SAJC 2022 Chapter 2 Inequalities Tutor copy Final version

SAJC 2022 JC2 H1 Mathematics Chapter 2: Inequalities Page 1 of 15 (Pure Mathematics) Chapter 2: Inequalities Objectives At the end of the chapter, you should be able to:  solve quadratic inequalities. Content 2.1 Definition of Inequality 2.1.1 Basic Rules of Inequalities 2.2 Solving Quadratic Inequalities 2.3 Solving Simultaneous Inequalities References 1. New Additional Mathematics, Ho Soo Thong (Msc, Dip Ed), Khor Nyak Hiong (Bsc, Dip Ed) 2. New Syllabus Additional Mathematics (7th Edition), Shinglee Publishers Ptd Ltd

SAJC 2022 JC2 H1 Mathematics Chapter 2: Inequalities Page 2 of 15 2.1 Definition of Inequality An inequality is a mathematical statement involving any of ‘<’ , ‘>’ , ‘≤’ and ‘≥’. Examples: 2 6x , 2sin 4cos 3xx , 3 25xx x . 2.1.1 Basic Rules of Inequalities Let ,,ab c . Basic Rules Comments Examples If ab , then (i) a c b c   (ii) a c b c   Addition and subtraction of the same number on both sides of the inequality does not change the inequality sign. Given 36 , then 3 2 6 2   ; 3 1 6 1   . If ab and 0c , then (i) ac bc (ii) ab cc Multiplication and division of the same positive number at both sides of the inequality does not change the inequality sign. Given 35 , then 3 2 5 2   ; 35 44  . If ab and 0c , then (i) ac bc (ii) ab cc Multiplication and division of the same negative number at both sides of the inequality always change the inequality sign. Given 37 , then 3 ( 2) 7 ( 2)     ; 37 33  . If ab and bc , then ac . Inequalities are transitive. If 32 and 2 < 5, then 35 . If a and b are both positive, 0 ab , then 11 ab Taking reciprocal of the two positive numbers in an inequality always changes the inequality sign. If 2 < 5, then 11 25 . If and a b c d , then a c b d   Note: and does not implya b c d a c b d   If 12 and 35 , then 1 3 2 5   . But 12 and 35 ,  1 3 2 5   . ab > 0  either ( a > 0 and b > 0 ) or ( a < 0 and b < 0 ) ab < 0  either ( a > 0 and b < 0 ) or ( a < 0 and b > 0 )

SAJC 2022 JC2 H1 Mathematics Chapter 2: Inequalities Page 3 of 15 Common Mistakes Comments Counter Examples 3 and 4xy 1xy   Do not subtract inequalities in this manner as it may lead to a sign change. 2 3 and 0 4 but 2 30 4    5 31 3 1 5 x x      Do not cross multiply unless the terms are positive, as it may lead to a sign change. Let 2x     > 5 but 3 3 21 5 2 1 15       2 42xx  Do not square both sides as the terms may have different signs, as it may lead to a sign change.   2 2 4 2 but 24     Q: What do x and y represent in the table directly above? A: They represent unknown constants (which might be positive or negative). Q: What does it mean to “solve an inequality”? A: It is to find the range of values of the quantity you are interested in (usually x) which satisfy the given inequality. Example 1 Linear Inequality Solve the inequality 6 3 2x . Solution: 6 3 2 3 2 6 34 4 3 113 x x x x x         

SAJC 2022 JC2 H1 Mathematics Chapter 2: Inequalities Page 4 of 15 0 2.2 Solving Quadratic Inequalities Consider the quadratic inequality   2 3 0xx   . The range of values of x which satisfies the inequality can be found from the quadratic curve   23y x x   as follows: For   2 3 0xx   (i.e. y is negative), we choose the interval for which the curve is below the x-axis.   2 3 0 3 2x x x      For   2 3 0xx   (i.e. y is positive), we choose the interval for which the curve is above the x-axis.   2 3 0 3 or 2x x x x      Note that   2 3 0 2 or 3x x x x      . For   2 3 0xx   (i.e. 0y ), we choose the interval for which the curve is on and below the x-axis.   2 3 0 3 2x x x      For   2 3 0xx   (i.e. 0y ), we choose the interval for which the curve is on and above the x-axis.   2 3 0 3 or 2x x x x      A guide to solving quadratic inequalities: 1. Collect all the terms to one side of the inequality so that the other side is zero. 2. Find the x-intercepts of the graph by factorization or by using the quadratic formula. 3. Sketch the quadratic function and identify the required region. 4. Solve.   23y x x   -3 2 -6 y x

SAJC 2022 JC2 H1 Mathematics Chapter 2: Inequalities Page 5 of 15 Example 2 Find the exact solution set of the following inequalities. (a)   2 19x (b) 0232 2  xx (c) 6)4)(3(  xx (d) xx 212  (e) xx 472  (f) 0522  xx Solution: (a)        2 2 2 19 1 3 0 1 3 1 3 0 ( 4)( 2) 0 x x xx xx             2 or 4xx  The solution set is { : 2 or 4}x x x   or    , 2 4,    Common mistake: Note that ( 4)( 2) 0xx   does NOT mean 4 or 2xx  . (b)      22 3 2 0 2 1 2 0 2 (1 2 ) 0 xx xx xx          1 or 22xx  The solution set is 1{ : or 2} 2x x x   or 1( , ] [2, )2    4 2 x 2 1 2 x

SAJC 2022 JC2 H1 Mathematics Chapter 2: Inequalities Page 6 of 15 (c) 6)4)(3(  xx 2 2 7 12 6 0 7 6 0 xx xx        ( 6)( 1) 0xx   61 x   The solution set is { : 6 1 }xx    or [ 6, 1] (d) xx 212  0122  xx 02)1( 2 x 0)2()1( 22 x 0)21)(21(  xx 1 2 1 2 x    The solution set is { :1 2 1 2}xx     x 1 6 x 12 12 Alternatively, we can solve it this way: Let 2 2 1 0xx   . Then        2 2 2 4 1 1 2(1) 12 x        xx 212  0122  xx    1 2 1 2 0 ( 1 2)( 1 2) 0 xx xx               (Then sketch the curve on the left and write the answer)

SAJC 2022 JC2 H1 Mathematics Chapter 2: Inequalities Page 7 of 15 (e) xx 472  0742  xx 22( 2) 2 7 0x    03)2( 2 x Since   2 20x , 2( 2) 3 3 0x    for all real values of x. Thus, all real values of x will satisfy the inequality given in the question. The solution set is (f) 0522  xx 22( 1) 1 5 0x    04)1( 2 x Since   2 10x , 2( 1) 4 4 0x    for all real values of x. Thus, there is no real value of x that satisfy the inequality given in the question. The solution set is  2.3 Solving Simultaneous Inequalities In inequalities, a x b means xa and xb . “and” in this context has the meaning of “intersection” which is represented by  . The simultaneous inequalities 0 < x < 4 and x < 2 give 0 < x < 2. “or” in inequalities has the meaning of “union” which is represented by  . The simultaneous inequalities 0 < x < 4 or x < 2 give x < 4. 0 2 4 x Alternatively, we can solve it this way: Let 2 4 7 0xx   . Discriminant      2 4 4 1 7 12 0     Since coefficient of 2 0x  and discriminant < 0, therefore 2 4 7 0xx   for all real values of x. Thus, all real values of x will satisfy the inequality given in the question. The solution set is Alternatively, we can solve it this way: Let 2 2 5 0xx   . Discriminant    22 4 1 5 16 0    Since coefficient of 2 0x  and discriminant < 0, therefore 2 2 5 0xx   for all real values of x. Thus, there is no real value of x that satisfy the inequality given in the question. The solution set is 

SAJC 2022 JC2 H1