4 (ii) f(x) x = 3 f(x) = |2x – 6| f(x) = 5
|2x – 6| = 4 4x – 7 = 5
2x – 6 = ±4 (1,4) (5,4) 4x = 12
x = 3
2x – 6 = ‒4
2x = 2 (ii) Domain = 2 N x N 3
x = 1
0 (3, 0) x
atau/ or
Julat/ Range: h(x) M 0
2x – 6 = 4
2x = 10
x = 5
4
Matematik Tambahan Tingkatan 4 Bab 1
(d) Rajah di bawah menunjukkan sebuah graf bagi (e) Rajah di bawah menunjukkan graf fungsi f(x) yang
fungsi mx + c yang memetakan kepada y dengan ditakrifkan oleh f : x → |x – k| dengan keadaan k ialah
keadaan m dan c ialah pemalar. pemalar.
The diagram below shows the graph of linear function The diagram below shows the graph of function f(x)
mx + c which maps onto y where m and c are constants. which is defined by f : x → |x – k| where k is a constant.
y f (x)
f (x) = |x – k|
(1, 4)
(3, k) 0 P (2, 0) x
(4, 1)
0 x Cari nilai k dan seterusnya, tentukan nilai objek
yang dipetakan kepada 3 di bawah fungsi f.
Find the value of k and hence, determine the values of
(i) Nyatakan jenis fungsi ini. objects which mapped onto 3 under function f.
State the type of this function.
(ii) Tunjukkan bahawa m = ‒1 dan c = 5. f(x) = |x – k|
Show that m = –1 and c = 5.
(iii) Seterusnya, cari nilai k.
Hence, find the value of k.
CONTOH f(2) = 0
y = mx + c |2 – k| = 0 → f (x) = |x – 2|
2 – k = 0
k = 2
(i) Fungsi satu kepada satu f(x) = 3
One-to-one function |x – 2| = 3
x – 2 = ±3
(ii) 4 = m(1) + c 1 = m(4) + c
x – 2 = ‒3
m + c = 4 …➀ 4m + c = 1 …➁ x = ‒1 atau/ or
➁ – ➀: 4m – m = 1 – 4 x–2=3
3m = ‒3 x=5
m = ‒1 (Ditunjukkan/ Shown)
m = ‒1 ↷ ➀ ‒1 + c = 4
c = 5 (Ditunjukkan/ Shown)
(iii) k = (‒1)(3) + 5
= ‒3 + 5
= 2
(f) Fungsi f memetakan x kepada ax + b dengan keadaan a dan b ialah pemalar.
Function f maps x onto ax + b where a and b are constants.
(i) Diberi bahawa imej bagi 3 dan 7 masing-masing ialah 1 dan 5. Ungkapkan fungsi f dalam sebutan x.
Given that the images of 3 and 7 are 1 and 5 respectively. Express function f in terms of x.
(ii) Lakarkan graf fungsi f dan seterusnya, nyatakan
Sketch the graph of function f and hence, state
(a) julat bagi domain 2 < x < 7, (b) domain bagi julat 1 < f(x) < 5.
the range of domain 2 < x < 7, the domain for range 1 < f(x) < 5
(i) f(3) = 1 f (7) = 5 (ii) f(x)
3a + b = 1 …➀ 7a + b = 5 …➁ f (x) = x – 2
➁ – ➀: 7a – 3a = 5 – 1 5
4a = 4
a=1 1
0 23
a = 1 ↷ ➀, 3(1) + b = 1 –2 x
3+b=1 7
b = ‒2
(a) Julat/ Range: 0 < f (x) < 5
\ f (x) = x – 2 (b) Domain: 3 < x < 7
5
Matematik Tambahan Tingkatan 4 Bab 1
MODUL PP Praktis ke Arah S P M
Kertas 1
Jawab semua soalan./ Answer all questions.
Bahagian A / Section A
1 Rajah 1 menunjukkan hubungan antara set P dan 3 Diberi bahawa f : x → p + qx.
set Q. Given that f : x → p + qx.
(a) Ungkapkan f ‒1(x) dalam sebutan p dan q.
Diagram 1 shows the relation between set P and set Q. Express f –1(x) in terms of p and q.
ep [3 markah/ marks]
(b) Cari nilai p dan nilai q sekiranya f ‒1(17) = ‒4 dan
fq
f(6) = ‒13.
r Find the values of p and q if f –1(17) = –4 and
g
f(6) = –13.
s
[4 markah/ marks]
Set P Set Q
Jawapan/ Answer:
Rajah 1/ Diagram 1 Rajah 1/ Diagram 1 Rajah 1/ Diagram 1 (a) Jadikan/ Let k = f ‒1(x)
Nyatakan jenis hubungan dan julat bagi hubungan di
CONTOH
atas. f(k) = x
State the type of relation and the range for the above
relation. p + qk = x
[2 markah/ marks] qk = x – p
Jawapan/ Answer: x – p
q
Jenis hubungan: Hubungan banyak kepada satu k =
Type of relation: Relation of many-to-one
x – p
⇒ f ‒1(x) = q
Julat/ Range: {p, r}
2 Diberi fungsi f : x → 2x + 5 dan fg : x → 17 – 2x. (b) f ‒1(17) = ‒4
Given the functions f : x → 2x + 5 and fg : x → 17 – 2x. 17q– p = ‒4
(a) Ungkapkan fungsi gf. 17 – p = ‒4q ……➀
Express the function of gf.
[3 markah/ marks] f(6) = ‒13
(b) Sekiranya gf (c2 – 1) = 3 – 6c, cari nilai-nilai c p + 6q = ‒13 ……➁
yang mungkin. ➀ + ➁: 17 + 6q = ‒4q + (‒13)
If gf (c2 – 1) = 3 – 6c, find the possible values of c. 6q + 4q = ‒13 – 17
[3 markah/ marks] 10q = ‒30
q = ‒3
Jawapan/ Answer:
(a) f[g(x)] = 2g(x) + 5
17 – 2x = 2g(x) + 5
2g(x) = ‒2x + 17 – 5
‒2x + 12 q = ‒3 ↷ ➀, 17 – p = ‒4(‒3)
g(x) = 2 p = 17 – 12
= ‒x + 6 =5
gf (x) = g[f (x)] \ p = 5, q = ‒3
= g(2x + 5)
= ‒(2x + 5) + 6
= ‒2x – 5 + 6
= ‒2x + 1
(b) gf (c2 – 1) = 3 – 6c
‒2(c2 – 1) + 1 = 3 – 6c
‒2c2 + 2 + 1 = 3 – 6c
2c2 – 6c = 0
c2 – 3c = 0
c(c – 3) = 0
c = 0 atau/ or 3
18
Matematik Tambahan Tingkatan 4 Bab 1
Bahagian B / Section B
4 Rajah 2 menunjukkan pemetaan y kepada x oleh fungsi g : y → py + q dan pemetaan y kepada z oleh fungsi
8 ‒ 2q.
h : y → 2y + q, y ≠
Diagram 2 shows the mapping of y onto x by function g : y → py + q and the mapping of y onto z by function h : y → 8 q,
y ≠ – q2. 2y +
xyz (a) Cari nilai p dan nilai q.
Find the values of p and q.
44 4 [3 markah/ marks]
(b) Tentukan fungsi yang memetakan x kepada y. [3 markah/ marks]
3 Determine the function which maps x onto y. [2 markah/ marks]
33 2 (c) Cari fungsi yang memetakan x kepada z.
1 Find the function which maps x onto z.
22 0 g(3) = 2
3p – 4 = 2
–1 3p = 6
p = 2
1 1 –2
Rajah 2/ Diagram 2
CONTOH
Jawapan/ Answer:
(a) h(3) = 4
2(38) + q = 4
8 = 6 + q
4
6 + q = 2
q = ‒4
\ p = 2, q = ‒4
(b) g(y) = x ⇒ g‒1(x) = y (c) g‒1(x) = y = x+4 h(y) = 8 4 = y 4 2
2 2y ‒ –
Jadikan/ Let k = g‒1(x) ( ) x+4
hg‒1(x) = h 2
g(k) = x
4
2k – 4 = x =
+4
2k = x + 4 x 2 ‒ 2
k = x + 4 ⇒ g‒1(x) = x + 4
2 2
= 8
x + 4 ‒ 4
= 8x ⇒ hg‒1(x) = 8 , x ≠ 0
x
5 Diberi bahawa h : x → 3x – 2 dan k : x → 5 – 3x.
Given that h : x → 3x – 2 and k : x → 5 – 3x.
(a) (i) Hitung nilai bagi k(5).
Calculate the value of k(5).
(ii) Cari nilai bagi m jika h(m + 2) = 1 k(5).
5
1
Find the value of m if h(m + 2) = 5 k(5).
(iii) Tentukan fungsi kh(x).
Determine the function kh(x).
(b) Seterusnya, lakar graf y = kh(x) bagi 0 N x N 5 dan nyatakan julat bagi y. [5 markah/ marks]
Hence, sketch the graph of y = kh(x) for 0 N x N 5 and state the range of y. [3 markah/ marks]
19
Matematik Tambahan Tingkatan 4 Bab 1
Jawapan/ Answer:
(a) (i) k(5) = 5 – 3(5)
= 5 – 15
= ‒10
(ii) h(m + 2) = 51k(5)
3(m + 2) – 2 = 15(‒10)
3m + 6 – 2 = ‒2
3m = ‒6
m = ‒2
(iii) kh(x) = k[h(x)]
= k(3x – 2)
= 5 – 3(3x – 2)
= 5 – 9x + 6
= –9x + 11
(b) CONTOHx 0 1 234 5
y 11 2 –7 –16 –25 –34
y
15
10
kh(x) = –9x + 11
5
0 x
–5
–10 1 23 4 5 67
–15
–20
–25
–30
–35
Julat bagi y/ The range of y: ‒34 N y N 11
20
Matematik Tambahan Tingkatan 4 Bab 1
Kertas 2
Jawab semua soalan./ Answer all questions.
Bahagian A / Section A
1 Jia Xiong bekerja di sebuah kilang kereta mainan dan dia mengambil 20 saat untuk menghasilkan sebuah kereta
mainan. Setiap hari, dia menggunakan 15 minit untuk menyediakan stesen kerjanya sebelum memulai kerjanya.
Jia Xiong works at a toy car factory and he takes 20 seconds to produce one toy car. Every day, he used 15 minutes to set up his
work station before starting his work.
(a) Dengan x sebagai bilangan kereta mainan yang terhasil dalam sehari, ungkapkan fungsi bagi jumlah masa yang
diambil oleh Jia Xiong untuk menghasilkan kereta mainan dalam sehari dalam sebutan x.
With x as the number of toy cars produced in a day, express a function for the total time taken by Jia Xiong to produce toy
cars in a day in terms of x.
[1 markah/ mark]
(b) Diberi kerja Jia Xiong dalam sehari ialah menghasilkan sekurang-kurangnya 1 400 buah kereta mainan. Hitung
masa minimum, dalam jam, Jia Xiong perlu untuk menyiapkan kerja hariannya.
Given Jia Xiong’s job in a day is to produce at least 1 400 toy cars. Calculate the minimum time, in hours, that Jia Xiong needs
to complete his daily job.
[2 markah/ marks]
(c) Ungkapkan fungsi yang memetakan masa kepada bilangan kereta mainan yang terhasil dalam sehari. Seterusnya,
hitung bilangan kereta mainan yang terhasil jika Jia Xiong bekerja selama 10 jam dan 30 minit hari ini.
Express the function which maps the time to the number of toy cars produced in a day. Hence, calculate the number of toy
cars produced if Jia Xiong works for 10 hours and 30 minutes today.
[5 markah/ marks]
Jawapan/ Answer:
(a) f(x) = 15(60) + 20x
= 20x + 900
CONTOH
(b) f(x) = 20x + 900
f(1 400) = 20(1 400) + 900
= 28 000 + 900
= 28 900 saat/ seconds
= (28 900 ÷ 60 ÷ 60) jam/ hours
= 8316 jam/ hours
(c) Jadikan/ Let f ‒1(x) = k
20k + 900 = x
20k = x – 900
k = x –29000 x – 900
⇒ f ‒1(x) = 20
10 jam/ hours 30 minit/ minutes = 10(60 × 60) + 30(60)
= 36 000 + 1 800
= 37 800 saat/ seconds
f ‒1(37 800) = 37 800 ‒ 900
20
= 36 900
20
= 1 845 buah kereta mainan/ toy cars
21