Jawapan Modul Aktiviti Pintar Bestari Matematik Tambahan Tingkatan 4

89 Pages·2017·0.69 MB·Indonesian

JAWAPAN
BAB 1: FUNGSI C
1.1 1. Hubungan satu kepada satu
A 2. Hubungan banyak kepada banyak
1. (a) Bahagi dengan 3
A B 3. Hubungan banyak kepada satu
3 1 D
6 2
1. (a) Hubungan satu kepada banyak
12 4 (b) {2, 3, 4, 5, 6}
15 5 (c) 7
17 6 (d) 5 dan 6

(b) {(3, 1), (6, 2), (12, 4), (15, 5)} 2. (a) Hubungan banyak kepada banyak
(b) {(a, 1), (a, 2), (b, 5), (c, 3), (c, 5), (d, 3)}
(c) Set B
(c) {1, 2, 3, 5}
6
5
4 3. (a) (i) 36
2 (ii) { (2, 4), (4, 16), (6, 36), (8, 64), (y, y2)}
1 (b) y = –5 atau 5
Set A
3 6 12 15 17
2. (a) P Tambah 5 Q 1.2
A
7 12
Jenis hubungan Gambar rajah anak panah
8 13
9 14
a x
10 15 Satu kepada satu b y

c z
(b) { (7, 12), (8, 13), (9, 14), (10, 15)}
(c) Set B a x
Satu kepada
y
15
banyak b z
14
13
12
Set A a x
7 8 9 10 Banyak kepada
b y
B satu c z
1. (a) {3, 6, 9, 12}
(b) {15, 30, 45, 60, 70}
(c) 3, 6, 9, 12 a x
Banyak kepada
(d) 15, 30, 45, 60} b y
(e) 45 banyak c z
(f) 12
(g) {15, 30, 45, 60}
B
2. (a) {3, 4, 5, 6, 7}
1. Tatatanda fungsi: f : x → –x
(b) {10, 20, 30, 40, 50}
(a) –3, –4, –5, –k
(c) 3, 4, 5, 6, 7
(b) –10
(d) 10, 30, 40, 50
(e) 50 2. Tatatanda fungsi: f : x → x – 3
(f) Tiada (a) 33, 44, 55, 68, k
(g) {10, 30, 40, 50} (b) k = 73
© Sasbadi Sdn. Bhd. (139288-X) 1 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
C 2. f(x) = mx + n
1. f(4) = 3(4) – 5 = 7 f(1) = m(1) + n = 5
m + n = 5 ...... ➀
f(–10) = 3(–10) – 5 = –35
f(6) = m(6) + n = 15
2. f(3) = 32 + 4 = 13 6m + n = 15 ...... ➁
f(–7) = (–7)2 + 4 = 53 ➁ – ➀: 5m = 10
m = 2
D
Gantikan m = 2 ke dalam ➀.
1. 3x + 5 = 11 2. 5 – 4x = 9
2 + n = 5
3x = 6 –4x = 4
n = 3
x = 2 x = –1
4
3x + 5 = –1 5 – 4x = –11 3. g(x) = – 10
x
3x = –6 –4x = –16
x = –2 x = 4 Apabila g(x) = 8
4
E – 10 = 8
x
1. (a) f(–4) = 2(–4) + 6 4
= 18
= –8 + 6 x
= –2 x 1
=
4 18
(b) 2x + 6 = 10
4
2x = 4 x =
18
x = 2
2
x =
2. (a) f(2) = 10 9
3(2) – p = 10
–p = 10 – 6 1.3
–p = 4 A
p = –4
1. fg(x) = f(2x + 8)
(b) f(x) = 3x – (–4) = (2x + 8) + 3
f(x) = 3x + 4 = 2x + 11
f(x) = –5 gf(x) = g(x + 3)
3x + 4 = –5 = 2(x + 3) + 8
3x = –9 = 2x + 6 + 8
x = –3 = 2x + 14
3. (a) f(3) = g(2x) gf(6) = 2(6) + 14
2(3) + 4 = 2x – 2 = 12 + 14
10 = 2x – 2 = 26
12 = 2x
x = 6 2. fg(x) = f(4x + 3)
= 2(4x + 3) – 5
(b) f(x) + g(x) = 11
= 8x + 6 – 5
(2x + 4) + (x – 2) = 11
= 8x + 1
3x + 2 = 11
3x = 9 gf(x) = g(2x – 5)
x = 3 = 4(2x – 5) + 3
= 8x – 20 + 3
F = 8x – 17
1. f(x) = ax + b
fg(2) = 8(2) + 1
f(1) = a(1) + b = –3 = 16 + 1
a + b = –3 ...... ➀ = 17
f(4) = a(4) + b = 6
4a + b = 6 ...... ➁
➁ – ➀: 3a = 9
a = 3
Gantikan a = 3 ke dalam ➀.
3 + b = –3
b = –6
© Sasbadi Sdn. Bhd. (139288-X) 2 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
3. hg(t) = h(t2 + 3) 5. f 2(x) = gf(x)
= 4(t2 + 3) – 5 f(2 + 3x) = g(2 + 3x)
= 4t2 + 12 – 5 2 + 3(2 + 3x) = 4 – 5(2 + 3x)
= 4t2 + 7 2 + 6 + 9x = 4 – 10 – 15x
8 + 9x = –6 – 15x
gh(t) = g(4t – 5)
24x = –14
= (4t – 5)2 + 3
= 16t2 – 40t + 25 + 3 x = – 14
= 16t2 – 40t + 28 24
7
hg(2) = 4(2)2 + 7 x = –
12
= 16 + 7
= 23
C
4. fg(x) = f(x + 2) 1. fg(x) = 37 – 8x
= –(x + 2)2 – 3(x + 2) 2g(x) – 7 = 37 – 8x
= –(x2 + 4x + 4) – 3x – 6 2g(x) = 44 – 8x
= –x2 – 4x – 4 – 3x – 6 g(x) = 22 – 4x
= –x2 – 7x – 10
2. fg(x) = 12x + 2
gf(x) = g(–x2 – 3x) 3g(x) + 5 = 12x + 2
= (–x2 – 3x) + 2 3g(x) = 12x – 3
= –x2 – 3x + 2 g(x) = 4x – 1
gf(3) = –(3)2 – 3(3) + 2 3. fg(x) = 4x2 – 4x + 9
= –9 – 9 + 2 2g(x) – 1 = 4x2 – 4x + 9
= –16 2g(x) = 4x2 – 4x + 10
g(x) = 2x2 – 2x + 5
B
1. fg(x) = 11 D
f(4 – 2x) = 11 1
1. (a) fg(x) = x + 7
(4 – 2x) + 3 = 11 2
4 – 2x + 3 = 11 1
g(x) + 3 = x + 7
–2x = 4 2
x = –2 1
g(x) = x + 4
2
2. gf(2) = 25
(b) gf(x) = g(x + 3)
g[2(2) + 3] = 25
1
g(7) = 25 = (x + 3) + 4
2
5(7) + h = 25
1 3
35 + h = 25 = x + + 4
2 2
h = –10
1 11
= x +
3. f 2(x) = 4 2 2
f(2x – 4) = 4
2. (a) fg(x) = 2x2 + 7
2(2x – 4) − 4 = 4
2g(x) + 1 = 2x2 + 7
4x – 8 − 4 = 4
2g(x) = 2x2 + 6
4x = 16
g(x) = x2 + 3
x = 4
g(x + 3) = (x + 3)2 + 3
4. gf(x) = 6x – 3
= x2 + 6x + 9 + 3
g(ax + 4) = 6x – 3
= x2 + 6x + 12
5 − 2(ax + 4) = 6x – 3
5 − 2ax − 8 = 6x – 3 (b) g(x) = x2 + 3
−2ax − 3 = 6x – 3 g(–1) = (–1)2 + 3
−2ax = 6x = 4
−2a = 6
3. fg(2) = 43
a = −3
2g(2) + 9 = 43
2g(2) = 34
g(2) = 17
© Sasbadi Sdn. Bhd. (139288-X) 3 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
E 3. f 2(x) = 4x – 4
f(b – ax) = 4x – 4
1. fg(x) = –2x – 6
b – a(b – ax) = 4x – 4
f(x + 5) = –2x – 6
b – ab + a2x = 4x – 4
Katakan y = x + 5
Maka, a2 = 4
Maka, x = y – 5
a = 2 (a (cid:2) 0)
f(y) = –2(y – 5) – 6
dan
f(y) = –2y + 10 – 6
f(y) = 4 – 2y b – ab = –4
b – 2b = –4
Maka, f(x) = 4 – 2x
–b = –4
2. fg(x) = 2x2 + 9 b = 4
f(x2 + 4) = 2x2 + 9
4. (a) fg(4) = 2
Katakan y = x2 + 4 f(4 – 4) = 2
Maka, x2 = y – 4 f(0) = 2
a(0) + b = 2
f(y) = 2(y – 4) + 9 b = 2
f(y) = 2y – 8 + 9
gf(2) = 4
f(y) = 2y + 1
g(2a + b) = 4
Maka, f(x) = 2x + 1 g(2a + 2) = 4
(2a + 2) – 4 = 4
3. fg(x) = 25x + 34 2a = 6
f(5x + 6) = 25x + 34 a = 3
Katakan y = 5x + 6 (b) f(x) = 3x + 2
5x = y – 6
gf(–3) = g[3(–3) + 2]
Maka, 25x = 5y – 30
= g(–7)
f(y) = (5y – 30) + 34 = –7 – 4
f(y) = 5y – 30 + 34 = –11
f(y) = 5y + 4
fg(–5) = f(–5 – 4)
Maka, f(x) = 5x + 4 = f(–9)
= 3(–9) + 2
F
= –25
1. fg(–2) = 7
6
f [2(–2) + 3] = 7 5. fg(x) = + 4
x + 1
f(–1) = 7
a(–1) + 3 = 7 3 6
= + 4
a(–1) = 4 g(x) + 1 x + 1
a = –4
6 + 4x + 4
=
f(x) = –4x + 3 x + 1
gf(2) = g[–4(2) + 3] 4x + 10
=
= g(–5) x + 1
= 2(–5) + 3 g(x) + 1 x + 1
=
= –7 3 4x + 10
3x + 3 4x + 10
2. fg(x) = gf(x) g(x) = –
4x + 10 4x + 10
f(a – 2x) = g(2x + b)
2(a – 2x) + b = a – 2(2x + b) g(x) = –x – 7 , x ≠ – 5
2a – 4x + b = a – 4x – 2b 4x + 10 2
2a – a = –2b – b
a = –3b
© Sasbadi Sdn. Bhd. (139288-X) 4 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
1.4 Maka, f –1(x) = x + 3
4
A
2 + 3
1. f –1(14) = m f –1(2) =
4
f f –1(14) = f(m)
5
14 = f(m) =
4
14 = 2m + 6
1
2m = 8 = 1
4
m = 4
2. Katakan f –1(x) = y
f –1(22) = n
f f –1(22) = f(n) Maka, f f –1(x) = f(y)
22 = f(n) x = f(y)
22 = 2n + 6 1
x = y + 3
2n = 16 2
n = 8 y = 2(x – 3)
Maka, f –1(x) = 2(x – 3)
2. f –1(6) = m
f f –1(6) = f(m) f –1(4) = 2(4 – 3)
6 = f(m) = 2
1
6 = m + 4
2 3. Katakan f –1(x) = y
1 Maka, f f –1(x) = f(y)
m = 2
2 x = f(y)
m = 4 2y + 3
x =
4
f –1(9) = n
4x = 2y + 3
f f –1(9) = f(n)
4x – 3
9 = f (n) y =
2
1
9 = n + 4
2 Maka, f –1(x) = 4x – 3
1 2
n = 5
2
4(6) – 3
n = 10 f –1(6) =
2
21
3. f –1(8) = m =
2
f f –1(8) = f(m)
1
8 = f(m) = 10
2
8 = 10 – 2m
2m = 2
4. Katakan f –1(x) = y
m = 1
Maka, f f –1(x) = f(y)
f –1(0) = n x = f(y)
f f –1(0) = f(n) 5
x =
0 = f(n) y + 2
0 = 10 – 2n xy + 2x = 5
2n = 10 xy = 5 – 2x
n = 5 5 – 2x
y =
x
B
Maka, f –1(x) = 5 – 2x, x ≠ 0
1. Katakan f –1(x) = y x
Maka, f f –1(x) = f(y)
5 – 2(2)
x = f(y) f –1(2) =
2
x = 4y – 3
1
x + 3 =
y = 2
4
© Sasbadi Sdn. Bhd. (139288-X) 5 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
C (a) hg–1(x) = h (cid:2)x + 1(cid:3)
3
1. (a) fg(x) = 2x – 2
(cid:2)x + 1(cid:3)
g(x) + 3 = 2x – 2 = 2 + 3
3
g(x) = 2x – 5
2x + 2
= + 3
g –1(x) = y 3
x = g(y)
2x + 11
=
x = 2y – 5
3
x + 5
y = 2 (b) gh–1(x) = g (cid:2)x – 3(cid:3)
2
x + 5
g –1(x) = (cid:2)x – 3(cid:3)
2 = 3 – 1
2
f g–1(x) = f (cid:2)x + 5(cid:3) 3x – 11
2 =
2
x + 5
= + 3
2
x + 11 Praktis Formatif: Kertas 1
=
2 1. (a) Julat = {a, b, d}
(b) g –1f (x) = g –1(x + 3) (b) Hubungan banyak kepada satu
= (x + 3) + 5 2. (a) {(–1, 5), (0, 2), (1, 5)}
2
(b) {–1, 0, 1}
x + 8
=
2
3. f(6) = 10
6 – 2m = 10
2. (a) Katakan g –1(4) = y
2m = –4
4 = g(y)
m = –2
4
4 =
3y – 2 4. (a) f(x) = x
3y – 2 = 1 2x – 4 = x
3y = 3 x = 4
y = 1
(b) f(3h – 1) = 3h
Maka, g–1(4) = 1 2(3h – 1) – 4 = 3h
6h – 2 – 4 = 3h
(b) Katakan f –1(x) = y 6h – 6 = 3h
x = f(y) 3h = 6
x = y + 8 h = 2
y = x – 8
f –1(x) = x – 8 5. (a) 3
(b) f(2) = | 2 – 4(2) |
f –1g(x) = f –1 (cid:2) 4 (cid:3) = | –6 |
3x – 2
= 6
= 4 – 8, x ≠ 2 (c) –1 (cid:3) x (cid:3) 2
3x – 2 3
6. hk(x) = 2mx + p
3. Katakan h–1(x) = y h(2x – 1) = 2mx + p
x = h(y) m(2x – 1) + 3 = 2mx + p
x = 2y + 3 2mx – m + 3 = 2mx + p
x – 3 –m + 3 = p
y =
2 m = 3 – p
x – 3
Maka, h–1(x) =
2 7. (a) Fungsi f
Katakan g–1(x) = y (b) g –1(c) = b
x = g(y)
8. g f (2) = 8
x = 3y – 1
g(2 × 2) = 8
x + 1
y = g(4) = 8
3
k(4) + h = 8
x + 1
Maka, g–1(x) = 4k + h = 8
3
h = 8 – 4k
© Sasbadi Sdn. Bhd. (139288-X) 6 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
9. (a) Katakan g–1(x) = y (ii) Berdasarkan rajah yang diberi,
x = g(y) gf(x) = 12x – 5
x = 2y + 4 Maka, g(3x – 2) = 12x – 5
2y = x – 4
Katakan u = 3x – 2.
y = x – 4 u + 2
Maka, x =
2
3
Maka, g–1(x) = x – 4 (cid:2)u + 2(cid:3)
g(u) = 12 – 5
2
3
(b) fg(x) = 4x2 + 16x + 10 = 4u + 8 – 5
f(2x + 4) = 4x2 + 16x + 10 = 4u + 3
Katakan y = 2x + 4 Maka, g(x) = 4x + 3
y – 4
x = (b) fg(x) = 5x + 14
2
f(4x + 3) = 5x + 14
f(y) = 4(cid:2)y – 4(cid:3)2 + 16(cid:2)y – 4(cid:3) + 10 3(4x + 3) – 2 = 5x + 14
2 2
12x + 9 – 2 = 5x + 14
= (y – 4)2 + 8(y – 4) + 10
7x = 7
= y2 – 8y + 16 + 8y – 32 + 10
x = 1
= y2 – 6
Maka, f(x) = x2 – 6 2. (a) Diberi f(x) = 4x – 5.
Katakan y = f–1(x)
10. (a) Katakan g–1(x) = y
f(y) = x
x = g(y)
4y – 5 = x
x = 3y – 6
4y = x + 5
3y = x + 6
x + 5
y =
x + 6
y = 4
3
x + 5
x + 6 Maka, f–1(x) =
Maka, g–1(x) = 4
3
Lakaran graf f–1(x):
(cid:2)2 (cid:3)
(b) g2 p = 12
3 f –1(x)
(cid:4) (cid:2)2 (cid:3) (cid:5)
g 3 p – 6 = 12
5
3
4
g (2p – 6) = 12
3(2p – 6) – 6 = 12 x
6p – 18 – 6 = 12 –5 0
6p = 36
Domain x ialah semua nilai nyata.
p = 6
(b) f–1g(x) = f –1 (cid:2)x – 2(cid:3)
4
Praktis Formatif: Kertas 2
(cid:2)x (cid:3)
– 2 + 5
1. (a) (i) Fungsi yang memetakan set B kepada 4
=
set A ialah f –1(x). 4
x + 12
Berdasarkan rajah yang diberi, =
16
f(x) = 3x – 2.
(c) hg(x) = x – 8
Katakan y = f –1(x)
Maka, f(y) = x h(cid:2)x – 2(cid:3) = x – 8
3y – 2 = x 4
x
3y = x + 2 Katakan u = – 2
4
x + 2
y = Maka, x = 4u + 8
3
x + 2 h(u) = (4u + 8) – 8
Maka, f –1(x) =
3 = 4u
Maka, h(x) = 4x
© Sasbadi Sdn. Bhd. (139288-X) 7 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
FOKUS KBAT
(a) Bukan fungsi.
Kerana hubungan yang memetakan set B kepada set A ialah hubungan satu kepada banyak.
(b) f(x) = kx2 – 5x
f(6) = –18
k(62) – 5(6) = –18
36k – 30 = –18
36k = 12
1
k =
3
18x
(c) g–1(x) =
x – 1
Katakan g(x) = y
Maka, g–1(y) = x
18y
= x
y – 1
18y = xy – x
xy – 18y = x
y(x – 18) = x
x
y =
x – 18
x 1
g(x) = dan f(x) = x2 – 5x
x – 18 3
(cid:2)1 (cid:3)
gf(x) = g x2 – 5x
3
1
x2 – 5x
3
=
(cid:2)1 (cid:3)
x2 – 5x – 18
3
x2 – 15x
=
x2 – 15x – 54
= x(x – 15) , x ≠ –3, 18
(x + 3)(x – 18)
© Sasbadi Sdn. Bhd. (139288-X) 8 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
JAWAPAN
BAB 2: PERSAMAAN KUADRATIK 3. –3x2 + 2x + 8 = 0
3x2 – 2x – 8 = 0
2.1
(3x + 4)(x – 2) = 0
A
3x + 4 = 0 atau x – 2 = 0
1. a = 2, b = –4, c = 5
x = – 4 x = 2
2. a = –3, b = 4, c = –7 3
3. a = 1, b = –6, c = 0 B
1. 3x2 + 10x + 6 = 0
B
1. Ya. Kerana kuasa tertinggi x ialah 2. x2 + 10x + 2 = 0
3
2. 5(x + 3) = x – 2 x2 + 10x = –2
5x + 15 = x – 2 3
4x + 17 = 0 x2 + 10x + (cid:2)5(cid:3)2 = –2 + (cid:2)5(cid:3)2
3 3 3
Bukan. Kerana kuasa tertinggi x bukan 2. (cid:2)x + 5(cid:3)2 = 7
3 9
3. 2 + x – 3 = 0
x2 x + 5 = ± 7
2x−2 + x – 3 = 0 3 9
Bukan. Kerana kuasa tertinggi x bukan 2. x = –53 ± 79
= –0.7847 atau –2.5486
C
1. Gantikan x = 3 ke dalam mx2 – 7x + 3 = 0. 2. –3x2 + 12x – 5 = 0
m(3)2 – 7(3) + 3 = 0 3x2 – 12x + 5 = 0
9m – 21 + 3 = 0 x2 – 4x + 5 = 0
9m = 18 3
m = 2 x2 – 4x = – 5
3
2. x = 2: a(2)2 – 5(2) + c = 0 x2 – 4x + (cid:2)4(cid:3)2 = – 5 + (cid:2)4(cid:3)2
2 3 2
4a + c = 10 …… ➀
(x – 2)2 = 7
x = 3: a(3)2 – 5(3) + c = 0 3
9a + c = 15 …… ➁ x – 2 = ± 7
➁ – ➀: 5a = 5 dan 4(1) + c = 10 3
a = 1 c = 6 x = 2 ± 7
3
= 3.5275 atau 0.4725
2.2
A C
1. x2 + 4x – 5 = 0 1. a = 1, b = 5, c = –6
(x + 5)(x – 1) = 0 –5 ± 52 – 4(1)(–6)
x =
x + 5 = 0 atau x – 1 = 0 2(1)
x = –5 x = 1 = –5 ± 7
2
2. 2x2 + x – 10 = 0 = –6 atau 1
(2x + 5)(x – 2) = 0
2x + 5 = 0 atau x – 2 = 0 2. a = –1, b = 4, c = –2
x = – 5 x = 2 x = –4 ± 42 – 4(–1)(–2)
2 2(–1)
= –4 ± 8
–2
= 0.5858 atau 3.4142
© Sasbadi Sdn. Bhd. (139288-X) 1 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4
3. a = 4, b = –6, c = 1 3. –2x2 – 6x + 15 = 0
–(–6) ± (–6)2 – 4(4)(1) x2 + 6 x – 15 = 0
x =
2 2
2(4)
x2 + 3x – 15 = 0
= 6 ± 20 2
8 x2 – (–3)x + (cid:2)– 15(cid:3) = 0
= 0.1910 atau 1.3090 2
Maka, HTP = –3
4. a = –2, b = 5, c = 8
15
HDP = –
x = –5 ± 52 – 4(–2)(8) 2
2(–2)
4. 5x2 + 9x – 25 = 0
= –5 ± 89
–4 x2 + 9 x – 25 = 0
5 5
= –1.1085 atau 3.6085
x2 + 9 x – 5 = 0
D 5
1. (x – 2)(x + 4) = 0 x2 – (cid:2)– 9(cid:3) x + (–5) = 0
5
x2 + 4x – 2x – 8 = 0
x2 + 2x – 8 = 0 Maka, HTP = –9
5
2. (cid:2)x – 1(cid:3)(x – 3) = 0 HDP = –5
2
G
x2 – 3x – 1x + 3 = 0
2 2 1. (a) x2 – 7x – k = 0
2x2 – 6x – x + 3 = 0
HTP = 7
2x2 – 7x + 3 = 0
Katakan r ialah punca yang satu lagi.
3. (x + 5)(x + 4) = 0 Maka, 6 + r = 7
x2 + 4x + 5x + 20 = 0 r = 1
x2 + 9x + 20 = 0 Jadi, punca yang satu lagi ialah 1.
(b) HDP = –k
E Maka, –k = 6 × 1
1. x2 – (p + q)x + pq = 0 k = –6
x2 – (–1)x + (–2) = 0
x2 + x – 2 = 0 2. x2 + kx + 8 = 0
x2 – (–k)x + 8 = 0
2. x2 – (p + q)x + pq = 0
x2 – 4x + (–5) = 0 HTP : m + (m – 2) = –k
x2 – 4x – 5 = 0 2m – 2 = –k
k = 2 – 2m
3. x2 – (p + q)x + pq = 0 HDP : m(m – 2) = 8
x2 – 3x + (cid:2)– 7(cid:3) = 0 m2 – 2m = 8
2 2 m2 – 2m – 8 = 0
x2 – 3x – 7 = 0 (m – 4)(m + 2) = 0
2 2 m = 4 atau –2
2x2 – 3x – 7 = 0
Apabila m = 4, k = 2 – 2(4)
F = –6
1. x2 + 5x + 4 = 0 Apabila m = –2, k = 2 – 2(–2)
x2 – (–5)x + 4 = 0 = 6
Maka, HTP = –5 Nilai k yang mungkin ialah –6 dan 6.
HDP = 4
3. x2 + 5x + 3 = 0
2. x2 – 8x – 20 = 0 x2 – (–5)x + 3 = 0
x2 – 8x + (–20) = 0 α + β = –5 dan αβ = 3
Maka, HTP = 8
HDP = –20
© Sasbadi Sdn. Bhd. (139288-X) 2 PINTAR BESTARI SPM Matematik Tambahan Tingkatan 4