FullyWorkedSolutions f4 2021

Additional Mathematics Form 4 Chapter 10 Index Numbers

(b) Pollution index in the P14 × 100 = 108 ⇒ x + 114 = 108 SPM Year-End Assessment
year 2016 based on the P10 P10 100
Town
Bandar year 2015 x + 114 = 110080P10 Paper 1 / Kertas 1 3 + 1 3 + 1
Indeks pencemaran pada tahun 3 – 1 3 + 1
M 2016 berasaskan tahun 2015 114 = 108 100x Section A / Bahagian A
N 100 105
P 105 1 2x+ 1. x2 = 4(mx – 4) = ×
x2 – 4mx + 16 = 0
100 105x + 11 970 = 108x = 3 + 23 + 1
(4m)2 – 4(1)(16) = 0 3–1
115 x = 3990 16(m2 – 4) = 0

\ P14 = 3990 + 114 = RM4 104 16(m – 2)(m + 2) = 0 = 4 + 23
m = –2, 2 2
I– = 105(50) + 100(30) + 115(20) = 2 + 3
100
10 550 (b) P12 × 100 = 118
= 100 P10 (b) Let the length of the rectangle = y m

= 105.5 3 422 × 100 = 118 2. (a) log2(3 − 2x) − log2 x2 = log2 5 Andaikan panjang salah satu sisi segi empat tepat
P10 P10 = =ym
(c) –I (2017 based on the year 2015) 3 422 × 100 log2 3(3 –x22x)4 = log2 5
118 3(3 –x22x)4 = 5 (23 – 2 )y = (1 + 6)
(2017 berasaskan tahun 2015)
= 2 900 5x2 + 2x – 3 = 0 1 + 6
114 y = 23 – 2
= 105.5 × 100
P14
= 120.27 P10 × 100 = 121 (5x – 3)(x + 1) = 0 1 + 6 23 + 2
23 – 2 23 + 2
P14 x = 3 , x –1 = ×
2 900 5 x
Q2017 × 100 = 120.27 × 100 = 121 ≠ 3 23 + 2 + 218 + 12
Q2015 = 5 [(23)2 – (2)2]
∴ P14 = 121 × 2 900 =
Q2017 100
1 200 × 100 = 120.27
(b) (2x – 1) log 3 = log 15
120.27 × 1 200 = RM3 509 = 23 + 2 + 62 + 23
Q2017 = 100 12 – 2
2x = log 15 + 1
log 3
= 1 443.24 110 = 43 + 72
(c) h= 100 × 110 = 121 x = 1.7325 10
Maximum number of treated students
Bilangan maksimum murid yang dirawat 3. Midpoint of AB = 11 +2(–7), 6 + 2 2 = 1403 + 1702
= 1 443 Titik tengah AB 2
106(2) + 108(3) + 2 3 1702
5
(d) (i) 121(6) + 121(w) = 116.4 6–2 = (–3, 4) = +
2+3+6+w 1 – (–7)
HOTS Challenge mAB =

121w + 1262 = 1280.4 + 116.4w = 1 1 21 25. (a) a—31 – b—32 a—32 + a—31 b—23 + b—34
2
1. (a) Let x = P12 = expenses for biils in the 4.6w = 18.4 = a + a—23 b—23 + a—31 b—34 – a—23 b—32 – a—31 b—34 – b2
year 2012
w=4 m = –2 = a – b2
Biarkan x = P12 = perbelanjaan untuk bil-bil
pada tahun 2012 (ii) P14 × 100 = 116.4 The equation: y – 4 = –2[x – (–3)]
P10 Persamaan y – 4 = –2x – 6

Let x = price in the year 2008 P17 × 100 = 130 y = –2x – 2 (b) 22x + 3 × 5x – 1 = 23x × 52x
Biarkan x = harga pada tahun 2008 P14
4. 1 + k 22x × 23 × 5x × 5–1 = 23x × 52x
then x + 15 = price in the year 2010 P17 P17 P14 (a) p = 1 – k
maka x + 15 = harga pada tahun 2010 P10 × 100 = P14 × P10 × 100 23 × 5–1 = 23x × 52x ÷ 22x ÷ 5x
1
P12 × 100 = 105 ⇒ P12 = 105 = 130 × 116.4 × 100 = 1+ 3 8 = 2x × 5x
P10 P10 100 100 100 1– 1 5
3
x = 105 = 151.32 3 + 1 8 = 10x
P10 100 5

P10 = 100x = 3
105 3 – 1

3

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A52

(c) 8x + 1 + 20(22x) = (23)x + 1 + (22 × 5)(22x) Additional Mathematics Form 4 Year-End Assessment Additional Mathematics Form 4 Year-End Assessment (iii) S∞ = –243 = –145.8
2x – 316x + 2 2x – 3(24)x + 2
(c) Let O→P = k(3~i – 4~j ), 10. y 2
(3k)2 + (–4k)2 = 15 3
23x + 3 + (22 + 2x × 5) 1– 31– 24
2= x – 3 + 4x + 8 25k2 = 15
5k = 15 3 14. (a) 4= 20
k=3 x–1
22x + 2(2x + 1 + 5) O3 x x–1=5
= 25x + 5 \ O→P = 3(3~i – 4~j ) 2
= 9~i – 12~j x=6
= 22x + 2 – 5x – 5(2x + 1 + 5)
= 2–3x – 3(2x + 1 + 5) (b) (i) y = x2 – 1

6. Y = mX + c 11. Let the point P(x, y) x2 = y + 1
x = y + 1
Katakan titik P(x, y) f −1 = x + 1 for / bagi x > –1.

2y = mx2 + c 1 – 1.6 1 PA = 2
0.2 – 0.5 3 PB 1
At (0.2, 1), 1 = 1 2(0.2)2 + c 8. (a) Range of f(x) is f(x)  .
Julat bagi
(x – 2)2+ (y – 1)2 = 2(x – 3)2+ (y + 4)2 209
1 = 2(0.04) + c 16
Hence, domain of f −1 is x  1 . (x2 – 4x + 4) + (y2 – 2y + 1) = 4[x2 – 6x + 9 + y2 (ii) f 2(x) =
Maka, domain bagi 3 + 8y + 16]
c = 0.92 209
16
3x2 + 3y2 – 20x + 34y + 95 = 0 f(x2 – 1) =

When / Apabila x = 0, then / maka x2 = 0. (b) Its vertex is 1 1 , –112 and it is a minimum 12. 2x2 – 7x – 15 . 0 (x2 – 1)2 – 1= 209
When / Apabila x2 = 0, 2y = c = 0.92 point. 3 (2x + 3)(x – 5) . 0 16

7. (a) A→B = A→O + O→B Verteks ialah 1 1 , –112 dan ialah titik minimum. (x2 – 1)2 = 209 +1
= –7~i + 3~j + (–2~i + 9~j) 3 16
= –9~i + 12~j
= 225
| A→B | = (–9)2 + 122 16
= 15 x
–b ± b2 – 4ac – 3 5 (x2 – 1) = ± 15
Unit vector parallel to A→B is 2a 2 4
Vektor unit yang selari dengan A→B ialah 9. (a) x =
115(–9~i + 12~j) = – 53~i + 54~j 3 x2 = 15 + 1 , – 15 + 1
–(–3) ± (–3)2 – 4(2)(–4) \ x ,– 2 , x.5 4 4
= 2(2)
Section B / Bahagian B x2 = 19 , x2 ≠ – 11
4 4
3 ± 41 19
= 4 x = ± 12

= 2.35, –0.85 13. (a) 120(2a + 9d) = 400

2a + 9d = 80 15. (a) ex(2ex + 1) = 10

(b) O→A = kO→B (b) y = x2 + ax + 5 5 [(a + 10d) + (a + 14d)] = 425 Let / Andaikan y = ex
q~i – ~j = k(11~i + 10~j) 2 2a + 24d = 170
q~i – ~j = 11k~i + 10k~j = x2 + ax + 1 a 22 – 1 a 22 + 5 y(2y + 1) = 10
2 2
15d = 90 2y2 + y – 10 = 0

= 1x + a 22 – 1 a 22 + 5 d =6 (2y + 5)(y – 2) = 0
2 2
–1 = 10k a = 13 y =– 5 , y =2
2
k = – 1
10 Compare with y = (x + 2)2 + b, (b) JG/GP : ar = 162 ex ≠ – 5 , ex = 2
ar4 = – 48 2
q = 111– 1102 Bandingkan dengan x = ln 2 = 0.6931
a
2 = 2 (i) r3 = – 8
27
= – 11 a=4 (b) log5 4y +1 = log4 42
10 2 y –5
r= – 3
4y + 1
–1 a 22 +5=b (ii) a= 162 log5 y–5 = 2
2
b = –1 4 22 + 5 2
2 – 3 4y + 1 = 52
y–5
=1
= –243 4y + 1 = 25y – 125

21y = 126

y=6

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A53

Additional Mathematics Form 4 Year-End Assessment Additional Mathematics Form 4 Year-End Assessment

Paper 2 / Kertas 2 2. (a) When / Apabila x = 0, y = 2, 4. =(a2)(7521++×A135B)××(555 +–– 3) = 7 + 15 7. (a) Distance between X and Z
Section A / Bahagian A p – e0 = 2 AB 3
p–1 =2 3 Jarak di antara X dan Z
p =3
= XZ

1. y + 4x = 23 ...........................a 2(7 + 15)(5 – 3) = XO + OZ
xy – 3x = 16 ..........................b = (5)2 – (3)2 1 1 2  1 8 2
(b) When / Apabila y = 0, = –2 + 5
0 = 3 – ex
From / Dari a : y = 23 – 4x .c ex = 3 = 2(75 – 73 + (5 × 3)(5) – (5 × 3)(3) = 1 9 2
x = ln 3 2 3

Substitute c into b: = 75 – 73 + 53 – 35 = 92 × 32
Gantikan c ke dalam b, = 45 – 23 = 90
x(23 – 4x) – 3x – 16 = 0 \ (ln 3, 0) = 310 units
23x – 4x2 – 3x – 16 = 0 (c)
(b) BC2 = (45 – 23)2 + (5 + 3)2
4x2 – 20x + 16 = 0 y = 80 – 1615 + 12 + 5 + 215 + 3
x2 – 5x + 4 = 0 = 100 – 1415
2 y=x (b) XZ = 1 9 2
(x – 4)(x – 1) = 0 ln 3 f 3
x = 1, 4
O f –1 = 31 3 2
ln 3 2 1

x 5. a2y + 2 – x × b(3 – x) – y = a–1 × b–6 = 3OY (shown / tertunjuk)
2y + 2 – x = –1
When / Apabila x = 1, y = 23 – 4(1) = 19 3. (a) (i) u = log3 x 2y – x = –3 ................ a (c) Unit vector in the direction of ZX
When / Apabila x = 4, y = 23 – 4(4) = 7 x = 3u Vektor unit dalam arah Z→X
\ A(1, 19), B(4, 7) 3 – x – y = –6
x + y = 9 ................. b
1 –9
a + b : 3y = 6 310 –3
Midpoint of AB = 11 + 4 , 19 + 72 9 y=2 1 2=
Titik tengah AB 2 2 x
(ii) log3 1 2 = log3 9 – log3 x Substitute y = 2 into b: 1 2= – 3
= log3 32 – u Gantikan y = 2 ke dalam b, 10
= 1 5 , 132 x+2=9
2 = 2 log3 3 – u
x=7 1
mAB = 19 – 7 = 2(1) – u – 10
1–4
=2–u

= 12 (iii) logx 27 = log3 27 (d) OX + rOY = sOZ
–3 log3 x
1 –1 2 r1 3 2 s1 8 2
= –4 log3 33 (–4) 2 + 1 = 5
u 2
–4 × m = –1 = 6. (a) α+ β = – =2 −1 + 3r = 8s ................a
2 + r = 5s ....................b
m = 1 = 3 log3 3 αβ = 3
4 u 2 b,–12 + 3r 8
a ÷ + r = 5
= 3(1) (α + β)2 = α2 + 2αβ + β2
u −5 + 15r = 16 + 8r
Equation of the perpendicular bisector of α2 + β2 = (α + β)2 – 2αβ
AB: 3 3 7r = 21
Persamaan pembahagi dua sama serenjang bagi = u = (2)2 – 21 2 2
AB. r=3

(b) (log4 y)2 + (log4 y2) = 8 =1 Substitute r = 3 into a, / Gantikan r = 3
(log4 y)2 + 2 log4 y = 8 ke dalam a,
– 13 = 1 1x – 5 2 (b) The quadratic equation is −1 + 3(3) = 8s
4 2
y Persamaan kuadratik ialah 8s = 8
s=1
y – 13 = 1 x – 5 Let v = log4 y x2 – 1 1 + 1 2 x + 1 1 × 1 2 = 0
4 8 v 2 + 2v – 8 = 0 α2 β2 α2 β2
(v – 2)(v + 4) = 0
= 1 + 99 x2 – 1 β2 + α2 2 x + 1 1 2 = 0 Section B / Bahagian B
4 8 v = 2, –4 α2 β2
y x α2 β2 a
+
3 4 3 4x2 –1  x + 1 =0 8. (a) y = x b

When / Apabila log4 y = 2, y = 42 = 16 1 3 22 1 3 22 y(x + b) = a
2 2
When / Apabila log4 y = –4, y = 4-4 = 1 –944xx++ 4 by = –xy + a
256 x2 – 94 = 0 1
9x2 = 0 y = 1– b 2xy + a
b

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A54

Additional Mathematics Form 4 Year-End Assessment Additional Mathematics Form 4 Year-End Assessment

xy 0.8 2.4 4.0 5.04 5.7 (ii) 4y = x + 28 ⇒ y = 1 x + 7 (b) (i) a = 1, d = 1, n = n (c) O→R = O→P + P→R
y 8.0 6.0 4.0 2.8 1.9 4 = 10~i – 24~j + 143P→Q
h is the y-intercept of line BC, so h Sn = n [2(1) + (n – 1)(1)]
2 cm 2
y 2 cm =7 = 10~i – 24~j + 143(26~i + 39~j )
9 = n (n + 1) Sn = n [2a + (n – 1)d]
8 CD parallel to the y-axis, so k = 4 2 2
7
6 h ialah pintasan-y bagi garis BC jadi h = 18~i – 12~j
5
4 = 7. (ii) Tn = 1 + (n – 1)(1) Last term Unit vector of O→R
3 =n Sebutan terakhir Vektor unit bagi O→R
2 CD selari dengan paksi-y, jadi k = 4.
1
0 1 2 3 4 5 6 xy (b) (i) Area of ∆BDC Sn < 100 = 18~i – 12~j
| 18~i – 12~j |
Luas ∆BDC n (1 + n) < 100 Sn = n [a + T2]
u= 07u 2 2
1 0 4 4 18~i – 12~j
2 7 7 8 n2 + n – 200 < 0 = 182 + 122

= 1 |(0 + 32 + 28) – (28 + 28 + 0)| n< –1 ± 12 – 4(1)(–200) 18~i – 12~j
2 2(1) 468
=
= 2 unit2 Since / Sejak n . 0, n = 13.65
Thus, there will be 13 rows.
(ii) Area of ABCD Maka, terdapat 13 baris. = 18~i – 12~j
613
Luas ABCD
u= u2
1 2 0 4 4 Total number of cups in the triangle = 133~i – 123~j
2 –3 7 8 7 –3 Jumlah bilangan cawan di dalam segi
tiga
= 1 |(14 + 0 + 28 – 12) – = 123(1 + 13)
2 = 91

(0 + 28 + 32 + 14)| Section C / Bahagian C

(b) From the graph, – 1 = 9–7 = 22 unit2 12. (a) (i) AC = 73
Daripada graf b 0 – 1.6 sin 45° sin 52°

= –1.25 Area of BDC : Area of ABCD The number of cups not being used AC = 73 × sin 45°
Luas BDC : Luas ABCD Bilangan cawan yang tidak digunakan sin 52°
a b = 0.8 = 2 : 22 = 100 – 91
b = 1 : 11 =9
= 9 = 65.51 cm

a = 9(0.8) (ii) ∆BAC = 180° – 45° – 52°
= 83°
= 7.2 1 11. (a) | 5~i – 12~j | = 52 + (–12)2 = 13
2 | O→P | = 26 = 2 × 13 GH2 = 332 + 302 – 2(33)(30) cos 83°
(c) (i) When y = 5.2, xy = 3.05 10. (a) (i) ar2 = ................... a O→P = 2(5~i – 12~j ) = 10~i – 24~j GH = 41.81 cm

Apabila x = 3.05 ar5 = – 1 .................b
5.2 16

= 0.5865 (i) b÷ a : r3 = – 1 | 1O→2Q~i | + 5~j | = 122 + 52 = 13 (b) Area of ∆ABC = 1 (73)(65.51) sin 83°
8 | = 39 = 3 × 13 Luas ∆ABC 2

(ii) When / Apabila xy = 2.8, y = 5.5 r=– 1 = 2 373.29 cm2
x(5.5) = 2.8 2
x = 0.5091 O→Q = 3(12~i + 5~j) = 36~i + 15~j (c) (i) The length of two sides, AB and
a 1– 1 22 1 AC, and a non-included acute
(ii) 2 = 2 angle, /ABC are given.

9. (a) (i) At C, 4(2x) = x + 28 a=2 (b) | P→Q | = | P→O + O→Q | The side opposite to the non-
Pada 8x = x + 28 included angle, AC is shorter than
7x = 28 (iii) Tn = 2 1– 21 n – 1 = | (–10~i + 24~j) + (36~i + 15~j) | the other side, AB. (AC , AB)
x=4 = | 2266~i2 ++ 3399~j2|
2 = Panjang dua sisi, AB dan AC, dan satu
y = 2(4) sudut tirus yang tidak termasuk, /ABC
=8 = 2 197 diberi. Sisi yang bertentangan dengan
sudut yang tidak termasik, AC lebih
\ C(4, 8) = 1313 pendek daripada sisi lain, AB. (AC , AB)

\λ = 13

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A55

Additional Mathematics Form 4 Year-End Assessment Additional Mathematics Form 4 Year-End Assessment

(ii) Another possible triangle: ABD 14. (a) (i) 120 × 100 = 90 (b) (i) Composite index, I (c) Q2020 × 100 = 140 ⇒ Q2020 = 140
where AD = AC x x= 133.33 Indeks gubahan, I Q2018 Q2018 100
Segi tiga lain yang mungkin: ABD
dengan AD = AC y = 125(1) + 80(1) + 150(1) + 80(1) Q2019 × 100 = 108.75 ⇒ Q2019 = 108.75
140 4 Q2018 Q2018 100
(ii) × 100 = 125
y = 175 = 108.75

A (iii) z = 150 × 100 (ii) 108.75 = 578 × 100 I 20/19 = Q2020 × 100
65.51 cm 100 Q2018 Q2019
C
73 cm = 150 Q2018 = 531.49 = Q2020 × Q2018 × 100
Q2018 Q2019
45° The corresponding cost in 2018
BD = RM531.49 = 140 × 100 × 100
Kos sepadan pada tahun 2018 100 108.75
(b) I= 90(3) + 125(1) + 150(2) + 110(5) + 75(4) = RM531.49
= 103 3+1+2+5+4
13. (a) = 128.7

S

12.8 cm (c) Q2019 × 100 = 103
600 Q2019 = RM618
85°
PT R \ The cost in the year 2019 = RM618
Kos pada tahun 2019
28° 6.2 cm
Q

sin 85° = ST (d) Q2019 × 100 = 103 ⇒ Q2019 = 103
12.8 Q2018 Q2018 100

ST = 12.75 cm Q2018 × 100 = 115 ⇒ Q2018 = 115
Q2017 Q2017 100

(b) (i) QS = 12.8 Q2019
sin 85° sin 28° Q2017
Price index, I = × 100
QS = 27.16 cm Indek harga,

(ii) /SQR = 180° – 85° – 28° = Q2019 × Q2018 × 100
Q2018 Q2017
= 67°
∠SQR = ∠PSQ 103 115
= 100 × 100 × 100

RS2 = 27.162 + 6.22 – 2(27.16)(6.2) = 118.45
cos 67°
RS = 25.39 cm 0.80
x
sin /QRS sin 67° 15. (a) (i) × 100 = 125
27.16 25.39 x = 0.64
(c) =

sin /QRS = sin 67° × 27.16 (ii) y = 2.00 × 100
25.39 2.50
= 80
/QRS = 79.96°

(d) Area of triangle QRS (iii) z × 100 = 150
0.60 z= 0.90
Luas segi tiga QRS

= 1 (27.16)(6.2) sin 67°
2

= 77.50 cm2

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A56