Page 7 - Modul A+1 Matematik Tambahan Tingkatan 4

6. Selesaikan setiap yang berikut.
Solve each of the following. TP 3
BAB 1 Contoh/Example (a) Fungsi f ditakrifkan oleh f (x) = |9m + 2x|.

Cari nilai-nilai m dengan keadaan
Fungsi h ditakrifkan oleh h : x → 4x – 12. Function f is defined by f (x) = |9m + 2x|.
2
Function h is defined by h : x → 4x – 12. Find the values of m such that
2
(i) Cari imej bagi –1. (i) f (3) = 7
Find the image of –1. (ii) f (m) = 2
(iii) f (m) = m + 5
(ii) Diberi imej bagi p ialah p + 2, cari nilai-nilai yang
mungkin bagi p. (i) |9m + 2(3)| = 7
Given the image of p is p + 2, find the possible values 9m + 6 = 7 atau 9m + 6 = –7
of p. 9m = 1 9m = –13
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9m = 1 9 m = – 13
9
9
(i) h(–1) = 4(–1) – 12 = –8 (ii) |9m + 2m| = 2
2
11m = 2 atau 11m = –2
(ii) h(p) = p + 2 2 2
4p – 12 = p + 2 11m = 11 11m = – 11
2
4p – p – 14 = 0 (iii) |9m + 2m| = m + 5
2
p = 2 atau/or p = –1.75 11m = m + 5 atau 11m = –m – 5
11m – m = 5 11m + m = –5
10m = 5 12m = –5
1 1m = 1 m = – 5
2 12
(b) Diberi fungsi f (x) = |x + 8|, cari (c) Diberi fungsi f (x) = |1 + 2x|. Nyatakan
Given the function f (x) = |x + 8|, find Given the function f (x) = |1 + 2x|. State
(i) f (–10), (i) objek-objek bagi 15,
(ii) nilai-nilai x dengan keadaan f (x) = 5. the objects of 15,
the values of x such that f (x) = 5. (ii) domain bagi 0  f (x)  15.
the domain for 0  f (x)  15.
(i) |–10 + 8| = |–2| = 2
(ii) |x + 8| = 5 (i) |1 + 2x| = 15
x + 8 = 5 atau x + 8 = –5 1 + 2x = –15 atau 1 + 2x = 15
x = –3 x = –13 x = –8 x = 7
(ii) –8  x  7

(d) Rajah di bawah menunjukkan gambar rajah anak (e) Diberi satu fungsi f : x → 6 , x ≠ m.
panah yang mewakili fungsi h(x) = s – tx, dengan 1 – 2x
keadaan s dan t ialah pemalar. Cari nilai s dan nilai t. Given a function f : x → 6 , x ≠ m.
The diagram below shows an arrow diagram which (i) Nyatakan nilai m. 1 – 2x
represents the function h(x) = s – tx, where s and t are State the value of m.
constants. Find the values of s and t. (ii) Cari nilai-nilai k dengan keadaan f (k) = –k.
Find the values of k such that f (k) = –k.
h
1 1
(i) Oleh sebab x ≠ , maka m =
–13 2 2
5 6
(ii) = –k
–16 1 – 2k

6 = –k(1 – 2k)
6 6 = –k + 2k 2
0 = 2k – k – 6
2
k = –1.5 atau k = 2
s – t(5) = –13
s – 5t = –13 …
s – t(6) = –16
s – 6t = –16 …
Persamaan serentak
 – :
–t = –3
–t = 3
s – 5(3) = –13
s = –13 + 15
s = 2

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01_Modul A+ MateTam Tg4.indd 4 08/10/2021 11:18 AM