Page 20 - 1202 Bank Soalan Matematik Tambahan Tingkatan 5

1
1
5
2
Apabila t = —, (b) f (x) = x + 2 —– = ——– = – —
3
5

5
2
gf (x) = x + 4x + 2
αβ
5
5
1 3 21
1 3 22
2
v = 2 — 10 – 3 — gf (x) = g(x + 2) – —
50 Katakan y = x + 2 Maka, persamaan ialah
v = —– x = y – 2 1 2
2
3 2 x + —x – — = 0
5
5
Maka, halaju maksimum zarah ialah g(y) = (y – 2) + 4(y – 2) + 2 5x + x – 2 = 0
2
= y – 4y + 4 + 4y – 8 + 2
2
50
—– m s . 2
–1
3 = y – 2 5. (a) Titik tengah AC = (3, 1)
2
∫
(b) s = v dt (i) g(x) = x – 2 y
2
g(2) = 2 – 2 = 2
∫
s = (20t – 6t ) dt (ii) fg(x) = 9 A(1, 4)
2
2
s = 10t – 2t + c f (x – 2) = 9 D(h, k)
3
2
2
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Apabila t = 0, s = 0 x – 2 + 2 = 9
2
10(0) – 2(0) + c = 0 x = 9 B(–3, 0) O x
2
3
c = 0 x = ±3
Oleh itu, s = 10t – 2t . 3. (a) x + 3 = t(x + 1) C(5, –2)
2
3
2
Apabila t = 2, s = 10(2) – 2(2) 3 x – tx + 3 – t = 0 h – 3
2
2
2
s = 24 b – 4ac , 0 ——– = 3
2
2
2
Apabila t = 3, s = 10(3) – 2(3) 3 (–t) – 4(1)(3 – t) , 0 h = 9
2
2
3
k + 0
s = 36 t + 4t – 12 , 0 ——– = 1
2
3
Jumlah jarak yang dilalui (t – 2)(t + 6) , 0 2
= s – s 2 –6 , t , 2 k = 2
3
= 36 – 24 (b) y = 2x ∴ h = 9, k = 2
2
2 – (–2)
4
= 12 m y = mx + c (b) m = ———– = — = 1
(c) Apabila zarah melalui titik P, (mx + c) = 2x PC 9 – 5 4
2
s = 0 m x + 2mcx + c – 2x = 0 Kecerunan  dengan DC = –1
2
2 2
10t – 2t = 0 m x + (2mc – 2)x + c = 0 Maka, persamaan ialah
2 2
2
2
3
2t (5 – t) = 0 b – 4ac = 0 y – 4 = –(x – 1)
2
2
t = 0 atau t = 5 (2mc – 2) – 4m c = 0 y = –x + 5
2
2 2
Maka, t = 5 s. 4m c – 8mc + 4 – 4m c = 0 6. (a) 2 2x + 2 + 1 = 5(2 )
2 2
2 2
x
(d) Apabila zarah bertukar arah –8mc = –4 2 ⋅2 + 1 – 5(2 ) = 0
2x 2
x
–4
pergerakannya, m = —–– 2 ⋅ 2 – 5(2 ) + 1 = 0
2
2x
x
v = 0 –8c 4(2 ) – 5(2 ) + 1 = 0
x
2x
1
2t(10 – 3t) = 0 m = —– (4(2 ) – 1)(2 – 1) = 0
x
x
2c
10
t = 0 atau t = —– 4. (a) h(t) = 20t – 5t 2 4(2 ) = 1 atau 2 = 2 0
x
x
3
x
10
Maka, t = —– s. = 5t(4 – t) 2 = 2 –2 x = 0
3 x = –2
xy
h(t) (b) (i) log ABB
4
1 log xy
PENTAKSIRAN SPM = — ——–– 2
2 1
2
20 log 4
2
Kertas 1 h(t) = 20t – 5t 2 1
= —(log x + log y)
4
2
2
Bahagian A 1
1. (a) y = —(p + q)
4
16y
0 4 t (ii) log ——
3 8 x 2
y = x 16y
–1
y = f (x) log ——
Apabila t = 2, 2 x 2
3
2 h(2) = 20(2) – 5(2) 2 = ————
1
= 20 = —(log 16 + log y – 2 log x)
0 < h(t) < 20 3 2 2 2
1 1
5
y = f(x) (b) (α + 1) + (β + 1) = — = —(4 + q – 2p)
3
2
x 5
0 1 2 3 4 α + β = — – 2 7. (a) M = M e –0.2t
2
0
1
1
0 < x < 2 = — M = —M 0
2
2
(b) f : (x, y) → (x – y, x + 2y) (α + 1)(β + 1) = –1 —M = M e –0.2t
1
f : (2, 3) → (–1, 8) αβ + (α + β) + 1 = –1 2 0 0
1
f : (–1, 8) → (–9, 15) αβ = –1 – 1 – — –0.2t ln e = ln 0.5
ln 0.5
∴ B(–1, 8), C(–9, 15) 2 t = ———
5
(c) f (x) = 3 – 4x = – — –0.2
2
Katakan y = 3 – 4x = –3 1 1 t = 3.47 tahun
6 = 4x Jika — dan — adalah punca-punca, (b) ———— × ———
36 + 4AB 6
AB 6 – 2
α
β
3
x = — — + — = ——–– AB 6 + 2 AB 6 – 2
α + β
1
1
2
3
36AB 6 – 72 + 24 – 8AB 6
∴ f (–3) = — α β — αβ = —————————
–1
1
2
2
2
28AB 6 – 48
2. (a) y = f (x) + 1 = —— = ——–——
5
–2 < y < 3 – — 2
2
–2 < f (x) + 1 < 3 = – — = 14AB 6 – 24
1
–3 < f (x) < 2 5
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