Download Additional maths study guide and more Study notes Mathematics in PDF only on Docsity!
Title Secondary School Additional Mathematics Material Version 10
Author AprilDolphin
Date 10/10/
Page
Number
Topic
2 Surds Manipulation
7 Quadratic Functions and Quadratic Equations
10 Polynomial and Partial Fractions Decomposition
26 Exponents and Logarithms
32 Binomial Expansion and Binomial Theorem
37 Coordinate Geometry of Circles
42 Trigonometry
54 Differentiation of Algebraic Functions
65 Differentiation of Exponential & Logarithmic Functions
69 Differentiation of Trigonometric Functions
71 Application of Differentiation – Tangent and Normal Lines
75 Integration of Algebraic Functions
82 Integration Leading to Logarithmic Functions & Integration of Exponential
Functions
84 Integration of Trigonometric Functions
85 Definite Integrals – Integration as Method to find Area under Curve
Title Surds Manipulation
Author -
Date 31/12/
Basic Surds Rules to Understand before Proceeding
Given the following expression can be written in the following form
2
It can be rewritten as the following
Example 1.
√ 8 can be decomposed into the following
√ 2 × 4
Since 4 = 2
2
, we can rewrite in the following manner: 2 √
Law of Surds (Only involving square roots)
𝑎 × √𝑏 = √𝑎𝑏
𝑏 × 𝑐
2
Rules of rationalizing the denominator in surds calculation and manipulation
Rule 1
If expression is in the following form
Multiply by the denominator to both numerator and denominator to get the following
×
(a) 11 √
Rewrite as 11 √
7 × 4 − 5
9 × 7
Once again can be rewritten as 11 √ 7 + 6
We then proceed to simplify the expression as
(b) ( 4 √ 3 − √ 2 )(√ 3 − 5 √ 2 )
Expand the expression into the following
1.2 Rationalize the denominator of the following
(a)
12
√
3
(b)
2 − √
7
3 + 4 √
7
(c)
1
3 −√ 5
Solutions
(a)
12
√ 3
×
√ 3
√ 3
12 √ 3
3
(b)
2 − √
7
3 + 4 √
7
×
3 − 4 √
7
3 − 4 √
7
( 2 −√ 7 )( 3 − 4 √ 7 )
( 3 + 4 √
7 )( 3 − 4 √
7 )
2
(c)
1
3 − √
5
×
3 +√ 5
3 + √
5
3 +√ 5
3
2
− 5
2.3[Equations Involving Surds]
Step by Step Instructions for solving such equations
- Rearrange the equation (if necessary) such that all square-roots are on the
left-hand-side and all non-square-roots are on the right-hand-side.
- Square both sides and then use the usual method you learn during
Elementary Mathematics to solve equation (i.e. Rearrange and then use
Quadratic Formula, etc.)
- Substitute answers back into the original questions and reject values that
don’t make sense (i.e. square root of negative values, etc.)
- Use the remaining values as answers. I typically leave the answers in surd
form for Additional Mathematics paper unless the paper specifies something
else.
(a) √
(b) 5 √
(a) (√ 7 𝑥 + 5 )
2
2
2
2
2
After using the quadratic formula, we get the following
5
2
√ 41
2
5
2
√ 41
2
(b) 5 √ 5 𝑥 + 9 − 4 𝑥 − 3 = 0
Rearrange the equation such that square roots are on one side and non-square-
roots are on the other side.
2
2
2
2
2
2
27
16
Title Quadratic Functions and Quadratic Equation
Date 3/1/
Author -
Finding minimum point or maximum point of a Quadratic Function by completing
the square method
Step 1. Determine if the Quadratic Function in question has a minimum point or a
maximum point, it can be done after equating the quadratic function to zero and
rearranging in the following form
2
If 𝑎 > 0 the quadratic function in question has a minimum point
If 𝑎 < 0 the quadratic function in question has a maximum point
Step 2. Apply completing the square method
- Factor the value 𝑎 out of the equation
Given an equation 2 𝑥
2
2
- Add (
𝑏
2 𝑎
2
into the bracket and deduct (
𝑏
2
4 𝑎
) from outside the bracket
2 [𝑥
2
2
] − 18 −
2
2
- In this case, x-coordinate value of the minimum point is −
𝑏
2 𝑎
which is
5
2
The y coordinate value in this case is −
61
2
Basic Concepts of Discriminant to Understand Before Proceeding
Any quadratic equation being rearranged in the following form 𝑎𝑥
2
the values has of 𝐷 = 𝑏
2
− 4 𝑎𝑐 as the discriminant value.
Condition of Discriminant Consequences and Interpretation
𝐷 > 0 Quadratic Equation has two real and distinct roots.
𝐷 = 0 Quadratic Equation has real and equal roots.
𝐷 ≥ 0 Quadratic Equation has real roots in general.
𝐷 < 0 Quadratic Equation has unreal roots. (No real roots)
Discriminant Manipulation
Question 1.
Find the range of values of 𝑘 for which the expression 3 𝑥
2
- 6 𝑥 + 𝑘 is always positive
for all real values of 𝑥.
[When you see the following keywords: “Always Positive”, “Always Negative”, “No Real
Roots”, it implies the quadratic graph in question will not have contact with the 𝑥 −axis,
thus having no real roots and implying 𝑏
2
− 4 𝑎𝑐 < 0 ]
Knowing such information, we can proceed to find values of 𝑘 that can satisfy the
question demands of having no real roots at all.
2
2
2
Question 2.
The expression
2
- 6 𝑥 + 𝑘 = 5 has two distinct solutions for 𝑥
(a) Show that 𝑘 satisfy 𝑘
2
(b) Find the set of possible value(s) of 𝑘
[When you see the following keywords: “pass through 𝑥-axis at two distinct points”,
“has two distinct solutions” or anything similar, it means the quadratic graph in question
has two real and distinct roots and thus, 𝑏
2
− 4 𝑎𝑐 > 0 .]
2(a)
2
2
2
2
2
2
2
2
2
− 2 𝑘 − 24 < 0 (Shown)
2(b)
2
Title Polynomials and Partial Fractions Decomposition
Author AprilDolphin
Date 24/9/
A polynomial is any form of algebraic expression that can be expressed in the form as
shown below.
𝑛
𝑛
𝑛− 1
𝑛− 1
𝑛− 2
𝑛− 2
2
2
1
0
Where the degree of polynomial is characterized by the highest power of within the
polynomial
In any case, an expression or an algebraic function can only be called a polynomial if
- The expression or function in question only has non-negative integer exponents.
Polynomial Type Example Name of Polynomial Degree of polynomial
0
Constant 0
3 𝑥 + 6 Linear 1
2
3
− 9 𝑥 + 5 Cubic 3
4
3
5
2
Examples of non-polynomials Justification on why it is not a polynomial
2
− 2
The expression contains negative exponents of the
unknown 𝑥 in the third term (𝑥
− 2
12
2
7
The expression contains fractional exponents of
unknown 𝑥 in the second term ( 4 𝑥
2
7 ).
Addition and Subtraction of Polynomials
To add and subtract polynomials together, we just simply combine like terms together
and add them or subtract them in the following manner:
Example 1:
Given 𝑃
3
2
2
− 6 𝑥 + 7 , evaluate 𝑃
and 𝑃
3
2
2
3
2
3
2
2
3
2
Multiplication of Polynomials
To multiply polynomials, we have to do it as in the following example.
Example 2:
Expand and simplify (𝑥 + 1 )(𝑥
2
3
2
2
3
Division of Polynomials by Long Division
Example 3:
Evaluate
3
2
÷
, listing down the dividends and, the
remainder if any.
2
3
2
3
2
2
2
Hence the dividend is 𝑥
2
− 𝑥 + 1 and remainder is 4
Remainder Theorem as Follows:
The remainder theorem states that if a polynomial 𝑃
is divided by 𝑎𝑥 − 𝑏, the
remainder of the division is given as 𝑅 = 𝑃 (
𝑏
𝑎
), where 𝑅 is the remainder. Specifically, if
𝑃(𝑥) is divided by 𝑥 − 𝑏, the remainder of the division is given as 𝑅 = 𝑃(𝑏).
Factor Theorem as Follows:
The factor theorem states that if 𝑎𝑥 − 𝑏 is indeed a factor of polynomial 𝑃
, the
remainder of the division is given as 𝑅 = 𝑃 (
𝑏
𝑎
) = 0. Specifically, if 𝑥 − 𝑏 is a factor of
, the remainder of the division is given as 𝑅 = 𝑃
Let’s see how the remainder theorem above works in practice:
Example 4:
Let 𝑃
3
2
- 11 𝑥 − 7. Find the remainder when 𝑃
is divided by
(a) 𝑥 − 1
(b) 2 𝑥 + 3
For example 4(a), we will substitute 𝑥 = 1 into the polynomial, and get the following
results.
3
2
Hence, remainder 𝑅 = 34
For example 4(b), we will substitute 𝑥 = −
3
2
into the polynomial, and get the following
results
3
2
Hence, remainder 𝑅 =
𝑎 + 2 𝑏 = 8 [Equation 2]
𝑎 = 8 − 2 𝑏 [Equation 2b]
Substituting Equation 2b into Equation 1
Factoring of Cubic Polynomials and Solving of Cubic Equations
Common identities for dealing with cubic polynomial factoring at secondary level
3
3
2
2
3
3
2
2
Example 7.
Factorise the following completely
(a) 𝑥
3
(b) 64 𝑥
3
Example 7(a)
3
2
2
2
Example 7(b)
3
)[(
2
2
]
2
Example 8.
Solve the following equations
(a) 9 𝑥
3
2
(b) 𝑥
3
2
Using FX-97 SGX to help find solutions to cubic equations for Example 8(a)
Action Taken Resulting Screenshot
Press [Menu] [5] [2]
When prompted for polynomial degree,
press 3 (cubic polynomials have a degree
of 3.)
Using 9 𝑥
3
2
− 23 𝑥 + 4 = 0 as an
example, we will have to key in 9 into the
3
field and press [=] button for the
cursor to jump to the next field which is
the 𝑥
2
field.
Press 3 into the 𝑥
2
field and press [=] for
the cursor to jump to the 𝑥 field
After being done with this step, we are going to use long division of polynomial to find
out the quadratic polynomial which is also a factor of 9 𝑥
3
2
Description of Steps Method as seen
Long Division 3 𝑥
2
3
2
3
2
2
2
Since we found out the quadratic factor which is 3 𝑥
2
- 5 𝑥 − 1 , we can use quadratic
formula to solve the quadratic equation:
2
− 5 ±
√ 5
2
− 4
( 3
)( − 1
)
2
( 3
)
Hence the solutions to the cubic equation are 𝑥 =
(− 5 + √
37 )
6
(− 5 − √
37 )
6
OR 𝑥 =
4
3
In the case of Example 8(b), we need to first rearrange the cubic equation such that the
right-hand-side is exactly 0.
3
2
3
2
3
2
3
2
After ensuring the right-hand-side is 0, we proceed to key in the equation into the
calculator
Action taken Resulting screenshot
Press [Menu], [5], [2], [3] in order
Key in 1 into the field labelled as 𝑥
3
and
press [=] button
Key in − 6 into the field labelled as 𝑥
2
and
press [=] button
Key in 11 into the field labelled as 𝑥 and
press [=] button
Key in − 6 into the bottom field and press
[=] button
Hence the solution to the cubic equation is 𝑥 = 1 , 𝑥 = 2 𝑂𝑅 𝑥 = 3
Partial Fractions Decomposition
While we have learnt about how to combine multiple algebraic fractions, what if we
want to reverse the process of combining multiple algebraic fractions together? (i.e. the
act of separating already combined algebraic fraction into their original algebraic
fractions) Such a technique of separation does exist and is known as partial fraction
decomposition.
Applications of Partial Fraction Decomposition
The benefit of learning partial fractions decomposition may not seem immediate but it
serves as a rather strong basis for studying calculus at higher levels Mathematics –
where you may be required to decompose algebraic fraction into partial fractions to
make integrals much easier to evaluate.
Concept of Proper and Improper Algebraic Fractions given
𝑷
( 𝒙
)
𝑸(𝒙)
Naming & Characteristics Examples
Proper Algebraic Fractions has:
Degree of denominator greater than
numerator
3
2
4
2
2
Improper Algebraic Fraction has:
Degree of numerator greater than or
equal to denominator
2
2
3
2
Partial Fraction Common Decomposition
Types at O Level Additional Mathematics
Resulting Decompositions
2
2
2
2
2
2
Example 9:
Express the following in partial fractions
(a)
5 𝑥+ 3
𝑥
( 𝑥+ 1
)
(b)
𝑥+ 6
𝑥(𝑥− 3 )
(c)
7 𝑥− 1
(𝑥− 1 )(𝑥+ 2 )
(d)
𝑥+ 2
( 2 𝑥− 3
)( 𝑥− 4
)
9(a)
Method Used: Substitution Approach
Substituting 𝑥 = 0 would get us the following
Substituting 𝑥 = − 1 would get us the following
[(
] = 5
∴ The partial fraction is decomposed as
2
𝑥+ 1
3
𝑥
9(b)
Substituting 𝑥 = 0 would get us the following
Substituting 𝑥 = 3 would get us the following
∴ The partial fraction is decomposed as
3
𝑥− 3
2
𝑥
