Module & More Add Math Tg 5

∫ 3. (a) (i) 6 g(x)dx = 14, ∫ 4 g(x)dx = –4 ∫ = 4 – 2 (y – 2)2dy y
–3 0 0
y
(y – 2)3
18 3
 4 2 x = (y – 2)2
=4– 0

= 4 – + 8 4 2
3
x y =– 1 x + 2
4 2
–3 0 46 = 3 unit2

–4 x

∫ ∫ 0 g(x)dx = 2 6 g(x)dx ∫ (ii) Isipadu = 2πx2dy 0 246
–3 4
Volume 0
∫ ∴ 0 g(x)dx = 12
–3 ∫ = π 2(y – 2)4dy
0
∫ 4 g(x)dx = 12 – 4 = 8
(y – 2)5
–3 5
 4 =π 2
∫ ∫(ii) 4 g(x)dx = 6 g(x)dx = 4 0
04
= π 3524 = 32 π unit3
∫ ∴ 0 g(x)dx = 18 – 4 = 14 5
–3
BAB 4: PILIH ATUR DAN GABUNGAN

∴2∫ 0 g(x)dx = 28 1. (a) (i) 4C1 × 5C1 = 20 cara
(ii) 4 baju 5 seluar
–3

∫ (iii) 6 g(x)dx = 2
0
2 hitam 2 hitam
∫ ∴ 6 g(x)dx = 6
Kes 1: Jika pakai baju hijam
4 Case 1: It she wears black blouse

∫ 0 g(x)dx = 18 – 6 = 12 Bilangan cara = 2C1 × 3C1 = 6
–3 Kes 2: Jika pakai seluar hitam
Case 2: It she wears black skirt
(b) (i) y

x = 2 – y2 x = 3 + y2 Bilangan cara = 2C1 × 2C1 = 4
1 Kes 3: Jika bukan baju hitam dan seluar

x hitam
0 1 23 4 5 Case 3: It not black blouse and black skirt.
-1
Bilangan cara = 2C1 × 3C1 = 6

Number of ways

Jumlah cara = 6 + 4 + 6 = 16

Total of ways

∫ luas = 1 [3 + y2 – (2 – y 2)]dy (b) (i) 6! = 120 cara/ways
3!
area –1 (ii)  Jika mesti bersama/If must together

∫ = 1 (1 + 2y2)dy = 4!
–1

 4=y+ 2 y3 1 M63!!e−sti4d! i=as9in6gkan/Must be separated
3 –1

= 1 + 2 4 – –1 – 2 4 (iii) M _ _ _ _ K
3 3
4!
= 3 1 unit2 3! = 4 cara/ways
3
(iv) 1 cara/way
∫ (ii) Isipadu = π 5 y2dx
2 2. (a) (i) Bilangan cara = (8 – 1)!3!
Volume 3

∫ = π 5(x – 3)dx Number of ways = 30240
2
3 (ii) Jika duduk bersama = (9 – 1)!2!

 4 π x2 5 It sit next to each other
= 2 2 – 3x 3
Jika tidak ada halangan = (10 – 1)!
= π 1225 – 152 – 1 9 – 924
2 2 If no restriction

= π [8 – 6] = π unit3 ∴Tidak duduk bersama = 9! – 8!2!
2
Not sit next to each other = 282240

(iii) Bilangan cara = 10C8(8 – 1)!
Number of ways = 226800

(c) (i) luas  = 1 [2](4) = 4 unit2 (b) (i) 0, 1, 3, 4, 5, 7, 8
2
area 6 6 5 4 = 720 cara/ways

luas berlorek (ii) 4 3 2 1 = 24 cara/ways

shaded area

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(iii) 4 6 6 6 = 864 cara/ways (ii) P[X ≥ 5] = 0.2 + 0.3 + 0.2

(iv) 5 6 5 4 = 600 cara/ways = 0.7

(b)
3, 4, 5, 7, 8
P[X = r]

(c) (i) 4 merah/red, 3 biru/blue, 4 hijau/green, 0.3
0.2
3 kuning/yellow 0.1

4C1 × 3C1 × 4C1 × 3C1 = 144 0 3579 r
(ii) Kes 1: 2 merah, 2 berlainan warna
2 red, 2 different colour

4C2 × 10C2 = 270
Kes 2: 3 merah, 1 berlainan warna
3 red, 1 different colour

4C3 × 10C1 = 40 2. (a) (i) p = 3 n = 10
Kes 3: 4 merah/red = 4C4 = 1 100
Jumlah/Total = 270 + 40 + 1

= 311 P[X = 5] = 10C5(0.03)5(0.97)5
= 5.26 × 10–6
(iii) Jika tiada guli hijau dan merah

If no green and red marble (ii) P[X ≥ 2] = 1 – P[X < 2]

6C4 = 15 = 1 – P[X = 0] – P[X = 1]
Jika tiada guli merah sahaja
= 1 – 10C0(0.03)0(0.97)10 –
If no red marble only
10C1(0.03)(0.97)9
10C4 = 210 = 0.0345
Sekurang-kurangnya 1 hijau
At least 1 green (b) n = 2500, p = 0.03

210 – 15 = 195 µ = np = 2500 × 0.03

(d) (i) nC2 = 253 = 75

n! σ2 = npq = 75(0.97)
– 2)!
(n = 253 σ = 8.53

n(n – 1) = 506 3. (a) P[X = x] = k(1 – x)2

n2 – n – 506 = 0 X –1 0 1 2

(n + 22)(n – 23) = 0

n = 23 P[X = x] 4k k 0 k

(ii) Bilangan salam antara 3 graduan.
Number of handshakes among 3 graduates.
4k + 2k = 1
B=il3aCn2g=an3salam/Number of handshakes
k = 1
= 253 – 3 6

= 250 (b) P[X < 0] + P[X > 1]

(e) m + n = 26 ...➀ = 4 + 1 = 5
6 6 6
nC2 + mC2 = 160
(c)
n! + m! = 160
(n – 2)!2! (m – 2)!2! P (X = x)

n(n – 1) + m(m – 1) = 320 4
n2 + m2 – (n + m) = 320 6
(n + m)2 – 2mn = 320 + 26
262 – 346 = 2mn 3
mn = 165 6
m(26 – m) = 165
m2 – 26m + 165 = 0 2
(m – 11)(m – 15) = 0 6
n = 15,
m = 11 1
m = 11 ; 15 6

0 –1 0 1 2 x

4. (a) (i) p = 0.6 n = 8 q = 0.4

BAB 5: TABURAN KEBARANGKALIAN P[X = 5] = 8C5(0.6)5(0.4)3
= 0.2787

(ii) P[X ≥ 5] = P[X = 5] + P[X = 6] + P[X = 7] +

1. (a) (i) P[X = 3] = P[X = 7] = 0.3 = p P[X = 8]

1 – 0.3 – 0.2 – 0.3 = q = 8C5(0.6)5(0.4)3 + 8C6(0.6)6(0.4)2 +

q = 0.2 8C7(0.6)7(0.4) + 8C8(0.6)8
= 0.5941

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(b) µ = 200 × 0.6 = 120 60° – A = 30°

(c) P[X = n]= nCn(0.6)n = 0.07776 A = 30°

n = log 0.07776 = 5 = π 2 Ί๶3
log 0.6 6
60°
5. (a) (i) X ~ N(45, 8.32) y 1

Z=1= X – 45 30° x
8.3 30°

X = 45 + 8.3
= 53.3

(ii) –2 = X – 45 (b) kot A + tan A = sek A kosek A
8.3 cot A + tan A = sec A cosec A
Sebelah kiri
X = 45 – 2(8.3)
= 45 – 16.6 LHS = 1 + tan A
= 28.4 tan A

(b) P[X > 40] =

= P X–µ > 408–.3454 1 + tan2 A = sek2A
σ tan A tan A

= PZ > –5 4 = 1 · kos A
8.3 kos2 tan A
A
= 0.7266
(c) P[X > n] = 0.3589 = sek A kosek A

n – 45 = sec A cosec A
8.3
P3 Z > 4 = 0.3589 = Sebelah kanan

n – 45 = 8.3(0.361) RHS
n = 48
6. (a) X ~ N(142, σ2) 2. (a) cos A + sin A = sin A + cos A
P[139 < X < 145] = 0.95 1 − tan A 1 − cot A

LHS = cos2 A A + sin sin2 A
cos A – sin A − cos A

P 139 – 142 < Z < 145 – 142 4 = 0.95 (sebelah kiri)
σ σ
= cos2 A – sin2 A
P– 3 3 4 = 0.95 cos A − sin A
σ < Z< σ
(cos A + sin A)(cos A – sin A)
= cos A − sin A
PZ > 3 4 = 0.05
σ 2 = sin A + cos A = RHS (sebelah kanan)

= 0.025 (b) (i) 4 sin2x + 8 cos x – 7 = 0

σ3 = 1.96 4[1 – cos2x] + 8 cos x – 7 = 0

4 cos2x – 8 cos x + 3 = 0

σ = 3 = 1.53 (2 cosx – 1)(2 cosx – 3) = 0
1.96 cos x = x
1 cos = 3 y
141 – 142 2 2
(b) P[X < 141] = PZ < 1.53 4
∴x = 60°, 300°
2 Ί๵3
= 0.2567 (ii) Jika x = θ 60° x
2 1
(c) P[X > 140.5]

= PZ > 140.5 – 142 4 2θ = 60°, 300°
1.53
θ = 120°, 600° (Tidak diterima)
= 0.8366
(Rejected)
25
Bilangan orang perempuan = 0.8366 3. (a) 2 sin x = 4 cos x – 1
tan x
Number of girls

= 30 2 sin2 x = 4 cos x – 1
cos x
BAB 6: FUNGSI TRIGONOMETRI
2[1 – cos2x] = 4 cos2x – cos x

1. (a) tan ( π – A) = kot π 0 = 6 cos2x – cos x – 2
3 3
6 cos2x – cos x – 2 = 0 (Tertunjuk)

tan (60° – A) = kot 60° (b) (3 cos x – 2)(2 cos x + 1) = 0

= 1 cos x = 2 , – 1
3 3 2

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y y 5. (a) 2 sin2x + 2 = 7 cos x
48.19° x 2 sin2x – 7 cos x + 2= 0
2 2[1 – cos2x] – 7 cos x + 2 = 0
2 cos2x + 7 cos x – 4 = 0
60° x (2 cos x – 1)(cos x + 4) = 0
1

cos x = 1 atau/or cos x = –4
2

x = 48.19°, 311.81° atau/or x = 120°, 240° ∴x = 60°, 300°

∴x = 48.19°, 120°, 240°, 311.81° = π , 5π
3 3
(c) y

(b) (i) y = 3 sin bx

y = 4 kos x – 1 a = 3 b → 2π
y = 4 cos x – 1
1→π

x ∴b = 2
2
01 (ii) 3 sin 2x = 1.4

sin 2x = 1.4
3

2π sin x tan x = 2π – x x = 27.82°, 152.18°
x A(0.486, 1.4)
4 kos x – 1= 2 – π =y B(2.656, 1.4)
atau A(0.155π, 1.4)
x B(0.845π, 1.4)
π
4 cos x – 1 = 2 – =y

2 penyelesaian/solution BAB 7: PENGATURCARAAN LINEAR

4. (a) tan 2x = 5 sin 2x 1. (a) x + y ≤ 5
3x + 2y ≤ 12
sin 2x = 5 sin 2x (b)
kos 2x
y
sin 2x = 5 sin 2x
cos 2x 3x + 2y ഛ 12
sin2x = 5 sin 2x · cos 2x 7
sin 2x[1 – 5 cos 2x] = 0 (Tertunjuk/Proved)
6
(b) sin 2x(1 – 5 cos 2x) = 0
5
sin 2x = 0 x+yഛ5

2x = 0°, 180°, 360° 4

540°, 720° 3
R
x = 0°, 90°, 180°, 270°, 360°
2
cos 2x = 1
5 1

2x = 78.46°, 281.54° 0 12345
438.46°, 641.54°
x = 39.23°, 140.77°, 219.23°, 320.77°
(c)

y

y = tan 2x

x

y = x1 (c) (i) P = 30x + 25y
0 S Sx Titik optimal/Optimal point = (2, 3)
Keuntungan maksimum/Maximum profit
2
y = tan 2x P = 30(2) + 25(3)
= RM 135
5x sin 2x = 1 (ii) Jika/If x = 1 , 0 ≤ y ≤ 4
2. (a) 6x + 3y ≤ 36 atau 2x + y ≤ 12
5 sin 2x = 1 y ≤ 4x
x x ≤ 4y
x ≤ 4
= tan 2x
1
y = x

2 penyelesaian/solutions

J6

(b) y 3. (a) vA = 3 + 2t – t2

y р 4x ddvtA = 2 – 2t = 0
8 t = 1

6 6x + 3y р 36 vmax = 3 + 2(1) – (1)2 = 4 m s–1
(b) Di R vA = 0 = 3 + 2t – t2
xр4 0 = (3 – t)(1 + t)
4
∴t = 3
R
sA = 3t + t2 – 1 t3
2 3

0 24 x t = 3, sA = 3(3) + 9 – 1 (27) = 9 m
3

(c) Untung/Profit P = 30x + 10y ∴RP = 9 m

Titik optimum/ Optimum point = (4, 4) (c) vB = –5
sB = –5t + c
Untung/Profit P = 60(4) + 20(4) Apabila t = 0 sB = 30 m

= RM320 ∴sB = –5t + 30

BAB 8: KINEMATIK GERAKAN LINEAR Apabila t = 3, sB = –5(3) + 30
= 15 m

1. (a) v = 2t(6 – t) ∴Jarak di antara A dan B
Distance between A and B
= 12t – 2t2
= (15 – 9) m

ddvt = 12 – 4t = 0 = 6 m
t = 3
(d) 3 + 2t – t2 = –5

∴vmak = 2(3)(6 – 3) t2 – 2t – 8 = 0
= 18 m s–1
(t + 2)(t – 4) = 0

(b) v = 0, 2t(6 – t) = 0 ∴ t = 4

t = 0, 6 4. (a) (i) a = –10

∴t = 6 v = –10t + c

(c) v t = 0 v = 30

18 ∴v = –10t + 30

Apabila tinggi maksimum

When it is the maximum height

0 67 t v = 0 ∴10t = 30

t = 3

(ii) s = –5t2 + 30t

(d) Jumlah jarak/Total distance Apabila s = 0 5t[–t + 6] = 0

∫ ∫ = 6(12t – 2t2)dt + 7(12t – 2t2)dt t = 6
06
∴v = –10(6) + 30
3 4 3 4 = 2 6 2 7
6t2 – 3 t3 0 + 6t2 – 3 t3 6 = –30 m s–1

2 (b) s = –5(9) + 90
3
= 72 + –6  = 45 m s

= 78 2 m 45 m t = 3
3

2. (a) s = t(t – 4)2 0 36 t

ds = v = t(2)(t – 4) + (t – 4)2
dt

= (t – 4)[2t + t – 4] V
30
= (t – 4)(3t – 4)

Apabila t = 3 v = (3 – 4)(9 – 4)

= –5 m s–1

(b) v = 0 = (t – 4)(3t – 4) 0 t(s)
–30 36
4
∴t =4 s dan/and 3 s

(c) v = (t – 4)(3t – 4)

dv = a = (t – 4)3 + (3t – 4) (c) Jumlah jarak = 1 (3)(30) 2
dt 2
Total distance ×
Apabila/When t = 4
= 90 m
a = (4 – 4)3 + (3(4) – 4)

= 8 m s–2

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