Page 20 - Modul A+1 Matematik Tambahan Tingkatan 4

2
BAB 1 (ii) gf(x) = 24x + 15 2. (a) 2x + px – 2x + 8 = 0
2
g[8x – 5] = 24x + 15 ( p – 2) – 4(2)(8) > 0
2
Kertas 2/Paper 2 y + 5 p – 4p + 4 – 64 > 0
g( y) = 24   + 15 p – 4p – 60 > 0
2
8
Bahagian A/Section A g( y) = 3(y + 5) + 15 ( p + 6)( p – 10) > 0
1. (a) (i) f(x) = x + 1 g( y) = 3y + 30 p < –6 atau/or p > 10
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(ii) (2, 5), (–2, 5), (–3, 10) \ g(x) = 3x + 30 α β –k
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(b) (iii) fg(x) = 18x + 1 (b) 2 + = 2
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y 8[3x + 30] – 5 = 18x + 1 αβ = 2
y = x 24x + 240 – 5 – 18x – 1 = 0 4 2
6x + 234 = 0 α + β = –k
x = –39 αβ = 4
(b) f(x) = 8x – 5 –k = –( p – 2)
6 2
gf(x) = 24x + 15 –2k = –p + 2
g[8x – 5] = 24x + 15
x p = 2 + 2k
0 6 Katakan/Let 8x – 5 = y
2
y + 5 3. (a) f(x) = –(x – h) + 2k
x = 8 –2 = –h + 2k … 1
2
∴ Bukan fungsi. Hubungan satu g(y) = 24 [ y + 5 ] + 15 2k = –1
kepada banyak. 8 k = – 1
Not a function. One to many = 3y + 15 + 15 2
relation. = 3y + 30 h = 2 – 1 + 2
 
2
–1
f (x) = x + 5 2 2
Bahagian B/Section B 8 h = 1
–1
2. (a) (i) n(2) = 1 – 2 g (x) = x – 30 h = ±1
3
2
= –1 x – 30 (b) f(x) = –(x – 1) – 1
2 f g (x) = 3 + 5
–1 –1
(ii) m(k + 2) = n(2) 8 f(x)
3
2 x – 30 + 15
k + 2 + 3 = (–1) = 24 x
3
17 x – 15 0 1 (1, –1)
k = – 3 = 24 –1
–2
–1 –1
(iii) nm(x) = 1 – (x + 3) ∴ f g (x) = (gf) (x)
–1
= 1 – x – 3
= –2 – x BAB 2
(b) y (c) f(x) = (x – 1) + 1
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Kertas 2/Paper 2
4. (a) Katakan α dan β ialah punca-
6 Bahagian A/Section A punca bagi x + px + q = 0.
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b
5 1. (a) (i) h + k = Let α and β are the roots of
a x + px + q = 0.
2
4 hk = – c a α + β = –p ..... 1
3 c c a α – β = 23 ..... 
b = × b
a
2 1 αβ = q = 6
1 = –hk h + k 1 + : 2α = 23 – p ..... 3
1 – : 2β = –p – 23 ..... 4
x = –hk 3 × 4:
–2 –1 0 1 2 3 h + k
–1 a + b + c a b 4ab = –(23 – p)(p + 23)
2
(ii) = + + 1 4(6) = –(12 – p )
c c c 2
0  y  5 = –1 + h + k + 1 24 = –12 + p
2
(c) nm(x) = –2 – x hk –hk p = 36
Katakan/Let –2 – x = y = –1 – h – k + hk p = ±6
–2 – y = x hk q = 6
Jadi/Hence, (nm) (x) = –2 – x = 1 –  1 + h + k  (b) Apabila/When p > 0, p = 6
–1
–1
∴ (nm) (x) = nm(x) h + k hk Daripada/From 3,
– a
b
1
(b) 1 + = = ×  
3. (a) (i) Katakan/Let 8x – 5 = y k h hk a c α = 23 – 6
y + 5 b 2
x = = –
8 c = 3 – 3
1
1
–1
f (x) = x + 5    = 1 = – a Daripada/From 4,
8 k h hk c –6 – 23
b a β = 2
2
x + x – = 0
c c = –3 – 3
cx + bx – a = 0
2
MG-1
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