Secondary School Additional Mathematics Notes Compilation-13

Title Secondary School Additional Mathematics Materials Version 13 Author AprilDolphin – Lim Wang Sheng Date 15/5/2026 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 72 Applications of Differentiation – Tangent & Normal Lines 76 Applications of Differentiation – Rates of Change 83 Integration of Algebraic Functions 90 Integration Leading to Logarithmic Functions and Integration of Exponential Functions 92 Integration of Trigonometric Functions 93 Definite Integrals – Area Between Function Curve and 𝑥-axis

Title Surds Manipulation Author - Date 31/12/2022 Basic Surds Rules to Understand before Proceeding Given the following expression can be written in the following form 𝑔√𝑎𝑏2 It can be rewritten as the following 𝑔𝑏√𝑎 Example 1.1 √8 can be decomposed into the following √2 × 4 Since 4 = 22, we can rewrite in the following manner: 2√2 Law of Surds (Only involving square roots) √𝑎 × √𝑏 = √𝑎𝑏 𝑎√𝑏 × 𝑐√𝑑 = 𝑎𝑐(√𝑏𝑑) √𝑎2 = 𝑎 √𝑎 √𝑏 = √𝑎 𝑏 𝑚√𝑎 + 𝑛√𝑎 = √𝑎(𝑚 + 𝑛) 𝑚√𝑎 − 𝑛√𝑎 = √𝑎 (𝑚 − 𝑛) Rules of rationalizing the denominator in surds calculation and manipulation Rule 1 If expression is in the following form 𝑎 𝑔√𝑏 Multiply by the denominator to both numerator and denominator to get the following 𝑎 𝑔√𝑏 × 𝑔√𝑏 𝑔√𝑏

Rule 2. If the expression is in the following form or show some near resemblance to the following form 𝑎 + 𝑔√𝑏 𝑐 − 𝑑√𝑝 Find the conjugate value of the denominator whereby the sign in between 𝑐 − 𝑑√𝑝 is flipped to positive and multiply conjugate value to both numerator and denominator to get the following 𝑎 + 𝑔√𝑏 𝑐 − 𝑑√𝑝 × 𝑐 + 𝑑√𝑝 𝑐 + 𝑑√𝑝 Rule 3. If the expression is in the following form or show some near resemblance to the following form 𝑎 + 𝑔√𝑏 𝑐 + 𝑑√𝑝 Find the conjugate value of the denominator whereby the sign in between 𝑐 + 𝑑√𝑝 is flipped to negative and multiply conjugate value to both numerator and denominator to get the following 𝑎 + 𝑔√𝑏 𝑐 + 𝑑√𝑝 × 𝑐 − 𝑑√𝑝 𝑐 − 𝑑√𝑝 Question 1. 1.1 Simplify the following expression (a) 11√7 + 6√28 − 5√63 (b) (4√3 − √2)(√3 − 5√2)

(a) 11√7 + 6√28 − 5√63 Rewrite as 11√7 + 6√7 × 4 − 5√9 × 7 Once again can be rewritten as 11√7 + 6(2)√7 − 5(3)√7 We then proceed to simplify the expression as (11 + 12 − 15)√7 = 8√7 (b) (4√3 − √2)(√3 − 5√2) Expand the expression into the following 4√3(√3) − √2 (√3) − 5√(2)(4√3) + 5√2(√2) 4(3) − √6 − 20√6 + 5(2) = 12 + 10 − √6 − 20√6 = 22 − 21√6 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 = 4√3 (b) 2−√7 3+4√7 × 3−4√7 3−4√7 = (2−√7)(3−4√7) (3+4√7)(3−4√7) = 6 − 3√7 − 8√7 + 28 9 − 42(7) = 6 + 28 − 3√7 − 8√7 −103 = (34 − 11√7) −103 = −34 + 11√7 103 = 11√7 103 − 34 103 (c) 1 3−√5 × 3+√5 3+√5 = 3+√5 32−5 = 3 + √5 4 = 3 4 + √5 4

2.3[Equations Involving Surds] Step by Step Instructions for solving such equations 1. 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. 2. 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.) 3. Substitute answers back into the original questions and reject values that don’t make sense (i.e. square root of negative values, etc.) 4. 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)√7𝑥 + 5 = 𝑥 + 1 (b) 5√5𝑥 + 9 − 4𝑥 − 3 = 0 (a) (√7𝑥 + 5) 2 = (𝑥 + 1)2 7𝑥 + 5 = 𝑥2 + 2𝑥 + 1 0 = 𝑥2 + 2𝑥 + 1 − 7𝑥 − 5 0 = 𝑥2 − 5𝑥 − 4 After using the quadratic formula, we get the following 𝑥 = 5 2 − √41 2 𝑂𝑅 𝑥 = 5 2 + √41 2 (b) 5√5𝑥 + 9 − 4𝑥 − 3 = 0 Rearrange the equation such that square roots are on one side and non-square- roots are on the other side. 5√5𝑥 + 9 = 4𝑥 + 3 (5√5𝑥 + 9) 2 = (4𝑥 + 3)2 25(5𝑥 + 9) = (4𝑥)2 + 2(3)(4𝑥) + 9 125𝑥 + 225 = 16𝑥2 + 24𝑥 + 9 0 = 16𝑥2 + 24𝑥 − 125𝑥 + 9 − 225 0 = 16𝑥2 − 101𝑥 − 216 𝑥 = − 27 16 𝑂𝑅 𝑥 = 8

After substituting the value 𝑥 = − 27 16 into 5√5𝑥 + 9 − 4𝑥 − 3 = 0 5√5 (− 27 16) + 9 − 4 (− 27 16) − 3 = 7.5 Literally implies that 7.5 = 0 (Reason for rejecting 𝑥 = − 27 16, substituting the value into the question yield illogical results.) After substituting value 𝑥 = 8 into 5√5𝑥 + 9 − 4𝑥 − 3 = 0 5√5(8) + 9 − 4(8) − 3 = 0 When we worked out the left side of the equation, we get 0 = 0, which is within mathematical logic, therefore, we accept this as the only valid answer and thus 𝑥 = 8 2.4 [Equality of Surds] (a) Given that 𝑎 + 𝑏√2 = (3 − √2) 2 − 8 1−√2 Find the value for 𝑎 and 𝑏 (3 − √2) 2 − 8 1 − √2 × 1 + √2 1 + √2 = 9 − 2(3)(√2) + 2 − [8(1 + √2)] (1 − √2)(1 + √2) = 9 − 6√2 + 2 − [8(1 + √2)] (1 − √2)(1 + √2) = 11 − 6√2 − 8+8√2 12−2 = 11 − 6√2 + 8 + 8√2 = 19 + 2√2 19 + 2√2 = 𝑎 + 𝑏√2 Therefore 𝑏 = 2 and 𝑎 = 19

Title Quadratic Functions and Quadratic Equation Date 3/1/2023 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 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 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 1. Factor the value 𝑎 out of the equation Given an equation 2𝑥2 − 10𝑥 − 18 = 0 2(𝑥2 − 5𝑥) − 18 = 0 2. Add ( 𝑏 2𝑎) 2 into the bracket and deduct ( 𝑏2 4𝑎) from outside the bracket 2 [𝑥2 − 5𝑥 + ( 10 2(2)) 2 ] − 18 − 102 4(2) 2 (𝑥 − 5 2) 2 − 61 2 3. In this case, x-coordinate value of the minimum point is − 𝑏 2𝑎 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 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has the values has of 𝐷 = 𝑏2 − 4𝑎𝑐 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 3𝑥2 + 6𝑥 + 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 𝑏2 − 4𝑎𝑐 < 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. 𝑏2 − 4𝑎𝑐 = 62 − 4(3)(𝑘) 62 − 4(3)(𝑘) 36 − 12𝑘 < 0 −12𝑘 < −36 𝑘 > 3 Question 2. The expression (𝑘 + 3)𝑥2 + 6𝑥 + 𝑘 = 5 has two distinct solutions for 𝑥 (a) Show that 𝑘 satisfy 𝑘2 − 2𝑘 − 24 < 0 (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, 𝑏2 − 4𝑎𝑐 > 0.] 2(a) (𝑘 + 3)𝑥2 + 6𝑥 + (𝑘 − 5) = 0 62 − 4(𝑘 + 3)(𝑘 − 5) > 0 36 − 4(𝑘2 + 3𝑘 − 5𝑘 − 15) > 0 36 − 4𝑘2 − 12𝑘