KSSM Form 4 Additional Mathematics Notes (Chapter 1-5) PDF

Preview

Summary

Project Leader: Lim Jun Hao Associate Editor: Lai Zhi Jun Associate Project Editor: Siaw Jia Qi Senior Managing Editor: Lim Jun Hao Production Coordinator: Ooi Ming Yang Cover Design: Ooi Hian Gee Cover image: Pixabay About the cover: The cover images comprise of dictionary, notes and an ancient cl...

Description

Project Leader: Lim Jun Hao Associate Editor: Lai Zhi Jun Associate Project Editor: Siaw Jia Qi Senior Managing Editor: Lim Jun Hao Production Coordinator: Ooi Ming Yang Cover Design: Ooi Hian Gee Cover image: Pixabay About the cover: The cover images comprise of dictionary, notes and an ancient clock. The illustrations mean well-management of time and notes-taking habit can help us in achieving our goals. We shall also always make inquiry to the right person whenever we have any doubts during learning process. A collaboration work by Faculty of Science, University of Malaya Applied Mathematics undergraduate students.

Form 4 Additional Mathematics / Lim Jun Hao, Lai Zhi Jun, Ooi Hian Gee, Ooi Ming Yang, Siaw Jia Qi Font used: Times New Roman Copyright © 2020 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. The work is published in electronic form only. For information on obtaining permission for use of material in this work, please email us at [email protected].

i

Table of Contents Chapter 1 Functions 1.1 Functions ................................................................................................................................ 1 1.2 Composite Functions.............................................................................................................. 7 1.3 Inverse Functions ............................................................................................................... 12 Summative Exercises ................................................................................................................ 18

Chapter 2 Quadratic Functions 2.1 Quadratic Equations and Inequalities ............................................................................... 19 2.2 Types of Roots of Quadratic Equations ............................................................................. 24 2.3 Quadratic Functions ........................................................................................................... 26 Summative Exercises ............................................................................................................... 38

Chapter 3 Systems of Equations 3.1 Systems of Linear Equations in Three Variables .............................................................. 39 3.2 Simultaneous Equations Involving One Linear Equation & One Non-Linear Equation . 45 Summative Exercises ............................................................................................................... 49

Chapter 4 Indices, Surds and Logarithms 4.1 Laws of Indices.................................................................................................................... 50 4.2 Laws of Surds ..................................................................................................................... 52 4.3 Laws of Logarithms ........................................................................................................... 55 4.4 Applications of Indices, Surds and Logarithms ................................................................ 62 Summative Exercises ............................................................................................................... 63

Chapter 5 Progressions 5.1 Arithmetic Progressions ..................................................................................................... 64 5.2 Geometric Progressions ..................................................................................................... 71 Summative Exercises ............................................................................................................... 82

Solutions References

ii

CHAPTER 1 FUNCTIONS

Chapter 1 Functions 1.1 Functions 1.1.1 What is a function? Let’s take an example. If the radius of a circle is 2 cm, then the circumference, C of the circle is C = 2πr = 4π cm. If the radius of a circle is 4 cm, then the circumference, C of the circle is C = 2πr = 8π cm. If the radius of a circle is 5 cm, then the circumference, C of the circle is C = 2πr = 10π cm. From this example, we know that the radius of a circle will affect the circumference of the circle; therefore, the circumference of a circle is a function of radius, such that C = 2πr. Look at the graph of y = x − 2 on the right. The relation between the value of 3 on the x-axis and the value of 1 on the y-axis can be written as 3 → 1. We can say that 3 is mapped to 1, and likewise for 5 → 3 and 8 → 6. Every point (x, y) on the line corresponds to the mapping of x → y where the values of x on the x-axis are mapped to the values of y on the y-axis. A relation from x-axis to y-axis is called a function if each element in the x-axis is related to exactly one element in the y-axis. Hence, this is a function since each element from X is related to only one element in Y.

(8,6)

(5,3) (3,1)

A relation can be represented by an arrow diagram as shown below. 3

1

5

3

8

6

Set X Set Y 1. f is a function from set X to set Y . 2. The element in set X is known as object, whereas the element in set Y is known as image. Note that the term “preimage” is in common usage in math instead of the term “object”. 3. Example, the element 3 in set X and the element 1 in set Y correspond to each other whereby 3 is the object of 1 and 1 is the image of 3. 4. Any element x in set X that is mapped to one element y in set Y by a function is written in function notation as 𝒇 ∶ 𝒙 → 𝒚 or 𝒇(𝒙) = 𝒚 , where x is the object and y is the image. 5. A function maps an object to an image only. Therefore, the relation is called a function if the relation is either one-to-one relation or many- to-one relation. 1

CHAPTER 1 FUNCTIONS

EXAMPLE 1 Which of the following is a function? Explain. (a)

Solution: This is a function since each object is mapped to only one image although the element 1 has no object. This relation is known as one-to-one function.

12

8

20

6

8

12 1

(b)

a

5

b

4

c

10

(c)

3 9

-3

1

1

Solution: This is a function since each object has only one image although the element 10 has no object. This relation is known as many-to-one relation.

Solution: This is not a function since one object has two images. (Does not satisfy the condition of a function) This relation is known as one-to-many relation.

-1 (d) 2

1

4

2

Solution: This is not a function since each object has more than one image. (Does not satisfy the condition of a function)

4

2

CHAPTER 1 FUNCTIONS To determine whether a graph represents a function or not, we use vertical line test.

The vertical line intersects more than 1 point of the graph. Hence the graph does not represent a function.

The vertical line intersects only 1 point of the graph. Hence the graph represents a function.

1.1.2 Introducing asymptotes The asymptote of a curve is a line such that the distance between the line and the curve approaches zero as one or both of the x or y coordinates tend to infinity. However, the asymptote of a curve will not touch the curve. y

Graph of 𝑦 =

1 𝑥−3

Vertical asymptote x=3

x

From the graph, we know that: 1. When the x-value approaches 3 from the left ( 3− ), the corresponding y-value approaches negative infinity (−ꚙ). 2. When the x-value approaches 3 from the right ( 3+ ), the corresponding y-value approaches infinity (ꚙ). Function is undefined when denominator = 0, because division by zero makes an operation undefined. 𝑥−3=0 𝑥=3

The function is undefined at x = 3, therefore the graph will not touch the line x = 3. Is there a way to calculate the vertical asymptote? Take note that the function is undefined when denominator is 0. In this example, the denominator of the function is x−3. We will have the function undefined when x−3 = 0, therefore x = 3.

3

CHAPTER 1 FUNCTIONS

1.1.3 Introducing the absolute-value function y

The absolute value |x| is expressed as: |x| = Graph of y=|x| x

x −x

when x ≥ 0 when x < 0

For instance, |1| = 1; |−1| = 1.

1.1.4 How to determine the domain and the range of a discrete function? Let’s take back Example 1(a). a) 12

8

20

6

8

12

1. The domain of a function is the set of all possible input values. 2. The codomain of a function is the set into which all of the output function is forced to fall. 3. The range of a function is the set of all possible output values.

1 Set X Domain

Set Y Codomain

In this example, Domain = {12, 20, 8} Codomain = {8, 6, 12, 1} Range = {8, 6, 12}

1.1.5 How to determine the domain and the range of a continuous function? y

𝑓 (𝑥) = 𝑠𝑖𝑛 𝑥

x

Range

Domain In this example, the domain is −5 ≤ 𝑥 ≤ 6 whereas the range is −1 ≤ 𝑓(𝑥) ≤ 1.

4

CHAPTER 1 FUNCTIONS

EXAMPLE 2 Sketch the graph of f : x → |2x–4| in the domain −3 ≤ 𝑥 ≤ 3.

Solution: Step 1: You need to have 4 points in order to construct a graph of an x |2𝑥 − 4| absolute-value function. -3 10 The four points are: 0 4 (a) vertex, 2 0 (b) y-intercept, 3 2 (c) left endpoint of the graph, (d) right endpoint of the graph. Step 2: You need to plot the points onto a graph paper and indicate the x-axis and y-axis. Step 3: Connect the points in the shape of V with vertex as the turning point. The vertex can be obtained by equating the algebraic expression inside the modulus sign with 0 and solving it. In this example, 2x – 4 = 0 2x = 4 x=2 when x = 2, f(2) = |2(2) – 4| = 0 Hence the coordinate of the vertex is (2,0).

(-3,10)

(0,4)

𝑓(𝑥) = |2𝑥 − 4| (3,2)

(2,0)

This is the vertex.

5

CHAPTER 1 FUNCTIONS

1.1.6 How to determine the image of a function when its domain is given and vice versa? Note: For a function, for example f : x → 2x + 2, if x = 1, then f (1) = 2(1) + 2 = 4; if x = 2, then f (2) = 2(2) + 2 = 6; if x = u, then f (u) = 2(u) + 2 = 2u + 2.

EXAMPLE 3 A function f is defined by f : x → 2x + 2, find (a) the image of 2, (b) the object that have the image 4. Solution: (a) f : x → 2x + 2 f (x) = 2x + 2 When x = 2, f (2) = 2(2) + 2 =6 Hence, the image of 2 is 6.

(b) When f (x) = 4 2x + 2 = 4 2x = 2 x=1 Hence the object that has the image 4 is 1.

EXAMPLE 4

2𝑥 + 𝑎 A function f is defined by f : x → for all values of x, except x = b, where a is a constant. 𝑥−2 (a) State the value of b. (b) Find the value of a given that the value 5 is mapped to itself under f. Solution: (a) Function is undefined when: denominator = 0 x–2=0 x=2 It is given that function is undefined when x = b, therefore b = 2.

(b) 5 is mapped to itself, therefore 𝑓 (5) = 5 ( ) 2 5 +𝑎 =5 5−2 10 + 𝑎 = 5(3) 𝑎=5

6

CHAPTER 1 FUNCTIONS

1.2 Composite Functions 1.2.1 What are composite functions? x

f

f (x)

g

gf (x)

For example, let x be a tree. In order to produce furniture, the tree (x) undergoes cutting process (function f) to obtain wood (f (x)). The wood (f (x)), is then manufactured (Function g) to produce furniture (gf (x)). fg(x) is read as “f composed with g of x” . fg(x) = f[g(x)]. gf(x) is read as “g composed with f of x” . gf(x) = g[f(x)]. Note: 𝒇𝟐 (𝒙) = 𝒇𝒇(𝒙).

1.2.2 How to determine composite functions? EXAMPLE 5 Two functions f and g are defined by f (x) = x + 1 and g (x) = 𝑥 2 . Find the following composite functions: (𝑏) 𝑔2 (𝑐)𝑓𝑔 (𝑑)𝑔𝑓 (a) 𝑓 2 Solution: (a) 𝑓 2 (𝑥) = 𝑓𝑓 (𝑥) = 𝑓[𝑓(𝑥)] = 𝑓(𝑥 + 1) = (𝑥 + 1) + 1 =𝑥+2 2( ) (b) 𝑔 𝑥 = 𝑔𝑔(𝑥) = 𝑔[𝑔(𝑥)] = 𝑔(𝑥 2 ) = (𝑥 2 )2 = 𝑥4

f (x) = x + 1 f (x + 1) = (x + 1) + 1

(c) 𝑓𝑔(𝑥) = 𝑓[𝑔(𝑥)] = 𝑓(𝑥 2 ) = 𝑥2 + 1 (d) 𝑔𝑓(𝑥) = 𝑔[𝑓(𝑥)] = 𝑔(𝑥 + 1) = (𝑥 + 1)2 = 𝑥 2 + 2𝑥 + 1

7

CHAPTER 1 FUNCTIONS

1.2.3 How to determine the objects or the images of composite functions? EXAMPLE 6

Two functions f and g are defined by 𝑓(𝑥) = 𝑥 + 2 and 𝑔(𝑥) = 𝑥 2 , find 𝑔𝑓(𝑥). Then, find the value of 𝑔𝑓 (2). Solution: 𝑔𝑓 (𝑥) = 𝑔[𝑓 (𝑥)] = 𝑔(𝑥 + 2) = (𝑥 + 2)2 = 𝑥 2 + 4𝑥 + 4

𝑓(𝑥) = 𝑥 + 2

𝑔𝑓 (2) = 22 + 4(2) + 4 = 16

EXAMPLE 7 Two functions f and g are defined by f (x) = x − 2 and g (x) = 𝑥 2 + 2. Find the values of x when gf (x) = 6. Solution: 𝑔𝑓 (𝑥) = 𝑔[𝑓 (𝑥)] = 𝑔(𝑥 − 2) = (𝑥 − 2)2 + 2 = (𝑥 2 − 4𝑥 + 4) + 2 = 𝑥 2 − 4𝑥 + 6

𝑔𝑓(𝑥) = 6 𝑥 − 4𝑥 + 6 = 6 𝑥 2 − 4𝑥 = 0 (𝑥)(𝑥 − 4) = 0 𝑥 = 0, 𝑥 = 4 2

Tip: The word “values” indicates that there will be at least 2 values of x

8

CHAPTER 1 FUNCTIONS

1.2.4 How to determine a function when the composite function and one of the functions are given? EXAMPLE 8 (a) A function f is defined as 𝑓 ∶ 𝑥 → 𝑥 + 2. Find the function g such that 𝑓𝑔 ∶ 𝑥 → 3𝑥 + 4. (b) A function f is defined as 𝑓 ∶ 𝑥 → 𝑥 + 2. Find the function g such that 𝑔𝑓 ∶ 𝑥 → 𝑥 + 5. Solution: (a) (Case where the function determined is situated “inside” the composite function) 𝑓𝑔 ∶ 𝑥 → 3𝑥 + 4 𝑓𝑔 (𝑥) = 3𝑥 + 4 𝑓𝑔 (𝑥) = 𝑔(𝑥) + 2 3𝑥 + 4 = 𝑔(𝑥) + 2 𝑔(𝑥) = 3𝑥 + 2

Note: For a function, for example f : x → 2x + 2, if x = 1, then f (1) = 2(1) + 2 = 4; if x = 2, then f (2) = 2(2) + 2 = 6; if x = u, then f (u) = 2(u) + 2 = 2u + 2.

(b) (Case where the function determined is situated “outside” the composite function) 𝑔𝑓 (𝑥) = 𝑥 + 5 𝑔(𝑥 + 2) = 𝑥 + 5 𝑔(𝑢) = (𝑢 − 2) + 5 𝑔(𝑢) = 𝑢 + 3 𝑔(𝑥) = 𝑥 + 3 How to change 𝒈(𝒙 + 𝟐) to 𝒈(𝒙)? 𝐿𝑒𝑡 𝑥 + 2 = 𝑢 𝑥 = 𝑢−2

9

CHAPTER 1 FUNCTIONS

1.2.5 How to determine the domain of a composite function? If ƒ and g are functions, the composite function (“ƒ composed with g”) is defined by ƒ(g(x)). The domain of this composite function consists of the numbers x in the domain of g for which g(x) lies in the domain of ƒ. Similarly, if ƒ and g are functions, the composite function (“g composed with f”) is defined by g(f(x)). The domain of this composite function consists of the numbers x in the domain of f for which f(x) lies in the domain of g.

EXAMPLE 9 Two functions f and g are defined by 𝑓(𝑥) = √𝑥 and 𝑔(𝑥) = 𝑥 − 2. Find 𝑔𝑓(𝑥) and its domain.

Solution: 𝑓𝑔(𝑥) = 𝑓(𝑥 − 2) = √𝑥 − 2

Notice that 𝑔(𝑥) = 𝑥 − 2 is defined for all real x but 𝑔(𝑥) belongs to the domain of ƒ only if x ≥ 0. The domain of the composite function 𝑔𝑓 (𝑥) = √𝑥 − 2 is 𝑥 ≥ 2 . 𝑥−2≥0

1.2.6 Further examples and applications on composite functions EXAMPLE 10 1

A function f is defined as 𝑓: 𝑥 → 𝑥 , 𝑥 ≠ 0. Find 𝑓 2 , 𝑓 4 , 𝑓 5 and 𝑓 20 .

Solution: 1 𝑓 (𝑥 ) = 𝑥

𝑓 2 (𝑥) = 𝑓𝑓 (𝑥) 1 = 𝑓( ) 𝑥 1 = 1 𝑥 =𝑥

𝑓 4 (𝑥) = (𝑓 2 )2 (𝑥) = 𝑓 2 [𝑓 2 (𝑥)] 𝑓 2 (𝑥) = 𝑥 = 𝑓 2 (𝑥) =𝑥 5( ) 𝑓 𝑥 = 𝑓[𝑓 4 (𝑥)] = 𝑓 (𝑥) 1 = 𝑥 𝑓 20 (𝑥) = 𝑓 4 𝑓 4 𝑓 4 𝑓 4 𝑓 4 (𝑥) =𝑥

10

CHAPTER 1 FUNCTIONS

EXAMPLE 11 1

In a grocery store, the discounted price, q of a product A is denoted as 𝑞(𝑥) = 2 𝑥, where x is the original price of the product. Upon checking out, the product will be charged service tax, p and is denoted by 𝑝(𝑞) = 1.2𝑞. How much does product A costs if the original price of the product is RM20? Solution:

pq The question is to find 𝑝𝑞(20). 1 𝑝𝑞(𝑥) = 𝑝 ( 𝑥) 2 1 = 1.2 ( 𝑥) 2 = 0.6𝑥

𝑝𝑞(20) = 0.6(20) = 12 Therefore product A costs RM12.

11

CHAPTER 1 FUNCTIONS

1.3 Inverse Functions 1.3.1 What are inverse functions? Let’s take an example of ice and water. f Ice

Water 𝑓 −1

Water undergoes freezing process to become ice. Freezing process is denoted by the function f. Ice undergoes melting process to become water again. Melting process is the inverse of freezing process and is denoted by the inverse function 𝑓 −1 .

1.3.2 What are the characteristics of an inverse function?

1) A function f that maps set X to set Y has an inverse function, 𝑓 −1 if f is a one-to-one function. (Inverse function, 𝑓 −1 is valid when each element in set Y is mapped onto one and only one element in set X and the function f is valid) 2) The inverse of an inverse function gives back the original function. Example 𝒇𝒇−𝟏 (𝒙) = 𝒙 and 𝒇−𝟏 𝒇(𝒙) = 𝒙 3) If two functions f and g are inverses of each other: a) The domain of f = the range of g. b) The domain of g = the range of f. c) The graph of function f is the reflection of graph of function g across the line y = x. 4) If (a,b) lies on the graph f , where a and b are real numbers, then (b,a) lies on the graph 𝒇−𝟏 since graph f is the reflection of graph 𝑓 −1 at line y = x. Note: 𝒇−𝟏 (𝒙) ≠

𝟏 𝒇(𝒙)

1.3.3 How to determine whether the function is one-to-one? A function ƒ(x) is one-to-one on a domain D if 𝑓(𝑥1 ) ≠ 𝑓(𝑥2 ) whenever 𝑥1 ≠ 𝑥2 in D. For example, 𝑓 (𝑥) = 𝑥 3 is one-to-one on any values of x in the domain D because 𝑥1 3 ≠ 𝑥2 3 whenever 𝑥1 ≠ 𝑥2 in D. 𝑓(𝑥) 𝑓(𝑥2 ) 𝑥1

𝑓(𝑥1 )

𝑓 (𝑥) = 𝑥 3 𝑥2

x

12

CHAPTER 1 FUNCTIONS

1.3.4 How to determine whether the graph of function has an inverse function? We use horizontal line test to test whether the graph of function has an inverse function. y

y

𝑦 = 𝑥2

𝑦 = 𝑥+2

x

x 𝑦 = 𝑥 2 is a function. However, it does not have an inverse function since horizontal line cuts the graph of function at least 2 points.

𝑦 = 𝑥 + 2 is a function. It has an inverse function since horizontal line cuts the graph of function at only 1 point.

EXAMPLE 12 Given that 𝑓 (3) = 5 and 𝑓 (1) = 3. Determine 𝑓 −1 (5) and 𝑓 −1 (3).

Solution: 𝑓 (3) = 5 therefore 𝑓 −1 (5) = 3 𝑓 (1) = 3 therefore 𝑓 −1 (3) = 1

EXAMPLE 13 Given that 𝑓: 𝑥 →

Solution: 𝐿𝑒𝑡 𝑎 = 𝑓 −1 (3) 𝑓(𝑎) = 3

1

𝑥

, 𝑥 ≠ 0. Find 𝑓 −1 (3).

𝑓 (𝑎) =

1 𝑎

𝑓 (𝑥) = 𝑦 𝑓 −1 (𝑦) = 𝑥

13

CHAPTER 1 FUNCTIONS To determine whether two functions f and g are inverses of each other, we need to make sure both the domain for 𝑓𝑔(𝑥) and 𝑔𝑓(𝑥) are the same.

EXAMPLE 14 Determine whether the function 𝑓 (𝑥) =

𝑥−3 3 + 2𝑥 has an inverse of 𝑔(𝑥) = − . 𝑥+2 𝑥−1

Solution: Check the domain of the function 𝑓(𝑥). Function is undefined when denominator = 0 𝑥+2= 0 𝑥 = −2 Therefore the domain of the function 𝑓(𝑥) is 𝑥 < −2 and 𝑥 > −2.

Check the domain of the function 𝑔(𝑥). Function is undefined when denominator = 0 𝑥−1= 0 𝑥=1 Therefore the domain of the function 𝑔(𝑥) is 𝑥 < 1 and 𝑥 > 1.

Check the domain of the function 𝑓𝑔(𝑥). 3 + 2𝑥 Notice that 𝑔(𝑥) = − is defined for all real x except 𝑥 = 1 but 𝑔(𝑥) belongs to the domain of 𝑥−1 ƒ only if 𝑥 ≠ −2. The domain of the composite function 𝑓𝑔(𝑥) = 𝑥 is 𝑥 < −2, −2 < 𝑥 < 1, 𝑥 > 1.

Check the domain of the function 𝑔𝑓(𝑥). 𝑥−3 Notice that 𝑓 (𝑥) = is defined for all real x except 𝑥 = −2 but 𝑓(𝑥) belongs to the domain of 𝑔 𝑥+2 only if 𝑥 ≠ 1. The domain of the composite function 𝑔𝑓 (𝑥) = 𝑥 is 𝑥 < −2, −2 < 𝑥 < 1, 𝑥 > 1.

14

CHAPTER 1 FUNCTIONS

𝑥−3 ) 𝑥+2 𝑥−3 3 + 2 (𝑥 + 2) =− 𝑥−3 𝑥+2−1

𝑔𝑓 (𝑥) = 𝑔 (
...