HEBAT SPM 2019 - addmaths

Practice Progressions
1 Janjang 1



Practice Linear Law
2 Hukum Linear 10



Practice Integration
3 Pengamiran 21



Practice Vectors
4 Vektor 29



Practice Trigonometric Functions
5 Fungsi Trigonometri 39



Practice Permutations and Combinations
6 Pilih Atur dan Gabungan 46



Practice Probability
7 Kebarangkalian 51



Practice Probability Distributions
8 Taburan Kebarangkalian 57



Practice Motion along a Straight Line
9 Gerakan pada Garis Lurus 66



Practice Linear Programming
10 Pengaturcaraan Linear 75




Forecast Paper 86


Answers A1 – A20

PRACTICE Linear Law
2 Hukum Linear





A nalysis of PAPER ‘11 ✓ ‘12 ✓ ‘13 ✓ ‘14 ✓ ‘15 ✓ ‘16 ✓ ‘17 ✓
SPM
1
Questions 2 ✓ ✓ ✓ ✓ ✓ ✓ ✓
FORMULAE


y – y
1. Gradient (m) of straight lines passing through points A(x , y ) and B(x , y ): m = 2 1
1 1 2 2
x – x
2 1
y – y
Kecerunan garis lurus (m) yang melalui dua titik A(x , y ) dan B(x , y ): m = 2 1
1 1 2 2
x – x
2 1
2. Linear equation : y = mx + c
Persamaan linear : y = mx + c
PAPER 1


Lines of Best Fit The diagram shows the line of best fit obtained
2.1 Garis Lurus Penyuaian Terbaik by plotting y against x. The line passes through
(2, 2) and (6, 10). Find the equation of the line
of best fit.
1.
y
Rajah menunjukkan garis lurus penyuaian
(4, 16) terbaik yang diperoleh dengan memplot y
melawan x. Garis itu melalui (2, 2)
dan (6, 10). Cari persamaan garis lurus
penyuaian terbaik itu.
4
x 3. y
O
Graph of y against x
The diagram above shows the line of best fit 40 Graf y melawan x
obtained by plotting y against x.
Rajah di atas menunjukkan garis lurus
penyuaian terbaik yang diperoleh dengan 30
memplot y melawan x.
(a) Find the equation of the line of best fit.
Cari persamaan garis lurus penyuaian 20
terbaik itu.
(b) Determine the value of y when x = 4.5.
Tentukan nilai y apabila x = 4.5. 10

2. y
0 x
(6, 10) 10 20 30
The diagram above shows the line of best fit
obtained by plotting the values of y against x.
(2, 2) Determine the gradient and the y-intercept of
x the line. Hence, find the equation of the line of
O
best fit.

© Penerbitan Pelangi Sdn. Bhd. 10

Additional Mathematics Form 5 Practice 2 Linear Law

Rajah di atas menunjukkan satu garis lurus 8. log y
2
penyuaian terbaik yang diperoleh dengan
B(3, 8)
memplot nilai y melawan x. Tentukan kecerunan
dan pintasan-y garis itu. Seterusnya, cari
persamaan garis lurus penyuaian terbaik itu.
A(1, 2)
Application of Linear Law to Non- O x
2.2 linear Relations The diagram above shows the graph of log y
2
Mengaplikasikan Hukum Linear kepada against x where points A(1, 2) and B(3, 8) lie
Hubungan Tak Linear on the graph. Given the values of x and y are
related by the equation y = 2 px + q , where p and
4. Convert the equation y = 3x + 2x to the linear q are constants. Find the values of p and of q.
3
form of Y = mX + c. State the quantities to be Rajah di atas menunjukkan graf log y
2
plotted on the Y-axis and the X-axis. melawan x dengan titik A(1, 2) dan B(3, 8)
Tukar persamaan y = 3x + 2x kepada bentuk terletak pada graf itu. Diberi nilai x dan y
3
linear Y = mX + c. Nyatakan kuantiti yang dikaitkan dengan persamaan y = 2 px + q , dengan
diplotkan pada paksi-Y dan paksi-X. p dan q ialah pemalar. Cari nilai p dan nilai q.

5. Express the equation y = 4x q + 1 in the linear 9. log y
form Y = mX + c. State the quantities to be 10
plotted on the Y-axis and the X-axis. (2, 2.2)
Ungkapkan persamaan y = 4x q + 1 dalam bentuk
linear Y = mX + c. Nyatakan kuantiti yang 1
diplotkan pada paksi-Y dan paksi-X.
x
O
6. Explain how a straight line can be drawn from The variables x and y are related by the
h x
the equation y = —– + kx, where h and k are equation y = ab , where a and b are constants.
x 2 A line of best fit is obtained by plotting log y
constants. 10
against x as shown in the diagram above. Find
Terangkan bagaimana satu garis lurus boleh the values of a and of b.
h
dilukis daripada persamaan y = —– + kx, Pemboleh ubah x dan y dihubungkan oleh
x 2 x
dengan h dan k ialah pemalar. persamaan y = ab , dengan a dan b ialah
pemalar. Satu garis lurus penyuaian terbaik
diperoleh dengan memplot log y melawan x
7. 10
log 10 y seperti yang ditunjukkan dalam rajah di atas.
Cari nilai a dan nilai b.
3
10. The variables x and y are related by the
k
equation y = hAB + —, where h and k are
x
x
x + 1 — 3
2
O 4 constants. By plotting xy against x , a straight
line passing through the points (8, 4) and
The straight line in the diagram above is (12, 18) is obtained. Determine the value of h
the line of best fit when the graph of log y and of k.
10
against (x + 1) is plotted. Find the value of y Diketahui pemboleh ubah x dan y dihubungkan
when x is 7. k
x
oleh persamaan y = hAB + —, dengan h dan k
Garis lurus dalam rajah di atas ialah garis x 3
lurus penyuaian terbaik apabila graf log y — 2
10 ialah pemalar. Dengan memplot xy melawan x ,
melawan (x + 1) diplotkan. Cari nilai y
satu garis lurus yang melalui titik (8, 4) dan
apabila x ialah 7.
(12, 18) diperoleh. Tentukan nilai h dan nilai k.
11 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Practice 2 Linear Law

11. log 10 y xy
(5, 8)
B (3,5)

(3, 2) A
1
x 2 —
O O x
Based on the line of best fit in the diagram (a) Find the coordinates of point A.
above, form an equation that relates the Cari koordinat titik A.
variable x to the variable y. (b) Express y in terms of x.
Ungkapkan y dalam sebutan x.
Berdasarkan garis lurus penyuaian terbaik (c) Find the value of y when x = 2.
dalam rajah di atas, dapatkan satu persamaan Cari nilai y apabila x = 2.
yang menghubungkan pemboleh ubah x
dengan pemboleh ubah y. 14. y

12. log y
10
(2, 8)

(n, 8)

(1, 1)
(0, 2) x
0
0 log x
10
Diagram (a) / Rajah (a)
y
The diagram above shows a straight line graph x
plotted which represents non-linear equation
y = ax , where a is a constant.
3
Rajah di atas menunjukkan graf garis lurus
yang diplotkan bagi mewakili persamaan tak x
linear y = ax , dengan keadaan a ialah pemalar. 0
3
(a) Based on the graph, express the equation (0, p)
y = ax in linear form.
3
Berdasarkan graf tersebut, ungkapkan Diagram (b) / Rajah (b)
3
persamaan y = ax dalam bentuk linear.
2
Diagram (a) shows part of a curve y = ax +
(b) Find
bx, where a and b are constants.
Cari
Rajah (a) menunjukkan sebahagian lengkung
(i) log a,
2
10 bagi graf y = ax + bx, dengan a dan b adalah
(ii) n.
pemalar.
(a) Calculate the values of a and b.
13. The following diagram shows the graph of xy Hitung nilai a dan nilai b.
1 (b) If the equation is reduced to the linear
against —. Point A lies on the xy-axis and the
x y
length of AB is 5 units. HOTS form x = mx + c, a straight line graph
is obtained as shown in Diagram (b). Find
Rajah berikut menunjukkan graf bagi xy the value of p.
1
melawan —. Titik A terletak pada paksi-xy dan Jika persamaan tersebut ditukar kepada
x y
bentuk linear, = mx + c, satu graf garis
panjang AB ialah 5 unit. x
lurus diperoleh seperti Rajah (b). Cari
nilai p.
© Penerbitan Pelangi Sdn. Bhd. 12

Additional Mathematics Form 5 Practice 2 Linear Law

15. log y linear yang menghubungkan pemboleh ubah
10
x kepada pemboleh ubah y diberi oleh
A log y = m log x + c. Lakar satu garis lurus
10 10
untuk persamaan itu.
C (2,3) y
(1, 10)
B(4,0)
log x
O 10
The diagram above shows the graph of log y (5, 2)
10
against log x. Point A lies on the log y-axis
10 10 x
and point B lies on the log x-axis. HOTS O
10
Given that the distance of AC = distance of BC.
Rajah di atas menunjukkan graf log y melawan 18. Two variables, x and y are related by the
10 equation y = p , where p is a positive constant.
–x
log x. Titik A terletak pada paksi-log y dan
10 10
titik B terletak pada paksi-log x. Diberi jarak Dua pemboleh ubah, x dan y, dihubungkan
10 –x
AC = jarak BC. oleh persamaan y = p , dengan keadaan p
(a) Find the coordinates of point A. ialah pemalar positif. HOTS
–x
Cari koordinat titik A. (a) Sketch the graph of y = p .
–x
(b) State log y in terms of log x and hence, Lakarkan graf y = p .
10 10 –x
express y in terms of x. (b) Convert y = p to the linear form. Sketch
Nyatakan log y dalam sebutan log x dan the straight line graph and explain how the
10 10
seterusnya, ungkapkan y dalam sebutan x. value of p can be obtained given that the
(c) Find the value of y when x = 100. gradient is –2.
–x
Cari nilai y apabila x = 100. Tukar y = p kepada bentuk linear. Lakar
graf garis lurus dan terangkan bagaimana
1
16. – nilai p boleh diperoleh diberi bahawa
y
(4, 61) kecerunannya ialah –2.
19. The variables x and y are related by the equation
3
y = px + qx, where p and q are constants.
2
The equation is reduced to the linear form of
(1, 13)
1
– Y = mX + c as shown in the table below. HOTS
O x
Pemboleh ubah x dan y dihubungkan oleh
Variables x and y are related by the equation 2 3
x persamaan y = px + qx, dengan p dan q ialah
y = ———, where a and b are constants. pemalar. Persamaan itu ditukar kepada bentuk
a + bx
1 linear Y = mX + c seperti yang ditunjukkan
A line of best fit is obtained by plotting —
y dalam jadual di bawah.
1
against —. Find the values of a and of b. 2
x y
X Y =
Pemboleh ubah x dan y dihubungkan oleh x
x
persamaan y = ———, dengan a dan b ialah 1 5
a + bx
pemalar. Satu garis lurus penyuaian terbaik 3 13
1 1 (a) State the quantities to be plotted on the
diperoleh dengan memplot — melawan —.
y x X-axis.
Cari nilai a dan nilai b.
Nyatakan kuantiti yang diplot pada paksi-X.
17. The following diagram shows part of the graph (b) Hence, find the value of
of y against x. The linear equation that relates Seterusnya, cari nilai bagi
the variable x to the variable y is given by (i) p, (ii) q.
log y = m log x + c. Sketch a straight line (c) The line of best fit Y = mX + c passes
10 10
graph for the equation. HOTS through point (5, n). Find the value of n.
Rajah berikut menunjukkan sebahagian Garis lurus penyuaian terbaik Y = mX + c
daripada graf y melawan x. Persamaan melalui titik (5, n). Cari nilai n.
13 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Practice 2 Linear Law

PAPER 2


1. The table below shows some experimental data of two related variables x and y.
Jadual di bawah menunjukkan beberapa data eksperimen bagi dua pemboleh ubah x dan y.

x 1 1.6 2 2.5 3 3.4
y 7.1 14.9 24.0 42.1 70.8 99.9

3
(a) Draw a graph of y against x using a scale of 2 cm to 5 units on the x -axis and 2 cm to 10 units on the
3

y-axis.
3
Lukis graf y melawan x dengan menggunakan skala 2 cm kepada 5 unit pada paksi-x dan 2 cm
3
kepada 10 unit pada paksi-y.
(b) Form an equation to show the relation between x and y.
Bentuk satu persamaan untuk menunjukkan hubungan antara x dengan y.
(c) Calculate the approximate value of y when x = 8.
Hitung anggaran nilai y apabila x = 8.
–3
2. An experiment was carried out to study the changes of the concentration, K mole dm , of a solution with
respect to time, t seconds. The results were recorded in the table below.
–3
Satu eksperimen dijalankan untuk mengkaji perubahan kepekatan, K mol dm bagi satu larutan terhadap
masa, t saat. Keputusan eksperimen direkod dalam jadual di bawah.

t (s) 100 200 300 400 500 600
–3
K(mole dm ) 0.048 0.038 0.031 0.027 0.023 0.0206
a
It is expected that the values of P and t are related by the equation K = , where b is a constant and
bt + 1
a is the initial concentration.
a
Nilai P dan nilai t dihubungkan oleh persamaan K = , dengan keadaan b ialah pemalar dan a ialah
bt + 1
kepekatan awal.
1
(a) Based on the table above, construct a table for the values of —.
K
1
Berdasarkan jadual di atas, bina satu jadual bagi nilai-nilai —.
K
1 1
(b) Plot — against t, using a scale of 2 cm to 100 units on the t-axis and 2 cm 5 units on —-axis.
K K
Hence, draw the line of best fit.
1
Plot — melawan t, menggunakan skala 2 cm kepada 100 unit pada paksi-t dan 2 cm kepada 5 unit
K
1
pada paksi-—. Seterusnya, lukis garis lurus penyuaian terbaik.
K
(c) Find the values of a and b.
Cari nilai a dan b.

3. The table below shows some of the experimental values of the variables x and y. Given the variables x and
x 3
y are related by an equation y = hk , where h and k are constants.
Jadual di bawah menunjukkan beberapa nilai eksperimen bagi pemboleh ubah x dan y. Diberi pemboleh
x 3
ubah x dan y adalah dihubungkan oleh persamaan y = hk , dengan h dan k ialah pemalar.
x 1.4 1.6 1.8 2.0 2.2 2.4
y 33.5 85.5 284.8 1 280.1 8 023.0 72 511.8



© Penerbitan Pelangi Sdn. Bhd. 14

Additional Mathematics Form 5 Practice 2 Linear Law

3
3
(a) Plot a graph of log y against x using a scale of 2 cm to 2 units on the x -axis and 2 cm to 0.5 unit
10
on the log y-axis. Hence, draw a line of best fit.
10
Plot graf log y melawan x dengan menggunakan skala 2 cm kepada 2 unit pada paksi-x dan 2 cm
3
3

10
kepada 0.5 unit pada paksi-log y. Seterusnya, lukis satu garis lurus penyuaian terbaik.
10
(b) By using the graph obtained in (a), determine the values of h and of k.
Dengan menggunakan graf yang diperoleh di (a), tentukan nilai h dan nilai k.
4. The table below shows some of the experimental values of variables x and y.
Jadual di bawah menunjukkan beberapa nilai eksperimen bagi pemboleh ubah x dan y.
x 1.0 1.5 2.0 2.5 3.0 3.5
y 5.0 6.6 10.0 17.8 30.2 69.1
2
2
(a) Plot a graph of log y against x using a scale of 2 cm to 2 units on the x -axis and 2 cm to 0.2 unit
10
to the log y-axis. Hence, draw a line of best fit.
10
2
2
Plot graf log y melawan x dengan menggunakan skala 2 cm kepada 2 unit pada paksi-x dan 2 cm
10
kepada 0.2 unit pada paksi-log y. Seterusnya, lukis satu garis lurus penyuaian terbaik.
10
(b) Express y in terms of x.
Ungkapkan y dalam sebutan x.
(c) Calculate the value of x when y = 1 000.
Hitung nilai x apabila y = 1 000.
5. The table below shows the values of two variables, x and y, obtained from an experiment. Variables x and
– x
y are related by the equation y = pk , where p and k are constants.
Jadual di bawah menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada satu
– x
eksperimen. Pemboleh ubah x dan y dihubungkan oleh persamaan y = pk , dengan keadaan p dan k
adalah pemalar.
x 1.0 1.5 2.0 2.5 3.0 3.5
y 31.62 16.60 9.55 4.57 2.19 1.26
(a) Plot log y against x, using a scale of 2 cm to 0.5 unit on the x-axis and 2 cm to 0.2 unit on the
10
log y-axis. Hence, draw the line of best fit.
10
Plot log y melawan x, dengan menggunakan skala 2 cm kepada 0.5 unit pada paksi-x dan 2 cm
10
kepada 0.2 unit pada paksi-log y. Seterusnya, lukis garis lurus penyuaian terbaik.
10
(b) Use your graph in (a) to find the value of
Gunakan graf anda dari (a) untuk mencari nilai
(i) p,
(ii) k,
(iii) y when x = 0.3.
y apabila x = 0.3.

6. The table below shows some experimental values of two variables x and y which are related by an equation
a
y = , where a and b are constants.
b 3x 2 + 1
Jadual di bawah menunjukkan beberapa nilai eksperimen bagi dua pemboleh ubah x dan y yang
a
dihubungkan oleh persamaan y = , dengan a dan b ialah pemalar.
b 3x 2 + 1
x 0.1 0.2 0.4 0.5 0.8 0.9
y 1.867 1.512 0.662 0.356 0.024 0.007





15 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Practice 2 Linear Law

(a) Explain how a straight line can be drawn from the equation. Hence, by using a scale of 2 cm to 0.5
unit on both axes, draw a line of best fit.
Terangkan bagaimana satu garis lurus penyuaian terbaik boleh dilukis daripada persamaan itu.
Seterusnya, dengan menggunakan skala 2 cm kepada 0.5 unit pada kedua-dua paksi, lukis satu garis
lurus penyuaian terbaik.
(b) By using the graph obtained in (a), determine the values of a and of b.
Dengan menggunakan graf yang diperoleh di (a), tentukan nilai a dan nilai b.

7. The table below shows some of the experimental values of two variables x and y. HOTS
Jadual di bawah menunjukkan beberapa nilai eksperimen bagi pemboleh ubah x dan y.

x 1.6 2.1 2.5 4.6 5.0 6.0
y 5.01 0.83 0.51 0.16 0.14 0.11

1
(a) Plot a graph of against (x + 1) and draw a line of best fit by using a scale of 2 cm to 1 unit on the
y
1
(x + 1)-axis and 2 cm to 2 units on the -axis.
y
1
Plot graf — melawan (x + 1) dan lukis satu garis lurus penyuaian terbaik dengan menggunakan skala
y
1
2 cm kepada 1 unit pada paksi-(x + 1) dan 2 cm kepada 2 unit pada paksi- .
y
(b) By using the graph obtained in (a),
Dengan menggunakan graf yang diperoleh di (a),
(i) find the value of y when x = 0.5,
cari nilai y apabila x = 0.5,
1
(ii) find the gradient and the —-intercept for the graph and hence, express y in terms of x.
y
1
cari kecerunan dan pintasan-— untuk graf itu dan seterusnya, ungkapkan y dalam sebutan x.
y
(c) Calculate the value of y when x = 25.
Hitung nilai y apabila x = 25.

2
2
8. It is found that the values of variables x and y are related by the equation py + qx = x. The following table
shows the corresponding values of x and y. HOTS
2
2
Didapati bahawa nilai-nilai pemboleh ubah x dan y adalah dihubungkan oleh persamaan py + qx = x.
Jadual berikut menunjukkan beberapa nilai yang sepadan bagi x dan y.
x 0.4 0.7 0.9 1.2 1.3 1.4
y 1.10 1.35 1.41 1.34 1.30 1.19

y 2
(a) Plot a graph of —– against x by using a scale of 2 cm to 0.2 unit on the x-axis and 2 cm to 0.5 unit
x
y 2
on the -axis. Hence, draw a line of best fit.
x
y 2
Plot graf —– melawan x dengan menggunakan skala 2 cm kepada 0.2 unit pada paksi-x dan 2 cm
x 2
kepada 0.5 unit pada paksi- y . Seterusnya, lukis satu garis lurus penyuaian terbaik.
x
(b) One of the values of y has been wrongly recorded. Identify the incorrect value of y and determine its
correct value.
Salah satu daripada nilai y salah direkod. Kenal pasti nilai y itu dan tentukan nilai yang betul.
(c) From the graph, find the values of p and of q.
Daripada graf itu, cari nilai p dan nilai q.


© Penerbitan Pelangi Sdn. Bhd. 16

Additional Mathematics Form 5 Practice 2 Linear Law








1
PAPER
shows the straight line MN obtained by plotting
1. y against x .
2
xy
SPM
CLONE Pemboleh ubah x dan y dihubungkan oleh
2016 (4, k)
persamaan xy = 9x – 3x . Rajah di atas
3
menunjukkan garis lurus MN yang diperoleh
2
dengan memplot y melawan x .
(a) Express the equation xy = 9x – 3x in its
3
1 x 2
O linear form used to obtain the straight line
graph shown in the diagram above.
3
The variables x and y are related by the equation Ungkapkan persamaan xy = 9x – 3x
p
y = — + 3x, where p is a constant. A straight dalam bentuk linear yang digunakan
x
2
line is obtained by plotting xy against x , as untuk memperoleh graf garis lurus seperti
ditunjukkan dalam rajah di atas.
shown in the diagram above. Find the value of
(b) State
p and of k.
Nyatakan
Pemboleh ubah x dan y dihubungkan oleh
p (i) the gradient of the straight line MN,
persamaan y = — + 3x, dengan keadaan p ialah
x kecerunan bagi garis lurus MN,
pemalar. Satu garis lurus diperoleh dengan (ii) the coordinates of M.
2
memplotkan xy melawan x , seperti ditunjukkan koordinat M.
dalam rajah di atas. Cari nilai p dan nilai k. [3 marks / 3 markah]
[3 marks / 3 markah]
4. The variables x and y are related by the equation
2. y + x SPM m
SPM CLONE y = x + , where m is a constant. A straight
CLONE 2013 x 2
2015
4 1
line is obtained by plotting (y – x) against ,
x 2
as shown in the diagram below.
x 2 Pembolehubah x dan y dihubungkan oleh
O 6
persamaan y = x + m dengan keadaan m ialah
The diagram above shows the straight line x 2
2
graph obtained by plotting ( y + x) against x . pemalar. Satu garis lurus diperoleh dengan
Express y in terms of x. 1
memplot (y – x) melawan seperti yang
Rajah di atas menunjukkan graf garis lurus yang x 2
diperoleh dengan memplot ( y + x) melawan x . ditunjukkan dalam rajah di bawah.
2
Ungkapkan y dalam sebutan x. (y – x)
[3 marks / 3 markah]
3.
k
y ( – , )
SPM 2 5t
CLONE
2014
M – 1
O x 2
x 2 Express k in terms of t and m.
O N
Ungkapkan k dalam sebutan t dan m.
[3 marks / 3 markah]
The variables x and y are related by the
3
equation xy = 9x – 3x . The diagram above
17 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Practice 2 Linear Law

2
PAPER
1. The table below shows the values of two variables, x and y, obtained from an experiment. The variables x
SPM p
CLONE and y are related by the equation y = + 2kx, where p and k are constants.
2013
x
SPM
CLONE Jadual di bawah menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada suatu
2016
p
eksperimen. Pemboleh ubah x dan y dihubungkan oleh persamaan y = + 2kx, dengan p dan k ialah
x
pemalar.
x 1 2 3 4 5 6
y 8.50 4.70 3.63 3.25 3.14 3.17
2
(a) Based on the table, construct a new table for the values of x and xy.
2
Berdasarkan jadual itu, bina jadual baru untuk nilai-nilai x dan xy. [2 marks / 2 markah]
2
2
(b) Plot xy against x , using a scale of 2 cm to 5 units on the x -axis and 2 cm to 2 units on the xy-axis.
Hence, draw a line of best fit.
2
2
Plot xy melawan x , dengan menggunakan skala 2 cm kepada 5 unit pada paksi-x dan 2 cm kepada
2 unit pada paksi-xy. Seterusnya, lukis satu garis lurus penyuaian terbaik. [3 marks / 3 markah]
(c) Use the graph in (b) to find the value of
Gunakan graf di (b) untuk mencari nilai
(i) p,
(ii) k.
[5 marks / 5 markah]
2. The following table shows some of the experimental values of two variables x and y. The variables x and
SPM p
CLONE y are related by the equation — = qx + 1, where p and q are constants.
2015 y
Jadual yang berikut menunjukkan beberapa nilai eksperimen bagi dua pemboleh ubah x dan y. Pemboleh
p
ubah x dan y dihubungkan oleh persamaan — = qx + 1, dengan p dan q ialah pemalar.
y
x 0.1 0.2 0.3 0.4 0.5 0.6

y 4.012 2.498 1.754 1.429 1.178 1.021

1
(a) Based on the table, construct a new table for the values of — .
y
1
Berdasarkan jadual itu, bina jadual baru untuk nilai-nilai — . [1 mark / 1 markah]
y
1
(b) Plot — against x, using a scale of 2 cm to 0.1 unit on both axes. Hence, draw a line of best fit.
y
1
Plot — melawan x, dengan menggunakan skala 2 cm kepada 0.1 unit pada kedua-dua paksi.
y
Seterusnya, lukis satu garis lurus penyuaian terbaik. [3 marks / 3 markah]
(c) Use the graph in (b) to find the value of
Gunakan graf di (b) untuk mencari nilai
(i) y when x = 0.24,
y apabila x = 0.24,
(ii) p,
(iii) q.
[6 marks / 6 markah]



© Penerbitan Pelangi Sdn. Bhd. 18

Additional Mathematics Form 5 Practice 2 Linear Law

3. The table below shows the values of two variables, x and y, obtained from an experiment. Variables x and
SPM h
CLONE y are related by the equation y = , where h and k are constants.
2014 x
k
Jadual di bawah menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada suatu
h
eksperimen. Pemboleh ubah x dan y dihubungkan oleh persamaan y = , dengan h dan k ialah pemalar.
k x
x 4 6 8 10 12 14
y 3.69 3.07 2.54 2.10 1.73 1.42

(a) Based on the table, construct a table for the values of log y.
10
Berdasarkan jadual itu, bina jadual untuk nilai-nilai log y. [1 mark / 1 markah]
10
(b) Plot log y against x, using a scale of 2 cm to 2 units on the x-axis and 2 cm to 0.1 unit on the
10
log y-axis. Hence, draw a line of best fit.
10
Plot log y melawan x, dengan menggunakan skala 2 cm kepada 2 unit pada paksi-x dan 2 cm
10
kepada 0.1 unit pada paksi-log y. Seterusnya, lukis satu garis lurus penyuaian terbaik.
10
[3 marks / 3 markah]
(c) Use the graph in (b) to find the value of
Gunakan graf di (b) untuk mencari nilai
(i) y when x = 5,
y apabila x = 5,
(ii) h,
(iii) k.
[6 marks / 6 markah]

4. Table below shows the values of two variables, x and y which are obtained from an experiment. The
SPM pq
CLONE variable x and y are related by the equation y – p = —, where p and q are constants.
2010 x
Jadual di bawah menunjukkan nilai-nilai bagi dua pembolehubah, x dan y, yang diperoleh daripada satu
pq
eksperimen. Pemboleh ubah x dan y dihubungkan oleh persamaan y – p = —, dengan keadaan p dan
x
q ialah pemalar.
x 1.5 2.0 3.5 4.5 5.0 6.0
y 4.5 5.25 5.5 6.3 6.34 6.5


(a) Plot xy against x by using a scale 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, 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 marks / 4 markah]
(b) Using the graph in 4(a), find
Menggunakan graf di 4(a), cari
(i) the value of p and q,
nilai p dan q,
(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
dijalankan.
[6 marks / 6 markah]





19 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Practice 2 Linear Law

h k

5. The variables u and v are related by an equation — = 1 – —, where h and k are constants. The following
v u
SPM
CLONE table shows the corresponding values of u and v.
2011
h k
Pemboleh ubah u dan v adalah dihubungkan oleh persamaan — = 1 – —, dengan h dan k ialah pemalar.

v u
Jadual yang berikut menunjukkan nilai-nilai u dan v yang sepadan.
u 50 40 25 20 15 12.5
v 12.5 13.3 16.1 20.1 30.0 50.2
1 1
(a) Based on the table, construct a new table for the values of — and —.
u v
1 1
Berdasarkan jadual itu, bina satu jadual baru untuk nilai-nilai — dan —. [2 marks / 2 markah]
u v
1 1
(b) Plot — against — by using a scale of 2 cm to 0.01 unit on both axes. Hence, draw the line of best fit.
v u
1 1
Plot — melawan — dengan menggunakan skala 2 cm kepada 0.01 unit pada kedua-dua paksi.
v u
Seterusnya, lukis garis lurus penyuaian terbaik. [3 marks / 3 markah]
(c) Use the graph obtained in (b) to find the value of
Gunakan graf di (b) untuk mencari nilai
(i) h,
(ii) k.
[5 marks / 5 markah]






The table below shows the values of variables x and y which are obtained from an experiment. The variables
x and y are related by the equation y = px + r , where p and r are constants.
px
Jadual di bawah menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada satu
eksperimen. Pemboleh ubah x dan y dihubungkan oleh persamaan y = px + r , dengan keadaan p dan r adalah
px
pemalar.
x 1 2 3 4 5 6
y 5.9 4.2 4.3 4.6 5.1 5.9

(a) Plot xy against x by using a scale of 2 cm to 5 units on both axes. Hence, draw the line of best fit.
2
Plotkan xy melawan x dengan menggunakan skala 2 cm kepada 5 unit pada kedua-dua paksi. Seterusnya,
2
lukiskan garis lurus penyuaian terbaik.
(b) Based on your graph, find the value of
Berdasarkan graf anda, cari nilai bagi
(i) p,
(ii) r,
(iii) y when x = 35.
y apabila x = 35.







© Penerbitan Pelangi Sdn. Bhd. 20

Additional Mathematics Form 5 Answers







2. (a) RM1 875
PRACTICE Progressions (b) Year / Tahun 2007
1 Janjang (c) RM15 607
1
3. (a) —
PAPER 1 2
(b) (i) 9th rectangle / Segi empat tepat ke-9
1. (a) 4 (b) 860 (ii) 6 400 cm 2
2. (a) 9; 6 (b) 153 1 1
4. (a) a = —r p, d = – —r p
3. d = 5 2 9
(b) (i) 15p cm
4. (a) k = 4 (b) T = 69 (ii) 125p cm
17
(c) n = 51
5. (a) d = –16 cm (b) 84 cm
5. 27, 30, 33 (c) 408 cm
1
1
6. (a) a = 100; d = –4 (b) –380 6. (a) r = —, a = 81; r = – —, a = 162
3
3
7. –737 1 n – 1
1
(b) T = 162 – — 2 ; T = 3 5 – n
8. (a) 19 (b) 1 007 n 3 n
1
(c) 121—
9. 12 2
10. (a) a = –5 (b) m = 14 7. (a) 6
11. 456 cm (b) 4.713 cm
2 8. (a) The length of the rectangles form a geometric
12. (a) a = 81, r = — (b) 243 1
3 progression with common ratio —;
1 2
13. (a) 8 (b) 788— Panjang segi empat tepat masing-masing
8
3 membentuk janjang geometri dengan nisbah
1
14. (a) 546— (b) 48 sepunya —;
4 2
5 41 The height of the rectangles form a geometric
15. (a) —––– (b) 426—–
1 024 64 progression with common ratio 2.
3 64 Tinggi segi empat tepat masing-masing
16. (a) r = – —, a = – —–
2 81 membentuk janjang geometri dengan nisbah
33 sepunya 2.
(b) –102—–
64 (b) (i) n = 4
(ii) 5 125 cm
1 3
17. (a) a = —, r = 2 (b) 255—
4 4
2
18. (a) a = 729, r = – —
3
2 PAPER 1
(b) 437— 8
5 1. (a) 4 (b) —
3 5 3
19. (a) — (b) — 2. (a) m = 0, m = 1, m = –1
2 6 3
20. (a) r = 2 (b) —
2
(b) T = 24 576
14 3. Amirul won the prize.
4 6
21. (a) — (b) —– Amirul memenangi hadiah itu.
9 11
4. 567
22. RM1 023
5. a = 99, d = –11
27 13
23. q = 6. T = — – 3n
p 2 n 2
7. (a) Arithmetic progression / Janjang aritmetik
PAPER 2 (b) 40°
(c) 360°
1. (a) a = –21, d = 6 8. Syarikat Arif; RM102 598
(b) 45
(c) 32
A1 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

2
PAPER 18. (a) y
1. RM2 950
2. (a) (i) 164 cm (ii) 4563 cm
(b) n = 46, Green
y = p –x
1
O x
1. (a) 29 (b) 210
(c) 30 (b) log y = (–log p)x,
10
10
2. (a) n = 6 (b) 500 cm log y
10
(c) 12.58°
PRACTICE Linear Law
2 Hukum Linear
O x
y = (– log p)x
PAPER 1 10
1. (a) y = 3x + 4 (b) 17.5
Given m = –log p, compare with m = –2 to
10
2. y = 2x – 2 obtain the value of p. Hence, p = 100.
3. 1.21, 2; y = 1.21x + 2 Diberi m = –log p, bandingkan dengan m = –2
10
y y untuk memperoleh nilai bagi p. Maka, p = 100.
4. — = 3x + 2; —, x 2
2
x
x
19. (a) x 2
5. log y = (q + 1) log x + log 4; log y, log x
10 10 10 10 10 (b) (i) p = 4 (ii) q = 1
6. Plotting a graph of x y against x 3 (c) n = 21
2
2
Memplot satu graf x y melawan x 3
PAPER 2
7. y = 0.001
1. (a)
8. p = 3, q = –1 y
9. a = 10, b = 3.981 Graph of y against x 3
110 Graf y melawan x 3
10. h = 3.5, k = –24
11. y = 10 3x 2 – 7 100
12. (a) log y = log a + 3 log x
10
10
10
(b) (i) log a = 2 90
10
(ii) n = 2
4 1 80
13. (a) A(0, 1) (b) y = —– + —
x
3x
2
(c) 5 70
6
14. (a) a = 3, b = –2 60
(b) p = –2
15. (a) A(0, 6) 50
(b) log y = – 3 log x + 6, y = 1 000 000
10 2 10 3 40
(c) 1 000 x 2
16. a = 16, b = –3 30
17. y
log 10 20
1 10
5
0 x 3
5 10 15 20 25 30 35 40
log x 3
O 1 10 (b) y = 2.425x + 5
(c) 1 247
© Penerbitan Pelangi Sdn. Bhd. A2

Additional Mathematics Form 5 Answers

2. (a) t(s) 100 200 300 400 500 600 4. (a) log 10 y
Graph of log y against x 2
1 20.83 26.32 32.26 37.04 43.48 48.54 2.0 Graf log y melawan x 2
10
K 10
(b) 1.8
1
Graph of against t 1.6
—
60 K
1
Graf melawan t 1.4
—
55 K
50 1.2
45 1.0
40
0.8
35
0.6
30
0.4
25
20 0.2

15 0 x 2
2 4 6 8 10 12 14
10
(b) y = 10 0.102x 2 + 0.6
5
(c) 4.851
0 t(s)
100 200 300 400 500 600 700 800 5. (a) log y
Graph of log y against x
2.2 y melawan x
10
(c) a = 0.067 Graf log 10
b = 0.0037 2.0
1.9
3. (a) 1.8
log y
10
Graph of log y against x 3 1.6
10
Graf log y melawan x 3
5.0 10 1.4
4.5 1.2
1.0
4.0
0.8
3.5
0.6
3.0
0.4
2.5
0.2
2.0 x
0 0.3 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
1.5 (b) (i) p = 117.49
(ii) k = 3.631
1.0
(iii) y = 79.43
0.5 0.7

0 x 3
2 4 6 8 10 12 14
(b) h = 5.01, k = 2.02





A3 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

6. (a) Plot a graph of log y against (3x + 1) 8. (a) – 2
2
y
10
Lukis graf log y melawan (3x + 1) x y 2
2
10 Graph of – against x
4.5 x
log y 4.25 y 2
10 Graf – melawan x
x
Graph of log y against (3x + 1)
2
2.0 10 4.0
2
Graf log 10 y melawan (3x + 1)
1.5 1.3 3.5 3.3
1.0 3.0
0.5 2.5
0 3x 2 + 1 2.0
0.5 1.0 1.5 2.0 2.5 3.0 3.5
–0.5 1.5
–1.0 1.0
–1.5 0.5
–2.0 0 x
0.2 0.4 0.6 0.8 1.0 1.2 1.4
–2.5 (b) 1.10; 1.15 (c) p = 0.235; q = 0.5385



(b) a = 20; b = 10 PAPER 1
2
1. p = 1, k = 13 2. y = – 2 x – x + 4
7. (a) 3. (a) y = –3x + 9 3
2
– 1 y (b) (i) –3 (ii) M(0, 9)
1
Graph of – against (x + 1) 10t
12 y 4. k =
1
Graf – melawan (x + 1) PAPER 2 m
y
10
1. (a) x 2 1 4 9 16 25 36
8 xy 8.50 9.40 10.89 13.00 15.70 19.02
(b) xy
6
20
4
18
2
16
0 x + 1
1 2 3 4 5 6 7
–1.8 14
–2
12
–4
10
–6 –5
8
8.2
(b) (i) –0.56 6
1
(ii) 2, –5; y = ———–
2x – 3 4
(c) 0.0213
2
x 2
0 5 10 15 20 25 30 35 40
(c) (i) p = 8.2 (ii) k = 0.15



© Penerbitan Pelangi Sdn. Bhd. A4

Additional Mathematics Form 5 Answers

2. (a) x 0.1 0.2 0.3 0.4 0.5 0.6 4. (a) x 1.5 2.0 3.5 4.5 5.0 6.0
1 y
— 0.249 0.400 0.570 0.700 0.849 0.979 4.5 5.25 5.5 6.3 6.34 6.5
y
xy 6.75 10.50 19.25 28.35 31.70 39.00
(b) 1 xy
–
y Graph of xy against x
1

Graph of – against x Graf xy melawan x
1.0 1 y 40

Graf – melawan x
y
35
0.9
30
0.8
25
0.7
20
0.6
15
0.5 10
0.4 5
x
0.3 0 1 2 3 4 5 6 7
–5 –4
0.2
(b) (i) p = 51.41, q = –0.078
(ii) 6
0.1
5. (a) 1 1
0 x — —
0.1 0.2 0.3 0.4 0.5 0.6 u v
(c) (i) 2.17 (ii) p = 10 0.02 0.08
(iii) q = 15 0.025 0.075
0.04 0.062
0.05 0.05
3. (a) x 4 6 8 10 12 14
0.067 0.033
log y 0.57 0.49 0.40 0.32 0.24 0.15
10
0.08 0.02
(b) log y (b)
10
0.8 1 – v
1
0.74 0.10 Graph of – against – u 1
v
0.7 1 1
Graf – melawan – u
v
0.6 0.09
0.53 0.08
0.5
0.4 0.07 0.0665
0.3 0.06
0.05
0.2
0.1 0.04
x 0.03
0 2 4 5 6 8 10 12 14
0.02
(c) (i) 3.39
(ii) h = 5.5 0.01
(iii) k = 1.1 0.058
0 1 – u
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08


A5 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

11
(c) (i) h = 10 18. —–
(ii) k = 10 40
19. 9 unit 2
3
20. 14— unit 2
(a) xy 4
3
40 21. — unit 2

4
35
22. 3 unit 2
30 3
23. 72— unit 2
4
25
24. a = 3
20
25. p = 4
15
26. q = 3
10 1
27. 4—p unit 3
2
5 1
28. 91—–p unit 3
2
x 15
0 5 10 15 20 25 30 35 40 29. 2p unit 3
(b) (i) p = 0.84 (ii) r = 4.2
2
(iii) y = 5.832 30. 6—p unit 3
5
31. (a) P(1, 1)
1
2
PRACTICE Integration (b) 1 12 unit
3 Pengamiran 169 3

(c)
320 p unit
1
PAPER 1 32. 75—p unit 3
5
1
3
1. (a) x + c (b) – —– + c PAPER 2
x 4
4 1 1 1. k = 3
3
2. (a) x – — + c (b) —x – —x – 6x + c
2
3
x
3
2
1 4 2. (a) E(2, 12)
3. (a) —–(4x – 3) + c (b) x – — + c
3
8
2
32 x (b) 4— unit 2
3 3
4. y = —x – 2x + 9
2
2 (c) f = 2
1
5. y = 8x – —x – 3
4
2 3. (a) A(2, 4), B(5, 1)
6. 2(4x + 5) + c, where c is a constant. (b) (i) 3 unit (ii) 6—p unit 3
3
4
2
dengan c ialah suatu pemalar. 5
2
7. k = 6, y = 3x – 8x – 2 4. (a) 2— unit 2 (b) k = 3
2
3
2
8. a = –1, n = –4 (c) 10—–p unit 3
15
9. (a) 9 (b) 123
5. (a) h = –2, k = 2
10. (a) 274.5 (b) 61.5
2
5 (b) (i) 10 unit 2 (ii) 8p unit 3
11. —
2 3
1
3 6. 9—p unit 3
12. m = – , 6 3
2
13. a = 4
14. k = 3
15. (a) 7 (b) 5 PAPER 1
2
16. (a) –4 (b) m = 2 1. 2,
3
17. a = 2, b = –3 2. (a) p = –1 (b) 16
q = 5
© Penerbitan Pelangi Sdn. Bhd. A6

Additional Mathematics Form 5 Answers

1 14. (a) –5i + 12j
3. p = – —, m = 6 ~ ~
2
(b) 13 units / unit
PAPER 2 1
1 (c) —–(–5i + 12j )
2
1. (a) a = –9, b = –6 (b) unit 13 ~ ~
4
(c) –1.2 15. (a) 7i + 7j ~
~
(b) 5 units / unit
2. (a) Q(2, –10) (b) 11 
(c) –10 12 16. (a) k = –4 (b) k = ± 91
21
3. (b) 4 000p cm 3 17. (a) h = 15 (b) h = —–
5
29
18. 9i – —j
~ 6 ~
19. p = 5, q = 2
(a) A(–2, 0), B(2, 0) (b) 21 1 unit 2
3 20. p = 4, q = 3
(c) k = 4
PAPER 2
1. (a) (i) – 4a + 10b (ii) 4a + 2b
PRACTICE Vectors ~ ~ ~ ~
4 Vektor (iii) 8b ~
1
2. (a) (i) 6i + 3j (ii) –—(2i + j)
~
~
PAPER 1 ~ AB 5 ~
(b) (i) 5i + 7j
~
1. (a), (b) ~
(ii) collinear / segaris

Q → →
4

3. (a) RA = 3x – 2y, PB = 3x + —y

~
~
→ → ~ 3 ~
1
→ –PQ

P –2PQ 2 (b) (i) RC = 3hx – 2hy ~
~
→ 2

(ii) RC = 3(1 – k)x – —(1 + 2k)y
~ 3 ~
3 2
(c) h = —, k = —
3 5 5
2. k = –2, h = — →
2 4. (a) (i) RS = 2a – 10b

15 8 → ~ ~
3. h = – —–, k = — 8 20

4 3 (ii) PT = —a + —–b
3 ~
3 ~
→ →
4. (a) AD (b) DC (b) h = 3
→ →
(c) AD 5. (a) (i) SQ = x – 3y
~
~
→ 2
→ → (ii) PT = —x + y
5. (a) AB = 5a + 5b, AC = 6a + 6b ~ 3 ~ ~
~
~
~
(b) collinear / segaris
3
2
(b) h = —, k = —
6. h = 2, 14 3 2
7. 18.76 units / unit 6. (a) (i) q – p
5 5 ~ ~
8. (a) 3x + —y (b) —y – x n 3

~ 2 ~ 2 ~ ~ (ii) p + q
3 + n ~ 3 + n ~
9. h = 6 (b) n = 4
10. (a) 3x + 2y (b) x + 2y
~ ~ ~ ~ → →


7. (a) RS = 2a – 10b , PT = 4a + 5b ~
~
~
~
–1
4
2
11. (a) –i – 11j (b) 1 2 (b) h = —, k = —
~
~
5
5
–11
(c) 50 units / unit
–6
12. (a) –6i + 4j (b) 1 2
~ ~ 4 8. (a) (i) 6a – 5b
~ ~
13. (a) 8i + 3j (ii) —–a + 2b
18
~
~
(b) ABB 5 ~ ~
73 units / unit
A7 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers
→ 1 18
2
1
(b) (i) TS = — —–a + 2b

h 5 ~ ~
→ 18 1
1
2
(ii) TS = k – —– a + 3b ~ (a) (i) –x + y ~ (ii) 2 (x + y)

~
~
5 ~
~
~
~
2 (iii) –x + 2y
(c) h = —, k = 9 → → → →
3 (b) RW = 1 OP or / atau OP = 2RW
9. (a) (i) 2x + 3y ~ (ii) –2x + y ~ 2
~
~
(b) (i) 2mx + 3 my (ii) (2 – 2n)x + ny
~ ~ ~ ~ Trigonometric Functions
5
→ 1 → → → PRACTICE
(c) QB = AC or / atau AC = 3 QB Fungsi Trigonometri
3
PAPER 1
1. (a) (b)
PAPER 1
→ y y
1. (a) SU
(b) 1 (q – p)
5 ~ ~ x
2. (a) 3x – 2y –85° x
~ ~ –225°
3
(b) – x – 1 y
2 ~ 2 ~
Fourth quadrant Second quadrant
3. m = 3(n – 6) Sukuan keempat Sukuan kedua
n + 2 (c) (d)
4. (a) –3a (b) 6a – 5b
~ ~ ~ y y
5. (a) 10 units / unit (b) –8
6. m = 5 x – – π rad
7
4
8
–4
1
7. (a) 5 1 2 (b) k = 8 – – π rad x

9
3
8. (a) 5 unit Third quadrant First quadrant
(b) (i) b – a Sukuan ketiga Sukuan pertama
~ ~
(ii) 2b – a 2. (a) Second quadrant / Sukuan kedua
~ ~
9. –10i – j (b) First quadrant / Sukuan pertama
~ ~ (c) Fourth quadrant / Sukuan keempat
(d) Fourth quadrant / Sukuan keempat
2
PAPER
3 1
1. (a) (i) 6x – 9y (ii) 2x + 6y 3. (a) – (b)
~ ~ ~ ~ 2 2
(b) m = 3 5 12
8 4. (a) —– (b) —–
13
13
n = 3 5 12
4 (c) – —– (d) – —–
(c) h = 3 13 5
5. (a) h (b) ABBBBB 2
1 – h
2. (a) (i) a – 2b (ii) 1 (a – 6b) h
~ ~ 4 ~ ~ (c) ———— (d) h
ABBBBB 2
1 – h
(b) 3 37 units / unit
4 6. (a) 8.44°, 171.56°
(c) m = 2 , n = 1 (b) 192.91°, 347.09°
5 10
(c) 62.80°, 297.20°
3. (a) (i) –8a + 6b ~ (ii) 16a + 6b (d) 151.81°, 208.19°
~
~
~
(b) k = 1 (e) 60.80°, 240.80°
3 (f) 146.26°, 326.26°
4. (a) (i) –10u + 15v
~
~
(ii) 4u + 9v 7. (a) 7.37°, 82.63°, 187.37°, 262.63°
~ ~ (b) 53.91°, 96.09°, 233.91°, 276.09°
(b) PT : TR = 1 : 2 (c) 337.03°
© Penerbitan Pelangi Sdn. Bhd. A8

Additional Mathematics Form 5 Answers

(d) 5°, 65°, 185°, 245° PAPER 2
(e) 150°, 240°
1. (a), (b)
8. (a) y
y
y = 2 sin 2x y = sin 2x
2 1
x x
0 90° 180° 270° 360° 0 – π π 3π 2π
–
2 2
1
x

y = – 1 – –
π
–2 –1 2
5 solutions / 5 penyelesaian
(b) 1
2
y y = – cos 2x 2. (a) (i), (ii)
1
y = – kos 2x y
0.5 2
y = 2 – x y = 1 – sin 2x
2 p
x
0 90° 180° 270° 360° 1
–0.5 0 x
π 2π
(c) 3 solutions / 3 penyelesaian
y
y = |tan x| 3. (b) & (c)
y
2
1
x x
0 90° 180° 270° 360° O 1 π π
2 y = 1 + 1 π x
9. (a) y = 2 + sin 2x (b) y = utan 1 xu – 1 2 solutions / 2 penyelesaian
(c) y = cos x – 2 2
y = kos x – 2 4. (a) 19° 28ʹ, 90°, 160° 32ʹ, 270°
10. 146.31°, 153.43°, 326.31°, 333.43° (b) y
11. (a) 0 rad, p rad, 2p rad 3 y = 3 – sin 2x
1 5 7 11
(b) —p rad, —p rad, —p rad, —–p rad 2
6 6 6 6 2x
y = 2 –
1 3 p
(c) —p rad, —p rad
2 2 0 π π x
1 3 2
(d) —p rad, —p rad
2 2
(e) p rad –3
120 56
12. (a) – (b) 3 solutions / 3 penyelesaian
169 65
16 3 5. (a), (b)
13. (a) —– (b) ——
63 ABB y
10
4
1
119
2
y = 2 | kos 2x|
(c) 1—– (d) —–– y = 2 | cos 2x| y = – + – x
2
5π
63 169 2
169
14.
119 1
7 11
15. (a) —p rad, —–p rad
6 16
2
4
(b) 0 rad, —p rad, p rad, —p rad, 2p rad 0 x
3 3 – π π 3π 2π
–
1 1 5 3 2 2
(c) —p rad, —p rad, —p rad, —p rad
23p
9p
11p
3p
33p
7p
6 2 6 2 —– rad, —– rad, —– rad, —–– rad, —–– rad, —–– rad,
(d) p rad. 40 10 10 40 20 8
(e) 0 rad, 2p rad 8p
—– rad
5
A9 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

4. (b) y
PAPER 1 2
1. 45°, 75°58ʹ
17
2. – —– O 2π x
7
3. 30°, 90°, 150°, 270° –2
1
4. (a) — (b) 2kABBBBB
1 – k
2
k 4 solutions / 4 penyelesaian
1
1 – r
5. (a) ABBBBB (b) —–——–
2
2rABBBBB
2
1 – r
6. (a) (i) n = 3
(i) q = 2 (b) 0°, 65.91°, 114.09°, 180°, 245.91°
(b) 1 solution / 1 penyelesaian (c) y
PAPER 2 2 y = 1 + sin x
1. (b), (c)
y y = 2 – x
1 p
y = | – tan 2 |
θ
x
x
y = 1 – –– 0 π 2π
π
3 solutions / 3 penyelesaian
x
0 π π 3π π
– – –
PRACTICE Permutations and Combinations
4 2 4
4 solutions / 4 penyelesaian 6 Pilih Atur dan Gabungan
2. (a) (ii) 52°15ʹ, 127°46ʹ, PAPER 1
232°15ʹ, 307°46ʹ
(b) y 1. 24
2. 5 040
1 y = – cos 2x 3. (a) 24 (b) 15 120
y = – kos 2x
4. 40 320
x 5. (a) 48 (b) 48
0 π π 3π 2π
– –
2 2 6. 120
–1 7. 1 080
8. (a) 1 440 (b) 840
9. (a) 120 (b) 72
3. (a), (b)
10. (a) 362 880
y (b) 80 640
3 11. (a) 479 001 600 (b) 103 680
y = tan 2x − 1 12. (a) 210 (b) 462
2
13. (a) 34 650 (b) 369 600
1
14. (a) 35 (b) 35
x
0 π π 3π π 5π 3π 7π π 15. (a) 21 (b) 35
2
8
8
8
4
8
– 1 — — — — — — —
4
2x
y = − — 16. 840
– 2 π − 1
17. 4
– 3
18. (a) 210 (b) 80
19. (a) 1 400 (b) 350
3 solutions / 3 penyelesaian
20. (a) 140 (b) 371
© Penerbitan Pelangi Sdn. Bhd. A10

Additional Mathematics Form 5 Answers

PAPER 2 14. (a) 2 (b) 40
7 77
1. (a) 40 320 (b) 1 440 5
15.
2. (a) 17 280 (b) 1 555 200 18
11 1
3. (a) 72 (b) 48 16. (a) (b)
(c) 36 8 12 16
4. (a) 15 120 (b) 42 17. —–
75
(c) 1 008
18. 0.07
5. (a) 1 716 (b) 462
(c) 308 19. 0.243
74
20. —––
231
PAPER 1 PAPER 2
1. (a) 40 320 (b) 4 320
1
3
2. (a) 560 (b) 480 1. (a) — (b) —–
8 16
3. (a) 210 (b) 217 12 13
2. (a) —– (b) —–
25 25
25 107
3. (a) —– (b) —––
53 212
1. (a) 5 005 (b) 6 435 4. (a) 0.504 (b) 0.054
2. 240 1 4
5
5
3. (a) 1 716 (b) 315 5. (a) — (b) —
43
58
6. (a) —– (b) —––
67 134
PRACTICE Probability
7 Kebarangkalian
PAPER 1
13 36
PAPER 1 1. (a) (b)
25 125
1
1. — 2. 2544
2 3125
1
2. — 3. (a) 1 (b) 5
4 3 9
2
3. — 4. (a) 21 (b) 19
5 40 40
4. n = 5
1 3
5. (a) (b)
56 4
6. y = 10 1 3
1. (a) (b)
256 16 4 8
7. (a) (b) 81
625 3 125 2.
125
8. (a) 0.0625 (b) 0.1406
9. (a) 0.7 (b) 0.1
10. 1 PRACTICE Probability Distributions
8
Taburan Kebarangkalian
11
11. —–
13
2 PAPER 1
12.
3 1. 0.02842
2 1
13. (a) (b)
3 30 2. 4, 2.4, 1.549
7
(c) 3. 0.1115
10
4. 0.6563
A11 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

5. 0.6230
6. (a) 0.2416 (b) n = 3
PAPER 1
7. (a) 0.2787 (b) 0.0006554
1. (a) {(E, E), (E, F), (F, E), (F, F)}
8. P(X = 0) = 0.216, (b) 0, 1, 2
P(X = 1) = 0.432,
2
P(X = 2) = 0.288, 2. (a) — (b) 20
3
P(X = 3) = 0.064
3. (a) 1.2 (b) 37.556
P(X = x)
4. (a) (i) 0
0.5 (ii) 1
(b) 0.4772
0.4 5. (a) 1.697 (b) 0.9978
6. 0.4202
0.3 36
7. (a) e + f – —– (b) p = 0.5879
49
0.2
PAPER 2
1. (a) 0.2150
0.1 (b) (i) 0.5769 (ii) 65
2. (a) (i) p = 0.2 (ii) 0.6778
0.0 x (b) (i) 0.2595 (ii) 50.10
0 1 2 3
3. (a) (i) 0.3115 (ii) 0.6329
1
9. n = 90, k = (b) (i) 0.1239 (ii) k = 3.29
3
10. (a) –2 (b) 19.5
11. (a) 0.1151 (b) 0.3829
12. (a) 0.8944 (b) 0.3384 1. (a) (i) 0.1762 (ii) 0.8658
(b) (i) 0.6304 (ii) 3.774 kg
13. 0.8427
2. (a) (i) n = 24 (ii) 0.03077
14. (a) 1 (b) 0.1359 (b) (i) 0.1587
15. (a) 2.28% (b) 53.28% (ii) 1 417 or / atau 1 418
16. (a) 2.5 (b) 0.9938
PRACTICE
17. (a) k = –1.264 (b) x = 56.208 Motion along a Straight Line
18. 60 9 Gerakan pada Garis Lurus

PAPER 2
PAPER 1
1. (a) 0.6083
(b) (i) 2.275% (ii) 0.02275 1. (a) 13 m
(b) 4 m
2. (a) (i) p = 0.025, n = 3000
(ii) 0.0000022 2. (a) 30 m
(b) (i) 0.9772 (ii) 53 g (b) t = 1.732
3. (a) (i) n = 3 040 (ii) 0.00385 3. (a) 12 m
(b) 29.02% (b) 50 m
4.
4. (a) (i) 0.3456 (ii) 300 t (s) 0 1 2 3 4 5
(b) (i) 0.0099 (ii) m = 2.3428
s (m) 0 3 4 3 0 –5
5. (a) (i) 0.01622 (ii) 375; 9.682
(b) (i) 0.2149 (ii) 0.3962
t = 0 t = 1
6. (a) 0.09935 t = 5 t = 2
(b) (i) 0.02275 (ii) 64 –5 –4 –3 –2 –1 0 1 2 3 4 s (m)
7. (a) (i) 0.3364 (ii) 0.8840 t = 4 t = 3
(b) 253
5. (a) t = 1, 3
8. (a) 0.9192 (b) 2 m to the left of O. / 2 m ke kiri O.
(b) (i) 32 (ii) 246.8




© Penerbitan Pelangi Sdn. Bhd. A12

Additional Mathematics Form 5 Answers

6. (a) 22. (a) –17 m s –1
s (m) –1
2
s = t – 4t + 5 (b) –15.5 m s
17
3
1
3
2
23. (a) s = —t – —t + 4t
2 3
2
(b) 18— m to the right of O.
3
2
5 18— m ke kanan O.
3
1 t (s) 24. (a) 6 m to the right of O.
O 2 6 6 m ke kanan O.
(b) 20 m (b) 9 m to the right of O.
9 m ke kanan O.
7. (a) v = 2t – 4 (b) v = 1 – 4t – 9t 2
(c) v = t + 8t (d) v = t – 3t 2 PAPER 2
2
1
(e) v = 10t – —
4 1. (a) 27
2
8. v = 6t – 6t –1
(b) –24.5 m s
v (m s –1 ) (c) –36 m s –2
2
v = 6t – 6t
36 2. (a) –15 m s
–1
1
(b) –16— m s –1
3
(c) 36 m to the left of O. / 36 m ke kiri O.
3. (a) h = 6
0 (b) t = 1, 3
–1.5 0.5 1 3 t (s) (c) 8 m
–1
2
9. (a) v = 6t – 3 (b) –3 m s –1 4. (a) 16 m s (b) 36 m
(c) 21 m s , 147 m s –1 (c) 6 s (d) t > 4
–1
4
10. (a) 2t – 9 (b) –1 m s –1 5. (a) (i) –1 m s –1 (ii) 3 m
(c) t = 4.5
(b) v
11. (a) s = 2t + 3t
2
(b) s = t – 3t + 2t 15
2
3
5
(c) s = 6t – 2t – —t
3
2
2
(d) s = 2t – 7t
3
1 1
2
(e) s = —t – —t 3
4 3
12. (a) 24 m to the left of O. 8
24 m ke kiri O.
(b) 13.5 m to the right of O.
13.5 m ke kanan O.
13. (a) 105 m to the right of O.
105 m ke kanan O.
(b) 29 m t
0 2 4 7
14. (a) 9 m s –1 (b) 20 m
(3, –1)
15. (a) a = 10t (b) a = 4 – 6t 3  t  7
(c) a = 6(2t – 7) 2
6. (a) –8 m s
–2
16. (a) a = 36t – 10 (b) a = –24t (b) 16 m s –1
(c) a = 18t – 4 2
(c) 90— m
–2
17. (a) 14 m s (b) 4 m s –2 3
18. (a) 20 m s (b) 8 m s –2 1
–2
7. (a) — , t , 3
19. (a) –6 m s –2 2
–2
(b) –12 m s , 12 m s –2 17 1
(b) —– m to the right of O at t = —;
20. (a) 24 m s (b) 18 m s –2 24 2
–2
17 1
21. (a) v = t – 12t + 12 —– m ke kanan O pada t = —;
2
2
24
(b) –24 m s –1
A13 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

1
4— m to the left of O at t = 3.
2 PRACTICE Linear Programming
1 10 Pengaturcaraan Linear
4— m ke kiri O pada t = 3.
2
(c) –2 m s –1
11 PAPER 1
(d) 5—– m
12
1.
y
8. (a) p = 3 (b) 3 m s –1
(c) y = 2x – 3
v (m s )
–1
x
O
2
3 v = –3t + 6t
t (s)
O 1 2 3
2. y
–9
y = 5 – 2x
(d) 8 m x
O
3.
2
PAPER y
1. (a) 1 (b) 0 , t , 6 2y = x + 10
1
(c) t = 12 (d) 108— m
3 y = x – 3
2. (a) 12 m s –1 (b) t = 4 O x
1 x + y = 8
(c) 12.25 m s –1 (d) 51— m
3
3. (a) –1 m s –1
(b) 22 m 4. y , x + 2, y + x , 6
(c) Particle X = 2 m
Particle Y = 2 m 5. 2y , 3x + 4
6. y – 4x  2, y + x . –2
7. 3y  2x + 9, y + x  3
1. (a) (i) 4 m s –1 (ii) t = 4 s 8. y
(b) (i) v y = x + 4

10 y = 10
4



t O 4 x
0 2 4 5 x = 4 2y + 5x = 20
9.
y
–5
8 3y – 4x = 12
(ii) 13 m
4
1 5 5y + 8x = 40
–1
2. (a) 10 m s (b) 9 m
8 6
x
27 9 –3 O 5
(c) s (d) s
4 2

© Penerbitan Pelangi Sdn. Bhd. A14

Additional Mathematics Form 5 Answers

10. y 2. (a) x + y > 50;
y = 8 y – x > 10;
8 (6, 8) 4x + 3y  300
(b)
y
y = 2x – 4
3 100
x
O 2 2y + 3x = 6 90
11. y  6, 2y + x > 8, y > 2x – 12 80
4x + 3y = 300
12. (a) x , y (b) y  2x y – x = 10
(c) 2x + 3y  30 70
13. (a) x  50 (b) y > 20
(c) x + y > 100 60
R
14. (a) x + y  200 (b) x  60
(c) y > 50 50
15. (a) x + y  80 000 (b) x  2y y = 40
(c) y > 40 000 40
16. The objective function is a set of lines which is 30
parallel to 20x + 35y.
Fungsi objektif adalah satu set garis lurus yang selari
dengan 20x + 35y. 20
17. 54
10
x + y = 50
PAPER 2
1. (a) x + 2y  100; 0 10 20 30 40 50 60 70 x
3
2y > —x; (c) (i) 30 (ii) 10  x  30
5
x  30 3. (a) 4x + 5y  200;
(b) y > 8;
y y > x – 10
50 (b)
y
x + 2y = 100 k = 2x + y
45
x = 30 40
y = 40
40
35
35 (30, 35) 4x + 5y = 200
30
30
25
R
25 20 y = x – 10
R
20
15
16x + 15y = 360
15 10
y = 8
10
5
3
5 2y = – x 26
5 0 x
5 10 15 20 25 30 35
0 x
5 10 15 20 25 30 35
(c) (i) 26 units / unit
(c) (i) 95 (ii) 15 units / unit
(ii) x = 20

A15 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

4. (a) 8x + 18y > 720; 40
5x + 8y  800; III : y > 100 x
8y  5x (b)
(b) y
y
100 x
100 x = 35
90
90 4x + 5y = 500
80 5x + 8y = 800
70 80
60 8y = 5x 70
50 (80, 50)
60
40 = 720 Region R
50
8x + 18y
30
20 40
R y = 40 x
10 30 100
k = 10x + 30y k = 40x + 30y
x
0 20 40 60 80 100 120 140 160 20
y = 14
(c) (i) maximum / maksimum = 120 10
minimum = 40 x + y = 45
(ii) RM2 300 0 x
5. (a) x + y  9; 10 20 30 40 50 60 70 80 90 100
1 (c) (i) 14
y > —x;
2 (ii) RM4 310
5x + 2y  30
(b) y

9 PAPER 2
k = 1200x + 600y
1. (a) x + 2y  60; y  2x
8 (b) The number of badminton rackets is not more
than 5 times the number of tennis rackets.
Bilangan raket badminton tidak boleh melebihi
7
5 kali bilangan raket tenis.
5x + 2y = 30 (c)
6 y
30 y = 2x
5 (4, 5)
k = x + y
25
4
R x + y = 9
20
3
15
2 R
(42, 9) x = 5y
10
1
1
y = –x 5 x + 2y = 60
2
x
0 1 2 3 4 5 6 7
0 x
(c) (i) 4 lorries / buah lori 10 20 30 40 50 60
(ii) 7 800 kg
(d) (i) 10
6. (a) I : 80x + 100y  10 000 or 4x + 5y  500 (ii) 51
II : x + y > 45


© Penerbitan Pelangi Sdn. Bhd. A16

Additional Mathematics Form 5 Answers

2. (a) x + y . 40; (b) y
6x + 5y  900;
3y > 5x 1000
(b)
y 900
800
180 400x + 200y = 200 000
700
160
600
140 y = 500
500
120
3y = 5x 400
100 k = 5x + 3y R
300 100x + 50y = 30 000
80 x = 2y
200
60 6x + 5y = 900
100
40
x
20 k = 5x + 3y 0 100 200 300 400 500
x + y = 40 (c) (i) 400 pairs / pasang
0 x
20 40 60 80 100 120 140 160 (ii) RM50 000
2. (a) x + y  240;
(c) 150 , P  625
x  40;
y > 3x
3. (a) x + y  100; (b)
y  2x; y
3x + 5y > 200 x = 40
240
(b) y (40, 200)
200 y = 3x
80
y = 2x
160 R
70
(34, 66)
120 x + y = 240
60
k = 60x + 100y 80
50
k = 80x + 40y
40
40
R x
30 0 40 80 120 160 200 240
22 (c) (i) 40 wooden chairs / kerusi kayu
20 200 plastic chairs / kerusi plastik
x + y = 100
(ii) RM11 200
10 3x + 5y = 200
x = 30
0 x
20 40 60 80 100 120 PAPER 1
1. (a) 16
(c) (i) 22 units / unit
(b) 4
(ii) RM8 640
2. (a) p = –1
(b) 0
5x – 4 4 – 5x
3. (a) g(x) = or g(x) =
1. (a) 400x + 200y  200 000; 5 –2 2
(b) p = – —
100x + 50y > 30 000; 2
x  2y


A17 © Penerbitan Pelangi Sdn. Bhd.

Additional Mathematics Form 5 Answers

4. b = –5, and c = –3 5. (a) (0, 2)
(b) (12, –6)
5. h  –3, h > 5 (c) 2y = 3x + 4
p 2
6. q = — 6. (a) mode = 62.5
8
2
7. x = – — Students
3 Pelajar
8. x = –3 18
9. 3m + n – 3
16
10. n = 18
14
1
11. s = —
∞ 2
12
1
12. p = —, and q = 16
6 10
3 7
13. y = – —x + —
4 8 8
19
14. (10, —)
2 6
15. h = 7
7i + 2j 4
~
~
16. ————
ABB 2
53
17. –19
Mass
(2x – 1) (4x – 5)
2
18. —–—————— 0 Berat
(x – 1) 2 50.5 55.5 60.5 65.5 70.5 75.5
dy
19. —– = –8 62.5
dx
20. 43 (b) 62.72
21. h = 3
7. (a) log x 0.18 0.30 0.48 0.60 0.65 0.78
22. k = 2
log y 0.66 0.84 1.08 1.26 1.33 1.52
23. 19.47°, 90°, 160.53°
24. (a) q = 1.712 rad log y
10
(b) 34.30 Graph of log y against log x
10
25. (a) x = 5 1.8 Graf log y melawan log x 10
10
10
(b) 14
1.6
PAPER 2
5
1. x = 2, – — 1.4 (0.65, 1.33)
4
or 1.2
7
y = 3; – —
2
1.0
4u – 2
2. (a) t = ———–
3
(b) 12 0.8 (0.3, 0.84)
3. (a) h = 2 or k = –6 0.6
3
2
(b) y = x – 3x + 7
0.4 0.42
4. 5
2
0.2
0 log x
10
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
(b) (i) m = 2.63
π π 3 π 2 π
– 1 2 2 (ii) n = 1.4
2
4 solutions / 4 penyelesaian
© Penerbitan Pelangi Sdn. Bhd. A18

Additional Mathematics Form 5 Answers

→ 14. (a) x + y > 30
8. (a) (i) AB = 12y – 8x y  2x
→ ~ ~
(ii) OQ = 8x + 16y 80x + 60y  3 600
~
→ ~
(b) (i) OP = h(8x + 16y) '(b) y
~
→ ~ 60
(ii) OP = (8 – 8k)x + 12ky y = 2x
~
~
4
3
(c) h = — , k = — 50
7 7
4
9. (a) —p or 4.189 rad 40
3
(b) 50.27 cm 30
(c) 134.05 cm 2
(d) 122.5 cm 2 20 80x + 60y = 3600
10. (a) (i) a = 2 y = 12
b = –8 10 x = 18
32
(ii) — unit 2
3 0 x
20
10
(b) 27p k = 80x + 60y 30 40 50 60
–10
11. (a) (i) 8
(ii) 1.88
(b) (i) 0.2373 –20
(ii) 0.1035 x + y = 30
–30
12. (a) x = 110, y = RM3.40, z = RM49.50
(b) 109.85
(c) RM24 716.25
(d) 115.34 (c) (i) Maximum x = 36, Minimum x = 18
(ii) 2 000
13. (a) 3
(b) 4 15. (a) 48.85°
(c) t , –1 and t . 3 (b) 13.75
59
(d) — m (c) 17.66°
3 (d) 41.08 cm 2































A19 © Penerbitan Pelangi Sdn. Bhd.

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© Penerbitan Pelangi Sdn. Bhd. A20