MMoodduull 336600°°
JAWAPAN 4
Tingkatan
KSSM
Matematik
Tambahan
(Dwibahasa)
PERCUMA
Jawapan
Langkah
Bab 1 Penyelesaian
1.1 Fungsi Lengkap
Function (b) h(x) = |2x – 6|
(i) x = 1 atau/ or x = 5
(ii) f (x) x = 3 f(x) = |2x – 6|
1 (a) A B (1,4)
–5
(5,4)
CONTOH 1
2 –3 0 (3, 0) x
3 –2
5 –1 Julat/ Range: h(x) M 0
(c) (i) x 2 3
5
(b) 0 2 3 4 f(x) 1
A
(ii) Domain/ Domain = 2 N x N 3
(d) (i) Fungsi satu kepada satu
One-to-one function
B (ii) 4 = m(1) + c 1 = m(4) + c
–2 0 2
m + c = 4 …➀ 4m + c = 1 …➁
(c) x
2x – 1 ➁ – ➀: 4m – m = 1 – 4
7 3m = ‒3
45 m = ‒1 (Ditunjukkan/ Shown)
33 m = ‒1 ↷ ➀ ‒1 + c = 4
c = 5 (Ditunjukkan/ Shown)
2 (iii) 2
(e) x = ‒1 atau/ or x = 5
2 (a) Hubungan antara set A dan set B ialah hubungan (f) (i) f (x) = x – 2
banyak kepada satu, maka hubungan ini ialah suatu (ii) f (x)
fungsi. f(x) = x – 2
The relation between set A and set B is many-to-one relation,
so this relation is a function. 5
(b) Hubungan antara set A dan set B ialah hubungan
banyak kepada banyak, maka hubungan ini bukan
suatu fungsi. Unsur b mempunyai dua imej, iaitu v
dan w. 1 x
The relation between set A and set B is many-to-many relation, 0 23 7
so this relation is not a function. The element b has two –2
images, which are v and w.
(c) Hubungan antara set A dan set B bukan suatu fungsi
kerana unsur 3 tidak dipetakan kepada mana-mana (a) 0 < f (x) < 5
(b) 3 < x < 7
imej.
The relation between set A and set B is not a function because
the element 3 is not mapped to any images. 5 (a) (i) Fungsi banyak kepada satu
Function of many-to-one relation
3 Domain Kodomain Julat (ii) f (0) = 1
Domain Codomain Range 02 + c = 1
{x, y, z} c = 1 (Ditunjukkan/ Shown)
(a) {3, 5, 7, 9} {3, 5, 9}
(b) {1, 2, 3, 4, 5, 6} \ f (x) = x2 + 1
(c) ‒5 N x N 20 {2, 3, 4} {2, 3}
(d) ‒10 N f (x) N 30 ‒10 N f (x) N 30 k = 10
{‒4, ‒1, 0, 2, 4} {‒4, ‒2, 2, 8, 10} {‒4, ‒2, 2, 8, 10}
(b) k = 7 1
(c) (i) a =1, b = 3
4 (a) f(x) = ‒2x + 3 (ii) 13 N y N 1
(i) Julat/ Range = ‒1 < f (x) < ‒7
(ii) Domain/ Domain = 0 < x < 2 (d) (i) ‒20
(ii) ‒2
(iii) 4
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Matematik Tambahan Tingkatan 4 Jawapan
(e) (i) f(k) = 0 7 (a) Keuntungan daripada jualan produk boleh ditakrifkan
|k + 3| = 0 oleh fungsi k(p) = 15p dengan p ialah bilangan produk
k + 3 = 0 yang terhasil.
k = ‒3 (Ditunjukkan/ Shown)
(ii) f(1) = 4, f(‒7) = 4 The revenue from the sales of products can be defined by
the function k(p) = 15p where p is the number of products
\ Fungsi banyak kepada satu/ Many-to-one function produced.
(f) (i) g(2) = 4 kp(t) = k[p(t)]
= k(2t2 – 5t)
a(2 – p)2 + 4 = 4 = 15(2t2 – 5t)
= 30t2 – 75t
a(2 – p)2 = 0
(2 – p)2 = 0
2–p=0 Jumlah keuntungan yang diperoleh:
Total revenue which can be earned:
p = 2 (Ditunjukkan/ Shown)
g(3) = 3 kp(75) = 30(75)2 – 75(75)
= 168 750 – 5 625
CONTOH a(3 – 2)2 + 4 = 3
= 163 125
a+4=3
a = ‒1 (Ditunjukkan/ Shown) Maka, jumlah keuntungan sebanyak RM163 125 boleh
(ii) x = 0 atau/ or x = 4 diperoleh Ali daripada jualan produk yang terhasil
1.2 Fungsi Gubahan selepas kilang beroperasi selama 75 minit.
Thus, a total revenue of RM163 125 can be earned by Ali from
Composite Functions
the sales of products produced after the factory has operated
1 (a) f (x) = 2x ‒ 3, g(x) = x + 1, gf : x → 2x – 2 for 75 minutes.
(b) Perimeter boleh ditakrifkan oleh:
(b) f (x) = 2x ‒ 6, g(y) = 15 , gf : x → 15 The perimeter can be defined by:
2y – 4x ‒ 13
1 p(x) = 2x + 2(2x + 1)
(c) f (y) = 5 4 , g(x) = 2 x , fg : x → 20 – 5x = 2x + 4x + 2
3y – – 4x – 10
4 = 6x + 2
2 (a) (i) f g : x → x2 + x + 4 Fungsi kos pagar/ Function of the cost of fence:
(ii) gf : x → x2 – 3x + 2 cp(x) = c[p(x)]
(iii) g2 : x → x + 4 = c(6x + 2)
(b) (i) gf : x → 3x + 3 = 10(6x + 2)
(ii) f 2 : x → 9x – 8 = 60x + 20
1 Dengan/ With x = 12,
6x –
(c) (i) f g : x → 1 Kos pagar/ Cost of the fence:
→ x + 4 cp(12) = 60(12) + 20
x – 2
(ii) gf : x = 720 + 20
= 740
3 (a) 35 \ Kos pagar ialah RM740./ The cost of the fence is RM740.
(b) 7
(c) ‒7 (c) Biar d sebagai jarak perjalanan, t sebagai masa dan p
sebagai penggunaan petrol.
Let d be the travel distance, t be the time and p be the
4 (a) 1 consumption of petrol.
(b) 2
(c) ‒33 atau/ or ‒1 d(t) = 80t = 4t p(d) = 200d = d
60 3 1 000 5
5 (a) g(x) = 3x + 1 pd(t) = p[d(t)] pd(90) = 4(90)
15
‒2x + 29 ( ) 4t
(b) f : x → 3 =p 3 = 24
(c) fg(x) = 4x + 26 4t
1 = 3
x 5
6 (a) (i) – 4t
= 15
(ii) x
(iii) x 1 \ Petrol yang telah digunakan dalam 90 minit ialah
– 24 m./ The petrol consumed in 90 minutes is 24 m.
(b) (i) – x 1
(d) Isi padu air selepas t minit:
(ii) x Volume of water after t minutes:
(iii) x –x 1 V(t) = 20 + 15t
1 Tinggi air di dalam bekas:
– Height of water in the container:
(iv) – x 1
V
(c) (i) x h(V) = 12 × 4
(ii) xx +– 1 = V
1 48
(iii) x
(iv) xx +– 1
1
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Matematik Tambahan Tingkatan 4 Jawapan
Tinggi air selepas t minit: Domain 0NxN4 3 N x N 19
Height of water after t minutes: Domain 3 N y N 19 0NyN4
hV(t) = h[V(t)] Julat 5 12
Range
= h(20 + 15t)
(b)
= 20 + 15t
48
y –4 3 4
20 + 15(5) y
hV(5) = 48
= 1.9792 15
\ Bekas belum diisi penuh dengan air. Tinggi air dalam y = f (x)
bekas ialah 1.9792 cm selepas 5 minit. 10 y=x
The container is not fully with water yet. The height of
water in the container is 1.9792 cm after 5 minutes.
CONTOH1.3 Fungsi Songsang 5
Inverse Functions –5 0
–5
1 (a) 2 y = f –1(x)
(b) 2
(c) 4 x
(d) 5 5 10 15
(e) 8
2 (a) Ada, kerana fungsi ini ialah fungsi satu dengan satu.
Has, because this function is a one-to-one function.
(b) Tidak ada, kerana fungsi ini ialah fungsi banyak dengan (c) Domain –2 N x N 2 –4 N x N 12
Domain –4 N y N 12 –2 N y N 2
satu. –1 –4
Does not have, because this function is a many-to-one function. Julat 52
Range
(c) Tidak ada, kerana fungsi ini ialah fungsi banyak dengan
y8
satu.
Does not have, because this function is a many-to-one function.
(d) Ada, kerana fungsi ini ialah fungsi satu dengan satu.
Has, because this function is a one-to-one function.
(e) Ada, kerana fungsi ini ialah fungsi satu dengan satu. y
Has, because this function is a one-to-one function.
3 (a) Fungsi f dan g ialah fungsi songsang antara satu sama 10 y=x
lain. y = f (x)
y = f –1(x) x
Functions f and g are the inverse function of each other. 5 5 10
(b) Fungsi f dan g bukan fungsi songsang antara satu sama –5 0
lain. –5
Functions f and g are not the inverse function of each other. Domain –1 N x N 3 –4 N x N 8
Domain –4 N y N 8 –1 N y N 3
(c) Fungsi f dan g ialah fungsi songsang antara satu sama
lain. Julat
Range
Functions f and g are the inverse function of each other.
(d) Fungsi f dan g bukan fungsi songsang antara satu sama
lain.
Functions f and g are not the inverse function of each other.
(e) Fungsi f dan g ialah fungsi songsang antara satu sama
lain.
Functions f and g are the inverse function of each other.
4 (a) y3 4 7 12 19
y y=x
20 y = f (x) 5 (a) (‒7, 2)
(b) (‒4, –3)
(c) (3, 0)
15 6 (a) f ‒1(x) = x – 11
10 (b) g‒1(x) = x
5 y = f –1(x) 5
(c) h‒1(x) = 1 + 4x , x ≠ 0
x
(d) f ‒1(x) = 1 + 3x , x ≠ 0
x
0 x (e) h‒1(x) = x –5
5 10 15 20 3
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Matematik Tambahan Tingkatan 4 Jawapan
Praktis ke Arah S P M (b) g(y) = x ⇒ g‒1(x) = y
Kertas 1
Jadikan/ Let k = g‒1(x)
Bahagian A/ Section A g(k) = x
2k – 4 = x
1 Jenis hubungan: Hubungan banyak kepada satu 2k = x + 4
Type of relation: Relation of many-to-one
Julat/ Range: {p, r} k =x + 4 ⇒ g‒1(x) = x+4
2 2
x + 4 8 4
(c) g‒1(x) = y = 2 h(y) = 2y – 4=y –
2
2 (a) f[g(x)] = 2g(x) + 5 ( ) x +4
hg‒1(x) = 2
17 – 2x = 2g(x) + 5 h
2g(x) = ‒2x + 17 – 5 = 4
–2x + 12 =
g(x) = x +4
2 x 28 ‒ 2
+6
CONTOH = ‒x
gf (x) = g[f (x)] +4–4
= g(2x + 5) = 8 ⇒ hg‒1(x) = 8 , x ≠ 0
= ‒(2x + 5) + 6 x x
= ‒2x – 5 + 6
= ‒2x + 1 5 (a) (i) k(5) = 5 – 3(5)
= 5 – 15
(b) gf (c2 – 1) = 3 – 6c = ‒10
‒2(c2 – 1) + 1 = 3 – 6c
‒2c2 + 2 + 1 = 3 – 6c (ii) h(m + 2) = 1 k(5)
5
2c2 – 6c = 0
1
c2 – 3c = 0 3(m + 2) – 2 = 5 (‒10)
c(c – 3) = 0 3m + 6 – 2 = ‒2
c = 0 atau/ or 3
3m = ‒6
3 (a) Jadikan/ Let k = f ‒1(x) m = ‒2
(iii) kh(x) = k[h(x)]
f(k) = x = k(3x – 2)
p + qk = x = 5 – 3(3x – 2)
qk = x – p = 5 – 9x + 6
k = x–p = –9x + 11
q
(b) x 0 1 2 3 4 5
⇒ f ‒1(x) = x – p y 11 2 –7 –16 –25 –34
q
(b) f ‒1(17) = ‒4 y
17q– p = ‒4 15
17 – p = ‒4q ……➀
10
f(6) = ‒13 kh(x) = –9x + 11
p + 6q = ‒13 ……➁
5
➀ + ➁: 17 + 6q = ‒4q + (‒13)
6q + 4q = ‒13 – 17 x
10q = ‒30
q = ‒3
0
q = ‒3 ↷ ➀, 17 – p = ‒4(‒3) –5 1 23 456 7
–10
p = 17 – 12 –15
–20
=5 –25
–30
\ p = 5, q = ‒3 –35
Bahagian B/ Section B
4 (a) h(3) = 4
2(38) + q = 4
8 = 6 + q
4
6 + q = 2
q = ‒4
g(3) = 2 Julat bagi y/ The range of y: ‒34 N y N 11
3p – 4 = 2 J4
3p = 6
p = 2
\ p = 2, q = ‒4
Matematik Tambahan Tingkatan 4 Jawapan
Kertas 2 VA2 = VO2 + AO2
( ) 3 2
Bahagian A/ Section A VA2 = 5 VA AO2
+
1 (a) f(x) = 15(60) + 20x
= 20x + 900 AO2 = VA2 – 9 VA2
25
16
(b) f(x) = 20x + 900 = 25 VA2
f(1 400) = 20(1 400) + 900 AO = 4 VA
= 28 000 + 900 5
= 28 900 saat/ seconds
= (28 900 ÷ 60 ÷ 60) jam/ hours Luas/ Area = 9π
= 8316jam/ hours π(AO)[AO + VA] = 9π
(AO)(AO + VA) = 9
( )( ) 4 VA 4 VA + VA =9
5 5
(c) Jadikan/ Let f ‒1(x) = k ( )( )
CONTOH 4 VA 9 VA =9
20k + 900 = x 5 5
20k = x – 900 x – 900 3256VA2 = 9
k = x –29000 20
⇒ f ‒1(x) = VA2 = 25
4
10 jam/ hours 30 minit/ minutes
= 10(60 × 60) + 30(60) VA = 5
= 36 000 + 1 800 2
= 37 800 saat/ seconds
AB = AO + OB
2r = 2AO
37 800 – 900 r = AO
20 ( )( ) 4 5
f ‒1(37 800) = = 5 2
= 36 900 = 2
20
( )( ) 3 5
= 1 845 buah kereta mainan/ toy cars VO = 5 2
= 3
2
Bab 2
Isi padu/ Volume:
( ) 1 1 3
2.1 Persamaan dan Ketaksamaan Kuadratik 3 π(AO2)(VO) = 3 π(22) 2
Quadratic Equations and Inequalities = 2π cm3
1 (a) x = 7.0000 atau/ or ‒1.0000 (c) (i) v = 180 v + 10 = 180
(b) x = 0.3723 atau/ or ‒5.3723 t1 t2
(c) y = 2.6861 atau/ or ‒0.1861 t1 = 18v0 180
t2 = v + 10
2 (a) x = 4.6458 atau/ or ‒0.6458 (ii) 180 t1 – t2 = 15 min
(b) x = 3.6794 atau/ or ‒0.6794 v 180
(c) z = 3.0000 atau/ or ‒3.0000 ‒ v + 10 = 1
4
180(v + 10) ‒ 180v
3 (a) (i) Jumlah luas permukaan = 210 v(v + 10) = 1
4
Total surface area
4(180v + 1 800 – 180v) = v2 + 10v
2(2x + 3)(2x + 3) + 4(2x)(2x + 3) = 210
7 200 = v2 + 10v
2(4x2 + 12x + 9) + 4(4x2 + 6x) = 210
8x2 + 24x + 18 + 16x2 + 24x – 210 = 0 v2 + 10v – 7 200 = 0
(v + 90)(v – 80) = 0
24x2 + 48x – 192 = 0 v = ‒90 atau/ or 80
x2 + 2x – 8 = 0
(Ditunjukkan/ Shown) \ Oleh sebab v > 0, v = 80 km/j.
(ii) x2 + 2x – 8 = 0
Because v > 0, v = 80 km/h.
(x + 4)(x – 2) = 0 (iii) 15 jam 48 minit – 12 jam = 3.8 jam
x = ‒4 atau/ or 2 15 hours 48 minutes – 12 hours = 3.8 hours
\ Oleh sebab panjang tidak boleh bernilai negatif,
180 + 180 = 3.8 jam/ hours
maka x = 2. v v + 10
Because length cannot be negative value, thus x = 2. 180(v + 10) + 180v = 19
v(v + 10) 5
(b) sin ÐVAB = 3
5 5(180v + 1 800 + 180v) = 19(v2 + 10v)
VO = 3 1 800v + 9 000 = 19v2 + 190v
VA 5 19v2 ‒ 1 610v – 9 000 = 0
VO = 3 VA v = ‒(‒1 610) ± (‒1 610)2 ‒ 4(19)(‒9 000)
5
2(19)
1 610 ‒ 1 810 1 610 + 1 810
= 38 atau/ or 38
= ‒5.263 atau/ or 90
\ v > 0, maka/ then v = 90 km/j/ km/h.
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