Module & More Add Math Tg 5

7. (a) kos/ cos 23° = sin(90° − 23°) (f) sek/sec 420° + kosek/cosec (−120°)

= sin 67° = 1 + 1
(–120°)
= 0.9205 kos/ cos 420° sin

(b) tan 23° = kot/ cot = (90 – 23°) = 1 + 1
kos/ cos 60° –sin 60°
= kot/ cot 67° = 1 y
tan 67°
= 1 – 2 2
1 13
= 2.3559 2 60Њ √3
x
 2 = 2 – 2 = 2 3 – 1 60Њ 1
33
= 0.4245

(c) sek/ sec 23° = kosek/cosec (90° – 23°)
= kosek/cosec 67°
9. (a)
1
= sin 67° π π 3 5 3 7
4 2 4 4 2 4
1 x 0 π π π π π 2π
0.9205
= 0° 45° 90° 135° 180° 225° 270° 315° 360°

= 1.0864 kos/ 1 0.71 0 –0.71 –1 –0.71 0 0.71 1
cos x
8. (a) tan 225° = tan 45°

=1

y kos/ cos x

–1 225Њ 1
x
45Њ
–1 0

√2

(b) sek/ sec (−240°) = kos/ 1
cos(–240°)

= 1 y 0 x
–kos/ cos
60° 90Њ 180Њ 270Њ 360Њ

= 1 = –2 2

– 12 √3 60Њ 0 x
–1
–240Њ

–1

(c) – kosek/ cosec (330°) = – 1
sin 330°
(b)
= – 1
–sin 30° π π 3 5 3 7
0 4 2 4 π π 4 π 2 π 4 π 2π
= 1 =2 y
1 x

20 √3 x 0° 45° 90° 135° 180° 225° 270° 315° 360°
30Њ –1

2 tan x 0 1 ∞ –1 0 1 ∞ –1 0

tan x

y

(d) – sin (−300°) 4
= – sin 60°
= – 23 2 √3 2
x
60Њ
01

0 x

(e) sin 315° + kos/ cos 135° y 90Њ 180Њ 270Њ 360Њ

= – 12 +  – 12  √2 1 x –2
1 45Њ –1
= – 22
45Њ √2
–1

–4

J34

10. (a) y y = 3 kos 2x 11. (a) y = a kos bx + c
4
3 3 a = −2
4 y 4 cos 2x
= 1 1 kala dalam 2π
2

0␲ x 1 1 cycle in 2π
2␲ 2
␲x
– 3 b = 3 ; c = 0
4 2
1
(b) a = −2 ; b = 2 kala/cycle ; c = 0

(b) y y = 1+ kos 2x (c) a = 2 ; b = 1 kala/cycle ; c = 1

y = 1+ cos 2x 12. (a) y y = tan x
2
1
1 x
0 ␲x
–1 0 ␲ 2␲

–2

(c) y y = tan 2x–1 π(1 – tan x) = 2x

1 – tan x = 2x
π
2x
0 1 – π = tan x
–1
Maka/Hence y = 1 – 2x
π

Apabila/When x = 0 y = 1

x = π y = –1

(d) y Terdapat 3 penyelesaian

3 There are 3 solution
2
1 y = 3│kos 1 x│ (b) (i) y y = │2 sin 2x│
2
0 1
–1 y = 3│cos 2 x│ 2
–2
–3 ␲ x 0 x
2␲ ␲ 2␲
(e) y
–2
1
0 (ii) 2|2 sin 2x| – 3 = –2x
–1 π
–2
|2 sin 2x| = 3 – x
(f) y 2 π

2 ∴ y= 3 – x
2 π
0 x
␲ 2␲ Apabila/When x = 0, y = 3
2

3 x = 2π y= 3 –2=– 1
4 2 2
y = –│sin x│–1
5 penyelesaian/solutions

(c) (i) y = –2 sin 3 x 0 ≤ x ≤ 2π
2

y y = –2 sin 3 x
2
2 y=k

␲ 2␲ x x
y = 2 – │tan x│ 0 2␲
–2 y = k

(ii) y = k = −2 atau/or 0 < k ≤ 2

J35

(d) (i) y ␲ y = –2 kos 3 x (c) Sebelah kiri / Left side = sin A + kos/ cos A
2x 4 kos/ cos sin A
2 y = 3 A
y cos 4 x
= –2 sin2 A + kos2 A 1
kos A sin A kos A sin A
= =

0 x sin2 A + cos2 A 1
2␲ cos A sin A cos A sin A
= =

–2 = sek A kosek A (sebelah kanan)

(ii) π = –2 kos 3 x = sec A cosec A (right side)
2x 4
(d) Sebelah kiri / Left side
∴y= π = π  1  kot2 A + kos2 A = kosek2 A − 1 + kos2 A
2x 2 x
cot2 A + cos2 A = cosec2 A − 1 + cos2 A
2 penyelesaian/solutions
= kosek2 A – (1 − kos2 A)
(e) (i) y y = 2 sin 2x+1
cosec2 A – (1 − cos2 A)

= kosek2 A – sin2 A (sebelah kanan)

3 cosec2 A – sin2 A (right side)

2 (e) Sebelah kanan / Right side
= (kosek A + kot A)2
1
(cosec A + cot A)2
0 ␲ ␲ 3␲ x
= kosek2 A + 2 kosek A kot A + kot2 A
–1 2 2
cosec2 A + 2 cosec A cot A + cot2 A

–2 = 1 + 2 kos A + kos2 A
sin2A sin2A sin2A
(ii) 2 sin 2x + 2 kos x + 1 = 0
2 sin 2x + 1 = –2 kos x 1 + 2 cos A + cos2 A
∴ y = –2 kos x sin2A sin2A sin2A
2 penyelesaian/solutions
= 1 + 2 kos A + kos2 A = (1 + kos A)2
sin2A sin2A
1 r2
13. (a) Sebelah kanan: sek2A = kos2A = x2 1 + 2 cos A + cos2 A = (1 + cos A)2
sin2A sin2A
1 r2
Right: sec2A = cos2A = x2 = (1 + kos A)2 = (1 + kos A)2
1 – kos2A kos A)(1 + kos
Sebelah kiri: 1 + tan2 A = 1+ y2 = (x2 + y2) (1 – A)
x2 x2
(x2 + y2) (1 + cos A)2 = (1 + cos A)2
y2 x2 1 – cos2A (1 – cos A)(1 + cos A)
Left: 1 + tan2 A = 1 + x2 =
1 + kos A
Tetapi/But x2 + y2 = r2 = 1 – kos A (sebelah kiri)

Maka/Hence 1 + tan2A = r2 = sebelah kanan 1 + cos A (left side)
x2 1 – cos A
1 + tan2A = sek2A
(b) Sebelah kanan: kosek2A = 1 = r2 15. (a) kos (A + B) = 1 – tan A + tan B
sin2A y2 sin (A – B) tan A – tan B
1
Right: cosec2A = sin2A = r2 sebelah kanan/Right side
y2
sin A sin B
Sebelah kiri: 1 + kot2 A = 1 + x2 = (x2 + y2) 1– cos A cos B
y2 y2
(x2 + y2) =
x2 y2 sin A sin B
Left: 1 + cot2 A = 1 + y2 = cos A – cos B

Tetapi/But x2 + y2 = r2 r2 = sebelah kanan = cos A cos B – sin A sin B
Maka/Hence 1 + kot2 A = y2 cos A cos B

1 + kot2 A = kosek2 A. sin A cos B – cos A sin B
cos A cos B
14. (a) Sebelah kiri: kos A = kos A = 1
1 – sin2A kos2A kos A = cos (A + B) = Sebelah kiri/Left side
sin (A – B)
cos A cos A 1
Left: 1 – sin2A = cos2A = cos A
(b) sin (A – B) = tan A – tan B
= sek/ sec A kos A kos B
= sebelah kanan/right side
Sebelah kanan/Right side
(b) Sebelah kiri / Left side
= 1 − 2(1 − kos2 A) = sin A – sin B
cos A cos B
1 − 2(1 − cos2 A)
= sin A cos B – cos A sin B
= 1 − 2 + 2 kos2 A cos A cos B

1 − 2 + 2 cos2 A = sin (A – B) = Sebelah kiri/Left side
cos A cos B
= 2 kos2 A − 1 (sebelah kanan)

2 cos2 A − 1 (right side)

J36

(c) sin (A + B) + sin (A – B) = tan A (c) Sebelah kiri / Left side
sin (A + B) – sin (A – B) tan B
11 – kos 2A = 1 – (1 – 2 sin2A)
Sebelah kiri/Left side + kos 2A 1 + (2 kos2A – 1)

sin A cos B + cos A sin B + = 2 sin2A = tan 2A
2 kos2A
= sin A cos B – cos A sin B
sin A cos B + cos A sin B – (d) Sebelah kanan / Right side

sin A cos B + cos A sin B 2 sin θ kos θ
2 2
= 2 sin A cos B sin θ =
2 cos A sin B 1 + kos θ θ – 1)
1 + (2 kos2 2
sin A cos B
= cos A × sin B θ
2
= tan A × cot B 2 sin θ
2
tan A = θ = tan
tan B 2
= = Sebelah kanan/Right side 2 kos

16. (a) Jika/If A = B (e) Sebelah kiri / Left side

Maka, kos (A + A) = kos A kos A − sin A sin A sin A + kos A = sin2A + kos2 A = sin 1 A
kos A sin A sin A kos A A kos
Hence, cos (A + A) = cos A cos A − sin A sin A

kos 2A = kos2 A – sin2 A = 2 = 2
2 sin A kos A sin 2A
cos 2A = cos2 A – sin2 A

Oleh kerana kos2 A = 1 − sin2 A = 2 kosek 2A

Since cos2 A = 1 − sin2 A 2 cosec 2A

Maka/Hence kos/ cos 2A = 1 − sin2 A – sin2 A

= 1 − 2 sin2 A 18. (a) sin (132° − 72°)
= sin 60°
Jika/If sin2 A = 1 − kos2 A
= 3
sin2 A = 1 − cos2 A 2

Maka/Hence kos 2A = kos2 A – (1 − kos2 A)
cos 2A = cos2 A – (1 − cos2 A)
(b) tan (120° + 45°)
= 2 kos2 A − 1 /2 cos2 A − 1
(tan 120° + tan 45°)
Jika/If A = θ , kos θ = kos2 θ – sin2 θ = 1 – tan 120° tan 45°
2 2 2
3+
cos θ = cos2 θ – sin2 θ = (– (– 1)
2 2 1– 3)
θ
= 2 kos2 2 −1 = 1– 3
1+ 3
2 cos2 θ −1
2 4–2 3
θ = –2
= 1 − 2 sin2 2
= 3 – 2
(b) Jika/If A = B
(c) kos (45° − 30°)
sin (A + A) 2 sin A kos A
tan (A + A) = kos (A + A) = kos2 A – sin2 A = kos 45° kos 30° + sin 45° sin 30°

2 sin A kos A = 1 1 21 3 2 + 1 1 21 1 2
2 2 2 2
kos2 A
tan 2A = kos2 A – sin2 A = 2 tan A = ( 3 + 1)
1 – tan2A 2 2

kos2 A θ 19. (a) y
2
Jika A = θ , maka tan θ = 2 tan
2
1 – tan2 θ 51Њ5Ј x
2 51Њ5Ј

17. (a) Sebelah kanan / Right side x = 180° – 51° 5’, 180° + 51.5’
(kos A + sin A)2 = kos2 A + 2 kos A sin A + sin2 A = 128° 55’, 231° 5’
= 1 + 2 sin A kos A (b) y
= 1 + sin 2A (sebelah kiri)
(b) Sebelah kiri / Left side

kos2 A + sin2 A

kos2 A 51Њ34Ј x

kos2 A – sin2 A 51Њ34Ј

kos2 A

= 1 = sek 2A
kos 2A

J37

2 tan x = 2.52 sin 2x = 0.782
tan x = 1.26 3
x = 51° 34’, 180° + 51.34’
= 51° 34’, 231° 34’ = 51° 27’, 128° 33’
(c) y
x = 77° 11’ , 192° 50’

(h) y

20Њ55Ј x 38Њ25Ј x
20Њ55Ј 38Њ25Ј

sin 2x = –0.357 2 kos(x – 25°) = 1.567
2x = 200° 55’, 339° 5’
560° 55’, 699° 5’ kos(x – 25°) = 0.7835
x = 100° 28’, 169° 33’, 280° 28’, 349° 33’
(d) x – 25° = 38° 25’, 321° 35’

y x = 63° 25’, 346° 35’

2 0. (a) (i) kos (A + B) = kos A kos B – sin A sin B

=  – 53 2 – 11232 –  4 2 – 1532 y
5
44Њ36Ј x = 56 45
44Њ36Ј 65 –3 A

(ii) sin 2A = 2 sin A kos A –12 x
–5
= 2 4 2– 53 2 =– 24
5 25 13

kos 2x = 0.712 (iii) tan (A – B) = tan A – tan B
2x = 135° 24’, 224° 36’ =
495° 24’, 584° 36 = 1 + tan A tan B
x = 67°42’. 112° 18’
247° 42’, 292° 18’ – 34 –  5 2
(e) 12

y 1 +  – 43 2 5 2
12

65Њ3Ј x – 3 15
16

65Њ3Ј (b) (i) tan (A – B) = tan A – tan B

1 + tan A tan B

5 – (–1)
12
tan (x + 30) = 2.15 =
 5 2(–1)
x + 30 = 65° 3’, 245° 3’ 1 + 12

x = 35° 3’, 215° 3’ 17

(f) y = 7y

(ii) tan 2A = 2 tan A 13
1 – tan2 A

42Њ55Ј 42Њ55Ј x 2 5 2 √2 5
12 B x
1 A
12
= 25 –1
144
1–

sin(2x – 10°) = 0.681 = 1 1 = 120
119 119
(2x – 10°) = 42° 55’, 137° 5’

402° 55’, 497° 5’ (iii) kosek (A + B)

2x = 52° 55’, 147° 5’ 1
sin (A + B)
412° 55’, 507° 5’ =

x = 26° 28’, 73° 32’, 206° 28’

253° 32’ sin (A + B) = sin A kos B + kos A sin B

(g) y  5 2– 1 2+  12 2 1 )
= 13 2 13 2

51Њ27Ј 51Њ27Ј x = 7
13 2

= kosek (A + B) = 13 2
7

J38

(c) (i) kos 2A = 2 kos2A – 1 y x kos x = 0 2 = sin x kos x
p √1–p2 5 0.8
cos 2A = 2 cos2A – 1 4
A sin 2x = 5 =
= 2p2 – 1

(ii) tan A = 1 – p2 1 x = 90°, 270° 2x = 53° 8’, 126° 52’
p
413° 8’, 486° 52’
(iii) kos 4A = kos 2(2A)
cos 4A = cos 2(2A) x = 26° 34’, 63° 26’

= 2 kos22A –1 206° 34’, 243° 26’

= 2[2p2 – 1]2 – 1 (d) 4 tan 2x = 9 tan x

= 2[4p4 – 4p2 + 1] – 1 8 tan x = 9 tan x
1 – tan2x
= 8p4 – 8p2 + 1
8 tan x = 9 tan x – 9 tan3 x
–1
(d) (i) sin A = 1 + p2 9 tan3x = tan x

tan x[9 tan2x – 1] = 0

(ii) sin 2A = 2 sin A kos A tan x = 0 tan x = ± 1
–1 –p 3
= 2( 1 + p2 )( 1 + p2 )
x = 0, 180°, 360° x = 18° 27’, 161° 34’, 198° 27’,

2p 341° 34’
1 + p2
= (e) 5 sin x kos x – 5 sin x – 2 kos x + 2 = 0

A 5 sin x[ kos x – 1] – 2[kos x – 1] = 0
2
(iii) tan A (kos x – 1)(5 sin x – 2) = 0
2
2tan y kos x = 1 sin x = 2
–p 5
tan A = A
2 A
1 – tan2 x x = 0°, 360°, 23° 35’, 156° 25’
√1+p2 (f) 3 kos x = 5 sin x
A
2
1p = 2tan –1 3 = tan x
5
A x = 30° 58’ 210° 58’
1 – tan2 2

1 – tan2 A = 2ptan A (g) 3 sin x + 3 kos x + 1 x = 0
2 2 1 – kos
sin x

tan2 A + 2p tan A – 1= 0 3[sin x + kos x][sin x – kos x] + 1 = 0
2 2
3[sin2x – kos2x] + 1 = 0

A –2p ± 4p2 – 4(–1) 3[sin2x – (1 – sin2x)] + 1 = 0
tan 2 2
= 3[2 sin2x – 1] + 1 = 0

= –2p ± 6 sin2x – 2 = 0

4p2 + 4 sin2x = 1
2 3

= –p ± p2 + 1 sin x = ± 1
x = 35° 16’, 144° 44’ 3

21. (a) kot x + kos x = 0 = 215° 16’, 324° 44’
ksions
x + kos x = 0 (h) 3 sin2x = 8 sin x kos x + 3 kos2x
x
3 sin2x – 8 sin x kos x – 3 kos2x = 0
kos x[1 + sin x] = 0
(3 sin x + kos x)(sin x – 3 kos x) = 0
kos x = 0
3 sin x = –kos x tan x = 3,
sin x = –1
tan x = – 31 , x = 71° 34’, 251° 34’
x = 90°, 270°

(b) 2 kos2x – 1 = sin x x = 161° 34’, 341° 33’

2[1 – sin2 x] – 1 = sin x (i) sin2x = 1 – 2 kos2x

2 sin2 x + sin x – 1 = 0 = 1 – 2[1 – sin2x]

(2 sin x – 1)(sin x + 1) = 0 = –1 + 2 sin2x

sin x = 1 sin2x = 1
2
sinx = ± 1
sin x = –1
x = 90°, 270°

x = 30°, 150°, 270° 22. (a) (i) Q(–x, y) P(x, y)

(c) 2 sek2 x = 5 tan x 4y
␪
2 = 5 sin x R O xS
kos2 x kos x
x2 + y2 = 16
2 kos x = 5 sin x kos2 x

kos x[2 – 5 sin x kos x] = 0

J39

Maka OP = 4 4. (a) kos(180° – x)

luas berlorek = kos 180° kos x + sin 180° sin x

= luas semibulatan – luas PQRS = –kos x = – 1 – k2 y
=
1 π(4)2 – (2x)(y) (b) kosek 2x = 1
2 sin 2x
y x 1 k
sin θ = 4 , kos θ = 4
= 2k 1 x x
1 – k2 √1–k2
∴ luas berlorek = 8π – 2(4 sin q)(4 kos q)

= 8π – 16 sin 2q Kertas 2

(ii) Apabila/When q = 30°. 1 1
2 2
Luas/Area = 8p − 16 sin 2(30°) 1. (a) kot2 x = kos 2x +
1 – kos2 x
= 8p − 16 3
2 kot2 x
Sebelah kiri = sin2 x
= 8p − 8 3
(b) (i) P(x, y) Left side

OP = 5 = kos2 x ÷ 1
sin2 x sin2 x
OP2 = 52
= kos2 x
x2 + y2 = 52 Tetapi kos 2x = 2kos2 x – 1

x2 + y2 = 25 kos 2x + 1
2
QOP = 180° – 2θ kos2 x =

(ii) luas berlorek = luas sektor – luas ∆QOP 1 1
2 2
= (180° – 2θ) × p(5)2 – 1 (2x)y = kos 2x +
360° 2
(90° – θ) 1 1 3 y
= 25 180° π – xy (b) 2 kos 2x + 2 = 4 2

x = 5 kos θ, y = 5 sin θ kos 2x = 1 60Њ
2 01
luas berlorek = 5 (90° – θ)p – 25 sin θ kos θ x
36
(iii) Apabila θ = 15° 2x = 60°, 300°, 420°, 660°

5 25 x = 30°, 150°, 210°, 330°
36 2
luas berlorek = [90° – 15°]p – sin 30° (c) (i) y= 1 kos2x + 1
2 2
= 32.72 – 6.25

= 26.47 unit2 y y 1 2x 1
2 2
= kos +

Praktis SPM 6 y = 1 cos 2x + 1 y = 2x
2 2 3␲
1
Kertas 1 1

1. sin 2x + sin x = 0 2 x
2 sin x kos x + sin x = 0 2␲
sin x[2 kos x + 1] = 0 y 0␲

sin x = 0 ; 2 x 1 kos 2x + 1 ) = 2x
60Њ 3π( 2 2
1

x = 0°, 180°, 360°, 1 1 2x
2 2 3π
kos x = – 21 kos 2x + =

x = 120°, 240° y = 2x x = 0, y = 0
3π
2. sin (x – 45°) y 2
3
= sin x kos 45° – kos x sin 45° –12 x x = π, y =
x
= sin x cos 45° – cos x sin 45° 3 penyelesaian/solutions
–5
= 1  –5 2 – 1  –12 2 13 2. (a) tan(x + 45°) + tan(x – 45°) = 2 tan 2x
2 13 2 13
tan x + 1 tan x – 1
= 7 Sebelah kiri = 1 – tan x + 1 + tan x
13 2
Left side

3. 3 kos x – kosek x + 2 = 0 = (1 + tan x)2 – (1 – tan x)2
1 1 – tan2 x
3 kos x – kos x +2=0
1 + 2 tan x + tan2 x – 1 +

3 kos2x + 2 kos x – 1 = 0 = 2 tan x – tan x
1 – tan2 x
[3 kos x – 1][kos x + 1] = 0

kos x = 1 ; –1 = 2(2 tan x)
3 1 – tan2 x

x = 70° 32’, 289° 28‘, 180° = 2 tan 2x

J40

(b) 2 tan 2x = 1 (b) (i) & (ii)
2
y y = tan x
1
tan 2x = 4

2x = 14° 2’, 194° 2’,

274° 2’, 554° 2’ 0 ␲ x
2␲
x = 7° 1’, 97° 1’, 137° 1’, 277° 1’

(c) y = 1 [tan(x + 45°) + tan(x – 45°)] + 1 y = – x
2 ␲

y = 1 [2 tan 2x] +1 kot x [sek x – 1) = – x
2 2 π

= tan 2x + 1 ∴seytak=nx––2x πx1 x
π
y = tan x = –

1 x = 0, , y = 0
0 ␲x

x = π, y = –1
3 penyelesaian/solutions

3. (a) (i) sek A – 1 = tan A Sudut KBAT
tan A 2
(a) sin(A – B) = 3
Sebelah kiri sin(A + B) 5
Left side
5[sin A kos B – kos A sin B]

1 – 1 1 – kos A = 3[sin A kos B + kos A sin B]
= kos sin A
A = 2 sin A kos B = 8 kos A sin B

sin A sin A kos B = 4 kos A sin B

kos A 1 – [1 – 2 sin2 A ] (b) (i) tan B = 1
= 2 2

2 sin A kos A ksions AA = 4 tan B
2 2
1
A = 4 2 2 = 2
sin 2 A
= 2
A = tan tan A = 2
2
kos 2 tan A
1 – tan2 A
2 2 21 = sek 45° – 1 (ii) tan 2A =
tan 45°
(ii) tan 2(2) 4
1–4 –3
1 = =
kos 45°
= –1 = 2 –1

J41

BAB Pengaturcaraan Linear (c) y

7 Linear Programming x ≥ –1
R x≤y
1. (a) y
x + 2y ≥ 4

0x

0 x 4. (a) y
y
x ≥ 1 – 2y
(b)
R
0 –x ≥ –4
x

0 x –2 ≤ y
2y d 4x – 1
(b) y

2. (a) y = 4 x + y ≤ 60
y > 4

(b) m = 5− (−2) = – 78
−4 −4
x + 4y ≤ 100
y = – 87 x + c R
0 x

c = 3
2

y = – 87 x + 3 (c) y
2
y ≥ –3x – 15
y ≤ – 87 x + 3
2

3. (a) y

y≤5 2y + x ≥ –12
0R x

2 d x + 2y

y ≥ 4x – 12

R x 5. (a) y ≤ 3x + 3
0

x 4 y ≥ – 43 x + 4

(b) y y ≥ 1 x −1
3x – 2y t –5 2
x + y 10

(b) x ≥ 0
x ≤ 2

x y ≥ 2 x
3
0R
2y +3x + 2 ! 0 y ≤ 2x + 2
(c) y + x ≥ 4
3x + 2y ≤ 12
x + 2y ≤ 8

J42

6. (a) Pada (−2, 2) ialah titik minimum, maka (b) Keuntungan/Profit K = 3x + 6y
Titik optimum ialah titik persilangan.
At (−2, 2) is a minimum point, hence
The optimum point is the intersection point.
k = 2 + 2(−2) = −2
Pada (4, 5) ialah titik maksimum, maka y = 80 dan/and x − 3y = 300
Nilai integer x terdekat ialah 46.
At (4, 5) is a maximum point, hence
The value of the nearest integer x is 46.
k = 5 + 2(4) = 13
(b) Pada (3, 1) ialah titik minimum. Maka, Maka, untung maksimum K

k = 2(1) + 3 = 5. Hence, the maximum profit K
Tidak wujud titik maksimum kerana garis
= 3(46) + 6(80)
lurus boleh dianjak merentas rantau yang = RM618
tidak ada had. (c) Untuk y = x, titik maksimum = (60, 60).

At (3, 1) is a minimum point. Hence, k = 2(1) + 3 = 5. For y = x, maximum point = (60, 60).
There is no maximum point because the straight
Jenis A dan B masing-masing ialah 60 unit.
lines can be moved across an unlimited region.
Type A and type B are 60 units respectively.
7. (a) I 4x + 5y ≤ 120
II 3x + 5y ≤ 150 (b) (a) I x + y ≤ 100
III 600x + 1 500y ≤ 30 000 II y ≥ 20
atau 2x + 5y ≤ 100 III y ≤ 2x
(b) (i) Jika perbelanjaan untuk penyuntingan (b)
dan penulisan ialah x, maka x ≤ 5 000.
y
If the cost for editing and writing is x, then
x ≤ 5 000. 100
y = 2x
(ii) Jika perbelanjaan lain ialah y, maka
y ≤ 1 000. 80

If the other expenses is y, then y ≤ 1 000. 60

(iii) Kos 150 ribu naskhah ialah = 150 ×2 000
20
40 R
= RM15 000.

Kos untuk percetakan dan pengangkutan 20 (80, 20) y = 20

Z ≥ 15 000. 150 ×2 000 x + y = 100
20
The cost of 150 thousand copies is 0 20 40 60 80 100 x
= RM15 000.

The cost for printing and transportation (c) (i) x maksimum / x maximum = 80
Z ≥ 15 000. (ii) 20 ≤ y ≤ 50
(iii) Andaikan P = 1 200x + 600y
8. (a) (i) Ketaksamaan bagi setiap kekangan. Yuran maksimum / Maximum fees
= 1 200(80) + 600(20)
The inequalities for each constraint. = RM108 000

I y ≤ 80

II x − 3y ≤ 30

III 3x + 2y ≤ 300

(ii)

y y=x Praktis SPM 7
3x + 2y = 300
Kertas 2
100
80 y = 80 1. (a) I : 10x + 30y ≥ 30
x + 3y ≥ 3
60 II : y ≤ x
40 R III : x ≤ 9
(b) y

10 x = 9

20 x – 3y = 30 8

0 20 40 60 80 100 x 6 y=x

–20 R
4

(iii) (a) Daripada graf, julat bilangan B ialah 2 (3, 3) x + 3y = 12

From the graph, the range of the number of 0 2 4 6 8 10 12 x
B is

10 ≤ y ≤ 60

J43

(c) (i) Julat ialah 4 ≤ x ≤ 9. (b)

The range is 4 ≤ x ≤ 9. 3x + 4y = 400 y x – y = 15
100
(ii) K = 120x + 180y 80
Kos maksimum / Maximum cost 60
K = 120(9) + 180(9)
= RM2 700 40
2. (a) 50x + 60y ≤ 750
5x + 6y ≤ 75 20
75x + 45y ≤ 15 × 60
5x + 3y ≤ 60 –20 0 x
(b) –20 20 40 60 80 100 120 140 160 180

y

x + y = 40 3x + 4y = 360

12 (c) (i) Titik minimum = (20, 20)
10
8 Minimum point
6
4 Harga / Price = 30(20) + 40(20)
2
= RM1 400
-2 0
Baki = RM4000 − RM1 400 = RM2 600

(ii) Peruntukan yang tinggal = RM3 600
Allocation left = RM3 600

Maka/Hence

30x + 40y ≤ 3600

x 3x + 4y ≤ 360
2 4 6 8 10 12 14 16 18 20
Bilangan pen maksimum = 60
5x + 3y = 60
5x + 6y = 75 Maximum number of pens

(c) (i) y = 2x, Sudut KBAT

Dalam rantau, x maksimum = 4 dan y (a) x + y ≤ 20
80x + 50y ≤ 1 200
maksimum = 8 8x + 5y ≤ 120
In the region, x maximum = 4 and y maximum = (b) y
8
25
(ii) U = 60x + 30y, apabila x = 9, y = 5 akan 8x + 5y = 120

mendapat maksimum 20

U = 60(9) + 30(5)

= RM 690 15

3. (a) x + y ≥ 40

30x + 40y ≤ 4 000 10

3x + 4y ≤ 400

x − y ≤ 15 5

–5 0 x
5 10 15 20

x + y = 20

(c) (i) Julat / Range 0 ≤ x ≤ 11
(ii) Lukis x = 2y, maksimum kambing y = 5

J44

BAB (ii) Apabila t = 2 s, v(2) = (2 − 2)(3(2) − 4)

8 Kinematik Gerakan Linear = 0 m s−1 (zarah berhenti seketika)

Kinematics of Linear Motion (the particle stops instantaneously)

(iii) Apabila t = 5 s, v(5) = (2 − 5)(3(5) − 4)

= −33 m s−1 (zarah bergerak ke arah kiri)

1. (a) (i) Jumlah jarak = jumlah semua perjalanan (the particle moves to the left)

Total distance = total all the journey (b) (i) Apabila t = 0 s, v(0) = 1 (0)2 – 2(0)
2
= (2 + 3 + 7) km

= 12 km = 0 m s−1 (zarah tidak bergerak)

(ii) Titik asal Cik Nor ialah rumahnya dan (the particle is not moving)

titik akhir juga rumahnya. Maka sesaran = (ii) Apabila t = 2 s, v(2) = 1 (2)2 – 2(2)
2
0 km.
= −2 m s−1 (zarah bergerak ke arah kiri)
Cik Nor’s original point is her house and the end

point is also her house. Hence, displacement = (the particle moves to the left)
0 km.
(iii) Apabila t = 5 s, v(5) = 1 (5)2 – 2(5)
(b) (i) Jumlah jarak = jumlah semua perjalanan 2
1
Total distance = total all the journey = 2 2 m s−1

= (5 + 8) m

= 13 m (zarah bergerak ke arah kanan)

(ii) Titik asal zarah ialah di O dan titik akhirnya (the particle moves to the right)

ialah 3 m ke kiri O. (c) (i) Apabila t = 0 s, v(0) = 1 (2(0) − 3)2 − 4
2
Maka sesaran = −3 m. 1
2
The original point of the particle is at O and its = m s−1
end point is 3 m to the left of O.
(zarah bergerak ke arah kanan)
So, displacement = −3 m.
(the particle moves to the right)
2. (a) (i) Apabila t = 0 s, s(0) = 4 – 2(0) − 02 (ii) Apabila t = 1 s, v(1)
= 1
=4m = – 3 21 m s−1 2 (2(1) − 3)2 − 4

(zarah berada ke kanan O)
the particle is at the right of O)

(ii) Apabila t = 1 s, s(1) = 4 – 2(1) − 12 (zarah bergerak ke arah kiri)

=1m (the particle moves to the left)

(zarah berada ke kanan O) (iii) Apabila t = 2 s, v(2) = 1 (2(2) − 3)2 − 4
(the particle is at the right of O) = –3 12 m s−1 2

(iii) Apabila t = 2 s, s(2) = 4 – 2(2) − 22

= −4m (zarah bergerak ke arah kiri)

(zarah berada di titik O) (the particle moves to the left)
(the particle is at O)
4. (a) (i) Apabila t = 0 s, a(0) = 2 − 5(0) + 02
(b) (i) Apabila t = 1 s, s(1) = 2 + 4(1) − 2(1)2
= 2 m s−2
= 4 m (zarah berada ke kanan O)
(the particle is at the right of O) (halaju zarah bertambah)

(ii) Apabila t = 5 s, s(5) = 2 + 4(5) − 2(5)2 (the velocity is increasing)

= −28 m (zarah berada ke kiri O) (ii) Apabila t = 1 s, a 1  = 2 − 51 1  + 1 1 2
(the particle is at the left of O) 2 2 2 2

(iii) Apabila t = 1 s, s( 1 ) = 2 + 4( 1 ) − 2( 1 )2 = − 1 m s−2
2 2 2 2 4

= 3 1 m (zarah berada ke kanan O) (halaju zarah berkurang)
2
(the particle is at the right of O) (the velocity is decreasing)

(c) (i) Apabila t = 0 s, s(0) = (1 − 0)(3(0) + 4) (iii) Apabila t = 3 s, a(3) = 2 − 5(3 ) + (3)2

= 4 m (zarah berada ke kanan O) = −4 m s−2

(the particle is at the right of O) (halaju zarah berkurang)

1 1 1 1 (the velocity is decreasing)
3 3 3 3
(ii) Apabila t = s, s( ) = (1 − )(3( )+ 4) (b) (i) Apabila t = 1 s, a(1) = −2(1) + 6

1 = 4 m s−2
3
= 3 m (zarah berada ke kanan O) (halaju zarah bertambah)

(the particle is at the right of O) (the velocity is increasing)

(iii) Apabila t = 1.5 s, (ii) Apabila t = 3 s, a(3) = −2(3) + 6

s(1.5) = (1 − 1.5)(3(1.5) + 4) = 0 m s−2

= −4 1 m (zarah berada ke kiri O) (halaju zarah adalah malar)
4
(the velocity is constant)

(the particle is at the left of O) (iii) Apabila t = 4 s, a(4) = −2(4) + 6

3. (a) (i) Apabila t = 0 s, v(0) = (2 − 0)(3(0) − 4) = −2 m s−2

= −8 m s−1 (zarah bergerak ke arah kiri) (halaju zarah berkurang)

(the particle moves to the left) (the velocity is decreasing)

J45

(c) (i) Apabila t = 2 s, a(2) = −(2)2 + 4(2) (halaju zarah adalah malar)

= 4 m s−2 (the velocity is constant)

(halaju zarah bertambah) (iii) Apabila t = 6 s, a(6) = −(6)2 + 4(6)

(the velocity is increasing) = −12 m s−2

(ii) Apabila t = 4 s, a(4) = −(4)2 + 4(4) (halaju zarah berkurang)

= 0 m s−2 (the velocity is decreasing)

5. (a)

Lakaran graf sesaran- Jumlah jarak Jarak yang dilalui
masa
dilalui dalam saat ketiga
Displacement-time graph Total distance travelled
Distance travelled in the
s(t) third second

(a) s(t) = 3t − 4, Dari graf, Jarak dalam saat ketiga
0 ≤ t ≤ 3
jumlah jarak Distance in the third second
s(t) = 3t − 4
Graf ini ialah graf garis From the graph, total = s(3) − s(2)
lurus dengan kecerunan distance
3 dan titik pintasan-y = 3(3) − 4 − [3(2) − 4)]
ialah −4. =4+5=9m
5 =3m
This is a straight line graph
with the gradient of 3 and 03 t
the y-interception point is −4. –4

(b) s(t) = (t − 2)(t − 5), s(t) Dari graf, Jarak dalam saat ketiga
0≤t≤6
jumlah jarak total Distance in the third second

s(t) = (t − 2)(t − 5) From the graph, = s(3) − s(2)
distance = (3 − 2)(3 − 5) –
(2 − 2)(2 − 5)
= t2 − 7t + 10 10 = 10 + 2 41 + 2 14 + 4 = −2 m
= 18 12 m =2m
Graf ini ialah graf

minimum dengan punca- 7
2
punca 2 dan 5. t
02 5
This is a minimum graph
with the roots of 2 and 5. –2 14

Titik minimum

ds(t) = 2t − 7= 0
dt

t= 7 , s = – 2 41
2

(c) s(t) = −t2 + 4t − 3, s(t) Dari graf, jumlah jarak Jarak dalam saat ketiga
0 ≤ t ≤ 3 1
0 123 From the graph, total Distance in the third second

–3 distance = s(3) − s(2)

s(t) = −t2 + 4t − 3 =3+1+1 = (1 − 3)(3 − 3) −

= (1 − t)(t − 3) =5m (1 − 2)(2 − 3)
=1m
Graf ini ialah graf t

maksimum dengan

punca-punca 1 dan 3

This is a maximum graph

with the roots of 1 and 3.

Titik maksimum

ds(t) = −2t +4 = 0
dt

t = 2, s = 1

J46

6. (a) (i) Untuk sesaran positif, s > 0 (ii) Untuk halaju malar, pecutan mesti 0.
For a uniform velocity, the acceleration must be 0.
For positive displacement

s(t) = t2 − 3t > 0 Jadi, a = d2s = 2t − 4 = 0
dt2
t(t − 3) > 0

t > 3 t = 2

(ii) Apabila s = 10 Apabila t = 2, halaju malar
When t = 2, the velocity is uniform.
t2 − 3t = 10
v = 22 − 4(2) − 12 = −16 m s−1
t2 − 3t −10 = 0
ds
(t − 5)(t + 2) = 0 (b) dt = −3t2 + 6t + 9

t = 5 (i) Untuk sesaran maksimum atau minimum,

s(t) For maximum or minimum displacement,

ds =0
dt

−3t2 + 6t + 9 = 0

t2 − 2t − 3 = 0

0 3t (t + 1)(t − 3) = 0

t = 3

Apabila t = 3,

(b) (i) Untuk sesaran positif, s > 0. s = −33 + 3(3)2 + 9(3)

For positive displacement = 27 m (ke kanan O)

s(t) = t2 − 3t – 4 > 0 Untuk menentukan sesaran maksimum

(t − 4)(t + 1) > 0 atau minimum.

t > 4 To determine maximum or minimum

(ii) Apabila s = – 245 displacement.
4t2 −12t – 16 = −25
d2s = −6t + 6
4t2 −12t + 9 = 0 dt2
d2s
(2t − 3)(2t − 3) = 0 Apabila t = 6, dt2 = −30 < 0

Maka, s = 27 m ialah maksimum.
Hence, s = 27 m is maximum
t = 3
2 (ii) Untuk halaju malar, pecutan mesti 0
s(t)
For a uniform velocity, the acceleration must

be 0.

Jadi, a = d2s = −6t + 6 =0
dt2
t = 1
t
4 Apabila t = 1, halaju malar.
–4 When t = 1, the velocity is uniform.

ds v = −3(1)2 + 6(1) + 9
dt
7. (a) = t2 − 4t − 12 = 12 m s−1

(i) Untuk sesaran maksimum atau minimum, 8. (a) (i) ds = v = −6t2 + 12t
dt
For maximum or minimum displacement, Apabila t = 0, v = 0 m s−1 v

ds = 0 (ii) dv = −12t + 12
dt dt

t2 − 4t −12 = 0 Apabila t = 0, a = 12 m s−2

(t + 2)(t − 6) = 0 Apabila t = 2, 0 2t
a = −12(2) + 12
t = 6

Apabila t = 6, = −12 m s−2

s = (6)3 − 2(6)2 − 12(6) + 4 (iii) Apabila halaju maksimum atau minimum,
3
dv = 0.
= −68 m (ke kiri O) dt

Untuk menentukan sesaran maksimum When velocity is maximum or minimum,

atau minimum. dv = 0.
dt
To determine maximum or minimum
−12t + 12 = 0
displacement.
t = 1
d2s = 2t − 4
dt2 Untuk menentukan sama ada maksimum
d2s
Apabila t = 6, dt2 =8>0 atau minimum, kita guna

Maka, s = −68 ialah minimum. To determine whether the velocity is maximum
or minimum, we use
Hence, s = −68 is minimum

J47

d2v = −12 (< 0 maksimum) Maka / Hence,
dt2
v = 1 3 2 − 31 3  + 10
Maka, apabila t = 1, halaju maksimum = 2 2

−6(1)2 + 12 = 7 34 m s–1

= 6 m s−1 (ke arah kanan / to the right)

(b) (i) ds = 9t2 − 15t + 6 v 10. (a) (i) a = dv = 4 − t
dt dt

Apabila t = 0, v = 6 m s−1 dv = (4 − t) dt

(ii) ddvt = 18t − 15 6 ∫dv = ∫(4 − t) dt

Apabila t = 0, a = −15 m s−2 2 v = 4t – 1 t2 + c
3 2
Apabila t = 2,
1t A−5pa=b4il(a2/W) –he21n t = 2, v = −5. Maka/Hence
a = 18(2) − 15 (2)2 + c
= 21 m s−2 – 1
4

(iii) Apabila halaju maksimum atau minimum, c = −11

dv = 0. v = 4t – 1 t2 − 11
dt 2

18t − 15 = 0 Apabila/When a = 2, 4 − t = 2

t = 5 t = 2 1
6 2
Untuk menentukan sama ada maksimum v = 4(2) – (2)2 − 11

atau minimum, kita guna v = −5 m s−1

To determine whether the velocity is maximum

or minimum, we use (ii) s = ∫v dt

d2v = 18 ( > 0 minimum) =∫[ –1 t2 + 4t −11] dt
dt2 2
5
Maka, apabila t= 6 , halaju mininum = –1
6
9 5 22 − 15 5 2 + 6 = t3 + 2t2 − 11t + c
6 6
= – 41 m s−1 ( ke arah kiri / to the left) sA=pa–b61ilat/3W+h2etn2 t = 0, s = 0, c = 0
− 11t

9. (a) (i) a = dv = 12 − 4t Apabila halaju maksimum atau minimum.
dt
When the velocity is maximum or minimum.
dv = (12 − 4t) dt
dv = 0, 4 − t = 0
∫dv = ∫12t − 4t dt dt

v = 12t − 2t2 + c t = 4

Apabila t = 0, v = −2. Maka, c = −2 s = –1 (4)3 + 2(4)2 − 11(4)
6
v = 12t − 2t2 − 2 = – 22 23 m

Apabila t = 2, v = 12(2) − 2(2)2 – 2

= 14 m s−1 (b) (i) a = dv = 6t
dt
(ii) Apabila pecutan ialah sifar, dv = (6t) dt

When the acceleration is zero

12 − 4t = 0 ∫dv = ∫(6t) dt

t = 3 v = 3t2 + c

Maka / Hence, Apabila/When t = 1, v = 4. Maka/Hence

v = 12(3) − 2(3)2 − 2 4 = 3(1)2 + c

= 16 m s−1 c = 1
dv
(b) (i) a = dt = 2t − 3 v = 3t2 + 1

dv = (2t − 3)dt tA p=ab13ila a = 2, 6t = 2

∫dv = ∫2t − 3 dt v = 3 1 22 + 1
3
v = t2 − 3t + c

Apabila t = 0, v = 10. Maka, c = 10

v = t2 − 3t + 10 v = 1 13 m s−1

Apabila t = 2, v = (2)2 − 3(2) + 10

= 8 m s−1 (ii) s = ∫v dt
= ∫[3t2 + 1] dt
(ii) Apabila pecutan ialah sifar,

When the acceleration is zero

2 t − 3 = 0 = t3 + t + c
3 Apabila t = 0, s = 0, c = 0
t = 2

J48

s = t3 + t (iii) Apabila balik ke O, s = 0

Apabila halaju maksimum atau minimum. 4t2 – 1 t3 = 0
3
When the velocity is maximum or minimum.
dv
dt = 0, 6t = 0 t2[4 – 1 t] = 0 v (8, 108
3
t = 0
t = 0 t = 12
Jadi/So s = 0
∫(iv) s = 6(8t – t2)dt
11. (a) v = 2t(6 − t) 0

Luas dari 3 hingga 6 ialah = [4t2 – 1 t3]6
3 0
The area from 3 to 6 is O 68 t

∫ s1 = 6(12t − 2t2)dt = 72 m
3
= − 2 t3]6 = [6[6]2 − 2 (6)3] − [6(3)2 − (b) (i) t = 0 v = 3
[6t2 3 3 3 2

2 (3)3] ∫ = a + bt
3 b
v = at + 2 t2 + c
= 36 m

Luas dari t = 6 ke t = 8 t = 0 v = 3 c= 3
2 2
The area from t = 6 to t = 8.

∫ s2 = 8 (12t − 2t2)dt v = at + b t2 + 3
6 2 2
= − 2 t3]8 = [6[8]2 − 2 (8)3] − [6(6)2 −
[6t2 3 6 3 7
2
2 (6)3] = –29 31 Apabila t = 2 v =
3
27 = 2a + 2b + 3
Jumlah jarak/Total distance = 36 + 2 9 13 2

= 6 5 31 m a + b = 1

v = 0, t = 3

(b) a = dv = (2 − t)2 0 = 3a + 9b + 3
dt 2 2

dv = (2 − t)2 dt –3 = 6a + 9b

∫dv = ∫(2 − t)2 dt –1 = 2a + 3b = 2(1 – b) + 3b

(2 – t)3 –3 = b, a = 4
–3
v = + c ∴ 4t – 3 t2 + 3
2 2
Apabila/When t = 2, maka/hence
(ii) s = 2 4t – 3 t2 + 3 dt
c = 0, 1 2 2
∫  2 2
v = (2 – t)3 3 4 = 1 3 1
–3 2t2 – 2 t3 + 2 t

Jarak/Distance = 3(2(2)2 – 4 + 3) – 2 – 1 + 3 24
2 2
∫ s = 4 (2 – t)3 dt
2 –3 = 4 m

=[ (2 – t)4 ]4 = 4 m (c) (i) v = 6t – kt2
3(4) 2 3
dv
12. (a) (i) a = 8 – 2t a = dt = 6 – 2kt

v = 8t – t2 + c Apabila t = 2, a = 2

t = 0 v = 0 c = 0 2 = 6 – 2k(2)

v = 8t – t2 –4 = –4k

Apabila vmak, 8 – 2t = 0 k = 1
t = 4
v = 6t – t2
∴ v = 8(4) – 42 = 16 m s–1
(ii) ddvt = 6 – 2t = 0
(ii) s = 4t2 – t3 +c
3 t = 3

t = 0, s = 0 c = 0 Apabila t = 3

s = 4t2 – 1 t3 v = 6(3) – 32
3
= 18 – 9 = 9
Untuk smak,8 t – t2 = 0
[8 – t] = 0 vmak = 9 m s–1

t = 0 ; t = 8 (iii) s = 3t2 – t3 +c
3
Apabila t = 8, s = 4(8)2 – 1 (8)3
3 t = 0; s = 0, c = 0

= 8 5 13 m s = 3t2 – t3
3

J49

Apabila kembali ke O ∫(d) s = 3( 5 t2 – 8t)dt
2
When return to 0. 0

s = 0 = t23 – t  = [ 5 t3 – 4t2]3
3 6 0

∴t=9s = 13.5 m

(d) (i) s = 0 ; v = –10 2. (a) P Q
O A
a = p – 3t2 13 m

v = pt – t3 + c

t = 0 vP = 6t2 + 12 SQ = 2t3 – t
ddstQ = 6t2 – 1
v = –10

c = –10

v = pt – t3 – 10 dsQ
dt
s = pt2 – 1 t 4 – 10t + c Apabila t = 0 = –1
2 4
p 1 ∴vQ = –1 m s–1
t = 1 s1 = 2 – 4 – 10 + c
∫ 1
t = 2 s2 = 4 p – 4 – 20 + c (b) s1 = 6 (6t2 – 1)dt vQ
2 0
0
 1
s2 – s1 = 3 p – 1 3 34 = 4 41 0
2 = (2t3 – t) 6

32 p = 18 = ( 2 – 1 ) 13 t
6 6 Ίෆ6
6

p = 12 =  –2 = 2 m
36 36
(ii) v = 12t – t3 – 10

a = 12 – 3t2 = 0 3
∫ s2 = 1 (6t2 – 1)dt
t = 2 6

vmak = 12(2) – 23 – 10 = 3(2t3 – t)4 31
= 6 m s–1

6

Praktis SPM 8 = 51 – 3 6 2 – 1 4
6 6
Kertas 1
= 51 – 3– 2 4
1. (a) v = at2 – 8t 36

a= dv = 2at – 8 = 51 + 2
dt 36

Apabila t = 4, a = 12 Jumlah jarak = 51 + 4
12 = 8a – 8 36
Total distance

a = 5 = 51.54 m
2
(c) sp = 2t3 + 12t + c
(b) v Apabila t = 0 sp = –13
sp = 2t3 + 12t – 13
Apabila berselisah

sp = sQ

2t3 + 12t – 13 = 2t3 – t

0 8 16 t 13t = 13
55
t = 1

v= 5 t2 – 8t sp = 2 + 12 – 13 = 1 m
2 sQ = 2 – 1 = 1 m dari A

v = t[ 5 t – 8] 3. (a) Q P
2
B 0 6m A

∴ 8 <t< 16 Sp = 6 + 6t – 6t2
5 5 Bila t = 0, Sp = 6 m
SQ = 4t2 – 6t – 10
(c) 5 t2 – 8t = 0 Bila t = 0, SQ = –10 m
2 ∴ jarak AB = 10 + 6 = 16 m
(b) Sp = SQ
t[ 52 t – 8] = 0 6 + 6t – 6t2= 4t2 – 6t – 10
16 10t2 – 12t – 16 = 0
t = 5 s

J50

5t2 – 6t – 8 = 0 Sudut KBAT

(5t + 4)(t – 2) = 0 v (a) v = 36t – 18t2
Apabila berhenti
∴ t =2

(c) vp = 6 – 12t 6 v = 0 = 36t – 18t2
Jarak oleh P 0 = 18t[2 – t]
1 1 1 3 01
= 2 (6)( 2 ) + 2 ( 2 9 2 t=0;2 AB

)(18) t ∴t = 2 s ke B

= 3 + 27 = 30 = 15 m (b) a = dv = 36 – 36t
2 2 2 dt

4. (a) v = t3 – 5t2 + 4t Apabila t = 2, a = 36 – 72
dv
a = dt = 3t2 – 10t + 4 = –36 m s–2

Apabila t = 0, a = 4 m s–2 (c) vmak apabila dv = 0, t = 1
dt
(b) 3t2 – 10t + 4 < 12 v
v maks = 36(1) – 18(1)2 0
3t2 – 10t – 8 < 0 = 18 m s–1 (8, 1

(3t + 2)(t – 4) < 0 ∫(d) SAB = 2(36t – 18t2)dt

0 ≤ t < 4 0

(c) v = 0 = t[t2 – 5t + 4] = [18t2 – 6t3]2

t[(t – 4)(t – 1) = 0 0 t
2
t = 0 t = 1; 4 = 24 m

∴ t = 1 atau 4 (e) Apabila s = 18t2 – 6t3 = 0

∫ ∫(d) s = 1 t3 – 5t2 + 4t dt + 4 t3 – 5t2 + 4t dt 6t2[3 – t] = 0
01
t = 0 ; 3

 4  4 = t4 – 5 t3 + 2t2 1 + t4 – 5 t3 + 2t2 4 ∴ t = 3
4 3 0 4 3 1

= 1 1 – 5 + 22 + 1 256 – 320 + 322
4 3 4 3

= 7 + –1 1 14  v
12

= 1 1 65 m

01 4t

J51

Kertas PRA SPM (b) 32x + 3 + 6(3x) – 1 = 0
Kertas 1/Paper 1
Bahagian A/Section A 27·32x + 6(3x) – 1 = 0
1. (a) (i)
(9·3x – 1)(3·3x + 1) = 0

3x = 1 3x = – 31 = –(3)–1
9

Merah / Red ➤1 3x = 3–2 Tidak diterima

Putih / White ➤4 x = –2 Rejected

Hitam / Black ➤ 5 4. (a) 2x – 3y + 1 = 0 ...➀

x – 2y = 1

x = 1 + 2y ...➁

(ii) Satu kepada satu 2[1 + 2y] – 3y + 1 = 0

One to one 2 + 4y – 3y + 1 = 0

(b) (i) k = 0 y = –3

(ii) f(1) = 8 +b= 163 ...➀ x = 1 – 6 = –5
a
∴Titik persilangan (–5, –3)
f(2) = 8 +b = 5 ...➁
2a 6 Intersection point

Persamaan J3 (y + 3) = 2(x + 5)
Equation J3 y = 2x + 7
➀ – ➁ 4 = 8
a 6 (b) y = 2x + 7

2a = 6 J2 : y = 2 x + 1 y
3 3
a = 3

b= 13 – 8 = – 36 J1 : y = 1 x – 1 A7
6 3 2 2
B 1
= – 21 AB : BC 3

7 − 1 2 :  1 – − 1 22 x
3 3 2 0
8 1 1
(iii) f(h) = 3h – 2 = 6 230 : 5 =8:1 C – 1
6 2

38h = 4 5. (a) 2, 1.5, ...
6
3
h = 4 a=2,r= 4
2. (a) (i) x2 + px + 3q = 0
a 2
α + 2α = –p S= 1–r = – 3 =8m
∞ 4
3α = –p 1

α(2α) = 3q Tidak sampai tempat makanan.

2 –p 2 = 3q Not reach the food.
3
(b) x = y y = x2
2p2 = 27q 2 x 2

q = 2p2 a=2
27
a+d=x⇒d=x–2

1 2 (ii) x2 + px + 3q = x+ p 2– p2 + 3q a + 11d = y 2 + 11(x – 2) = x2
2 4 2

1 2=x+ p 2– p2 + 2p2 2 + 11x – 22 = x2
2 4 9 2

= 1x + p 22 p2 2(11x – 20) = x2
2 36
– x2 – 22x + 40 = 0

∴ Titik pusingan –p –p2 (x – 20)(x – 2) = 0
2 36
1 2 Turning point , x = 20 ; 2

y = 200, 2

(b) x(x – 2) > x + 4 2, 20, 200, x = 2, y = 2

x2 – 2x – x – 4 > 0 ∴x = 20, y = 200 tidak diterima
6. (a) α, α + 4
x2 – 3x – 4 > 0
α + α + 4 = 2k + 4
(x + 1)(x – 4) > 0 2α + 4 = 2k + 4

Julat x /Range of x

∴x < –1, x > 4 α = k

3. (a) log2 8 + 2 log2 p = 5 α(α + 4) = 3k + 2

3 + log2p2 = 5 k(k + 4) = 3k + 2

log2p2 = 2 k2 + k – 2 = 0

p2 = 22 (k – 1)(k + 2) = 0

p = 2 k = 1 ; –2

J52

(b) Jika k = 1 x2 – 6x + 5 = 0 ∫10. (a) (i) 4 f(x)dx
0
(x –1)(x – 5) = 0 = 6 + 4 = 10

Jika k = –2 x2 – 4 = 0 ∫ (ii) 0 f(x)dx
–2
( a) PPP→→→QRS === x = ±2
Q→R = = –4
7. –2~i + ~j

2~i + 5~j (iii) luas Q/ Area Q = 6 × 4 – 10

= 14 unit2

4~i ++7P→~jR (b) d2y = 1 – x2
Q→P dx2
= –21 +  2 2 =  4 2
5 4 dy =x– 1 x3 + c
dx 3
Q→S = Q→P + P→S = –212 +  4 2 =  6 2
7 6 Apabila x = 1, dy = –1
Q→R = 4(~i + ~j ) Q→S = dx
6(~i + ~j )
6 Q→R 1 – 1 + c= –1
= 4 3

Q→S = 3 Q→R c = –1 – 2 = –5
2 3 3

dy 1 5
Oleh sebab titik Q ialah titik sepunya dan dx =x– 3 x3 – 3

Q→S = 3 Q→R. Q, R dan S adalah segaris. y= x2 – 1 x 4 – 5 x + c1
2 is a common point and Q→S = 2 12 3
Since 3 Q→R.
point Q 2 Q, (1, 4); 4 = 1 – 1 – 5 +c
2 12 3
R and S are collinear.

(b) ~v q = 5~i + 36~~jj , c1 = 21
~v r = a~i + 4
6 3
5 = a ∴y= x2 – 1 x4 – 5 x + 21
2 12 3 4

6a = 15 11. (a) np2 = 56

a = 5 n! = 56
2 – 2)!
(n
8. (a) (i) y = 2x2 − 8
n(n – 1) = 56
y = – 8 +2
x2 x2 n2 – n – 56 = 0

Y = y , X = 1 , m = −8 , c = 2 (n – 8)(n + 7) = 0
x2 x2
n = 8

(ii) k = −8(2) + 2 (b) (i) 9! = 362 880

= −14 (ii) 27!2! ! = 1 260

(b) y = axn 12. (a) P[X = 1] + P[X = 2] + P[X = 3] +

log y = log a + n log x P[X = 4] + P[X = 5] = 1

n = 2 log a = 1.3

a ≈ 20 1 + 2 + 3 + 4 + 5 = 1
2k 2k 2k 2k 2k
9. (a) y = 1 = 1
(x + 2)(x – 3) (x2 – x – 6) 15 15
2k = 1 k = 2
dy = –(x2 – x – 6)–2(2x – 1)
dx
(b)
= 1 – 2x x 1234 5
(x2 – x – 6)2

dy + y2(2x – 1) P(X = x) 1234 5
dx 15 15 15 15 15

= 1 – 2x + (x (2x – 1)
(x2 – x – 6)2 + 2)2(x – 3)2
(c) P[2 < x ≤ 5]
0
= (x + 2)2(x – 3)2 = 0 = 1 + 4 + 1 = 12 = 4
5 15 3 15 5
(b) V = 100 – 2t – t2
(i) Apabila t = 0, V = 100 cm3 Bahagian B/Section B

(ii) ddvt = –2 – 2t dv = –2 –2(4) 13. (a) sin B = sin 30°
Apabila t = 4, dt 10 7

= –10 cm3 s–1 B = 134° 25’

J53

(b) P Panjang lengkok BD = 90° + 11.33’ × 2p(1)
360°
B Length of arc BD
7 cm
= 1.77 cm
7 cm
A 10 cm 30° C CB = 2.52 – 0.52 = 2.45 cm
Perimeter = 2.05 + 1.77 + 2.45

= 6.27 cm

(c) ∠APC = 45°35’ ; ∠PAC = 104°25’ (b) luas berlorek = luas trapezium OABC – sektor

PC2 = 72 + 102 – 2(7)(10) kos/ cos 104°25’ ODC – sektor ADB

PC = 13.56 cm Shaded area = area of trapezium OABC – sector ODC
– sector ADB

(d) Luas/ Area ABC = 1 [1 + 1.5]2.45 – 78°27’ × p(1.5)2 –
2 360°
= 1 (10)(7) sin(150° – 134°35’)
2 10316°03°3’ × p(1)2

= 9.4 cm2

14. (a) tan A = 3 = 0.64 cm2

tan(A + B) = tan A + tan B =1 Kertas 2/Paper 2
1 – tan A tan B

3 + tan B = 1 – 3 tan B Bahagian A/Section A
(1 + 3)tan B = 1 – 3
1. x + 2y – z = 4 ...➀
...➁
tan B = (1 – 3)(1 – 3) 2x + y + z = –2 ...➃
(1 + 3)(1 – 3) ...➂
➀ + ➁ 3x + 3y = 2
...➄
1–2 3+3 x + 2y + z = 2
1–3
= = 3–2 ➀ + ➂ 2x + 4y = 6

(i) y = a sin bx + c x + 2y = 3

1 3[3 – 2y] + 3y = 2
2
a =3 ,b = , c = −1 9 – 6y + 3y = 2

(ii) y 7 = 3y

3 y = 7
3
7
2 x = 3 – 2 3 2

1 = –5
x 3

0 S S 2S –5 + 14 – 4 = z
–1 2 3 3

x(a sin bx + c) = π z = –1

a sin bx + c = π 2. (a) & (b)
x
f (x)
y = π
x
(–1, 5) 5
Apabila/When x = π, y = 1 (–3, 3) 4 f(x) = 3 – 2x
3 f –1(x)
Apabila/When x = π , y = 2 2
2 1

2 penyelesaian/ solutions x

15. (a) Kos/ cos θ = 0.5 –3 –2 –1–1 0 1 2 3 4 5
2.5 –2 (5,–1)
–3 (3,–3)
θ = 78°27’

Panjang lengkok CD = 78°27’ × 2p(1.5)
360°
Length of arc CD (c) Domain bagi f−1(x)

= 2.05 cm Domain for f −1(x)

−3 ≤ x ≤ 5

3. (a) y = x2 + (1 – p)x + 2p

(1 – p)2 + 2p – 1–p 2
4 2
 2 y = x +
– (1
O D – p)2 = 6
0.5 cm T 2.5 cm 2p 4
A

1.5 cm 1 cm 8p – (1 – 2p + p2) = 24
P B
Q p2 – 10p + 25 = 0
C
(p – 5)(p – 5) = 0

p = 5

J54

(b) (i) L(x) = 2x(4 – x) 6. (a) A→B = A→D + D→B

= 8x – 2x2 = A3~→i ~Di+++2~~jjD→C+ 2~i – ~j
=
(ii) dLd(xx) = 8 – 4x = 0
x = 2 A→C =

Apabila x = 2 L(2) = 8(2) – 2(2)2 = 9~i ~i++23~j ~j+ 8~i + ~j
=
= 16 – 8

= 8 cm2 (b) A→C = 3A→B
Ax E B

2x A, B dan C segaris
G A, B and C collinear
4
F AB : BC = 2 : 1

D4 C (c)

D

4. (a) (i) y = x + 4 h

y = 4 + 3x – x2 A B1 C
2
x + 4 = 4 + 3x – x2
x2 – 2x = 0 ∆ADC = 1 ACh = 24
x (x – 2) = 0 2
x = 0 ; 2
∴∆ADB = 2 × 24 = 16 unit2
A(0, 4) B(2, 6) 3

2(4 + 3x – x2)dx – 1 7. (a) y = a x + b
2 x
0
(ii) Luas/Area = [4 + 6]2 y x = ax + b

∫ 3 – x3 2 – 10
  = 2 3
4x + x2 0

= 11 13 – 10 = 1 13 unit2 (b)

x 12 3 4 5
13.9
∫(b) Isipadu = 1 y2dx0 y x 5.9 7.9 10.0 12
Volume
= 1(4 + 3x – x2)2 dx
∫ 0 y

∫= 1(16 + 24x + x2 – 6x3 + x4)dx
0

 4= 1 – 3 + 1 1 14
16x + 12x2 + 3 x3 2 x4 5 x5 0 12
10
= 2 7 310 unit3
∴ isipadu sebenar = 13 3610 unit3 8
6
5. (a) (i) p = 0.01, n = 1000 4
2
m = np = 1000(0.01)

= 10

σ = npq = 1000(0.01)(0.99)

= 3.15

(ii) P[x ≥ 1] = 1 – P[X = 0]

= 1 – 5C0(0.01)0(0.99)5

= 0.049 0 1234 x

(b) (i) m = 200, σ = 40 5
P[ 150 ≤ z ≤ 180]
(c) (i) b = 4
150 – 200 180 – 200
= P 40 < z< 40 4 a = 14 – 4 = 2
5
= P – 54 < z < – 21 4
(ii) Apabila/When x = 2.5

= 0.2029 y x = 9
(ii) P[X > 230] = y = 9 = 5.69
P[Z > 230 – 200 ] 2.5
40
3
= P[Z > 4 ]

= 0.2266 Bahagian B/Section B

Bilangan pekerja = 8 8. (a) 7 tan x = 4 tan (45° – x)
Number of wokers 0.2266
= 4[1 – tan x]
1 + tan x
 35

J55

7 tan x + 7 tan2 x = 4 – 4 tan x (b) PAQ = 90°
7 tan2x + 11 tan x – 4 = 0 y – 5 y+2
mPA = x + 1 , mQA = x–3

tan x = –11 ± 112 – 4(7)(–4) y ( y – 5 )( y + 2 ) = –1
14 16°56′ x + 1 x – 3

tan x = –11 + 233 x y2 – 3y – 10 = –(x2 – 2x – 3)
14
16°56′ y2 + x2 – 2x – 3y – 13 = 0
y
x = 16° 56’, 196°56’ Jika/If x = 5

–11 – 233 y2 + 25 – 10 – 3y – 13 = 0
tan 14
x= y2 – 3y + 2 = 0

61°56′ x (y – 1)(y – 2) = 0
61°56′
x = 118°4’, 298°4’ y = 1 ; 2

(b) y (5, 1) ialah R

2 y = 1 + kos x 10. (a) (i) S2n = S3n – S2n

1 y = kos x (2a + (2n – 2S2n = S3n [2a + (3n – 1)d]
2[ 22n 1)d] = 3n
2

y = 0.5 a = 12, d = 3

0 S x 4n(24 + 6n – 3) = 3n[24 + 9n – 3]
2S
84 + 24n = 63 + 27n

–1 y = sin 2x 21 = 3n

n = 7

(i) 4 penyelesaian/solutions (ii) Tn = a + (n – 1)d = 48
12 + 3(n – 1) = 48
(ii) 4 penyelesaian/solutions
36
(iii) sin 2x = 1 + kos/ cos x (n – 1) = 3 = 12

2 penyelesaian/solutions n = 13

9. (a) (i) 5x = 2 – y (b) ar – a = 20

y = –5x + 2 ar3 – ar = 15

Titik tengah PR = (2, 3) a(r – 1) = 20 ...➀

∴Persamaan QS / Equation QS ar(r2 – 1) = 15 ...➁

y – 3= –5(x – 2) ➁ ÷ ➀

y= –5x + 13 ...➀ r(r + 1) = 15
20
(ii) mPR = 5–1 = 4 = 2
–1 – 5 –6 –3 4r2 + 4r = 3

Persamaan QR,/ Equation QS, 4r2 + 4r – 3 = 0

y – 1= 3 (x – 5) (2r + 3)(2r – 1) = 0
2
r = – 3 ; 1
3 15 2 2
y= 2 x – 2 +1
3 1
3 13 ∴a[– 2 –1] = 20 atau a( 2 – 1) = 20
2 2
y= x – ...➁ a = –8 a = –40

(iii) (a) ➀ = ➁ –5x + 13 = 3 x – 13 11. (a) COB 0.8π – π
2 2 = 2π – 3

–10x + 26 = 3x – 13 = 13π = 156°
15
39 = 13x

x = 3 BC2 = 82 + 82 – 2(8)(8) kos 156°

y = –2 BC = 15.65 cm

∴Q(3, –2) (b) Perimeter = 6 π(8) + (8) + 8 + π (8)
5 3
3 2+ x = 2 –2 + y = 3
x = 1 2 = 54.54 cm

S = (1, 8) y = 8 (c) Luas = 1 (8)2 13π 2 – 1 (8)2 π 2
2 15 2 3
Area B
  ( b) la ureaas == 2121 13π
–1 3 5 1 –1 = 323 15 – π 4
5 –2 1 8 5 3
0.8 S
|2 + 3 + 40 + 5 – 15 + = 256π = 17 π cm2 A O
15 15
10 –1 + 8| 60°

= 1 |52| = 26 unit2 C
2

J56

Bahagian C/Section C 14. (a) P20 × 100 = n
P18
12. ( a) v= 2(3t2 – 2t – 1)
dv = a = 12t – 4
dt 24 × 100 = n = 120
20
Apabila/When t = 0 a = –4 m s–2
3(120) + 4(125) + 130 m + 110
(b) Apabila/When dv = 0 t= 1 (b) (8 + m) = 125
dt 3

∴v = 2[31 1 2 – 21 1  – 1] 970 + 130 m = 1000 + 125 m
3 3
5 m = 30

= –2 23 m s–1 m = 6

(c) v = 0 3t2 – 2t – 1 = 0 (c) 110 × 120 = 132
100

(3t + 1)(t – 1) = 0 110 × 130 = 143
100
t = 1
132 × 3 + 125 × 4 + 143 × 6 + 110
(d) jumlah jarak/total distance I = 14

∫  ∫ = 1(6t2 – 4t – 2)dt + 2(6t2 – 4t – 2)dt = 133.14
01
P21
= (2t3 – 2t2 – 2t]1 + 2t3 – 2t2 – 2t 2 (d) P18 × 100 = 143

  3 40 1

= |–2| + 6 = 8 m P18 = 650 × 100 = RM454.55
143
13. (a) 25x + 50y ≤ 50 000

x + 2y ≤ 2 000 ... ➀ 15. (a) AM2 = 52 + 42
... ➁ AM = 6.4 cm
150x + 50y ≤ 75 000 (b) AR2 = 62 + PR2
= 36 + 102 + 82
6x + 2y ≤ 3 000 AR = 14.14 cm
(c) MR2 = 42 + 62 + 52
3x + y ≤ 1 500 MR = 8.77 cm
6.42 = 77 + 200 – 2 77 × 200 kos ARM
y = 2x

(b)

y

2000 y = 2x kos ARM = 77 + 200 – 41
2 77 × 200
3x + y = 1500 cos ARM

ARM = 18°2’

(d) MP2 = 36 + 41

1000 (200, 900) MP = 77 = 8.77

PR = 100 + 64 = 12.81

R (300, 600) x + 2y = 2000 12.81 + 8.77 + 8.77
S = 2

x = 15.17

0 1000 2000 ∆PMR =

15.17(15.17 − 8.77)2(15.17 − 12.81)

(c) (i) Gred/ Grade A = 300 , Gred/ Grade B = 600 = 38.29 cm2

(ii) p = 4x + 2y DC

2y = −4x + p

y = −2x + p 8 cm M
2
A BS T R
p = 4(200) + 2(900) 5 cm

= 800 + 1 800 6 cm

= RM2 600

P 10 cm Q

J57

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

BAB 1: SUKATAN MEMBULAT

Tahap Penguasaan Tafsiran

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian
masalah rutin yang kompleks

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian
masalah bukan rutin secara kreatif

1. Selesaikan setiap yang berikut.

Solve each of the following. TP 5

R 1
Q

PO S

Bentuk PQRSOP yang ditunjukkan dalam rajah terdiri daripada sebuah sektor bulatan ROS dengan pusat O LEMBARAN PBD

menyambung dengan satu sektor POQ daripada satu bulatan berpusat O. Diberi bahawa panjang lengkok

RS = 4.5 cm, ∠ROS = 0.55 radian dan POS ialah satu garis lurus dengan panjang 13 cm. Cari

The shape of PQRSOP as shown in the diagram consists of a sector ROS of a circle with centre O joined to a sector POQ
of a different circle also with centre O. Given that the arc length RS = 4.5 cm, ∠ROS = 0.55 radians and POS is a straight
line of length 13 cm.
Find

(a) panjang OQ

the length of OQ.

(b) perimeter bagi PQRSOP.

the perimeter of PQRSOP.

(c) luas rantau berlorek itu.

the area of the shaded region.

B1

LEMBARAN PBD   Matematik Tambahan  Tingkatan 5  Lembaran PBD

2. Selesaikan setiap masalah yang berikut.

Solve each of the following problems. TP 6

(a)

P

O
R
S

2T

Rajah menunjukkan dua sektor, POS daripada bulatan berpusat O dengan jejari 18 cm dan PRT dengan
pusat R dan ORT ialah garis lurus. Diberi bahawa ∠POR ialah 1.5 radian dan OR = 8 cm. Cari

The diagram shows two sectors, POS of a circle with centre O and radius 18 cm and PRT with centre R and ORT is a
straight line. Given that ∠POR is 1.5 radians and OR = 8 cm. Find

(i) jejari sektor PRS

the radius of the sector PRS.

(ii) perimeter rantau berlorek.

the perimeter of the shaded region.

(iii) luas rantau berlorek itu.

the area of the shaded region.

B2

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

(b) Rajah menunjukkan sebuah kotak kek yang digunakan oleh sebuah kedai.

Kotak berbentuk prisma tegak dengan tinggi h cm. Luas keratan rentas adalah r
berbentuk sebuah sektor bulatan dengan jejari r cm bercangkum pada sudut 1.2

1.2 radian. Isi padu kotak ialah 216 cm3. h

The diagram shows a cake box which is used by a shop. The box is a right

prism of height h cm. The cross-sectional area is a sector of a circle with radius
r cm and subtended at an angle of 1.2 radians. The volume of the box is
216 cm3.
1152
(i) Tunjukkan bahawa luas permukaan kotak, S cm2 diberi oleh 1.2r2 + r .

Show that the surface area of the box, S cm2 is given by 1.2r2 + 1152 .
r

(ii) Sekeping kek dengan keratan rentas ialah satu sektor bulatan dengan panjang lengkok 6.5 cm 3

dan jejari 5 cm akan dimasuk ke dalam kotak. Bolehkah kek ini dapat dimasukkan ke dalam

kotak jika tinggi kotak ialah 10 cm? Tunjukkan langkah pengiraan anda.

A piece of cake whose cross section is a sector of a circle with arc length 6.5 cm and radius 5 cm is to be put
inside the box. Can this cake be put in the box if the height of the box is 10 cm? Show your working steps.

LEMBARAN PBD

B3

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

BAB 2: PEMBEZAAN

Tahap Penguasaan Tafsiran

4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah
rutin yang mudah.

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah
rutin yang kompleks.

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah
bukan rutin secara kreatif.

1. Selesaikan setiap yang berikut.

4 Solve each of the following. TP 4
(a) Nilaikan had had x – 4 . x2 – a2
x → 16 x – 16 (b) Cari pemalar a dengan keadaan had x + a = −6

x → –4

Evaluate the limit lim x – 4 . Find the constant a such that lim x2 – a2 = −6
x → 16 x – 16 x → –4 x+a


LEMBARAN PBD 2. Selesaikan setiap masalah yang berikut.

Solve each of the following problems. TP 5

(a) Sebuah tangga dengan panjang 12 m bersandar pada dinding mencancang. Hujung atas menggelongsor
ke bawah dinding manakala hujung bawah tangga menggelongsor jauh dari dinding sepanjang tanah
dengan kadar 0.5 m min−1. Apakah kadar hujung atas tangga menurun apabila jarak kaki tangga pada
tanah mengufuk ialah 4 m dari dinding?

A ladder of 12 m length leans against a vertical wall. Its top slides down the wall while the bottom moves away
from the wall along the level ground at a rate of 0.5 m min−1. At what rate is the top of the ladder falling when the
foot of the ladder on the ground is 4 m from the wall?

B4

(b) Dua buah keretapi meninggalkan stesen pada   Matematik Tambahan  Tingkatan 5  Lembaran PBD 

masa yang sama. Satu ke arah utara di atas (c) Jika f(x) = ax2 + bx + c. Diberi bahawa f(2) = 26,
f ’ (2) = 23 dan f’’(2) = 14. Cari
landasan dengan halaju 60 km j−1. Satu lagi
If f(x) = ax2 + bx + c. Given that f(2) = 26, f' (2) = 23
ke arah timur di atas landasan dengan halaju and f ''(2) = 14. Find

45 km j−1. Berapakah laju kedua-dua keretapi (i) a, b dan c.

itu bergerak antara satu sama lain, dalam a, b and c.

km j−1 ketika keretapi ke utara itu ialah (ii) f(−1)

120 km dari stesen. 5

Two trains leave a station at the same time. One
travels north on the track at a velocity of 60 km j−1.
The other train travels east on the track at a velocity
of 45 km h−1. How fast are they travelling away from
each other, in km h−1 when the northbound train is
120 km from the station.

3. Selesaikan setiap masalah yang berikut. LEMBARAN PBD

Solve each of the following problems. TP 6

(a) Sebuah bekas berbentuk silinder mempunyai isi padu 28π m3. Bahan untuk membuat penutup bekas
berharga RM5 meter persegi dan bahan digunakan membuat sekeliling dan tapak berharga RM2
setiap meter persegi. Apakah dimensi bekas supaya harga kos bekas adalah minimum?

A storage container of a right circular cylinder has a volume of 28π m3. The material that is used to make the cover
of the container costs RM5 per square and the material that is used to make the surrounding and the base costs RM2
for every square metre. What is the dimension of the container so that the cost of the container is minimum?

B5

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

(b) Diberi y = 4 , cari
x
4
Given y = x , find

(i) perubahan hampir dalam y apabila x menokok dari 4 ke 4.05.

the approximate change in y when x increases from 4 to 4.05

(ii) peratus perubahan dalam y, dalam sebutan p jika x berubah dengan 2p%.

the percentage change in y, in terms of p if x changes by 2p%.

6

LEMBARAN PBD BAB 3: PENGAMIRAN

Tahap Penguasaan Tafsiran

4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengamiran dalam konteks penyelesaian masalah
rutin yang mudah.

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengamiran dalam konteks penyelesaian masalah
rutin yang kompleks.

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengamiran dalam konteks penyelesaian masalah
bukan rutin secara kreatif.

1. Selesaikan setiap yang berikut.

Solve each of the following. TP 4

Cari pengamiran tak tentu bagi setiap yang berikut. (b) ∫(2 −42x)3dx

Find the indefinite integral for each of the following.

(a) ∫x(2 − x)(x − 1)dx

B6

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

2. Selesaikan setiap masalah yang berikut.

Solve each of the following problems. TP 5

Fungsi kecerunan lengkung ialah (4x – 2) dan lengkung itu melalui titik A(−1, 3). Cari

The gradient function of the curve is (4x − 2) and the curve passes through A(−1, 3). Find

(a) persamaan normal pada A.

the equation of the normal at A.

(b) persamaan lengkung itu.

the equation of the curve.

7

3. Selesaikan setiap masalah yang berikut. LEMBARAN PBD

Solve each of the following problems. TP 6

∫ ∫(a) Diberi 6 g(x)dx = 14 dan 4 g(x)dx = −4, cari
–3 0

∫ ∫ Given6 = 14 4 g(x)dx = –4, find
–3 g(x)dx and 0

∫ ∫ ∫(i) 4 g(x)dx jika/if 0 g(x)dx = 6 2g(x)dx
−3 –3 4

∫ ∫ ∫(ii) 0 2g(x)dx jika/if 0 g(x)dx = 6 g(x)dx
−3 4 4

∫ ∫(iii) 0 g(x)dx jika/if 6 g(x)dx = 2
−3 0

B7

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

(b) Rajah menunjukkan dua lengkung, x = 3 + y2 (c) Rajah menunjukkan lengkung x = (y − 2)2.
dan x = 2 − y2. Cari

The diagram shows two curves, x = 3 + y2 and The diagram shows the curve x = (y − 2)2. Find
x = 2 − y2.
y
y 3

x = 2 – y2 x = 3 + y2 2
1
1
0 1 23 4 5 x x
-1
0 246

8 Cari/ Find (i) luas dibatasi oleh lengkung dan garis lurus

(i) luas dibatasi oleh dua lengkung, y = 1 dan y = – 12 x + 2.
y = −2.
the area bounded by the curve and the line
the area bounded by the two curves, y = 1 and
y = −2. y = – 21 x + 2.

(ii) isi padu dijanakan apabila luas dibatasi (ii) Isi padu dijanakan oleh rantau yang dibatasi

oleh lengkung x = 3 − y2 dan garis x = 5 oleh lengkung, paksi-x dan paksi-y apabila
diputarkan melalui 180° pada paksi-x
dalam sebutan π. diputarkan lengkap pada paksi-y dalam

the volume generated when area enclosed by sebutan π.
the curve x = 3 − y2 and the line x = 5 is rotated
through 180° about the x-axis in terms of π. the volume generated by the region bounded by the
LEMBARAN PBD curve , the x-axis and the y-axis when it is rotated
completely around the y-axis in terms of π.

B8

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

BAB 4: PILIH ATUR DAN GABUNGAN

Tahap Penguasaan Tafsiran

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam konteks
penyelesaian masalah rutin yang kompleks.

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam konteks
penyelesaian masalah bukan rutin secara kreatif.

1. Selesaikan setiap yang berikut.

Solve each of the following. TP 5

(a) Emma mempunyai 4 baju dan 5 seluar. 9

Emma has 4 blouses and 5 skirts.

(i) Berapa cara dia boleh memakai baju dan seluarnya jika tidak ada halangan.

How many ways can she wears a blouse to a skirt if there is no restriction.

(ii) Jika 2 daripada 4 baju ialah hitam dan 2 daripada 5 pasang seluarnya juga hitam. Berapa cara dia

boleh memakai baju dan seluar itu jika dia tidak memakai baju hitam dengan seluar hitam pada

masa yang sama.

If 2 of the 4 blouses are black and 2 of the 5 skirts are black. How many ways can she wear blouse and skirt
if she cannot wear her black blouse with her black skirt at the same time.

(b) LEMBARAN PBD

Rajah menunjukkan tiga biji guli biru yang sama dan tiga biji guli yang berlainan warna.
Cari bilangan cara menyusun semua guli pada satu baris jika

The diagram shows three identical blue marbles and three different coloured marbles.
Find the number of ways to arrange all the marbles in a row if

(i) tidak ada halangan./there is no restriction.
(ii) tiga biji guli biru yang sama mest diasingkan.

three identical blue marbles must be separated

(iii) mesti bermula dengan guli merah dan berakhir dengan guli kuning.

It must start with the red marble and ends with a yellow marble.

(iv) Berapa bilangan cara untuk memilih empat guli dengan warna berlainan.

How many ways to choose four different colour marbles.

B9

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

2. Selesaikan setiap yang berikut.

Solve each of the following. TP 6

10 (a) 10 kawan dalam satu kumpulan dijemput ke suatu jamuan hari jadi dan duduk pada satu meja bulat.

Berapa cara untuk menyusun semua jika

10 friends of a group are invited to a birthday party and to be seated at a round table.
How many ways to arrange all of them if

(i) tiga kawan tertentu mesti duduk sebelah menyebelah.

three particular friends must sit next to each other.

(ii) dua orang itu tidak boleh duduk bersama.

two of them cannot sit next to each other.

(iii) Berapa cara untuk menyusun mereka jika meja itu hanya boleh muat 8 orang sahaja.

How many ways to arrange them if the table can only accommodate 8 of them.

LEMBARAN PBD (b) Suatu kod empat digit akan dibentuk daripada digit-digit dari 0, 1, 3, 4, 5, 7 dan 8. Cari bilangan cara

membentuk kod jika

A four digit code is to be formed from the digits 0, 1, 3, 4, 5, 7 and 8. Find the number of ways to form the codes if

(i) kod tidak boleh bermula dengan 0 dan ulangan tidak dibenarkan.

the code cannot start with 0 and repetitions is not allowed.

(ii) kod mengandungi digit ganjil sahaja dan ulangan tidak dibenarkan.

the code contains only odd digits and repetitions is not allowed.

(iii) 0 digunakan sekali sahaja manakala digit lain boleh diulang.

0 is used once only and the rest of the digits can be repeated.

(iv) nilai kod lebih daripada 3 000 dan ulangan tidak dibenarkan.

the value of the code is more than 3 000 and repetitions is not allowed.

B10

  Matematik Tambahan  Tingkatan 5  Lembaran PBD LEMBARAN PBD

(c) Sebuah bakul mengandungi 4 biji merah, 3 biji biru, 4 biji hijau dan 3 biji kuning guli dengan setiap
guli ditulis dengan nombor berlainan. Berapa cara memilih 4 biji guli jika

A basket contains 4 red, 3 blue, 4 green and 3 yellow marbles each with a different number written on it. How
many ways to choose 4 marbles if

(i) semua mempunyai warna berlainan.

all have different colours.

(ii) sekurang-kurangnya 2 ialah merah.

at least 2 are red.

(iii) tiada guli merah dipilih tetapi sekurang-kurangnya satu guli ialah hijau.

there is no red marble chosen but at least one marble is green.

11

(d) (i) Pada suatu jamuan graduan, setiap graduan bersalam antara satu sama lain. Terdapat 253 salam.
Berapa graduan yang hadir dalam jamuan itu?

At a graduation party, every graduate shakes hands with each other. There are 253 handshakes. How many
graduates are present in the party?

(ii) Dari bahagian (i), jika 3 daripada graduan tidak bersalam antara satu sama lain tetapi bersalam
dengan semua yang lain, berapa salam dibuat?

From part (i), if 3 of them did not shake hands with each other but shake hands with the rest of them, how
many handshakes are made?

B11

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

(e) Terdapat n orang pengakap lelaki dan m orang pengakap perempuan. Jika dua orang lelaki dan dua
orang perempuan dipilih, bilangan gabungan ialah 160 dan jumlah orang pengakap ialah 26. Hitung
nilai n dan m jika n > m.

There are n boys scouts and m girls scouts. If two boy scouts and two girl scouts are chosen, the number of
combinations is 160 and the total number of scouts is 26. Calculate the value of n and of m if n > m.

12

LEMBARAN PBD BAB 5: TABURAN KEBARANGKALIAN

Tahap Penguasaan Tafsiran

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang taburan kebarangkalian dalam konteks
penyelesaian masalah rutin yang kompleks.

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang taburan kebarangkalian dalam konteks
penyelesaian masalah bukan rutin secara kreatif.

Selesaikan setiap yang berikut.

Solve each of the following. TP 5

1. (a) X ialah pemboleh ubah rawak diskret dengan taburan kebarangkalian berikut

X is a discrete random variable with the following probability distribution.

X=r 3 5 7 9
P(X = r) 0.3 0.2 p q

Diberi bahawa P(X = 3) = P(X = 7), cari

It is given that P(X = 3) = P(X = 7), cari

(i) nilai p dan nilai q

the value of p and of q.

   (ii) P(X ≥ 5).

(b) Seterusnya, lukis graf taburan kebarangkalian itu.

Then, draw a probability distribution graph

B12

  Matematik Tambahan  Tingkatan 5  Lembaran PBD LEMBARAN PBD



13

2. Didapati bahawa 3 daripada 100 pesakit mengalami kesan sampingan daripada suatu ubat tertentu. Jika
10 pesakit dipilih secara rawak dari kumpulan pesakit itu,

It is found that 3 out of 100 patients show side effects from a certain drug. If 10 patients are chosen at random from the
group of patients,

(a) cari kebarangkalian bahawa

find the probability that

(i) tepat separuhnya menunjukkan kesan sampingan daripada ubat tertentu itu.

exactly half of them show the side effects of the drug.

(ii) sekurang-kurangnya 2 menunjukkan kesan sampingan.

at least 2 of them show the side effects.

(b) Jika 2 500 pesakit menggunakan ubat tersebut, cari min dan sisihan piawai bilangan pesakit yang
menunjukkan kesan sampingan.

If there are 2 500 patients using the drug, find the mean and standard deviation of the number of patients who show
side effects.

B13

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

Selesaikan setiap yang berikut.

Solve each of the following. TP 6

  3.  Satu pemboleh ubah diskret X mempunyai fungi kebarangkalian P(X = x) = k(1 − x)2 untuk X = {−1, 0, 1, 2}.
A discrete random variable X has the probability function P(X = x) = k(1 − x)2 for X = {−1, 0, 1, 2}.

(a) Tunjukkan bahawa k = 1 .
6
Show that k = 1
6 .

(b) Cari P(X < 0) atau P(X > 1).
LEMBARAN PBD Find P(X < 0) or P(X > 1).

14 5(c) Lukis graf taburan kebarangkalian itu.

Draw the probability distribution graph.

4. Dalam sebuah sekolah, 60% daripada pelajar mempunyai akses internet di rumah mereka.

In a school, 60% of the students have internet access at home.

(a) Satu kumpulan 8 orang pelajar dipilih secara rawak. Cari kebarangkalian bahawa

A group of 8 students are chosen at random. Find the probability that

(i) 3 daripadanya tidak ada akses internet di rumah.

3 of them do not have internet access at home.

(ii) sekurang-kurangnya 5 daripada mereka ada akses internet di rumah.

at least 5 of them have internet access at home.

(b) Cari bilangan min pelajar yang mempunyai akses internet jika 200 orang pelajar dipilih secara rawak.

Find the mean of the number of students that have internet access if 200 students are chosen at random.

(c) Cari bilangan pelajar yang dipilih jika kebarangkalian bahawa semua ada akses internet di rumah ialah
0.07776.

Find the number of students chosen if the probability that all have internet access at home is 0.07776.

B14

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

5. Satu set data X mempunyai min 45 dan sisihan piawai 8.3 bertaburan normal.

A set of data X has a mean of 45 and standard deviation of 8.3 is normally distributed.

(a) Cari nilai X jika diberi jaraknya dari min ialah

Find the value of X given its distance from the mean is

(i) +1 sisihan piawai dari min.

+1 standard deviation from the mean.

(ii) −2 sisihan piawai dari min.

−2 standard deviation from the mean.

(b) Cari kebarangkalian bagi data yang dipilih secara rawak adalah lebih daripada 40.

Find the probability of the data chosen at random is greater than 40.

(c) Cari nilai n jika kebarangkalian data yang dipilih secara rawak lebih daripada n ialah 0.3589.

Find the value of n if the probability of the data chosen at random greater than n is 0.3589.
5 15

LEMBARAN PBD
  6. Tinggi sekumpulan perempuan bertaburan normal dengan min 142 cm. Jika 95% orang perempuan
mempunyai tinggi di antara 139 cm dan 145 cm, hitung

The heights of a group of girls are normally distributed with mean 142 cm. If 95% of the girls have heights between
139 cm and 145 cm, calculate

(a) sisihan piawai bagi kumpulan perempuan ini.

the standard deviation of this group of girls.

(b) kebarangkalian seorang perempuan dipilih secara rawak daripada kumpulan ini mempunyai tinggi
kurang daripada 141 cm.

the probability that a girl chosen at random from this group has a height less than 141 cm.

(c) bilangan orang perempuan dalam kumpulan ini jika 25 orang mempunyai tinggi lebih daripada
140.5 cm.

the number of girls in this group if 25 of the girls has heights more than 140.5 cm.

B15

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

BAB 6: FUNGSI TRIGONOMETRI

Tahap Penguasaan Tafsiran

4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi trigonometri dalam konteks penyelesaian
masalah rutin yang mudah.

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi trigonometri dalam konteks penyelesaian
masalah rutin yang kompleks.

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang fungsi trigonometri dalam konteks penyelesaian
masalah bukan rutin secara kreatif.

1. Selesaikan setiap yang berikut. Selesaikan setiap yang berikut.

Solve each of the following. TP 5
16LEMBARAN PBDSolve each of the following. TP 4

6(a) Jikatan(π−A)=kotπ,cari A untuk0≤A≤2π. 2. (a) Buktikan bahawa kos A + 1 sin A =
3 3 1 – tan A – kot A
π π
If tan ( 3 − A) = cot 3 , find A for 0 ≤ A ≤ 2π. sin A + kos A.

(b) Buktikan bahawa kot A + tan A = sek A kosek Prove that cos A + sin A = sin A + cos A
1 – tan A 1 – cot A
A.
Prove that cot A + tan A = sec A cosec A. (b) (i) Selesaikan persamaan 4 sin2 x + 8 kos
x − 7 = 0 untuk 0° ≤ x ≤ 360°.

Solve the equation 4 sin2 x + 8 cos x − 7 = 0
for 0° ≤ x ≤ 360°.

(ii) Seterusnya, cari penyelesaian bagi

persamaan 4 sin2 θ + 8 kos θ −7=0
2 2

untuk 0° ≤ x ≤ 360°.
Hence, find the solution of the equation

4 sin2 θ + 8 cos θ − 7 = 0 for 0° ≤ x ≤ 360°.
2 2

B16

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

Selesaikan setiap yang berikut.

Solve each of the following. TP 6

3. (a) Tunjukkan bahawa persamaan 2 sin x = 4 kos x − 1 boleh diungkapkan dalam bentuk 6 kos2 x −
tan x
kos x −2 = 0.
2 sin x = 4 cos x − 1 can be expressed in the form of 6 cos2x − cos x − 2 = 0.
Show that the equation tan x

(b) Seterusnya, selesaikan persamaan 2 sin x = 4 kos x − 1 untuk 0° ≤ x ≤ 360°.
tan x
Hence, solve the equation 4 cos x − 1
2 sin x = tan x for 0°≤ x ≤ 360°.


(c) Lakar graf y = 4 kos x − 1 untuk 0 ≤ x ≤ 2π. Seterusnya, cari bilangan penyelesaian bagi persamaan 17
2π sin x tan x + x = 2π untuk 0 ≤ x ≤ 2π.

Sketch the graph y = 4 cos x − 1 for 0 ≤ x ≤ 2π. Hence, find the number of solution for the equation 2π sin x tan x + x
= 2π for 0 ≤ x ≤ 2π.

LEMBARAN PBD

B17

LEMBARAN PBD   Matematik Tambahan  Tingkatan 5  Lembaran PBD

4. (a) Tunjukkan bahawa persamaan tan 2x = 5 sin 2x boleh ditulis dalam bentuk (1 − 5 kos 2x) sin 2x = 0.

Show that the equation tan 2x = 5 sin 2x can be written in the form (1 − 5 cos 2x) sin 2x = 0.

(b) Seterusnya, selesaikan persamaan tan 2x = 5 sin 2x untuk 0° ≤ x ≤ 360°.

Hence, solve the equation tan 2x = 5 sin 2x for 0° ≤ x ≤ 360°.

(c) Lakar graf y = tan 2x untuk 0 ≤ x ≤ π. Seterusnya, pada paksi sama, lakarkan satu graf sesuai untuk
menyelesaikan persamaan 5x sin 2x = 1 dan nyatakan bilangan penyelesaian untuk 0 ≤ x ≤ π.

Sketch the graph y = tan 2x for 0 ≤ x ≤ π. Hence, on the same axes, draw a suitable graph to solve the equation
5x sin 2x = 1 and state the number of solutions for 0 ≤ x ≤ π.

18

B18

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

5. (a) Selesaikan persamaan 2 sin2 x + 2 = 7 kos x untuk 0 ≤ x ≤ 2π. Beri jawapan dalam radian.

Solve the equation 2 sin2 x + 2 = 7 cos x for 0 ≤ x ≤ 2π. Give your answer in radians.

(b) Rajah menunjukkan sebahagian graf y = 3 sin bx. Garis lurus y = 1.4 menyilang lengkung pada A dan
B. Cari

The diagram shows part of the graph y = 3 sin bx. The line y = 1.4 cuts the curve at A and B. Find

y y = 1.4
a
πx
0
–a 19

(i) nilai a dan nilai b.

the value of a and of b.

(ii) koordinat A dan B dalam radian.

the coordinates of A and B in radians.

LEMBARAN PBD

B19

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

BAB 7: PENGATURCARAAN LINEAR

Tahap Penguasaan Tafsiran

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pengaturcaraan linear dalam konteks
penyelesaian masalah bukan rutin secara kreatif.

Selesaikan setiap yang berikut.

Solve each of the following problems. TP 6

1. Sebuah syarikat pengeluaran coklat menghasilkan dua jenis coklat, A dan B. Amaun tepung susu dan

tepung koko digunakan dijadualkan di bawah.

20 A chocolate manufacturing company produces two types of chocolates, A and B. The amount of milk and cocoa powder
needed are shown in the table below.

Jenis coklat Tepung susu (unit) Tepung koko

Type of chocolate Milk powder (units) Cocoa powder

A 1 3

LEMBARAN PBD B12

Syarikat mempunyai sejumlah 5 unit tepung susu dan 12 unit tepung koko.

The company has a total of 5 units of milk powder and 12 units of cocoa powder.

(a) Tulis semua ketaksamaan yang memuaskan semua kekangan di atas, selain daripada x ≥ 0 dan y ≥ 0.

Write all the inequalities that satisfy the above constraints, other than x ≥ 0 and y ≥ 0.

(b) Pada skala yang sesuai, lukis dan lorek rantau R yang memuaskan semua kekangan tersebut.

On a suitable scale, draw and shade the region R which satisfies the above constraints.

(c) Gunakan graf / Using graph
(i) cari keuntungan maksimum syarikat sehari jika setiap unit coklat A dan B memberi untung masing-
masing sebanyak RM30 dan RM50.

find the maximum profit per day of the company if each unit of chocolate A and B gives a profit of RM30 and
RM50 respectively.

(ii) julat bilangan coklat B dihasilkan jika syarikat ingin menghasilkan 1 unit coklat A.

the range of the number of chocolate B that is produced if the company wants to produce 1 unit of chocolate A.

B20

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

2. Seorang tukang kayu membuat meja dan kerusi. Setiap meja boleh dijual dengan keuntungan RM30 21

manakala sebuah kerusi memberi untung RM10. Tukang kayu menggunakan 36 jam seminggu bekerja

dan sebuah meja memerlukan 6 jam dan sebuah kerusi memerlukan 3 jam. Tukang kayu dikehendaki

membuat sekurang-kurangnya empat kali lebih banyak kerusi daripada meja ke atas keperluan pelanggan.

Meja memerlukan selebih-lebihnya empat kali lebih besar ruang simpanan daripada kerusi dan ruang yang

ada hanya boleh muat selebih-lebihnya 4 buah meja seminggu.

A carpenter makes tables and chairs. Each table can be sold for a profit of RM30 and each chair for a profit of RM10. The
carpenter can afford to spend up to 36 hours per week working and takes 6 hours to make a table and 3 hours to make a
chair. The carpenter must make at most four times as many chairs than tables from the demand of customers. Tables take
up at most four times storage space as chairs and there is room for at most 4 tables a week.

(a) Tulis semua ketaksamaan yang memuaskan semua kekangan di atas, selain daripada x ≥ 0 dan y ≥ 0.

Write all inequalities that satisfy the above constraints, other than x ≥ 0 and y ≥ 0.

(b) Pada skala yang sesuai, lukis dan lorek rantau R yang memuaskan semua kekangan tersebut.

On a suitable scale, draw and shade the region R which satisfies the above constraints.

(c) Gunakan graf, cari keuntungan maksimum syarikat seminggu.

Use the graph, find the maximum profit of the company per week.

LEMBARAN PBD

B21

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

BAB 8: KINEMATIK GERAKAN LINEAR

Tahap Penguasaan Tafsiran

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang kinematik gerakan linear dalam konteks
penyelesaian masalah rutin yang kompleks.

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang kinematik gerakan linear dalam konteks
penyelesaian masalah bukan rutin secara kreatif.

Selesaikan setiap yang berikut.

Solve each of the following problems. TP 5

22 1. Sebuah kereta bergerak pada satu jalan lurus dari 2. Suatu zarah bergerak pada satu garis lurus
lampu isyarat O. Halajunya, v m s−1, diberi oleh
dengan sesaran, s m selepas t saat dari titik tetap
2t(6 − t), dengan t ialah masa, dalam saat, selepas
O, diberi oleh s = t(t − 4)2. Cari
melalui lampu isyarat. (Anggap pergerakan ke
A particle moves in a straight line so that its
kanan ialah positif). Cari displacement, s m after t seconds from a fixed point O is
given by s = t(t − 4)2. Find
A car moves along a straight road from a traffic lights
O. Its velocity, v m s−1, is given by 2t(6 − t), where t (a) halaju apabila t = 3.
is the time, in seconds, after passing the traffic lights.
(Assume motion to the right is positive). Find the velocity when t = 3.

(a) halaju maksimum kereta. (b) nilai-nilai t apabila zarah berhenti seketika.

the maximum velocity of the car. the values of t when the particle stops
instantaneously.
(b) masa apabila kereta berhenti seketika.
(c) pecutan selepas 4 saat.
the time when the car stops instantaneously.
the acceleration after 4 seconds.
(c) Lakar graf halaju-masa bagi 0 ≤ t ≤ 7.
LEMBARAN PBD
Sketch the velocity-time graph for 0 ≤ t ≤ 7.

(d) jumlah jarak yang dilalui dalam 7 saat pertama.

the total distance travelled in the first 7 seconds.

B22

  Matematik Tambahan  Tingkatan 5  Lembaran PBD 

Selesaikan setiap masalah yang berikut.

Solve each of the following problems. TP 6

3. Rajah menunjukkan kedudukan dan arah pergerakan dua objek, A dan B pada satu garis lurus dan melalui

titik tetap, P dan Q pada masa yang sama. Jarak di antara P dan Q ialah 30 m. Halaju bagi A, vA = 3 + 2t − t2,
dengan keadaan t ialah masa, dalam saat, selepas melalui P manakala B bergerak dengan halaju tetap

−5 m s−1. Objek A berhenti seketika pada titik R. Anggapkan pergerakan ke kanan ialah positif, cari
The diagram shows the position and direction of motion of two objects, A and B on a straight line passing two fixed points,
mveslo−1c.itOybojef cAt,AvAst=op3s+in2stta−ntt2a, nwehoeurselyt
P and Q at the same time. The distance between P and Q is 30 m. The is the time, in
seconds, after passing P while B moves with a constant velocity of −5 at R. Assume

motion to the right is positive, find

AB 23

P 30 m Q

(a) halaju maksimum A.

the maximum velocity of A.

(b) jarak R dari P.

the distance of R from P

(c) jarak, dalam m, di antara A dan B apabila A berada di titik R.

the distance, in m, between A and B when A is at the point R.

(d) masa apabila kedua-dua objek mempunyai halaju yang sama.

the time when the two objects have the same velocity.

LEMBARAN PBD

B23

  Matematik Tambahan  Tingkatan 5  Lembaran PBD

4. Sebiji batu dilontar tegak ke atas dari tanah dengan halaju awal 30 m s−1. Anggapkan pecutan graviti ialah

−10 m s−2. Cari
A stone is thrown vertically upwards from the ground with an initial velocity of 30 m s−1. Assuming that the gravitational
acceleration is −10 m s−2. Find

(a) (i) masa yang diambil untuk mencapai tinggi maksimum.

the time taken to reach the maximum height.

(ii) halaju apabila batu menghentam tanah.

the velocity when the stone hits the ground.

(b) Lakar graf kedudukan-masa dan graf halaju-masa bagi pergerakan tersebut.

Sketch the displacement-time graph and the velocity-time graph for the motion.

(c) Cari jumlah jarak yang dilalui oleh batu sehingga batu kembali ke tanah lagi.

24 Find the total distance travelled by the stone until it returns to the ground.

LEMBARAN PBD

B24

JAWAPAN LEMBARAN PBD

BAB 1: SUKATAN MEMBULAT r2h = 360 ...➀ r
Luas permukaan S h 1.2
1. (a) OS(0.55) = 4.5
Surface area S

OS = 8.18 cm = 1 r2(1.2) × 2 + 2rh + (1.2r)h
2
∴OQ = 13 – 8.18 R

= 4.82 cm Q 4.5 cm = r2(1.2) + 3.2rh

(b) Panjang lengkok PQ P 0.55 rad S = 1.2r2 + 3.2r 360 2
O 13 cm r2
Arc length PQ
= 1.2r2 + 1152
= 4.82 (π – 0.55) r

= 12.49 cm

Perimeter = 12.49 + (8.18 – 4.82) + 4.5 + 13 (ii) Jika/If h = 10 cm
r2(10) = 360
= 33.35 cm Kek

(c) luas berlorek = 1 (4.82)2(π – 0.55) r = 6 cm 6.5 6 cm
shaded area = 2 5
sudut θ = 6.5 cm 5 cm
30.10 cm2 angle T 1.2
A O
  2. (a) (i) = 1.3 rad B 5 cm
P
Maka, CAO = (π – 1.3) rad
18 cm
Hence

O 1.5 CO = 5
8R S
sin(π – 1.53)180°4 sin 1.2 × 180° 4
π
T 1.5
PR2  2π
= 182 + 82 – 2(18)(8) kos × 180° CO = 5.17 cm

PR = 19.17 cm Oleh sebab OC < 6 cm

(ii) panjang lengkok PS = 18(1.5) Maka, kek dapat disimpan dalam kotak.

arc length PS = 27 cm ...➀ Since OC < 6 cm
Hence, the cake can be put in the box.

sin1O8RP sin 1π.5 × 180°2 BAB 2: PEMBEZAAN

= 19.17

ORP = 69.49° 1. (a) had x – 4
x → 16 x – 16
∴PRT = 180° – 69.49°
= had ( x – 4)( x + 4)
= 110.51° x → 16 (x – 16)( x + 4)

panjang lengkok PT = 110.51° × 2π(19.17) [x – 16]
arc length PT = 360° 16)( x +
= had
36.97 cm ...➁ (x – 4)
x → 16

Perimeter = 27 + 36.97 + 9.17 = 1
8
= 73.14 cm x2 – aa2) =
(x +
(iii) luas POR = 1 1.5 (b) had –6
2 π
 2 area POR x → –4
(18)(8) sin × 180°

= 71.82 cm2 had (x + a)(x – a) = –6
(a + x)
x → –4
luas sektor POS =
area of sector POS 1 (18)2(1.5) –4 – a = –6
2
a = 2
= 243 cm2
Luas sektor PRT = 2. (a) dx = 0.5 m/min
area of sector 110.51° dt
PRT 360° × π(19.17)2
x2 + y2 = 122
12
= 354.4 cm2 x2 = 144 – y2 y x
1
Luas rantau berlorek
x = (144 – y2)2
Shaded area 1
dx = 1 [144 – – 2 [–2y]
354.4 – [243 – 71.82] dy 2
y2]

= 183.22 cm2 dx –y
dy 144 – y2
(b) (i) Isipadu = 216 =

Volume –y
144 – y2
1 r2(1.2)h = 216 ∴ dx = · dy
2 dt dt

J1

Jika x = 4 m y2 = 144 – 16 (b) (i) y = 4x–1

If y = 128 dd yx = –4 , x=4
x2
= 8 2 δx = 0.05

0.5 = –8 2 · dy δy = dy · δx
16 dt dx

21 = –2 2· dy = –4 (0.05)
dt x2

ddyt = –1 m min–1 A = –164 (0.05)
–4 2
= –0.0125

(b) VA = 60t 60 km j–1 (ii) δx × 100% = 2p%
VB = 45t 60 km h–1 x
AB2 = (60t)2 + (45t)2
δx × 100% = –4δx × 100%
AB = 5625t2 B x yx2
= 75t 45 km j–1
45 km h–1 = –4δx × 100%

Apabila A ialah 120 km dari stesen. 4 · x2
When A is 120 km from the station. x

60t = 120 = –δx × 100%
x
t = 2 jam

d(AdtB) = 75 km/j = –2p%

(c) (i) f(x) = ax2 + bx + c BAB 3: PENGAMIRAN

f(2) = 4a + 2b + c = 26 ...➀ 1. (a) ∫x(2 – x)(x – 1)dx
...➁ = ∫(2x – x2)(x – 1)dx
f’(x) = 2ax + b

f’(2) = 4a + b = 23

f’’(x) = 2a

f’’(2) = 14 = 2a ∫= (3x2 – 2x – x3)dx

a = 7 1
4
b = 23 – 28 = –5 = x3 – x2 – x 4 + c

4(7) + 2(–5) + c = 26 4
– 2x)3
28 – 10 + c = 26 (b) ∫(2 dx

c = 8 4
8(1 –
(ii) f(x) = 7x2 – 5x + 8 = ∫ x)3 dx

f(–1) = 7 + 5 + 8 = 20 1
2
3. (a) Kos C/Cost C = 5πr2 + 2πr2 + 2πrh(2) ∫ = (1 – x)–3dx

= 7πr2 + 4πrh 1  (1 – x)–2 4
2 –2(–1)
Isipadu = 28π = πr2h r = + c
Volume

∴h= 28 = 1 +c
r2 4(1 – x)2
h
28
∴ C = 7πr2 + 4πr( r2 ) 2. (a) dy = 4x – 2
= 7πr2 + dx
112π dy
Apabila x = –1, dx = –6
r
When
dC = 14πr – 11r22 π = 0
dr Persamaan normal/Equation of normal

r3 = 11142ππ = 8 y = 1 x + c
r = 2 6

3 = 1 (–1) + c
6

dd2rC2 = 14π + 224π 169 = c
r3
1 19
Apabila/When r = 2 d2C = 14π + 28π ∴y = 6 x + 6
dr2
(b) y = ∫(4x – 2)dx
= 42π > 0
y = 2x2 – 2x + c
∴ Harga kos ialah minimum apabila r = 2 cm. Pada A(–1, 3)
The cost is minimum when r = 2 cm. 3 = 2 + 2 + c c = –1
∴y = 2x2 – 2x – 1
h= 28 = 7 cm
4

J2