Page 11 - Modul A+1 Matematik Tambahan Tingkatan 4

5
3. Diberi f (x) = , cari setiap yang berikut.
x
5
Given f (x) = , find each of the following. TP 3
BAB 1 Contoh/Example (a) f (x) (b) f (x)
x
5
3
f (x)
2
5
5
5
5
2 2 1
3
5
5
x
x
x
x
f (x) = f   =   = 5 ÷ f (x) = f f f = f 4   = 5 f (x) = f 2   = 5
2
x
5
x
x
= 5 × = x x
5
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(c) f (x) (d) f (x) (e) f (x)
4
8
13
5
f (x) = f f = f (x) = x f (x) = f f = f (x) = x f (x) = f f f f = f 12   = 5 x
4 4
8
4 4 4 1
13
4
2 2
4
2
x
4. Seorang penjaja sate memperoleh keuntungan harian, dalam RM, mengikut fungsi f : x → 2x – 100 dengan x ialah bilangan
4
sate yang dijual setiap hari.
A satay hawker earns a daily profit, in RM, according to the function f : x → 2x – 100 , where x is the number of satay sold per day.
4
TP 4

(a) Pada satu minggu, penjaja sate tersebut menjual (b) Berapakah bilangan minimum sate yang perlu dijual
3 580 sate dalam 5 hari. Kira purata keuntungan supaya penjaja sate itu tidak rugi?
harian bagi minggu itu. What is the minimum number of satay that needs to be sold
In a week, the satay hawker sold 3 580 satay in 5 days. so that the satay hawker does not suffer loss?
Calculate the average daily profit for the week. 2x – 100
4  0
3 580 ÷ 5 = 716 2x  100
x  50
Purata keuntungan harian
= 2(716) – 100 Bilangan minimum sate yang perlu dijual ialah
4
50 cucuk supaya dia tidak rugi.
= RM333
Uji Kendiri 1.2
1. Tuliskan fungsi berulang f , f , f , f dan f bagi 2. Luas permukaan riak air di permukaan kolam, L, dalam
4
3
50
25
2
4
2
2
fungsi f (x) = , x ≠ 0. cm , diberi oleh fungsi L( j) = πj dengan j ialah jejari
x riak air, dalam cm. Jejari riak air bertambah mengikut
Write the iterated function f , f , f , f and f for the 1
3
2
4
50
25
2
4 fungsi j(t) = t , t > 0, dengan t ialah masa, dalam
4
function f (x) = , x ≠ 0. KBAT Mengaplikasi
x saat. Cari luas permukaan riak air di permukaan kolam
f (x) = f f (x) = 4 = x selepas 4 saat. 2
2
The surface area of water ripple on a pool, L, in cm , is
4
  given by the function L( j) = πj , where j is the radius of the
2
x
f (x) = f f (x) = 4 water ripple, in cm. The radius of the water ripple increases
3
2
1
x according to the function j(t) = t , t > 0, where t is the time,
2
4
f (x) = f f (x) = x in seconds. Find the surface area of water ripple on the pool
2 2
4
f (x) = 4 after 4 seconds.
25
x
f (x) = x 1 2 1
50
t =
Lj(t) = π   16 πt 4
2
4
Lj(4) = 1 π(4) = 16π cm 2
4
16
8
01_Modul A+ MateTam Tg4.indd 8 08/10/2021 11:18 AM