Jawapan Pelangi Analysis Tingkatan 4 & 5 Matematik Tambahan (Dwibahasa)

Jawapan

Tingkatan 4 Fungsi (c) f(x) = x

BAB Functions 2 3 = x
5x –
1 2 = x(5x − 3)

Masteri SPM 5x2 − 3x − 2 = 0

(5x + 2)(x − 1) = 0
x = – 25 , x = 1
Kertas 1

1. Isi padu kon/Volume of cone, 6. k[p(x)] = k(mx − 3)

I = 1 πj2(4) = 3(mx − 3) + 5
3
4 = 3mx − 9 + 5
3
= πj2 = 3mx − 4

4 4 3mx + 2ℎ − m = 3mx − 4
3 3
I : j → πj2 atau/or I(j) = πj2 2ℎ − m = −4

2ℎ = m – 4
2. Domain = {0, 1, 2, 3, 4, 5} m− 4
ℎ = 2
Julat/Range = {0, 1, 2, 3, 4}

3. Domain = {0°, 90°, 180°, 270°, 360°} 7. (a) Daripada graf/From the graph,
Julat/Range = {−1, 0, 1}
f(−3) = 6

(b) f(x) = 2

4. (a) f(4) = |3(4) − 2| |4x + 6| = 2
= |12 − 2|
= 10 4x + 6 = 2 atau/or 4x + 6 = −2

4x = −4 4x = −8

(b) x = t, f(t) = 0 x = −1 x = −2

|3t − 2| = 0 (c) −2  x  −1

t = 2 (d) x  −3 atau/or x  0
3
8. A B C D
(c) 0  f(x)  10 ✓
✓✓
5. (a) f(x) tidak tertakrif apabila ✓

f(x) is undefined when ✓

5x − 3 = 0,

x = p = 3 9. (a) 0 (b) 2 (c) 0 (d) −2
5
(b) Domain f ialah x  , x ≠ 53. 10. (a) Fungsi f tidak mempunyai fungsi
3 songsang kerana f bukan fungsi satu
Domain of f is x  , x ≠ 5 . dengan satu.
Function f does not have an inverse function
Julat f ialah y  , y ≠ 0. because f is not a one-to-one function.

Range of f is y  , y ≠ 0.

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(b) Fungsi g mempunyai fungsi songsang 14. (a) Untuk kadar denyutan nadi Encik Salleh
kerana g ialah fungsi satu dengan satu. menjadi 100 denyutan per minit, dia
mesti berehat selama 6 minit.
Function g has an inverse function because For Mr Salleh’s pulse rate to be 100 pulses
g is a one-to-one function. per minute, he must rest for 6 minutes.

11. (a) Katakan/Let y = g(x) (b) 10 minit/minutes

g–1(y) = x

y = 4x − 6 Kertas 2

4x = y + 6 1. (a) h(x) = 4x2 − 4x + 1
y+6 h(5) = 4(5)2 − 4(5) + 1
x = 4 m = 100 − 20 + 1
= 81
∴ g−1(x) = x+6 g(y) = z
4 g(81) = k = 81
k = 9
(b) g2(x) = g(4x − 6)
= 4(4x − 6) − 6 (b) gh(x) = g[h(x)]
= 16x − 24 − 6
= 16x − 30 = g(4x2 − 4x + 1)
= 4x2 − 4x + 1
  g2 5p = 50 = (2x − 1)2
2 = 2x − 1
  16
5p − 30 = 50
2
40p = 80
2. (a) (i) g(5) = 1 − 2(5)
p = 2 = 1 − 10
= –9
12. (a) Katakan/Let y = g(x)

g–1(y) = x (ii) f(2 − h) = 1 g(5)
3
y = 5x + 1 1
3(2 − h) + 2 = 3 (–9)
5x = y − 1
y–1
x = 5 6 − 3h + 2 = −3

x–1 3h = 11
5
∴ g−1(x) = h = 11
3

(b) f (5x + 1) = 25x2 + 10x − 2 (iii) fg(x) = f(1 − 2x)

Katakan/Let 5x + 1 = u, then x = u – 1 = 3(1 − 2x) + 2
5
= 3 − 6x + 2

    f(u)= 25 u – 1 2 + 10 u – 1 −2 = 5 – 6x
5 5
(b) x = −1, y = |5 − 6(−1)|
= (u − 1)2 + 2(u − 1) − 2 = 11

= u2 − 2u + 1 + 2u − 2 − 2 x = 3, y = |5 − 6(3)|
= |−13|
= u2 − 3 = 13

∴ f(x) = x2 − 3

13. (a) Julat bagi f ialah f(x)  −3. y = 0, |5 − 6x| = 0
Range of f is f(x)  −3.
x = 5
(b) Julat bagi f  −1 ialah f  −1(x)  −2. 6
Range of f  −1 is f  −1(x)  −2.

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y y = |5 – 6x| (ii) Fungsi yang memetakan set A
kepada set C ialah gf  −1(x).
13 Julat/Range: Function which maps set A to set C
11 0  y  13 is gf  −1(x).

gf  −1(x) = g(x – 1)
5

–1 O 5 x = 2(x − 1)2 + 5(x − 1) + 2
6 3

3. (a) (i) 1 − 2k = 0 = 2(x2 – 2x + 1) + 5(x − 1)

1 +2
2
k = = 2x2 – 4x + 2 + 5x – 5 + 2

t = 1 – 2(0) = 2x2 + x − 1

= 1 (b) fg(x) = 18 – 2x

(ii) x = 4, y = |1 − 2(4)| f(2x2 + 5x + 2) = 18 – 2x

= |−7| 2x2 + 5x + 2 + 1 = 18 – 2x

=7 2x2 + 7x – 15 = 0

(2x – 3)(x + 5) = 0

0  y  7 2x – 3 = 0 atau/or x + 5 = 0
3
(b) (i) Katakan/Let y = 1 – 2x x = 2 x = –5

x = 1 – y 5. (a) Katakan/Let y = 4x – 3
2
4x = y + 3
x = 1 (1 y) y + 3
2 – x = 4

∴ f  −1(x) = 1 (1 – x) ∴ f  −1 (x) = x + 3
2 4

  1
gf  −1(x) = g 2 (1 – x)  (b) f  −1 x
f  −1g (x) = 2 + 1
 = 41 1
2 (1 – x) + 2  x + 1 + 3
2
= 2(1 − x) + 1 = 4
2
1 x
= 2 – 2x + 2 = 2 + 4
= 5 – 2x 4
x
2 = 8 + 1

(ii) gh(x) = f  −1(x)

4h(x) + 1 = 1 (1 − x) (c) hg(x) = 3x + 9
2 2
1 1 1 Katakan/Let y = x + 1
4h(x) = 2 − 2 x − 2 2
x = 2y – 2
= − 12 x
h(y) = 3(2y – 2) + 9
h(x) = − 18 x = 6y – 6 + 9
= 6y + 3
4. (a) (i) Fungsi yang memetakan set A
kepada set B ialah f  −1(x). / Function Maka/Hence, h(x) = 6x + 3
which maps set A to set B is f  −1(x).

Katakan/Let y = x + 1

x = y – 1

Maka/Hence, f  −1(x) = x – 1

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6. (a) Katakan/Let x = 5 8. (a) Dalam ujian garis mengufuk, garis
2 mengufuk akan memotong graf pada
2x = 5 dua titik, maka f bukan hubungan satu
dengan satu dan tidak mempunyai fungsi
2x − 5 = 0 songsang.

∴ p = 2, q = –5 In the horizontal line test, the horizontal
line will cut the graph at two points,
(b) (i) Domain f  −1(x) = julat f(x), iaitu therefore f is not a one-to-one relation and
x  0. does not have an inverse function.
Domain of f  −1(x) = range of f(x), that
is x  0. (b) f  2(x) = f [ f(x)]
= f 4 + x2 
(ii) f  2(x) = f [ f(x)]
f  1  1
= 2x – 5
= 4 + 4 + x2 22
= 1
2 1  1
2x – 5 – 5
= (4 + 4 + x2)2

= 8 + x2

= 1 (c) (i) Sebarang nilai yang lebih atau sama
2 – 5(2x – 5)
dengan sifar.
2x – 5
Any value that is more than or equal
2x – 5
= 27 – 10x to zero.

7. (a) 1 (ii) Katakan/Let y = 4 + x2

(b) (i) f (x)  –5 y2 = 4 + x2

x2 = y2 − 4

(ii) Katakan/Let y = (x − 1)2 − 5 x = y2 – 4

y + 5 = (x − 1)2 Maka/Hence, g−1(x) = x2 – 4

x − 1 = y + 5

x = 1 + y + 5 9. (a) Katakan/Let y = f(x)

Maka/Hence, f  −1(x) = 1 + x + 5 f –1(y) = x

(iii) y y = 2x + 3
3
y=x
y = f –1(x) y – 3 = 2x
(–5, 1) x 3
y = f(x)
O 3y – 9 = 2x

x = 3y – 9
2
(1, –5)

(iv) Apabila/When f(x) = f  −1(x), Maka/Hence, g(x) = 3x – 9
2

y = x

(x − 1)2 − 5 = x

x2 − 2x + 1 − 5 = x

x2 − 3x − 4 = 0

(x − 4)(x + 1) = 0

x = 4 atau/or x = −1

(diabaikan/rejected)

∴ x = 4

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(b) fg(x) = f [ g(x)] (b) (i) Kos/Cost, x = 120

= f  3x – 9  Harga jualan sebelum potongan,
2 Sales price before reduction,

2 3x – 9 f(120) = 34(120)
2 = RM160
= +3
3

= x – 3 + 3 (ii) Harga jualan sebelum potongan,

= x Sales price before reduction,

gf (x) = g[f(x)] f(x) = 360

= g 2x + 3 4x = 360
3 3

3 2x + 3 – 9 Kos/Cost, x = 3(360)
3 4
= 2
= RM270
2x + 9 – 9
= 2x 2 2. (a) (i) A(r) = πr2
= 2 (ii) V(A) = 10A

= x (b) V(r) = VA(r)
= V(πr2)
(c) fg(x) = gf(x) = x, dengan g(x) ialah = 10πr2

fungsi songsang bagi f(x). (c) V(r) = 90π
10πr2 = 90π
fg(x) = gf(x) = x, with g(x) being the inverse r2 = 9
r = 3
function of f(x).

Soalan KBAT

1. (a) Harga jualan − kos = keuntungan

Sales price − cost = profit

0.9f(x) − x = 0.2x

0.9f(x) = 1.2x

9  f(x) = 12 x
10 10

f(x) = 4 x
3

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Jawapan

Tingkatan 4 4. 2x2 + 9x + 8 = 0

BAB Fungsi Kuadratik α + β = – 29
8
2 Quadratic Functions 2

αβ = =4

Masteri SPM 2α + 2β = 2(α + β)

Kertas 1   = 2 – 29

1. (a) f  (x) = x2 + 3x + p = –9

    32 32 (2α)(2β) = 4αβ
= x2 + 3x + 2– 2 +p = 4(4)
= 16
32 32
Maka/Hence, x2 – (–9)x + 16 = 0
    = x + 2 – 2 + p

  = x + 3 2 9 x2 + 9x + 16 = 0
2 4
+p–

(b) p – 9 = 9 5. (a) 2x2 + px – 9 = 0
4
45 2(4)2 + p(4) – 9 = 0
4
p = 32 + 4p – 9 = 0

2. (x + h)2 = 9 4p + 23 = 0

(–6 + h)2 = 9 4p = –23
p = – 243
h2 – 12h + 36 = 9

h2 – 12h + 27 = 0 (b) − ab = – p2 = –8
p = 16
(h – 9)(h – 3) = 0

h = 9 atau/or h = 3

3. px2 – 4x + q = 0 6. (a) x2 + (n + 5)x – n2 = 0

α + 3α = − –p4 = 4 Katakan punca-puncanya ialah α dan
p – α.
4 Let’s say the roots are α and –α.
4α = p

1 Maka/Hence, α + (–α) = –(n + 5)
p
α = ...........a –n – 5 = 0

(α)(3α) = q n = –5
p
q Hasil darab punca / Product of roots,
p
3α2 = ...........b (α)(–α) = –n2

Gantikan a dalam b/Substitute a into b, = –(–5)2

  3 1 2 q = –25
p p
=

3 = q
p2 p

p = 3
q

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(b) hx2 – 9kx + 4h = 0 Dua punca nyata dan berbeza:

Dua punca sama/Two equal roots: Two distinct real roots:

b2 – 4ac = 0 b2 – 4ac  0

(– 9k)2 – 4(h)(4h) = 0 (–3)2 – 4(q – 3)(–1)  0

81k2 – 16h2 = 0 9 + 4(q – 3)  0

16h2 = 81k2 9 + 4q – 12  0

h2 = 81 4q  3
k2 16 3
q  4
h
= 81 = 9 11. (2 – p)x2 – x + 6 = 0
k 16 4

h : k = 9 : 4 Tiada punca nyata:

7. 8 + 7x – x2  0 y No real roots:
 x2 – 7x – 8  0 y = x2 – 7x – 8
b2 – 4ac  0

Jika/If x2 – 7x – 8 = 0, (–1)2 – 4(2 – p)(6)  0
(x – 8)(x + 1) = 0 –1 O 8
x 1 – 48 + 24p  0
x
24p  47
m
x = 8 atau/or x = –1 p  47
24
Maka/Hence, x  –1 atau/or x  8

8. 2x2 + 7x  4 y 12. f(x) = hx2 – 3x – 2
2x2 + 7x – 4  0 y = 2x2 + 7x – 4
Tiada punca nyata:

Apabila/When No real roots:

2x2 + 7x – 4 = 0 –4 O1 b2 – 4ac  0
2
(2x – 1)(x + 4) = 0 (–3)2 – 4(h)(–2)  0

x = 1 atau/or x = –4 9 + 8h  0
2 h  – 89

Maka/Hence, –4  x  1 13. (a) x(x – 6) = 5
2 x2 – 6x = 5
x2 – 6x – 5 = 0
9. f(x) = x2 + 2mx – 2m + 3
(b) – –16 = 6
Tiada punca nyata/No real roots:

b2 − 4ac  0 y

(2m)2 – 4(1)(3 – 2m)  0 (c) b2 – 4ac
= (–6)2 – 4(1)(–5)
4m2 + 8m – 12  0 y = m2 + 2m – 3 = 36 + 20
= 56 > 0
m2 + 2m – 3  0

Jika/If m2 + 2m – 3 = 0, –3 O 1

(m + 3)(m – 1) = 0 Dua punca nyata dan berbeza.

m = –3 atau/or m = 1 Two distinct real roots.

Maka/Hence, –3 < m < 1 14. (a) f(x) mempunyai titik maksimum, maka
∴ h = –3, k = 1 pekali x2  0.
f(x) has maximum point, therefore the
10. y = (q – 3)x2 – x + 6 coefficient of x2  0.

y = 2x + 7 h + 1  0
h  –1
(q – 3)x2 – x + 6 = 2x + 7
Maka/Hence, h = –2
(q – 3)x2 – 3x – 1 = 0

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(b) f(x) = (h + 1)x2 – 6x + k Kertas 2

= –x2 – 6x + k 1. (a) x(x – 2) = 4p – 5

Dua punca sama/Two equal roots: x2 – 2x = 4p – 5

b2 − 4ac = 0 x2 – 2x + 5 – 4p = 0

(–6)2 – 4(–1)(k) = 0 α ≠ β

36 + 4k = 0

4k = –36 ⇒ Dua punca nyata dan berbeza:

k = –9 Two distinct real roots:

15. (a) (5, –49) b2 − 4ac  0
(b) x = 5
(c) –2  x  12 (–2)2 – 4(1)(5 – 4p)  0

4 – 20 + 16p  0

16p – 16  0

16. (a) f(x) = –(x – 1)2 + 2p 16p  16
2p = 10
p = 5 p  1

dan/and q = 1 (b) Untuk/For x2 – 2x + 5 – 4p = 0,
α + β = 2 ...................a
(b) f(x) = –(x – 1)2 + 10 αβ = 5 – 4p ...............b

Di pintasan-y, x = 0. Untuk/For 2x2 + qx – 8 = 0,
At y-intercept, x = 0. α3 + β – q2
3 =
∴ n = –(0 – 1)2 + 10
= –1 + 10 α + β = – 32q .............c
= 9 – 28 = –4
 α3  β  =
17. Katakan x = lebar jalur kayu 3
Let’s say x = width of wooden strip αβ
9 = –4

Panjang jalur kayu pendek = 10.5 – 2x αβ = –36 ...............d
Length of shorter wooden strip

Jumlah luas permukaan jalur kayu Membanding a dan c,
Total surface area of wooden strips
Comparing a and c,
= 2(14x) + 2(10.5 – 2x)(x)
= 28x + 21x – 4x2 –  32q = 2
= 49x – 4x2 q = – 43

Maka/Hence, 49x – 4x2 = 55 Membanding b dan d,

4x2 – 49x + 55 = 0 Comparing b and d,

(4x – 5)(x – 11) = 0 5 – 4p = –36

x = 1.25 atau/or x = 11 (diabaikan/ 4p = 41
ignored)
41
4
Lebar jalur kayu = 1.25 cm p =
Width of wooden strip
Maka/Hence, p = 41 , q = – 43
4

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2. (a) f(x) = 3(x – p)2 + 3q Soalan KBAT
3q = –27
q = –9 1. Biar lebar = y, panjang = x
Let width = y, length = x
(b) p = 2+8 =5
2 ∴y=x−7

f(x) = 3(x – 5)2 – 27 Dengan Theorem Pythagoras,
By Pythagoras’ Theorem,
Apabila/When x = 0,
f(0) = 3(0 – 5)2 – 27 y2 + x2 = 752
= 48 (x − 7)2 + x2 = 5 625
x2 − 14x + 49 + x2 = 5 625
Apabila/When x = 9, 2x2 − 14x − 5 576 = 0
f(9) = 3(9 – 5)2 – 27 x2 − 7x − 2 788 = 0
= 21
Menggunakan formula kuadratik,
f(x) Using quadratic formula,

48 x = −(−7) ± (−7)2 − 4(1)(−2 788)
f(x) = 3(x – 5)2 – 27 2(1)

21

O 2 5 89 x 7 ± 49 + 11 152
2
–27 =

(c) g(x) = –3(x – 5)2 + 27 = 7 ± 11 201
2

= 7 ± 105.83
2

= 56.42 atau/or −49.42 (diabaikan/ignored)

∴ y = 56.42 − 7 = 49.42

Luas tanah/Area of plot of land
= (56.42)(49.42)
= 2 788.3 m2

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Jawapan

Tingkatan 4 Sistem Persamaan Gantikan b = −4 dan c = 6 ke dalam b,

BAB Systems of Equations Substitute b = −4 and c = 6 into b,

3 a + 2(−4) − 3(6) = −25

a = 1

Masteri SPM Maka/Hence, a = 1, b = −4, c = 6

Kertas 1 3. 8k – 2m + n = 7 .................a
13k + m − 4n = −4  ..............b
1. x + 2y + z = 14 ..................a −6k + 5m + n = 14 ...............c
3x + y + z = 18 ..................b
5x – 2y + 3z = 32 ..................c a − c : 14k – 7m = −7 ..............d
a × 4 : 32k − 8m + 4n = 28 ..............e
b – a : 2x – y = 4 ................d e + b : 45k – 7m = 24 ..............f

a × 3 : 3x + 6y + 3z = 42 ..............e f − d : 31k = 31

e – c : −2x + 8y = 10 ..............f k = 1

d × 8 : 16x – 8y = 32 ..............g Gantikan k = 1 ke dalam d,

f + g : 14x = 42 Substitute k = 1 into d,

x = 3 14(1) – 7m = −7

Gantikan x = 3 ke dalam d, m = 3
Substitute x = 3 into d,
Gantikan k = 1 dan m = 3 ke dalam a,
2(3) – y = 4
y = 2 Substitute k = 1 and m = 3 into a,

Gantikan x = 3 dan y = 2 ke dalam a, 8(1) – 2(3) + n = 7

Substitute x = 3 and y = 2 into a, n = 5

3 + 2(2) + z = 14 Maka/Hence, k = 1, m = 3, n = 5

z = 7 4. x – y + 1 = 0 ........................a
x2 + x – y = 3 .......................b
Maka/Hence, x = 3, y = 2, z = 7

2. 4a + b + c = 6 .......................a Daripada/From a, y = x + 1 ..............c
a + 2b – 3c = −25 ..................b
a + 3b – 2c = −23 ..................c Gantikan c ke dalam b,

Substitute c into b,

c – b : b + c = 2 ..................d x2 + x – (x + 1) = 3

b × 4: 4a + 8b – 12c = −100 ............e x2 − 4 = 0

e – a : 7b − 13c = −106 ............f (x − 2)(x + 2) = 0

d × 7 : 7b + 7c = 14 ...............g x = 2, −2

g − f : 20c = 120 Apabila/When x = 2, y = 2 + 1 = 3

c = 6 Apabila/When x = −2, y = −2 + 1 = −1

Gantikan c = 6 ke dalam d, Maka/Hence, x = 2, y = 3
Substitute c = 6 into d, atau/or x = −2, y = −1

b + 6 = 2
b = −4

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5. 2x + y = 1 ..................a Oleh sebab/Since x > 0,
x2 – xy = 2 .................b ∴ x = 11

Daripada/From a, y = 1 – 2x ............c Apabila/When x =11, y = 11 – 4 = 7

Gantikan c ke dalam b, Maka, dua integer itu ialah 7 dan 11.
Substitute c into b, Therefore, the two integers are 7 and 11.

x2 – x(1 − 2x) = 2 Kertas 2

x2 – x + 2x2 = 2

3x2 – x – 2 = 0 1. p + 3q + r = −2  .................a
2p + 7q − 3r = –2 .................b
(3x + 2)(x − 1) = 0 −2p + q – 5r = 18 .................c
x = – 32 , 1


Apabila/When x = – 32 , y = 1 − 2– 23  = 7 b + c : 8q − 8r = 16
3
q – r = 2 .................d

Apabila/When x = 1, y = 1 − 2(1) = −1 a × 2 : 2p + 6q + 2r = −4.........e
b − e : q – 5r = 2 ...................f
Maka/Hence, x = 1, y = −1
d − f : 4r = 0
atau/or x = – 32 , y = 7
3 r = 0

6. x – 2y = 5 ....................a Gantikan r = 0 ke dalam d,
x2 + y2 = 10 ..................b Substitute r = 0 into d,

Daripada/From a, x = 2y + 5 .............c q – 0 = 2
q = 2
Gantikan c ke dalam b,
Gantikan r = 0 dan q = 2 ke dalam a,
Substitute c into b,
Substitute r = 0 and q = 2 into a,
(2y + 5)2 + y2 = 10
p + 3(2) + 0 = −2
4y2 + 20y + 25 + y2 = 10
p = −8
5y2 + 20y + 15 = 0
Maka/Hence, p = −8, q = 2, r = 0
y2 + 4y + 3 = 0

(y + 1)(y + 3) = 0 2. 2x – 3y – 5z = 39 .................a
4x + 6y + z = 0 ...................b
y = −1, −3 −3x + 8y – 5z = 13 .................c

Apabila/When y = −1, x = 2(−1) + 5 = 3 a – c : 5x – 11y = 26 .............d
Apabila/When y = −3, x = 2(−3) + 5 = −1

Maka/Hence, x = 3, y = −1 b × 5 : 20x + 30y + 5z = 0 ...............e
atau/or x = −1, y = −3
a + e : 22x + 27y = 39 .............f

7. Katakan dua integer positif itu ialah x dan y. d × 22 : 110x − 242y = 572 ...........g
Let the two positive integers be x and y.
f × 5 : 110x + 135y = 195 ...........h
x − y = 4 ..........................a
x2 + y2 = 170 ....................b h − g : 377y = −377

y = −1

Daripada/From a, y = x − 4 ................c Gantikan y = −1 ke dalam d,

Gantikan c ke dalam b, Substitute y = −1 into d,

Substitute c into b, 5x – 11(−1) = 26

x2 + (x – 4)2 = 170 x = 3

x2 + x2 – 8x + 16 = 170 Gantikan x = 3 dan y = −1 ke dalam b,

2x2 – 8x – 154 = 0 Substitute x = 3 and y = −1 into b,

x2 – 4x – 77 = 0 4(3) + 6(−1) + z = 0

(x + 7)(x − 11) = 0 z = −6

x = −7, 11 Maka/Hence, x = 3, y = −1, z = −6

11 © Penerbitan Pelangi Sdn. Bhd.

3. x + 2y + 4z = 1 .................a 5. 2x + 5y – 18 = 0 .............a
8x – 6y + 3z = 13 ...............b
2x – 10y − 5z = 3 .................c x2 – 3y2 – 4x + 6y + 19 = 0 .............b

a × 2: 2x + 4y + 8z = 2 .................d Daripada/From a, x = 18 − 5y ............c
d – c : 14y + 13z = −1 ...............e 2
a × 8 : 8x + 16y + 32z = 8 .................f
Gantikan c ke dalam b,
Substitute c into b,

   
f – b : 22y + 29z = −5 ...............g 18 − 5y 2 18 − 5y
e × 11 : 154y + 143z = −11 .............h 2 2
g × 7 : 154y + 203z = −35 .............i − 3y2 – 4

+ 6y + 19 = 0

i − h : 60z = −24 324 − 180y + 25y2
4 – 3y2 – 36
z = −0.4 + 10y + 6y + 19 = 0

Gantikan z = −0.4 ke dalam e, 324 – 180y + 25y2 – 12y2 – 144

Substitute z = −0.4 into e, + 40y + 24y + 76 = 0

14y + 13(−0.4) = −1 13y2 – 116y + 256 = 0

y = 0.3 (y − 4)(13y − 64) = 0 64
13
Gantikan y = 0.3 dan z = −0.4 ke dalam a, y = 4,

Substitute y = 0.3 and z = −0.4 into a, 18 − 5(4)
2
x + 2(0.3) + 4(−0.4) = 1 Apabila/When y = 4, x = = −1

x = 2 64
13
Maka/Hence, x = 2, y = 0.3, z = −0.4 Apabila/When y = ,

4. 6x – 4y = 3.................a 18 − 5 64  = – 1433
13
 2x +  1y = 7.................b x =
2

Daripada/From a, y = 6x − 3 ..........c Maka/Hence, x = −1, y = 4
4
b × xy : 2y + x = 7xy ................d atau/or x = – 1433 , y = 64
13

Gantikan c ke dalam d, 6. x + 2y = 6.................a
Substitute c into d,
  3 –  6y = 5.................b
    x
2 6x − 3 + x = 7x 6x − 3
4 4 Daripada/From a: x = 6 – 2y.........c
b × xy : 3y – 6x = 5xy .............d
12x – 6 + 4x = 42x2 – 21x

42x2 – 37x + 6 = 0 Gantikan c ke dalam d,
Substitute c into d,
(3x − 2)(14x − 3) = 0
3y – 6(6 – 2y) = 5(6 – 2y)y
x = 2 , 3 3y – 36 + 12y = 30y – 10y2
3 14 10y2 − 15y – 36 = 0
  Apabila/When x = 2
2 , y = 6 3 −3 = 1
3 4
4 y = −(−15) ± (−15)2 − 4(10)(−36)
2(10)
  Apabila/When x = 6 3 − 3
3 , y = = – 73 15 ± 1 665
14 14 = 20
4

Maka/Hence, x = 2 , y = 1 = 15 + 1 665 , 15 – 1 665
3 4 20 20
– 73
atau/or x = 3 , y = = 2.79, −1.29
14

© Penerbitan Pelangi Sdn. Bhd. 12

Apabila/When y = 2.79, x = 6 – 2(2.79) d × 5 : 20a + 35c = 150 ................g
f × 7: 77a + 35c = 378 ................h
= 0.42

Apabila/When y = −1.29, x = 6 – 2(−1.29) h − g : 57a = 228

= 8.58 a = 4

Maka/Hence, x = 0.42, y = 2.79 Gantikan a = 4 ke dalam d,
atau/or x = 8.58, y = −1.29 Substitute a = 4 into d,

7. 4x + y = 1 .................a 4(4) + 7c = 30
x2 – 2y2 − 7y = 6 .................b c = 2

Daripada/From a: y = 1 − 4x .............c Gantikan a = 4 dan c = 2 ke dalam b,

Substitute a = 4 and c = 2 into b,

Gantikan c ke dalam b, 5(4) + b + 2 = 27

Substitute c into b, b = 5

x2 – 2(1 − 4x)2 – 7(1 − 4x) = 6 Maka/Therefore,
harga sepinggan nasi lemak = RM4
x2 – 2(1 – 8x + 16x2) – 7 + 28x = 6 price of a plate of nasi lemak

x2 – 2 + 16x − 32x2 – 7 + 28x = 6

31x2 – 44x + 15 = 0 harga sepinggan mi goreng = RM5
price of a plate of fried noodles
−(−44) ± (−44)2 − 4(31)(15)
x = harga secawan kopi = RM2
2(31) price of a cup of coffee

= 44 ± 76 9. Katakan/Let
62 a = bilangan peserta dari sekolah teknik A
number of participants from technical school A
= 44 + 76 , 44 – 76
62 62
b = bilangan peserta dari sekolah teknik B
= 0.85, 0.57 number of participants from technical school B

Apabila/When x = 0.85, y = 1 – 4(0.85) c = bilangan peserta dari sekolah teknik C
number of participants from technical school C
= −2.40

Apabila/When x = 0.57, y = 1 – 4(0.57) a + b + c = 36 ................a

= −1.28 a = b + c

Maka/Hence, x = 0.85, y = −2.40 a – b – c = 0 ..................b
atau/or x = 0.57, y = −1.28
a + c = 2b

8. Katakan/Let a – 2b + c = 0 ..................c
a = harga sepinggan nasi lemak
price of a plate of nasi lemak a + b : 2a = 36

b = harga sepinggan mi goreng a = 18
price of a plate of fried noodles
a – c : 3b = 36
c = harga secawan kopi
price of a cup of coffee b = 12

Gantikan a = 18 dan b = 12 ke dalam a,

Substitute a = 18 and b = 12 into a,

3a + 2b + 4c = 30 ..................a 18 + 12 + c = 36

c = 6

5a + b + c = 27 ..................b Maka/Therefore,
bilangan peserta dari sekolah teknik A = 18
2b = a + 3c
number of participants from technical school A
a – 2b + 3c = 0 ....................c
bilangan peserta dari sekolah teknik B = 12
a + c : 4a + 7c = 30 ..................d number of participants from technical school B
b × 2 :1 0a + 2b + 2c = 54 ..................e
c + e : 11a + 5c = 54 ..................f bilangan peserta dari sekolah teknik C = 6
number of participants from technical school C

13 © Penerbitan Pelangi Sdn. Bhd.

10. (x + 2) + x + y = 40 Luas/Area = 2 280 m2
1
y = 38 – 2x ..................a x(y + 16) − 2 (16)(30) = 2 280

Dengan theorem Pythagoras, xy + 16x – 240 = 2 280

By using the Pythagoras’ theorem, xy + 16x – 2 520 = 0 ..................b
x2 + y2 = (x + 2)2 ..................b
Gantikan a ke dalam b,
Gantikan a ke dalam b,
Substitute a into b,
Substitute a into b,
x(85 – x) + 16x – 2 520 = 0
x2 + (38 – 2x)2 = (x + 2)2
85x – x2 + 16x – 2 520 = 0
x2 + 1 444 – 152x + 4x2 = x2 + 4x + 4
x2 – 101x + 2 520 = 0
4x2 – 156x + 1 440 = 0
(x − 45)(x − 56) = 0
x2 − 39x + 360 = 0
x = 45, 56
(x − 15)(x − 24) = 0
Apabila/When x = 45, y = 85 − 45 = 40
x = 15, 24

Apabila/When x = 15, y = 38 – 2(15) = 8 Apabila/When x = 56, y = 85 – 56 = 29

Apabila/When x = 24, y = 38 – 2(24) = −10 Maka/Hence, x = 45, y = 40
atau/or x = 56, y = 29
Oleh sebab/Since x > 0, y > 0,

maka/hence x = 15, y = 8 Soalan KBAT

Luas segi tiga/Area of triangle = 1 (15)(8) 1. p + q + r = 48 000 ..................a
2
p + r – q = 12 000
= 60 cm2 p – q + r = 12 000 ..................b

11. x – 3y = 5 0.035p + 0.05q + 0.04r = 2 000
x = 3y + 5 ..................a
x2 + y2 = 5 ...........................b 35p + 50q + 40r = 2 000 000

Gantikan a ke dalam b, 7p + 10q + 8r = 400 000 ........c

Substitute a into b, a – b : 2q = 36 000

(3y + 5)2 + y2 = 5 q = 18 000

9y2 + 30y + 25 + y2 = 5 b × 8 : 8p – 8q + 8r = 96 000 ..........d

10y2 + 30y + 20 = 0 d – c : p − 18q = −304 000 ......e

y2 + 3y + 2 = 0 Gantikan q = 18 000 ke dalam e,

(y + 1)(y + 2) = 0 Substitute q = 18 000 into e,

y = −1, −2 p − 18(18 000) = −304 000

Apabila/When y = −1, x = 3(−1) + 5 = 2 p = 20 000

Apabila/When y = −2, x = 3(−2) + 5 = −1 Gantikan p = 20 000 dan q = 18 000 ke

Maka/Hence, A(−1, −2) dan/and B(2, −1). dalam a,

Substitute p = 20 000 and q = 18 000 into a,

12. 162 + 302 = 34 20 000 + 18 000 + r = 48 000

r = 10 000

Perimeter = 190 m Maka/Therefore,
kos membeli saham P/cost of buying shares P
x + y + 34 + (x − 30) + (y + 16) = 190 = RM20 000

2x + 2y + 20 = 190 kos membeli saham Q/cost of buying shares Q
= RM18 000
y = 85 − x ...a
kos membeli saham R/cost of buying shares R
= RM10 000

© Penerbitan Pelangi Sdn. Bhd. 14

2. Perimeter = 340 m

x + y + 50 + (x – 30) = 340

y = 320 – 2x ...........a

Dengan theorem Pythagoras,

By using Pythagoras’ theorem,

(x – 30)2 + (x – 50)2 = y2

x2 – 60x + 900 + x2 − 100x + 2 500 = y2

2x2 – 160x + 3 400 = y2 ......b

Gantikan a ke dalam b,

Substitute a into b,

2x2 – 160x + 3 400 = (320 – 2x)2

2x2 – 160x + 3 400 = 102 400 – 1 280x + 4x2

2x2 – 1 120x + 99 000 = 0

x2 – 560x + 49 500 = 0

(x − 110)(x − 450) = 0

x = 110, 450

Apabila/When x = 110, y = 320 – 2(110)

= 100

Apabila/When x = 450, y = 320 – 2(450)

= −580

Oleh sebab/Since x  0, y  0,

∴ Luas/Area = 1 (110 + 50)(110 – 30)
2
= 6 400 m2


15 © Penerbitan Pelangi Sdn. Bhd.

Jawapan

Tingkatan 4 Indeks, Surd dan Logaritma 5. Katakan/Let, x = 0.0234234… a

BAB Indices, Surds and Logarithms Maka/Thus, 10x = 0.234234… b 

4 dan/and 10 000x = 234.234…  c 

Masteri SPM c – b, 9 990x = 234
234
x = 9 990

Kertas 1 13
555
1. 3k + 3k + 3k = 3m =

3(3k) = 3m Jadi/Thus, 0.0234234… = 13
555
31 + k = 3m

1 + k = m 6. u = 2gh
= 2 × 10 × 16
k = m – 1 = 320
= 64 × 5
2. 2y + 3x = 5 + 8x = 64 × 5
= 85
2y + 3x = 5 + (23)x

2y × 23x = 5 + 23x

n × m = 5 + m

mn = 5 + m

mn – m = 5 7. (a) a > 0, a ≠ 1

m(n – 1) = 5 (b) log2 pq = 3
1
m = 5
n – 1 log2 p

3. 34493pq+–21 = 1 = 3 log2 p
= log2 p3

49p + 2 = 343q – 1 Maka/Thus, pq = p3
(72)p + 2 = (73)q – 1
(7)2p + 4 = (7)3q – 3 pq – p3 = 0

2p + 4 = 3q – 3 p(q – p2) = 0

p = 3q – 7 p = 0 (Ditolak kerana p ≠ 0
2 Rejected because p ≠ 0)

4. 11 – 2 = p atau/or
p  11 + 2 
q – p2 = 0
p2 = 11 – 2 11 + 2  q = p2

= 11 – 2

= 9

p = 3 atau/or p = –3

© Penerbitan Pelangi Sdn. Bhd. 16

8. log5 27 = log5 27 – log5 p2 12. (a) loga 25 = loga 52
p2 logp 27
= 2 loga 5

= logp 5 – 2 log5 p = 2k

logp 33 (b) log5 125a2 = log5 125 + log5 a2
logp 5
 = –2 1 = log5 53 + 2 log5 a
logp 5
 =3 + 2 1
3x 2 loga 5
= y – y
2
= 3x – 2 = 3 + k
y
= 3k + 2
k
9. loga b5a2b – x2 = 2
1   13. log2
23x = log2 23x – log2  37 + 23x
(5a2b – x2)b = a2  37 + 23x
1
5a2b – x2 = a2b
= 3x – log2  (7 + 23x)3
x2 = 5a2b – a2b

= 4a2b = 3x – 1 log2  (7 + 23x)
3
x = 2ab

10. log1m p + logp m = 5 Maka/Thus, a = 3 dan/and b = – 31 .

logp m + logp m = 5 Kertas 2

2 logp m = 5   
1. (a)  p 1 + 2 2 12 4
5 3
2 q3 p3 – p3q3 + q3
logp m =
12 112 14 22 122

5 = p3p3 – p3p3q3 + p3q3 + p3q3 – p3q3q3

m = p2 24

2 + q3q3

p = m5 22 14 22 14
= p – p3q3 + p3q3 + p3q3 – p3q3 + q2

11. logn 864 – log3n 3n = 2 = p + q2

logn 864 – logn 3n = 2 (b) 32x + 2 × 5x – 1 = 27x × 52x
logn 3n 32x + 2 52x
27x = 5x – 1

logn 864 – logn 3n = 2 32x + 2
33x
1 = 52x – x + 1

logn n3 32 – x = 5x + 1

logn 864 – logn 3n = 2 32 × 3–x = 5x × 51
1
39x = 5x × 5
3

logn 864 – 3 logn 3n = 2 9 ×
5
logn 864 – logn (3n)3 = 2 = 3x 5x

logn 864 – logn 27n3 = 2 15x = 9
864 5
logn 27n3 = 2

logn 32 = 2
n3

32 = n2
n3
n5 = 32

n = 2

17 © Penerbitan Pelangi Sdn. Bhd.

2. (a) 112x – 11x = 0 4. 5 + 3 x2 + 3x + 5 – 3  = 0
(11x)2 – 11x = 0
dengan keadaan/where
Katakan/Let u = 11x a = 5 + 3, b = 3, c = 5 – 3

Maka/Then, u2 – u = 0 Dengan rumus kuadratik/By quadratic formula,

u(u – 1) = 0 –b ± b2 – 4ac
2a
u = 0 atau/or u – 1 = 0 x =
11x = 0 u = 1
Tiada penyelesaian = –3 ± 32 – 45 + 3 5 – 3 
11x = 1
No solution 11x = 110 25 + 3 
∴ x = 0
= –3 ± 9 – 4(5 – 9)
(b) 27 – x – x + 1 = 0 25 + 3 

27 – x = x + 1 = –3 ± 25
25 + 3 
27 – x 2 = x + 1 2

4(7 – x) = x + 1 = –3 ± 5
25 + 3 
28 – 4x = x + 1

5x = 27 ∴ x = –3 + 5
25 + 3 
x = 27
5
2
3. (a) m = 2 + 1× 2 + 1 = 25 + 3 
2 – 1  2 + 1 
= 1 × 5 – 3
2 + 22 + 1 5 + 3  5 – 3 
= 2 – 1 
5 – 3
= 3 + 22 = 5 – 9 

(b) m1 = 2 – 1 = 5 – 3
2 + 1  –4

= 2 – 1 × 2 – 1 = 3 – 1 5
2 + 1  2 – 1  4 4

2 – 22 + 1 atau/or
2 – 1 
= x = –3 – 5
25 + 3 
= 3 – 22
= –8
m – m1 = 3 + 22 – 3 – 22  25 + 3 
= 3 + 22 – 3 + 22 
= 42 = –4 × 5 – 3
5 + 3  5 – 3 

= –45 – 3 
5 – 9 

= –45 – 3 
–4

= –  3 + 5

Maka/Thus,

x = 3 – 1 5  atau/or x = –  3 + 5
4 4

© Penerbitan Pelangi Sdn. Bhd. 18

5. (a) 1 + ex – 1 = ex b – a, –10q = 20

ex – 1 = ex – 1 q = –2

ex – 1 = (ex – 1)2 Gantikan/Substitute q = –2 ke dalam/into a,

ex – 1 = (ex)2 – 2ex + 1 p + 2(–2) = 5

(ex)2 – 3ex + 2 = 0 p – 4 = 5

Katakan/Let y = ex p = 9 Tanda
ketaksamaan
y2 – 3y + 2 = 0
8. (a) 0.7x  5 perlu
(y – 1)(y – 2) = 0 disongsangkan
log10 0.7x  log10 5 kerana log10 0.7
y = 1 atau/or y = 2 x log10 0.7  log10 5 adalah negatif.
The inequality
ex = 1 ex = 2 log10 5 sign must be
log 0.7 reversed because
ln ex = ln 1 ln ex = ln 2 x  log10 0.7 is
negative.
x = 0 x = 0.6931 x  – 4.51

(b) 3x = 5x – 1 Oleh kerana x merupakan integer,
x = –5.
log10 3x = log10 5x – 1 Since x is an integer, x = –5.

x log10 3 = (x – 1) log10 5

x log10 3 = x log10 5 – log10 5

log10 5 = x log10 5 – x log10 3 (b) Gantikan/Substitute t = 0

x (log10 5 – log10 3) = log10 5 M(0) = 7.0 × 20

x = log10 5 = 7.0
log10 5 – log10
3 Maka, jisim awal koloni bakteria ialah

= 3.151 7 gram.

Thus, the initial mass of the bacterial

6. (a) a = ln b ⇔ ea = b colony is 7 grams.

∴ e3 ln eb = eln (eb)3 Apabila jumlah jisim bakteria melebihi

= (eb)3 800 g,
= e3 × b3
= e3 × (ea)3 When the total mass of the bacteria exceeds
= e3 × e3a
800 g,

M(t)  800

= e3 + 3a 7.0 × 2t  800
800
(b) ex = 15 2t  7

ln ex = ln 15   log10 2t  log10 800
7
x ln e = ln 15

x = ln 15   t log10 2  log10 800
7
= 2.708

7. 11p – 2 × 121q = 1 331  t  log10 800
11p – 2 × (112)q = 113 7
11p – 2 × 112q = 113
log10 2
11p – 2 + 2q = 113
t  6.84

Maka/Thus, p – 2 + 2q = 3 Jadi, 7 hari diperlukan bagi jumlah
jisim bakteria untuk melebih 800 g.
p + 2q = 5 ......a
Therefore, 7 days are needed for the total
log5 (p – 8q) = 2 mass of the bacteria to exceed 800 g.
p – 8q = 52
p – 8q = 25  ....................b

19 © Penerbitan Pelangi Sdn. Bhd.

Soalan KBAT 2. PQ

3 cm 1.5 cm E 1.5 cm
C
1. x + y = 16 + 2 ..........a A 5 cm 2.5 3.5 cm
x + z = 18 – 82 ........b 1 cm cm
y + z = 22 – 2 ..........c
D B 2 cm

a – b: y – z = –2 + 92 ..........d Dalam segi tiga ADB/In the triangle ADB,

c + d: 2y = 20 + 82 AB = 5 cm, AD = 1 cm

y = 10 + 42 ..............e Maka/Thus, DB2 = AB2 – AD2

Gantikan/Substitute e ke dalam/into a: = 52 – 12

x + 10 + 42 = 16 + 2 = 24

x = 6 – 32 DB = 24

Gantikan/Substitute e ke dalam/into c: = 26 cm

10 + 42 + z = 22 – 2 Dalam segi tiga BEC/In the triangle BEC,
BC = 2 + 1.5, BE = 6 – 1.5 – 2
z = 12 – 52 = 3.5 cm = 2.5 cm
EC2 = BC2 – BE2
Maka/Thus, = 3.52 – 2.52
x = 6 – 32 = 12.25 – 6.25
y = 10 + 42 = 6
z = 12 – 52 EC = 6 cm

Oleh kerana/Since PQ = DB + EC,

Maka/Thus, PQ = 26 + 6

= 36 cm

© Penerbitan Pelangi Sdn. Bhd. 20

Jawapan

Tingkatan 4 Janjang 4. T6 = 9

BAB Progressions a + (6 − 1)d = 9

5 a + 5d = 9

a = 9 − 5d .........a

Masteri SPM S8 = −12

Kertas 1 28 [2a + (8 − 1)d] = −12
4(2a + 7d) = −12
p + 3
1. (a) Sn = 2 (a + k) 2a + 7d = −3 .........b

n Gantikan a ke dalam b,
2
= (a + l) Substitute a into b,

∴ n = p + 3 2(9 – 5d) + 7d = −3

p = n – 3 18 – 10d + 7d = −3

(b) Suatu janjang mestilah mempunyai −3d = −21
sekurang-kurangnya 2 sebutan.
A progression must have at least 2 terms. d = 7

∴ p + 3  2 a = 9 – 5(7)
p  −1 = −26

2. (a) d = 21 – 17 5. Fara: a = 150, d = 50
= 4 Hasliza: a = 470, d = 30

(b) T5 = 17 TnF = TnH
150 + (n – 1)(50) = 470 + (n – 1)(30)
a + (5 − 1)(4) = 17
150 + 50n – 50 = 470 + 30n – 30
a + 16 = 17 20n = 340
n = 17
a = 1
Nilai wang dalam akaun Fara pada masa itu
S13 = 13 [2(1) + (13 − 1)(4)] Amount of money in Fara's account at that time
2
13 = 150 + (17 – 1)(50)
= 2 (2 + 48) = RM950

= 325 6. (a) p = −1, 0, 1

3. Tn = Sn – Sn−1 (b) a = T1

= n (9 − 5n) − n − 1 [9 − 5(n − 1)] = 5r1 – 1
2 2 3

= 9n − 5n2 − n − 1 (14 − 5n) = 5r 0
2 2 3

= 9n − 5n2 − 14n − 5n2 − 14 + 5n = 5
2 2 3

= –10n + 14
2

= −5n + 7

21 © Penerbitan Pelangi Sdn. Bhd.

7. 18 + h + k = 14 10. (a) 5n = 5
k
k = −h − 4 ............a

h = k ....................b nk = 25
18 h 25
k = n
Gantikan a ke dalam b,
25
Substitute a into b, Sebutan pertama/First term = n

h = –h – 4 Nisbah sepunya/Common ratio = n
18 h 5

h2 = −18h – 72 25

h2 + 18h + 72 = 0 (b) S∞ = n

(h + 6)(h + 12) = 0 1 – n
5
h = −6, h = −12

Apabila/When h = −6, k = −(−6) – 4 25

=2 = n
5–n
Apabila/When h = −12, k = −(−12) – 4

=8 5

8. (a) 2xx+– 6 = 1 = 125
3 2 5n – n2

4x – 12 = x + 3 11. a = −12, r = 8 = – 23
−12
3x = 15

x = 5 z = T5

(b) T9 = x + 3   = −12 − 23 5 − 1

  a1 9–1 5 + 3 = − 64
2 27
=

  a 1 = 8   S∞ = −12
256
1 – – 32
a = 2 048 = − 356

9. (a) T4 = S4 – S3 12. a = 800, = 1.05

= [2(54+1) – 10] – [2(53+1) – 10]
= 6 240 – 1 240
= 5 000

(b) Sn = 2(5n+1) – 10 Bilangan bulan dari Januari 2015 hingga
April 2016,
= 2(5n)(51) – 10 Number of months from January 2015 to April
2016,
= 10(5n) – 10

= 10(5n − 1) n = 12 + 4 = 16

= 10(5n − 1)(5 − 1) S16 = 800(1.0516 – 1)
= 1.05 – 1
(5 − 1) Kaedah Alternatif
40(5n − 1) a = S1 = 2(51 + 1) – 10 = RM18 926

5−1 = 40
Also/Juga,
∴ a = 40, r = 5 T4 = 5 000 Pemilik itu memperoleh keuntungan
40(r 4 − 1) = 5 000 The owner makes a profit
r3 = 125
r = 5 = RM25 000 – RM18 926
= RM6 074

© Penerbitan Pelangi Sdn. Bhd. 22

13. Untuk Syarikat Alfa/For Alpha Company, Gantikan a ke dalam b,

a = 2 800 × 12 = 33 600, r = 1.08 Substitute a into b,

(33 600)(1.088 – 1) 2(6 – 4d) + 7d = 15
1.08 – 1
S8 = = 357 391 12 – 8d + 7d = 15

Untuk Syarikat Beta/For Beta Company, −d = 3
a = 3 500 × 12 = 42 000, r = 1.05
d = −3

S8 = (42 000)(1.058 – 1) = 401 063 Gantikan d = −3 ke dalam a,
1.05 – 1 Substitute d = –3 into a,

Syarikat Beta akan dipilih. a = 6 – 4(−3)
Beta Company will be chosen. =18

Jumlah simpanan/Total savings (b) Tn = 1 − n
= RM401 063 × 30%
18 + (n − 1)(−3) = 1 – n
= RM120 319
18 – 3n + 3 = 1 – n

2n = 20

Kertas 2 n = 10

5(12)(12 – 17) (c) Hasil tambah dari sebutan ke-9 hingga
2
1. (a) S12 = sebutan ke-20

= −150 Sum from 9th term to 20th term

(b) a = S1 = S20 – S8

= 5(1)(1 – 17) = 20 [2(18) + (20 − 1)(−3)] − 60
2 2

= −40 = −270

S12 = −150 3. (a) 3r – (2p – 1) = (4p – 5) – 3r

122 [2(−40) 3r – 2p + 1 = 4p – 5 – 3r

+ (12 − 1)d] = −150 6r = 6p – 6
−80 + 11d = −25
r = p − 1

11d = 55 (b) a = 2p – 1
= 2(r + 1) – 1
d = 5 = 2r + 1

(c) Tn  0

−40 + (n – 1)(5)  0 d = 3r – (2r + 1)
= r − 1
(n – 1)(5)  40

n – 1  8 S10 = 10 [2(2r + 1) + (10 − 1)(r − 1)]
2
n  9
= 5(4r + 2 + 9r – 9)
p = 10
= 5(13r – 7)
2. (a) T5 = 6
= 65r − 35
a + (5 − 1)d = 6

a + 4d = 6

a = 6 – 4d ..........a

S8 = 60

82 [2a + (8 − 1)d] = 60
4(2a + 7d) = 60

2a + 7d = 15 .........b

23 © Penerbitan Pelangi Sdn. Bhd.

4. a = 4500, d = −300 6. (a) T2 = 14
ar = 14 ..............a
Tn = 600

4500 + (n – 1)(−300) = 600 S3 = 114

(n – 1)(−300) = −3900 a + ar + ar2 = 114 ............b

n – 1 = 13 Gantikan a ke dalam b,
Substitute a into b,
n = 14

S14 = 14 (4500 + 600) a + 14 + 14r = 114
2
= 35 700 14
Daripada a/From a, r = a

Jumlah kos/Total cost = RM0.80 × 35 700 a + 14 + 14 14  = 114
a
= RM28 560 a2 + 14a + 196 = 114a

5. (a) a = 6k, r = 4k = 2 a2 – 100a + 196 = 0
6k 3
(a – 2)(a – 98) = 0
S∞ = 45
a = 2, a = 98

6k = 45 Apabila/When a = 2, r = 14 =7
2 2
1 – 3
14 1
Apabila/When a = 98, r = 98 = 7
6k = 15

k = 2.5 (b) Sn = 39 216

(b) a = 6(2.5) = 15, r = 2 2(77n–−11) = 39 216
3

Tn  1 2(7n6− 1) = 39 216

  15 2 n–1 1 7n – 1 = 117 648
3


 2 n–1 1 7n = 117 649
15
3  7n = 76

  log10 2 n–1 log10 1 n = 6
3 15
 98

  (n − 1) log10 2  log10 1 (c) S∞ = 1
3 15 7
1 –

n − 1  6.68 1
3
n  7.68 = 114

Nilai minimum n/Minimum value of n
=8

© Penerbitan Pelangi Sdn. Bhd. 24

7. (a) T4 = 8T7 Soalan KBAT

ar4−1 = 8(ar7−1)
aarr36 = 8
1 1. (a) Katakan/Let D = diameter
r3 = 8
T1 = πD
r3 = 1
8 T2 = π(D + 4)
= πD + 4π
r = 1
2 T3 = π(D + 8)
= πD + 8π

Bahagian tali paling panjang T2 – T1 = πD + 4π – πD = 4π
= sebutan pertama T3 – T2 = πD + 8π – (πD + 4π) = 4π

Longest part of rope = first term Janjang yang terbentuk ialah janjang
aritmetik kerana mempunyai beza
a = 896 cm, Sn = 1785 cm sepunya 4π.

    (896) 1 – 1 n The progression formed is an arithmetic
2 progression as it has a common difference
= 1785 of 4π.
1
1 – 2 (b) T13 = 3T3

    1 n a + (13 − 1)(4π) = 3[a + (3 − 1)(4π)]
1792 1 – 2 = 1785
a + 48π = 3(a + 8π)
   1 –1 n 1785
2 = 1792 a + 48π = 3a + 24π

 1 n 1785 2a = 24π
1792
2 = 1– a = 12π

 1 n 1 2πr = 12π
256
2 = r = 6 cm

   1n 18 Jejari bulatan terkecil ialah 6 cm.
2 Radius of the smallest circle is 6 cm.
2 =

n = 8

 (b) T8 = (896) 1 8–1
2

= 7 cm

25 © Penerbitan Pelangi Sdn. Bhd.

Soalan KBAT dalam kod QR pada m.s. 66

1. Dari 40 tahun hingga 56 tahun:
From 40 years old to 56 years old:

56 − 40 + 1 = 17 tahun/years

Bagi Syarikat A/For Company A:
a = 42 000, d = 4800

S17 = 17 [2(42 000) + (17 − 1)(4800)]
2

= RM1 366 800

Bagi Syarikat B/For Company B:
a = 36 000, r = 1.06

S17 = 36 000(1.0617 − 1)
1.06 – 1

= RM1 015 664

Syarikat A lebih memanfaatkan Irwan.
Company A benefits Irwan more.

© Penerbitan Pelangi Sdn. Bhd. 26

Jawapan

Tingkatan 4 Hukum Linear 4. xy = ax2 – bx
xy2 = ax – b  dan/and
BAB Linear Law = y a − b 1
−b = 4r x
6 b = −4r .......a x3

Masteri SPM a = 1 − r

r = 1 − a .......b

Kertas 1 Gantikan b ke dalam a,

(a)   yx 6 2 Darab dengan x Substitute b into a,
 x 3x Multiply by x
1. = + b = −4(1 − a)

2 b = −4 + 4a
 3x
y = 6 + a = b + 4
4
2
Kecerunan/Gradient = 3 –6 – 0 = – 53 , c = −6
0 – (–10)
(b) A(0, 6) 5. m=

2. (a) ky = px2 + 15 Dalam bentuk/In the form Y = mX + c,
x
p y + x = – 53 x3 + (−6)
y = k x2 + 15 y = – 35 x3 − x − 6
kx
p
xy = k x3 + 15
k
6. y = 3x + k
Pintasan-Y/Y-intercept, x2
1k5 = 5
x2y = 3x3 + k

k = 3 k = −9

p 5 – 3 Gantikan (p, 2) ke dalam Y = 3X – 9,
3 0 – 6
(b) Kecerunan/Gradient, = Substitute (p, 2) into Y = 3X – 9,

p = – 31 2 = 3(p) – 9
3
11 = 3p

p = −1 p = 11
3
3. (a) y = 64nx
log4 y = log4 64nx 7. y = x + b
log4 y = log4 64 + log4 nx x
log4 y = log4 nx + log4 64   y – x = b
log4 y = x log4 n + log4 43 1
log4 y = (log4 n)x + 3 x

  Gantikan k , 4n ke dalam Y = bX,
3
  Substitute
∴ h = 3 k , 4n into Y = bX,
3
11 – 3   4n = b k
(b) m = 4–0 = 2 3

∴ log4 n = 2 k = 12n
b
n = 42 = 16

27 © Penerbitan Pelangi Sdn. Bhd.

8. Kecerunan/Gradient, m = 4 – (–2) = – 35 Gantikan (6, 2) ke dalam Y = – 31 X + c,
0 – 10 Substitute (6, 2) into Y = – 13 X + c,
2 = – 13 (6) + c
Dalam bentuk/In the form Y = mX + c, c = 4

  xy = – 53 1 +4 Pintasan-Y/Y-intercept,
x2 log10 k = 4
k = 104
yx = – 53x2 + 4 = 10 000

yx = –3 + 20x2
5x2

1y = –3 + 20x2
5x3

y = 5x3 3 Kertas 2
20x2 –

9. y = 3 1. (a) x + 1 2 3 4 5 6 7
2x – 12 xy 29.0 24.0 19.0 14.0 9.0 4.0

1y = 2x – 12 xy
3
40
1y = 2x – 4 Graf xy melawan (x + 1)
3 39

Graph of xy against (x + 1)

∴ h = −4 30

Gantikan (k, 0) ke dalam Y = 2 X – 4, 20
3
2 17
Substitute (k, 0) into Y = 3 X – 4,
10

0 = 2 k – 4 O x+1
32 k = 3 1 2 3 4 4.4 5 6 7

4 (b) (i) Apabila/When x = −1, x + 1 = 0
dan/and xy = 39
k = 6
(−1)y = 39
10. y = k y = −39
 nx
k (ii) Apabila/When x = 3.4, x + 1 = 4.4
log10 y = log10  nx dan/and xy = 17

log10 y = log10 k – log10 nx (3.4)y = 17
y = 5
1
(iii) Pintasan-xy/xy-intercept = 39
log10 y = log10 k – log10 xn

log10 y = log10 k – 1 log10 x Kecerunan graf/Gradient of graph,
n
log10 y = – 1n log10 x + log10 k 4 – 39
m= 7–0 = −5

Kecerunan/Gradient, – n1 = 7–2 Persamaan/Equation:
–9 – 6 xy = −5(x + 1) + 39

– 1n = – 13 2. (a)

n = 3 log10 x 0.2 0.3 0.5 0.7 0.8 1.1
log10 y −0.14 −0.07 0.07 0.21 0.28 0.49

© Penerbitan Pelangi Sdn. Bhd. 28

log10 y (c) (i) Apabila/When x = 3.6, x2y = 24
0.4
Graf log10 y melawan log10 x Maka/Thus, (3.6)2y = 24
Graph of log10 y against log10 x
y = 1.85

0.2 (ii) y = h – hk
x x2
0.16
x2y = hx – hk
O
0.2 0.4 0.63 0.8 1.0 1.2 log10 x h = kecerunan graf/gradient of graph
–0.2
0.6 30.5 – 8
–0.28 5–0
=

= 4.5

(b) (i) Apabila/When x = 4.3, log10 x = 0.63 (iii) −hk = pintasan-x2y/x2y-intercept
−4.5k = 8
Maka/Thus, log10 y = 0.16 k = −1.78

y = 1.445

(ii) Apabila/When y = 1, log10 y = 0 4. (a) 1 0.1 0.2 0.3 0.4 0.5 0.6
x
Maka/Thus, log10 x = 0.4
x = 2.512 y −4.2 −3.0 −1.8 −0.6 0.6 1.8

(iii) Pintasan-log10 y = −0.28 y
log10 y-intercept
2 Graf y melawan 1
x
1
Kecerunan graf/Gradient of graph, Graph of y against x

m = 0.49 – (–0.28) = 0.7 0.25 1
1.1 – 0 0.5 0.6 x
O 0.1 0.2 0.3 0.4

Persamaan/Equation: –2
log10 y = 0.7 log10 x − 0.28
–2.4

3. (a) x 1 2 3 4 5 6 –4

x2y 12.5 17.0 21.5 26.1 30.5 34.9 –5.4
(b)
(b) (i) Apabila/When x = 4, 1 = 0.25
x2y x

35 Maka/Thus, y = −2.4
Graf x2y melawan x
Graph of x2y against x (ii) px + xy = q

30 xy = q – px

  y = q 1 −p
x
25
−p = pintasan-y/y-intercept
24 −p = −5.4
  p = 5.4
20

15 (iii) q = kecerunan graf/gradient of graph

10 = 1.8 – (–5.4)
0.6 – 0
8
= 12
5

O x
1 2 3 3.6 4 5 6

29 © Penerbitan Pelangi Sdn. Bhd.

5. (a) 6. (a)
x 123456 log10 x −0.4 −0.3 −0.2 −0.1 0.1 0.2
log10 y 0.14 0.34 0.54 0.74 1.14 1.34
log10 y 0.59 0.51 0.43 0.35 0.27 0.19
(b)
log10 y
log10 y
0.7 Graf log10 y melawan x
Graph of log10 y against x 1.4
0.67

0.6

1.2

0.5

0.4 Graf log10 y melawan log10x 1.0
Graph of log10 y against log10x
0.39 0.94

0.3 0.8

0.6

0.2

0.4

0.1

0.2

O x
1 2 3 3.5 4 5 6

–0.4 –0.3 –0.2 –0.1 O 0.1 0.2 log10 x

(b) (i) Apabila/When x = 3.5,

log10 y = 0.39 (c) (i) y =  r x
y = 2.455 p
 r x
(ii) y = a log10 y = log10 p
log10 y = bx
log10 a log10 y = log10  r x – log10 p
bx
log10 y = log10 a – log10 bx 1

log10 y = log10 a − x log10 b log10 y = log10 xr – log10 p

log10 y = (−log10 b)x + log10 a log10 y = 1 log10 x − log10 p
r

Pintasan-log10 y/log10 y-intercept, Pintasan-log10 y/log10 y-intercept,
log10 a = 0.67 −log10 p = 0.94
a = 4.677 log10 p = −0.94
p = 0.115
(iii) Kecerunan graf/Gradient of graph,

–log10 b = 0.67 – 0.27 (ii) Kecerunan graf/Gradient of graph,
0 – 5
1 = 0.94 – 0.14
−log10 b = −0.08 r 0 – (–0.4)

log10 b = 0.08 1 = 2
r
b = 1.202

r = 0.5

© Penerbitan Pelangi Sdn. Bhd. 30

7. (a) log3 y

1 4.0 6.0 8.0 10.0 12.0 14.1 B(5, 3)
x

y2 4.0 10.0 16.0 22.0 28.0 34.0 A(2, 1)
x
O x

(b) ∴ A(2, 1) dan/and B(5, 3)

y2 (b) Kecerunan/Gradient,
x
30 y2 m= 3–1 = 2
20 x melawan 5–2 3
Graf 1
x dYa=lam23 XY+=c,32
y2
Graph of x against 1 Gantikan (2, 1) ke X + c,
x Substitute (2, 1) into

12.5 1 = 2 (2) + c
3
10

1 c = − 31
O 10 12 14 x
2 4 6 6.8 8

–8 Daripada/From y = pqx,

1 log3 y = log3 pqx
x
(c) (i) Apabila/When x = 0.147, = 6.8 log3 y = log3 p + log3 qx
y2
dan/and x = 12.5 x
log3 y = log3 p + log3 q2
x
∴ y2 = 12.5 log3 y = log3 p + 2 log3 q
0.147
  log3 y = 1 q x + log3 p
y2 = 1.8375 2 log3

y = 1.356 Pintasan-Y/Y-intercept,

(ii) Pintasan-Y/Y-intercept, log3 p = − 13
c = −8

Kecerunan graf/Gradient of graph, p = 3– 31

m = 28 – (–8) = 1
12 – 0 3 3

= 3 = 0.693

Dalam bentuk/In the form Y = mX + c, Kecerunan/Gradient,

  y2 = 3 1 −8 12 log3 q = 2
x x 3

y2 = 3 – 8x 4
3
  y = 3 – 8x log3 q =

4

Soalan KBAT q = 33

1. (a) Untuk titik A/For point A, x = 2, = 334
log3 y = log3 3 = 1
= 4.327
Untuk titik B/For point B, x = 5,

log3 y = log3 27 = 3

31 © Penerbitan Pelangi Sdn. Bhd.

2. (a) 1 0.5 1.0 1.5 2.0 2.5 3.0 (ii) Pintasan-Y/Y-intercept,
M
c = 60

T 54 48 42 36 30 24 Kecerunan graf/Gradient of graph,
M
m = 60 – 30 = −12
0 – 2.5
T
M (°C kg–1) Dalam bentuk/In the form Y = mX + c,

60 Graf T melawan 1   MT = –121
M M M
T 1 + 60
Graph of M against M

41 T = −12 + 60M

40

T = 60M − 12

20

O 1.51.6 1 (kg–1)
M
0.5 1.0 2.0 2.5 3.0

(b) (i) M = 625 g
= 0.625 kg

1 = 1
M 0.625

= 1.6

Daripada graf/From the graph,

MT = 41
0.6T25 = 41
T = 25.625°C

© Penerbitan Pelangi Sdn. Bhd. 32

Jawapan

Tingkatan 4 Geometri Koordinat PA : PB = 1 : 2

BAB Coordinate Geometry   P =
2(–7) + 1(8) , 2(–1) + 1(5)
7 1+2 1+2

Masteri SPM   = –6 , 3
3 3

Kertas 1 = (−2, 1)

4. (a) 7

1. 4AB = 7TB 3 A(8, 11) (b) PN : NQ = 1 : 5
AB : TB = 7 : 4 T(x, y)
AT : TB = 3 : 4 4 Katakan koordinat Q ialah (x, y).
B(–6, 4)
Let coordinates of Q be (x, y).

  5(31) 1(x)
  T = + 5 5(–2) + 1(y)
4(8) + 3(–6) , 4(11) + 3(4) + , 1+5 = (4, 0)
3+4 3+4
 15 +
6
  = 14 , 56 x, –10 + y = (4, 0)
7 7 6

= (2, 8) 156+ x = 4 –10 + y = 0
6
 
2. 1(h)2++21(10) , 1(–4) + 2(p) = (2p, k) 15 + x = 24 –10 + y = 0
2+1
x = 9 y = 10
 h –4 + 2p
+ 20 , 3 = (2p, k) ∴ Q(9, 10)
3

h + 20 = 2p –4 + 2p = k 5. Garis selari/Parallel lines
3 3 ⇒ kecerunan sama/same gradient

h + 20 = 6p ...a −4 + 2p = 3k ∴ p – 4 = 2
p = 6
p = 3k + 4 ...b
2
6. (a) k x + hy – 4 = 0
Gantikan b ke dalam a,
Substitute b into a, hy = −kx + 4

  h + 20 = 6 y = – hk x + 4
3k + 4 h
2
h + 20 = 9k + 12 Kecerunan/Gradient = − hk

h = 9k − 8 (b) m1 × m2 = −1
− hk × m2 = −1
3. Katakan kedua-dua semut bertemu di P(x, y).
Let both ants meet at P(x, y). ∴ Kecerunan/Gradient = h
k
Oleh sebab halaju B = 2 × halaju A
Since velocity of B = 2 × velocity of A   7. (a) M = –8 + 0, 0 + 6
2 2
∴ Jarak dilalui B = 2 × jarak dilalui A
Distance covered by B = 2 × distance covered = (−4, 3)
 by A

33 © Penerbitan Pelangi Sdn. Bhd.

(b) m1 = 6–0 = 3 9. 15 0 2 5 = 8
0 – (–8) 4 2k –3 1 k

∴ m2 = – 13 = − 43 21|[(5)(−3) + (0)(1) + (2)(k)] − [(k)(0)
+ (−3)(2) + (1)(5)]| = 8
4
1
Gantikan m2 = − 43 dan (−4, 3) ke dalam 2 |(−15 + 0 + 2k) − (0 − 6 + 5)| = 8
y = mx + c,
− 34 |2k − 14| = 16

Substitute m2 = and (−4, 3) into 2k – 14 = 16 2k – 14 = –16
y = mx + c, 2k = 30 2k = −2
k = 15 k = –1
3 = − 34 (−4) + c
c = − 37 10. 1h 4 k h = 0
2 –2 1 3 –2

Persamaan garis lurus ialah 21|[(ℎ)(1) + (4)(3) + (k)(−2)]

Equation of straight line is − [(−2)(4) + (1)(k) + (3)(ℎ)]| = 0

y = − 43 x − 7 . |(ℎ + 12 − 2k) − (−8 + k + 3ℎ)| = 0
3
|20 − 2ℎ − 3k| = 0
8. (a) 3y – x = 18
20 – 2h – 3k = 0
3y = x + 18
3k = 20 – 2h
1
y = 3 x + 6 k = 20 – 2h
3
1 11. Katakan d = jarak terdekat
m1 = 3 Let d = shortest distance

m × m1 = −1 1 × ×
2
m × 1 = –1 Luas/Area of ∆ABC = AB d
3
12 × (9 − 0)2 =+ (121 08− 8)–521× d
∴ m = −3 9 0
1 8
(b) Gantikan y = −3x + 26 ke dalam
3y – x = 18, 11.4018d = |(0 + 5 + 72) − (40 − 9 + 0)|
Substitute y = −3x + 26 into 3y – x = 18,
11.4018d = |46|
3 (−3x + 26) – x = 18
11.4018d = 46
–9x + 78 – x = 18
d = 4.0345 m
–10x = –60
12. (x − 4)2 + (y − 0)2 = 7
x = 6 (x – 4)2 + y2 = 72
x2 – 8x + 16 + y2 = 49
y = –3(6) + 26 x2 + y2 – 8x – 33 = 0
= 8

∴ E(6, 8)

© Penerbitan Pelangi Sdn. Bhd. 34

13. AP : PB = 1 : 3

AP = 1
PB 3

3AP = PB

3(x − 0)2 + [y − (−2)]2 = (x − 5)2 + (y − 3)2

9[x2 + (y + 2)2] = (x – 5)2 + (y – 3)2

9(x2 + y2 + 4y + 4) = x2 – 10x + 25 + y2 – 6y + 9

9x2 + 9y2 + 36y + 36 = x2 + y2 – 10x – 6y + 34

8x2 + 8y2 + 10x + 42y + 2 = 0

4x2 + 4y2 + 5x + 21y + 1 = 0

14. [x − (−5)]2 + (y − 3)2 = (x − 10)2 + (y − 8)2

(x + 5)2 + (y – 3)2 = (x – 10)2 + (y – 8)2

x2 + 10x + 25 + y2 – 6y + 9 = x2 – 20x + 100 + y2 – 16y + 64

30x + 10y – 130 = 0

3x + y – 13 = 0

Kertas 2

1. (a) Daripada/From 2y + x + 2 = 0  
4(–3) + 1(x) , 4(8) + 1(y) = (–6, 2)
2y = –x – 2 1+4 1+4
y = – 21x – 1
–125+ x = –6 32 + y = 2
–12 + x = –30 5
– 21 , 1 x = –18
mBC = mAB = –  – 21 =2 32 + y = 10

y = –22

Persamaan garis lurus AB ialah ∴ D(–18, –22)

Equation of the straight line AB is 2. (a) 2x – 3y + 15 = 0
y – 8 = 2[x – (–3)]
y – 8 = 2x + 6 3y = 2x + 15
y = 2x + 14
y = 2 x + 5
3
(b) Gantikan y = 2x + 14 ke dalam
2y + x + 2 = 0, mAB = 2
Substitute y = 2x + 14 into 2y + x + 2 = 0, 3

2(2x + 14) + x + 2 = 0 ∴ k × mAB = −1

4x + 28 + x + 2 = 0 k × 2 = – 1
3
5x = –30 k = – 32

x = –6

y = 2(–6) + 14
= 2
∴ B(–6, 2)

35 © Penerbitan Pelangi Sdn. Bhd.

(b) Gantikan y = – 23 x ke dalam (c) Gantikan y = 3x – 15 ke dalam
2x – 3y + 15 = 0,
y = – 13x + 1, – 31
Substitute y = – 32 x into 2x – 3y + 15 = 0, Substitute y = 3x – 15 into y = x + 1,

  2x – 3 – 23x + 15 = 0 3x – 15 = – 13x + 1
130 x = 16
123x = −15 24
5
x = − 1303 x =

  y = – 23 – 3103   24
y=3 5 – 15

= 45 = – 53
13
  24 – 35
  ∴S 5 ,
∴ A − 3130 , 45
13

(c) mBC = mOA = – 32 (d) Luas PQRS

Persamaan garis lurus BC ialah Area of PQRS

Equation of the straight line BC is = 2 × luas segi tiga PRS
2 × area of triangle PRS
y – 1 = – 32 (x – 14)
y – 1 = – 23x + 21 = 2 × 1 –3 24 6 –3
y = – 32x + 22 5

2 2 – 35 3 2

    =  9+ 72 + 12 – 48 – 18 – 9  
5 5 5 5

3. (a) mSR = mPQ = 3 =  1451 – (–3) 

Persamaan garis lurus SR ialah =  1556 
Equation of the straight line SR is
= 31.2 unit2
y – 3 = 3(x – 6)
y – 3 = 3x – 18 4. (a) (i) 2y + x = 14
y = 3x – 15 2y = –x + 14
y = – 21 x + 7
(b) mPS = –  1 = – 31
mPQ mCD = – 21

Persamaan garis lurus PS ialah

Equation of the straight line PS is

y – 2 = – 13[x – (–3)] ∴ mAD = 2  mAD × mCD = –1
y – 2 = – 31x – 1
y = – 31x + 1 Persamaan garis lurus AD ialah
Equation of the straight line AD is

y – (–1) = 2[x – (–5)]
y + 1 = 2x + 10
y = 2x + 9

© Penerbitan Pelangi Sdn. Bhd. 36

(ii) Gantikan y = 2x + 9 ke dalam 5. (a) (i) 1 0 12 2 –6 0
2y + x = 14, 2 0 6 10 4 0
Substitute y = 2x + 9 into 2y + x = 14,
= 1 |(0 + 120 + 8 + 0) – (0 + 12
2(2x + 9) + x = 14 2
– 60 + 0)|
4x + 18 + x = 14
1
5x = –4 = 2 |128 – (–48)|

x = – 54 = 1 |176|
2
  y = 2 – 45 + 9
= 88 unit2

37 (ii) DF : EF = 5 : 3
5
= 5 F(7, 8)
2 E(1, 5) 3
  – 54 , 37 D(x, y)
∴ 5
D ∴ DE : EF = 2 : 3

(b) Di/At C, x = 4, 2y + 4 = 14

 3x + 2(7)
2+3
2y = 10 , 3y + 2(8) = (1, 5)
2+3
y = 5

∴ C(4, 5) 3x +5 14 = 1 3y + 16 = 5
3x = −9 5
KC = 6 x = –3
3y = 9
(x − 4)2 + (y − 5)2 = 6
y = 3
(x – 4)2 + (y – 5)2 = 62

x2 – 8x + 16 + y2 – 10y + 25 = 36 ∴ D(–3, 3)

x2 + y2 – 8x – 10y + 5 = 0 (b) (x − 1)2 + (y − 5)2 = 3

(x – 1)2 + (y – 5)2 = 32

x2 – 2x + 1 + y2 – 10y + 25 = 9

x2 + y2 – 2x – 10y + 17 = 0

6. (a) [x − (−2)]2 + (y − 5)2 = (x − 4)2 + [y − (−4)]2

(x + 2)2 + (y – 5)2 = (x – 4)2 + (y + 4)2

x2 + 4x + 4 + y2 – 10y + 25 = x2 – 8x + 16 + y2 + 8y + 16

12x – 18y – 3 = 0

4x – 6y – 1 = 0

(b) x – 2y = 2   x = 2 – 27 + 2
x = 2y + 2 .....a

Gantikan a ke dalam 4x – 6y – 1 = 0, = –5

 
Substitute a into 4x – 6y – 1 = 0, Kedudukan tiang lampu = –5, – 27
Location of lamp post
4(2y + 2) – 6y – 1 = 0

8y + 8 – 6y – 1 = 0 ∴ Jarak/Distance

2y = –7   = [−5 − (−2)]2 + – 72 – 5 2

y = – 27 = 9.014 km

37 © Penerbitan Pelangi Sdn. Bhd.

7. (a) 10 k 8 0 = 38
20 5 9 0

1 |[(0)(5) + (k)(9) + (8)(0)] − [(0)(k) + (5)(8) + (9)(0)]| = 38
2
|(0 + 9k + 0) – (0 + 40 + 0)| = 76

|9k − 40| = 76

9k – 40 = 76 9k – 40 = –76
9k = –36
9k = 116
k = 116 k = –4

9

Oleh sebab/Since k  0, k = –4

(b) (i) PR : RQ = 5 – (–4) : 8 – 5

=9:3

=3:1

(ii) Katakan koordinat T ialah (x, y).
Let coordinates of T be (x, y).

PT = 2RT

[x − (−4)]2 + (y − 5)2 = 2(x − 5)2 + (y − 8)2

(x + 4)2 + (y – 5)2 = 4[(x – 5)2 + (y – 8)2]

x2 + 8x + 16 + y2 – 10y + 25 = 4(x2 – 10x + 25 + y2 – 16y + 64)

x2 + y2 + 8x – 10y + 41 = 4x2 – 40x + 100 + 4y2 – 64y + 256

3x2 + 3y2 – 48x – 54y + 315 = 0

x2 + y2 – 16x – 18y + 105 = 0

Soalan KBAT

1. (a) 11 k p 1 = 38
29 2 12 9 38
21 |[(1)(2) 38
+ (k)(12) + (p)(9)] – [(9)(k) + (2)(p) + (12)(1)]| =

1 |(2 + 12k + 9p) – (9k + 2p + 12)| =
2

|3k + 7p – 10| = 76

3k + 7p – 10 = 76 atau/or 3k + 7p – 10 = –76

7p = 86 – 3k 7p = –66 – 3k

p = 86 – 3k (diabaikan kerana nilai p dan nilai k sepatutnya
7 positif/ ignored as the value of p and of k should be

positive)

© Penerbitan Pelangi Sdn. Bhd. 38

(b) Apabila/When k = 10, p = 86 – 3(10) = 8
7

mPS = 9 – (–3) = –3
1–5

∴ −3 × m = −1
m = 1

3

Persamaan garis lurus yang melalui Q(8, 12) dan berserenjang dengan PS:

Equation of the straight line that passes through Q(8, 12) and is perpendicular to PS:

y – 12 = 1 (x – 8)
3

y – 12 = 1 x – 8
3 3
1 28
y = 3 x + 3

2. (a) 3PK = 2PQ

PK = 2
PQ 3

PK : PQ = 2 : 3

∴ PK : KQ = 2 : 1

Katakan koordinat P ialah (x, y).
Let coordinates of P be (x, y).

 1(x) + 21(2), 1(y) + 2(8) = (0, 6)
2 + 2 + 1

x + 4 = 0 y + 16 = 6
3 3

x + 4 = 0 y + 16 = 18

 x = –4 y = 2

∴ P(–4, 2)

mPQ = 8–2 = 1
2 – (–4)

mPR = −  1 = –1
mPQ

Persamaan garis lurus PR ialah
Equation of the straight line PR is

y – 2 = –1[x – (–4)]
y – 2 = –x – 4
y = –x – 2

39 © Penerbitan Pelangi Sdn. Bhd.

(b) Katakan koordinat R ialah (x, y).

Let coordinates of R be (x, y).

1 –4 x 2 –4 = 26
22 y 8 2

21 |[(−4)(y) + (x)(8) + (2)(2)] − [(2)(x) + (y)(2) + (8)(−4)]| = 26
1
2 |−4y + 8x + 4 − 2x − 2y + 32| = 26

|6x – 6y + 36| = 52

6x – 6y + 36 = 52     atau/or 6x – 6y + 36 = –52

6x – 6y = 16 6x – 6y = –88

3x – 3y = 8 ......a 3x – 3y = –44 ......b

Titik R berada pada garis lurus PR. Jadi, gantikan y = –x – 2 ke dalam a dan b.
Point R lies on the straight line PR. So, substitute y = –x – 2 into a and b.

3x – 3(–x – 2) = 8    atau/or 3x – 3(–x – 2) = –44

3x + 3x + 6 = 8 3x + 3x + 6 = –44
6x = –50
6x = 2 x = – 560
1
x = 3

Oleh sebab/Since x  0, x = 1
3
y = – 31 – 2

= –2 1
3

  ∴ R 1 , –2 1
3 3

© Penerbitan Pelangi Sdn. Bhd. 40

Jawapan

Tingkatan 4 Vektor 3. P9:Q = kA9:B + 3C9:D

BAB Vectors 222~~~aaa + h~b = kk(k(~a~a+––655)k~a~b~b)+++(6–3~a5(2k+~a+3+3~b)~b~b)
+ h~b =
8 + h~b =

Masteri SPM Bandingkan kedua-dua belah persamaan,

Kertas 1 Comparing both sides of the equation,

2 = k + 6   dan/and    h = –5k + 3

1. A9:B = 4~x + 6~y dan/and B9:C = 14~x + 21~y k = –4
A9:B = 2(2~x + 3~y) ...a B9:C = 7(2~x + 3~y)
Gantikan k = –4 ke dalam h = –5k + 3,
1 B9:C = 2~x + 3~y ...b Substitute k = –4 into h = –5k + 3,
7
h = –5(–4) + 3
= 23

Gantikan b ke dalam a, 4. (a) A9:F + F9:C + B9:F = A9:F + F9:C + C9:E

Substitute b into a, = A9:F + F9:E

  A9:B = 2 1 B9:C = A9:E
7
(b) B9:C = B9:O + O9:C
A:B = 2 B9:C = A9:F + A9:B
7

Maka, A, B dan C adalah segaris. = ~y + ~x 1
Therefore, A, B and C are collinear. 5
Vektor unit/Unit = (~x + ~y)
2. Jika 9M:N dan S9:T adalah selari, maka vector
9M:N = λ9S:T .
If M9:N and 9S:T are parallel, then 9M:N = λS9:T . 5. (a) P9:U = 1 U9:Q
3
1
(p – 44))~~aa + 3~b = λ–(5–λ5~a~a++qqλ~b~b) = 3 (6~x)
(p – + 3~b =
= 2~x
Bandingkan kedua-dua belah persamaan,
Comparing both sides of the equation, (b) P9:Q = P9:U + U9:Q
= 2~x + 6~x
p – 4 = –5λ ....a  dan/and 3 = qλ = 8~x
Q9:S = Q9:P + P9:S
λ = 3 ......b
q

Gantikan b ke dalam a, =−8~x + 10~y
Substitute b into a, U9:T = U9:Q + Q9:T

  p – 4 = –5 3 = 6~x + 1 Q9:S
q 2

p = – 1q5 + 4 = 6~x + 1 (–8~x + 10~y)
2

= 2~x + 5~y

41 © Penerbitan Pelangi Sdn. Bhd.

6. B9:C = B9:A + A9:C (b) L9:M = L9:O + O9M:

= –––2(33mm~~m~++–~n(~n1+)++m~k()+m~~n k+~n k~n ) = –21 + –182
=
= =  16

B9:D = B9:A + A9:D –3

= 4–m(~3m~+ –3~n~n) + (7m~ + 2~n)  L9:M   = 162 + (–3)2
=
= 16.28 unit/units

B, C dan D adalah segaris, 9. (a) O9:P = λO9:Q

B, C and D are collinear,   7k  λ –1 
B9:C = λB9:D 2
=

–– 22mm~~ + (1 + kk))~~nn == λ4(λ4m~m~ ++33λ~n~n) 7 = –λ   dan/and  k = 2λ
+ (1 + λ = –7

Bandingkan kedua-dua belah persamaan, Gantikan λ = –7 ke dalam k = 2λ,
Comparing both sides of the equation, Substitute λ = –7 into k = 2λ,

–2 = 4λ  dan/and  1 + k = 3λ ∴ k = 2(–7)
λ = – 12 = –14
Gantikan λ = – 12 ke dalam 1 + k = 3λ,
(b) P9:Q = P9:O + O9:Q
Substitute λ = – 12 into 1 + k = 3λ,
= –7k + –21
  1 + k = 3 – 12
  =
k = – 25 –8
–k + 2

 P9:Q  = 10

7. (a) A9:B = A9:O + O9:B (–8)2 + (–k + 2)2 = 10

= –(2~i 8+~j3~j) + (8~i – 5~j) 64 + k2 – 4k + 4 = 100
= 6~i –
k2 – 4k – 32 = 0

(b)  A9:B  = 62 + (–8)2 = 10 (k + 4)(k – 8) = 0

k = –4, k = 8

Vektor unit/Unit vector = 6~i – 8~j 10. A9:B = –2~e + q~f
10 A9:O + O9:B = ––(r26q(~i3–~+i 6–2)~~~ijj)+++(r2qq~q(ir~++i +22q)2~~j~jj)
p~–j)~i++(3(1~i2+–1p2~)j~)j ==
= 3 ~i – 4 ~j (–4~i –
5 5

    8. (a) O9:K = 5 2
–3 dan/and O9:L = 1 rq – 6 = –1  dan/and 12 – p = 2q + 2

2O9:K = –4O9:L + O9:M q = 5 .....a 10 – p = 2q .....b
r
    2 5 2 + O9:M
–3 = –4 1 Gantikan a ke dalam b,

    O9:M = 5 2 Substitute a into b,
–3 1
2 + 4   10 – p = 2 5
r
  = 18
–2 r = 10 p
10 –
∴ M(18, –2)

© Penerbitan Pelangi Sdn. Bhd. 42

Kertas 2 (c) C9:F = 3 C9:D ⇒ F9:D = 5 C9:D
8 8

1. (a) (i) AB : BD = 3 : 1  F9:D   = 5  C9:D  
8
∴ AB : AD = 3 : 4 ⇒ A9:B = 3 A9:D
4 5  A9:C 2 +  A9:D 2
dan/and B9:E = B9:A + A9:E = 8

= −A9:B + A9:E = 5 (16 × 3)2 + (16 × 2)2
8
A9:F = A9:B + B9:F
= 36.06 unit/units
3 hB9:E
= 4 (16~x) + 2. (a) (i) PS : SR = 1 : 2

= 1(122~x + 1h2(h–)1~x2~x+ +202h0~y~y) ∴ PS : PR = 1 : 3 ⇒ P9:S = 1 P9:R
= – S9:Q = S9:P + P9:Q 3

(ii) AC : CE = 4 : 1 = – 31 (6~b) + 10~a

∴ AC : AE = 4 : 5 ⇒ A9:C = 4 A9:E = 10~a – 2~b
5 (ii) Q9:R = Q9:P + P9:R
dan/and C9:D = C9:A + A9:D
= −A9:C + A9:D
= −10~a + 6~b

A9:F = A9:C + C9:F QT : TR = 3 : 2

= 4 (20~y) + kC9:D ∴ QT : QR = 3 : 5 ⇒ Q9:T = 3 Q9:R
5 5
P9:T = P9:Q + Q9:T
= 16~y + k(–16~y + 16~x)
= 16k~x + (16 – 16k)~y = 10~a + 3 (–10~a + 6~b)
5
18
(b) (12 – 12h)~x + 20h~y = 4~a + 5 ~b
= 16k~x + (16 – 16k)~y
(b) (i) P9:V = mP9:T
Bandingkan kedua-dua belah persamaan,
  = m 4~a + 158~b
Comparing both sides of the equation,

12 – 12h = 16k .....a = 4m~a + 158m~b

20h = 16 – 16k .....b

Gantikan a ke dalam b, (ii) PS : SR = 1 : 2 ⇒ S9:R = 2 P9:R
Substitute a into b, R9:V = R9:S + S9:V 3

20h = 16 – (12 – 12h) = – 23 (6~b) + nS9:Q

8h = 4
1
h = 2 = –4~b + n(10~a – 2~b)
= 10n~a – (4 + 2n)~b
Gantikan h = 1 ke dalam a,
2
1
Substitute h = 2 into a,

  12 – 12 1 = 16k
2
6 = 16k
3
k = 8

43 © Penerbitan Pelangi Sdn. Bhd.

(c) P9:V = P9:R + R9:V Gantikan a ke dalam b,

4m~a + 18 m~b = 6~b + 10n~a – (4 + 2n)~b Substitute a into b,
5
18 6h = 14 – 2(12h)
5
4m~a + m~b = 10n~a + (2 – 2n)~b 30h = 14
7
Bandingkan kedua-dua belah persamaan, h = 15

Comparing both sides of the equation, Gantikan h = 7 ke dalam a,
15
4m = 10n dan/and 18 m = 2 – 2n ...b 7
 n = 2 5 15
5 m ...a Substitute h = into a,

Gantikan a ke dalam b,   k = 12 7
15
Substitute a into b, 28
= 5
  158m = 2
2 – 2 5 m (c) A, E dan F adalah segaris.

158m = 2 – 54m A, E and F are collinear.
252 m = A9:F = λA9:E
m = 2 p~x + 10~y = λ(12~x + 6~y)
5
11 5 p = 12λ   dan/and 10 = 6λ
11
Gantikan m = ke dalam a, λ = 5
3
5   Maka/Thus, 5
Substitute m = 11 into a, p = 12 3 = 20

  n = 2 5 4. (a) (i) A9:B = 4 D9:C ⇒ D9:C = 3 A9:B
5 11 3 4

= 2 A9:C = A9:D + D9:C
11
3
3. (a) (i) B9:C = B9:A + A9:C = 5~q + 4 (8~p)

= –21~x + 14~y = 6~p + 5~q
(ii) B9:C = B9:A + A9:C
(ii) BE : EC = 3 : 4
∴ BE : BC = 3 : 7 ⇒ B9:E = 37B9:C = –8~p + 6~p + 5~q
= –2~p + 5~q
A9:E = A9:B + B9:E
= 21~x + 73(–21~x + 14~y) (b) B, E dan D adalah segaris, maka
B, E and D are collinear, then
= 12~x + 6~y
A9:D = A9:C + C9:D B9:E = kB9:D = k B9:A + A9:D 

(b) Juga/Also, A9:E = A9:B + B9:E
hA9:E =
11k~x44~~yy++(+1k4k(~~xx–––2k22)k~y~y~y) nA9:C = 88(8~~pp–++8kkk() –B9~p:8A~+p++5kA95~q:D~q) 
h (12~x + 66h~y~y) ==
12h~x + n6(n6~p~p + 55n~q~q) ==
+

Bandingkan kedua-dua belah persamaan, Bandingkan kedua-dua belah persamaan,

Comparing both sides of the equation, Comparing both sides of the equation,

12h = k ...a dan/and 6h = 14 – 2k ...b 6n = 8 – 8k  ......a dan/and

5n = 5k
  n = k  ......b

© Penerbitan Pelangi Sdn. Bhd. 44

Gantikan b ke dalam a, Bandingkan kedua-dua belah persamaan,
Substitute b into a,
Comparing both sides of the equation,

6n = 8 – 8n 43 m – 2n = 0 dan/and 130m + 4n = 4 ...b

14n = 8 4
4 3
n = 7 m = 2n

5. (a) (i) A9:R = A9:P + P9:R n = 2 m .....a
3

= –2~x + 4~y Gantikan a ke dalam b,

(ii) PA : AQ = 1 : 3 Substitute a into b,

∴ PA : PQ = 1 : 4 ⇒ P9:Q = 4P9:A 130m + 4 2 m = 4
3

Q9:R = Q9:P + P9:R 6m = 4
= –4(2~x) + 4~y 2
= –8~x + 4~y m = 3
QB : BR = 5 : 1
Gantikan m = 2 ke dalam a,
3
2
∴ BR : QR = 1 : 6 Substitute m = 3 into a,

B9:R = 1 Q9:R   n = 2 2
6 3 3

= 1 (–8~x + 4~y) = 4
6 9

= – 34 ~x + 2 ~y Soalan KBAT
3

(b) B9:R = – 34 (2~i – ~j) + 2 (–~i + 3~j) 1. O9:Q = O9:P + P9:Q
3 = O9:P +  P9:R + R9:Q 

= – 83 ~i + 4 ~j – 2 ~i + 2~j
3 3
Diberi OPQR ialah segi empat selari, maka
= – 130~i + 130~j R9:Q = O9:P.
Given OPQR is a parallelogram, then R9:Q = O9:P .
  
 B9:R  = – 130 2 + 10 2 O9:Q = O9:P + P9:R + O9:P
3

= 4.714 unit/units     121 = 2O9:P + 12
–1
(c) P9:C + C9:R = P9:R
mP9:B + nA9:R =     2O9:P =2 12
  m P9:R + R9:B  + nA9:R = 4~y 11 – –1
4~y
4~y + 4 ~x – 2 ~y + n(–2~x + 4~y) = 4~y   2O9:P =–10
m 3 3 12

  m 4 ~x + 130~y − 2n~x + 4n~y = 4~y   O9:P = –5
3 6

4 m~x + 130m~y − 2n~x + 4n~y = 4~y Kedudukan Pei Rong/Position of Pei Rong
3 = (–5, 6)

   
4 m – 2n ~x + 130m + 4n ~y = 4~y
3

45 © Penerbitan Pelangi Sdn. Bhd.

2. (a) Halaju paduan A/Resultant velocity of A, 3. O9:A = 6~x, O9:B = 10~x, O9:C = 4~y, O9:D = 6~y
AE : ED = 1 : 3
  5 1
vA = 5~i + 2 ~j + (~i – 2~j) = 6~i + 2 ~j

Halaju paduan B/Resultant velocity of B, ∴ AE : AD = 1 : 4

  8 2 (a) (i) O9:E = O9:A + A9:E
vB = 7~i + 3 ~j + (~i – 2~j) = 8~i + 3 ~j 1 A9:D
4
Katakan vB adalah n kali ganda vA, = 6~x + A9:D = 9A:O + 9O:D
Let vB be n times vA,
= 6~x + 1 (–6~x + 6~y)
vB = nvA 4

  8~i +2 1 = 9 ~x + 3 ~y
3 ~j = n 6~i + 2 ~j 2 2

8~i + 2 ~j (ii) B9:E = B9:O + O9:E
n = 6~i + 3  
= –10~x + 9 ~x + 3 ~y
1 ~j 2 2
2
– 121~x 3
24~i + 2~j = + 2 ~y
3
= (b) B9:C = B9:O + O9:C
12~i + ~j
2 = –10~x + 4~y
= –2(5~x – 2~y)
= 48~i + 4~j B9:E – 121~x 3
36~i + 3~j Juga/Also, = + 2 ~y

= 4(12~i + ~j ) = – 21 (11~x – 3~y)
3(12~i + ~j ) B9:C dan B9:E tidak dapat dihubungkaitkan
4
= 3 dalam bentuk B9:C = λB9:E, maka B, E

dan C tidak segaris. Shamsuddin tidak

Halaju paduan kereta mainan B adalah akan dihalang oleh tiang bendera E.
4 9BB9::CC =anλdB9:EB9:E, socaBn,nEotabned
3 kali ganda halaju paduan kereta related in the form
C are not collinear.
mainan A.
4 Shamsuddin will not be blocked by flagpole
3
The resultant velocity of toy car B is E.
times the resultant velocity of toy car A.

(b) (i) Halaju paduan C,
Resultant velocity of C,

vC = (4~i – 10~j) + (~i – 2~j)
= 5~i – 12~j

(ii) Vektor unit/Unit vector

= 5~i – 12~j
 52 + (–12)2

= 153~i – 1132~j

© Penerbitan Pelangi Sdn. Bhd. 46

Jawapan

Tingkatan 4 Penyelesaian Segi Tiga 2. (a) (i) AC 2 = 122 + 82 – 2(12)(8) kos 84°
= 187.93
BAB Solution of Triangles AC = 13.71 cm

9

Masteri SPM (ii) ∠ABC = 180° – 84°

= 96°

Kertas 2 sin ∠BAC = sin 96°
5 13.71

1. (a) (i) 12 (6)(4.5)sin ∠BCD = 13 sin ∠BAC = sin 96° × 5
13.71

13.5 sin ∠BCD = 13 = 0.3627
∠BAC = 21.27°
sin ∠BCD = 0.96296

∠BCD = 180° − 74.36° ∠ACB = 180° – 96° – 21.27°

= 105.64° = 62.73°

(ii) (b) (i) Luas/Area of ∆ADC
1
BD2 = 62 + 4.52 – 2(6)(4.5) kos 105.64° = 2 (12)(8)sin 84°

= 70.81 = 47.74 cm2

BD = 70.81

= 8.41 cm (ii) Katakan jarak terdekat dari D ke

(iii) sin ∠CDB = sin 105.64° AC = h
4.5 8.41 Let the shortest distance from D to
AC = h
sin 105.64°
sin ∠CDB = 8.41 × 4.5 12 × 13.71 × h = 47.74

= 0.51527 h = 6.96 cm
∠CDB = 31.02°
3. (a) (i) ∠ADC = 180° – 34° – 81°
(b) AB2 = 142 – 4.52
= 175.75 = 65°
AB = 175.75
= 13.26 cm sinAC65° = 9
sin 34°

∠CBD = 180° – 105.64° – 31.02° AC = 9 × sin 65°
= 43.34° sin 34°

= 14.59 cm

∠ABD = 90° – 43.34° (ii)
= 46.66°
14.592 = 112 + 52 – 2(11)(5) kos ∠ABC

Luas/Area of ∆ABD kos ∠ABC = 112 + 52 – 14.592
1 2(11)(5)
= 2 (13.26)(8.41) sin 46.66°
= –0.6079
= 40.55 cm2
∠ABC = 127.44°

47 © Penerbitan Pelangi Sdn. Bhd.

(iii) Luas/Area of ∆ABC (ii) AC 2 = 4.12 + 92 – 2(4.1)(9) kos 136°
= 150.90
= 1 (11)(5) sin 127.44° AC = 150.90
2 = 12.28 cm
AE = 12.28 – 9.06
= 21.83 cm2 = 3.22 cm

Luas/Area of ∆ACD
1
= 2 (9)(14.59) sin 81°

= 64.85 cm2 (iii) ∠CED = 180° – 95° – 38°

Luas sisi empat/Area of quadrilateral = 47°
= 21.83 + 64.85
= 86.68 cm2 Luas/Area of ∆CDE
1
= 2 (5.6)(9.06) sin 47°

(b) (i) CЈ = 18.55 cm2
5 cm
AЈ 14.59 cm (b) (i) DЈ

BЈ 5.6 cm 38° CЈ
EЈ
(ii) ∠CʹBʹAʹ = 180° – 127.44°

= 52.56° (ii) ∠CʹEʹDʹ = 180° – 47°

C/CЈ

= 133°

A/AЈ 127.44° ∠CʹDʹEʹ = 180° – 133° – 38°
B
BЈ = 9°

4. (a) (i) sinCE95° = 5.6 D/DЈ
sin 38°
95°

CE = 5.6 × sin 95° 47° 38° C/CЈ
sin 38° 47° EЈ
E
= 9.06 cm

5. (a) (i) 92 = 72 + 122 – 2(7)(12) kos ∠ACD

kos ∠ACD = 72 + 122 – 92
2(7)(12)

= 0.6667

∠ACD = 48.19°

(ii) ∠CAB = ∠ACD = 48.19°

sin B48C.19° = 12
sin 105°

BC = 12 × sin 48.19°
sin 105°

= 9.26 cm

(b) (i) DЈ D 7 cm C

9 cm 9 cm 12 cm

105°
AB

© Penerbitan Pelangi Sdn. Bhd. 48

(ii) 122 = 72 + 92 – 2(7)(9) kos ∠ADC

kos ∠ADC = 72 + 92 – 122
2(7)(9)

= –0.1111

∠ADC = 96.38°

∠ADDʹ = 180° – 96.38°
= 83.62°

∠DʹAD = 180° – 2(83.62°)

= 12.76°

Luas/Area of ∆DDʹA

= 1 (9)(9) sin 12.76°
2

= 8.95 cm2

Soalan KBAT

1. (a) sinAE50° = 15
sin 35°

AE = 15 × sin 50°
sin 35°

= 20.03 cm

(b) EC = 152 + 152
= 152 cm

∠AEB = 180° – 35° – 50°
= 95°

sinAB95° = 15
sin 35°

AB = 15 × sin 95°
sin 35°

= 26.05 cm

AC = 26.052 + 152
= 30.06 cm

30.062 = 20.032 + 152 2 – 2(20.03)152  kos ∠AEC
20.032 + 152 2 – 30.062
kos ∠AEC = 2(20.03)152 

= –0.0617

∠AEC = 93.54°

Luas satah/Area of plane ACE
1
= 2 (20.03)152  sin 93.54°

= 212.04 cm2

49 © Penerbitan Pelangi Sdn. Bhd.

(c) Katakan panjang terdekat dari E ke AC = h

Let the shortest length from E to AC = h

12 × 30.06 × h = 212.04
h = 14.11 cm

2. (a) sin ∠CAB = sin 82°
16 24

sin ∠CAB = sin 82° × 16
24

= 0.66018

∠CAB = 41.31°

(b) ∠ABC = 180° – 82° – 41.31°

= 56.69°
AC 2 = 242 + 162 – 2(24)(16) kos 56.69°

= 410.24
AC = 410.24
= 20.25 cm

(c) VA = 142 + 92
= 277 cm

VC = 122 + 92
= 15 cm

20.252 = 277 2 + 152 – 2277 (15) kos ∠AVC

kos ∠AVC = 277 2 + 152 – 20.252
2277 (15)

= 0.1841

∠AVC = 79.39°

Luas satah/Area of plane VAC = 1 (15)277  sin 79.39°
2

= 122.69 cm2

© Penerbitan Pelangi Sdn. Bhd. 50