JC and Polytechnic Mathematics Material Compilation - Pure Mathematics

Title Junior College (H2) and Polytechnic Mathematics Material Compilation – Pure Mathematics Author AprilDolphin Date 16/2/2025 Topic Page Graphs and Functions – Conic Sections 2 Inequalities – Involving Quadratic, Cubic and Rational Functions 5 Inequalities – Involving Modulus Functions 11 Complex Numbers – Basic Operations 17 Complex Numbers – Argument & Modulus 21 Calculus – Differentiation – Involving Inverse Trigonometric Functions 27 Calculus – Differentiation – Implicit Differentiation 29 Calculus – Differentiation – Parametric Differentiation 33 Calculus – Integration – Integration by Substitution 36 Calculus – Integration – Integration by Parts 39 Calculus – Differentiation Equations – Basic Concepts & Separation of Variables 42

Title Graphs and Functions – Conic Sections – Hyperbola and Ellipse Author AprilDolphin Date 21/12/2024 Note This article assume you already understand the properties of circle and parabola as taught in secondary school Additional Mathematics and Elementary Mathematics respectively. In general, all conic sections can be represented using the equation below, 𝐴𝑥2 + 𝐵𝑥 + 𝐶𝑦2 + 𝐷𝑦 + 𝐸 = 0 Different types of conic sections appears when we begin changing the value of 𝐴, 𝐵, 𝐶, 𝐷 and 𝐸 If the conic section is an Ellipse, it has the following equation required in general: (Make sure the right-hand side is exactly 1 before proceeding!) (𝑥 − ℎ)2 𝑎2 + (𝑦 − 𝑘)2 𝑏2 = 1 For an ellipse to look vertically stretched, 𝑎 < 𝑏 For an ellipse to look horizontally stretched, 𝑎 > 𝑏

For an ellipse to have an exactly circular shape, 𝑎 = 𝑏, where 𝑎 represents the radius of a circle. (Circle is a special case of an ellipse.) Properties of an ellipse. The point where the centre lies is 𝐶(ℎ, 𝑘) The coordinates of vertices from up, down, left and right can be represented as the following Upper Vertex (Top of Ellipse) 𝑈(ℎ, 𝑘 + 𝑏) Lower Vertex (Bottom of Ellipse) 𝐷(ℎ, 𝑘 − 𝑏) Left Vertex (Left side of Ellipse) 𝐿(ℎ − 𝑎, 𝑘) Right Vertex (Right side of Ellipse) 𝑅(ℎ + 𝑎, 𝑘) The lines of symmetry of an ellipse are represented by 𝑥 = ℎ, 𝑦 = 𝑘 If the conic section is a Hyperbola, it has the following equation in general: (Make sure the right-hand side is exactly 1 before proceeding!) Hyperbola can be left and right opening as seen with the below diagram Properties of left-and-right-opening hyperbola as follows: Can be reduced to the following equation, (𝑥−ℎ)2 𝑎2 − (𝑦−𝑘)2 𝑏2 = 1, where RHS is exactly 1

Hyperbola can also be up-and-down-opening as seen with the below diagram Properties of up-and-down-opening hyperbola as follows: Can be reduced to the following equation, (𝑦−𝑘)2 𝑏2 − (𝑥−ℎ)2 𝑎2 = 1, where RHS is exactly 1 Further Properties of Hyperbola Left-and-Right-Opening Up-and-Down-Opening Centre 𝐶(ℎ, 𝑘) 𝐶(ℎ, 𝑘) Vertices Left Vertex, 𝐿(ℎ − 𝑎, 𝑘) Right Vertex, 𝑅(ℎ + 𝑎, 𝑘) Upper Vertex, 𝑈(ℎ, 𝑘 + 𝑏) Bottom Vertex,𝐵(ℎ, 𝑘 − 𝑏) Line of symmetry 𝑥 = ℎ 𝑦 = 𝑘 𝑥 = ℎ 𝑦 = 𝑘 Asymptotes 𝑦 = 𝑘 + 𝑏 𝑎 (𝑥 − ℎ) 𝑦 = 𝑘 − 𝑏 𝑎 (𝑥 − ℎ) 𝑦 = 𝑘 + 𝑏 𝑎 (𝑥 − ℎ) 𝑦 = 𝑘 − 𝑏 𝑎 (𝑥 − ℎ)

Title Inequalities Involving Quadratic, Cubic and Rational Functions Author AprilDolphin Date 3/2/2025 Flashback from Secondary School Elementary Mathematics Description of Rules Demonstration Addition & Subtraction Rule: The act of adding any numbers to both sides of any inequalities doesn’t affect the sign. 𝑥 > 8 𝑥 + 4 > 8 + 4 𝑥 ≤ 9 𝑥 + 1 ≤ 9 + 1 Multiplication & Division of Positive Numbers Rule: The act of multiplying or dividing of positive numbers to both sides of any inequalities doesn’t affect the sign. 5𝑥 + 8 ≥ 9 5𝑥 5 ≥ 1 5 𝑥 ≥ 1 5 Multiplication & Division of Negative Numbers Rule: The act of multiplying and dividing of negative numbers to both sides of any inequalities will flip the inequality sign (i.e. ≤ changes to ≥ and ≥ changes to ≤) 2𝑥 > 6 After dividing both sides by −2 , we get −𝑥 < −3 𝑥 3 ≥ 5 After multiplying both sides by −3, we get −𝑥 ≤ −5

Common Mistakes Made by Inexperienced Students when Solving Inequalities at H2 Level Math. Mistake Number 1. Cross Multiplication 𝑥 − 6 2𝑥 + 3 > 1 𝑥 − 6 > 2𝑥 + 3 Issue: It is incorrect to cross multiply this way as it is currently not known if the sign of 2𝑥 + 3 is positive or negative. Mistake Number 2. Square Rooting Both Sides of Inequality 𝑥2 ≥ 9 𝑥 ≥ ±3 Issue: It is not correct if you treat the inequality sign as if it were an equal sign for the purpose of square rooting both sides in this way. Correct Steps to Deal with Inequalities at H2 Mathematics as follows (The steps are to be followed with discretion, as not all steps are required for all question types.) 1. Moving all the terms of inequalities from RHS to LHS, making sure the RHS of the inequality is 0 before proceeding to step 2. 2. Add up everything on LHS (If LHS now has multiple algebraic fractions, add them up into a state where the LHS shares a common denominator.) 3. Factorize the LHS [If LHS cannot be factored, apply quadratic formula to any quadratic portions of the inequality.] 4. If the quadratic portion has no real roots (such as 𝑥2 + 𝑥 + 1), complete the square as such portion will be either always positive or always negative. 5. For each linear factor, find the critical value that satisfy the inequality 6. Mark all critical values on the number line, 𝑛 critical values divide the number line into 𝑛 + 1 interval 7. Do a sign test on each interval to determine whether the function is positive or negative in that interval. 8. Complete the sign diagram 9. Determine the solution set, noting that any values that will cause the denominator to be zero must be excluded as division by zero is undefined.

Q1: Solve the inequality 𝑥3 ≥ 𝑥(3𝑥 + 10). Step 1. Move terms from RHS to LHS of the inequality. 𝑥3 − 𝑥(3𝑥 + 10) ≥ 0 Step 2. Add up everything on LHS of inequality 𝑥3 − 3𝑥2 − 10𝑥 ≥ 0 Step 3. Take out 𝑥 by factoring. 𝑥(𝑥2 − 3𝑥 − 10) ≥ 0 Step 4. Further factorize to get the following. 𝑥(𝑥 + 2)(𝑥 − 5) ≥ 0 Note: to get sign test result on the bottom right of this page, substitute 𝑥 on LHS with sign test value. Example if your sign test interval is less than −2, find a value that is less than −2, in this case, I use −3 and substitute 𝑥 with −3 to get (−3)[(−3) + 2][−3 − 5] = −24 Since the inequality in step 4 indicates the inequality has to be ≥ 0, we look out for set of values that will produce positive or zero. Which get us the following as a result. −2 ≤ 𝑥 ≤ 0 OR 𝑥 ≥ 5 Sign Test Result -3 -24 (Negative) -1 6 (Positive) 1 -12 Negative 6 48 (Positive) 0 -2 5

Q2: Solve the inequality 𝑥3 < 4𝑥2 − 𝑥 Step 1. Move terms from RHS to LHS 𝑥3 − 4𝑥2 + 𝑥 < 0 Step 2. Factorize the LHS of the inequality 𝑥(𝑥2 − 4𝑥 + 1) < 0 Step 3. Use quadratic formula (which I will not further elaborate here) to deal with quadratic portion (Since it cannot be factorized). 𝑥2 − 4𝑥 + 1 = 0 𝑥 = 2 ± √3 Final Inequality: [𝑥 − (2 + √3)] [𝑥 − (2 − √3)](𝑥) < 0 𝑥 < 0 OR 2 − √3 < 𝑥 < 2 + √3 Sign Test Result -2 -26 (Negative) 0.1 0.061 (Positive 3 -6 (Negative) 4 4 (Positive) 0 2 − √3 2 + √3

Q3. Solve the inequality 3𝑥−5 𝑥−1 ≥ 2 𝑥 Step 1: Move all terms from RHS to LHS of inequality 3𝑥 − 5 𝑥 − 1 − 2 𝑥 ≥ 0 Step 2: Add up everything on the LHS of inequality 𝑥(3𝑥 − 5) 𝑥(𝑥 − 1) − 2(𝑥 − 1) 𝑥(𝑥 − 1) ≥ 0 3𝑥2 − 5𝑥 − 2𝑥 + 2 𝑥(𝑥 − 1) ≥ 0 3𝑥2 − 7𝑥 + 2 𝑥(𝑥 − 1) ≥ 0 Step 3. Factorize any quadratic portion of the LHS of inequality. (3𝑥 − 1)(𝑥 − 2) 𝑥(𝑥 − 1) ≥ 0 Step 4. Complete sign diagram and exclude any value that will cause the denominator to be 0. 𝑥 ≠ 0 𝐴𝑁𝐷 𝑥 ≠ 1 Therefore: 𝑥 < 0, 1 3 ≤ 𝑥 < 1 𝑂𝑅 𝑥 ≥ 2 Sign Test Result -1 6 (Positive) 0.25 -2.333 (Negative) 0.5 3 (Posi