MMoodduull 336600°°
JAWAPAN 5
Tingkatan
KSSM
Matematik
Tambahan
(Dwibahasa)
PERCUMA
Jawapan
Bab 1 (b) 0.5237 rad = 0.5237 × 180°
π
Radian = 30°
1.1 Radian Perimeter = 40(0.5237) + 2(40) sin 30°
= 20.948 + 20.7068 2
1 (a) 3.385 × 180°
π = 41.65 cm
CONTOH
= 193.92° π
(c) 87° = 87° × 180°
60
2π 180°
(b) 13 × π = 1.5186 rad
= 120° 180° 87°
π 2
(c) 5.175 × Perimeter = 15(1.5186) + 2(15) sin
= 296.47° = 22.7795 + 20.6506
π = 43.43 cm
2 (a) 15° × 180°
= 0.2618 rad 5 (a) Sudut bagi tiga sektor rim tayar:
The angle for three sectors of the rim of tyre:
π
(b) 60° × 180° 2π × 3 = 2(3.142) × 3
= 1.0473 rad 5 5
π = 3.7704 rad
(c) 170° × 180°
= 2.9674 rad Panjang lengkok/ Length of arc:
22 × 3.7704 rad = 11 × 3.7704
2
= 41.47 cm
1.2 Panjang Lengkok Suatu Bulatan (b) θmin = 2π – θmaj
Arc Length of a Circle
1 (a) AB = 2(3.142)(8) × 64° 7θmin = 14.665
360°
7(2π – θmaj) = 14.665
= 8.937 cm 2(3.142) – θmaj = 2.095
(b) AB = 9(1.484 rad) θmaj = 6.284 – 2.095
= 13.356 cm = 4.189 rad
2 (a) 40 = r(5.25 rad) θmin = 2.095 × 180°
3.142
40
r = 5.25 = 120.02°
= 7.62 cm Panjang perentas/ Length of chord:
(b) 7= 2(3.142)(r) × 40° 2(7) sin 120.02° = 2(7)(0.8661)
360° 2
= 12.125 cm
7(360°)
r = 40°(2)(3.142) Perimeter tembereng major:
Perimeter of major segment:
= 10.025 cm
12.125 + 7(4.189) = 12.125 + 29.323
3 (a) 11 = 8θ = 41.448 cm
θ = 11 1.3 Luas Sektor Suatu Bulatan
8
Area of Sector of a Circle
= 1.375 rad
2(3.142)(13) 3.142
360° 180°
( ) 1 1
(b) 10 = × θ (a) Luas/ Area = 2 × 122 × 95° ×
θ = 10(360°) = 1 × 122 × 1.6583 rad
2(3.142)(13) 2
= 44.068° = 119.4 cm2
3.142
180°
[ ] 1 ×
π (b) Luas/ Area = 2 × 152 × (360° – 302°)
4 (a) 20° = 20° × 180°
= 0.3491 rad 3.142
180°
( )=1 × 152 × 58° ×
2
Perimeter = 21(0.3491) + 2(21) sin 20° = 1 × 152 × 1.0124 rad
= 7.3313 + 7.2932 2 2
= 113.9 cm2
= 14.62 cm
J1
Matematik Tambahan Tingkatan 5 Jawapan
(c) Luas/ Area = 1 × 42 × (6.284 – 0.4364) rad 1.4 Aplikasi Sukatan Membulat
2
Application of Circular Measures
1
= 2 × 42 × 5.8476 rad (a) ( i) 124 × ÐBOC = 5.25π
ÐBOC = 0.75π
= 46.78 cm2
(ii) ÐAOB = π – 0.75π
= 0.25π
2 (a) L = 291.852
1
2 × r2 × 1.206 rad = 291.852 Isi padu/ Volume:
r2 = 484 (b) (i) L21ua×s7/2A×re0a.2=52π8×5 6 = 36.75π cm3
( ) –
r = 22 cm
(b) L = 157.1 1 × 172 × 55° × 3.142
2 180°
3.142 = 285 – 138.728
180°
( ) 1 × r2 × 45° × = 157.1 = 146.272 cm2
2
CONTOH
1 × r2 × 0.7855 rad = 157.1 (ii) Perimeter: 55° × 3.142
2 180°
( )
r2 = 400 13.9 + (30 – 17) + 17 × +
r = 20 cm (30 – 17 kos/ cos 55°)
= 13.9 + 13 + 16.32 + 20.25
3 (a) L = 39.15 = 63.47 m
1
2 × 52 × θ = 39.15 Praktis ke Arah S P M
θ = 3.132 rad
(b) L = 33.504 Kertas 1
21
× 82 × θ = 33.504
θ = 1.047 rad
Bahagian A/ Section A
30 1 (a) Lilitan bulatan/ Circumference of circle:
1 32
4 (a) 1π rad × 180° = 30° 2πj = 3.142 × 2
16 1π 2×
1
LABC = LOABC – LOAC = 100.544 cm
= 1 × 72 × (0.5237 rad – sin 30°) Perimeter semibulatan/ Perimeter of semicircle:
2
2(32) + πj = 64 + 3.142(32)
= 0.5807 cm2 = 64 + 100.544
(b) 0.7855 rad × 180° = 45° = 164.544 cm
π
Perimeter kawasan berlorek:
LABC = LOABC – LOAC Perimeter of shaded region:
= 1 × 182 × (0.7855 rad – sin 45°) 100.544 + 164.544 = 265.088 cm
2
(b) Luas semibulatan/ Area of semicircle:
= 12.70 cm2 1
π 2 × 322 × π = 512π cm2
(c) 88° × 180° = 1.5361 rad Luas bulatan/ Area of circle:
162 × π = 256π cm2
LABC = LOABC – LOAC
= 1 × 62 × (1.5361 rad – sin 88°) Luas kawasan berlorek:
2
Area of shaded region:
= 9.661 cm2 512π – 256π = 256π cm2
5 (a) 90° = 90° × 3.142 Q R
1 2 180° 2 (a) (i) θmaj = 2π – θmin
= 1.571 rad 78.55
5
Luas/ Area PQR: rθmin =
( )
P = 15.71 cm
1 × 15.24 2 × (1.571 rad – sin 90°)
2 2
2r2(r2πrππ–––1rθr5θθm.7mmin1ia)nj = 78.55
= 1 × (7.62)2 × (1.571 – 1) = 78.55
2 = 78.55
= 78.55
= 16.577 cm2
Jumlah luas/ Total area: r = 78.55 + 15.71
16.577 × 2 × 16 = 530.5 cm2 2(3.142)
(b) θmin = 2π – θmaj
= 15 cm
7.5θmin = 10.47 (ii) (15)θmin = 15.71
θmin = 1.0473 rad
2(3.142) – θmaj = 1.396 (b) Perimeter:
θmaj = 6.284 – 1.396
1.0473 × 180°
2 × 3.142
( )
= 4.888 rad 78.55 + 2(15) sin
Isi padu piza yang tinggal: = 78.55 + 30 sin(29.99°)
The volume of the remaining pizza: = 78.55 + 14.996
1 × 7.52 × 4.888 rad × 0.5 = 68.74 cm3 = 93.55 cm
2
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Matematik Tambahan Tingkatan 5 Jawapan
Kertas 2 (ii) Luas tembereng yang dipotong pada bulatan
besar/ Area of the cut segment at the big circle:
Bahagian B/ Section B
12 × 92 × (2.0947 rad – sin 120°) = 40.5(1.2287)
1 (a) PQ = QR = PR = 49.7613 cm2
⇒ ∠RPQ = ∠PQR = ∠QRP = 60°
RS = 43.3 Luas semibulatan kecil/ Area of small semicircle:
2r sin 60° = 43.3 1
2 × 4.52 × 3.142 = 31.8128 cm2
r = 2 43.3 Luas bulatan besar/ Area of big circle:
sin 60° 3.142 × 92 = 254.502 cm2
= 25 cm Luas kawasan berlorek/ Area of shaded region:
254.502 – 49.7613 – 31.8128 = 172.93 cm2
(b) Panjang lengkok/ Length of arc RPS:
25 × 120° × 3.142 = 25 × 2.0947 rad
180°
CONTOH
= 52.367 cm Bab 2
Lilitan bulatan/ Circumference of circle: Had dan Hubungannya dengan Pembezaan
2(3.142) × 25 = 157.1 cm
Limit and Its Relation to Differentiation
Perimeter: 2.1
157.1 + 2(52.367) = 157.1 + 104.734 100 100
x 0
= 261.834 cm 1 (a) haxd→/ 0lim = (tidak tertakrif/ undefined)
(c) Luas tembereng/ Area of segment PROS:
1 × 252 × (2.0947 rad – sin 120°) (b) had/ lim 3 = 0 3 4
2 x+4 +
x→0
1 3
= 2 × 252 × 1.2287 = 4
= 383.96 cm2 (c) had/ lim –3 + √x + 9
x
Luas kawasan berlorek/ Area of shaded region: x → 0
3.142(252) – 2(383.96) = 1 963.75 – 767.922
( )( )
= 1 195.828 cm2 = had/ lim –3 + √x + 9 –3 – √x + 9
x –3 – √ x + 9
x → 0
2 (a) 2rk = rb = had/ lim 9 – (x + 9)
x → 0 x(–3 – √ x + 9 )
Luas kawasan berlorek/ Area of shaded region = 60.75π –x 1
πrb2 – πrk2 = 60.75π
= had/ lim
π(rb2 – rk2) = 60.75π x → 0 1 x(–3 – √ x + 9 )
( )rb2 – rb 2 = had/ lim 1
2 x → 0 3 + √ x + 9
= 60.75
rb2 – rb2 = 60.75 = 1
4 3 + √0 + 9
1
3rb2 = 3+ 3
4
= 60.75 = 1
6
rb2 = 81
rb = 9 2 (a) had/ lim x2 – 2x – 3
x–3
∴ Diameter: 9 × 2 = 18 cm x → –3
= had/ lim (x + 1)(x – 3)
(b) (i) O x → –3 x–3
9 cm 4.5 cm = had/ lim x + 1
P
Q x → –3
= ‒3 + 1
= ‒2
kos/ cos ∠POQ = 4.5 (b) had/ lim 3 + x
2 9
x → 4
=3+4
∠POQ kos–1/ cos–1 =7
2 = 0.5 (c) had/ lim 3x2 – 12
∠POQ = 2(60°) x → –2 x + 2
= had/ lim 3(x – 2)(x + 2)
= 120° × 3.142 x → –2 x+2
180°
= 2.0947 rad = had/ lim 3(x – 2)
x → –2
= 3(–2 – 2)
= 3(–4)
= –12
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Matematik Tambahan Tingkatan 5 Jawapan
(d) had/ lim x – 3 2.2 Pembezaan Peringkat Pertama
x → 3 √ 5x – 6 – 3
The First Derivative
( )( ) 2
= had/ lim x–3 √ 5x – 6 + 3 ( ) 1 (a) d 2x10 = (10)2x10 – 1
√5x – 6 – 3 √ 5x – 6 + 3 dx 5 51
x → 3
= had/ lim (x – 3)(√5x – 6 + 3) = 4x9
(5x – 6) – 9 √x 1
x → 3 (b) 3 ⇔ 1 x 2
3
= had/ lim (x – 3)(√ 5x – 6 + 3)
x → 3 5x – 15 ( ) ( )( ) ddx 1 x 1 = 1 1 x 1 – 1
3 2 2 3 2
= had/ lim (x – 3)(√5x – 6 + 3) = 1 x– 1
5(x – 3) 6 2
x→3
= had/ lim √ 5x – 6 + 3 =1
6√x
x → 3 5
CONTOH dy 2x2 – 1
(c) dx = 5
= √5(3) – 6 + 3
5
= 2x
3+3 5
= 5
(d) dy =0
6 dx
= 5
63x6 – 1
(e) had/ lim x2 – 9x (e) f ′(x) = 48
x → 0 x 3x5
x(x – 9) 4
= had/ lim x =
x → 0
= had/ lim x – 9 2 (a) 3√ x – 2x 1 – 2x
x → 0
⇔ 3x 2
=0–9 d 1 – 2x) = d (3x 1 ) – d (2x)
= –9 dx dx 2 dx
(3x 2
3 (a) y = 3x – 5 …… ➀ = 3 –2
2√x
y + δy = 3(x + δx) – 5
( ) ( ) d 3x2 d 3x2 d (2x)
= 3x + 3δx – 5 …… ➁ (b) dx 2 + 2x = dx 2 + dx
➁ – ➀: δy = 3δx = 3x + 2
dy d
δy = 3δx (c) dx = dx (5x2 – 7x + 1)
δx δx
=3 = d (5x2) – d (7x) + d (1)
dx dx dx
ddyx δy
= had/ lim δx = 10x – 7
δx → 0 (d) dy = d (25x3 + 19x2 + 28x – 100)
dx dx
= had/ lim 3
δx → 0
= d (25x3) + d (19x2) + d (28x) – d (100)
dx dx dx dx
= 3
(b) y = x2 + 2 = f (x) = 75x2 + 38x + 28
δy = f (x + δx) – f (x) (e) y = 3x2 – 7x + 1 ⇔ y = 3x – 7 + 1
= [(x + δx)2 + 2] – (x2 + 2) x x
= x2 + 2xδx + (δx)2 + 2 – x2 – 2
= 2xδx + (δx)2 ( ) dy d 1
dx = dx 3x – 7 + x
δy = 2xδx + (δx)2 = d (3x) – d (7) + d (x–1)
δx δx dx dx dx
= 2x + δx = 3 + (‒1)x‒1 – 1
dy = had/ lim δy = 3 – 1
dx δx → 0 δx x2
= had/ lim 2x + δx 3 (a) f ′(x) = 3x2 + 4x
δx → 0
f ′(2) = 3(2)2 + 4(2)
= 3(4) + 8
= 2x = 12 + 8
= 20
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Matematik Tambahan Tingkatan 5 Jawapan
(b) f ′(x) = 18x2 – x +7 5 (a) Biar/ Let u = (2x + 5)2 dan/ and v = 4x2 + 2
4
u = (2x + 5)2 v = 4x2 + 2
f ′(2) = 18(2)2 – 2 1 + 7
4
2 du = 2(2x + 5)⋅(2) dv = 8x
dx dx
= 18(4) – 1 +7 = 4(2x + 5)
2
= 72 – 1 +7 dy = (2x + 5)2(8x) + (4x2 + 2)[4(2x + 5)]
2 dx
= 78 1 = (2x + 5)2(8x) + 8(2x2 + 1)(2x + 5)
2
= 8(2x + 5)[(2x + 5)(x) + (2x2 + 1)]
5
(c) f ′(x) = 3x – 3 = 8(2x + 5)(2x2 + 5x + 2x2 + 1)
5 = 8(2x + 5)(4x2 + 5x + 1)
3
f ′(2) = 3(2) – (b) Biar/ Let u = x3 dan/ and v = (5 + 6x)3
CONTOH = 6 – 5 u = x3 v = (5 + 6x)3
3
du = 3x2 dv = 3(5 + 6x)2⋅(6)
= 4 1 dx dx
3 = 18(5 + 6x)2
4 (a) Biar/ Let u = x3 + 8 dy = x3⋅18(5 + 6x)2 + (5 + 6x)3⋅3x2
dx
y = u5 u = x3 + 8 = 3x2(5 + 6x)2[6x + (5 + 6x)]
dy = 5u4 du = 3x2 = 3x2(5 + 6x)2(12x + 5)
du dx
(c) Biar/ Let u = 3x2 + 1 dan/ and v = (2x + 5)3
dy dy du
dx = du × dx u = 3x2 + 1 v = (2x + 5)3
= 5u4 × 3x2 du = 6x dv = 3(2x + 5)2(2)
dx dx
= 15x2u4 = 6(2x + 5)2
= 15x2(x3 + 8)4
(b) y= 2 ⇔ y = 2(5�2 + 7)‒1 dy = (3x2 + 1)⋅6(2x + 5)2 + (2x + 5)3⋅6x
5x2 + dx
7
= 6(2x + 5)2[(3x2 + 1) + (2x + 5)⋅x]
Biar/ Let u = 5x2 + 7
= 6(2x + 5)2(3x2 + 1 + 2x2 + 5x)
y = 2u–1 u = 5x2 + 7 = 6(2x + 5)2(5x2 + 5x + 1)
dy = –2u–2 du = 10x 6 (a) Biar/ Let u = (4 – x)3 dan/ and v = x – 1
du dx
dy = dy × du u = (4 – x)3 v=x–1
dx du dx
du = 3(4 – x)2⋅(‒1) dv
= –2u–2 × 10x dx = ‒3(4 – x)2 dx = 1
= –20xu–2
20x
= – (5x2 + 7)2 dy (x ‒ 1)⋅(‒3)(4 ‒ x)2 ‒ (4 ‒ x)3(1)
dx (x ‒ 1)2
4 ⇔ y – 1)– 1 =
2
(c) y = = 4(3� (4 ‒ x)2[‒3x + 3 ‒ (4 ‒
√3x – 1 (x ‒ 1)2 x)]
=
Biar/ Let u = 3x – 1 (4 ‒ x)2(–2x ‒ 1)
(x ‒ 1)2
y = 4u– 1 u = 3x – 1 =
2
dy = –2u– 3 du = 3 (b) Biar/ Let u = (3 – x)4 dan/ and v = x2
2 dx
du u = (3 – x)4
dy dy du v = x2
dx du dx
= × du = 4(3 – x)3⋅(‒1) dv = 2x
dx = ‒4(3 – x)3 dx
= –2u– 3
2 ×3
= –6u– 3 dy = x2⋅(‒4)(3 ‒ x)3 ‒ (3 ‒ x)4(2x)
2 dx (x2)2
= – (√ 6 1)3 2x1(3 ‒ x)3[‒2x ‒ (3 ‒
3x – x43
= x)]
= 2(3 ‒ x)3(‒2x ‒ 3 + x)
x3
= 2(3 ‒ x)3(‒x ‒ 3)
x3
J5