Page 17 - Modul A+1 Matematik Tambahan Tingkatan 4

Bahagian B/Section B
4. (a) Rajah di bawah menunjukkan hubungan antara set 5. (a) Rajah di bawah menunjukkan graf bagi fungsi f : x
BAB 1 KLON A, set B dan set C. KLON → |2x + 1| untuk domain –3  x  4.
The diagram below shows the graph of the function f : x
The diagram below shows the relation among set A, set B
SPM M
SPM M
SP
SP
and set C. KBAT Mengaplikasi → |2x + 1| for domain –3  x  4. KBAT Mengaplikasi
2
pq(x) = x – 2x + 1 f(x)
A C (–3, 5) 9
B
x
2
Diberi set A dipetakan kepada set B oleh fungsi x + 1 0 4
dan dipetakan kepada set C oleh pq(x) = x – 2x + 1. Nyatakan/State
2
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2
Given that set A maps to set B by the function x + 1 and (i) objek bagi 9,
maps to set C by pq(x) = x – 2x + 1. the object of 9,
2
(i) Tulis fungsi yang memetakan set A kepada set (ii) imej bagi 2,
B dengan menggunakan tatatanda fungsi. the image of 2,
Write down the function that maps set A to set B by (iii) domain bagi 0  f (x)  5.
using the function notation. the domain of 0  f (x)  5.
(ii) Cari fungsi yang memetakan set B kepada set [3 markah/marks]
C. (b) Diberi fungsi f : x → 3x – 7, cari
Find the function which maps set B to set C. Given the function f : x → 3x – 7, find KBAT Mengaplikasi
[3 markah/marks] (i) f (x),
–1
(b) Diberi fungsi s(x) = x – 2 dan rs(x) = 3x + x + 1, cari 3
2
Given the functions s(x) = x – 2 and rs(x) = 3x + x + 1, find (i) nilai k dengan keadaan f 2   = –1.
2k
2
 
KBAT Mengaplikasi the value of k such that f 2 3 = –1.
(i) s (x) 2k [3 markah/marks]
–1
(ii) r(x) [3 markah/marks] 5x – 3
(c) Diberi fungsi m(x) = x + 2, lakarkan graf bagi m (x). (c) Diberi p(x) = 7 + x , x ≠ t, nyatakan nilai t.
–1
Given the function m(x) = x + 2, sketch the graph of Seterusnya, tentukan p(t).
m (x). [2 markah/marks] Given that p(x) = 5x – 3 , x ≠ t, state the value of t. Hence,
–1
7 + x
2
(a) (i) q : x → x + 1 determine p(t). [2 markah/marks]
2
(ii) p  x + 1  = x – 2x + 1
2
2
2
Katakan x + 1 = y, x = –1 (a) (i) 4
y
(ii) 5
2


p(y) =  2 y – 1 – 2  2 – 1 + 1 (iii) |2x + 1| = 5 atau 2x + 1 = –5
y

2x + 1 = 5
4
4
4
= − + 1 − + 2 + 1 x = 2 x = –3
y 2 y y Maka, –3  x  2.
4
8
= – + 4
y 2 y (b) (i) Katakan 3x – 7 = y
4
8
p : x → – + 4 x = y + 7
x
x
3
2
(b) (i) Katakan x – 2 = y, x = y + 2 Maka, f (x) = x + 7
–1
Maka, s (x) = x + 2 3
–1
3
(ii) rs(x) = 3x + x + 1 (ii) f 2   = –1
2
2k
r(x – 2) = 3x + x + 1 3
2
   
Katakan x – 2 = y, x = y + 2 f 3 2k – 7 = –1
2
   
r(y) = 3(y + 2) + (y + 2) + 1 3 3 3 – 7 – 7 = –1
= 3(y + 4y + 4) + y + 2 + 1 2k
2
2
= 3y + 12y + 12 + y + 2 + 1 27 − 21 − 7 = –1
2k
2
= 3y + 13y + 15 27
2
r(x) = 3x + 13x + 15 2k = – 1 + 28
(c) y 2k = 1
m(x)
y = x 1
k =
–1
m (x) 2
2
(c) 7 + t = 0
x t = –7
–2 0 2 Nilai bagi p(t) adalah tidak tertakrif.
–2
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