Page 13 - Modul A+Matematik Tamb Tg 4_Pan Asia Publications

J awapan

Bab 1 (ii) a + b + c = a + b + 1 5. (a) (i) CQ = x
c c c 3 4
Kertas 2 / Paper 2 = –1 + h + k + 1 QC = 3x
hk –hk 4
1. (a) (i) n(2) = 1 – 2 –1 – h – k + hk NQ = 3 – QC
= –1 = hk
2 1 + h + k = 3 – 3x
(ii) m(k + 2) = n(2) = 1 –  hk  4
3
2 1 1 h + k b – a 12 – 3x
k + 2 + 3 = (–1) (b) k + = hk = ×   = 4
3
h
a
c
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17 b AB 4
k = – 3 = – c 5 = 3
(iii) nm(x) = 1 – (x + 3) 1 1 1 = – a 20
k
h
= 1 – x – 3    = hk c AB = 3
a
b
2
2
= –2 – x x + x – = 0 BC = AB + AC 2
2
20
(b) 2 c c BC =   2 + 5 2
2
y cx + bx – a = 0 3
2
2. (a) 2x + px – 2x + 8 = 0 = 625
1 ( p – 2) – 4(2)(8) > 0 9
2
2
x p – 4p + 4 – 64 > 0 BC = 25
2
–2 –1 0 1 2 3 p – 4p – 60 > 0 3
–1 ( p + 6)( p – 10) > 0 BN = BC – NC
–2 \ Julat nilai p / The range of BN = 25 – 3
values of p 3
–3 p < –6, p > 10 = 16
β
–4 (b) α + = –k 3
2 4 2 PN = 4 – x
–5 αβ 2 BN 4
4 = 2 4 – x 16 4
–6 α + β = –k PN = 4   = (4 – x)
3
3
αβ = 4 PQ = NQ + PN
0  y  –5 –( p – 2)
–k = 2 = 12 – 3x + 4 (4 – x)
2. (a) Katakan / Let 8x – 5 = y –2k = –p + 2 4 3
y + 5 36 – 9x + 16(4 – x)
x = p = 2 + 2k =
8 12
x + 5 3. (a) f(x) = –(x – h) + 2k
2
f (x) = 100 – 25x
–1
8 –2 = –h + 2k … 1 = 12
2
(b) gf(x) = 24x + 15 2k = –1 25
g[8x – 5] = 24x + 15 k = – 1 = 12 (4 – x)
2
g( y) = 24  y + 5  + 15 1 25x
 
2
8 h = 2 – + 2 (ii) Luas / Area = 12 (4 – x)
g( y) = 3(y + 5) + 15 h = 1 2
2
2
g( y) = 3y + 30 h = ±1 = 25 (4x – x )
\ g(x) = 3x + 30 12
2
(b) f(x) = –(x – 1) – 1 = – 25 (x – 4x)
2
(c) fg(x) = 18x + 1 12
8[3x + 30] – 5 = 18x + 1 f(x) = – 25 [(x – 2) – 4]
2
24x + 240 – 5 – 18x – 1 = 0 12
6x + 234 = 0 0 1 x = – 25 (x – 2) + 25
2
x = –39 –1 (1, –1) 12 3
–2
(b) Apabila / When x = 2,
Bab 2 Luas maksimum / Maximum area
= 25(2) (4 – 2)
2
Kertas 2 / Paper 2 (c) f(x) = (x – 1) + 1 25 12
2
2
4. (a) D = 6 + 3x – x – (x – 5x + 2) = 3 unit 2
1. (a) (i) h + k = b D = 6 + 3x – 2x + 5x – 2
2
a 2 6. (a) Luas / Area =
hk = – c D = 4 + 8x – 2x L = 11.5x + 11.5x + (10 – 2x)x
a (b) D = –2(x – 4x) + 4 = 23x + 10x – 2x 2
2
c
c = × a D = –2[(x – 2) – 4] + 4 = 33x – 2x 2
2
b a b D = –2(x – 2) + 8 + 4 33x – 2x  45
2
2
1 D = –2(x – 2) + 12 2
2
= –hk h + k (b) 2x – 33x + 45  0
Panjang maksimum = 12 unit (2x – 3)(x – 15)  0
–hk apabila x = 2 1.5  x  15
=
h + k Maximum length = 12 units when x = 2 Minimum x = 1.5 m
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12_Modul A+ MateTam Tg4_Jawapan K2_Final.indd 188 3/9/20 6:41 PM