66 Solutions Chapter 2 Inequalities in One Unknown Skills Assessment (P.54) 1. 21 + 3x 2 15 ` Exercise 2A (P.62) 1. (a) 3x 2 15 - 21 3x 2 -6 x 2 -2 01x18 (b) -10 G x G -4 (c) Graphical representation: 3 -5 G x 1 4 1 1 (d) -2 3 1 x G - 2 –2 0 2. (a) Graphical representation: 2. 2x - 14 G 16 2x G 16 + 14 2x G 30 x G 15 0 3 5 (b) Graphical representation: Graphical representation: 0 –4 15 0 7 (c) Graphical representation: 3. (a) 1 (b) G (c) H 0 2 6 (d) Graphical representation: (d) 2 0 3 (e) Graphical representation: 0 (f) 1 Graphical representation: 0 02 NSS TM 5A (E)-2P-KJ.indd 66 2010/7/20 4:39:36 PM 67 Chapter 2: Inequalities in One Unknown 3. Solving x - 3 2 1: x 2 4 .......................................... (i) ` Solving x + 2 2 9: ` x 2 7 ......................................... (ii) According to the question, x must satisfy (i) and (ii). Therefore x 2 7. According to the question, x must satisfy (i) and (ii). Therefore x G -3. Graphical representation: –3 0 Graphical representation: 0 7. Rewrite the compound inequality as -5 1 3x - 2 and 3x - 2 1 4 we have -3 1 3x and 3x 1 6 -1 1 x and x12 7 Therefore -1 1 x 1 2. 4. Solving 7x 1 16 + 3x: ` 4x 1 16 x 1 4 ............................................... (i) Solving 9x 2 7x - 10: 2x 2 -10 ` x 2 -5 ........................................... (ii) According to the question, x must satisfy (i) and (ii). Therefore -5 1 x 1 4. Graphical representation: Graphical representation: –1 0 2 8. Rewrite the compound inequality as 1 - 2x 2 7 and 7 2 2x - 7 we have -6 2 2x and 14 2 2x -3 2 x and 7 2 x Therefore x 1 -3. Graphical representation: –5 0 4 –3 5. Solving 3x H 16 - 5x: ` 8x H 16 x H 2 ............................................... (i) Solving 13 + 4x G 17: 4x G 4 ` x G 1 ..................................... (ii) [No values of x can satisfy both (i) and (ii).] ` This compound inequality has no solutions. 6. Solving 2x - 8 H 5x + 1: ` -9 H 3x x G -3 ..................................... (i) Solving 11 - 2x G 21 - 4x: 2x G 10 ` x G 5 ..................................... (ii) 02 NSS TM 5A (E)-2P-KJ.indd 67 0 9. Rewrite the compound inequality as 15 + 2x G 8x - 3 and 8x - 3 G 2x - 9 we have 18 G 6x and 6x G -6 3Gx and x G -1 Therefore the compound inequality has no solutions. 10. Rewrite the compound inequality as x - 2 H 16 - 5x and 16 - 5x H 11 we have 6x H 18 and 5 H 5x xH3 and 1Hx Therefore the compound inequality has no solutions. 2010/7/20 4:39:38 PM 68 Solutions 14 + 3x 2 2 3x 2 -12 x 2 -4 ......................................... (i) 11. 9 1 11 - 2x 2x 1 2 x 1 1 ................................................... (ii) According to the question, x must satisfy (i) and (ii). Therefore -4 1 x 1 1. Graphical representation: 1 2 0 and 15. Solving 3x + 2 1 6 - x: 4x 1 4 x 1 1 ........................................ (i) ` x Solving 2 - 5 G 2 + x: Graphical representation: x -7 G 2 –4 12. 0 ` x H -14 .................................. (ii) According to the question, x must satisfy (i) and (ii). Therefore -14 G x 1 1. 1 2x + 1 H 2 2x H 1 Graphical representation: x H 21 ............................................ (i) and 3 - 3x H 8 - 2x -x H 5 x G -5 ........................................ (ii) [No values of x can satisfy both (i) and (ii).] ` This compound inequality has no solutions. 13. No, they are incorrect. 3x + 4 G -5x and -5x G 5 8x G -4 and x H -1 x G - 21 ` -1 G x G - 21 2x - 1 1 3 2x 1 4 x 1 2 ............................................ (i) 14. and – 14 10x - 7 2 4x + 11 6x 2 18 x 2 3 ............................................ (i) 16. and x G 21 ......................................... (ii) According to the question, x must satisfy (i) and (ii). Therefore x G 21 . x+9 H3 4 x + 9 H 12 x H 3 ............................................. (ii) According to the question, x must satisfy (i) and (ii). Therefore x 2 3. Graphical representation: 4x + 1 G1 3 4x + 1 G 3 4x G 2 01 0 3 17. According to the question, we have 25t 2 3 000 ................................................... (i) 25t 1 5 000 .................................................. (ii) From (i): 25t 2 3 000 t 2 120 From (ii): 25t 1 5 000 t 1 200 ` The possible range of values of t is ( 120 1 t 1 200. 02 NSS TM 5A (E)-2P-KJ.indd 68 2010/7/20 4:39:41 PM Chapter 2: Inequalities in One Unknown 18. Solving Graphical representation: 3 - 4x G 1: 4 3 - 4x G 4 -4x G 1 0 x H - 41 ................................... (i) ` Solving 21. (a) 2 (x + 3) + 2 G 3: 3 we have 3(v - 9) 1 12v 3v - 27 1 12v -9v 1 27 v 2 -3 Therefore -3 1 v G 2. 2(x + 3) G 3 2x G -3 3 x G - 2 ....................... (ii) [No values of x can satisfy both (i) and (ii).] ` This compound inequality has no solutions. 19. Rewrite the compound inequality as and and and and 22. (a) 2x - 4 1 6 2x 1 10 x15 x15 Graphical representation: 0 2 5 and 3 6v G 2 (v + 6) and 12v G 3(v + 6) and 12v G 3v + 18 and 9v G 18 and vG2 (b) The possible values of v are -2, -1, 0, 1 and 2. 1 (6 - 3x) G 31 (2x - 4) and 31 (2x - 4) 1 2 4 3(6 - 3x) G 4(2x - 4) 18 - 9x G 8x - 16 34 G 17x xH2 ` 2Gx15 4 Rewrite the compound inequality as 3 (v - 9) 1 6v 2 2 (x + 3) G1 3 ` 69 3(0.1k - 0.5) G 1.35 0.3k - 1.5 G 1.35 0.3k G 2.85 k G 9.5 ....................... (i) and 0.4(k + 3) G 1.3(2k - 5) 0.4k + 1.2 G 2.6k - 6.5 7.7 G 2.2k k H 3.5 .......................... (ii) According to the question, k must satisfy (i) and (ii). Therefore 3.5 G k G 9.5. (b) From (a), the maximum value of k = 9.5 20. 4(x - 2) H 12 - x 4x - 8 H 12 - x 5x H 20 x H 4 ........................................... (i) and x - 10 G 6(3 - x) x - 10 G 18 - 6x 7x G 28 x G 4 ........................................... (ii) According to the question, x must satisfy (i) and (ii). Therefore x = 4. 02 NSS TM 5A (E)-2P-KJ.indd 69 the minimum value of k = 3.5 23. According to the question, we have * 6x - 30c 2 90c ............................................. (i) 6x - 30c 1 162c .......................................... (ii) From (i): 6x - 30c 2 90c 6x 2 120c x 2 20c From (ii): 6x - 30c 1 162c 6x 1 192c x 1 32c ` The possible range of x is 20c 1 x 1 32c. 2010/7/20 4:39:45 PM 70 Solutions 24. According to the question, we have * 2 (2x + 10 + 25) 1 160 ................................ (i) 25 (2x + 10) H 1 200 ................................... (ii) From (i): 2(2x + 10 + 25) 1 160 2x + 35 1 80 2x 1 45 Exercise 2B (P.70) 1. (a) (b) x G -6 or x H 10 (c) 45 x1 2 From (ii): 25(2x + 10) H 1 200 2x + 10 H 48 2x H 38 x H 19 x 1 -4 or x 2 0 x 1 21 or x H 3 4 1 (d) x G -3 2 or x 2 - 5 2. (a) Graphical representation: 45 ` 19 G x 1 2 a x is an integer. ` The maximum value of x is 22. 0 4 6 (b) Graphical representation: 25. According to the question, we have * 2 (x + 5 + x) H 50 ........................................ (i) –4 2 (x + 5 + x) G 100 ...................................... (ii) From (i): 2(x + 5 + x) H 50 2x + 5 H 25 2x H 20 x H 10 From (ii): 2(x + 5 + x) G 100 2x + 5 G 50 2x G 45 x G 22.5 ` The possible range of values of x is AB + AC 2 BC 0 (c) Graphical representation: –5 0 (d) Graphical representation: 0 10 G x G 22.5. 26. (a) –2 13 (e) Graphical representation: (b) AB + BC 2 AC (c) From (a) and (b), 6 + x 2 17 ............................................ (i) * 6 + 17 2 x ........................................... (ii) From (i): 6 + x 2 17 x 2 11 From (ii): 6 + 17 2 x x 1 23 ` The solutions of the compound –3 (f) 0 2 Graphical representation: –4 –3 –2 –1 0 1 2 3 4 inequality are 11 1 x 1 23. 02 NSS TM 5A (E)-2P-KJ.indd 70 2010/7/20 4:39:48 PM Chapter 2: Inequalities in One Unknown 3. Solving 3x - 7 2 8: ` 71 6. Solving 5 + 3x G 9 - x: 3x 2 15 x 2 5 ........................................ (i) ` 4x G 4 x G 1 ........................................ (i) Solving 5x + 1 1 11: 5x 1 10 ` x 1 2 ....................................... (ii) According to the question, x must satisfy (i) or (ii). Therefore x 2 5 or x 1 2. Solving 5x - 1 H 3x + 13: 2x H 14 ` x H 7 ....................................... (ii) According to the question, x must satisfy (i) or (ii). Therefore x G 1 or x H 7. Graphical representation: Graphical representation: 0 2 4. Solving 5x - 8 2 2: ` 7 7. Solving 2x + 5 H 4x - 3: 5x 2 10 x 2 2 ........................................ (i) Solving 4x 2 -1: ` 0 1 5 x 2 - 41 ......................................... (ii) According to the question, x must satisfy (i) or (ii). Therefore x 2 - 41 . ` 8 H 2x x G 4 ........................................ (i) Solving 6x + 7 G -3x - 2: 9x G -9 ` x G -1 .................................... (ii) According to the question, x must satisfy (i) or (ii). Therefore x G 4. Graphical representation: Graphical representation: 0 –1 4 4 0 8. Solving 8x - (3x - 2) H 7: 5. Solving 3x - 17 1 4: ` 3x 1 21 x 1 7 ...................................... (i) ` 5x + 2 H 7 5x H 5 x H 1 ............................ (i) Solving 14 + 4x 2 -3x: 7x 2 -14 ` x 2 -2 ................................. (ii) Since all values of x can satisfy either (i) or (ii), any real number is a solution of this compound inequality. i.e. x is any real number. Solving 3x - 11 G 13 - 3x: 6x G 24 ` x G 4 ..................................... (ii) Since all values of x can satisfy either (i) or (ii), any real number is a solution of this compound inequality. i.e. x is any real number. Graphical representation: Graphical representation: –4 –3 –2 –1 0 1 2 3 4 02 NSS TM 5A (E)-2P-KJ.indd 71 –4 –3 –2 –1 0 1 2 3 4 2010/7/20 4:39:51 PM 72 Solutions 9. Solving 4(x - 3) 1 1 - x: Solving 21 (6x + 4) H 11: 4x - 12 1 1 - x 5x 1 13 13 x 1 5 .................................. (i) ` Solving 7x H 3(x - 2): 7x H 3x - 6 4x H -6 3 x H - 2 .......................................... (ii) ` Since all values of x can satisfy either (i) or (ii), any real number is a solution of this compound inequality. i.e. x is any real number. 6x + 4 H 22 6x H 18 ` x H 3 ................................ (ii) According to the question, x must satisfy (i) or (ii). Therefore x 2 2. Graphical representation: 0 12. (a) 2 No. Graphical representation: (b) No. 13. Solving 4x - 1 2 –4 –3 –2 –1 0 1 2 3 4 16x - 4 2 x + 1 15x 2 5 4x + 5 10. Solving - 3 H 5: 4x + 5 G -15 4x G -20 x G -5 ................................. (i) ` Solving 7 - (3x + 8) 1 2x: 7 - 3x - 8 1 2x -1 1 5x x 2 - 51 ........................ (ii) ` x+1 : 4 ` x 2 31 Solving 3(x + 1) 2 x + 5: 3x + 3 2 x + 5 2x 2 2 ` x21 Therefore x 2 31 . Graphical representation: According to the question, x must satisfy (i) or (ii). 0 Therefore x G -5 or x 2 - 51 . Graphical representation: 1 3 14. Solving 3(2x + 1) 1 2x + 1: 6x + 3 1 2x + 1 4x 1 -2 –5 – 10 5 1 11. Solving 5 (7x - 4) 2 2: ` 7x - 4 2 10 7x 2 14 x 2 2 ................................. (i) 02 NSS TM 5A (E)-2P-KJ.indd 72 ` Solving x 1 - 21 19 - 3x H 2x - 4: 5 19 - 3x H 10x - 20 39 H 13x ` xG3 Therefore x G 3. 2010/7/20 4:39:55 PM 73 Chapter 2: Inequalities in One Unknown Graphical representation: 17. (a) Solving 10m - 5 + 6 1 16: 7 10m - 5 1 10 7 0 3 15. Solving 2x + 3 2 9 - x: ` 3x 2 6 x 2 2 ........................................ (i) ` x Solving 5 - 21 G Solving 2x - 5 G 5 - 3x 5x G 10 ` x G 2 ..................................... (ii) Since all values of x can satisfy either (i) or (ii), any real number is a solution of this compound inequality. i.e. x is any real number. –4 –3 –2 –1 0 1 2 3 4 x 16. Solving 2 + 2 H ` Graphical representation: 0 3 Solving 2 (x - 3) 1 3 - x: 3(x - 3) 1 2(3 - x) 3x - 9 1 6 - 2x 5x 1 15 ` x 1 3 .................................. (ii) Since all values of x can satisfy either (i) or (ii), any real number is a solution of this compound inequality. i.e. x is any real number. Graphical representation: 8 (b) The maximum value of m is 8. 18. (a) Solving 30 (2k - 1) - 1 G -2: 8 30(2k - 1) - 1 G -16 60k - 31 G -16 60k G 15 x+5 : 3 3x + 12 H 2(x + 5) 3x + 12 H 2x + 10 x H -2 ..................................... (i) 2m + 5 G 3: 7 2m + 5 G 21 2m G 16 ` mG8 Therefore m G 8. 5 - 3x : 10 Graphical representation: 10m - 5 1 70 10m 1 75 m 1 7.5 k G 41 ................. (i) ` k-4 Solving 21 a 8 + 8 k 2 4: k-4 +424 16 k-4 20 16 k-420 k 2 4 .................. (ii) ` According to the question, k must satisfy (i) or (ii). Therefore k G 41 or k 2 4. Graphical representation: –4 –3 –2 –1 0 1 2 3 4 0 1 4 02 NSS TM 5A (E)-2P-KJ.indd 73 4 2010/7/20 4:40:00 PM 74 Solutions (b) The numbers that can satisfy the compound inequality in (a) are: -2, 0, 81 , 41 , 8. From the sketch, y G 0 when -4 G x G 3. ` The required solutions are -4 G x G 3. Graphical representation: 19. According to the question, we have 10 + x 1 12 or 10 + x 2 65 ` x 1 2 or x 2 55 20. (a) –4 According to the question, we have 4 000 + 0.1x 1 6 000 0.1x 1 2 000 x 1 20 000 or 4 000 + 0.1x 2 18 000 0.1x 2 14 000 x 2 140 000 ` The range of sales revenue of those to be interviewed is less than $20 000 or 0 3 2. The corresponding quadratic function is y = (x - 2)(3x + 1). When y = 0, we have (x - 2)(3x + 1) = 0 ` The x-intercepts of the graph of the quadratic function are 2 and - 31 . Since the coefficient of x2 is 3 (2 0), the parabola opens upward. Sketch: greater than $140 000. y (b) Graphical representation of the range of values of x in (a): 0 20 000 y = (x – 2)(3x + 1) 140 000 –1 0 3 x 2 Exercise 2C (P.79) From the sketch, 1. The corresponding quadratic function is y = (x - 3)(x + 4). When y = 0, we have (x - 3)(x + 4) = 0 ` The x-intercepts of the graph of the quadratic function are 3 and -4. Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. Sketch: y 02 NSS TM 5A (E)-2P-KJ.indd 74 ` The required solutions are x 1 - 31 or x 2 2. Graphical representation: –1 0 3 y = (x – 3)(x + 4) –4 y 2 0 when x 1 - 31 or x 2 2. 0 x 3 2 3. The corresponding quadratic function is y = 2x(x - 5). When y = 0, we have 2x(x - 5) = 0 ` The x-intercepts are 0 and 5. 2010/7/20 4:40:02 PM Chapter 2: Inequalities in One Unknown Since the coefficient of x 2 is 2 (2 0), the parabola opens upward. Sketch: Graphical representation: –4 y y = 2x(x – 5) 0 5 From the sketch, y H 0 when x G 0 or x H 5. ` The required solutions are x G 0 or x H 5. y = x2 - 2x - 8. When y = 0, we have x2 - 2x - 8 = 0 i.e. (x - 4)(x + 2) = 0 ` The x-intercepts are 4 and -2. Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. Sketch: y y = x2 – 2x – 8 Graphical representation: 0 4 5. The corresponding quadratic function is x 0 75 5 x 0 –2 4 2 4. The inequality can be written as x - 16 1 0. The corresponding quadratic function is y = x2 - 16. When y = 0, we have x2 - 16 = 0 i.e. (x + 4)(x - 4) = 0 ` The x-intercepts are -4 and 4. Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. Sketch: y 0 Graphical representation: 0 4 6. The corresponding quadratic function is x 4 From the sketch, y 1 0 when -4 1 x 1 4. ` The required solutions are -4 1 x 1 4. 02 NSS TM 5A (E)-2P-KJ.indd 75 x 2 4. –2 y = x2 – 16 –4 From the sketch, y 2 0 when x 1 -2 or x 2 4. ` The required solutions are x 1 -2 or y = 3x2 - 17x + 10. When y = 0, we have 3x2 - 17x + 10 = 0 i.e. (x - 5)(3x - 2) = 0 ` The x-intercepts are 5 and 23 . Since the coefficient of x2 is 3 (2 0), the parabola opens upward. 2010/7/20 4:40:04 PM 76 Solutions Sketch: From the sketch, 7 y 1 0 when 0 1 x 1 8 . y y = 3x2 – 17x + 10 7 ` The required solutions are 0 1 x 1 8 . Graphical representation: 0 x 5 2 3 0 7 8 2 8. The inequality can be written as 4x + 25x H 0. From the sketch, y G 0 when 23 G x G 5. ` The required solutions are 23 G x G 5. Graphical representation: The corresponding quadratic function is y = 4x2 + 25x. When y = 0, we have 4x2 + 25x = 0 i.e. x(4x + 25) = 0 25 ` The x-intercepts are 0 and - 4 . 0 2 Since the coefficient of x 2 is 4 (2 0), the parabola opens upward. Sketch: 5 3 2 7. The inequality can be written as 8x - 7x 1 0. The corresponding quadratic function is y = 8x2 - 7x. When y = 0, we have 8x2 - 7x = 0 i.e. x(8x - 7) = 0 7 ` The x-intercepts are 0 and 8 . Since the coefficient of x 2 is 8 (2 0), the parabola opens upward. Sketch: y 2 y = 4x + 25x 0 – 25 4 x From the sketch, 25 y H 0 when x G - 4 or x H 0. y y = 8x2 – 7x 25 ` The required solutions are x G - 4 or x H 0. 0 x 7 8 Graphical representation: – 25 4 02 NSS TM 5A (E)-2P-KJ.indd 76 0 2010/7/20 4:40:07 PM 77 Chapter 2: Inequalities in One Unknown 9. The corresponding quadratic function is Sketch: 2 y = 7x - x . When y = 0, we have 7x - x2 = 0 i.e. -x(x - 7) = 0 ` The x-intercepts are 0 and 7. Since the coefficient of x2 is -1 (1 0), the parabola opens downward. Sketch: y x 0 –3 2 y = 6 – x – x2 y x 0 7 From the sketch, y 2 0 when -3 1 x 1 2. ` The required solutions are -3 1 x 1 2. Graphical representation: y = 7 x – x2 –3 From the sketch, y G 0 when x G 0 or x H 7. ` The required solutions are x G 0 or x H 7. Graphical representation: 0 7 10. The corresponding quadratic function is y = 6 - x - x2. When y = 0, we have 6 - x - x2 = 0 i.e. -(x + 3)(x - 2) = 0 ` The x-intercepts are -3 and 2. Since the coefficient of x2 is -1 (1 0), the parabola opens downward. 0 2 11. The corresponding quadratic function is y = 4 + 3x - x2. When y = 0, we have 4 + 3x - x2 = 0 i.e. -(x + 1)(x - 4) = 0 ` The x-intercepts are -1 and 4. Since the coefficient of x2 is -1 (1 0), the parabola opens downward. Sketch: y y = 4 + 3x – x2 –1 0 x 4 From the sketch, y H 0 when -1 G x G 4. ` The required solutions are -1 G x G 4. 02 NSS TM 5A (E)-2P-KJ.indd 77 2010/7/20 4:40:08 PM 78 Solutions Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. Sketch: Graphical representation: –1 0 4 y 2 12. The corresponding quadratic function is y = x + 7x + 15 y = 2 - 3x - 2x2. When y = 0, we have 2 - 3x - 2x2 = 0 i.e. -(x + 2)(2x - 1) = 0 0 ` The x-intercepts are -2 and 21 . Since the coefficient of x2 is -2 (1 0), the parabola opens downward. Sketch: a y 2 0 for all values of x. ` x can be any real number. 14. The corresponding quadratic function is y y = 2 – 3x – 2x2 –2 x x 0 1 2 From the sketch, y = 3x2 - 8x + 6. When y = 0, we have 3x2 - 8x + 6 = 0 Since the discriminant of 3x2 - 8x + 6 = 0 is (-8)2 - 4(3)(6) = -8 1 0, y = 3x2 - 8x + 6 has no x-intercepts. Since the coefficient of x2 is 3 (2 0), the parabola opens upward. Sketch: y y 1 0 when x 1 -2 or x 2 21 . y = 3x2 – 8x + 6 ` The required solutions are x 1 -2 or x 2 21 . Graphical representation: 0 –2 0 1 2 x a y G 0 is not true for all values of x. ` The inequality has no solutions. 13. The corresponding quadratic function is y = x2 + 7x + 15. When y = 0, we have x2 + 7x + 15 = 0 Since the discriminant of x2 + 7x + 15 = 0 is 72 - 4(1)(15) = -11 1 0, y = x2 + 7x + 15 has no x-intercepts. 02 NSS TM 5A (E)-2P-KJ.indd 78 2010/7/20 4:40:09 PM Chapter 2: Inequalities in One Unknown 15. 5x 1 2(2x - 1) + x2 5x 1 4x - 2 + x2 0 1 x2 - x - 2 2 x -x-220 The corresponding quadratic function is y = x2 - x - 2. When y = 0, we have x2 - x - 2 = 0 i.e. (x + 1)(x - 2) = 0 ` The x-intercepts are -1 and 2. Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. Sketch: Sketch: y y = x2 – 4x + 4 2 From the sketch, y = 0 when x = 2. ` The required solution is x = 2. Graphical representation: y = x2 – x – 2 –1 x 0 y 0 x 0 79 2 2 17. From the sketch, y 2 0 when x 1 -1 or x 2 2. ` The required solutions are x 1 -1 or x 2 2. Graphical representation: x(x - 1) 2 (2x - 5)(x + 1) x2 - x 2 2x2 - 3x - 5 0 2 x2 - 2x - 5 2 x - 2x - 5 1 0 The corresponding quadratic function is y = x2 - 2x - 5. When y = 0, we have x2 - 2x - 5 = 0 2 x= –1 16. 0 2 2x(x + 6) H (3x + 2)(x + 2) 2x2 + 12x H 3x2 + 8x + 4 0 H x2 - 4x + 4 2 x - 4x + 4 G 0 The corresponding quadratic function is y = x2 - 4x + 4. When y = 0, we have x2 - 4x + 4 = 0 i.e. (x - 2)2 = 0 ` x = 2 (repeated) Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. 02 NSS TM 5A (E)-2P-KJ.indd 79 = -(-2) ! (-2) - 4 (1)(-5) 2 (1) 2 ! 24 (or 1 ! 2 6) Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. Sketch: y y = x2– 2 x – 5 2 – 24 2 0 x 2 + 24 2 2010/7/20 4:40:11 PM 80 Solutions 19. (a) From the sketch, y 1 0 when 2 + 24 2 - 24 1x1 . 2 2 ` The required solutions are The corresponding quadratic function is y = 2x2 - 9x - 21. When y = 0, we have 2x2 - 9x - 21 = 0 2 2 - 24 2 + 24 1x1 2 2 (or 1 - 6 1x11+ x= = 6 ). Graphical representation: 0 2 – 24 2 - (-9) ! (-9) - 4 (2)(-21) 2 (2) 9 ! 249 4 Since the coefficient of x2 is 2 (2 0), the parabola opens upward. Sketch: y 2 + 24 2 y = 2 x2– 9 x – 21 18. 4x - 5 G (2x + 1)(2x - 1) 4x - 5 G 4x2 - 1 0 G 4x2 - 4x + 4 2 x -x+1H0 The corresponding quadratic function is y = x2 - x + 1. When y = 0, we have x2 - x + 1 = 0 Since the discriminant of x2 - x + 1 = 0 is (-1)2 - 4(1)(1) = -3 1 0, y = x2 - x + 1 has no x-intercepts. Since the coefficient of x 2 is 1 (2 0), the parabola opens upward. Sketch: 9 – 249 4 0 x 9 + 249 4 From the sketch, y 2 0 when x 1 x2 9 - 249 or 4 9 + 249 . 4 ` The required solutions are x1 9 - 249 9 + 249 or x 2 . 4 4 Graphical representation: y y = x2 – x + 1 9 – 249 0 4 9 + 249 4 (b) From the sketch in (a), we have 0 a y 2 0 for all values of x. ` x can be any real number. Graphical representation: x y H 0 when x G xH 9 - 249 or 4 9 + 249 . 4 ` The required solutions are xG 9 - 249 9 + 249 or x H . 4 4 –4 –3 –2 –1 0 1 2 3 4 02 NSS TM 5A (E)-2P-KJ.indd 80 2010/7/20 4:40:16 PM 81 Chapter 2: Inequalities in One Unknown 20. (a) -4x2 G 49 - 28x 4x2 - 28x + 49 H 0 The corresponding quadratic function is y = 4x2 - 28x + 49. When y = 0, we have 4x2 - 28x + 49 = 0 i.e. (2x - 7)2 = 0 Since the coefficient of x2 is -1 (1 0), the parabola opens downward. Sketch: y y = 1 + 3 x – x2 7 ` x = 2 (repeated) Since the coefficient of x2 is 4 (2 0), the parabola opens upward. Sketch: 3 – 13 2 y y = 4x2 – 28x + 49 x 0 3 + 13 2 From the sketch, y H 0 when 3 - 13 3 + 13 GxG . 2 2 ` The required solutions are 3 - 13 3 + 13 GxG . 2 2 x 0 7 2 Graphical representation: a y H 0 for all values of x. ` x can be any real number. 3 – 13 0 2 Graphical representation: 3 + 13 2 (b) From the sketch in (a), we have –4 –3 –2 –1 0 1 2 3 4 (b) 21. (a) y 1 0 when x 1 2 -4x 2 49 - 28x 4x2 - 28x + 49 1 0 The corresponding quadratic function is y = 4x2 - 28x + 49. a y 1 0 is not true for all values of x. ` The inequality has no solutions. The corresponding quadratic function is y = 1 + 3x - x2. When y = 0, we have 1 + 3x - x2 = 0 x2 - 3x - 1 = 0 2 x= = 02 NSS TM 5A (E)-2P-KJ.indd 81 -(-3) ! (-3) - 4 (1)(-1) 2 (1) 3 ! 13 2 3 - 13 3 + 13 or x 2 . 2 2 ` The required solutions are x1 22. (a) 3 - 13 3 + 13 or x 2 . 2 2 x + 7 G (x - 2)(3x + 4) x + 7 G 3x2 - 2x - 8 0 G 3x2 - 3x - 15 2 x -x-5H0 The corresponding quadratic function is y = x2 - x - 5. When y = 0, we have x2 - x - 5 = 0 2 x= = -(-1) ! (-1) - 4 (1)(-5) 2 (1) 1 ! 21 2 2010/7/20 4:40:20 PM 82 Solutions Since the coefficient of x2 is 1 (2 0), the parabola opens upward. Sketch: y y = x2– x – 5 x 0 1 – 21 2 1 + 21 2 2 2. x + 13x + 36 G 0 (x + 4)(x + 9) G 0 ` x + 4 H 0 and x + 9 G 0, i.e. x H -4 and x G -9, this compound inequality has no solutions. or x + 4 G 0 and x + 9 H 0, i.e. x G -4 and x H -9, the solutions of this compound inequality are -9 G x G -4. ` The solutions of the quadratic inequality x2 + 13x + 36 G 0 are -9 G x G -4. Graphical representation: From the sketch, y H 0 when x G –9 1 - 21 1 + 21 or x H . 2 2 ` The required solutions are 1 – 21 2 0 1 + 21 2 (b) From the sketch in (a), we have 1 - 21 1 + 21 y 1 0 when 1x1 . 2 2 ` The required solutions are 1 - 21 1 + 21 1x1 . 2 2 0 3. A table is created below. 1 - 21 1 + 21 xG or x H . 2 2 Graphical representation: –4 3x + 4 x-7 (3x + 4)(x - 7) x 1 - 43 - - + x = - 43 0 - 0 - 43 1 x 1 7 + - - x=7 + 0 0 x27 + + + From the above table, x G - 43 or x H 7 when (3x + 4)(x - 7) H 0. ` The solutions of (3x + 4)(x - 7) H 0 are Exercise 2D (P.90) x G - 43 or x H 7. 1. (x - 2)(x - 6) 2 0 ` x - 2 2 0 and x - 6 2 0, i.e. x 2 2 and x 2 6, the solutions of this compound inequality are x 2 6. or x - 2 1 0 and x - 6 1 0, i.e. x 1 2 and x 1 6, the solutions of this compound inequality are x 1 2. ` The solutions of the quadratic inequality (x - 2)(x - 6) 2 0 are x 1 2 or x 2 6. Graphical representation: –4 0 3 7 Graphical representation: 0 02 NSS TM 5A (E)-2P-KJ.indd 82 2 6 2010/7/20 4:40:25 PM 83 Chapter 2: Inequalities in One Unknown 4. 5x2 + 8x 1 4 5x2 + 8x - 4 1 0 i.e. The inequality can be written as (x + 2)(5x - 2) 1 0. A table is created below. x+2 x 1 -2 - 6. ` The critical points are x = - 17 and x = 3. 5x - 2 (x + 2)(5x - 2) - 20x + 3 G 7x2 0 G 7x2 - 20x - 3 2 7x - 20x - 3 H 0 Factorizing 7x2 - 20x - 3 H 0, we have (7x + 1)(x - 3) H 0 i.e. + interval 1 interval 2 interval 3 –1 3 7 x = -2 0 - 0 -2 1 x 1 25 + - - x = 25 + 0 0 x = - 17 and x = 3. x 2 25 + + + In interval 2: a- 17 1 x 1 3 k 7x2 - 20x - 3 = 0 when x = - 17 or x = 3. i.e. The (partial) solutions of the inequality are Take x = 0, then [7(0) + 1](0 - 3) = -3 1 0. From the above table, -2 1 x 1 25 Therefore the required solutions are x G - 17 when (x + 2)(5x - 2) 1 0. or x H 3. ∴ The solutions of 5x2 + 8x 1 4 are -2 1 x 1 25 . 7. Graphical representation: 6x2 - 5x - 4 1 0 (2x + 1)(3x - 4) 1 0 ` - 21 1 x 1 43 –2 0 2 5 Graphical representation: 2 5. Factorizing 2x - 8x 1 0, we have 2x(x - 4) 1 0 ` The critical points are x = 0 and x = 4. i.e. interval 1 interval 2 interval 3 0 4 2x2 - 8x = 0 when x = 0 or x = 4. In interval 2: (0 1 x 1 4) Take x = 1, then 2(1)(1 - 4) = -6 1 0. Therefore the required solutions are 0 1 x 1 4. – 1 2 4 3 2 8. -2x + 5x + 3 G 0 2x2 - 5x - 3 H 0 (2x + 1)(x - 3) H 0 ` x G - 21 or x H 3 Graphical representation: – 02 NSS TM 5A (E)-2P-KJ.indd 83 0 1 0 2 3 2010/7/20 4:40:29 PM 84 9. Solutions x(x - 2) 2 4(x - 2) x(x - 2) - 4(x - 2) 2 0 (x - 2)(x - 4) 2 0 ` x 1 2 or x 2 4 13. (a) Graphical representation: 0 10. 2 2 (b) For the quadratic equation x + kx - 3k = 4 0 to have no real roots, the discriminant of the equation = k2 - 4(1)(-3k) 1 0. k2 + 12k 1 0 k(k + 12) 1 0 ` -12 1 k 1 0 2 3x G -2(7x + 4) 2 3x + 14x + 8 G 0 (x + 4)(3x + 2) G 0 ` -4 G x G - 23 14. Substituting x = 3 into (x - k)(x - 2k) 1 2, Graphical representation: –4 11. For the quadratic equation x2 + kx - 3k = 0 to have two distinct real roots, the discriminant of the equation = k2 - 4(1)(-3k) 2 0. k2 + 12k 2 0 k(k + 12) 2 0 ` k 1 -12 or k 2 0 we have (3 - k)(3 - 2k) 1 2 9 - 3k - 6k + 2k2 - 2 1 0 2k2 - 9k + 7 1 0 (k - 1)(2k - 7) 1 0 –2 0 3 x2 + 14 H 6x x2 - 6x + 14 H 0 x2 - 6x + 14 = x2 - 6x + 9 + 5 = (x - 3)2 + 5 a For all real values of x, (x - 3)2 + 5 H 0 holds. ` The solutions of x2 + 14 H 6x can be any real number. Graphical representation: 7 ` 11k1 2 15. According to the question, we have x(x - 7) 2 98 x2 - 7x - 98 2 0 (x + 7)(x - 14) 2 0 ` x 1 -7 (rejected) or x 2 14 ` The range of values of the larger number x is x 2 14. 16. (x - 2)(2 - x) H 0 (x - 2)(x - 2) G 0 (x - 2)2 G 0 ` x=2 –4 –3 –2 –1 0 1 2 3 4 12. 4x2 1 4x - 3 4x2 - 4x + 3 1 0 4x2 - 4x + 3 = 4(x2 - x) + 3 Graphical representation: 2 = 4 a x - x + 41 k + 2 0 2 = 4 a x - 21 k + 2 2 a For all real values of x, 4 a x - 21 k + 2 1 0 does not hold. ` 4x2 1 4x - 3 has no solutions. 2 02 NSS TM 5A (E)-2P-KJ.indd 84 2010/7/20 4:40:32 PM 85 Chapter 2: Inequalities in One Unknown 20. Let the smaller number be x, then the larger 3x (x - 3) 1 1 - 2x 2 17. number is x + 1. According to the question, we have x2 + (x + 1)2 G 25 x2 + x2 + 2x + 1 - 25 G 0 2x2 + 2x - 24 G 0 x2 + x - 12 G 0 (x + 4)(x - 3) G 0 ` -4 G x G 3 a x is a positive integer. ` The possible values of the smaller number 3x(x - 3) 1 2 - 4x 2 3x - 9x + 4x - 2 1 0 2 3x - 5x - 2 1 0 (3x + 1)(x - 2) 1 0 ` - 31 1 x 1 2 Graphical representation: –1 0 are 1, 2 and 3. 2 3 21. Let the length of the rectangular field along the 18. (3x + 2)(3x - 4) - (3x - 4)(2x + 3) H 0 (3x - 4)(3x + 2 - 2x - 3) H 0 (3x - 4)(x - 1) H 0 river be x m, then the width of the field is a 100 - x k m, i.e. a 50 - x k m. 2 2 The area of the field must be at least 800 m2. ` x G 1 or x H 43 i.e. Graphical representation: x a 50 - 2 k H 800 x 50x - 21 x2 - 800 H 0 0 19. 1 4 3 (x + 3)(2x - 1) 1 6x2 + 5x + 1 2x2 + 6x - x - 3 1 6x2 + 5x + 1 0 1 4x2 + 4 4x2 + 4 2 0 2 a For all real values of x, 4x + 4 2 0 holds. ` The solutions of (x + 3)(2x - 1) 1 6x2 + 5x + 1 can be any real number. Graphical representation: –4 –3 –2 –1 0 1 2 3 4 02 NSS TM 5A (E)-2P-KJ.indd 85 x2 - 100x + 1 600 G 0 (x - 20)(x - 80) G 0 ` 20 G x G 80 ` The maximum length of the field along the river will be 80 m. 22. (a) Total surface area of the cylinder = [2rr2 + 2rr(r - 2)] cm2 = (4rr2 - 4rr) cm2 (b) According to the question, we have 4rr2 - 4rr 1 48r r2 - r 1 12 r2 - r - 12 1 0 (r + 3)(r - 4) 1 0 ` -3 1 r 1 4 a r 2 0 and r - 2 2 0, i.e. r 2 2 ` 21r14 2010/7/20 4:40:35 PM 86 Solutions 23. (a) According to the question, we have m(36 - m) 2 320 36m - m2 - 320 2 0 m2 - 36m + 320 1 0 (m - 16)(m - 20) 1 0 ` 16 1 m 1 20 ` The possible values of m are 17, 18 (b) Solving 17 - 4x 1 x + 2: 15 1 5x ` x 2 3 ............................. (i) Solving 23x - 3 H 5x: 18x H 3 x H 61 ........................... (ii) ` According to the question, x must satisfy (i) and (ii). Therefore x 2 3. and 19. (b) When m = 17, the number of seats in the hall = 17(36 - 17) = 323 When m = 18, the number of seats in the hall = 18(36 - 18) = 324 When m = 19, the number of seats in the hall = 19(36 - 19) = 323 ` The maximum number of seats in the Graphical representation: 0 2. (a) 3 Rewrite the compound inequality as 2x - 2 G 6 and 6 G 1 - 5x 2x G 8 and 5x G -5 x G 4 and x G -1 ` x G -1 Graphical representation: hall is 324. –1 Supp. Exercise 2 (P.95) 1. (a) Solving 5x H 16 - 3x: 8x H 16 ` x H 2 ....................................... (i) Solving 9x G 35 + 4x: 5x G 35 ` x G 7 ...................................... (ii) According to the question, x must satisfy (i) and (ii). Therefore 2 G x G 7. Graphical representation: 0 02 NSS TM 5A (E)-2P-KJ.indd 86 2 7 0 (b) Rewrite the compound inequality as 25 - 7x 1 3 + x and 3 + x G 12 - 8x 22 1 8x and 9x G 9 x 2 11 4 and xG1 ` The compound inequality has no solutions. 3. (a) 7 + 2x H -3 2x H -10 x H -5 ................................. (i) and 8x - 9 1 7 8x 1 16 x 1 2 ................................... (ii) According to the question, x must satisfy (i) and (ii). Therefore -5 G x 1 2. 2010/7/20 4:40:37 PM Chapter 2: Inequalities in One Unknown Graphical representation: (c) Solving 3x - 4 H 15 - 2x: 5x H 19 19 x H 5 ............................ (i) ` –5 0 2 Solving 2x + 11 1 -1 2x 1 -12 x 1 -6 ................................ (i) (b) and 4x + 3 25 2 4x + 3 2 10 4x 2 7 87 x-5 1 3: 2 x-516 ` x 1 11 .............................. (ii) According to the question, x must satisfy (i) or (ii). Therefore x can be any real number. Graphical representation: 7 x 2 4 ................................. (ii) [No values of x can satisfy both (i) and (ii).] ` This compound inequality has no (d) Solving 2x + 7 2 10 - x: solutions. 4. (a) Solving 3x - 2 1 4: 3x 1 6 ` x 1 2 ............................... (i) Solving -2x 1 -20: 2x 2 20 ` x 2 10 ................................. (ii) According to the question, x must satisfy (i) or (ii). Therefore x 1 2 or x 2 10. Graphical representation: 0 –4 –3 –2 –1 0 1 2 3 4 2 3x 2 3 x 2 1 ............................... (i) ` x Solving 3 - 1 G x: -1 G 23 x ` 3 x H - 2 .......................... (ii) According to the question, x must satisfy (i) or (ii). 3 Therefore x H - 2 . Graphical representation: 10 –3 2 0 (b) Solving 5 + 3x G -7: ` 3x G -12 x G -4 ............................ (i) Solving 13x - 8 G 11x: 2x G 8 ` x G 4 ............................ (ii) According to the question, x must satisfy (i) or (ii). Therefore x G 4. Graphical representation: 0 02 NSS TM 5A (E)-2P-KJ.indd 87 4 2010/7/20 4:40:40 PM 88 Solutions 5. (a) The corresponding quadratic function is y = (x - 2)(x + 8). When y = 0, we have (x - 2)(x + 8) = 0 ` The x-intercepts of the graph of the quadratic function are 2 and -8. Since the coefficient of x2 is 1 (2 0), the parabola opens upward. Sketch: y From the sketch, y 1 0 when -15 1 x 1 0. ` The required solutions are -15 1 x 1 0. Graphical representation: –15 (c) 0 3x2 G 18x 3x - 18x G 0 x2 - 6x G 0 The corresponding quadratic function is y = x2 - 6x. When y = 0, we have x2 - 6x = 0 i.e. x(x - 6) = 0 ` The x-intercepts of the graph of the quadratic function are 0 and 6. Since the coefficient of x2 is 1 (2 0), the parabola opens upward. Sketch: 2 y = (x – 2)(x + 8) x 0 –8 2 From the sketch, y 2 0 when x 1 -8 or x 2 2. ` The required solutions are x 1 -8 or x 2 2. Graphical representation: y y = x2 – 6x –8 0 2 x 0 6 (b) The corresponding quadratic function is y = x(x + 15). When y = 0, we have x(x + 15) = 0 ` The x-intercepts of the graph of the quadratic function are 0 and -15. Since the coefficient of x2 is 1 (2 0), the parabola opens upward. Sketch: From the sketch, y G 0 when 0 G x G 6. ` The required solutions are 0 G x G 6. Graphical representation: y y = x(x + 15) –15 02 NSS TM 5A (E)-2P-KJ.indd 88 0 0 6 x 2010/7/20 4:40:42 PM 89 Chapter 2: Inequalities in One Unknown (d) The corresponding quadratic function is Sketch: 2 y = 2x - 7x - 15. When y = 0, we have 2x2 - 7x - 15 = 0 i.e. (2x + 3)(x - 5) = 0 y y = x2– x – 1 ` The x-intercepts of the graph of the 3 1– 5 2 quadratic function are - 2 and 5. x 0 1+ 5 2 Since the coefficient of x2 is 2 (2 0), the parabola opens upward. Sketch: From the sketch, y y 2 0 when x 1 y = 2x2 – 7x – 15 1- 5 1+ 5 or x 2 . 2 2 ` The required solutions are x 1 –3 2 x 0 or x 2 5 1- 5 2 1+ 5 . 2 Graphical representation: From the sketch, 1– 5 2 3 y H 0 when x G - 2 or x H 5. 3 ` The required solutions are x G - 2 or x H 5. Graphical representation: (b) 0 1+ 5 2 2x2 - 3x G 1 2x2 - 3x - 1 G 0 The corresponding quadratic function is y = 2x2 - 3x - 1. When y = 0, we have 2x2 - 3x - 1 = 0 2 –3 0 5 2 x= = 6. (a) The corresponding quadratic function is y = x2 - x - 1. When y = 0, we have x2 - x - 1 = 0 2 -(-1) ! (-1) - 4 (1)(-1) x= 2 (1) = -(-3) ! (-3) - 4 (2)(-1) 2 (2) 3 ! 17 4 ` The x-intercepts of the graph of the quadratic function are 3 ! 17 . 4 Since the coefficient of x2 is 2 (2 0), the parabola opens upward. 1! 5 2 ` The x-intercepts of the graph of the quadratic function are 1! 5 . 2 Since the coefficient of x2 is 1 (2 0), the parabola opens upward. 02 NSS TM 5A (E)-2P-KJ.indd 89 2010/7/20 4:40:46 PM 90 Solutions a y 2 0 for all values of x. ` x can be any real number. Sketch: y Graphical representation: y = 2 x2– 3 x – 1 3 – 17 4 –4 –3 –2 –1 0 1 2 3 4 x 0 3 + 17 4 (b) The corresponding quadratic function is y = x2 - 6x + 10. When y = 0, we have x2 - 6x + 10 = 0 Since the discriminant of x2 - 6x + 10 = 0 is (-6)2 - 4(1)(10) = -4 1 0, y = x2 - 6x + 10 has no x-intercepts. Since the coefficient of x2 is 1 (2 0), the parabola opens upward. Sketch: From the sketch, y G 0 when 3 - 17 3 + 17 GxG . 4 4 ` The required solutions are 3 - 17 3 + 17 GxG . 4 4 Graphical representation: y y = x2 – 6x + 10 3 – 17 0 3 + 17 4 4 7. (a) The corresponding quadratic function is y = x2 + 4x + 6. When y = 0, we have x2 + 4x + 6 = 0 Since the discriminant of x2 + 4x + 6 = 0 is 42 - 4(1)(6) = -8 1 0, 2 y = x + 4x + 6 has no x-intercepts. Since the coefficient of x2 is 1 (2 0), the parabola opens upward. Sketch: y y = x2 + 4x + 6 0 02 NSS TM 5A (E)-2P-KJ.indd 90 0 x a y G 0 is not true for all values of x. ` The inequality has no solutions. (c) 4x2 + 11 H 12x 4x2 - 12x + 11 H 0 The corresponding quadratic function is y = 4x2 - 12x + 11. When y = 0, we have 4x2 - 12x + 11 = 0 Since the discriminant of 4x2 - 12x + 11 = 0 is (-12)2 - 4(4)(11) = -32 1 0, y = 4x2 - 12x + 11 has no x-intercepts. Since the coefficient of x2 is 4 (2 0), the parabola opens upward. x 2010/7/20 4:40:48 PM Chapter 2: Inequalities in One Unknown 91 Graphical representation: Sketch: y y = 4x2 – 12x + 11 0 (c) 5x2 - 17x + 6 1 0 (x - 3)(5x - 2) 1 0 ` x - 3 2 0 and 5x - 2 1 0, i.e. x 2 3 2 and x 1 5 , this compound inequality x 0 16 has no solutions. a y H 0 for all values of x. ` x can be any real number. or x - 3 1 0 and 5x - 2 2 0, i.e. x 1 3 2 and x 2 5 , the solutions of this Graphical representation: compound inequality are 25 1 x 1 3. ` The solutions of the quadratic inequality –4 –3 –2 –1 0 1 2 3 4 8. (a) (x + 3)(x - 6) G 0 ` x + 3 H 0 and x - 6 G 0, i.e. x H -3 and x G 6, the solutions of this compound inequality are -3 G x G 6. or x + 3 G 0 and x - 6 H 0, i.e. x G -3 and x H 6, this compound inequality has no solutions. ` The solutions of the quadratic inequality (x + 3)(x - 6) G 0 are -3 G x G 6. Graphical representation: 5x2 - 17x + 6 1 0 are 25 1 x 1 3. Graphical representation: 0 2 3 5 (d) 4(3x - 1) G x(3x - 1) 0 G (3x - 1)(x - 4) (3x - 1)(x - 4) H 0 ` 3x - 1 G 0 and x - 4 G 0, i.e. x G 1 and x G 4, the solutions of this 3 compound inequality are x G 31 . or 3x - 1 H 0 and x - 4 H 0, i.e. –3 0 6 2 (b) 2x - 32x 2 0 2 x - 16x 2 0 x(x - 16) 2 0 ` x 1 0 and x - 16 1 0, i.e. x 1 0 and x 1 16, the solutions of this compound inequality are x 1 0. or x 2 0 and x - 16 2 0, i.e. x 2 0 and x 2 16, the solutions of this compound inequality are x 2 16. ` The solutions of the quadratic inequality 2x2 - 32x 2 0 are x 1 0 or x 2 16. 02 NSS TM 5A (E)-2P-KJ.indd 91 x H 31 and x H 4, the solutions of this compound inequality are x H 4. ` The solutions of the quadratic inequality 4(3x - 1) G x(3x - 1) are x G 31 or x H 4. Graphical representation: 01 3 4 2010/7/20 4:40:52 PM 92 Solutions 9. (a) x2 + 8 H 2x x2 - 2x + 8 H 0 (x - 1)2 + 7 H 0 a For all real values of x, (x - 1)2 + 7 H 0 holds. ` x can be any real number. 12. According to the question, we have 15t - 5t2 2 10 -5t + 15t - 10 2 0 t2 - 3t + 2 1 0 (t - 1)(t - 2) 1 0 ` 11t12 ` The range of values of t is 1 1 t 1 2 2 Graphical representation: such that the height of the ball from the ground is more than 10 m. –4 –3 –2 –1 0 1 2 3 4 13. (b) 2 4x 1 4x - 5 2 4x - 4x + 5 1 0 2 4(x - x) + 5 1 0 2x2 + (k - 4)x = 2(k - 1) 2x + (k - 4)x - 2(k - 1) = 0 For the quadratic equation to have two distinct real roots, the discriminant of the equation = (k - 4)2 - 4(2)[-2(k - 1)] 2 0. k2 - 8k + 16 + 16k - 16 2 0 k2 + 8k 2 0 k(k + 8) 2 0 ` k 1 -8 or k 2 0 2 2 4 a x - 21 k + 4 1 0 a For all real values of x, 2 4 a x - 21 k + 4 1 0 does not hold. ` The inequality has no solutions. 10. When the length of the rectangle is x cm, its width is (x - 6) cm and its perimeter is [2(x + x - 6)] cm. ` 60 G 2(x + x - 6) G 80 i.e. 30 G 2x - 6 and 2x - 6 G 40 36 G 2x and 2x G 46 x H 18 and x G 23 ` The possible range of values of x is 18 G x G 23. 14. (a) Solving 3x + 4 H -1 - 2x: 5x H -5 ` x H -1 ............................ (i) Solving 8 - x 1 4x - 2: 10 1 5x ` x 2 2 ................................ (ii) According to the question, x must satisfy (i) and (ii). Therefore x 2 2. Graphical representation: 11. According to the question, we have 9 9 c + 32 1 -37.9 or 5 c + 32 2 674.1 5 9 c 1 -69.9 or 5 9 c 2 642.1 5 c 1 -38.8 or c 2 356.7, cor. to 1 d.p. ` The possible range of the temperature in 0 (b) Rewrite the compound inequality as 3 - 6x G 2x + 1 and 2x + 1 G 9 2 Celsius scale is below -38.8cC or above 356.7cC when mercury is not in liquid state. ` 02 NSS TM 5A (E)-2P-KJ.indd 92 2 1 G 8x 2 and 2x G 8 1 x H 16 and xG4 1 GxG4 16 2010/7/20 4:40:55 PM Chapter 2: Inequalities in One Unknown and 3x + 4 2 x - 2 Graphical representation: 0 1 5x - 9 2 x - 1 4x 2 8 x 2 2 ..................................... (i) (c) 2x 2 -6 x 2 -3 ................................ (ii) According to the question, x must satisfy (i) and (ii). Therefore -3 1 x G 2. 4 16 93 (b) If x is a positive integer, the possible values of x are 1 and 2. 8x and 6x + 2 G 3 - 2 16. (a) 10x G -4 3 Rewrite the compound inequality as x+22 6 x G - 5 .............................. (ii) 4x + 8 2 3x + 2 x 2 -6 ................................ (i) [No values of x can satisfy both (i) and (ii).] ` This compound inequality has no and solutions. x H 6x - 60 60 H 5x x G 12 ........................................... (i) and 3x + 4 G 4(x - 2) 3x + 4 G 4x - 8 12 G x x H 12 ................................. (ii) According to the question, x must satisfy (i) and (ii). Therefore x = 12. Graphical representation: 0 15. (a) 3x + 2 H 2x + 1 4 3x + 2 H 8x + 4 -2 H 5x x H x - 10 6 (d) 12 2x - 1 3x G4- 2 3 2(2x - 1) G 24 - 3(3x) 4x - 2 G 24 - 9x 13x G 26 x G 2 ................................ (i) x G - 25 .............................. (ii) According to the question, x must satisfy (i) and (ii). Therefore -6 1 x G - 25 . (b) The maximum value of x = - 1 The minimum value of x = - 5 17. (a) Solving 2(x + 8) 2 30 - 5x: 2x + 16 2 30 - 5x 7x 2 14 ` x 2 2 ............................ (i) Solving 2x + 5 1 x: 3 2x + 5 1 3x 51x ` x 2 5 .............................. (ii) According to the question, x must satisfy (i) or (ii). Therefore x 2 2. Graphical representation: 0 02 NSS TM 5A (E)-2P-KJ.indd 93 3x + 2 4 2 2010/7/20 4:40:59 PM 94 Solutions (b) Solving 3(3x - 1) 2 4: Graphical representation: 9x - 3 2 4 9x 2 7 ` 7 x 2 9 ......................... (i) 3x Solving 2x - 21 G 2 : –5 0 18. (a) x G 21 2 ` x G 1 ............................. (ii) According to the question, x must satisfy (i) or (ii). Therefore x can be any real number. Solving 3(3x + 2) 1 5(x - 2): 9x + 6 1 5x - 10 4x 1 -16 ` x 1 -4 ...................... (ii) According to the question, x must satisfy (i) or (ii). Therefore x can be any real number except -4. 1 4 (b) 0, - 2 and 5 cannot satisfy the compound inequality in (a). 19. (a) Solving 12 a 81 x - 1k 1 -x - 22: 3 x - 12 1 -x - 22 2 5 x 1 -10 2 x 1 -4 .................... (i) 5 1 Solving 24 x+ 1 2 4: 0 (d) Solving 2(3x - 4) G 2x - 13: 6x - 8 G 2x - 13 4x G -5 ` Therefore x 1 - 45 or x 2 1. ` Graphical representation: –4 x 1 - 45 ...................... (i) Solving 3(x + 2) 2 13 - 4x: 3x + 6 2 13 - 4x 7x 2 7 ` x 2 1 ............................ (ii) According to the question, x must satisfy (i) or (ii). –4 –3 –2 –1 0 1 2 3 4 Solving 4x - (3x - 1) 2 -3: 4x - 3x + 1 2 -3 ` x 2 -4 ................ (i) Solving 2(1 - 2x) 2 6 + x: 2 - 4x 2 6 + x -4 2 5x ` Graphical representation: (c) 4 4 5 x G - 4 ...................... (i) x 6 Solving 6 - a x + 5 k 1 5 : 30 - 5x - 6 1 x 24 1 6x ` x 2 4 .................... (ii) According to the question, x must satisfy (i) or (ii). x + 24 2 30 ` x 2 6 ........................... (ii) According to the question, x must satisfy (i) or (ii). Therefore x 1 -4 or x 2 6. Graphical representation: –4 0 6 (b) From (a), x 1 -4 or x 2 6. ` -4 G t G 6 5 Therefore x G - 4 or x 2 4. 02 NSS TM 5A (E)-2P-KJ.indd 94 2010/7/20 4:41:06 PM 95 Chapter 2: Inequalities in One Unknown 20. (a) Rewrite the compound inequality as 3x - 2 G 5x + 6 and 5x + 6 G 2x + 15 -8 G 2x and 3x G 9 x H -4 and xG3 ` -4 G x G 3 ..................................... (i) or 1-x 20 2 1-x20 12x x 1 1 ........................................ (ii) According to the question, x must satisfy (i) or (ii). Therefore x G 3. (b) 4x H 3 - 4x2 4x2 + 4x - 3 H 0 The corresponding quadratic function is y = 4x2 + 4x - 3. When y = 0, we have 4x2 + 4x - 3 = 0 i.e. (2x - 1)(2x + 3) = 0 ` The x-intercepts of the graph of the 3 quadratic function are 21 and - 2 . Since the coefficient of x2 is 4 (2 0), the parabola opens upward. Sketch: (b) Graphical representation of the solutions y in (a): y = 4x2 + 4x – 3 0 3 –3 x 0 1 2 2 21. (a) The corresponding quadratic function is y = 11x - 24 - x2. When y = 0, we have 11x - 24 - x2 = 0 i.e. -(x - 3)(x - 8) = 0 ` The x-intercepts of the graph of the quadratic function are 3 and 8. Since the coefficient of x2 is -1 (1 0), the parabola opens downward. Sketch: From the sketch, 3 y H 0 when x G - 2 or x H 21 . ` The required solutions are 3 x G - 2 or x H 21 . Graphical representation: y x 0 8 3 –3 2 0 1 2 y = 11x – 24 – x2 From the sketch, y 2 0 when 3 1 x 1 8. ` The required solutions are 3 1 x 1 8. Graphical representation: 0 02 NSS TM 5A (E)-2P-KJ.indd 95 3 8 2010/7/20 4:41:08 PM 96 Solutions 22. (a) (c) The corresponding quadratic function is y = 9x2 - 12x + 4. When y = 0, we have 9x2 - 12x + 4 = 0 (3x - 2)2 = 0 From the sketch in (a), y = 0 when x = 23 . ` The solution of 9x2 - 12x + 4 G 0 is x = 23 . x = 23 (repeated) Graphical representation: Since the coefficient of x2 is 9 (2 0), the parabola opens upward. Sketch: 0 y y = 9x2 – 12x + 4 2 3 (d) From the sketch in (a), a y 1 0 is not true for all values of x. ` 9x2 - 12x + 4 1 0 has no solutions. 23. x 0 2 3 From the sketch, y H 0 when x H 23 or x G 23 . ` The solutions of 9x2 - 12x + 4 H 0 can be any real number. Graphical representation: –4 –3 –2 –1 0 1 2 3 4 12x(x - 1) G x + 4 12x2 - 12x G x + 4 12x2 - 13x - 4 G 0 The corresponding quadratic function is y = 12x2 - 13x - 4. When y = 0, we have 12x2 - 13x - 4 = 0 i.e. (3x - 4)(4x + 1) = 0 ` The x-intercepts of the graph of the quadratic function are 43 and - 41 . Since the coefficient of x2 is 12 (2 0), the parabola opens upward. Sketch: (b) From the sketch in (a), y y 2 0 when x 2 23 or x 1 23 . y = 12x2 – 13x – 4 ` x can be any real number except 23 . Graphical representation: 0 2 3 –1 4 0 x 4 3 From the sketch, y G 0 when - 41 G x G 43 , and x is an integer. ` The required solutions are 0 and 1. 02 NSS TM 5A (E)-2P-KJ.indd 96 2010/7/20 4:41:12 PM Chapter 2: Inequalities in One Unknown 24. (a) (2x - 1)(x + 2) G (3x + 2)(x - 1) 2x2 + 3x - 2 G 3x2 - x - 2 0 G x2 - 4x 2 x - 4x H 0 The corresponding quadratic function is y = x2 - 4x. When y = 0, we have x2 - 4x = 0 i.e. x(x - 4) = 0 ` The x-intercepts of the graph of the quadratic function are 0 and 4. Since the coefficient of x2 is 1 (2 0), the parabola opens upward. Sketch: (b) 9x2 + 16 2 24x 9x2 - 24x + 16 2 0 (3x - 4)2 2 0 ` 3x - 4 2 0, i.e. x 2 43 , 4 or 3x - 4 1 0, i.e. x 1 3 . ` The required solutions are any real number except 43 . Graphical representation: 0 4 3 y y = x2 – 4x 0 From the sketch, y H 0 when x G 0 or x H 4. ` The required solutions are x G 0 or x H 4. (b) From the sketch in (a), when 0 1 x 1 4, y = x2 - 4x 1 0, i.e. (2x - 1)(x + 2) 2 (3x + 2)(x - 1). ` The required solutions are 0 1 x 1 4. 25. (a) 2 x + 10x + 25 G 0 2 (x + 5) G 0 ` x + 5 H 0 and x + 5 G 0, i.e. x H -5 and x G -5, the solution of this compound inequality is x = -5. ` The required solution is x = -5. Graphical representation: –5 02 NSS TM 5A (E)-2P-KJ.indd 97 (c) 4x2 + 49 1 28x 4x2 - 28x + 49 1 0 (2x - 7)2 1 0 a For all real values of x, (2x - 7)2 1 0 does not hold. ` The inequality has no solutions. 26. (a) (x - 2)(3x + 5) + (2 - x)(4x + 7) 1 0 (x - 2)(3x + 5) - (x - 2)(4x + 7) 1 0 (x - 2)[(3x + 5) - (4x + 7)] 1 0 (x - 2)(-x - 2) 1 0 (x - 2)(x + 2) 2 0 ` x - 2 2 0 and x + 2 2 0, i.e. x 2 2 and x 2 -2, the solutions of this compound inequality are x 2 2. or x - 2 1 0 and x + 2 1 0, i.e. x 1 2 and x 1 -2, the solutions of this compound inequality are x 1 -2. ` The required solutions are x 2 2 or x 4 97 x 1 -2. Graphical representation: –2 0 2 0 2010/7/20 4:41:14 PM 98 Solutions (b) From (a), 30. (a) the required solutions are -2 G x G 2. Graphical representation: –2 27. (a) 0 2 x(x + 1) G 3(x + 1) (x + 1)(x - 3) G 0 ` -1 G x G 3 ........................ (i) and 4x2 + 4x - 3 1 0 (2x - 1)(2x + 3) 1 0 ` 3 - 2 1 x 1 21 .......... (ii) According to the question, we have 15 + 0.005 85(x - 20) 1 15(1 + 0.5%) i.e. 0.005 85x 1 0.192 ` x 1 33, cor. to the nearest integer .. (i) or 15 + 0.005 85(x - 20) 2 15(1 + 2%) i.e. 0.005 85x 2 0.417 ` x 2 71, cor. to the nearest integer.. (ii) According to the question, x must satisfy (i) or (ii), and must satisfy x 2 20 at the same time. i.e. 20 1 x 1 33 or x 2 71. (b) Graphical representation of the range in (a): According to the question, x must satisfy (i) and (ii). 0 20 33 71 Therefore -1 G x 1 21 . (b) The maximum value of x is 0. 28. Let the number of the $2 coins be x, then the number of the $5 coins is 36 - x. According to the question, we have 2x + 5(36 - x) H 90 and 2x + 5(36 - x) G 120 2x + 180 - 5x H 90 and 2x + 180 - 5x G 120 -3x H -90 and -3x G -60 x G 30 and x H 20 ` The maximum number of the $2 coins is 30 and the minimum number is 20. 31. Let the larger odd number be 2x + 3, then the smaller odd number is 2x + 1. According to the question, we have (2x + 1)2 + (2x + 3)2 1 100 2 4x + 4x + 1 + 4x2 + 12x + 9 1 100 8x2 + 16x + 10 1 100 8x2 + 16x - 90 1 0 4x2 + 8x - 45 1 0 (2x + 9)(2x - 5) 1 0 9 5 ` - 2 1 x 1 2 ......................................... (i) and 2x + 1 2 0 2x 2 -1 29. According to the question, we have t t 50: 60 2 45 and 50: 60 1 100 - 30 5 t 2 45 and 6 5 t 1 70 6 t 2 54 and ` 54 1 t 1 84 t 1 84 ` x 2 - 21 ...................................... (ii) According to the question, x must satisfy (i) and (ii). 5 ` - 21 1 x 1 2 The possible values of x are 0, 1 and 2. ` The possible values of the larger odd number are 3, 5 and 7. 02 NSS TM 5A (E)-2P-KJ.indd 98 2010/7/20 4:41:18 PM Chapter 2: Inequalities in One Unknown 32. (a) If 2x2 - 4x + k H 0, the discriminant of the quadratic equation 2x2 - 4x + k = 0 is: D = (-4)2 - 4(2)(k) G 0 16 - 8k G 0 16 G 8k kH2 ` The minimum value of k is 2. According to the question, x must satisfy (i) and (ii). Therefore -1 G x G 0 or 4 G x G 5. 35. (a) 3 = 21 x $ 2 x cm2 3 2 = 4 x cm2 (b) Total cost of the ornament 2 = $ b 0.5 # 12x + 2 # 4 x # 8 l 3 2 = $ (6x + 4 3 x ) (c) According to the question, we have * 2 6x + 4 3 x H 50 6x + 4 3 x2 G 300 6x + 4 3 x2 H 50 4 3 x2 + 6x - 50 H 0 2 3 x2 + 3x - 25 H 0 -2 G 2 cos i G 2 i.e. -2 G 3x - 1 G 2 -2 G 3x - 1 and 3x - 1 G 2 -1 G 3x and 3x G 3 and ` xG y2 - y - 6 G 0 (y - 3)(y + 2) G 0 ` -2 G y G 3 or x H 2 -2 G x - 4x - 2 G 3. 2 -2 G x - 4x - 2 2 x - 4x H 0 x(x - 4) H 0 ` x G 0 or x H 4 .......... (i) and 02 NSS TM 5A (E)-2P-KJ.indd 99 -3 + 9 + 200 3 4 3 i.e. x H 2.3, cor. to 1 d.p. .............. (i) 6x + 4 3 x2 G 300 4 3 x + 6x - 300 G 0 2 3 x2 + 3x - 150 G 0 and 2 ` (b) From the result in (a), we have -3 - 9 + 200 3 4 3 i.e. x G -3.2, cor. to 1 d.p. xG1 ` - 31 G x G 1 34. (a) 2 ` Area of one triangular face of the ornament 33. a -1 G cos i G 1 x H - 31 2 3 (b) If -x - 4x + k G 8, Integrated Questions (cross-chapters) x - a 2x k cm Height of the triangle = = 2 x cm 2 then x2 + 4x - k + 8 H 0. The discriminant of the quadratic equation x2 + 4x - k + 8 = 0 is: D = 42 - 4(1)(-k + 8) G 0 16 + 4k - 32 G 0 4k G 16 kG4 ` The maximum value of k is 4. 99 -3 - 9 + 1 200 3 GxG 4 3 -3 + 9 + 1 200 3 4 3 i.e. -7.0 G x G 6.2, cor. to 1 d.p. ..... (ii) a x must satisfy (i), (ii) and x 2 0. ` 2.3 G x G 6.2 2 x - 4x - 2 G 3 2 x - 4x - 5 G 0 (x - 5)(x + 1) G 0 ` -1 G x G 5 ........ (ii) 2010/7/20 4:41:22 PM 100 Solutions 2 36. Let the length of BC be l cm. ` It is given that the perimeter of ABCD is 20 cm, and AB = AD, BC = DC. 10 (r + 2) - 100 (r + 2) - 4 (r + 2) $ 50 (r - 2) 2 (r + 2) GlG 2 ` AB + BC = 21 (20) cm 10 (r + 2) + 100 (r + 2) - 4 (r + 2) $ 50 (r - 2) 2 (r + 2) ` 1.3 G l G 8.7, cor. to 1 d.p. ` The maximum length of BC is 8.7 cm. = 10 cm ` Length of AB = (10 - l ) cm A Public Exam Questions D B 37. 2 1 x 1 6 38. x 1 -3 or x 2 2 39. -8 1 x G -5 40. x 1 3; 0, 1, 2 C Lively Maths Problem Join AC. a 3ABC , 3ADC (SSS) ` +ABC = +ADC a +ABC + +ADC = 180c ` +ABC = +ADC = 90c Length of AC = 2 BC + AB 2 41. (a) Area of the kite = 21 x $ (76 - x) cm2 2 2 x = c 38x - 2 m cm 2 2 = l + (10 - l ) cm a +ABC = 90c ` AC is a diameter. Area of the shaded region 2 2 l + (10 - l ) F - 21 l (10 - l ) $ 2 3 cm2 = )r < 2 2 2 2 l + (100 - 20l + l ) F - 10l + l 2 2 cm2 = (r < 4 (b) According to the question, we have x 2 38x - 2 H 720 76x - x2 H 1 440 x - 76x + 1 440 G 0 (x - 36)(x - 40) G 0 ` 36 G x G 40 2 2 rl - 10rl + 50r - 10l + l 2 m cm2 =c 2 According to the question, we have 2 rl - 10rl + 50r - 10l + l 2 G 50 2 rl 2 - 10rl + 50r - 20l + 2l 2 G 100 (r + 2)l 2 - (10r + 20)l + 50r - 100 G 0 (r + 2)l 2 - 10(r + 2)l + 50(r - 2) G 0 02 NSS TM 5A (E)-2P-KJ.indd 100 2010/7/20 4:41:26 PM Chapter 2: Inequalities in One Unknown Revision Test (P.105) 1. 5. 2 9x - 6 2 2x - 4x 2 0 2 2x - 13x + 6 2 2x - 13x + 6 1 0 (x - 6)(2x - 1) 1 0 x 2 - 21 ..................................... (i) 3x - 15 2 -4 7 and 3x - 15 2 -28 3x 2 -13 9x - 6 2 x(2x - 4) C -4x 1 2 B 101 ` 6. A 1 1x16 2 Let the length of BC be x cm, then AB = (x + 4) cm 13 x 2 - 3 ........................... (ii) 2 2 2 2 According to the question, x must satisfy (i) and (ii). AC = x + (x + 4) cm According to the question, we have x20 Therefore x 2 - 21 . and x + (x + 4) 2 0 x2 + (x + 4) 2 1 106 x2 + (x + 4)2 1 106 2 2 x + x + 8x + 16 - 106 1 0 2x2 + 8x - 90 1 0 x2 + 4x - 45 1 0 (x + 9)(x - 5) 1 0 -9 1 x 1 5 ` 01x15 i.e. 0 cm 1 length of BC 1 5 cm and 2. A -2 G 13 - 7x 7x G 15 15 x G 7 ........................................ (i) or x G 7 - 12x 13x G 7 7 x G 13 ....................................... (ii) According to the question, x must satisfy (i) or (ii). 15 Therefore x G 7 . 3. x 2 1 ........................................... (i) D HKCEE Questions 7. A 8. D and 2x 1 3x - 4 41x x 2 4 .......................................... (ii) and 3x - 4 1 5 3x 1 9 x 1 3 .................................. (iii) According to the question, x must satisfy (i), (ii) and (iii). Therefore the compound inequality has no 9. A solutions. 4. A From the graph, y = 0 when x = 3 and there does not exist other values of x such that y 1 0. ` The solution of f(x) G 0 is x = 3. 02 NSS TM 5A (E)-2P-KJ.indd 101 2010/7/20 4:41:29 PM