FullyWorkedSolutions f4 2021

PTER Additional Mathematics Form 4 Chapter 1 Functions

CHA 1 Functions x2 + 5 = 21 (ii) f(x) = 4x – 7
x2 = 16 4x – 7 = –19
Fungsi x = 4 or / atau –4 4x = –12
x = –3
7. (a) –2 0 3 4
x (b) (i) h(x) = 6x – 5
h(2) = 6(2) – 5
1. (a) A function. Each object has one image Range / Julat = {1, 2, 3, 4, 5} f(x) 5 3 0 1 =7
only.
Satu fungsi. Setiap objek hanya mempunyai satu Function notation: f(x) (ii) f(x) = 6x – 5
imej. Tatatanda fungsi 5 6x – 5 = x
f : x → x or f(x) = x 5x = 5
(b) Not a function. The object 3 has two x=1
images in the codomain. 3
Bukan fungsi. Objek 3 mempunyai dua imej (c) g(x) = px2 – qx
dalam kodomain. (c) Domain = {1, 2, 4, 5} 1 x g(3) = p(3)2 – q(3)
Codomain / Kodomain = {4, 7, 10, 13, 16} –2 0 34
(c) Not a function. Object 2 has two images 9p – 3q = 12
in the codomain. Range / Julat = {4, 7, 13, 16} The range is 0 < f(x) < 5. 3p – q = 4 ––– 
Bukan fungsi. Objek 2 mempunyai dua imej Julat ialah
dalam kodomain. Function notation: g(4) = p(4)2 – q(4)
Tatatanda fungsi (b) 0 1 16p – 4q = 28
x –4 –3 4p – q= 7 ––– 
f : x → 3x + 1 or f(x) = 3x + 1
From : q = 3p – 4 ––– 
2. (a) Not a function. When tested with a 4. (a) Domain = {–4, –2, 0, 2, 3} f(x) 2 0 6 8 Dari
vertical line, the line cuts the graph at Codomain = {0, 2, 4, 8}
two points. Kodomain f(x) Substitute  into :
Bukan satu fungsi. Apabila diuji dengan garis Range / Julat = {0, 2, 4, 8} Gantikan  ke dalam :
mencancang, garis itu memotong graf pada dua 8 4p – (3p – 4) = 7
titik. (b) The domain of f is 0 < x < 3. 6
Domain bagi f ialah 4p – 3p + 4 = 7
(b) A function. When tested with a vertical The codomain of f is 0 < f(x) < 4. –4 –3 2 x p=3
line, the line cuts the graph at one Kodomain bagi f ialah 01
point. The range of f is 0 < f(x) < 4. Substitute p = 3 into :
Satu fungsi. Apabila diuji dengan garis Julat bagi f ialah The range is 0 < f(x) < 8. Gantikan p = 3 ke dalam :
mencancang, garis itu memotong graf pada Julat ialah q = 3(3) – 4
satu titik. 5. (a) f(–4) = 3(–4) + 7
= –5 (c) –3 0 4 2 =5
(c) A function. When tested with a vertical x 5 6
line, the line cuts the graph at one f(5) = 3(5) + 7 (d) h(x) = 4 – x
point. = 22 f(x) 19 4 0 4 – x = 7
Satu fungsi. Apabila diuji dengan garis 4–x=7
mencancang, garis itu memotong graf pada (b) f(3) = (3)2 + 1 f(x) x = –3
satu titik. = 10 19 and / and
4 – x = –7
3. (a) Domain = {2, 3, 4, 5} f(–2)= (–2)2 + 1 x = 11
=5
9. (a) fg(x) = f(4x – 5)
Codomain / Kodomain = { 1 , 1 , 1 , 1 , 0} 6. (a) 5x + 7 = –1 6 = 4x – 5 + 2
2 3 4 5 4 = 4x – 3
5x = –8 ∴ fg : x → 4x – 3
8 x
Range / Julat = { 1 , 1 , 1 , 1 } x = – 5 –3 0 –45 2 gf(x) = g(x + 2)
2 3 4 5 = 4(x + 2) – 5
5x + 7 = –3 = 4x + 8 – 5
Function notation: The range is 0 < f(x) < 19. = 4x + 3
Tatatanda fungsi 5x = –10
Julat ialah ∴ gf : x → 4x + 3
x = –2
1 1 8. (a) (i) f(x) = 4x – a
f : x → x or f(x) = x (b) x2 + 5 = 9 f(–1) = 4(–1) – a
x2 = 4
(b) Domain = {1, 4, 9, 16, 25} x = 2 or / atau –2 –4 – a = –11
Codomain / Kodomain = {1, 2, 3, 4, 5} a=7

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A2

Additional Mathematics Form 4 Chapter 1 Functions Additional Mathematics Form 4 Chapter 1 Functions

(b) fg(x) = f (x2) 12. (a) f [g(x)] = 3x + 9 So, f(y) = (y – 2)2 + 4(y – 2) + 4 (ii) V(5) = 17.6(5)
= 2x2 – 3 g(x) + 4 = 3x + 9 = y2 – 4y + 4 + 4y – 8 + 4 = 88
g(x) = 3x + 5 = y2 Hence, the volume of water in the
∴ fg : x → 2x2 – 3 container after 5 seconds is 88 cm3.
∴ g : x → 3x + 5 ∴ f(x) = x2
gf(x) = g(2x – 3) Maka, isi padu air di dalam bekas selepas
= (2x – 3)2 (b) f [g(x)] = 5x + 8 (ii) gf(x) = 2x2 – 3x + 4 5 saat ialah 88 cm3.
3g(x) – 6 = 5x + 8 f(x) + 2 = 2x2 – 3x + 4
= 4x2 – 12x + 9 3g(x) = 5x + 14 f(x) = 2x2 – 3x + 2 (c) (i) gf(x) = g(x – 3 000)
∴ gf : x → 4x2 – 12x + 9 g(x) = 5x + 14 = 0.05(x – 3 000)
3 (c) ff(x) = 4x + 6
(c) fg(x) = f(–5x) f(px + q) = 4x + 6 gf(x) is the commission received
= (–5x)2 – 4(–5x) + 3 ∴ g : x → 5x + 14 by Azlan, that is 5% on sales over
= 25x2 + 20x + 3 3 p(px + q) + q = 4x + 6 RM3 000.
p2x + pq + q = 4x + 6
∴ fg : x → 25x2 + 20x + 3 gf(x) ialah komisen yang diterima oleh
13. (a) f [g(x)] = 7x – 4 Comparing: / Bandingkan: Azlan, iaitu 5% pada jualan yang melebihi
gf(x) = g(x2 – 4x + 3) f(2x + 3) = 7x – 4
= –5(x2 – 4x + 3) p2 = 4 RM3 000.
= –5x2 + 20x – 15 Let / Biar y = 2x + 3
x= y–3 p = ±√4 (ii) gf(6 499) = 0.05(6 499 – 3 000)
∴ gf : x → –5x2 + 20x – 15 2 = 2 or / atau –2 = 174.95

10. (a) f 2(x)= f [ f(x)]  So / Jadi, f(y) = 7 y – 3 – 4 When p = 2, Salary / Gaji = 2 000 + 174.95
= f(7x – 8) 2 = 2 174.95
= 7(7x – 8) – 8 Apabila
= 49x – 56 – 8 pq + q = 6 Hence, his salary is RM2 174.95
= 49x – 64 2q + q = 6
3q = 6 Maka, gajinya ialah RM2 174.95
(b) f 2(x) = f [ f(x)] q=2
= f(x2 + 3) = 7y – 29 16. (a) f(2) = 3 (b) f –1(11) = 6
= (x2 + 3)2 + 3 2 (c) f –1(3) = 2 (d) f –1(9) = 5
= x4 + 6x2 + 9 + 3
= x4 + 6x2 + 12 ∴ f : x → 7x – 29 When p = –2,
2
11. (a) (i) fg(x) = f(2 – x2) Apabila 17. (a) Does not have inverse function. When
= 3(2 – x2) + 2 (b) f [g(x)] = 3x2 – 5 pq + q = 6 tested with a horizontal line, the line
= 6 – 3x2 + 2 f(x2 – 9) = 3x2 – 5 cuts the graph at two points.
= 8 – 3x2 –2q + q = 6 Tidak mempunyai fungsi songsang. Apabila diuji
Let / Biar y = x2 – 9 –q = 6 dengan garis mengufuk, garis itu memotong graf
fg(2) = 8 – 3(2)2 x2 = y + 9 q = –6 pada dua titik.
= –4 x = √y + 9
(d) g(x) = 5x – 6 (b) Has inverse function. When tested with
gf(x) = g(3x + 2) So / Jadi, f(y) = 3(√y + 9)2 – 5 3x a horizontal line, the line cuts the graph
= 2 – (3x + 2)2 = 3(y + 9) – 5 h(x) = x–2 at one point.
= 2 – (9x2 + 12x + 4) = 3y + 27 – 5 Mempunyai fungsi songsang. Apabila diuji
= –9x2 – 12x – 2 = 3y + 22 gh(x) = h(x) dengan garis mengufuk, garis itu memotong
graf pada satu titik.
gf(4) = –9(4)2 – 12(4) – 2 ∴ f : x → 3x + 22 g x 3x  = 3x
= –194 –2 x–2

(ii) fg(x) = 5 5 x 3x  – 6 = 3x 4–x
8 – 3x2 = 5 –2 x–2 3
3x2 = 3  18.
x2 = 1 14. (a) (i) fg(x) = f(x – 6) 15x 3x (a) fg(x) = f
x = 1 or / atau –1 = 3(x – 6) + 8 x–2 x–2
= 3x – 18 + 8 – 6 = = 4 – 3 4 – x 
= 3x – 10 3
15x – 6(x – 2) = 3x = 4 – (4 – x)
(ii) fg(–2) = 3(–2) – 10
= –16 15x – 6x + 12 = 3x =x

(b) (i) f g(x) = x2 + 4x + 4 6x = –12
f(x + 2) = x2 + 4x + 4
x = –2 gf(x) = g(4 – 3x)
Let y = x + 2
x=y–2 15. (a) (i) V[h(t)] = V(1.4t) = 4– (4 – 3x)
= 3
88 3x
= 7 (1.4t) 3

= 17.6t =x

∴ V : t → 17.6t

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A3

Additional Mathematics Form 4 Chapter 1 Functions Additional Mathematics Form 4 Chapter 1 Functions

Since fg(x) = gf(x) = x, f and g are inverse x =y–1 Since / Oleh sebab f (y) = x, ∴ g : x → 4x + 4
functions of each other. 4 y + 4 x–1
Oleh sebab fg(x) = gf(x) = x, f dan g ialah fungsi x = 4(y – 1) f (y) = 3
songsang antara satu sama lain.
x = 4y – 4 (e) Let / Biar y = h–1(x)
5
Substitute y with x, y = x –2

19. (a) Since / Oleh sebab g–1(y) = x, Gantikan y dengan x, 5
g–1(y) = 4y – 4 x + 4 x
x 1 5 10 f (x) = 3 x = y+2
=
y –5 –1 –0.5 Substitute y with x, ∴ f : x → x + 4 5
Gantikan y dengan x, 3 y+2
y g–1(x) = 4x – 4
∴ g–1 : x → 4x – 4 Since / Oleh sebab h(y) = x,
(–0.5, 10) (b) Let / Biar y = f –1(x) 5
h(y) = +
y=x x +6 y 2
y = 5
f –1 (d) Let / Biar y = h(x)
x–1 Substitute y with x,
y = 2x + 1 5y = x + 6
Gantikan y dengan x,
(–5, 1) 2xy + y = x – 1 x = 5y – 6 5
x h(x) = +
0f (10, –0.5) y + 1 = x – 2xy x 2
(1, –5) Since / Oleh sebab f(y) = x,
y + 1 = x(1 – 2y) f(y) = 5y – 6 ∴ h : x → 5
y+1 +
x = 1 – 2y x 2

Domain of f –1: –5 < x < –0.5 Since / Oleh sebab h–1(y) = x, Substitute y with x, 22. (a) (i) g(4) = 2(4) + p
Gantikan y dengan x, p–4
20. (a) Let / Biar y = f(x) h–1(y) = y+1 f(x) = 5x – 6
1 – 2y 5= 8+p
y = 3x – 1 p–4
Substitute y with x, ∴ f : x → 5x – 6
3x = y + 1 5(p – 4)= 8 + p
y +1 Gantikan y dengan x,
x = 3 (c) Let / Biar y = g–1(x) 5p – 20 = 8 + p
x+1
h–1(x) = 1 – 2x y = x–7 4p = 28
3
Since / Oleh sebab f –1(y) = x, : x+1 p=7
y +1 ∴ h–1 x → 1 – 2x
f –1(y) = 3 3y = x – 7 2x + 7
3
x = 3y + 7 (ii) g(x) =

Substitute y with x, (e) Let / Biar y = h(x) Since / Oleh sebab g(y) = x, Let / Biar y = g(x) g–1(y) = x
g(y) = 3y + 7
Gantikan y dengan x 3 2x + 7
x +1 x–1 y= 3
f –1(x) = 3 y=

∴ f –1 : x → x +1 xy – y = 3 Substitute y with x, 3y = 2x + 7
3 Gantikan y dengan x,
xy = 3 + y g(x) = 3x + 7 x = 3y – 7
2
(b) Let / Biar y = g(x) x= 3 +y
y
y = 7 – 2x 3 ∴ g : x → 3x + 7 ∴ g –1(x) = 3x – 7
y 2
2x = 7 – y x= + 1
7–y
x= 2 Since / Oleh sebab h–1(y) = x, (d) Let / Biar y = g–1(x) (iii) g –1(x) = 3x – 7
2
3 y = x +4 3x – 7
Since / Oleh sebab g–1(y) = x, h–1(y) = y +1 x –4 4= 2
7–y
g–1(y) = 2 Substitute y with x, xy – 4y = x + 4 8 = 3x – 7

Gantikan y dengan x, xy – x = 4y + 4 3x = 15
4y + 4
Substitute y with x, h–1(x) = 3 +1 x = y–1 x=5
x
Gantikan y dengan x,
7–x
g–1(x) = 2 ∴ h–1 : x → 3 + 1 Since / Oleh sebab g(y) = x, (b) (i) Let / Biar y = g(x)
x 4y + 4
∴ g–1 : x → 7–x g(y) = y–1 y = 8x – 3
2
21. (a) Let / Biar y = f –1(x) 8x = y + 3
y+3
y = 3x – 4 Substitute y with x, x = 8

(c) Let / Biar y = g(x) 3x = y + 4 Gantikan y dengan x, Since /Oleh sebab g –1(y) = x,
4x + 4
y = x + 1 x = y+4 g(x) = x–1 g –1(y) = y+3
4 3 8

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A4

Additional Mathematics Form 4 Chapter 1 Functions Additional Mathematics Form 4 Chapter 1 Functions

Substitute y with x, 4. (a) Let / Biar y = g(x) (b) fg : x → 16x2 + 8x – 5 f(y) = 91 y – 6 2 + 12 Substitute y with x,
5
Gantikan y dengan x, y = 3x – 9 f(x) = fgg-1(x) Gantikan y dengan x,

g –1(x) = x+3 3x = y + 9 = fg[g-1(x)] = 9y – 54 + 60 f(x) = 9x + 6
8 y+9 5 5
x= 3 = fg1 x – 1 2 9y + 6 9x + 6
4 5 5
x+3 = \ f : x →
8
 (ii) Since / Oleh sebab g-1(y) = x, x – 1 22 x – 1
hg–1(x) = h = 161 4 + 81 4 2 – 5

= 5 + 9 161x2 – 2x + 12 10. (a) The coloured
x+3 3 16 dice thrown
g-1(y) = y = + 2(x – 1) – 5 Warna dadu The number obtained
yang dilontarkan Nombor yang diperoleh
8 Substitute y with x, = x2 – 2x + 1 + 2x – 2 – 5

= 40 Ganti y dengan x. = x2 – 6
x+3
Yellow 1
(iii) h(10) = 5 = 1 g-1(x) = x + 9 Kuning 4
10 2 3 Blue
7. (a) 3 Biru
 g–1h(10) = g–1 1 (b) g2(x) = gg(x) Orange
2 = g(3x – 9) (b) f(2) = 1 – 3(2) Jingga
= 3(3x – 9) – 9 = –5
1 + 3 = 9x – 27 – 9 =5
2 = 9x – 36
= (c) When / Apabila f(x) = 2, (b) Relation many to one. The relation is a
8 1 – 3x = 2
function.
= 7 g2 4p  = 9 4p  – 36 Hubungan banyak dengan satu. Hubungan itu
16 3 3 ialah fungsi.

SPM Practice 1 24 = 12p – 36 1 – 3x = 2 and 1 – 3x = –2

Paper 1 12p = 60 3x = -1 dan 3x = 3 11. (a) f : x → 5x – 1 m = 5(4) – 1
1 f(x) = 5x – 1 18 = 20 – 1
1. (a) The function that maps a to b is g. p=5 x= – 3 x=1 18 ≠ 19
Fungsi yang memetakan a kepada b ialah g. f(4) = 5(4) – 1
\ Domain is / Domain ialah – 1 < x < 1.
(b) f -1(c) = b 5. m(x) = px + 3 3 The value of m is wrong because when
h(x) = 4x – 1
mh(x) = m(4x – 1) 8. (a) g(x) = 3x – 1 x = 4, the value of f(x) is not 18.
= p(4x – 1) + 3 Nilai m adalah salah kerana apabila x = 4, nilai
2. g(x) = mx – n h(x) = 6x f(x) bukan 18.
g(2) = m(2) – n
= 2m – n hg(x) = h(3x – 1)

Comparing / Bandingkan: = 6(3x – 1) (b) f(x) = 5x – 1
4px + q = 4px – p + 3
= 18x – 6 g–1f(x) = 2 + 3
q = –p + 3 1 x
fg(2) = 1 p=3–q (b) hg(x) = 3 g(x)

5(2m – n) + 1 = 1 1 g–1f(x) = 2 + 3x
3 x
10m – 5n + 1 = 1 18x – 6 = (3x – 1)

10m = 5n 6. (a) Let / Biar y = g(x) 3(18x – 6) = 3x – 1 g–1f(x) = g–1 (5x – 1)

m = 1 n y = 4x + 1 54x – 18 = 3x – 1 2 + 3x = g–1(5x – 1)
2 x
4x = y – 1 51x = 17
y–1
3. (a) f(x) = x x = 4 x = 1 Let y = 5x – 1,
3
8x – 4 = x 5x = y + 1

7x = 4 Since / Oleh sebab g-1(y) = x, 9. (a) g : x → 5x + 6 x= y+1
4 y–1 5
x = 7 g-1(y) = 4 (b) fg(x) = 9x + 12 31 y + 1 2
g(x) = 5x + 6 5
g -1(y) = 2 +

(b) f(x) = 8x – 4 Substitute y with x, Let / Biar y = g(x) y+1

f(3 – h) = 8(3 – h) – 4 Ganti y dengan x, y = 5x + 6 5
x–1
4h = 24 – 8h – 4 g-1(x) = 4 5x = y – 6 13 + 3y 5
5 y+1
4h = 20 – 8h y–6 = ×
5
12h = 20 x= 13 + 3y
5 y+1
h = 3 =

g -1(x) = 13 + 3x
x+1

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A5

Additional Mathematics Form 4 Chapter 1 Functions Additional Mathematics Form 4 Chapter 1 Functions

Paper 2 3. (a) f(x) = px – q (d) x–7 = 31 x – 2 2 (b) Let y = f(x)
1. g(x) = 2x – 3 2 2
y = 3x
2(x – 7)= 2[3x – 6] 5–x
fg(x) = f(2x – 3)
x –4 –3 –2 –1 0 1 1 2 34 = p(2x – 3) – q 2x = –1 1 5y – xy = 3x
3 = 2px – 3p – q 2
x=– 5y = 3x + xy

y 16 13 10 7 4 0 2 5 8 5y = x(3 + y)

y fg(2) = 4p – 3p – q x = 5y y
p – q = 5 ………… a 3+
HOTS Challenge
16 gf(x) = g(px – q) Since f -1(y) = x,
= 2(px – q) – 3
= 2px – 2q – 3 1. (a) f(x) = 3388.50x f -1(y) = 5y
f(800) = (3388.50)(800) 3+y
8 1– 1 2 2p1– 1 2 = 2 710 800 Rp
4 4
4 gf = – 2q – 3 Substitute y with x,

x –3 = – 1 p – 2q – 3 (b) Let y = f(x) f -1(x) = 5x
2 3+x
–4 0 131 4 y = 3 388.50x
1 y
– 2 p – 2q = 0 x = 3 388.50 Let y = g(x)

0 < f(x) < 16 1 Since f -1(y) = x, y = 2x – 3
2
4 p + 2q = 0 y 2x = y + 3
x 388.50
2. (a) f(x) = 3x + ,x ≠0 p + 4q = 0 …………… b f -1(y) = 3 x = y + 3
+ 4 2
f(–1) = 3(–1) = –3 – 4= –7 From / Dari a : p = q + 5 ………… c Substitute y with x,
(–1)
f -1(x) = x Since g-1(y) = x,
f(6) = 3(6) + 4 = 18 2 Substitute c into b: 3 388.50
6 3 y +3
Gantikan c ke dalam b: g-1(y) = 2
The inverse function obtained converts
(b) 1 1 2 = 31 1 2 + 4 q + 5 + 4q = 0 5q = –5 Substitute y with x,
3 3 money in Indonesian Rupiah (Rp) to
f 1 1 2 5q + 5 = 0 q = –1
3 Malaysian Ringgit (RM).
= 1 + 12 When / Apabila q = –1, Fungsi songsang yang diperoleh menukar wang g-1(x) = x +3
p = –1 + 5 Rupiah Indonesia (Rp) kepada Ringgit Malaysia 2
= 13 p=4 (RM).

(c) f(x) = 3x + 4 (c) fg(x) = f(2x – 3)
= 8 x
4 (b) f 2(x) = ff(x) = 16x + 5 2. (a) gf(x) = g1 mx 2 = 3(2x – 3)
3x + x f 3(x) = ff 2(x) = 64x + 21 n–x 5 – (2x – 3)
f 4(x) = ff 3(x) = 256x + 85
3x2 + 4 = 8x 9x – 15 = 21 mx 2 –m = 6x – 9
5–x n–x 8 – 2x
3x2 – 8x + 4 = 0 (c) f–1(x) = x–1
4 = 2mx – m
(x – 2)(3x – 2) = 0 n–x
2 g1 x – 1 2 g(x) = fg(x)
x = 2 or / atau x = 3 gf –1(x) = 4 2mx – m(n – x)
= n–x 6x – 9
2 2x – 3 = 8 – 2x
3 21 x – 1 2 –
The possible values of x is 2 and . = 4 3 3mx
n – mn 4x2 – 16x + 15 = 0
Nilai-nilai x yang mungkin ialah 2 dan 2 . x–7 = – x (2x – 5)(2x – 3) = 0
3 2
= Comparing / Bandingkan:
5–x =n–x
(d) f(x) = 3x + 4 f–1g(x) = f–1(2x – 3) x = 5 or 3
–13 x n =5 2 2
and / dan
3x + 4 = = 2x –3 – 1 3mx – mn = 9x – 15
x 4
3m = 9
3x2 + 4 = –13x = x–2 m=3
2
3x2 + 13x + 4 = 0

(3x + 1)(x + 4) = 0 gf–1(x) is not equal to f–1g(x) because
gf –1(x) tidak sama dengan f –1g(x) kerana
3x + 1 = 0 or / atau x + 4 = 0

x = – 1 or / atau x = –4 x–7 ≠ x–2
3 2 2

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A6

PTER Additional Mathematics Form 4 Chapter 2 Quadratic Functions

CHA 21 Quadratic Functions = 8 + √112 or / atau x= 8 – √112 x2 – – 4 x – 1 =0
2 2 5 5
Fungsi Kuadratik x 4 1
x2 + 5 x – 5 =0
= 9.29 = –1.29 (rejected)
(tidak diterima)
∴ x = 9.29 5x2 + 4x – 1 = 0

1. (a) x2 – 4x = 3 (b) a = 3, b = –±5,(c–=5)12 – 4(3)(1) 4. (a) Let / Biar α = 1 and / dan β = 6. (ii) Sum of roots:
x = –(–5) Sum of roots, α + β = 1 + 6
x2 – 4x + – 4 2 = 3 + – 4 2 2(3) Hasil tambah punca = 7 Hasil tambah punca:
2 2 (α + β)2 + (α + β)2
Product of roots, αβ = 1 × 6 = 42 + 42
x2 – 4x + (–2)2 = 3 + (–2)2 = 5 ± 13 Hasil tambah punca = 6 = 32
6
(x – 2)2 = 7 x2 – (α + β)x + (αβ) = 0 Product of roots:
x = 5 + 13 or / atau x = 5 – 13 x2 – 7x + 6 = 0
x – 2 = ±√7 6 6 Hasil darab punca:
x = 4.6458 or / atau = 1.4343 = 0.2324
(α + β)2 × (α + β)2
x = –0.6458 = 42 × 42
= 256
(b) x2 – 6x – 1 = 0 (c) a = –1, b =(88,)2c –= 4–(9–1)(–9) (b) Let / Biar α = –1 and / dan
x2 – 6x = 1 x = –8 ± 1 x2 – 32x + 256 = 0
β = 4 .

x2 – 6x + – 6 2 = 1 + – 6 2 2(–1) Sum of roots, α + β= –1 + 1 (b) 4x2 – px – 4 = 0
2 2 Hasil tambah punca = 4
= –8 ± 28 p
x2 – 6x + (–3)2 = 1 + (–3)2 –2 3 x2 – 4 x – 1 = 0
– 4
(x – 3)2 = 10
x – 3 = ±10 x = –8 + 28 or / atau x = –8 – 28 1 Product of roots = –1
–2 –2 4 Hasil darab punca
x = 6.1623 or / atau Product of roots, αβ = –1 ×
Hasil darab punca =
= 1.3542 = 6.6458 – 1 ∴ q– 1  = –1
4 4
x = –0.1623

2 3. (a) 3x(x + 2) = –x + 12 x2 – (α + β)x + (αβ) = 0 q=4
3
(c) x2 + 3x – =0 3x2 + 6x = –x + 12 3 1 p
4 4 4
x2 + 3x = 2 3x2 + 7x – 12 = 0 x2 – – x – =0 Sum of roots =
3
3 2 3 x2 + 7 x – 4 = 0 x2 + 3 x – 1 =0 Hasil tambah punca
 2 2 3  2 2 3 4 4
x2 + 3x + = + 7 p
x2 + 3 x = 4 ∴ q + – 1  = 4
4x2 + 3x – 1 = 0 4
x + 3 2 = 35
2 12 72 72
   x27 3 3 4 – 1 = p
3 35 + 3 x + 2 =4+ 2 5. (a) x2 – 4x – 5 = 0 4 4
x + 2 =± 12 p
b 15 = 4
7 7 2 7 2 α +β= – a 4
x = 0.2078 or / atau x2 + 3 x +  6 = 4 +  6
x = –3.2078 =4 p = 15

x + 7 2 = 193 αβ = c (c) Let the roots of the quadratic equation
6 36 a
x2 – kx + 8 = 0 be α and 2α.
2. (a) a = 2, b = –±5,(c–=5)–29– 4(2)(–9) x + 7 = ± 193 = –5 Biar punca persamaan kuadratik
x = –(–5) 6 36
2(2) (i) Sum of roots: x2 – kx + 8 = 0 menjadi α dan 2α.
x = 1.149 or / atau
5 ± 97 Hasil tambah punca:
= 4
x = –3.482 1 + 1 = β+α Product of roots = 8
α β αβ
x = 5 + 97 or / atau x = 5 – 97 Hasil darab punca
4 4 (b) x2 = 4(2x + 3) = – 4
x2 = 8x + 12 5 ∴ 2α2 = 8
= 3.7122 = –1.2122 α2 = 4
x2 – 8x – 12 = 0 Product of roots: α = ±4
= 2 or –2
–(–8) ± √(–8)2 – 4(1)(–12) Hasil darab punca:
2(1)
x =  1  1  = 1
α β αβ
= 8 ± √112 1
2 = – 5

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A7

Additional Mathematics Form 4 Chapter 2 Quadratic Functions Additional Mathematics Form 4 Chapter 2 Quadratic Functions

Sum of roots = k (b) For two different real roots: When (3m + 1)(m – 5) = 0,
Hasil tambah punca Bagi dua punca nyata yang berbeza:
∴ α + 2α = k x b2 – 4ac . 0 m = – 1 or m = 5
25 (p)2 – 4(1)(4) . 0 3
k = 3α 3 p2 – 16 . 0
k = 3(2) or k = 3(–2) (p – 4)(p + 4) . 0
For (3x – 2)(–x + 5) . 0, the range of When / Apabila (p – 4)(p + 4) = 0, m
= 6 = –6 2 p = 4 or p = –4
the values of x is 3 , x , 5. – 1 5
6. (a) x2 + 4x + 3 > 0 3
(x + 1)(x + 3) > 0
Untuk (3x – 2)(–x + 5) . 0, julai nilai x ialah ∴ – 1 , m , 5
When / Apabila (x + 1)(x + 3) = 0, 2 3
x = –1 or / atau x = –3 3 , x , 5.

x 7. (a) x2 – 8x + 16 = 0 p 11. (a) x(2x – 5) = p – 4
–3 –1
–4 4 2x2 – 5x – p + 4 = 0

Discriminant, b2 – 4ac ∴ p , –4 or p . 4 (i) For two different real roots,
Pembezalayan
= (–8)2 – 4(1)(16) Bagi dua punca nyata yang berbeza,
=0
9. (a) For two equal real roots: b2 – 4ac . 0

Bagi dua punca nyata yang sama: (–5)2 – 4(2)(–p + 4) . 0

For (x + 1)(x + 3) > 0, the range of the ∴ two equal real roots b2 – 4ac = 0 25 – 8(–p + 4) . 0
Dua punca nyata yang sama
values of x is x < –3 or x > –1. (2)2 – 4(q)(–3) = 0 25 + 8p – 32 . 0
Untuk (x + 1)(x + 3) > 0, julai nilai x ialah (b) 6x2 – 5x – 1 = 0
x < –3 atau x > –1. 4 + 12q = 0 8p – 7 . 0

12q = –4 8p . 7
7
q=– 1 p . 8
3
(b) (x + 3)(x – 5) < 0 Discriminant, b2 – 4ac (ii) For two equal real roots,
Pembezalayan (b) For two equal real roots: Bagi dua punca nyata yang sama,
When / Apabila (x + 3)(x – 5) < 0, = (–5)2 – 4(6)(–1) Bagi dua punca nyata yang sama: b2 – 4ac = 0
x = –3 or / atai x = 5 = 49 . 0 b2 – 4ac = 0
(q – 1)2 – 4(1)(1) = 0
x ∴ two different real roots q2 – 2q + 1 – 4 = 0 Hence / Maka, p = 7
–3 5 Dua punca nyata yang berbeza q2 – 2q – 3 = 0 8
(q – 3)(q + 1) = 0
For (x + 3)(x – 5) < 0, the range of the q = 3 or / atau q = –1 (b) (q + 1)x2 – 8x + p = 0
values of x is –3 < x < 5.
Untuk (x + 3)(x – 5) < 0, julai nilai x ialah (c) 2(x2 + 2) = –x For two equal real roots,
–3 < x < 5. 2x2 + 4 = –x
Bagi dua punca nyata yang sama,
(c) –x2 + 5x + 6 , 0 2x2 + x + 4 = 0
(x + 1)(–x + 6) , 0 b2 – 4ac = 0
Discriminant, b2 – 4ac
When / Apabila (x + 1)(–x + 6) = 0, Pembezalayan 10. (a) 2x2 + 6x + (–3 + m) = 0 (–8)2 – 4(q + 1)(p) = 0
x = –1 or / atau x = 6 = (1)2 – 4(2)(4)
= –31 , 0 No real roots: 64 – 4p(q + 1) = 0

∴ no real roots Tiada punca nyata: 64 – 4pq – 4p = 0
Tiada punca nyata
b2 – 4ac , 0 16 – pq – p = 0

(6)2 – 4(2)(–3 + m) , 0 pq + p = 16

(6)2 – 8(–3 + m) , 0 p(q + 1) = 16

36 + 24 – 8m , 0 p = 16
q+1
60 – 8m , 0

x 8. (a) For two different real roots: –8m , –60 (c) 3x2 – 5x + 2p + 3 = 0
–1 6
8m . 60
Bagi dua punca nyata yang berbeza: 15 No real roots,
m . 2
For (x + 1)(–x + 6) , 0, the range of the b2 – 4ac . 0 Tida punca nyata,

values of x is x , –1 or x . 6. (–5)2 – 4(4)(3p – 1) . 0 (b) No real roots: b2 – 4ac , 0
Untuk (x + 1)(–x + 6) , 0, julai nilai x ialah Tiada punca nyata
x , –1 atau x . 6. 25 – 16(3p – 1) . 0 b2 – 4ac , 0 (–5)2 – 4(3)(2p + 3) , 0
(–3m – 1)2 – 4(1)(12m + 4) , 0
25 – 48p + 16 . 0 9m2 + 6m + 1 – 48m – 16 , 0 25 – 12(2p + 3) , 0
9m2 – 42m – 15 , 0
41 – 48p . 0 3m2 – 14m – 5 , 0 25 – 24p – 36 , 0
(3m + 1)(m – 5) , 0
(d) –3x2 + 17x – 10 . 0 –48p . –41 –24p – 11 , 0
(3x – 2)(–x + 5) . 0
48p , 41 –24p , 11
41
When / Apabila (3x – 2)(–x + 5) = 0, p, 48 p . – 11
2 24
x = 3 or / atau x = 5

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A8

Additional Mathematics Form 4 Chapter 2 Quadratic Functions Additional Mathematics Form 4 Chapter 2 Quadratic Functions

12. (a) has two different real roots (b) does not have real roots (c) f(x) = 2x2 + 10x + 7 (c) (i) In the form f(x) = a(x – h)2 + k
tidak mempunyai punca nyata
mempunyai dua punca nyata yang berbeza b2 – 4ac , 0  = 2 x2 + 5x + 7 Dalam bentuk f(x) = a(x – h)2 + k
(2h)2 – 4(2h – 3)(–1) , 0 2 h = –2, k = –9
b2 – 4ac . 0 4h2 + 4(2h – 3) , 0 ∴ f(x) = a(x + 2)2 – 9
4h2 + 8h – 12 , 0
(8)2 – 4(k)(6) . 0 h2 + 2h – 3 , 0  = 2 x2 + 5x +  5 2 + 7 –  5 2 = a(x2 + 4x + 4) – 9
(h – 1)(h + 3) , 0 2 2 2 = ax2 + 4ax + 4a – 9
64 – 24k . 0 When / Apabila (h – 1)(h + 3) = 0,
h = 1 or h = –3
–24k . –64  = x 5 2 7 25
2 2 4
24k , 64 2 + + – Comparing / Bandingkan:
8 ax2 + bx + c = ax2 + 4ax + 4a – 9
k , 3  = x 5 2 11 ∴ b = 4a and / dan c = 4a – 9
2 4
2 + –

(b) has two different real roots = 2x + 5 2 – 11 f(–1) = a(–1)2 + 4a(–1) + 4a – 9
mempunyai dua punca nyata yang berbeza 2 2 –7 = a – 4a + 4a – 9
b2 – 4ac . 0 –7 = a – 9
(k – 3)2 – 4(1)(1) . 0 h Minimum value / Nilai minimum a=2
k2 – 6k + 9 – 4 . 0 –3 1
k2 – 6k + 5 . 0 b = 4(2)
(k – 5)(k – 1) . 0 \ –3 , h , 1 =– 11 =8
2
c = 4(2) – 9
Axis of symmetry / Paksi simetri = –1

When (k – 5)(k – 1) = 0, 15. (a) f(x) = x2 – 6x – 3 x = – 5
k = 5 or k = 1 2
– 6 2 – 6 2
= x2 – 6x + 2 – 3 – 2 Minimum point / Titik minimum

= x2 – 6x + (–3)2 – 3 – (–3)2 = – 5 , – 11  (ii) x = –2
2 2

15 k = (x – 3)2 – 3 – 9 –3 + 1
2
\ k , 1 or / atau k . 5 = (x – 3)2 – 12 16. (a) (i) x= 17. (a) a . 0, then the shape of the graph is
.
Minimum value / Nilai minimum =– 2
= –12 2 a . 0, maka bentuk graf ialah .

13. (a) has two equal real roots Axis of symmetry / Paksi simetri = –1 Discriminant, b2 – 4ac = (–4)2 – 4(1)(–5)
mempunyai dua punca nyata yang sama x=3
b2 – 4ac = 0 (ii) p = –1 Pembezalayan = 16 + 20
(–2)2 – 4(1)(p – 5) = 0
4 – 4(p – 5) = 0 (iii) (–1, –8) = 36 . 0
4 – 4p + 20 = 0
–4p = –24 Minimum point / Titik minimum (b) (i) f(x) = –x2 + 8x + k2 ∴ f(x) = x2 – 4x – 5 has two different
p =6 = (3, −12) = –[x2 – 8x – k2] real roots.

(b) f(x) = –x2 + 2x – 8  = – 8 2 – 8 2 f(x) = x2 – 4x – 5 mempunyai dua punca
x2 – 8x + – 2 – k2 – 2
= –[x2 – 2x + 8] nyata yang berbeza.

(b) has two equal real roots  =– x2 – 2x + – 2 2 + 8 – – 2 2 = –[x2 – 8x + (–4)2 – k2 – (–4)2] f(x) = x2 – 4x – 5 2 2
2 2 = –[(x – 4)2 – k2 – 16]
mempunyai dua punca nyata yang sama = –(x – 4)2 + k2 + 16 = x2 – 4x + – 4 – 5 – – 4
2 2
b2 – 4ac = 0 = –[x2 – 2x + (–1)2 + 8 – (–1)2]
= –[(x – 1)2 + 8 – 1] = x2 – 4x + (–2)2 – 5 – (–2)2
(4)2 – 4(p –3)(–5) = 0 = –[(x – 1)2 + 7]
= –(x – 1)2 – 7 = (x – 2)2 – 5 – 4
16 + 20(p – 3) = 0 Therefore, the equation of axis of
symmetry is x = 4. = (x – 2)2 – 9
16 + 20p – 60 = 0 Maka, persamaan paksi simetri ialah
x = 4.
20p = 44 Maximum value / Nilai maksimum ∴ The minimum point is (2, –9).
11 = –7 Titik minimum ialah
p = 5
The axis of symmetry is x = 2.
14. (a) does not have real roots Axis of symmetry / Paksi simetri (ii) k2 + 16 = 25
x=1 Paksi simetri ialah
k2 = 9
tidak mempunyai punca nyata When / Apabila f(x) = 0, x2 – 4x – 5 = 0
k = ±√9
b2 – 4ac , 0 Maximum point / Titik maksimum k = 3 or k = –3 (x – 5)(x + 1) = 0
= (1, −7)
(4)2 – 4(8)(h) , 0 x = 5 or / atau x = –1

16 – 32h , 0 ∴ The x-intercepts are (5, 0) and (–1, 0).
Pintasan-x ialah (5, 0) dan (–1, 0).
–32h , –16

32h . 16

h . 1
2

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A9

Additional Mathematics Form 4 Chapter 2 Quadratic Functions Additional Mathematics Form 4 Chapter 2 Quadratic Functions

When / Apabila x = 0,  = –2 x – 7 2 – 2– 49 18. (a) (i) 5x2 – 9x – 2 . 0
f(0) = (0)2 – 4(0) – 5 4 16 (x – 2)(5x + 1) . 0
x
= –5 4 4
 = –2 x – 7 2 81 When / Apabila (x – 2)(5x + 1) = 0, – 3 3
∴ The y-intercept is (0, –5). 4 16
Pintasan-y ialah – x=2 ∴ For / Untuk 9k2 – 16 , 0,

or / atau x = – 1 – 4 k 4 .
5 3 3
When / Apabila = –2x 7 2 81 , ,
x = –2, f(–2) = (–2)2 – 4(–2) – 5 – 4 + 8

=4+8–5 ∴ The maximum point is 1 3 , 10 1 . x (c) (i) h(x) = –x2 + 6x + 1
=7 4 8 –1 1
∴ (–2, 7) = –(x2 – 6x – 1)
4
Titik maksimum ialah  =– x2 – 6x +  –6 2 – 1 –  –6 2
2 2
When / Apabila The axis of symmetry is x = 1 3 . ∴ For / Untuk 5x2 – 9x – 2 . 0,
x = 6, f(6) = (6)2 – 4(6) – 5 Paksi simetri ialah 4 1
x , – 5 or / atau x . 2. = –[x2 – 6x + (–3)2 – 1 – (–3)2]
= 36 – 24 – 5
=7 = –[(x – 3)2 – 10]
∴ (6, 7)
When / Apabila f(x) = 0, –2x2 + 7x + 4 = 0 (ii) 5x2 – 9x – 2 , 0 = –(x – 3)2 + 10
f(x) (x – 2)(5x + 1) , 0
(2x + 1)(4 – x) = 0 1 a = –1 , 0, so the graph has a
2 maximum point.
x =– or / atau x=4 When / Apabila (x – 2)(5x + 1) = 0, a = –1 , 0 jadi, graf mempunyai titik
maksimum.
(–2, 7) x=2 (6, 7) ∴ The x-intercepts are – 1 , 0 x = 2 or / atau x = – 1
Pintasan- x ialah 2 5 The maximum value is 10 when
x = 3.
x and / dan (4, 0). Nilai maksimum ialah 10 apabila x = 3.
5
–1 0 f(x) = x2 – 4x – 5 1 2 x Hence, the maximum height is 10 m.
–5 5 Maka, tinggi maksimum ialah 10 m.
When /Apabila x = 0, –
f(0)= –2(0)2 + 7(0) + 4
∴ For / Untuk 5x2 – 9x – 2 , 0,
=4 1
5
(2, –9) ∴ The y-intercept is (0, 4). – , x , 2. (ii) h(x) = 0
Pintasan-y ialah
–x2 + 6x + 1 = 0
(b) (i) x2 + 4x + 4 , 11x + 12 –b ± √b2 – 4ac
x2 – 7x – 8 , 0 x= 2a
(b) a , 0, then the shape of the graph is (x – 8)(x + 1) , 0
. When / Apabila x = 5, When / Apabila (x – 8)(x + 1) = 0, = –6 ± √62 – 4(–1)(1)
f(5) = –2(5)2 + 7(5) + 4 x = 8 or / atau x = –1 2(–1)
a , 0, maka bentuk graf ialah .
= –50 + 35 + 4
Discriminant, b2 – 4ac = (7)2 – 4(–2)(4) = –11 = –6 ± √40
∴ (5, –11) –2
Pembezalayan
= 49 + 32 x = –6 + √40 or x = –6 – √40
–2 –2
= 81 . 0 When / Apabila x = –1, x
f(–1) = –2(–1)2 + 7(–1) + 4 –1 8 = –0.16 (ignore) = 6.16

∴ f(x) = –2x2 + 7x + 4 has two different = –2 – 7 + 4 ∴ For / Untuk x2 – 7x – 8 , 0, (diabaikan)
real roots. = –5 –1 , x , 8.
f(x) = –2x2 + 7x + 4 mempunyai dua punca ∴ (–1, –5) Hence, the horizontal distance is
(ii) 3kx – x2 = 4 6.16 m.
nyata yang berbeza. f(x) x2 – 3kx + 4 = 0 Maka, jarak mengufuk ialah 6.16 m.
No real roots / Tiada punca nyata,
f(x) = –2x2 + 7x + 4 (1 3 , 10 1 ) b2 – 4ac , 0
4 8 (–3k)2 – 4(1)(4) , 0
 =–2 x2 – 7 x – 2 9k2 – 16 , 0 SPM Practice 2
    =–2 2 (3k – 4)(3k + 4) , 0
f(x) = –2x2 + 7x + 4
– 7 2 – 7 2 4 x Paper 1
2 2
x2 – 7 x + 2 –2– 2 1 4 1. (a) p = 1 The graph has minimum point, p . 0
2 – 2 0 Graf mempunyai titik minimum, p . 0
(5, –11) When / Apabila (3k – 4)(3k + 4) = 0,

–2  = x2 – 7 x + – 7 2 – 2 – – 7 2 (–1, –5) k = 4 or / atau k = – 4 (b) f(x) = px2 + 6x + q
2 4 4 3 3 = x2 + 6x + q
x = 134

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A10

Additional Mathematics Form 4 Chapter 2 Quadratic Functions Additional Mathematics Form 4 Chapter 2 Quadratic Functions

For two equal real roots, x2 – (Sum of new roots / Hasil tambah punca 8. f(x) , 0 10. (4x)(3x) – 3x2 > (x2 + 32)
baharu)x + (Product of of new roots / Hasil 10 + 3x – x2 , 0 12x2 – 3x2 > x2 + 32
Untuk dua punca yang sama, darab punca baharu) = 0 8x2 – 32 > 0
b2 – 4ac = 0 –x2 + 3x + 10 , 0 8(x2 – 4) > 0
x2 – 9 x + 81 = 0 (x + 2)(5 – x) , 0
(6)2 – 4(1)(q) = 0 2 16 8(x – 2)(x + 2) > 0
36 – 4q = 0 When / Apabila (x + 2)(5 – x) = 0, x = –2
– 4q = –36 16x2 – 72x + 81 = 0 or / atau x = 5 When / Apabila 8(x – 2)(x + 2) = 0, x = 2
q=9 or / atau x = –2

2. 4x2 + 7x < 2 5. f(x) = –x2 – 2kx + 4k – 5 x

4x2 + 7x – 2 < 0 f(x) , 0, the function does not –2 5

(4x – 1)(x + 2) < 0 have real roots. x
–2 2
When / Apabila (4x – 1)(x + 2) = 0, f(x) , 0, fungsi tidak f(x)
For / Untuk 8(x – 2)(x + 2) > 0, x < –2
x = 1 or / atau x = –2 mempunyai punca nya \ For / Untuk (x + 2)(5 – x) , 0, x , –2 (ignore) or / (diabaikan) atau x > 2.
4 or / atau x . 5 Hence / Maka, x > 2
b2 – 4ac , 0 x
(–2k)2 – 4(–1)(4k – 5) , 0 0

x 4k2 + 16k – 20 , 0 9. (a) x2 + (p + 4)x – p2 = 0
–2 1
k2 + 4k – 5 , 0 a = 1, b = p + 4, c = –p2
4
(k + 5)(k – 1) , 0 Let the roots be / Katakan
For / Untuk (4x – 1)(x + 2) 0, –2 x 1 . 11. At intersection points,
\ < < < 4 root 1 / punca 1 = a
root 2 / punca 2 = –a Pada titik persilangan,
3. (x + p)2 = 49 When / Apabila (k + 5)(k – 1) = 0, (p – 3)x2 – x + 8 = 4x + 6
k = –5, k = 1
Substitute x = 4 into the equation: b (p – 3)x2 – x – 4x + 8 – 6 = 0
Sum of roots / Hasil tambah punca = – a (p – 3)x2 – 5x + 2 = 0
Gantikan x = 4 ke dalam persamaan:

(4 + p)2 = 49 k + (–a) = – p + 4
–5 1 1
4 + p = ± √49 a Two intersection points,
4 + p = ±7 For / Untuk (k + 5)(k – 1) , 0, –5 , k , 1.
Hence / Maka, m = –5, n = 1 p + 4= 0 Dua titik persilangan,

4 + p = 7 or 4 + p = –7 6. x2 + 30 , 11x p = –4 b2 – 4ac . 0
x2 – 11x + 30 , 0
p = 3 atau p = –11 (x – 6)(x – 5) , 0 (–5)2 – 4(p – 3)(2) . 0

When / Apabila (x – 6)(x – 5) = 0, x = 6 Product of roots = c 25 – 8(p – 3) . 0
or / atau x = 5 Hasil darab punca = a
4. 16x2 – 24x + 9 = 0 25 – 8p + 24 . 0
–
a = 16, b = –24, c = 9 b p2 –8p + 49 . 0
a 1
Sum of roots, a + b = – –8p . –49
Hasil tambah punca = –p2
(–24) 8p , 49
16 = –(–4)2
= – p, 49
= –16 8
3
= 2

ab = c (b) mx2 – 9nx + m = 0 12. hx2 – 5x + k = 0
a a = h, b = –5, c = k

9 x Two equal roots, b
16 56 a
= Dua punca sama, Sum of roots =– (–5)
For / Untuk (x – 6)(x – 5) , 0, 5 , x , 6. Hasil tambah punca =–
b2 – 4ac = 0 h

Sum of new roots, 3a + 3b = 3(a + b) (–9n)2 – 4(m)(m) = 0

Hasil tambah punca baharu = 31 3 2 81n2 – 4m2 = 0 = 5
2 h
7. (a) f(x) = x2 + 6x + 1 6 22 + h – 1 6 22 4m2 = 81n2
2 2
= 9 m2 = 81 Product of roots = c
2 = x2 + 6x + (3)2 + h – (3)2 n2 4 Hasil darab punca a

Product of new roots, 3a × 3b = 9(ab) = (x + 3)2 + h – 9 1 m 22 = 1 9 22 = k
n 2 h
Hasil darab punca baharu 91 9 2
= 16 (b) h – 9 = 7
h = 16 \m:n=9:2
Given roots b and 3b,
= 81 Diberi punca-punca b dan 3b,
16

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A11

Additional Mathematics Form 4 Chapter 2 Quadratic Functions Additional Mathematics Form 4 Chapter 2 Quadratic Functions

Sum of roots / Hasil tambah punca = b + 3b When / Apabila m = 3 n, Paper 2
= 4b 2
3 4 1. (a) x(x – 8) = 3h – 15 f(x)
2 n k (0, 44) x=5
Product of roots / Hasil darab punca = b(3b) . x2 – 8x = 3h – 15

= 3b2 3n . 4 x2 – 8x – 3h + 15 = 0
2 k
Comparing: / Bandingkan: Two different real roots, (8, 12)

5 = 4b ............................... 3n . 8 Dua punca nyata berbeza, 03 7 x
h k
b2 – 4ac . 0 (5, –6) f(x) = 2(x – 5)2 – 6
k 8
h = 3b2 .............................. n . 3k (–8)2 – 4(1)(–3h + 15) . 0

From / Dari , b = 5 ...... 64 – 4(–3h + 15) . 0
4h
14. (a) f(x) = 4[2h – (x – 3)2] 64 + 12h – 60 . 0
= 8h – 4(x – 3)2
Substitute  into , = –4(x – 3)2 + 8h 12h + 4 . 0 (c) Equation of reflected curve,
Persamaan bagi lengkung terpantul,
Gantikan  ke dalam , 12h . – 4 f(x) = –[2(x – 5)2 – 6]
1 = –2(x – 5)2 + 6
k 31 5 22 h . – 3
h 4h 3. (a) (i) x2 – 7x + 12 = 0
= 8h = h + 7 (x – 4)(x – 3) = 0
7h = 7 x = 4 or x = 3
k = 31 25 2 h =1 (b) From x(x – 8) = 3h – 15, sum of roots Since / Oleh sebab p . q, p = 4
h 16h2 is 8 and product of roots is –3h + 15. and / dan q = 3.
Dari x(x – 8) = 3h – 15, hasil tambah punca
k = 75 (b) f(x) = 4[2h – (x – 3)2] (ii) If x2 – 7x + 12 . 0,
h 16h2 = 4[2 – (x – 3)(x – 3)] ialah 8 dan hasil darab punca ialah –3h + 15. (x – 4)(x – 3) . 0
= 4[2 – (x2 – 6x + 9)] a + b = 8...................  When / Apabila (x – 4)(x – 3) = 0,
16h2k = 75h = 8 – 4(x2 – 6x + 9) ab = –3h + 15.......  x = 4 or / atau x = 3
= 8 – 4x2 + 24x – 36
16hk = 75 = –4x2 + 24x – 28 For / Untuk x2 + kx + 3 = 0, x
34
h = 75 a + b = –k
16k 2 2 \ For / Untik x2 – 7x + 12 . 0,
x , 3 or / atau x . 4.
a + b = –2k .......... 
(b) p – 2 = 4 – 2
13. (mx)2 – 3nx + 1 = 0 f(x) = –4x2 + 24x – 28 1 a 21 b 2 = 3 =2
m2x2 – 3nx + 1 = 0 2 2
b2 – 4ac = (24)2 – 4(–4)(–28) q+4=3+4
b2 – 4ac = 0 = 576 – 448 ab =3 =7
= 128 . 0 4
(–3n)2 – 4(m2)(1) = 0 ab = 12 ............  The new roots are 2 and 7.
Two different real roots Punca-punca baharu ialah 2 dan 7.
9n2 – 4m2 = 0 Dua punca nyata berbeza  = , Sum of roots = 2 + 7
Hasil tambah punca = 9
4m2 = 9n2 –2k = 8 Product of roots = 2 × 7
Hasil darab punca = 14
m2 = 9 n2 k = –4 \ New equation: x2 – 9x + 14 = 0
4
Persamaan baharu:
m=± 9 n2  = ,
4 –3h + 15 = 12
3 3 15. (a) Quadratic function, the highest power
m= 2 n or – 2 n (Rejected) is 2. –3h = –3
Fungsi kuadratik, kuasa tertinggi ialah 2. h =1
(Tidak diterima) n = –2

kx2 – 4x + m = 0 (No roots / Tiada punca) (b) (i) f(x) = ax2 + bx + c 2. (a) p = 3 + 7
SOR / HTP 2
= –p + p = 0 =5
b2 – 4ac , 0 \b=0

(–4)2 – 4(k)(m) , 0 3q = –6
q = –2
16 – 4km , 0

–4km , –16 (ii) f(x) = ax2 + bx + c (b) f(x) = 2(x – 5)2 – 6
When / Apabila x = 8, f(8) = 2(8 − 5)2 − 6
4km . 16 c = 12
a
km . 4 POR / HDP = When / Apabila x = 0, f(0) = 2(0 − 5)2 − 6
= 44
m . 4 c =c
k a
3 4
Substitute m = 2 n into the range of m . k a=1
Gantikan
ke dalam julat

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A12

Additional Mathematics Form 4 Chapter 2 Quadratic Functions Additional Mathematics Form 4 Chapter 2 Quadratic Functions

4. (a) x2 + 3(x – p) = 0 HDP = (–3a)(–3B) (b) Area of the right-angled triangle PQR q – 9 = 6q + 1
x2 + 3x – 3p = 0 = 9aB 4p 4p
= 9(3)(–2) Luas bagi segi tiga bersudut tegak PQR
For roots q and –2q, = 27(–2) 1 9
Bagi punca-punca q dan –2q. = –54 = 2 (12.32 – 2)[2(12.32) + 1] q – 4(2) = 6q + 1
Sum of roots = q + (–2q) 4(2)
Hasil tambah punca = –q New quadratic equation
Persamaan kuadratik baharu = 132.30 cm2 q – 9 = 6q + 1
Product of roots = q(–2q) = x2 – (HTP)x + HDP 8 8
Hasil darab punca = –2q2 = x2 – (–3)x + (–54)
\ x2 + qx – 2q2 = 0 = x2 + 3x – 54 2. (a) f(x) = px2 – 3x + q q– 9 = 3q + 1
8 4 8
     =px2– 3 x + q
p p q = 5
4 4
2 2
– 3 – 3 q=5
3 p + q p
=p x2 – p x + p –
Compare with / Bandingkan dengan 2 2
x2 + 3x – 3p = 0, q = 3 (b) a = 3 dan B = –2 (b) f(x) = px2 – 3x + q
and / dan 3 q = 2x2 – 3x + 5
–2q2 = –3p SOR = 3 + 2 = p x2 – p x + – 3 2 + p – – 3 2
HTP a B 2p 2p
2q2 = 3p
2(3)2 = 3p = 3B + 2a = px – 3 2 + q – 9 2x2 – 3x + 5 > 4
2p p 2x2 – 3x + 1 > 0
3p = 18 aB 4p2 (x – 1)(2x – 1) > 0
p=6 3(–2) + 2(3)
= 3(–2) =0 = px – 3 2 + q – 9
(b) p – 3 = 6 – 3 2p 4p
=3 When / Apabila (x – 1)(2x – 1) = 0,
1 3 21 2 2 1
p+1=6+1 POR = a B  Compare with f(x) 3 2+ 6q4p+1, x = 1 or / atau x = 2
=7 HDP = 4
6 Bandingkan dengan = p x–
The new roots are 3 and 7.
Punca-punca baharu ialah 3 dan 7. aB 2p = 4
6 p=2
Sum of roots = 3 + 7 = (3)(–2) x
Hasil tambah punca = 10 11
= –1 and for minimum value 2
Product of roots = 3 × 7 dan bagi nilai minimum
Hasil darab punca = 21 New quadratic equation \ For (x – 1)(2x – 1) > 0, x< 1
Persamaan kuadratik baharu or x > 1. 2
\ New equation: x2 – 10x + 21 = 0
Persamaan baharu: = x2 – (SOR / HTP)x + (POR / HDP)

= x2 – 0x – 1

= x2 – 1

HOTS Challenge

5. (a) x2 – x – 6 = 0 1. (a) (2x + 3)2 = (x – 2)2 + (2x + 1)2
(x – 3)(x + 2) = 0
x = 3 or x = –2 4x2 + 12x + 9 = x2 – 4x + 4 + 4x2 + 4x + 1
a = 3 and B = –2 because / kerana a . B.
4x2 + 12x + 9 = 5x2 + 5

x2 – 12x – 4 = 0

New roots / Punca-punca baharu x = – (–12) ± √(–12)2 – 4(1)(–4)
= –3a and –3B 2

HTP = –3a + (–3B) = 12 ± √160
= –3a – 3B 2
= –3(a + B)
= –3(3 – 2) x = 12 + √160 or x = 12 – √160
= –3 2 atau 2

= 12.32 = –0.32

(rejected)

\ x = 12.32 (tidak diterima)

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A13

PTER Additional Mathematics Form 4 Chapter 3 Systems of Equations

CHA 31 Systems of Equations Substitute  into , Substitute  into ,
Gantikan  ke dalam ,
Sistem Persamaan Gantikan  ke dalam , 14(4 – z) + 19z = 91
56 – 14z + 19z = 91
5 5y + 6z + 12  – 4y + 3z = 9
2 5z = 35
25y + 30z + 60 – 8y + 6z = 18 z =7

17y + 36z = –42 ..... Substitute z = 7 into ,
Gantikan z = 7 ke dalam ,
1. (a) Yes. This is because the three equations (b) x – 2y + 5z = –22 ......  From / Dari , y + 2z = –2 y =4–7
have three variables and the highest x – 6y + 3z = –12........  y = –2 – 2z ......
power of each variable is 1. 3x – 4y + z = 19 .........  = –3
Ya. Ini kerana ketiga-tiga persamaan mempunyai Substitute  into ,
tiga pemboleh ubah dan kuasa tertinggi bagi  – : 4y + 2z = –10 Gantikan  ke dalam , Substitute y = –3 and z = 7 into ,
setiap pemboleh ubah ialah 1. 2y + z = –5 ............  17(–2 – 2z) + 36z = –42 Gantikan y = –3 dan z = 7 ke dalam ,

(b) No. This is because the highest power  × 3: 3x – 18y + 9z = –36 ..........  –34 – 34z + 36z = –42 x – 2(–3) – 4(7) = –17
of the equation x2 + 4y + z = 9 is 2. : 3x – 4y + z = 19 2z = –8 x + 6 – 28 = –17
Bukan. Ini kerana kuasa tertinggi bagi  – : –14y + 8z = –55 ..........  z = –4 x=5
persamaan x2 + 4y + z = 9 ialah 2.
Substitute z = –4 into , \ x = 5, y = –3, z = 7
(c) No. This is because there are only two Gantikan z = –4 ke dalam ,
variables in each linear equation.  × 7: 14y + 7z = –35 ..........  y = –2 – 2(–4)
Bukan. Ini kerana terdapat dua pemboleh ubah
sahaja dalam setiap persamaan linear. : –14y + 8z = –55 = –2 + 8 4. (a) x + 2y – 3z = 22 .............
=6 –2x – 4y + 6z = –34 ............
 + : 15z = –90 5y – z = 28 ..............
Substitute y = 6 and z = –4 into ,
z = –6 Gantikan y = 6 dan z = –4 ke dalam ,  × 2: 2x + 4y – 6z = 44.......... 
2x – 5(6) – 6(–4) = 12  + : 0 + 0 + 0 = 10
2. (a) x + y + z = –2 …………  Substitute z = –6 into ,
2x – 3y + 4z = 14 …………  2x – 30 + 24 = 12 0 = 10
5x + 2y – 3z = 10 …………  Gantikan z = –6 ke dalam , 2x = 18
x =9 \ Since 0 ≠ 10, the system has no
2y – 6 = –5 solution.
Oleh sebab 0 ≠ 10, sistem itu tiada
2y = 1 penyelesaian.
1
 × 2: 2x + 2y + 2z = –4.......... y= 2 (b) 2x + 5y – z= –9 ................ 
3x – 4y + 6z = –9 ........ 
: 2x – 3y + 4z = 14 Substitute y= 1 and z = –6 into ,
2 –17x – 31y + z = 72 ........ 
 – : 5y – 2z = –18........ 1 \ x = 9, y = 6, z = –4
2
 × 5:10x – 15y + 20z = 70.......... Gantikan y = dan z = –6 ke dalam ,
 × 2: 10x + 4y – 6z = 20..........
 – : –19y + 26z = 50.......... x – 2( 1 ) + 5(–6) = –22 (b) x – 2y – 4z = –17 ............... 
2 3x + 7y + z = 1 .................. 
x – 1 – 30 = –22
4x + 6y + 3z = 23 ................ 
x – 31 = –22  × 6: 12x + 30y – 6z = –54 ... 
From / Dari , x = 2y + 4z – 17....... 
 × 19: 95y – 38z = –342...... x=9  + : 15x + 26y = –63 ... 
Substitute  into ,
 × 5: –95y + 130z = 250........ \x = 9, y= 1 , z = –6 Gantikan  ke dalam ,  + : –15x – 26y = 63 ..... 
2
 + : 92z = –92 3(2y + 4z – 17) + 7y + z = 1  + : 0+0=0
6y + 12z – 51 + 7y + z = 1
z = –1 3. (a) 2x – 5y – 6z = 12 .......  13y + 13z = 52 0=0
x + 3y + 8z = –5 .......  y + z = 4............. 
Substitute z = –1 into , 5x – 4y + 3z = 9 ......... 
Substitute  into ,
Gantikan z = –1 ke daman , Gantikan  ke dalam , \ Since 0 = 0, the system has infinite
4(2y + 4z – 17) + 6y + 3z = 23 solutions.
5y – 2(–1) = –18
8y + 16z – 68 + 6y + 3z = 23 Oleh sebab 0 = 0, sistem itu mempunyai
5y + 2 = –18 From / Dari , x = 5y + 6z + 12  14y + 19z = 91......... 
2 penyelesaian tak terhingga.
5y = –20 From / Dari , y + z = 4
y = 4 – z ............. 
y = –4 Substitute  into ,

Substitute y = –4 and z = –1 into , Gantikan  ke dalam , 5. (a) Let / Biar x = price of a loaf of bread
Gantikan y = –4 dan z = –1 ke dalam , harga bagi sebuku roti
x – 4 – 1 = –2 5y + 6z + 12 + 3y + 8z = –5
2 y = price of a packet of biscuits
x – 5 = –2 harga bagi sepaket biskut
x=3 5y + 6z + 12 + 6y + 16z = –10
z = price of a box of milk
11y + 22z = –22
harga bagi sekotak susu
\ x = 3, y = –4, z = –1 y + 2z = –2...........

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A14

Additional Mathematics Form 4 Chapter 3 Systems of Equations Additional Mathematics Form 4 Chapter 3 Systems of Equations

3x + 5y + 2z = 41 .............  From / Dari , 6. (a) x – 5y = 9 ................. 7. (a) 3 – 4 = –5...................... 
2x + 8y + 3z = 56 .............  x y
4x + 7y + 5z = 70 .............  2x = 2.80 – y – 3z x2 – 4y2 + 6xy = –12..............
2x + y = –5...................... 
2.80 – y – 3z From / Dari , x = 9 + 5y ……
41 – 5y – 2z x= 2 ............. 3y – 4x
3 Substitute  into , From / Dari , xy = –5
From / Dari , x = .... 
Gantikan  ke dalam ,
Substitute  into , Substitute  into , 3y – 4x = –5xy..... 
(9 + 5y)2 – 4y2 + 6y(9 + 5y) = –12
Gantikan  ke dalam , Gantikan  ke dalam ,
81 + 90y + 25y2 – 4y2 + 54y + 30y2 + 12 = 0 From / Dari , y = –5 – 2x............. 
41 – 5y – 2z  2.80 –y – 3z  + 2y + 4z = 3.30
2 3  2 51y2 + 144y + 93 = 0
+ 8y + 3z = 56 Substitute  into ,
17y2 + 48y + 31 = 0
82 – 10y – 4z + 24y + 9z = 168 2.80 – y – 3z + 4y + 8z = 6.60 Gantikan  ke dalam ,
(y + 1)(17y + 31) = 0
14y + 5z = 86 .......  3y + 5z = 3.80 31 3(–5 – 2x) – 4x = –5x(–5 – 2x)
y = –1 or / atau y = – 17
30y + 50z = 38 .......  –15 – 6x – 4x = 25x + 10x2
Substitute  into ,
Substitute the values of y into , 10x2 + 35x + 15 = 0
Gantikan  ke dalam , Gantikan nilai-nilai y ke dalam ,
Substitute  into , When y = –1, x = 9 + 5(–1) 2x2 + 7x + 3 = 0
41 – 5y – 2z
4 3  + 7y + 5z = 70 Gantikan  ke dalam , =4 (2x + 1)(x + 3) = 0
1
164 – 20y – 8z + 21y + 15z = 210 3 2.80 –y – 3z  + 4y + z = 4.30 x =– 2 or / atau x = –3
2
y + 7z = 46.....  Substitute the values of x into ,
8.40 – 3y – 9z + 8y + 2z = 8.60 When / Apabila
Gantikan nilai-nilai x ke dalam ,
From / Dari , y = 46 – 7z ……  5y – 7z = 0.20 31 5– 31 
y = – 17 , x = 9 + 17 1 = –5 – 2– 1 
Substitute  into , 50y – 70z = 2 .........  When / Apabila x = – 2 , y 2
Gantikan  ke dalam , = –4
14(46 – 7z) + 5z = 86 38 – 50z = – 2
From /Dari , y = 30 ……  17
644 – 98z + 5z = 86 When / Apabila x = –3, y = –5 – 2(–3)
–93z = –558 Substitute  into , \ x = 4, y = –1; x = – 2 , y = – 31 =1
z=6 Gantikan  ke dalam , 17 17

Substitute z = 6 into , 5038 3–050z – 70z = 2 (b) 3x + 2y = 9 .............  \ x= – 1 , y= –4; x = –3, y =1
Gantikan z = 6 ke dalam , x2 + y2 + 2xy = 4 .............  2
y = 46 – 7(6) 1 900 – 2 500z – 2 100z = 60
–4 600z = –1 840 From / Dari , x = 9 – 2y (b) x – y = 3 ................. 
=4 z = 0.40 Substitute  into , 3 4 3
…… 8 6
Substitute y = 4 and z = 6 into , x – y = 3 ................. 
Gantikan y = 4 dan z = 6 ke dalam ,
3x + 5(4) + 2(6) = 41 Gantikan  ke dalam , From / Dari ,

3x + 20 + 12 = 41 Substitute z = 0.40 into ,  9 – 2y 2 + y2 + 2y 9 – 2y  =4 3x – 4y
3x = 9 3 3 12
x= 3 Gantikan z = 0.40 ke dalam , 81 – 36y + 4y2 18y – 4y2 =3
9 3
Therefore, the prices of a loaf of bread, y= 38 – 50(0.40)   + y2 +   =4 x = 36 + 4y .........
a packet of biscuits and a box of milk 30 3
are RM3, RM4 and RM6 respectively. 81 – 36y + 4y2 + 9y2 + 3(18y – 4y2) = 36
Maka, harga bagi sebuku roti, sepaket biskut = 0.60 8y – 6x
dan sekotak susu masing-masing ialah RM3, 81 – 36y + 4y2 + 9y2 + 54y – 12y2 – 36 = 0 From / Dari , xy =3
RM4 dan RM6.
Substitute y = 0.60 and z = 0.40 into , y2 + 18y + 45 = 0 8y – 6x = 3xy ....... 
(b) Let / Biar x = price of a curry puff Gantikan y = 0.60 dan z = 0.40 ke dalam ,
harga karipap 2x + 0.60 + 3(0.40) = 2.80 (y + 3)(y + 15) = 0

y = price of a doughnut 2x + 0.60 + 1.20 = 2.80 y = –3 or / atau y = –15 Substitute  into ,
harga donat 2x = 1.00 Gantikan  ke dalam ,
x = 0.50 Substitute the values of y into ,
z = price of a nugget Gantikan nilai-nilai y ke dalam , 8y – 6 36 + 4y  = 3y 36 + 4y 
harga nuget 3 3
9 – 2(–3) 8y – 72 – 8y = 36y + 4y2
2x + y + 3z = 2.80 ............... When / Apabila y = –3, x = 3
x + 2y + 4z = 3.30 ............... Therefore, the prices of a curry puff, a 4y2 + 36y + 72 = 0
3x + 4y + z = 4.30 ............... =5
doughnut and a nugget are RM0.50, 9 – 2(–15) y2 + 9y + 18 = 0
When / Apabila y = –15, x = 3
RM0.60 and RM0.40 respectively. (y + 3)(y + 6) = 0
Maka, harga bagi sebiji karipap, sebiji donat
dan seketul nuget masing-masing ialah RM0.50, = 13 y = –3 or / atau y = –6
RM0.60 dan RM0.40.
\ x = 5, y = –3; x = 13, y = –15 Substitute the values of y into ,
Gantikan nilai-nilai y ke dalam ,

27 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 28

A15

Additional Mathematics Form 4 Chapter 3 Systems of Equations Additional Mathematics Form 4 Chapter 3 Systems of Equations

When / Apabila y = –3, x = 36 + 4(–3) \ The solutions are (3, –2) and (5, 2). Gantikan c ke dalam b, (d) Perimeter of ACDF,
3 Penyelesaian ialah (3, –2) dan (5, 2). (18 – y)2 + y2 = 170 4(y – 4) + 2(2x) = 116
4y – 16 + 4x = 116
=8 324 – 36y + y2 + y2 = 170 4y + 4x = 132
2y2 – 36y + 154 = 0 x + y = 33 ............
When / Apabila y = –6, x = 36 + 4(–6) 9. (a) x cm y2 – 18y + 77 = 0
3 (y – 11)(y – 7) = 0 Area of ACDF,
y cm y = 11 or / atau y = 7 Luas ACDF,
=4
Substitute the values of y into c, (y – 4)(2x + y – 4) = 480
\ x = 8, y = –3; x = 4, y = –6 Let / Biar x = length of the rectangular Gantikan nilai-nilai y ke dalam c, y(2x + y – 4) – 4(2x + y – 4) = 480
board When / Apabila y = 11, x = 18 – 11 2xy + y2 – 4y – 8x – 4y + 16 = 480
8. (a) 23 4 5 panjang papan segi empat tepat
–4 –2 0 2 =7 y2 – 8y – 8x + 2xy = 464 .....b
x y = width of the rectangular When / Apabila y = 7, x = 18 – 7
–3 –2 –0.65 or 2 board From / Dari , x = 33 – y ……… c
y (For/ Untuk or 5 or 5 4.65 or 3 lebar papan segi empat tepat = 11
2x – y = 8) Substitute c into b,
2x + 2y = 50 ....................  Therefore, the value of the two numbers Gantikan c ke dalam b,
y (For/ Untuk xy = 126 ..................  are 7 and 11. y2 – 8y – 8(33 – y) + 2y(33 – y) = 464
x2 + y2 – xy Maka, nilai dua nombor itu ialah 7 dan 11. y2 – 8y – 264 + 8y + 66y – 2y2 = 464
= 19)
(c) 3y = –(x + 4).......  –y2 + 66y – 728 = 0
 ÷ 2: x + y = 25.............  x(2 + y) + y = –8 y2 – 66y + 728 = 0
2x + xy + y = –8 .............. b (y – 52)(y – 14) = 0
y From / Dari , x= 126 .. 
5 y y = 52 or y = 14
4
3 Substitute  into , From / Dari , y = – x+4 ……… c Substitute y = 52 into c,
2 (5, 2) 3 Gantikan y = 52 ke dalam c,
1 Gantikan  ke dalam , x = 33 – 52

x 126 + y = 25 Substitute c into b, = –19 (Negative, rejected)
0 246 y Gantikan c ke dalam b, (Negatif, tidak diterima)
–1
–2 (3, –2) 126 + y2 = 25y 2x + x1– x + 4 2 1– x + 4 2 Substitute y = 14 into c,
–3 3 3 Gantikan y = 14 ke dalam c,
–4 y2 – 25y + 126 = 0 + = –8 x = 33 – 14

(y – 18)(y – 7) = 0 x1 –x – 42 1 –x – 42 = 19
3 3
y = 18 or y = 7 2x + + = –8 \ x = 19, y = 14

Substitute the values of y into , 6x + x(–x – 4) + (–x – 4) = –24

Gantikan nilai-nilai y ke dalam , 6x – x2 – 4x – x – 4 = –24
126
When / Apabila y = 18, x = 18 = 7 –x2 + x + 20 = 0
x2 – x – 20 = 0
When / Apabila y = 7, x = 126 = 18
7 (x – 5)(x + 4) = 0
x = 5 or / atau x = –4 SPM Practice 3
Therefore, the length of the rectangle is
18 cm and the width of the rectangle Substitute the values of x into c, Paper 1
is 7 cm.
Maka, panjang segi empat tepat ialah 18 cm dan Gantikan nilai-nilai x ke dalam c,
lebar segi empat tepat ialah 7 cm. (5 + 4)
When / Apabila x = 5, y = – 3 1. 4x – y – 6 = 0..................... 
3x2 + 4y2 – 8xy = 27 .......... b
= –3
From / Daripada , y = 4x – 6 ..........c
(b) Let the two numbers be x and y. When / Apabila x = –4, y =– (–4 + 4)
Biar dua nombor ialah x dan y. 3
x + y = 18 .......... 
x2 + y2 = 170 ........  =0

From / Dari , x = 18 – y ……… Therefore, the coordinates of
intersection points of the straight line
Substitute  into , and the curve are (5, −3) and (−4, 0).
Maka, koordinat titik persilangan antara garis

lurus dan lengkung ialah (5, –3) dan (–4, 0).

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A16

Additional Mathematics Form 4 Chapter 3 Systems of Equations Additional Mathematics Form 4 Chapter 3 Systems of Equations

Substitute  into , Therefore, the lengths of sides of the land Substitute  into , Substitute z = 5 into ,
are 18 m, 24 m and 30 m.
Gantikan  ke dalam , Oleh itu, panjang sisi tanah ialah 18 m, 24 m, dan Gantikan  ke dalam , Gantikan z = 5 ke dalam ,
30 m.
3x2 + 4(4x – 6)2 – 8x(4x – 6) = 27 31 9 – 7z 2 + z = –1 y = 9(5) – 51
3. x – 5y = 3 ............ 5 2
3x2 + 4(16x2 – 48x + 36) – 32x2 + 48x = 27 2x2 + 3xy + 4y2 = 18 .......... 27 – 21z + 5z = –5

3x2 + 64x2 – 192x + 144 – 32x2 + 48x = 27 From / Daripada , x = 5y + 3 ………  16z = 32 = –3

35x2 – 144x + 117 = 0 z=2 Substitute y = –3 and z = 5 into ,
Gantikan y = –3 dan z = 5 ke dalam ,
(x – 3)(35x – 39) = 0 x – 3(–3) + 5(5) = 36

x = 3 or x = 39 Substitute z = 2 into , x + 9 + 25 = 36
atau 35 x=2
Gantikan z = 2 ke dalam ,
\ x = 2, y = –3, z = 5
Substitute the values of x into , Substitute  into , y = 9 – 7(2)
Gantikan nilai-nilai x ke dalam , Gantikan  ke dalam , 5
When / Apabila x = 3,
2(5y + 3)2 + 3y(5y + 3) + 4y2 = 18 = –1
y = 4(3) – 6 2(25y2 + 30y + 9) + 15y2 + 9y + 4y2 = 18
=6 Substitute y = –1 and z = 2 into , 6. x + 2y – z = 17 .............. 
50y2 + 60y + 18 + 15y2 + 9y + 4y2 = 18 x–1+2=6 –2x + y + 6z = –8 ............. 
When / Apabila x = 3395, y = 433592 – 6 69y2 + 69y = 0 5x – 3y + 9z = –7 ............. 
y2 + y = 0 x=5
= –11359 y(y + 1) = 0  × 2: 2x + 4y – 2z = 34.......... 
\ x = 3, y = 6; x = 3359, y = –13195 y = 0 or y = –1 \ x = 5, y = –1, z = 2 : –2x + y + 6z = –8
 + :
2. Perimeter, Substitute the values of y into , 5y + 4z = 26.......... 
x + x + 6 + y = 72 Gantikan nilai-nilai y ke dalam ,
2x + y = 66 ..............  When / Apabila y = 0, x = 5(0) + 3 5. x – 3y + 5z = 36 ..............  × 5: –10x + 5y + 30z = –40 ........ 
2x – 4y + z = 21 ..............
By Pythagoras’ theorem, =3  × 2: 10x – 6y + 18z = –14 ........ 
Daripada Teorem Pythagoras, When / Apabila y = –1, x = 5(–1) + 3 3x + 5y – 6z = –39.............
y2 = x2 + (x + 6)2  + : –y + 48z = –54 ........ 
y2 = x2 + x2 + 12x + 36 = –2
y2 = 2x2 + 12x + 36 ............  From / Daripada , x = 3y – 5z + 36 ……  : 5y + 4z = 26
\ x = 3, y = 0; x = –2, y = –1  × 5: –5y + 240z = –270 ..... 
Substitute  into ,  + :
4. x + y + z = 6 ....................  Gantikan  ke dalam , 244z = –244
3x – 2y – 4z = 9......................  2(3y – 5z + 36) – 4y + z = 21 z = –1
2x + 5y + 3z = 11.................... 
6y – 10z + 72 – 4y + z = 21 Substitute z = –1 into ,
2y – 9z = –51.........  Gantikan z = –1 ke dalam ,
5y + 4(–1) = 26
From / Daripada , y = 66 – 2x .....  Substitute  into ,
Gantikan  ke dalam , 5y = 30
3(3y – 5z + 36) + 5y – 6z = –39 y=6
9y – 15z + 108 + 5y – 6z = –39
From / Daripada , x = 6 – y – z ………  Substitute y = 6 and z = –1 into ,
14y – 21z = –147 Gantikan y = 6 dan z = –1 ke dalam ,
Substitute  into , Substitute  into , 2y – 3z = –21....... x + 2(6) – (–1) = 17
Gantikan  ke dalam , Gantikan  ke dalam ,
3(6 – y – z) – 2y – 4z = 9 From / Daripada , y = 9z – 51 ........ x + 12 + 1 = 17
(66 – 2x)2 = 2x2 + 12x + 36 18 – 3y – 3z – 2y – 4z = 9 2 x=4
4 356 – 264x + 4x2 = 2x2 + 12x + 36
2x2 – 276x + 4 320 = 0 5y + 7z = 9............  \ x = 4, y = 6, z = –1

x2 – 138x + 2 160 = 0 Substitute  into ,
(x – 120)(x – 18) = 0
Substitute  into , Gantikan  ke dalam ,
x = 120 or / atau x = 18 Gantikan  ke dalam ,
2(6 – y – z) + 5y + 3z = 11 21 9z – 51 2 – 3z = –21 7. 2x + 3y = 1............... 
Substitute x = 18 into , 12 – 2y – 2z + 5y + 3z = 11 2
y = 66 – 2(18) 4 – 3 = –8................ 
3y + z = –1 ........ 9z – 51 – 3z = –21 x y
= 66 – 36
= 30 6z = 30 From / Daripada , 2x = 1 – 3

x = 120 rejected because perimeter is only z=5
72 m.
120 tidak diterima kerana perimeter hanya 72 m. From / Daripada , y = 9 – 7z ………  x = 1 – 3y .......
5 2

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A17

Additional Mathematics Form 4 Chapter 3 Systems of Equations Additional Mathematics Form 4 Chapter 3 Systems of Equations

Substitute  into , y 10. 2x + y = p + 1 \ Another solution are x = – 3 and
Gantikan  ke dalam , 3 2 y = 6. 2
p
4 – 3 = –8 25 –2 + 1 = 2 + 1 Penyelesaian yang satu lagi ialah x = – 3
y 20 dan y = 6. 2
 1 – 3y 2 15 –1 = p + 1
2 10 2
p
8 – 3 = –8 5 –2 = 2 11. Let / Katakan x = first number
1 – 3y y nombor pertama
–4 –3 –2 –1 0 p = –4
8y –5 y = second number
1 – 3y –3 = –8y x nombor kedua
–10
12 34 – 3 + 6 = q – 1 z = third number
1 3 nombor ketiga
8y – 3(1 – 3y) = –8y (1 – 3y)
8y – 3 + 9y = –8y + 24y2 –3 + 2 = q – 1 x + y + z = –6.................. a
17y – 3 = –8y + 24y2 3x + y – 2z = –6.................. b
24y2 – 8y = 17y – 3 –1 = q – 1 5x + 5y + 2z = –24 ............... c

24y2 – 25y + 3 = 0 The intersection points are (–1, 10) and q=0 From / Daripada a, x = –6 – y – z …… d
(–2, 13)
–(–25) ± √(–25)2 – 4(24)(3) Titik-titik persilangan ialah (–1, 10) dan (–2, 13). 2x + y = p + 1 Substitute d into b,
2(24) 3 2 Gantikan d ke dalam b:
y= 9. y 3(–6 – y – z) + y – 2z = –6
2x + 3 = –2 + 1 –18 – 3y – 3z + y – 2z = –6
x –4 –3 –2 –1 0 1 2 3 4
= 25 ± √337 2x + y = –1 –2y – 5z = 12...........e
48 Value of 3
y for the Substitute d into c,
= 25 + √337 or / atau 25 – √337 equation 6x + y = –3............ a Gantikan d ke dalam c:
48 48 Nilai y bagi –4.8 –4.4 –4.0 –3.6 –3.2 –2.8 –2.4 –2.0 –1.6 5(–6 – y – z) + 5y + 2z = –24
persamaan 3 6 –30 – 5y – 5z + 5y + 2z = –24
= 0.903 or / atau 0.138 2x – 5y = x + y = q – 1
16 –3z = 6
When / Apabila y = 0.903, 3 + 6 = –1 z = –2
Value of x y
1 – 3(0.903) y for the Substitute z = –2 into e,
x = 2 equation 3 + 6x = –x Gantikan z = –2 ke dalam e:
Nilai y bagi 3.25 –2.00 –5.75 –8.00 –8.75 –8.00 –5.75 –2.00 3.25 y –2y – 5(–2) = 12
persamaan
x = –0.855 3x2 – 4y 3y + 6x = –xy .......... b –2y + 10 = 12
= 35 –2y = 2
When / Apabila y = 0.138, From / Daripada a: y = –3 – 6x …… c y = –1

x = 1 – 3(0.138) Substitute y = –1 and z = –2 into d,
2 Gantikan y = –1 dan z = –2 ke dalam d:
Substitute c into b:
x = 0.293 Gantikan c ke dalam b: x = –6 – (–1) – (–2)
3(–3 – 6x) + 6x = –x(–3 – 6x) x = –6 + 1 + 2
8. y x = –3
–9 – 18x + 6x = 3x + 6x2
x –4 –3 –2 –1 0 1 2 3 4 4 1 2 34 x –12x – 9 = 3x + 6x2 \ First number / Nombor pertama: –3
2 0 = 6x2 + 15x + 9 Second number / Nombor kedua : –1
Value of y for Third number / Nombor ketiga : –2
the equation –4 –3 –2 –1 0 6x2 + 15x + 9 = 0
Nilai y bagi 19 16 13 10 7 4 1 –2 –5 –2 2x2 + 5x + 3 = 0
persamaan –4
3x + y = 7 –6 (x + 1)(2x + 3) = 0
–8
Value of y for x+1=0 2x + 3 = 0
the equation –10
Nilai y bagi 25 18 13 10 9 10 13 18 25 x = –1 2x = –3
persamaan 3
x2 – y = –9 x = –1 , x= – 2

When / Apabila x = – 3 ,
2
61– 3 2
y = –3 – 2

The intersection points are (3, –2) and y=6
(–2.5, –4.2)
Titik-titik persilangan ialah (3, –2) dan (–2.5, –4.2).

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A18

Additional Mathematics Form 4 Chapter 3 Systems of Equations Additional Mathematics Form 4 Chapter 3 Systems of Equations

12. Let / Katakan x = price of greeting card Substitute y = 5 and z = 18 into d, Substitute g into f, 170x = 805 – 360y – 285z
harga sekeping kad ucapan Gantikan g ke dalam f:
Gantikan y = 5 dan z = 18 ke dalam d: x = 805 – 360y – 285z ............. d
y = price of flowers 9(50 – z) + 10z = 480 170
harga sejambak bunga x = 572 – 18y – 25z 450 – 9z + 10z = 480
16
z = price of bracelets 572 – 18(5) – 25(18) z = 30 Substitute d into ,
harga seutas gelang tangan x = 16 Gantikan d ke dalam :
Substitute z = 30 into g,
x =2 Gantikan x = 30 ke dalam g: 425 1 805 – 360y – 285z 2 = 1 415
y = 50 – 30 170
16x + 18y + 25z = 572 ......... a \ Price of greeting cards / Harga kad ucapan: y = 20 +180y + 475z
23x + 19y + 17z = 447 .........  RM2
18x + 20y + 21z = 514 ........ c Substitute y = 20 and z = 30 into d, 425(805 – 360y – 285z) = 240 550
Price of flowers / Harga bunga : RM5 Gantikan y = 20 dan z = 30 ke dalam d: + 30 600y + 80 750z
Price of bracelets / Harga gelang tangan: x = 100 – y – z
From / Daripada a, RM18 x = 100 – 20 – 30 342 125 – 153 000y – 121 125z
x = 50 + 30 600y + 80 750z = 240 550
16x = 572 – 18y – 25z –122 400y – 40 375z = –101 575
\ x = 50, y = 20 and / dan z = 30. 122 400y + 40 375z = 101 575
x= 572 – 18y – 25z …… d 13. Let / Katakan x = number of tables 4 896y + 1 615z = 4 063....... 
16 bilangan meja \ Number of table / Bilangan meja: 50
Number of wooden chairs: 20
Substitute d into , y = number of wooden chairs Bilangan kerusi kayu: Substitute d into ,
Gantikan d ke dalam : bilangan kerusi kayu Number of plastic chairs: 30 Gantikan d ke dalam :
Bilangan kerusi plastik:
23 572 – 18y – 25z 2 + 19y + 17z = 447 z = number of plastic chairs
16 bilangan kerusi plastik HOTS Challenge

13156 – 414y – 575z + 304y + 272z = 7152 250x + 70y + 50z = 15 400 1. Let x = price of a pen before discount 2551 805 – 360y – 285z 2
x + y + z = 100 harga pen sebelum diskaun 170
110y + 303z = 6 004 .e x–y–z=0 = 1 015
y = price of an eraser before discount
x + y + z = 100 ................a harga pemadam sebelum diskaun + 630y + 190z
x – y – z = 0 ....................
Substitute d into c, 250x + 70y + 50z = 15 400 ..........c z = price of a ruler before discount 255(805 – 360y – 285z) = 172 550
Gantikan d ke dalam c: harga pembaris sebelum diskaun + 107 100y + 32 300z
From / Daripada a, x = 100 – y – z …… d
18 572 – 18y – 25z 2 + 20y + 21z = 514 For Ali, Bagi Ali, 205 275 – 91 800y – 72 675z = 172 550
16 Substitute d into , 2(0.85x) + 4(0.90y) + 3(0.95z) = 8.05 + 107 100y + 32 300z
10296 – 324y – 450z + 320y + 336z = 8224 Gantikan d ke dalam :
1.70x + 3.60y + 2.85z = 8.05 15 300y – 40 375z = –32 725
–4y – 114z = –2072 100 – y – z – y – z = 0 170x + 360y + 285z = 805........ 
100 – 2y – 2z = 0 612y – 1 615z = –1 309 ..... f
4y + 114z = 2072 ..f 2y + 2z = 100 For Bala,/ Bagi Bala,
y + z = 50 ......e 5(0.85x) + 2(0.90y) + 5(0.95z)= 14.15
From / Daripada e, From /Daripada ,
Substitute d into c, 4.25x + 1.80y + 4.75z = 14.15
110y + 303z = 6004 Gantikan d ke dalam c: 425x + 180y + 475z = 1 415 ....  1 615z = 4 063 – 4 896y

110y = 6004 – 303z 250(100 – y – z) + 70y + 50z = 15400 z= 4 063 – 4 896y .......... g
25000 – 250y – 250z + 70y + 50z = 15400 1 615
y= 6004 – 303z …… g
110 –180y – 200z = –9600 Substitute g into f,
180y + 200z = 9600 Gantikan g ke dalam f:
Substitute g into f,
Gantikan g ke dalam f: 18y + 20z = 960 612y – 1 6151 4 063 – 4 896y 2 = –1 309
9y + 10z = 480 ....f 1 615
4y + 114z = 2072
From / Daripada e, y = 50 – z ……… g 612y – (4 063 – 4 896y) = –1 309
4 6004 – 303z 2 + 114z = 2072
110 612y – 4 063 + 4 896y = –1 309

24016 – 1212z + 12540z = 227920 5 508y = 2 754

11328z = 203904 y = 0.50

z = 18

Substitute z = 18 into g, For Chong, / Bagi Chong, Substitute y = 0.50 into g,
3(0.85x) + 7(0.90y) + 2(0.95z)= 10.15 Gantikan y = 0.50 ke dalam g:
Gantikan x = 18 ke dalam g:
2.55x + 6.30y + 1.90z = 10.15 4 063 – 4 896(0.50)
y = 6004 – 303z 255x + 630y + 190z = 1 015 ....  z= 1 615
110
6004 – 303(18) = 1.00
= 110
y

y =5 From / Daripada ,

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A19

Additional Mathematics Form 4 Chapter 3 Systems of Equations PTER

Substitute y = 0.50 and z = 1.00 into , Substitute  into , CHA 41 Indices, Surds and Logarithms
Gantikan y = 0.50 dan z = 1.00 ke dalam : Gantikan  ke dalam :
170x + 360(0.50) + 285(1.00) = 805 Indeks, Surd dan Logaritma
(6 – y)2 + y2 – y(6 – y) = 21
170x = 340 36 – 12y + y2 + y2 – 6y + y2 = 21 12 1 2(f) 2p + 3 –4p 1
x = 2.00 2
3y2 – 18y + 15 = 0 1. (a) (32)4 × 93 ÷ 813
Therefore, the prices of a pen, an eraser and y2 – 6y + 5 = 0 9
a ruler before discount are RM2.00, RM0.50 12
and RM1.00 respectively. (y – 5)(y – 1) = 0 (22)p 1
Oleh itu, harga pen, pemadam dan pembaris masing- y = 5 or / atau 38 × (32)3 ÷ (34)3 1 2= 2p × 23 – 2
masing sebelum diskaun ialah RM2.00, RM0.50 dan y=1
RM1.00. 28 32
When / Apabila y = 5, x = 6 – 5
2. Let x and y be the values of the two =1 = 38 × 33 ÷ 33 1 2= 8(2p) – 1 11
numbers. 32
Biar x dan y ialah nilai dua nombor itu. When / Apabila y = 1, x = 6 – 1 = 38 + 2 – 8 2 (22p) 2
x + y = 6 ................. =5 3 3
x2 + y2 – xy = 21 ...............
Therefore, the values of the two numbers = 36 1 2= 8(2p) – 1 (2p)
From / Daripada , x = 6 – y ………  are 1 and 5. 3
Oleh itu, nilai-nilai dua nombor itu ialah 1 dan 5.
(b) 1 24 3 ÷ 1 2 24 = 18 – 21 (2p)
2 3
3
9 23
3
=1 4 23 ÷ 1 2 24 = (2p)
9 3
21 × 5x 21
= 1 2 23 ÷ 1 2 24 2. (a) = 625
3 3
5x = 625–1
1 2= 2 3–4 5x = (54)–1
3 5x = 5–4
x = –4
1 2= 2 –1 or 3
3 2

(c) 2a3b4 × 5ab2 (b) (23 – 5x)(8x – 4) = 1
= 2 × 5 × a3 × a × b4 × b2
= (2 × 5)a3 + 1 × b4 + 2 (23 – 5x)[(23)x – 4] = 20
= 10a4b6
(23 – 5x)(23x – 12) = 20

2 = 23 – 5x +(3x – 12) 0

2–9 – 2x = 20

1 –9 – 2x = 0
3
(d) 21m3n2 ÷ m5n 2x = –9 9
2
21m3n2 x= –

= 1
3
m5n (c) 9x – 3 = 81
9x – 3 = 92
= 63m3 – 5n2 – 1 x–3=2
x=5
= 63m–2n

= 63n 3. After the 6th fission, the number of amoeba
m2 Selepas pembelahan ke-6, bilangan ameba
= 18 × 26
(e) 4p5q2 × 6pq3 = 2 × 9 × 26
3p3q4 = 21 + 6 × 32
= 2732
4×6
3 Initial stage: 18
1 2= p q5 + 1 – 3 2 + 3 – 4 Peringkat awal
After the 1st fission: 18 × 21
= 8p3q Selepas pembelahan pertama
After the 2nd fission: 18 × 2 × 2 = 18 × 22
Selepas pembelahan ke-2

37 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 38

A20

Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms

4. (a) Let / Biar x = 2.333… ———— 1 (d) 3 2 = 0.65863… (b) √12 – √75 + √108 (b) 3 = 3 × 2√3
7 = √4 × 3 – √25 × 3 + √36 × 3 2√3 2√3 2√3
1 × 10: 10x = 23.33…—— 2 = 2√3 – 5√3 + 6√3
= (2 – 5 + 6)√3
2 – 1: 9x = 21 Surd. Its value is a non-recurring = 3√3 = 3 × 2 × √3
decimal. 2 × 2 × √3 × √3
x = 21 (c) √2 + √3 + √12 – √48
9 Surd. Nilainya ialah perpuluhan tak berulang. = √2 + √3 + √4 × 3 – √16 × 3
= √2 + √3 + 2√3 – 4√3 6√3
= 7 or 2 1 = √2 + (1 + 2 – 4)√3 = 12
3 3 = √2 – √3
6. (a) √14 × √5
Hence / Maka, 2.333… = 7 or 2 1 = √14 × 5 = √3
3 3 = √70 2

(b) Let / Biar x = 0.1666… ————— 1 (b) √2 × √27 (c) √6 = 6
= √2 × 27 √10 10
1 × 10: 10x = 1.666…——— 2 = √54

1 × 100: 100x = 16.66…——— 3 3
5
3 – 2: 90x = 15 =

x = 15 (c) √126 ÷ √3 (d) √18 + √8 – √2 √3
90 3 √5
=
1 126
= 6 = 3 = √9 × 2 + √4 × 2 – √2 √3 √5
3 √5 √5
Hence / Maka, 0.1666… = 1 = ×
6 3√2
= √42 = 3 + 2√2 – √2 √15
5
(c) Let / Biar x = 0.001111… ———— 1 (d) √56 × √3 = √2 + 2√2 – √2 =
√7
1 × 100: 100x = 0.1111…—— 2 = (1 + 2 – 1)√2

1 × 1 000: 1 000x = 1.111…—— 3

3 – 2: 900x = 1 = 56 × 3 = 2√2 10. (a) 7 = 7 × 2 + 3√2
7 2 – 3√2 2 – 3√2 2 + 3√2
x = 1
900 = √24 (e) 3√10 – √5 × √2 = 7(2 + 3√2)
22 – (3√2)2
Hence / Maka, 0.001111… = 1 7. (a) √208 = 3√10 – √5 × 2
900 = √2 × 2 × 2 × 2 × 13
= 2 × 2 × √13 = 3√10 – √10 = 7(2 + 3√2)
(d) Let / Biar x = 0.141414… ————— 1 = 4√13 4 – 18
= (3 – 1)√10
1 × 100: 100x = 14.1414…————2 = 2√10 = 7(2 + 3√2)
–14
2 – 1: 99x = 14

x = 14 (f) √5(8 – √5) = 2 + 3√2
99 = 8√5 – √5 × 5 –2
(b) 2√72 = 8√5 – 5
Hence / Maka, 0.141414… = 14 = 2 × √2 × 2 × 2 × 3 × 3 =– 2 + 3√2
99 = 2 × 2 × 3 × √2 2
= 12√2
5. (a) √72 = 8.48528… (g) (1 + 2√3)(7 – 4√3) √3 √3 √5 + √3
Surd. Its value is a non-recurring (c) 3√24 = (1)(7) – (1)(4√3) + (2√3)(7) – (2√3) (b) √5 – √3 = √5 – √3 × √5 + √3
= 3√2 × 2 × 2 × 3 (4√3)
decimal. = 2 × 3√3 = 7 – 4√3 + 14√3 – 24 = √3(√5 + √3)
Surd. Nilainya ialah perpuluhan tak berulang. = 23√3 = 10√3 – 17 (√5)2 – (√3)2

(b) √81 = 9 = √3(√5 + √3)
Not a surd. Its value is an integer. 5–3
8. (a) 7√2 – 3√2 9. 2 2
Bukan surd. Nilainya ialah satu integer. = (7 – 3)√2 (a) √12 = √4 × 3 √15 + √9)
= 4√2 2
(c) – 121 = – 11 = –2.2 =
25 5
Not a surd. Its value is a terminating = 2 √15 + 3
2√3 = 2
decimal.
= 1 × √3
Bukan surd. Nilainya ialah perpuluhan tak √3 √3

berulang. √3
3
=

39 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 40

A21

Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms

(c) 3√2 – √3 12. (a) (i) AC = (√125)2 + (√500)2 14. (a) 0.6990 (f) Let log1 0.25 = x
2√2 – √3 2
= √125 + 500 (b) –0.3010
= √625 1 1 2x = 0.25
= 3√2 – √3 × 2√2 + √3 = 25 cm (c) 2.3802 2
2√2 – √3 2√2 + √3
(d) –1.1073 1 1 2x = 1
2 4
= (3√2 – √3)(2√2 + √3) 1 2(e) log104 = –0.3522
(2√2)2 – (√3)2 9 1 1 2x = 1 1 22
Perimeter 2 2
= (3√2)(2√2) + (3√2)(√3) – (√3)(2√2) – (√3) (√3) 1 2(f) 1 = –2.0969
(2√2)2 – (√3)2 = √125 + √500 + 25 log10 125 x=2
= √5 × 5 × 5 + √2 × 2 × 5 × 5 × 5 + 25
= 6 × 2 + 3 × √6 – 2 × √6 – 3 = 5√5 + 10√5 + 25 15. (a) x = antilog 2.831 17. (a) x = 2–3
4×2–3 = (15√5 + 25) cm x = 677.6
1
= 12 + 3 √6 – 2 √6 – 3 (b) x = antilog 0.0842 = 8
8–3 x = 1.214
(ii) Area / Luas = 1 × √125 × √500
= 9 + √6 2 (c) x = antilog 3 1
5 x = 1 000
= 1 × 5√5 × 10√5 (b) x = 42
1– 1 2 (d) x = antilog (–1.45) = √4
x = 0.03548 =2
√2 = 1 × 50 × 5
11. (i) n= 1+ 1 2

√2 = 125 cm2 16. (a) Let log2 256 = x (c) x = 1 1 23
2x = 256 3
1 1 2x = 28
= 11 – √2 2 ÷ 11 + √2 2 (b) Let h = height of the cuboid x =8 = 1
27
Biar h = tinggi kuboid
= √2 – 1 ÷ √2 + 1 Volume of cuboid (b) Let log3 1 = x (d) 64 = x2
√2 √2 = Area of base × Height 27
Isi padu kuboid = Luas tapak × Tinggi 1 82 = x2
= √2 – 1 × √2 27 x =8
√2 √2 + 1 3x =

= √2 – 1 × √2 – 1 32(4 – √7) = (√14 – √2)2 × h 3x = 3–3 (e) log10 x = –4
√2 + 1 √2 – 1
h = 32(4 – √7) x = –3 x = 10–4
(√14 – √2)2
= (√2 – 1)2 = 1
(√2)2 – 12 32(4 – √7) 10 000
= 14 – 2√28 + 2 (c) Let log15 1 = x
2 – 2√2 + 1 15x = 1 or 0.0001
= 2–1 15x = 150
32(4 – √7) x=0
= 3 – 2√2 = 16 – 2√4 × 7 (f) loge x = 0

(ii) n + 1 = 32(4 – √7) (d) Let log8 √8 = x x = e0
n 16 – 4√7 8x = √8 =1

= (3 – 2√2) + 1 = 32(4 – √7) 1 18. (a) 2 + log10 4
3 – 2√2 4(4 – √7)
8x = 82 = 2 × 1 + log10 4
= (3 – 2√2) + 1 × 3 + 2√2 =8 1 = 2 log10 10 + log10 4
3 – 2√2 3 + 2√2 2
1 x= = log10 102 + log10 4
81
= (3 – 2√2) + 3 + 2√2 13. (a) log3 = –4 (e) Let log√3 9 = x = log10 (102)(4)
9–8 (√3)x = 9 = log10 400
2 1
= (3 – 2√2) + (3 + 2√2) 1 2(b) log 4 9 = 2
81
=6 (√3)x = (√3)4
x=4 9 12
(c) 7–2 = 1 (b) log3 4 + log3 5
49
1 2= log3 9 12
(d) ay = x 4 × 5

= log3 27
5

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A22

Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms

(c) logx p + 3 logx q – 1 logx r (c) 1+ 1 log7 196 – log7 14 (b) log8 4 = log2 4 (e) log16 x2y8 = log2 x2y8
3 2 log2 8 log2 16
1
1 = 1 + log7 (142)2 – log7 14 log2 x2 + log2 y8
= log2 22 = log2 24
= logx p + logx q3 – logx r3 = 1 + log7 14 – log7 14 log2 23

= logx pq3 =1 = 2 log2 2 = 2 log2 x + 8 log2 y
3 log2 2 4
1
r3

(d) 3 loga x + 4 loga y – loga x2y (d) 2 logx xy – 2 logx y + 3 = 2(1) = 1 m + 2n
= loga x3 + loga y4 – loga x2y = 2(logx x + logx y) – 2 logx y + 3 3(1) 2
= 2 logx x + 2 logx y – 2 logx y + 3
x3 × y4 =2+3 = 2 23. (i) P = 6.9(1.010)t
x2y =5 3
1 2= loga log10 P = log10 6.9(1.010)t
log3 243 log10 P = log10 6.9 + t log10 1.010
= loga xy3 (c) log81 243 = log3 81

(e) loga m + 3 loga n – loga mn (e) 1 logx x – logx 4 x3 = log3 35 (ii) (a) t = 2030 – 2019 = 11 years / tahun
= loga m + loga n3 – loga mn 2 √y y log3 34 P = 6.9(1.010)11 = 7.698 billion / bilion

= 1 (logx x– 1 logx y) = 5 log3 3 \ The population in the year 2030
logx 2 – logx y) 4 log3 3 is 7.698 billion.
1 2= loga mn3 2 1 (3 x Populasi pada tahun 2030 ialah 7.698
mn – 4 bilion.

= loga n2 = 1 (411)lo–gx14ylogx y – 3 (1) = 5(1)
4 4(1)
2
(f) 1 log2 m – 3 log2 n + 2 log2 q + = 5 (b) In the year 2019
2 4 Pada tahun 2019
1 1 P2019 = 6.9(1.010)0
= log2 m2 – log2 n3 + log2 q2 =– 4 = 6.9
22. (a) log2 xy3 1
q2 √m 3√2 t = 0 as it starts on that year, that is 2019
n3 = log2 x + log2 y3 – log2 23 t = 0 kerana ia bermula pada tahun ini, iaitu tahun 2019
1 2= log2
log10 14 = log2 x + 3 log2 y – 1 log2 2
20. (a) log3 14 = log10 3 3
1
19. (a) 2 log2 4 + log2 5 – log2 10 = 2.402 = m + 3n – 3
= log2 42 + log2 5 – log2 10
4x 1 Double of that in 2019,
3√y4
16 × 5 log10 0.06 (b) log2 = log2 22 + log2 x – log2 (y4)3 Dua kali ganda pada 2019,
10 log10 15
1 2= log2 (b) log15 0.06 = =2 log2 2 + log2 x – 4 log2 y 2 × 6.9 = 6.9(1.010)t
3 13.8 = 6.9(1.010)t
= log2 8 = –1.039
4 13.8
= log2 23 = 2 + m – 3 n (1.010)t = 6.9

= 3 log2 2 (c) log0.7 9 = log10 9 (c) logxy 4x = log2 4x =2
log10 0.7 log2 xy
=3 t log10 1.010 = log10 2

3 = –6.160 log2 22 + log2 x t= log10 2
4 log2 x + log2 y log10 1.010
(b) log4 24 – log4 + log4 2 =

1 2= log4 24 ÷ 3 ×2 21. (a) log9 27 = log3 27 = 2+m = 69.66
4 log3 9 m+n
≈ 70
1 2= log4 4 2 = log3 33 1 2019 + 70 = 2089
24 × 3 × log3 32
(d) log2 2√xy2 = log2 2 + log2 (xy2)2 \ In the year 2089, the population
will be double of that in 2019.
= log4 64 = 3 log3 3 =1 + 1 log2 (xy2)
2 log3 3 2 Pada tahun 2089, populasi akan dua kali
= log4 43
=1 + 1 (log2 x + log2 y2) ganda pada 2019.
= 3 log4 4 2
= 3(1)
=3 2(1) = 1 + 1 (log2 x) + 1 (2 log2 y)
2 2
= 3
2 = 1 + 1 m + n
2

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A23

Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms

26. (a) Let its side is x cm. SPM Practice 4 (b) log√p Xp4 = logp Xp4 8. (a) t > 0 and / dan t ≠ 1
logp √p
Biar sisi ialah x cm. Paper 1 (b) log2 x = 3
(Changing base / Tukar asas) logxy 2
x2 = 8
1. log7 4x3 = logx 4x3 = (logp X + 4 logp p) ÷ 1 logp p log2 x = 3
x = √8 logx 7 2 1
= √4 × 2 = 2(q + 4)
= 2√2 logx 4 + logx x3 log2 xy
q
Perimeter = 4 × 2√2 = = 2q + 8 log2 x = 3 log2 xy
= 8√2 cm
= logx 22 + 3 logx x 5. 16p + 3 = 1 log2 x = 3(log2 x + log2 y)
q 64h – 2
log2 x = 3 log2 x + 3 log2 y

(b) Let its mass after t days is m pg. = 2 logx 2 + 3 24(p + 3) = 20 3 log2 x – log2 x = -3 log2 y
Biar jisimnya selepas t hari ialah m pg. q 26(h – 2)
Thus / Maka, m = 2(3)t 2 log2 x = -3 log2 y
32lolgo2gx2–x32
= 2p + 3 2 = 24(p + 3) – 6(h – 2) 0 – = log2 y
q = log2 y
If / Jika m = 15 pg, 15 = 2(3)t 4p + 12 – 6h + 12 = 0

15 = 3t 1 6h = 4p + 24 y = x– 2
2 3
2. (ap)9 = a(a2)q
log10 15 = log10 3t Comparing / Bandingkan: h = 2 p + 4
2 3

log10 7.5 = t log10 3 p = 1 , q = 9 6. loga 256 – log√a 2a = 1 9. √27 – 4√5 – √20 + 5√3
2 = √9 × 3 – 4√5 – √4 × 5 + 5√3
t = log10 7.5 loga 256 – loga 2a = loga a = 3√3 – 4√5 – 2√5 + 5√3
log10 3 loga √a = 8√3 – 6√5
3. From / Dari 2n = 3m,
= 1.834 n log10 2 = m log10 3 loga 256 – loga 2a = loga a

It takes 2 days. 1 loga a
2
Ia mengambil 2 hari. From / Dari 3m = 6p, 10. (a) 2x = 64
m log10 3 = p log10 6 2x = 26
1 2(c)MV =P 1 + r nt m log10 3 = p(log10 2 + log10 3) loga 256 – 2 loga 2a = loga a x=6
n
m log10 3 loga 256 – loga (2a)2 = loga a 2
p = log10 2 + log10 3 (b) 2 × 5x = 625
1 25 431.20 =3 0.05 (1)t 256
500 1 + 1 loga 4a2 = loga a

5 431.20 = 1.05t m log10 3 256 2 × 5x = 2 × (54)–1
3 500 4a2
= m log10 3 = a 5x = 5–4
n
5 431.20 + log10 3 x = –4
3 500
1 2ln = ln 1.05t 64 = a3

= m log10 3 a=4 11. log6 64 – log6 36 + 2 log2 8
= 4 log6 6 – log6 62 + 2 log2 23
ln 5 431.20 – ln 3 500 = t ln 1.05 1log10 m 12 = 4 – 2 log6 6 + 2(3 log2 2)
n =4–2+6
t = ln 5 431.20 – ln 3 500 m 3 +
ln 1.05 =8
= 7. 2m + 2m = 2n
=9 1m + n2 2(2m) = 2n
n 21 + m = 2n
It takes 9 years. 1+m=n
Ia mengambil 9 tahun. = mn m=n–1 12. log2 25 – log2 4√3 + log2 60
m+n
1

= log2 52 – log2 34 + log2 (22 × 3 × 5)

1 24. 1 = logp X–1 = 2 log2 5 – 1 log2 3 + log2 22 + log2 3 + log2 5
(a) logp x 4
1
= –logp X = 2b – 4 a + 2 log2 2 + a + b

= –q = 3b + 3 a + 2
4

45 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 46

A24

Additional Mathematics Form 4 Chapter 4 Indices, Surds and Logarithms PTER

Paper 2 3. √a + √b = √a + 2√ab + b CHA 51 Progressions
√a + √b 22 = √a+ 2√ab + b 22
1. 3n + 1 – 3n + 3n – 1 = 3(3n) – (3n) + 3n Janjang
3 √a 22 + 2 √a 2√b 2 + √b 22 = a + 2√ab + b
3n3 1 2 a + 2√ab + b = a + 2√ab + b
= – 1 + 3 (shown / tertunjuk)

= 3n(6 + 1) Compare / Bandingkan √6 + 2√5 1. (a) d1 = 1 – 1 = – 1 3. (a) T3 = a + (3 – 1)d = –5
3 with / dengan √a + 2√ab + b, 3 2 6 a + 2d = –5 ……… 1

= 7(3n – 1) a + b = 6 ..........................a 1 1 1 T8 = a + (8 – 1)d= 15
ab = 5 ...............................b 4 3 12 a + 7d = 15 ……… 2
27(3n + 1 – 3n + 3n – 1) = 7(3n2) d2 = – =–
27[7(3n – 1)] = 7(3n2) From / Dari a, b = 6 – a ……. c
33(3n – 1) = 3n2 d3 = 1 – 1 =– 1 2 – 1, 5d = 20
3n + 2 = 3n2 Substitute c into b, 5 4 20
Gantikan c ke dalam b, d=4
n + 2 = n2
a(6 – a) = 5 Not an arithmetic progression because Substitute d = 4 into 1,
n2 – n – 2 = 0 6a – a2 = 5
a2 – 6a + 5 = 0 Bukan satu janjang aritmetik kerana Gantikan d = 4 ke dalam a,
(n – 2)(n + 1) = 0 (a – 1)(a – 5)= 0 d1 ≠ d2 ≠ d3.
a + 2(4) = –5
n = –1 or / atau 2 a = 1 or / atau a = 5
a = –13
Substitute a = 1 into b,
Gantikan a = 1 ke dalam b, (b) d1 = 3 – (–2) = 5 \The first term is –13 and the common
b=6–1 d2 = 8 – 3 = 5 difference is 4.
d3 = 13 – 8 = 5
=5 Sebutan pertama ialah –13 dan beza sepunya
An arithmetic progression because
Substitute a = 5 into c, Satu janjang aritmetik kerana ialah 4.
Gantikan a = 5 ke dalam c, d1 = d2 = d3 = 5.
HOTS Challenge b=6–5 (b) T2 = a + d = x ………… 1
T5 = a + 4d = 2y ………… 2
1. Amount / Amaun r 2nt =1 (c) d1 = (–x – 2y) – (–y) = –x – y 2 – 1, 3d = 2y – x
= 5 000(1 + 0.06)8 n d2 = –y – x = –x – y
MV = P 1 + Hence / Maka, √6+ 2√5 = √1 + √5 d3 = x – (2x + y) = –x – y d = 2y – x
= RM7 969.24 3
= 1 + √5 An arithmetic progression because 2y – x
Satu janjang aritmetik kerana Substitute d = 3 into 1,
d1 = d2 = d3 = –x – y.
2. A = 390A0 Gantikan d = 2y – x ke dalam a,
3
2y – x
 2R = log10 A a + 3 = x
A0 a = x
2. (a) a = 3 – 2y – x
390A0 d=7–3=4 3
A0
 2= log10 T8 = 3 + 7(4) = 4x – 2y
= 31 3
= log10 390

= 2.6 \ The first term is 4x – 2y and the
3
(b) a = –10 2y – x
d = –13 – (–10) = –3 common difference is 3 .

T16 = –10 + 15(–3) Sebutan pertama ialah 4x – 2y dan beza
= –55 3
2y – x.
sepunya ialah 3

(c) a = 2p 4. (a) a = –7,
d = 3 – (–7) = 10,
d = 3p – 2p = – p Tn = 93
2 2

1 2T21 = 2p + 20 – p = 2p – 10p = –8p 93 = –7 + (n – 1)(10)
2

n = 93 + 7 + 1
10

= 11

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A25

Additional Mathematics Form 4 Chapter 5 Progressions Additional Mathematics Form 4 Chapter 5 Progressions

(b) a = 4p, (d) a = –15, d = –12 – (–15) = 3 S19 = 129[2(–22) + (19 – 1)(8)] (c) Sn – 1 = 4(n – 1) – (n – 1)2
d = 11p – 4p = 7p, = 950 = (n – 1)(4 – n + 1)
Tn = 67p 1 2S15 = 15 [2(–15) + (15 – 1)(3)] = (n – 1)(5 – n)
2 S20 → 40 = S40 – S19
= 5 360 – 950 Sn – 2 = 4(n – 2) – (n – 2)2
67p = 4p + (n – 1)(7p) = 90 = 4 410 = (n – 2)(4 – n + 2)
= (n – 2)(6 – n)
n = 67p – 4p +1 8. (a) a = 15, d = 23 – 15 = 8, Tn = 255
7p Tn – 1 = Sn – 1 – Sn – 2
255 = 15 + (n – 1)(8) = (n – 1)(5 – n) – (n – 2)(6 – n)
= 10 10. (a) a = 21, d = 15 – 21 = –6, Sn = 21 = (5n – n2 – 5 + n) –
255 – 15 (6n – n2 – 12 + 2n)
(2p) + (3p + 15) n = 8 + 1 Sn = n [2(21) + (n – 1)(–6)] = 21 = 7 – 2n
2 2 21
5. (a) 4p + 9 = = 31 n
2
8p + 18 = 5p + 15 (48 – 6n) =

3p = –3 S31 = 321[15 + 255] 24n – 3n2 – 21 = 0 12. (a) 41, 37, 33, …
= 4 185
p = –1 n2 – 8n + 7 = 0 a = 41, d = 37 – 41 = –4
(i) T8 = 41 + (8 – 1)(–4)
(n – 1)(n – 7) = 0
= 13
(b) 4p = (2p – 1) + (5p + 4) (b) a = –52, d = –48 – (–52) = 4, Tn = 0 \ n ≠ 1, n = 7
8p = 7p + 3 2
0 = –52 + (n – 1)(4) (b) a = –36, d = –32 – (–36) = 4, Sn = –168
p=3 (ii) Tn = 25
0 + 52 n 41 + (n – 1)(–4) = 25
6. (a) a + d = 51 ……… 1 n = 4 +1 Sn = 2 [2(–36) + (n – 1)(4)] = –168

a + 4d = 33 ……… 2 = 14 n (–76 + 4n) = –168 4n = 20
2
2 – 1, 3d = –18 –38n + 2n2 + 168 = 0 n =5

d = –6 S14 = 124[–52 + 0] n2 –19n + 84 = 0 \ 5 rows
= –364
(n – 12)(n – 7) = 0 5 baris

Substitute d = –6 into 1, \ n = 12 or n = 7 (iii) (a) Tn . 0
Gantikan d = –6 ke dalam a,
a + (–6)= 51 9. (a) S10 = 120[2(–22) + (10 – 1)(8)] 41 + (n – 1)(–4) . 0
= 140
a = 57 (c) a = –10, d = –15 – (–10) = –5, Sn = –595 4n , 45
S2 = T1 + T2
Tn = a + (n – 1)d , 0 = –22 + (–14) n n , 11.25
57 + (n – 1)(–6) , 0 = –36 2
Sn = [2(–10) + (n – 1)(–5)] = –595 \ 11 rows are possible.
57 – 6n + 6 , 0 S3 → 10 = S10 – S2 n = –595
6n . 63 = 140 – (–36) 2 (–15 – 5n) 11 baris yang mungkin.
n . 10.5 = 176
– 125n – 5 n2 + 595 = 0 (b) T11 = 41 + (11 – 1)(–4)
\ n = 11 2 =1

n2 + 3n – 238 = 0 (c) S11 = 121(41 + 1)
= 231
(n – 14)(n + 17) = 0

T11 = 57 + (11 – 1)(–6) = –3 \ n ≠ –17, n = 14

7. (a) a = 4, d = 6 – 4 = 2 (b) S15 = 125[2(–22) + (15 – 1)(8)] 11. (a) S10 = 1 + 4(10)2 T11 = S11 – S10 13. (a) r1 = –27 = –3
= 510 = 401 = 485 – 401 r2 = –3
1 2S10 = 10 [2(4) + (10 – 1)(2)] 99 =
2 S11 = 1 + 4(11)2 = 84 –3
= 485
= 130 S4 = 4 [2(–22) + (4 – 1)(8)] r3 = –3 = –3
2 1

(b) a = –23, d = –16 – (–23) = 7 = –40 A geometric progression because r1 = r2

1 2S14 = 14 [2(–23) + (14 – 1)(7)] S5 → 15 = S15 – S4 (b) S5 = 2 – 5(5) + (5)2 = 2 = r3 = –3.
2 = 510 – (–40) S4 = 2 – 5(4) + (4)2 = –2 Satu janjang geometri kerana r1 = r2 = r3 = –3.
= 550 \ T5 = S5 – S4 = 2 – (–2) = 4
= 315

(c) a = 23, d = 17 – 23 = –6 (c) S40 = 420[2(–22) + (40 – 1)(8)] S6 = 2 – 5(6) + (6)2 = 8
= 5 360 T6 = S6 – S5 = 8 – 2 = 6
1 2S16 = 16 [2(23) + (16 – 1)(–6)] \ d = T6 – T5 = 6 – 4 = 2
2

= –352

49 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 50

A26

Additional Mathematics Form 4 Chapter 5 Progressions Additional Mathematics Form 4 Chapter 5 Progressions

(b) r1 = 2x = 2x = 2x 16. (a) a = 20, (b) T4 = 1 a = 32 19. (a) a = 27, r = 9 = 1 , Tn = 1
20 1 2 27 3 6 561
r = –4 = – 1 , a = 64
2x + 1 (2x)(2) 20 5
r2 = 2x = 2x =2 4 1 21 = 27 1 n–1
15 625 3
Tn =– T4 = ar3 = 32 6 561
64r3 = 32
2x + 2 (2x)(22) 4 1 1 21n–1 1
2x + 1 (2x)(2) 15 625 5 3 177 147
r3 = = =2 1 2– – n–1 =
= 20
r3 = 1
Not a geometric progression because 1 1 2 1 1
78 125 5 3 177 147
1 2– = – n–1 (n – 1) log10 = log10

r1 ≠ r2 and r1 ≠ r3. 1 2r = 1 1
Bukan satu janjang geometri kerana r1 ≠ r2 dan 3

r1 ≠ r3. 1 2 1 2–1 7= – 1 n–1 2 n = 12
5 5
n=8 \ The first term is 64 and the common 3 1 2 4S12 = 1 12
3
loga x 4 4 loga x 4 1 2ratio is 1 1 27 1 –
loga x 3 3 loga x 3 2
(c) r1 = = = 3.

(b) a = 18, Sebutan pertama ialah 64 dan nisbah sepunya 1 – 1
3
loga x 3 3 loga x 3 r = 27 = 3 , 1 2ialah1 1
r2 = loga x 2 = 2 loga x = 2 18 2 2 = 40.50
3.

loga x 2 2 loga x Tn = – 2 187 (c) T3 = ar2 = 4x3 …… 1 (b) a = 2, r = – 1 , Tn = 1
loga x loga x 16 T8 = ar7 = –128x8 …… 2 2 128
r3 = = =2
1 22 187 3 n–1
16 2
= 18 1 21 1 n–1
2
128
Not a geometric progression because r1 1 2243 3 n–1 2 ÷ 1, r5 = –32x5 =2–
2 r = –2x
≠ r2 ≠ r3. 32 = 1 2– 1 n–1 1
Substitute r = –2x into 1, 2 256
Bukan satu janjang geometri kerana r1 ≠ r2 ≠ r3. 1 2 1 235 3 n–1 Gantikan r = –2x ke dalam a, =
2 2 a(–2x)2 = 4x3
6k2 = 1– 21 8
2k a=x
14. (a) r= = 3k n=6 \ The first term is x and the common = 2
ratio is –2x.
(b) r = – 2 ÷ 2 = – 1 n =9
9 3 3 Sebutan pertama ialah x dan nisbah sepunya
(c) a = –12, 231 – 1– 1 294
ialah –2x. 2
(c) r= e6x = e6x – 8x = e–2x r = 24 = –2, S9 =
e8x –12 1– 1 2
1 1 – 2
1 2(d) 1 2 x (ln p) Tn = –3 072
ln p 2 x 1 –3 072 = (–12)(–2)n – 1
2
r= ln px = x ln p = 256 = (–2)n – 1 216 2 = 171
324 3 128
(–2)8 = (–2)n – 1 18. (a) a = 324, r = = r,1

15. (a) a = 1 , r = –1 ÷ 1 = –4 n =9 Sn = a(1 – rn) (c) a = – 4, r = 8 = –2, Tn = 2 048
4 4 1–r –4

1 2T5 = 1 17. (a) T5 = ar4 = –48 …… 1 32431 1 2 254 2 048 = (–4)(–2)n – 1
4 T7 = ar6 = –12 …… 2 3
(–4)4 = 64 – (–2)n – 1 = –512

2 2 3 S5 = 2 = (–2)9
3 3 2 3
(b) a = , r = 1 ÷ = 2 ÷ 1, r2 = 1 1 – n = 10
4
2 3 2 187 1 = 844 – 4[1 – (–2)10]
1 3 21 2 28 128 r = 2 = 1 – (–2)
T9 = = S10
1.6
(c) a = 1, r = 52x Substitute r = 1 into 1, (b) a = 0.4, r = 0.4 =4 r.1 = 1 364
T8 = (1)(52x)7 = 514x 2
1 Sn = a(rn – 1)
Gantikan r = 2 ke dalam a, r–1

(d) a = 2, r = –4k2 = –2k2 a1 1 24 = –48 S7 = 0.4(47 – 1)
2 2 4–1

T7 = (2)(–2k2)6 = 128k12 a = –768 = 2 184.4

\ The first term is –768 and the common
1
(e) a = p, r= p2 ÷p = p ratio is 2 .
5 5
Sebutan pertama ialah –768 dan nisbah
1 2T6 = (p) p 5 p6 1
5 = 3 125 2
sepunya ialah .

51 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 52

A27

Additional Mathematics Form 4 Chapter 5 Progressions Additional Mathematics Form 4 Chapter 5 Progressions

20. (a) a= –2, r = 1 22. (a) a= –5, r =– 1 (b) 0.00931931931… (ii) Tn , 37
4 5 = 0.00931 + 0.00000931 95(0.95)n – 1 , 37
–5 + 0.00000000931 + …
The sum from the 2nd term to 6th term 1 2S∞ = (0.95)n – 1 , 37
1 95
Hasil tambah dari sebutan ke-2 hingga sebutan 1– – 5 = 0.00931
1 – 0.001
ke-6. 1 2(n – 1)(log10 0.95) 37
25 , log10 95
= S6 – S1 = – 6 0.00931
= 0.999
(–2)31 1 1 264 1 2n – 1 . log10 37
– 4 2 1 95
(b) a = 4, r = 4 = 2 931
= 1 – (–2) = 99 900 log10 0.95
4
1– S∞ = 4 n – 1 . 18.38

= – 1 365 – (–2) 1 – 1 (c) 3.0111… n . 19.38
512 2
= 3 + (0.01 + 0.001 + 0.0001 + …) \ after the 20th bounce

= – 341 =8 =3 + 0.01 selepas lantunan ke-20
512 1 – 0.1

(b) a = 1, r = 3 (c) a = –8, = 3 + 0.01 (iii) S∞ = 95
0.9 – 0.95
The sum from the 4th term to 7th term r = – 4 ÷ (–8) = 1 1
5 10
Hasil tambah dari sebutan ke-4 hingga sebutan = 3910 or / atau 271 = 1 900
90
ke-7. S∞ = –8 =– 80
9
= S7 – S3 1 – 1 Total distance covered
10 Jumlah jarak dilalui
1(37 – 1) 1(33 – 1) 25. (a) a = 36 000, r = 1 + 0.07 = 1.07 = 2(1 900) + 100
= 3–1 – 3–1 = 3 900 m
(i) T5 = 36 000(1.07)4
= 1 093 – 13 23. (a) S3 = – 13 = RM47 188.66
8
= 1 080
a(1 – r3) = – 13 …… 1 (ii) Tn = 12 × RM4 200 = RM50 400
1–r 8 36 000(1.07)n – 1 . 50 400 SPM Practice 5
(c) a = 2, r = –2
a =– 8 …… 2 (1.07)n – 1 . 1.4
The sum from the 3rd term to 7th term 1–r 5 Paper 1
log10 1.4
Hasil tambah dari sebutan ke-3 hingga sebutan 1 21 ÷ 2, (1 – r3) = – 13 ÷ – 8 n–1 . log10 1.07
8 5
ke-7. n – 1 . 4.97 1. (a) Compare with the formula

= S7 – (T1 + T2) 1 – r3 = 65 n . 5.97 Sn = n (a + l).
64 2
= 2[1 – (–2)7] – [2 + 2(–2)] \ in the 6th year n
1 – (–2) 1 Bandingkan dengan rumus Sn = 2 (a + l).
r3 = – 64 dalam tahun ke-6 Hence / Maka, m = r
= 86 – (–2)

= 88 \ r= – 1 (iii) Annual salary: 36 000, 38 520, 41 (b) k > 1
4
–12 3 1 216.40, …
21. (a) a = 8, r = 8 =– 2 , –1 Substitute r= – 4 into 2,
Gaji tahunan:
1
Since r < –1, the sum to infinity Gantikan r = – 4 ke dalam b, Savings / Simpanan: 3 600, 3 852, 4 2. (a) T4 = a + 3d = 7
cannot be determined. T5 = a + 4d = 10
Oleh sebab r , –1, hasil tambah hingga a 8 121.64, … Solving by simultaneous equation,
ketakterhinggaan tidak dapat ditentukan. 5 Selesai dengan persamaan serentak,
1– 1 2 =– r = 1.07, a = 3 600, n = 10 d=3
4
1 – S10 = 3 600(1.0710 – 1) (b) Substitute d = 3 into equation a.
1.07 – 1 Gantikan d = 3 ke dalam persamaan a.
1 a = –2 a = –2
(b) a = –2, r = – 2 –1 = RM49 739.21 23
. \ r = – 1 , a = –2 S23 = 2 [2(–2) + (23 – 1)(3)]
4 = 713
Since r . –1, the sum to infinity can be (b) a = 100 × 0.95 = 95, r = 0.95
(i) T5 = 95(0.95)4
determined. 24. (a) 0.222… = 0.2 + 0.02 + 0.002 + …
= 77.38 m
Oleh sebab r . –1, hasil tambah hingga = 0.2
1 – 0.1
ketakterhinggaan dapat ditentukan.
–2 4
1 2S∞ = =– 3 = 0.2
1 0.9
1– – 2
2
= 9

53 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 54

A28

Additional Mathematics Form 4 Chapter 5 Progressions Additional Mathematics Form 4 Chapter 5 Progressions

3. (a) (7 – 2x) = 1  120 minutes 2 hours = 120 minutes 11. Perimeter: 12 cm, 28 cm, 44 cm, … (c) Tr  0
(x – 1) 2 minit 2 jam = 120 minit a = 12, d = 28 − 12 = 16 a + (r – 1)d  0

14 – 4x = x – 1 ∴ Mat qualified for the state level run. Sn = 10.12 × 100 −45 + (r – 1)(3)  0
Mat layak untuk larian peringkat negeri. n [2(12) + (n − 1)(16)] = 1 012
5x = 15 2 n(12 + 8n – 8) = 1 012 (r – 1)  45
3
x=3 r  16

(b) T5 = ar5 – 1 = (x − 1) 4n + 8n2 = 1 012 ∴ r = 17

1 24 5rn – 1 n + 2n2 = 253
2 4
a1 a = 2 7. Tn = arn – 1 = , r ≠ 1 2n2 + n – 253 = 0
= 32
(a) r ≠ 1, so k = 1 (n – 11)(2n + 23) = 0

4. m, 3, n, … (b) a = 5 ∴ n = 11, n ≠ – 23 3. a = 3 500, d = −300, Tn = 500
4 2
(a) n = 3 Tn = a + (n – 1)d
3m Paper 2 500 = 3 500 + (n – 1)(−300)
300(n – 1) = 3 000
mn = 9 9 8. Tn = Sn – Sn – 1 n – 1 = 10
n
a = m= = n [15 – 7n] − n – 1 [15 – 7(n −1)] 1. (a) a = −12, d = −15 − (−12) = −3 n = 11
2 2
r = n (b) (i) S6 = 6 [2(−12) + (6 – 1)(−3)]
3 = 1 [15n – 7n2 – (n – 1)(22 – 7n)] 2
2 Sn = n (a + l)
= −117 2
(b) S∞ = a = 1 [15n – 7n2 – (22n – 7n2 – 22 + 7n)]
1– r 2 (ii) S7 = 7 [2(−12) + (7 – 1)(−3)] S11 = 121(3 500 + 500)
2 = 22 000
9 = 1 [−14n + 22] = −147
n 2
= n
1 – 3 = 11 – 7n (c) T7 = S7 – S6
= −147 – (−117) Total cost / Jumlah kos = 22 000 × RM0.50
= 9 × 3 9. 27 + 27r + 27r2 = 21 = RM11 000
n – = −30
3 n
9r2 + 9r + 2 = 0
27
= n(3 – n) (3r + 1)(3r + 2) = 0 2. Sn = 3n(n – 31) 4. (a) (i) a = 5, d = 2, n = 48
2
r = – 1 , r = – 2 Tn = a + (n − 1)d
3 3 3(10)(10 – 31) T48 = 5 + (48 – 1)(2)
5. Let the smallest sector = T1 = a When / Apabila r = – 1 , p = –9, q = 3 (a) S10 = 2
3 = 99 cm
Katakan sektor terkecil = −315

and the largest sector = T5 = 9a When / Apabila r = – 2, p = –18, q = 12 (ii) Sn = n (a + l)
3 2
dan sektor terbesar (b) T1 = S1 = 3(1)(1 – 31)
2
S5 = 360° S48 = 48 (5 + 99)
5 (a + 9a) = 360 Sn = n [a + l] 10. T3 = a + 2d = 22 .....................a = −45 2
2 10a = 144 2 T17 = a + 16d = −20 .....................b‚
3(2)(2 – 31) = 2 496 cm
b − a, 14d = −42 S2 = 2
a = 14.4 d = −3

Angle of the largest sector = 9 × 14.4 Substitute d = −3 into a, = −87
Sudut sektor terbesar Gantikan d = –3 ke dalam a,
a + 2(−3) = 22 T2 = S2 – S1 (b) Tn × (1.6 × 100) = 14 880
= 129.6° = −87 – (−45)
a = 28 Tn = 14 880
= −42 1.6 × 100
T8 = 28 + 7(−3) = 7
6. a = 5, r = 9 d = T2 – T1 5 + (n – 1)(2) = 93
8 = −42 – (−45) n = 45
=3
S10 = a[r10 – 1] ∴ The 45th coloured rectangle has an
r–1 ∴ first term = −45, common difference area of 14 880 cm2. The colour is
yellow.
= 51 9 210 – 1 =3
8 Sebutan pertama = –45, beza sepunya = 3 Segi empat tepat berwarna ke-45
9 –1 mempunyai keluasan 14 880 cm2. Warna
8
= 89.89 minutes kuning

55 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 56

A29

Additional Mathematics Form 4 Chapter 5 Progressions PTER

5. a = RM111.00, 2. Common difference: CHA 61 Linear Law

l = Tn = RM1 321.25, Beza sepunya Hukum Linear
Sn = RM74 477.00
r−q=q−1

(a) Sn = n (a + l) 1 + r = 2q
2
q= 1+ r
2
74 477.00 = n (111.00 + 1 321.25)
2 1. (a) Non-linear relation 3. (a)
n = 104 Common ratio: Hubungan tak linear
Nisbah sepunya In y
(b) Linear relation
∴ There are 104 lucky draw winners. r +1 = q Hubungan linear 11
Terdapat 104 pemegang cabutan bertuah. q 1
(c) Non-linear relation
(b) Tn = a + (n − 1)d q2 = r + 1 Hubungan tak linear
1 321.25 = 111.00 + (104 – 1)d
d = RM11.75 q = √r + 1 2. (a) log10 y 10
2.3
∴ The constant amount is RM11.75. 1 + r = √r + 1 2.5
Amaun yang malar ialah RM11.75. 2 9

(1 + r)2 = r + 1 8
4
(1 + r)2 – 4(1 + r) = 0
(c) T6 = 111.00 + (6 – 1)(11.75) x2
= RM169.75 (1 + r)(1 + r – 4) = 0 0 7

(1 + r)(r – 3) = 0 (b) xy

r = −1 or 3 6

(d) (i) Sn = n [2a + (n − 1)d] ∴ r ≠ −1, r = 3
2

S5 = 5 [2(111.00) + (5 – 1)(11.75)] 5
2
= RM672.50
3. …, 71, 84, … 4
a = 71, d = 84 – 71 = 13 x
(ii) S20 – S9 = 220[2(111.00) + (20 – 1) 0
(11.75)] − 9 [2(111.00) 1 hour 3
2 (c)
+ (9 – 1)(11.75)] Tn  71 + 60 = 60 minutes
a + (n – 1)d  131 1 jam In y
= 4 452.50 − 1 422 = 60 minit
= RM3 030.50 2

71 + (n – 1)(13)  131

n  131 – 71 + 1 1
13 x

n  5.615 0 In x 0 1 2 34 5

HOTS Challenge T6 = 71 + (6 – 1)(13)
= 136
1. 84, 91, 98, …, 252
a = 84, d = 7, Tn = 252 The tourist has to wait [136 – (71 + 60)] m = 11.3 – 9
Tn = a + (n − 1)d = 5 minutes. 0 – 2.5
252 = 84 + (n – 1)(7) Pelancong itu perlu menunggu [136 – (71 + 60)]
n – 1 = 24 = 2.3
n = 25 = 5 minit. –2.5

= –0.92

c = 11.3

\ ln y = –0.92x + 11.3

57 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 58

A30

Additional Mathematics Form 4 Form 4 Chapter 6 Linear Law Additional Mathematics Form 4 Chapter 6 Linear Law

(b) Ί๶y (b) m = 6–3 6. (a) (i) m= 0–e (b) xy = kx3 + p
20 – 10 e2 – 0 Y = mX + c
15 \ Y = xy, X = x3, m = k, c = p
10 = 3 =– e
10 e2
5
c =0 = –e –1 y1 2(c) 1
01 x
16 q = 3 p c =e =2 5 +n
10
\ y = –e–1 ln x + e x2

(ii) (a) 3e = –e–1 ln x + e Y = mX + c 1
q e–1 ln x = –2e y
3.2 x2 \Y= x , X = 5 , m = 2, c = n
23 4 6 ln x = –2e2
x = e–2e2 x2
5 5
(b) y = –e–1 ln e3 + e (d) log10 y = log10 q + (–rt) log10 5
4 y = –3e–1 + e
log10 y = (–r log10 5)t + log10 q
m = 17 – 1 2
3.8 – 0.6 Y = mX + c
1.5
16 7–1 \ Y = log10 y, X = t, m = –r log10 5,
= 3.2 p (b) m= 3–1 c = log10 q

=5 0 10 12 20 =3

(e) y2 = 21x2 – 12x

c = –2 At (1, 1), 1 = 3(1) + c y2 = 21x – 12
c = –2 x
\ √y = 5x2 – 2 Y = mX + c
(x – √y ) = 3x – 2
(c) 5. (a) √y = 2 – 2x \ Y= y2 , X = x, m = 21, c = –12
√y = 2(1 – x) x
log10 q y y = 4(1 – x)2
3 10 (f) ln e y = ln rx t
1.65
4–2 y = ln r + t ln x ln e = 1, y ln e = y
8 6–2
(c) m = Y = c + mX
6
2 1 \ Y = y, X = ln x, m = t, c = ln r
5.6 2
=

1 1 8. From / Daripada (a), 5 = p(e0) + 2
2 p=3
At (2, 2), 2 = (2) + c
In / Dalam (b), Y = mX + c
0 2 4 6 8 log10 p 4 c=1 y = 3ex + 2
2
log2 y = 1 log2 x +1 Hence, m = 3 and c = 2
0.39 x 2 Maka, m = 3 dan c = 2
m = 3.45 – 1.8 0 0.1 0.2 0.3 0.4 0.5
8.4 – 3.4 log2 y – log2 √x = 1 Y = 3X + 2
q = 3(2) + 2
= 1.65 log2 y =1
5 √x =8
y = 21
= 0.33 (b) m = 6.6 – 9.6 √x 14 = 3(r) + 2
0.2 – 0 3r = 12
c = 0.7 y = 2√x
= –15 r =4
\ log10 q = 0.33 log10 p + 0.7
c = 9.6

4. (a) (i) 1.5 \ y = –15x + 9.6 7. (a) y = 4x – 9
(ii) 5 x2
(iii) 12
(c) From the graph Y = mX + c
Daripada graf
(i) 5.6 \ Y= y , X = x, m = 4, c = –9
(ii) 0.39 x2

59 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 60

A31

Additional Mathematics Form 4 Form 4 Chapter 6 Linear Law Additional Mathematics Form 4 Chapter 6 Linear Law

9. (a) y = p(qx – 1) (b) yqx = p SPM Practice 6 5. Y = mX + c
log10 y = log10 p + (x – 1) log10 q ln yqx = ln p m = 4 – (–5)
log10 y = (log10 q)(x – 1) + log10 p Paper 1 6–0
Y = mX + c ln y + x ln q = ln p =3
ln y = (–ln q)x + ln p 1. From Diagram (a) / Daripada Rajah (a) 2
Y = mX + c
Y = px – q c = −5
5
3 – 1 5 m = 2–1 1 = – 1 (1) + c From the graph / Daripada graf,
3 –1 – 1 2
m= 1 = 1(1) +c q = 5t x 3 1x122
3 y 2
1– c = 2 = – 1 c = 3 y = px – q = + (−5)
3 2 2
=1 x2 q x = 3 –5
y x y 2x2
–ln q = – 1 ln p = 3 x3 =p–
2 2
log10 q= 1 log10 p = 2 y = p – qX x = 3 – 10x2
q = 10 3 1 3 x3 y 2x2

2 q = e2 p = e2 p=t+2 y = 2x2
p= q +2 x 3 – 10x2
p = 103
5
10. (a) (i) V = kt n q = 5p – 10 y= 2x3
ln V = n ln t + ln k 3 – 10x2

(ii) ln t 0 0.69 1.10 1.39 1.61 2. y = 3x2 − q 6. y= x a
ln V 3.91 3.57 3.36 3.22 3.11 xy = 3x3 – qx + 3b
−q = 1 Y = mX + c
q = −1 1 x + 3b
y = a
13 – 1 = 3
(iii) (a) ln k = y-intercept In V h–0 1 = 1 x + 3b Y = mX + c
pintasan-y 3.9 y a a
3h = 12
= 3.91 h=4 Hence / Maka,
k = e3.91
3.7 3. Y = mX + c 1 1
= 49.90 a 2
3.52
=
3.5
(b) n = gradient m= 9–1 a=2
3–8
kecerunan

= 3.76 – 3.20 = – 8 3b = −6
0.3 – 1.44 5 a
3.3 Substitute (8, 1) into
8 3b
= –0.49 1 = – 5 (8) + c the equation 2 = −6
Gantikan (8, 1) ke dalam
(c) When / Apabila t = 2.2, 3.1 c= 69 persamaan b = −4
ln 2.2 = 0.79 5
From the graph,
0 0.2 0.340.4 0.6 0.8 1.0 1.2 1.4 In t y = – 8 x2 + 69 7. Y = mX + c
Daripada graf, 1.6 x3 5 5
ln V = 3.52
ln V = e3.52 y = – 8 x5 + 69 x3 m = 3.6 – 1.4
55 1.5 – 2.6
= 33.78 = −2

(d) When / Apabila V = 42, 4. y = x2 + r 3.6 = −2(1.5) + c Substitute (1.5, 3.6)
x c = 6.6
into the equation
ln 42 = 3.74 y − x2 = r1 1 2 Gantikan (1.5, 3.6) ke dalam
x persamaan

From the graph, ln t = 0.34 Substitute ( p , 4h) into the (a) 1 = −2x + 6.6
equation 3 y2
Daripada graf t = e0.34 4h = r1 p 2 Gantikan ( p , 4h) ke dalam y2 = 1
3 persamaan 3 –2x + 6.6
= 1.40 rp
12
h =

(iv) When t = 0, V is the initial volume of the liquid.
Apabila t = 0, V ialah isi padu awal cecair.

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A32

Additional Mathematics Form 4 Form 4 Chapter 6 Linear Law Additional Mathematics Form 4 Chapter 6 Linear Law

(b) y2 = 1 Paper 2 2. (a) xy
x2 1 40
–2(1.3) + 6.6 1. (a) xy 2 4 9 16 25 36
2.4 4.0 5.6 7.8 10.4 35
=1 1 0.67 0.50 0.33 0.25 0.20 0.17
4 x 1.30 2.16 2.59 2.83 2.94 30

y= ± 1 1 (b) 25
2 y
0.50 xy 21.5

12 20

8. (a) Y = mX + c (b) 10 15

m= 3–1 y1 8 10
16 – 8
Substitute (8, 1) into 3.5 5
= 1 3.0 6 x
4 the equation 2.5
Gantikan (8, 1) ke dalam 2.0 0 123456
1 = 1 (8) + persamaan 1.5
4 1.0
c 0.5 4

c = −1 0 0.1 0.2 0.3 0.4 0.5 0.6 1x 2
x2
log2 y = 1 x – 1
4 0 5 10 15 20 25 30 35

(b) y = 21x – 1
4
21x
4 (c) y = 3px + q k√h
4x
y= 2 q (b) y – h = x
4
y = 1 1214x2 Compare with y = p(2qx) xy = 3px2 +
2 Bandingkan dengan y = p(2qx)
xy – xh = k√h
p=1 From the graph,
2 xy = hx + k√h
Daripada graf,
1 From the graph,
q = 4 (i) 3p = gradient
Daripada graf,
= 9 – 1.8
30 – 0 (i) h = gradient / kecerunan

= 0.24 = 32.5 – 5
5 – 1.2
9. (a) A = Ao ekt (c) px = qy + xy p = 0.08 = 7.24
ln A = ln Ao ekt
ln A = ln Ao + ln ekt px = y(q + x) q
ln A = ln Ao + kt ln e 4
ln A = kt + ln Ao 1 = q+x (ii) = xy-intercept / pintasan-xy k√h = xy-intercept
y px
q pintasan-xy
1= q 1 g1xr2ap+h1p, 4 = 1.8
Fryom p k√7.24 = −4
(b) ln 3Ao = kt + ln Ao the q = 7.2
ln 3Ao – ln Ao = kt k = −1.49
Daripada graf,
(ii) From the graph, y is wrongly
1ln 3Ao 2 = kt (i) 1 = 1 -intercept recorded when x = 3.5.
Ao p y Daripada graf, y salah dicatat apabila
3. (a) x = 3.5.
ln 3 = kt = 3.8 xy = 21.5
x 1.5 3.5y = 21.5
k = ln 3 p = 0.26 xy 6.8 2.0 3.0 3.5 4.5 6.0 y = 6.14
t 10.5 18.0 19.3 29.5 39.5
q
(ii) p = gradient / kecerunan

q = 1.0 – 3.3
p 0.56 – 0.1
–2.3
3.8q = 0.46

q = −1.32

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A33

Additional Mathematics Form 4 Form 4 Chapter 6 Linear Law Additional Mathematics Form 4 Chapter 6 Linear Law

4. (a) 5. (a) (ii) a will be the gradient and HOTS Challenge
b is the pv-intercept.
1 0.64 0.49 0.25 0.16 0.09 0.04 v2 4 6 8 a adalah kecerunan dan b ialah
x2 v2p 24.84 45.60 65.88 71.68 pintasan-pv
1. (a) (i)
y2 6. (a) R = kvµ
x 15.98 12.39 6.77 4.79 3.20 2.00 v2 p log10 R = log10 k + µ log10 v x 50 100 150 200 250
y-axis = log10 R, x-axis = log10 v, 110 144 179 214
(b) 86 gradient / kecerunan = µ, y 74
y-intercept / pintasan-y = log10 k x
y2 80
x (b)
log10 v 0.70 1.00 1.18 1.30 1.40 y y =x
70 log10 R 1.70 2.08 2.25 2.48 2.58 x x
16
log10 R
200
2.5
14 60
2.35
150
2.0
12 50
1.5
100

10 40

8 30 50
x
6 20
0 50 100120.5 150 200 250

4 10 1.0 (ii) The equation formed from the
2 0.5 situation:
0 v Persamaan ini terbentuk daripada keadaan
1 2468 0 0.2 0.4 0.6 0.8 0.94 1.0 1.2 1.4 log10 v ini
0 0.1 0.2 0.3 0.4 0.5 0.6 x2 y = x(ax + b)

(b) From the graph, y = ax + b
(8, 71.68) is not correctly recorded. x
Daripada graf, (8, 71.68) tidak dicatatkan (c) From the graph,
1 dengan betul. Daripada graf, From the graph,
x2 When / Apabila v = 8, v2p = 86 Daripada graf,
(c) (i) When / Apabila x = 1.82, = 0.30, log10 k = 0.85 a = gradient / kecerunan
64p = 86 k = 100.85
from the graph, y2 = 8 p = 1.34 = 200 – 100
daripada graf, x k = 7.079 230 – 85

y2 = 8 × 1.82 µ = 2.2 – 1.0 = 0.6897
= 14.56 1.10 – 0.12
a b y
y = 3.82 (c) p= v2 + v = 1.224 b = x -intercept = 40
pintasan
v2p = a + bv 1
(d) (i) 2 log10 v= 0.6
(ii) From the graph, From the graph, log10 v = 1.2
Daripada graf, (b) y = x2
Daripada graf,
y =x
gradient / kecerunan a = v2p-intercept / pintasan = 5 From the graph,log10 R = 2.35 x

= 9.2 – 1 = 23.4, b = gradient /kecerunan = 55 – 5 = 10 Daripada graf, R = 102.35 (i) x = 120.5
0.35 – 0 5–0 (ii) The rectangle will become a square
R = 233.9
of side of length of this value of x,
y2 (d) p= a + b (ii) R = 100 that is 120.5 m.
x v2 v
-intercept / pintasan = 1 log10 R = log10 100 = 2.0 Segi empat tepat itu akan menjadi sebuah
segi empat sama dengan panjang sisi nilai
y2 = 23.41 1 2 +1 pv = a1 1 2 + b From the graph,log10 v = 0.94
x x2 v x, iaitu 120.5 m.
Daripada graf, v = 100.94
(i) pv is plotted on the y-axis.
√y = 23.4 + x pv diplot pada paksi-y. v = 8.710
x

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A34

PTER Additional Mathematics Form 4 Chapter 7 Coordinate Geometry

CHA 71 5(–10) + 2(q) 5(3q) + 2(p) 4(–9) + 1(–4)
5+2 5+2 4+1
Coordinate Geometry 1 2(c) , = (–6, 2p) (ii) =k

Geometri Koordinat –50 + 2q = –6, 15q + 2p = 2p –40 =k
7 7 5

–50 + 2q = –42 15q + 2p = 14p k = –8

q=4 15(4) = 12p 6. (a) QR = 3 PR → QR = 3
5 PR 5
1. (a) y 1(4) + 3(–2) 1(7) + 3(–1) p=5
1+3 1+3
4 1 2(b) P = ,

A = 1– 1 12 5. (a) B(10, 11) Hence / Maka, PQ : QR = 2 : 3
2
, n

B(4, 7) m M(1, 5) Let / Biar R(x, y),
A(–5, 1)
2(y) + 3(–5)
2+3
2 P1 3 1 2 2(x) + 3(–4) , = (2, –1)
O 2 B4 2+3
x
(b) y A x nn(–5), m(11) + n(1) 2x – 12 = 2, 2y – 15 = –1
P1 m+n 5 5
4 1P 1 2m(10) + = (1, 5)
2 A(–2, –1) m+
2x – 12 = 10 2y – 15 = –5
11m + n x = 11 y =5
m+n
1 210m – 5n , = (1, 5)
m+n
3(1) + 2(–1) 3(8) + 2(5)
3+2 3+2
1 2(c) P = , 10m – 5n =1 \ The coordinates of traffic light R are
m+n (11, 5).
= 1 1 , 3542
5 10m – 5n = m + n Koordinat bagi lampu isyarat R ialah (11, 5).
10m – m = n + 5n
O 2 B4 B(1, 8)
9m = 6n
2 7. (a) m1 = 3 y = –3x + 2
m2 = –3
(c) P m 6
3 n = 9
y
\ m1 ≠ m2, the two lines are not parallel.
2 CP D A(–1, 5) = 2 kedua-dua garis tidak selari.
1 3

O 246 x \m:n =2:3 (b) y = 5x + 3 y = 5x – 3
m1 = 5 m2 = 5
1(4) + 3(p) 1(q) + 3(3)
1+3 1+3
1 24. (a) , = (4, 1) (b) Q(–9, 9) n

2. (a) 2 : 7 4 + 3p = 4, q+9 =1 A(k, 6) \ m1 = m2, the two lines are parallel.
(b) 5 : 4 4 4 m kedua-dua garis adalah selari.
(c) 8 : 1
4 + 3p = 16 q+9=4 P(–4, –6)
p =4 q = –5 1 4 1
2(–1) + 1(5) 2(10) + 1(1) 1 2m(–9) + (c) y= 3 x – 3 m2 = 3
2+1 2+1 m+
1 23. , n(–4) m(9) + n(–6)
(a) P = n , m+n = (k, 6) m1 = 1
3
3(q) + 2(7) 3(p) + 2(2q)
3+2 3+2
= (1, 7) 1 2(b) , = (1, –2) 9m – 6n
m+n
1 2–9m – 4n , = (k, 6) \ m1 = m2, the two lines are parallel.
m+n kedua-dua garis adalah selari.
B(–1, 10) 3q + 14 = 1, 3p + 4q = –2
1 5 5 9m – 6n
P (i) m+n =6
3q + 14 = 5 3p + 4q = –10

q = –3 3p + 4(–3) = –10 9m – 6n = 6m + 6n 8. (a) 2x + ky = 1 2y + x = 12
2 2
p = 3 9m – 6m = 6n + 6n ky = –2x + 1 2y = –x + 12

A(5, 1) 3m = 12n y = – 2 x + 1 y = – 1 x + 6
k k 2
m = 12
n 3 m1 = – 2 m2 = – 1
k 2
= 4
1
\m:n =4:1

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A35

Additional Mathematics Form 4 Chapter 7 Coordinate Geometry Additional Mathematics Form 4 Chapter 7 Coordinate Geometry

The two lines are parallel, hence 1 2m1m2 1 The equation is / Persamaan ialah (iii) 1
= –5 5 1 2
Kedua-dua garis adalah selari, maka y – (–5) = 2 (x – 3)
= –1
– 2 =– 1 1 3 T(h, k) Q(–3, 2) P(3, 14)
k 2 2 2
\ Both the lines are perpendicular. y + 5 = x – 2h + 3 2k + 14
Kedua-dua garis adalah berserenjang. 3 3
k =4 1 13 = –3 , =2
2 2
y = x – 2h + 3 = –9 2k + 14 = 6

(b) kx + 4y = 3 8y – x = 6 (b) y – 3x = 8 3y = x – 9 (b) 2y – 5x = 6 h = –6 k = –4
y = 3x + 8
4y = –kx + 3 8y = x + 6 m1 = 3 1 2y = 5x + 6
3
y = – k x + 3 y = 1 x + 3 y = x – 3 5 \ T(–6, –4)
4 4 8 4 2
1 y = x + 3
k 1 m2 = 3
m1 = – 4 m3 = 8 5 14. (a) Area of DPQR
1 = 2
The two lines are parallel, hence 1 2m1m2 = 3 3 m1 Luas DPQR

Kedua-dua garis adalah selari, maka =1 m1m2 = –1 = 1  ––32 5 4 –3 
2 0 8 –2
k 1 5
– 4 = 8 \ Both the lines are not perpendicular. 2 m2 = –1 = 1 |(0 + 40 – 8) – (–10 +0– 24)|
Kedua-dua garis tidak berserenjang. 2
k = – 1 m2 = – 2
2 5 1
11. (a) 5y = –kx + 3 2y = 5x + 6 Equation of the road is = 2 |32 – (–34)|

9. (a) 6x – 2y – 1 = 0 y = – k x + 3 y= 5 x + 3 Persamaan jalan lurus ialah = 33 unit2
5 5 2 2
2y = 6x – 1 y – (–4) = – 5 [x – (–1)]
k 5
y = 3x – 1 m1 =– 5 m2 = 2 y + 4 = – 2 x – 2 (b) Area of DABC
2 5 5
Both the lines are perpendicular, hence Luas DABC

Gradient / Kecerunan = 3 Kedua-dua garis adalah berserenjang, maka y = – 2 x – 22 = 1  32 5 6 3 
5 5 2 –2 0 2

Equation of the straight road is 1– k 2 1 5 2 = –1 = 1 |(–6 +0 + 12) – (10 – 12 + 0)|
5 2 2
Persamaan bagi jalan lurus ialah 13. (a) (i) 2y = –x + 1
y – 7 = 3[x – (–2)] k=2 1 1 1
y – 7 = 3x + 6 y = – 2 x + 2 = 2 |6 – (–2)|
y = 3x + 13
(b) ky = –3x + 4 4y = –x + 8 mQR =– 1 = 4 unit2
2
(b) 3x – 4y + 5 = 0 y = – 3 x + 4 y= – 1 x + 2 mPT = 2
k k 4
4y = 3x + 5 (c) Area of DPQR
m1 = – 3 m2 = – 1
3 5 k 4 The equation of PT is Luas DPQR
y = 4 x + 4 Persamaan PT ialah
= 1  –86 1 5 –8 
Both the lines are perpendicular, hence y – 14 = 2(x – 3) 2 –3 2 6
3 y = 2x + 8
Gradient / Kecerunan = 4 Kedua-dua garis adalah berserenjang, maka = 1 |(24 +2 + 30) – (6 – 15 – 16)|
2
Equation of the side is 1– 3 2 1– 1 2 = –1 1 1
k 4 2 2
Persamaan sisi ialah 3 (ii) – x + = 2x + 8 = 1 |56 – (–25)|
k =– 4 2
y – (–3) = 3 (x – 8) 5 15
4 – 2 x = 2 = 40.5 unit2
12. (a) 6x + 3y = 9
y + 3 = 3 x – 6 3y = –6x + 9 x = –3 15. (a) Area of quadrilateral PQRS
4 y = –2x + 3
m1 = –2
y = 3 x – 9 y =– 1 (–3) + 1 =2 Luas sisi empat PQRS
4 2 2
= 1  03 1 5 3 0 
10. (a) y + 5x – 2 = 0 5y – x + 1 = 0 2 1 8 10 3
m1m2 = –1 \ Q(–3, 2)
y = –5x + 2 5y = x – 1 1
m1 = –5 –2m2 = –1 = 2 |(0 +8 + 50 + 9) – (3 +5 + 24 + 0)|
1 1
y = 5 x – 5 m2 = 1 1
2 2
1 = |67 – 32|
5
m2 = = 17.5 unit2

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A36

Additional Mathematics Form 4 Chapter 7 Coordinate Geometry Additional Mathematics Form 4 Chapter 7 Coordinate Geometry

(b) Area of quadrilateral PQRS 17. (a) Area of ∆ABC (ii) mPQ = 1 – (–5) = 2 3x2 + 3y2 – 28x – 60y = 0
3–0 + 168
Luas sisi empat PQRS Luas ∆ABC
(b) 2AP = 3PB
= 1  08 5 0 4 8  = 1  18 –4 6 8  m2 =– 1 AP = 3
2 7 –2 –3 0 2 –8 –0.5 1 2 PB 2

= 1 |(56 – 10 + 0 + 0) – (0 + 0 – 8 – 24)| = 1 |(–64 + 2+ 6) – (–4 – 48 – 4)| y – (–3) = – 1 (x – 1) 2√(x – 5)2 + (y – 2)2 = 3√(x – 1)2 + [(y – 7)]2
2 2 2

= 1 |46 – (–32)| = 1 |–56 – (–56)| y =– 1 x – 5 4(x2 – 10x + 25 + y2 = 9(x2 – 2x + 1 + y2
2 2 2 2 – 4y + 4) – 14y + 49)

= 39 unit2 =0 (iii) Area of ∆PON 4x2 – 40x + 100 9x2 – 18x + 9 +
+4y2 – 16y + 16 9y2 – 126y + 441
(c) Area of quadrilateral PQRS Luas ∆PON =
Luas sisi empat PQRS
Therefore, points A, B and C are = 1  00 1 3 0  5x2 + 5y2 + 22x – 110y = 0
collinear. 2 –3 1 0 + 334
= 1  ––23 –4 7 2 –2  Maka, titik-titik A, B dan C adalah segaris.
2 –1 2 9 –3 1
= 2 |(1) – (–9) |

= 1 |(2 + 8 + 63 + 6) – (12 –7+4– 18)| (b) Area of ∆ABC = 5 unit2 21. (a) (i) 3AP = 2PB
2 9[(x – 2)2 + (y – 4)2] = 4[(x – 5)2 + (y – 1)2]
Luas ∆ABC

= 1 |79 – (–9)| = 1  13 –12 6 3  OP = √32 + 12 = √10
2 2 14 –4 1
Let h = shortest distance from N to 9(x2 – 4x + 4 + y2 4(x2 – 10x + 25 + y2
= 44 unit2 1 – 8y + 16) = – 2y + 1)
= 2 |(42 + 48 + 6) – (–12 + 84 – 12)| line OP

16. (a) Area of pentagon FGHKL = 1 |96 – 60| Biar h = jarak terdekat dari N ke garis OP 9x2 – 36x + 36 + 9y2 = 4x2 – 40x + 100 +
2 – 72y + 144 4y2 – 8y + 4
Luas pentagon FGHKL = 18 unit2 1 h(OP) = 5
2
1  03 1 6 5 4 0  5x2 + 5y2 + 4x – 64y + 76 = 0
= 2 1 3 8 8 3 Therefore, points A, B and C are not 1 h(√10) = 5
collinear. 2
1 Maka, titik-titik A, B dan C tidak segaris. h = 10 = 3.16 units (ii) intersects at x-axis ⇒ y = 0
= 2 | (0 + 3 + 48 + 40 + 12) – √10 menyilang paksi-x
(3 + 6 + 15 + 32 + 0)|
1 5x2 + 5(0)2 + 4x – 64(0) + 76 = 0
= 2 |103 – 56| (c) Area of ∆ABC
Luas ∆ABC 5x2 + 4x + 76 = 0

= 23.5 unit2 19. (a) AP = 10 b2 – 4ac = (4)2 – 4(5)(76)
= –1 504 , 0
= 1  –31 4 –2 1  √(x – 6)2 + [y – (–8)]2 = 10
2 3 –9 –3 There are no real roots. Therefore,
(b) Area of hexagon MNPQRS (x – 6)2 + (y + 8)2 = 100 the locus of point P does not
intersect the x-axis.
Luas heksagon MNPQRS = 1 |(3 – 36 + 6) – (–12 – 6 – 9)| x2 – 12x + 36 + y2 + 16y + 64 = 100 Tiada punca nyata. Maka, lokus titik P
2 tidak menyilang paksi-x.
= 1  –41 –4 2 11 6 3 –41  x2 + y2 – 12x + 16y = 0
2 –3 1 7 8 10 1
= 2 |–27 – (–27)| (b) AP = 8
1 √(x – (–7)2 + (y – 7)2 = 8
= 2 |(3 – 4 + 14 + 88 + 60 + 12) – =0 (x + 7)2 + (y – 7)2 = 64
(-16 – 6 + 11 + 42 + 24 – 10)|
Therefore, points A, B and C are x2 + 14x + 49 + y2 – 14y + 49 = 64 (b) (i)
= 1 |173 – (45)| collinear. x2 + y2 + 14x – 14y + 34 = 0 √(x – 3)2 + (y – 2)2 = √(0 – 3)2 + (–2 – 2)2
2 Maka, titik-titik A, B dan C adalah segaris. x2 – 6x + 9 + y2 – 4y + 4 = 25

= 64 unit2 x2 + y2 – 6x – 4y – 12 = 0

(c) Area of heptagon TUVWXYZ 18. (a) (i)

Luas heptagon TUVWXYZ 1 2 20. (a) 2AP = PB AP = 1 (ii) x2 + y2 – 6x – 4y – 12 = 0 ……a
PB 2
= 1  –56 –1 4 5 8 7 2 –6   Substitute x = 0, y = p into equation
2 –2 –2 3 4 10 9 5 Q(0, –5) N(x, y) P(3, 1) a:
2√(x – 4)2 + (y – 6)2 = √(x – 2)2 + [(y – (–6)]2 Gantikan x = 0, y = p ke dalam a:
= 1 |(12 + 2 + 12 + 20 + 80 + 63 + 10) –
2 (–5 – 8– 10 + 24 + 28 + 20 – 54)| 2(0) + 1(3) 2(–5) + 1(1) 4(x2 – 8x + 16 + y2 x2 – 4x + 4 + y2 + p2 – 4p – 12 = 0
3 3 – 12y + 36) 12y + 36 (p – 6)(p + 2) = 0
1 2N = , =
p = 6 or p = –2
= 1 |199 – (–5)| = (1, –3) 4x2 – 32x + 64 + 4y2 x2 – 4x + 4 + y2
2 – 48y + 144 + 12y + 36 Hence / Maka, p = 6
=
= 102 unit2

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A37

Additional Mathematics Form 4 Chapter 7 Coordinate Geometry Additional Mathematics Form 4 Chapter 7 Coordinate Geometry

mPQ = 1 Area of triangle PQR (c) MN2 + LN2 = ML2
2p 20 + (–6 – 0)2 + (–4 + x)2 = (–x + 2)2 + (0 + 2)2
SPM Practice 7 Luas segi tiga PQR
20 + 36 + x2 – 8x + 16 = 4 – 4x + x2 + 4
Paper 1 y – 6x = 1 = 1 –5 6 2 –5 –8x + 4x = –64
2 5k 2 2 –1 7 2 –4x = –64
y x = 16
2 + x = 1 = 1 (5 + 42 + 4) – (12 – 2 – 35)
2 \ L(–16, 0)
1. (a) py = 4x – 8 1– 5k 2
6
y = 1 4 2 x – 8 = 1 51 – (–25)
p p 2
Area ∆LMN
x-intercept = – 8 mRS = – y-intercept = 38 Luas ∆LMN
p x-intercept
pintasan-x =– 2 1 –16 –4 –2 –16
PQ = √[6 – (–5)]2 + (–1 – 2)2 = 2 0 –6 –2 0
1– 5k 2 = √112 + (–3)2
∴– 8 = –r 6 = √130 = 1 (96 + 8 + 0) – (0 + 12 + 32)
p 2
8 12
p= r = 5k 1
2
Area of triangle PQR = 38 = 104 – 44

4 8 mPQ × mRS = −1 Luas segi tiga PQR = 30 unit2
p p
(b) y = 1 2 x – 1 1 21 12 2 = −1 1 (PQ)(RS) = 38
2p 5k 2
4 12 = −10pk 1
∴m = p 2 √130 (RS) = 38
= 6.6656 m
m= k–0 = 4 p = – 12 RS 2. (a) Area of ∆OPQ = 12
0 – (–q) p 10k
Luas ∆OPQ
=– 6
k= 4 5k Paper 2 10 h –4 0 = 12
qp 1. (a) (i) 20 8 –4 0
Coordinates
4q 4. 4AC = 3AB L(–x, 0) P(x, y) 1 [–4h – (–32)] = +12
p AC = 3 M(–2, –2) 2 are arrange in
∴k= AB 4 −4h + 32 = 24
anticlockwise
32 − 4h = 24
direction
2. (a) 2 × h = −1 4h = 8 Koordinat disusun
dalam arah lawan
31 h =2 jam.
C(x, y) B(–9, –2)
h= – 1 A(5, 5)
2
3(–9) + 1(5) 3(–2) + 1(5) N(–4, –6)
3+ 1 3 +1
1 , 2 = (x, y) (b) (i)

(b) y = 2x − 7 x= –27 + 5 , y= –6 + 5 MN = √[–4 – (–2)]2 + (–6 – (–2)]2 m P(2, 8)
4 4 A(1, 6)
y = – 1x + 8 = √4+ (–4)2
2 = – 22 = – 1 n
4 4 = √20
2x − 7 = – 1 x + 8
2 11 PM = MN
5 = – 2 [x – (–2)2] + [y – (–2)]2 = 20
2 x = 15 Q(–4, –4)
(x + 2)2 + (y + 2)2 = 20
x=6 ∴ C1– 11 , – 1 2 x2 + 4x + 4 + y2 + 4y + 4 – 20 = 0  m(–4) + n(2) , m(–4) + nn(8)  = (1, 6)
2 4 m +n m +
y = 2(6) − 7 x2 + y2 + 4x + 4y – 12 = 0
=5  –4m + 2n , –4m + 8n  = (1, 6)
(b) substitute x = q and y = 2 into the m+ n m+ n
5. Let RS = shortest distance from R to the equation: –4m + 2n = 1
∴ A(6, 5) other side of the road gantikan x = q dan y = 2 ke dalam persamaan: m+n
Katakan RS = jarak terpendek dari R ke seberang q2 + 22 + 4q + 4(2) – 12 = 0
jalan raya. q2 + 4q = 0 –4m + 2n = m + n
q(q + 4) = 0
3. 5 + 6py − 3x = 0 R(2, 7) q = 0 , q = –4 5m = n

6py = 3x − 5 5 m =1
1 6p n5
y = 2p x –
P(–5, 2) PA : AQ = 1 : 5

S Q(6, –1)

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A38

Additional Mathematics Form 4 Chapter 7 Coordinate Geometry Additional Mathematics Form 4 Chapter 7 Coordinate Geometry

(ii) RA = 2RQ 4. (a) (i) mKL = mBC = 4 – 2 = 2 KL and BC (c) Area of ∆ABC 4[x2 + (y − 6)2] = (x – 3)2 + (y + 1)2
6 – 5 4(x2 + y2 − 12y + 36) = x2 – 6x + 9 + y2 + 2y + 1
√ (x − 1)2 + (y − 6)2 = 2√ [x − (–4)]2 + [y − (–4)]2 are parallel. Luas ∆ABC 4x2 + 4y2 − 48y + 144 = x2 – 6x + 9 + y2 + 2y + 1
KL dan BC 3x2 + 3y2 + 6x – 50y + 134 = 0
(x − 1)2 + (y − 6)2 = 4[(x + 4)2 + (y + 4)2] Equation of KL is adalah selari. = 1 2 5 6 2
Persamaan KL ialah 2 7 2 4 7 (ii) cuts the x-axis ⇒ y = 0
x2 − 2x + 1 + y2 = 4(x2 + 8x + 16 + y2 + 8y y − 7 = 2(x − 2) memotong paksi-x
− 12y + 36 + 16) y − 7 = 2x − 4 = 1 [(4 + 20 + 42) – (35 + 12 + 8)]
2 3x2 + 3(0)2 + 6x – 50(0) + 134 = 0
x2 + y2 − 2x – 12y + 3 = 4x2 + 4y2 + 32x + 32y + 128 y = 2x + 3 3x2 + 6x + 134 = 0
= 1 (66 – 55)
3x2 + 3y2 + 34x = 0 2 b2 – 4ac = 62 – 4(3)(134)
+ 44y + 91 = 36 – 1 608
(ii) mPK = mAB = 11 unit2 = –1 572  0
2
7 – 4 ∴ The locus does not cut the x-axis.
= 2 – 6 Area of ∆KLP Lokus itu tidak memotong paksi-x.

3. (a) = – 3 Luas ∆KLP
4
√[x − (–3)]2 + (y − 8)2 = √(x − 9)2 + [y − (–2)]2 = 4 × Area of ∆ABC / Luas ∆ABC

(x + 3)2 + (y − 8)2 = (x – 9)2 + (y + 2)2 m2 × mPK = −1 The equation of =4 × 11
2
x2 + 6x + 9 + y2 x2 – 18x + 81 + y2 + 1− 3 2 the perpendicular = 22 unit2
− 16y + 64 = 4y + 4 m2 × 4 = −1
bisector of PK is HOTS Challenge
24x – 20y – 12 = 0 m2 = 4 an equation of a 5. (a) (i) y − 3x = 6
3
6x – 5y – 3 = 0 straight line that x = 0, y = 6 ⇒ R(0, 6)
Equation of perpendicular 1. (a) m n
is perpendicular y = 0,0 − 3x = 6
bisector of PK is
to PK and passing x = −2 ⇒ T(–2, 0) P(–2, 6) Q(0, 4h) R(4h + 1, –h)
(b) (i) y = −2x + 9 ...................a Persamaan pembahagi dua through the
6x − 5y − 3 = 0 .............b S = 10 + (–2) , 6 + 02 = (−1, 3)
sama serenjang PK ialah midpoint of PK. 2 2
Persamaan bagi m(4h + 1) + n(–2) = 0
y – 2 = 4 (x – 5) pembahagi dua sama m+ n
Substitute a into b: 3 serenjang PK ialah satu (ii) Area of quadrilateral OTSU
Gantikan a ke dalam b: 3y – 6 = 4x – 20 persamaan garis lurus 4mh + m – 2n = 0
6x − 5(−2x + 9) − 3 = 0 yang berserenjang kepada Luas bagi sisi empat OTSU
3y = 4x – 14 PK dan melalui titik m(4h + 1) = 2n
6x + 10x − 45 − 3 = 0 tengah PK. 1 0 0 –1 –2 0
16x = 48 2 0 2 3 00
x =3 = m = 2 ……… a
n 4h +
= 1 0 − (−2 − 6) 1
2
(b) y = 2x + 3 ..................................a m(–h) + n(6)
Substitute x = 3 into a: 3y = 4x − 14 ..............................b = 4 unit2 m+n = 4h
Gantikan x = 3 ke dalam a: a × 3: 3y = 6x + 9 ................... c
y = −2(3) + 9 (b)
–mh + 6n = 4mh + 4nh
=3 c − b: 0 = 2x + 23 S(–1, 3)
1 U(0, 2) 4mh + mh = 6n – 4nh
∴ (3, 3) 2x = −23
m = 6 – 4h ……… b
x = – 23 3 V(x, y) n 5h
2
C11, 3 2 a = b:
(ii) 5 Substitute x = – 23 into a: 1 3(–1) + 1(x) , 3(3) + 1(y) 2 = (0, 2) 2 = 6 – 4h
2 a: 3+1 3 + 1
y = −2x + 9 ke dalam 4h + 1 5h
Gantikan x = – 23 –3 + x 9 + y
= −2(1) + 9 2 4 =0 , 4 = 2 10h = (4h + 1)(6 – 4h)
10h = 24h – 16h2 + 6 – 4h
=7 y = 21– 23 2 + 3 −3 + x = 0 9+y=8 16h2 – 10h – 6 = 0
2 8h2 – 5h – 3 = 0
6x – 5y – 3 = 0 = −20 x=3 y = −1 (8h + 3)(h – 1) = 0
6(1) − 5y − 3 = 0
∴ S 1– 23 , –202 ∴ V(3, −1)
5y = 3 2
y= 3 (c) (i) h = – 3 (ignore / diabaikan) , h = 1
5 8
AR = 1 AV
Road PQ passes through mosque 2 (b) m = 6 – 4(1) = 2
Jalan raya PQ melalui masjid n 5(1) 5
2AR = AV

C11, 3 2. 2√ (x − 0)2 + (y − 6)2 = √ (x − 3)2 + [y − (–1)]2 ∴ PQ : QR = 2 : 5
5

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A39

PTER Additional Mathematics Form 4 Chapter 8 Vectors

CHA 81 Vectors 6. (a) E→F = 3 m~ , ⇒ m~ = 2 E→F 7. (a) U→V = 1 4 26~r 2W
2 3 6 4V
Vektor F→G E→F 2 U
= 51 2 = 4~r
3 4W
10 E→F 5V
= 3 V→W = 1 2 26~r U
6
Hence, E→F and F→G are parallel. Points W
1. = 2~r 3
E, F and G are collinear as F is the U 1V
Quantity Scalar Vector Justification
Kuantiti Skalar Vektor Justifikasi common point. (b) U→V = 1 5 26~r
Maka, →EF dan →FG adalah selari. Titik-titik E, 9
(a) 10 kg 7 7 No direction / Tiada arah
F dan G adalah segaris kerana F ialah titik = 130~r
7 7 Possesses direction and magnitude
(b) 50 N Mempunyai arah dan magnitud sepunya.
7
(c) 5% No direction / Tiada arah (b) H→I = –2~n, ⇒ ~n =– 1 H→I V→W = 1 4 26~r
2 9
Possesses direction and magnitude I→K H→I2
(d) 1 m s–1 due north Mempunyai arah dan magnitud = –41– 1 = 8 ~r
ke utara 2 3
No direction / Tiada arah
(e) 45°C = 2H→I (c) U→V = 1 1 26~r
4
Hence, H→I and I→K are parallel. Points H,
2. (a) (c) Magnitude / Magnitud I and K are collinear as I is the common = 3 ~r
= 4 units point. 2
~a m s–2 Maka, →HI dan →IK adalah selari. Titik-titik
30° \ 4 units due west H, I dan K adalah segaris kerana I ialah titik V→W = 1 3 26~r
4 unit ke arah barat 4
sepunya.
(b) N = 9 ~r
K 2
(d) Magnitude / Magnitud R→S = –3~q , 1 R→S
(c) ⇒ ~q = – 3
= √32 + 32
56 m = √18 T→U 1 R→S2 8. (a) m + 3 = 0
= 3√2 units 3 m = –3
H = –51–
\ 3√2 units due south-east
(c) 3√2 unit ke arah tenggara = 5 R→S 2–n =0
3 n =2

60 km h–1 Hence, R→S and T→U are parallel but

(ii) (a) 5 units due west points R, S, T and U are not collinear (b) 4 – 2m= 0
5 unit ke arah barat 2m = 4
as there is no common point. m =2
Maka, →RS dan →TU adalah selari tetapi titik-titik

3. (i) (a) Magnitude / Magnitud (b) √5 units due north-west R, S, T dan U tidak segaris kerana tidak ada titik
= 5 units √5 unit ke arah barat laut
sepunya. n+1 =0
\ 5 units due east n = –1
5 unit ke arah timur (c) 4 units due east (d) V→W = 25~p, ⇒ ~p = 215V→W
4 unit ke arah timur (c) (5m + 6)~a = (2n + 4)~b
(b) Magnitude / Magnitud X→Y = 81215V→W2
= √22 + 12 (d) 3√2 units due north-west = 285V→W 5m + 6 = 0 2n + 4 = 0
= √5 units 3√2 unit ke arah barat laut Hence, V→W and X→Y are parallel but
5m = –6 2n = –4
\ √5 units due south-east points V, W, X and Y are not collinear 6 n = –2
√5 unit ke arah tenggara 4. (a) U→V and ~t as there is no common point. m = – 5
Maka, V→W dan X→Y adalah selari tetapi titik-titik
5. (a) – 3 ~r (b) 3~r V, W, X dan Y tidak segaris kerana tidak ada titik
2
sepunya.
(c) 2~r

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A40

Additional Mathematics Form 4 Chapter 8 Vectors Additional Mathematics Form 4 Chapter 8 Vectors

C→D 3E→F 2F→G 1 (e) (c) (i) A→C = A→D + D→C (c) (i) (a) –4~i + 3~j
1 29. = 3 = ~r + ~s
(a) (i) – + = 23~u~u – 3 ~u + 2(~u) –2 p~ + 3q~ (b) 1–342

(ii) uC→D – 3E→F + 2F→Gu = 33u(~u4u) 3q~ –2 p~ (ii) D→B = D→A + A→B (ii) u~t u
= = –~r + ~s = √(–4)2 + 32
= √25 units
= 12 units C→X C→B 1 B→A = 5 units
2
1 F→G 3 E→F 1 D→C (iii) = + (d) (i) (a) 3~i – ~j
4 2 8
(b) (i) + – 11. (a) c~ 1 (b) 1–312
2
1 2= 1 3 1 1 = –~r – ~s (ii) u~uu
4 2 3 8 = √32 + (–1)2
(~u) + ~u – (–2~u) b~ c~ +31– ~b (iv) X→D = X→A + A→D = √10 units
1–3 ~b c~
= 1 ~u + 1 ~u + 1 ~u = – 1 ~s + ~r
4 2 4 2
= ~u
u u(ii) F→G E→F D→C (d) O→B = ~b and / dan 2OM = MB,
1 + 3 – 1 = u~uu hence / maka O→M = 1 O→B 1
4 2 8 (b) ~a A→M = M→C 3 = 3 ~b
2a~ + d~ (e) (i) (a) 3~j
= 4 units d~

(c) (i) G→F + 3C→D – 6E→F A→O + O→M = M→O + O→C (b) 1 0 2
3
–O→A + O→M = –O→M + O→C
1 2= –~u + – 1 (ii) u~vu
3(2~u) 6 3 ~u ~d –~a + 1 ~b = – 1 ~b + ~c = √32
3 3
= 3~u 12. (a) D→H 2a~ = 3 units
(c) C→A
(ii) uG→F + 3C→D – 6E→Fu = 33(u~u4u) (b) F→C 2 ~b = ~a + ~c
= (d) D→A 3
3 (f) (i) (a) –3~i – 3~j
~b = 2 (~a + ~c )
(b) 1––332
= 12 units O→M 1
3 (ii) uw~u
10. (a) 13. (a) C = ~b = √(–3)2 + (–3)2

2 p~ – ~q 45° = 1 [ 3 (~a + ~c )] = √18
2 ~p – q~ B 3 2 = 3√2

Displacement = 1 (~a + ~c )
Sesaran 2

(b) 3_ A 14. (a) (i) (a) –3~i – 2~j 12~i – 5~j
2 1 cm represents 200 km √122 + (–5)2
– p~ + 2 ~q 1 cm mewakili 200 km (b) 1––232 15. (a) ~a^ =

2 ~q – _3 ~p (ii) u~r u = 1123~i – 153~j
2 = √(–3)2 + (–2)2
V = √13 units
(c) (b) (i) P→V = ~a + ~b a~ + ~b (b) Unit vector in the direction of P→Q
P a~ b~ Vektor unit dalam arah P→Q
2 ~p + 2 ~q 2~q Q

2 p~ W (b) (i) (a) –3~i + 2~j = –3~i – 3~j
√(–3)2 + (–3)2
(ii) P→W = ~a + ~b + ~c c~ (b) 1–232
(d) ~a + ~b + c~ V = –3~i – 3~j
P ~a + b~ (ii) u~s u 3√2
– 3 ~p = √(–3)2 + 22
= √13 units = – 1 ~i – 1 ~j
√2 √2
U
–3~p – 4 ~q (iii) U→Q = –~b + ~a – b~ – ~b + ~a
– 4 q~
P ~a Q

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A41

Additional Mathematics Form 4 Chapter 8 Vectors Additional Mathematics Form 4 Chapter 8 Vectors

(c) Unit vector in the direction of R→S 17. (a) P→Q = P→O + O→Q At t = 2 s, the position vectors of both 4h – 1 = 0
Vektor unit dalam arah →RS the particles P and Q are the same. 1
= –2–~(~ii~i ––+~3j~~jj+) + (3~i – 2~j ) Hence, P meets Q after 2 s. h = 4
= 3~i – 2~j Pada t = 2 s, kedudukan vektor bagi kedua-dua
1 2= 1 4 = zarah P dan Q adalah sama. Maka, P bertemu Q 31 2 2 1–452
√42 + 62 6 selepas 2 s. (ii) (b) 3~v + w~ = 1 +

1 2= 1 1 4 2 = 2 (b) P→R = P→O + O→R R→S 3 = 1 6 2 + 1–452
2√13 6 5 3
√13 = –(~i + ~j ) + (6~i + k~j ) (b) (i) = × 10~r 2T 1
3 = –~i – ~j + 6~i + k~j 1 7 2
= 5~i + (k – 1)~j
√13 = 6~r =

1 –8 3S
√(–8)2 + 152 15
1 2(d)~^r = S→T 2 R u3~v + w~ u = √12 + 72
5 = √50
1 2– 8 (c) P→Q = µP→R = × 10~r
17
= 15 2~i – 3~j = µ[5~i + (k – 1)~j ] = 4~r = 5√2

17 By comparing the like terms, (ii) uR→Su = 6u~r u Unit vector in the direction of
=6×4
16. (a) 2u – 3p + 4r 2 = 5µ –3 = (k – 1)µ = 24 units 3~v + w~
Vektor unit dalam arah 3~v + w~
2 1 2– –3 = 2
= 2+(74~(i–+2~i7~–j )6–~j )3(–8~i + 2~j ) µ = 5 (k – 1) 5 = 1 1 1 2
5√2 7
15 = k –1 u S→T u = 4u~r u
2 =4×4
= 14~i + 14~j + 24~i – 6~j – 8~i – 24~j k= – 13 = 16 units The required vector is
= 30~i – 16~j 2 Vektor diperlukan

= 1 30 2 18. (a) Initial position: 3√2 × 1 1 1 2 = 3 1 1 2
–16 Kedudukan awal: 5√2 7 5 7
(c) (i) ~u =deln~vgawn lheiarleahlpiesmaalcaor nstant
(b) 5s + 2t – u + p
1 23
= –15––25(7(30~i7~~~iii~i––––+376407~~3jj~j~)j~j–)–+8+22~i0((~––i+81–~20i ~~8ji+~j–2~4j ~)j ) O→P = 1 2 2 1 8 2 = l1 2 2 5
= 4 + 1 = 21
= p 3
O→Q = 11822 5
= 12ll2

1 –20 2 Velocities of P and Q: By comparing / Bandingkan, SPM Practice 8
–43 Halaju P dan Q:
= 8 = 2l p + 3 = l
1 1 2 l=4 p+3=4 Paper 1
~v P = 1
2 p=1 1. PQ = kPR
(c) 4r + 2q + 3 s
~vQ = 1––322
= 4+(–322(~i3~–i 6~j ) + 2(3~i + 5~j ) displacement (ii) (a) = 1 8 2 PO + OQ = k(PO + OR)
– 6~j ) After 2 seconds, 4
Selepas 2 saat, = time taken × velocity ~u −1 m 2 + 1 2 2 = k−1 m 2 + 1 n 2
sesaran 5 3 5 –7
= masa diambil × halaju 8 2 k1–452
= –8~i – 24~j + 6~i + 10~j + 2~i – 4~j h1 4 2 – 1 1 2 – = 0 2 –m n–m
1 –2 2 = k1 –12 2
= –18~j
1 2 1 2P: 2 1 2 1 28h – 2 + 5k
1 2 4h – 1 – 4k

1 0 2 displacement = = =0 −2 = −12k Compare each
–18 sesaran
= 1 component
6 Bandingkan setiap
1 2 1 2Q: –2 –4 8h – 2 + 5k = 0 ……… 1 k = komponen
–3 –6

(d) – 1 t–u–q displacement = 2 = 4h – 1 – 4k = 0 ……… 2 1
= 2 sesaran 6
– 2 × 2: 8h – 2 – 8k = 0 ……… 3 2 – m = (n – m)
1
2 (–10~i – 4~j ) – (7~i + 7~j ) O→Pt 1 2 2 1 2 2 1 4 2 3 – 1: –13k = 0 12 – 6m = n – m
4 2 6
– (3~i + 5~j ) = 2 = + = k=0 5m = 12 – n

= 5~i + 2~j – 7~i – 7~j – 3~i – 5~j Substitute k = 0 into 2, m = 1 (12 – n)
5
= –5~i – 10~j O→Qt = 11822 + 1––642 = 1 4 2 Gantikan k = 0 ke dalam 2,
6
= 2

= 1 –5 2
–10

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A42

Additional Mathematics Form 4 Chapter 8 Vectors Additional Mathematics Form 4 Chapter 8 Vectors

2. (a) BF + FD + FA = BD + DC F→A = D→C 8. (a) OA =  3 , OB =  b , r =  3 , s =  12  Paper 2
= BC a 5 –1 4

(b) AB = AO + OB RS AB = 3 3  + c 12  1. (a) (i) CR = CA + AR
= −̰a + ̰b –1 4
= −6̰y + 2̰x
= AO + OB =  –3  +  b  = 2̰x − 6̰y
–a 5
Unit vector in the direction of AB Q
Vektor unit dalam arah ~q  9  +  12c  =  –3 + b  (ii) AB = 3 × AR
1 O P –3 4c –a + 5 = 3 × 2̰x
= 2 (−a̰ + ̰b) ~p = 6̰x
9 + 12c = –3 + b –3 + 4c = –a + 5
= – 1 a̰ + 1 ̰b b – 12c = 12 = 1
2 2 3
3 –2 2 2 2 2 4c = 8 – a CT CB
6. (a) 3 = 5 + OC b – 3(8 – a) = 12
1
 –6 2  4 2 b = 36 – 3a = 3 CA + AB 
9 10
3. (a) PU = 1 UQ = 1 (4~p) = 2~p OC = − = 1 (−6̰y + 6̰x)
2 2 3
 –10 2
(b) UV = UP + PV = –2~p + 1 (PR ) = –1 9. PR = PO + OR = 2̰x − 2̰y
2 = −5̰p + (24̰q – 3̰p)
1 ∴ Coordinates of C = (−10, −1) = −8̰p + 24q̰
= –2~p + 2 PS + SR 2 Koordinat bagi C = (–10, –1) (b) CT = 2(2̰i) − 2(−̰i + 4̰j)
= 6̰i − 8̰j
= –2~p + 1 (6~s + PQ ) (b) AC = AO + OC PQ = PO + OQ
= –2~p + 2 + = −5̰p + 15q̰
3~s + 1 (PU UQ ) = – –2 2 +  –10 2
2 3 –1
1 −5̰p + 15q̰ CT  = √62 + 82
= –2~p + 3~s + 2 (6~p)  –8 2 PQ = −8̰p + 24q̰
–4 PR = 10 units
=
= ~p + 3~s 5(−p̰ + 3̰q )
= √(–8)2 + (–4)2 = 8(−p̰ + 3̰q) (c) AT = AC + CT
= 6̰y + 2̰x − 2̰y
4.  4 2 = m 2 2 = √80 =5 A constant = 2̰x + 4̰y
3 – 8 Satu pemalar
= 4√5 units CS = pCR
k 1 PQ = 5 . Thus, Q lies on PR. CA + AS = p(2̰x − 6̰y)
PR 8 Maka, Q berada pada PR. −6̰y + (1 − q)AT = 2p̰x − 6p̰y
4 = 2m Compare each component −6̰y + (1 − q) (2̰x + 4̰y) = 2p̰x − 6p̰y
m =2 Bandingkan setiap komponen −6̰y + 2̰x + 4̰y − 2q̰x − 4q̰y = 2p̰x − 6p̰y
(2 − 2q)̰x − (2 + 4q)̰y = 2p̰x − 6p̰y
3 = m(k – 1) 7. (a) Magnitude of the vector = √(–3)2 + 52 10. ̰r = ̰s
̰c + p(̰c + 2d̰ ) = 2̰c + d̰ + q̰c 2 − 2q = 2p ...............................a
3 = 2(k – 1) Magnitud bagi vektor = √34 units ̰c + p̰c + 2p̰d = 2̰c + ̰d + q̰c 2 + 4q = 6p ...............................b
a × 2, 4 − 4q = 4p ...................c
3 =k–1 (1 + p − 2 − q)̰c = (1 − 2p)d̰
2 1 − 2p = 0
k = 5 1 –3
2 5
(b) Unit vector = √34  2
Vektor unit 2p = 1
5. (a) OR = OP + PR c + b, 6 = 10p
= ̰p + 2̰q 1
 2– 3 p = 2 p = 6
(b) RS + OQ = 2OP √34 10
RS = 2OP − OQ 1+p−2−q =0
=5 =3
√34 1+ 1 −2−q =0 5

2 q =–1
2

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A43

Additional Mathematics Form 4 Chapter 8 Vectors Additional Mathematics Form 4 Chapter 8 Vectors

Substitute p = 3 into a, 3. 6u D 2u E (ii) OR = OQ + QR PR = 2 PB uP→Bu = uB→Pu
Gantikan p = 5 dalam a, C 3
3 ke O = 3 + kQA 2
5 3v 4 OB = 3 × 2√61
A
2 − 2q = 21 3 2 = 3 (12̰y) + kQO + OA = 4√61 units
5 4
4 = 9̰y  + k(−9̰y + 15̰x) 3
2q = 5 6v
B = 15k̰x + (9 – 9k)̰y 5. (a) OQ = OA + AQ
q= 2 = 6̰a + (2̰c + 3b̰ – 6̰a)
5 = 600 = 6ṵ , = 800 = 8̰u , = 3̰b + 2̰c
OD 100 ̰u OE 100 ̰u

2. (a) Boat P: / Bot P: OA = 450 ̰v = 3̰v (b) (5 – 5h)̰x + 12h̰y = 15k̰x + (9 – 9k)̰y (b) (i) AC = AO + OC
150 = −6̰a + 3c̰
5 – 5h = 15k ...............................a
12̰i 12̰j2 900 12h = 9 – 9k ...............................b
150
Resultant velocity = (̰i +̰j ) + + AB = ̰v , DC : CB =1:5 From / Daripada a, 5h = 5 – 15k
3̰i + 32̰j h = 1 – 3k ..........c
Halaju paduan = (ii) PQ = PA + AQ
From / Daripada b, 4h = 3 – 3k ........d = −(3̰b – 2̰a) + (2̰c + 3̰b– 6a̰ )
= 31̰i + 12̰j2 m s–1 (a) (i) OC = OD + DC = −4̰a + 2̰c

= 6̰u + 1 DB d − c, 3h = 2 2
6 3
Boat Q: / Bot Q: 1 h= (c) PQ = –4a̰ + 2̰c
= 6̰u + 6 (DO + OB ) AC –6a̰ + 3̰c
12̰i 21̰j2 2
Resultant velocity = (3̰i + 2̰j) + + 1 Substitute h = 3 into c, = 2(–2a̰ + ̰c)
Halaju paduan = 5̰i + 25̰j 6 Gantikan p = ke dalam c, 3(–2̰a + ̰c )
= 6ṵ + (–6ṵ + 9̰v ) 2
3
= 5ṵ + 3 ̰v 2 = 2 A constant
= 51̰i + 21̰j2 m s–1 2 3 = 1 – 3k 3 Pemalar

∴ The resultant velocity of boat Q is (ii) EC = ED + DC 3k = 1 ∴ PQ is parallel to / selari dengan AC.
3
5 1
3 times the resultant velocity of = –2̰u + 1 (–6̰u + 9̰v ) k = 9
6
boat P.
5 = –3̰u + 3 ̰v 1 6. (a) (i) OD = OA + AD
Halaju paduan bagi bot Q ialah 3 kali 2 (c) OP = 3 OA  = a̰ + mAB
= ̰a + mAO + OB
ganda halaju paduan bagi bot P (b) EA = EO + OA = 1 15̰x  = a̰ + m(−̰a + b̰ )
= –8̰u + 3̰v 3 = (1 − m)̰a + m̰b
1
(b) (i) Resultant velocity of boat R 3 = 3 × 15 × 2 (ii) OD = OC + CD
Halaju paduan bot R 2 = 3b̰ – 9a̰ + nOB
EC = –3̰u + ̰v = 10 units = 3̰b – 9a̰ + nb̰
= 1̰i – 23̰j2 + 12̰i + 12̰j2 = −9̰a + (3 + n)̰b
EA –8ṵ + 3̰v OB  = 12̰y
= (3̰i –̰j ) m s–1 = 12 × 1 (b) (1 − m)̰a + m̰b = −9a̰ + (3 + n)̰b
= 1 1 –6ṵ + 3̰v 2 ≠ constant / pemalar = 12 units
2 –8̰u + 3̰v 1 − m = −9
m = 10
(ii) Unit vector in the direction of Hence, the LRT will not pass through
m=3+n
boat R building C BP = √102 + 122 10 = 3 + n
= √244
Vektor unit dalam arah bot R Maka, LRT itu tidak akan melalui bangunan C = √4× 61 n=7
= 2√61 units
= 3̰i –̰j 4. (a) (i) OR = OP + PR ∴ OD = −9a̰ + 10b̰

√32 + (–1)2

= 3̰i –̰j = 1 OA + hPB
3
√10
= 1 (15̰x ) + h1PO + OB 2
= √310̰i – √110̰j 3
= 5̰x + h(−5̰x + 12̰y )
= (5 – 5h)̰x + 12h̰y

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A44

Additional Mathematics Form 4 Chapter 8 Vectors PTER

HOTS Challenge 2. Given OQ = ̰c + ̰d, OR = 2̰c + 3d̰ , OS = 4̰c – d̰ , CHA 91 Solution of Triangles
Diberi
Penyelesaian Segi Tiga
1. PQ = SR

C PO + OQ = SO + OR

−OP + (̰c + ̰d ) = −(4̰c – d̰ ) + (2̰c + 3̰d )

−OP + (̰c + d̰ ) = −2̰c + 4d̰ S R 1. (a) (i) BC = 5 (ii) ∠E = 180° – 135° – 17.14°
sin 110° sin 34° = 27.86°
D E OP = 3̰c  − 3̰d
~d
OT = 5̰c BC = 5 × sin 110° FG 12
sin 34° sin 27.86° = sin 135°
O ~a A B OQ = ̰c + d̰
PQ = 8.40 cm
12
(a) BD = BO + OD ST = SO + OT CA 5 FG = sin 135° × sin 27.86°
= −2̰a + d̰ = −(4̰c – ̰d ) + 5̰c sin (180° – 110° –34°) sin 34°
= ̰c + d̰ (ii) = = 7.93 cm

(b) AE = AB + BE ∴ OQ = ST CA = 5 × sin 36° (b) (i) sin ∠IHJ = sin 94°
sin 34° 8.6 13.7
2
= a̰ + 3 BD OS = 4̰c – d̰ = 5.26 cm sin 94°
13.7
= + 2 (−2a̰ + d̰ ) QT = QO + OT GE 13 1 2∠IHJ = sin–1 × 8.6
3 = −(̰c + ̰d ) + 5̰c sin 48° sin 75°
a̰ = 4̰c – d̰ (b) (i) = = 38.77°

= – 1 a̰ + 2 d̰ GE = 13 × sin 48° (ii) ∠HJI = 180° – 94° – 38.77°
3 3 sin 75° = 47.23°
∴ OS = QT
AC = AO + OC = 10.00 cm
= −a̰ + 2d̰
Since the opposite pairs of the vectors of (ii) ∠E = 180° – 75° – 48° HI = 13.7
the quadrilateral are equal, hence OQTS is = 57° sin 47.23° sin 94°
1– 1 2 2 a parallelogram.
AE : AC = 3 ̰a + 3 ̰d : (−a̰ + 2̰d ) HI = 13.7 × sin 47.23°
Oleh sebab sisi yang bertentangan vektor bagi sisi sin 94°
= 1 (−̰a + 2d̰ ) : (−a̰ + 2d̰ ) empat adalah sama, maka OQTS ialah segi empat FG = 13
3 selari. sin 57° sin 75° = 10.08 cm

= 1 :1 FG = 13 × sin 57° sin ∠LMN sin 62°
3 sin 75° (c) (i) 4.5 = 7.8
=1:3
= 11.29 cm
sin 62°
7.8
1 2∠LMN = sin–1 × 4.5

(c) (i) ∠R = 180° – 95° – 62° = 30.62°
= 23°

(ii) QR as it has the greatest opposite ∠MNL = 180° – 62° – 30.62°
angle. = 87.38°
QR kerana ia mempunyai sisi bertentangan
yang paling panjang. (ii) LM = 7.8
sin 87.38° sin 62°

QR = 12 LM = 7.8 × sin 87.38°
sin 95° sin 23° sin 62°

12 = 8.82 cm
sin 23°
QR = × sin 95°

= 30.59 cm 3.

2. (a) (i) sin G = sin 135° 8.4 cm 6.6 cm
5 12 40°

sin 135°
12
1 2∠G = sin–1 × 5

= 17.14°

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A45

Additional Mathematics Form 4 Chapter 9 Solution of Triangles Additional Mathematics Form 4 Chapter 9 Solution of Triangles

(b) sin 40° (c) 5.22 = 10.42 + 6.32 – 2(10.4)(6.3) cos q
5.8
5.6 cm I 1 2∠RQP = sin–1 × 7.4 10.42 + 6.32 – 5.22
7.5 cm 2(10.4)(6.3)
5.8 cm G1 = 55.10° cos q =

Since 5.8 , 7.4, = 0.9219

42° G 44° or ∠RQP = 180° – 55.10° there is two q = 22.79°
8 cm 2 possible answers
H = 124.9
for ∠RQP
Oleh sebab 5.8 , 7.4,
(d) 3.42 = 5.22 + 6.82 – 2(5.2)(6.8) cos q
terdapat dua jawapan

yang mungkin bagi cos q = 5.22 + 6.82 – 3.42
2(5.2)(6.8)
sin G = sin 44° /RQP
7.5 5.6
(ii) PS = 7.4 = 0.8727
19 cm sin 44° sin (180° – 80° – 40°) sin 80°
15 cm 5.6
37° 1 2∠HG1I = sin–1 × 7.5 q = 29.22°

= 68.49° PS = 7.4 × sin 60° 8. (a) (i) ∠DBC is the largest angle as it has
sin 80° the longest opposite side in the
∠HG2I = 180° – 68.49°
= 111.51° = 6.507 km triangle.

4. (a) F 6.5 cm ∆G1HI is an acute-angled triangle. 6. (a) EG2 = EF2 + FG2 – 2(EF)(FG) cos 113° ∠DBC ialah sudut terbesar kerana
D1 ∆G1HI ialah segi tiga bersudut tirus. = 5.62 + 8.12 – 2(5.6)(8.1) cos 113° mempunyai sisi bertentangan yang paling
6.9 cm = 132.42
∠G1HI = 180° – 44° – 68.49° panjang di dalam segi tiga.
= 67.51° EG = 11.51 cm
cos ∠DBC = 8.42 + 7.72 – 9.62
2(8.4)(7.7)

68° D2 G1H = 5.6 (b) cos 36° = DC ∠DBC = 73.06°
E sin 67.51° sin 44° 20
(ii) AD2 = 15.52 + 8.42 – 2(15.5)(8.4)
sin D sin 68° 5.6 DC = 20 cos 36° cos 24°
6.9 6.5 sin 44°
= G1H = × sin 67.51° AC = 3 × DC AD = 8.54 cm
2
sin 68° = 7.448 cm AC = 8.54 + 9.6 = 18.14 cm
6.5
1 2∠ED1F = sin–1 × 6.9 = 79.81° = 3 × 20 cos 36°
2
∠ED2F = 180° – 79.81° = 100.19° 5. (a) (i) ∠B = 180° – 70° – 33°
= 77° = 24.27 cm (b) (i)

In triangle D1EF, AB = 110 AB2 = AC2 + BC2 – 2(AC)(BC) cos 36° AD = DC
sin 33° sin 77° = 24.272 + 202 – 2(24.27)(20)
Dalam segi tiga D1EF, cos 36° = √15.82 + 14.22 – 2(15.8)(14.2) cos 43°
∠EFD1 = 180° – 68° – 79.81° = 32.19° 110 = 203.64 = 11.10 cm
sin 77°
ED1 6.5 AB = × sin 33° AB = 14.27 cm 11.32 + 14.22 – 11.102
sin 32.19° sin 68° 2(11.3)(14.2)
= = 61.49 m (ii) cos ∠ABD=

ED1 = 6.5 × sin 32.19° sin D sin 142° (c) BT 2 = AB2 + AT2 – 2(AB)(AT) cos 42° ∠ABD = 50.04°
sin 68° 110 138 = 162 + 312 – 2(16)(31) cos 42°
(ii) = = 479.80
= 3.735 cm 1
BT = 21.90 m 9. (a) Area of ∆DEF = 2 (6)(10) sin 145°
Luas ∆DEF
sin 142°
138
1 2∠D = sin–1
In triangle D2EF, × 110 = 17.21 cm2

Dalam segi tiga D2EF, = 29.39° 7. (a) 122 = 112 + 102 – 2(11)(10) cos q (b) ∠G = 180° – 33° – 123°
∠EFD2 = 180° – 68° – 100.19° = 11.81°

ED2 = 6.5 ∠CAD = 180° – 142° – 29.39° cos q = 112 + 102 – 122 = 24°
sin 11.81° sin 68° = 8.61° 2(11)(10)
Area of ∆GHI = 1 (16)(18) sin 24°
6.5 Luas ∆GHI 2
ED2 = sin 68° × sin 11.81° = 0.35
(b) (i) ∠QRP = 40° Alternate angle = 58.57 cm2
Sudut berselang-seli q = 69.51°
= 1.435 cm
sin Q = sin 40° (b) 19.92 = 16.62 + 11.52 – 2(16.6)(11.5) cos q
7.4 5.8 (c) ∠J = 180° – 52° – 31° = 97°

cos q = 16.62 + 11.52 – 19.92 JL = 8.4
2(16.6)(11.5) sin 52° sin 97°

= 0.0309

q = 88.23°

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A46

Additional Mathematics Form 4 Chapter 9 Solution of Triangles Additional Mathematics Form 4 Chapter 9 Solution of Triangles

JL = 8.4 × sin 52° (c) sin Q = sin 68° (ii) 16.422 = 52 + 142 – 2(5)(14) cos ∠BCD 3. (a) sin ∠QSR = sin 65°
sin 97° 16 18 20.5 21
cos ∠BCD = 52 + 142 – 16.422
= 6.67 cm sin Q = sin 68° × 16 2(5)(14) sin 65°
18 sin ∠QSR = 21 × 20.5
∠BCD = 110.3°
1 ∠Q = 55.5° ∠QSR = 62.22°
Area of ∆JKL = 2 (8.4)(6.67) sin 31°
Luas ∆JKL ∠R = 180° – 68° – 55.5°
= 14.43 cm2 (iii) Area of ABCD / Luas ABCD QS 21
= 56.5° (b) sin (180° – 65° – 62.22°) = sin 65°
= 1 (9)(16.42) sin 79° + 1 (5)(14)
10. (a) s = 1 (20 + 14 + 14) = 24 Area of ∆PQR = 1 (16)(18) sin 56.5° 22 QS = 21 × sin 52.78°
2 Luas ∆PQR 2 sin 110.3° sin 65°

Area of ∆STU / Luas ∆STU = 120.08 cm2 = 105.36 cm2 = 18.45 cm

= √24(24 – 20)(24 – 14)(24 – 14) (b) (i) (c) VQ2 = 122 + 52
= 97.98 cm2 VQ = 13 cm
12. (a) BD2 = 152 + 82 = BЈ
289
(b) s = 1 (32 + 37 + 14) = 41.5
2 VS2 = 52 + 102
VB = √289 + 62 VS = 11.18 cm
Area of ∆VWX / Luas ∆VWX = 18.03 cm
DЈ
= √41.5(41.5 – 32)(41.5 – 37)(41.5 – 14) 5 cm s = 1 (13 + 11.18 + 18.45) = 21.315
= 220.9 cm2 2
VC = √62 + 152 CЈ
1 = 16.16 cm Area ∆QVS /Luas ∆QVS
2 (ii) ∠DʹCʹBʹ = 180° – 110.3°
11. (a) Area of ∆STU = (US)(UT) sin U = 1 (18.03 + 8 + 16.16) = 21.10 = 69.7° = √21.315(21.315 – 13)(21.315 – 11.18)(21.315 – 18.45)
Luas ∆STU sin x 2
1 s = 71.74 cm2
53.97 = 2
(5)(21.8) Area of ∆VBC
Luas ∆VBC
sin x = 0.990275 2. (a) (i) 1 (4.8)(4.2) sin ∠BCD = 8.64 4. (a) (i) AV = √82 + 52
2 = √89 cm (shown / Tertunjuk)
x = sin–1 0.990275 = √21.10(21.10 – 18.03)(21.10 – 8)(21.10 – 16.16)
= 64.75 cm2 ∠BCD = 59.00° (ii) AB = √82 + 62
= 82°
(from scientific calculator) = 10 cm

(b) Area of ∆PQR / Luas ∆PQR (b) 10.22 = 6.82 + 5.12 – 2(6.8)(5.1) cos (daripada kalkulator saintifik) (b) (i) BV = √62 + 52
= √61 cm
= √s(s – a)(s – b)(s – c) where/ dengan ∠ABC ∠BCD = 180° – 59.00°
= 121.0° AV2 = AB2 + BV2 − 2(AB)(BV) cos
s = 1 (a + b + c) cos ∠ABC = 6.82 + 5.12 – 10.22 ∠VBA
2 2(6.8)(5.1)
(ii) BD2 = 4.82 + 4.22 – 2(4.8)(4.2) cos 89 = 100 + 61 – 2(10)√61 cos
s = 1 (3.8 + 1.8 + 3) = 4.3 ∠ABC = 117.28° 121.0°
2 ∠VBA
BD = 7.839 cm 100 + 61 – 89
Area of ∆PQR (b) 10.22 = 6.82 + 5.12 – 2(6.8)(5.1) cos ∠ABC
Luas ∆PQR cos ∠VBA = 2(10)√61
cos ∠ABC = 6.82 + 5.12 – 10.22 (iii) sin ∠CBD = sin 121.0°
= √4.3(4.3 – 3.8)(4.3 – 1.8)(4.3 – 3) 2(6.8)(5.1) 4.8 7.839 ∠VBA = 62.55°
= 2.643 cm2
∠ABC = 117.28° ∠CDB = 31.66°

Alternative Method (b) ∠ABD = 90° – 31.66°
= 58.34°
1.82 = 32 + 3.82 – 2(3)(3.8) cos Q (ii) Area of ∆VBA / Luas ∆VBA
AB2 = 8.62 – 4.22 1
cos Q = 32 + 3.82 – 1.82 SPM Practice 9 = 2 (10)√61 sin 62.55°
2(3)(3.8) AB = 7.505
= 34.65 cm2
∠Q = 27.63° Paper 2 1
2
1 Area / Luas ∆ABD = (7.505)(7.839) ST 11
2 sin 45° sin 36°
Area of ∆PQR = (3)(3.8) sin 27.63° 1. (a) (i) BD = 9 sin 58.34° 5. (a) =
Luas ∆PQR sin(180° – 79° – 31°) sin 31°
= 2.643 cm2
BD = 9 × sin70° = 25.04 cm2 ST = 11 × sin 45°
sin 31° sin 36°

= 16.42 cm = 13.23 cm

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A47

Additional Mathematics Form 4 Chapter 9 Solution of Triangles Additional Mathematics Form 4 Chapter 9 Solution of Triangles

(b) PS = 11 AʹDʹ = 1 600 D 18 m C ∴ The distance between the ship and
– 36° sin 36° sin(180° – 86° – 36°) sin 86° the buoy when the ship is at a
sin (180° – 45°) h position that is the closest to the
buoy is 242 km/h
PS = 11 × sin 99° AʹDʹ = 1 600 × sin 58° 21 m
sin 36° sin 86° Jarak di antara kapal dan boya apabila
kapal itu berada di kedudukan yang paling
= 18.48 cm = 1 360 km B
dekat dengan boya ialah 242 km/j
QS = √112 + 18.482 7. (a) (i) AC2 = 11.12 + 8.92 – 2(11.1) 1 × h × 21 = 72.08
(8.9) cos 60° 2
= 21.51 cm
AC = 10.18 cm
QT = √112 + 112 h = 6.865 m (ii) ∠ABC = 180° – 40° – 74°
= 66°
= 15.56 cm (e) s = 1 (21 + 25 + 9.77) = 27.885
(ii) ∠ABC = 180° − 60° 2 AC2 =2522 +3772 –2(252)(377)cos66°
s = 1 (13.23 + 21.51 + 15.56) = 25.15 = 120° AC = 358 km
2 Area of ABCD
Area of coloured plane BC 10.18 Luas ABCD
sin(180° – 120° – 35°) = sin 120° 40 = 358 Distance
= √27.885(27.885 – 21)(27.885 – 25) t Speed = Time
Luas bagi satah berwarna
10.18 (27.885 – 9.77) + 72.08
= √25.15(25.15 – 13.23)(25.15 – 21.51)(25.15 – 15.56) BC = sin 120° × sin 25° = 100.17 + 72.08 t = 8.95 hours / jam
= 172.25 m2
= 102.30 cm2 = 8 hours 57 minutes
(c) Let h = the shortest length from T to QS
Biar h = panjang terdekat dari T ke QS = 4.968 cm 8 jam 57 minit

QT (b) Area of ABCD HOTS Challenge ∴ The ship will reach port C at 0642
h hours the next day.
Luas ABCD
21.51 cm Kapal akan tiba di pelabuhan C pada jam
= 1 (11.1)(8.9) sin 60° + 1 (4.968)
2 2 0642 keesokan harinya.

S (10.18) sin 35° 1. (a) AB = 252 2. (a) (i) AD2 = 552 + 802 – 2(55)(80) cos 40°
sin 74° sin 40° AD = 51.81 m
1 = 57.28 cm2
102.30 = 2 × 21.51 × h = 252 sin 74°
AB sin 40° ×

h = 9.512 cm sin ∠DBC sin 100° = 377 km (ii) ∠BDC = ∠ABD = 40°
18 21
6. (a) (i) BC2 = 1 4002 + 1 6002 – 2(1 400) 8. (a) =
(1 600) cos 15°
sin ∠DBC = sin 100° × 18 (b) (i) Let h = perpendicular distance from BC = 80
BC = 438.9 km 21 B to AC sin 40° sin 42°
Biar h = jarak serenjang dari B ke AC
∠DBC = 57.58° BC = 80 × sin 40°
sin 42°

(ii) AD = 1 600 (b) ∠DBA = 180° − 100° − 57.58° = 76.85 m
sin(180° – 94° – 36°) sin 94° = 22.42°
(b) ∠DBC = 180° – 40° – 42°
AD = 1 600 × sin 50° AD2 = 212 + 252 – 2(21)(25) cos 22.42° C = 98°
sin 94° AD = 9.77 m

= 1 229 km 252 km Area of ABCD
h
(c) ∠CDB = ∠DBA = 22.42° Luas ABCD
(b) (i)
A 40° B = 1 (55)(80) sin 40° + 1 (76.85)(80)
CЈ Area of ∆BCD = 1 (18)(21) sin 22.42° 2 2
Luas = 2 377 km
1 600 km 72.08 m2 sin 98°

AЈ 36° sin 40° = h Opposite side = 4 458 m2
377 sin θ = Hypotenuse
94° (d) Let h = perpendicular distance from C
to BD h = 377 × sin 40°
D DЈ Biar h = jarak serenjang dari C ke BD
= 242 km
(ii) ∠CʹDʹAʹ = 180° − 94°
= 86°

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A48

PTER Additional Mathematics Form 4 Chapter 10 Index Numbers

CHA 110 Index Numbers Index number, I = Q2015 × 100 (b) Q2017 × 100 = 125
Nombor indeks Q2010 Q2015
Nombor Indeks
= Q2015 × Q2019 × 100 360 × 100 = 125
Q2019 Q2010 x
360 × 100
100 150 x= 125
125 100
= × × 100 = 288

1. (a) Let / Biar Q0 = sales in the year 2011 Accident index, I = Q2012 × 100 = 75 = 120 Q2017
jualan pada tahun 2011 Indeks kemalangan Q2016 Q2015
× 100 = 140
Q1 = sales in the year 2018 Q2012 Q2004 Q2004
jualan pada tahun 2018 460 × 100 =75 (b) Q2000 × 100 = 116 ⇒ Q2000 = 116 y × 100 = 140
Q2012 = 100 125
Q1 75 × 460
Sales index, I = Q0 × 100 100 Q2006 × 100 = 125 ⇒ Q2006 = 125 y = 140 × 125
Indeks jualan Q2000 Q2000 100 100
= 345

= 560 × 100 Hence, the number of road accidents = 175
320
occured in the year 2012 was 345. Q2000 × 100 = 100 × 100 Q2009
Q2006 125 Q2008
= 175 Maka, bilangan kemalangan jalan raya berlaku (c) × 100 = 110

pada tahun 2012 ialah 345. = 80

Hence, there is an increase of sales of (c) Let / Biar Q2000 = price of the terrace in Q2004 Q2004 Q2000 2.31 × 100 = 110
75% from the year 2011 to the year 2000 Q2006 Q2000 Q2006 x
2018. × 100 = × × 100 2.31 × 100
harga rumah teres pada x = 110
Maka, keuntungan menokok 75% dari tahun 2011
ke tahun 2018. tahun 2000 = 116 × 100 × 100 = RM2.10
Q1984 = price of the terrace in 100 125
Q2009
2. (a) Let / Biar Q0 = price in the year 2014 1984 = 92.8 Q2008 × 100 = 105
harga pada tahun 2014
harga rumah teres pada (c) Q1996 × 100 = 140 ⇒ Q1996 = 140 y × 100 = 105
Q1 = price in the year 2016 Q1990 Q1990 100 0.60
harga pada tahun 2016 tahun 1994

Price index, I = Q2000 × 100 = 150 Q2000 Q2000 y = 0.60 × 105
Indeks harga Q1984 Q1996 Q1996 100
Q1 × 100 = 105 ⇒ = 105
Price index, I= Q0 × 100 100 = RM0.63
Indeks harga
6 750 139 500 × 100 =150 z = 1.60 × 100
Q0 Q1984 Q2000 Q2000 Q1996 1.20
125 = × 100 139 500 × 100 Q1990 × 100 = Q1996 × Q1990 × 100
150
Q0 = RM5 400 Q1984 = = 133.33

= RM93 000 = 105 × 140 × 100 (d) For item A: / Untuk barangan A:
Hence, the price of the terrace in the 100 100
year 1984 was RM93 000. Q2008 Q2008
Hence, the price of the motorcycle in = 147 Q2005 × 100 = 120 ⇒ Q2005 = 120
the year 2014 was RM5 400. Maka, harga rumah teres pada tahun 1984 ialah 100
Maka, harga motosikal pada tahun 2014 ialah RM93.000.
RM5 400. Q2010 × 100 = 150 ⇒ Q2010 = 150
Q2005 Q2005 100
(b) Let / Biar Q2016 = number of accidents in
2016 4. (a) Price index x = Q2010 × 100
bilangan kemalangan pada Item Indeks harga Q2008
tahun 2016 Q2019 Q2019 150
3. (a) Q2010 × 100 = 150 ⇒ Q2010 = 100 Barangan = Q2010 × Q2005 × 100
Q2012 = number of accidents in A Q2005 Q2008
2012 Q2019 Q2019 125 4.90 × 100 = 140
bilangan kemalangan pada Q2015 × 100 = 125 ⇒ Q2015 = 100 B 3.50
tahun 2012 = 150 × 100 × 100
8.10 100 120
6.00 × 100 = 135
= 125

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Additional Mathematics Form 4 Chapter 10 Index Numbers Additional Mathematics Form 4 Chapter 10 Index Numbers

For item B: / Untuk barangan B: 5. (a) I– = ∑Iiwi (d) 136y + 552 = 120y + 600
∑wi Food
Q2008 × 100 = 110 ⇒ Q2008 = 110 Makanan Price index in the year 2007 16y = 48
Q2005 Q2005 100
= 124(1) + 140(1) + 132(1) + 116(1) A based on the year 2002 60 y =3
4 B Indeks harga pada tahun 2007 50
Q2010 × 100 = 140 ⇒ Q2010 = 140 C (b) (i) h = × 100
Q2008 Q2008 100 D berasaskan tahun 2002
= 512 = 120
4
Q2010 1.75 × 100 = 125
y = Q2005 × 100 = 128 1.40 100 × 100 = 125
k
Q2010 Q2008 (b) I– = ∑Iiwi 2.80 100 = 140
= Q2008 × Q2005 × 100 ∑wi 2.00 × k = 100 × 100
125
6.00
= 140 × 110 × 100 = 120(2) + 95(8) + 110(6) + 140(4) 4.00 × 100 = 150 = 80
100 100 2+8+6+4
p × 100 = 110
= 154 = 2 220 2.40 × 100 = 80 40
20 3.00
p = 110 × 40
(e) For gold 916: / Untuk emas 916: = 111 100
I– 125 × 35 + 140 × 20 + 150 × 15 + 80 × 30
Q2016 Q2016 125 = 35 + 20 + 15 + 30 = 44
Q2002 Q2002 100
× 100 = 125 ⇒ = I– = 100 × 4 + 120 × 2 + 145 × 4 11 825
4+2+4 100
Q2018 Q2018 6. (a) = (ii) ICndoemkspgousbiatehainn,dI–ex, I– = 132
Q2002 Q2002
× 100 = 140 ⇒ = 140 = 1 220 = 118.25
100 10

x = Q2018 × 100 = 122 7. (a) x(1) + 120 × 3 + 130 × 2 + 114 × 4 = 114.6 120 × 1 + 160 × 3 + 125 × 4 + 110 × m = 132
Q2016 1+3+2+4 1+3+4+m

= Q2018 × Q2002 × 100 I– = 128 × 1 + 110 × 1 + 125 × 1 + 90 × 1 x + 1 076 = 114.6 1 100 + 110m = 132
Q2002 Q2016 4 10 8+m
(b)
1 100 + 110m = 1 056 + 132m
140 100 453 x + 1 076 = 1 146
= 100 125 × 100 = 4 22m = 44
× x = 70 m =2

= 112 = 113.25 112 × 4 + 130 × 1 + 120 × 2 + x(3)
4+1+2+3
(b) = 119 (c) (i) 0.70 × 100 = 175
x
For gold 999: / Untuk emas 999: (c) 3x + 818
10
Q2016 × 100 = 135 ⇒ Q2016 = 135 Price index in the year 2004 = 119 x= 0.70 × 100
Q2002 Q2002 100 175
Component based on the year 2000 3x + 818 = 1 190
Q2018 Q2018 Komponen Indeks harga pada tahun 2004 = 0.40
Q2016 Q2016 120 3x = 372
× 100 = 120 ⇒ = 100 berasaskan tahun 2000 5.50
4.00
Q2018 E 105 x = 124 y = × 100
Q2002 70
y = × 100 × 100 = 150 Q2008 = 137.5
Q2013
F 36 100 = 90 8. (a) x= × 100
40
= Q2018 × Q2016 × 100 × z
Q2016 Q2002 2.50
60 = 34.20 × 100 × 100 = 120
50 30
120 135 G × 100 = 120 z = 120 × 2.50
100 100 100
= × × 100 125 = 114
100 ICnodemkspgousbitaehainn,dI–ex, I– = 120
= 162 H × 100 = 125 =3

I–= 150 × 4 + 90 × 2 + 120 × 6 + 125 × 3 136 × y + 114 × 2 + 108 × 3 = 120 (ii)
4+2+6+3 y+2+3 175 × 15 + 125 × 30 + 137.5 ×

1 875 136y + 552 I– = 24 + 150 × 33 + 120 × 12
15 y+5 15 + 30 + 24 + 33 + 12
= = 120
= 16 065
= 125 114

= 140.92

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Additional Mathematics Form 4 Chapter 10 Index Numbers Additional Mathematics Form 4 Chapter 10 Index Numbers

SPM Practice 10 115(3) + 135(p) + (ii) Q2015 × 100 = 150 5. (a) (i) h = Q2014 × 100
125(7) + 126(4) Q2011 Q2000
(b) 3 + p + 7 + 4 = 126.7
Paper 2 Q2015 × 100 = 150 = 16.10 × 100
135p + 1724 = 1773.8 + 126.7p 8 14.00
8.3p = 49.8 150 × 8
1. (a) Q2010 × 100 = 120 p=6 Q2015 = 100 = 115
Q2008
= 12 sen
7.80 × 100 = 120
Q2008 (c) P19 × 100 = 126.7 86 × 100 (ii) Q2012 × 100 = 120
7.80 × 100 P15 Number of meat balls = 12 Q2000
Q2008 = 120 Bilangan bebola daging
= 716.67 Q2012
= RM6.50 P19 × 100 = 126.7 14.00 × 100 = 120
12.00
126.7 × 12.00 Maximum number of meat balls 120 × 14.00
P19 = 100 Bilangan maksimum bebola daging Q2012 = 100
= 716
(b) Price index of P= 120 × 90 = 15.20 = RM16.80
Indeks harga P = 108 100
The cost of biscuit in 2019
Price index of Q = 108 Kos membuat biskut pada tahun 2019 (b) (i) I–= 110(m) + 120(3) + 125(2) = 116
Indeks harga Q = RM15.20 m+3+2
4. (a) Q2012 × 100 = 130
Q2010 110m + 610 = 116m + 580
Price index of R = 125 \ the selling price of biscuits in 2019 6m = 30
Q2012 × 100 = 130 m=5
Indeks harga R \ harga jualan biskut pada tahun 2019 3.60
110
Price index of S = 150 × 100 = 160 ×15.20 Q2012 = 130 × 3.60
Indeks harga S = 165 100 100
(ii) Q2012 × 100 = 116
= RM24.32 = RM4.68 Q2000

108(30) + 108(20) + 3. (a) (i) p = 120 (b) Percentage of usage of M 69.60 × 100 = 116
Peratus penggunaan M Q2000
I– = = 100% – 10% – 30% – 40%
(c) (i) 125(40) + 165(10) Q2015 = 20% Q2000 = 69.60 × 100
100 Q2013 116
12 050 (ii) × 100 = 120
= 100 = RM60.00
I– 120(10) + 90(30) + 140(20) + 130(40)
= 120.5 2.16 × 100 = 120 = 100
q
2.16 × 100 11 900 Q2014 Q2014 143
(ii) 22.16 × 100 = 120.5 q= 120 = 100 (c) Q2000 × 100 = 143 ⇒ Q2000 = 100
Q2008 Q2008 =
22.16 × 100 = 1.80 = 119 Q2012 Q2012 110
120.5 Q2000 × 100 = 110 ⇒ Q2000 = 100

= RM18.39 I– = 150(60) + 120(30) + 125(10) Q2012 × 100 = 119 Q2014 × 100 = Q2014 × Q2000 × 100
100 Q2010 Q2012 Q2000 Q2012
(b)

= 13 850 23 800 × 100 = 119 = 143 × 100 × 100
100 Q2010 Q2010 = 100 110
90 23 800 × 100
2. (a) (i) x = 100 × 150 = 135 = 138.5 119 = 130

y = 105 × 120 = 126 = RM20 000
100
(c) (i) Q2015 100 = 150 Q2015 = 150 (c) I– = 119 125 6. (a) I– = 140(50) + 120(30) + k(20) = 128
Q2011 × ⇒ Q2011 100 × 100 100

(ii) P19 × 100 = 135 Q2015 Q2015 138.5 = 148.75 20k + 10 600 = 12 800
P15 Q2013 Q2013 100
× 100 = 138.5 ⇒ = 20k = 2 200

5.40 × 100 = 135 Q2013 100 = Q2013 Q2015 100 Percentage of changes in production k = 110
P15 P15 = Q2011 Q2015 Q2011 cost
5.40 × 100 × × ×
135 Peratus perubahan dalam kos pengeluaran
= 100 × 150 × 100 = 48.75%
138.5 100 \ 48.75% increase
= RM4.00
= 108.3 Menokok 48.75%

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