ࡱ> gifqG 3bjbjَ /]BBBBBBBVVVV8dV2.../Ai$ B[BB..-2B.B.VVBBBBGBB.t`>-ϏVVCHAPTER: INEQUALITIES Contents 1 Basic Concepts and Results Solving Inequalities Involving Factors 2.1 Factorisation 2.2 Completing The Square 3 Solving Inequalities Involving Equal Signs 4 Solving Inequalities Involving Modulus 4.1 Breaking the modulus/Squaring Both Sides 4.2 Substitution 4.3 Graphical Method 5 Miscellaneous Examples ( 1 Basic Concepts and Results Basic Results Recall the following familiar axioms. Let a, b ( ( a > b ( a - b > 0. Clearly, a > b ( b < a a < b ( a - b < 0. a > 0 and b > 0 ( a + b > 0 a > 0 and b > 0 ( ab > 0 Note: The above axioms holds also for ( and ( . Let a, b ( (. Then one and only one of the following relations is true: a > b, a = b, a < b. The above results can be used to prove the following basic properties. Let a, b, c ( (. Then, a > b and b > c ( a > c. a > b ( a + c > b + c c ( 0. a > b and c > 0 ( ac > bc. a > b and c < 0 ( ac < bc. a > b and c > 0 (  eq \f(a,c)  >  eq \f(b,c)  a > b and c < 0 (  eq \f(a,c)  <  eq \f(b,c)  ( 2 Solving Inequalities Involving Factors For inequalities without modulus, the first step is to set one side of the inequality to zero and then factorise or complete the square on the other side. ( 2.1 Fractorisation Let a, b ( (. Then the following is true: 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.  eq \f(a,b) > 0 (b ( 0) ( either a > 0 and b > 0 or a < 0 and b < 0.  eq \f(a,b) < 0 (b ( 0) ( either a > 0 and b < 0 or a < 0 and b > 0. Note: and means it must exist in both cases. or means it may exist in either case. Example 2.1 Find the solution sets of the following inequalities: a) (x -3)(x - 5) > 0 b) x2 - 3x + 2 > 0 [a) x < 3 or x > 5 b) x < 1 or x > 2] Solution Example 2.2 Find the solution set of the inequality  eq \f(x - 1,3x + 4) > 0 , x ( (  eq \f(4,3)  [x < (  eq \f(4,3)  or x > 1] Solution Note: Sometimes, the range of values of x which satisfy the inequality is given using set notation. A set is just a collection of objects. So as in the above example, the solution can also be expressed as {x ( (: x < (  eq \f(4,3)  or x > 1}. This is read as the solution is the set of real numbers, x, such that x is less than (  eq \f(4,3) or x is larger than 1. Example 2.3 For which values of x is (x + 3)(2x - 1)(2 - x) < 0 ? [- 3 < x <  eq \f(1,2) or x > 2 ] Solution Example 2.4 Find the solution set of the inequality  eq \f(x + 1,x - 1) >  eq \f(6,x)  ( x ( 1 and x ( 0). [x < 0 or 1 < x < 2 or x >3] Solution Note: Short-cut to determine signs on the number line if all factors are linear. Step 1: Look at the signs of the highest power of x in the expression. If positive, start with + on the extreme RHS of the number line. If negative, start with ( on the extreme RHS of the number line. Step 2: Put the signs alternating + or ( after the first sign. ( 2.2 Completing The Square There are some quadratic expressions which cannot be factorized. In such cases, we use the method of completing the square. Example 2.5 Find the solution set of the inequality x2 - 4x + 1 > 0. [x < 2 -  eq \r(,3) or x > 2 +  eq \r(,3)  ] Solution Example 2.6 Find the solution set of the inequality 2x2 - 3x - 1 < 0. [  eq \f(1,4) (3 -  eq \r(,17) ) < x <  eq \f(1,4) (3 +  eq \r(,17) ) ] Solution Example 2.7 Solve the inequality x2 3x + 5 < 0 [empty set] Solution Example 2.8 (ACJC 97/1/5) Solve  eq \f(x - 1,2x - 1) <  eq \f(x,4 - 3x)  [ eq \f(1,2) < x <  eq \f(4,3) ] Solution ( 3 Solving Inequalities Involving Equal Signs When solving inequalities with equal signs, the critical values in the answer must be tested with the original expression to check if the values are applicable as a solution. Example 3.1 (JJC 00/1/3) Solve the inequality  eq \f(3x + 2,x - 1) ( 1 [x ( - 3/2 or x > 1] Solution Example 3.2 Solve  eq \f(x - 1,x + 1)  (  eq \f(x,6) . [x < - 1, 2 ( x ( 3] Solution ( 4 Solving Inequalities Involving Modulus Definition: The modulus function on the set of real numbers, denoted by ( (, is defined as (x( = {  eq \a\ac( x if x ( 0, (x if x ( 0)  For example, (6( = 6, (- 8(= 8, (0(= 0 (x( is called the modulus of x or the absolute value of x. E.g. The graph of a modulus function, y = (x(, is as follows:  Let a, b ( (. Then the following is true: a2 > b2 ( (a(>(b(. For example, (3)2 > (2)2 , (-3)2 > (2)2 , (3)2 > (-2)2 , (-3)2 > (-2)2 . (a(< b (b > 0) ( - b < a < b. For example, (a(< 2 ( - 2 < a < 2 (a(> b (b > 0) ( a < -b or a > b. For example, (a(> 3 ( a < -3 or a > 3 There are various methods to solve inequalities with modulus: Break the modulus using (a(< b ( - b < a < b where b > 0 (a(> b ( a < -b or a > b where b > 0 Squaring both sides when both sides are positive for all real values of x Substituting y = (x( Using definition of modulus to separate into cases for x > 0 and x < 0 Graphical method ( 4.1 Breaking the modulus/Squaring Both Sides When using the method of breaking the modulus, you must be careful to combine the solutions in the final answer. For the method of squaring both sides, do not expand, instead use a2 ( b2 = (a + b)(a ( b). Example 4.1 Solve the inequality ( eq \f(x -5,x + 3) ( > 2 [ ( 11 < x < (  eq \f(1,3) , x ( (3] Solution Example 4.2 Solve the inequality |x ( 2| < 2x (x ( 0) [ x >  eq \f(2,3) ] Solution Example 4.3 (SAJC 00/1/1ii) Solve the inequality 3 ( (2x + 5( ( 13 [ -9 ( x ( -4 or - 1 ( x ( 4] Solution ( 4.2 Substitution Example 4.4 (HCJC Prelim 97/1/2) Find the range of values of x for which  eq \f(x2 - 12,x)  > 1. Deduce the range of values of x for which  eq \f(x2 - 12,(x() > 1. [ - 3 < x < 0 or x > 4, x < -4 or x > 4] Solution Example 4.5 (CJC Promo 00/13) Solve the inequality  eq \f(1,x - 1)  -  eq \f(2,x - 2)  < 0. Hence solve  eq \f(1,(x( - 1)  -  eq \f(2,(x( - 2)  < 0. [Ans: - 1< x < 1, x (0, x < -2 or x > 2] Solution Example 4.6 Solve the inequality  eq \f(5 ( 2|x|,4 ( x2)  ( 1. [x = ( 1, x < ( 2 or x > 2] Solution ( 4.3 Using Definition of Modulus to separate into Cases for x > 0 and x < 0 Example 4.7 (AJC 00/1/5) Solve the inequality  eq \f(4,x + 1) < x 2. Hence, or otherwise, solve the inequality  eq \f(4,(x( + 1) < x 2. [x > 3] Solution ( 4.4 Graphical Method In this method, you need to sketch the modulus curve and the other curve on the same diagram. Example 4.8 Sketch the graph of y = |x + 2| and hence solve the inequality |x + 2| > 2x + 1, x ( (. [x < 1] ( STOP. THINK. Can we use the method of squaring both sides for this question? Why? Solution Example 4.9 (SRJC Prelim 97/1/2) Sketch, on a single diagram, the graphs of y = (3x + 2( and y = - x + 1. Hence, solve the inequality (3x + 2(> - x + 1. [x < -  eq \f(3,2)  or x > - eq \f(1,4) ] Solution Example 4.10 (YJC 99/1/14aiii) Sketch y = | x + 4 | and y = | x2 ( 4x |. Hence, solve | x + 4 | < | x2 ( 4x |. [x<  eq \f(5 ( \r(,41),2) or x >  eq \f(5 + \r(,41),2) ] Solution ( 5 Miscellaneous Examples Example 5.1 Solve the inequality 0 <  eq \f(x(2 + x),2x + 1)  < 1 [- 2 < x < -1 or 0 < x < 1] Solution Example 5.2 (SAJC Prelim 97/1/4) Solve the inequality (2x - 3( < x2 . [x < -3 or x > 1] Solution Example 5.3 (VJC Prelim 97/1/3) Express x2 + x + 1 in the form (x + A)2 + B, giving the values of the constants A and B. Deduce that x2 + x + 1 > 0 for all real values of x. Solve the inequality  eq \f(x + 3,2 - x2)  < 1. [x < -  eq \r(,2)  or x >  eq \r(,2) ] Solution Example 5.4 (SRJC 98/1/3) Solve the following: (i)  eq \f(x,x ( 1)  >  eq \f(1,x)  (ii)  eq \f(|x|,|x| ( 1)  >  eq \f(1,|x|)  [(i) x< 0 or x > 1 (ii) x< ( 1 or x > 1] Solution Example 5.5 (NJC 01/1/13 modified) Solve this inequality using graphical method: | 1 ( 2x2 | ( 8x ( 1 > 0. [x < 0, x > 2 +  eq \r(,5) ] Solution Summary (Inequalities) Basic Results The following are axioms or statements which do not require proof. Let a, b ( ( a > b ( a - b > 0. Clearly, a > b ( b < a a < b ( a - b < 0. a > 0 and b > 0 ( a + b > 0 a > 0 and b > 0 ( ab > 0 Note: The above axioms holds also for ( and ( . Theorem Let a, b, c ( (. Then, a > b and b > c ( a > c. a > b ( a + c > b + c c ( 0. a > b and c > 0 ( ac > bc. a > b and c < 0 ( ac < bc. a > b and c > 0 (  eq \f(a,c)  >  eq \f(b,c)  a > b and c < 0 (  eq \f(a,c)  <  eq \f(b,c)  (b > 0) (a(< b ( - b < a < b. For example, (a(< 2 ( - 2 < a < 2 (b > 0) (a(> b ( a < -b or a > b. For example, (a(> 3 ( a < -3 or a > 3 Guidelines for solving inequalities: For inequalities without modulus, the first step is to set one side of the inequality to zero and then factorise or complete the square on the other side. When solving inequalities with equal signs, the critical values in the answer must be evaluated first with the original expression to test if the value is applicable as a solution. Let a, b ( (. Then the following is true: a2 > b2 ( (a(>(b(. For example, (3)2 > (2)2 , (-3)2 > (2)2 , (3)2 > (-2)2 , (-3)2 > (-2)2 . (a(< b (b > 0) ( - b < a < b. For example, (a(< 2 ( - 2 < a < 2 (a(> b (b > 0) ( a < -b or a > b. For example, (a(> 3 ( a < -3 or a > 3 There are various methods to solve inequalities with modulus: Break the modulus using (a(< b ( - b < a < b where b > 0 (a(> b ( a < -b or a > b where b > 0 Squaring both sides when both sides are positive for all real values of x Substituting y = (x( Using definition of modulus to separate into cases for x > 0 and x < 0 Graphical method  Teach a child what to think and you make him a slave to your knowledge. Teach a child how to think and you make all knowledge his slave. ( 3 Solving Inequalities in Two Variables ( 3.1 The Geometrical Method The line ax + by = c divides the xy-plane into two half-planes. One of them consists of the points (x, y) satisfying the inequality ax + by > c whereas the other consists of points satisfying inequality ax + by < c.  y x + 3y > 6 2 x + 3y < 6 x 0 6 x + 3y = 6 If we are given a system of simultaneous inequalities, then Draw the respective straight lines (or curves if the inequalities are non-linear) and find the appropriate region which satisfy their respective inequalities. The intersection of all the appropriate region is the set of all the points which satisfy all the inequalities. Example 3.1 Indicate on a graph the set of points which satisfy the inequalities. x + 3 ( 2y, x + 4 ( 3y, x ( y. Solution By taking the intersection of the three regions shown, we have the common region shown which is the set of points where the three given inequalities are satisfied simultaneously. Example 3.2 Given that x + 2y < 1 and x - y < 3, show geometrically that x <  EQ \f(7,3) . Solution Example 3.3 Given that x + y < 1 and y - 3x < 5, show geometrically that y < 2. Solution ( 3.2 The Analytical Method For this method, we need the following result. Let a, b, c, d ( (. Then a > b and c > d ( a + c > b + d. Example 3.4 Given that x + 2y < 1 and x - y < 3, prove that x < EMBED Unknown Solution Example 3.5 Given that x + y < 1 and y - 3x < 5, prove that y < 2. Solution Example 3.6 Given x < 2y + 1 and y < 3x, prove that y > - eq \f(3,5)  and x > -  eq \f(1,5)  Solution PAGE 15 @ggh !>?st:;=>YZij%&(*+X     C D jUmH jU j j j j j j jx>*>*>*5>*R@gz!3Ifg-H & F hhhh  & F h@gz!3Ifg-H-.Hc'(XŽzrolgdMM            Z[  1  G   x  Mz  &-.Hc'(X 5 w _ ` & F h & F h 5 w _ ` g p q r s t u v w x y z { | } ~        ! " # $ ~{lz{4         7L0   " # + , h k % & g o    / X Y            ! " c d p q  5Aij j j j-H*>*65 j j jUX g p q r s t u v w x y z { | } ~         ! 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