Page 16 - Modul A+1 Matematik Tambahan Tingkatan 4

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Soalan Berformat SPM

Kertas 1 Paper 1 BAB 1

Bahagian A/Section A
1. (a) Rajah di bawah menunjukkan fungsi gubahan hk 2. (a) Diberi fungsi k : x → 2x + 9, cari
yang memetakan p kepada r. Given the function k : x → 2x + 9, find KBAT Mengaplikasi
KLON KLON
SPM M The diagram below shows the composite function hk SPM M (i) nilai x apabila k(x) = 3x,
SP
SP
which maps p to r. KBAT Mengaplikasi the value of x when k(x) = 3x,
hk (ii) nilai a dengan keadaan k(a + 9) = a.
the value of a such that k(a + 9) = a
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[4 markah/marks]
x – m
(b) Diberi bahawa q(x) = n ialah fungsi songsang
bagi fungsi k(x) = 2x + 9. Tentukan nilai bagi 2m – n.
x – m
p q r Given that q(x) = n is the inverse function of the
function k(x) = 2x + 9. Determine the value of 2m – n.
Nyatakan
State [2 markah/marks]
(i) fungsi yang memetakan q kepada r.
the function that maps q to r. (a) (i) 2x + 9 = 3x
(ii) k h (r) –x = –9
–1 –1
[2 markah/marks] x = 9
(b) Diberi fungsi f (x) = 2mx dan g(x) = 3x + n, dengan
keadaan m dan n ialah pemalar. Ungkapkan n dalam (ii) 2(a + 9) + 9 = a
sebutan m supaya fungsi gubahan gf memetakan 2 2a + 27 = a
kepada dirinya sendiri. a = –27
Given the functions f(x) = 2mx and g(x) = 3x + n, where m (b) k : x ˜ 2x + 9
and n are constants. Express n in terms of m such that the 2x + 9 = y
composite function gf maps 2 to itself. x = y – 9
2
KBAT Mengaplikasi –1 x – 9
[3 markah/marks] k (x) = 2
Jadi, m = 9 dan n = 2.
(a) (i) h 2m – n = 2(9) – 2
(ii) p = 16
(b) gf (2) = 2
3(2m(2)) + n = 2
12m + n = 2
n = 2 – 12m

3. (a) Diberi f : x → 1 – 2mx, g : x → 5x – n dan fg : x → m x + n. Cari nilai-nilai m dan n.
2
2
KLON Given f : x → 1 – 2mx, g : x → 5x – n and fg : x → m x + n. Find the values of m and n. KBAT Mengaplikasi
SPM M [4 markah/marks]
SP
(b) Seterusnya, dengan menggunakan nilai m yang negatif, ungkapkan gf(x) menggunakan tatatanda fungsi.
Hence, by using the negative value of m, express gf(x) by using function notation. [2 markah/marks]
(a) fg(x) = 1 – 2m(5x – n) (b) m = –10 Ú n = 1
= 1 – 10mx + 2mn 21
m x + n = –10mx + 2mn + 1 f(x) = 1 – 2(–10)x
2
Bandingkan pekali x: = 1 + 20x
m = –10m 1
2
m + 10m = 0 g(x) = 5x – 21
2
m = 0 atau m = –10 gf(x) = 5(1 + 20x) – 1
Bandingkan pemalar: 21
1
n = 2mn + 1 = 5 + 100x – 21
Jika m = 0 ⇒ n = 1 104
Jika m = –10 ⇒ n = 2(–10)n + 1 = 100x + 21
n = –20n + 1
21n = 1
n = 1
21

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