2024 JPJC Chapter 3 Equation and Inequalies

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2023 Chapter 3 Equations and Inequalities / Pg 1 Chapter 3: Equations and Inequalities Content Outline Include:  formulating an equation, a system of linear equations, or inequalities from a problem situation  solving an equation exactly or approximately using a graphing calculator  solving a system of linear equations using a graphing calculator  solving inequalities of the form f ( ) g( ) x x > 0 where f(x) and g(x) are linear or quadratic expressions that are either factorisable or always positive  concept of |x| and use of relations x a b a b x a b       and or x a b x a b x a b       in the course of solving inequalities  solving inequalities by graphical methods References  Further Pure Mathematics by Brian and Mark Gaulter (Oxford)  My First Step in using TI-84Plus CE for H1 and H2 Math by Dr Spario Soon (Interactive Math Exploration Centre) Prerequisites  Basic rules for manipulating inequalities  Ability to sketch graphs of quadratic, cubic and quartic functions and use the graphing calculator to find the x-intercepts of these graphs  Determine the equations of vertical, horizontal asymptotes and restrictions on the possible values of x and y 1. Properties of Inequalities (Recap from O Level Mathematics) 1.1 Addition & Subtraction (a) Any quantity can be added to or subtracted from both sides of an inequality without changing the inequality sign. If a b , then a c b c   and a – c > b – c For example, If x > 3, then x + 7 > 3 + 7 and x – 8 > 3 – 8 (b) Inequalities of the same kind can be added together . If a > b and c > d , then a + c > b + d For example, If x > 3 and y > 4, then x + y > 3 + 4

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2023 Chapter 3 Equations and Inequalities / Pg 2 (c) NEVER SUBTRACT inequalities. For example, If x > 2 & y > 0, then x y may not be greater than 2 Counter example: For 3x and 4y  , 3 4x y   1 ≯2 1.2 Multiplication & Division (a) Both sides of an inequality can be multiplied or divided by any positive quantity without changing the inequality sign. If c > 0 and a b , then ac > bc and a b c c For example: If x > 3, then x × 2 > 3 × 2 and 3 2 2 x  (b) Both sides of an inequality can be multiplied or divided by any negative quantity but the inequality sign must be reversed. If d < 0 and a > b , then ad < bd and a b d d For example: If x > 3, then 2 2(3)x  and 3 2 2 x   Note: Greater care must be taken when dealing with logarithmic terms. When solving the inequality, ln 5 .(Since )ln 0. ln 0.2 ln 5 ln 0.2 2 0 n n   When solving the inequality, ln 5.(Since )ln 2 ln 2 ln 5 ln 2 0 n n   Recall: ln x > 0 if x > 1 and ln x < 0 if 0 < x < 1 so ln 2 > 0 but ln 0.2 < 0 0 y x lny x 1

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2023 Chapter 3 Equations and Inequalities / Pg 3 1.3 Inequalities involving Modulus Function Recall definition of x , if 0 if 0 x xx x x   Hence 0x  for all real values of x. Then, definition of x a :   if 0 if 0 x a x ax a x a x a        Hence 0x a  for all real values of x. x a b  x a b  0b x a b b x a b a b x a b            For example, 2 5 5 2 5 3 7 x x x          or or x a b x a b x a b x a b x a b             For example, 2 5 2 5 or 2 5 3 or 7 x x x x x           0b No solution for x x   For 0a a x y O For 0a a x y O x y y = x y = – x O Graph of y x Graph of y x a 

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2023 Chapter 3 Equations and Inequalities / Pg 4 2 Solving inequalities of the form ( )( )...... 0x a x b− − > (Including cases involving <,  and ≤) Example 1: (Recap from Additional Mathematics) Solve the inequality ( 5)( 2) 0x x+ − > Method 1: Graphical Sketch Sketch the graph of the LHS of the inequality with the graphic calculator (GC). From the question, we need the LHS of the inequality to be ‘> 0’ (i.e. positive). Indicate with a hollow circle on the sketch to show that 5x and 2 are not required in the answer. Hence, the answer is x < 5 or x > 2. Method 2: Sign Test The above problem could also be solved with a line diagram. Equate each factor to zero to find the critical values of x. Hence, critical values of x are – 5 and 2 To determine the sign in each zone: i) choose a value of x in any one zone ii) substitute the value into the LHS of the inequality to determine the sign (+/– ) iii) the signs will then alternate between each zone unless there are repeated factors. Hence, x < 5 or x > 2. 2 -5 – 5 2 + – + E.g. Choose 10x = − , then ( 5)( 2)x x+ − has Choose 0x = , then ( 5)( 2)x x+ − has Choose 10x = , then ( 5)( 2)x x+ − has

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2023 Chapter 3 Equations and Inequalities / Pg 5 3 Solving inequalities of the form f( ) 0g( ) x x  (including cases involving < , ≥ and ≤) 3.1 Solving inequalities where f( x) and g(x) are linear or are factorisable (By Sign Test) Use of sign test □ Factorise f(x) and g(x) completely into linear factors. □ Equate each factor to zero to find the critical values of x. □ Use the sign test to determine the regions that correspond to the inequality. Note 1: The sign test must be properly presented. Note 2: For quadratic factors which are difficult to factorise, we use completing the square on the factor, or the quadratic formula to find the roots. Example 2: Without using calculator, solve the inequality 2 2 3 01 x x x    . Solution: 2 2 3 0 11 x x xx        1 3 01 x x x    (check signs of zones with a value in each region) Hence, 1 ≤ x < 1 or x ≥ 3. Note: 1x because 2 2 3 1 x x x    is undefined when 1x . Example 3: Solve the inequality 2 3 07 10 x x x    algebraically. Solution: 3 0( 5)( 2) x x x    (Note 5x and 2x ) The answer is 5 2 or 3x x     –5 –2 3 – 1 1 3

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2023 Chapter 3 Equations and Inequalities / Pg 6 3.2 Solving inequalities where f( x) and g(x) are linear or are factorisable (By Graphical Method) Use of graphical method □ Sketch graph using GC. □ Read required range of x. Example 4: Sketch the graph of 2 2 3 1 x xy x    . Hence, solve the inequality 2 2 3 01 x x x    . Solution: 2 2 3 0 11 x x xx      Hence, 1 ≤ x < 1 or x ≥ 3 Note 1: Notice that the vertical asymptote 1x is not shown on the GC. You need to be mindful of the presence of vertical asymptote. Note 2: For inequalities involving  and , we need to be careful on whether x can be equal to the critical values. In the above example, 2 2 3 01 x x x    , 1x as this value of x will result in the denominator being zero. Hence, the answer is 1 ≤ x < 1 or x ≥ 3. x y 0 1x 1y x  1 3 3

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2023 Chapter 3 Equations and Inequalities / Pg 7 Example 5: Solve the inequality 2 3 07 10 x x x    . Solution:    2 3 07 10 3 02 5 x x x x x x       Using GC, OR 5x   2x The answ