# Nota Ulangkaji SPM Matematik Tingkatan 4 dan Tingkatan 5

## Tingkatan 4

### Bab 1 Bentuk Piawai

1. Angka Bererti
2. Bentuk Piawai
3. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

### Bab 2 Ungkapan dan Persamaan Kuadratik

1. Ungkapan Kuadratik
2. Pemfaktoran Ungkapan Kuadratik
3. Persamaan Kuadratik
4. Punca Persamaan Kuadratik
5. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

### Bab 3 Set

1. Set
2. Subset, Set Semesta, dan Set Pelengkap
3. Operasi ke atas Set
4. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

### Bab 5 Garis Lurus

1. Kecerunan Garis Lurus
2. Kecerunan Garis Lurus dalam Sistem Koordinat Cartesan
3. Pintasan
4. Persamaan Garis Lurus
5. Garis Selari
6. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

### Bab 6 Statistik III

1. Selang Kelas
2. Mod dan Min bagi Data Terkumpul
3. Histogram Selang Kelas Sama Saiz
4. Poligon Kekerapan
5. Kekerapan Longgokan
6. Sukatan Serakan
7. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

### Bab 7 Kebarangkalian I

1. Ruang Sampel
2. Peristiwa
3. Kebarangkalian suatu Peristiwa
4. Soalan Model SPM
1. Soalan pendek (Kertas 1)

### Bab 8 Bulatan III

1. Tangen kepada Bulatan
2. Sudut di antara Tangen dengan Perentas
3. Tangen Sepunya
4. Soalan Model SPM
1. Soalan pendek (Kertas 1)

### Bab 9 Trigonometri II

1. Nilai Sin θ, Kos θ, dan Tan θ (0o ≤ θ ≤ 360o)
2. Graf Sinus, Kosinus, dan Tangen
3. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

### Bab 10 Sudut Dongakan dan Sudut Tunduk

1. Sudut Dongakan dan Sudut Tunduk
2. Menyelesaikan Masalah yang Melibatkan Sudut Dongakan dan Sudut Tunduk
3. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

### Bab 11 Garis dan Satah dalam Tiga Dimensi

1. Sudut di antara Garis dengan Satah
2. Sudut di antara Dua Satah

## Tingkatan 5

### Bab 1 Asas Nombor

1. Nombor dalam Asas Dua, Asas Lapan, dan Asas Lima
2. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

### Bab 2 Garf Fungsi II

1. Graf bagi beberapa Fungsi (Bahagian 1)
2. Graf bagi beberapa Fungsi (Bahagian 2)
3. Penyelesaian Persamaan dengan Kaedah Graf
4. Rantau Ketaksamaan dalam Dua Pemboleh Ubah
5. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

### Bab 5 Ubahan

1. Ubahan Langsung
2. Ubahan Songsang
3. Ubahan Tercantum
4. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

### Bab 6 Kecerunan dan Luas di bawah Graf

1. Kuantiti yang diwakili oleh Kecerunan Graf
2. Kuantiti yang diwakili oleh Luas di Bawah Graf
3. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

### Bab 7 Kebarangkalian II

1. Kebarangkalian Suatu Peristiwa
2. Kebarangkalian Peristiwa Pelengkap
3. Kebarangkalian Peristiwa Bergabung
4. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

1. Bearing

### Bab 9 Bumi sebagai Sfera

1. Longitud
2. Latitud
3. Kedudukan Tempat
4. Jarak pada Permukaan Bumi
5. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

### Bab 10 Pelan dan Dongakan

1. Unjuran Ortogon
2. Pelan dan Dongakan
3. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

# SPM Practice 2 (Question 8 – 10)

Question 8:
The variables x and y are related by the equation $y=\frac{p}{{3}^{x}}$, where k is a constant.
Diagram below shows the straight line graph obtained by plotting ${\mathrm{log}}_{10}y$ against x.

1. Express the equation$y=\frac{p}{{3}^{x}}$ in its linear form used to obtain the straight line graph shown in Diagram above.
2. Find the value of p.

Solution:

Question 9:
Variable x and y are related by the equation ${y}^{2}=p{x}^{q}$ . When the graph lg y against lg x is drawn, the resulting straight line has a gradient of -2 and an vertical intercept of 0.5 . Calculate the value of p and of q.

Solution:

Question 10:
Variable x and y are related by the equation $y=\frac{c}{d-x}$. When the graph y against xy is drawn the resulting line has gradient 0.25 and an intercept on the y-axis of 1.25. Calculate the value of c and of d.

Solution:

# SPM Practice 2 (Question 6 & 7)

Question 6:
The variables x and y are related by the equation $y=p{x}^{3}$, where p is a constant. Find the value of p and n.

Solution:

Question 7:
Diagram A shows part of the curve $y=a{x}^{2}+bx$ .  Diagram B shows part of the straight line obtained when the equation is reduced to the linear form. Find
(a) the values of a and b,
(b) the values of p and q.

Diagram A

Diagram B

Solution:

# SPM Practice 2 (Question 4 & 5)

Question 4:
The diagram shows part of the straight line graph obtained by plotting $\frac{y}{x}$ against $\sqrt{x}$.

Given its original non-linear equation is $y=px+q{x}^{\frac{3}{2}}$.  Calculate the values of p and q.

Solution:

Question 5:
The diagram below shows the graph of the straight line that is related by the equation $\frac{x}{y}=\frac{2}{x}+3x$.

Find the values of p and k.

Solution:

# (G) Sum to Infinity of Geometric Progressions (Part 2)

(H) Recurring Decimal
Example of recurring decimal:
$\begin{array}{l}\frac{2}{9}=0.2222222222222.....\\ \\ \frac{8}{33}=0.242424242424.....\\ \\ \frac{41}{333}=0.123123123123.....\end{array}$

Recurring decimal can be changed to fraction using the sum to infinity formula:
$\overline{)\text{}S\infty =\frac{a}{1-r}\text{}}$

Example (Change recurring decimal to fraction)
Express each of the following recurring decimals as a fraction in its lowest terms.
(a) 0.8888 ...
(b) 0.171717...
(c) 0.513513513 ….

Solution:
(a)
0.8888 = 0.8 + 0.08 + 0.008 +0.0008 + ….. (recurring decimal)
$\begin{array}{l}GP,\text{}a=0.8,\text{}r=\frac{0.08}{0.8}=0.1\\ {S}_{\infty }=\frac{a}{1-r}\\ {S}_{\infty }=\frac{0.8}{1-0.1}\\ {S}_{\infty }=\frac{0.8}{0.9}\\ {S}_{\infty }=\frac{8}{9}\to \overline{)\begin{array}{l}\text{check using calculator}\\ \text{}\frac{8}{9}=0.888888....\end{array}}\end{array}$

(b)

0.17171717 …..
= 0.17 + 0.0017 + 0.000017 + 0.00000017 + …..
$\begin{array}{l}GP,\text{}a=0.17,\text{}r=\frac{0.0017}{0.17}=0.01\\ {S}_{\infty }=\frac{a}{1-r}\\ {S}_{\infty }=\frac{0.17}{1-0.01}=\frac{0.17}{0.99}=\frac{17}{99}\to \overline{)\begin{array}{l}\text{remember to check the}\\ \text{answer using calculator}\end{array}}\end{array}$

(c)
0.513513513…..
= 0.513 + 0.000513 + 0.000000513 + …..
$\begin{array}{l}GP,\text{}a=0.513,\text{}r=\frac{0.00513}{0.513}=0.001\\ {S}_{\infty }=\frac{a}{1-r}\\ {S}_{\infty }=\frac{0.513}{1-0.001}=\frac{0.513}{0.999}=\frac{513}{999}=\frac{19}{37}\end{array}$

# 3. The nth Term of Geometric Progressions (Part 2)

(D) The Number of Term of a Geometric Progression
Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term

Example 2:
Find the number of terms for each of the following geometric progressions.
(a) 2, 4, 8, ….., 8192
(b) $\frac{1}{4},\text{}\frac{1}{6},\text{}\frac{1}{9},\text{}.....,\text{}\frac{16}{729}$
(c) $-\frac{1}{2},\text{1},\text{}-\text{2},.....,\text{64}$

Solution:
(a)
$\begin{array}{l}2,\text{}4,\text{}8,.....,\text{}8192←\left(\text{Last term is given)}\\ a=2\\ r=\frac{{T}_{2}}{{T}_{1}}=\frac{4}{2}=2\\ {T}_{n}=8192\\ a{r}^{n-1}=8192←\left(Then\text{th term of GP,}{T}_{n}=a{r}^{n-1}\right)\\ \left(2\right){\left(2\right)}^{n-1}=8192\\ {2}^{n-1}=4096\\ {2}^{n-1}={2}^{12}\\ n-1=12\\ n=13\end{array}$

(b)
$\begin{array}{l}\frac{1}{4},\text{}\frac{1}{6},\text{}\frac{1}{9},\text{}.....,\text{}\frac{16}{729}\\ a=\frac{1}{4},r=\frac{\frac{1}{6}}{\frac{1}{4}}=\frac{2}{3}\\ {T}_{n}=\frac{16}{729}\\ a{r}^{n-1}=\frac{16}{729}\\ \left(\frac{1}{4}\right){\left(\frac{2}{3}\right)}^{n-1}=\frac{16}{729}\\ {\left(\frac{2}{3}\right)}^{n-1}=\frac{16}{729}×4\\ {\left(\frac{2}{3}\right)}^{n-1}=\frac{64}{729}\\ {\left(\frac{2}{3}\right)}^{n-1}={\left(\frac{2}{3}\right)}^{6}\\ \therefore n-1=6\\ n=7\end{array}$

(c)
$\begin{array}{l}-\frac{1}{2},\text{1},\text{}-\text{2},.....,\text{64}\\ a=-\frac{1}{2},\text{}r=\frac{-2}{1}=-2\\ {T}_{n}=64\\ a{r}^{n-1}=64\\ \left(-\frac{1}{2}\right){\left(-2\right)}^{n-1}=64\\ {\left(-2\right)}^{n-1}=64×-2\\ {\left(-2\right)}^{n-1}=-128\\ {\left(-2\right)}^{n-1}={\left(-2\right)}^{7}\\ n-1=7\\ n=8\end{array}$

(E) Three consecutive terms of a geometric progression
If e, f and g are 3 consecutive terms of GP, then
$\overline{)\text{}\frac{g}{f}=\frac{f}{e}\text{}}$
Example 3:
If p + 20,   p − 4, p −20 are three consecutive terms of a geometric progression, find the value of p.

Solution:
$\begin{array}{l}\frac{p-20}{p-4}=\frac{p-4}{p+20}\\ \left(p+20\right)\left(p-20\right)=\left(p-4\right)\left(p-4\right)\\ {p}^{2}-400={p}^{2}-8p+16\\ 8p=416\\ p=52\end{array}$

# SPM Practice Question 13 – 15

Question 13:
An arithmetic progression consists of 10 terms. The sum of the last 5 terms is 5 and the fourth term is 9. Find the sum of this progression.

Solution:

Question 14:
The sum of the first 6 terms of an arithmetic progression is 39 and the sum of the next 6 terms is –69. Find
(a) The first term and the common difference.
(b) The sum of all the terms from the 15th term to the 25th term.

Solution:

Question 15:
An arithmetic progression has 9 terms. The sum of the first four terms is 24 and the sum of all the odd number terms is 55. Find
(a) The first term and common difference,
(b) The seventh term.

Solution:

# SPM Practice Question 4 – 6

Question 4:
It is given that -7, h, k, 20,… are the first four terms of an arithmetic progression. Find the value of h and of k.

Solution:

Question 5:
51, 58, 65,... 191 are the first n terms of an arithmetic progression. Find the value of n.

Solution:

Question 6:
Find the number of the multiples of 8 between 100 and 300.

Solution:

# The nth Term of Arithmetic Progression (Example 3 & 4)

Example 3:
The volume of water in a tank is 75 litres on the first day. Subsequently, 15 litres of water is added to the tank everyday.
Calculate the volume, in litres, of water in the tank at the end of the 12th day.

Solution:
Volume of water on the first day = 75l
Volume of water on the second day = 75 + 15 = 90l
Volume of water on the third day = 90 + 15 = 105l
75, 90, 105, …..
AP, a = 75, d = 90 – 75 = 15
Volume of water on the 12th day,
T12 = a + 11d
T12 = 75 + 11 (15)
T12 = 240l

Example 4:
The first three terms of an arithmetic progression are 72, 65 and 58.
The nth term of this progression is negative.
Find the least value of n.

Solution:
72, 65, 58
AP, a = 72, d = 65 – 72 = –7
The nth term is negative,
Tn < 0
a + (n – 1) d < 0
72 + (n – 1) (–7) < 0
(n – 1) (–7) < –72
n – 1 > –72/ –7
n – 1 > 10.28
n > 11.28
n must be integer, n = 12, 13, 14, ….

Therefore, the least value of n = 12.

# 9.2.1 First Derivative for Polynomial Function (Examples)

Example:
Find $\frac{dy}{dx}$ for each of the following functions:
(a) y = 12
(b) y = x4
(c) y = 3x
(d) y = 5x3

Solution:
(a)
= 12
$\frac{dy}{dx}=0$

(b)
= x4
$\frac{dy}{dx}$ = 4x3

(c)
= 3x
$\frac{dy}{dx}$ = 3

(d)
= 5x3
$\frac{dy}{dx}$ = 3(5x2) = 15x2

(e)

$\begin{array}{l}y=\frac{1}{x}={x}^{-1}\\ \frac{dy}{dx}=-{x}^{-1-1}=-\frac{1}{{x}^{2}}\end{array}$

(f)

$\begin{array}{l}y=\frac{2}{{x}^{4}}=2{x}^{-4}\\ \frac{dy}{dx}=-4\left(2{x}^{-4-1}\right)=-8{x}^{-5}=-\frac{8}{{x}^{5}}\end{array}$

(g)

$\begin{array}{l}y=\frac{2}{5{x}^{2}}=\frac{2{x}^{-2}}{5}\\ \frac{dy}{dx}=-2\left(\frac{2{x}^{-2-1}}{5}\right)=-\frac{4{x}^{-3}}{5}=-\frac{4}{5{x}^{3}}\end{array}$

(h)

$\begin{array}{l}y=\text{3}\sqrt{x}=3{\left(x\right)}^{\frac{1}{2}}\\ \frac{dy}{dx}=\frac{1}{2}\left(3{x}^{\frac{1}{2}-1}\right)=\frac{3}{2}{x}^{-\frac{1}{2}}=\frac{3}{2\sqrt{x}}\end{array}$

(i)
$\begin{array}{l}y=4\sqrt{{x}^{3}}=4{\left({x}^{3}\right)}^{\frac{1}{2}}=4{x}^{\frac{3}{2}}\\ \frac{dy}{dx}=\frac{3}{2}\left(4{x}^{\frac{3}{2}-1}\right)=6{x}^{\frac{1}{2}}=6\sqrt{x}\end{array}$

# Change of Base of Logarithms – Example 5

Example 5
Given that ${\mathrm{log}}_{2}3=1.585$   and ${\mathrm{log}}_{2}5=2.322$  , evaluate the following.
(a) ${\mathrm{log}}_{8}15$
(b) ${\mathrm{log}}_{5}0.6$
(c) ${\mathrm{log}}_{15}30$
(d) ${\mathrm{log}}_{16}45$

Solution: