TMJC H2 Chapter 4 Equations & Inequalities Extra Practice Solution 2024

Chapter 4 Equations & Inequalities TMJC 2024 Page 1 of 6 H2 Mathematics (9758) Chapter 4 Equations and Inequalities Extra Practice Solutions 1 2009/VJC Prelim/I/2 Let the price (per kg) for crab, lobster and bamboo clam for his first visit be c, l, b. 3.20c + 1.50l + 7b = 277.50 5.60c + 1.20 (1.1l) + 6.50b = 347 4.50c + 2 (1.12 l) + 6.50b = 395.18 From GC, c = 36.20, l = 79.9983, b = 5.95 Required price: $36.20, $79.9983 x 1.12 = $96.80 and $5.95 respectively. 2 2010/SAJC Prelim/I/1 Let 32 nu an bn cn d= + + + 1 63u = : ( ) ( ) 23 1 1 (1) 63 63 a b c d a b c d + + + = + + + = 2 116u = : ( ) ( ) 23 2 2 (2) 116 8 4 2 116 a b c d a b c d + + + = + + + = 3 171u = : ( ) ( ) 23 3 3 (3) 171 27 9 3 171 a b c d a b c d + + + = + + + = 4 234u = : ( ) ( ) 23 4 4 (4) 234 64 16 4 234 a b c d a b c d + + + = + + + = Using GC, 1, 5, 61, 6a b c d= =− = = . 32 5 61 6nu n n n = − + + Hence 50u = ( ) ( ) ( ) 32 50 5 50 61 50 6 115556− + + = Qn 3 2010/NJC Prelim/I/1 Let the unit digit be z. Let the tenth digit be y. Let the hundredth digit be x. 15x y z+ + = (100 10 ) (100 10 ) 594 99 99 594x y z z y x x z+ + − + + =  − = 4 5 4 5y z x x y z+ = + − + + = Using GC , 8, 5, 2x y z= = = . Thus the number is 852

Chapter 4 Equations & Inequalities TMJC 2024 Page 2 of 6 4 2018/RI Promo/1 32f ( )x x ax bx c= + + + f ( 1) 24 25 a b c− =  − + = f ( 2) 36 4 2 44a b c− =  − + = f (1) 0 1 abc=  + + =− By GC, 2, 13, 10a b c= =− = 32f ( ) 2 13 10x x x x = + − + 5 2015/I/2 (i) ( )1212 11 1 1 xxy x x x − − ++= = = − +− − − (ii) From the graph, Using GC, 1.73 0.414 or 1.73xx−    . y x O 1 2 1.73 0.414

Chapter 4 Equations & Inequalities TMJC 2024 Page 3 of 6 6 2008/II/1 Modified 23 g( ) f ( ) 0.5 1 e sin 0.53 x xx x x x x − + + −  Alternative graph: From the graph, the solution set is  : 1.96 1.56xx −   . x y O x y O y = 0.5

Chapter 4 Equations & Inequalities TMJC 2024 Page 4 of 6 7 2007/H1 A level/I/5(part) Graphical Method 1: Using intersection of 2 graphs 22 3 2 2 3 11 or 2 x x x xx + +  + −  Graphical Method 2: Draw 1 graph ( )( ) 2 2 2 3 2 2 3 2 1 0 1 2 1 0 11 or 2 x x x xx xx xx + +  + + −  + −  −  Algebraic Method: ( )( ) 2 2 2 3 2 2 3 2 1 0 1 2 1 0 x x x xx xx + +  + + −  + −  11 or 2xx −  Let cosx = , 1cos 1 or cos 2 −  For cos 1 − , 180 , 540 =   For 1cos 2  , 0 60 or 300 420      180 or 540 or 0 60 or 300 420 .    =  =        x θ 60° 180° 540° 300° 420° 1 0 −1 x = cos θ (-0.75,0.875) -1.5 (-1, 1) (0.5, 4) -1 0.5 x y -1 0.5 x −1 0.5 + − +

Chapter 4 Equations & Inequalities TMJC 2024 Page 5 of 6 8 2018/SAJC P