F
Additional Mathematics
2. Determine whether each of the following graphs is a function using the vertical line test.
Tentukan sama ada setiap graf yang berikut ialah fungsi atau bukan dengan menggunakan ujian garis mencancang.
y
(a)
y
x
0
x
0
Not a function. When tested with a vertical line,
the line cuts the graph at two points.
A function. When tested with a vertical line, the line
cuts the graph at one point.
y
(b)
y
(c)
x
0
x
0
A function. When tested with a vertical line, the
line cuts the graph at one point.
A function. When tested with a vertical line, the
line cuts the graph at one point.
3. State the domain, codomain and range for each of the following. Then, write the function by using function
notation.
Nyatakan domain, kodomain dan julat bagi setiap yang berikut. Kemudian, tulis setiap fungsi dengan menggunakan tatatanda fungsi.
Domain
= {2, 4, 6, 8}
Codomain
Kodomain
= {8, 16, 24, 28, 32}
Range
Julat
= {8, 16, 24, 32}
f
x
(a) Domain
= {2, 3, 4, 5}
y
2
4
6
8
8
16
24
28
32
X
Y
Range
Julat
= {1, 1, 1, 1}
2 3 4 5
Codomain
Kodomain
= {1, 2, 3, 4, 5}
Range
Julat
= {1, 2, 3, 4, 5}
Function notation:
Tatatanda fungsi
f : x → x or f(x) = x
y
1
–
2
1
–
3
2
3
4
5
1
–
4
1
–
5
X
0
Y
Function notation:
Tatatanda fungsi
f : x → 1 or f(x) = 1
x
x
Function notation: f : x → 4x or f(x) = 4x
Tatatanda fungsi
(b) Domain
= {1, 4, 9, 16, 25}
Codomain
Kodomain
= { 1 , 1 , 1 , 1 , 0}
2 3 4 5
f
x
y
Domain
= {1, 2, 4, 5}
x
1
4
9
16
25
1
2
3
4
5
Codomain
Kodomain
= {4, 7, 10, 13, 16}
1
2
4
5
4
7
10
13
16
X
Y
Range
X
Y
x
f
f
y
Julat
= {4, 7, 13, 16}
Function notation: f : x → 3x + 1 or f(x) = 3x + 1
Tatatanda fungsi
F
Additional Mathematics
4. State the domain, codomain and range for each of the following.
Nyatakan domain, kodomain dan julat bagi setiap yang berikut.
(i)
(ii)
f(x)
f(x)
19
8
6
4
0
–2
2
–4 –3 –2 –1 0
1
2
3
x
x
2 3
–16
The domain of f is –2 x
3.
The codomain of f is –16
f(x)
19.
The range of f is –16 f(x) 19.
Domain = {–4, –2, –1, 2, 3}
Codomain = {2, 4, 6, 8}
Range = {2, 4, 6, 8}
(a)
–8
(b)
f(x)
f(x)
8
4
6
3
4
2
–4 –3 –2 –1 0
1
2
3
x
0
1
3
x
The domain of f is 0 x 3.
The codomain of f is 0
f(x)
The range of f is 0 f(x) 4.
Domain = {–4, –2, 0, 2, 3}
Codomain = {0, 2, 4, 8}
Range = {0, 2, 4, 8}
4.
5. For each of the following functions, find the image for the object x given.
Untuk setiap fungsi berikut, cari imej bagi objek x yang diberikan.
(a) f(x) = 3x + 7; x = –4, x = 5
f(x) = 9x – 4; x = 2, x = –1
f(2) =
=
f(–1) =
=
9(2) – 4
14
9(–1) – 4
–13
f(–4) =
=
f(5) =
=
3(–4) + 7
–5
3(5) + 7
22
(b) f(x) = x2 + 1; x = 3, x = –2
f(3) =
=
f(–2) =
=
(3)2 + 1
10
(–2)2 + 1
5
6. For each of the following functions, find the object x based on the image given.
Untuk setiap fungsi berikut, cari objek x bagi imej yang diberikan.
(a) f(x) = 5x + 7; f(x) = –1, f(x) = –3 (b) f(x) = x2 + 5; f(x) = 9, f(x) = 21
f(x) = 2x – 1; f(x) = 3, f(x) = 5
2x – 1 = 3
2x = 4
x= 2
2x – 1 = 5
2x = 6
x= 3
5x + 7 = –1
5x = –8
8
x=–
5
x2 + 5 = 9
x2 = 4
x = 2 or –2
5x + 7 = –3
5x = –10
x = –2
x2 + 5 = 21
x2 = 16
x = 4 or –4
F
Additional Mathematics
7. Sketch the graphs of the following functions f(x) for the given domains. Then, state the ranges.
Lakar graf bagi fungsi f(x) berikut untuk domain yang diberikan. Seterusnya, nyatakan julatnya.
(a) f(x) = x – 3 , –2
f(x) = 1 – 4x , –1
3
x
x
–1
0
1
4
3
f(x)
5
1
0
11
x
–2
0
3
4
f(x)
5
3
0
1
f(x)
5
f(x)
11
3
5
1
1
–1 0 1
–
4
The range is 0
4
x
–2
x
3
The range is 0
0
3 4
f(x)
5.
x
2
11.
f(x)
(b) f(x) = 2x + 6 , –4
1
x
(c) f(x) = 5x – 4 , –3
x
–4
–3
0
1
x
–3
0
4
5
2
f(x)
2
0
6
8
f(x)
19
4
0
6
f(x)
f(x)
8
19
6
6
4
2
–4 –3
The range is 0
f(x)
0
1
x
4
–
5
8.
8. Solve each of the following problems.
Selesaikan setiap masalah berikut.
A function is defined by f : x → x – 8. Find
Suatu fungsi ditakrifkan oleh f : x → x – 8. Cari
(i) f(–3)
(ii) the value of x if f(x) = –5.
nilai x jika f(x) = –5.
f(x) = x – 8
f(–3) = –3 – 8
= –11
0
–3
The range is 0
(i)
x
(ii)
f(x) = –5
x – 8 = –5
x =3
f(x)
19.
2
x
Additional Mathematics
F
(a) Given f : x → 4x – a, where a is a constant and f(–1) = –11. Find
Diberi f : x → 4x – a, dengan keadaan a ialah pemalar dan f(–1) = –11. Cari
(i) the value of a.
nilai a.
(ii) the value of x when f(x) = –19.
nilai x apabila f(x) = –19.
(i)
f(x)
f(–1)
–4 – a
a
=
=
=
=
4x – a
4(–1) – a
–11
7
(ii)
f(x)
4x – 7
4x
x
=
=
=
=
4x – 7
–19
–12
–3
(ii)
f(x)
6x – 5
5x
x
=
=
=
=
6x – 5
x
5
1
(b) A function is defined by h : x → 6x – 5. Find
Suatu fungsi ditakrifkan oleh h : x → 6x – 5. Cari
(i) the image of object 2.
imej bagi objek 2.
(ii) the object which maps onto itself.
objek yang memeta kepada dirinya sendiri.
(i)
h(x) = 6x – 5
h(2) = 6(2) – 5
=7
(c) Given g : x → px2 – qx, where p and q are constants. If g(3) = 12 and g(4) = 28, find the values of p and q.
Diberi g : x → px2 – qx dengan keadaan p dan q ialah pemalar. Jika g(3) = 12 dan g(4) = 28, cari nilai p dan q.
g(x)
g(3)
9p – 3q
3p – q
=
=
=
=
px2 – qx
p(3)2 – q(3)
12
4 –––
g(4) = p(4)2 – q(4)
16p – 4q = 28
4p – q = 7 –––
From
Substitute
4p – (3p – 4)
4p – 3p + 4
p
into
=7
=7
=3
:
Substitute p = 3 into
q = 3(3) – 4
=5
: q = 3p – 4 –––
(d) Given the function h : x → 4 – x . Find the value of x when h(x) = 7.
Diberi fungsi h : x → 4 – x . Cari nilai x apabila h(x) = 7.
h(x)
4–x
4–x
x
=
=
=
=
4–x
7
7
–3
and
4 – x = –7
x = 11
:
F
Additional Mathematics
11. For each pair of the following functions, find
Bagi setiap pasangan fungsi yang berikut, cari
(i) fg(2) and gf(4), / fg(2) dan gf(4),
(ii) the value of x when fg(x) = 5. / nilai x apabila fg(x) = 5.
(a) f : x → 3x + 2, g : x → 2 – x2
f : x → 5x, g : x → 4 – 2x
(i)
fg(x) =
=
=
=
f [ g(x)]
f(4 – 2x)
5(4 – 2x)
20 – 10x
fg(2) = 20 – 10(2)
=0
PA
g(2) =
=
fg(2) =
=
=
P A
gf(x) =
=
=
=
(i)
g[f(x)]
g(5x)
4 – 2(5x)
4 – 10x
gf(4) = 4 – 10(4)
= –36
gf(x) =
=
=
=
f(4) = 5(4)
= 20
gf(4) = g(20)
= 4 – 2(20)
= –36
fg(x) =
20 – 10x =
10x =
x=
g(3x + 2)
2 – (3x + 2)2
2 – (9x2 + 12x + 4)
–9x2 – 12x – 2
gf(4) = –9(4)2 – 12(4) – 2
= –194
(ii)
(ii)
f(2 – x2)
3(2 – x2) + 2
6 – 3x2 + 2
8 – 3x2
fg(2) = 8 – 3(2)2
= –4
AP
4 – 2(2)
0
f(0)
5(0)
0
fg(x) =
=
=
=
5
5
15
1.5
fg(x) =
8 – 3x2 =
3x2 =
x2 =
x=
5
5
3
1
1 or –1
12. Find the function g based on the functions f and fg given.
Cari fungsi g berdasarkan fungsi f dan fg yang diberikan.
(a) f(x) = x + 4, fg(x) = 3x + 9
f(x) = x + 3, fg(x) = 2x – 5
f [g(x)] = 2x – 5
g(x) + 3 = 2x – 5
g(x) = 2x – 8
f [g(x)] = 3x + 9
g(x) + 4 = 3x + 9
g(x) = 3x + 5
∴ g : x → 3x + 5
∴ g : x → 2x – 8
(b) f(x) = 3x – 6, fg(x) = 5x + 8
f [g(x)]
3g(x) – 6
3g(x)
g(x)
=
=
=
=
5x + 8
5x + 8
5x + 14
5x + 14
3
∴ g : x → 5x + 14
3
13. Find the function f based on the functions g and fg given.
Cari fungsi f berdasarkan fungsi g dan fg yang diberikan.
(a) g(x) = 2x + 3, fg(x) = 7x – 4
g(x) = x + 2, fg(x) = 4x – 9
f [g(x)] = 4x – 9
f(x + 2) = 4x – 9
Let y = x + 2
x=y–2
So, f(y) = 4(y – 2) – 9
= 4y – 8 – 9
= 4y – 17
Substitute y with x, f(x) = 4x – 17
∴ f : x → 4x – 17
(b) g(x) = x2 – 9, fg(x) = 3x2 – 5
f [g(x)] = 7x – 4
f(2x + 3) = 7x – 4
f [g(x)] = 3x2 – 5
f(x2 – 9) = 3x2 – 5
Let y = 2x + 3
x= y–3
2
Let y = x2 – 9
x2 = y + 9
x = √y + 9
So, f(y) = 7 y – 3 – 4
2
7y – 29
=
2
So, f(y) =
=
=
=
∴f:x→
7x – 29
2
3(√y + 9)2 – 5
3(y + 9) – 5
3y + 27 – 5
3y + 22
∴ f : x → 3x + 22
F
Additional Mathematics
14. Solve the following problems.
Selesaikan masalah-masalah berikut.
Given the functions f : x → 4x – 1 and g : x → (x – 3)2. Find
Diberi fungsi f : x → 4x – 1 dan g : x → (x – 3)2. Cari
(i) fg(x),
(ii) fg(3).
(i)
fg(x) =
=
=
=
=
f[(x – 3)2]
4(x – 3)2 – 1
4(x2 – 6x + 9) – 1
4x2 – 24x + 36 – 1
4x2 – 24x + 35
(ii) fg(3) = 4(3)2 – 24(3) + 35
= 36 – 72 + 35
= –1
(a) Given f : x → 3x + 8 and g : x → x – 6. Find
Diberi f : x → 3x + 8 dan g : x → x – 6. Cari
(i) fg(x),
(ii) fg(–2).
(i)
fg(x) =
=
=
=
f(x – 6)
3(x – 6) + 8
3x – 18 + 8
3x – 10
(ii) fg(–2) = 3(–2) – 10
= –16
(b) A function g is defined by g : x → x + 2. Find the function f for each of the following composite functions.
Fungsi g ditakrifkan oleh g : x → x + 2. Cari fungsi f bagi setiap fungsi gubahan berikut.
(i)
fg : x → x2 + 4x + 4,
(ii) gf : x → 2x2 – 3x + 4.
(i)
f g(x) = x2 + 4x + 4
f(x + 2) = x2 + 4x + 4
(ii)
Let y = x + 2
x=y–2
gf(x) = 2x2 – 3x + 4
f(x) + 2 = 2x2 – 3x + 4
f(x) = 2x2 – 3x + 2
So, f(y) = (y – 2)2 + 4(y – 2) + 4
= y2 – 4y + 4 + 4y – 8 + 4
= y2
∴ f(x) = x2
(c) Given f(x) = px + q and f 2(x) = 4x + 6. Find the values of p and q.
Diberi f(x) = px + q dan f 2(x) = 4x + 6. Cari nilai-nilai p dan q.
ff(x) =
f(px + q) =
p(px + q) + q =
p2x + pq + q =
4x
4x
4x
4x
+
+
+
+
6
6
6
6
Comparing:
p2 = 4
p = ±√4
= 2 or –2
When p
pq + q
2q + q
3q
q
=
=
=
=
=
2,
6
6
6
2
When p
pq + q
–2q + q
–q
q
=
=
=
=
=
–2,
6
6
6
–6
3x
(d) Functions g and h are defined as g : x → 5x – 6 and h : x →
. Find the value of x if gh(x) = h(x).
x–2
3x
Fungsi g dan h ditakrifkan oleh g : x → 5x – 6 dan h : x →
. Cari nilai x jika gh(x) = h(x).
x–2
g(x) = 5x – 6
3x
h(x) =
x–2
gh(x) = h(x)
3x
g 3x
=
x–2
x–2
5
3x – 6 = 3x
x–2
x–2
3x
15x
–6=
x
–2
x–2
15x – 6(x – 2) =
15x – 6x + 12 =
6x =
x=
3x
3x
–12
–2
Additional Mathematics
F
15. Solve each of the following.
Selesaikan setiap yang berikut.
5
The temperature in Fahrenheit, F, can be converted to Celsius, C, using the function C(F) = (F – 32).
9
9
The temperature in Kelvin, K, can be converted to Fahrenheit, F, using F(K) = (K – 273) + 32. Suraya
5
obtained a result of temperature of 258 K in an experiment. Convert the temperature to Celsius, in °C.
5
Suhu dalam Fahrenheit, F, boleh ditukar kepada Celsius, C, dengan menggunakan fungsi C(F) = 9 (F – 32). Suhu dalam Kelvin, K,
9
boleh ditukar kepada Fahrenheit, F, dengan menggunakan fungsi F(K) = 5 (K – 273) + 32. Suraya memperoleh hasil suhu 258 K
dalam suatu eksperimen. Tukar suhu tersebut ke Celsius, dalam °C.
9
(258 – 273) + 32
5
=5
F(258) =
C[F(258)] = C(5)
5
= (5 – 32)
9
= –15
Hence, 258 K can be converted to –15°C.
(a) The height of water, h cm, in a cylindrical container is increasing with respect to time, t s, and
its function is h(t) = 1.4t. The volume of water, V cm3, in the container is given by the function
88
V(h) =
h.
7
Ketinggian air, h cm, dalam suatu bekas berbentuk silinder semakin meningkat mengikut masa, t s, dan fungsinya ialah
88
h(t) = 1.4t. Isi padu air, V cm3, dalam bekas tersebut diberi oleh fungsi V(h) = 7 h.
(i) State the volume of water, V, as a function of time, t.
Nyatakan isi padu air, V, sebagai fungsi masa, t.
(ii) Find the volume of water, in cm3, in the container after 5 seconds.
Cari isi padu air, dalam cm3, dalam bekas tersebut selepas 5 saat.
(i)
V[h(t)] = V(1.4t)
88
=
(1.4t)
7
= 17.6t
∴ V : t → 17.6t
(ii) V(5) = 17.6(5)
= 88
Hence, the volume of water in the container after 5 seconds is 88 cm3.
(c) Azlan works in a furniture store. Every month, he receives a basic salary of RM2 000 plus a 5%
commission on sales over RM3 000. Let x represents his sales per month.
Azlan bekerja di sebuah kedai perabot. Setiap bulan, dia menerima gaji asas sebanyak RM2 000 dan komisen 5% pada jualan
yang melebihi RM3 000. Let x mewakili jualannya setiap bulan.
(i) Given f(x) = x – 3 000 and g(x) = 0.05x, write the function of gf(x) and explain its meaning.
Diberi f(x) = x – 3 000 dan g(x) = 0.05x, tulis fungsi gf(x) dan terangkan maksudnya.
(ii) If Azlan’s sales are RM6 499, find his salary.
Jika jualan Azlan ialah RM6 499, cari gajinya.
(i)
gf(x) = g(x – 3 000)
= 0.05(x – 3 000)
gf(x) is the commission received by Azlan, that is 5% on sales over RM3 000.
(ii) gf(6 499) = 0.05(6 499 – 3 000)
= 174.95
Salary = 2 000 + 174.95
= 2 174.95
Hence, his salary is RM2 174.95
F
Additional Mathematics
18. Verify that f and g are inverse functions of each other.
Sahkan f dan g ialah fungsi songsang antara satu sama lain.
(a) f(x) = 4 – 3x, g(x) =
f(x) = x – 8, g(x) = x + 8
fg(x) = f(x + 8)
= (x + 8) – 8
=x
gf(x) = g(x – 8)
= (x – 8) + 8
=x
4–x
3
fg(x) = f 4 – x
3
=4–3 4–x
3
= 4 – (4 – x)
=x
Since fg(x) = gf(x) = x, f and g are inverse functions
of each other.
gf(x) = g(4 – 3x)
4 – (4 – 3x)
3
3x
=
3
=x
=
Since fg(x) = gf(x) = x, f and g are inverse
functions of each other.
19. Sketch graphs of f and f –1 on the same plane. Then, state the domain of f –1.
Lakar graf bagi f dan f –1 pada satah yang sama. Seterusnya, nyatakan domain bagi f –1.
(a) f : x → –
f : x → 4x, domain: 0
3
x
x
0
1
2
3
y
0
4
8
12
x
1
5
10
y
–5
–1
–0.5
y=x
f
(12, 3)
–1
x
Domain of f –1:
–5
x
–0.5
f –1
Domain of f :
0 x 12
–1
0
10
(–0.5, 10)
y=x
f
x
y
y
(3, 12)
5
, domain: 1
x
(–5, 1)
0
Range of f
x
(10, –0.5)
f
(1, –5)
20. Find the inverse function of each of the following functions.
Cari fungsi songsang bagi setiap fungsi berikut.
(a) f(x) = 3x – 1
f(x) = 4x + 5
Let y = f(x)
y = 4x + 5
4x = y – 5
y–5
x =
4
Since f –1(y) = x,
y–5
f –1(y) =
4
Substitute y with x,
x–5
f –1(x) =
4
∴ f –1 : x →
x–5
4
Let y
y
3x
x
=
=
=
=
f(x)
3x – 1
y+1
y+1
3
(b) g(x) = 7 – 2x
Let y
y
2x
x
=
=
=
=
g(x)
7 – 2x
7–y
7–y
2
Since f –1(y) = x,
f –1(y) = y + 1
3
Since g–1(y) = x,
g–1(y) = 7 – y
2
Substitute y with x,
f –1(x) = x + 1
3
∴ f –1 : x → x + 1
3
Substitute y with x,
g–1(x) = 7 – x
2
∴ g–1 : x → 7 – x
2
Additional Mathematics
(c) g(x) =
x
+1
4
Let y = g(x)
x
y=
+1
4
x
=y–1
4
x = 4(y – 1)
x = 4y – 4
Since g–1(y) = x,
g–1(y) = 4y – 4
Substitute y with x,
g–1(x) = 4x – 4
∴ g–1 : x → 4x – 4
F
(d) h(x) =
x–1
2x + 1
Let
y = h(x)
x–1
y=
2x + 1
2xy + y = x – 1
y + 1 = x – 2xy
y + 1 = x(1 – 2y)
y+1
x=
1 – 2y
Since h–1(y) = x,
y+1
h–1(y) =
1 – 2y
Substitute y with x,
x+1
h–1(x) =
1 – 2x
x+1
∴ h–1 : x →
1 – 2x
(e) h(x) =
3
x–1
Let y = h(x)
3
y=
x–1
xy – y = 3
xy = 3 + y
3+y
x=
y
3
x=
+1
y
Since h–1(y) = x,
3
h–1(y) =
+1
y
Substitute y with x,
3
h–1(x) =
+1
x
3
∴ h–1 : x →
+1
x
21. Find the function f, g or h based on the given inverse function.
Cari fungsi f, g dan h berdasarkan fungsi songsang yang diberikan.
(a) f –1(x) = 3x – 4
4+x
f –1(x) =
2
Let y = f –1(x)
4+x
y=
2
2y = 4 + x
x = 2y – 4
Since f (y) = x,
f (y) = 2y – 4
Substitute y with x,
f (x) = 2x – 4
Let y
y
3x
x
=
=
=
=
f –1(x)
3x – 4
y+4
y+4
3
Since f (y) = x,
f (y) = y + 4
3
Substitute y with x,
f (x) = x + 4
3
∴f :x→ x+4
3
(b) f –1(x) =
x+6
5
Let y = f –1(x)
y = x+6
5
5y = x + 6
x = 5y – 6
Since f(y) = x,
f(y) = 5y – 6
Substitute y with x,
f(x) = 5x – 6
∴ f : x → 5x – 6
∴ f : x → 2x – 4
(c) g–1(x) =
x–7
3
Let y = g–1(x)
y = x–7
3
3y = x – 7
x = 3y + 7
Since g(y) = x,
g(y) = 3y + 7
Substitute y with x,
g(x) = 3x + 7
∴ g : x → 3x + 7
(d) g–1(x) =
x+4
x–4
Let
y = g–1(x)
y = x+4
x–4
xy – 4y = x + 4
xy – x = 4y + 4
x = 4y + 4
y–1
Since g(y) = x,
g(y) = 4y + 4
y–1
Substitute y with x,
g(x) = 4x + 4
x–1
∴ g : x → 4x + 4
x–1
(e) h–1(x) =
5
–2
x
Let y = h–1(x)
5
y =
–2
x
5
=y+2
x
x = 5
y+2
Since h(y) = x,
h(y) = 5
y+2
Substitute y with x,
h(x) = 5
x+2
∴h:x→ 5
x+2
F
Additional Mathematics
22. Solve each of the following problems.
Selesaikan masalah-masalah berikut.
Given the function f : x → 4x – 7 and g : x → x2 – 5, find
Diberi fungsi f : x → 4x – 7 dan g : x → x2 – 5, cari
(i)
f –1(x),
(ii) gf –1(x),
(iii) gf –1(5)
(i)
Let y = f(x)
y = 4x – 7
4x = y + 7
y+7
x=
4
(ii) gf –1(x) = g x + 7
4
2
(iii) gf –1(5) = 5 + 14(5) – 31
16
64
=
16
=4
= x+7
4
Since f –1(y) = x,
y+7
f –1(y) =
4
Substitute y with x,
x+7
f –1(x) =
4
2
–5
2
= x + 14x + 49 – 5
16
2
= x + 14x + 49 – 80
16
2
x
+
14x
– 31
=
16
(a) Given g : x → 2x + p , p ≠ 4 and g(4) = 5, find
p–4
2x + p
Diberi g : x →
, p ≠ 4 dan g(4) = 5, cari
p–4
(i) the value of p / nilai p,
(ii) g –1(x),
(iii) the value of x if g –1(x) = 4 / nilai x jika g –1(x) = 4.
(i)
g(4) =
5=
5(p – 4)
5p – 20
4p
p
=
=
=
=
2(4) + p
p–4
8+p
p–4
8+p
8+p
28
7
2x + 7
3
Let y = g(x)
g–1(y) = x
2x + 7
y=
3
3y = 2x + 7
3y – 7
x=
2
(ii) g(x) =
∴ g –1(x) =
(iii) g –1(x) =
4=
8=
3x =
x=
3x – 7
2
3x – 7
2
3x – 7
15
5
3x – 7
2
5
(b) It is given g : x → 8x – 3 and h : x → , x ≠ 0. Find
x
5
Diberi bahawa g : x → 8x – 3 dan h : x → , x ≠ 0. Cari
x
(i) g –1(x),
(ii) hg–1(x),
(iii) g –1h(10).
(i)
Let y = g(x)
y = 8x – 3
8x = y + 3
y+3
x =
8
Since g –1(y) = x,
y+3
g –1(y) =
8
Substitute y with x,
x+3
g –1(x) =
8
(ii) hg –1(x) = h x + 3
8
=
5
x+3
8
= 40
x+3
(iii) h(10) =
5
1
=
10
2
g –1h(10) = g –1
1
2
1
+3
2
=
8
7
=
16
Additional Mathematics
F
(c) Given the function h(x) = ax + b, where a
0, a and b are constants and h2(x) = 49x – 40, find
Diberi fungsi h(x) = ax + b, dengan keadaan a 0, a dan b ialah pemalar dan h2(x) = 49x – 40, cari
(i) the value of a and of b, / nilai a dan b,
(ii) h–1(–3),
(iii) the value of x if (h–1)2(x) = 1 / nilai x jika (h–1)2(x) = 1.
(i)
Given h(x) = ax + b
h2(x) = hh(x)
= h(ax + b)
= a(ax + b) + b
= a2x + ab + b
Compare with 49x – 40,
a2 = 49
a = 49
=7
ab + b
7b + b
8b
b
=
=
=
=
–40
–40
–40
–5
(ii) h(x) = 7x – 5
(iii)
x+5
+5
7
=1
7
Let y = h(x)
y = 7x – 5
7x = y + 5
y+5
x=
7
x+5
+5
7
x+5
7
x+5
x
Since h–1(y) = x,
y+5
h–1(y) =
7
Substitute y with x,
x+5
h–1(x) =
7
–3 + 5
7
2
=
7
∴ h–1(–3) =
3x – 6
mx + 6
and f (x) =
. Find
7 + 4x
3 + nx
3x – 6
mx + 6
Diberi bahawa f –1(x) =
dan f (x) =
. Cari
7 + 4x
3 + nx
(d) It is given that f –1(x) =
(h–1)2(x) = 1
276
O
(i) the values of m and n, / nilai m dan n,
(ii) the value of x if f(4x + 3) = 7. / nilai x jika f(4x + 3) = 7.
(i)
3x – 6
7 + 4x
Let
y = f –1(x)
3x – 6
y=
7 + 4x
7y + 4xy = 3x – 6
7y + 6 = 3x – 4xy
7y + 6 = x(3 – 4y)
7y + 6
x=
3 – 4y
f –1(x) =
Since f(y) = x,
7y + 6
f(y) =
3 – 4y
Substitute y with x,
7x + 6
f(x) =
3 – 4x
Compare with
mx + 6
f (x) =
3 + nx
7x + 6 mx + 6
=
3 – 4x
3 + nx
∴ m = 7, n = –4
7x + 6
3 – 4x
7(4x + 3) + 6
3 – 4(4x + 3)
28x + 21 + 6
3 – 16x – 12
28x + 27
–9 – 16x
28x + 27
28x + 27
90
9
–
14
(ii) f(4x + 3) = 7, f(x) =
7=
7=
7=
7(–9 – 16x) =
–63 – 112x =
–140x =
x=
=7
=2
= 14
=9
F
Additional Mathematics
SPM
Paper
P A
5. Given the functions m : x → px + 3, h : x → 4x – 1 and
mh(x) = 4px + q, express p in terms of q.
1
630
Diberi fungsi m : x → px + 3, h : x → 4x – 1 dan
mh(x) = 4px + q, ungkapkan p dalam sebutan q.
[3]
Ans: p = 3 – q
1. The diagram shows the composite function fg that maps a
630
to c.
Rajah di bawah menunjukkan fungsi gubahan fg yang
memetakan a kepada c.
fg
6. It is given the functions
630
a
b
Diberi fungsi
g : x → 4x + 1
fg : x → 16x2 + 8x – 5
c
Find / Cari
(a) g–1(x),
(b) f(x).
State / Nyatakan
(a) the function that maps a to b,
fungsi yang memetakan a kepada b,
(b) f –1(c).
[3]
x–1
4
(b) f(x) = x2 – 6
Ans: (a) g –1(x) =
[2]
Ans: (a) g
(b) f –1(c) = b
7. The diagram shows the graph of the function
f : x → |1 – 3x| for the domain –1 x 3.
630
2. Given the functions f(x) = 5x + 1 and g(x) = mx – n, where
630
m and n are constants, express m in terms of n such that
fg(2) = 1.
Rajah di bawah menunjukkan graf bagi fungsi
f : x → |1 – 3x| untuk domain –1 x 3.
f(x)
8
Diberi fungsi f(x) = 5x + 1 dan g(x) = mx – n, dengan keadaan
m dan n ialah pemalar, ungkapkan m dalam sebutan n dengan
keadaan fg(2) = 1.
[3]
1
Ans: m = 2 n
Diberi fungsi f : x → 8x – 4. Cari
x
(b) the image of 2,
nilai x jika f(x) memeta kepada dirinya sendiri,
(b) the value of h if f(3 – h) = 4h.
4
Ans: (a) x = 7
5
(b) h = 3
3
objek bagi 8,
(a) the value of x if f(x) maps onto itself,
nilai h jika f(3 – h) = 4h.
0
State / Nyatakan
(a) the object of 8,
3. It is given the function f : x → 8x – 4. Find
630
(–1, 4)
(c)
[4]
imej bagi 2,
the domain of 0 f(x) 2.
domain bagi 0 f(x) 2.
[3]
Ans: (a) 3
(b) 5
(c) – 1
3
x
1
4. It is given the function g : x → 3x – 9. Find
630
Diberi fungsi g : x → 3x – 9. Cari
8. Given the functions g : x → 3x – 1 and h : x → 6x, find
(a) g –1(x),
4p
(b) the value of p if g2 3 = 24.
4p
nilai p jika g 2 3 = 24.
Ans: (a) g –1(x) =
(b) p = 5
x+9
3
[4]
Diberi fungsi g : x → 3x – 1 dan h : x → 6x, cari
(a) hg(x),
(b) the value of x if hg(x) = 1 g(x).
3
1
nilai x jika hg(x) = 3 g(x).
Ans: (a) hg(x) = 18x – 6
(b) x = 1
3
[4]
F
Additional Mathematics
9. The diagram shows the relation between set P, set Q and
set R.
630
Rajah di bawah menunjukkan hubungan antara set P, set Q
dan set R.
P
fg : x → 9x + 12
12. Given the function g : x → 2x – 7, find
Diberi fungsi g : x → 2x – 7, cari
(a) g(6),
(b) the value of p if 2g–1(p) = g(6).
nilai p jika 2g–1(p) = g(6).
R
Q
2
It is given that set P maps to set Q by the function
5x + 6 and maps to set R by fg : x → 9x + 12.
Diberi bahawa set P dipetakan kepada set Q oleh fungsi 5x + 6
dan dipetakan kepada set C oleh fg : x → 9x + 12.
(a) Write the function which maps set P to set Q by using
the function notation.
Tulis fungsi yang memetakan set P kepada set Q dengan
menggunakan tatatanda fungsi.
(b) Find the function which maps set Q to set R.
Cari fungsi yang memetakan set Q kepada set R.
[4]
Ans: (a) g : x → 5x + 6
9x + 6
(b) f : x →
5
1. In the diagram, function f maps set P to set Q and function
g maps set Q to set R.
Dalam rajah di bawah, fungsi f memetakan set P kepada set Q
dan fungsi g memetakan set Q kepada set R.
f
Rajah di bawah menunjukkan fungsi g : x → x – 2p, dengan
keadaan p ialah pemalar.
g
6
x – 2p
2x + 5
6x + 11
P
Q
R
Find/ Cari
(a) in terms of x, the function
[5]
(b) the value of x such that fg(x) = 3x + 4.
nilai x dengan keadaan fg(x) = 3x + 4.
Ans: (a)
(b)
x–5
(i) f (x) =
; (ii) g(x) = 3x – 4
2
7
1
x=
=2
3
3
2. The diagram shows the mapping from set X to set Y defined
as g(x) = px + 3 and the mapping from set Y to set Z
60
defined as h(y) =
, y ≠ –q.
y+q
Rajah di bawah menunjukkan pemetaan dari set X ke set Y yang
ditakri an sebagai g(x) = px + 3 dan pemetaan dari set Y ke
set Z yang ditakri an sebagai h(y) = 60 , y ≠ –q.
y+q
nilai p,
(b) g–1(x),
(c) the value of x if g–1(x) = 5.
nilai x jika g–1(x) = 5.
[5]
p = –2
g–1(x) = x – 4
x=9
15
X
Y
Z
[4]
(b) the function that maps set X to set Z.
fungsi yang memetakan set X kepada set Z.
[3]
[4]
27 – 3x
5
h
11
nilai-nilai p dan q,
Diberi fungsi f(x) = 9 – 5x dan g(x) = 3x, cari
(a) the value of fg(–1),
nilai bagi fg(–1),
(b) gf –1(x).
(b) gf –1(x) =
g
4
Find/ Cari
(a) the values of p and q,
11. Given the functions f(x) = 9 – 5x and g(x) =3x, find
Ans: (a) 24
[3]
–1
10
Find / Cari
(a) the value of p,
Ans: (a)
(b)
(c)
g
x
dalam sebutan x, fungsi
(i) which maps set Q to set P,
yang memetakan set Q kepada set P,
(ii) g(x),
10. The diagram shows the function g : x → x – 2p, where p is
a constant.
x
[4]
Ans: (a) 5
(b) p = –2
Ans: (a) p = 2, q = –7
60
(b) hg(x) =
,x≠2
2x – 4
Additional Mathematics
37 5
SPM
3 SH
D
F R
P
7
I
-
WLR W DW
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6L
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Additional Mathematics
W
6L
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→
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Additional Mathematics
D
D
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Additional Mathematics
276
OOH
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F
Additional Mathematics
QF
Q
2. Solve the following quadratic equations by using the formula method. Give your answers correct to four
decimal places.
Selesaikan persamaan kuadratik berikut dengan menggunakan kaedah rumus. Berikan jawapan anda betul kepada empat tempat
perpuluhan.
(a) 2x2 – 5x – 9 = 0
x – 6x + 7 = 0
2
a = 2, b = –5, c = –9
a = 1, b = –6, c = 7
–(–5) ± (–5)2 – 4(2)(–9)
2(2)
5
±
97
=
4
5
+
97 or x = 5 – 97
x =
4
4
= 3.7122
= –1.2122
x =
(–6) ± (–6) – 4(1)(7)
2(1)
6± 8
=
2
2
x =–
6+ 8
2
= 4.4142
x =
or
6– 8
2
= 1.5858
x=
(b) 3x2 – 5x + 1 = 0
(c) –x2 + 8x – 9 = 0
a = 3, b = –5, c = 1
x =
a = –1, b = 8, c = –9
–(–5) ± (–5) – 4(3)(1)
2(3)
2
= 5 ± 13
6
5
+
13
x =
6
= 1.4343
x =
or x = 5 – 13
6
= 0.2324
–8 ± (8)2 – 4(–1)(–9)
2(–1)
= –8 ± 28
–2
–8
+
28
x =
–2
= 1.3542
or
x = –8 – 28
–2
= 6.6458
3. Solve the following problems.
Selesaikan masalah-masalah berikut.
(a) The product of 3x and (x + 2) is (–x + 12). Find (b)
the values of x. Give your answers correct to
A
x cm
B
(2x + 3) cm
three decimal places.
Hasil tambah 3x dan (x + 2) ialah (–x + 12). Cari nilainilai x. Berikan jawapan anda betul kepada tiga tempat
The diagram shows two squares, A and B. If the
perpuluhan.
area of the square A is equal to the perimeter
of the square B, find the value of x. Give your
3x(x + 2) = –x + 12
answer correct to three significant figures.
2
3x + 6x = –x + 12
Rajah di atas menunjukkan dua buah segi empat sama,
2
3x + 7x – 12 = 0
A dan B. Jika luas segi empat sama A bersamaan dengan
x2 + 7 x – 4 = 0
perimeter segi empat sama B, cari nilai x. Berikan jawapan
3
anda betul kepada tiga angka bererti.
x2 + 7 x = 4
3
x2 = 4(2x + 3)
7 2
7 2
x2 = 8x + 12
x2 + 7 x + 3 = 4 + 3
2
3
x – 8x – 12 = 0
2
2
x2 + 7 x + 7
3
6
2
x+ 7
6
2
=4+ 7
6
2
= 193
36
7
x+
= ± 193
6
36
x = 1.149 or x = –3.482
–(–8) ± √(–8)2 – 4(1)(–12)
2(1)
8
±
112
√
=
2
x = 8 + √112 or x = 8 – √112
2
2
= 9.29
= –1.29 (rejected)
x =
∴ x = 9.29
F
Additional Mathematics
QF
Q
4. Form quadratic equations by using the given roots.
Bentukkan persamaan kuadratik dengan menggunakan punca-punca yang diberikan.
(b) –1 and / dan 1
4
(a) 1 and / dan 6
2 and / dan –5
Let α = 1 and β = 6.
Let α = 2 and β = –5.
Sum of roots, α + β
= 2 + (–5)
= –3
Product of roots, αβ
= 2 × (–5)
= –10
Sum of roots, α + β =
=
Product of roots, αβ =
=
P A
1+6
7
1×6
6
x – (α + β)x + (αβ) = 0
x2 – 7x + 6 = 0
2
x2 – (α + β)x + (αβ) = 0
x2 – (–3)x + (–10) = 0
x2 + 3x – 10 = 0
PA
Let α = –1 and β = 1 .
4
Sum of roots, α + β = –1 + 1
4
=– 3
4
Product of roots, αβ = –1 × 1
4
=– 1
4
x2 – (α + β)x + (αβ) =
x2 – – 3 x – 1 =
4
4
3
2
x + x– 1=
4
4
4x2 + 3x – 1 =
AP
(x – 2)(x + 5) = 0
x2 + 5x – 2x – 10 = 0
x2 + 3x – 10 = 0
0
0
0
0
5. Solve the following problems.
Selesaikan masalah-masalah berikut.
If α and β are the roots of the quadratic equation 2x2 – x – 15 = 0, form new quadratic equations using the
roots
Jika α dan β ialah punca-punca bagi persamaan kuadratik 2x2 – x – 15 = 0, bentukkan persamaan kuadratik yang baru
menggunakan punca-punca
(i) 2 and / dan 2
(ii) (α + 3) and / dan (β + 3)
α
β
2x2 – x – 15 = 0
α+β =–
= 1
2
c
a
= – 15
2
αβ =
b
a
(i)
Sum of roots:
2 + 2 = 2β + 2α
α
β
αβ
2(α + β)
=
αβ
2 1
2
=
– 15
2
2
=–
15
Product of roots:
2 2 = 4
α β
αβ
4
– 15
2
8
=–
15
=
x2 – – 2 x + – 8 = 0
15
15
2
2
x +
x– 8 =0
15
15
15x2 + 2x – 8 = 0
(ii) Sum of roots:
(α + 3) + (β + 3) = α + β + 6
= 1 +6
2
= 13
2
Product of roots:
(α + 3)(β + 3) = αβ + 3α + 3β + 9
= αβ + 3(α + β) + 9
= – 15 + 3 1 + 9
2
2
=3
x2 – 13 x + 3 = 0
2
2x2 – 13x + 6 = 0
F
Additional Mathematics
QF
Q
(a) If the roots of the quadratic equation x2 – 4x – 5 = 0 are α and β, form new quadratic equations using
the roots
Jika punca-punca bagi persamaan kuadratik x2 – 4x – 5 = 0 ialah α dan β, bentukkan persamaan kuadratik baru menggunakan
punca-punca
(i) 1 and / dan 1
(ii) (α + β)2 only / sahaja
α
β
x2 – 4x – 5 = 0
b
α+β =–
a
=4
c
a
= –5
αβ =
(i)
Sum of roots:
1 + 1 = β+α
α
β
αβ
=– 4
5
(ii) Sum of roots:
(α + β)2 + (α + β)2
= 42 + 42
= 32
Product of roots:
1 1 = 1
α β
αβ
=– 1
5
Product of roots:
(α + β)2 × (α + β)2
= 42 × 42
= 256
x2 – – 4 x – 1 = 0
5
5
x2 + 4 x – 1 = 0
5
5
x2 – 32x + 256 = 0
5x2 + 4x – 1 = 0
(b) It is given that the roots of the quadratic equation 4x2 – px – 4 = 0 are q and – 1 . Find the values of
4
p and q.
Diberi punca-punca bagi persamaan kuadratik 4x2 – px – 4 = 0 ialah q dan – 1 . Cari nilai-nilai p dan q.
4
4x2 – px – 4 = 0
p
x2 – x – 1 = 0
4
Product of roots = –1
∴ q – 1 = –1
4
q =4
Sum of roots =
p
4
p
∴ q+ – 1 =
4
4
p
4– 1 =
4
4
15 = p
4
4
p = 15
(c) Find the possible values of k if one of the roots of the quadratic equation x2 – kx + 8 = 0 is twice the
other.
Cari nilai-nilai k yang mungkin jika salah satu punca bagi persamaan kuadratik x2 – kx + 8 = 0 adalah dua kali ganda
punca yang satu lagi.
Let the roots of the quadratic equation
x2 – kx + 8 = 0 be α and 2α.
Product of roots = 8
∴ 2α2 = 8
α2 = 4
α=± 4
= 2 or –2
Sum of roots
∴ α + 2α =
k =
k =
=
=k
k
3α
3(2)
6
or k = 3(–2)
= –6
F
Additional Mathematics
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Q
6. Find the range of the values of x which satisfy the following quadratic inequalities.
Cari julat nilai x yang memenuhi ketaksamaan kuadratik berikut.
2x2 + 3x – 5 0
(x – 1)(2x + 5)
0
Write in the form of (x – a)(x – b), where
a and b are the roots
Method 1: Graph sketching
When (x – 1)(2x + 5) = 0, x = 1 or x = – 5
2
–5
2
Method 2: Number line
Point testing:
x = –3
(–3 – 1)(–6 + 5)
x
1
x
For (x – 1)(2x + 5)
0, the range of the values of
x is x
– 5 or x 1.
2
–5
2
x=0
(0 – 1)(0 + 5)
0
–5
2
–5
2
x
x=2
(2 – 1)(4 + 5)
0
1
1
0
1
x
x
For (x – 1)(2x + 5)
0, the range of the values of
x is x
– 5 or x
1.
2
Methohd 3: Table
The range of the values of x
– 5
2
x
– 5
2
x
1
1
x
(x – 1)
–
–
+
(2x + 5)
–
+
+
(x – 1)(2x + 5)
+
–
+
For (x – 1)(2x + 5)
(a) x2 + 4x + 3
0, the range of the values of x is x
0
x2 + 4x + 3
(x + 1)(x + 3)
0
0
– 5 or x
2
1.
(b) (x + 3)(x – 5)
0
(x + 3)(x – 5)
0
When (x + 3)(x – 5)
0, x = –3 or x = 5
When (x + 1)(x + 3) = 0, x = –1 or x = –3
–3
For (x + 1)(x + 3)
of x is x
–3 or x
(c) –x2 + 5x + 6
–x2 + 5x + 6
(x + 1)(–x + 6)
–1
x
0, the range of the values
–1.
0
0
0
When (x + 1)(–x + 6) = 0, x = –1 or x = 6
–1
For (x + 1)(–x + 6)
of x is x
–1 or x
6
–3
5
For (x + 3)(x – 5)
of x is –3
x
5.
0, the range of the values
(d) –3x2 + 17x – 10
0
–3x2 + 17x – 10
(3x – 2)(–x + 5)
0
0
When (3x – 2)(–x + 5) = 0, x = 2 or x = 5
3
x
0, the range of the values
6.
x
2
3
For (3x – 2)(–x + 5)
of x is 2
x
5.
3
5
x
0, the range of the values
F
Additional Mathematics
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Q
9. Find the value of q if each of the following quadratic equations has two equal real roots.
Cari nilai q jika setiap persamaan kuadratik berikut mempunyai dua punca nyata yang sama.
(a) qx2 + 2x – 3 = 0
x2 – 6x + q = 0
(b) x2 + (q – 1)x + 1 = 0
For two equal real roots:
b2 – 4ac = 0
(2)2 – 4(q)(–3) = 0
4 + 12q = 0
12q = –4
q=– 1
3
For two equal real roots:
b2 – 4ac = 0
2
(–6) – 4(1)(q) = 0
36 – 4q = 0
–4q = –36
q=9
For two equal real roots:
b2 – 4ac = 0
(q – 1)2 – 4(1)(1) = 0
q2 – 2q + 1 – 4 = 0
q2 – 2q – 3 = 0
(q – 3)(q + 1) = 0
q = 3 or q = –1
10. Find the range of values of m if each of the following quadratic equations has no real roots.
Cari julat nilai m jika setiap persamaan kuadratik berikut tidak mempunyai punca nyata.
(a) 2x2 + 6x – (3 – m) = 0
x2 – 3x + 7m = 0
No real roots:
b2 – 4ac
2
(–3) – 4(1)(7m)
9 – 28m
–28m
28m
m
2x2 + 6x + (–3 + m) = 0
No real roots:
b2 – 4ac
0
2
(6) – 4(2)(–3 + m)
0
(6)2 – 8(–3 + m)
0
36 + 24 – 8m
0
60 – 8m
0
–8m
–60
8m
60
15
m
2
0
0
0
–9
9
9
28
(b) x2 – (3m + 1)x + 12m + 4 = 0
No real roots:
b2 – 4ac
(–3m – 1) – 4(1)(12m + 4)
9m2 + 6m + 1 – 48m – 16
9m2 – 42m – 15
3m2 – 14m – 5
(3m + 1)(m – 5)
When (3m + 1)(m – 5) = 0,
m = – 1 or m = 5
3
0
0
0
0
0
0
2
–
∴ – 1
3
5
1
3
m
m
5
11. Solve the following problems.
Selesaikan masalah-masalah berikut.
(a) It is given the quadratic equation x(2x – 5) = p – 4, where p is a constant, has two different real roots.
Diberi persamaan kuadratik x(2x – 5) = p – 4, dengan keadaan p ialah pemalar, mempunyai dua punca nyata yang berbeza.
(i) Find the range of values of p.
Cari julat nilai p.
(ii) State the value of p if the quadratic equation has two equal real roots.
Nyatakan nilai p jika persamaan kuadratik itu mempunyai dua punca nyata yang sama.
x(2x – 5) = p – 4
2x2 – 5x – p + 4 = 0
(i)
For two different real roots,
b2 – 4ac
(–5)2 – 4(2)(–p + 4)
25 – 8(–p + 4)
25 + 8p – 32
8p – 7
8p
p
0
0
0
0
0
7
7
8
(ii) For two equal real roots,
b2 – 4ac = 0
Hence, p = 7
8
Additional Mathematics
F
QF
Q
12. Find the range of values of k if the graphs of the following quadratic functions intersect the x-axis at two
points.
Cari julat nilai k jika graf bagi fungsi kuadratik berikut menyilang paksi-x pada dua titik.
(a) f(x) = kx2 + 8x + 6
f(x) = x2 – 3x – k
The graph has
two different real roots.
b2 – 4ac 0
2
(–3) – 4(1)(–k) 0
9 + 4k 0
4k –9
k – 9
4
has two different real
b2 – 4ac
2
(8) – 4(k)(6)
64 – 24k
–24k
24k
k
(b) f(x) = x2 + (k – 3)x + 1
roots
0
0
0
–64
64
8
3
has two different real
b2 – 4ac
2
(k – 3) – 4(1)(1)
k2 – 6k + 9 – 4
k2 – 6k + 5
(k – 5)(k – 1)
roots
0
0
0
0
0
When (k – 5)(k – 1) = 0,
k = 5 or k = 1
1
∴k
5
1 or k
k
5
13. Find the value of p if the graphs of the following quadratic functions touch the x-axis at one point only.
Cari nilai p jika graf bagi fungsi kuadratik berikut menyentuh paksi-x pada satu titik sahaja.
(a) f(x) = x2 – 2x + (p – 5)
f(x) = x – 4x – p
(b) f(x) = (p – 3)x2 + 4x – 5
2
The graph has
two equal real roots.
b2 – 4ac = 0
2
(–4) – 4(1)(–p) = 0
16 + 4p = 0
4p = –16
p = –4
has two equal real roots
b2 – 4ac = 0
2
(–2) – 4(1)(p – 5) = 0
4 – 4(p – 5) = 0
4 – 4p + 20 = 0
–4p = –24
p =6
has two equal real roots
b2 – 4ac = 0
2
(4) – 4(p –3)(–5) = 0
16 + 20(p – 3) = 0
16 + 20p – 60 = 0
20p = 44
p = 11
5
14. Find the range of values of h if the following quadratic functions do not intersect the x-axis.
Cari julat nilai h jika graf bagi fungsi kuadratik berikut tidak menyilang paksi-x.
(a) f(x) = 8x2 + 4x + h
f(x) = x2 + 5x – h
The graph does not
have real roots.
b2 – 4ac
0
(5)2 – 4(1)(–h)
0
25 + 4h
0
4h
–25
h
– 25
4
does not have real roots
b2 – 4ac
0
2
(4) – 4(8)(h)
0
16 – 32h
0
–32h
–16
32h
16
1
h
2
(b) f(x) = (2h – 3)x2 + 2hx – 1
does not have real roots
b2 – 4ac
0
2
(2h) – 4(2h – 3)(–1)
0
4h2 + 4(2h – 3)
0
4h2 + 8h – 12
0
h2 + 2h – 3
0
(h – 1)(h + 3)
0
When (h – 1)(h + 3) = 0,
h = 1 or h = –3
–3
∴ –3
h
1
1
h
F
Additional Mathematics
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Q
15. Express each of the following quadratic functions in the form of f(x) = a(x – h)2 + k. Then, state the minimum
or maximum value, axis of symmetry and minimum or maximum point.
Ungkapkan setiap fungsi kuadratik berikut dalam bentuk f(x) = a(x – h)2 + k. Kemudian, nyatakan nilai minimum atau maksimum,
paksi simetri dan titik minimum atau maksimum.
Quadratic function
Fungsi kuadratik
f(x) = x2 – 2x + 5
a 0, shape
which passes
through minimum
point
f(x) = a(x – h)2 + k
f(x) = x2 – 2x + 5
2
2
= x2 – 2x + – 2 + 5 – – 2
2
2
= x2 – 2x + (–1)2 + 5 – (–1)2
= (x – 1)2 + 5 – 1
Add and subtract
= (x – 1)2 + 4
coefficient of x 2
Minimum /
Axis of
maximum
symmetry
value
Nilai minimum / Paksi simetri
maksimum
Minimum /
maximum
point
Titik minimum /
maksimum
Minimum
value
=4
Minimum
point
= (1, 4)
x=1
2
(a) f(x) = x2 – 6x – 3
(b) f(x) = –x2 + 2x – 8
f(x) = x2 – 6x – 3
2
= x2 – 6x + – 6 – 3 – – 6
2
2
2
2
2
= x – 6x + (–3) – 3 – (–3)
= (x – 3)2 – 3 – 9
= (x – 3)2 – 12
2
f(x) = –x2 + 2x – 8
= –[x2 – 2x + 8]
= – x2 – 2x + – 2
2
=
=
=
=
2
+8– – 2
2
2
Minimum
value
= –12
x=3
Minimum
point
= (3, −12)
Maximum
value
= –7
x=1
Maximum
point
= (1, −7)
Minimum
value
= – 11
2
x=– 5
2
–[x2 – 2x + (–1)2 + 8 – (–1)2]
–[(x – 1)2 + 8 – 1]
–[(x – 1)2 + 7]
–(x – 1)2 – 7
(c) f(x) = 2x2 + 10x + 7 f(x) = 2x2 + 10x + 7
= 2 x2 + 5x + 7
2
= 2 x2 + 5x + 5
2
2
+ 7 – 5
2
2
=2 x+ 5
2
2
+ 7 – 25
2
4
=2 x+ 5
2
2
– 11
4
=2x+ 5
2
2
– 11
2
2
Minimum
point
= – 5 , – 11
2
2
Additional Mathematics
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QF
Q
16. Solve the following problems.
Selesaikan masalah-masalah berikut.
Given the quadratic function f(x) = x2 + 6x – 5 has
a minimum point (h, k), find
Diberi fungsi kuadratik f(x) = x2 + 6x – 5 mempunyai titik
minimum (h, k), cari
(i) the values of h and k,
nilai-nilai h dan k,
(ii) the equation of the axis of symmetry,
persamaan paksi simetri,
(iii) the minimum value of f(x).
nilai minimum f(x).
(i) f(x) = x2 + 6x – 5
2
2
= x2 + 6x + 6 – 5 – 6
2
2
= x2 + 6x + (3)2 – 5 – (3)2
= (x + 3)2 – 5 – 9
= (x + 3)2 – 14
∴ h = –3 and k = –14
(ii) x = –3
(iii) –14
(a) The diagram shows the graph of quadratic
function f(x) = (x – p)2 – 8.
Rajah di bawah menunjukkan graf bagi fungsi kuadratik
f(x) = (x – p)2 – 8.
f(x)
–3
0
1
x
Find / Cari
(i) the equation of the axis of symmetry,
persamaan paksi simetri,
(ii) the value of p,
nilai p,
(iii) the coordinates of the minimum point.
koordinat bagi titik minimum.
(i)
x = –3 + 1
2
2
=–
2
= –1
(ii) p = –1
(iii) (–1, –8)
(b) A quadratic function is given by f(x) = –x2 + 8x (c) The quadratic function f(x) = ax2 + bx + c has the
+ k2, where k is a constant. Find
minimum point (–2, –9) and f(–1) = –7. Find
Suatu fungsi kuadratik diberi oleh f(x) = –x2 + 8x + k2,
Fungsi kuadratik f(x) = ax2 + bx + c mempunyai titik
dengan keadaan k ialah pemalar. Cari
minimum (–2, –9) dan f(–1)= –7. Cari
(i) the equation of the axis of symmetry,
(i) the values of a, b and c,
persamaan paksi simetri,
nilai-nilai a, b dan c,
(ii) the possible values of k if the quadratic
(ii) the equation of the axis of symmetry.
persamaan paksi simetri.
function f(x) has a maximum value of 25.
nilai-nilai k yang mungkin jika fungsi kuadratik f(x)
mempunyai nilai maksimum 25.
(i) In the form f(x) = a(x – h)2 + k
h = –2, k = –9
(i) f(x) = –x2 + 8x + k2
∴ f(x) = a(x + 2)2 – 9
= a(x2 + 4x + 4) – 9
= –[x2 – 8x – k2]
= ax2 + 4ax + 4a – 9
2
2
8
8
Comparing:
= – x2 – 8x + –
– k2 – –
2
2
ax2 + bx + c = ax2 + 4ax + 4a – 9
= –[x2 – 8x + (–4)2 – k2 – (–4)2]
∴ b = 4a and c = 4a – 9
= –[(x – 4)2 – k2 – 16]
= –(x – 4)2 + k2 + 16
f(–1) = a(–1)2 + 4a(–1) + 4a – 9
Therefore, the equation of axis of symmetry
is x = 4.
(ii) k2 + 16 =
k2 =
k=
k=
25
9
±√9
3 or k = –3
–7 = a – 4a + 4a – 9
–7 = a – 9
a= 2
b = 4(2)
=8
c = 4(2) – 9
= –1
(ii) x = –2
F
Additional Mathematics
QF
17. Sketch the graph of each of the following quadratic functions.
Lakarkan graf bagi setiap fungsi kuadratik berikut.
f(x) = x2 – x – 6 for / untuk –3
a
x
4
0, then the shape of the graph is
.
Discriminant, b – 4ac = (–1) – 4(1)(–6)
= 1 + 24
= 25 0
∴ f(x) = x2 – x – 6 has two different real roots.
2
When x = 0, f(0) = (0)2 – 0 – 6
= –6
∴ The y-intercept is (0, –6).
2
When x = –3, f(–3) = (–3)2 – (–3) – 6
=9+3–6
=6
Range: –3
∴ (−3, 6)
When x = 4, f(4) = (4)2 – 4 – 6
= 16 – 4 – 6
=6
∴ (4, 6)
f(x) = x2 – x – 6
2
2
= x2 – x + – 1 – 6 – – 1
2
2
2
= x– 1 –6– 1
2
4
2
1
25
= x–
–
2
4
∴ The minimum point is 1 , –6 1 .
2
4
The axis of symmetry is x = 1 .
2
f(x)
x=
(–3, 6)
–2
Discriminant, b2 – 4ac = (–4)2 – 4(1)(–5)
= 16 + 20
= 36 0
∴ f(x) = x – 4x – 5 has two different real roots.
2
2
–5– – 4
2
2
= x2 – 4x + (–2)2 – 5 – (–2)2
= (x – 2)2 – 5 – 4
= (x – 2)2 – 9
∴ The minimum point is (2, –9).
(4, 6)
x
3
f(x) = x2 – x – 6
–6
∴ The x-intercepts are (3, 0) and (–2, 0).
(a) f(x) = x2 – 4x – 5 for / untuk –2 x
6
a
0, then the shape of the graph is
.
1
2
0
When f(x) = 0, x2 – x – 6 = 0
(x – 3)(x + 2) = 0
= 3 or x = –2
f(x) = x2 – 4x – 5
= x2 – 4x + – 4
2
x
(
1
1
, –6 )
2
4
When x = 0, f(0) = (0)2 – 4(0) – 5
= –5
∴ The y-intercept is (0, –5).
When x = –2, f(–2) = (–2)2 – 4(–2) – 5
=4+8–5
=7
∴ (–2, 7)
When x = 6, f(6) = (6)2 – 4(6) – 5
= 36 – 24 – 5
=7
∴ (6, 7)
f(x)
(–2, 7)
x=2
(6, 7)
The axis of symmetry is x = 2.
When f(x) = 0, x – 4x – 5 = 0
(x – 5)(x + 1) = 0
x = 5 or x = –1
2
∴ The x-intercepts are (5, 0) and (–1, 0).
–1 0
–5
5
f(x) = x2 – 4x – 5
(2, –9)
x
4
Q
F
Additional Mathematics
QF
(b) f(x) = –2x2 + 7x + 4 for / untuk –1 x
a 0, then the shape of the graph is
Q
5
.
When f(x) = 0, –2x2 + 7x + 4 = 0
(2x + 1)(4 – x) = 0
x = – 1 or x = 4
2
1
∴ The x-intercepts are (– , 0) and (4, 0).
2
Discriminant, b2 – 4ac = (7)2 – 4(–2)(4)
= 49 + 32
= 81 0
∴ f(x) = –2x2 + 7x + 4 has two different real roots.
When x = 0, f(0) = –2(0)2 + 7(0) + 4
=4
f(x) = –2x2 + 7x + 4
= –2 x2 – 7 x – 2
2
∴ The y-intercept is (0, 4).
– 7
2
7
2
= –2 x – x +
2
2
= –2 x2 – 7 x + – 7
2
4
2
2
–2– – 7
4
= –2 x – 7
4
2
– 2 – 49
16
= –2 x – 7
4
2
– 81
16
= –2 x – 7
4
2
– 7
2
–2–
2
When x = 5, f(5) = –2(5)2 + 7(5) + 4
= –50 + 35 + 4
= –11
∴ (5, –11)
2
2
When x = –1, f(–1) = –2(–1)2 + 7(–1) + 4
= –2 – 7 + 4
= –5
∴ (–1, –5)
f(x)
(1
+ 81
8
3
1
, 10 )
4
8
f(x) = –2x2 + 7x + 4
4
∴ The maximum point is 1 3 , 10 1 .
4
8
3
The axis of symmetry is x = 1 .
4
–1
2
(–1, –5)
0
x
4
x = 13
4
(5, –11)
18. Solve the following problems.
Selesaikan masalah-masalah berikut.
(i) Given f(x) = 3x2 – 27, find the range of values of x such that f(x) is always positive.
Diberi f(x) = 3x2 – 27, cari julat nilai x dengan keadaan f(x) sentiasa positif.
(ii) Given f(x) = 3x2 – 9, find the range of values of x such that f(x) –6x.
Diberi f(x) = 3x2 – 9, cari julat nilai x dengan keadaan f(x) –6x.
(i)
f(x)
3(x2 – 9)
3(x – 3)(x + 3)
0
0
0
(ii)
When 3(x – 3)(x + 3) = 0,
x = 3 or x = –3
f(x)
3x2 – 9
3x2 + 6x – 9
x2 + 2x – 3
(x – 1)(x + 3)
–6x
–6x
0
0
0
When (x – 1)(x + 3) = 0,
x = 1 or x = –3
–3
∴ For 3x – 27
2
0, x
3
–3 or x
x
3.
∴ For 3x2 – 9
–3
1
–6x, –3
x
x
1.
F
Additional Mathematics
QF
Q
(a) Given f(x) = 5x2 – 9x – 2, find the range of values (b) (i) Find the range of values of x for
of x such that f(x) is always
(x + 2)2
11x + 12.
Cari julat nilai x bagi (x + 2)2 11x + 12.
Diberi bahawa f(x) = 5x2 – 9x – 2, cari julat nilai x dengan
(ii) Given the quadratic equation x(3k – x) = 4
keadaan f(x) sentiasa
does not have any real roots, find the range
(i) positive / positif,
of values of p.
(ii) negative / negatif.
Diberi persamaan kuadratik x(3k – x) = 4 tidak
mempunyai punca nyata, cari julat nilai p.
(i)
5x2 – 9x – 2
0
(x – 2)(5x + 1)
0
(i)
x2 + 4x + 4
11x + 12
When (x – 2)(5x + 1) = 0, x = 2 or x = – 1
5
x2 – 7x – 8
0
(x – 8)(x + 1)
0
1
4
–1
∴ For 5x2 – 9x – 2
(ii)
5x2 – 9x – 2
(x – 2)(5x + 1)
0, x
When (x – 8)(x + 1) = 0, x = 8 or x = –1
x
– 1 or x
5
2.
0
0
∴ For x2 – 7x – 8
When (x – 2)(5x + 1) = 0, x = 2 or x = – 1
5
–
–1
1
5
∴ For 5x2 – 9x – 2
2
0, – 1
5
(ii)
x
0, –1
2.
8.
x
3kx – x2 = 4
x2 – 3kx + 4 = 0
No real roots, b2 – 4ac
(–3k)2 – 4(1)(4)
9k2 – 16
(3k – 4)(3k + 4)
x
x
8
0
0
0
0
When (3k – 4)(3k + 4) = 0, k = 4 or k = – 4
3
3
–
4
3
∴ For 9k2 – 16
x
4
3
0, – 4
3
k
4.
3
(c) Hashim threw a ball. The height of the ball is given by the function h(x) = –x2 + 6x + 1, where h is the
height of the ball from the ground, in m, and x is the horizontal distance of the ball from Hashim’s
position, in m. Find
Hashim membaling sebiji bola. Tinggi bola itu diberi oleh fungsi h(x) = –x2 + 6x + 1, dengan keadaan h ialah tinggi bola
dari permukaan tanah, dalam m, dan x ialah jarak mengufuk bola dari kedudukan Hashim, dalam m. Cari
(i) the maximum height, in m, of the ball thrown by Hashim,
tinggi maksimum, dalam m, bola yang dibaling oleh Hashim,
(ii) the horizontal distance between Hashim and the ball, in m, when the ball touches the ground.
jarak mengufuk di antara Hashim dan bola, dalam m, apabila bola itu menyentuh permukaan tanah.
(i)
h(x) = –x2 + 6x + 1
= –(x2 – 6x – 1)
2
2
= – x2 – 6x + –6 – 1 – –6
2
2
= –[x2 – 6x + (–3)2 – 1 – (–3)2]
= –[(x – 3)2 – 10]
= –(x – 3)2 + 10
a = –1
0, so the graph has a maximum point.
The maximum value is 10 when x = 3.
Hence, the maximum height is 10 m.
(ii)
h(x) = 0
–x2 + 6x + 1 = 0
–b ± √b2 – 4ac
x=
2a
–6 ± √62 – 4(–1)(1)
=
2(–1)
–6 ± √40
=
–2
–6 + √40
–6 – √40
x =
or
x =
–2
–2
= –0.16 (ignore)
= 6.16
Hence, the horizontal distance is 6.16 m.
F
Additional Mathematics
QF
Q
SPM
Paper
P A
7. A quadratic function is defined as f(x) = x2 + 6x + h,
where h is a constant.
1
630
Suatu fungsi kuadratik ditakri an sebagai f(x) = x2 + 6x + h,
dengan keadaan h ialah pemalar.
(a) Express f(x) in the form (x + m)2 + n, where m and n
are constants.
Ungkapkan f(x) dalam bentuk (x + m)2 + n, dengan
keadaan m dan n ialah pemalar.
(b) If the minimum value of f(x) is 7, find the value of h.
Jika nilai minimum f(x) ialah 7, cari nilai h.
[4]
Ans: (a) (x + 3)2 + h – 9
(b) h = 16
1. It is given the graph of a quadratic function
630
f(x) = px2 + 6x + q, where p and q are constants, has a
minimum point.
Diberi graf bagi fungsi kuadratik f(x) = px2 + 6x + q, dengan
keadaan p dan q ialah pemalar, mempunyai satu titik minimum.
(a) If p is an integer such that –2 p 2, state the value
of p.
Jika p ialah integer dengan keadaan –2 p 2, nyatakan
nilai p.
(b) Using the answer in 1(a), find the value of q when the
graph touches the x-axis at one point only.
Menggunakan jawapan di 1(a), cari nilai q apabila graf
tersebut menyentuh paksi-x pada satu titik sahaja.
[3]
Ans: (a) p = 1
(b) q = 9
8. Find the range of values of x such that the quadratic
function f(x) = 10 + 3x – x2 is negative.
630
Cari julat nilai x dengan keadaan fungsi kuadratik
f(x) = 10 + 3x – x2 adalah negatif.
[3]
2. Find the range of values of x for 4x2 + 7x
630
Cari julat nilai x untuk 4x2 + 7x
Ans: –2
x
Ans: x
2.
[2]
9. (a) Given one of the roots of the quadratic equation
x2 + (p + 4)x – p2 = 0, where p is a constant, is the
negative of the other, find the value of product of
roots.
630
3. Given the quadratic equation (x + p)2 = 49, where p is
630
a constant, find the values of p if one of the roots of the
equation is 4.
Diberi satu daripada punca-punca persamaan kuadratik
x2 + (p + 4)x – p2 = 0, dengan keadaan p ialah pemalar,
adalah negatif kepada yang satu lagi, cari nilai hasil darab
punca.
[2]
(b) Given the quadratic equation mx2 – 9nx + m = 0,
where m and n are constants, has two equal roots,
find m : n.
Diberi persamaan kuadratik mx2 – 9nx + m = 0, dengan
keadaan m dan n ialah pemalar, mempunyai dua punca
yang sama, cari m : n.
[2]
Ans: (a) –16
(b) 9 : 2
Diberi persamaan kuadratik (x + p)2 = 49, dengan keadaan
p ialah pemalar, cari nilai-nilai p jika satu daripada punca
persamaan tersebut ialah 4.
[2]
Ans: p = 3 or p = –11
4. The quadratic equation 16x2 – 24x + 9 = 0 has roots α and
β. Form a quadratic equation with the roots 3α and 3β.
630
Persamaan kuadratik 16x2 – 24x + 9 = 0 mempunyai punca α
dan β. Bentukkan persamaan kuadratik dengan punca-punca
3α dan 3β.
[3]
Ans: 16x2 – 72x + 81= 0
5. The quadratic function f(x) = –x – 2kx + 4k – 5, where
k is a constant, is always negative when m k n. Find
the values of m and n.
2
630
Fungsi kuadratik f(x) = –x2 – 2kx + 4k – 5, dengan keadaan
k ialah pemalar, adalah sentiasa negatif apabila m k n.
Cari nilai-nilai m dan n.
[3]
Ans: m = –5, n = 1
6. Find the range of values of x for x2 + 30
11x.
11x.
[3]
Ans: 5
x
6
5
2.
1
4
Cari julat nilai x untuk x2 + 30
–2 or x
10. Encik Samad has a rectangular land with a dimension 4x m
in length and 3x m in width. Three square regions of the
land are planted with different types of vegetables. Each
region has a side length of x m. Find the range of values
of x if the region that is not planted with vegetables is at
least (x2 + 32) m2.
630
Encik Samad mempunyai sebidang tanah berbentuk
segi empat tepat yang berukuran 4x m panjang dan
3x m lebar. Tiga kawasan berbentuk segi empat sama ditanam
dengan pelbagai jenis sayur. Setiap kawasan itu mempunyai
panjang sisi x m. Cari julat nilai x jika kawasan yang tidak ditanam
dengan sayuran adalah sekurang-kurangnya (x2 + 32) m2.
[3]
Ans: x 2
Additional Mathematics
11. Given the curve y = (p – 3)x2 – x + 8, where p is a constant,
630
intersects the straight line y = 4x + 6 at two points, find
the range of values of p.
Diberi lengkung y = (p – 3)x2 – x + 8, dengan keadaan p ialah
pemalar, menyilang garis lurus y = 4x + 6 pada dua titik, cari
julat nilai p.
[3]
49
Ans: p
8
Paper
13. It is given 5 and h + 1 are the roots of the quadratic
equation x2 + (k – 1)x – 5 = 0, where h and k are constants.
Find the values of h and k.
y
3
–4
0
y = f(x)
6
x
State / Nyatakan
(a) the roots of the equation when f(x) = 0,
punca-punca persamaan tersebut apabila f(x) = 0,
(b) the equation of the axis of symmetry of the curve.
persamaan paksi simetri lengkung itu.
[2]
Ans: (a) –4 and 6
(b) x = 1
15. It is given the quadratic function f(x) = x2 – 4x + 3 can
be expressed in the form f(x) = (x – 2)2 + m, where m is a
constant.
Diberi fungsi kuadratik f(x) = x2 – 4x + 3 boleh diungkapkan
dalam bentuk f(x) = (x – 2)2 + m, dengan keadaan m ialah
pemalar.
(a) Find the value of m.
Cari nilai m.
(b) Sketch the graph of the function f(x).
Lakar graf bagi fungsi f(x).
[5]
Ans: (a) m = –1
(b) Refer to Answer Section
2
Diberi α dan β ialah punca-punca bagi persamaan kuadratik
x(x – 8) = 3h – 15, dengan keadaan h ialah pemalar.
(a) Find the range of values of h if α ≠ β.
Cari julat nilai h jika α ≠ β.
[3]
α
β
(b) Given
and
are the roots of another quadratic
2
2
2
equation x + kx + 3 = 0, where k is a constant, find
the values of h and k.
α
β
Diberi
dan
ialah punca-punca suatu persamaan
2
2
kuadratik yang lain, x2 + kx + 3 = 0, dengan keadaan k
ialah pemalar, cari nilai-nilai h dan k.
[4]
1
Ans: (a) h –
3
(b) h = 1, k = –4
Diberi persamaan kuadratik hx2 – 5x + k = 0, dengan keadaan
h dan k ialah pemalar, mempunyai punca-punca β dan 3β.
Ungkapkan h dalam sebutan k.
[3]
75
Ans: h =
16k
Rajah di bawah menunjukkan graf bagi suatu fungsi kuadratik
y = f(x).
Q
1. It is given α and β are the roots of the quadratic equation
x(x – 8) = 3h – 15, where h is a constant.
12. Given the quadratic equation hx2 – 5x + k = 0, where h and
k are constants, has roots β and 3β, express h in terms of k.
14. The diagram shows the graph of a quadratic function
y = f(x).
QF
630
630
Diberi 5 dan h + 1 ialah punca-punca bagi persamaan kuadratik
x2 + (k – 1)x – 5 = 0 dengan keadaan h dan k ialah pemalar.
Cari nilai-nilai h dan k.
[3]
Ans: h = –2, k = –3
F
2. The curve of the quadratic function f(x) = 2(x – p)2 + 3q
intersects the x-axis at points (3, 0) and (7, 0). The straight
line y = –6 touches the minimum point of the curve.
630
Lengkung bagi fungsi kuadratik f(x) = 2(x – p)2 + 3q menyilang
paksi-x pada titik (3, 0) dan (7, 0). Garis lurus y = –6
menyentuh titik minimum lengkung tersebut.
(a) Find the values of p and q.
Cari nilai-nilai p dan q.
[2]
(b) Hence, sketch the graph of f(x) for 0 x 8.
Seterusnya, lakar graf f(x) untuk 0 x 8
[3]
(c) If the curve is reflected about the x-axis, write the
equation of the curve.
Jika lengkung tersebut dipantulkan pada paksi-x, tulis
persamaan bagi lengkung tersebut.
[1]
Ans: (a) p = 5, q = –2
(b) Refer to Answer Section
(c) f(x) = –2(x – 5)2 + 6
3. The quadratic equation x2 – 7x + 12 = 0 has two roots p
and q, where p q.
Persamaan kuadratik x2 – 7x + 12 = 0 mempunyai dua punca
p dan q, dengan keadaan p q.
(a) Find / Cari
(i) the values of p and q,
nilai-nilai p dan q,
(ii) the range of values of x if x2 – 7x + 12 0.
julat nilai x jika x2 –7x + 12 0.
[4]
(b) By using the values of p and q in 3(a)(i), form a
quadratic equation which has the roots p – 2 and q + 4.
Dengan menggunakan nilai-nilai p dan q di 3(a)(i), bentuk
satu persamaan kuadratik yang mempunyai punca-punca
p – 2 dan q + 4.
[3]
Ans: (a) (i) p = 4, q = 3; (ii) x 3 or x 4
(b) x2 – 9x + 14 = 0
Additional Mathematics
37 5
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Additional Mathematics
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Additional Mathematics
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Additional Mathematics
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Additional Mathematics
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S
Additional Mathematics
2. Solve the following systems of linear equations in three variables by using the elimination method.
Selesaikan sistem-sistem persamaan linear dalam tiga pemboleh ubah berikut dengan menggunakan kaedah penghapusan.
x – 3y + z = 12
x + 4y + 3z = 8
4x – 5y – 2z = 4
x – 3y + z = 12 ………
x + 4y + 3z = 8 ……..…
– : 7y + 2z = –4 ………
x + 4y + 3z
4x – 5y – 2z
× 4: 4x + 16y + 12z
– :
21y + 14z
=
=
=
=
Choose two
equations and then
eliminate variable x
8 ……...…
4 ……...…
32 …….…
28 …….…
× 3: 21y + 6z = –12 ………..
– :
8z = 40
z= 5
(a)
Choose another two
equations and then
eliminate variable x
Eliminate
variable y
Substitute z = 5 into
7y + 2(5) = –4
7y = –14
y = –2
,
Substitute y = –2 and z = 5 into
x – 3(–2) + 5 = 12
x + 6 + 5 = 12
x + 11 = 12
x=1
,
\ x = 1, y = –2, z = 5
x + y + z = –2 …………
2x – 3y + 4z = 14 …………
5x + 2y – 3z = 10 …………
× 2:
:
–
:
2x + 2y + 2z = –4 ….........…
2x – 3y + 4z = 14
5y – 2z = –18 …………
× 5: 10x – 15y + 20z = 70 …….........
× 2:
10x + 4y – 6z = 20 ….....……
– :
–19y + 26z = 50 ……..……
× 19:
× 5:
+ :
95y – 38z
–95y + 130z
92z
z
=
=
=
=
–342 …….….
250 …………
–92
–1
Substitute z = –1 into
5y – 2(–1) = –18
5y + 2 = –18
5y = –20
y = –4
,
Substitute y = –4 and z = –1 into
x – 4 – 1 = –2
x – 5 = –2
x=3
,
\ x = 3, y = –4, z = –1
(b) x – 2y + 5z = –22 ..............
x – 6y + 3z = –12 ...............
3x – 4y + z = 19.................
–
:
:
4y + 2z = –10
2y + z = –5 ....................
× 3: 3x – 18y + 9z = –36 ..................
3x – 4y + z = 19
– :
–14y + 8z = –55 ..................
× 7:
:
+
:
14y + 7z =
–14y + 8z =
15z =
z=
–35 ..................
–55
–90
–6
Substitute z = –6 into
2y – 6 = –5
2y = 1
y= 1
2
,
Substitute y = 1 and z = –6 into
2
x – 2( 1 ) + 5(–6) = –22
2
x – 1 – 30 = –22
x – 31 = –22
x= 9
\ x = 9, y = 1 , z = –6
2
,
S
Additional Mathematics
3. Solve the following systems of linear equations in three variables by using the substitution method.
Selesaikan sistem-sistem persamaan linear dalam tiga pemboleh ubah berikut dengan menggunakan kaedah penggantian.
3x – 2y + 4z = 12
x + 5y – 3z = –19
2x + 7y – z = –15
3x – 2y + 4z = 12 ..........
x + 5y – 3z = –19.........
2x + 7y – z = –15.........
From
Express x in
terms of y and z
2y – 4z + 12
,x=
............
3
Substitute
into
,
Substitute z = 4 into
13(4) – 69
y =
17
= –1
2y – 4z + 12
+ 5y – 3z = –19
3
2y – 4z + 12 + 15y – 9z = –57
17y – 13z = –69 ......
Substitute
2
into
,
2y – 4z + 12
+ 7y – z = –15
3
4y – 8z + 24 + 21y – 3z = –45
25y – 11z = –69 .....
Substitute
into ,
13z – 69
25
– 11z =
17
325z – 1 725 – 187z =
138z =
z =
Have two variables
only, y and z
–69
–1 173
552
4
,
Substitute y = –1 and z = 4 into
3x – 2(–1) + 4(4) = 12
3x + 18 = 12
3x = –6
x = –2
,
\ x = –2, y = –1, z = 4
From
, 17y – 13z = –69
13z – 69
y=
.....
17
Express y in
terms of z
(a) 2x – 5y – 6z = 12 ...........
x + 3y + 8z = –5 ...........
5x – 4y + 3z = 9 .............
From
,x=
5y + 6z + 12
.....................
2
Substitute
into ,
5y + 6z + 12
+ 3y + 8z =
2
5y + 6z + 12 + 6y + 16z =
11y + 22z =
y + 2z =
–5
–10
–22
–2 .................
Substitute
into ,
5y + 6z + 12
5
– 4y + 3z = 9
2
25y + 30z + 60 – 8y + 6z = 18
17y + 36z = –42 ............
From
, y + 2z = –2
y = –2 – 2z ..............
Substitute
into
17(–2 – 2z) + 36z =
–34 – 34z + 36z =
2z =
z=
,
–42
–42
–8
–4
Substitute z = –4 into
y = –2 – 2(–4)
= –2 + 8
=6
,
Substitute y = 6 and z = –4 into
2x – 5(6) – 6(–4) = 12
2x – 30 + 24 = 12
2x = 18
x =9
\ x = 9, y = 6, z = –4
,
S
Additional Mathematics
(b)
x – 2y – 4z = –17 .........
3x + 7y + z = 1 .............
4x + 6y + 3z = 23 ...........
From
, x = 2y + 4z – 17 .......................
Substitute
into
Substitute
into ,
14(4 – z) + 19z = 91
56 – 14z + 19z = 91
5z = 35
z= 7
,
3(2y + 4z – 17) + 7y + z =
6y + 12z – 51 + 7y + z =
13y + 13z =
y+z=
1
1
52
4 ...................
Substitute z = 7 into
y =4–7
= –3
Substitute
into ,
4(2y + 4z – 17) + 6y + 3z = 23
8y + 16z – 68 + 6y + 3z = 23
14y + 19z = 91 ..............
From
,
Substitute y = –3 and z = 7 into
x – 2(–3) – 4(7) = –17
x + 6 – 28 = –17
x= 5
, y+z=4
y = 4 – z ....................
,
\ x = 5, y = –3, z = 7
4. Solve the following systems of linear equations.
Selesaikan sistem-sistem persamaan linear berikut.
(i)
x – 3y + 4z = 11
2x – 6y + 8z = 14
y – 6z = –25
(ii)
4x – y + 2z = –4 ............
3x + 2y + z = 18 ............
15x – y + 7z = 6 ..............
× 2: 8x – 2y + 4z = –8 .............
+ : 11x + 5z = 10 .............
–
: 11x + 5z = 10 ............
– :
0+0 =0
0=0
\ Since 0 = 0, the system has infinite solutions.
x – 3y + 4z = 11 ...............
2x – 6y + 8z = 14 ...............
y – 6z = –25 .............
× 2: 2x – 6y + 8z = 22 ...............
– : 0+0+0 =8
0=8
\ Since 0 ≠ 8, the system has no solution.
(a)
x + 2y – 3z = 22 ............
–2x – 4y + 6z = –34 ..........
5y – z = 28 ............
× 2: 2x + 4y – 6z = 44 ..........
+ : 0 + 0 + 0 = 10
0 = 10
\ Since 0 ≠ 10, the system has no solution.
4x – y + 2z = –4
3x + 2y + z = 18
15x – y + 7z = 6
(b)
2x + 5y – z = –9 ..........
3x – 4y + 6z = –9 ..........
–17x – 31y + z = 72 ..........
× 6: 12x + 30y – 6z = –54 ...........
+ :
15x + 26y = –63 ...........
+ :
–15x – 26y = 63 .............
+ : 0+0 =0
0=0
\ Since 0 = 0, the system has infinite solutions.
S
Additional Mathematics
5. Solve the following problems.
Selesaikan masalah-masalah berikut.
A transport company provides bus, taxi and van service to send students to school and adults to work. If
a person takes the bus, another person takes the taxi and a third person takes the van, then the total fare
collected by the transport company is RM25. If a person takes the bus, 2 people take the taxi and 3 other
people take the van, then the total fare collected is RM49. Finally, if 2 people take the bus, a person takes
the taxi and another person takes the van, then the fare collected is RM32. Find the bus, taxi and van fares
for one person.
Sebuah syarikat pengangkutan menyediakan perkhidmatan bas, teksi dan van untuk menghantar pelajar ke sekolah dan orang
dewasa pergi bekerja. Jika seorang menaiki bas, seorang lagi menaiki teksi dan orang ketiga menaiki van, maka jumlah tambang
yang akan dikutip oleh syarikat pengangkutan tersebut ialah RM25. Jika seorang menaiki bas, 2 orang menaiki teksi dan 3 orang
lain menaiki van, maka jumlah tambang yang dikutip ialah RM49. Akhir sekali, jika 2 orang menaiki bas, seorang menaiki teksi
dan seorang lagi menaiki van, tambang yang dikutip ialah RM32. Cari tambang bas, teksi dan van untuk satu orang.
Let x = bus fare
y = taxi fare
z = van fare
From
x + y + z = 25 .......
x + 2y + 3z = 49 .......
2x + y + z = 32 .......
From
, x = 25 – y – z .........
, y = 24 – 2z ........
Substitute
into
24 – 2z + z = 18
z=6
,
Substitute z = 6 into
y = 24 – 2(6)
= 12
,
Substitute
into ,
25 – y – z + 2y + 3z = 49
y + 2z = 24 .......
Substitute y = 12 and z = 6 into
x + 12 + 6 = 25
x=7
,
Substitute
into ,
2(25 – y – z) + y + z = 32
50 – 2y – 2z + y + z = 32
y + z = 18......
Therefore, the bus, taxi and van fares for
one person are RM7, RM12 and RM6
respectively.
(a) Customer A buys 3 loaves of bread, 5 packets of biscuits and 2 boxes of milk with RM41. Customer
B buys 2 loaves of bread, 8 packets of biscuits and 3 boxes of milk with RM56. Customer C buys 4
loaves of bread, 7 packets of biscuits and 5 boxes of milk with RM70. Find the price of a loaf of bread,
a packet of biscuits and a box of milk.
Pelanggan A membeli 3 buku roti, 5 paket biskut dan 2 kotak susu dengan RM41. Pelanggan B membeli 2 buku roti, 8 paket
biskut dan 3 kotak susu dengan RM56. Pelanggan C membeli 4 buku roti, 7 paket biskut dan 5 kotak susu dengan RM70. Cari
harga bagi sebuku roti, sepaket biskut dan sekotak susu.
Let x = price of a loaf of bread
y = price of a packet of biscuits
z = price of a box of milk
3x + 5y + 2z = 41 ...................
2x + 8y + 3z = 56 ...................
4x + 7y + 5z = 70 ...................
From
,x=
41 – 5y – 2z
....
3
Substitute
into ,
41 – 5y – 2z
2
+ 8y + 3z = 56
3
82 – 10y – 4z + 24y + 9z = 168
14y + 5z = 86..............
Substitute
into ,
41 – 5y – 2z
4
+ 7y + 5z = 70
3
164 – 20y – 8z + 21y + 15z = 210
y + 7z = 46 ............
From
, y = 46 – 7z ……
Substitute
into
14(46 – 7z) + 5z =
644 – 98z + 5z =
–93z =
z =
,
86
86
–558
6
Substitute z = 6 into
y = 46 – 7(6)
=4
,
Substitute y = 4 and z = 6 into
3x + 5(4) + 2(6) = 41
3x + 20 + 12 = 41
3x = 9
x=3
,
Therefore, the prices of a loaf of bread, a
packet of biscuits and a box of milk are
RM3, RM4 and RM6 respectively.
S
Additional Mathematics
6. Solve the following simultaneous equations.
Selesaikan persamaan-persamaan serentak berikut.
2x – 3y = 5 ..........
x2 – 4y2 – y = 3 ..........
From
,x=
Substitute
5 + 3y
....
2
into
y =
,
5 + 3y
– 4y2 – y =
2
25 + 30y + 9y2
– 4y2 – y =
4
25 + 30y + 9y2 – 16y2 – 4y =
–7y2 + 26y + 13 =
2
–b ± b2 – 4ac
Method of quadratic formula
2a
–26 ± (26)2 – 4(–7)(13)
=
2(–7)
–26 ± 1 040
=
–14
Make x as the
subject of formula
3
3
12
0
–26 + 1 040
–14
= −0.4464
y =
Solve the
quadratic
equation
or
–26 – 1 040
–14
= 4.1606
y =
Substitute the values of y into ,
5 + 3(–0.4464)
When y = −0.4464, x =
2
= 1.8304
Quadratic equations can be solved using the method
of completing the square, formula or factorisation.
Persamaan kuadratik boleh diselesaikan dengan kaedah
penyempurnaan kuasa dua, rumus atau pemfaktoran.
5 + 3(4.1606)
2
= 8.7409
When y = 4.1606, x =
\ x = 1.8304, y = –0.4464; x = 8.7409, y = 4.1606
(a)
x – 5y = 9............
x2 – 4y2 + 6xy = –12 ........
From
, x = 9 + 5y ……
Substitute
into ,
(9 + 5y)2 – 4y2 + 6y(9 + 5y)
81 + 90y + 25y2 – 4y2 + 54y + 30y2 + 12
51y2 + 144y + 93
17y2 + 48y + 31
(y + 1)(17y + 31)
y = –1 or y
(b)
=
=
=
=
=
=
–12
0
0
0
0
– 31
17
Substitute the values of y into ,
When y = –1, x = 9 + 5(–1)
=4
When y = – 31 , x = 9 + 5 – 31
17
17
=– 2
17
\ x = 4, y = –1; x = – 2 , y = – 31
17
17
3x + 2y = 9 .............
x2 + y2 + 2xy = 4 .............
From
,x=
9 – 2y
……
3
Substitute
into ,
9 – 2y 2
9 – 2y
+ y2 + 2y
=
3
3
81 – 36y + 4y2
18y – 4y2
+ y2 +
=
9
3
2
2
2
81 – 36y + 4y + 9y + 3(18y – 4y ) =
81 – 36y + 4y2 + 9y2 + 54y – 12y2 – 36 =
y2 + 18y + 45 =
(y + 3)(y + 15) =
y = –3 or y =
4
4
36
0
0
0
–15
Substitute the values of y into
9 – 2(–3)
When y = –3, x =
3
=5
9 – 2(–15)
When y = –15, x =
3
= 13
\ x = 5, y = –3; x = 13, y = –15
,
Additional Mathematics
(c)
S
x – 4y = 1 ............
x2 – 5y2 – 2xy = 12 ..........
From
, x = 4y + 1 ……
–6 ± √(6)2 – 4(3)(–11)
2(3)
–6 ± √168
=
6
y =
Substitute
into ,
(4y + 1)2 – 5y2 – 2y(4y + 1) = 12
16y2 + 8y + 1 – 5y2 – 8y2 – 2y – 12 = 0
3y2 + 6y – 11 = 0
–6 + √168
6
= 1.1602
y =
or
–6 – √168
6
= –3.1602
y =
Substitute the values of y into ,
When y = 1.1602, x = 4(1.1602) + 1
= 5.6408
When y = –3.1602, x = 4(–3.1602) + 1
= –11.6408
\ x = 5.6408, y = 1.1602;
x = –11.6408, y = –3.1602
(d)
3x + 8y = –1 .............
x2 + 2xy – y = 7 ...............
From
,y=
–1 – 3x
……
8
Substitute
into ,
–1
–
3x
–1 – 3x
x2 + 2x
–
=
8
8
2
–2x – 6x
–1 – 3x
x2 +
–
=
8
8
8x2 + (–2x – 6x2) – (–1 – 3x) =
8x2 – 2x – 6x2 + 1 + 3x – 56 =
2x2 + x – 55 =
(x – 5)(2x + 11) =
x = 5 or x =
(e)
Substitute the values of x into
–1 – 3(5)
When x = 5, y =
8
= –2
7
,
–1 – 3 – 11
2
When x = – 11 , y =
8
2
= 31
16
7
56
0
0
0
– 11
2
\ x = 5, y = –2; x = – 11 , y = 31
2
16
3x – 2y = 4 ..............
2x2 – xy + y2 = 5 ..............
From
,x=
2y + 4
……
3
into ,
2y + 4 2
2y + 4
2
–y
+ y2
3
3
4y2 + 16y + 16
2y2 + 4y
2
–
+ y2
9
3
2(4y2 + 16y + 16) – 3(2y2 + 4y) + 9y2
8y2 + 32y + 32 – 6y2 – 12y + 9y2 – 45
11y2 + 20y – 13
–20 ± √(20)2 – 4(11)(–13)
2(11)
–20 ± √972
=
22
y =
Substitute
=5
=5
= 45
=0
=0
–20 + √972
22
= 0.5080
y =
–20 – √972
22
= –2.3262
or y =
Substitute the values of y into ,
2(0.5080) + 4
When y = 0.5080, x =
3
= 1.6720
2(–2.3262) + 4
When y = –2.3262, x =
3
= –0.2175
\ x = 1.6720, y = 0.5080;
x = –0.2175, y = –2.3262
Additional Mathematics
S
7. Solve the following simultaneous equations.
Selesaikan persamaan-persamaan serentak berikut.
10 + 2 = 3 ...............
x
y
x + 3y = 11 .............
From
,
10y + 2x
=3
xy
10y + 2x = 3xy ……
Substitute the values of y into
When y = 2, x = 11 – 3(2)
=5
From
, x = 11 – 3y .................
When y = 11 , x = 11 – 3 11
9
9
= 22
3
\ x = 5, y = 2; x = 22 , y = 11
3
9
Substitute
into
10y + 2(11 – 3y) =
10y + 22 – 6y =
9y2 – 29y + 22 =
(y – 2)(9y – 11) =
y=
(a)
,
3y(11 – 3y)
33y – 9y2
0
0
2 or y = 11
9
3 – 4 = –5 ............
x
y
2x + y = –5 ............
3y – 4x
= –5
xy
3y – 4x = –5xy ..........
From
,
From
, y = –5 – 2x .................
Substitute
into
3(–5 – 2x) – 4x =
–15 – 6x – 4x =
10x2 + 35x + 15 =
2x2 + 7x + 3 =
(2x + 1)(x + 3) =
x =
(b)
,
,
–5x(–5 – 2x)
25x + 10x2
0
0
0
– 1 or x = –3
2
Substitute the values of x into
When x = – 1 , y = –5 – 2 – 1
2
2
= –4
,
When x = –3, y = –5 – 2(–3)
=1
\ x = – 1 , y = –4; x = –3, y = 1
2
x
y
–
= 3 ..............
4
3
8 – 6 = 3 ..............
x
y
From
,
From
,
3x – 4y
=3
12
36 + 4y
x=
..........
3
Substitute the values of y into
36 + 4(–3)
When y = –3, x =
3
=8
8y – 6x
=3
xy
8y – 6x = 3xy ..................
When y = –6, x =
Substitute
into
36 + 4y
8y – 6
=
3
8y – 72 – 8y =
4y2 + 36y + 72 =
y2 + 9y + 18 =
(y + 3)(y + 6) =
y=
,
36 + 4y
3
36y + 4y2
0
0
0
–3 or y = –6
3y
36 + 4(–6)
3
=4
\ x = 8, y = –3; x = 4, y = –6
,
S
Additional Mathematics
8. Solve the following simultaneous equations using graph representation.
Selesaikan persamaan-persamaan serentak berikut menggunakan perwakilan graf.
(a)
3x – y = 6
x2 + 2y2 – xy = 8
2x – y = 8
x2 + y2 – xy = 19
x
0
1
2
3
x
2
3
4
5
y (For/ Untuk
3x – y = 6)
–6
–3
0
3
y (For/ Untuk
2x – y = 8)
–4
–2
0
2
y (For/ Untuk
x2 + 2y2 – xy = 8)
±2
–1.64
or 2.14
–1 or
2
0.5
or 1
–3
or 5
–2
or 5
–0.65 or
4.65
2
or 3
y (For/ Untuk
x2 + y2 – xy = 19)
y
y
5
3
4
2
(2.55, 1.7)
3
1
0
1
2
3
x
(5, 2)
2
1
–1
(1.55, –1.35)
–2
0
–3
–1
–4
–2
–5
–3
–6
–4
\ The solutions are (1.55, –1.35) and (2.55, 1.7).
2
4
6
(3, –2)
\ The solutions are (3, –2) and (5, 2).
x
Additional Mathematics
S
9. Solve the following problems.
Selesaikan masalah-masalah berikut.
The straight line x + y = 2 and the curve x2 + y2 + xy = 7 intersect at two different points. Find the coordinates
of the points of intersection.
Garis lurus x + y = 2 dan lengkung x2 + y2 + xy = 7 bersilang pada dua titik yang berbeza. Cari koordinat bagi titik-titik
persilangan tersebut.
x + y = 2 .........
x2 + y2 + xy = 7 .........
From
Substitute the values of y into
When y = 3, x = 2 – 3
= –1
When y = –1, x = 2 – (–1)
=3
, x = 2 – y .....
Substitute
into ,
(2 – y)2 + y2 + y(2 – y) =
4 – 4y + y2 + y2 + 2y – y2 =
y2 – 2y – 3 =
(y – 3)(y + 1) =
y = 3 or y =
7
7
0
0
–1
,
Therefore, the coordinates of the points of
intersection are (3, –1) and (–1, 3).
(a) The perimeter of a rectangular board is 50 cm and its area is 126 cm2. Find the length and width of
the board.
Perimeter bagi sekeping papan yang berbentuk segi empat tepat ialah 50 cm dan luasnya ialah 126 cm2. Cari panjang dan
lebar papan itu.
x cm
y cm
Let x = length of the rectangular board
y = width of the rectangular board
2x + 2y = 50 .................
xy = 126 ...............
÷ 2: x + y = 25 ..........
126
From , x =
.........
y
Substitute
into ,
126
+ y = 25
y
126 + y2 = 25y
2
y – 25y + 126 = 0
(y – 18)(y – 7) = 0
y = 18 or y = 7
Substitute the values of y into
126
When y = 18, x =
18
=7
126
When y = 7, x =
7
= 18
,
Therefore, the length of the rectangle is 18 cm and
the width of the rectangle is 7 cm.
(b) The sum of two numbers is 18 and the sum of squares of the two numbers is 170. Find the value of
the two numbers.
Hasil tambah dua nombor ialah 18 dan hasil tambah kuasa dua bagi dua nombor itu ialah 170. Cari nilai dua nombor itu.
Let the two numbers be x and y.
x + y = 18 .......
x2 + y2 = 170 .....
From
, x = 18 – y ...
Substitute
into
(18 – y)2 + y2
324 – 36y + y2 + y2
2y2 – 36y + 154
y2 – 18y + 77
(y – 11)(y – 7)
y
,
= 170
= 170
=0
=0
=0
= 11 or y = 7
Substitute the values of y into
When y = 11, x = 18 – 11
=7
When y = 7, x = 18 – 7
= 11
,
Therefore, the value of the two numbers are 7
and 11.
S
Additional Mathematics
(c) The straight line 3y = –(x + 4) intersects the curve x(2 + y) + y = –8 at points A and B. Find the
coordinates of intersection points of the straight line and the curve.
Garis lurus 3y = –(x + 4) menyilang lengkung x(2 + y) + y = –8 pada titik A dan B. Cari koordinat titik-titik persilangan
antara garis lurus dan lengkung itu.
3y = –(x + 4) ...........
x(2 + y) + y = –8
2x + xy + y = –8 ...................
From
,y=–
Substitute the values of x into
(5 + 4)
When x = 5, y = –
3
= –3
(–4 + 4)
When x = –4, y = –
3
=0
x+4
.............
3
Substitute
into ,
x
+
4
2x + x –
+ – x+4
3
3
–x
–
4
–x
–4
2x + x
+
3
3
6x + x(–x – 4) + (–x – 4)
6x – x2 – 4x – x – 4
–x2 + x + 20
x2 – x – 20
(x – 5)(x + 4)
x
= –8
= –8
=
=
=
=
=
=
,
Therefore, the coordinates of intersection points
of the straight line and the curve are (5, −3) and
(−4, 0).
–24
–24
0
0
0
5 or x = –4
(d) The diagram shows the plan of a playground. ABEF is a square and BCDE is a rectangle.
Rajah di bawah menunjukkan pelan sebuah taman permainan. ABEF ialah sebuah segi empat sama dan BCDE ialah sebuah
segi empat tepat.
A
B
C
(y – 4) m
F
E
2x m
D
It is given AF = CD, the perimeter of the rectangle ACDF is 116 m and its area is 480 m2. Find the
values of x and y.
Diberi AF = CD, perimeter segi empat tepat ACDF ialah 116 m dan luasnya ialah 480 m2. Cari nilai-nilai x dan y.
Perimeter of ACDF,
4(y – 4) + 2(2x) = 116
4y – 16 + 4x = 116
4y + 4x = 132
x + y = 33 ..............
Area of ACDF,
(y – 4)(2x + y – 4) =
y(2x + y – 4) – 4(2x + y – 4) =
2xy + y2 – 4y – 8x – 4y + 16 =
y2 – 8y – 8x + 2xy =
From
, x = 33 – y ...............
Substitute
into ,
y2 – 8y – 8(33 – y) + 2y(33 – y)
y2 – 8y – 264 + 8y + 66y – 2y2
–y2 + 66y – 728
y2 – 66y + 728
(y – 52)(y – 14)
y
480
480
480
464 ........
Substitute y = 52 into ,
x = 33 – 52
= –19 (Negative, rejected)
Substitute y = 14 into
x = 33 – 14
= 19
\ x = 19, y = 14
,
=
=
=
=
=
=
464
464
0
0
0
52 or y = 14
S
Additional Mathematics
SPM
Paper
P A
6. Solve the following system of linear equations in three
variables by using the elimination method.
2
1. Solve the following simultaneous equations.
630
Selesaikan persamaan serentak berikut.
4x – y – 6 = 0,
Ans: x = 3, y = 6; x =
3x2 + 4y2 – 8xy = 27
39
19
, y = –1
35
35
[5]
2. Danial planted cabbages on a plot of land in the shape
of a right-angled triangle. The longest side of the land is
y metres. The lengths of the other two sides are x metres
and (x + 6) metres. He fenced the land with 72 metres of
barbed wire. Find the length, in metres, of each side of the
land.
630
Danial menanam kubis di sebidang tanah yang berbentuk segi
tiga bersudut tegak. Sisi yang paling panjang tanah itu ialah
y meter. Panjang dua sisi yang lain ialah x meter dan (x + 6)
meter. Dia menggunakan kawat berduri sepanjang 72 meter
untuk memagar tanah itu. Cari panjang, dalam meter, setiap
sisi tanah itu.
[7]
Ans: 18 m, 24 m and 30 m
3. Solve the following simultaneous equations.
630
7. Solve the following simultaneous equations.
Selesaikan persamaan serentak berikut.
1
x + y = –3, x2 + 9y = 7
5
Ans: x =
34
9
, y = –4 ; x = –5, y = –2
5
25
[5]
p
q
13
+
=
and 2 – 4 = –16, find the
72
2
3
p
q
possible values of p and q.
8. Given that
p
q
13
2 4
+ =
dan – = –16, cari nilai-nilai p dan
p q
2
3
72
q yang mungkin.
[5]
1
1
13
13
Ans: p = , q = ; p = –
,q=
4
6
72
16
Diberi
9. Solve the simultaneous equations h – 3k = –1 and
k + 2hk – 5h = 0. Give your answer correct to three
decimal places.
Selesaikan persamaan serentak berikut.
x – 5y = 3,
Selesaikan sistem persamaan linear dalam tiga pemboleh ubah
berikut dengan menggunakan kaedah penghapusan.
x + 2y – z = 17
–2x + y + 6z = –8
5x – 3y + 9z = –7
[5]
Ans: x = 4, y = 6, z = –1
2x2 + 3xy + 4y2 = 18
[5]
Ans: x = 3, y = 0; x = –2, y = –1
Selesaikan persamaan serentak h – 3k = –1 dan
k + 2hk – 5h = 0. Berikan jawapan anda betul kepada tiga
tempat perpuluhan.
[5]
4. Solve the following system of linear equations in three
variables by using the substitution method.
Selesaikan sistem persamaan linear dalam tiga pemboleh ubah
berikut dengan menggunakan kaedah penggantian.
x+y+z=6
3x – 2y – 4z = 9
2x + 5y + 3z = 11
Ans: h = 5.915, k = 2.305; h = 0.086, k = 0.362
10. Solve the following simultaneous equations.
Selesaikan persamaan serentak berikut.
y – 3x + 2 = 0,
x2 – 4y2 – 5y + 3 = 0
Give your answer correct to four significant figures.
[5]
Ans: x = 5, y = –1, z = 2
Berikan jawapan anda betul kepada empat angka bererti.
[5]
Ans: x = 0.8409, y = 0.5227; x = 0.1019, y = –1.694
5. Solve the following system of linear equations in three
variables by using the substitution method.
Selesaikan sistem persamaan linear dalam tiga pemboleh ubah
berikut dengan menggunakan kaedah penggantian.
x – 3y + 5z = 36
2x – 4y + z = 21
3x + 5y – 6z = –39
[5]
Ans: x = 2, y = –3, z = 5
11. Find the coordinates of the intersection points of the
straight line x – 1 y = 6 and curve y2 – 2x + 3y = –6.
4
Cari koordinat titik-titik persilangan antara garis lurus
1
x – y = 6 dan lengkung y2 – 2x + 3y = –6.
4
3 1
Ans: (6 , 1 ) and (5, –4)
8 2
[5]
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Additional Mathematics
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2. Determine the value of x in each of the following equations.
Tentukan nilai x dalam setiap persamaan berikut.
2x = 64
2x = 2 6
x=6
Compare the powers of both sides of
the equation when the bases are the
same.
Bandingkan kuasa kedua-dua belah
persamaan apabila asasnya adalah sama.
(b) (23 – 5x)(8x – 4) = 1
(23 – 5x)[(23)x – 4] =
(23 – 5x)(23x – 12) =
23 – 5x +(3x – 12) =
2–9 – 2x =
–9 – 2x =
2x =
x=
1
1
(a) 2 × 5x = 2
625
5x = 625–1
5x = (54)–1
5x = 5–4
x = –4
(c)
2
20
20
20
0
–9
– 9
2
0
9x – 3
9x – 3
x–3
x
=
=
=
=
81
92
2
5
3. Solve each of the following problems.
Selesaikan setiap masalah berikut.
A sample of radioactive material with an initial mass of 250 g decay from time to time such that its mass is
p
halved every hour. Determine its mass after 10 hours in the form of k a where k, a, b and p are integers.
b
Satu sampel bahan radioaktif dengan jisim permulaan 250 g mereput dari semasa ke semasa, dengan jisimnya diseparuhkan setiap
a p
jam. Tentukan jisimnya selepas 10 jam dalam bentuk k
dengan keadaan k, a, b dan p ialah integer. 276 SSO L
b
After 1 hour, mass = 250 × 1
2
After 2 hours, mass = 250 × 1 × 1 = 250 1
2
2
2
10
1
After 10 hours, mass = 250
2
2
Amoeba is a unicellular organism which has the ability to alter its shape. It carries out binary fission to
increase its number, that is, splits into two halves. In a certain colony, at the initial stage, there are 18
amoebae. Find the number of amoebae in the colony after the 6th fission in the form of 2m3n where m and
n are positive integers. 276 SSO L
Ameba ialah sejenis organisma unisel yang sentiasa berubah bentuknya. Ameba menjalankan proses pembelahan dedua untuk
menambah bilangannya, iaitu membahagi kepada dua. Dalam satu koloni tertentu, didapati ada 18 ameba pada awalnya.
Cari bilangan ameba dalam koloni tersebut selepas pembelahan ke-6 dalam bentuk 2m3n dengan keadaan m dan n ialah integer
positif.
After the 6th fission, the number of amoebae =
=
=
=
18 × 26
2 × 9 × 26
21 + 6 × 32
2732
Initial stage: 18
After the 1st fission: 18 × 21
After the 2nd fission: 18 × 2 × 2 = 18 × 22
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Additional Mathematics
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L
(a) 2.333…
L
(b) 0.1666…
Let x = 2.333… ——————
× 10: 10x = 23.33… ——
– : 9x = 21
x = 21
9
7
=
or 2 1
3
3
7
Hence, 2.333… =
or 2 1
3
3
(c) 0.001111…
Let x = 0.1666…
× 10: 10x =
× 100: 100x =
– :
90x =
———————
1.666… ———
16.66… ———
15
x = 15
90
= 1
6
Hence, 0.1666… = 1
6
(d) 0.141414…
Let x = 0.001111…
× 100:
100x =
× 1 000: 1 000x =
– :
900x =
——————
0.1111… ——
1.111… ——
1
x= 1
900
Hence, 0.001111… = 1
900
Let x = 0.141414… ————————
× 100: 100x = 14.1414…
———
– :
99x = 14
x = 14
99
Hence, 0.141414… = 14
99
5. Determine whether each of the following is a surd or not. Give your reason.
Tentukan sama ada setiap yang berikut ialah surd atau bukan. Beri alasan anda.
(i)
(ii) √9 = 3
√81 = 4.32674…
3
Surd. Its value is a non-recurring
decimal.
(a) √72
= 8.48528…
Surd. Its value is
a non-recurring
decimal.
Not a surd. Its value
is an integer.
Not all the roots are surds.
Tidak semua punca kuasa
ialah surd.
(b) √81
3 2
(c) – 121
(d)
25
7
=9
11
Not a surd. Its value
=–
= –2.2
= 0.65863…
5
is an integer.
Not a surd. Its value
Surd. Its value is
is a terminating
a non-recurring
decimal.
decimal.
6. Write each of the following as single surd.
Tulis setiap yang berikut sebagai surd tunggal.
(i)
√3 × √7 = √3 × 7
= √21
√a × √b = √ab
(ii) √20 ÷ √10 = 20
10
√a ÷ √b =
a
b
= √2
(a) √14 × √5
= √14 × 5
= √70
(b) √2 × √27
= √2 × 27
= √54
(c) √126 ÷ √3
= 126
3
= √42
(d)
√56 × √3
√7
= 56 × 3
7
= √24
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7. Simplify each of the following.
Permudahkan setiap yang berikut.
(i)
√12 = √2 × 2 × 3
= 2√3
(ii) 5√18 = 5 × √2 × 3 × 3
= 5 × 3 × √2
= 15√2
n√a and n√a are different.
n√a dan n√a adalah berbeza.
(a) √208
= √2 × 2 × 2 × 2 × 13
= 2 × 2 × √13
= 4√13
(b) 2√72
= 2 × √2 × 2 × 2 × 3 × 3
= 2 × 2 × 3 × √2
= 12√2
(c)
√24
3
= √2 × 2 × 2 × 3
3
= 2 × √3
3
= 2 √3
3
8. Simplify the following expressions.
Permudahkan ungkapan berikut.
3√5 + 4√5
= (3 + 4)√5
= 7√5
3√5 and 4√5 are similar surds.
Factorisation is carried out
with √5 as the common factor.
(a) 7√2 – 3√2
= (7 – 3)√2
= 4√2
3√5 dan 4√5 ialah surd serupa.
Pemfaktoran dilakukan dengan
√5 sebagai faktor sepunya.
(b) √12 – √75 + √108
= √4 × 3 – √25 × 3 + √36 × 3
= 2√3 – 5√3 + 6√3
= (2 – 5 + 6)√3
= 3√3
(c) √2 +
= √2
= √2
= √2
= √2
(d) √18 + √8 – √2
3
= √9 × 2 + √4 × 2 – √2
3
3
2
√
=
+ 2√2 – √2
3
= √2 + 2√2 – √2
= (1 + 2 – 1)√2
= 2√2
(e) 3√10 – √5 × √2
= 3√10 – √5 × 2
= 3√10 – √10
= (3 – 1)√10
= 2√10
(f)
(g) (1 + 2√3)(7 – 4√3)
= (1)(7) – (1)(4√3) + (2√3)(7) – (2√3)(4√3)
= 7 – 4√3 + 14√3 – 24
= 10√3 – 17
√5(8 – √5)
= 8√5 – √5 × 5
= 8√5 – 5
√3 + √12 – √48
+ √3 + √4 × 3 – √16 × 3
+ √3 + 2√3 – 4√3
+ (1 + 2 – 4)√3
– √3
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Additional Mathematics
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9. Simplify the following expressions by rationalising the denominators.
Permudahkan ungkapan berikut dengan menisbahkan penyebut.
√3
2
2
=
×
√3
√3
√3
= 2√3
3
(b)
(a)
Multiply with same surd
as the denominator.
3
3
2√3
=
×
2√3
2√3
2√3
3 × 2 × √3
=
2 × 2 × √3 × √3
=
6√3
12
=
√3
2
(c)
2
2
=
√12
√4 × 3
2
=
2√3
1
√3
=
×
√3
√3
√
3
=
3
√6
6
=
10
√10
3
=
5
√3
=
√5
√3
√5
=
×
√5
√5
√15
=
5
10. Simplify the following expressions by rationalising the denominators using conjugate surds.
Permudahkan ungkapan berikut dengan menisbahkan penyebut melalui surd konjugat.
5
5
2 – √3
×
=
2 + √3
2 + √3
2 – √3
5(2 – √3)
= 2
(a + b)(a – b) = a2 – b2
2 – (√3)2
5(2 – √3)
=
4–3
= 5(2 – √3)
2 – √3 is conjugate surd
(a)
for 2 + √3.
2 – √3 ialah surd konjugat
bagi 2 + √3.
(b)
√3
√3
√5 + √3
=
×
√5 – √3
√5 – √3
√5 + √3
√3(√5 + √3)
=
(√5)2 – (√3)2
√3(√5 + √3)
5–3
√15 + √9)
=
2
√15 + 3
=
2
=
(c)
7
7
2 + 3 √2
=
×
2 – 3√2
2 – 3√2
2 + 3 √2
7(2 + 3√2)
= 2
2 – (3√2)2
= 7(2 + 3√2)
4 – 18
7(2
+ 3√2)
=
–14
= 2 + 3 √2
–2
2
+ 3√2
=–
2
3√2 – √3
3 √2 – √3
2√2 + √3
=
×
2√2 – √3
2 √2 – √3
2√2 + √3
(3√2 – √3)(2√2 + √3)
=
(2√2)2 – (√3)2
(3√2)(2√2) + (3√2)(√3) – (√3)(2√2) – (√3) (√3)
=
(2√2)2 – (√3)2
6 × 2 + 3 × √6 – 2 × √6 – 3
=
4×2–3
12 + 3 √6 – 2 √6 – 3
=
8–3
9 + √6
=
5
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11. Selesaikan setiap yang berikut.
Solve each of the following.
1
1
Given p =
and q =
, simplify the following
3 – √2
3 + √2
expressions by rationalising the denominators.
1
1
Diberi p =
dan q =
, permudahkan ungkapan
3 – √2
3 + √2
berikut dengan menisbahkan penyebut.
(i) p + q
(ii) p – q
(i)
p+q=
=
=
=
=
(ii) p – q =
=
=
=
=
1
1
+
3 – √2 3 + √2
(3 + √2)
(3 – √2)
+
(3 – √2)(3 + √2)
(3 + √2)(3 – √2)
(3 + √2) + (3 – √2)
32 – (√2)2
3 + √2 + 3 – √2
9–2
6
7
1
1
–
3 – √2 3 + √2
(3 + √2)
(3 – √2)
–
(3 – √2)(3 + √2) (3 + √2)(3 – √2)
(3 + √2) – (3 – √2)
32 – (√2)2
3 + √2 – 3 + √2
9–2
2√2
7
1
1–m
and n =
, express the following
1+m
√2
expressions in the simplest form.
Given m =
1
1–m
dan n =
, ungkapkan ungkapan berikut
1+m
√2
dalam bentuk yang paling ringkas.
Diberi m =
(i) n
(ii) n + 1
n
1
√2
1
1+
√2
1
1
1–
÷ 1+
√2
√2
√2 – 1
√2 + 1
÷
√2
√2
√2 – 1
√2
×
√2
√2 + 1
√2 – 1
√2 – 1
×
√2 + 1
√2 – 1
(√2 – 1)2
(√2)2 – 12
2 – 2√2 + 1
2–1
3 – 2√2
1–
(i)
n=
=
=
=
=
=
=
=
1
(ii) n + 1 = (3 – 2√2) +
n
3 – 2√2
1
3 + 2√ 2
= (3 – 2√2) +
×
3 – 2√2
3 + 2√ 2
3 + 2√2
= (3 – 2√2) +
9–8
= (3 – 2√2) + (3 + 2√2)
=6
12. Selesaikan masalah berikut.
Solve the following problems.
A square has an area of 216 m2. Write the length, in m, of the side of the square in surd form in its simplest
form. 276 SSO L
Sebuah segi empat sama mempunyai luas 216 m2. Tulis panjang, dalam m, sisi segi empat sama tersebut dalam bentuk surd yang
paling ringkas.
The length of the side = √216
= √2 × 2 × 2 × 3 × 3 × 3
= 2 × 3 × √2 × 3
= 6√6 m
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Additional Mathematics
V
V W
logarithmic functions.
V
/
W
Rajah di bawah menunjukkan graf fungsi eksponen dan
fungsi logaritma.
y=a
y
x
RI R
L
a
y
a
• log
a
a
y
• loga
a
x
x
H
RI E
N
: F
V
W W W
W
W
F
functions are re ection of one another in the straight
y
. The exponential and logarithmic functions
are inverse functions of one another.
VW
V
≠ 1
RI R
L
LP
c
• loga
b
c
c ≠ 1
Kita dapat lihat bahawa fungsi eksponen dan fungsi logaritma
berpantulan pada garis lurus y = x. Fungsi eksponen dan
fungsi logaritma adalah fungsi songsang antara satu sama lain.
W
untuk sebarang nombor nyata n
c
• loga
V
LP
dengan keadaan a, x dan y ialah nombor positif dan a ≠ 1
(1, 0)
W
unde ned.
L
y
y
y = loga x
L
N P
• loga xy
y=x
(0, 1)
,
VW
V
≠ 1 and
dengan keadaan a, b dan c ialah nombor positif, a ≠ 1 dan
c ≠ 1
V
(common logarithms) and ln = loge (natural
V
e is a constant.
W
(logaritma biasa dan
keadaan e ialah pemalar.
Logaritma bagi nombor negatif dan sifar adalah tak tertakrif.
e
(logaritma jati dengan
13. Convert the following equations in the form of index to the form of logarithm or vice versa.
Tukarkan persamaan dalam bentuk indeks kepada persamaan dalam bentuk logaritma atau sebaliknya.
(i)
103 = 1 000
log10 1 000 = 3
(a) 3–4 = 1
81
1
log3
= –4
81
(ii) log2 64 = 6
ax = N
26 = 64
loga N = x
(b)
1
2
4
81
log 4
81
= 2
9
2 = 1
9
2
(c) log7 1 = –2
49
7–2 = 1
49
(d) loga x = y
ay = x
14. Use a calculator to determine the value of each of the following.
Gunakan kalkulator untuk menentukan nilai bagi setiap yang berikut.
(a) 1og10 5
log10 23
= 1.3617
3
(b) 1og10 0.5
= 0.6990
Tekan
= –0.3010
(c) 1og10 240
= 2.3802
(e) log10 2
3
2
(d) 1og10 0.0781
= –1.1073
= log10 4
9
= –0.3522
(f)
log10 1
5
3
1
125
= –2.0969
= log10
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15. Use a calculator to determine the value x.
Gunakan kalkulator untuk menentukan nilai x.
log10 x = 1.23
x = antilog 1.23
x = 16.98
3
Tekan
(a) 1og10 x = 2.831
x = antilog 2.831
x = 677.6
(b) 1og10 x = 0.0842
x = antilog 0.0842
x = 1.214
(c) 1og10 x = 3
x = antilog 3
x = 1 000
(d) 1og10 x = –1.45
x = antilog (–1.45)
x = 0.03548
16. Without using a calculator, find the value of each of the following.
Tanpa menggunakan kalkulator, cari nilai bagi setiap yang berikut.
(a) log2 256
log2 32
Let log2 32
2x
2x
x
=
=
=
=
x
32
25
5
Since the bases are
the same, equate
the indices.
(c) log15 1
Let log15 1
15x
15x
x
Let log2 256
2x
2x
x
(d) log8 √8
=x
=1
= 150
=0
=x
= 256
= 28
=8
(e) log√3 9
Let log8 √8 = x
8 x = √8
(f)
log 1 0.25
2
Let log√3 9 = x
Let log 1 0.25 = x
2
(√3)x = 9
1
2
1
2
1
2
1
2
(√3)x = (√3)4
x=4
8 =8
x = 1
2
x
(b) log3 1
27
Let log3 1 = x
27
3x = 1
27
3x = 3–3
x = –3
x
= 1
4
x
= 1
2
x =2
x
17. Determine the value of x for each of the following equations without using a calculator.
Tentukan nilai x bagi setiap persamaan berikut tanpa menggunakan kalkulator.
(a)
log3 x = 4
x = 34
= 81
Convert to the
form of index
(c) log 1 x = 3
3
x = 1
3
1
=
27
3
(d) logx 64
64
82
x
=
=
=
=
2
x2
x2
8
log2 x = –3
x = 2–3
= 1
8
(e) lg x = –4
log10 x = –4
x = 10–4
1
=
10 000
or 0.0001
(b)
log4 x = 1
2
1
x = 42
= √4
=2
(f)
ln x = 0
loge x = 0
x = e0
=1
= 0.25
2
S
Additional Mathematics
,
L
L
18. Express each of the following expressions as a single logarithm.
Ungkapkan setiap ungkapan berikut sebagai logaritma tunggal.
log5 4 – 2 log5 4 + 1 log5 64
2
1
= log5 4 – log5 42 + log5 (82) 2
= log5 4 – log5 16 + log5 8
= log5 4 × 8
16
= log5 2
Pastikan asas logaritma adalah sama
before laws of logarithms is applied
Pastikan pekali setiap sebutan logaritma ialah
hukum logaritma digunakan
sebelum
(a) 2 + log10 4
= 2 × 1 + log10 4
= 2 log10 10 + log 10 4
= log10 102 + log10 4
= log10 (102)(4)
= log10 400
(b) log3 9 + log3 12
4
5
9
12
= log3
×
4
5
= log3 27
5
(c) logx p + 3 log x q – 1 logx r
3
(d) 3 loga x + 4 loga y – loga x2y
= loga x3 + loga y4 – loga x2y
(e) loga m + 3 log a n – loga mn
= loga m + loga n3 – loga mn
(f)
3
4
= loga x ×2 y
xy
= loga xy3
3
= loga mn
mn
= loga n2
1
= logx p + logx q3 – logx r 3
= logx
pq3
1
r3
1 log m – 3 log n + 2 log q
2
2
2
2
1
= log2 m 2 – log2 n3 + log2 q2
2
= log2 q √3m
n
19. Evaluate the following expressions.
Nilaikan ungkapan berikut.
log3 18 + log 3 4 – 3 log3 2
= log3 18 + log3 4 – log3 23
= log3 18 × 4
8
= log3 9
= log3 32
= 2 log3 3
log3 3 = 1
=2
(c) 1 + 1 log7 196 – log7 14
2
1
= 1 + log7 (142) 2 – log7 14
= 1 + log7 14 – log7 14
=1
(a) 2 log2 4 + log2 5 – log2 10
= log2 42 + log2 5 – log2 10
= log2 16 × 5
10
= log2 8
= log2 23
= 3 log2 2
=3
(b) log4 24 – log4 3 + log4 2
4
3
= log4 24 ÷
×2
4
= log4 24 × 4 × 2
3
= log4 64
= log4 43
= 3 log4 4
=3
(d) 2 logx xy – 2 logx y + 3
x3
(e) 1 logx x – logx 4
y
2
= 2(logx x + logx y) – 2 logx y + 3
√y
1
1
= 2 logx x + 2 logx y – 2 logx y + 3
= (logx x – logx y)
2
2
=2+3
– 1 (3 logx x – logx y)
4
=5
1
= (1) – 1 logx y – 3 (1)
2
4
4
+ 1 logx y
4
1
=–
4
Additional Mathematics
S
,
L
L
20. Evaluate each of the following. State the answer correct to 4 significant figures.
Nilaikan setiap yang berikut. Nyatakan jawapan betul kepada 4 angka bererti.
(a) log3 14
log5 2
log10 2
=
log10 5
= 0.4307
Can also
ln 2
use
ln 5
(b) log15 0.06
log10 14
log10 3
= 2.402
(c) log0.7 9
log10 0.06
log10 15
= –1.039
=
log10 9
log10 0.7
= –6.160
=
=
21. Evaluate the following without using a calculator.
Tentukan nilai berikut tanpa menggunakan kalkulator.
(a) log9 27
log64 4
=
=
=
=
=
log4 4
log4 64
1
log4 64
1
log4 43
1
3 log4 4
1
3
=
=
=
=
log4 4 = 1
=
(b) log8 4
log3 27
log3 9
log3 33
log3 32
3 log3 3
2 log3 3
3(1)
2(1)
3
2
=
=
=
=
=
(c) log81 243
log2 4
log2 8
log2 22
log2 23
2 log2 2
3 log2 2
2(1)
3(1)
2
3
=
=
=
=
=
log3 243
log3 81
log3 35
log3 34
5 log3 3
4 log3 3
5(1)
4(1)
5
4
22. Given log2 x = m and log2 y = n, express each of the following in terms of m and n.
Diberi log2 x = m dan log2 y = n, ungkapkan setiap berikut dalam sebutan m dan n.
1
3
(a) log2 3xy = log2 x + log2 y3 – log2 2 3
√2
= log2 x + 3 log2 y – 1 log2 2
3
1
= m + 3n –
3
2
log2 x = log2 x2 – log2 y
y
= 2 log2 x – log2 y
= 2m – n
1
(b) log2 34x4 = log2 22 + log2 x – log2 (y4) 3
√y
= 2 log2 2 + log2 x – 4 log2 y
3
= 2 + m – 4n
3
1
(d) log2 2√xy2 = log2 2 + log2 (xy2) 2
= 1 + 1 log2 (xy2)
2
= 1 + 1 (log2 x + log2 y2)
2
= 1 + 1 (log2 x) + 1 (2 log2 y)
2
2
= 1 + 1m + n
2
log2 4x
log2 xy
log2 22 + log2 x
=
log2 x + log2 y
= 2+m
m+n
(c) logxy 4x =
log2 x2y8
log2 16
log2 x2 + log2 y8
=
log2 24
2 log2 x + 8 log2 y
=
4
= 1 m + 2n
2
(e) log16 x2y8 =
S
Additional Mathematics
,
L
L
23. Solve the following problems.
Selesaikan masalah berikut.
The total savings, RMs, of Encik Zaidi after n years is given by the formula s = 20 000(1.032)n.
Jumlah simpanan, RMs, Encik Zaidi selepas n tahun diberi oleh rumus s = 20 000(1.032)n. 276 SSO L
(i) Convert the given formula to an equation in the form of logarithm.
Tukarkan rumus yang diberi kepada persamaan dalam bentuk logaritma.
(ii) After how many years does Encik Zaidi’s total savings exceed RM30 000 for the first time?
Selepas berapa tahunkah jumlah simpanan Encik Zaidi akan melebihi RM30 000 buat pertama kalinya?
(i)
s = 20 000(1.032)n
s
= (1.032)n
N = ax
20 000
s
log1.032
=n
loga N = x
20 000
(ii) 20 000(1.032)n
n
n
n
n
PA
P A
AP
s = 20 000(1.032)n
log10 s = log10 20 000(1.032)n
log10 s = log10 20 000 + n log10 1.032
30 000
log1.032 30 000
20 000
log1.032 1.5
log10 1.5
log10 1.032
12.87
After 13 years, Encik Zaidi’s total savings exceeds RM30 000 for the first time.
Given that the world population can be estimated by using the formula P = 6.9(1.010)t, where t is the
number of years after 2019 and P is the world population in billions of people. 276 SSO L
Diberi populasi dunia boleh dianggarkan dengan rumus P = 6.9(1.010)t, dengan keadaan t ialah bilangan tahun selepas 2019 dan
P ialah populasi dunia dalam bilion.
(i)
Convert the given formula to an equation in the form of logarithm.
Tukarkan rumus yang diberi kepada persamaan dalam bentuk logaritma.
(ii) Estimate / Anggarkan
(a) the population in the year 2030,
populasi pada tahun 2030,
(b) by what year will the population be double of that in 2019.
pada tahun apa populasi akan dua kali ganda populasi pada 2019.
(i)
P = 6.9(1.010)t
log10 P = log10 6.9(1.010)t
log10 P = log10 6.9 + t log10 1.010
(ii) (a) t = 2030 – 2019 = 11 years
P = 6.9(1.010)11 = 7.698 billion
\ The population in the year 2030
is 7.698 billion.
(b) In the year 2019
P2019 = 6.9(1.010)0
= 6.9
t = 0 as it starts on that year, that is 2019
Double of that in 2019,
2 × 6.9 = 6.9(1.010)t
13.8 = 6.9(1.010)t
(1.010)t = 13.8
6.9
=2
t log10 1.010 = log10 2
log10 2
log10 1.010
= 69.66
≈ 70
t =
2019 + 70 = 2089
\ In the year 2089, the population
will be double of that in 2019.
S
Additional Mathematics
SSOL
SL
L
L,
H
I,
L H 6
,
L
L
L P
L
Textbook
pg. 122 – 123
26. Solve each of the following.
Selesaikan setiap yang berikut.
(a)
pH value is defined as pH = –log10 [H+] where [H+]
is the concentration of hydrogen ions. Liquid A has
the concentration of hydrogen ions of 1 × 10–5 moles
per litre. Determine pH value of liquid A and hence
276
O L
state whether it is acidic or alkaline.
+
Nilai pH ditakrifkan sebagai pH = –log10 [H ] dengan keadaan
[H+] ialah kepekatan ion hidrogen. Cecair A mempunyai kepekatan
ion hidrogen 1 × 10–5 mol per liter. Tentukan nilai pH cecair A
dan seterusnya nyatakan adakah cecair A berasid atau beralkali.
1
2
3
4
5
6
Acidic
Berasid
pH =
=
=
=
=
7
8
9
Neutral
–log10 (1 × 10–5)
–(log10 1 + log10 10–5)
–[log10 1 + (–5) log10 10]
–[0 + (–5)]
5
Hence, liquid A is acidic.
10
11
12
Alkaline
Beralkali
13
14
A
B
Area
Luas
8 cm2
D
C
The diagram shows a square ABCD with an
area of 8 cm2. Determine the perimeter, in cm,
of the square ABCD. Write your answer in surd
276
SSO L
form.
Rajah di atas menunjukkan sebuah segi empat sama yang
mempunyai luas 8 cm2. Tentukan perimeter, dalam cm,
segi empat sama ABCD itu. Tulis jawapan anda dalam
bentuk surd.
Let its side is x cm.
x2 = 8
x = √8
= √4 × 2
= 2√2
Perimeter = 4 × 2√2
= 8√2 cm
(b) A bacterial colony B with the initial mass of (c) Ahmad deposited RM3 500 in a fixed deposit
2 picogram (pg) will increase every day, that is
account with interest rate of 5% per annum
3 times the mass of the previous day. Find the
and compounded yearly. Determine the number
time, in the nearest day, that will be taken by
of years taken for his account to become
the bacterial colony B to become 15 pg.
RM5 431.20.
276
O L
Ahmad menyimpan RM3 500 dalam akaun simpanan tetap
dengan kadar faedah 5% setahun dan dikompaun setiap
Suatu koloni bakteria B dengan jisim asal 2 pikogram (pg)
tahun. Tentukan tempoh, dalam tahun, yang diambil untuk
akan bertambah setiap hari, iaitu 3 kali jisimnya pada hari
amaun dalam akaunnya menjadi RM5 431.20.
sebelumnya. Cari masa, dalam hari yang terhampir, yang
akan diambil oleh koloni bakteria B untuk menjadi 15 pg.
nt
MV = P 1 + r
n
Let its mass after t days is m pg.
(1)t
Thus, m = 2(3)t
5 431.20 = 3 500 1 + 0.05
If m = 15 pg,
15 = 2(3)t
1
5 431.20 = 1.05t
15 = 3t
3 500
2
5
431.20
15
ln
= ln 1.05t
log10
= log10 3t
2
3 500
log10 7.5 = t log10 3
ln 5 431.20 – ln 3 500 = t ln 1.05
log10 7.5
t = ln 5 431.20 – ln 3 500
t=
ln 1.05
log10 3
=9
= 1.834
It takes 2 days.
It takes 9 years.
S
Additional Mathematics
SPM
Paper
630
2. Given 23x = m, 2y = n and 23x + y = 5 + 8x, express m in
terms of n.
630
Diberi 23x = m, 2y = n dan 23x + y = 5 + 8x, ungkapkan m dalam
sebutan n.
[3]
5
Ans: m =
n–1
3. Given 2n = 3m = 6p, express p in terms of m and n.
Diberi 2n = 3m = 6p, ungkapkan p dalam sebutan m dan n.
mn
m+ n
[3]
630
[4]
2
Ans: h = p + 4
3
[3]
Selesaikan persamaan:
loga 256 – log√a 2a = 1
3 , ungkapkan y dalam sebutan x.
logxy 2
[3]
Ans: (a) t > 0 and t ≠ 1
2
–—
(b) y = x 3
9. Simplify:
Permudahkan:
√27 – 4√5 – √20 + 5 √3
[2]
Ans: 8√3 – 6√5
10. Determine the value of x for each of the following equations.
Tentukan nilai x dalam setiap persamaan berikut.
(a) 2x = 64
2
(b) 2 × 5x =
625
[4]
log6 64 – log6 36 + 2 log2 8
[3]
Ans: 8
Diberi log2 3 = a dan log2 5 = b, ungkapkan berikut dalam
sebutan a dan b.
4
log2 25 – log2 √3 +log2 60
[3]
3
Ans: 3b + a + 2
4
13. Simplify the expression by rationalising the denominator.
6. Solve the equation:
630
[1]
12. Given log2 3 = a and log2 5 = b, express the following in
terms of a and b.
16p + 3
= 1, express h in terms of p.
64h – 2
p+3
Diberi 16h – 2 = 1, ungkapkan h dalam sebutan p.
64
3
, express y in terms of x.
logxy 2
Diberi log2 x =
Nilaikan:
Ans: (a) 3q
2q + 3
(b)
q
5. Given
(b) Given log2 x =
11. Evaluate:
Diberi log p 5 = q, ungkapkan p dalam sebutan q.
(a) logp 125
(b) log5 25p3
Diberi S = logt R, nyatakan syarat-syarat bagi t.
Ans: (a) x = 6
(b) x = –4
4. Given logp 5 = q, express p in terms of q.
630
L
8. (a) Given S = logt R, state the conditions of t.
1
Diberi logx 2 = p dan logx 7 = q, ungkapkan log7 4x3 dalam
sebutan p dan q.
[3]
2p + 3
Ans:
q
Ans:
L
P A
1. Given logx 2 = p and logx 7 = q, express log7 4x3 in terms
630
of p and q.
630
,
[3]
Ans: a = 4
Permudahkan ungkapan dengan menisbahkan penyebut.
3√7
2√3 – √7
6√21 + 21
Ans:
5
[3]
14. Convert the equation in the form of index to the form of
logarithm and vice versa.
7. Given 2m + 2 m = 2 n, express m in terms of n.
630
Diberi 2m + 2m = 2n, ungkapkan m dalam sebutan n.
[2]
Ans: m = n – 1
Tukarkan persamaan dalam bentuk indeks kepada bentuk
logaritma dan sebaliknya.
(a) 103 = 1 000
(b) log2 64 = 6
[2]
Ans: (a) log10 1 000 = 3
(b) 26 = 64
Form 4 Answers
37 5
SPM
3 SH
,
L H 6
R
L P
P
R
R
R
R
R
R
R
R
R
R √
R
R √
R
R
R
R
R
R
R
R
R
)R
R
R
R
R
R
R
R
R
R
R
≠1
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
)R
R
R
R
R
R
R
- R
- R
R
R
R
R
√ – 4√ – √ + 5√
= √ × – 4√ – √ × + 5√
= 3√ – 4√ – 2√ + 5√
= 8√ – 6√
R
R
×
×
×
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
R
R
R
R
R
R
R
R
R
R
√
√
R
R
×
R
3√
2√ 3 – √
R
3√
2√
6√
R
R
OOH
2√ 3 + √
2√ 3 + √
2√ 3 + √
6√ ×
276
×
H
50
R
R
R
© Penerbitan Pelangi Sdn. Bhd.
√
×
√
√
√
+ 2√
√ 6 + 2√
√
√
√
√ + 2√
√ + 2√
+ 2√
+ 2√
√ + 2√
×
R
3√
2√ 3 – √
√
R S H
R
R
R
R
R
)R
6
H
6
H
6
H
+H FH
R
R
R
R
√ 6 + 2√ 5 = √ 1 + √ 5 = 1 + √
R
Additional Mathematics
2. For each of the following arithmetic progressions, determine the term stated in bracket.
Bagi setiap janjang aritmetik berikut, tentukan sebutan yang dinyatakan dalam kurungan.
(a) 3, 7, 11, 15, … [8th term/ sebutan ke-8]
ln x, ln 3x, ln 9x, ln 27x, … [5 term/ sebutan ke-5]
a = ln x
d = ln 3x – ln x = ln 3x = ln 3
x
T5 = ln x + 4 ln 3
Tn = a + (n – 1)d
= ln 34x = ln 81x
th
(b) –10, –13, –16, –19, … [16th term/ sebutan ke-16]
a=3
d=7–3=4
T8 = 3 + 7(4)
= 31
(c) 2p, 3p , p, p , … [21th term/ sebutan ke-21]
2
2
a = –10
d = –13 – (–10) = –3
a = 2p
d = 3p – 2p = – p
2
2
p
T21 = 2p + 20 –
= 2p – 10p = –8p
2
T16 = –10 + 15(–3)
= –55
3. Determine the first term and the common difference of each of the following.
Tentukan sebutan pertama dan beza sepunya bagi setiap yang berikut.
In an arithmetic progression, the 4th and 8th terms are 16 and 56 respectively.
Dalam suatu janjang aritmetik, sebutan ke-4 dan ke-8 masing-masing ialah 16 dan 56.
Tn = a + (n – 1)d
T4 = a + (4 – 1)d
a + 3d
T8 = a + (8 – 1)d
a + 7d
–
= 16
= 16 …………
= 56
= 56 …………
, 4d = 40
d = 10
Substitute d = 10 into ,
a + 3(10) = 16
a = –14
\ The first term is –14 and the common difference
is 10.
(a) In an arithmetic progression, the 3rd and 8th terms (b) In an arithmetic progression, the 2nd and 5th
are –5 and 15 respectively.
terms are x and 2y respectively.
Dalam suatu janjang aritmetik, sebutan ke-3 dan ke-8
Dalam suatu janjang aritmetik, sebutan ke-2 dan ke-5
masing-masing ialah –5 dan 15.
masing-masing ialah x dan 2y.
T3 = a + (3 – 1)d = –5
a + 2d = –5 ………
T8 = a + (8 – 1)d = 15
a + 7d = 15 ………
– ,
5d = 20
d =4
Substitute d = 4 into ,
a + 2(4) = –5
a = –13
\ The first term is –13 and the common
difference is 4.
T2 = a + d = x …………
T5 = a + 4d = 2y …………
– ,
3d = 2y – x
d = 2y – x
3
Substitute d = 2y – x into ,
3
2y
–
x
a+
=x
3
a = x – 2y – x
3
4x
–
2y
=
3
\ The first term is 4x – 2y and the common
3
difference is 2y – x .
3
Additional Mathematics
4. Find the number of terms for each of the following arithmetic progressions.
Cari bilangan sebutan bagi setiap janjang aritmetik berikut.
(a) –7, 3, 13, …, 93
43, 31, 19, …, –29
a = 43,
d = 31 – 43 = –12,
Tn = –29
Tn = a + (n – 1)d
–29 = 43 + (n – 1)(–12)
n = –29 – 43 + 1
–12
=7
(b) 4p, 11p, 18p, …, 67p
a = –7,
d = 3 – (–7) = 10,
Tn = 93
a = 4p,
d = 11p – 4p = 7p,
Tn = 67p
93 = –7 + (n – 1)(10)
n = 93 + 7 + 1
10
= 11
67p = 4p + (n – 1)(7p)
n = 67p – 4p + 1
7p
= 10
5. Each of the following shows three consecutive terms of an arithmetic progression. Find the value of p.
Setiap yang berikut menunjukkan tiga sebutan berturutan dalam suatu janjang aritmetik. Cari nilai p.
(a) …, 2p, 4p + 9, 3p + 15, …
…, p + 4, 3p, 3p + 8, …
3p = (p + 4) + (3p + 8)
2
T
+ Tn + 1
6p = 4p + 12
Tn = n – 1
,
2
2p = 12
Tn is known as the
p=6
arithmetic mean of
(b) …, 2p – 1, 4p, 5p + 4, …
4p + 9 = (2p) + (3p + 15)
2
8p + 18 = 5p + 15
3p = –3
p = –1
4p = (2p – 1) + (5p + 4)
2
8p = 7p + 3
p=3
Tn – 1 and Tn + 1
6. Solve the following problems.
Selesaikan masalah berikut.
–35, –31, –27, … are the first three terms in an
arithmetic progression. Find the smallest value of n
such that the nth term is positive. Hence, determine
the term.
–35, –31, –27, … ialah tiga sebutan pertama dalam suatu
janjang aritmetik. Cari nilai n yang terkecil dengan keadaan
sebutan ke-n adalah positif. Seterusnya, tentukan sebutan ini.
a = –35, d = –31 – (–35) = 4 and Tn
Tn = a + (n – 1)d
–35 + (n – 1)(4)
(n – 1)(4)
(n – 1)
n
n
0
0
35
35
4
8.75 + 1
9.75
\ n = 10
T10 = –35 + (10 – 1)(4)
=1
0 (positive)
(a) Given 51 and 33 are the 2nd and 5 th terms of
an arithmetic progression respectively. Find
the smallest value of n such that the nth term is
negative. Hence determine the term.
Diberi 51 dan 33 masing-masing ialah sebutan ke-2 dan
ke-5 bagi suatu janjang aritmetik. Cari nilai n yang terkecil
dengan keadaan sebutan ke-n adalah negatif. Seterusnya,
tentukan sebutan ini.
a+d
a + 4d
– , 3d
d
= 51 ………
= 33 ………
= –18
= –6
Substitute d = –6 into
a + (–6) = 51
a = 57
Tn = a + (n – 1)d
57 + (n – 1)(–6)
57 – 6n + 6
6n
n
\ n = 11
,
0
0
0
63
10.5
T11 = 57 + (11 – 1)(–6) = –3
Additional Mathematics
7. Find the sum of the first n terms for each of the following arithmetic progressions.
Cari hasil tambah n sebutan pertama bagi setiap janjang aritmetik berikut.
1, 6, 11, 16, … (the first 12 terms)
(12 sebutan pertama)
a = 1, d = 6 – 1 = 5
Sn = n [2a + (n – 1)d]
2
S12 = 12 [2(1) + (12 – 1)(5)]
2
= 342
(a) 4, 6, 8, 10, … (the first 10 terms)
(10 sebutan pertama)
a = 4, d = 6 – 4 = 2
S10 = 10 [2(4) + (10 – 1)(2)]
2
= 130
(c) 23, 17, 11, 5, … (the first 16 terms)
(16 sebutan pertama)
a = 23, d = 17 – 23 = –6
S16 = 16 [2(23) + (16 – 1)(–6)]
2
= –352
1
[2a + (n – 1)d]
2
where / dengan keadaan
Sn =
Sn = sum of first nth terms / hasil tambah n sebutan pertama
a = first term / sebutan pertama
d = common difference / beza sepunya
(b) –23, –16, –9, –2, … (the first 14 terms)
(14 sebutan pertama)
a = –23, d = –16 – (–23) = 7
S14 = 14 [2(–23) + (14 – 1)(7)]
2
= 315
(d) –15, –12, –9, –6, … (the first 15 terms)
(15 sebutan pertama)
a = –15, d = –12 – (–15) = 3
S15 = 15 [2(–15) + (15 – 1)(3)]
2
= 90
8. Find the sum of the terms in each of the following arithmetic progressions.
Cari hasil tambah sebutan dalam setiap janjang aritmetik berikut.
2, –3, –8, …, –38
a = 2, d = –3 – 2 = –5, Tn = –38
–38 = 2 + (n – 1)(–5)
n = –38 – 2 + 1
–5
=9
To determine the number of
terms, use Tn = a + (n – 1)d.
(a) 15, 23, 31, 39, …, 255
S9 = 9 [2 + (–38)]
2
= –162
Use Sn = n [a + l ]
2
where l = last term
(b) –52, –48, –44, …, 0
a = 15, d = 23 – 15 = 8, Tn = 255
a = –52, d = –48 – (–52) = 4, Tn = 0
255 = 15 + (n – 1)(8)
n = 255 – 15 + 1
8
= 31
0 = –52 + (n – 1)(4)
n = 0 + 52 + 1
4
= 14
S31 = 31 [15 + 255]
2
= 4 185
S14 = 14 [–52 + 0]
2
= –364
Additional Mathematics
9. The sequence –22, –14, –6, … is an arithmetic progression. Find the sum from the
Jujukan –22, –14, –6, … ialah suatu janjang aritmetik. Cari hasil tambah dari
2nd term to 8th term
sebutan ke-2 hingga sebutan ke-8
a = –22, d = –14 – (–22) = 8
S8 = 8 [2(–22) + (8 – 1)(8)] = 48
2
S1 = –22
S8 = T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8
S1 = T1
‘From 2nd term to 8th term’ means T2 is
included in the calculation, so
‘Dari sebutan ke-2 hingga sebutan ke-8’
bererti T2 diambil kira dalam pengiraan, jadi
The sum from the 2 term to 8 term
Sm → n = Sn – Sm – 1
= S8 – S1
= 48 – (–22)
= 70
nd
th
(a) 3rd term to 10th term
sebutan ke-3 hingga sebutan ke-10
S10 = 10 [2(–22) + (10 – 1)(8)]
2
= 140
S2 = T1 + T2
= –22 + (–14)
= –36
S3 → 10 = S10 – S2
= 140 – (–36)
= 176
S2 → 8 = T2 + T3 + T4 + T5 + T6 + T7 + T8
= S8 – S2 – 1
(b) 5th term to 15th term
sebutan ke-5 hingga sebutan ke-15
(c) 20th term to 40th term
sebutan ke-20 hingga sebutan ke-40
S15 = 15 [2(–22) + (15 – 1)(8)]
2
= 510
S40 = 40 [2(–22) + (40 – 1)(8)]
2
= 5 360
S4 = 4 [2(–22) + (4 – 1)(8)]
2
= –40
S19 = 19 [2(–22) + (19 – 1)(8)]
2
= 950
S20 → 40 = S40 – S19
= 5 360 – 950
= 4 410
S5 → 15 = S15 – S4
= 510 – (–40)
= 550
10. For each of the following arithmetic progressions, the sum of the first n terms is given in the bracket. Find the
value of n.
Bagi setiap janjang aritmetik yang berikut, hasil tambah sebutan n pertama diberikan dalam kurungan. Cari nilai n.
(a) 21, 15, 9, …
12, 18, 24, …
[390]
a = 12, d = 18 – 12 = 6, Sn = 390
Sn = n [2(12) + (n – 1)(6)] = 390
2
n (18 + 6n) = 390
2
9n + 3n2 – 390 = 0
n2 + 3n – 130 = 0
(n + 13)(n – 10) = 0
\ n ≠ –13, n = 10
(b) –36, –32, –28, …
[–168]
a = –36, d = –32 – (–36) = 4, Sn = –168
Sn = n [2(–36) + (n – 1)(4)] = –168
2
n (–76 + 4n) = –168
2
–38n + 2n2 + 168 = 0
n2 –19n + 84 = 0
(n – 12)(n – 7) = 0
\ n = 12 or n = 7
[21]
a = 21, d = 15 – 21 = –6, Sn = 21
Sn = n [2(21) + (n – 1)(–6)] = 21
2
n (48 – 6n) = 21
2
24n – 3n2 – 21 = 0
n2 – 8n + 7 = 0
(n – 1)(n – 7) = 0
\ n ≠ 1, n = 7
(c) –10, –15, –20, …
[–595]
a = –10, d = –15 – (–10) = –5, Sn = –595
Sn = n [2(–10) + (n – 1)(–5)] = –595
2
n (–15 – 5n) = –595
2
15
– n – 5 n2 + 595 = 0
2
2
n2 + 3n – 238 = 0
(n – 14)(n + 17) = 0
\ n ≠ –17, n = 14
Additional Mathematics
11. Solve each of the following.
Selesaikan setiap yang berikut.
The sum of the first n terms of an arithmetic
progression is given by Sn = 3 – 8n. Find the first term
and the common difference.
Hasil tambah n sebutan pertama suatu janjang aritmetik diberi
oleh Sn = 3 – 8n. Cari sebutan pertama dan beza sepunya.
a = T1 = S1 = 3 – 8(1) = –5
Tn = Sn – Sn – 1
T2 = S2 – S1
= [3 – 8(2)] – (–5)
= –8
d = T2 – T1
= –8 – (–5)
= –3
(a) The sum of the first n terms of an arithmetic
progression is given by Sn = 1 + 4n2. Find the 11th
term.
Hasil tambah n sebutan pertama suatu janjang aritmetik
diberi oleh Sn = 1 + 4n2. Cari sebutan ke-11.
S10 = 1 + 4(10)2
= 401
T11 = S11 – S10
= 485 – 401
= 84
S11 = 1 + 4(11)2
= 485
(b) The sum of the first n terms of an arithmetic (c) The sum of the first n terms of an arithmetic
progression is given by Sn = 2 – 5n + n2. Find the
progression is given by Sn = 4n – n2. Find Tn – 1 in
th
5 term and the common difference.
terms of n. 276 SSO
Hasil tambah n sebutan pertama suatu janjang aritmetik
Hasil tambah n sebutan pertama suatu janjang aritmetik
diberi oleh Sn = 2 – 5n + n2. Cari sebutan ke-5 dan beza
diberi oleh Sn = 4n – n2. Cari Tn – 1 dalam sebutan n.
sepunya.
Sn – 1 = 4(n – 1) – (n – 1)2
2
S5 = 2 – 5(5) + (5) = 2
= (n – 1)(4 – n + 1)
S4 = 2 – 5(4) + (4)2 = –2
= (n – 1)(5 – n)
\ T5 = S5 – S4 = 2 – (–2) = 4
Sn – 2 = 4(n – 2) – (n – 2)2
= (n – 2)(4 – n + 2)
S6 = 2 – 5(6) + (6)2 = 8
= (n – 2)(6 – n)
T6 = S6 – S5 = 8 – 2 = 6
Tn – 1 = Sn – 1 – Sn – 2
\ d = T6 – T5 = 6 – 4 = 2
= (n – 1)(5 – n) – (n – 2)(6 – n)
= (5n – n2 – 5 + n) – (6n – n2 – 12 + 2n)
= 7 – 2n
12. Solve each of the following.
Selesaikan setiap yang berikut.
Chong borrows RM8 000 from a financial institution and promises to return the amount borrowed in 12
instalments with a total interest of RM400. The payment of the instalments is such that every subsequent
instalment is RM50 more than the previous one. How much is the starting and last instalments?
Chong meminjam sejumlah RM8 000 daripada sebuah institusi kewangan dan berjanji akan membayar balik dalam 12 ansuran
dengan faedah sejumlah RM400. Pembayaran ansuran dibuat dengan keadaan setiap ansuran berikut adalah RM50 lebih daripada
ansuran sebelumnya. Berapakah bayaran ansuran pertama dan terakhir? 276 SSO
S12 = 8 000 + 400 = 8 400,
n = 12, d = 50
8 400 = 12 [2a + (12 – 1)(50)]
2
1 400 = 2a + 550
2a = 850
a = 425
Total to pay
= Amount borrowed
+ interest incurred
Hence, the first instalment is RM425.
T12 = 425 + (12 – 1)(50)
= 975
Hence, the last instalment is RM975.
Additional Mathematics
13. Determine whether each of the following sequences is a geometric progression. Give your justification.
Tentukan sama ada setiap jujukan yang berikut ialah janjang geometri atau bukan. Berikan justifikasi anda.
x – 2k, 2x – 4k, 4x – 8k, 8x – 16k, …
r1 = 2x – 4k = 2(x – 2k) = 2
x – 2k
x – 2k
4x
–
8k
4(x
– 2k) = 2
r2 =
=
2x – 4k
2(x – 2k)
r3 = 8x – 16k = 8(x – 2k) = 2
4x – 8k
4(x – 2k)
A geometric progression because r1 = r2 = r3 = 2.
(a) 1, –3, 9, –27, …
r1 = –27 = –3
9
9
r2 =
= –3
–3
r3 = –3 = –3
1
A geometric progression because r1 = r2 = r3 = –3.
(b) 20, 2x, 2x + 1, 2x + 2, …
(c) loga x, loga x2, loga x3, loga x 4, …
x
x
loga x 4
4 loga x
r1 = 20 = 2 = 2x
r1 =
=
= 4
3
2
1
3
loga x
3 loga x
x+1
x
3
2
(2
)(2)
loga x
3 loga x
r2 = x =
=2
r2 =
=
= 3
2
2x
2
loga x 2
2 loga x
x+2
x
2
2
(2
)(2
)
2
r3 = x + 1 =
=2
loga x
2 loga x
r3 =
=
=2
2
(2x)(2)
loga x
loga x
Not a geometric progression because r1 ≠ r2 and
Not a geometric progression because r1 ≠ r2 ≠ r3.
r1 ≠ r3.
14. For each of the following geometric progressions, determine the common ratio.
Bagi setiap janjang geometri berikut, tentukan nisbah sepunya.
(a) 2k, 6k2, 18k3, 54k4, …
–3, 12, –48, 192, …
r=
r = 12 = –4
–3
(b) 2 , – 2 , 2 , – 2 , …
3
9 27
81
r=– 2 ÷ 2 =– 1
9
3
3
12
–48
192
=
=
= –4
–3
12
–48
(c) e8x, e6x, e4x, e2x, …
6x
r = e8x = e6x – 8x = e–2x
e
2
r = 6k = 3k
2k
1
x
1
x
(c) 1, 52x, 54x, …
[8th term/ sebutan ke-8]
a = 1, r = 52x
T8 = (1)(5 ) = 5
2x 7
14x
(a)
1 , –1, 4, …
4
[5th term/ sebutan ke-5]
a = 1 , r = –1 ÷ 1 = – 4
4
4
1
T5 =
(–4)4 = 64
4
(d) 2, –4k2, 8k4, …
[7th term/ sebutan ke-7]
2
a = 2, r = –4k = –2k2
2
T7 = (2)(–2k2)6 = 128k12
x
1 x (ln p)
1
x
ln p 2
r=
= 2
= 1
ln px
x ln p
2
15. For each of the following geometric progressions, determine the term stated in bracket.
Bagi setiap janjang geometri berikut, tentukan sebutan yang dinyatakan dalam kurungan.
1, 2, 4,…
3 15 75
[5th term/ sebutan ke-5]
a = 1, r = 2 ÷ 1 = 2
3
15 3 5
4
T5 = ar5 – 1 = 1 2 = 16
3 5
1 875
1
(d) ln p x, ln p 2 , ln p 4 , ln p 8 , …
(b) 2 , 1, 3 , …
3
2
[9th term/ sebutan ke-9]
a = 2, r = 1 ÷ 2 = 3
3
3 2
8
2
3
T9 =
= 2 187
3 2
128
2
3
(e) p, p , p , …
5 25
[6th term/ sebutan ke-6]
2
a = p, r = p ÷ p = p
5
5
6
5
p
p
T6 = (p)
=
5
3 125
Additional Mathematics
16. Find the number of terms for each of the following geometric progressions.
Cari bilangan sebutan bagi setiap janjang geometri berikut.
PA
P A AP
1 = 1 n–1
729
3
n–1
1
log10
= log10 1
729
3
1
log10
= (n – 1) log10 1
729
3
1
log10
729
n–1 =
=6
log10 1
3
\ n=7
12, 4, 4 , …, 4
3
243
4
a = 12, r =
= 1, T = 4
12 3 n 243
Tn = ar n – 1
4 = 12 1 n – 1
243
3
1 = 1 n–1
729
3
1 6 = 1 n–1
3
3
n =7
4
(a) 20, – 4, 4 , …, –
5
15 625
a = 20,
r = –4 = – 1,
20
5
4
Tn = –
15 625
n–1
4
–
= 20 – 1
15 625
5
n–1
1
1
–
= –
78 125
5
n–1
1
7
–
= – 1
5
5
n =8
(b) 18, 27, 81 , …, 2 187
2
16
a = 18,
r = 27 = 3 ,
18 2
Tn = – 2 187
16
2 187 = 18 3 n – 1
16
2
243 = 3 n – 1
32
2
3 5 = 3 n–1
2
2
n =6
(c) –12, 24, –48, …, –3 072
a = –12,
r = 24 = –2,
–12
Tn = –3 072
–3 072 = (–12)(–2)n –
256 = (–2)n – 1
(–2)8 = (–2)n – 1
n =9
1
17. Each of the following shows some information of a geometric progression. Determine the first term and the
common ratio of the geometric progression.
Setiap yang berikut menunjukkan maklumat suatu janjang geometri. Tentukan sebutan pertama dan nisbah sepunya janjang geometri
tersebut.
(a) –48 and –12 are the 5th term and 7th term
respectively
1 and –128 are the 2nd term and 9th term respectively
–48 dan –12 masing-masing ialah sebutan ke-5 dan ke-7
1 dan –128 masing-masing ialah sebutan ke-2 dan ke-9
Tn = arn – 1
T2 = ar = 1 ……………
T9 = ar8 = –128 ………
÷
, r7 = –128
r = –2
Substitute r = –2 into ,
a(–2) = 1
a =– 1
2
\ The first term is – 1 and the common ratio is –2.
2
T5 = ar4 = –48 ……
T7 = ar6 = –12 ……
, r2 = 1
4
r = 1
2
Substitute r = 1 into ,
2
4
a 1 = –48
2
a = –768
\ The first term is –768 and the common ratio
is 1 .
2
÷
Additional Mathematics
(b) 32 is the fourth term and it is half of the first (c) 4x3 and –128x8 are the 3 rd term and 8th term
term.
respectively.
32 ialah sebutan ke-empat dan adalah separuh daripada
4x3 dan –128x8 masing-masing ialah sebutan ke-3 dan ke-8.
sebutan pertama.
T3 = ar2 = 4x3 ……
T8 = ar7 = –128x8 ……
T4 = 1 a = 32
2
a = 64
÷ , r5 = –32x5
3
r = –2x
T4 = ar = 32
Substitute r = –2x into ,
64r3 = 32
a(–2x)2 = 4x3
r3 = 1
a =x
2
1
The
first
term is x and the common ratio is
\
r= 1 3
–2x.
2
\ The first term is 64 and the common ratio
1
is 1 3 .
2
18. Find the sum of the first n terms for each of the following geometric progressions.
Cari hasil tambah n sebutan pertama bagi setiap janjang geometri berikut.
(a) 324, 216, 144, 96, …
(the first 5 terms)
(5 sebutan pertama)
3, –6, 12, –24, …
(the first 6 terms)
(6 sebutan pertama)
a = 3, r = –6 = –2
3
n
a(1
–
r
)
Sn =
1–r
6
S6 = 3[1 – (–2) ]
1 – (–2)
= –63
r
1
a = 324, r = 216 = 2
324 3
n
a(1
–
r
)
Sn =
1–r
5
324 1 – 2
3
S5 =
1– 2
3
= 844
(b) 0.4, 1.6, 6.4, 25.6, …
(the first 7 terms)
(7 sebutan pertama)
r
a = 0.4, r = 1.6 = 4
0.4
n
a(r
–
1)
Sn =
r–1
7
S7 = 0.4(4 – 1)
4–1
= 2 184.4
1
19. Find the sum of the terms in each of the following geometric progressions.
Cari hasil tambah sebutan dalam setiap janjang geometri berikut.
1, 5, 25, …, 78 125
a = 1, r = 5, Tn = 78 125
To determine the number
Tn = ar n – 1
of terms, use Tn = arn – 1
78 125 = 1(5)n – 1
log10 78 125 = (n – 1) log10 5
log10 78 125
log10 5
n–1=7
n=8
n–1=
n
Sn = a(r – 1)
r–1
8
1(5
– 1)
S8 =
5–1
= 97 656
Use this formula as
r=5 1
1
6 561
a = 27, r = 9 = 1 , Tn = 1
27 3
6 561
(a) 27, 9, 3, …,
1
6 561
1 n–1
3
(n – 1) log10 1
3
n
n–1
= 27 1
3
1
=
177 147
1
= log10
177 147
= 12
27 1 – 1
3
S12 =
1
1–
3
= 40.50
12
r
1
Additional Mathematics
(b) 2, –1, 1 , …, 1
2
128
a = 2, r = – 1 , Tn = 1
2
128
1 = 2 – 1 n–1
128
2
n
–
1
– 1
= 1
2
256
8
= – 1
2
n =9
21– – 1
2
S9 =
1
1– –
2
= 171
128
(c) – 4, 8, –16, …, 2 048
a = – 4, r = 8 = –2, Tn = 2 048
–4
2 048 = (–4)(–2)n – 1
(–2)n – 1 = –512
= (–2)9
n = 10
10
S10 = – 4[1 – (–2) ]
1 – (–2)
= 1 364
9
20. Calculate the sum of specific number of consecutive terms in a geometric progression.
Hitung hasil tambah sebutan tertentu yang berturutan bagi suatu janjang geometri.
Given the first term of a geometric progression is 1
3
and its common ratio is 1 . Calculate the sum from
the 3rd term to 8th term. 3
1
Diberi sebutan pertama bagi suatu janjang geometri ialah dan
3
1
nisbah sepunya ialah . Hitung hasil tambah dari sebutan ke-3
3
hingga sebutan ke-8.
a=r= 1
3
The sum from the 3rd term to 8th term
= S8 – S2
1 1– 1
3
= 3
1
1–
3
3
280
=
– 4
6 561 9
= 364
6 561
Sm → n = Sn – Sm – 1
8
1 1– 1
3
– 3
1
1–
3
2
(a) Given the first term of a geometric progression is
–2 and its common ratio is 1 . Calculate the sum
4
from the 2nd term to 6th term.
Diberi sebutan pertama bagi suatu janjang geometri ialah
1
–2 dan nisbah sepunya ialah . Hitung hasil tambah dari
4
sebutan ke-2 hingga sebutan ke-6.
a = –2, r = 1
4
The sum from the 2nd term to 6th term
= S6 – S1
6
(–2) 1 – 1
4
=
1– 1
4
1
365
=–
– (–2)
512
= – 341
512
– (–2)
(b) Given the first term of a geometric progression (c) Given the first term of a geometric progression is
is 1 and its common ratio is 3. Calculate the sum
2 and its common ratio is –2. Calculate the sum
from the 4th term to 7th term.
from the 3rd term to 7th term.
Diberi sebutan pertama bagi suatu janjang geometri ialah
Diberi sebutan pertama bagi suatu janjang geometri ialah
1 dan nisbah sepunya ialah 3. Hitung hasil tambah dari
2 dan nisbah sepunya ialah –2. Hitung hasil tambah dari
sebutan ke-4 hingga sebutan ke-7.
sebutan ke-3 hingga sebutan ke-7.
a = 1, r = 3
The sum from the 4th term to 7 th term
= S7 – S3
7
3
= 1(3 – 1) – 1(3 – 1)
3–1
3–1
= 1 093 – 13
= 1 080
a = 2, r = –2
The sum from the 3rd term to 7 th term
= S7 – (T1 + T2)
7
= 2[1 – (–2) ] – [2 + 2(–2)]
1 – (–2)
= 86 – (–2)
= 88
Additional Mathematics
21. For each of the following geometric progressions, determine the sum to infinity if exist.
Bagi setiap janjang geometri berikut, tentukan hasil tambah hingga ketakterhinggaan jika wujud.
(i)
– 1 , –1, –2, …
2
a = – 1 , r = –1 ÷ – 1 = 2 1
2
2
Since r 1, the sum to infinity cannot be
determined.
In a geometric progression, the sum to infinity can
be determined when –1 < r < 1.
Dalam suatu janjang geometri, hasil tambah hingga
ketakterhinggan dapat ditentukan apabila –1 < r < 1.
(a) 8, –12, 18, …
a = 8, r = –12 = – 3
8
2
(ii) 18, 12, 8, …
a = 18, r = 12 = 2
1
18 3
Since r
1, the sum to infinity can be
determined.
S∞ = a
1–r
18
=
1– 2
3
= 54
(b) –2, 1, 1 , …
2
–1
Since r < –1, the sum to infinity cannot be
determined.
a = –2, r = – 1
–1
2
Since r
–1, the sum to infinity can be
determined.
–2
S∞ =
=– 4
3
1
1– –
2
22. Find the sum to infinity for each of the following geometric progressions.
Cari hasil tambah hingga ketakterhinggaan bagi setiap janjang geometri yang berikut.
3, 3 , 3 , …
2 4
a = 3, r = 3 ÷ 3 = 1
2
2
3
S∞ = a =
=6
1–r
1– 1
2
(a) –5, 1, – 1 , …
5
a = –5, r = – 1
5
–5
S∞ =
1– – 1
5
25
=–
6
(b) 4, 2, 1, …
a = 4, r = 2 = 1
4 2
4
S∞ =
1– 1
2
=8
(c) –8, – 4 , – 2 , …
5
25
a = –8,
r = – 4 ÷ (–8) = 1
5
10
–8
80
S∞ =
=–
9
1– 1
10
23. Solve the following problems.
Selesaikan masalah berikut.
The sum of the first two terms of a geometric progression is 9 and the sum to infinity is 8 . Find the possible
40
35
values of the first term and the common ratio of the geometric progression.
9
8
Hasil tambah dua sebutan pertama suatu janjang geometri ialah
dan hasil tambah hingga ketakterhinggaan ialah . Cari nilai40
35
nilai yang mungkin bagi sebutan pertama dan nisbah sepunya janjang geometri tersebut.
a + ar = 9
40
a(1 + r) = 9 ……
40
a = 8 ……
1 – r 35
, (1 + r)(1 – r) = 9 ÷ 8
40 35
1 – r2 = 63
64
r2 = 1
64
1
1
or r = –
\ r=
8
8
÷
Substitute the values of r into
a= 9 ÷ 1+ 1 = 1
40
8
5
a= 9 ÷ 1+ – 1 =
40
8
1
1
\ r= ,a= ;r=–
8
5
9
35
1, a = 9
8
35
,
Additional Mathematics
(a) The sum of the first three terms of a geometric progression is – 13 and the sum to infinity is – 8 . Find
8
5
the first term and the common ratio of the geometric progression.
13
8
Hasil tambah tiga sebutan pertama suatu janjang geometri ialah –
dan hasil tambah hingga ketakterhinggaan ialah – .
8
5
Cari sebutan pertama dan nisbah sepunya janjang geometri tersebut.
S3 = – 13
8
a(1 – r3) = – 13 ……
1–r
8
a = – 8 ……
1–r
5
, (1 – r3) = – 13 ÷ – 8
8
5
65
3
1–r =
64
3
r =– 1
64
1
\ r=–
4
÷
Substitute r = – 1 into
4
a
8
=–
5
1– – 1
4
a = –2
,
\ r = – 1 , a = –2
4
24. Express each of the following recurring decimals as a fraction in its simplest form.
Ungkapkan setiap perpuluhan berulang berikut sebagai satu pecahan dalam bentuk yang paling ringkas.
0.727272…
= 0.72 + 0.0072 + 0.000072 + …
Use S∞ = a
= 0.72
1–r
1 – 0.01
a = 0.72, r = 0.01
0.72
=
0.99
= 8
11
(a) 0.222…
= 0.2 + 0.02 + 0.002 + …
= 0.2
1 – 0.1
= 0.2
0.9
2
=
9
PA
P A
AP
Let/ Katakan x = 0.727272… ……
× 100, 100x = 72.7272… ……
– ,
99x = 72
x= 8
11
(b) 0.00931931931…
= 0.00931 + 0.00000931
+ 0.00000000931 + …
= 0.00931
1 – 0.001
= 0.00931
0.999
= 931
99 900
(c) 3.0111…
= 3 + (0.01 + 0.001
+ 0.0001 + …)
= 3 + 0.01
1 – 0.1
= 3 + 0.01
0.9
1
=3
or 271
90
90
25. Solve each of the following.
Selesaikan setiap yang berikut.
The value of a new car is RM50 000. If the value of the car depreciates 5% each year, find
Nilai sebuah kereta baru ialah RM50 000. Jika nilai kereta itu menyusut 5% setiap tahun, cari
(i) the value of the car in the 8th year, / nilai kereta itu pada tahun ke-8,
(ii) the number of years when the value of the car is / bilangan tahun apabila nilai kereta itu ialah
(a) RM30 000
(b) half of the original price / separuh daripada nilai asalnya
a = 50 000,
r = 1 – 0.05 = 0.95
(i) T8 = 50 000(0.95)8 – 1
= RM34 916.86
276
O
(ii) (a)
Tn = 30 000
(ii) (b)
Tn = 50 000 ÷ 2
n–1
n–1
50
000(0.95)
= 25 000
50 000(0.95)
= 30 000
n–1
n–1
(0.95)
= 0.5
(0.95)
= 0.6
log10 0.5
log10 0.6
n–1=
n–1=
log10 0.95
log10 0.95
=
13.5
= 9.96
n = 14.5 years
n = 10.96 years
Additional Mathematics
(a) Faiz joined a new company in the year 2018 and he was offered an annual salary of RM36 000 for the
O
year 2018 with an increase of 7% yearly thereafter. 276
Faiz menyertai sebuah syarikat baru pada tahun 2018 dan ditawarkan gaji tahunan RM36 000 bagi tahun 2018 dengan
kenaikan 7% setiap tahun berikutnya.
(i) Determine his annual salary in the 5th year.
Tentukan gaji tahunannya pada tahun ke-5.
(ii) In which year he will start receive a monthly salary of more than RM4 200?
Pada tahun yang keberapa dia akan mula menerima gaji bulanan yang lebih daripada RM4 200?
(iii) If he saves 10% of his annual salary every year, how much is the savings after 10 years in the
company?
Jika dia menyimpan 10% daripada gaji tahunannya setiap tahun, berapakah simpanannya selepas berkhidmat 10 tahun
dengan syarikat tersebut?
a = 36 000, r = 1 + 0.07 = 1.07
(i) T5 = 36 000(1.07)4
= RM47 188.66
(ii) Tn = 12 × RM4 200 = RM50 400
36 000(1.07)n – 1 50 400
(1.07)n – 1
1.4
log10 1.4
n–1
log10 1.07
n–1
4.97
n
5.97
\ in the 6th year
(iii) Annual salary: 36 000, 38 520, 41 216.40, …
Savings: 3 600, 3 852, 4 121.64, …
r = 1.07, a = 3 600, n = 10
10
S10 = 3 600(1.07 – 1)
1.07 – 1
= RM49 739.21
(b) A ball is dropped from a tall building of height 100 m. Each time it hits the ground, it bounces 95% of
O
the previous height. 276
Sebiji bola dijatuhkan daripada sebuah bangunan yang setinggi 100 m. Setiap kali menyentuh tanah, bola itu akan melantun
95% daripada ketinggian sebelumnya.
(i) What is the height, in m, of the ball after the 5th bounce?
Berapakah ketinggian, dalam m, bola itu selepas lantunan ke-5?
(ii) After which bounce the height of the ball is less than 37 m?
Selepas lantunan yang keberapa ketinggian bola itu kurang daripada 37 m?
(iii) Find the total distance covered, in m, when the ball stops to bounce.
Cari jumlah jarak yang dilalui, dalam m, apabila lantunan berhenti.
a = 100 × 0.95 = 95, r = 0.95
(i)
T5 = 95(0.95)
= 77.38 m
4
(ii)
T
95(0.95)
n
n–1
(0.95)n – 1
37
37
37
95
log10 37
95
log10 37
95
n–1
log10 0.95
n–1
18.38
n
19.38
\ after the 20th bounce
(n – 1)(log10 0.95)
95
1 – 0.95
= 1 900
(iii) S∞ =
Total distance covered
= 2(1 900) + 100
= 3 900 m
Additional Mathematics
SPM
Paper
1
1. It is given 4, 8, a, b and c are five consecutive terms of a
geometric progression.
630
Diberi 4, 8, a, b dan c adalah lima sebutan berturutan suatu
janjang geometri. Cari nilai c.
[2]
Ans: c = 64
P A
6. Mat took 5 minutes to complete the first kilometre of 10 km
run. He could not sustain his stamina for each subsequent
1
kilometre, he took
more time compared to the time
8
he took for the previous kilometre. The participants who
finished the run more than two hours are not qualified for
the state level run. Did Mat qualified? Show calculation to
support your answer.
630
Mat mengambil masa 5 minit untuk menghabiskan kilometer
pertama dalam suatu acara larian 10 km. Dia tidak dapat
mengekalkan staminanya bagi setiap kilometer berikutnya, dia
mengambil 1 lebih masa berbanding dengan masa yang diambil
8
untuk kilometer sebelumnya. Peserta-peserta yang menamatkan
larian melebihi dua jam tidak layak untuk acara larian peringkat
negeri. Adakah Mat layak? Tunjukkan kiraan untuk menyokong
jawapan anda.
[4]
Ans: Mat qualified for the state level run.
2. In an arithmetic progression, the sum of the first four terms
is 2 and the sixth term is –10. Find the first term and the
common difference of the progression.
630
Dalam suatu janjang aritmetik, hasil tambah empat sebutan
pertama ialah 2 dan sebutan keenam ialah –10. Cari sebutan
pertama dan beza sepunya janjang itu.
[3]
Ans: a = 5, d = –3
x–2
are three
2
consecutive terms of a geometric progression with a
1
common ratio of . Find
2
3. It is given that (x – 1), (7 – 2x) and
630
Diberi bahawa (x – 1), (7 – 2x) dan x – 2 ialah tiga sebutan
2
berturutan bagi suatu janjang geometri dengan nisbah sepunya
1 . Cari
2
(a) the value of x,
nilai x,
(b) the first term if (x – 1) is the 5 th term of the
progression.
sebutan pertama jika (x – 1) ialah sebutan ke-5 janjang itu.
[4]
Ans: (a) x = 3
(b) 32
4. It is given that m, 3 and n are the first three terms of a
geometric progression. Express in terms of n
630
Diberi bahawa m, 3 dan n ialah tiga sebutan pertama bagi suatu
janjang geometri. Ungkapkan dalam sebutan n
(a) the first term and the common ratio of the progression,
sebutan pertama dan nisbah sepunya janjang itu,
(b) the sum to infinity of the progression.
hasil tambah sebutan hingga ketakterhinggaan janjang itu.
[4]
9
n
27
Ans: (a) a = , r =
(b)
n
3
n(3 – n)
5. A circle is divided to 5 sectors in such a way that the angles
of the sectors are in an arithmetic progression. Given that
the angle of the largest sector is 9 times the angle of the
smallest sector, find the angle of the largest sector.
Suatu bulatan dibahagikan kepada 5 sektor dengan keadaan
sudut-sudut bagi kesemua sektor adalah suatu janjang aritmetik.
Diberi bahawa sudut sektor terbesar adalah 9 kali sudut sektor
terkecil, cari sudut sektor terbesar.
[3]
Ans: 129.6°
7. It is given that the nth term of a geometric progression is
5r n – 1
Tn =
, r ≠ k. State
4
630
Diberi bahawa sebutan ke-n bagi suatu janjang geometri ialah
n–1
Tn = 5r , r ≠ k. Nyatakan
4
(a) the value of k,
nilai k,
(b) the first term of the progression.
sebutan pertama bagi janjang itu.
[2]
Ans: (a) k = 1
5
(b)
4
8. It is given that the sum of the first n terms of an arithmetic
n
progression is Sn = [15 – 7n]. Find the nth term.
2
630
Diberi bahawa hasil tambah n sebutan pertama bagi suatu
janjang aritmetik ialah Sn = n [15 – 7n]. Cari sebutan ke-n.
2
[3]
Ans: 11 – 7n
9. The nth term of an arithmetic progression, Tn, is given by
1
Tn = (6 + 5n). Find
3
Sebutan ke-n suatu janjang aritmetik, Tn, diberi oleh
Tn = 1 (6 + 5n). Cari
3
(a) the 15th term,
sebutan ke-15,
(b) the common difference.
beza sepunya.
[4]
Ans: (a) 27
5
(b)
3
Additional Mathematics
10. In an arithmetic progression, the 3rd term and the 17th term
are 22 and –20 respectively. Find the 8th term.
(c)
Dalam suatu janjang aritmetik, sebutan ke-3 dan sebutan ke-17
masing-masing ialah 22 dan –20. Cari sebutan ke-8.
[3]
Ans: 7
11. Su has a wire with a length of 10.12 m. Su divided the wire
into several pieces of different lengths. Each piece is bent
to form a square. The diagram shows the first three squares
formed by Su.
630
Su mempunyai seutas dawai dengan panjang 10.12 m. Su
membahagikan dawai itu kepada beberapa bahagian yang
berlainan panjangnya. Setiap bahagian dibengkokkan untuk
membentuk satu segi empat sama. Rajah di bawah menunjukkan
tiga segi empat sama yang pertama yang dibentuk oleh Su.
nilai r, diberi bahawa sebutan ke-r adalah sebutan positif
pertama bagi janjang itu.
[2]
Ans: (a) –315
(b) a = –45, d = 3
(c) r = 17
3. At a certain day, a breeder has 3 500 chickens in his farm. He
delivers 300 chickens every day to a wholesaler. The breeder
feeds the chickens before delivering. If the cost to feed a
chicken is RM0.50 per day, calculate the total cost until his
remaining chickens are 500.
630
Pada suatu hari yang tertentu, seorang penternak mempunyai
3 500 ekor ayam di ladangnya. Dia menghantar 300 ekor
ayam setiap hari kepada seorang pemborong. Penternak itu
akan memberi makan dahulu kepada ayamnya sebelum
penghantarannya. Jika kos memberi makan seekor ayam ialah
RM0.50 sehari, hitung jumlah kos sehingga bilangan ayamnya
berbaki 500 ekor.
[6]
Ans: RM11 000
11 cm
7 cm
11 cm
3 cm
the value of r, given that the rth term is the first
positive term of the progression.
7 cm
3 cm
How many squares can be formed by Su?
Berapa buah segi empat sama yang boleh dibentuk oleh Su?
[3]
Ans: 11
Paper
2
4. The diagram shows part of a rectangular wall painted red,
R, black, B and yellow, Y and so on in that order. The height
of the wall is 1.6 m. The side length of the first coloured
rectangle is 5 cm and the side length of each subsequent
coloured rectangle increases by 2 cm.
630
Rajah di bawah menunjukkan sebahagian daripada dinding
berbentuk segi empat tepat yang dicat dengan warna merah,
R, hitam, B dan kuning, Y secara berselang seli. Tinggi dinding
ialah 1.6 m. Panjang sisi segi empat tepat berwarna yang
pertama ialah 5 cm dan panjang sisi bagi setiap segi empat tepat
berwarna yang berikutnya bertambah sebanyak 2 cm.
1. Given an arithmetic progression.
Diberi suatu janjang aritmetik:
–12, –15, –18, …
(a) State the first term and the common difference.
Nyatakan sebutan pertama dan beza sepunya.
[2]
R B
(b) Find the sum of
Cari hasil tambah bagi
(i) the first six terms, / enam sebutan pertama,
(ii) the first seven terms, / tujuh sebutan pertama.
(c)
[4]
Use the answer in (b), find the 7th term.
Gunakan jawapan dalam (b), cari sebutan ke-7.
[1]
Ans: (a) a = –12, d = –3
(b) (i) –117; (ii) –147
(c) –30
2. The sum of the first n terms of an arithmetic progression, Sn
3n(n – 31)
is given by Sn =
. Find
2
630
Hasil tambah n sebutan pertama bagi suatu janjang aritmetik,
Sn diberi oleh Sn = 3n(n – 31) . Cari
2
(a) the sum of the first 10 terms,
hasil tambah 10 sebutan pertama,
[1]
(b) the first term and the common difference,
sebutan pertama dan beza sepunya,
[3]
Y
R
B
Y
R
It is given that the total number of the coloured rectangles
is 48.
Diberi jumlah bilangan segi empat tepat berwarna ialah 48.
(a) Find/ Cari
(i) the side length, in cm, of the last coloured
rectangle,
panjang sisi, dalam cm, bagi segi empat tepat
berwarna yang terakhir,
(ii) the total length, in cm, of the painted wall.
jumlah panjang, dalam cm, dinding yang dicat.
[4]
(b) Which coloured rectangle has an area of 14 880 cm2?
Hence, state the colour of that particular rectangle.
Segi empat tepat berwarna yang keberapa mempunyai
keluasan 14 880 cm2? Seterusnya, nyatakan warna bagi
segi empat tepat berkenaan.
[3]
Ans: (a) (i) 99 cm; (ii) 2 496 cm
(b) 45th, yellow
Form 4 Answers
37 5
SPM
3 R H
R
P
3 SH
Let the smallest sector =
and the largest sector =
= 360°
10
) = 360
Angle of the largest sector = 9 ×
= 129.6°
+ (4 − 1)
10
10
10
= −10
= −10 ......................
+ 10 = −20 ......................
−
6
= −21
= −3
= −3 into
+ 3(−3) = 1
= 10
= 89.89 minutes
120 minutes
∴ Mat quali ed for the state level run.
≠1
≠ 1, so
]−
−1)]
− 1)
[−14
−
∞
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
= −20 .....................
−
6
= −42
= −3
= −3 into
+ 2(−3) = 22
= 3 500,
= −300,
= 500
500 = 3 500 + ( – 1)(−300)
300( – 1) = 3 000
– 1 = 10
= 28 + 7(−3) = 7
(3 500 + 500)
Perimeter: 12 cm, 28 cm, 44 cm, …
= 28 − 12 = 16
=
− 1)(16)] =
– 8) =
=
10.12 × 100
1 012
1 012
1 012
= 22 000
Total cost = 22 000 × RM0.50
= RM11 000
− 1)
– 253 = 0
+ 23) = 0
≠–
∴
3 SH
= −12,
= −15 − (−12) = −3
× 100) = 14 880
14 880
× 100
×
[2(−12) + (6 – 1)(−3)]
= −117
[2(−12) + (7 – 1)(−3)]
= −147
= −147 – (−117)
= −30
10
∴7
coloured rectangle has an area of 14 880 cm
The colour is yellow.
= RM111.00,
5
= RM74 477.00
3(10)(10 – 31)
74 477.00 =
= −315
(111.00 + 1 321.25)
= 104
∴ There are 104 lucky draw winners.
− 1)
1 321.25 = 111.00 + (104 – 1)
5
= −45
∴ The constant amount is RM11.75.
= −87
= 111.00 + (6 – 1)(11.75)
5
= −87 – (−45)
= −42
(d) (i)
= −42 – (−45)
− 1)
[2(111.00) + (5 – 1)(11.75)]
∴ rst term = −45, common difference = 3
−45 + (
∴
© Penerbitan Pelangi Sdn. Bhd.
0
0
0
= RM672.50
20
20 [2(111.00) + (20 – 1)(11.75)]
−
[2(111.00) + (9 – 1)(11.75)]
= 4 452.50 − 1 422
= RM3 030.50
Form 4 Answers
276
OOH
H
71 + 60
− 1)
Common difference:
−T T−1
T
T
The tourist has to wait [136 – (71 + 60)] = 5 minutes.
Common ratio:
T
T
T
T
√
√
)=
– 4) =
– 3) =
=
∴
0
0
0
−1 or 3
≠ −1,
© Penerbitan Pelangi Sdn. Bhd.
L
Additional Mathematics
3. Plot a graph of the variables for each of the following tables and draw the line of best fit in each case. Hence,
determine the equation of the line of best fit.
Plot graf bagi nilai pemboleh ubah dalam setiap jadual berikut dan lukis garis lurus penyuaian terbaik dalam setiap kes berikut.
Seterusnya, tentukan persamaan garis lurus penyuaian terbaik tersebut.
(a)
x
2
4
6
8
10
x2y
24.96
45.12
64.44
85.12
105.0
Plot a graph of x2y against x.
Plot graf x2y melawan x.
x
1
2
3
4
5
ln y
10.4
9.51
8.55
7.62
6.69
Plot a graph of ln y against x.
Plot graf ln y melawan x.
In y
x 2y
11
100
10
80
56
9
60
2.5
8
40
5.6
7
20
0
2.3
2
4
6
8
10
x
6
5
4
From the graph,
gradient, m = 96 – 40
9 – 3.4
56
=
5.6
= 10
3
2
1
y-intercept, c = 6
\ x2y = 10x + 6
0
To draw a line of best fit, make sure
Bagi melukis satu garis lurus penyuaian terbaik, pastikan
garis lurus itu melalui seberapa banyak titik yang mungkin.
bilangan titik yang terletak di kiri dan kanan garis lurus itu
adalah hampir sama dan jaraknya adalah hampir sama dari
garis lurus itu.
GeoGebra – lines of best t
XX S
VIDEO
X
1
2
m = 11.3 – 9
0 – 2.5
= 2.3
–2.5
= –0.92
c = 11.3
\ ln y = –0.92x + 11.3
3
4
5
x
L
Additional Mathematics
(b)
x2
0
1
2
3
4
√y
–2
3
8
13
18
Plot a graph of √y against x2.
Plot graf √y melawan x2.
(c)
log10 p
2
4
6
8
log10 q
1.34
2.13
2.55
3.32
Plot a graph of log10 q against log10 p.
Plot graf log10 q melawan log10 p.
y
log10 q
3
15
1.65
10
2
16
5
1
5
3.2
0
1
2
3
4
x2
0
2
m = 17 – 1
3.8 – 0.6
= 16
3.2
=5
m = 3.45 – 1.8
8.4 – 3.4
1.65
=
5
= 0.33
c = –2
c = 0.7
\ √y = 5x – 2
2
4
6
log10 p
8
\ log10 q = 0.33 log10 p + 0.7
4. The graphs below show the lines of best fit of two variables as given in the respective graphs.
Graf berikut menunjukkan garis penyuaian terbaik bagi dua pemboleh ubah seperti ditunjukkan dalam graf masing-masing.
From the graph,
Daripada graf,
(a) determine the value of
tentukan nilai
(i) s when / apabila t = 3.8,
(ii) s when / apabila t = 7.2,
(iii) t when / apabila s = 21.5,
(b) obtain the equation of the line of best fit.
dapatkan persamaan garis lurus penyuaian terbaik.
(a) (i) 19.5
(ii) 7
(iii) 3.2
(b) m = 33 – 14.5 = –3.56
0 – 5.2
c = 33
Equation of the line of best fit: s = –3.56t + 33
s
30
25
20
19.5
15
10
7
5
0
2
3.2
4
6
8
10
t
L
Additional Mathematics
From the graph,
Daripada graf,
(a) determine the value of
tentukan nilai
(i) q when / apabila p = 5,
(ii) q when / apabila p = 16.5,
(iii) p when / apabila q = 3.6,
(b) obtain the equation of the line of best fit.
dapatkan persamaan garis lurus penyuaian terbaik.
q
6
5
4
(a) (i) 1.5
(ii) 5
(iii) 12
2
1.5
6–3
20 – 10
= 3
10
c =0
q= 3p
10
0
(b) m =
10
12
20
p
5. The tables below show the corresponding values of the related variables. Answer the questions given.
Jadual berikut menunjukkan nilai-nilai sepadan bagi pemboleh ubah yang berkait. Jawab soalan yang diberi.
t
5
10
15
20
25
s
34
39
44
49
55
(c) Based on the graph obtained, determine
(a) Plot a graph of s against t.
Berdasarkan graf yang diperolehi, tentukan
Plot graf s melawan t.
(i)
the value of s when t = 13,
(b) Draw the line of best fit and determine its
nilai s apabila t = 13,
equation.
(ii) the value of t when s = 46.
Lukis garis lurus penyuaian terbaik dan tentukan
nilai t apabila s = 46.
persamaannya.
(a)
(b) m = 39 – 29
10 – 0
=1
c = 29
\ s = t + 29
s
60
50
(c) From the graph
(i) 42
(ii) 17
42
40
30
20
10
0
5
10
15
17
20
25
t
L
Additional Mathematics
x
0.1
0.2
0.3
0.4
0.5
y
7.8
6.6
5.1
4.0
2.1
(a) Plot a graph of y against x.
Plot graf y melawan x.
(c) Based on the graph obtained, determine
(b) Draw the line of best fit and determine its
Berdasarkan graf yang diperolehi, tentukan
equation.
(i) the value of y when x = 0.27,
Lukis garis lurus penyuaian terbaik dan tentukan
nilai y apabila x = 0.27,
persamaannya.
(ii) the value of x when y = 3.8.
nilai x apabila y = 3.8.
(a)
(b) m = 6.6 – 9.6
0.2 – 0
= –15
c = 9.6
y
10
\ y = –15x + 9.6
8
(c) From the graph
(i) 5.6
(ii) 0.39
6
5.6
4
2
0
0.39
0.1
0.2
0.3
0.4
0.5
x
6. Solve each of the following.
Selesaikan setiap yang berikut.
The diagram shows the line of best fit obtained by plotting y2 against √x .
Find
Rajah di sebelah menunjukkan garis lurus penyuaian terbaik yang diperolehi dengan
memplotkan y2 melawan √x . Cari
(i) the equation of the line of best fit,
persamaan garis lurus penyuaian terbaik tersebut,
(ii) the exact value(s) of / nilai tepat
(a) y when / apabila x = 25,
(b) x when / apabila y = 16 .
5
(i)
m = 13 – 9
10 – 0
2
=
5
c =9
\ y2 = 2 √x + 9
5
(ii) (a) x = 25, √x = 5
y2 = 2 (5) + 9
5
= 11
y = ±√11
y2
(10, 13)
9
0
(ii) (b) y = 16 , y2 = 256 ,
5
25
256 = 2 √x + 9
25
5
2 √x = 31
5
25
√x = 31
10
x = 9.61
x
L
Additional Mathematics
(a) The diagram shows the line of best fit obtained by plotting y against
ln x. Find
Rajah di sebelah menunjukkan garis lurus penyuaian terbaik yang diperolehi
dengan memplotkan y melawan ln x. Cari
(i) the equation of the line of best fit,
persamaan garis lurus penyuaian terbaik tersebut,
(ii) in terms of e, the exact value of
dalam sebutan e, nilai tepat
(a) x when / apabila y = 3e,
(b) y when / apabila x = e3.
(i)
m = 02 – e
e –0
= – e2
e
–1
= –e
c =e
\ y = –e –1 ln x + e
(ii) (a)
y
(0, e)
(e 2, 0)
0
3e = –e –1 ln x + e
e ln x = –2e
ln x = –2e2
x = e –2e2
–1
(b) y = –e –1 ln e3 + e
y = –3e –1 + e
(b) The diagram shows part of the line of best fit obtained by plotting
(x – √y ) against x. The line passes through the points (1, 1) and
(3, 7). Express y in terms of x.
Rajah di sebelah menunjukkan sebahagian daripada satu garis lurus penyuaian
terbaik yang diperolehi dengan memplotkan (x – √y ) melawan x. Garis lurus
tersebut melalui titik-titik (1, 1) dan (3, 7).
)x – y )
(3, 7)
(1, 1)
x
0
m = 7–1
3–1
=3
At (1, 1), 1 = 3(1) + c
c = –2
(x – √y ) = 3x – 2
√y = 2 – 2x
√y = 2(1 – x)
y = 4(1 – x)2
(c) The diagram shows part of the line of best fit obtained by plotting
log2 y against log2 x. The line passes through the points (2, 2) and
(6, 4). Express y in terms of x.
Rajah di sebelah menunjukkan sebahagian daripada satu garis lurus penyuaian
terbaik yang diperolehi dengan memplotkan log2 y melawan log2 x. Garis lurus
tersebut melalui titik-titik (2, 2) dan (6, 4). Ungkapkan y dalam sebutan x.
log2 y
(6, 4)
(2, 2)
0
m = 4–2
6–2
1
=
2
At (2, 2), 2 = 1 (2) + c
2
c=1
In x
log2 y = 1 log2 x + 1
2
log2 y – log2 √x = 1
log2 y = 1
√x
y = 21
√x
y = 2√x
log2 x
L
Additional Mathematics
8. Solve the following problems.
Selesaikan masalah berikut.
y
y
y
y = px2 + q
x
0
(a)
(r, 14)
y = pex + 2
(2.5, 10)
(0.5, 6)
y
(2, q)
(0, 5)
x2
0
x
0
(a)
(b)
ex
0
(b)
Diagram (a) shows part of the graph of y = px + q.
The non-linear relation is reduced to a linear relation
which is represented by the graph in diagram (b).
Find the value of p and of q.
Rajah (a) menunjukkan sebahagian graf bagi y = px2 + q.
Hubungan tak linear tersebut ditukarkan kepada hubungan
linear yang diwakili oleh graf dalam rajah (b). Cari nilai bagi
p dan q.
Diagram (a) shows part of the graph of y = pex + 2.
The non-linear relation is reduced to a linear relation
which is represented by the graph in diagram (b).
Find the values of p, q and r.
Rajah (a) menunjukkan sebahagian graf bagi y = pex + 2.
Hubungan tak linear tersebut ditukarkan kepada hubungan
linear yang diwakili oleh graf dalam rajah (b). Cari nilai bagi
p, q dan r.
In (b), Y = mX + c
y = mx2 + c
From (a), 5 = p(e0) + 2
p=3
2
10 – 6
2.5 – 0.5
=2
In (b), Y = mX + c
y = 3ex + 2
Hence, m = 3 and c = 2
m =
At (0.5, 6), 6 = 2(0.5) + c
c=5
Y = 3X + 2
q = 3(2) + 2
=8
Equation: y = 2x2 + 5
Hence, p = 2 and q = 5.
14 = 3(r) + 2
3r = 12
r =4
9. Solve each of the following.
Selesaikan setiap yang berikut.
The variables x and y are related by the equation y = p(q2x), where p and q are constants. When log2 y is plotted
against x, a straight line passing through the point 1 , 3 and with gradient 4 is obtained. Find the value of
4
p and of q.
Pemboleh ubah x dan y dihubungkan oleh persamaan y = p(q2x), dengan keadaan p dan q ialah pemalar. Apabila log2 y diplotkan
1
melawan x, satu garis lurus yang melalui titik
, 3 dan berkecerunan 4 diperolehi. Cari nilai p dan nilai q.
4
y = p(q2x)
log2 y = log2 p + 2x log2 q
log2 y = (2 log 2 q)x + log2 p
m = 2 log2 q
log2 q
q
q
=4
=2
= 22
=4
Y = mX + c
Y = 4X + log2 p
3 = 4 1 + log2 p
4
log2 p = 2
p = 22
p =4
L
Additional Mathematics
(a) The variables x and y are related by the equation (b) The variables x and y are related by the equation
y = p(q x – 1), where p and q are constants. When
yqx = p, where p and q are constants. When
log10 y is plotted against (x – 1), a straight line
ln y is plotted against x, a straight line passing
5
1
through the points (1, 1) and (–1, 2) is obtained.
passing through the points 1,
and
, 1 is
3
3
Determine the exact value of p and of q.
obtained. Find the value of p and of q.
Pemboleh ubah x dan y dihubungkan oleh persamaan
Pemboleh ubah x dan y dihubungkan oleh persamaan
yqx = p, dengan keadaan p dan q ialah pemalar. Apabila
y = p(qx – 1), dengan keadaan p dan q ialah pemalar. Apabila
ln y diplotkan melawan x, satu garis lurus yang melalui
log10 y diplotkan melawan (x – 1), satu garis lurus yang
titik-titik (1, 1) dan (–1, 2) diperolehi. Tentukan nilai tepat
5
1
p dan q.
melalui titik-titik 1,
dan
, 1 diperolehi. Cari nilai
3
3
p dan nilai q.
yqx = p
x–1
ln yqx = ln p
y = p(q )
ln y + x ln q = ln p
log10 y = log10 p + (x – 1) log10 q
ln y = (–ln q)x + ln p
log10 y = (log10 q)(x – 1) + log10 p
Y = mX + c
Y = mX + c
5 –1
3
m=
1– 1
3
=1
5 = 1(1) + c
3
c = 2
3
log10 q = 1
q = 10
log10 p = 2
3
m = 2–1
–1 – 1
=– 1
2
–ln q = – 1
2
1 = – 1 (1) + c
2
3
c =
2
ln p = 3
2
1
3
q = e2
p = e2
2
p = 10 3
SSOL
L
L
LR R
L H
Textbook
pg. 166 – 169
L H
10. Solve each of the following problems.
Selesaikan setiap masalah yang berikut.
The table shows the experimental values of the distance travelled, s metres, and the time taken, t seconds,
by a particle. 276 SSO L
Jadual di bawah menunjukkan nilai-nilai daripada eksperimen mengenai jarak dilalui, s meter dan masa diambil, t saat, oleh suatu
zarah.
t
2
4
6
8
10
s
3.9
20
48
88
141
It is known that s and t are related by the equation s = pt2 + qt, where p and q are constants to be determined.
Diketahui bahawa s dan t dihubungkan oleh persamaan s = pt2 + qt, dengan keadaan p dan q ialah pemalar yang perlu ditentukan.
(i) Convert the non-linear equation to a linear equation.
Tukarkan persamaan tak linear tersebut kepada persamaan linear.
s
s
(ii) Use a scale of 2 cm to 2 s on the t-axis and 2 cm to 2 m s–1 on the -axis, plot against t and draw the
t
t
line of best fit.
s
s
Gunakan skala 2 cm kepada 2 s pada paksi-t dan 2 cm kepada 2 m s–1 pada paksi- , plot melawan t dan lukis garis lurus
t
t
penyuaian terbaik.
(iii) From the graph obtained, determine the values of p and q.
Daripada graf yang diperolehi, tentukan nilai p dan q.
L
Additional Mathematics
(iv) Explain the significance of the gradient of the straight line.
Terangkan keertian kecerunan garis lurus itu.
(v) Without further calculation, find
Tanpa membuat sebarang pengiraan lanjutan, cari
(a) the initial velocity / halaju awal,
(b) the acceleration / pecutan.
(i)
s = pt2 + qt
s
= pt + q
t
(ii)
By plotting s against t, a straight line will be obtained.
t
t
2
4
6
8
10
s
t
1.95
5
8
11
14.1
2 cm
s
t
2 cm
14
x
12
Persamaan tak linear ialah persamaan yang mengandungi
pemboleh ubah x yang kuasanya bukan
10
7.5
8
6
5
4
For Y = mX + c
Bagi Y = mX + c
x and y
Y
Y boleh diungkapkan dalam sebutan x dan y.
2
X
y
X tidak mempunyai sebutan y.
0
2
4
6
8
10
t
c
x and y
c ialah pemalar (tanpa sebutan x dan y).
(iii) p = gradient
= 12.5 – 5
9–4
= 7.5
5
= 1.5
q = y-intercept
= –1
(iv) The gradient of the straight line is the rate of change of velocity of the particle which is a constant. That
is the acceleration which is 1.5 m s–2.
(v) (a) When t = 0, s = –1 m s–1
t
\ The initial velocity of the particle is –1 m s–1.
(b) The gradient of the straight line represents the acceleration. Hence, the acceleration is 1.5 m s–2.
L
Additional Mathematics
(a) Suhaimi measured the volume of a type of liquid, in ml, every day. The readings are recorded and
displayed in the table below. The volume of the liquid, V, and the number of days, t, are related by the
equation V = kt n, where k and n are constants. 276 SSO L
Suhaimi mengukur isi padu, dalam ml, sejenis cecair setiap hari. Bacaan isi padu direkodkan dan dipaparkan dalam jadual di
bawah. Isi padu cecair itu, V, dan bilangan hari, t, dihubungkan oleh persamaan V = kt n, dengan keadaan k dan n ialah pemalar.
(i)
t
1
2
3
4
5
V
50
35.50
28.87
25.00
22.36
Convert the equation V = kt n to linear form.
Tukarkan persamaan V= kt n kepada bentuk linear.
(ii) Plot ln t against ln V on the graph paper provided and draw the line of best fit.
Plot ln t melawan ln V pada kertas graf yang disediakan dan lukis garis lurus penyuaian terbaik.
(iii) From the graph drawn in (a)(ii), determine the value of
Daripada graf yang dilukis di (a)(ii), tentukan nilai
(a) k
(b) n
(c) V when / apabila t = 2.2
(d) t when / apabila V = 42
(iv) Explain the significance of the value of V when t = 0.
Terangkan keertian nilai V apabila t = 0.
(i)
(ii)
V = kt n
ln V = n ln t + ln k
ln t
0
0.69
1.10
1.39
1.61
ln V
3.91
3.57
3.36
3.22
3.11
(iii) (a) ln k = y-intercept
= 3.91
k = e3.91
= 49.90
(b) n = gradient
= 3.76 – 3.20
0.3 – 1.44
= –0.49
In V
3.9
3.7
3.52
3.5
3.3
(c) When t = 2.2,
ln 2.2 = 0.79
From the graph,
ln V = 3.52
ln V = e3.52
= 33.78
3.1
0
0.2
(d) When V = 42,
ln 42 = 3.74
From the graph, ln t = 0.34
t = e0.34
= 1.40
(iv) When t = 0, V is the initial volume of the liquid.
0.34
0.4
0.6
0.8
1.0
1.2
1.4
1.6
In t
L
Additional Mathematics
(b) The pressure at the opening of the pipe, P, and the corresponding velocities of the flow of water, v, are
recorded. The equation relating P and v is v = kP q, where k and q are constants. The readings are as
follows. 276 SSO L
Tekanan pada bukaan paip, P, dan halaju aliran air yang sepadan, v, direkodkan. Persamaan yang menghubungkan P dan v ialah
v = kPq, dengan keadaan k dan q ialah pemalar. Bacaan adalah seperti berikut.
(i)
P
10.6
13.3
17.2
24.7
29.3
v
9.8
11.0
12.4
14.9
16.3
Use a scale of 2 cm to 0.2 unit on the log10 P-axis and 2 cm to 0.2 unit on the log10 v-axis, plot log10 v
against log10 P and hence draw the line of best fit.
Gunakan skala 2 cm kepada 0.2 unit pada paksi-log10 P dan 2 cm kepada 0.2 unit pada paksi-log10 v, plot log10 v melawan
log10 P dan seterusnya lukis garis lurus penyuaian terbaik.
(ii) From the graph of (b)(i), determine the values of k and q.
Daripada graf dalam (b)(i), tentukan nilai k dan q.
(iii) Hence, find the value of
Seterusnya, cari nilai
(a) v when P = 9.2
v apabila P = 9.2
(b) P when v = 7.6
P apabila v = 7.6
(i)
log10 P
1.03
1.12
1.24
1.39
1.47
log10 v
0.99
1.04
1.09
1.17
1.21
(ii) v = kP q
log10 v = q log10 P + log10 k
log10 k = y-intercept
= 0.5
k = 100.5
= 3.162
q = gradient
= 1.16 – 0.64
1.36 – 0.30
= 0.49
(iii) (a) When P = 9.2,
log10 9.2 = 0.96
From the graph,
log10 v = 0.96
v = 100.96
= 9.12
(b) When v = 7.6,
log10 7.6 = 0.88
From the graph,
log10 P = 0.78
P = 100.78
= 6.03
2 cm
log10 v
2 cm
1.4
1.2
1.0
0.96
0.8
0.6
0.4
0.2
0
0.78
0.2
0.4
0.6
0.8
1.0
1.2
1.4
log10 P
L
Additional Mathematics
SPM
Paper
P A
1
4. The variables x and y are related by the equation
r
y = x2 + x , where r is a constant. The diagram shows a
1
straight line graph obtained by plotting (y – x2) against x .
630
1. The diagram shows the straight line graph obtained by
plotting (y – x) against x2.
r
Pemboleh ubah x dan y dihubungkan oleh persamaan y = x2 + ,
x
dengan keadaan r ialah pemalar. Rajah di bawah menunjukkan
graf garis lurus yang diperoleh dengan memplotkan (y – x2)
1
melawan .
x
630
Rajah di bawah menunjukkan graf garis lurus yang diperoleh
dengan memplotkan (y – x) melawan x2.
(y – x)
9
(y – x 2)
p
( , 4h(
3
–3
x2
O
1
x
O
Express y in terms of x.
Express h in terms of p and r.
Ungkapkan y dalam sebutan x.
Ungkapkan h dalam sebutan p dan r.
[3]
Ans: y = 3x2 + x + 9
Ans: h =
2. The variables x and y are related by the equation
q
y = 3x2 – x , where q is a constant. A straight line is
obtained by plotting xy against x3, as shown in the diagram.
630
Pemboleh ubah x dan y dihubungkan oleh persamaan
q
y = 3x2 – , dengan keadaan q ialah pemalar. Suatu garis
x
lurus diperoleh dengan memplotkan xy melawan x3, seperti
ditunjukkan pada rajah di bawah.
[3]
rp
12
x
1
5. The diagram shows the graph of y against 2 .
x
630
Rajah di bawah menunjukkan graf
x
1
melawan 2 .
y
x
x
y
(6, 4)
xy
O
(h, 13)
1
x2
–5
1
O
Based on the diagram, express y in terms of x.
Berdasarkan rajah, ungkapkan y dalam sebutan x.
x3
Find the value of h and of q.
Ans: y =
Cari nilai h dan nilai q.
2x3
3 – 10x2
[3]
[3]
Ans: h = 4, q = –1
y
3. The variables x and y are related so that when 3 is plotted
x
2
against x , a straight line passing through the points (3, 9)
and (8, 1) is obtained. Express y in terms of x.
Pemboleh ubah x dan y dihubungkan dengan keadaan apabila
y
diplotkan melawan x2, suatu garis lurus yang melalui titikx3
titik (3, 9) dan (8, 1) diperolehi. Ungkapkan y dalam sebutan x.
[3]
8
69 3
Ans: y = – x5 +
x
5
5
a
6. The equation y =
, where a and b are constants, can
x + 3b
be converted to an equation of straight line. This straight
1
line has a gradient of
and passes through the point
2
(0, –6). Determine the values of a and b.
Persamaan y = a , dengan keadaan a dan b ialah pemalar,
x + 3b
boleh ditukarkan kepada persamaan garis lurus. Garis lurus ini
mempunyai kecerunan 1 dan melalui titik (0, –6). Tentukan
2
nilai a dan nilai b.
[3]
Ans: a = 2, b = –4
L
Additional Mathematics
1
is
y2
plotted against x, a straight line passing through the points
(1.5, 3.6) and (2.6, 1.4) is obtained.
Paper
7. The variables x and y are related such that when
Pemboleh ubah x dan y dihubungkan dengan keadaan apabila
1 diplotkan melawan x, satu garis lurus melalui titik-titik
y2
(1.5, 3.6) dan (2.6, 1.4) diperolehi.
(a) Express y2 in terms of x.
Ungkapkan y2 dalam sebutan x.
(b) Find the values of y when x = 1.3
Cari nilai-nilai y apabila x = 1.3
[4]
1
2
Ans: (a) y =
–2x + 6.6
1
(b) y = ±
2
1. The table shows the values of two variables, x and y,
obtained from an experiment. Variables x and y are related
by the equation px = qy + xy, where p and q are constants.
630
Jadual berikut menunjukkan nilai-nilai bagi dua pemboleh ubah,
x dan y, yang diperoleh daripada suatu eksperimen. Pemboleh
ubah x dan y dihubungkan oleh persamaan px = qy + xy, dengan
keadaan p dan q ialah pemalar.
x
y
log2 y
(16, 3)
x
(a) Express log10 y in terms of x.
Ungkapkan log10 y dalam sebutan x.
(b) Show that y = p(2qx) where p and q are constants. State
the value of p and of q.
Tunjukkan bahawa y = p(2qx) dengan keadaan p dan q
ialah pemalar. Nyatakan nilai p dan nilai q.
[3]
1
Ans: (a) log2 y = x – 1
4
1
1
(b) p = , q =
2
4
9. A particular species of fish is put into a pond. After t weeks,
the number of these fish, A, in the pond is given by the
equation A = Ao ekt, where Ao is the initial number of fish
and k is a constant.
Sejenis spesies ikan tertentu dimasukkan ke dalam satu kolam.
Selepas t minggu, bilangan ikan ini, A, di dalam kolam diberikan
oleh persamaan A = Ao ekt, dengan keadaan Ao ialah bilangan
ikan pada permulaan dan k ialah satu pemalar.
(a) Reduce the equation to linear form.
Tukarkan persamaan itu kepada bentuk linear.
(b) Express k in terms of t, when the number of these fish
is thrice that of the initial number.
Ungkapkan k dalam sebutan t, apabila bilangan ikan di
dalan kolam adalah tiga kali bilangan pada permulaan.
[4]
Ans: (a) ln A = kt + ln Ao
ln 3
(b) k =
t
2.0
3.0
4.0
5.0
6.0
Berdasarkan jadual di atas, bina satu jadual bagi nilai-nilai
1 dan 1 .
y
x
[2]
1
1
(b) Plot against , using a scale of 2 cm to 0.1 unit on
y
x
1
1
-axis and 2 cm to 0.5 unit on -axis. Hence, draw
x
y
the line of best fit.
Plot 1 melawan 1 , dengan menggunakan skala 2 cm
y
x
kepada 0.1 unit pada paksi- 1 dan 2 cm kepada 0.5 unit
x
pada paksi- 1 . Seterusnya, lukis garis lurus penyuaian
y
terbaik.
[3]
(c) Using the graph in 1(b), find the value of
Menggunakan graf di 1(b), cari nilai
(i) p,
(ii) q.
[5]
Ans: (a), (b) Refer to Answer Section
(c) (i) 0.26; (ii) –1.32
Rajah di bawah menunjukkan graf suatu garis lurus dengan
memplotkan log10 y melawan x.
(8, 1)
1.5
2.018 0.768 0.464 0.386 0.353 0.340
(a) Based on the above table, construct a table for the
1
1
values of and .
y
x
8. The diagram shows the graph of a straight line with log10 y
plotted against x.
O
2
2. The table shows the values of two variables, x and y,
obtained from an experiment. Variables x and y are related
q
by the equation y = 3px + , where p and q are constants.
4x
630
Jadual berikut menunjukkan nilai-nilai bagi dua pemboleh ubah,
x dan y, yang diperoleh daripada suatu eksperimen. Pemboleh
ubah x dan y dihubungkan oleh persamaan y = 3px + q ,
4x
dengan keadaan p dan q ialah pemalar.
x
1
2
3
4
5
6
y
2
1.2
1.33
1.4
1.56
1.73
(a) Based on the above table, construct a table for the
values of x2 and xy.
Berdasarkan jadual di atas, bina satu jadual bagi nilai-nilai
x2 dan xy.
[2]
(b) Plot xy against x2, using a scale of 2 cm to 5 units on
the x2-axis and 2 cm to 2 units on the xy-axis. Hence,
draw the line of best fit.
Plot xy melawan x 2, dengan menggunakan skala 2 cm
kepada 5 unit pada paksi-x2 dan 2 cm kepada 2 unit pada
paksi-xy. Seterusnya, lukis garis lurus penyuaian terbaik.
[3]
L
Additional Mathematics
(c)
Using the graph in 2(b), find the value of
Menggunakan graf di 2(b), cari nilai
(i) p,
(ii) q.
[5]
Ans: (a), (b) Refer to Answer Section
(c) (i) 0.08; (ii) 7.2
3. The table shows the values of two variables, x and y,
obtained from an experiment. The variables x and y are
k√h
related by the equation y – h =
, where h and k are
x
constants.
630
Jadual berikut menunjukkan nilai-nilai bagi dua pemboleh ubah,
x dan y, yang diperoleh daripada suatu eksperimen. Pemboleh
ubah x dan y dihubungkan oleh persamaan y – h = k√h ,
x
dengan keadaan h dan k ialah pemalar.
x
1.5
2.0
3.0
3.5
4.5
6.0
y
4.50
5.25
6.00
5.51
6.56
6.58
(a) Plot xy against x, using a scale of 2 cm to 1 unit on the
x-axis and 2 cm to 5 units on the xy-axis. Hence, draw
the line of best fit .
Plot xy melawan x, dengan menggunakan skala 2 cm
kepada 1 unit pada paksi-x dan 2 cm kepada 5 unit
pada paksi-xy. Seterusnya, lukis garis lurus penyuaian
terbaik.
[4]
(b) Using the graph in 3(a), find
Menggunakan graf di 3(a), cari
(i) the value of h and of k,
nilai h dan nilai k,
(ii) the correct value of y if one of the values of y has
been wrongly recorded during the experiment.
nilai y yang betul jika satu daripada nilai-nilai y telah
tersalah catat semasa eksperimen.
[6]
Ans: (a) Refer to Answer Section
(b) (i) h = 7.24, k = –1.49; (ii) 6.14
4. The table shows the values of two variables, x and
y, obtained from an experiment. A straight line will be
y2
1
obtained when a graph of
against 2 is plotted.
x
x
630
Jadual berikut menunjukkan nilai-nilai bagi dua pemboleh ubah,
x dan y, yang diperoleh daripada suatu eksperimen. Satu garis
2
1
lurus akan diperoleh apabila graf y melawan 2 diplotkan.
x
x
x
1.25
1.43
2.00
2.50
3.34
5.00
y
4.47
4.21
3.68
3.46
3.27
3.16
(a) Based on the above table, construct a table for the
1
y2
values of 2 and .
x
x
Berdasarkan jadual di atas, bina satu jadual bagi nilai-nilai
1 dan y2 .
x2
x
[2]
y2
1
(b) Plot
against 2 , using a scale of 2 cm to 0.1 unit on
x
x
1
y2
the 2 -axis and 2 cm to 2 units on the
-axis. Hence,
x
x
draw the line of best fit.
2
Plot y melawan 12 , dengan menggunakan skala 2 cm
x
x
kepada 0.1 unit pada paksi- 12 dan 2 cm kepada 2 unit
x
2
pada paksi- y . Seterusnya, lukis garis lurus penyuaian
x
terbaik.
[3]
(c) Using the graph in 4(b),
Menggunakan graf di 4(b),
(i) find the value of y when x = 1.82,
cari nilai y apabila x = 1.82,
(ii) express y in terms of x.
ungkapkan y dalam sebutan x.
[5]
Ans: (a), (b) Refer to Answer Section
(c) (i) 3.82; (ii) y = 23.4 + x
x
5. The table shows the values of the variables v and p which
a
b
are related by the equation p = 2 + , where a and b are
v
v
constants.
Jadual di bawah menunjukkan nilai-nilai bagi pemboleh ubah v
dan p yang dihubungkan oleh persamaan p = a2 + b , dengan
v
v
keadaan a dan b ialah pemalar.
v
2
4
6
8
p
6.21
2.85
1.83
1.12
(a) Plot v2p on the y-axis against v on the x-axis, using a
scale of 2 cm to 2 units on the x-axis and 2 cm to 10
units on the y-axis. Hence, draw the line of best fit.
Plot v 2p pada paksi-y melawan v pada paksi-x, dengan
menggunakan skala 2 cm kepada 2 unit pada paksi-x dan
2 cm kepada 10 unit pada paksi-y. Seterusnya, lukis satu
garis lurus penyuaian terbaik.
[4]
(b) Based on the graph in 5(a), determine which point is
not correctly recorded. Give the appropriate value of p.
Berdasarkan graf di 5(a), tentukan titik manakah yang
salah direkodkan. Berikan nilai yang sesuai bagi p.
[2]
(c) Use your graph to estimate the value of a and of b.
Gunakan graf anda untuk menganggarkan nilai a dan
nilai b.
[2]
(d) In another method of finding the values of a and b
1
from a straight line graph, v is plotted on the x-axis.
In this case without drawing a second graph,
Dalam satu cara lain untuk menentukan nilai a dan nilai b
1
daripada graf garis lurus, diplotkan pada paksi-x. Dalam
v
kes ini, tanpa melukiskan graf kedua,
(i) state the variable that should be plotted on the
y-axis,
nyatakan pemboleh ubah yang akan diplotkan pada
paksi-y,
(ii) explain how the value of a and of b can thus be
obtained.
terangkan bagaimana nilai a dan nilai b boleh
diperolehi.
[2]
Ans: (a)
(b)
(c)
(d)
Refer to Answer Section
(8, 71.68), p = 1.34
a = 5, b = 10
(i) pv is plotted on the y-axis.
(ii) a will be the gradient and b is the
pv-intercept.
Form 4 Answers
37 5
SPM
L H
P
3 SH
= −5
−
+ (−5)
S
S
−
−
−
= −1
+H FH
= −6
6
= −6
= −4
= −2
3.6 = −2(1.5) +
−
6
= −2
6
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
F
6
)
= −1
W H
WH FH W
H W
S
= −1.32
R
R
R
R
R
R
R
xy
R
12
R
10
R
R
8
6
3 SH
4
2
0
1
y
5
10
F
3.5
)
3.0
W H
H W
2.5
2.0
WH FH W
1.5
1.0
0.5
0
0.1
0.2
© Penerbitan Pelangi Sdn. Bhd.
0.3
0.4
0.5
0.6
1
x
15
20
25
30
35
x2
Form 4 Answers
xy
40
y2
x
35
16
30
14
25
12
21.5
20
10
15
8
10
6
5
0
1
2
3
4
5
6
4
x
2
√
0
√
)
√
W H
F
H W
0.1
0.2
0.3
0.4
0.5
0.6
1
x2
: H
W H
×
√
)
H
√
WH FH W
= −4
= −1.49
W H
)
W H
H W
HF
H
WH FH W
√
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
Y
Y
H W
WH FH W
Y
OR
v2 p
OR
86
80
log10 R
70
2.5
2.35
60
2.0
50
1.5
40
1.0
30
0.5
20
0
10
F
0
)
2
W H
: H Y
Y
WF
4
HFW
6
HF
8
)
0.2
0.4
0.6
0.8
W H
v
H
Y
Y
F
Y
Y
)
Y
Y
Y
W H
Y
Y
Y
Y
W H
)
W H
WH FH W
H W
Y
)
WWH
W H
HW H
H W
W H Y WH FH W
© Penerbitan Pelangi Sdn. Bhd.
Y
Y
Y
0.94
1.0
1.2
1.4
log10 v
Form 4 Answers
276
OOH
H
7 H HT W
)
y
x
W H
H
W H W W
H W
y
x =x
200
WH FH W
150
100
50
0
50
120.5
100
150
200
250
x
7 H HFW
W
H
H
HF
W W
H
T
H
H
H W
© Penerbitan Pelangi Sdn. Bhd.
G
Additional Mathematics
3. Find the coordinates of point P which divides the straight line AB in the ratio AP : PB.
Cari koordinat titik P yang membahagikan garis lurus AB mengikut nisbah AP: PB.
(a) A(5, 1), B(–1, 10); AP : PB = 2 : 1
A(–3, 2), B(7, 17); AP : PB = 2 : 3
P = 2(7) + 3(–3) , 2(17) + 3(2)
2+3
2+3
= (1, 8)
B(7, 17)
P = 2(–1) + 1(5) , 2(10) + 1(1)
2+1
2+1
= (1, 7)
B(–1, 10)
1
3
P
P
2
2
A(–3, 2)
A(5, 1)
(b) A(–2, –1), B(4, 7); AP : PB = 1: 3
(c) A(–1, 5), B(1, 8); AP : PB = 3 : 2
P = 1(4) + 3(–2) , 1(7) + 3(–1)
1+3
1+3
1
= – ,1
2
B(4, 7)
P = 3(1) + 2(–1) , 3(8) + 2(5)
3+2
3+2
1
34
=
,
5 5
B(1, 8)
2
3
P
3
1
P
A(–1, 5)
A(–2, –1)
4. For each of the following, point Q divides the straight line AB according to the ratio AQ : QB. Find the value of
p and of q.
Bagi setiap yang berikut, titik Q membahagi garis lurus AB mengikut nisbah AQ : QB. Cari nilai p dan nilai q.
(a) A(p, 3), Q(4, 1), B(4, q); AQ : QB = 1 : 3
A(10, p), Q(2p, –3), B(4, q); AQ : QB = 2 : 1
1(4) + 3(p) , 1(q) + 3(3) = (4, 1)
1+3
1+3
4 + 3p = 4,
q+9 =1
4
4
4 + 3p = 16
q+9=4
p =4
q = –5
2(4) + 1(10) , 2(q) + 1(p) = (2p, –3)
2+1
2+1
2(4) + 1(10) = 2p,
2(q) + 1(p) = –3
3
3
18 = 6p
2q + p = –9
p =3
2q + 3 = –9
q = –6
A(10, p)
2
1 Q(2p, –3)
B(4, q)
(b) A(7, 2q), Q(1, –2), B(q, p); AQ : QB = 3 : 2
3(q) + 2(7) , 3(p) + 2(2q) = (1, –2)
3+2
3+2
3q + 14 = 1,
3p + 4q = –2
5
5
3q + 14 = 5
3p + 4q = –10
q = –3
3p + 4(–3) = –10
p= 2
3
(c) A(q, p), Q(–6, 2p), B(–10, 3q); AQ : QB = 5 : 2
5(–10) + 2(q) , 5(3q) + 2(p) = (–6, 2p)
5+2
5+2
–50 + 2q
15q + 2p = 2p
= –6,
7
7
–50 + 2q = –42
15q + 2p = 14p
q =4
15(4) = 12p
p=5
G
Additional Mathematics
5. Solve each of the following.
Selesaikan setiap yang berikut.
Point M(3, 6) divides a straight line joining point P(–5, 2) and point Q(5, 7) in the ratio of m : n. Find the
ratio m : n.
Titik M(3, 6) membahagikan garis lurus yang menyambungkan titik P(–5, 2) dan titik Q(5, 7) dengan nisbah m : n. Cari nisbah m : n.
n
5m – 5n
m+n
5m – 5n
5m – 5n
5m – 3m
2m
m
n
Q(5, 7)
M(3, 6)
m
=3
Equate the x-coordinate
= 3(m + n)
= 3m + 3n
= 3n + 5n
= 8n
= 8
2
= 4
1
\m:n =4:1
P(–5, 2)
m(5) + n(–5) , m(7) + n(2) = (3, 6)
m+n
m+n
5m – 5n , 7m + 2n = (3, 6)
m+n
m+n
(a) Point M(1, 5) divides a straight line joining point A(–5, 1) and point B(10, 11) in the ratio of m : n. Find
the ratio m : n.
Titik M(1, 5) membahagikan garis lurus yang menyambungkan titik A(–5, 1) dan titik B(10, 11) dengan nisbah
m : n. Cari nisbah m : n.
n
m
10m – 5n
m+n
10m – 5n
10m – m
9m
m
n
B(10, 11)
=1
=m+n
= n + 5n
= 6n
= 6
9
= 2
3
\m:n =2:3
M(1, 5)
A(–5, 1)
m(10) + n(–5) , m(11) + n(1) = (1, 5)
m+n
m+n
10m – 5n , 11m + n = (1, 5)
m+n
m+n
(b) Point A(k, 6) divides a straight line joining point P(–4, –6) and point Q(–9, 9) in the ratio of m : n, find
Titik A(k, 6) membahagikan garis lurus yang menyambungkan titik P(–4, –6) dan titik Q(–9, 9) dengan nisbah m : n, cari
(i) m : n,
(ii) the value of k. / nilai bagi k.
Q(–9, 9) n
A(k, 6)
(i)
9m – 6n = 6
m+n
9m – 6n = 6m + 6n
9m – 6m = 6n + 6n
3m = 12n
m = 12
n
3
4
=
1
\m:n=4:1
(ii)
4(–9) + 1(–4) = k
4+1
–40 = k
5
k = –8
m
P(–4, –6)
m(–9) + n(–4) , m(9) + n(–6) = (k, 6)
m+n
m+n
–9m – 4n , 9m – 6n
m+n
m+n
= (k, 6)
G
Additional Mathematics
7. Determine whether each of the following pairs of straight lines are parallel.
Tentukan sama ada setiap pasangan garis lurus yang berikut adalah selari atau tidak.
y = 2x + 5
m1 = 2
,
(a) y = 3x – 4
m1 = 3
y – 2x = 8
y = 2x + 8
m2 = 2
2y = 10x – 6
y = 5x – 3
m2 = 5
3x + y = 2
y = –3x + 2
m2 = –3
\ m1 ≠ m2, the two lines are not parallel.
\ m1 = m2, the two lines are parallel.
(b) y – 5x = 3
,
y = 5x + 3
m1 = 5
,
(c) 3y = x – 4
y = 1x – 4
3
3
1
m1 =
3
\ m1 = m2, the two lines are parallel.
,
y = 1x + 5
3
m2 = 1
3
\ m1 = m2, the two lines are parallel.
8. Given that each of the following pairs of straight lines are parallel, find the value of k.
Diberi setiap pasangan garis lurus yang berikut adalah selari, cari nilai k.
3x + ky = 2, 4y + x – 8 = 0
4y + x – 8 = 0
4y = –x + 8
Write in
y = – 1x + 2
gradient form.
4
m2 = – 1
4
3x + ky = 2
ky = –3x + 2
y = – 3x + 2
k
k
m1 = – 3
k
(a) 2x + ky = 1, 2y + x = 12
2x + ky = 1
ky = –2x + 1
y = – 2x + 1
k
k
2
m1 = –
k
The two lines are parallel, hence
m1 = m2
– 3 =– 1
k
4
k = 12
(b) kx + 4y = 3, 8y – x = 6
2y + x = 12
2y = –x + 12
y = – 1x + 6
2
m2 = – 1
2
The two lines are parallel, hence
– 2 =– 1
k
2
k=4
kx + 4y = 3
4y = –kx + 3
y = – kx + 3
4
4
k
m1 = –
4
8y – x = 6
8y = x + 6
y = 1x + 3
8
4
1
m3 =
8
The two lines are parallel, hence
– k = 1
4
8
k=– 1
2
9. Solve each of the following.
Selesaikan setiap yang berikut.
The equation of a straight line AB is 4x + y = 6. Find the equation of the straight line
CD that is parallel to AB and passes through point P(-2, 5).
Persamaan garis lurus AB ialah 4x + y = 6. Cari persamaan garis lurus CD yang selari dengan AB dan
melalui titik P(–2, 5).
4x + y = 6
y = –4x + 6
Gradient = –4
Write in
gradient
form.
Equation of the straight line CD is
y – 5 = –4[x – (-2)]
y – 5 = –4x – 8
y = –4x – 3
y
D B
P(–2, 5)
C
4x + y = 6
O
A
x
G
Additional Mathematics
11. Given that each of the following pairs of straight lines are perpendicular, find the value of k.
Diberi setiap pasangan garis lurus yang berikut adalah berserenjang, cari nilai k.
4x + ky = 1, 2y + x = 8
4x + ky = 1
ky = –4x + 1
y = – 4x + 1
k
k
4
m1 = –
k
Both the lines are perpendicular, hence
m1m2 = –1
2y + x = 8
2y = –x + 8
Write in
y = – 1x + 4
gradient
2
form.
m2 = – 1
2
(a) kx + 5y = 3, 2y – 5x = 6
5y = –kx + 3
y= – kx+ 3
5
5
k
m1 = –
5
– 4
k
– 1 = –1
2
2 = –1
k
k = –2
(b) 3x + ky = 4, 4y + x – 8 = 0
2y = 5x + 6
y= 5x + 3
2
m2 = 5
2
Both the lines are perpendicular, hence
– k 5 = –1
5 2
k=2
ky = –3x + 4
y= – 3x + 4
k
k
3
m1 = –
k
4y = –x + 8
y= – 1x + 2
4
m2 = – 1
4
Both the lines are perpendicular, hence
– 3 – 1 = –1
k
4
k=– 3
4
12. Solve each of the following.
Selesaikan setiap yang berikut.
The equation of a straight road AB is x + 3y = 10. Find the equation of a straight
road that is perpendicular to AB and passes through point P(–3, 4).
Persamaan satu jalan lurus AB ialah x + 3y = 10. Cari persamaan satu jalan lurus yang berserenjang
dengan AB dan melalui titik P(–3, 4).
x + 3y = 10
y = – 1 x + 10
3
3
m1 = – 1
3
Both the roads are perpendicular, so
m1m2 = –1
– 1 m2 = –1
3
m2 = 3
y
A
P(–3, 4)
B
O
x
Equation of the road is
y – 4 = 3[x – (–3)]
y – 4 = 3x + 9
y = 3x + 13
(a) Akmal wants to draw a straight line that is perpendicular to a straight line 6x + 3y = 9 and passes through
(3, –5) using a dynamic software. What is the equation he should key in?
Akmal ingin melukis satu garis lurus yang berserenjang dengan satu garis lurus 6x + 3y = 9 dan melalui (3, –5) dengan
menggunakan perisian dinamik. Apakah persamaan yang patut ditaipkan?
6x + 3y = 9
3y = –6x + 9
y = –2x + 3
m1 = –2
m1m2 = –1
–2m2 = –1
m2 = 1
2
The equation is
y – (–5) = 1 (x – 3)
2
y+5= 1 x– 3
2
2
1
13
y= x–
2
2
G
Additional Mathematics
(b) Hidayah’s house is located at (–1, –4). The straight road from Hidayah’s house
to road AB is perpendicular to the road AB. It is given that the equation of the
road AB is 2y – 5x = 6. Find the equation of the straight road from Hidayah’s
house to road AB.
Rumah Hidayah terletak di (–1, –4). Jalan lurus dari rumah Hidayah ke jalan AB adalah
berserenjang dengan jalan AB. Diberi bahawa persamaan jalan AB ialah 2y – 5x = 6. Cari
persamaan jalan lurus dari rumah Hidayah ke jalan AB.
2y – 5x = 6
2y = 5x + 6
y = 5x + 3
2
m1 = 5
2
m1m2 = –1
5 m = –1
2 2
m2 = – 2
5
y
B
(–1, –4)
A
x
O
Equation of the road is
y – (–4) = – 2 [x – (–1)]
5
y+4=– 2 x– 2
5
5
2
22
y=– x–
5
5
13. Solve the following problems.
Selesaikan masalah berikut.
y
In the diagram, QR is a straight line with an equation of x + 2y – 6 = 0.
Dalam rajah di sebelah, QR ialah garis lurus yang mempunyai persamaan x + 2y – 6 = 0.
(i) Find the equation of the straight line PQ.
Cari persamaan garis lurus PQ.
(ii) Find the coordinates of point Q.
Cari koordinat titik Q.
(iii) The straight line PQ is extended to a point T such that PQ : QT = 2 : 1.
Find the coordinates of T.
Garis lurus PQ dipanjangkan ke satu titik T dengan keadaan PQ : QT = 2 : 1. Cari koodinat T.
(i)
x + 2y – 6 = 0
2y = –x + 6
y = – 1x + 3
2
mQR = – 1
2
– 1 (mPQ) = –1
2
mPQ = 2
Write in
gradient
form.
QR and PQ are
perpendicular to
each other.
y = – 1 (–2) + 3 = 4
2
\ Q(–2, 4)
1
T(h, k)
O
R
x
2k + 16 = 4
3
2k + 16 = 12
k = –2
: – 1 x + 3 = 2x + 8
2
– 5x = 5
2
x = –2
2
Q(–2, 4)
P(4, 16)
2(h) + 1(4) , 2(k) + 1(16) = (–2, 4)
2+1
2+1
2h + 4 = –2
3
2h + 4 = –6
h = –5
(ii) y = – 1 x + 3 ............
2
y = 2x + 8 ...............
into
Q
(iii) Let T(h, k),
The equation of PQ is
y – 16 = 2(x – 4)
y – 16 = 2x – 8
y = 2x + 8
Substitute
P(4, 16)
\ T(–5, –2)
Compare the x-coordinate
and y-coordinate
G
Additional Mathematics
14. Find the area of the triangles with the vertices given.
Cari luas segi tiga yang mempunyai bucu yang diberikan.
(a) P(–3, –2), Q(5, 0), R(4, 8)
A(3, 1), B(7, 6), C(4, 8)
Area of ∆ABC
3
7
4
3
= 1
6
8
1
2 1
1
= |(18 + 56 + 4) – (7 + 24 + 24)|
2
= 1 |78 – 55|
2
= 11.5 unit2
Area of ∆PQR
–3
5
= 1
0
2 –2
INFO
S
.
K R
RQ RK
L
(b) A(3, 2), B(5, –2), C(6, 0)
I
4
8
–3
–2
= 1 |(0 + 40 – 8) – (–10 + 0 – 24)|
2
= 1 |32 – (–34)|
2
= 33 unit2
I
Q
(c) P(–8, 6), Q(1, –3), R(5, 2)
Area of ∆ABC
3
5
6
3
= 1
2
–2
0
2
2
= 1 |(–6 + 0 + 12) – (10 – 12 + 0)|
2
= 1 |6 – (–2)|
2
= 4 unit2
Area of ∆PQR
–8
1
5
–8
= 1
6
–3
2
6
2
= 1 |(24 + 2 + 30) – (6 – 15 – 16)|
2
= 1 |56 – (–25)|
2
= 40.5 unit2
15. Find the area of each of the following quadrilaterals PQRS.
Cari luas bagi setiap sisi empat PQRS yang berikut.
(a) P(0, 3), Q(1, 1), R(5, 8), S(3, 10)
P(2, 3), Q(9, 8), R(–2, 11), S(–5, 4)
Area of quadrilateral PQRS
2
9
–2
–5
2
= 1
8
11
4
3
2 3
= 1 |(16 + 99 – 8 – 15) – (27 – 16 – 55 + 8)|
2
= 1 |92 – (–36)|
I
2
S
I
.
K R
Q
= 64 unit2
INFO
RQ RK
L
(b) P(8, 0), Q(5, 7), R(0, –2), S(4, –3)
Area of quadrilateral PQRS
8
5
0
4
= 1
7
–2
–3
2 0
Area of quadrilateral PQRS
0
1
5
3
0
= 1
1
8
10
3
2 3
= 1 |(0 + 8 + 50 + 9) – (3 + 5 + 24 + 0)|
2
= 1 |67 – 32|
2
= 17.5 unit2
8
0
= 1 |(56 – 10 + 0 + 0) – (0 + 0 – 8 – 24)|
2
= 1 |46 – (–32)|
2
= 39 unit2
(c) P(–2, 3), Q(4, –1), R(7, 2), S(2, 9)
Area of quadrilateral PQRS
–4
7
2
–2
= 1 –2
2 –3
–1
2
9
–3
= 1 |(2 + 8 + 63 + 6) – (12 – 7 + 4 – 18)|
2
= 1 |79 – (–9)|
2
= 44 unit2
G
Additional Mathematics
16. Find the area of each of the following polygons.
Cari luas bagi setiap poligon yang berikut.
(a) F(0, 3), G(1, 1), H(6, 3), K(5, 8), L(4, 8)
A(2, 3), B(–3, –2), C(5, 0), D(8, 5), E(6, 9)
Area of pentagon ABCDE
–3
5
8
6
2
= 1 2
2 3
–2
0
5
9
3
1
= |(–4 + 0 + 25 + 72 + 18) – (–9 – 10 + 0 + 30 + 18)|
2
I
= 1 |111 – 29)|
S
I
2
.
K R
Q
= 41 unit2
RQ RK
L
Area of pentagon FGHKL
0
1
6
5
4
0
= 1
1
3
8
8
3
2 3
1
= | (0 + 3 + 48 + 40 + 12) – (3 + 6 + 15 + 32
2 + 0)|
= 1 |103 – 56|
2
= 23.5 unit2
INFO
(b) M(–1, 4), N(–4, –3), P(2, 1), Q(11, 7), R(6, 8), (c) T(–6, 5), U(–1, –2), V(4, –2), W(5, 3), X(8, 4),
S(3, 10)
Y(7, 10), Z(2, 9)
Area of hexagon MNPQRS
–1
–4
2
11
6
3
–1
= 1
–3
1
7
8
10
4
2 4
1
= |(3 – 4 + 14 + 88 + 60 + 12) – (-16 – 6 + 11
2 + 42 + 24 – 10)|
Area of heptagon TUVWXYZ
–6
–1
4
5
8
7
2
–6
= 1
–2
–2
3
4
10
9
5
2 5
1
= |(12 + 2 + 12 + 20 + 80 + 63 + 10) – (–5 – 8
2 – 10 + 24 + 28 + 20 – 54)|
= 1 |173 – (45)|
2
= 64 unit2
= 1 |199 – (–5)|
2
= 102 unit2
17. Determine whether the points A, B and C are collinear.
Tentukan sama ada titik A, B dan C adalah segaris.
(a) A(8, 1), B(–4, –8), C(6, –0.5)
A(6, 4), B(0, 1), C(–4, –1)
Area of DABC
6
0
–4
6
= 1
1
–1
4
2 4
1
= |(6 + 0 – 16) – (0 – 4 – 6)|
2
= 1 |–10 – (–10)|
2
=0
Collinear: lying in the
same straight line
Segaris: terletak di atas
garis lurus yang sama.
Therefore, points A, B and C are collinear.
(b) A(3, 1), B(–12, 14), C(6, –4)
Area of ∆ABC
8
–4
6
8
= 1
–8
–0.5
1
2 1
1
= |(–64 + 2 + 6) – (–4 – 48 – 4)|
2
= 1 |–56 – (–56)|
2
=0
Therefore, points A, B and C are collinear.
(c) A(1, –3), B(4, 3), C(–2, –9)
Area of ∆ABC
–12
6
3
= 1 3
2 1
14
–4
1
= 1 |(42 + 48 + 6) – (–12 + 84 – 12)|
2
= 1 |96 – 60|
2
= 18 unit2
Area of ∆ABC
4
–2
1
= 1 1
2 –3
3
–9
–3
= 1 |(3 – 36 + 6) – (–12 – 6 – 9)|
2
= 1 |–27 – (–27)|
2
=0
Therefore, points A, B and C are not collinear.
Therefore, points A, B and C are collinear.
G
Additional Mathematics
18. Solve each of the following.
Selesaikan setiap yang berikut.
(a)
y
y
P(3, 1)
x
O
B(4, 7)
N
A(–6, 1)
O
x
In the diagram, AB is a straight line.
Dalam rajah di atas, AB ialah garis lurus.
(i) Find the equation of a perpendicular bisector
of AB.
Cari persamaan pembahagi dua sama serenjang AB.
(ii) The perpendicular bisector cuts the x-axis at
point C. Find the coordinates of point C.
Pembahagi dua sama serenjang tersebut menyilang paksi-x
pada titik C. Cari koordinat titik C.
(iii) Find the area of ∆ABC.
Cari luas ∆ABC.
(i)
7–1
4 – (–6)
= 3
5
m2 = – 5
3
mAB =
\ C 7, 0
5
1
Q(0, –5)
The equation of the perpendicular bisector is
y – 4 = – 5 [x – (–1)]
3
y – 4 = – 5x – 5
3
3
5
y=– x+ 7
3
3
,
The diagram shows a straight line PQ. Given N is
a point such that PN = 2QN. Find
Rajah di atas menunjukkan garis lurus PQ. Diberi N ialah
satu titik dengan keadaan PN = 2QN. Cari
(i) the coordinates of N,
koordinat titik N,
(ii) the equation of the straight line which is
perpendicular to PQ and passing through N,
persamaan garis lurus yang berserenjang kepada PQ
dan melalui titik N,
(iii) the area of ∆PON. Hence or otherwise, find
the shortest distance from N to the line OP.
luas ∆PON. Seterusnya atau dengan cara lain, cari
jarak terdekat dari N ke garis OP. 276 DO L
(i)
Midpoint of AB = –6 + 4 , 1 + 7
2
2
= (–1, 4)
(ii) y = 0
Q(0, –5)
0 = – 5x + 7
3
3
0 = –5x + 7
x= 7
5
2
P(3, 1)
N(x, y)
N = 2(0) + 1(3) , 2(–5) + 1(1)
3
3
= (1, –3)
(ii) mPQ = 1 – (–5) = 2
3–0
m2 = – 1
2
y – (–3) = – 1 (x – 1)
2
y = – 1x – 5
2
2
(iii) Area of ∆PON
0
1
= 1
–3
2 0
3
1
0
0
= 1 |(1) – (–9) |
2
= 5 unit2
OP = √32 + 12 = √10
(iii) Area of ∆ABC
7
–6
–6
4
1
5
=
2 1
7
0
1
1
7
49
=
–42 + 0 +
– 4+
+0
2
5
5
= 136 unit2
5
Let h = shortest distance from N to line OP
1 h(OP) = 5
2
1 h(√10) = 5
2
h = 10 = 3.16 units
√10
G
Additional Mathematics
21. Solve each of the following.
Selesaikan setiap yang berikut.
Given that A(7, 4) and B(1, -2) are two fixed points and P(x, y) is a moving point such that AP : PB = 3 : 1,
Diberi bahawa A(7, 4) dan B(1, –2) ialah dua titik tetap dan P(x, y) ialah suatu titik yang bergerak dengan keadaan AP : PB = 3 : 1,
(i) find the equation of the locus of P, / cari persamaan lokus bagi titik P,
(ii) determine whether the locus of P intersects the x-axis. / tentukan sama ada lokus P menyilang paksi-x.
(i)
AP
PB
AP
(x – 7)2 + (y – 4)2
x2 – 14x + 49 + y2 – 8y + 16
x2 – 14x + 49 + y2 – 8y + 16
8x2 + 8y2 – 4x + 44y – 20
2x2 + 2y2 – x + 11y – 5
= 3
(ii) intersects at x-axis ⇒ y = 0
1
2x2 + 2(0)2 – x + 11(0) – 5 = 0
= 3PB
2x2 – x – 5 = 0
= 9[(x – 1)2 + [y – (–2)]2]
b2 – 4ac = (–1)2 – 4(2)(–5)
= 9(x2 – 2x + 1 + y2 + 4y + 4)
= 41 0 (two different roots)
= 9x2 – 18x + 9 + 9y2 + 36y + 36
Therefore, the locus of P intersects the
=0
x-axis.
=0
(a) Given A(2, 4) and B(5, 1) are two fixed points and P(x, y) is a moving point such that AP : PB = 2 : 3,
Diberi A(2, 4) dan B(5, 1) ialah dua titik tetap dan P(x, y) ialah suatu titik yang bergerak dengan keadaan AP : PB = 2 : 3,
(i) find the equation of the locus of point P,
cari persamaan lokus bagi titik P,
(ii) determine whether the locus of point P intersects the x-axis.
tentukan sama ada lokus titik P menyilang paksi-x.
(i)
3AP = 2PB
9[(x – 2) + (y – 4)2] = 4[(x – 5)2 + (y – 1)2]
9(x2 – 4x + 4 + y2 – 8y + 16) = 4(x2 – 10x + 25 + y2 – 2y + 1)
2
9x – 36x + 36 + 9y2 – 72y + 144 = 4x2 – 40x + 100 + 4y2 – 8y + 4
5x2 + 5y2 + 4x – 64y + 76 = 0
2
(ii) intersects at x-axis ⇒ y = 0
5x2 + 5(0)2 + 4x – 64(0) + 76 = 0
5x2 + 4x + 76 = 0
b2 – 4ac = (4)2 – 4(5)(76)
= –1 504 0
There are no real roots. Therefore, the locus of point P does not intersect the x-axis.
(b) A cow moves at the end of a tight rope around the pole A as shown in the
diagram. The cow passes through B(0, –2) and C(0, p). Find
Seekor lembu bergerak di hujung tali yang tegang mengelilingi tiang A seperti dalam rajah di
sebelah. Lembu itu melalui B(0, –2) dan C(0, p). Cari
(i) the equation of the path of the cow,
persamaan laluan lembu itu,
(ii) the value of p.
nilai p.
(i)
√(x – 3)2 + (y – 2)2 = √(0 – 3)2 + (–2 – 2)2
x2 – 6x + 9 + y2 – 4y + 4 = 25
x2 + y2 – 6x – 4y – 12 = 0
C(0, p)
A(3, 2)
B(0, –2)
(ii) x2 + y2 – 6x – 4y – 12 = 0 ……………
Substitute x = 0, y = p into equation
p2 – 4p – 12 = 0
(p – 6)(p + 2) = 0
p = 6 or p = –2
Hence, p = 6
:
G
Additional Mathematics
SPM
Paper
1
P A
4. A straight line passes through A(5, 5) and B(–9, –2).
The point C divides the line segment AB such that
4AC = 3AB. Find the coordinates of C.
630
1. The equation of a straight line is ax + by – 9 = 0, where a
and b are constants. Find in terms of a and b,
Satu garis lurus melalui A(5, 5) dan B(–9, –2). Titik
C membahagi tembereng garis AB dengan keadaan
4AC = 3AB. Cari koordinat C.
[3]
11
1
Ans: C –
,–
2
4
630
Persamaan suatu garis lurus ialah ax + by – 9 = 0. Cari dalam
sebutan a dan b,
(a) the gradient of the straight line,
kecerunan garis lurus itu,
(b) the gradient of the straight line which is perpendicular
to the line ax + by – 9 = 0.
kecerunan garis lurus yang berserenjang dengan garis
ax + by – 9 = 0.
[2]
a
b
Ans: (a) –
(b)
b
a
5. The diagram shows the position of three buildings P, Q and
R at the side of a main road drawn on a Cartesian plane,
such that P and Q lie on the same side of the straight main
road.
630
Rajah di bawah menunjukkan kedudukan tiga buah bangunan
P, Q dan R di tepi jalan raya utama yang dilukis pada suatu
satah Cartes, dengan keadaan P dan Q terletak di tepi jalan raya
utama lurus yang sama.
2. The diagram shows two straight lines drawn on a Cartesian
plane.
630
Rajah di bawah menunjukkan dua garis lurus yang dilukis pada
suatu satah Cartes.
y (m)
y
R(2, 7)
y = 2x – 7
A
P(–5, 2)
y = hx + 8
x
O
x (m)
O
Nadiah wants to cross the main road from building R to the
opposite side of the road where the buildings P and Q are
located. Find the shortest distance, in m, that she can take
to cross the main road. Give your answer correct to four
decimal places.
Both the straight lines are intersecting at point A and
perpendicular to each other.
Kedua-dua garis lurus itu bersilang di titik A dan berserenjang
antara satu sama lain.
(a) State the value of h.
Nyatakan nilai h.
(b) Find the coordinates of A.
Cari koordinat A.
[3]
1
Ans: (a) –
(b) A(6, 5)
2
3. The following information refers to the equation of two
straight lines, PQ and RS.
630
Maklumat berikut adalah merujuk kepada persamaan dua garis
lurus, PQ dan RS.
PQ : 5 + 6py – 3x = 0
y
6x
RS : –
=1
2
5k
Q(6, –1)
Nadiah hendak melintas jalan raya utama itu dari bangunan R
ke seberang jalan raya yang bertentangan di mana terletaknya
bangunan P dan Q. Cari jarak terdekat, dalam m, yang dia boleh
lalui untuk melintas jalan raya utama itu. Beri jawapan anda
betul kepada empat tempat perpuluhan.
[4]
Ans: 6.6656 m
Paper
2
1. In the diagram, PRS is a triangle. Side PR intersects the
y-axis at point Q.
630
Dalam rajah di bawah, PRS ialah sebuah segi tiga. Sisi PR
bersilang dengan paksi-y pada titik Q.
where k and p are constants.
y
dengan keadaan k dan p ialah pemalar.
It is given that the straight lines PQ and RS are
perpendicular to each other. Express p in terms of k.
Diberi bahawa garis lurus PQ dan RS adalah berserenjang
antara satu sama lain. Ungkapkan p dalam sebutan k.
[3]
6
Ans: p = –
5k
P
Q(0, 4)
S(–12, 4)
R(2, 3)
O
x
G
Additional Mathematics
(a) Given PQ : QR = 3 : 1, find
PQ is a straight road such that the distance from mosque A
and mosque B to any point on the road is always equal.
Diberi PQ : QR = 3 : 1, cari
(i) the coordinates of P,
koordinat P,
(ii) the equation of the straight line PS,
persamaan garis lurus PS,
(iii) the area, in unit2, of triangle PRS.
luas, dalam unit2, segi tiga PRS.
[7]
(b) Point T moves such that its distance from point R
is always thrice its distance from point S. Find the
equation of the locus T.
Titik T bergerak dengan keadaan jaraknya dari titik
R adalah sentiasa tiga kali jaraknya dari titik S. Cari
persamaan lokus T.
[3]
10
44
2
Ans: (a) (i) P(–6, 7); (ii) y = – x – ; (iii) 24 unit
19
19
(b) 8x2 + 8y2 + 220x – 66y + 1 427 = 0
2. The diagram shows a triangle OPQ.
630
Rajah di bawah menunjukkan segi tiga OPQ.
y
P(h, 8)
4. In the diagram, A(2, 7), B(6, 4) and C(5, 2) are the midpoints
of the straight lines KL, LP and PK respectively such that
KABC forms a parallelogram.
x
O
Q(–4, –4)
(a) It is given that the area of the triangle OPQ is
12 unit2. Calculate the value of h.
Diberi bahawa luas segi tiga OPQ ialah 12 unit2. Hitung
nilai h.
[2]
(c) Point A(1, 6) lies on the straight line PQ.
Titik A(1, 6) terletak pada garis lurus PQ.
(i) Find PA : AQ.
Cari PA : AQ.
(ii) Point R moves such that RA = 2RQ. Find the
equation of the locus R.
Titik R bergerak dengan keadaan
= 2RQ. Cari
persamaan lokus R.
[4]
Ans: (a) h = 2
(b) (i) 1 : 5
(ii) 3x2 + 3y2 + 34x + 44y + 91 = 0
3. The diagram shows the locations of mosque A and mosque
B drawn on a Cartesian plane.
630
Rajah di bawah menunjukkan kedudukan bagi masjid A dan
masjid B yang dilukis pada suatu satah Cartes.
Q
y
Mosque A
Masjid A
(–3, 8)
O
P
x
Mosque B
Masjid B
(9, –2)
PQ ialah jalan raya lurus dengan keadaan jarak dari masjid
A dan masjid B ke mana-mana titik pada jalan raya tersebut
adalah sentiasa sama.
(a) Find the equation of PQ.
Cari persamaan bagi PQ.
[3]
(b) Another straight road, ST with an equation of
y = –2x + 9 is to be built.
Satu lagi jalan lurus, ST dengan persamaan y = –2x + 9
akan dibina.
(i) A signboard is to be built at the crossroad of the
two roads. Find the coordinates of the signboard.
Papan tanda akan dibina di persimpangan kedua-dua
jalan raya itu. Cari koordinat bagi papan tanda itu.
(ii) Which of the two roads passes through mosque
3
C 1,
?
5
Antara dua jalan raya itu, yang manakah melalui
masjid C 1, 3 ?
5
[4]
Ans: (a) 6x – 5y – 3 = 0
(b) (i) (3, 3) ; (ii) PQ
Dalam rajah di bawah, A(2, 7), B(6, 4) dan C(5, 2) masingmasing ialah titik tengah bagi garis lurus KL, LP dan PK dengan
keadaan BC membentuk sebuah segi empat selari.
y
L
A(2, 7)
K
B(6, 4)
C(5, 2)
O
x
P
(a) Find/ Cari
(i) the equation of the straight line KL,
persamaan garis lurus KL,
(ii) the equation of the perpendicular bisector of
line PK.
persamaan pembahagi dua sama serenjang garis PK.
[6]
(b) The straight line KL is extended and cuts the
perpendicular bisector of straight line PK at point S.
Find the coordinates of point S.
Garis lurus KL dipanjangkan sehingga bersilang dengan
pembahagi dua sama serenjang garis lurus PK di titik S.
Cari koordinat titik S.
[2]
(c) Find the area of triangle ABC and of triangle KLP.
Cari luas segi tiga ABC dan segi tiga KLP.
[2]
Ans: (a) (i) y = 2x + 3 ; (ii) 3y = 4x – 14
23
(b) S –
, –20
2
11
(c) Area of ∆ABC =
unit2
2
Area of ∆KLP = 22 unit2
Form 4 Answers
37 5
SPM
RR G
H
HRPH
=3
= 3
P
3 SH
−9=0
= −D + 9
9
=– D
D
3(–9) + 1(5) , 3(–2) + 1(5) = ( ,
3+1
3+1
= –27 + 5
∴ Gradient = – D
– D
(b)
=–
P = −1
P =
=–
=–
D
–
∴
× K = −1
K=–
,–
= shortest distance from
P(–5, 2)
−7=–
S
= 15
Area of triangle
–5 6
=
–1 7
=6
= 2(6) − 7
=5
∴ (6, 5)
S
=
– 6 =1
–
=
S
(5 + 42 + 4) – (12 – 2 – 35)
=
51 – (–25)
= √ [6 – (–5)] + (–1 – 2)
=√
6S
= √ 130
+ (–3)
Area of triangle
√ 130
= 38
) = 38
) = 38
= 6.6656 m
6
3 SH
L
L
–
L
6
P(x, y)
3
Q(0, 4)
1
P = −1
P
–5
=
=1
P =–
=–
S
–
Q(6, –1)
= 38
5 + 6S − 3 = 0
6S = 3 − 5
P
to the other side of the road
R(2, 7)
=2 −7
=–
=
= –6 + 5
,
= −1
12 = −10S
S=–
10
=– 6
R(2, 3)
) + 3(2) ,
1+3
+6 =0
+6=0
= −6
∴ (−6, 7)
) + 3(3) = (0, 4)
1+3
+9 =4
,
+ 9 = 16
=7
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
LL
P
= 4 – (–6)
–12 – 7
= – 10
19
− 4 = – 10 − (−12)]
19
− 4 = − 10 – 120
19
19
10
=−
–
19
19
(iii) Area of ∆PRS
–6
–12
=
7
3
LL
−2
3
–6
7
(−18 + 8 − 84) − (14 − 36 − 24)
=
(−94) − (− 46)
L
=3
− 3) = 3√ − (–12)]
− 4)
− 3) = 9[(
− 4) ]
−4
− 6 + 9 = 9(
− 8 + 16)
−4
− 6 + 9 = 9 + 216 + 1 296
+ 9 − 72
+ 220 − 66 + 1 427 = 0
0
= 12
0
∴ (3, 3)
1, 3
= −2 + 9
= −2(1) + 9
=7
6 –5 –3=0
6(1) − 5 − 3 = 0
=3
= 3
= 24
= 24
=8
=2
Road
L
− (–2)]
Substitute = 3 into
= −2(3) + 9
=3
[–4K – (–32)] = +12
−4K + 32
32 − 4K
K
K
√ − 9)
( – 9)
– 18
0
0
= −2 + 9 ...................
6 − 5 − 3 = 0 .............
LL
= 12
=
=
=
=
=
Substitute
into
6 − 5(−2 + 9) − 3 = 0
6 + 10 − 45 − 3 = 0
16 = 48
=3
− 2)
− 2)
Area of ∆
0 K –4
0
–4
− 6) = 2√ − (–4)]
− (–4)]
− 6) = 4[(
]
− 12 + 36 = 4(
+ 16 +
+ 16)
− 2 – 12 + 37 = 4
+ 32 + 32
+ 34
+ 91 = 0
− (–3)]
− 8)
+ 3)
− 8)
+ 6 + 9 + − 16 + 64
– 20 – 12
6 –5 –3
= 24 unit
(a)
+3
− 1)
− 1)
√
=
√
√
m P(2, 8)
A(1, 6)
n
P(–4) + Q , P(–4) + Q
P Q
P Q
–4P
Q , –4P
Q
P Q
P Q
–4P
Q
P Q
–4P
Q
P
P
Q
= (1, 6)
= (1, 6)
=1
=P
=Q
=
Q
= 4–2 =2
6–5
Equation of
is
− 7 = 2( − 2)
−7=2 −4
=2 +3
L
P
=P
LL
P
=P
= 7–4
2–6
=– 3
Q(–4, –4)
1, 3 .
passes through mosque
T
T
P × P = −1
P × − 3 = −1
P =
3
Equation of perpendicular bisector of
–2=
– 5)
3
3 – 6 = 4 – 20
3 = 4 – 14
© Penerbitan Pelangi Sdn. Bhd.
is
Form 4 Answers
= 2 + 3 ..................................
3 = 4 − 14 ..............................
× 3: 3 = 6 + 9 ...................
−
:
L
2√
0 = 2 + 23
= −23
= – 23
3
= – 23 into
Substitute
+3
3
= −20
=
∴ The locus does not cut the -axis.
7
=
[(4 + 20 + 42) – (35 + 12 + 8)]
=
(66 – 55)
=
27
= 3
8–T = 3
–6 – S
16 – 2T = –18 – 3S
–2T + 3S = –34
=4×
(b) Area of ∆
0 –6
=
= 22 unit
−3 =6
= 0, = 6 ⇒ (0, 6)
= 0, 0 − 3 = 6
= −2 ⇒ (–2, 0)
0 + (–2) 6 + 0
=
,
= (−1, 3)
(ii) Area of quadrilateral
0 0 –1 –2
=
0
3
0
[(–6T
=
(–6T – 4S
(c) Area of rectangle
0
0
× 3:
−
=0
=3
−6T − 4S
−6T
−2T
−6T
104
+ 24
− 4S
+ 3S
+ 9S
= 2 × Area of ∆
= 2(−3T −2S
= 104
= 80 ...................................
= −34 .................................
= −102 ...............................
: 13S = −182
S = −14
Substitute S = −14 into
−2T + 3(−14) = −34
−2T = 8
T = −4
S(–1, 3)
1 U(0, 2)
∴ (−14, −4)
V(x, y)
3(–1) + 1( , 3(3) + 1(
3+1
3+1
–3 +
=0
,
S) – (–24 + 8S)]
= –3T – 2S
0 − (−2 − 6)
3
0
S
T
=
= 4 unit
−3 +
H
= 8–4 =–
–6 – 0
3
P
Area of ∆
= 4 × Area of ∆
=
OOH
P
unit
L
=0
– 4D = 6 – 4(3)(134)
= 36 – 1 608
= –1 572 0
6
7
= √ − 3)
− (–1)]
= ( – 3)
= –6 +9+
= –6 +9+
=0
+ 3(0) + 6 – 50(0) + 134 = 0
3 + 6 + 134 = 0
– 23 , –20
(c) Area of ∆
− 6)
− 6) ]
− 12 + 36)
− 48 + 144
+ 6 – 50 + 134
(ii) cuts the -axis ⇒
= 2 – 23 + 3
∴
− 0)
= (0, 2)
9+
=2
9+
=8
= −1
=P
P
= 3
=4
Equation of
3 –
∴ (3, −1)
is
= 3
= –4
3 –
=1
(–8)
(–4)
–
3
=1
3
© Penerbitan Pelangi Sdn. Bhd.
V
Additional Mathematics
2. Draw and label each of the following vectors.
Lukis dan tandakan setiap vektor berikut.
(a) a m s–2 represents acceleration making an angle
~
of 30° with the horizontal line.
a m s–2 mewakili pecutan pada 30° dengan garis mengufuk.
~
→
QR represents a force of 70 N due east.
→
QR mewakili daya 70 N ke arah timur.
70 N
a~ m s–2
Q
R
30°
→
v represents a velocity of 60 km h–1 due south.
(b) HK represents a displacement of 56 m due north- (c) ~
v mewakili halaju 60 km h–1 ke arah selatan.
~
east.
→
HK mewakili sesaran 56 m ke timur laut.
N
60 km h–1
K
56 m
H
3. (i)
(ii)
Determine the magnitude and direction of each vector in
the diagram.
Tentukan magnitud dan arah bagi setiap vektor pada rajah di sebelah.
Hence draw the corresponding negative vector and state
the magnitude and direction respectively.
Seterusnya lukis vektor negatif yang sepadan dan nyatakan magnitud
serta arah masing-masing.
N/U
e~
a~
(a)
(b)
b~
(d) M
(c)
N
(i)
B
(a) a
~
e
~
Magnitude = √22 + 22
= √8
= 2√2 units
Pythagoras' theorem
Teorem Pythagoras
c2 = a2 + b2
c
a
Magnitude
= 5 units
\ 5 units due east
b
\ 2√2 units due north-west
(b) b
~
(c)
Magnitude
= √22 + 12
= √5 units
→
AB
→
(d) MN
Magnitude
= 4 units
\ √5 units due south-east
\ 4 units due west
Magnitude
= √32 + 32
= √18
= 3√2 units
\ 3√2 units due south-east
(ii)
N/U
– ~e
(a)
– e : 2√2 units due south-east
~
–a
~
(a) – a
~
(b)
–b
~
(b) – b : √5 units due north-west
~
→
(c) – AB : 4 units due east
(d) M
(c)
N
: 5 units due west
B
A
→
(d) –MN : 3√2 units due north-west
A
Additional Mathematics
4. State the pairs of vectors which are equal.
Nyatakan pasangan vektor yang sama.
5. Express vectors below in terms of r .
~
Ungkapkan vektor berikut dalam sebutan ~r .
F
P
Q
(a)
~s
U
w
~
~r
Example
~r
V
~t
E
(b)
(c)
V
– 1r
2~
→
Answer/ Jawapan : UV and t
~
(a) – 3 r
2~
(b) 3 r
~
(c) 2 r
~
6. Prove that the following pairs of vectors are parallel. Then, determine whether the points involved are
collinear.
Buktikan bahawa pasangan vektor berikut adalah selari. Kemudian, tentukan sama ada titik-titik yang terlibat adalah segaris atau tidak
→
→
AB = 3k , CD = –1.2 k
~
~
→
→
AB = 3k ⇒ k = 1 AB
~
~ 3
→
→
1
CD = –1.2 AB
3
→
= – 2 AB
5
→
→
Hence, AB and CD are parallel but points A, B, C and
D are not collinear as there is no common point.
a and b
Vector ~
~ are parallel if and only if
a = kb
~
~ where k is a constant.
a dan ~
b adalah selari jika dan hanya
Vektor ~
jika ~
a = k~
b dengan keadaan k ialah pemalar.
→
→
→
→
(b) HI = –2n , IK = –4n
(a) EF = 3 m , FG = 5m
~
~
~
~
2
→
→
→
→
HI = –2n , ⇒ n = – 1 HI
EF = 3 m , ⇒ m = 2 EF
~
~
~
~
2
2
3
→
→
→
→
1
2
IK = –4 – HI
FG = 5 EF
2
3
→
→
10
EF
=
= 2 HI
3
→
→
→
→
Hence, HI and IK are parallel. Points H, I and K
Hence, EF and FG are parallel. Points E, F and
are collinear as I is the common point.
G are collinear as F is the common point.
→
→
→
→
(c) RS = –3 q , TU = –5 q
(d) VW = 25 p , XY = 8 p
~
~
~
~
→
→
→
→
1
RS = –3q , ⇒ q = – RS
VW = 25p , ⇒ p = 1 VW
~
~
~
~ 25
3
→
→
→
→
TU = –5 – 1 RS
XY = 8 1 VW
3
25
→
→
= 5 RS
= 8 VW
3
25
→
→
→
→
Hence, RS and TU are parallel but points R, S, T
Hence, VW and XY are parallel but points V, W,
and U are not collinear as there is no common
X and Y are not collinear as there is no common
point.
point.
V
Additional Mathematics
→
7. U, V and W are three collinear points. Point V divides UW in the ratio of p : q and UW = 6 r . Determine the
~
→
→
vectors UV and VW in terms of r when
~
→
→
→
U, V dan W ialah tiga titik segaris. Titik V membahagikan UW dengan nisbah p : q dan UW = 6 ~r . Tentukan vektor UV dan VW dalam
sebutan ~r apabila
U, V and W are collinear points means these points are lying on a single straight line. The vectors connecting these points are parallel.
U, V dan W ialah titik-titik yang segaris bermaksud titik-titik tersebut terletak pada satu garis lurus. Vektor-vektor yang menyambungkan titiktitik tersebut adalah selari.
(a) p = 4 and / dan q = 2
p = 2 and / dan q = 3
→
→
→
2
2 6r
UV =
VW
UV =
2+3
2+3 ~
= 12 ~r
5
→
VW =
3 6r
2+3 ~
= 18 ~r
5
→
VW =
3
2
V
→
VW = 2 6~r
6
= 2~r
U
→
3
UW
2+3
(b) p = 5 and / dan q = 4
2
4
W
V
U
(c) p = 1 and / dan q = 3
→
UV = 5 6~r
9
10
r
=
3~
→
VW = 4 6~r
9
8
= ~r
3
W
→
UV = 4 6~r
6
= 4~r
4
5
W
V
U
→
UV = 1 6~r
4
3
= ~r
2
→
VW = 3 6~r
4
9
= ~r
2
3
1
U
W
V
8. In the following equations, m and n are constants. Vectors a and b are not parallel and non-zero. Determine
~
~
the values of m and n in each of the following cases.
a dan ~
b tidak selari dan bukan sifar. Tentukan nilai m dan n dalam
Dalam persamaan berikut, m dan n ialah pemalar. Vektor-vektor ~
setiap kes berikut.
(3m – 2) a = (n + 5) b
~
~
3m – 2 = 0
3m = 2
m= 2
3
n+5 =0
n = –5
(b) (4 – 2m) a = (n + 1) b
~
~
4 – 2m = 0
2m = 4
m =2
n+1 =0
n = –1
(a) (m + 3) b = (2 – n) a
~
~
If ~
a and b
~ are not parallel
and non-zero, and ha
~ = kb
~,
then h = k = 0.
a dan ~
b tidak selari dan
Jika ~
a = k~
b,
bukan sifar, dan h ~
maka h = k = 0.
m+3=0
m = –3
2–n =0
n =2
(c) (5m + 6) a – (2n + 4) b = 0
~
~
(5m + 6) a = (2n + 4) b
~
~
5m + 6 = 0
5m = –6
m =– 6
5
2n + 4 = 0
2n = –4
n = –2
V
Additional Mathematics
10.
p
~
q
~
The diagram shows vectors p and q . Draw the resultant vectors for the following.
~
~
Rajah di atas menunjukkan vektor-vektor p dan q . Lukis vektor paduan bagi berikut.
~
~
(a) 2p – q
~ ~
p + 3q
~
~
(b) – 3 p + 2q
~
2~
3q
~
p +3q
~ ~
2q
~
2p–q
~ ~
p
~
(c) 2p + 2q
~
~
(d) –3p – 4q
~
~
2q
~
2 p + 2q
~
~
3
– _ p~ + 2 q~
2
–q
~
2p
~
3
– _ p~
2
(e) –2p + 3q
~
~
–3 p
~
–2 p + 3 q
~
~
2p
~
3q
~
–3 p – 4 q
~
~
–4 q
~
–2 p
~
11. Use the given vectors to draw the resultant vectors in the space provided using parallelogram law.
Gunakan vektor yang diberi untuk melukis vektor paduan di ruang yang disediakan dengan menggunakan hukum segi empat selari.
(a) c + 1 b
~ 3~
+
~r ~s
(b) 2a + d
~ ~
r~
c~
s~
b~
r~
r~ + s~
2 a~ + d~
1
c~ + – b~
3
1
– b~
3
s~
a~
d~
d~
c~
2 a~
12. The diagram shows an octagon ABCDEFGH. Find the resultant vector for each of the following.
Rajah di bawah menunjukkan sebuah oktagon ABCDEFGH. Cari vektor paduan bagi setiap yang berikut.
B
C
D
A
H
G
E
F
→
→
→
→
BC + CD + DE = BE
→
→
→
→
(a) DE + EF + FG + GH =
→
→
→
(b) DC + ED + FE =
→
FC
→
→
→
CD – ED + EA =
→
CA
(c)
→
→
→
→
(d) DE + EF – GF – AG =
→
DH
→
DA
V
Additional Mathematics
13. Solve the following problems.
Selesaikan masalah berikut.
Two forces F1 = 40 N and F2 = 20 N are acting on a moving object. Draw vector diagrams to represent the
following situations.
Suatu objek yang sedang bergerak dikenakan dua daya F1 = 40 N dan F2 = 20 N. Lukis gambar rajah vektor untuk mewakili situasi
berikut.
(i) F1 and F2 are in the same direction
F1 dan F2 dalam arah yang sama
(ii) F1 and F2 are in the opposite direction
F1 dan F2 dalam arah yang bertentangan
Hence, determine the magnitude of the force and the direction of the movement of the object.
Seterusnya, tentukan magnitud daya dan arah pergerakan objek tersebut.
(i)
(ii)
F1
F1
F2
F2
Magnitude = F1 + F1
= 40 + 20
= 60 N
Magnitude = F1 + (–F2)
= 40 + (–20)
= 20 N
The magnitude of the force acted on the object is
60 N in the same direction as F1 and F2.
The magnitude of the force acted on the object is
20 N in the same direction as F1.
(a) Displacement is a vector quantity that shows the (b)
change in position of an object. An aeroplane
flies 200 km north from airport A to airport B.
Then, it flies to airport C which is 400 km northwest of the airport B. Draw a vector diagram to
represent the path taken by the aeroplane and
the displacement of the aeroplane.
Sesaran ialah kuantiti vektor yang menunjukkan perubahan
kedudukan suatu objek. Sebuah kapal terbang bergerak
200 km ke utara dari lapangan terbang A ke lapangan
terbang B. Kemudian, kapal terbang itu bergerak ke
lapangan terbang C yang berada di 400 km barat laut
lapangan terbang B. Lukis satu gambar rajah vektor untuk
mewakili perjalanan kapal terbang itu dan sesaran kapal
terbang itu.
T
S
c~
P
R
U
a~
Q
V
b~
The diagram shows a cube PQRSTUVW.
According to the diagram, express in terms of a ,
~
b and c , vector
~
~
Rajah di atas menunjukkan sebuah kubus PQRSTUVW.
a,
Berdasarkan rajah tersebut, ungkapkan dalam sebutan ~
b
c
dan
,
vektor
~
~
→
(i) PV
→
(ii) PW
→
(iii) UQ
V
(i)
C
W
→
PV = a + b
~ ~
a~ + b~
P
b~
a~
Q
45°
B
Displacement
a~ + b~ + c~
→
(ii) PW = a + b + c
~ ~ ~
P
A
FP
S V
V
P
→
(iii) UQ = –b + a
~ ~
U
– b~ + a~
– b~
P
a~
Q
a~ + b~
W
c~
V
V
Additional Mathematics
14. The diagram shows some vectors drawn on the Cartesian plane.
Rajah di sebelah menunjukkan beberapa vektor dilukis pada satah Cartes.
(i) Express each of the following vectors in the form of
Ungkapkan setiap vektor berikut dalam bentuk
(a) x i + y j
~
~
x
(b)
y
(ii) Hence, find the magnitude of each of the vectors.
Seterusnya, cari magnitud bagi setiap vector tersebut.
q
~
(i)
(a) 2 i + j
~ ~
2
(b)
1
~r
q
~
y
~t
~s
w
~
v
~
x
O
u
~
(a) r
~
(b) s
~
(i)
(a) –3 i – 2 j
~
~
(b) –3
–2
(i)
(a) –3 i + 2 j
~
~
(b) –3
2
(ii)
r
~
= √(–3)2 + (–2)2
= √13 units
(ii)
s
~
= √(–3)2 + 22
= √13 units
(ii)
q = √22 + 12
~
= √5 units
√x2 + y2
(c)
t
~
(d) u
~
(e) v
~
(f)
w
~
(i)
(a) –4 i + 3 j
~
~
–4
(b)
3
(i)
(a) 3 i – j
~ ~
(b) 3
–1
(i)
(a) 3 j
~
(b) 0
3
(i)
(a) –3 i – 3 j
~
~
–3
(b)
–3
(ii)
t
~
= √(–4)2 + 32
= √25 units
= 5 units
(ii)
u
~
= √32 + (–1)2
= √10 units
(ii)
v
~
= √32
= 3 units
(ii)
w
~
= √(–3)2 + (–3)2
= √18
= 3√2
15. Determine unit vector for each of the following.
Tentukan vektor unit bagi setiap berikut.
(i)
(a) a = 12 i – 5 j
~
~
~
→
AC = 3 i + 4 j
~
~
Unit vector in the
→
direction of AC
→
= AC
→
AC
3i + 4j
= ~2 ~2
√3 + 4
= 3i + 4j
5~ 5~
12 i – 5 j
~ ~
~ = √122 + (–5)2
= 12 i – 5 j
13 ~ 13 ~
a^
x~i + y j
~
√x2 + y2
(ii) s = 5
~
–12
1
5
~ = √52 + (–12)2 –12
5
13
=
– 12
13
^
s
(c)
→
RS = 4
6
Unit vector in the
→
direction of RS
1
4
=
2
2
√4 + 6 6
2
13
√
4 =
= 1
3
2√13 6
√13
→
(b) PQ = –3 i – 3 j
~
~
Unit vector in the
→
direction of PQ
–3 i – 3 j
~ ~
=
√(–3)2 + (–3)2
–3 i – 3 j
~ ~
=
3√2
=– 1 i – 1 j
√2 ~ √2 ~
(d) r = –8
~
15
1
–8
^
~r = √(–8)2 + 152 15
– 8
17
=
15
17
V
Additional Mathematics
16. Given the vectors p = –8 i + 2 j , q = 3 i + 5 j , r = –2 i – 6 j , s = 3 , t = –10 and u = 7 , find in the form of
~
~
~
~
~
~
–6
–4
7
x i + y j and x for each of the following.
~ ~
y
3
–10
7
Diberi vektor-vektor p =–8~i + 2 j , q = 3~i + 5 j , r = –2~i – 6 j , s =
,t=
dan u =
, cari dalam bentuk x~i + y j dan
–6
–4
7
~
~
~
~
x
bagi setiap berikut.
y
p+q
= (–8 i + 2 j ) + (3 i + 5 j )
~
~
~
~
= –8 i + 3 i + 2 j + 5 j
~
~
~
~
= –5 i + 7 j
~
~
= –5
7
(b) 5s + 2t – u + p
= 5(3 i – 6 j ) + 2(–10 i – 4 j )
~
~
~
~
– (7 i + 7 j ) + (–8 i + 2 j )
~
~
~
~
= 15 i – 30 j – 20 i – 8 j
~
~
~
~
– 7i – 7j – 8i + 2j
~
~
~
~
= –20 i – 43 j
~
~
–20
=
–43
components in the vectors involved.
Tambah atau tolak komponen yang
sepadan dalam vektor-vektor yang
terlibat.
scalar involved.
Darab setiap komponen dengan skalar
yang terlibat.
(c) 4r + 2q + 2 s
3
= 4(–2 i – 6 j ) + 2(3 i + 5 j )
~
~
~
~
+ 2 (3 i – 6 j )
~
3 ~
= –8 i – 24 j + 6 i + 10 j
~
~
~
~
+ 2i – 4j
~
~
= –18 j
~
= 0
–18
(a) 2u – 3p + 4r
= 2(7 i + 7 j ) – 3(–8 i + 2 j )
~
~
~
~
+ 4(–2 i – 6 j )
~
~
= 14 i + 14 j + 24 i – 6 j
~
~
~
~
– 8 i – 24 j
~
~
= 30 i – 16 j
~
~
30
=
–16
(d) – 1 t – u – q
2
= – 1 (–10 i – 4 j ) – (7 i + 7 j )
~
~
~
~
2
– (3 i + 5 j )
~
~
= 5i + 2j – 7i – 7j – 3i – 5j
~
~
~
~
~
~
= –5 i – 10 j
~
~
= –5
–10
17. It is given that the position vectors of P, Q and R are i + j , 3 i – 2 j and 6 i + k j respectively.
~ ~ ~
~
~
~
Diberi bahawa vektor kedudukan P, Q dan R masing-masing ialah ~i + j , 3~i – 2 j dan 6~i + k j .
~
~
~
→
(a) Find the vector of PQ.
→
→
Find the vector of RQ in term of k.
Cari vektor PQ .
→
RQ
Cari vektor
dalam sebutan k.
→
→ →
PQ = PO + OQ
→ → →
→
→
= –( i + j ) + (3 i – 2 j )
– OR + OQ
RQ = RO + OQ
~ ~
~
~
= –i – j + 3i – 2j
= –(6 i + k j ) + (3 i – 2 j )
~ ~
~
~
~
~
~
~
= 2i – 3j
= –6 i – k j + 3 i – 2 j
~
~
~
~
~
~
= –3 i – (k + 2) j
~
~
→
(b) Find the vector of PR in term of k.
→
Cari vektor PR dalam sebutan k.
→ → →
PR = PO + OR
= –( i + j ) + (6 i + k j )
~ ~
~
~
= –i – j + 6i + kj
~ ~
~
~
= 5 i + (k – 1) j
~
~
→
→
(c) If PQ = µ PR, find the values of µ and k.
→
→
Jika PQ = µ PR , cari nilai µ dan k.
→
→
PQ = µ PR
2 i – 3 j = µ[5 i + (k – 1) j ]
~
~
~
~
By comparing the like terms,
2 = 5µ
–3 = (k – 1)µ
2
µ=
–3 = (k – 1) 2
5
5
– 15 = k –1
2
k = – 13
2
V
Additional Mathematics
18. Solve each of the following.
Selesaikan setiap yang berikut.
The current of a river is flowing parallel to its bank with a velocity of 1.25 km h–1. A swimmer is swimming
276 SSO
at 2.5 km h–1 perpendicularly to the river bank across the river. Calculate
–1
Arus sungai mengalir selari dengan tebing sungai dengan halaju 1.25 km h . Seorang perenang berenang secara berserenjang kepada
tebing sungai itu dengan halaju 2.5 km h–1. Hitung
(i) the magnitude of the resultant velocity of the swimmer in km h–1,
magnitud halaju paduan perenang itu dalam km h–1,
(ii) the time taken, in hour, if the width of the river is 200 m.
masa yang diambil, dalam jam, jika lebar sungai itu ialah 200 m.
1.25 km h–1
Assuming the direction of the water current is along the positive x-axis and the
direction of the swimmer is along the positive y-axis.
The magnitude of the water current = 1.25 km h–1
The magnitude of the swimmer = 2.5 km h–1
Let v km h–1 represents the resultant velocity of the swimmer.
v = 1.25~i + 2.5 j
~
~
(i) |v | = √1.252 + 2.52
~
= 2.795 km h–1
2.5 km h–1
v km h–1
(ii) Time taken = Displacement
Velocity
200 m = 0.2 km
= 0.2
2.795
= 0.07156 hour
(a) In this question, i and j are the unit vectors, in metre, due east and due north respectively. The initial
~
~
positions of particles P and Q relative to the origin O are 2 i + 4 j and 8 i + 12 j respectively. Particles P
~
~
~
~
dan Q move from their initial position at the same time with constant velocities in m s–1. Velocities of
particles P and Q are i + j and –2 i – 3 j respectively. Show that after 2 seconds, both the particles P and
~ ~
~
~
Q will meet.
276 SSO
Dalam soalan ini, ~i dan j masing-masing ialah vektor unit, dalam meter, ke timur dan ke utara. Kedudukan awal zarah P dan
~
zarah Q relatif kepada titik asalan O masing-masing ialah 2~i + 4 j dan 8~i + 12 j . Zarah P and zarah Q bergerak pada masa
~
~
yang sama dari kedudukan awal dengan halaju tetap dalam m s–1. Halaju zarah P dan halaju zarah Q masing-masing ialah
~i + ~j dan –2~i – 3~j . Tunjukkan selepas 2 saat, kedua-dua zarah P dan Q akan bertemu.
Initial position:
→
OP = 2
4
→
OQ = 8
12
After 2 seconds,
Velocities of P and Q:
→
OP t = 2 = 2 + 2 = 4
4
2
6
1
v =
~P
1
v = –2
~Q
–3
P: displacement = 2 1 =
1
Q: displacement = 2 –2 =
–3
2
2
–4
–6
displacement
= time taken × velocity
→
OQt = 2 = 8 + –4 = 4
12
–6
6
At t = 2 s, the position vectors of both the particles P and Q are the same. Hence, P meets Q after 2 s.
V
Additional Mathematics
→
(b) Given R, S and T are three collinear points. If RS : ST = 3 : 2 and RT = 10 ~r .
→
Diberi R, S dan T adalah tiga titik segaris. Jika RS : ST = 3 : 2 dan RT = 10~r .
→
→
(i) Find RS and ST in terms of ~r .
→
→
Cari RS dan ST dalam sebutan ~r .
→
→
(ii) If ~r = 4 units, find RS and ST .
→
→
Jika ~r = 4 unit, cari RS dan ST .
(i)
→
RS = 3 × 10~r
5
= 6~r
→
ST = 2 × 10~r
5
= 4~r
2
3
T
S
R
8 , = 2 and = –5 .
(c) It is given that u =
v
w
~
~
p+3 ~
1
4
8
2
–5
Diberi bahawa ~
u=
, v=
dan w
~= 4 .
p+3 ~
1
(i)
(ii)
→
RS = 6 ~r
=6×4
= 24 units
→
ST = 4 ~r
=4×4
= 16 units
276
SSO
If u is parallel with v , find the value of p.
~
~
Jika ~
u selari dengan ~
v , cari nilai p.
(ii) With the value of p obtained, determine
Dengan nilai p yang diperolehi, tentukan
(a) the value of h and of k such that hu – v – kw = 0.
~ ~
~
v
nilai h dan nilai k dengan keadaan hu
–
–
kw
=
0.
~ ~
~
(b) the vector in the same direction and parallel to 3v + w and has a magnitude of 3√2.
~ ~
vektor dalam arah yang sama dan selari dengan 3v
~+w
~ dan mempunyai magnitud 3√2.
(i)
u = l v where l is a constant
~
~
8
=l 2
p+3
1
= 2l
l
By comparing,
8 = 2l
p+3=l
p+3=4
l=4
p=1
(ii) (a) u = 8
~
4
h 8 – 2 – k –5 = 0
4
1
4
8h – 2 + 5k = 0
4h – 1 – 4k
8h – 2 + 5k = 0 ………
4h – 1 – 4k = 0 ………
× 2: 8h – 2 – 8k = 0 ………
– :
–13k = 0
k =0
Substitute k = 0 into ,
4h – 1 = 0
h = 1
4
(ii) (b) 3v + w = 3 2 + –5
~ ~
1
4
6
–5
=
+
3
4
1
=
7
3v + w = √12 + 72
~ ~
= √50
= 5√2
Unit vector in the direction of 3v + w
~ ~
1
1
=
5√2 7
The required vector is
3√2 ×
1
5√2
1 = 3 1
7
5 7
=
3
5
21
5
Additional Mathematics
V
(d) In the diagram, M and R are the midpoints of PQ and OM respectively. The lines OM and PN intersect at
→
→
R and ON = 1 OQ. If OP = p and OQ = q , 276 SSO
~
~
3
Dalam rajah di sebelah, M dan R masing-masing ialah titik tengah bagi PQ dan OM. Garis OM dan garis PN bersilang di R dan
→
→
1
ON = OQ. Jika OP = p dan OQ = q ,
P
3
~
~
(i) find, in terms of p and q , in its simplest form,
~
~
M
cari, dalam sebutan p dan q , dalam bentuk termudah,
~
~
→
→
→
R
(a) OM
(b) OR
(c) PR
O
Q
(ii) find the ratio PR : PN in its simplest form.
N
cari nisbah PR : PN dalam bentuk termudah.
→
→
→
→
→
→
(i) (a) OM = OQ + QM = q + 1 QP = q + 1 ( QO + OP ) = q + 1 (–q + p ) = 1 p + 1 q
~ 2
~ 2
~ 2 ~ ~
2~ 2~
→
→
(b) OR = 1 OM = 1 1 p + 1 q = 1 p + 1 q
2
2 2~ 2~
4~ 4~
→ → →
1
1
3
1
(c) PR = PO + OR = –p +
p+ q =– p+ q
~
4~ 4~
4~ 4~
→ →
→
→
(ii) PN = PO + ON = –p + 1 OQ = –p + 1 q
~ 3
~ 3~
3
1
PR : PN = – p + q : –p + 1 q = 1 –3p + q : 1 –3p + q = 1 : 1 = 3 : 4
~ 3~
~ ~ 3
~ ~
4~ 4~
4
4 3
(e) Points P and Q have the position vectors p and q respectively. A point R lies on the line OP such that
~
~
→
→
OR = 1 OP and a point S lies on line OQ such that OS = 1 OQ. Express QR and SP in terms of p and q .
~
~
2
3
O V
The lines QR and SP intersect at T. 276
Titik P dan titik Q mempunyai vektor kedudukan p dan q masing-masing. Suatu titik R berada pada garis OP dengan keadaan
~
~
→
→
1
1
OR = OP dan suatu titik S berada pada garis OQ dengan keadaan OS = OQ. Ungkapkan OQ dan SP dalam sebutan p dan
2
3
~
q . Garis QR dan garis SP bersilang di T.
~
→
→
→
(i) Given that PT = l SP, express OT in terms of l, p and q .
~
~
→
→
→
Diberi bahawa PT = l SP , ungkapkan OT dalam sebutan l, p dan q .
~
~
→
→
→
(ii) Given that QT = µQR, express OT in terms of µ, p and q .
~
~
→
→
→
Diberi bahawa QT = µ QR, ungkapkan OT dalam sebutan µ, p dan q .
~
~
Hence, calculate the value of l and of µ. With the values obtained, find the position vector of T.
Seterusnya, hitung nilai bagi l dan µ. Dengan nilai-nilai yang diperolehi, cari vektor kedudukan bagi T.
→
→
→
→
OP = p , OQ = q , OR = 1 p and OS = 1 q
~
~
2~
3~
→
→
→
QR = QO + OR = –q + 1 p = 1 p – q
~ 2~ 2~ ~
→
→
→
SP = SO + OP = – 1 q + p = p – 1 q
3~ ~ ~ 3~
→
→
→
(i) OT = OP + PT
→
= p + l SP
~
= p +l p – 1q
~
~ 3~
Q
= (1 + l)p – 1 l q
~ 3 ~
→
→
→
(ii) OT = OQ + QT
→
= q + µ QR
~
= q + µ( 1 p – q )
~
2~ ~
= 1 µp + (1 – µ)q
~
2 ~
S
O
Hence, (1 + l)p – 1 l q = 1 µp + (1 – µ)q
~ 3 ~ 2 ~
~
1
1 + l = µ ———————
2
– 1 l = 1 – µ ——————
3
From , l = 1 µ – 1 ———
2
Substitute
into ,
– 1 1µ – 1
3 2
– 1µ + 1
2
5µ
2
µ
T
R
P
l= 1 4 –1
2 3
=– 3
5
=1–µ
= 3 – 3µ
=2
= 4
5
→
\ OT = 1 – 3 p – 1 – 3 q = 2 p + 1 q
5 ~ 3 5 ~ 5~ 5~
V
Additional Mathematics
SPM
Paper
P A
4
2
Given p =
and q =
where k is a constant, find
3
k–1
~
~
the value of k.
1
→
→
m →
2
n
1. It is given OP =
, OQ =
and OR =
, where m
5
3
–7
630
and n are constants. Express m in terms of n, if points P, Q
and R lie on a straight line.
→
→
→
Diberi OP = m , OQ = 2 dan OR = n , dengan keadaan
5
3
–7
m dan n ialah pemalar. Ungkapkan m dalam sebutan n, jika
titik-titik P, Q dan R terletak pada satu garis lurus.
[3]
1
Ans: m = (12 – n)
5
2 , dengan keadaan k ialah
Diberi p = 4 dan q =
~
~
3
k–1
pemalar, cari nilai k.
[3]
5
Ans: k =
2
→ →
→
5. The diagram shows the vectors OP , OQ and OR drawn on
630
a square grid.
→ →
→
Rajah di bawah menunjukkan vektor-vektor OP, OQ dan OR
dilukis pada grid segi empat sama.
2. The diagram shows a regular hexagon with centre O.
630
Rajah di bawah menunjukkan sebuah heksagon sekata dengan
pusat O.
F
A
E
D
O
B
R
C
Q
→ → →
(a) Express BF + FD + FA as a single vector.
q
~
→ → →
Ungkapkan BF + FD + FA sebagai satu vektor tunggal.
→
→
a , OB = ~
b and the length of each side
(b) Given OA = ~
of the hexagon is 2 units, find the unit vector in the
→
a and ~
b.
direction of AB, in terms of ~
→
→
→
that RS + OQ = 2 OP.
→
BC
1
1
(b) – ~
a+ ~
b
2
2
Pada rajah yang sama, tanda and label titik S dengan
→ →
→
keadaan RS + OQ = 2 OP.
[3]
Ans: (a) p + 2 q
~
~
(b) Refer to the diagram above
3. Given ~
a = –2 ~i + 3 j , ~
b = 2 ~i – 7 j and ~c = – ~i + 4 j .
~
~
~
If l~
a + µb
=
2
c
,
find
the
values of l and µ.
~
~
4. The diagram shows a trapezium ABCD.
Rajah di bawah menunjukkan trapezium ABCD.
A
q
~
D
O
→
Ungkapkan OR dalam bentuk mp + n q , dengan keadaan
~ ~
m dan n ialah pemalar.
(b) On the same diagram, mark and label the point S such
Ans: (a)
630
P
p
~
→
(a) Express OR in the form mp + nq , where m and n are
~
~
constants.
→
→
Diberi OA = ~a , OB = ~b dan panjang setiap sisi heksagon
→
itu ialah 2 unit, cari vektor unit dalam arah AB , dalam
sebutan ~a dan ~
b.
[3]
Diberi bahawa ~a = –2~i + 3 j , ~b = 2~i – 7 j dan ~c = –~i + 4 j .
~
~
~
Jika λa~ + µb~ = 2~c , cari nilai λ dan µ.
[3]
1
5
Ans: l = – , µ = –
4
4
S
6. A(–2, 3) and B(2, 5) lie on a Cartesian plane. It is given
→
→
→
that 3 OA = 2 OB + OC . Find
630
A(–2, 3) dan B(2, 5) terletak pada satah Cartes. Diberi bahawa
→
→ →
3 OA = 2 OB + OC. Cari
(a) the coordinates of C,
koordinat C.
(b)
→
AC .
[4]
B
p
~
Ans: (a) (–10, –1)
C
(b) 4√5 units
V
Additional Mathematics
7. (a) Calculate the magnitude of the vector
630
→
x and
It is given that AR : RB = 1 : 2, BT : TC = 2 : 1, AR = 2~
→
AC = 6y
~.
–3
.
5
Hitung magnitud bagi vektor –3 .
5
–3
(b) Hence, state the unit vector of
.
5
Seterusnya, nyatakan vektor unit bagi –3 .
5
→
→
Diberi AR : RB = 1 : 2, BT : TC = 2 : 1, AR = 2x~ dan AC = 6y .
~
x and ~
y,
(a) Express in terms of ~
Ungkapkan dalam sebutan ~x dan y ,
→
→~
(i) CR ,
(ii) CT .
[3]
→
(b) Given ~
x = 2~i and ~
y = –~i + 4~j , find CT .
→
Diberi ~x = 2~i dan y = –~i + 4 j , cari CT .
~
~
[2]
→
→
→
→
(c) Given CS = p CR and ST = q AT, where p and q are
constants, find the values of p and q.
→
→
→
→
Diberi CS = pCR dan ST = q AT, dengan keadaan p dan
q ialah pemalar, cari nilai p dan q.
[5]
Ans: (a) (i) 2~
x – 6y (ii) 2~
x – 2y
~
~
(b) 10 units
3
2
(c) p = , q =
5
5
[3]
Ans: (a) √34 units
3
–
√34
(b)
5
√34
8. Given the points P(5, 3), Q(2, –4) and R(–1, –2),
Diberi titik-titik P(5, 3), Q(2, – 4) dan R(–1, –2),
→
x
(a) write RP in the form of
,
y
→
tulis RP dalam bentuk x ,
y
→ →
(b) find RP – RQ in the form of a i + b j .
~
~
→ →
cari RP – RQ dalam bentuk a~i + b j .
~
[3]
6
5
(b) 3 i + 7 j
~
~
2. The diagram shows the position and the direction of boats
P, Q and R on a river.
630
Rajah di bawah menunjukkan kedudukan dan arah bot P, Q dan
R di suatu sungai.
Ans: (a)
R
Q
→
→
→
9. Given OP = 5 p , OQ = 15 q and OR = 24 q – 3 p , show that
~
~
~ ~
Q lies on the line PR.
→
→
→
Diberi OP = 5p , OQ = 15q dan OR = 24q – 3p , tunjukkan
~
~
~ ~
bahawa Q berada pada garis PR.
[3]
Ans: Refer to Answer Section
10. Given ~r = ~c + p(~c + 2d
d + q~c , where
~) and ~s = 2~c + ~
c~ and ~
d are non-parallel and non-zero. If ~r = ~s , find the
values of p and q, where p and q are constants.
Diberi ~r = ~c + p( ~c + 2d
d + q ~c , dengan
~ ) dan ~s = 2 ~c + ~
keadaan ~c dan ~
tidak
selari
dan
bukan
sifar.
Jika ~r = ~s , cari
d
nilai p dan q, dengan keadaan p dan q ialah pemalar.
[3]
1
1
Ans: p = , q = –
2
2
Paper
2
1. The diagram shows a triangle ABC.
630
Rajah di bawah menunjukkan sebuah segi tiga ABC.
C
T
S
A
R
B
P
Riverbank
Tebing sungai
Both boats P and Q move in the direction of the water
1
–1
current which flow with a velocity of w
~ = (2~i + 2 ~j ) m s .
The velocity of boat P is p = (~i + j ) m s–1 and the velocity
~
~
of boat Q is q = (3~i + 2 j ) m s–1.
~
~
Kedua-dua bot P dan bot Q bergerak mengikut arah arus
1
–1
air yang mengalir dengan halaju w
~ = (2 ~i + 2 ~j ) m s .
Halaju bot P ialah p = (~i + j ) m s–1 dan halaju bot Q ialah
~
~
q = (3~i + 2 j ) m s–1.
~
~
(a) Determine how many times the resultant velocity of
boat Q compare to the resultant velocity of boat P.
Tentukan berapa kali ganda halaju paduan bot Q
berbanding halaju paduan bot P.
[4]
(b) At the half way, boat R changes its velocity to
3
–1
~r = (~i – 2 ~j ) m s . Find
Pada separuh perjalanan, bot R menukar halajunya ke
3
–1
~r = (~i – 2 ~j ) m s . Cari
(i) the resultant velocity of boat R,
halaju paduan bot R,
(ii) the unit vector in the direction of boat R.
vektor unit dalam arah bot R.
[3]
5
Ans: (a)
times
3
3
1
(b) (i) (3~i – j ) m s–1 (ii)
i –
j
~
√10 ~ √10 ~
V
Additional Mathematics
3. The diagram shows a triangle ABC. The straight line AT
630
intersects with the straight line BC at point S. Point R lies
on the straight line AT.
Rajah di bawah menunjukkan segi tiga ABC. Garis lurus AT
bersilang dengan garis lurus BC di titik S. Titik R terletak pada
garis lurus AT.
B
T
S
R
C
A
→
→
1→ →
It is given that BS = BC , AC = 6x
.
~ and AB = 12y
3
~
→
→ →
→
Diberi bahawa BS = 1 BC , AC = 6x~ dan AB = 12y
~
3
(a) Express in terms of ~
x and / or y :
~
Ungkapkan dalam sebutan ~x dan / atau y :
→
→ ~
(i) BC ,
(ii) AS .
[3]
→
→
→
(b) Given AR = m AS and BR = n(x
y ), where m and n
~ – 8~
are constants, find the value of m and of n.
→
→
→
Diberi AR = m AS dan BR = n(x~ – 8y ), dengan keadaan
~
m dan n ialah pemalar, cari nilai m dan nilai n.
(c)
[5]
→
Given AT = kx
+
9y
,
where
k
is
a
constant,
find
the
~
~
value of k.
→
Diberi AT = kx~ + 9y , dengan keadaan k ialah pemalar,
~
cari nilai k.
[2]
Ans: (a) (i) 6x
–
12y
(ii)
2x
+
8y
~
~
~
~
1
(b) m = , n = 1
2
9
(c) k =
4
4. The diagram shows triangles OAQ and OPB where P lies on
630
OA and Q lies on OB. The straight lines AQ and PB intersect
at the point R.
Rajah di bawah menunjukkan segi tiga OAQ dan OPB dengan
keadaan P terletak pada OA dan Q terletak pada OB. Garis
lurus AQ dan garis lurus PB bersilang pada titik R.
A
P
R
B
O
Q
→
→
It is given that OA = 15 ~
x , OB = 12 y , OP : PA = 1 : 2,
~→
→
→
→
OQ : QB = 3 : 1, PR = h PB and QR = k QA, where k and h
are constants.
→
→
Diberi bahawa OA = 15 ~x , OB = 12 y , OP : PA = 1 : 2,
→
→
→ ~→
OQ : QB = 3 : 1, PR = h PB dan QR = k QA, dengan keadaan
k dan h ialah pemalar.
→
(a) Express OR in terms of
→
Ungkapkan OR dalam sebutan
(i) h, ~
x and / dan y ,
~
(ii) k, ~
x and / dan y .
~
[4]
(b) Hence, find the value of h and of k.
Seterusnya, cari nilai h dan nilai k.
[4]
(c)
x = 2 units, y = 1 unit and OA is
Given ~
~
→
perpendicular to OB, calculate PR .
Diberi |x~ | = 2 unit, | y | = 1 unit dan OA berserenjang
~
→
kepada OB, hitung PR .
[2]
Ans: (a) (i) (5 – 5h)x
+
12hy
~
~
y
(ii) 15kx
~ + (9 – 9k)~
2
1
(b) h = , k =
3
9
(c)
4√61
units
3
→
→
5. In the diagram, O is the origin and OA = 6a
~, OB = 9b
~ and
→
OC = 3~c .
→
Dalam rajah di bawah, O ialah titik asalan dan OA = 6a~ ,
→
→
OB = 9b~ dan OC = 3~c .
P
C
B
Q
A
O
→
The point P lies on CB and AP = 3 ~
b – 2~
a . The point Q lies
→
on AB and AQ = 2~c + 3b
~ – 6a
~.
→
Titik P berada pada CB dan AP = 3b~ – 2a~. Titik Q berada
→
pada AB dan AQ = 2~c + 3b~ – 6a~.
(a) Find, in terms of ~
a and ~
b , the position vector of Q in
its simplest form.
Cari, dalam sebutan ~a dan ~b , vektor kedudukan Q dalam
bentuk teringkas.
[3]
(b) Find, in terms of ~
a and ~c , in its simplest form,
Cari, dalam sebutan ~a dan ~c , dalam bentuk teringkas,
→
(i) AC ,
→
(ii) PQ .
[4]
(c) Hence, based on the answers in (b), make a conclusion
→
→
regarding PQ and AC .
Seterusnya, berdasarkan jawapan pada (b), buat satu
→
→
kesimpulan mengenai PQ dan AC .
[3]
c
Ans: (a) 3b
+
2
~
~
(b) (i) –6a
~ + 3~c
(ii) –4a
~ + 2~c →
→
(c) PQ is parallel to AC .
Form 4 Answers
37 5
SPM
9H R V
D
S̰
P
3 SH
T̰
−
P
−
−
P
P
Q
Q
R
P
−2 = −12
P
P
P
P
Q
S
Q
Q
P
Q
q
~
P
P
O
p
~
D
Q
−
D
→
→
∴ RRUGL D
= −a̰
L
GLU F LR R
–8
b̰
√ (–8)
√ 80
√
λ
−2λ + 2µ = −2 ......................
3λ − 7µ = 8 ...........................
)UR
= (−10, −1)
b̰
F RU L
(−a̰ b̰
a̰
R
, −λ + µ = −1
µ = λ – 1 ......
D
0D L G R
L
6
L
L R
3λ − 7(λ – 1) = 8
−4λ + 7 = 8
4λ = −1
λ=–
L
F RU
√
√
F RU
L
√
√
√
−1
D
→
→
P
P
P
P
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
F
̰i
= −5S
̰
= −8S
̰
T̰
T̰
= −5S
̰
T̰
−5S
̰
−8S
̰
−6̰ + (1 − T
−6̰ + (1 − T x̰
̰
−6̰
x̰
̰ − 2qx̰ − 4T̰
(2 − 2T x̰ − (2 + 4T ̰
̰
→
→
→
S
S
T̰
T̰
6
L
L R
S
2 − 2T
. Thus,
L
R
T
.
̰r ̰s
̰c S ̰c
̰d
̰c d̰ qc
̰
̰c pc̰
pd̰
̰c d̰ qc
̰
S − 2 − T ̰c = (1 − 2S d̰
T
D
RD
5
D
RFL
̰i
̰i
1 − 2S
S
S
̰
RD
5
−2−T
D
RFL
̰i
L
5
D
+8
L
© Penerbitan Pelangi Sdn. Bhd.
R
L
L
RD
̰
̰i ̰
LL
̰x
̰x − 2̰
=√
̰i
̰
L
√
̰i) − 2(−ḭ
− 8̰
RFL
̰i
×
× ̰x
̰x
(−6̰
̰
∴ The resultant velocity of boat
U
D
RFL R RD .
= −6̰
̰x
̰x − 6̰
̰
̰
̰i
3 SH
̰i
̰
̰i
T
LL
̰
̰
̰i
S−2−T
L
̰i
̰
̰
F RU L
̰i ̰
GLU F LR R
̰i ̰
√
√
̰i
√
̰
RD
T
→
2 − 2T
S ...............................
T
S ...............................
× 2, 4 − 4T
S ...................
̰S
8
D
S
S x̰ − 6̰
px̰ − 6S̰
px̰ − 6S̰
px̰ − 6S̰
px̰ − 6S̰
T̰
T̰
5(−S̰
8(−S̰
8
x̰ − 2̰
̰
̰
x̰
T
Form 4 Answers
D
L
= −12̰
̰x
6
̰x
L
L R
̰
LL
̰x
̰
̰x + 8̰
̰
F
Q ̰x − 8̰ ) = −12̰ P
nx
̰ − 8Q̰ = −12̰ P ̰x + 8̰
nx
̰ − 8Q̰
mx
̰ + (8P
̰
̰x
×
Q
P .................................
−8Q = 8P − 12 ....................
L
6
L
L R
−8(2P) = 8P
P
P
̰
×
U
L
=√
=√
= √ ×
= 2√
Q
F
×
is a constant.
L
→
kx̰
̰x + 8̰
̰
× 2√
.................
9 = 8 .................
)UR
D
4√
8
∴
D
L
L
LL
̰x
(−5x
̰
̰x
̰
̰
̰
̰
̰x
kx̰
̰
(−9̰
)UR
−
b̰
= −6a̰
F̰
= −(3b̰
= −4a̰
a̰
̰c
a̰
̰c
b̰
a̰
̰c
̰c
a̰ ̰c
a̰ ̰c
̰x
̰
kx̰
...............................
...............................
)UR
̰c
̰c
a̰
̰a
F
LL
L
̰a
b̰
8
̰x
→
∴
̰
D
L
...............
...............
LL
L SDUD
̰a
a̰
a̰
= (1
R
.
P
P
P(−a̰ ̰b
− P a̰ mb̰
̰b
a̰
b̰
a̰
= −9a̰
Q
nb̰
Q b̰
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
(b) (1 − P a̰
mb̰ = −9a̰
Q b̰
276
OOH
H
1 − P = −9
P
P
Q
D
= −9a̰
∴
D
C
Q
Q
b̰
L D F
E
d
~
D G
a
~
O
D
A
B
= −2a̰ d̰
8
a̰
=√
=√
= 3√
(−2a̰
a̰
a̰
L
= −a̰
F
L
d̰
d̰
a̰
R
F RU L
(−a̰
GLU F LR R
d̰
d̰ : (−a̰
d̰ ) : (−a̰
d̰
d̰
√
√
L
̰c
d̰
̰c
d̰
̰c
d̰
√
S
F
−
̰c
8
̰c
d̰
P
Q
d̰
= −(4c̰ d̰
̰c d̰
−5 + 15 = 16 + 8
L R
̰c
d̰
d̰ ) = −2c
̰
d̰
̰c − 3d̰
̰c
−1 + 3 = 8 ................
....................
6
L
−1 + 3(3) = 8
8 =8
d̰ ) = −(4c̰
̰c
−
8
R
̰c
∴
̰c d̰
= −(c̰ d̰
̰c ̰d
̰c
∴
6L F
D
© Penerbitan Pelangi Sdn. Bhd.
RSSR L
F
SDLU R
F RU R
is a parallelogram.
DGUL D UD DU
S
Additional Mathematics
(b)
E
(c)
P
12 cm
G
75°
Q
13 cm
95°
62°
48°
R
In the diagram, PQR is a triangle.
F
Dalam rajah di atas, PQR ialah sebuah segi tiga.
In the diagram, EFG is a triangle. Find the length
of
Dalam rajah di atas, EFG ialah sebuah segi tiga. Cari
panjang bagi
(i) GE,
(ii) FG.
(i)
(i)
Find ∠R.
Cari ∠R.
(ii) Which side of the triangle is the longest in
length? Give a reason for your answer and
hence find the length.
Sisi segi tiga yang manakah paling panjang? Berikan
alasan bagi jawapan anda dan seterusnya cari
panjangnya.
GE
= 13
sin 48°
sin 75°
GE = 13 × sin 48°
sin 75°
= 10.00 cm
(i)
∠R = 180° – 95° – 62°
= 23°
(ii) QR as it has the greatest opposite angle.
(ii) ∠E = 180° – 75° – 48°
= 57°
QR
= 12
sin 95°
sin 23°
QR = 12 × sin 95°
sin 23°
= 30.59 cm
FG
= 13
sin 57°
sin 75°
FG = 13 × sin 57°
sin 75°
= 11.29 cm
2. Solve each of the following.
Selesaikan setiap yang berikut.
(a)
4.8 cm
A
67°
E
B
12 cm
5 cm
135°
6.1 cm
F
C
In the diagram, ABC is a triangle. Calculate
Dalam rajah di atas, ABC ialah sebuah segi tiga. Hitung
(i) ∠C,
(ii) the length of AC.
(i)
sin C = sin 67°
Using the sine rule
4.8
6.1
∠C = sin–1 sin 67° × 4.8
6.1
= 46.41°
(ii) ∠B = 180° – 46.41° – 67°
= 66.59°
AC
= 6.1
sin 66.59°
sin 67°
AC = 6.1 × sin 66.59°
sin 67°
= 6.08 cm
G
In the diagram, EFG is a triangle. Calculate
Dalam rajah di atas, EFG ialah sebuah segi tiga. Hitung
(i) ∠G,
(ii) the length of FG.
panjang FG.
(i)
sin G = sin 135°
5
12
–1 sin 135°
∠G = sin
×5
12
= 17.14°
(ii) ∠E = 180° – 135° – 17.14°
= 27.86°
FG
12
=
sin 27.86° sin 135°
12
FG =
× sin 27.86°
sin 135°
= 7.93 cm
S
Additional Mathematics
4. Solve each of the following.
Selesaikan setiap yang berikut.
Given a triangle ABC such that AB = 6.2 cm, AC = 4.8 cm and ∠ABC = 43°, find the possible values of
Diberi sebuah segi tiga ABC dengan keadaan AB = 6.2 cm, AC = 4.8 cm dan ∠ABC = 43°, cari nilai yang mungkin bagi
(i) ∠BCA
(ii) the lengths of BC / panjang BC
B
43°
C2
C1
6.2 cm
Ambiguous case
4.8 cm
A
(i)
sin C = sin 43°
6.2
4.8
∠BC1A = sin–1 sin 43° × 6.2 = 61.75°
4.8
∠BC2A = 180° – 61.75° = 118.25°
(ii) In triangle ABC1,
∠BAC1 = 180° – 43° – 61.75° = 75.25°
BC1
= 4.8
sin 75.25° sin 43°
BC1 = 4.8 × sin 75.25° = 6.806 cm
sin 43°
In triangle ABC2,
∠BAC2 = 180° – 43° – 118.25° = 18.75°
BC2
= 4.8
sin 18.75° sin 43°
BC2 = 4.8 × sin 18.75° = 2.262 cm
sin 43°
(a) Given a triangle DEF such that ∠FED = 68°, (b) In a triangle GHI, HI = 7.5 cm, GI = 5.6 cm and
DF = 6.5 cm and EF = 6.9 cm, find the possible
∠IHG = 44°. Find the possible values for ∠HGI
lengths of ED.
and hence find the length of the remaining side
Diberi sebuah segi tiga DEF dengan keadaan ∠FED = 68°,
of the acute-angled triangle GHI.
DF = 6.5 cm dan EF = 6.9 cm, cari nilai yang mungkin
Dalam sebuah segi tiga GHI, HI = 7.5 cm, GI = 5.6 cm
bagi panjang ED.
dan ∠IHG = 44°. Cari nilai yang mungkin bagi ∠HGI dan
seterusnya cari panjang sisi yang tertinggal bagi segi tiga
F
bersudut tirus GHI.
6.5 cm
D1
6.9 cm
G1
7.5 cm
68° D
2
G2
E
sin D = sin 68°
6.9
6.5
∠ED1F = sin–1 sin 68° × 6.9 = 79.81°
6.5
∠ED2F = 180° – 79.81° = 100.19°
In triangle D1EF,
∠EFD1 = 180° – 68° – 79.81° = 32.19°
ED1
= 6.5
sin 32.19° sin 68°
ED1 = 6.5 × sin 32.19°
sin 68°
= 3.735 cm
In triangle D2EF,
∠EFD2 = 180° – 68° – 100.19° = 11.81°
ED2
= 6.5
sin 11.81° sin 68°
ED2 = 6.5 × sin 11.81°
sin 68°
= 1.435 cm
I
5.6 cm
44°
H
sin G = sin 44°
7.5
5.6
∠HG1I = sin–1 sin 44° × 7.5
5.6
= 68.49°
∠HG2I = 180° – 68.49°
= 111.51°
∆G1HI is an acute-angled triangle.
∠G1HI = 180° – 44° – 68.49°
= 67.51°
G 1H
= 5.6
sin 44°
sin 67.51°
G1H = 5.6 × sin 67.51°
sin 44°
= 7.448 cm
S
Additional Mathematics
5. Solve the following problems.
Selesaikan masalah yang berikut.
S
In the diagram, PQRS is a quadrilateral and SQ is the diagonal. Find the length of
Dalam rajah di sebelah, PQRS ialah sebuah sisi empat dan SQ ialah pepenjuru. Cari panjang bagi
(i) RQ
(ii) PQ
(i)
∠R = 180° – 46° – 30°
= 104°
25 cm
128°
(ii) ∠PSQ = 180° – 128°
2
= 26°
RQ =
25
sin 30°
sin 104°
25
RQ =
× sin 30°
sin 104°
= 12.88 cm
R
30°
P
46°
Q
PQ =
25
sin 26° sin 128°
25
PQ =
× sin 26°
sin 128°
= 13.91 cm
(a)
(b)
B
70°
A
R
110 m
33°
7.4 km
C
142°
138 m
Q
S
D
80°
5.8 km
40°
P
The diagram shows a playground ABCD, in the
shape of a quadrilateral, where AC is a footpath
along the diagonal. Calculate
Rajah di atas menunjukkan sebuah taman permainan ABCD,
dalam bentuk suatu sisi empat, dengan keadaan AC ialah
lorong jalan kaki di sepanjang pepenjuru, Hitung
(i) the length of AB/ panjang AB
(ii) ∠CAD
The diagram shows five straight paths. The
positions P, Q, R and S will be planted with trees.
Calculate
Rajah di atas menunjukkan lima jalan lurus. Kedudukan P,
Q, R dan S akan ditanam dengan pokok. Hitung
(i) ∠RQP
(ii) the distance between P and S
jarak di antara P dan S
(i)
(i)
∠B = 180° – 70° – 33°
= 77°
∠RQP = sin–1 sin 40° × 7.4
5.8
= 55.10°
Since 5.8
or ∠RQP = 180° – 55.10°
= 124.9°
sin D = sin 142°
110
138
∠D = sin–1 sin 142° × 110
138
= 29.39°
∠CAD = 180° – 142° – 29.39°
= 8.61°
Alternate angle
sin Q = sin 40°
7.4
5.8
AB = 110
sin 33°
sin 77°
AB = 110 × sin 33°
sin 77°
= 61.49 m
(ii)
∠QRP = 40°
(ii)
7.4,
there is two
possible answers
for ∠RQP
PS
= 7.4
sin (180° – 80° – 40°)
sin 80°
PS = 7.4 × sin 60°
sin 80°
= 6.507 km
S
Additional Mathematics
(c) Ahmad (A) stands 16 metres from a bridge (B) and 31 metres from
a tree (T). Calculate the distance of BT.
Ahmad (A) berdiri 16 meter dari sebuah jambatan (B) dan 31 meter dari sebatang
pokok (T). Hitung jarak BT.
B
T
16 m
31 m
42°
A
BT = AB + AT – 2(AB)(AT) cos 42°
= 162 + 312 – 2(16)(31) cos 42°
= 479.80
BT = 21.90 m
2
2
2
7. Find the value of θ, in degree, in each of the following triangles.
Cari nilai θ, dalam darjah, bagi setiap segi tiga berikut.
(a)
6.2 cm
R
S
P
θ
7.3 cm
12.9 cm
12 cm
10 cm
Q
RQ2 = PR2 + PQ2 – 2(PR)(PQ) cos θ
12.92 = 6.22 + 7.32 – 2(6.2)(7.3) cos θ
2
2
2
cos θ = 6.2 + 7.3 – 12.9
2(6.2)(7.3)
U
θ
T
11 cm
122 = 112 + 102 – 2(11)(10) cos θ
2
2
2
cos θ = 11 + 10 – 12
2(11)(10)
= 0.35
θ = 69.51°
Using the cosine rule
= –0.8250
θ = 145.59°
(c)
(b)
V
16.6 cm
θ
19.9 cm
W
11.5 cm
A
θ
5.2 cm
C
5.22 = 10.42 + 6.32 – 2(10.4)(6.3) cos θ
2
2
2
cos θ = 10.4 + 6.3 – 5.2
2(10.4)(6.3)
= 0.9219
θ = 22.79°
(d) During a basketball match, a player runs from position A to position B
and then to position C as shown in the diagram. Find the value of θ, in
degree.
Dalam suatu pertandingan bola keranjang, seorang pemain berlari dari kedudukan A
ke kedudukan B dan kemudian ke kedudukan C seperti yang ditunjukkan dalam rajah
di sebelah. Cari nilai θ, dalam darjah.
3.42 = 5.22 + 6.82 – 2(5.2)(6.8) cos θ
2
2
2
cos θ = 5.2 + 6.8 – 3.4
2(5.2)(6.8)
= 0.8727
θ = 29.22°
B
6.3 cm
X
19.92 = 16.62 + 11.52 – 2(16.6)(11.5) cos θ
2
2
2
cos θ = 16.6 + 11.5 – 19.9
2(16.6)(11.5)
= 0.0309
θ = 88.23°
10.4 cm
B
5.2 m
θ
3.4 m
6.8 m
A
C
S
Additional Mathematics
9. Find the area of each of the following triangles.
Cari luas bagi setiap segi tiga berikut.
(a)
A
36°
16 cm
D
145°
6 cm
13 cm
C
10 cm
F
E
B
Area of ∆DEF = 1 (6)(10) sin 145°
2
= 17.21 cm2
Area of ∆ABC = 1 (16)(13) sin 36°
2
= 61.13 cm2
(b)
(c)
H
123°
16 cm
I
8.4 cm
K
18 cm
33°
31°
52°
L
G
J
∠G = 180° – 33° – 123°
= 24°
∠J = 180° – 52° – 31° = 97°
JL
= 8.4
sin 52°
sin 97°
JL = 8.4 × sin 52°
sin 97°
= 6.67 cm
Area of ∆GHI = 1 (16)(18) sin 24°
2
= 58.57 cm2
Area of ∆JKL = 1 (8.4)(6.67) sin 31°
2
= 14.43 cm2
10. Find the area of each of the following triangles.
Cari luas bagi setiap segi tiga berikut.
Area of ∆PQR = √s(s – a)(s – b)(s – c) where s = 1 (a + b + c)
2
1
s = (22 + 17 + 15) = 27
Heron’s formula
2
Area of ∆PQR
= √27(27 – 22)(27 – 17)(27 – 15)
= 127.3 cm2
P
15 cm
Q
17 cm
22 cm
R
(a)
(b)
S
14 cm
T
37 cm
V
14 cm
20 cm
s = 1 (20 + 14 + 14) = 24
2
Area of ∆STU
= √24(24 – 20)(24 – 14)(24 – 14)
= 97.98 cm2
U
X
32 cm
W
s = 1 (32 + 37 + 14) = 41.5
2
Area of ∆VWX
= √41.5(41.5 – 32)(41.5 – 37)(41.5 – 14)
= 220.9 cm2
S
Additional Mathematics
SSO
SL
L
R R 6 H 5 OH
L
R
L
H 5 OH
H R
L L
7
OH
Textbook
pg. 263 – 265
12. Solve each of the following.
Selesaikan setiap yang berikut.
HF = 5 cm
Use Pythagorean
triple: (3, 4, 5)
BF = √82 + 32
= 8.544 cm
BH = √82 + 42
= 8.944 cm
VB = √289 + 62
= 18.03 cm
VC = √62 + 152
= 16.16 cm
8 cm
G
H
3 cm
E
4 cm
F
V
D
A
6 cm
C
B
15 cm
s = 1 (18.03 + 8 + 16.16) = 21.10
2
Area of ∆VBC
= √21.10(21.10 – 18.03)(21.10 – 8)(21.10 – 16.16)
= 64.75 cm2
(b) The diagram shows a crane on a horizontal ground. AB is vertical.
O
Find ∠ABC. 276
Rajah di sebelah menunjukkan sebuah kren di atas tanah mengufuk. AB adalah
tegak. Cari ∠ABC.
10.22 = 6.82 + 5.12 – 2(6.8)(5.1) cos ∠ABC
2
2
2
cos ∠ABC = 6.8 + 5.1 – 10.2
2(6.8)(5.1)
∠ABC = 117.28°
C
D
Using Heron’s formula,
s = 1 (5 + 8.544 + 8.944) = 11.244
2
Area of ∆BFH
= √11.244(11.244 – 5)(11.244 – 8.544)(11.244 – 8.944)
= 20.88 cm2
(a) The diagram shows a pyramid with a horizontal rectangular base ABCD.
The vertex V is 6 cm vertically above D. It is given that AB = 15 cm and
BC = 8 cm. Find the area, in cm2, of the triangular plane VBC. 276 SSO
Rajah di sebelah menunjukkan sebuah piramid dengan tapak mengufuk berbentuk segi empat
tepat ABCD. Bucu V adalah 6 cm tegak di atas D. Diberi bahawa AB = 15 cm dan BC = 8 cm.
Cari luas, dalam cm2, bagi satah VBC yang berbentuk segi tiga.
BD2 = 152 + 82 = 289
B
A
In the diagram, ABCDEFGH is a cuboid with a horizontal rectangular base
EFGH. The plane ADEH is a rectangle which is parallel to the plane BCFG and
perpendicular to the base. The plane ABCD is parallel to the base. Find the area,
in cm 2, of the triangle BFH. 276 SSO
Dalam rajah di sebelah, ABCDEFGH ialah sebuah kuboid dengan tapak mengufuk EFGH yang
berbentuk segi empat tepat. Satah ADEH ialah segi empat tepat yang selari dengan satah BCFG dan
berserenjang dengan tapak. Satah ABCD adalah selari dengan tapak. Cari luas, dalam cm2, segi tiga
BFH.
C
5.1 m
B
6.8 m
A
10.2 m
8 cm
S
Additional Mathematics
12 cm
A
B
(c) The diagram shows a right prism placed on a horizontal surface. Triangle
AED is the uniform cross section of the prism where angle E is a right
5 cm
angle. The rectangular plane ABFE is perpendicular to the horizontal
F
E
rectangular plane CDEF and the rectangular surface ABCD is an inclined
9 cm
plane. Given the area of the triangle BCE is 65.77 cm2, calculate the
C
D
O
∠BEC. 276
Rajah di sebelah menunujukkan sebuah prisma tegak diletak pada suatu permukaan mengufuk. Segi tiga AED ialah keratan
rentas seragam prisma itu dengan keadaan sudut E ialah sudut tegak. ABFE adalah satah berbentuk segi empat tepat yang
berserenjang dengan satah mengufuk CDEF yang juga berbentuk segi empat tepat dan permukaan ABCD yang berbentuk segi
empat tepat ialah satah condong. Diberi bahawa luas segi tiga BCE ialah 65.77 cm2, hitung ∠BEC.
BE = √52 + 122
= 13 cm
Area of ∆BCE = 1 (BE)(EC) sin ∠BEC
2
65.77 = 1 (13)(15) sin ∠BEC
2
sin ∠BEC = 0.6746
∠BEC = 42.42°
EC = √122 + 92
= 15 cm
(d) The diagram shows a building. C is vertically above D. A
surveyor measures the angle of elevation of the top of the
building from two points, A and B. The distance between A
and B is 50 m. The angle of elevation obtained is shown in the
O
diagram. Calculate 276
Rajah di sebelah menunjukkan sebuah bangunan. C adalah tegak di atas D.
Seorang juruukur mengukur sudut dongak puncak bangunan dari dua titik,
A dan B. Jarak di antara A dan B ialah 50 m. Sudut dongak yang diperoleh
ditunjukkan dalam rajah di sebelah. Hitung
(i) the distance between B and C,
jarak di antara B dan C,
(ii) the height of the building.
tinggi bangunan tersebut.
(i)
∠ACB = 47° – 40°
= 7°
C
A
40°
B
50 m
47°
The exterior angle is the sum of two opposite interior angles.
BC = 50
sin 40°
sin 7°
BC = 50 × sin 40°
sin 7°
= 263.72 m
\ The distance between B and C is 263.72 m.
(ii) sin ∠CBD = CD
BC
sin 47° = CD
263.72
CD = 192.87 m
\ The height of the building is 192.87 m.
D
S
Additional Mathematics
SPM
Paper
P A
3. The diagram shows a quadrilateral PQRS on a horizontal
plane.
2
630
Rajah di bawah menujukkan sisi empat PQRS pada suatu satah
mengufuk.
1. The diagram shows a quadrilateral ABCD.
630
Rajah di bawah menunjukkan sebuah sisi empat ABCD.
V
B
A
31°
9 cm
14 cm
79°
D
5 cm
S
65°
C
Q
20.5 cm
R
VQSP is a pyramid such that PQ = 12 cm and V is 5 cm
vertically above P. Find
panjang, dalam cm, bagi BD,
(ii) ∠BCD,
(iii) the area, in cm2, of the quadrilateral ABCD.
luas, dalam cm2, bagi sisi empat ABCD.
[8]
Sketch a triangle D B C which has a different
shape from triangle DBC such that D'C'= DC,
D'B' = DB and ∠D B C = ∠DBC.
Lakarkan sebuah segi tiga D'B'C' yang mempunyai
bentuk berbeza daripada segi tiga DBC dengan
D'C' = DC, D'B' = DB dan ∠D'B'C' = ∠DBC.
(ii) Hence, state ∠D C B .
Seterusnya, nyatakan ∠D'C'B'.
[2]
Ans: (a) (i) 16.42 cm; (ii) 110.3°; (iii) 105.36 cm2
(b) (i) Refer to Answer Section
(ii) 69.7°
2. The diagram shows a cyclic quadrilateral PQRS.
VQSP ialah sebuah piramid dengan keadaan PQ = 12 cm dan
V adalah 5 cm tegak di atas P. Cari
(a) ∠QSR,
[2]
(b) the length, in cm, of QS,
panjang, dalam cm, bagi QS,
[3]
(c) the area, in cm2, of inclined plane QVS.
luas, dalam cm2, bagi satah condong QVS.
[5]
Ans: (a) 62.22°
(b) 18.45 cm
(c) 71.74 cm2
4. The diagram shows a sketch of a solid tetrahedron with V
vertically above C. The right-angled triangle ABC is its base.
Rajah di bawah menunjukkan lakaran sebuah pepejal
tetrahedron dengan V tegak di atas C. Segi tiga bersudut tegak
ABC ialah tapaknya.
Rajah di bawah menunjukkan sisi empat kitaran PQRS.
P
V
4 cm
9 cm
630
P
21 cm
(a) Find/ Cari
(i) the length, in cm, of BD,
(b) (i)
10 cm
80°
Q
7 cm
S
C
R
A
(a) Calculate/ Hitung
(i) the length, in cm, of PR,
panjang, dalam cm, bagi PR,
(ii) ∠PRS.
[6]
(b) Find/ Cari
(i) the area, in cm2, of ∆PQR,
luas, dalam cm2, bagi ∆PQR.
(ii) the shortest distance, in cm, from point Q to PR.
jarak terdekat, dalam cm, dari titik Q ke PR.
[4]
Ans: (a) (i) 10.40 cm; (ii) 57.74°
(b) (i) 31.02 cm2; (ii) 5.965 cm
B
Given that AC = 8 cm, CB = 6 cm and VC = 5 cm.
Diberi bahawa AC = 8 cm, CB = 6 cm dan VC = 5 cm.
(a) (i) Show that AV = √89 cm.
Tunjukkan AV = √89 cm.
(ii) Find the length of AB, in cm.
Cari panjang AB, dalam cm.
[4]
(b) Calculate/ Hitung
(i) ∠VBA,
(ii) the area of triangle VBA, in cm2.
luas segi tiga VBA, dalam cm2.
[6]
Ans: (a) (ii) 10 cm
(b) (i) 62.55°; (ii) 34.65 cm2
S
Additional Mathematics
Rajah di bawah menunjukkan prisma lutsinar dengan tapak
PQRS berbentuk segi empat tepat. Permukaan condong PQUT
ialah segi empat sama dengan sisi 11 cm dan permukaan
condong RSTU ialah segi empat tepat. TPS ialah keratan rentas
seragam bagi prisma itu. QST ialah satah berwarna di dalam
prima itu.
U
7. In the diagram, ABCD is a cyclic quadrilateral.
Dalam rajah di bawah, ABCD ialah sebuah sisi empat
kitaran.
A
11
.1
cm
5. The diagram shows a transparent prism with a rectangular
base PQRS. The inclined surface PQUT is a square with sides
11 cm and the inclined surface RSTU is a rectangle. TPS is
the uniform cross section of the prism. QST is a coloured
plane in the prism.
630
D
B
35°
60°
8.9 cm
C
(a) Calculate the length of
Hitung panjang bagi
Q
(i) AC,
(ii) BC.
R
T
[6]
(b) Find the area of quadrilateral ABCD, in cm2
P
Cari luas bagi sisi empat ABCD, dalam cm2.
S
[4]
Ans: (a) (i) 10.18 cm; (ii) 4.968 cm
(b) 57.28 cm2
It is given that ∠PST = 36° and ∠TPS = 45°. Find
Diberi bahawa ∠PST = 36° dan ∠TPS = 45°. Cari
(a) the length, in cm, of ST,
panjang, dalam cm, bagi ST,
[2]
(b) the area, in cm2, of the coloured plane,
luas, dalam cm2, bagi satah berwarna,
(c)
[6]
the shortest length, in cm, from point T to the straight
line QS.
8. The diagram shows the plan of a playground in the shape
of a trapezium. A straight path B to D has a length of 21 m.
Rajah di bawah menunjukkan pelan suatu taman permainan
dalam bentuk trapezium. Satu jalan lurus dari B ke D
mempunyai panjang 21 m.
panjang terdekat, dalam cm, dari titik T ke garis lurus QS.
[2]
Ans: (a) 13.23 cm
(b) 102.30 cm2
(c) 9.512 cm
6. The diagram shows the positions of four spots in a country,
A, B, C and D.
Rajah di bawah menunjukkan kedudukan empat lokasi dalam
sebuah negera, A, B, C dan D.
15°
A
km
C
0 km
1 60
21 m
A
B
Calculate/ Hitung
(a) ∠DBC,
[2]
(b) the length, in m, of AD,
panjang, dalam m, bagi AD,
[2]
(c)
the area, in m2, of triangle BCD,
[2]
94°
D
(b) (i)
25 m
luas, dalam cm2, bagi segi tiga BCD,
36°
(b) Calculate/ Hitung
(i) BC,
C
100°
B
00
14
18 m
D
(d) the perpendicular distance, in m, from C to BD,
jarak serenjang, dalam m, dari C ke BD,
[2]
(e) the area, in m2, of trapezium ABCD.
(ii) AD.
[6]
Sketch another triangle, ∆D A C such that
C D = CD, A C = AC and ∠C A D = ∠CAD.
Lukis segi tiga lain, ∆D A C dengan keadaan
C D = CD, A C = AC dan ∠C A D = ∠CAD.
(ii) Hence, find the length of A D .
Seterusnya, cari panjang A D .
[4]
Ans: (a) (i) 438.9 km; (ii) 1 229 km
(b) (i) Refer to Answer Section
(ii) 1 360 km
luas, dalam m2, bagi trapezium ABCD.
[2]
Ans: (a)
(b)
(c)
(d)
(e)
57.58°
9.77 m
72.08 m2
6.865 m
172.25 m2
Form 4 Answers
37 5
SPM
3 SH
6RO
R R 7
OH
Area of ∆
P
L
L
×
×
L
L
L ∠
L
L ∠
L
× L
L
×
∠
LL
∠
L
∠
L
× L
L
∠
LLL
L
L
L
B
V
Area ∆
=√
D
5 cm
LL
C
∠ ʹ ʹ ʹ
L
LL
∠
L ∠
L ∠
L
=√
=√
LL
=√
L
=√
=√
L
− 2(
L
×
∠
∠
∠
√
∠
∠
(b) (i)
∠
√
(ii) Area of ∆
Area of ∆
√
L
L
LL
L
L
P
L
× L
9 cm
10.40 cm
h
Q
L
7 cm
R
L
L
L
× L
© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
=√
L ∠
L
=√
L ∠
L
×
∠
V
= 180° − 100° − 57.58°
∠
=√
Q
∠
T
∠
Area of ∆
h
21.51 cm
L
S
L
×
L
18 m
D
C
h
L
21 m
B
LL
L
L
L
×
× L
×
V
L
C
=√
0 km
1 60
A
36°
94°
276
D D
LL
∠ ʹ ʹ ʹ = 180° − 94°
L
OOH
H
L
ʹ ʹ
L
ʹ ʹ
L
L
× L
L
× L
L
L
L
L
LL
C
∠
= 180° − 60°
252 km
h
L
L
L
A
× L
40°
L
L
SS
H
S H
θ
L
L
∴7
© Penerbitan Pelangi Sdn. Bhd.
B
377 km
L
L L
L
n
LL
L
L
H
H
Form 4 Answers
LL
L
∠
LL
SHH
D
H
H
∠
∠
L
L
L
L
∴7
L
L
× L
∠
L
L
© Penerbitan Pelangi Sdn. Bhd.
1
Additional Mathematics
2. Calculate each of the following.
Hitung setiap yang berikut.
A survey has been made on the number of road
accidents. The number of road accidents in the year
2010 was 13 260. Given the accident index in the
year 2010 based on the year 2008 is 130, calculate the
number of road accidents in the year 2008.
Satu kajian mengenai bilangan kemalangan jalan raya telah
dijalankan. Bilangan kemalangan jalan raya pada tahun 2010
ialah 13 260. Diberi indeks kemalangan pada tahun 2010
berasaskan tahun 2008 ialah 130, hitung bilangan kemalangan
jalan raya pada tahun 2008.
Let Q0 = number of road accidents in the year 2008
Q1 = number of road accidents in the year 2010
Accident index, I =
Q1
× 100
Q0
(a) The price of a motorcycle in the year 2016
was RM6 750. Given the price index of the
motorcycle in the year 2016 based on the year
2014 is 125, calculate the price of the motorcycle
in the year 2014.
Harga sebuah motosikal pada tahun 2016 ialah
RM6 750. Diberi indeks harga bagi motosikal itu pada
tahun 2016 berasaskan tahun 2014 ialah 125, hitung harga
motosikal itu pada tahun 2014.
Let Q0 = price in the year 2014
Q1 = price in the year 2016
Price index, I =
Q1
Q0
× 100
125 = 6 750 × 100
Q0
130 = 13 260 × 100
Q0
Q0 = RM5 400
Q0 = 10 200
Hence, the price of the motorcycle in the year
2014 was RM5 400.
Hence, the number of road accidents in the year 2008
was 10 200.
(b) The road accident index of a certain town in the (c) The price index of a double storey terrace in the
year 2012 based on the year 2016 was 75. If the
year 2000 based on the year 1984 was 150. If the
number of road accidents occurred in the year
price of the terrace was RM139 500 in the year
2016 was 460, find the number of road accidents
2000, find its price in the year 1984.
Indeks harga bagi sebuah rumah teres dua tingkat pada
occurred in the year 2012.
tahun 2000 berasaskan tahun 1984 ialah 150. Jika harga
Indeks kemalangan jalan raya bagi sebuah bandar pada
rumah teres itu ialah RM139 500 pada tahun 2000, cari
tahun 2012 berasaskan tahun 2016 ialah 75. Jika bilangan
harganya pada tahun 1984.
kemalangan jalan raya yang berlaku pada tahun 2016 ialah
460, cari bilangan kemalangan jalan raya pada tahun 2012.
Let Q2000 = price of the terrace in 2000
Let Q2016 = number of accidents in 2016
Q1984 = price of the terrace in 1984
Q2012 = number of accidents in 2012
Q
Price index, I = 2000 × 100 = 150
Q2012
Q1984
Accident index, I =
× 100 = 75
Q2016
139 500 × 100 =150
Q2012
Q1984
× 100 =75
460
Q1984 = 139 500 × 100
75
460
150
×
Q2012 =
100
= RM93 000
= 345
Hence, the number of road accidents occured in
the year 2012 was 345.
Hence, the price of the terrace in the year 1984
was RM93 000.
1
Additional Mathematics
3. Calculate the index number for each of the following situations.
Hitung nombor indeks bagi setiap situasi berikut.
The index numbers that show the change in the
number of primary students who wore spectacles in
the year 2014 based on the year 2010 and 2012 were
161 and 140 respectively. Find the index number of
the number of students in the year 2012 based on the
year 2010.
Nombor indeks yang menunjukkan perubahan bilangan murid
sekolah rendah yang memakai cermin mata pada tahun 2014
berasaskan tahun 2010 dan 2012 masing-masing ialah 161 dan
140. Cari nombor indeks bagi bilangan murid pada tahun 2012
berasaskan tahun 2010.
Q2014
× 100 = 161
⇒
Q2014
× 100 = 140
Q2012
⇒
Q2010
Index number, I =
=
Q2012
Q2010
Q2012
Q2014
Q2014
Q2010
= 161
100
Q2014
= 140
Q2012
100
× 100
×
Q2014
Q2010
× 100
= 100 × 161 × 100
140
100
(a) The price indices of a car in the year 2019 based
on the year 2010 and 2015 were 150 and 125
respectively. Find the price index of the car in
the year 2015 based on the year 2010.
Indeks harga bagi sebuah kereta pada tahun 2019
berasaskan tahun 2010 dan 2015 masing-masing ialah 150
dan 125. Cari indeks harga bagi kereta tersebut pada tahun
2015 berasaskan tahun 2010.
Q2019
× 100 = 150
Q2010
Q2019
Q2015
× 100 = 125
⇒
⇒
Q2019
= 150
Q2010
100
Q2019
Q2015
= 125
100
Q2015
× 100
Q2010
Q
Q
= 2015 × 2019 × 100
Q2019
Q2010
Index number, I =
= 100 × 150 × 100
125
100
= 120
= 115
(b) Price indices for a certain electric product in the (c) The price index for a bowl of curry mee in the
year 2004 and 2006 based on the year 2000 were
year 1996 based on the year 1990 was 140 and
116 and 125 respectively. Calculate the price
the price index for a bowl of curry mee in the
indices in the year 2000 dan 2004 if the year
year 2000 based on the year 1996 was 105. Find
2006 is used as base year.
the price index of a bowl of curry mee in the
Indeks harga bagi suatu barangan elektrik pada tahun 2004
year 2000 based on the year 1990.
dan 2006 berasaskan tahun 2000 masing-masing ialah 116
Nombor indeks bagi semangkuk mi kari pada tahun 1996
dan 125. Hitung indeks harga barangan elektrik itu pada
berasaskan tahun 1990 ialah 140 dan nombor indeks bagi
tahun 2000 dan 2004 jika tahun 2006 digunakan sebagai
semangkuk mi kari pada tahun 2000 berasaskan tahun
tahun asas.
1996 ialah 105. Cari indeks harga bagi semangkuk mi kari
pada tahun 2000 berasaskan tahun 1990.
Q2004
Q2004
= 116
× 100 = 116 ⇒
Q1996
Q1996
Q2000
Q2000
100
= 140
× 100 = 140 ⇒
Q
Q
100
Q2006
Q2006
1990
1990
= 125
× 100 = 125 ⇒
Q2000
Q2000
100
Q2000
Q2000
= 105
× 100 = 105 ⇒
Q1996
Q1996
100
Q2000
100
× 100 =
× 100
Q2006
125
Q2000
Q
Q
× 100 = 2000 × 1996 × 100
= 80
Q1990
Q1996
Q1990
Q2004
Q
Q
× 100 = 2004 × 2000 × 100
Q2006
Q2000
Q2006
= 116 × 100 × 100
100
125
= 92.8
= 105 × 140 × 100
100
100
= 147
1
Additional Mathematics
4. Solve each of the following.
Selesaikan setiap yang berikut.
The price index of a sport shoes in the year 2009 by
using the year 2006 as the base year was 130. If the
price of the sport shoes in the year 2006 was RM80,
find its price in the year 2009.
Indeks harga sejenis kasut sukan pada tahun 2009 dengan
menggunakan tahun 2006 sebagai tahun asas ialah 130.
Jika harga kasut sukan itu pada tahun 2006 ialah RM80, cari
harganya pada tahun 2009.
Let Q2009 = price of the sport shoes in 2009
Q2006 = price of the sport shoes in 2006
Price index, I =
(a) The table shows the prices of two items A and B
in the years 2005 and 2008.
Jadual di bawah menunjukkan harga bagi dua barangan A
dan B pada tahun 2005 dan 2008.
Price (RM)
Harga (RM)
Item
Barangan
2005
2008
A
3.50
4.90
B
6.00
8.10
Calculate the price indices of A and B in the year
2008 based on the year 2005.
Hitung indeks harga bagi A dan B pada tahun 2008
berasaskan tahun 2005.
Q2009
× 100 = 130
Q2006
Q2009
× 100 = 130
80
Q2009 = 130 × 80
100
Item
= RM104
A
Price index
4.90 × 100 = 140
3.50
B
8.10 × 100 = 135
6.00
(b) The table shows the prices of items P and Q in the years 2015 and 2017, and the price indices in the year
2017 based on the year 2015.
Jadual di bawah menunjukkan harga bagi barangan P dan Q pada tahun 2015 dan 2017, dan indeks harga pada tahun 2017
berasaskan tahun 2015.
Item
Barangan
Price/ Harga (RM)
2015
2017
Price index in the year 2017 based on the year 2015
Indeks harga pada tahun 2017 berasaskan tahun 2015
P
x
360
125
Q
125
y
140
Find the values of x and y. / Cari nilai x dan y.
Q2017
× 100 = 125
Q2015
360 × 100 = 125
x
x = 360 × 100
125
= 288
Q2017
× 100 = 140
Q2015
y × 100 = 140
125
y = 140 × 125
100
= 175
(c) The table shows the price per kg of items A, B and C in the year 2008 and 2009, and the price indices in
the year 2009 based on the year 2008.
Jadual di bawah menunjukkan harga per kg bagi barangan A, B dan C pada tahun 2008 dan 2009, dan indek harga pada tahun
2009 berasaskan tahun 2008.
Item
Price/ Harga
Barangan
2008
2009
Price index in the year 2009 based on the year 2008
Indeks harga pada tahun 2009 berasaskan tahun 2008
A
RM1.20
RM1.60
z
B
x
RM2.31
110
C
RM0.60
y
105
1
Additional Mathematics
Find the values of x, y and z. / Cari nilai x, y dan z.
Q2009
Q2009
× 100 = 110
× 100 = 105
Q2008
Q2008
2.31 × 100 = 110
y × 100 = 105
x
0.60
x = 2.31 × 100
y = 0.60 × 105
110
100
= RM2.10
= RM0.63
(d) The table shows the price indices of two
items A and B. Find the values of x and y.
Jadual di sebelah menunjukkan indeks harga bagi
dua barangan A dan B. Cari nilai x dan y.
Item
Barangan
A
B
For item A:
Q2008
× 100 = 120 ⇒
Q2005
Q2010
× 100 = 150 ⇒
Q2005
Q
x = 2010 × 100
Q2008
Q
Q
= 2010 × 2005 × 100
Q2005
Q2008
Q2008
= 120
Q2005
100
Q2010
= 150
Q2005
100
z = 1.60 × 100
1.20
= 133.33
Price index / Indeks harga
2008
2010
2010
(2005 = 100) (2005 = 100) (2008 = 100)
x
120
150
y
110
140
For item B:
Q2008
× 100 = 110 ⇒
Q2005
Q2010
× 100 = 140 ⇒
Q2008
Q
y = 2010 × 100
Q2005
Q
Q
= 2010 × 2008 × 100
Q2008
Q2005
= 150 × 100 × 100
100
120
= 140 × 110 × 100
100
100
= 125
= 154
Q2008
= 110
Q2005
100
Q2010
= 140
Q2008
100
(e) Price indices of two types of gold for certain years are given in the following table.
Indeks harga bagi dua jenis emas untuk tahun-tahun tertentu diberi dalam jadual berikut.
Price index / Indeks harga
Type of gold
Jenis emas
2016
(2002 = 100)
2018
(2002 = 100)
2018
(2016 = 100)
916
125
140
x
999
135
y
120
Calculate the value of x and of y.
Hitung nilai bagi x dan nilai bagi y.
For gold 916:
Q2016
× 100 = 125 ⇒
Q2002
Q2018
× 100 = 140 ⇒
Q2002
Q
x = 2018 × 100
Q2016
Q
Q
= 2018 × 2002 × 100
Q2002
Q2016
= 140 × 100 × 100
100
125
= 112
Q2016
= 125
100
Q2002
Q2018
= 140
Q2002
100
For gold 999:
Q2016
× 100 = 135 ⇒
Q2002
Q2018
× 100 = 120 ⇒
Q2016
Q
y = 2018 × 100
Q2002
Q
Q
= 2018 × 2016 × 100
Q2016
Q2002
= 120 × 135 × 100
100
100
= 162
Q2016
= 135
Q2002
100
Q2018
= 120
Q2016
100
Additional Mathematics
1
6. Solve each of the following.
Selesaikan setiap yang berikut.
Item
The table shows the price indices of four items in the year
2018 based on the year 2014 and their respective weightage.
Calculate the composite index in the year 2018 based on the
year 2014.
Jadual di sebelah menunjukkan indeks harga bagi empat barangan pada
tahun 2018 berasaskan tahun 2014 dan pemberatnya masing-masing. Hitung
indeks gubahan pada tahun 2018 berasaskan tahun 2014.
Price index Weightage
Indeks harga
Pemberat
Barangan
A
112.5
20
B
105
30
C
108
10
D
120
40
–
I = 112.5 × 20 + 105 × 30 + 108 × 10 + 120 × 40
20 + 30 + 10 + 40
11
280
=
100
= 112.8
(a)
Item
Barangan
Price index
Indeks harga
Weightage
Pemberat
P
100
4
Q
120
2
R
145
4
∑Iiwi
∑wi
The table shows the price indices of three items in
the year 2018 based on the year 2014 and their
respective weightage. Calculate the composite index
in the year 2018 based on the year 2014.
Jadual di sebelah menunjukkan indeks harga bagi tiga barangan
pada tahun 2018 berasaskan tahun 2014 dan pemberatnya
masing-masing. Hitung indeks gubahan pada tahun 2018
berasaskan tahun 2014.
–
I = 100 × 4 + 120 × 2 + 145 × 4
4+2+4
1
220
=
10
= 122
(b)
Price index in the year 1997
Ingredient
based on the year 1994
Bahan
Indeks harga pada tahun 1997
berasaskan tahun 1994
A
128
B
110
C
125
D
90
–
I = 128 × 1 + 110 × 1 + 125 × 1 + 90 × 1
4
453
=
4
= 113.25
The table shows the price indices of four types of
ingredients in the year 1997 based on the year 1994.
Calculate the composite index of the ingredients in
the year 1997 based on the year 1994.
Jadual di sebelah menunjukkan indeks harga bagi empat jenis
bahan pada tahun 1997 berasaskan tahun 1994. Hitung indeks
gubahan bagi bahan-bahan tersebut pada tahun 1997 berasaskan
tahun 1994.
1
Additional Mathematics
(c)
Component
Komponen
Price / Harga
(RM)
Weightage
Pemberat
2000
2004
E
70
105
4
F
40
36
2
G
50
60
6
H
100
125
3
Component
Komponen
E
F
G
H
Price index in the year 2004
based on the year 2000
Indeks harga pada tahun 2004
berasaskan tahun 2000
105 × 100 = 150
70
36 × 100 = 90
40
60 × 100 = 120
50
125 × 100 = 125
100
The table shows the prices of four components in the
years 2000 and 2004, and their respective weightage.
Calculate the composite index in the year 2004 based
on the year 2000.
Jadual di sebelah menunjukkan harga bagi empat komponen pada
tahun 2000 dan tahun 2004 serta pemberatnya masing-masing.
Hitung indeks gubahan pada tahun 2004 berasaskan tahun 2000.
Find the price index first before you
calculate the composite index.
Cari indeks harga terlebih dahulu sebelum
menghitung indeks gubahan.
–
I = 150 × 4 + 90 × 2 + 120 × 6 + 125 × 3
4+2+6+3
1
875
=
15
= 125
(d) The table shows the prices of four types of food in the years 2002 and 2007. The pie chart shows the
weightage for each type of food.
Jadual di bawah menunjukkan harga bagi empat jenis makanan pada tahun 2002 dan tahun 2007. Carta pai pula menunjukkan
pemberat bagi setiap jenis makanan.
Food
Makanan
Price / Harga
(RM)
2002
2007
A
1.40
1.75
B
2.00
2.80
C
4.00
6.00
D
3.00
2.40
D
30%
C
15%
A
35%
B
20%
Calculate the composite index in the year 2007 based on the year 2002.
Hitung indeks gubahan pada tahun 2007 berasaskan tahun 2002.
Food
Makanan
Price index in the year 2007
based on the year 2002
Indeks harga pada tahun 2007
berasaskan tahun 2002
A
1.75 × 100 = 125
1.40
B
2.80 × 100 = 140
2.00
C
6.00 × 100 = 150
4.00
D
2.40 × 100 = 80
3.00
–
I = 125 × 35 + 140 × 20 + 150 × 15 + 80 × 30
35 + 20 + 15 + 30
11
825
=
100
= 118.25
Additional Mathematics
1
7. Solve the following questions.
Selesaikan soalan-soalan berikut.
Barangan
Item
Price index
Indeks harga
Weightage
Pemberat
P
125
5
Q
x
3
R
110
2
The table shows the price indices of three items in the
year 1999 based on the year 1996 and their respective
weightage. Given the composite index in the year 1999
based on the year 1996 is 116. Find the value of x.
Jadual di sebelah menunjukkan indeks harga bagi tiga barangan pada
tahun 1999 berasaskan tahun 1996 dan pemberatnya masing-masing.
Diberi indeks gubahan pada tahun 1999 berasaskan tahun 1996 ialah
116. Cari nilai x.
125 × 5 + x(3) + 110 × 2 = 116
5+3+2
3x + 845 = 116
10
3x + 845 = 1 160
3x = 315
x = 105
(a)
Stationery
Alat tulis
Price index
Indeks harga
Weightage
Pemberat
K
x
1
L
120
3
M
130
2
N
114
4
The table shows the price indices of four types of
stationery in the year 2008 based on the year 2007 and
their respective weightage. Given the composite index in
the year 2008 based on the year 2007 was 114.6, find the
value of x.
Jadual di sebelah menunjukkan indeks harga bagi empat jenis alat tulis
pada tahun 2008 berasaskan tahun 2007 dan pemberatnya masingmasing. Diberi indeks gubahan pada tahun 2008 berasaskan tahun
2007 ialah 114.6, cari nilai x.
x(1) + 120 × 3 + 130 × 2 + 114 × 4 = 114.6
1+3+2+4
x + 1 076 = 114.6
10
x + 1 076 = 1 146
x = 70
(b)
Fruits
Buah-buahan
Price
index
Indeks harga
Weightage
Pemberat
watermelon
tembikai
112
4
papaya
betik
130
1
banana
pisang
120
2
pineapple
nanas
x
3
112 × 4 + 130 × 1 + 120 × 2 + x(3) = 119
4+1+2+3
3x + 818 = 119
10
3x + 818 = 1 190
3x = 372
x = 124
The table shows the price indices and the weightages of
four types of fruits in the year 2015 based on the year
2011. The composite index of all the four types of fruits in
the year 2015 based on the year 2011 was 119. Find the
value of x.
Jadual di sebelah menunjukkan indeks harga dan pemberat bagi empat
jenis buah-buahan pada tahun 2015 berasaskan tahun 2011. Indeks
gubahan bagi empat jenis buah-buahan tersebut pada tahun 2015
berasaskan tahun 2011 ialah 119. Cari nilai x.
1
Additional Mathematics
8. Solve each of the following.
Selesaikan setiap yang berikut.
The table shows the price indices of four items in the year 2008 based on the year 2004 and their respective
weightage.
Jadual di bawah menunjukkan indeks harga bagi empat barangan pada tahun 2008 berasaskan tahun 2004 dan pemberatnya
masing-masing.
Item / Barangan
Price index / Indeks harga
Weightage / Pemberat
A
120
1
B
114
y
C
x
5
D
130
2
The prices of item C in the year 2004 and the year 2008 were RM30 and RM42 respectively. Given the
composite index in the year 2008 based on the year 2004 is 128, find the value of x and of y.
Harga bagi barangan C pada tahun 2004 dan tahun 2008 masing-masing ialah RM30 dan RM42. Diberi indeks gubahan pada tahun
2008 berasaskan tahun 2004 ialah 128, cari nilai x dan nilai y.
–
Q
Composite index, I = 128
x = 2008 × 100
Q2004
120 × 1 + 114 × y + 140 × 5 + 130 × 2 = 128
1+y+5+2
42
=
× 100
30
114y + 1 080 = 128
y+8
= 140
114y + 1 080 = 128y + 1 024
14y = 56
y =4
(a) The table shows the price indices of three items in the year 2018 based on the year 2013 and their
respective weightage.
Jadual di bawah menunjukkan indeks harga bagi tiga barangan pada tahun 2018 berasaskan tahun 2013 dan pemberatnya
masing-masing.
Item / Barangan
Price index / Indeks harga
Weightage / Pemberat
A
136
y
B
x
2
C
108
3
The prices of item B in the year 2013 and 2018 were RM30 and RM34.20 respectively. The composite
index of the three items in the year 2018 based on the year 2013 was 120. Find the values of x and y.
Harga barangan B pada tahun 2013 dan 2018 masing-masing ialah RM30 dan RM34.20. Indeks gubahan bagi tiga barangan
itu pada tahun 2018 berasaskan tahun 2013 ialah 120. Cari nilai x dan y.
x =
Q2008
× 100
Q2013
= 34.20 × 100
30
= 114
–
Composite index, I = 120
136 × y + 114 × 2 + 108 × 3
y+2+3
136y + 552
y+5
136y + 552
16y
y
= 120
= 120
= 120y + 600
= 48
=3
1
Additional Mathematics
(b) The table shows the price indices of four components used to produce an electric item in the year 2006
based on the year 2004 and their respective weightage.
Jadual di bawah menunjukkan indeks harga bagi empat komponen yang digunakan untuk menghasilkan satu barang elektrik
pada tahun 2006 berasaskan tahun 2004 dan pemberatnya masing-masing.
Price / Harga
(RM)
Component
Komponen
(i)
Price index in the year 2006 based on the year 2004 Pemberat
Indeks harga pada tahun 2006 berasaskan tahun 2004
Weightage
2004
2006
Q
50
60
h
1
R
25
40
160
3
S
k
100
125
4
T
40
p
110
m
Find the values of h, k and p.
Cari nilai h, k dan p.
(ii) Given the composite index of all the components in the year 2006 based on the year 2004 is 132,
find the value of m.
Diberi indeks gubahan bagi semua komponen tersebut pada tahun 2006 berasaskan tahun 2004 ialah 132, cari nilai m.
(i)
h = 60 × 100
50
= 120
100 × 100 = 125
k
k = 100 × 100
125
= 80
p × 100 = 110
40
p = 110 × 40
100
= 44
–
Composite index, I = 132
(ii)
120 × 1 + 160 × 3 + 125 × 4 + 110 × m
1+3+4+m
1 100 + 110m
8+m
1 100 + 110m
22m
m
= 132
= 132
= 1 056 + 132m
= 44
=2
(c) The bar chart shows the weekly cost for five products P, Q, R, S and T of a company in the year 1998. The
table shows the prices and price indices for the five products.
Carta palang di bawah menunjukkan kos mingguan lima produk P, Q, R, S dan T bagi suatu syarikat pada tahun 1998. Jadual
pula menunjukkan harga dan indeks harga bagi lima produk tersebut.
Weekly cost (RM)
Kos mingguan (RM)
Product
Produk
33
30
24
15
12
0
(i)
P
Q
R
S
Product / Produk
T
Price / Harga
(RM)
Price index in the year
2013 based on the year
1998
Indeks harga pada tahun 2013
berasaskan tahun 1998
1998
2013
P
x
0.70
175
Q
2.00
2.50
125
R
4.00
5.50
y
S
6.00
9.00
150
T
2.50
z
120
Find the values of x, y and z.
Cari nilai bagi x, y dan z.
(ii) Find the composite index for the five products in the year 2013 based on the year 1998.
Cari indeks gubahan bagi lima produk tersebut pada tahun 2013 berasaskan tahun 1998.
1
Additional Mathematics
0.70 × 100 = 175
x
x = 0.70 × 100
175
= 0.40
(i)
–
I = 175 × 15 + 125 × 30 + 137.5 × 24 + 150 × 33 + 120 × 12
15 + 30 + 24 + 33 + 12
16
065
=
114
= 140.92
y = 5.50 × 100
4.00
= 137.5
z × 100 = 120
2.50
z = 120 × 2.50
100
=3
SPM
2
Paper
S
10%
1. The table shows the price indices and change in price
630
indices of four ingredients P, Q, R and S, which are used to
make a type of cake.
Jadual di bawah menunjukkan indeks harga dan perubahan
indeks harga bagi empat bahan P, Q R dan S, yang digunakan
untuk membuat sejenis kek.
Ingredient
Bahan
P
P A
Price index in 2010
based on 2008
Indeks harga pada
2010 berasaskan
2008
120
Change in price
index from 2010 to
2012
Perubahan indeks
harga dari 2010 ke
2012
10% decrease
Menyusut 10%
Q
108
No change
Tidak berubah
R
125
No change
Tidak berubah
S
150
10% increase
Menokok 10%
The diagram is a pie chart which represents the
percentage of usage of the ingredients used to make the
cake in 2008.
Rajah yang berikut adalah carta pai yang mewakili peratusan
penggunaan bahan-bahan yang digunakan untuk membuat kek
itu pada 2008.
R
40%
P
30%
Q
20%
(a) The price of ingredient P in 2010 was RM7.80. Find the
corresponding price in 2008.
Harga bahan P pada 2010 ialah RM7.80. Cari harga yang
sepadan pada 2008.
[2]
(b) Find the price indices of all the four ingredients in 2012
based on 2008.
Cari indeks harga bagi kesemua empat bahan itu pada
2012 berasaskan 2008.
[3]
(c) (i) Calculate the composite index for the cost of
making the cake in 2012 based on 2008.
Hitung indeks gubahan bagi kos membuat kek itu
pada 2012 berasaskan 2008.
(ii) Hence, find the cost of making the cake in 2008 if
the corresponding cost in 2012 was RM22.16.
Seterusnya, cari kos membuat kek itu pada 2008 jika
kos sepadan pada 2012 ialah RM22.16.
[5]
Ans: (a) RM6.50
(b) IP = 108, IQ = 108, IR = 125, IS = 165
(c) (i) 120.5; (ii) RM18.39
1
Additional Mathematics
2. The table shows the price indices in the years 2017 and
2019 based on the year 2015 of three materials P, Q and R
used in making a type of pen.
630
Jadual di bawah menunjukkan indeks harga bagi tahun 2017
dan 2019 berasaskan tahun 2015 untuk tiga jenis bahan P, Q
dan R yang digunakan untuk membuat sejenis pen.
Price index in the
year 2017 based on
the year 2015
Price index in the
year 2019 based on
the year 2015
Indeks harga
pada tahun 2017
berasaskan tahun
2015
Indeks harga
pada tahun 2019
berasaskan tahun
2015
P
125
x
Q
120
132
R
130
156
Material
Bahan
(a) The price of material P in the year 2015 was RM14.00
and its price in the year 2019 was RM24.50. Find
Harga bahan P pada tahun 2015 ialah RM14.00 dan
harganya pada tahun 2019 ialah RM24.50. Cari
(i) the value of x,
nilai x,
(ii) the price of material P in the year 2017.
harga bagi bahan P pada tahun 2017.
[3]
(b) The composite index for the production cost of the pen
in the year 2017 based on the year 2015 was 125. The
ratio of materials P, Q and R used was 1 : y : 3. Find
Indeks gubahan untuk kos pengeluaran pen itu pada tahun
2017 berasaskan tahun 2015 ialah 125. Nisbah bahanbahan P, Q dan R yang digunakan ialah 1 : y : 3. Cari
(i) the value of y,
nilai y,
(ii) the corresponding price of the pen in the year
2015 if the price of the pen in the year 2017 was
RM44.50.
harga sepadan bagi pen itu pada tahun 2015 jika
harga pen itu pada tahun 2017 ialah RM44.50.
[5]
(c)
Find the price index of material R in the year 2019
based on the year 2017.
Cari indeks harga bagi bahan R pada tahun 2019
berasaskan tahun 2017.
[2]
Ans: (a) (i) 175; (ii) RM17.50
(b) (i) 3; (ii) RM35.60
(c) 120
3. The table shows the prices and the price indices of three
types of ingredients R, S and T used in the production of a
type of meat balls.
630
Jadual di bawah menunjukkan harga dan indeks harga bagi
tiga jenis bahan R, S dan T yang digunakan dalam penghasilan
sejenis bebola daging.
Price (RM)
per kg
Ingredient Harga (RM)
Bahan
Price index in
2015 based on
2013
Indeks harga pada
2015 berasaskan
2013 2015
2013
per kg
Weightage
Pemberat
R
4.00
6.00
150
60
S
q
2.16
p
30
T
0.80
1.00
125
10
(a) The price of ingredient S is increased by 20% from 2013
to 2015.
Harga bahan S menokok sebanyak 20% dari 2013 hingga
2015.
(i) State the value of p.
Nyatakan nilai p.
(ii) Find the value of q.
Cari nilai q.
[3]
(b) Calculate the composite index for the cost of making
the meat balls in 2015 based on 2013.
Hitung indeks gubahan bagi kos membuat bebola daging
pada 2015 berasaskan 2013.
[2]
(c)
It is given that the composite index for the cost of
making meat balls increased by 50% from 2011 to
2015.
Diberi bahawa indeks gubahan bagi kos membuat bebola
daging meningkat sebanyak 50% dari 2011 hingga 2015.
(i) Calculate the composite index for the cost of
making the meat balls in 2013 based on 2011.
Hitung indeks gubahan bagi kos membuat bebola
daging pada 2013 berasaskan 2011.
(ii) The cost of making a meat ball was 8 sen in 2011.
Find the maximum number of meat balls that can
be produced using an allocation of RM86 in 2015.
Kos membuat sebiji bebola daging ialah 8 sen pada
2011. Cari bilangan maksimum bebola daging yang
boleh dihasilkan menggunakan peruntukan sebanyak
RM86 pada 2015.
[5]
Ans: (a) (i) 120; (ii) 1.80
(b) 138.5
(c) (i) 108.3; (ii) 716
Additional Mathematics
1
4. The table shows the information related to four ingredients
K, L, M and N used in the production of a type of bread.
630
Jadual di bawah menunjukkan maklumat berkaitan empat
bahan K, L, M dan N yang digunakan dalam pembuatan sejenis
roti.
Ingredient
Bahan
K
Change in price index
from 2010 to 2012
Perubahan indeks harga
dari 2010 ke 2012
20% increase
Menokok 20%
L
10% decrease
Menyusut 10%
M
40% increase
Menokok 40%
N
30% increase
Menokok 30%
Percentage
of usage (%)
Peratus
penggunaan
(%)
5. The table shows the price indices for the years 2012 and
2014 based on the year 2000 of three materials P, Q and R
used in making a type of handbag.
Jadual di bawah menunjukkan indeks harga bagi tahun 2012
dan 2014 berasaskan tahun 2000 untuk tiga jenis bahan P, Q
dan R yang digunakan untuk membuat sejenis beg tangan.
Price index in
the year 2012
based on the
year 2000
Price index in
the year 2014
based on the
year 2000
Indeks
harga pada
tahun 2012
berasaskan
tahun 2000
Indeks
harga pada
tahun 2014
berasaskan
tahun 2000
P
110
143
m
Q
120
h
3
R
125
115
2
Material
Bahan
10
30
40
The production cost for this bread was RM23 800 in the year
2012.
Weightage
Pemberat
(a) The price of material Q in the year 2000 was RM14.00
and its price in the year 2014 was RM16.10. Find
Kos pengeluaran bagi roti ini ialah RM23 800 pada tahun
2012.
(a) If the price of ingredient N in the year 2010 was
RM3.60, find its price in the year 2012.
Jika harga bahan N pada tahun 2010 ialah RM3.60, cari
harganya pada tahun 2012.
[2]
Harga bahan Q pada tahun 2000 ialah RM14.00 dan
harganya pada tahun 2014 ialah RM16.10. Cari
(i) the value of h,
nilai h,
(ii) the price of material Q in the year 2012.
harga bahan Q pada tahun 2012.
[3]
(b) Percentage of usage for several ingredients were given
in the table. Calculate the corresponding production
cost in the year 2010.
(b) The composite index for the production cost of the
handbag in the year 2012 based on the year 2000 was
116. Find
Indeks gubahan bagi kos pengeluaran beg tangan itu pada
tahun 2012 berasaskan tahun 2000 ialah 116. Cari
(i) the value of m,
nilai m,
(ii) the corresponding price of the handbag in the
year 2000 if the price of the handbag in the year
2012 was RM69.60.
harga sepadan bagi beg tangan itu pada tahun 2000
jika harga beg tangan itu pada tahun 2012 ialah
RM69.60.
[4]
Peratus penggunaan bagi beberapa bahan diberikan dalam
jadual di atas. Hitung kos pengeluaran yang sepadan pada
tahun 2010.
[5]
(c)
The production cost is expected to increase by 25%
from the year 2012 to the year 2015. Calculate the
percentage of changes in production cost from the year
2010 to the year 2015.
Kos pengeluaran dijangka meningkat sebanyak 25% dari
tahun 2012 ke tahun 2015. Hitung peratus perubahan
dalam kos pengeluaran dari tahun 2010 ke tahun 2015.
[3]
Ans: (a) RM4.68
(b) RM20 000
(c) 48.75% increase
(c)
Find the price index of material P for the year 2014
based on the year 2012.
Cari lndeks harga bagi bahan P pada tahun 2014
berasaskan tahun 2012.
[3]
Ans: (a) (i) 115; (ii) RM16.80
(b) (i) 5; (ii) RM60.00
(c) 130
Form 4 Answers
3
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© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
3
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© Penerbitan Pelangi Sdn. Bhd.
Form 4 Answers
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© Penerbitan Pelangi Sdn. Bhd.
SPM
Paper 1
[2 hours/ jam]
[80 marks/ markah]
Instruction: Answer all questions.
Arahan : Jawab semua soalan.
1.
Given the points A(1, 5), B(3, 2) and C(6, k) are
collinear, find
Diberi titik-titik A(1, 5), B(3, 2) dan C(6, k) adalah segaris, cari
the value of k,
nilai k,
the ratio of AB to AC in the form of m : n.
nisbah AB kepada AC dalam bentuk m : n.
[4 marks / 4 markah]
Answer/ Jawapan :
5–2
1–3
3
–
2
9
5
k
(a)
(b)
=
=
=
=
=
k–2
6–3
k–2
3
–2k + 4
–2k
5
–
2
m(6) + n(1)
,
m+n
m–
3.
Find the value of y in the equation
Cari nilai y bagi persamaan
log5 (4y + 1) – log5 (y – 5) = log4 16
[3 marks/ 3 markah]
Answer Jawapan :
(4y + 1)
log5
= log4 42
(y – 5)
log5
(4y + 1)
=2
(y – 5)
(4y + 1)
= 52
(y – 5)
4y + 1 = 25y – 125
21y = 126
y=6
5
+ n(5)
2
m+n
6m + n
=3
m+n
6m + n = 3m + 3n
3m = 2n
m
2
=
n
3
4.
= (3, 2)
A(1, 5)
m
\ AB : AC = m : m + n
=2:5
B(3, 2)
n
C(6, – 5 )
2
The second term of a geometric progression is 162 and
the fifth term is –48. Find
Sebutan kedua suatu janjang geometri ialah 162 dan sebutan
kelima ialah –48. Cari
(a) the common ratio,
nisbah sepunya,
(b) the first term,
sebutan pertama.
(c) the sum to infinity.
hasil tambah hingga ketakterhinggaan.
[5 marks/ 5 markah]
Answer/ Jawapan :
(a) ar = 162 ....................
ar4 = –48 ...................
÷
2.
Solve the equation ex(2ex + 1) = 10.
Selesaikan persamaan ex(2ex + 1) = 10.
[4 marks/ 4 markah]
Answer/ Jawapan :
Let y = ex,
–48
162
8
=–
27
2
r=–
3
: r3 =
162
2
–
3
= –243
(b) a =
y(2y + 1) = 10
2y2 + y – 10 = 0
(2y + 5)(y – 2) = 0
5
,y=2
2
5
ex ≠ – , ex = 2
2
x = ln 2
y=–
= 0.6931
–243
2
1– –
3
= –145.8
(c) S∞ =
Additional Mathematics
5.
7.
Solve the equation:
Selesaikan persamaan:
log2 (3 − 2x) − 2 log2 x = log2 5
[4 marks/ 4 markah]
Answer/ Jawapan :
log2 (3 − 2x) − log2 x2= log2 5
(3 – 2x)
x2
(3 – 2x)
x2
5x2 + 2x – 3
(5x – 3)(x + 1)
log2
= log2 5
=5
=0
=0
3
x = , x ≠ –1
5
\x=
The coordinates of point A are (1, 6) and the
coordinates of point B are (–7, 2). Find the equation
of the perpendicular bisector of straight line AB, giving
your answer in the form of y = mx + c.
Koordinat titik A ialah (1, 6) dan koordinat titik B ialah
(–7, 2). Cari persamaan pembahagi dua sama serenjang bagi
garis lurus AB, dengan memberi jawapan anda dalam bentuk
y = mx + c.
[4 marks / 4 markah]
Answer/ Jawapan :
1 + (–7) 6 + 2
Midpoint of AB =
,
2
2
= (–3, 4)
6–2
1 – (–7)
1
=
2
mAB =
3
5
m = –2
6.
The equation: y – 4 = –2[x – (–3)]
y – 4 = –2x – 6
y = –2x – 2
The function f is defined by f(x) = x2 – 1 for x 0.
Fungsi f ditakrifkan sebagai f(x) = x2 – 1 bagi x 0.
(a) Define the inverse function f −1 in a similar form.
Takrifkan fungsi songsang f −1 dalam bentuk yang sama.
(b) Solve the equation
Selesaikan persamaan
209
16
[7 marks / 7 markah]
f 2(x) =
Answer/ Jawapan :
(a) Let y = f(x), then x = f –1(y)
y = x2 – 1
x2 = y + 1
x= y+1
\ f −1(x) = x + 1 for x
(b)
–1
209
16
209
2
f (x – 1) =
16
209
2
2
(x – 1) – 1 =
16
209
2
2
(x – 1) =
+1
16
225
=
16
15
2
x –1=±
4
15
15
2
x =
+ 1, x2 = –
+1
4
4
19
11
x2 =
x2 ≠ –
4
4
f 2(x) =
\ x = ± 19
2
8.
In an arithmetic progression, the sum of the first ten
terms is 400 and the sum of the next five terms is 425.
Find the common difference and the first term.
Dalam suatu janjang aritmetik, hasil tambah sepuluh sebutan
pertama ialah 400 dan hasil tambah lima sebutan berikutnya
ialah 425. Cari beza sepunya dan sebutan pertama.
[4 marks / 4 markah]
Answer/ Jawapan :
S10 = 400
10
(2a + 9d) = 400
2
2a + 9d = 80 ............................................
S15 – S10 = 425
15
10
(2a + 14d) –
(2a + 9d) = 425
2
2
15a + 105d – 10a – 45d = 425
5a + 60d = 425
a + 12d = 85 ...................
From
: a = 85 – 12d ....................................
Substitute
into :
2(85 – 12d) + 9d = 80
170 – 24d + 9d = 80
15d = 90
d=6
a = 85 – 12(6)
= 13
Additional Mathematics
9.
1
—
2
—
Find in its simplest form, the product of a 3 – b 3 and
2
—
1 2
— —
12.
4
—
a3 + a3b3 + b3.
1
—
2
—
Cari dalam bentuk paling ringkas, hasil darab bagi a 3 – b 3
2
—
1 2
— —
salah satu sisi ialah (2 3 + 2 ) m. Cari tanpa menggunakan
kalkulator, panjang sisi yang satu lagi dalam bentuk surd yang
teringkas.
[3 marks/ 3 markah]
4
—
dan a 3 + a 3 b 3 + b 3 .
[2 marks / 2 markah]
Answer/ Jawapan :
1
—
2
—
2
—
1 2
— —
4
—
Answer/ Jawapan :
Let the length = y m
a3 – b3 a3 + a3b3 + b3
2 2
— —
1 4
— —
2 2
— —
1 4
— —
= a + a 3 b 3 + a 3 b 3 – a 3 b 3 – a 3 b 3 – b2
(2 3 + 2)y = (1 + 6)
= a – b2
10.
The area of a rectangle is (1 + 6) m2. The length of one
side is (2 3 + 2) m. Find without using a calculator,
the length of the other side in the simplest surd form.
Luas suatu segi empat tepat ialah (1 + 6 ) m2. Panjang bagi
y =
=
1
1+k
and p =
, express p in its
1–k
3
simplest surd form.
1
1+k
Diberi k =
dan p =
, ungkapkan p dalam bentuk
1–k
3
surd yang paling ringkas.
Given that k =
=
=
1+ 6
2 3+ 2
×
2 3– 2
2 3– 2
2 3 – 2 + ( 6)(2 3) – ( 6)( 2)
4(3) – 2
=
2 3– 2+6 2–2 3
10
=
5 2
10
=
2
2
3+1
3
3–1
3
3+1
3–1
3+1
×
3+1
13.
= 3+2 3+1
3–1
= 4+2 3
2
=2+ 3
11.
2 3+ 2
=
[2 marks / 2 markah]
Answer/ Jawapan :
1+k
p =
1–k
1
1+
3
=
1
1–
3
1+ 6
×5
Given that 2
= 2 × 5 , evaluate 10 .
Diberi bahawa 22x + 3 × 5x – 1 = 23x × 52x, nilaikan 10x.
[3 marks / 3 markah]
2x + 3
x–1
3x
2x
Answer/ Jawapan :
22x + 3 × 5x – 1 = 23x × 52x
2x
2 × 23 × 5x × 5–1 = 23x × 52x
23 × 5–1 = 23x × 52x ÷ 22x ÷ 5x
8
= 2x × 5x
5
8
= 10x
5
x
The position vectors of points A and B, relative to the
origin are 7i
and –2i~ + 9j respectively. Find the
~ – 3j
~
~
→
unit vector parallel to AB .
Vektor kedudukan bagi titik-titik A dan B, berhubung
dengan titik asalan masing-masing ialah 7i – 3j dan
~ ~
→
–2i + 9j . Cari vektor unit yang selari dengan AB .
~ ~
[3 marks/ 3 markah]
Answer/ Jawapan :
→
→
→
AB = AO + OB
= –7i~ + 3j + (–2i~ + 9j )
~
~
= –9i~ + 12j
~
→
| AB | = (–9)2 + 122
= 15
→
Unit vector parallel to AB is
1
3
4
(–9i + 12j ) = – ~i + j
~
15 ~
5
5~
Additional Mathematics
14.
Find the range of values of x for which 2x2 7x + 15.
Cari julat nilai x dengan keadaan 2x2 7x + 15.
[4 marks/ 4 markah]
Answer/ Jawapan :
2x2 – 7x – 15 0
(2x + 3)(x – 5) 0
–
\x
3
–
,x
2
3
2
5
17.
f(x) = 9 x –
x
5
Its vertex is
8x + 1 + 20(22x)
2x – 316x + 2
[3 marks/ 3 markah]
8
+ 20(2 )
(2 )
+ (2 × 5)(2 )
=
2x – 316x + 2
2x – 3(24)x + 2
=
1
, –11 and it is a minimum point.
3
ex + 1
for the domain
3
x 0. Determine the domain of inverse function, f −1.
ex + 1
Suatu fungsi ditakrifkan oleh f : x →
bagi domain x 0.
3
−1
Tentukan domain bagi fungsi songsang, f .
[2 marks/ 2 markah]
A function f is defined by f : x →
1
.
3
Range of f(x) is f(x)
3 x+1
– 11.
Answer/ Jawapan :
Answer/ Jawapan :
2x
2
3 by
Answer/ Jawapan :
Simplify
Ringkaskan
x+1
1
3
x
State the coordinates of the vertex and determine
whether it is a maximum point or a minimum point.
Nyatakan koordinat verteksnya dan tentukan sama ada verteks
itu ialah titik maksimum atau titik minimum.
[2 marks/ 2 markah]
18.
15.
The function f is defined for the domain –3
Fungsi f ditakrifkan bagi domain –3 x 3 oleh
2
2x
1
.
3
Hence, domain of f −1 is x
23x + 3 + (22 + 2x × 5)
2x – 3 + 4x + 8
22x + 2(2x + 1 + 5)
25x + 5
= 22x + 2 – 5x – 5(2x + 1 + 5)
=
= 2–3x – 3(2x + 1 + 5)
19.
The equation of a curve is y = x2 + ax + 5, where a is a
constant. Given the equation of the curve can also be
written as y = (x + 2)2 + b, find the value of a and of b.
Persamaan suatu lengkung ialah y = x2 + ax + 5, dengan
keadaan a ialah pemalar. Diberi persamaan lengkung itu juga
boleh ditulis sebagai y = (x + 2)2 + b, cari nilai a dan nilai b.
[3 marks/ 3 markah]
Answer Jawapan :
16.
The position vectors of A and B, relative to an origin O
are qi
~ – ~j and 11i
~ + 10j~ respectively, find the value of
the constant q if OAB are collinear.
Vektor kedudukan bagi A dan B, berhubung dengan suatu titik
asalan O masing-masing ialah q~
i – j dan 11i
, cari
~ + 10j
~
~
nilai pemalar q jika OAB adalah segaris.
[2 marks/ 2 markah]
Answer/ Jawapan :
→
→
OA = k OB
qi
)
~ – ~j = k(11i
~ + 10j
~
qi
~ – ~j = 11ki~ + 10kj
~
–1 = 10k
1
k=–
10
q = 11 –
=–
11
10
y = x2 + ax + 5
a
2
= x2 + ax +
= x+
a
2
2
–
a
2
–
2
a
2
2
+5
+5
Compare with y = (x + 2)2 + b,
a
=2
2
a=4
–
a
2
2
+5=b
b=–
=1
1
10
2
4
2
2
+5
Additional Mathematics
20.
→
The vector OP has a magnitude of 15 units and is
→
parallel to the vector 3i~ – 4j . Find the vector OP in
~
the same form.
→
Vektor OP mempunyai magnitud 15 unit dan selari dengan
→
vektor 3~
i – 4 j . Cari vektor OP dalam bentuk yang sama.
~
[3 marks/ 3 markah]
Answer Jawapan :
→
Let OP = k(3i~ – 4j ),
~
(3k)2 + (–4k)2 = 15
Solve the equation 2x2 – 3x – 4 = 0. Give your answers
correct to two decimal places.
Selesaikan persamaan 2x2 – 3x – 4 = 0. Berikan jawapan anda
betul kepada dua tempat perpuluhan.
[2 marks/ 2 markah]
Answer/ Jawapan :
2
x = –b ± b – 4ac
2a
2
= –(–3) ± (–3) – 4(2)(–4)
2(2)
3
±
41
=
4
= 2.35, –0.85
25k2 = 15
5k = 15
k=3
→
\ OP = 3(3i
)
~ – 4j
~
= 9i
–
12j
~
~
24.
21
Variables x and y are related such that when 2y is plotted
against x2, a straight line passing through the points
(0.2, 1) and (0.5, 1.6) is obtained. Find the value of
2y when x = 0.
Pemboleh ubah x dan y dihubungkan dengan keadaan apabila
2y diplotkan melawan x2, satu garis lurus yang melalui
titik-titik (0.2, 1) dan (0.5, 1.6) diperolehi. Cari nilai 2y apabila
x = 0.
[3 marks/ 3 markah]
Sketch the graph y = |2x – 3|. Indicate clearly the
intersections on the axes.
Lakarkan graf y = |2x – 3|. Tandakan dengan jelas titik
persilangan pada kedua-dua paksi.
[3 marks 3
]
Answer Jawapan :
y
3
Answer Jawapan :
Y = mX + c
2y = mx2 + c
O
1 – 1.6
(0.2)2 + c
0.2 – 0.5
1 = 2(0.04) + c
c = 0.92
3
2
x
At (0.2, 1), 1 =
When x = 0, then x2 = 0.
25.
When x2 = 0, 2y = c = 0.92
20
The function g is defined by g(x) =
for x
3.
x–1
–1
Find g (4).
20
Fungsi g ditakrifkan sebagai g(x) =
bagi x
3. Cari
x–1
–1
g (4).
[2 marks/ 2 markah]
Answer/ Jawapan :
Let y = g –1(4)
22.
Solve the equation 32x – 1 = 15.
Selesaikan persamaan 32x – 1 = 15.
g(y) = 4
[3 marks/ 3 markah]
Answer/ Jawapan :
log10 32x – 1 = log10 15
(2x – 1) log10 3 = log10 15
log10 15
2x =
+1
log10 3
x = 1.7325
20
=4
y–1
y–1=5
y=6
\ g –1(4) = 6
Additional Mathematics
Paper 2
[2 1 hours/ jam]
2
Section A
Bahagian A
[40 marks/ markah]
Instruction: Answer all questions in this section.
Arahan: Jawab semua soalan dalam bahagian ini.
1.
Diagram 1 shows part of the curve y = f(x), where
f(x) = p – ex and p is a constant.
Rajah 1 menunjukkan sebahagian lengkung y = f(x), dengan
keadaan f(x) = p – ex dan p ialah pemalar.
y
2
x
O
D D
Rajah 1
2. The line y + 4x = 23 intersects the curve xy – 3x = 16
at two points A and B. Find the equation of the
perpendicular bisector of the straight line AB.
Garis y + 4x = 23 menyilang lengkung xy – 3x = 16 pada dua
titik A dan B. Cari persamaan pembahagi dua sama serenjang
bagi garis lurus AB.
[7 marks/ 7 markah]
Answer/ Jawapan :
y + 4x = 23 .................................
xy – 3x = 16 ................................
From
: y = 23 – 4x .................
The curve crosses the y-axis at (0, 2).
Lengkung itu menyilang paksi-y pada (0, 2).
(a) Find the value of p.
Cari nilai p.
[2 marks/ 2 markah]
Substitute
into :
x(23 – 4x) – 3x – 16= 0
23x – 4x2 – 3x – 16 = 0
4x2 – 20x + 16 = 0
x2 – 5x + 4 = 0
(x – 4)(x – 1) = 0
x = 1, 4
(b) Find the coordinates of the point where the curve
crosses the x-axis.
Cari koordinat bagi titik dengan keadaan lengkung itu
menyilang paksi-x.
[3 marks/ 3 markah]
When x = 1, y = 23 – 4(1) = 19
When x = 4, y = 23 – 4(4) = 7
\ A(1, 19), B(4, 7)
(c) Copy the diagram above and on it sketch the
graph of y = f –1(x).
Salin rajah di atas dan lakar graf bagi y = f –1(x) padanya.
[3 marks/ 3 markah]
19 – 7
1–4
12
=
–3
= –4
mAB =
Answer/ Jawapan :
(a) When x = 0, y = 2,
p – e0 = 2
p – 1= 2
p= 3
–4 × m = –1
1
m=
4
(b) When y = 0,
0 = 3 – ex
ex = 3
x = ln 3
Equation of the perpendicular bisector of AB:
\ (ln 3, 0)
(c)
1 + 4 19 + 7
,
2
2
5
=
, 13
2
Midpoint of AB =
y – 13 =
1
5
x–
4
2
y – 13 =
1
5
x–
4
8
y
2
y=x
f
ln 3
O
f –1
ln 3
2
x
y=
1
99
x+
4
8
Additional Mathematics
3.
Given that u = log3 x, express in simplest form in
terms of u,
Diberi bahawa u = log3 x, ungkapkan dalam bentuk yang
paling ringkas dalam sebutan u,
(i) x,
9
(ii) log3
,
x
(iii) logx 27.
[5 marks/ 5 markah]
4.
Diagram 2 shows a right-angled triangle ABC.
Rajah 2 menunjukkan sebuah segi tiga bersudut tegak.
B
(b) Solve the equation (log4 y)2 + (log4 y2) = 8.
Selesaikan persamaan (log 4 y)2 + (log 4 y2) = 8.
[4 marks/ 4 markah]
Answer/ Jawapan :
(a) (i) u = log3 x
x = 3u
(ii) log3
9
=
x
=
=
=
=
(iii) logx 27 =
log3 9 – log3 x
log3 32 – u
2 log3 3 – u
2(1) – u
2–u
log3 27
log3 x
log3 33
u
3 log3 3
=
u
3(1)
=
u
3
=
u
=
=
=
=
=
D D
Rajah 2
The area of the triangle is (7 + 15) cm2.
Luas segi tiga itu ialah (7 + 15) cm2.
(a) Find the length of AB, in cm, in the form of
(a 5 + b 3), where a and b are constants.
Cari panjang AB, dalam cm, dalam bentuk (a 5 + b 3),
dengan keadaan a dan b ialah pemalar
[4 marks/ 4 markah]
(b) Express BC2 in the form of (c + d 5) cm2, where c
and d are constants.
Ungkapkan BC2 dalam bentuk (c + d 5) cm2, dengan
keadaan c dan d ialah pemalar.
[2 marks/ 2 markah]
(a)
log4 y
0
0
2, –4
1
× AB × ( 5 + 3) = 7 + 15
2
5– 3
2(7 + 15)
×
AB =
5– 3
5+ 3
=
2(7 + 15)( 5 – 3)
( 5)2 – ( 3)2
=
2(7 5 – 7 3 + ( 5 × 3)( 5) – ( 5 × 3)( 3)
2
=7 5–7 3+5 3–3 5
=4 5–2 3
When log4 y = 2, y = 42 = 16
When log4 y = –4, y = 4-4 =
( 5 + 3 ) cm C
Answer/ Jawapan :
(b) (log4 y)2 + (log4 y2) = 8
(log4 y)2 + 2 log4 y = 8
Let v
v 2 + 2v – 8
(v – 2)(v + 4)
v
A
1
256
(b) BC2 = (4 5 – 2 3)2 + ( 5 + 3)2
= 80 – 16 15 + 12 + 5 + 2 15 + 3
= 100 – 14 15
Additional Mathematics
Given that
Diberi bahawa
b3 – x
(ay + 1)2
1
=
×
x
a
by
ab6
Find the value of x and of y.
Cari nilai x dan nilai y.
[4 marks/ 4 markah]
a2y + 2 – x × b(3 – x) – y = a–1 × b–6
2y + 2 – x = –1
2y – x = –3 .....................
3 – x – y = –6
x + y = 9 .......................
Answer/ Jawapan :
(a) α + β = –
αβ =
: 3y = 6
y=2
Substitute y = 2 into
x+2=9
x=7
1
1
and 2 .
α2
β
1
1
persamaan kuadratik dengan punca 2 dan 2 .
α
β
[4 marks/ 4 markah]
(b) the quadratic equation of roots
Answer/ Jawapan :
+
If α and β are the roots of the equation 2x2 – 4x + 3 = 0,
find
Jika α dan β ialah punca bagi persamaan 2x2 – 4x + 3 = 0,
cari
(a) the value of α2 + β2,
nilai α2 + β2,
[2 marks/ 2 markah]
3
2
(–4)
=2
2
(α + β)2 = α2 + 2αβ + β2
:
α2 + β2 = (α + β)2 – 2αβ
3
= (2)2 – 2
2
=1
(b) The quadratic equation is
x2 –
1
1
1
1
+ 2 x+ 2 × 2 =0
α2
β
α
β
β2 + α2
1
x2 –
x+ 2 2 =0
α2 β2
α β
x2 –
1
3
2
2
x+
1
3
2
2
=0
4
4
x+
=0
9
9
9x2 – 4x + 4 = 0
x2 –
Additional Mathematics
Section B
Bahagian B
[40 marks/ markah]
Instruction: Answer any four questions from this section.
Arahan: Jawab mana-mana empat soalan daripada bahagian ini.
7.
Diagram 3, which is not drawn to scale, shows a
quadrilateral ABCD.
Rajah 3 yang tidak dilukis mengikut skala menunjukkan sebuah
sisi empat ABCD.
y
B(0, h)
4y = x +
(b) (i)
(ii) Area of ABCD
0
4
4
2
1 2
=
7
8
7
–3
2 –3
1
=
|(14 + 0 + 28 – 12) – (0 + 28 + 32 + 14)|
2
= 22 unit2
x
Area of BDC : Area of ABCD
= 2 : 22
= 1 : 11
D D
Rajah 3
The equation of the line BC is 4y = x + 28. Point C lies
on the straight line y = 2x. The line CD is parallel to the
y-axis.
Persamaan garis lurus BC ialah 4y = x + 28. Titik C berada
pada garis lurus y = 2x. Garis CD selari dengan paksi-y.
(a) Find / Cari
(i) the coordinates of C,
koordinat bagi C,
(ii) the value of h and of k.
nilai h dan nilai k.
[5 marks/ 5 markah]
(b) (i)
Show that the area of triangle BDC is 2 unit2.
Tunjukkan bahawa luas segi tiga BDC ialah 2 unit2.
(ii) Hence, find area of triangle BDC : area of
quadrilateral ABCD.
Seterusnya, cari luas segi tiga BDC : luas sisi empat
ABCD.
[5 marks/ 5 markah]
Answer/ Jawapan :
At C, 4(2x)
8x
7x
x
=
=
=
=
\ C(4, 8)
⇒
y=
8.
The table shows the values of variables x and y which
a
are related by the equation y =
, where a and b
x+b
are constants.
Jadual berikut menunjukkan nilai-nilai bagi pemboleh ubah x
a
dan y yang dihubungkan oleh persamaan y =
, dengan
x+b
keadaan a dan b ialah pemalar.
x
0.1
0.4
1.0
1.8
3.0
y
8.0
6.0
4.0
2.8
1.9
7D
Jadual 1
(a) Using a scale of 2 cm to 1 unit on the xy-axis and
2 cm to 1 unit on the y-axis, plot y against xy and
hence draw a line of best fit.
Dengan menggunakan skala 2 cm kepada 1 unit pada
paksi-xy dan 2 cm kepada 1 unit pada paksi-y, plot
y melawan xy dan seterusnya lukis satu garis lurus
penyuaian terbaik.
[3 marks/ 3 markah]
x + 28
x + 28
28
4
y = 2(4)
=8
(ii) 4y = x + 28
0
7
1
|(0 + 32 + 28) – (28 + 28 + 0)|
2
= 2 unit2
A(2, –3)
(a) (i)
4
8
=
y = 2x
28 C
D(k, 7)
O
Area of ∆BDC
4
1 0
=
7
2 7
1
x+7
4
h is the y-intercept of line BC, so h = 7
CD parallel to the y-axis, so k = 4
(b) Use your graph to estimate the values of a and b.
Gunakan graf anda untuk menganggarkan nilai a dan
nilai b.
[4 marks/ 4 markah]
(c) From the graph obtained, find the value of
Daripada graf yang diperolehi, cari nilai bagi
(i) x when y = 5.2 / x apabila y = 5.2
(ii) x when xy = 2.8 / x apabila xy = 2.8
[3 marks/ 3 markah]
Additional Mathematics
Answer/ Jawapan :
a
(a)
y=
x+b
y(x + b) = a
by = –xy + a
1
a
y= –
xy +
b
b
xy
0.8
2.4
4.0
5.04
5.7
y
8.0
6.0
4.0
2.8
1.9
2 cm
y
2 cm
9
8
7
6
5
4
3
2
1
0
(b) From the graph, –
a
=9
b
a = 9(0.8)
= 7.2
1
1
9–7
=
b
0 – 1.6
= –1.25
b = 0.8
2
3
4
5
6
(c) (i)
xy
When y = 5.2, xy = 3.05
3.05
x=
5.2
= 0.5865
(ii) When xy = 2.8, y = 5.5
x(5.5) = 2.8
x = 0.5091
Additional Mathematics
1
1
and –
as
2
16
the third and sixth terms respectively. Find
1
1
Diberi suatu janjang geometri dengan
dan –
2
16
masing-masing sebagai sebutan ketiga dan keenamnya.
Cari
(i) the common ratio,
nisbah sepunya,
(ii) the first term,
sebutan pertama,
(iii) the nth term in terms of n.
sebutan ke-n dalam sebutan n.
[5 marks/ 5 markah]
(ii)
9. (a) Given a geometric progression with
(b) Ros was given a certain number of cups to be
stacked up to take the shape of a triangle so that
the topmost row consists of 1 cup, the second row
has 2 cups, the third row has 3 cups and so on.
Ros diberikan beberapa biji cawan untuk disusun supaya
membentuk satu segi tiga dengan keadaan baris tertinggi
terdiri daripada 1 cawan, baris kedua mempunyai 2 cawan,
baris ketiga mempunyai 3 cawan dan seterusnya.
(i) Express the total number of cups required in
terms of n if the triangle formed consists of n
rows.
Ungkapkan jumlah bilangan cawan yang diperlukan
dalam sebutan n jika segi tiga yang dibentukkan
terdiri daripada n barisan.
(ii) Hence, determine the number of possible
rows if 100 cups are available and the number
of cups not being used in the formation of
such triangle.
Seterusnya, tentukan bilangan baris yang mungkin
jika 100 biji cawan diberikan dan bilangan cawan
yang tidak digunakan dalam pembentukan segi tiga
ini.
[5 marks/ 5 markah]
Answer/ Jawapan :
1
(a) ar2 =
................................
2
1
ar5 = –
............................
16
(i)
(ii) a –
: r3 = –
1
2
1
2
a=2
2
(iii) Tn = 2 –
(b) (i)
=
1
2
n
100
100
Sn = n [a + Tn)
2
0
–1 ± 12 – 4 (1)(–200)
2(1)
Since n
0, n = 13.65
Thus, there will be 13 rows.
Total number of cups in the triangle
13
=
(1 + 13)
2
= 91
The number of cups not being used
= 100 – 91
=9
10.
→
The vector OP has a magnitude of 26 units and is
→
parallel to the vector 5i
. The vector OQ has a
~ – 12j
~
magnitude of 39 units and is parallel to the vector
12i
.
~ + 5j
~
→
Vektor OP mempunyai magnitud 26 unit dan selari dengan
→
vektor 5 i – 12 j . Vektor OQ mempunyai magnitud 39 unit dan
~
~
selari dengan vektor 12 i + 5 j .
~
~
→
→
(a) Express OP and OQ in terms of ~i and j .
~
→
→
Ungkapkan OP dan OQ dalam sebutan i dan j .
~
~
[3 marks/ 3 markah]
→
(b) Given the magnitude of PQ is λ 13, find the
value λ.
→
Diberi magnitud PQ ialah λ 13, cari nilai λ.
[3 marks/ 3 markah]
(a) | 5i
| = 52 + (–12)2 = 13
~ – 12j
~
→
| OP | = 26 = 2 × 13
→
OP = 2(5i~ – 12j ) = 10i~ – 24j
~
~
n–1
a = 1, d = 1, n = n
n
Sn = [2(1) + (n – 1)(1)]
2
n
= (n + 1)
2
Sn
n
(1 + n)
2
n2 + n – 200
Last term
(c) Given R lies on PQ such that PR : RQ = 4 : 9, find
→
the unit vector of OR.
Diberi R berada pada PQ dengan keadaan PR : RQ = 4 : 9,
→
cari vektor unit bagi OR .
[4 marks/ 4 markah]
Answer/ Jawapan :
1
8
1
r=–
2
÷
Tn = 1 + (n – 1)(1)
=n
n
Sn = 2 [2a + (n – 1)d]
| 12i~ + 5j | = 122 + 52 = 13
~
→
| OQ | = 39 = 3 × 13
→
) = 36i
OQ = 3(12i
~ + 5j
~ + 15j
~
~
Additional Mathematics
→
→ →
(b) | PQ | = | PO + OQ |
= | (–10i~ + 24j ) + (36i
)|
~ + 15j
~
~
= | 26i
~ + 39j~ |
= 262 + 392
= 2 197
= 13 13
\λ = 13
→
→ →
(c) OR = OP + PR
4 →
= 10i
+
PQ
~ – 24j
~ 13
4
= 10i
+
(26i + 39j )
~ – 24j
~ 13 ~
~
= 18i
~ – 12j
~
(c) If the point C divides the line AB in the ratio of
CB : AB = 2 : 5, find the coordinates of C.
Jika titik C membahagikan garis AB dengan nisbah
CB : AB = 2 : 5, cari koordinat C.
[2 marks/ 2 markah]
(d) A point P moves such that its distance from B is
always twice its distance from the x-axis. Find the
equation of the locus of P.
Suatu titik P bergerak dengan keadaan jaraknya dari
B adalah sentiasa dua kali jaraknya dari paksi-x. Cari
persamaan lokus bagi P.
[3 marks/ 3 markah]
Answer/ Jawapan :
1 0
–4
6
0
2 0
2
–8
0
1
= |(0 + 32 + 0) – (0 + 12 + 0)|
2
= 10 unit2
(a) Area of ∆OAB =
→
Unit vector of OR
18i~ – 12j
~
=
| 18i~ – 12j |
~
18i~ – 12j
~
=
182 + 122
18i~ – 12j
~
=
468
18i~ – 12j
~
=
6 13
3
2
=
i –
j
13 ~
13 ~
2 – (–8)
–4 – 6
= –1
(b) Gradient of AB =
Gradient of perpendicular bisector, m = 1
–4 + 6 2 + (–8)
,
2
2
= (1, –3)
Midpoint of AB =
Equation:
y – (–3) = 1(x – 1)
y+3 =x–1
y =x–4
(c)
11.
Diagram 4 shows a triangle OAB where O is the origin.
Point C lies on the line AB.
Rajah 4 menunjukkan sebuah segi tiga OAB dengan keadaan O
ialah asalan. Titik C terletak pada garis AB.
A(–4, 2)
3
A(–4, 2)
2
C
B(6, –8)
3(6) + 2(–4) 3(–8) + 2(2)
,
3+2
3+2
= (2, –4)
C =
y
O
x
C
(d) Let a point on the x-axis be (x, 0).
PB : PX = 2 : 1
PB
2
=
PX
1
PB = 2PX
B(6, –8)
D D
Rajah 4
(a) Calculate the area of triangle OAB.
Hitung luas segi tiga OAB.
[2 marks/ 2 markah]
(b) Find the equation of the perpendicular bisector of
line AB.
Cari persamaan pembahagi dua sama serenjang bagi garis
AB.
[3 marks/ 3 markah]
(x – 6)2 + (y + 8)2
(x – 6)2 + (y + 8)2
2
x – 12x + 36 + y2 + 16y + 64
x2 – 3y2 – 12x + 16y + 100
=
=
=
=
2 (y – 0)2
4y2
4y2
0
Additional Mathematics
Section C
Bahagian C
[20 marks/ markah]
Instruction: Answer any two questions from this section.
Arahan: Jawab mana-mana dua soalan daripada bahagian ini.
12.
Diagram 5 shows a triangle ABC. It is given that
AH = 30 cm, AG = 33 cm and BG = 40 cm.
Rajah 5 menunjukkan sebuah segi tiga ABC. Diberi bahawa
AH = 30 cm, AG = 33 cm dan BG = 40 cm.
(c) (i)
The side opposite to the non-included
angle, AC is shorter than the other side, AB.
(AC
AB)
A
33 cm
G
The length of two sides, AB and AC, and a
non-included acute angle, ABC are given.
30 cm
H
(ii) Another possible triangle: ABD
where AD = AC
40 cm
B
A
45°
52°
D D
Rajah 5
C
73 cm
45°
B
(a) Calculate the length, in cm, of
Hitung panjang, dalam cm, bagi
(i) AC,
(ii) GH.
[5 marks/ 5 markah]
(b) Find the area of triangle ABC, in cm2.
Cari luas segi tiga ABC, dalam cm2.
[2 marks/ 2 markah]
(c) If ∠ACB is not given,
Jika ∠ACB tidak diberi,
(i) state the necessary conditions so that an
ambiguous case exists,
nyatakan syarat-syarat yang diperlukan supaya kes
berambiguiti wujud,
(ii) sketch another possible triangle with the
length of AC found in (a)(i).
lakarkan segi tiga lain yang mungkin dengan
panjang AC yang diperolehi dalam (a)(i).
[3 marks/ 3 markah]
13.
AC
73
=
sin 45°
sin 52°
73
× sin 45°
AC =
sin 52°
= 65.51 cm
(ii) ∆BAC = 180° – 45° – 52°
= 83°
GH2 = 332 + 302 – 2(33)(30) cos 83°
GH = 41.81 cm
1
(73)(65.51) sin 83°
2
= 2 373.29 cm2
(b) Area of ∆ABC =
C
D
Diagram 6 shows a trapezium PQRS.
Rajah 6 menunjukkan sebuah trapezium PQRS.
S
12.8 cm
P
85°
R
28°
D D
Rajah 6
Answer/ Jawapan :
(a) (i)
65.51 cm
6.2 cm
Q
It is given that QRS is an obtuse angle. Find
Diberi bahawa QRS ialah sudut cakah. Cari
(a) the shortest distance from S to the side PQ,
jarak terpendek dari S ke sisi PQ,
[1 mark/ 1 markah]
(b) the length, in cm, of
panjang, dalam cm, bagi
(i) QS,
(ii) RS,
[5 marks/ 5 markah]
(c)
QRS,
[2 marks/ 2 markah]
(d) the area of triangle QRS, in cm2.
luas segi tiga QRS, dalam cm2.
[2 marks/ 2 markah]
Additional Mathematics
Answer/ Jawapan :
(a)
(a) Find the value of
Cari nilai bagi
(i) x,
(ii) y,
(iii) z.
S
12.8 cm
85°
P
[3 marks/ 3 markah]
(b) Calculate the composite index for the five
components in the year 2019 based on the year
2018.
Hitung indeks gubahan bagi lima komponen itu pada
tahun 2019 berasaskan tahun 2018.
[2 marks/ 2 markah]
R
T
6.2 cm
28°
Q
ST
12.8
ST = 12.75 cm
sin 85° =
(c) The cost of the item in the year 2018 was RM600.
Calculate the cost of the item in the year 2019.
Kos barangan itu pada tahun 2018 ialah RM600. Hitung
harganya pada tahun 2019.
[2 marks/ 2 markah]
QS
12.8
=
sin 85° sin 28°
QS = 27.16 cm
(b) (i)
(ii)
∠SQR= ∠PSQ
SQR = 180° – 85° – 28°
= 67°
RS2 = 27.162 + 6.22 – 2(27.16)(6.2) cos 67°
RS = 25.39 cm
(c)
sin QRS
sin 67°
=
27.16
25.39
sin
(d) The cost of the item increases by 15% from the
year 2017 to 2018. Calculate the price index of the
item in the year 2019 based on the year 2017.
Kos barangan itu meningkat sebanyak 15% dari tahun
2017 hingga tahun 2018. Hitung indeks harga barangan
itu pada tahun 2019 berasaskan tahun 2017.
[3 marks/ 3 markah]
Answer/ Jawapan :
sin 67°
× 27.16
25.39
QRS = 79.96°
(a) (i)
QRS =
(ii)
1
(27.16)(6.2) sin 67°
2
= 77.50 cm2
(d) Area of triangle QRS =
14.
Table 2 shows the costs and weightages of five
components of an item in the beginning of the year
2019 based on the beginning of the year 2018.
Jadual 2 menunjukkan kos dan pemberat bagi lima komponen
suatu barangan pada awal tahun 2019 berasaskan awal tahun
2018.
Price
index
Cost in Cost in in 2019
2018
2019 based on
Component
Weightage
Kos pada Kos pada
2018
Komponen
Pemberat
2018
2019
(RM)
(RM)
Indeks harga
pada 2019
berasaskan
2018
A
x
120
90
3
B
140
y
125
1
C
100
150
z
2
D
110
100
110
5
E
90
120
75
4
7D
Jadual 2
120
× 100 = 90
x
x = 133.33
y
× 100 = 125
140
y = 175
150
× 100
100
= 150
(iii) z =
90(3) + 125(1) + 150(2) + 110(5) + 75(4)
3+1+2+5+4
= 103
(b) I =
(c)
Q2019
× 100 = 103
600
Q2019 = RM618
\ The cost in the year 2019 = RM618
(d)
Q2019
× 100 = 103
Q2018
⇒
Q2019
103
=
Q2018
100
Q2018
× 100 = 115
Q2017
⇒
Q2018
115
=
Q2017
100
Price index, I =
Q2019
× 100
Q2017
=
Q2019
Q
× 2018 × 100
Q2018
Q2017
=
103
115
×
× 100
100
100
= 118.45
Additional Mathematics
15.
Table 3 shows the prices and the price indices for the
four ingredients P, Q, R and S used in making a special
type of cake.
Jadual 3 menunjukkan harga dan indeks harga bagi empat
bahan P, Q, R dan S untuk membuat sejenis kek yang istimewa.
Ingredient
(a) (i)
(RM)
2019
2018
Indeks harga pada tahun 2019
berasaskan tahun 2018
P
0.80
x
125
Q
2.00
2.50
y
R
z
0.60
150
S
0.50
0.40
80
0.80
× 100 = 125
x
x = 0.64
2.00
× 100
2.50
= 80
(ii) y =
Price index in the
year 2019 based on
the year 2018
Price per kg
Harga per kg
Bahan
Answer/ Jawapan :
(iii)
(b) (i)
z
× 100 = 150
0.60
z = 0.90
Composite index, I
125(1) + 80(1) + 150(1) + 80(1)
4
= 108.75
=
7D
Jadual 3
(ii) 108.75 = 578 × 100
Q2018
(a) Find the value of
Cari nilai bagi
(i) x,
(ii) y,
(iii) z.
Q2018 = 531.49
The corresponding cost in 2018 = RM531.49
[3 marks/ 3 markah]
(b) (i) Calculate the composite index of the cost in
making the cake for the year 2019 based on
the year 2018.
Hitung indeks gubahan kos membuat kek itu pada
tahun 2019 berasaskan tahun 2018.
(ii) Hence, calculate the corresponding cost of
making the cake in the year 2018 if the cost
of making the cake in the year 2019 was
RM578.00.
Seterusnya, hitung kos yang sepadan untuk membuat
kek itu pada tahun 2018 jika kos untuk membuat kek
itu pada tahun 2019 ialah RM578.00.
[4 marks/ 4 markah]
(c) The cost of making the cake is expected to increase
by 40% from the year 2018 to the year 2020. Find
the expected composite index for the year 2020
based on the year 2019.
Kos membuat kek itu dijangkakan akan meningkat
sebanyak 40% dari tahun 2018 ke tahun 2020. Cari indeks
gubahan jangkaan bagi tahun 2020 berasaskan kepada
tahun 2019.
[3 marks/ 3 markah]
(c)
Q2020
× 100 = 140
Q2018
Q2019
× 100 = 108.75
Q2018
I 20/19 =
Q2020
140
=
Q2018
100
⇒
⇒
Q2020
× 100
Q2019
=
Q2020
Q
× 2018 × 100
Q2018
Q2019
=
140
100
×
× 100
100
108.75
= 128.7
Q2019
108.75
=
Q2018
100